Algebra
With Sessionwise Theory & Exercises
Algebra
With Sessionwise Theory & Exercises
Dr. SK Goyal
ARIHANT PRAKASHAN (Series), MEERUT
All Rights Reserved
© AUTHOR
No part of this publication may be re-produced, stored in a retrieval system or
by any means, electronic mechanical, photocopying, recording, scanning, web or
otherwise without the written permission of the publisher. Arihant has obtained
all the information in this book from the sources believed to be reliable and true.
However, Arihant or its editors or authors or illustrators don’t take any responsibility
for the absolute accuracy of any information published, and the damages or loss
suffered thereupon.
All disputes subject to Meerut (UP) jurisdiction only.
Administrative & Production Offices
Regd. Office
‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002
Tele: 011- 47630600, 43518550
Head Office
Kalindi, TP Nagar, Meerut (UP) - 250002 Tel: 0121-7156203, 7156204
Sales & Support Offices
Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati,
Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Nagpur & Pune.
ISBN : 978-93-25298-63-7
PO No : TXT-XX-XXXXXXX-X-XX
Published by Arihant Publications (India) Ltd.
For further information about the books published by Arihant, log on to
www.arihantbooks.com or e-mail at info@arihantbooks.com
Follow us on
PREFACE
‘‘THE ALGEBRAIC SUM OF ALL THE TRANSFORMATIONS OCCURRING IN A CYCLICAL
PROCESS CAN ONLY BE POSITIVE, OR, AS AN EXTREME CASE EQUAL TO NOTHING’’
MEANS IF YOU CONTINUOUSLY PUT YOUR EFFORTS ON AN ASPECT YOU HAVE VERY
GOOD CHANCE OF POSITIVE OUTCOME i.e. SUCCESS
It is a matter of great pride and honour for me to have received such an overwhelming response to the
previous editions of this book from the readers. In a way, this has inspired me to revise this book
thoroughly as per the changed pattern of JEE Main & Advanced. I have tried to make the contents more
relevant as per the needs of students, many topics have been re-written, a lot of new problems of new
types have been added in etcetc. All possible efforts are made to remove all the printing errors that had
crept in previous editions. The book is now in such a shape that the students would feel at ease while
going through the problems, which will in turn clear their concepts too.
A Summary of changes that have been made in Revised & Enlarged Edition
—
Theory has been completely updated so as to accommodate all the changes made in JEE Syllabus & Pattern in
recent years.
—
The most important point about this new edition is, now the whole text matter of each chapter has been
divided into small sessions with exercise in each session. In this way the reader will be able to go through the
whole chapter in a systematic way.
—
Just after completion of theory, Solved Examples of all JEE types have been given, providing the students a
complete understanding of all the formats of JEE questions & the level of difficulty of questions generally
asked in JEE.
—
Along with exercises given with each session, a complete cumulative exercises have been given at the end of
each chapter so as to give the students complete practice for JEE along with the assessment of knowledge
that they have gained with the study of the chapter.
—
Last 13 Years questions asked in JEE Main & Adv, IIT-JEE & AIEEE have been covered in all the chapters.
However I have made the best efforts and put my all Algebra teaching experience in revising this book.
Still I am looking forward to get the valuable suggestions and criticism from my own fraternity i.e. the
fraternity of JEE teachers.
I would also like to motivate the students to send their suggestions or the changes that they want to be
incorporated in this book.
All the suggestions given by you all will be kept in prime focus at the time of next revision of the book.
Dr. SK Goyal
CONTENTS
1. COMPLEX NUMBERS
LEARNING PART
Session 1
— Integral Powers of Iota (i)
— Switch System Theory
Session 2
— Definition of Complex Number
— Conjugate Complex Numbers
— Representation of a Complex Number in
Various Forms
Session 3
— amp (z)– amp (–z)=± p, According as amp (z)
is Positive or Negative
— Square Root of a Complex Number
— Solution of Complex Equations
— De-Moivre’s Theorem
— Cube Roots of Unity
1-102
Session 4
— nth Root of Unity
— Vector Representation of Complex Numbers
— Geometrical Representation of Algebraic
Operation on Complex Numbers
— Rotation Theorem (Coni Method)
— Shifting the Origin in Case of Complex
Numbers
— Inverse Points
— Dot and Cross Product
— Use of Complex Numbers in Coordinate
Geometry
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
2. THEORY OF EQUATIONS
LEARNING PART
Session 1
— Polynomial in One Variable
— Identity
— Linear Equation
— Quadratic Equations
— Standard Quadratic Equation
Session 2
— Transformation of Quadratic Equations
— Condition for Common Roots
Session 3
— Quadratic Expression
— Wavy Curve Method
— Condition for Resolution into Linear Factors
— Location of Roots (Interval in which Roots Lie)
103-206
Session 4
— Equations of Higher Degree
— Rational Algebraic Inequalities
— Roots of Equation with the
Help of Graphs
Session 5
— Irrational Equations
— Irrational Inequations
— Exponential Equations
— Exponential Inequations
— Logarithmic Equations
— Logarithmic Inequations
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
3. SEQUENCES AND SERIES
207-312
LEARNING PART
Session 5
Session 1
— Sequence
— Series
— Progression
— Mean
Session 2
— Arithmetic Progression
Session 3
— Geometric Sequence or Geometric
Progression
Session 4
— Harmonic Sequence or Harmonic Progression
Session 6
— Arithmetico-Geometric
Series (AGS)
— Sigma (S) Notation
— Natural Numbers
Session 7
— Application to Problems of Maxima and
Minima
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
4. LOGARITHMS AND THEIR PROPERTIES
LEARNING PART
Session 1
— Definition
— Characteristic and Mantissa
Session 2
— Principle Properties of Logarithm
313-358
Session 3
— Properties of Monotonocity of Logarithm
— Graphs of Logarithmic Functions
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
359-436
5. PERMUTATIONS AND COMBINATIONS
LEARNING PART
Session 5
Session 1
— Fundamental Principle of Counting
— Factorial Notation
Session 2
— Divisibility Test
— Principle of Inclusion and Exclusion
— Permutation
Session 3
— Number of Permutations Under Certain
Conditions
— Circular Permutations
— Restricted Circular Permutations
Session 4
— Combination
— Restricted Combinations
— Combinations from Identical Objects
Session 6
— Arrangement in Groups
— Multinomial Theorem
— Multiplying Synthetically
Session 7
— Rank in a Dictionary
— Gap Method
[when particular objects are never together]
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
437-518
6. BINOMIAL THEOREM
LEARNING PART
Session 1
— Binomial Theorem for Positive Integral Index
— Pascal’s Triangle
Session 2
— General Term
— Middle Terms
— Greatest Term
— Trinomial Expansion
Session 3
— Two Important Theorems
— Divisibility Problems
Session 4
— Use of Complex Numbers in Binomial Theorem
— Multinomial Theorem
— Use of Differentiation
— Use of Integration
— When Each Term is Summation Contains the
Product of Two Binomial Coefficients or
Square of Binomial Coefficients
— Binomial Inside Binomial
— Sum of the Series
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
519-604
7. DETERMINANTS
LEARNING PART
— System of Linear Equations
Session 1
— Definition of Determinants
— Expansion of Determinant
— Sarrus Rule for Expansion
— Window Rule for Expansion
— Cramer’s Rule
Session 2
— Minors and Cofactors
— Use of Determinants in Coordinate Geometry
— Properties of Determinants
Session 3
— Examples on Largest Value of a
Third Order Determinant
— Multiplication of Two Determinants of
— Nature of Solutions of System of Linear
Equations
— System of Homogeneous Linear Equations
Session 4
— Differentiation of Determinant
— Integration of a Determinant
— Walli’s Formula
— Use of S in Determinant
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
the Same Order
8. MATRICES
LEARNING PART
Session 1
— Definition
— Types of Matrices
— Difference Between a Matrix and a
Determinant
— Equal Matrices
— Operations of Matrices
— Various Kinds of Matrices
605-690
Session 2
— Transpose of a Matrix
— Symmetric Matrix
— Orthogonal Matrix
— Complex Conjugate (or Conjugate) of a Matrix
— Hermitian Matrix
— Unitary Matrix
— Determinant of a Matrix
— Singular and Non-Singular Matrices
Session 3
— Adjoint of a Matrix
— Inverse of a Matrix
— Elementary Row Operations
— Equivalent Matrices
— Matrix Polynomial
— Use of Mathematical Induction
Session 4
— Solutions of Linear Simultaneous Equations
Using Matrix Method
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
691-760
9. PROBABILITY
LEARNING PART
Session 1
— Some Basic Definitions
— Mathematical or Priori or Classical Definition
of Probability
— Odds in Favours and Odds Against the Event
Session 2
— Some Important Symbols
— Conditional Probability
Session 3
— Total Probability Theorem
— Baye’s Theorem or Inverse Probability
Session 4
— Binomial Theorem on Probability
— Poisson Distribution
— Expectation
— Multinomial Theorem
— Uncountable Uniform Spaces
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
761-784
10. MATHEMATICAL INDUCTION
LEARNING PART
PRACTICE PART
— Introduction
— JEE Type Examples
— Statement
— Chapter Exercises
— Mathematical Statement
785-836
11. SETS, RELATIONS AND FUNCTIONS
LEARNING PART
Session 1
— Definition of Sets
— Representation of a Set
— Different Types of Sets
— Laws and Theorems
— Venn Diagrams (Euler-Venn Diagrams)
Session 2
— Ordered Pair
— Definition of Relation
— Ordered Relation
— Composition of Two Relations
Session 3
— Definition of Function
— Domain, Codomain and Range
— Composition of Mapping
— Equivalence Classes
— Partition of Set
— Congruences
PRACTICE PART
— JEE Type Examples
— Chapter Exercises
SYLLABUS
Unit I Sets, Relations and Functions
numbers. Relation between AM and GM Sum upto n
terms of special series: ∑ n, ∑ n2, ∑n3. Arithmetico Geometric progression.
Sets and their representation, Union, intersection and
complement of sets and their algebraic properties, Power
set, Relation, Types of relations, equivalence relations,
functions, one-one, into and onto functions, composition
of functions.
Probability of an event, addition and multiplication
theorems of probability, Baye’s theorem, probability
distribution of a random variate, Bernoulli and Binomial
distribution.
JEE MAIN
Unit VIII Probability
Unit II Complex Numbers
Complex numbers as ordered pairs of reals,
Representation of complex numbers in the form a+ib and
their representation in a plane, Argand diagram, algebra
of complex numbers, modulus and argument (or
amplitude) of a complex number, square root of a
complex number, triangle inequality.
Unit III Matrices and Determinants
JEE ADVANCED
Algebra
Algebra of complex numbers, addition, multiplication,
conjugation, polar representation, properties of modulus
and principal argument, triangle inequality, cube roots of
unity, geometric interpretations.
Matrices, algebra of matrices, types of matrices,
determinants and matrices of order two and three.
Properties of determinants, evaluation of deter-minants,
area of triangles using determinants. Adjoint and
evaluation of inverse of a square matrix using
determinants and elementary transformations, Test of
consistency and solution of simultaneous linear
equations in two or three variables using determinants
and matrices.
Arithmetic, geometric and harmonic progressions,
arithmetic, geometric and harmonic means, sums of finite
arithmetic and geometric progressions, infinite geometric
series, sums of squares and cubes of the first n natural
numbers.
Unit IV Permutations and Combinations
Logarithms and their Properties
Fundamental principle of counting, permutation as an
arrangement and combination as selection, Meaning of
P(n,r) and C (n,r), simple applications.
Permutations and combinations, Binomial theorem for a
positive integral index, properties of binomial
coefficients.
Unit V Mathematical Induction
Matrices as a rectangular array of real numbers, equality
of matrices, addition, multiplication by a scalar and
product of matrices, transpose of a matrix, determinant of
a square matrix of order up to three, inverse of a square
matrix of order up to three, properties of these matrix
operations, diagonal, symmetric and skew-symmetric
matrices and their properties, solutions of simultaneous
linear equations in two or three variables.
Principle of Mathematical Induction and its simple
applications.
Unit VI Binomial Theorem and its
Simple Applications
Binomial theorem for a positive integral index, general
term and middle term, properties of Binomial coefficients
and simple applications.
Unit VII Sequences and Series
Arithmetic and Geometric progressions, insertion of
arithmetic, geometric means between two given
Quadratic equations with real coefficients, relations
between roots and coefficients, formation of quadratic
equations with given roots, symmetric functions of roots.
Addition and multiplication rules of probability,
conditional probability, independence of events,
computation of probability of events using permutations
and combinations.
CHAPTER
01
Complex Numbers
Learning Part
Session 1
● Integral Powers of Iota (i)
● Switch System Theory
Session 2
● Definition of Complex Number
● Conjugate Complex Numbers
● Representation of a Complex Number in Various Forms
Session 3
● amp ( z ) - amp ( - z ) = ± p , According as amp ( z ) is Positive or Negative
● Square Root of a Complex Number
● Solution of Complex Equations
● De-Moivre’s Theorem
● Cube Roots of Unity
Session 4
● nth Root of Unity
● Vector Representation of Complex Numbers
● Geometrical Representation of Algebraic Operation on Complex Numbers
● Rotation Theorem (Coni Method)
● Shifting the Origin in Case of Complex Numbers
● Inverse Points
● Dot and Cross Product
● Use of Complex Numbers in Coordinate Geometry
Practice Part
●
●
JEE Type Examples
Chapter Exercises
Arihant on Your Mobile !
Exercises with the
#L
symbol can be practised on your mobile. See inside cover page to activate for free.
2
Textbook of Algebra
The square of any real number, whether positive, negative
or zero, is always non-negative i.e. x 2 ³ 0 for all x Î R.
Therefore, there will be no real value of x , which when
squared, will give a negative number.
Thus, the equation x 2 + 1 = 0 is not satisfied for any real
value of x. ‘Euler’ was the first Mathematician to
introduce the symbol i (read ‘Iota’) for the square root of
- 1 with the property i 2 = - 1. The theory of complex
number was later on developed by Gauss and Hamilton.
According to Hamilton, ‘‘Imaginary number is that
number whose square is a negative number ’’. Hence, the
equation x 2 + 1 = 0
Þ
x2 = -1
or
x = ± -1
(in the sense of arithmetic, -1 has no meaning).
Symbolically, -1 is denoted by i (the first letter of the
word ‘Imaginary ’).
\ Solutions of x 2 + 1 = 0 are x = ± i.
Also, i is the unit of complex number, since i is present in
every complex number. Generally, if a is positive quantity,
then
-a ´ -a = ( -1) ´ a ´ ( -1) ´ a
= -1 ´ a ´ -1 ´ a
=i a ´i a
= i2a = - a
Remark
- a = i a ,where a is positive quantity. Keeping this result in
mind, the following computation is correct
- a - b = i a × i b = i 2 ab = - ab
where, a and b are positive real numbers.
But the computation, - a - b = ( - a)( - b) = |a||b| is wrong.
Because the property, a b = ab is valid only when atleast one
of a and b is non-negative.
If a and b are both negative, then a b = - a b .
y Example 1. Is the following computation correct?
If not, give the correct computation.
-2 -3 = ( -2) ( -3) = 6
Sol. No,
If a and b are both negative real numbers, then a b = - ab
Here, a = - 2 and b = - 3.
\
-2
- 3 = - ( - 2) ( - 3) = - 6
y Example 2. A student writes the formula
ab = a b . Then, he substitutes a = - 1 and b = - 1
and finds 1 = - 1. Explain, where he is wrong.
Sol. Since, a and b are both negative, therefore ab ¹ a b .
Infact a and b are both negative, then we have a b = - ab .
y Example 3. Explain the fallacy
- 1 = i ´ i = -1 ´ -1 = ( -1) ´ ( -1) = 1 = 1.
Sol. If a and b are both negative, then
a b = - |a | |b |
\
-1 ´ -1 = - | -1| | -1| = - 1
Session 1
Integral Powers of Iota (i ), Switch System Theory
Integral Powers of Iota ( i )
(i) If the index of i is whole number, then
i 0 = 1, i 1 = i , i 2 = ( -1 ) 2 = - 1,
i 3 = i × i 2 = - i, i 4 = (i 2 ) 2 = ( -1) 2 = 1
n
To find the value of i (n > 4 ) First divide n by 4.
Let q be the quotient and r be the remainder.
i.e.
4 ) n (q
- 4q
r
Þ
When, 0 £ r £ 3
\
n = 4q + r
i n = i 4 q + r = (i 4 ) q (i ) r = (1) q × (i ) r = i r
In general,
i 4n = 1, i 4n + 1 = i, i 4n + 2 = - 1,
i 4n + 3 = - i for any whole number n.
(ii) If the index of i is a negative integer, then
1 i
1
i
i -1 = = =
= - i, i -2 =
= - 1,
2
2
i i
-1
i
1
1 1
i
i -3 = = = i, i -4 = = = 1, etc.
3
4
i
i
i4 1
Chap 01 Complex Numbers
2
æ 1 + i 2 + 2i ö
æ1 + i ö
Sol. Q a 2 = ç
÷
÷ =ç
è 2 ø
2
ø
è
y Example 4. Evaluate.
(i) i 1998
(ii) i - 9999
æ 1 - 1 + 2i ö
=ç
÷ =i
ø
è
2
(iii) ( - -1 ) 4n +3 , n Î N
Sol. (i) 1998 leaves remainder 2, when it is divided by 4.
i.e.
4 ) 1998 (499
1996
2
i 1998 = i 2 = - 1
\
Aliter
i 2000
1
=
= -1
-1
i2
(ii) 9999 leaves remainder 3, when it is divided by 4.
i.e.
4 ) 9999 (2499
9996
3
i
i
1
1
\
i - 9999 = 9999 = 3 = 4 = = i
1
i
i
i
i 1998 =
1
= a (i )4 ´ 241 = a × (i 4 )241 = a
y Example 7. Dividing f (z ) by z - i , where i = -1, we
obtain the remainder i and dividing it by z + i , we get
the remainder 1 + i. Find the remainder upon the
division of f (z ) by z 2 + 1.
Sol. z - i = 0 Þ z = i
Remainder, when f (z ) is divided by (z - i ) = i
i.e.
f (i ) = i
K (i)
and remainder, when f (z ) is divided by (z + 1) = 1 + i
i.e.
f ( - i ) = 1 + i [Qz + i = 0 Þ z = - i ] K (ii)
Since, z 2 + 1 is a quadratic expression, therefore remainder
when f (z ) is divided by z 2 + 1, will be in general a linear
expression. Let g (z ) be the quotient and az + b (where a
and b are complex numbers) be the remainder, when f (z ) is
divided by z 2 + 1.
i
=i
1
i
i
(iii) 4n + 3 leaves remainder 3, when it is divided by 4.
i.e.,
4 ) 4n + 3 (n
4n
3
9999
=
i
10000
=
Then,
Aliter ( - -1 )4n + 3
= - (i 4 )n × i 3
= - (1)n ( - i ) = i
where i = -1 and n Î N .
Sol. Q 1 + i 2 + i 4 + i 6 + ... + i 2 n = 1 - 1 + 1 - 1 + ... + ( - 1)n
Case I If n is odd, then
1 + i 2 + i 4 + i 6 + ... + i 2 n = 1 - 1 + 1 - 1 + ... + 1 - 1 = 0
Case II If n is even, then
1 + i 2 + i 4 + i 6 + ... + i 2 n = 1 - 1 + 1 - 1 + ... + 1 = 1
y Example 6. If a =
value of a
1929
.
1+ i
2
[from Eq. (i)] K (iv)
ai + b = i
f ( - i ) = (i 2 + 1) g ( - i ) - ai + b = - ai + b
or
[from Eq. (ii)] …(v)
- ai + b = 1 + i
From Eqs. (iv) and (v), we get
1
i
b = + i and a =
2
2
Hence, required remainder = az + b
1
1
= iz + + i
2
2
= - (- i)
=i
= ( - i )4n + 3 = - i 4n + 3
y Example 5. Find the value of 1 + i 2 + i 4 + i 6 + ... + i 2n ,
K (iii)
f (i ) = (i + 1) g (i ) + ai + b = ai + b
or
and
( - -1 )4n + 3 = ( - i )4n + 3 = - (i )4n + 3
Now,
f (z ) = (z 2 + 1) g ( z ) + az + b
2
\
i 4n + 3 = i 3 = - i
\
a1929 = a × a1928 = a × (a 2 )964 = a (i )964
\
Aliter
i - 9999 =
3
The Sum of Four Consecutive
Powers of i (Iota) is Zero
If n Î I and i = - 1, then
i n + i n + 1 + i n + 2 + i n + 3 = i n (1 + i + i 2 + i 3 )
= i n (1 + i - 1 - i ) = 0
Remark
1.
, where i = - 1, then find the
2.
m
m- p + 1
r =p
r =1
å f( r ) = å
f ( r + p - 1)
m
m+ p + 1
r =-p
r =1
å f( r ) = å
f ( r - p - 1)
4
Textbook of Algebra
13
y Example 8. Find the value of
Sol. Q
å
(i n + i n + 1 )
Switch System Theory
n =1
( where ,i = - 1 )
13
å
(i n + i n + 1 ) =
n =1
13
13
n =1
n =1
(Finding Digit in the Unit’s Place)
å i n + å i n + 1 = ( i + 0) + ( i 2 + 0)
We can determine the digit in the unit’s place in
a b , where a, b Î N . If last digit of a are 0, 1, 5 and 6, then
13
ù
é 13 n
n +1
=0
ú
êQ å i = 0 and å i
=i -1 n=2
n=2
ú
ê
êë(three sets of four consecutive powers of i )úû
100
y Example 9. Find the value of
å
digits in the unit’s place of a b are 0, 1, 5 and 6
respectively, for all b Î N .
Powers of 2
in !
2 1 , 2 2 , 2 3 , 2 4 , 2 5 , 2 6 , 2 7 , 2 8 , 2 9 , ... the digits in unit’s place
of different powers of 2 are as follows :
n=0
( where , i = - 1).
Sol. n! is divisible by 4, " n ³ 4.
\
100
97
n=4
n =1
2, 4, 8, 6, 2, 4, 8, 6, 2,... (period being 4)
­ ­ ­ ­ ­ ­ ­ ­ ­
1 2 3 0 1 2 3 0 1 ... (switch number)
å in ! = å i (n + 3 )!
= i 0 + i 0 + i 0 + ... 97 times = 97
100
åi
\
n=0
n!
=
3
åi
n!
n=0
+
100
åi
…(i)
n!
n=4
= i 0 ! + i 1! + i 2 ! + i 3 ! + 97
1
1
2
[from Eq. (i)]
6
= i + i + i + i + 97 = i + i - 1 - 1 + 97
= 95 + 2i
4n + 7
y Example 10. Find the value of
4n + 7
å
ir = i1 + i 2 + i 3 +
r =1
y Example 12. What is the digit in the unit’s place of
( 5172)11327 ?
ir
Sol. Here, last digit of a is 2.
The remainder when 11327 is divided by 4, is 3. Then,
press switch number 3 and then we get 8.
r =1
( where ,i = - 1 ).
Sol.
å
4n + 7
å
ir = i - 1 - i +
r =4
4n + 4
å
ir
+3
r =1
= - 1 + 0 [(n + 1) sets of four consecutive powers of i ]
= -1
y Example 11. Show that the polynomial
x 4 p + x 4q + 1 + x 4r + 2 + x 4 s + 3 is divisible by
x 3 + x 2 + x + 1, where p , q, r , s Î N .
Sol. Let f ( x ) = x 4 p + x 4q + 1 + x 4r
and
+2
+ x 4s + 3
x 3 + x 2 + x + 1 = ( x 2 + 1) ( x + 1)
where
+2
+ i 4s + 3 = 1 + i + i 2 + i 3 = 0
[sum of four consecutive powers of i is zero]
f ( - i ) = ( - i )4 p + ( - i )4q + 1 + ( - i )4r + 2 + ( - i )4s + 3
= 1 + ( -i )1 + ( -i )2 + ( - i )3 = 1 - i - 1 + i = 0
and f ( - 1) = ( - 1)4 p + ( - 1)4q + 1 + ( - 1)4r
+2
Hence, the digit in the unit’s place of (5172)11327 is 8.
Powers of 3
3 1 , 3 2 , 3 3 , 3 4 , 3 5 , 3 6 , 3 7 , 3 8 , ... the digits in unit’s place of
different powers of 3 are as follows:
3, 9, 7, 1, 3, 9, 7, 1, ... (period being 4)
­ ­ ­ ­ ­ ­ ­ ­
1 2 3 0 1 2 3 0 ... (switch number)
The remainder when b is divided by 4, can be 1 or 2 or 3
or 0. Now, press the switch number and get the unit’s
place digit ( just above).
= ( x + i ) ( x - i ) ( x + 1),
i = -1
Now, f (i ) = i 4 p + i 4q + 1 + i 4r
(The remainder when b is divided by 4, can be 1 or 2 or 3 or 0).
Then, press the switch number and then we get the digit
in unit’s place of a b (just above the switch number) i.e.
‘press the number and get the answer’.
+ ( - 1)4s + 3
=1-1+1-1=0
Hence, by division theorem, f ( x ) is divisible by
x 3 + x 2 + x + 1.
y Example 13. What is the digit in the unit’s place
of
(143) 86 ?
Sol. Here, last digit of a is 3.
The remainder when 86 is divided by 4, is 2.
Then, press switch number 2 and then we get 9.
Hence, the digit in the unit’s place of (143)86 is 9.
Chap 01 Complex Numbers
Powers of 4
Powers of 8
4 1 , 4 2 , 4 3 , 4 4 , 4 5 ,... the digits in unit’s place of different
powers of 4 are as follows:
4, 6, 4, 6, 4, ... (period being 2)
­ ­ ­ ­ ­
8 1 , 8 2 , 8 3 , 8 4 , 8 5 , 8 6 , 8 7 , 8 8 ,... the digits in unit’s place of
different powers of 8 are as follows:
8, 4, 2, 6, 8, 4, 2, 6, ... (period being 4)
1 0 1 0 1 ... (switch number)
The remainder when b is divided by 2, can be 1 or 0. Now,
press the switch number and get the unit’s place digit
( just above the switch number).
y Example 14. What is the digit in unit’s place of
(1354 ) 22222 ?
Sol. Here, last digit of a is 4.
The remainder when 22222 is divided by 2, is 0. Then,
press switch number 0 and then we get 6.
Hence, the digit in the unit’s place of (1354 )22222 is 6.
­ ­ ­ ­ ­ ­ ­ ­
1 2 3 0 1 2 3 0 ... (switch number)
The remainder when b is divided by 4, can be 1 or 2 or 3
or 0.
Now, press the switch number and get the unit’s place
digit (just above the switch number).
y Example 16. What is the digit in the unit’s place of
(1008 ) 786 ?
Sol. Here, last digit of a is 8.
The remainder when 786 is divided by 4, is 2. Then, press
switch number 2 and then we get 4.
Hence, the digit in the unit’s place of (1008)786 is 4.
Powers of 7
7 1 , 7 2 , 7 3 , 7 4 , 7 5 , 7 6 , 7 7 , 7 8 , ... the digits in unit’s place of
different powers of 7 are as follows:
7, 9, 3, 1, 7, 9, 3, 1, ... (period being 4)
­ ­ ­ ­ ­ ­ ­ ­
1 2 3 0 1 2 3 0 ... (switch number)
Powers of 9
9 1 , 9 2 , 9 3 , 9 4 , 9 5 ,... the digits in unit’s place of different
powers of 9 are as follows:
9, 1, 9, 1, 9, ... ( period being 2)
­ ­ ­ ­ ­
1 0 1 0 1 ... (switch number)
(The remainder when b is divided by 4, can be 1 or 2 or 3
or 0). Now, press the switch number and get the unit’s
place digit ( just above).
The remainder when b is divided by 2, can be 1 or 0.
Now, press the switch number and get the unit’s place
digit (just above the switch number).
y Example 15. What is the digit in the unit’s place of
(13057 ) 941120579 ?
y Example 17. What is the digit in the unit’s place of
(2419 )111213 ?
Sol. Here, last digit of a is 7.
The remainder when 941120579 is divided by 4, is 3. Then,
press switch number 3 and then we get 3.
Sol. Here, last digit of a is 9.
The remainder when 111213 is divided by 2, is 1. Then,
press switch number 1 and then we get 9.
Hence, the digit in the unit’s place of (13057 )941120579 is 3.
5
Hence, the digit in the unit’s place of (2419 )111213 is 9.
6
Textbook of Algebra
#L
1
Exercise for Session 1
If (1 + i )2 n + (1 - i )2 n = - 2n + 1 (where, i = - 1) for all those n, which are
(a) even
(c) multiple of 3
2
(b) odd
(d) None of these
If i = - 1, the number of values of i n + i - n for different n Î I is
(a) 1
(c) 3
3
4
(b) 2
(d) 4
If a > 0 and b < 0, then a
b is equal to (where, i = -1)
(a) - a × b
(b) a × b i
(c) a × b
(d) None of these
Consider the following statements.
S1 : - 6 = 2i ´ 3i = ( - 4) ´ ( - 9) ( where, i = - 1)
S2 : ( - 4) ´ ( - 9) = ( - 4) ´ ( - 9)
S3 : ( - 4) ´ ( - 9) = 36
S4 : 36 = 6
Of these statements, the incorrect one is
(b) S 2 only
(d) None of these
(a) S1 only
(c) S 3 only
50
5
The value of
S
n=0
i ( 2n + 1) ! (where, i = - 1) is
(b) 47 - i
(d) 0
(a) i
(c) 48 + i
1003
6
The value of
S
r =-3
i r ( where i = - 1) is
(a) 1
(c) i
7
The digit in the unit’s place of (153)98 is
(a) 1
(c) 7
8
(b) - 1
(d) - i
(b) 3
(d) 9
The digit in the unit’s place of (141414)12121 is
(a) 4
(c) 3
(b) 6
(d) 1
Session 2
Definition of Complex Number, Conjugate Complex
Numbers, Representation of a Complex Number in
Various Forms
Definition of Complex Number
A number of the form a + ib, where a, b Î R and i = - 1, is
called a complex number. It is denoted by z i.e. z = a + ib.
A complex number may also be defined as an ordered pair
of real numbers; and may be denoted by the symbol (a, b).
If we write z = (a, b ), then a is called the real part and b is
the imaginary part of the complex number z and may be
denoted by Re (z ) and Im (z), respectively i.e., a = Re (z )
and b = Im (z ).
Two complex numbers are said to be equal, if and only if
their real parts and imaginary parts are separately equal.
Thus,
a + ib = c + id
Û
a = c and b = d
where, a, b, c , d Î R and i = - 1.
i.e.
Û
z1 = z2
Re (z 1 ) = Re (z 2 ) and Im (z 1 ) = Im (z 2 )
Important Properties of Complex Numbers
1. The complex numbers do not possess the property of order,
i.e., ( a + ib) > or < ( c + id ) is not defined. For example,
9 + 6 i > 3 + 2i makes no sense.
2. A real number a can be written as a + i × 0. Therefore, every
real number can be considered as a complex number, whose
imaginary part is zero. Thus, the set of real numbers (R) is a
proper subset of the complex numbers ( C ) i.e. R Ì C. Hence,
the complex number system is N Ì W Ì I Ì Q Ì R Ì C
3. A complex number z is said to be purely real, if Im ( z ) = 0; and
is said to be purely imaginary, if Re ( z ) = 0. The complex
number 0 = 0 + i × 0 is both purely real and purely imaginary.
4. In real number system, a2 + b2 = 0 Þ a = 0 = b.
But if z 1 and z 2 are complex numbers, then z 12 + z 22 = 0
does not imply z1 = z2 = 0.
For example, z 1 = 1 + i and z 2 = 1 - i
Here, z 1 ¹ 0, z 2 ¹ 0
But z 12 + z 22 = ( 1 + i ) 2 + ( 1 - i ) 2 = 1 + i 2 + 2i + 1 + i 2 - 2i
= 2 + 2i 2 = 2 - 2 = 0
However, if product of two complex numbers is zero, then
atleast one of them must be zero, same as in case of real
numbers.
If z 1z 2 = 0, then z 1 = 0, z 2 ¹ 0 or z 1 ¹ 0, z 2 = 0
or
z 1 = 0, z 2 = 0
Algebraic Operations on
Complex Numbers
Let two complex numbers be z 1 = a + ib and z 2 = c + id ,
where a, b, c , d Î R and i = - 1.
1. Addition z 1 + z 2 = (a + ib ) + (c + id )
= (a + c ) + i (b + d )
2. Subtraction z 1 - z 2 = (a + ib ) - (c + id )
= (a - c ) + i (b - d )
3. Multiplication z 1 × z 2 = (a + ib ) × (c + id )
= ac + iad + ibc + i 2 bd
= ac + i (ad + bc ) - bd
= (ac - bd ) + i (ad + bc )
(a + ib ) (c - id )
z
4. Division 1 =
×
z 2 (c + id ) (c - id )
[multiplying numerator and denominator by c - id
where atleast one of c and d is non-zero]
ac - iad + ibc - i 2 bd ac + i (bc - ad ) + bd
=
=
c 2 - i 2d 2
(c ) 2 - (id ) 2
(ac + bd ) + i (bc - ad ) (ac + bd )
(bc - ad )
=
=
+i
2
2
2
2
c +d
(c + d )
(c 2 + d 2 )
Remark
1+ i
1- i
= i and
= - i, where i =
1- i
1+ i
-1.
Properties of Algebraic Operations
on Complex Numbers
Let z 1 , z 2 and z 3 be any three complex numbers.
Then, their algebraic operations satisfy the following
properties :
Properties of Addition of Complex Numbers
(i) Closure law z 1 + z 2 is a complex number.
(ii) Commutative law z 1 + z 2 = z 2 + z 1
(iii) Associative law (z 1 + z 2 ) + z 3 = z 1 + (z 2 + z 3 )
8
Textbook of Algebra
(iv) Additive identity z + 0 = z = 0 + z , then 0 is called
the additive identity.
(v) Additive inverse - z is called the additive inverse of
z, i.e. z + ( - z ) = 0.
Let z, z 1 and z 2 be complex numbers. Then,
(i) (z ) = z
Properties of Multiplication
of Complex Numbers
Closure law z 1 × z 2 is a complex number.
Commutative law z 1 × z 2 = z 2 × z 1
Associative law (z 1 × z 2 ) z 3 = z 1 (z 2 × z 3 )
Multiplicative identity z × 1 = z = 1 × z , then 1 is
called the multiplicative identity.
(v) Multiplicative inverse If z is a non-zero complex
1
number, then is called the multiplicative inverse
z
1
1
of z i.e. z. = 1 = × z
z
z
(vi) Multiplication is distributive with respect to
addition z 1 (z 2 + z 3 ) = z 1 z 2 + z 1 z 3
(i)
(ii)
(iii)
(iv)
Conjugate Complex Numbers
The complex numbers z = (a, b ) = a + ib and
z = (a, - b ) = a - ib, where a and b are real numbers,
i = -1 and b ¹ 0, are said to be complex conjugate of each
other (here, the complex conjugate is obtained by just
changing the sign of i).
Note that, sum = (a + ib ) + (a - ib ) = 2a, which is real.
And product = (a + ib ) (a - ib ) = a 2 - (ib ) 2
= a 2 - i 2 b 2 = a 2 - ( -1) b 2
= a 2 + b 2 , which is real.
Geometrically, z is the mirror image of z along real axis on
argand plane.
Remark
Let z = - a - ib, a > 0, b > 0 = ( - a, - b) (III quadrant )
Imaginary axis
b
+z
-z
+z
-z
= 2 Re (z )
= 2 Im (z )
= 0 Þ z = - z Þ z is purely imaginary.
= 0 Þ z = z Þ z is purely real.
(vi) z 1 ± z 2 = z 1 ± z 2 Ingeneral,
z 1 ± z 2 ± z 3 ± ... ± z n = z 1 ± z 2 ± z 3 ± ... ± z n
(vii) z 1 × z 2 = z 1 × z 2
In general, z 1 × z 2 × z 3 ... z n = z 1 × z 2 × z 3 ... z n
æz ö z
(viii) ç 1 ÷ = 1 , z 2 ¹ 0
èz2 ø z2
(ix) z n = (z ) n
(x) z 1 z 2 + z 1 z 2 = 2 Re (z 1 z 2 ) = 2 Re (z 1 z 2 )
(xi) If z = f (z 1 , z 2 ), then z =f (z 1 , z 2 )
x-3 y -3
+
= i , where x , y ÎR and
3+ i
3-i
i = - 1, find the values of x and y.
y Example 18. If
Sol. Q
x -3 y -3
=i
+
3+i
3 -i
Þ
( x - 3) ( 3 - i ) + ( y - 3 ) ( 3 + i ) = i ( 3 + i ) ( 3 - i )
Þ ( 3x - xi - 9 + 3i ) + ( 3y + yi - 9 - 3i ) = 10i
Þ
(3x + 3y - 18) + i (y - x ) = 10i
On comparing real and imaginary parts, we get
3x + 3y - 18 = 0
…(i)
Þ
x +y = 6
and
...(ii)
y - x = 10
On solving Eqs. (i) and (ii), we get
x = - 2, y = 8
prove that (b + ia ) 5 = q + ip .
b
θ a
θ
(ii) z
(iii) z
(iv) z
(v) z
y Example 19. If (a + ib ) 5 = p + iq, where i = - 1,
P(z)
O
Properties of Conjugate
Complex Numbers
Real
axis
Q(z)
Then, z = - a + ib = ( - a, b) ( II quadrant). Now,
(i) If z lies in I quadrant, then z lies in IV quadrant and
vice-versa.
(ii) If z lies in II quadrant, then z lies in III quadrant and
vice-versa.
Sol. Q
\
(a + ib ) 5 = p + iq
(a + ib ) 5 = p + iq Þ (a - ib ) 5 = ( p - iq )
Þ
( - i 2a - ib ) 5 = ( -i 2 p - iq )
Þ
( - i )5 (b + ia ) 5 = ( - i ) (q + ip )
Þ
( - i ) (b + ia ) 5 = ( - i ) (q + ip )
\
(b + ia ) 5 = (q + ip )
[Qi 2 = - 1]
9
Chap 01 Complex Numbers
y Example 20. Find the least positive integral value of
n
æ 1-i ö
n, for which ç
÷ , where i = -1, is purely
è 1+ i ø
imaginary with positive imaginary part.
n
n
n
æ 1 + i 2 - 2i ö æ 1 - 1 - 2i ö
æ1 - i 1 - i ö
æ1 - i ö
Sol. ç
´
÷ =ç
÷
÷ =ç
÷ =ç
ø
2
2
è1+ i 1 -i ø
è1 +i ø
ø è
è
n
= ( - i )n = Imaginary
Þ
n = 1, 3, 5, ... for positive imaginary part n = 3.
y Example 21. If the multiplicative inverse of a
complex number is ( 3 + 4i ) / 19, where i = - 1, find
complex number.
Sol. Let z be the complex number.
æ 3 + 4i ö
Then, z × ç
÷ =1
è 19 ø
z=
or
=
19 ( 3 - 4i )
= ( 3 - 4i )
19
y Example 22. Find real q, such that
Þ
- 3 + ix 2y = x
2
- 3 - ix y = x
3 + 2 i sin q
,
1 - 2 i sin q
(ii) purely imaginary.
3 + 2i sin q
1 - 2i sin q
2
(1 + 4 sin q )
+i
8 sin q
1 + 4 sin 2 q
\
For
y = - 4, 1
y = - 4 , x2 = 1 Þ x = ± 1
For
y = 1, x 2 = - 4
\
x = ± 1, y = - 4
x = - 5 + 2 - 4 Þ x + 5 = 4i
Sol. Since,
Þ
( x + 5)2 = ( 4i )2 Þ x 2 + 10x + 25 = - 16
\
x 2 + 10x + 41 = 0
x
-
4
+ 9 x 3 + 35x
4
3
-x
+
3
(1 + 4 sin q )
2
\
pö
3 æ 3ö
æ
sin q = = ç ÷ = çsin ÷
è
4 è 2 ø
3ø
p
q = np ± , n Î I
3
3
2
-x+4
+ 35x
2
2
2
- 41x
+
+ 40x + 4
2
+ 40x + 164
- 160
-x+4
4x
-
\ x 4 + 9x
-x+4 x
2
- 10x
+
4x
= 0 or sin q = 0
2
…(i)
+ 10x + 41x
- x 3 - 6x 2 - x + 4
2
2
[impossible]
x 4 + 9 x 3 + 35x 2 - x + 4.
(8 sin q )
= ( x 2 + 10x + 41) ( x 2 - x + 4 ) - 160
\
q = n p, n Î I
(ii) For purely imaginary, Re (z ) = 0
(3 - 4 sin 2 q )
= 0 or 3 - 4 sin 2 q = 0
Þ
(1 + 4 sin 2 q)
or
…(ii)
4
From Eq. (ii), we get x 2 = y
é
ù
4
4
Then,
- + y = - 3 ê putting x 2 = - in Eq. (i)ú
y
y
ë
û
(i) For purely real, Im(z ) = 0
Þ
…(i)
-x y=4
and
Now,
x 2 + 10x + 41 x
(3 - 4 sin 2 q )
+ y + 4i
2
(3 + 2i sin q ) (1 + 2i sin q ) (3 - 4 sin 2 q ) + 8i sin q
=
(1 - 2i sin q ) (1 + 2i sin q )
(1 + 4 sin 2 q )
=
+ y + 4i
2
On comparing real and imaginary parts, we get
x 2 +y = -3
On multiplying numerator and denominator by conjugate
of denominator,
z=
2
y Example 24. If x = - 5 + 2 - 4, find the value of
where i = -1, is
Sol. Let z =
Sol. Given,
y 2 + 3y - 4 = 0 Þ (y + 4 ) (y - 1) = 0
19
( 3 - 4i )
´
( 3 + 4i ) ( 3 - 4i )
(i) purely real.
y Example 23. Find real values of x and y for which
the complex numbers - 3 + i x 2 y and x 2 + y + 4i ,
where i = - 1, are conjugate to each other.
= 0 - 160 = - 160
2
[from Eq. (i)]
y Example 25. Let z be a complex number satisfying
the equation z 2 - ( 3 + i ) z + l + 2 i = 0, where l ÎR and
i = -1. Suppose the equation has a real root, find the
non-real root.
Sol. Let a be the real root. Then,
a 2 - (3 + i ) a + l + 2i = 0
10
Textbook of Algebra
Þ
(a 2 - 3 a + l ) + i (2 - a ) = 0
On comparing real and imaginary parts, we get
a 2 - 3a + l = 0
Þ
2-a =0
From Eq. (ii), a = 2
Let other root be b.
Then,
a + b = 3 + i Þ 2+ b = 3 + i
\
b =1+i
Hence, the non-real root is 1 + i.
…(i)
…(ii)
Representation of a Complex
Number in Various Forms
Cartesian Form
(Geometrical Representation)
Every complex number z = x + iy , where x , y Î R and
i = - 1, can be represented by a point in the cartesian
plane known as complex plane (Argand plane) by the
ordered pair ( x , y ).
Modulus and Argument of a
Complex Number
Let
z = x + iy = ( x , y ) for all x , y Î R and i = - 1.
Imaginary axis
P (x, y)
y Example 26. Find the arguments of z 1 = 5 + 5i ,
z 2 = - 4 + 4 i , z 3 = - 3 - 3 i and z 4 = 2 - 2i ,
where i = - 1.
Sol. Since, z 1, z 2 , z 3 and z 4 lies in I, II, III and IV quadrants
respectively. The arguments are given by
5
arg (z1 ) = tan - 1
= tan - 1 1 = p / 4
5
arg (z 2 ) = p - tan - 1
4
p 3p
= p - tan -1 1 = p - =
-4
4
4
arg (z 3 ) = p + tan - 1
-3
p 5p
= p + tan -1 1 = p + =
-3
4
4
and arg (z 4 ) = 2p - tan - 1
-2
2
= 2p - tan -1 1 = 2p -
p 7p
=
4
4
Principal Value of the Argument
The value q of the argument which satisfies the inequality
-p < q £ p is called the principal value of the argument.
If z = x + iy = ( x , y ), " x , y Î R and i = - 1, then
æy ö
arg(z ) = tan - 1 ç ÷ always gives the principal value. It
èxø
depends on the quadrant in which the point ( x , y ) lies.
Y
r
y
Imaginary
axis
P (x, y)
θ
O
Argument of z will be q, p - q , p + q and 2p - q
according as the point z lies in I, II, III and IV
y
.
quadrants respectively, where q = tan - 1
x
x
Real axis
The length OP is called modulus of the complex number z
denoted by z ,
i.e.
OP = r = z = ( x 2 + y 2 )
and if ( x , y ) ¹ (0, 0 ), then q is called the argument or
amplitude of z,
æy ö
i.e. q = tan - 1 ç ÷ [angle made by OP with positive X-axis]
èxø
or arg (z ) = tan - 1 (y / x )
Also, argument of a complex number is not unique, since
if q is a value of the argument, so also is 2np + q, where
n Î I . But usually, we take only that value for which
0 £ q < 2 p. Any two arguments of a complex number differ
by 2np.
y
X′
θ
x
O
X
Real axis
Y′
(i) ( x , y ) Î first quadrant x > 0, y > 0.
æy ö
The principal value of arg (z ) = q = tan - 1 ç ÷
èxø
It is an acute angle and positive.
(ii) ( x , y ) Î second quadrant x < 0, y > 0.
The principal value of arg (z ) = q
æ y ö
÷
= p - tan - 1 çç
÷
è x ø
Chap 01 Complex Numbers
Y
(x, y)
y
p
p
p
p
p 3p
3p
p
or , , ,p- ,-p+ , ,4
4
4
4
4 4
4
4
Hence, the principal values of the arguments of z1, z 2 , z 3
p 3p
3p
p
and z 4 are , , , - , respectively.
4 4
4
4
or
θ
X′
x
X
Real axis
O
Remark
Y′
It is an obtuse angle and positive.
(iii) ( x , y ) Î third quadrant x < 0, y < 0.
æy ö
The principal value of arg (z ) = q = - p + tan -1 ç ÷
èxø
Y
Imaginary
axis
O
x
X′
θ
y
X
Real axis
1. Unless otherwise stated, amp z implies principal value of the
argument.
2. Argument of the complex number 0 is not defined.
3. If z1 = z2 Û z1 = z2 and arg ( z1 ) = arg ( z2 ).
4. If arg ( z ) = p /2 or - p /2, z is purely imaginary.
5. If arg ( z ) = 0 or p, z is purely real.
y Example 28. Find the argument and the principal
value of the argument of the complex number
2+ i
, where i = - 1.
z=
4i + (1 + i ) 2
Sol. Since, z =
(x, y)
Y′
Here, q = tan
The principal value of arg (z ) = q
æ y ö
÷
= - tan - 1 çç
÷
è x ø
2+i
4i + 1 + i 2 + 2i
=
2+i 1 1
= - i
6i
6 3
-1
1
3 = tan - 1 2
1
6
-
Hence, principal value of arg (z ) = - q = - tan -1 2.
Properties of Modulus
x
θ
O
=
\ arg (z ) = 2p - q = 2p - tan -1 2
Y
Imaginary
axis
2+i
4i + ( 1 + i ) 2
\ z lies in IV quadrant.
It is an obtuse angle and negative.
(iv) ( x , y ) Î fourth quadrant x > 0, y < 0.
X′
tan - 1 1, p - tan - 1 1, - p + tan - 1 1, - tan - 1 1
or
Imaginary
axis
11
y
X
Real axis
(i) z ³ 0 Þ z = 0, iff z = 0 and z > 0, iff z ¹ 0
(ii) - z £ Re (z ) £ z and - z £ Im (z ) £ z
(iii) z = z = - z = - z
(iv) zz = z
(x, y)
Y′
It is an acute angle and negative.
y Example 27. Find the principal values of the
arguments of z 1 = 2 + 2i , z 2 = - 3 + 3i , z 3 = - 4 - 4i
and z 4 = 5 - 5i , where i = -1.
Sol. Since, z 1, z 2 , z 3 and z 4 lies in I, II, III and IV quadrants
respectively. The principal values of the arguments are
given by
æ 3 ö
æ- 4ö
æ2ö
tan -1 ç ÷ , p - tan -1 ç
÷ , - p + tan -1 ç
÷,
è2ø
è- 4ø
è -3 ø
æ -5 ö
- tan -1 ç
÷
è 5 ø
2
(v) z 1 z 2 = z 1 z 2
In general, z 1 z 2 z 3 ... z n = z 1 z 2 z 3 ... z n
z1
z
(vi) 1 =
(z 2 ¹ 0 )
z2
z2
(vii) z 1 ± z 2 £ z 1 + z 2
In general, z 1 ± z 2 ± z 3 ± ... ± z n £ z 1 + z 2
+ z 3 + ... + z n
(viii) z 1 ± z 2 ³ z 1 - z 2
(ix) z n = z
(x)
n
z1 - z2 £ z1 + z2 £ z1 + z2
12
Textbook of Algebra
Thus, z 1 | + | z 2 is the greatest possible value of
z 1 + z 2 and | z 1 | - | z 2 | is the least possible value of
z1 + z2 .
(xi) z 1 ± z 2
2
= (z 1 ± z 2 ) (z 1 ± z 2 ) = z 1
2
+ z2
2
± (z 1 z 2 + z 1 z 2 )
or z 1
2
+ z2
2
± 2 Re (z 1 z 2 )
(xii) z 1 z 2 + z 1 z 2 = 2 z 1 z 2 cos(q 1 - q 2 ), where
q 1 = arg (z 1 ) and q 2 = arg (z 2 )
2
2
z
(xiii) | z 1 + z 2 |2 = z 1 + z 2 Û 1 is purely imaginary.
z2
(xiv) z 1 + z 2
2
+ z1 - z2
2
= 2 {z1
2
(xv) az 1 - bz 2 + bz 1 + az 2
2
2
2
+ z2 }
2
2
= (a + b ) ( z 1
2
2
+ z 2 ),
where a, b Î R
(xvi) Unimodular i.e., unit modulus
If z is unimodular, then z = 1. In case of unimodular,
let z = cos q + i sin q, q Î R and i = -1.
Remark
1. If f ( z ) is unimodular, then f ( z ) = 1and let
f ( z ) = cos q + i sin q, q ÎR and i = -1.
z
is always a unimodular complex number, if z ¹ 0.
2.
z
(xvii) The multiplicative inverse of a non-zero complex
number z is same as its reciprocal and is given by
z
1 z
.
=
=
2
z zz
z
y Example 29. If q i Î [0, p / 6], i = 1 , 2, 3, 4, 5 and
sin q 1 z 4 + sin q 2 z 3 + sin q 3 z 2 + sin q 4 z
3
+ sin q 5 = 2, show that < z < 1.
4
Sol. Given that,
sin q 1 z 4 + sin q 2 z 3 + sin q 3 z 2 + sin q 4 z + sin q 5 = 2
or 2 = sin q 1 z 4 + sin q 2 z 3 + sin q 3 z 2 + sin q 4 z + sin q 5
2 £ sin q 1 z
4
+ sin q 2 z
3
+ sin q 3 z
2
z 4 + sin q 2
+ sin q 4
Þ 2 £ sin q 1
z
4
z 3 + sin q 3 z 2
z + sin q 5
+ sin q 2
+ sin q 4 z + sin q 5
But given, q i Î[0, p / 6]
z
3
Þ
3£ z
4
3
+ z
+ z
2
Þ
3£ z + z + z
3
2
3< z + z
Þ
3<
2
+ z
3
+
1
1
z +
2
2
+ z
+ z
+ z
Þ
2
3
+ z
4
< z + z
4
+ z
4
2
+... + ¥
+ ... + ¥
z
[here, | z | < 1]
1- z
Þ
3- 3 z < z Þ 3 < 4 z
3
z >
\
4
3
Hence,
< z <1
4
[Q | z | < 1]
If z - 2 + i £ 2, find the greatest
y Example 30.
and least values of | z | , where i = -1.
Sol. Given that, z - 2 + i £ 2
…(i)
Q
z-2+i ³
z - 2-i
\
z-2+i ³ z - 5
[by property (x)]
…(ii)
From Eqs. (i) and (ii), we get
z - 5 £ z-2+i £2
\
z - 5 £2
Þ
-2£ z - 5 £2
Þ
5 -2£ z £ 5 +2
Hence, greatest value of z is 5 + 2 and least value of z
is 5 - 2.
y Example 31. If z is any complex number such
that z + 4 £ 3, find the greatest value of z + 1 .
+ sin q 4 z + sin q 5 [by property (vii)]
Þ 2 £ sin q 1
é 1ù
sin q i Î ê0, ú ,
ë 2û
1
i.e.
0 £ sinq i £
2
\ Inequality Eq. (i) becomes,
1
1
1
4
3
2£ z + z + z
2
2
2
\
Sol. Q z + 1 = (z + 4 ) - 3
= ( z + 4 ) + ( - 3) £ z + 4 + - 3
[by property (v)]
+ sin q 3 z
= z + 4 +3
2
[by property (ix)] …(i)
£3+3=6
\
z +1 £6
Hence, the greatest value of z + 1 is 6.
[Q z + 4 £ 3 ]
Chap 01 Complex Numbers
y Example 32. If z 1 = 1, z 2 = 2, z 3 = 3 and
9z 1z 2 + 4z 3z 1 + z 2 z 3 = 6, find the value of
z1 +z 2 +z 3 .
Sol. Q
z1 = 1
Þ
z1 z1 = 1
2
Þ z1
=1
z2 = 2 Þ z2
2
z1 z 2 z 3
Þ
z1
z2
z3
Þ
z1 + z 2 + z 3 = 1
=
- z1
2
× z2
çæ z1
è
2
2
2
= 4 + z1
+ 4 z2
2
z2
2
- 4 =0
2
- 4 ÷öø æçè1 - z 2 ö÷ø = 0
z2 ¹ 1
z1
2
[given]
=4
z1 = 2
Hence,
Properties of Arguments
[Q | z | = | z | ]
y Example 33. Prove that
1
1
z 1 + z 2 = (z 1 + z 2 ) + z 1 z 2 + (z 1 + z 2 ) - z 1 z 2 .
2
2
Sol. RHS =
2
+ 4 z2
2
\
é 1
ù
4
9
êQ z = z 1, z = z 2 and z = z 3 ú
2
3
ë 1
û
\
z1
2
But
z 3 + z 2 + z1 = 6
1 × 2 × 3 z1 + z 2 + z 3 = 6
z1
Þ
9
4
1
+
+
=6
z3
z 2 z1
Þ
[by property (iv)]
( z1 - 2z 2 ) ( z1 - 2z 2 ) = ( 2 - z1z 2 ) ( 2 - z1z 2 )
Þ
= 4 Þ z2 z2 = 4
and given 9z 1z 2 + 4z 3z 1 + z 2 z 3 = 6
Þ
2
= 4 - 2z1z 2 - 2z1z 2 + z1z1z 2 z 2
9
z3z3 =9 Þ
=z3
z3
Þ
= 2 - z1z 2
Þ z1z1 - 2z1z 2 - 2z 2 z1 + 4z 2 z 2
4
= z 2 and z 3 = 3 Þ | z 3 | 2 = 9
z2
Þ
2
z1 - 2z 2
Þ (z 1 - 2z 2 ) ( z1 - 2 z 2 ) = ( 2 - z1z 2 ) ( 2 - z1z 2 )
Þ
1
= z1
z1
Þ
Þ
13
(i) arg (z 1 z 2 ) = arg (z 1 ) + arg (z 2 ) + 2kp, k Î I
In general, arg (z 1 z 2 z 3 ... z n )
= arg (z 1 ) + arg (z 2 ) + arg (z 3 ) +... + arg (z n ) + 2kp,
k Î I.
æz ö
(ii) arg ç 1 ÷ = arg (z 1 ) - arg (z 2 ) + 2kp, k Î I
èz2 ø
æz ö
(iii) arg ç ÷ = 2 arg (z ) + 2kp, k Î I
èz ø
1
1
(z1 + z 2 ) + z1z 2 + (z1 + z 2 ) - z1z 2
2
2
(iv) arg (z n ) = n. arg (z ) + 2kp, k Î I , where proper value
of k must be chosen, so that RHS lies in ( -p, p ].
z1 + z 2 + 2 z1z 2
z + z 2 - 2 z1z 2
+ 1
2
2
æz ö
æz ö
(v) If arg ç 2 ÷ = q, then arg ç 1 ÷ = 2np - q, where n Î I .
èz2 ø
èz1 ø
2
2
1
= { z 1 + z2 + z 1 - z2 }
2
2
2
1
[ by property (xiv)]
= . 2 { z1 + z 2 }
2
= z1 + z 2 = LHS
(vi) arg (z ) = - arg (z)
17 p
7p
and arg (z 2 ) = , find
18
18
the principal argument of z 1z 2 and (z 1 / z 2 ).
y Example 35. If arg (z 1 ) =
Sol. arg (z1z 2 ) = arg (z1 ) + arg (z 2 ) + 2kp
y Example 34. z 1 and z 2 are two complex numbers,
z 1 - 2z 2
is unimodular, while z 2 is not
such that
2 - z 1 ×z 2
unimodular. Find | z 1 |.
Sol. Here,
Þ
Þ
z1 - 2z 2
2 - z1z 2
z1 - 2z 2
2 - z1z 2
=1
=1
z1 - 2z 2 = 2 - z1z 2
[by property (vi)]
17 p 7 p
+
+ 2kp
18
18
4p
=
+ 2kp
3
2p
4p
=
- 2p = 3
3
æ z1 ö
and arg ç ÷ = arg (z1 ) - arg (z 2 ) + 2kp
èz2 ø
17 p 7 p
10p
=
+ 2kp =
+ 2kp
18
18
18
5p
5p
=
+0=
9
9
=
[for k = - 1]
[for k = 0]
14
Textbook of Algebra
y Example 36. If z 1 and z 2 are conjugate to each
other, find the principal argument of ( - z 1z 2 ).
Sol. Qz1 and z 2 are conjugate to each other i.e., z 2 = z1, therefore, z1z 2 = z1z1 = z1
2
2
\arg ( - z1 z 2 ) = arg ( - z1 ) = arg [negative real number]
=p
(b) Trigonometric or Polar or
Modulus Argument Form of a
Complex Number
Let z = x + iy , where x , y Î R and i = -1, z is represented
by P ( x , y ) in the argand plane.
Y
Imaginary axis
y Example 37. Let z be any non-zero complex
number, then find the value of arg (z ) + arg (z ).
Sol. arg (z ) + arg (z ) = arg (zz )
2
= arg ( z ) = arg [positive real number]
By geometrical representation,
OP = ( x 2 + y 2 ) = z
and Arguments
(i) z 1 + z 2 = z 1 + z 2 Û arg (z 1 ) = arg (z 2 )
(ii) z 1 + z 2 = z 1 - z 2 Û arg (z 1 ) - arg (z 2 ) = p
Proof (i) Let arg (z 1 ) = q and arg (z 2 ) = f
z1 + z2 = z1 + z2
\
On squaring both sides, we get
Þ
z1
2
2
+ z2
= z1
2
+ z2
2
+ 2 z1 z2
+ 2 z 1 z 2 cos (q - f)
= z1
2
+ z2
2
+ 2 z1 z2
z1 + z2
Þ
z1
2
2
= z1
+ z2
2
+ z2
2
+ 2 z1
1. cos q + i sin q is also written as CiS q.
2. Remember
1 = cos 0 + i sin 0
Þ - 1 = cos p + i sin p
p
p
p
p
Þ - i = cos - i sin
i = cos + i sin
2
2
2
2
= z1
Þ
\
y Example 38. Write the polar form of -
- 2 z1 z2
(where, i = - 1).
z 2 cos (q - f)
2
+ z2
2
- 2 z1
z2
cos (q - f) = - 1
q - f = p or arg (z 1 ) - arg (z 2 ) = p
Remark
1. z1 - z2 = z1 + z2
2. z1 - z2 = z1 - z2
Û arg ( z1 ) = arg ( z2 )
Û arg ( z1 ) - arg ( z2 ) = p
3. z1 - z2 = z1 + z2
Û arg ( z1 ) - arg ( z2 ) = ±
z
and 1 are purely imaginary.
z2
ÐPOM = q = arg (z )
x = OP cos ( Ð POM ) = z cos (arg z )
y = OP sin ( Ð POM ) = z sin (arg z )
z = x + iy
z = z (cos (arg z ) + i sin (arg z ) )
z = r (cos q + i sin q )
z = r (cos q - i sin q )
where, r = z and q = principal value of arg (z ).
In DOPM,
and
Q
\
or
Remark
Þ
cos (q - f) = 1
\
q - f = 0 or q = f
\
arg (z 1 ) = arg (z 2 )
(ii) Q | z 1 + z 2 | = | z 1 | - | z 2 |
On squaring both sides, we get
2
X
M Real axis
x
O
(a) Mixed Properties of Modulus
2
y
θ
=0
z1 + z2
P (x, y)
1 i 3
2
2
æ 1
3ö
1 i 3
. Since, ç - , ÷ lies in III quadrant.
2 ø
2
2
è 2
- 3 /2
\ Principal value of arg (z ) = - p + tan - 1
- 1/2
p
2p
-1
= - p + tan
3 = -p +
=3
3
Sol. Let z = -
2
2
æ
3ö
æ 1ö
æ1 3ö
and z = ç - ÷ + ç ÷ = ç + ÷ = 1 =1
è 2ø
è4 4ø
2
ø
è
p , z1 z2
2
\ Polar form of z = z [cos (arg z ) + i sin(arg z )]
i.e. -
1 i 3 é
æ 2p ö ù
æ 2p ö
= ê cos ç ÷
÷ + i sin ç è 3 ø úû
ø
è
2
2
3
ë
Chap 01 Complex Numbers
(c) Euler’s Form
\
Also,
…(ii)
arg (z ) = q
æ arg (z ) ö
1 + i tan ç
÷
è 2 ø 1 + i tan (q / 2)
[from Eq. (ii)]
RHS =
=
æ arg (z ) ö 1 - i tan (q / 2)
1 - i tan ç
÷
è 2 ø
-i q
=
and if q, f Î R and i = - 1, then
-e
if
=e
1. e iq + 1 = e iq / 2× 2cos (q/ 2)
2. e
3.
e
e
iq
-1= e
iq
-1
iq
+1
iq / 2
× 2i sin (q/ 2)
= i tan ( q / 2)
[from Eq. (i)]
æ æ a - ib ö ö
2ab
y Example 41. Prove that tan ç i ln ç
÷÷ = 2
è è a + ib ø ø a - b 2
( where a, b ÎR + and i = - 1 ).
Sol. Q
æq+ fö
iç
÷
è 2 ø
æ q - fö
× 2i sin ç
÷
è 2 ø
æ q - fö
\
e iq - e if = 2 sin ç
÷
è 2 ø
q+f p
and arg (e iq - e if ) =
+
2
2
Remark
(ii) e
iq
cos q / 2 + i sin q / 2
e i q/ 2
= - i q/ 2
cos q / 2 - i sin q / 2 e
= e i q = z = LHS
æq+ fö
iç
÷
è 2 ø
æ q - fö
× 2 cos ç
÷
è 2 ø
æ q - fö
\
e iq + e if = 2 cos ç
÷
è 2 ø
æ q + fö
and arg (e i q + e if ) = ç
÷
è 2 ø
(i) e iq + e if = e
…(i)
Þ
z = 1 and arg (z ) = q
e +e
= 2 cos q and e i q - e - i q 2i sin q
iq
z =eiq
\
z = ei q
Let
z =1
Sol. Given,
If q Î R and i = - 1, then e i q = cos q + i sin q is known as
Euler’s identity.
Now,
e - i q = cos q - i sin q
15
Let
a - ib
a - ib
=
=1
a + ib
a + ib
a - ib
= e iq
a + ib
…(i)
By componendo and dividendo , we get
( a - ib ) - (a + ib ) e i q - 1 b
=
- i = i tan (q / 2)
(a - ib ) + (a + ib ) e i q + 1 a
[Qi = e ip / 2 ]
or
b
æq ö
tan ç ÷ = è2ø
a
…(ii)
(Remember)
æ
æ a - ib ö ö
LHS = tan çi ln ç
÷÷
è a + ib ø ø
è
(Remember)
= tan (i ln (e i q ))
(Remember)
\
5. If z - z0 = 1, then z - z0 = e iq
=-
y Example 39. Given that z - 1 = 1, where z is a point
z -2
on the argand plane, show that
= i tan (arg z ) ,
z
where i = -1.
\
z - 1 = e i q Þ z = e i q + 1 = e i q / 2 × 2 cos (q / 2)
\
…(i)
arg (z ) = q / 2
z - 2 1 + e iq - 2 e iq - 1
LHS =
=
= iq
= i tan (q / 2)
z
e +1
1 + e iq
[from Eq. (i)]
= i tan (arg z ) = RHS
y Example 40. Let z be a non-real complex number
æ arg (z ) ö
1 + i tan ç
÷
è 2 ø
lying on z = 1, prove that z =
æ arg (z ) ö
1 - i tan ç
÷
è 2 ø
( where, i = - 1 ).
[from Eq. (i)]
= tan (i × i q ) = - tan q
2 tan q / 2
=1 - tan 2 q / 2
4. If z = r e iq ; z = r , then arg ( z ) = q, z = r e- iq
Sol. Given, z - 1 = 1
[Q | z | = | z | ]
=
2 ( - b / a)
[from Eq. (ii)]
1 - ( - b / a )2
2ab
2
a - b2
= RHS
Applications of Euler’s Form
If x , y , q Î R and i = -1, then
let
z = x + iy
= z (cos q + i sin q )
= z e iq
[cartesian form]
[polar form]
[Euler’s form]
(i) Product of Two Complex Numbers
Let two complex numbers be
z 1 = | z 1 | e iq1 and z 2 = | z 2 | e iq2 ,
where q 1 , q 2 Î R and i = - 1
16
Textbook of Algebra
= - w (cos (arg w ) - i sin (arg w ))
= - w (cos ( - arg w ) + i sin ( - arg w ))
= - w (cos (arg w ) + i sin (arg w )) = - w
z 1 × z 2 = z 1 e iq1 × z 2 e iq2 = z 1 z 2 e i ( q1 + q2 )
\
= z 1 z 2 (cos (q 1 + q 2 ) + i sin (q 1 + q 2 ))
Thus,
z1 z2 = z1 z2
and arg (z 1 z 2 ) = q 1 + q 2 = arg (z 1 ) + arg (z 2 )
y Example 44. Express ( 1 + i ) - i , (where, i = -1) in the
form A + iB.
(ii) Division of Two Complex Numbers
Let two complex numbers be
z 1 = z 1 e iq1 and z 2 = z 2 e iq2 ,
Sol. Let A + iB = ( 1 + i )- i
On taking logarithm both sides, we get
loge ( A + iB ) = - i loge (1 + i )
where q 1 , q 2 Î R and i = - 1
i öö
æ æ 1
= - i loge ç 2 ç
+
÷÷
è è 2
2 øø
iq1
z1 e
z 1 i(q - q )
z1
=
=
e 1 2
i
q
2
z2
z2
z2 e
\
=
Thus,
z1
p
p öö
æ æ
= - i loge ç 2 ç cos + i sin ÷ ÷
è è
4
4 øø
(cos (q 1 - q 2 ) + i sin (q 1 - q 2 ))
z2
= - i loge ( 2 e i p / 4 ) = - i (loge 2 + loge e i p / 4)
ipö
i
p
æ1
= - i ç loge 2 + ÷ = - loge 2 +
ø
è2
4
2
4
z1
z1
=
, (z 2 ¹ 0 )
z2
z2
-
\ A + iB = e
æz ö
and arg ç 1 ÷ = q 1 - q 2 = arg (z 1 ) - arg (z 2 )
èz2 ø
=e
p/ 4
=e
p/ 4
(iii) Logarithm of a Complex Number
log e (z ) = log e ( z e
iq
) = log e z + log e (e iq )
\LHS = e 2 m i cot
-1
x
æ xi + 1 ö
2 m i q æ i cot q + 1 ö
÷ =e
÷
ç
ç
è xi - 1 ø
è i cot q - 1 ø
\
æ i (cot q - i ) ö
2 m i q æ cos q - i sin q ö
= e2 m iq ç
÷ =e
÷
ç
è i (cot q + i ) ø
è cos q + i sin q ø
=e
=e
2 mi q
æe - i q ö
× ç iq ÷
èe ø
2 mi q
æ
æ 1 öö
× cos ç loge ç ÷ ÷ + i e
è 2 øø
è
p/ 4
æ
æ 1 öö
sin ç loge ç ÷ ÷
è 2 øø
è
= sin ( i ( loge i + i arg i ))
= sin ( i ( loge 1 + (i p / 2)))
= sin ( i ( 0 + (i p / 2))) = sin ( - p / 2) = - 1
a = -1,b = 0
Aliter
Q
m
× (e - 2 i q )m
× e -2 m i q = e 0 = 1 = RHS
y Example 43. If z and w are two non-zero complex
numbers such that z = w and arg (z ) + arg (w ) = p,
prove that z = - w .
Sol. Let arg (w ) = q , then arg (z ) = p - q
\
× (cos ( loge 2- 1/ 2 ) + i sin ( loge 2- 1/ 2 ))
= 1 - ( - 1)2 = (1 - 1) = 0
m
=e
2 - 1/ 2
Now, cos( loge i i ) = 1 - sin 2 ( loge i i )
m
m
2miq
× e i log e
Sol. a + ib = sin ( loge i i ) = sin ( i loge i )
x = q , then cot q = x
m
p/ 4
find a and b, hence and find cos ( log e i i ).
y Example 42. If m and x are two real numbers and
m
2 m i cot -1 x æ xi + 1 ö
i = - 1 , prove that e
ç
÷ = 1.
è xi - 1 ø
Sol. Let cot
=e
y Example 45. If sin ( log e i i ) = a + ib , where i = - 1,
= log e z + iq = log e z + i arg (z )
So, the general value of log e (z )
= log e (z ) + 2npi ( -p < arg z < p ).
-1
i
p
log e 2 +
2
4
z = z (cos (arg z ) + i sin (arg z ))
= z (cos ( p - q ) + i sin ( p - q ))
= z ( - cos q + i sin q ) = - z (cos q - i sin q )
i i = ( e ip / 2 ) i = e - p / 2
ö
æ p
\ sin (loge i i ) = sin (loge e - p / 2 ) = sin ç - loge e ÷
ø
è 2
[given]
= sin ( - p / 2) = - 1 = a + ib
\
a = - 1, b = 0
and
cos ( loge i i ) = cos ( loge e - p / 2 )
ö
æ p
æ pö
= cos ç - loge e ÷ = cos ç - ÷ = 0
ø
è 2
è 2ø
y Example 46. Find the general value of log 2 ( 5i ),
where i = - 1.
Sol. log 2 5i =
=
loge 5i
1
{ loge 5i + i arg ( 5i ) + 2npi }
=
loge 2 loge 2
1
ip
{ loge 5 +
+ 2npi } , n Î I
loge 2
2
Chap 01 Complex Numbers
#L
1
Exercise for Session 2
If
1 - ix
= a - ib and a 2 + b 2 = 1, where a, b Î R and i = -1, then x is equal to
1 + ix
(a)
2a
2
(1 + a ) + b
(b)
2
2b
2
(1 + a ) + b
(c)
2
2a
2
(1 + b ) + a
2
2b
(d)
(1 + b ) 2 + a 2
n
2
æ1+ i ö
2æ
ö
-1 1
The least positive integer n for which ç
+ sin- 1 x ÷ ( where, x ¹ 0 , - 1 £ x £ 1and i = - 1), is
÷ = çsec
ø
è 1- i ø
x
pè
(a) 2
3
(b) 4
6
(b) 0
1/ 3
If ( x + iy )
If
(b) 4
1
2
(b) 1
(d) 2
(b) 0
3
+ 70x
(d) no value of x
b- a
is equal to
1- ab
(c) 1
(d) 2
2
- 204 x + 225, is
(c) 35
(b) 12
(d) 15
(c) 17
(d) 23
8p ö
8p
æ
The principal value of arg (z ), where z = ç1 + cos
( where, i = - 1) is given by
÷ + i sin
ø
è
5
5
p
5
If z1 = 2,
(b) -
4p
5
z 2 = 3, z 3 = 4 and z1 + z 2 + z
(b) 60
(c)
3
p
5
(b) 9 and 1
4p
5
(d)
= 5, then 4z 2 z 3 + 9z 3 z1 + 16 z1 z 2 is
(c) 120
(d) 240
If z - i £ 5 and z1 = 5 + 3i ( where, i = - 1), the greatest and least values of iz + z
(a) 7 and 3
14
(c) 2
If z1 - 1 £ 1, z 2 - 2 £ 2, z 3 - 3 £ 3, the greatest value of z1 + z 2 + z 3 is
(a) 24
13
1
2
If x = 3 + 4i ( where, i = - 1), the value of x 4 - 12x
(a) -
12
(d) 6
1
(c) x = æçn + ö÷,n ÎI
è
2ø
(b) x = 0
(b)
(a) 6
11
(c) 5
If a and b are two different complex numbers with | b | = 1, then
(a) - 45
10
(d) a 2 + b 2
The complex numbers sin x + i cos 2x and cos x - i sin 2x , where i = - 1, are conjugate to each other, for
(a) 0
9
(c) 4 a 2 - b 2
z -1
If
is purely imaginary, then z is equal to
z +1
(a) x = np,n Î I
8
(d) None of these
3
= a + ib , where i = - 1 and a 2 + b 2 = l a - 3, the value of l is
2 + cos q + i sin q
(a)
7
(c) 6
(b) 4 (a 2 - b 2 )
(a) 3
6
(d) 8
æx yö
= a + ib , where i = - 1, then ç + ÷ is equal to
èa b ø
(a) 4 a 2b 2
5
(c) 6
6
If z = (3 + 4i ) + (3 - 4i ) , where i = - 1, then Im(z ) equals to
(a) - 6
4
17
(c) 10 and 0
1
are
(d) None of these
æz ö
æz ö
If z1, z 2 and z 3, z4 are two pairs of conjugate complex numbers, then arg ç 1 ÷ + arg ç 2 ÷ equals to
è z4 ø
èz3 ø
(a) 0
(b)
p
2
(c) p
(d)
3p
2
Session 3
amp(z) — amp (—z) = ± π; According as amp (z) is Positive or
Negative, Square Root of a Complex Number, Solution of
Complex Equations, De-Moivre’s Theorem, Cube Roots of Unity
amp ( z ) - amp (- z ) = ± p ,
From Eq. (ii), we get
z1 = z 2 (cos ( p + arg (z 2 )) + i sin ( p + arg (z 2 )))
[from Eq. (i) and z1 = z 2 ]
= z 2 ( - cos (arg z 2 ) - i sin (arg z 2 )) = - z 2
[from Eq. (iii)]
\ z1 + z 2 = 0
According as amp ( z) is Positive
or Negative
Case I amp (z ) is positive.
If amp (z ) = q , we have
y Example 48.Let z and w be two non-zero complex
numbers, such that z = w and
amp (z ) + amp (w ) = p, then find the relation between
z and w .
Y
P
r
θ
O
r
P′
z
Sol. Given, amp (z ) + amp (w ) = p
amp (z ) - amp (w ) = p
Þ
Here,
z =w = w
X
– ( π – θ)
–z
If
amp (z ) = - q
We have,
amp ( -z ) = ÐP ¢OX = p - q
[here, OP = OP ¢]
\ amp (z ) - amp ( -z ) = - p
–z
O
Let
z = x + iy ,
where x , y Î R and i = - 1.
Suppose
…(i)
a2 - b2 = x
X
–θ
r
P
y Example 47. If z 1 = z 2 and arg (z 1 / z 2 ) = p, then
find the value of z 1 + z 2 .
…(i)
…(ii)
…(iii)
2
2
a +b =
…(iii)
2
2 2
2 2
(a - b ) + 4a b
=
a2 + b2 = z
2
(x + y 2 )
…(iv)
From Eqs. (ii) and (iv), we get
æ z +xö
÷, b = ±
a = ± çç
÷
è 2 ø
æz ö
Sol. Q arg ç 1 ÷ = p
èz2 ø
arg (z1 ) - arg (z 2 ) = p
z1 = z1 (cos ( arg z1 ) + i sin (arg z1 ))
z 2 = z 2 (cos (arg z 2 ) + i sin (arg z 2 ))
…(ii)
2ab = y
and
z
\
Þ
Q
and
( x + iy ) = a + ib
On comparing the real and imaginary parts, we get
π–θ
r
Square Root of a Complex Number
On squaring both sides, we get
( x + iy ) = (a 2 - b 2 ) + 2iab
Y
P'
amp (z ) > 0
z +w =0
and
Then,
amp ( - z ) = - ( ÐP ¢ OX ) = - ( p - q )
amp (z ) - amp ( -z ) = p
[here, OP = OP ¢]
\
Case II amp (z ) is negative.
[given | z | = | w | ]
or
æ z -xö
÷
ç
ç 2 ÷
ø
è
æ z + Re (z ) ö
÷ , b=±
a = ± çç
÷
2
è
ø
æ z - Re (z ) ö
ç
÷
ç
÷
2
è
ø
Chap 01 Complex Numbers
Now, from Eq. (i), the required square roots,
ì æ z + Re ( z )
z - Re ( z ) ö
ï± ç
÷ , if Im ( z ) > 0
+i
÷
ïï çè
2
2
ø
i.e. z = í
z - Re ( z ) ö
ï æç z + Re ( z )
÷ , if Im ( z ) < 0
-i
ï± ç
÷
2
2
ø
ïî è
Aliter
ìïæ 3 ö 2 æ i ö 2
3
i üï
×
= íç ÷ + ç ÷ + 2 ×
ý
è 2ø
2
2 ïþ
ïîè 2 ø
2
æ3 +i ö
æ3 +iö
= ç
÷
÷ = ±ç
è 2 ø
è 2 ø
(ii) Let z = - 5 + 12i
\
| z | = 13, Re (z ) = - 5, Im (z ) = 12 > 0
If ( x + iy ), where x , y Î R and i = - 1, then
(i) If y is not even, then multiply and divide in y by 2,
then ( x + iy ) convert in
x +y
æ
y 2 ö÷
- 1 = çx + 2 .
ç
4 ÷ø
è
y2
say a , b (a < b).
4
Take that possible factor which satisfy
(ii) Factorise: -
Q
æ
z =±ç
ç
è
2
or a + (ib) + 2iab
Remark
1+ iö
1. The square root of i is ± æç
÷, where i =
è 2 ø
-1.
1- iö
2. The square root of ( - i ) is æç
÷.
è 2 ø
y Example 49. Find the square roots of the following
(i) 4 + 3i
(ii) - 5 + 12i
(iii) - 8 - 15i
(iv) 7 - 24i (where, i = - 1)
Sol. (i) Let z = 4 + 3i
| z | = 5, Re (z ) = 4, Im (z ) = 3 > 0
æ
z =±ç
ç
è
z - Re (z ) ö
÷
÷
2
ø
ö
5
4
æ3 +iö
æ
ö
+i ç
÷÷ =±ç
÷
è 2 øø
è 2 ø
z + Re (z )
2
æ æ5 + 4ö
\ ( 4 + 3i ) = ± ç ç
÷
è 2 ø
è
Aliter
+i
æ 9ö
( 4 + 3i ) = 4 + 3 - 1 = 4 + 2 ç - ÷
è 4ø
=
= ( -5 +2
( - 36 )
= ( -5 +2
( - 9 ´ 4) )
= (- 9 + 4 + 2
and take their square root.
ì± (ai + b)
ì± (b - ia )
(iv) ( x + iy ) = í
and ( x - iy ) = í
±
(
+
i
or
a
b)
î
îor ± (a -ib)
Q
2
9 1
æ 9ö
- + 2 ç- ÷
è 4ø
2 2
+i
z - Re (z ) ö
÷
÷
2
ø
( - 5 + 12i ) = ( - 5 + 12 - 1 )
(iii) Finally, write x + iy = (ai)2 + b2 + 2iab
\
z + Re (z )
æ æ 13 - 5 ö
æ 13 + 5 ö ö
\ ( - 5 + 12i ) = ± ç ç
÷ +i ç
÷÷
è 2 øø
è è 2 ø
= ± (2 + 3i )
Aliter
x = (ai ) 2 + b2 , if x > 0 or x = a 2 + (ib)2 , if x < 0
2
19
( - 9 ´ 4 ))
= (3i ) 2 + 2 2 + 2 × 3i × 2
= ( 2 + 3i ) 2 = ± (2 + 3i )
(iii) Let
z = - 8 - 15i
\ z = 17, Re (z ) = - 8, Im (z ) = - 15 < 0
æ æ 17 - 8 ö
æ 17 + 8 ö ö
\ ( - 8 - 15i ) = ± ç ç
÷ -i ç
÷÷
è 2 øø
è è 2 ø
æ 3 - 5i ö
=±ç
÷
è 2 ø
Aliter ( - 8 - 15i ) = ( - 8 - 15 - 1 )
æ
æ
æ 225 ö ö
æ 25 9 ö ö
= ç- 8 - 2 ç´ ÷÷
÷ ÷ = ç- 8 - 2 çè 4 øø
è 2 2ø ø
è
è
æ 9 25
æ 25 9 ö ö
= ç - 2 ç´ ÷÷
è 2 2ø ø
è2 2
2
2
3 5i
æ 3 ö
æ 5i ö
= ç ÷ + ç ÷ -2×
×
è 2ø
è 2ø
2
2
2
æ 3 - 5i ö
æ 3 - 5i ö
= ç
÷
÷ =±ç
è 2 ø
è 2 ø
20
Textbook of Algebra
(iv) Let z = 7 - 24i
\ z = 25, Re (z ) = 7, Im (z ) = - 24 < 0
æ
Q z =±ç
ç
è
z + Re (z )
-i
2
=
z - Re (z ) ö
÷
÷
2
ø
æ æ 25 + 7 ö
æ 25 - 7 ö ö
\ (7 - 24i ) = ± ç ç
÷ -i ç
÷÷
è 2 øø
è è 2 ø
Aliter
(7 - 24 i ) = (7 - 24 -1 ) = 7 - 2 ( - 144 )
(16 ´ - 9 )
æ æx
=±ç ç
ç è
è
= ( 4 ) 2 + (3i ) 2 - 2 × 4 × 3i
= ( 4 - 3i ) 2 = ± ( 4 - 3i )
- 1).
z = x + (- x 4 - x
Sol. Let
= x + i (x 4 + x
\
z = x
2
+ (x 4 + x
= ( x 4 + 2x
\
2
2
- 1)
2
+ 1)
2
+ 1)
+ 1) = ( x
[Q -1 = i ]
2
Sol. Let
+ 1) 2
and
Re (z ) = x
Im (z ) = ( x + x
æ
z = ± çç
è
Q
\
+ 1) > 0
z + Re (z )
2
4
x + (- x - x
æ æx
=±ç ç
ç è
è
2
2
+i
z - Re (z ) ö
÷
÷
2
ø
æx
ç
è
2
+ 1 - x ö ö÷
÷
2
ø ÷ø
4
2
- 1ö
÷
ø
æ - ( x 2 + x + 1) ( x 2 - x + 1) ö
= x +2 ç
÷
4
è
ø
2
2
2
z = ( x + iy ) = x
2
2
- x + 1 ö ö÷
÷
2
ø ÷ø
2
…(i)
2
- y + 2ixy
2
z = (x + y )
x
æ- x - x
- 1) = x + 2 ç
4
è
éæ x
= x + 2 êç
êë è
æx
ç
è
æx
+ x + 1ö
÷´-ç
2
è
ø
2
2
- y 2 + 2ixy + ( x
2
+ y 2) = 0
On comparing the real and imaginary parts, we get
- x + 1öù
÷ú
2
ø úû
2
- y 2 + (x
2
+ y 2) = 0
and
2xy = 0
From Eq. (iii), let x = 0 and from Eq. (ii),
- y2 + y2 = 0
Þ
2
+ x + 1ö
÷ +i
2
ø
2
z = x + iy , where x , y Î R and i = - 1
x
Aliter
x + (- x 4 - x
2
ü
ï
ïï
ý
ï
ï
ïþ
Then, given equation reduces to
- 1)
+1+ xö
÷ +i
2
ø
- x + 1öù
÷ú
2
ø úû
Putting z = x + iy , where x , y Î R and i = - 1 in the given
equation and equating the real and imaginary parts, we
get x and y, then required solution is z = x + iy .
Þ
2
2
y Example 51. Solve the equation z 2 + z = 0.
z = ( x 2 + 1)
4
æx
+ x + 1ö
÷´-ç
2
ø
è
Solution of Complex Equations
y Example 50. Find the square root of
2
2
- x + 1ö
÷
2
ø
æ æ x 2 + x +1ö
æ x 2 - x +1ö ö
= ç ç
÷÷
÷ +i ç
ç è
2
2
ø ÷ø
è
ø
è
= 7 - 2 (16 ´ - 9 )
x + ( - x4 - x
éæ x
+ 2 êç
êë è
2
2
2
ìæ
2
2
ö æ
ö
ï ç æç x + x + 1 ö÷ ÷ + çi æç x - x + 1 ö÷ ÷
ïï çè è
2
2
ø ÷ø çè è
ø ÷ø
= í
ï
æ x 2 - x + 1ö
æ x 2 + x +1ö
÷
÷ ×i ç
ï+ 2 ç
2
2
ø
è
ø
è
ïî
= ± ( 4 - 3i )
= 16 - 9 - 2
æ x 2 + x + 1ö æ x
÷-ç
ç
2
ø è
è
2
- y
+ y =0
\
y = 0, 1
Þ
y = 0, ± 1
From Eq. (iii), let y = 0 and from Eq. (ii),
2
x
Þ
Þ
x
x
+ x
2
=0
2
+ x =0
2
+ x =0 Þ x =0
\ x + iy are 0 + 0 ×i, 0 + i, 0 - i
i.e. z = 0, i , - i are the solutions of the given equation.
…(ii)
…(iii)
Chap 01 Complex Numbers
y Example 52. Find the number of solutions of the
2
equation z 2 + z = 0.
Sol. Q
Þ
\
and
\
If
and
z2 + z
2
De-Moivre’s Theorem
Statements
= 0 or z 2 + z z = 0
z (z + z ) = 0
z =0
z + z = 0 Þ 2 Re (z ) = 0
Re (z ) = 0
z = x + iy
= 0 + iy , y Î R
i = -1
…(i)
[Q x = Re (z )]
z = (x 2 + y 2 )
(cos q + i sin q ) n = cos nq + i sin nq
and i = - 1, then cos n q + i sin n q is one of the values
Sol. Let z = x + iy , where x , y Î R and i = - 1
and
(ii) If q Î R, n Î I (set of integers) and i = - 1, then
(iii) If q Î R, n Î Q (set of rational numbers)
y Example 53. Find all complex numbers satisfying
2
the equation 2 z + z 2 - 5 + i 3 = 0, where i = - 1.
z 2 = ( x + iy )2 = x 2 - y 2 + 2ixy
(i) If q 1 , q 2 , q 3 , ..., q n Î R and i = - 1, then
(cos q 1 + i sin q 1 ) (cos q 2 + i sin q 2 )
(cos q 3 + i sin q 3 )... (cos q n + i sin q n )
= cos (q 1 + q 2 + q 3 + ...+ q n )
+ i sin (q 1 + q 2 + q 3 + ... + q n )
…(ii)
On combining Eqs. (i) and (ii), then we can say that the
given equation has infinite solutions.
Þ
of (cos q + i sin q ) n .
Proof
(i) By Euler’s formula, e iq = cos q + i sin q
LHS = (cos q 1 + i sin q 1 ) (cos q 2 + i sin q 2 )
(cos q 3 + i sin q 3 ) ... (cos q n + i sin q n )
= e iq1 × e iq2 × e iq3 ... e iqn = e i ( q1 + q2 + q3 + ... + qn )
Then, given equation reduces to
2 ( x 2 + y 2 ) + x 2 - y 2 + 2ixy - 5 + i 3 = 0
Þ
2
= cos (q 1 + q 2 + q 3 + ... + q n )
+ i sin (q 1 + q 2 + q 3 + ... + q n ) = RHS
2
(3x + y - 5) + i (2xy + 3 ) = 0 = 0 + i × 0
On comparing the real and imaginary parts, we get
3x 2 + y 2 - 5 = 0
and
2xy + 3 = 0
…(i)
…(ii)
On substituting the value of x from Eq. (ii) in Eq. (i), we get
2
Þ
æ
3ö
2
3 ç÷ +y -5=0
è 2y ø
9
+ y2 = 5
2
4y
or
4y 4 - 20y 2 + 9 = 0
Þ
2
(ii) If q 1 = q 2 = q 3 = ... = q n = q, then from the above
result (i), (cos q + i sin q ) (cos q + i sin q )
(cos q + i sin q ) ... upto n factors
= cos (q + q + q + ... upton times)
+ i sin (q + q + q + ... upto n times)
i.e., (cos q + i sin q ) n = cos nq + i sin nq
(iii) Let n =
p
, where p, q Î I and q ¹ 0, from above result (ii),
q
q
2
(2y - 9 ) (2y - 1) = 0
9
1
3
y 2 = , y 2 = or y = ±
,y = ±
2
2
2
3 3
1 1
or
y=,
,,
2 2
2 2
From Eq. (ii), we get
1
1
3
3
x=
,, ,2
6
6 2
\
z = x + iy
i
1
3i
1
3i
3
,,
,=
+
2
6
2
6
2 2
are the solutions of the given equation.
\
21
1
2
æ
æp ö
æ p öö
we have ç cos ç q ÷ + i sin ç q ÷ ÷
èq ø
è q øø
è
ææ p ö ö
ææ p ö ö
= cos ç ç q ÷ q ÷ + i sin ç ç q ÷ q ÷ = cos p q + i sin p q
èè q ø ø
èè q ø ø
æ pq ö
æ pq ö
Þ cos ç ÷ + i sin ç ÷ is one of the values of
è q ø
è q ø
(cos p q + i sin p q ) 1 /q
i
3
+
2
2
æ pq ö
æ pq ö
Þ cos ç ÷ + i sin ç ÷ is one of the values of
è q ø
è q ø
[(cos q + i sin q ) p ]1 /q
22
Textbook of Algebra
æ pq ö
æ pq ö
Þ cos ç ÷ + i sin ç ÷ is one of the values of
è q ø
è q ø
(cos q + i sin q ) p
/q
Other Forms of De-Moivre’s Theorem
1. (cos q - i sin q) n = cos n q - i sin n q, " n Î I
Proof (cos q - i sin q) n = (cos ( - q) + i sin ( - q)) n
= cos ( - nq) + i sin ( - nq) = cos nq - i sin nq
2. (sin q + i cos q) n = ( i ) n (cos nq - i sin n q), " n Î I
Proof (sin q + i cos q) n = ( i (cos q - i sin q)) n
= i n (cos q - i sin q) n = ( i ) n (cos nq - i sin n q)
[from remark (1)]
3. (sin q - i cos q) n = ( - i ) n (cos nq + i sin nq), " n Î I
Proof (sin q - i cos q) n = ( - i (cos q + i sin q)) n
= ( - i ) n (cos q + i sin q) n
= ( - i ) n (cos nq + i sin nq)
n
4. (cos q + i sin f) ¹ cos nq + i sin nf, " n Î I
[here, q ¹ f\ De-Moivre’s theorem is not applicable]
1
5.
= (cos q + i sin q) - 1
cos q + i sin q
= cos ( - q) + i sin ( - q) = cos q - i sin q
p
p
y Example 54. If z r = cos æç r ö÷ + i sin æç r ö÷ , where
è3 ø
è3 ø
i = - 1, prove that z 1 z 2 z 3 ... upto infinity = i.
æpö
Sol. We have, zr = cos ç r ÷ + i sin
è3 ø
æpö
ç r÷
è3 ø
p
æp p
ö
\ z1 z 2 z 3 ... ¥ = cos ç + 2 + 3 + ... + ¥ ÷
è3 3
ø
3
p
æp p
ö
+ i sin ç + 2 + 3 +... + ¥ ÷
è3 3
ø
3
æ p ö
æ p ö
÷
ç
÷
ç
æpö
æpö
3
= cos ç
÷ + i sin ç 3 ÷ = cos ç ÷ + i sin ç ÷
1
1
è2ø
ø
è
2
ç1 - ÷
ç1 - ÷
è
è
3ø
3ø
= 0 + i ×1 = i
y Example 55. Express
(cos q + i sin q )
(sin q + i cos q ) 5
= i (cos q + i sin q )- 5
(sin q + i cos q ) 5
=
(cos q + i sin q )4
i (cos q + i sin q )- 5
(cos q + i sin q )9
i
cos 9 q + i sin 9 q
= - i cos 9 q + sin 9 q
=
i
= sin 9 q - i cos 9 q
=
[polar form]
+ i sin (2np + q) )} p
=r
p /q
(cos (2np + q ) + i sin (2np + q ))
/q
,n ÎI
p /q
æ
öö
æp
ö
æp
ç cos ç (2np + q )÷ + i sin ç (2np + q )÷ ÷ ,
øø
èq
ø
èq
è
where, n = 0, 1, 2, 3, ..., q - 1
=rp
/q
y Example 56. Find all roots of x 5 - 1 = 0.
Sol. Q x 5 - 1 = 0 Þ x 5 = 1
\
x = (1)1 / 5 = (cos 0 + i sin 0)1 / 5 ,
where i = -1
= [cos (2np + 0) + i sin (2np + 0)]1/ 5
æ 2np ö
æ 2n p ö
= cos ç
÷ + i sin ç
÷,
è 5 ø
è 5 ø
where, n = 0, 1, 2, 3, 4
\ Roots are
æ 4p ö
æ 4p ö
æ 2p ö
æ 2p ö
1, cos ç ÷ + i sin ç ÷, cos ç ÷ + i sin ç ÷,
è 5 ø
è 5 ø
è 5 ø
è 5 ø
æ 8p ö
æ 8p ö
æ 6p ö
æ 6p ö
cos ç ÷ + i sin ç ÷, cos ç ÷ + i sin ç ÷
è 5 ø
è 5 ø
è 5 ø
è 5 ø
æ 6p ö
æ 6p ö
Now, cos ç ÷ + i sin ç ÷
è 5 ø
è 5 ø
4p ö
4p ö
æ
æ
= cos ç2p ÷
÷ + i sin ç2p è
ø
è
5 ø
5
æ 4p ö
æ 4p ö
= cos ç ÷ - i sin ç ÷
è 5 ø
è 5 ø
æ 8p ö
and cos ç ÷ + i sin
è 5 ø
æ 8p ö
ç ÷
è 5 ø
2p ö
2p ö
æ
æ
= cos ç2p ÷
÷ + i sin ç2p è
è
5 ø
5 ø
in a + ib
Sol. Q (sin q + i cos q )5 = (i )5 (cos q - i sin q ) 5
(cos q + i sin q )4
Let
a + ib = r (cos q + i sin q )
\ (a + ib )p / q = {r (cos (2np + q )
4
form, where i = - 1.
\
To Find the Roots of (a + ib ) p /q , where a , b Î R ;
p , q Î I , q ¹ 0 and i = - 1
æ 2p ö
æ 2p ö
= cos ç ÷ - i sin ç ÷
è 5 ø
è 5 ø
æ 2p ö
æ 2p ö
Hence, roots are 1, cos ç ÷ ± i sin ç ÷
è 5 ø
è 5 ø
æ 4p ö
æ 4p ö
and cos ç ÷ ± i sin ç ÷.
è 5 ø
è 5 ø
Remark
Five roots are 1, z1, z2, z1, z2 (one real, two complex and two
conjugate of complex roots).
Chap 01 Complex Numbers
y Example 57. Find all roots of the equation
x 6 - x 5 + x 4 - x 3 + x 2 - x + 1 = 0.
Sol. Q
1- x + x2 - x3 + x4 - x5 + x6 =0
[1 - ( - x )7 ]
= 0, 1 + x ¹ 0
1 - (- x )
1 + x 7 = 0, x ¹ - 1 or x 7 = - 1
Þ
1×
or
x = ( - 1)1 / 7 = (cos p + i sin p)1 / 7 , i = -1
\
= [cos (2n + 1) p + i sin (2n + 1) p ]1 / 7
æ (2n + 1)p ö
æ (2n + 1)p ö
= cos ç
÷
÷ + i sin ç
ø
è
ø
è
7
7
Cube Roots of Unity
3
Let z = (1)
Þ z =1 Þ z -1=0
2
Þ (z - 1) (z + z + 1) = 0 Þ z - 1 = 0 or z 2 + z + 1 = 0
- 1 ± (1 - 4 )
-1 ± i 3
2
2
- 1 + i 3 - 1 -i 3
Therefore, z = 1 ,
, where i = -1.
,
2
2
If second root is represented by w (omega), third root will
be w2 .
\
z = 1 or z =
=
\ Cube roots of unity are 1, w, w2 and w, w2 are called
non-real complex cube roots of unity.
Remark
1. w = w2, ( w) 2 = w
3. w = w
2
2. w = ± w2, w2 = ± w
=1
Aliter
Let
z = (1) 1 / 3 = (cos 0 + i sin 0 ) 1 / 3 , i = - 1
= [cos (2 np + 0 ) + i sin (2 np + 0 )]1 / 3
æ 2 np ö
æ 2 np ö
= cos ç
÷ + i sin ç
÷ , where, n = 0, 1, 2
è 3 ø
è 3 ø
Therefore, roots are
æ 2p ö
æ 2p ö
æ 4p ö
æ 4p ö
1, cos ç ÷ + i sin ç ÷ , cos ç ÷ + i sin ç ÷
è 3 ø
è 3 ø
è 3 ø
è 3 ø
or
1, e
2pi / 3
,e
(i) 1 + w + w2 = 0 and w 3 = 1
(ii) To find the value of wn (n > 3 ).
First divide n by 3. Let q be the quotient and r be the
remainder.
3 ) n (q
- 3q
r
i.e.
n = 3q + r , where 0 £ r £ 2
wn = w 3q + r = ( w 3 ) q × wr = wr
\
ì3, when n is a multiple of 3
(iii) 1 + wr + w2r = í
î0, when n is not a multiple of 3
(iv) Cube roots of - 1 are - 1, - w and - w2 .
Remark
QFor n = 3, x = - 1 but here x ¹ - 1
\
n¹ 3
3
Properties of Cube Roots of Unity
In general, w 3n = 1 , w 3n + 1 = w , w 3n + 2 = w 2
for n = 0, 1, 2, 4, 5, 6.
1/ 3
23
4pi / 3
If second root is represented by w, then third root will be w2
or if third root is represented by w, then second root will be w2 .
(v) a + b w + c w2 = 0 Þ a = b = c , if a, b, c Î R.
(vi) If a, b, c are non-zero numbers such that
a + b + c = 0 = a 2 + b 2 + c 2 , then a : b : c = 1 : w : w2 .
(vii) A complex number a + ib (where i = -1), for which
a : b = 1 : 3 or 3 : 1 can always be expressed in
terms of w or w2 .
For example,
(a) 1 + i 3 = - ( - 1 - i 3 )
[Q |1 : 3 | = 1 : 3 ]
æ-1-i 3ö
2
= -2 ç
÷ = -2w
2
ø
è
i ( 3 + i ) (- 1 + i 3 )
=
i
i
æ - 1 + i 3 ö æ2ö
=ç
[Q | 3 : 1| = 3 : 1]
÷ç ÷
2
ø èiø
è
2w
=
= -2i w
i
(viii) The cube roots of unity when represented on complex
plane lie on vertices of an equilateral triangle
inscribed in a unit circle, having centre at origin. One
vertex being on positive real axis.
(b)
3 +i =
Y
i
2π/3
X′
2π/3
ω
–1
1
O
2π/3
ω2
–i
Y′
X
24
Textbook of Algebra
Important Relations in Terms
of Cube Root of Unity
(i) a 2
(ii) a 2
(iii) a 3
(iv) a 3
(v) a 2
+ ab + b 2 = (a - bw ) (a - bw2 )
- ab + b 2 = (a + bw ) (a + bw2 )
+ b 3 = (a + b ) (a + bw ) (a + bw2 )
- b 3 = (a - b ) (a - bw ) (a - bw2 )
+ b 2 + c 2 - ab - bc - ca
= (a + bw + cw2 ) (a + bw2 + cw )
(vi) a 3 + b 3 + c 3 - 3abc
= (a + b + c ) (a + bw + cw2 ) (a + bw2 + cw )
y Example 58. If w is a non-real complex cube root of
unity, find the values of the following.
y Example 59. If a , b and g are the roots of
x 3 - 3x 2 + 3x + 7 = 0, find the value of
a - 1 b - 1 g -1
+
+
.
b -1 g -1 a -1
Þ
( x - 1 )3 + 8 = 0
Þ
( x - 1) 3 + 2 3 = 0
Þ ( x - 1 + 2) ( x - 1 + 2 w) ( x - 1 + 2 w2 ) = 0
( x + 1) ( x - 1 + 2 w) ( x - 1 + 2 w2 ) = 0
Þ
x = - 1, 1 - 2 w, 1 - 2 w2
\
a = - 1, b = 1 - 2 w, g = 1 - 2 w2
Þ
1999
(i) w
(ii) w
Then,
- 998
æ - 1+ i
(iii) ç
2
è
3ö
÷
ø
3n + 2
=
æ a + bw + gw 2 + dw 2 ö
÷ , where a, b, g , d ÎR
(v) ç
2
è b + aw + gw + dw ø
(vi) 1 × (2 - w) (2 - w 2 ) + 2 × ( 3 - w) ( 3 - w 2 ) + 3 ×
3 ´ 666 + 1
=w
(ii) w- 998 =
1
value of (z 101 + i 103 )105 .
Sol. Q
w998
æ - 1+i 3 ö
(iii) ç
÷
2
ø
è
=
w999
z=
3 +i 1
=
2
i
[Qi 2 = - 1]
Then, z 101 + i 103 = - i w2 - i = - i ( w2 + 1)
3n + 2
= w3n + 2 = w3n × w2 = ( w3 )n × w2
(iv) (1 + w) (1 + w2 ) (1 + w4 ) (1 + w8 ) ... upto 2n factors
= (1 + w) (1 + w2 ) (1 + w) (1 + w2 ) ... upto 2n factors
= ( - w2 ) ( - w) ( - w2 ) (- w) ... upto 2n factors
= ( w3 ) ( w3 )... upto n factors = 1 × 1× 1 × ...upto n factors
= (1)n = 1
= - i ( - w) = i w
Hence, (z 101 + i 103 )105 = (iw)105 = i 105 × w105 = i × 1 = i
æ3 i 3ö
y Example 61. If ç +
÷
2 ø
è2
w (a + b w + g w2 + d w2 )
( b w + a + g w2 + d w2 )
Sol. Q
S (n - 1) (n - w) (n - w2 ) = S (n 3
2
ì n ( n + 1) ü
=í
ý -n
2
þ
î
- 1) =
S n3 - S 1
= 3 25 ( x - iy ), where
æ 3 +iö
3 i 3
3
+
= 3ç
÷ =
2
2
i
è 2 ø
æi 3 + i 2 ö
÷
ç
2
ø
è
æ- 1 + i 3ö
= -i 3 ç
÷ = -i 3 w
2
ø
è
æ3 i 3 ö
\ç +
÷
2 ø
è2
=w
50
x , y ÎR and i = - 1, find the ordered pair of ( x , y ).
æ a + b w + g w2 + d w2 ö w (a + b w + g w2 + d w2 )
(v) ç
÷=
è b + a w2 + g w + d w ø ( b w + a w3 + g w2 + d w2 )
(vi)
æi 3 + i 2 ö
÷
ç
2
ø
è
\ z 101 = ( - iw)101 = - i 101 × w101 = - i w2 and i 103 = i 3 = - i
=w
= (1)n × w2 = w2
=
3+i
, where i = - 1, find the
2
æ- 1 + i 3ö
= -i ç
÷ = -iw
2
ø
è
=w
w
1 1
+ + w2 = w2 + w2 + w2 = 3 w2
w w
y Example 60. If z =
(4 - w)(4 - w 2 ) + K + K + (n - 1) ×(n - w)(n - w 2 )
Sol. (i) w
a - 1 b -1 g - 1
-2
- 2 w - 2 w2
+
+
+
=
+
b - 1 g - 1 a - 1 - 2 w - 2 w2
-2
, n Î N and i = - 1
(iv) (1 + w) (1 + w 2 ) (1 + w 4 ) (1 + w 8 )... upto 2n factors
1999
x 3 - 3x 2 + 3x + 7 = 0
Sol. We have,
50
= ( - i 3 w) 50 = i 50 × 325 × w50
æ- 1 - i 3ö
= - 1 × 325 × w2 = - 325 × ç
÷
2
ø
è
25
Chap 01 Complex Numbers
æ1 i 3 ö
25
= 325 ç +
÷ = 3 ( x - iy )
2
2
ø
è
1
x = ,y = 2
æ1
Þ Ordered pair is ç , è2
\
[given]
3
2
3ö
÷.
2 ø
f (x ) = 7x
3
+ ax + b
or
- 14 + a + 2b = 0 or a + 2b = 14
and on subtracting, we get
- a ( w - w2 ) = 0
Þ
Q f ( x ) is divisible by x - x + 1
#L
Exercise for Session 3
1
The real part of (1 - i )- i , where i = - 1 is
1
(b) - e - p / 4 sin æç loge 2ö÷
è2
ø
1
(a) e - p / 4 cos æç loge 2ö÷
è2
ø
2
1
cos æç loge 2ö÷
è2
ø
1
(d) e - p / 4 sin æç loge 2ö÷
è2
ø
- iq
The amplitude of e e , where q ÎR and i = - 1 is
(b) - sinq
(d) e sinq
(a) sinq
(c) e cos q
3
If z = i loge (2 - 3 ), where i = - 1, then the cos z is equal to
(a) i
4
(b) 2i
5
(b) e - p / 2
3
i
2
(b) ± (1 - 3i )
(c) ± (1 + 3 i )
(d) ± (3 - i )
3
(b) i
4
(c) -
3
i
4
(d) -
3
2
If 0 < amp (z ) < p, then amp (z ) - amp ( - z ) is equal to
(b) 2 amp (z )
(c) p
(d) - p
(c) z1 + z 2 = 0
(d) z1 = z 2
If z1 = z 2 and amp (z1) + amp (z 2 ) = 0, then
(a) z1 = z 2
9
(d) e p
(5 + 12i ) + (5 - 12i )
is equal to (where, i = - 1)
(5 + 12i ) - (5 - 12i )
(a) 0
8
(c) e - p
( - 8 - 6i ) is equal to (where, i = -1)
(a) -
7
(d) 2
If z = i i , where i = - 1, then z is equal to
(a) 1 ± 3i
6
(c) 1
i
(a) 1
(b) z1 = z 2
The solution of the equation z - z = 1 + 2i , where i = - 1, is
(a) 2 -
3
i
2
(b)
3
+ 2i
2
(c)
3
- 2i
2
…(i)
[Q w - w2 ¹ 0]
a=0
From Eq. (i), we get b = 7
\
2a + b = 7
2
p/4
- 7 - aw + b = 0
- 7 - aw2 + b = 0
On adding, we get
- 14 - a ( w + w2 ) + 2b = 0
and x 2 - x + 1 = ( x + w) ( x + w2 )
(c) e
- 7 w3 - aw + b = 0 and - 7 w6 - aw2 + b = 0
Þ
or
and
y Example 62. If the polynomial 7 x 3 + ax + b is
divisible by x 2 - x + 1, find the value of 2a + b .
Sol. Let
f ( - w) = 0 and f ( - w2 ) = 0
Then,
(d) - 2 +
3
i
2
26
Textbook of Algebra
10
The number of solutions of the equation z 2 + z = 0, is
(a) 1
(c) 3
11
(b) 2
(d) 4
æ ra ö
æ ra ö
If z r = cos ç 2 ÷ + i sin ç 2 ÷, where r = 1, 2, 3, ..., n and i = - 1, then lim z1z 2 z 3 K z n is equal to
èn ø
èn ø
n®¥
(b) e - ia / 2
(a) e ia
(c) e
ia / 2
(d) 3 e ia
n
12
13
æ 1 + sin q + i cos q ö
If q ÎR and i = - 1, then ç
÷ is equal to
è 1 + sin q - i cos q ø
np
np
(a) cos æç
- nqö÷ + i sin æç
- nqö÷
è 2
ø
è 2
ø
np
np
(b) cos æç
+ nqö÷ + i sin æç
+ nqö÷
è 2
ø
è 2
ø
np
np
(c) sin æç
- nqö÷
- nqö÷ + i cos æç
è 2
ø
è 2
ø
æ p
ö
æ p
ö
(d) cos çn æç + 2qö÷ ÷ + i sinçn æç + 2 qö÷ ÷
øø
øø
è è2
è è2
If i z 4 + 1 = 0, where i = - 1, then z can take the value
1+ i
2
1
(c)
4i
(a)
14
(d) i
If w ( ¹ 1) is a cube root of unity, then (1 - w + w2 ) (1 - w2 + w4 ) (1 - w4 + w8 ) ... upto 2n factors, is
(a) 2n
(c) 0
15
p
p
(b) cos æç ö÷ + i sin æç ö÷
è 8ø
è 8ø
(b) 22 n
(d) 1
If a, b and g are the cube roots of p ( p < 0), then for any x , y and z ,
1
(- 1 -i 3 ), i = -1
2
1
(c) (1 - i 3 ), i = -1
2
(a)
(b)
1
(1 + i
2
x a+ yb+z g
is equal to
xb+ y g +z a
3 ), i =
(d) None of these
-1
Session 4
nth Root of Unity, Vector Representation of Complex Numbers,
Geometrical Representation of Algebraic Operation on Complex Numbers,
Rotation Theorem (Coni Method), Shifting the Origin in Case of Complex
Numbers, Inverse Points, Dot and Cross Product, Use of Complex Numbers
in Coordinate Geometry
nth Root of Unity
Remark
1 × a × a2 × a3 ... an - 1 = ( - 1) n - 1 is the basic concept to be
understood.
Let x be the nth root of unity, then
x = (1) 1 /n = (cos 0 + i sin 0 ) 1 /n
= (cos (2kp + 0 ) + i sin (2kp + 0 ) 1 /n
[where k is an integer]
1 /n
= (cos 2kp + i sin 2kp)
æ 2kp ö
æ 2kp ö
\
x = cos ç
÷ + i sin ç
÷
è n ø
è n ø
where,
k = 0, 1, 2, 3, ..., n - 1
2p
2p
Let a = cos
+ i sin , the n, nth roots of unity are
n
n
a k (k = 0, 1, 2, 3, ... , n - 1) i.e, the n, nth roots of unity are
(c) If a is an imaginary nth root of unity, then other roots
are given by a 2 , a 3 , a 4 , ... , a n .
Imaginary
axis
Y
i
A2(α2)
r
θ
X′
–1
A1(α)
θ
θ
1, a, a 2 , a 3 , ... , a n - 1 which are in GP with common ratio
=
1
X Real axis
An-1(αn-1)
= e 2pi / n .
(a) Sum of n, nth roots of unity
1 + a + a 2 + a 3 + ... + a n - 1 =
A0
–i
Y′
1 × (1 - a n )
(1 - a )
(d) Q 1 + a + a 2 + ... + a n - 1 = 0
n -1
S ak = 0
Þ
1 - (cos 2 p + i sin 2 p)
1-a
k=0
n -1
n -1
æ 2 pk ö
æ 2 pk ö
or S cos ç
÷ + i S sin ç
÷ =0
k=0
k
=
0
è n ø
è n ø
1 - (1 + 0 )
=
=0
1-a
n -1
Remark
1 + a + a2 + a3 + ... + an - 1 = 0 is the basic concept to be
understood.
1 ´ a ´ a 2 ´ a 3 ´ ... ´ a n - 1 = a 1 + 2 + 3 + ... + (n - 1 )
2p
2p ö
æ
= ç cos
+ i sin ÷
è
n
n ø
= cos (n - 1) p + i sin (n - 1) p
= (cos p + i sin p)
n -1
= ( - 1)
n -1
(n - 1 ) n
2
æ 2 pk ö
cos ç
S
÷ =0
k=0
è
ø
n
n -1
and
(b) Product of n , nth roots of unity
(n - 1 ) n
=a 2
Þ
S sin æçè 2pk ö÷ø = 0
k=0
n
These roots are located at the vertices of a regular
plane polygon of n sides inscribed in a unit circle
having centre at origin, one vertex being on positive
real axis.
(e) x n - 1 = ( x - 1) ( x - a ) ( x - a 2 ) ... ( x - a n - 1 ).
28
Textbook of Algebra
Important Benefits
niq
ü
ì
iq
2 × 2i sin æ nq ö ï
ï
e
e
×
ç
÷
ìe iq { (e iq ) n - 1} ü
è 2 øï
ï
= Im í
ý = Im í
ý
iq
e -1
þ
î
ï e iq /2 × 2i sin æç q ö÷ ï
è2ø ï
ï
î
þ
ì æ nq ö
ü
æ nq ö
ïsin çè 2 ÷ø æç n + 1 ö÷iq ï sin çè 2 ÷ø
éæ n + 1ö ù
ï
ï
è
ø
× sin ê ç
×e 2
= Im í
÷ qú
ý=
ëè 2 ø û
ï sin æç q ö÷
ï sin æç q ö÷
è2ø
ïî
è2ø
þï
1. If 1, a1, a2, a3, ... , an - 1 are the n, nth root of unity, then
( 1) p + ( a1 ) p + ( a2 ) p + ... + ( an - 1 ) p
ì0, if p is not an integral multiple of n
=í
în, if p is an integral multiple of n
ì0 , if nis even
2. ( 1 + a1 ) ( 1 + a2 ) ... ( 1 + an - 1 ) = í
î1, if nis odd
3. ( 1 - a1 ) ( 1 - a2 ) ... ( 1 - an - 1 ) = n
4. z n - 1 = ( z - 1) ( z + 1)
( n - 2) / 2
P
r =1
æ z 2 - 2z cos 2rp + 1ö,
ç
÷
è
ø
n
if ‘n’ is even.
5. z n + 1 =
( n - 2) / 2
P
r =0
6. z n + 1 = ( z + 1)
æ 2
ç z - 2z cos
è
( n - 3) / 2
P
r =0
æ ( 2r + 1)
ç
è
n
pö
ö
÷ + 1÷, if n is even.
ø
ø
æ 2
æ ( 2r + 1) p ö + 1ö,
÷
ç z - 2z cos ç
÷
è
ø
è
ø
n
if ‘n’ is odd.
The Sum of the Following
Series Should be Remembered
(i) cos q + cos 2 q + cos 3 q + ... + cos nq
æ nq ö
sin ç ÷
è 2 ø
éæ n + 1ö ù
. cos ê ç
=
÷ qú
æqö
ëè 2 ø û
sin ç ÷
è2ø
(ii) sin q + sin 2 q + sin 3 q + ... + sin nq
æ nq ö
sin ç ÷
è 2 ø
éæ n + 1ö ù
=
× sin ê ç
÷ qú
æqö
ëè 2 ø û
sin ç ÷
è2ø
Proof
(i) cos q + cos 2 q + cos 3 q + ... + cos n q
= Re {e iq + e 2iq + e 3i q + ... + e niq }, where i = -1
ì iq niq /2
æ nq ö ü
× 2i sin ç ÷ ï
e ×e
ï
ìe { (e ) - 1} ü
è 2 øï
ï
= Re í
ý = Re í
ý
iq
iq /2
e -1
× 2i sin (q /2 ) ï
þ
î
ï e
ïî
ïþ
ì æ nq ö
ü
æ nq ö
ïsin çè 2 ÷ø æç n + 1 ö÷iq ï sin çè 2 ÷ø
éæ n + 1ö ù
ï
è
ø ï
× cos ê ç
×e 2
= Re í
÷ qú
ý=
q
q
ëè 2 ø û
ï sin æç ö÷
ï sin æç ö÷
è2ø
ïî
è2ø
ïþ
iq
iq n
(ii) sin q + sin 2 q + sin 3 q + ... + sin nq
= Im {e iq + e 2iq + e 3iq + ... + e niq }, where i = -1
Remark
2p
, we get
n
2p
4p
6p
( 2n - 2) p ö
1. 1 + cos æç ö÷ + cos æç ö÷ + cos æç ö÷ + ... + cos æç
÷ =0
è nø
è nø
è nø
è
ø
n
2p ö
4p ö
6p ö
( 2n - 2) p ö
æ
æ
æ
æ
2. sin ç ÷ + sin ç ÷ + sin ç ÷ + ... + sin ç
÷ =0
è nø
è nø
è nø
è
ø
n
For q =
y Example 63. If 1, w , w 2 , ..., wn - 1 are n, nth roots of
unity, find the value of (9 - w) (9 - w 2 ) ...(9 - wn -1 ).
x = (1)1/n
Sol. Let
Þ xn - 1 = 0
has n roots 1, w, w2 , ... , wn - 1
x n - 1 = ( x - 1) ( x - w) ( x - w2 ) K ( x - wn - 1 )
\
On putting x = 9 in both sides, we get
9n - 1
= (9 - w) (9 - w2 ) (9 - w3 ) ... (9 - wn - 1 )
9 -1
or
(9 - w) (9 - w2 ) ... (9 - wn - 1 ) =
9n - 1
8
Remark
x n -1
= ( x - w) ( x - w2 ) ... ( x - wn - 1 )
x -1
\ lim
x ®1
xn -1
= lim ( x - w) ( x - w2 ) ... ( x - wn - 1 )
x -1 x ®1
n = ( 1 - w) ( 1 - w2 ) ... ( 1 - wn - 1 )
Þ
æ 2p ö
æ 2p ö
If a = cos ç ÷ + i sin ç ÷ , where
è 7 ø
è 7 ø
y Example 64.
i = -1, find the quadratic equation whose roots
are a = a + a 2 + a 4 and b = a 3 + a 5 + a 6 .
Sol. Q
æ 2p ö
æ 2p ö
a = cos ç ÷ + i sin ç ÷
è7 ø
è7 ø
\ a 7 = cos 2p + i sin 2p = 1 + 0 = 1
or
a = (1)1 / 7
\ 1 , a , a 2 , a 3 , a 4 , a 5 , a 6 are 7, 7 th roots of unity.
\ 1 + a + a2 + a3 + a4 + a5 + a6 = 0
2
4
3
5
6
Þ (a + a + a ) + (a + a + a ) = - 1 or a + b = - 1
…(i)
29
Chap 01 Complex Numbers
and ab = (a + a 2 + a 4 ) (a 3 + a 5 + a 6 )
4
6
7
5
7
8
y Example 67. If n ³ 3 and 1, a 1 , a 2 , a 3 , ..., a n -1 are
7
9
10
= a + a + a +a + a + a + a + a + a
= a4 + a6 + 1 + a5 + 1 + a + 1 + a2 + a3
[Qa 7 = 1]
= (1 + a + a 2 + a 3 + a 4 + a 5 + a 6 ) + 2
[from Eq. (i)]
=0+2
=2
Therefore, the required equation is
x 2 - (a + b ) x + ab = 0 or x 2 + x + 2 = 0
y Example 65. Find the value of
10
é æ 2pk ö
æ 2pk ö ù
sin ç
S
÷ , where i = - 1.
÷ - i cos ç
ê
è 11 ø úû
è 11 ø
k =1 ë
11
x n = 1 or x n - 1 = 0
\
\
1 + a 1 + a 2 + a 3 + ... + a n - 1 = 0
or
a 1 + a 2 + a 3 + ... + a n - 1 = - 1
On squaring both sides, we get
2
+ ... + a 1a n - 1 + a 2 a 3 + ... + a 2 a n - 1
+ ... + a n - 2 a n - 1 ) = 1
or 12 + (a 1 )2 + (a 2 )2 + (a 3 )2 + ... + (a n - 1 )2
+2
é
æ 2p k ö
æ 2p k ö ù
ê cos çè 11 ÷ø + i sin çè 11 ÷ø ú
ë
û
S
= -i
k =1
0+2
üï
é
æ 2p k ö ù
æ 2p k ö
ê cos çè 11 ÷ø + i sin çè 11 ÷ø ú - 1ý
ë
û
ïþ
[sum of 11, 11th roots of unity]
= - i ( 0 - 1)
=i
ìï 10
= -iíS
k =0
îï
y Example 66. If a 0 , a 1 , a 2 , ..., a n - 1 are the n, nth
n -1
ai
roots of the unity, then find the value of S
.
i =0 2 - a i
Sol. Let x = (1)1/n Þ x n = 1 \
SS
0=
\
i =0
S loge ( x - a i )
i =0
nx
n -1
S
=
n
x -1
i =0
æ 1 ö
÷
ç
è x - ai ø
n (2)n - 1
2n - 1
n -1
Now,
S
i =0
1
(2 - a i )
n -1
n
n ×2
n
ai a j - 1
1£i < j £n -1
[ Q1 + a 1 + a 2 + ... + a n - 1 = 0]
ai a j = 1
Y
2 -1
=
1
2 × n × 2n - 1
[from Eq. (i)]
= - (n ) + n
(2 - a i )
2 -1
n
n
2 -1
P (x, y)
…(i)
y
θ
æ
ai
2 ö
= S ç- 1 +
÷
=
0
i
è
(2 - a i )
2 - ai ø
S 1 + 2 i S= 0
i =0
= -n +
S
i =0
SS
n -1
n -1
=-
=
a i a j + a 1 + a 2 + ... + a n - 1
1£i < j £n -1
SS
On putting x = 2, we get
n -1
SS
0 £i < j £n -1
If P is the point ( x , y ) on the argand plane corresponding
to the complex number z = x + iy , where x , y Î R and
i = - 1.
On differentiating both sides w.r.t. x, we get
n -1
ai a j = 1
Vector Representation of
Complex Numbers
n -1
loge ( x n - 1) =
[here, p is not a multiple of n]
On comparing the coefficient of x n - 2 both sides, we get
x - 1 = ( x - a 0 ) ( x - a 1 ) ( x - a 2 ) ... ( x - a n - 1 )
On taking logarithm both sides, we get
a i a j = 1 + 12
ai a j = 2
1£i < j £n -1
0=
P (x - ai )
SS
1£i < j £n -1
Aliter
Q x n - 1 = ( x - 1) ( x - a 1 ) ( x - a 2 ) ... ( x - a n -1 )
xn - 1 = 0
n -1
SS
1£i < j £n -1
\
n
=
2
11 û
10
or
x = (1)1/n
Sol. Let
2
S ésin æçè 2pk ö÷ø - i cos æçè 2pk ö÷ø ùú
k =1 ê
ë
1 £ i < j £ n -1
a 1 + a 2 + a 3 + ... + a 2n - 1 + 2 (a 1a 2 + a 1a 3
10
Sol.
the n, nth roots of unity, then find the value of
S S ai a j .
O
¾®
Then,
x
X
M
¾®
OP = x $i + y $j Þ OP = ( x 2 + y 2 ) = z
¾®
and arg (z ) = direction of the vector OP = tan - 1 (y / x ) = q
¾®
Therefore, complex number z can also be represented by OP.
30
Textbook of Algebra
Geometrical Representation
of Algebraic Operation on
Complex Numbers
(c) Product
(a) Sum
Let the complex numbers z 1 = x 1 + iy 1 =( x 1 , y 1 ) and
z 2 = x 2 + iy 2 = ( x 2 , y 2 ) be represented by the points P and
Q on the argand plane.
Let
z 1 = r1 (cos q 1 + i sin q 1 ) = r1 e i q1
\
z 1 = r1 and arg (z 1 ) = q 1
and
z 2 = r2 (cos q 2 + i sin q 2 ) = r2 e iq2
\
z 2 = r2 and arg (z 2 ) = q 2
Then, z 1 z 2 = r1 r2 (cos q 1 + i sin q 1 ) (cos q 2 + i sin q 2 )
= r1 r2 {cos (q 1 + q 2 ) + i sin (q 1 + q 2 )}
z 1 z 2 = r1 r2 and arg (z 1 z 2 ) = q 1 + q 2
\
Y
R(z1z2)
Y
R (z1 + z2)
r1 r2
Q (z2)
Q(z2)
P (z1)
X
O
θ1
Complete the parallelogram OPRQ. Then, the mid-points
of PQ and OR are the same. The mid-point of
æ x + x2 y1 + y2 ö
,
PQ = ç 1
÷.
è
2
2 ø
Hence, R = ( x 1 + x 2 , y 1 + y 2 )
Therefore, complex number z can also be represented by
¾®
OR
= ( x 1 + x 2 ) + i (y 1 + y 2 ) = ( x 1 + iy 1 ) + ( x 2 + iy 2 )
= z 1 + z 2 = (x 1 , y 1 ) + (x 2 , y 2 )
In vector notation, we have
¾®
¾®
¾®
¾®
¾®
z 1 + z 2 = OP + OQ = OP + PR = OR
(b) Difference
We first represent - z 2 by Q ¢, so that QQ ¢ is bisected at O.
Complete the parallelogram OPRQ ¢. Then, the point R
represents the difference z 1 - z 2 .
Q(z2)
P(z1)
O
X
R(z1 – z2)
Q ′( –z2)
θ2 θ1
O
P(z1)
r1
A
X
Let P and Q represent the complex numbers z 1 and z 2 ,
respectively.
\
OP = r1 , OQ = r2
ÐPOX = q 1 and ÐQOX = q 2
Take a point A on the real axis OX, such that OA = 1 unit.
Complete the ÐOPA
Now, taking OQ as the base, construct a DOQR similar to
OR OP
=
DOPA, so that
OQ OA
[since, OA = 1 unit]
i.e.
OR = OP × OQ = r1 r2
and
ÐROX = ÐROQ + ÐQOX = q 1 + q 2
Hence, R is the point representing product of complex
numbers z 1 and z 2 .
Remark
1. Multiplication by i
p
p
Since, z = r (cos q + i sin q) and i = æçcos + i sin ö÷
è
2
2ø
p
p
é
ù
æ
ö
æ
ö
iz = r êcos ç + q÷ + i sin ç + q÷ ú
\
è2
ø
è2
øû
ë
Y
X′
r2
Y′
¾®
¾®
We see that ORPQ is a parallelogram, so that OR = QP
We have in vectorial notation,
¾®
¾®
¾®
¾®
¾®
¾®
¾®
¾®
z 1 - z 2 = OP - OQ = OP + QO
= OP + PR = OR = QP
Hence, multiplication of z with i, then vector for z rotates a
right angle in the positive sense.
2. Thus, to multiply a vector by ( - 1) is to turn it through two
right angles.
3. Thus, to multiply a vector by (cos q + i sin q) is to turn it
through the angle q in the positive sense.
(d) Division
Let
\
and
z 1 = r1 (cos q 1 + i sin q 1 ) = r1 e iq1
z 1 = r1 and arg (z 1 ) = q 1
z 2 = r2 (cos q 2 + i sin q 2 ) = r2 e iq2
Chap 01 Complex Numbers
\
| z 2 | = r2 and arg (z 2 ) = q 2
z
r (cos q 1 + i sin q 1 )
Then, 1 = 1 ×
[z 2 ¹ 0, r2 ¹ 0 ]
z 2 r2 (cos q 2 + i sin q 2 )
z 1 r1
= [cos (q 1 - q 2 ) + i sin(q 1 - q 2 )]
z 2 r2
æz ö
z1
r
\
= 1 , arg ç 1 ÷ = q 1 - q 2
èz2 ø
z2
r2
Let P and Q represent the complex numbers z 1 and z 2 ,
respectively.
\ OP = r1 , OQ = r2 , ÐPOX = q 1 and ÐQOX = q 2
Let OS be new position of OP, take a point A on the real
axis OX, such that OA = 1 unit and through A draw a line
making with OA an angle equal to the ÐOQP and meeting
OS in R.
Then, R represented by (z 1 /z 2 ).
Y
¾®
¾®
Then, we have AC = z 3 - z 1 and AB = z 2 - z 1
¾®
and let
arg AC = arg (z 3 - z 1 ) = q
and
arg AB = arg (z 2 - z 1 ) = f
Let
ÐCAB = a
¾®
¾®
= arg (z 3 - z 1 ) - arg (z 2 - z 1 )
æz - z1 ö
= arg ç 3
÷
è z2 - z1 ø
or angle between AC and AB
æ affix of C - affix of A ö
= arg ç
÷
è affix of B - affix of A ø
z= z e
r2
O
r1
i (arg z )
æz - z1 ö
z 3 - z1
Similarly, ç 3
e
÷=
z2 - z1
è z2 - z1 ø
P(z1)
θ1
A
θ1 – θ2
X
S
Now, in similar DOPQ and DOAR.
r
OR OP
=
Þ OR = 1
r2
OA OQ
é
æ z - z1 ö ù
i ê arg ç 3
÷ú
è z 2 - z1 ø û
ë
z 3 - z 1 i ( Ð CAB ) AC i a
z 3 - z1
=
e
e
=
AB
z2 - z1
z2 - z1
or
R
Remark
since OA = 1 and ÐAOR = ÐPOR - ÐPOX = q 2 - q 1
Hence, the vectorial angle of R is - (q 2 - q 1 ) i.e., q 1 - q 2 .
Remark
1. Here, only principal values of the arguments are considered.
æ z - z2 ö
2. arg ç 1
÷ = q, if AB coincides with CD, then
è z3 - z4 ø
æ z - z2 ö
z1 - z2
is real. It follows that
arg ç 1
÷ = 0 or p, so that
z3 - z4
è z3 - z4 ø
z - z2
is real, then the points A, B, C, D are collinear.
if 1
z3 - z4
If q1 and q2 are the principal values of z1 and z2 , then q1 + q2 and
q1 - q2 are not necessarily the principal value of arg ( z1z2 ) and
arg ( z1 / z2 ).
D
P(z1)
θ
S(z4)
Rotation Theorem (Coni Method)
Let z 1 , z 2 and z 3 be the affixes of three points A, B and C
respectively taken on argand plane.
Y
C(z3)
α
B(z2)
α
φ
θ
A
R(z3)
B
Q(z2)
C
3. If AB is perpendicular to CD, then
æ z - z2 ö
z - z2
p
is purely imaginary.
arg ç 1
÷ = ± , so 1
2
z3 - z4
è z3 - z4 ø
4. It follows that, if z1 - z2 = ± k ( z3 - z4 ), where k is purely
imaginary number, then AB and CD are perpendicular to
each other.
A(z1)
O
¾®
ÐCAB = a = q - f = arg AC - arg AB
For any complex number z, we have
Q(z2)
θ2
31
X
32
Textbook of Algebra
y Example 68.
Complex numbers z 1 , z 2 and
z 1 - z 2 a ip / 3
= e
z 3 - z2 a
From Coni method,
z 3 are the vertices A, B, C respectively of an isosceles
p
3
z 3 - z 1 a ip / 3
From Coni method,
= e
z 2 - z1 a
ÐBAC =
and
right angled triangle with right angle at C. Show
that (z 1 - z 2 ) 2 = 2 (z 1 - z 3 ) (z 3 - z 2 ).
Sol. Since, ÐACB = 90° and AC = BC , then by Coni method
z1 - z 3 AC ip/ 2
=i
=
e
z 2 - z 3 BC
π/3
π/4
a
a
π/3
π/4
2
2
2
2
2
Þ
Þ
2
C(z3)
z1 - z 2 z 3 - z1
=
z 3 - z 2 z 2 - z1
(z1 - z 2 )(z 2 - z1 ) = (z 3 - z1 )(z 3 - z 2 )
z 12 + z 22 + z 23 = z1z 2 + z 2 z 3 + z 3z1
Remark
2
Þ z1 + z 2 - 2z1z 2 = 2(z1z 3 - z1z 2 - z 3 + z 2z 3 )
Therefore, (z1 - z 2 )2 = 2(z1 - z 3 ) (z 3 - z 2 )
CA = CB =
a
From Eqs. (i) and (ii), we get
z1 + z 3 - 2z1z 3 = -(z 2 + z 3 - 2z 2z 3 )
Aliter
π/3
B(z2)
C(z3)
Þ
z1 - z 3 = i ( z 2 - z 3 )
On squaring both sides, we get
( z 1 - z 3 ) 2 = -( z 2 - z 3 ) 2
Þ
Triangle with vertices z1, z2, z3, then
(i) ( z1 - z2 ) 2 + ( z2 - z3 ) 2 + ( z3 - z1 ) 2 = 0
1
BA
2
(ii) ( z1 - z2 ) 2 = ( z2 - z3 )( z3 - z1 )
(iv) å
(iii) å( z1 - z2 )( z2 - z3 ) = 0
B(z2)
π/4
π/4
\
or
and
\
…(i)
Let
z = r (cos q + i sin q ) = re iq
be a complex number representing a point P in the argand
plane.
C(z3)
ÐBAC = ( p / 4 )
z 2 - z1 BA (i p / 4 )
=
e
z 3 - z1 CA
z1 - z2
z1 - z 3
= 2e
Y
(i p/ 4 )
Ð CBA = (p / 4)
z 3 - z 2 CB (i p/ 4)
z - z2
1
or 3
=
e
=
e
z 1 - z 2 AB
z1 - z2
2
y Example 69. Complex numbers z 1 , z 2 , z 3 are the
vertices of A, B, C respectively of an equilateral
triangle. Show that
z 12 + z 22 + z 32 = z 1z 2 + z 2z 3 + z 3z 1 .
Q
AB = BC = CA = a
p
ÐABC =
3
φ
X′
(i p/ 4)
Q(ze if)
r
…(i)
On dividing Eq. (i) by Eq. (ii), we get
( z1 - z 2 ) 2 = 2 ( z1 - z 3 ) ( z 3 - z 2 )
Sol. Let
1
=0
( z1 - z2 )
Complex Number as a Rotating Arrow
in the Argand Plane
A(z1)
Q
…(ii)
A(z1)
B(z2)
A(z1)
…(i)
P(z)
θ
r
O
X
…(ii)
Y′
Then, OP = z = r and ÐPOX = q
Now, consider complex number z 1 = ze if
or
z 1 = re iq × e if = re i ( q + f )
[from Eq. (i)]
Clearly, the complex number z 1 represents a point Q in the
argand plane, when OQ = r and ÐQOX = q + f
¾®
Clearly, multiplication of z with e if rotates the vector OP
through angle f in anti-clockwise sense. Similarly,
¾®
multiplication of z with e - if will rotate the vector OP in
clockwise sense.
Chap 01 Complex Numbers
Shifting the Origin in Case
of Complex Numbers
Remark
1. If z1, z2 and z3 are the affixes of the
three points A, B and C, such that
AC = AB and ÐCAB = q. Therefore,
¾®
C(z3)
¾®
AB = z2 - z1 , AC = z3 - z1.
B(z2)
θ
¾®
Then, AC will be obtained by rotating A(z1)
¾®
AB through an angle q in anticlockwise sense and therefore,
¾®
¾®
AC = AB e i q
or
( z 3 - z1 ) = ( z2 - z1 ) e iq or
z 3 - z1
z2 - z1
Let O be the origin and P be a point with affix z 0 . Let a
point Q has affix z with respect to the coordinate system
passing through O. When origin is shifted to the point P
(z 0 ), then the new affix Z of the point Q with respect to
new origin P is given by Z = z - z 0 .
i.e., to shift the origin at z 0 , we should replace z by Z + z 0 .
= e iq
y
Y
Q
2. If A, B and C are three points in argand plane, such that
AC = AB and ÐCAB = q, then use the rotation about Ato find
e iq , but if AC ¹ AB, then use Coni method.
y Example 70. Let z 1 and z 2 be roots of the equation
z 2 + pz + q = 0, where the coefficients p and q may be
complex numbers. Let A and B represent z 1 and z 2 in
the complex plane. If Ð AOB = a ¹ 0 and OA = OB,
where O is the origin, prove that p 2 = 4 q cos 2 (a / 2).
¾®
¾®
Sol. Clearly, OB is obtained by rotating OA through angle a.
¾®
¾®
\
OB = OA e i a
Þ
z 2 = z1 e i a
Þ
z2
=eia
z1
A(z1)
O
Þ
z2
+ 1 = ( e i a + 1)
z1
( z1 + z 2 )
= e i a / 2 × 2 cos (a / 2)
z1
Þ
( z1 + z 2 )
z12
=
z2
× ( 4 cos 2 a / 2)
z1
Sol. In an equilateral triangle, the circumcentre and the
centroid are the same point.
z1 + z 2 + z 3
So,
z0 =
3
…(i)
\
z1 + z 2 + z 3 = 3z 0
To shift the origin at z 0 , we have to replace z1, z 2 , z 3 and z 0
by Z 1 + z 0 , Z 2 + z 0 , Z 3 + z 0 and 0 + z 0 .
Then, Eq. (i) becomes
( Z 1 + z 0 ) + ( Z 2 + z 0 ) + ( Z 3 + z 0 ) = 3( 0 + z 0 )
Þ
Z1 + Z 2 + Z 3 = 0
On squaring, we get
2
2
…(ii)
But triangle with vertices Z 1 , Z 2 and Z 3 is equilateral, then
2
2
2
Z 1 + Z 2 + Z 3 = Z 1Z 2 + Z 2 Z 3 + Z 3Z 1
[from Eq. (i)]
( - p )2 = 4 q cos 2 (a / 2)
éQz1 and z 2 are the roots of z 2 + pz + q = 0ù
ê
ú
ë\z1 + z 2 = - p and z1 z 2 = q
û
p 2 = 4 q cos 2 (a / 2)
equilateral triangle with z 0 as its circumcentre, then
changing origin to z 0 , show that Z 1 2+ Z 2 2 + Z 3 2 = 0,
2
(z1 + z 2 )2 = 4 z1z 2 cos 2 (a / 2)
or
y Example 71. If z 1 , z 2 and z 3 are the vertices of an
Z 1 + Z 2 + Z 3 + 2 (Z 1Z 2 + Z 2Z 3 + Z 3Z 1 ) = 0
On squaring both sides, we get
( z1 + z 2 ) 2
= e i a × ( 4 cos 2 a / 2)
2
z1
2
x
M
O
vertices.
B(z2)
or
X
P(z0)
where Z 1 , Z 2 , Z 3 are new complex numbers of the
…(i)
α
33
…(iii)
From Eqs. (ii) and (iii), we get
2
2
2
3 (Z 1 + Z 2 + Z 3 ) = 0
Therefore,
2
2
2
Z1 + Z 2 + Z 3 = 0
Inverse Points
(a) Inverse points with respect to a line Two points P
and Q are said to be the inverse points with respect to
the line RS. If Q is the image of P in RS, i.e., if the line
RS is the right bisector of PQ.
34
Textbook of Algebra
y Example 72. Show that z 1 , z 2 are the inverse points
with respect to the line z a + a z = b , if z 1 a + a z 2 = b .
Sol. Let RS be the line represented by the equation,
…(i)
z a +az =b
Let P and Q are the inverse points with respect to the line RS.
The point Q is the reflection (inverse) of the point P in the
line RS, if the line RS is the right bisector of PQ. Take any
point z in the line RS, then lines joining z to P and z to Q are
equal.
y Example 73. Show that inverse of a point a with
respect to the circle z - c = R (a and c are complex
R2
numbers, centre c and radius R) is the point c +
.
a -c
Sol. Let a¢ be the inverse point of a with respect to the circle
z - c = R, then by definition,
c
P
R
z - z1 = z - z 2 or z - z1
2
= z - z2
2
Þ
z1
A(z)
a¢ - c
a - c = R2
Þ
(a ¢ - c ) ( a - c ) = R 2
Þ
(a ¢ - c ) ( a - c ) = R 2
Þ
R
-
[Q arg z = - arg z ]
Þ arg (a ¢ - c ) + arg ( a - c ) = 0
Þ
arg {(a ¢ - c ) ( a - c )} = 0
\ (a ¢ - c ) ( a - c ) is purely real and positive.
By definition, a ¢ - c a - c = R 2
[QCP × CQ = r 2 ]
i.e., ( z - z1 ) ( z - z1 ) = ( z - z 2 ) ( z - z 2 )
…(ii)
Þ z ( z 2 - z1 ) + z ( z 2 - z1 ) + ( z1z1 - z 2 z 2 ) = 0
Hence, Eqs. (i) and (ii) are identical, therefore, comparing
coefficients, we get
P
a′
|z – c | = R
The points c , a, a ¢ are collinear.
We have, arg (a ¢ - c ) = arg(a - c )
= - arg ( a - c )
S
Q
i.e.,
a
[Q | z | = | z | ]
[Q(a ¢ - c )( a - c ) is purely real and positive]
R2
R2
a¢ - c =
Þ a¢ = c +
a -c
a -c
S
Dot and Cross Product
Q
z2
-b
a
a
=
=
z 2 - z1 z 2 - z1 z1z1 - z 2 z 2
So that,
z1a
az 2
=
z1( z 2 - z1 ) z 2 (z 2 - z1 )
=
z a + az 2 - b
-b
= 1
0
z1z1 - z 2 z 2
Let z 1 = x 1 + iy 1 º ( x 1 , y 1 ) and z 2 = x 2 + iy 2 º ( x 2 , y 2 ),
where x 1 , y 1 , x 2 , y 2 Î R and i = - 1, be two complex
numbers.
If ÐPOQ = q, then from Coni method,
z 2 iq
z2 - 0
Q(z2)
=
e
z1 - 0
z1
[by ratio and proportion rule]
z1a + az 2 - b = 0 or z1 a + az 2 = b
Þ
(b) Inverse points with respect to a circle If C is the
centre of the circle and P , Q are the inverse points
with respect to the circle, then three points C , P , Q are
collinear and also CP × CQ = r 2 , where r is the
radius of the circle.
Þ
C
P
Q
z 2 iq
z2z1
=
e
z1z1
z1
z2z1
z1
2
=
z2
z1
θ
P(z1)
O
e iq
z 2 z 1 = | z 1 || z 2 | e iq
z 2 z 1 = z 1 z 2 (cos q + i sin q )
…(i)
Þ
Re (z 2 z 1 ) = z 1 z 2 cos q
and
…(ii)
Im (z 2 z 1 ) = z 1 z 2 sin q
The dot product z 1 and z 2 is defined by,
z 1 ×z 2 = z 1 z 2 cos q
[from Eq. (i)]
= Re (z 1 z 2 ) = x 1 x 2 + y 1 y 2
Chap 01 Complex Numbers
and cross product of z 1 and z 2 is defined by
z 1 ´ z 2 = z 1 z 2 sin q
[from Eq. (ii)]
= Im (z 1 z 2 ) = x 1 y 2 - x 2 y 1
Hence, z 1 ×z 2 = x 1 x 2 + y 1 y 2 = Re (z 1 z 2 )
and z 1 ´ z 2 = x 1 y 2 - x 2 y 1 = Im (z 1 z 2 )
Remark
1. The distance of a point z from origin, z - 0 = z
2. Three points A( z1 ), B ( z2 ) and C ( z3 ) are collinear, then
AB + BC = AC
C(z3)
Results for Dot and Cross
Products of Complex Number
If z1 and z2 are perpendicular, then z1 × z2 = 0
If z1 and z2 are parallel, then z1 ´ z2 = 0
Projection of z1 on z2 = ( z1× z2 ) / z2
Projection of z2 on z1 = ( z1× z2 ) / z1
Area of triangle, if two sides represented by z 1 and z2 is
1
z1 ´ z2 .
2
6. Area of a parallelogram having sides z 1 and z2 is z1 ´ z2 .
7. Area of parallelogram, if diagonals represented by z1 and z2 is
1
z1 ´ z2 .
2
1.
2.
3.
4.
5.
y Example 74. If z 1 = 2 + 5i , z 2 = 3 - i , where i = -1, find
(i) z 1 × z 2
(ii) z 1 ´ z 2
(iii) z 2 × z 1
(iv) z 2 ´ z 1
(v) acute angle between z 1 and z 2 .
(vi) projection of z 1 on z 2 .
Sol. (i)
(ii)
(iii)
(iv)
(v)
z1×z 2 = x 1x 2 + y1y 2 = (2) (3) + (5) ( - 1) = 1
z1 ´ z 2 = x 1y 2 - x 2 y1 = (2) ( - 1) - (3) (5) = - 17
z 2 ×z1 = x 1 x 2 + y1y 2 = (2) (3) + (5) ( - 1) = 1
z 2 ´ z1 = x 2 y1 - x 1y 2 = (3) (5) - (2) ( - 1) = 17
Let angle between z1 and z 2 be q, then
z1 × z 2 = z1 z 2 cosq
Þ
B(z2)
A(z1)
i.e.
z1 - z2 + z2 - z3 = z1 - z3 .
y Example 75. Show that the points representing the
complex numbers ( 3 + 2i ), (2 - i ) and - 7i, where
i = - 1, are collinear.
Sol. Let z 1 = 3 + 2i , z 2 = 2 - i and z 3 = - 7i .
Then,
z 1 - z 2 = 1 + 3i = 10, z 2 - z 3 = 2 + 6i
= 40 = 2 10
and
z1 - z 3 = 3 + 9i = 90 = 3 10
z1 - z 2 + z 2 - z 3 = z1 - z 3
\
Hence, the points (3 + 2i ), (2 - i ) and - 7i are collinear.
(b) Equation of the
Perpendicular Bisector
If P (z 1 ) and Q (z 2 ) are two fixed points and R (z ) is
moving point, such that it is always at equal distance from
P (z 1 ) and Q (z 2 ).
P(z1)
R (z)
1 = ( 4 + 25) (9 + 1) cosq
æ 1 ö
\ q = cos - 1 ç
÷
è 290 ø
1
z ×z
1
(vi) Projection of z1 on z 2 = 1 2 =
=
z2
( 9 + 1)
10
\
cosq =
1
290
Use of Complex Numbers in
Coordinate Geometry
(a) Distance Formula
35
Q(z2)
i.e.
or
or
PR = QR
z - z1 = z - z2
z (z 1 - z 2 ) + z (z 1 - z 2 ) = z 1 z 1 - z 2 z 2
or
z (z 1 - z 2 ) + z (z 1 - z 2 ) = z 1
2
- z2
2
Hence, z lies on the perpendicular bisectors of z 1 and z 2 .
y Example 76. Find the perpendicular bisector of 3 + 4i
and - 5 + 6i, where i = - 1.
Sol. Let z1 = 3 + 4i and z 2 = - 5 + 6i
The distance between two points P (z 1 ) and Q (z 2 ) is given by
Q(z2)
If z is moving point, such that it is always equal distance
from z1 and z 2 .
i.e.
z - z1 = z - z 2
or
P(z1)
PQ = z 2 - z 1 = affix of Q - affix of P
z ( z1 - z 2 ) + z ( z1 - z 2 ) = z1 2 - z 2
2
Þ z (( 3 - 4i ) - ( - 5 - 6i )) + z (( 3 + 4i ) - ( - 5 + 6i )) = 25 - 61
Hence, ( 8 + 2i ) z + ( 8 - 2i ) z + 36 = 0
which is required perpendicular bisector.
36
Textbook of Algebra
(c) Section Formula
If R (z ) divides the joining of P (z 1 ) and Q (z 2 ) in the ratio
m 1 : m 2 (m 1 , m 2 > 0 ).
Q(z2)
(m2)
(m 1 )
R(z)
P(z1)
y Example 78. Let z 1 , z 2 and z 3 be three complex
numbers and a, b , c ÎR, such that a + b + c = 0 and
az 1 + bz 2 + cz 3 = 0, then show that z 1 , z 2 and z 3 are
collinear.
Sol. Given,
(i) If R (z ) divides the segment PQ internally in the ratio
m z + m2 z 1
of m 1 : m 2 , then z = 1 2
m1 + m2
(ii) If R (z ) divides the segment PQ externally in the ratio
m z - m2 z 1
of m 1 : m 2 , then z = 1 2
m1 - m2
m1
m2
Remark
z1 + z2
.
2
2. If z1 , z2 and z3 are affixes of the vertices of a triangle, then
z + z2 + z 3
.
affix of its centroid is 1
3
3. In acute angle triangle, orthocentre ( O), nine point centre ( N ),
OG 2
centroid ( G) and circumcentre ( C) are collinear and
= ,
GC 1
ON 1
= .
NG 1
4. If z1, z2, z 3 and z4 are the affixes of the vertices of a
parallelogram taken in order, then z1 + z 3 = z2 + z4 .
1. If R ( z ) is the mid-point of PQ, then affix of R is
y Example 77. If z 1 , z 2 and z 3 are the affixes of the
vertices of a triangle having its circumcentre at the
origin. If z is the affix of its orthocentre, prove that
z 1 + z 2 + z 3 - z = 0.
Sol. We know that orthocentre O, centroid G and circumcentre
C of a triangle are collinear, such that G divides OC in the
z1 + z 2 + z 3
and C is the
ratio 2 : 1. Since, affix of G is
3
origin. Therefore, by section formula, we get
O
G
…(i)
and
az1 + bz 2 + cz 3 = 0
Þ az1 + bz 2 - (a + b ) z 3 = 0
az + bz 2
or
z3 = 1
a+b
…(ii)
[from Eq. (i)]
It follows that z 3 divides the line segment joining z1 and z 2
internally in the ratio b : a. (If a, b are of same sign and
opposite sign, then externally.)
Hence, z1,z 2 and z 3 are collinear.
(d) Area of Triangle
P(z1)
2
a+b+c =0
R(z)
Q(z2)
1
z1 + z 2 + z 3
2 ×0 + 1 × z
=
2+1
3
Þ
z1 + z 2 + z 3 = z
Therefore, z1 + z 2 + z 3 - z = 0
Þ
If z 1 , z 2 and z 3 are the affixes of the vertices of a triangle,
z1 z1 1
1
then its area = | z 2 z 2 1 |
4
z3 z3 1
A(z1)
B(z2)
C(z3)
Remark
The area of the triangle with vertices z, wz and z + w z is
3
2
z ,
4
where w is the cube root of unity.
y Example 79. Show that the area of the triangle on
the argand plane formed by the complex numbers z, iz
1
2
and z + iz is z , where i = - 1.
2
1
Sol. Required area = |
4
C
1
= |
4
z
z
1
1|
iz
iz
z + iz z + iz 1
z
z
1
iz
iz
1|
z + iz z + iz 1
37
Chap 01 Complex Numbers
=
1
|
4
z
z
1
iz
-iz 1 |
z + iz z - iz 1
On applying R 3 ® R 3 - ( R 1
z
z
1
Area = | iz -iz
4
0
0
=
1
2izz
4
=
+ R 2 ), we get
1
1
1 | = ( - 1) ( - izz - izz )
4
-1
1
i
2
zz =
1
z
2
2
Aliter
We have, iz = z (cos( p / 2) + i sin( p / 2)) = ze (ip / 2 ) iz is the
vector obtained by rotating vector z in anti-clockwise
direction through ( p / 2). Therefore, OA ^ AB,
Imaginary axis
Y
Proof Equation of the straight line joining points having
affixes z 1 and z 2 is
z = tz 1 + (1 - t ) z 2 , where t Î R ~ {0 }
…(i)
Þ
z - z 2 = t (z 1 - z 2 )
z - z 2 = t (z 1 - z 2 )
and
or
z - z 2 = t (z 1 - z 2 )
From Eqs. (i) and (ii), we get
z - z2 z1 - z2
z - z2
z - z2
=
Þ
=
z - z2 z1 - z2
z1 - z2 z1 - z2
Þ
z - z2
z - z2
0
z1 - z2
z2
z1 - z2
z2
0 =0
1
…(ii)
Now, applying R 1 ® R 1 + R 3 and R 2 ® R 2 + R 3 , we get
z z 1
z1 z1 1 = 0
B
z2 z2 1
iz
z + iz
π/2
z
O
A
X Real axis
1
1
OA ´ AB = z iz
2
2
1
1
2
= z i z = z
2
2
Now, area of DOAB =
(e) Equation of a Straight Line
(i) Parametric form
Equation of the straight line joining the points having
affixes z 1 and z 2 is
z = tz 1 + (1 - t ) z 2 , where t Î R ~ {0 }
Proof
tz + (1 - t ) z 2
Q
z = tz 1 + (1 - t ) z 2 = 1
t + (1 - t )
Hence, z divides the line joining z 1 and z 2 in the ratio
1 - t : t . Thus, the points z 1 , z 2 , z are collinear.
(ii) Non-parametric form
Equation of the straight line joining the points having
affixes z 1 and z 2 is
z z 1
z1 z1 1 = 0
or z (z 1 - z 2 ) - z (z 1 - z 2 ) + z 1 z 2 - z 1 z 2 = 0
Aliter
Let P (z ) be an arbitrary point on the line, which pass
through A (z 1 ) and B (z 2 ).
\
Ð BAP = 0 or p
æ z - z1 ö
arg ç
\
÷ = 0 or p [by rotation theorem]
èz2 - z1 ø
z - z1
is purely real.
Þ
z2 - z1
æ z - z1 ö æ z - z1 ö
z - z1
z - z1
=
\
÷ Þ
ç
÷ =ç
z2 - z1 z2 - z1
èz2 - z1 ø èz2 - z1 ø
A(z1)
B(z2)
P(z)
z (z 1 - z 2 ) - z (z 1 - z 2 ) + z 1 z 2 - z 1 z 2 = 0
z z 1
z1 z1 1 = 0
z2 z2 1
or
Remark
z1
If z1, z 2 and z 3 are collinear, z2
z3
z2 z2 1
or z (z 1 - z 2 ) - z (z 1 - z 2 ) + z 1 z 2 - z 1 z 2 = 0
or
z1
1
1 =0
z3 1
z2
å z1( z2 - z3 ) = 0.
38
Textbook of Algebra
(iii) General form The general equation of a straight
line is of the form a z + a z + b = 0, where a is a
complex number and b is a real number.
(i) If the lines are perpendicular, then
(z3)
Sol. The equation of a straight line passing through points
having affixes z1 and z 2 is given by
…(i)
z (z1 - z 2 ) - z (z1 - z 2 ) + z1z 2 - z1z 2 = 0
On multiplying Eq. (i) by i (where, i = -1), we get
(z2)
(z1)
(z4)
zi ( z1 - z 2 ) - z i ( z1 - z 2 ) + i ( z1z 2 - z1z 2 ) = 0
Þ z { - i ( z1 - z 2 )} + z {i ( z1 - z 2 )} + i ( z1z 2 - z1z 2 ) = 0
(z 1 - z 2 ) (z 3 - z 4 ) ip / 2
=
e
z1 - z2
z3 -z4
Þ z { - i ( z1 - z 2 )} + z { - i ( z1 - z 2 )} + {i ( 2i Im (z1z 2 ))} = 0
Þ z { - i ( z1 - z 2 )} + z { - i ( z1 - z 2 )} + { ( - 2 Im (z1z 2 )} = 0
(z 1 - z 2 ) 2
e ip
Þ
Þ
(z 1 - z 2 ) 2
(z 3 - z 4 ) 2
=
e ip
(z 1 - z 2 ) (z 1 - z 2 ) (z 3 - z 4 ) (z 3 - z 4 )
(iv) Slope of the line a z + a z + b = 0
Let A (z 1 ) and B (z 2 ) be two points on the line
a z + a z + b = 0, then
a z1 + a z1 + b = 0
and
a z2 + a z2 + b = 0
\ a (z 1 - z 2 ) + a (z 1 - z 2 ) = 0
z1 - z2
a
[Remember]
=Þ
a
z1 - z2
Þ
(z 1 - z 2 ) (z 3 - z 4 )
=
( - 1)
(z 1 - z 2 ) (z 3 - z 4 )
Complex slope of AB = -
a
a
=-
coefficient of z
coefficient of z
Thus, the complex slope of the line a z + a z + b = 0 is
a
- .
a
Remark
The real slope of the line a z + az + b = 0 is
Re ( a)
Re (coefficient of z )
, i.e. .
Im ( a)
Im (coefficient of z )
Important Theorem
If a 1 and a 2 are the complex slopes of two lines on the
argand plane, then prove that the lines are
(i) perpendicular, if a 1 + a 2 = 0.
(ii) parallel, if a 1 = a 2 .
Proof Let z 1 and z 2 be the affixes of two points on one
line with complex slope a 1 and z 3 and z 4 be the affixes of
two points another line with complex slope a 2 . Then,
z 3 -z4
z - z2
and a 2 =
…(i)
a1 = 1
z1 - z2
z 3 -z4
z1 - z2
2
=
(z 3 - z 4 ) 2
z a + z a + b = 0,
Þ
where, a = - i (z1 - z 2 ), b = - 2 Im(z1z 2 )
Hence, the general equation of a straight line is of the form
a z + a z + b = 0,
where a is complex number and b is a real number.
z3 -z4
2
[from Eq. (i)]
Þ
a1 = - a2
\
a1 + a2 = 0
(ii) If the lines are parallel, then
z1 - z2
z -z4 0
= 3
e
z1 - z2
z3 -z4
(z 1 - z 2 ) 2
Þ
z1 - z2
2
=
(z 3 - z 4 ) 2
z3 -z4
2
Þ
(z 1 - z 2 ) 2
(z 3 - z 4 ) 2
=
(z 1 - z 2 ) (z 1 - z 2 ) (z 3 - z 4 ) (z 3 - z 4 )
Þ
(z 1 - z 2 ) (z 3 - z 4 )
=
(z 1 - z 2 ) (z 3 - z 4 )
Þ
a1 = a2
Remark
1. The equation of a line parallel to the line a z + az + b = 0 is
a z + az + l = 0, where l ÎR.
2. The equation of a line perpendicular to the line
a z + a z + b = 0 is a z - a z + i l = 0
where,
l ÎR and i = - 1
.
(v) Length of perpendicular from a given point on a
given line
The length of perpendicular from a point P (z 1 ) to the
line
a z + a z + b = 0 is given by
az 1 + a z 1 + b
2 a
.
39
Chap 01 Complex Numbers
Proof Let PM be perpendicular from P on the line
a z + a z + b = 0 and let the affix of M be z 2 , then
P(z1)
a - a¢ æ b ö
+ ç- ÷ = 0
a - a¢ è b ø
Þ
Þ
b (a - a ¢ ) - b ( a - a ¢ ) = 0
...(ii)
On subtracting Eq. (ii) from Eq. (i), we get
a ¢ b + ab + c = 0
M(z2)
PM = z 1 - z 2
a z +a z +b =0
and M(z 2 ) lies on a z + a z + b = 0, then
…(i)
a z2 + a z2 + b = 0
Since, PM perpendicular to the line (a z + a z + b = 0 ).
z1 - z2 æ a ö
Therefore,
+ ç- ÷ = 0
z1 - z2 è a ø
Þ
Þ
or
a z1 - a z2 - a z1 + a z2 = 0
a z1 + a z1 + b = 2 a z1 + a z2 - a z2 + b
= 2 a z 1 - az 2 + (az 2 + b )
= 2 a z1 - a z2 - a z2
[Qa z 2 + b = - a z 2 ]
= 2 a (z 1 - z 2 )
a z1 + a z1 + b = 2 a z1 - z2
= 2 a z1 - z2
= 2 a PM
\
PM =
[ Q|z | = |z | ]
Aliter
Equation of perpendicular bisector of PQ is
z ( a ¢ - a ) + z (a ¢ - a ) - a ¢ a ¢ + aa = 0
and given line
zb + z b + c = 0
Since, Eqs. (i) and (ii) are identical, we have
a ¢ - a a ¢ - a aa - a ¢ a ¢
=
=
=k
b
c
b
…(i)
…(ii)
[say]
a ¢ - a = b k , a ¢ - a = bk
Q
aa - a ¢ a ¢ = ck
ì æa¢ - a ö
Now,
a ¢ b + ab = ía ¢ ç
÷+a
î è k ø
1
= {a ¢ a ¢ - a a } =
k
Hence, a ¢b + ab + c = 0
and
æ a ¢ - a öü
÷ý
ç
è k øþ
1
( - ck ) = - c
k
(f) Circle
The equation of a circle whose centre is at point affix z 0
and radius r, is z - z 0 = r .
a z1 + a z1 + b
P(z)
r
2 a
C(z0)
y Example 80. Show that the point a ¢ is the reflection
of the point a in the line z b + z b + c = 0, if
a ¢b + ab + c = 0.
Sol. Since, a¢ is the reflection of point a through the line.
So, the mid-point of PQ
P
A(z)
Q
i.e.,
or
Þ
1. If the centre of the circle is at origin and radius r, then its
equation is z = r .
2. z - z0 < r represents interior of a circle z - z0 = r and
z - z0 > r represent exterior of the circle z - z0 = r.
3. r < z - z0 < R, this region is known as annulus.
a
B
(i) General Equation of a Circle
The general equation of the circle is
z z + a z + a z + b = 0,
a'
a + a¢
lies on z b + z b + c = 0
2
æ a + a¢ ö
æ a + a¢ ö
bç
÷ +c =0
÷ +bç
è 2 ø
è 2 ø
b (a + a ¢ ) + b (a + a ¢ ) + 2c = 0
Since, PQ ^ AB. Therefore,
Complex slope of PQ + Complex slope of AB = 0
Remark
where a is a complex number and b Î R, having centre at ( -a )
and radius =
…(i)
a
2
- b.
Proof The equation of circle having centre at z 0 and
radius r is
z -z0 =r
40
Textbook of Algebra
2
z -z0
Þ
(z - z 0 ) (z - z 0 ) = r 2
Þ
zz - zz 0 - z 0 z + z 0 z 0 = r 2
Þ zz + ( - z 0 ) z + ( - z 0 ) z + z 0
2
From Eqs. (i) and (ii), we get
(z - z 1 ) (z 2 - z 3 )
= Real
(z - z 2 ) (z 1 - z 3 )
=r2
Þ
Remark
-r2 =0
If four points z1, z2, z 3, z4 are concyclic, then
zz + az + az + b = 0
Þ
a = - z 0 and b = z 0
where,
2
z0
2
é ( z - z 3 ) ( z4 - z1 ) ù
arg ê 2
ú = p, 0.
êë ( z1 - z 3 ) ( z4 - z2 ) úû
or
(iii) Equation of Circle in Diametric Form
- b = | a |2 - b .
Remark
Rule to find the centre and radius of a circle whose equation is
given
1. Make the coefficient of z z equal to 1 and right hand side
equal to zero.
2. The centre of circle will be = ( - a) = ( - coefficient of z ).
If end points of diameter represented by A (z 1 ) and B (z 2 )
and P (z ) is any point on circle.
\
Ð APB = 90°
\ Complex slope of PA + Complex slope of PB = 0
P(z)
90°
3. Radius = (|a|2 - constant term)
B(z2)
y Example 81. Find the centre and radius of the circle
2 z z + ( 3 - i ) z + ( 3 + i ) z - 7 = 0, where i = -1.
Sol. The given equation can be written as
æ3 + i ö
7
æ3 + i ö
zz + ç
÷z +ç
÷z - =0
è 2 ø
2
è 2 ø
(ii)
2
7ö
æ9 1 7ö
+ ÷ = ç + + ÷= 6
÷
è 4 4 2ø
2ø
Equation of Circle Through Three
Non-Collinear Points
Let A (z 1 ), B (z 2 ), C (z 3 ) be three points on the circle and
P (z ) be any point on the circle, then
P(z)
θ
ter
ame
Di
A(z1)
æz - z1 ö æz - z2 ö
ç
÷ +ç
÷ =0
èz - z1 ø èz - z2 ø
Þ
æ3 + i ö
So, it represent a circle with centre at - ç
÷ and radius
è 2 ø
æ æ3 + i ö
= ç -ç
ç è 2 ÷ø
è
Hence, ( z - z 1 ) ( z - z 2 ) + ( z - z 2 ) ( z - z 1 ) = 0
which is required equation of circle in diametric form.
(iv) Other Forms of Circle
(a) Equation of all circles which are orthogonal to
z - z 1 = r1 and z - z 2 = r2 .
Let the circle be z - a = r cut given circles
orthogonally.
2
2
\
r 2 + r1 = a - z 1
and
2
On solving,
2
2
r2 - r1 = a (z 1 - z 2 ) + a (z 1 - z 2 ) + z 2
C(z3)
θ
…(i)
…(ii)
2
- z1
2
and let a = a + ib , i = - 1, a, b Î R
(b) Apollonius circle
B(z2)
Ð ACB = Ð APB
Using Coni method,
in DAPB,
2
r 2 + r2 = a - z 2
A(z1)
in DACB,
( z4 - z1 ) ( z2 - z3 )
=
( z4 - z2 ) ( z1 - z3 )
real [replacing z by z4 in Eq. (iii)]
-r2
Þ
zz + a z + a z + b = 0
where, b Î R represents a circle having centre at ( - a ) and
radius =
…(iii)
z 2 - z 3 BC iq
=
e
z 1 - z 3 CA
z 2 - z BP iq
=
e
z 1 - z AP
…(i)
…(ii)
z - z1
=k ¹1
z - z2
It is the circle with join of z 3 and z 4 as a diameter,
z + kz 2
z - kz 2
where z 3 = 1
, z4 = 1
1+k
1-k
for k = 1, the circle reduces to the straight line which
is perpendicular bisector of the line segment from z 1
to z 2 .
Chap 01 Complex Numbers
æz - z1 ö
(c) Circular arc arg ç
÷ =a
èz - z2 ø
z -a =
This is an arc of a circle in which the chord joining z 1
and z 2 subtends angle a at any point on the arc.
p
If a = ± , then locus of zis a circle with the join of
2
z 1 and z 2 as diameter. If a = 0 or p, then locus is a
straight line through the points z 1 and z 2 .
(d) The equation z - z 1
2
+ z - z2
2
= k, will represent a
z + z + 2a
2
1 2
{z + (z ) 2 }
2
where, a Î R (focus), directrix is z + z + 2a = 0.
or
z z - 4a (z + z ) =
(h) Equation of Ellipse
For ellipse
Imaginary axis
2
1
circle, if k ³ z 1 - z 2 .
2
P(z)
y Example 82. Find all circles which are orthogonal
to z = 1 and z - 1 = 4 .
z -a = k
Sol. Let
S′(z2)
Real axis
C S(z1)
…(i)
(where, a = a + ib and a, b, k Î R and i = - 1) be a circle
which cuts the circles
…(ii)
z =1
and
…(iii)
z -1 = 4
Orthogonally, then using the property that the sum of
squares of their radii is equal to square of distance between
centres. Thus, the circle (i) will cut the circles (ii) and (iii)
orthogonally, if
and
41
SP + S ¢ P = 2a
+ z - z 2 = 2a
Þ
z - z1
where, 2a > z 1 - z 2
[since, eccentricity < 1]
Then, point z describes an ellipse having foci at z 1 and z 2
and a Î R + .
k 2 + 1 = a -0
2
= aa
(i) Equation of Hyperbola
2
2
= ( a - 1) ( a - 1)
For hyperbola
k + 16 = a - 1
= aa - ( a + a ) + 1
Imaginary axis
\ 1 - ( a + a ) - 15 = 0 Þ a + a = -14
\
Þ
2a = - 14 Þ a = - 7
a = a + ib = - 7 + ib
Also,
k2 = a
Þ
k = (b 2 + 48)
2
P(z)
- 1 = ( - 7 )2 + b 2 - 1 = b 2 + 48
S′(z2)
C
S(z1)
Real axis
Therefore, required family of circles is given by
z + 7 - ib = ( 48 + b 2 ).
where,
(g) Equation of Parabola
Then, point z describes a hyperbola having foci at z 1 and
z 2 and a Î R + .
Now, for parabola
Imaginary axis
Examples on Geometry
P(z)
z + z + 2a = 0
M
N
A (O) S (a + i . 0)
SP = PM
SP - S ¢ P = 2a Þ z - z 1 - z - z 2 = 2a
[since, eccentricity > 1]
2a < z 1 - z 2
Real axis
y Example 83. Let z 1 = 10 + 6i , z 2 = 4 + 6i , where
i = - 1. If z is a complex number, such that the
argument of (z - z 1 ) / (z - z 2 ) is p / 4, then prove that
z - 7 - 9i = 3 2.
Sol. Q
æ z - z1 ö p
arg ç
÷=
èz - z2 ø 4
42
Textbook of Algebra
It is clear that z , z1, z 2 are non-collinear points. Always a
circle passes through z , z1 and z 2 . Let z 0 be the centre of the
circle.
sense through 180° and stretched three times. Then,
find the complex number represented by the new
vector.
A(z)
π/4
Sol. Q
O(z0)
r
π/2
r
(z2)B
or
[QOC = OB ]
4 3
+ i.
5 5
æ 4 3 ö
Therefore, required vector = 3 z ç - + i ÷
è 5 5 ø
æ 4 3 ö
= 15 ç - + i ÷ = - 12 + 9i
è 5 5 ø
Aliter
16 + 16i + 2i + 2i 2
=
2
14 + 18i
=
= 7 + 9i
2
and radius, r = OC = z 0 - z1 = 7 + 9i - 10 - 6i
= - 3 + 3i
= (9 + 9 ) = 3 2
Hence, required equation is
z - z0 = r
z - 7 - 9i = 3 2
y Example 84.If z - 2 + i £ 2, where i = - 1, then
find the greatest and least value of z .
Sol. Q Radius = 2 units
Imaginary axis
z1
4
Real axis
O
z
–3
Here, z1 = - 4 + 3i
Hence,3 z1 = - 12 + 9i
y Example 86.ABCD is a rhombus. Its diagonals AC
and BD intersect at the point M and satisfy BD = 2AC .
If the points D and M represents the complex numbers
1 + i and 2 - i, where i = - 1, respectively, find A.
Sol. Let
Q
A ºz
BD = 2AC or DM = 2 AM
B
C
M
O
A
2
(2 – i)
X
π/2
C(2, –1)
2
B
AC = CB = 2 units
\ Least value of z = OA = OC - AC = 5 - 2
and greatest value of z = OB = OC + CB = 5 + 2
Hence, greatest value of z is 5 + 2 and least value of z
is 5 - 2.
3
–4
Y
i.e.,
z = ( 4 ) 2 + ( - 3) 2 = 5
Þ
The unit vector in the direction of z1 is -
( z1 - z 0 ) = i ( z 2 - z 0 )
10 + 6i - z 0 = i ( 4 + 6i - z 0 )
16 + 2i = (1 - i ) z 0
(16 + 2i ) (1 + i )
×
z0 =
(1 - i ) (1 + i )
Þ
z = 4 - 3i
Let z1 be the new vector obtained by rotating z in the
clockwise sense through 180°, therefore
z1 = z e - ip = - z = - ( 4 - 3i ) = - 4 + 3i .
C(z1)
On applying rotation theorem in DBOC,
z1 - z 0 OC (ip / 2 )
=i
=
e
z 2 - z 0 OB
Þ
Þ
Þ
y Example 85. In the argand plane, the vector
z = 4 - 3i , where i = - 1, is turned in the clockwise
D(1 + i)
A(z)
Now, in DDMA,
Applying Coni method, we have
z - (2 - i )
AM ip / 2 1
= i
=
e
2
(1 + i ) - (2 - i ) DM
Þ
\
i
i
3
( - 1 + 2i ) = - - 1 or z = 1 - i
2
2
2
3
i
A º 1 - i or 3 2
2
[if positions of A and C interchange]
z -2+i =
43
Chap 01 Complex Numbers
If z ±
b
= a , then the greatest and least values of | z |
z
- a + (a 2 + 4 | b | )
a + (a 2 + 4 | b |)
and
,
2
2
respectively.
b
b
Proof
³ z z±
z
z
are
Þ
a³
or
-a£ z -
|b |
£a
|z |
z -
|b |
£a
|z |
Now,
\ Maximum and minimum values of z are
and
0 £ |z | £
and
|z | -
a + (a 2 + 4 | b| )
2
Sol. Here, b = 4 and a = 2.
\ Maximum and minimum values of z are
2 + ( 4 + 16)
- 2 + ( 4 + 16)
and
2
2
i.e. 1 + 5 and - 1 + 5, respectively.
|z | ³
- a + (a 2 + 4 | b | )
a + (a 2 + 4 | b | )
£ |z | £
2
2
Hence, the greatest value of | z | is
a + (a 2 + 4 | b | )
2
- a + (a 2 + 4 | b | )
.
and the least value of | z | is
2
1
Corollary For b = 1, z ±
=a
z
Then,
…(i)
y Example 89. If z ³ 3, then determine the least
1
value of z + .
z
…(ii)
Sol. Q
|b |
³ - a Þ |z |2 + a |z | - |b | ³ 0
|z |
- a + (a 2 + 4 | b | )
2
From Eqs. (i) and (ii), we get
\
- a + (a 2 + 4 )
a + (a 2 + 4 )
£ z £
2
2
- 2 + (4 + 4)
i.e., 1 + 2 and - 1 + 2, respectively.
2
4
= 2, find the maximum and
z
minimum values of z .
a - (a 2 + 4 | b| )
a + (a 2 + 4 | b | )
£ |z | £
2
2
or
2 + (4 + 4)
2
y Example 88. If z +
|z |2 - a |z | - |b | £ 0
Þ
\
Sol. Here, b = 1 and a = 2
|b |
|z |
z
y Example 87. Find the maximum and minimum
1
values of z satisfying z +
= 2.
z
z+
or
z -
z ³3 Þ
Q
\
1
³
z
z -
1
z
1
z
1
1
1
1
or £
³3
z
3
z
1
1 8
³3- = Þ
3 3
z
z -
1
z
³
8
3
From Eqs. (i) and (ii), we get
1
8
z+
³
3
z
\ Least value of z +
1
8
is .
z
3
=
z -
z -
…(i)
1
8
³
3
z
…(ii)
44
Textbook of Algebra
#L
1
Exercise for Session 4
If z1, z 2, z 3 and z4 are the roots of the equation z 4 = 1, the value of
(a) 0
2
3
(c) z k + 1 = z k + z k - 1
(d) z k = z k + 1
(b) 1
6
The value of
å
(c) n
(d) n 2
(c) - i
(b) 0
(d) i
If a ¹ 1 is any nth root of unity, then S = 1 + 3 a + 5 a 2 + ... upto n terms is equal to
2n
1- a
(b) -
2n
1- a
(c)
n
1- a
(d) -
n
1- a
If a and b are real numbers between 0 and 1, such that the points z1 = a + i ,z 2 = 1+ bi and z 3 = 0 form an
equilateral triangle, then
(b) a = b = 2 - 3
(d) None of these
3
If z = 2, the points representing the complex numbers - 1 + 5z will lie on
(a) a circle
(b) a straight line
(c) a parabola
(d) an ellipse
If z - 2 / z - 3 = 2 represents a circle, then its radius is equal to
(a) 1
9
-1
æ
æ 2 pk ö ö
æ 2 pk ö
÷ , where i = - 1, is
÷ - i cos ç
çsin çè
è 7 ø ÷ø
è
7 ø
(a) a = b = 2 + 3
(c) a = 2 - 3, b = 2 +
8
(d) 1 + i , i =
If 1, a1, a 2, a 3, ... , a n - 1 are n, nth roots of unity, then (1 - a1) (1 - a 2 ) (1 - a 3 ) ... (1 - a n - 1) equals to
(a)
7
-1
(b) z k + 1 = k z k
k =1
6
z i 3 is
(a) z k = k z k + 1
(a) - 1
5
i =1
If z1, z 2, z 3, ... , z n are n, nth roots of unity, then for k = 1, 2, 3, ... , n
(a) 0
4
(c) i , i =
(b) 1
4
S
(b)
1
3
(c)
3
4
(d)
2
3
If centre of a regular hexagon is at origin and one of the vertex on argand diagram is 1 + 2i , where i = - 1, its
perimeter is
(a) 2 5
10
(b) an ellipse
(d) a circle
(b) f
(d) None of these
(b) a circle
(c) an ellipse
(d) a pair of straight lines
If z, iz and z + iz are the vertices of a triangle whose area is 2 units, the value of z is
(a) 1
14
(c) a hyperbola
Locus of the point z satisfying the equation iz - 1 + z - 1 = 2, is (where, i = - 1)
(a) a straight line
13
(d) 6 5
If z - 2 - 3i + z + 2 - 6i = 4, where i = - 1, then locus of P (z ) is
(a) an ellipse
(c) line segment of points 2 + 3i and - 2 + 6i
12
(c) 4 5
If z is a complex number in the argand plane, the equation z - 2 + z + 2 = 8 represents
(a) a parabola
11
(b) 6 2
(b) 2
(c) 4
(d) 8
(c) 5 + 1
(d) 3 - 1
4
If z = 2, the greatest value of z is
z
(a) 5 - 1
(b) 3 + 1
Shortcuts and Important Results to Remember
1
z1 - z2
£ z1 + z2 £ z1 + z2
Thus, z1 + z2 is the greatest possible value of
z1 + z2 and z1 - z2 is the least possible value of
z1 + z2 .
æ z - z1 ö
12 If arg ç
÷ = a (fixed), then the locus of z is a segment
è z - z2 ø
of circle.
P(z)
P(z)
α
b
If z ±
= a, then the greatest and least values of z are
z
a + (a2 + 4| b|)
- a + (a2 + 4| b|)
and
, respectively.
2
2
z1 +
( z12
2
- z2 ) + z2 -
( z12
2
- z2 )
= z1 + z2 + z1 - z2
z1 + z2 = z1 + z2
Û arg ( z1 ) = arg ( z2 )
i.e. z1 and z2 are parallel.
z1 + z2 = z1 - z2 Û arg ( z1 ) - arg ( z2 ) = p
6
z1 + z2 = z1 - z2 Û arg ( z1 ) - arg ( z2 ) = ± p / 2
7 If z1 = z2 and arg ( z1 ) + arg ( z2 ) = 0, then z1 and z2 are
conjugate complex numbers of each other.
The equation z - z1
2
2
+ z - z2 = k, k Î R will
1
represent a circle with centre at ( z1 + z2 ) and radius is
2
1
1
2
2
provided
k³
z1 - z2 .
2 k - z1 - z2
2
2
9 Area of triangle whose vertices are z, iz and z + iz,
1
where i = - 1, is | z|2 .
2
10 Area of triangle whose vertices are z, w z and z + w z is
3
2
z , where w is cube root of unity.
4
11 arg ( z ) - arg (- z ) = p or - p according as arg ( z ) is
positive or negative.
α
A(z1)
B(z2)
æ z - z1 ö
If arg ç
÷ = ± p / 2 , the locus of z is a circle with z1
è z - z2 ø
and z2 as the vertices of diameter.
æ z - z1 ö
14 If arg ç
÷ = 0 or p, the locus of z is a straight line
è z - z2 ø
passing through z1 and z2 .
15 If three complex numbers are in AP, they lie on a straight
line in the complex plane.
16 If three points z1, z2 , z3 connected by relation
a z1 + b z2 + c z3 = 0, where a + b + c = 0, the three points
are collinear.
17 If z1, z2 , z3 are vertices of a triangle, its centroid
z0 =
z1 + z2 + z 3
3
orthocentre z =
and its area =
, circumcentre z1 =
å| z1|2 ( z2 - z3 ) ,
å z1( z2 - z3 )
å z1( z2 - z3 ) + å| z1|2 ( z2 - z3 )
å ( z1 z2 - z1 z2 )
z1 z1 1
1
| z2 z2 1 |.
4
z3 z3 1
18 If| z1| = n1,| z2 | = n2 ,| z 3| = n 3 , ...,| z m | = n m,
then
n12 n22 n 23
n2
+
+
+ ... + m = | z1 + z2 + z3 + ... + zm|.
z1 z2
z3
zm
JEE Type Solved Examples :
Single Option Correct Type Questions
n
This section contains 10 multiple choice examples.
Each example has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct.
On squaring both sides, we get
a2 b2 c 2
bc
ca ö
æ ab
+ 2 + 2 + 2ç
+
+
÷
2
è a1b1 b1c 1 c 1a1 ø
a1 b1 c 1
Ex. 1 If z1 and z 2 be the n th root of unity which subtend
right angled at the origin. Then, n must be of the form
l
= 1 + i 2 + 2i
(a) 4k + 1 (b) 4k + 2
(c) 4k + 3 (d) 4k
Sol. (d) The nth roots of unity is given by
æ 2r p ö
æ 2r p ö
2r pi /n
,
cos ç
÷ + i sin ç
÷ =e
è n ø
è n ø
2 pr 1 i / n
and z 2 = e
2 pr 2 i / n
p
æz ö
arg ç 1 ÷ = ±
èz2 ø
2
l
\
n = 4k , where k = ± (r1 - r 2 )
Ex. 2 If | z | =1 and w =
½ z ½
½× 1
(c)½
2
½z + 1½ | z + 1|
(d)
c2
c 12
+2
abc æ c 1 a1 b1 ö
ç + + ÷
a1b1c 1 è c
a
bø
2
b
+
2
b12
a12
+
+
c
2
+ 0 = 2i
c 12
b2
b12
+
c2
c 12
= 2i
Ex. 4 Let z and w be complex numbers. If Re( z ) = | z - 2|,
p
Re( w ) = | w -2 | and arg ( z - w ) = , the value of Im ( z + w ) is
3
l
1
3
(b)
2
3
(d)
(c) 3
4
3
z = x + iy , x, y Î R and i = -1
Re(z ) = | z - 2|
Q
x = ( x - 2) 2 + y 2
Þ
z -1
(where z ¹ -1), then Re( w ) is
z +1
(b) -
(a) 0
+
b12
a2
Sol. (d) Let
n = ± 4(r1 - r 2 )
b2
a12
\
(a)
Þ
+
a
Þ
2pr1 2pr 2
p
=±
n
n
2
Þ
a12
= 1 - 1 + 2i
, where 0 £ r1, r 2 < n ,
It is given that the line segment joining the points having
affixes z1 and z 2 , subtends a right angled at the origin.
Therefore,
a2
Þ
where r = 0, 1, 2, K, (n - 1)
So, let z1 = e
r1 ¹ r 2 .
a
b
c
+ + =1+i
a1 b1 c 1
Sol. (a) We have,
y 2 = 4( x - 1)
Þ
\ z = 1 + t 2 + 2ti , parametric form and let w = p + iq
1
Similarly,
| z + 1|2
\
2
| z + 1|2
w = 1 + s 2 + 2si
z - w = (t 2 - s 2 ) + 2i (t - s )
p
3
Þ
arg(z - w) =
æ 2 ö p
2
tan -1 ç
Þ
= 3
÷=
t +s
èt + s ø 3
Sol. (a) We have, | z | = 1.
[given]
Let
z = e i q , where q Î R and i = -1.
\
Then,
w=
z -1 eiq -1
æq ö
= i tan ç ÷
= iq
è2ø
z +1 e +1
Þ
(t + s ) =
Now,
z + w = 2 + t 2 + s 2 + 2i (t + s )
\
Re( w) = 0
Ex. 3 If a, b, c, a 1 , b1 and c 1 are non-zero complex
a b c
a
b
c
numbers satisfying
+ + = 1 + i and 1 + 1 + 1 = 0,
a1 b1 c 1
a
b
c
2
2
2
a
b
c
where i = -1, the value of
is
+
+
2
2
a 1 b1 c 12
\
l
(a) 2i
(b) 2 + 2i
(c) 2
(d) None of these
2
3
Im(z + w) = 2(t + s ) =
4
3
æ z - 3ö p
Ex. 5 The mirror image of the curve arg ç
÷= ,
è z -i ø 6
i = -1 in the real axis, is
l
Chap 01 Complex Numbers
æz + 3ö p
(a) arg ç
÷=
èz +iø 6
æz - 3ö p
(b) arg ç
÷=
èz +iø 6
æz +iö p
(c) arg ç
÷=
èz + 3ø 6
æz +iö p
(d) arg ç
÷=
èz - 3ø 6
æz - 3ö p
- arg ç
÷=
èz + i ø 6
[Q arg(z ) = - arg(z )]
Þ
æz + i ö p
arg ç
÷=
èz - 3ø 6
é
æ z1 ö
æz2 öù
êQ arg çè z ÷ø = - arg çè z ÷ø ú
2
1 û
ë
æz +iö p
Ex. 6 The mirror image of the curve arg ç
÷= ,
è z - 1ø 4
i = -1 in the line x - y = 0, is
l
æz + iö p
(a) arg ç
÷=
è z + 1ø 4
æ z + 1ö p
(b) arg ç
÷=
èz - iø 4
æz - iö p
(c) arg ç
÷=
è z + 1ø 4
æz + iö p
(d) arg ç
÷=
è z - 1ø 4
æz + 1ö p
- arg ç
÷=
èz - i ø 4
l
Ex. 7 If z +
(b) -1
(c) 1
(d) 2
1
1
1
2
2
Sol. (d) Q
+
+
= 2w =
a+w b+w c +w
w
and
1
1
and b is the last
= 1 and a = z 2017 +
2017
z
z
2
a+w
+
1
2
b+w
+
1
2
c +w
= 2w =
2
w2
Þ
Þ
…(i)
x å (b + x )(c + x ) = 2(a + x )(b + x )(c + x )
x 3 - (ab + bc + ca )x - 2abc = 0
a + w + w2 = 0 Þ a - 1 = 0
\
a =1
Third
root
is
1.
\
From Eq. (i), we get
1
1
1
+
+
=2
a+1 b+1 c +1
Ex. 9 If a, b and c are distinct integers and w( ¹1) is a
cube root of unity, then the minimum value of
| a + bw + cw 2 | + | a + bw 2 + c w |, is
l
(a) 2 3
(b) 3
(c) 4 2
(d) 2
2
Sol. (a) Let z = a + bw + c w . Then,
of a 2 + b 2 is
(b) 24
(c) 26
(d) 27
1
= 1 Þ z2 - z + 1 = 0
z
-( - 1 ) ± ( 1 - 4 )
z=
= - w, - w2
2
[w is cube root of unity]
2017
2017
z
= ( - w)
= - w,
Sol. (c) Qz +
and
1
It is clear that, w and w2 are the roots of the equation
1
1
1
2
+
+
=
a+x b+x c +x x
[Q arg( z ) = - arg(z )]
n
\
has last digit 6.
(a) -2
Þ
digit of the number 2 2 -1,, when the integer n >1, the value
(a) 23
n-4
Q Coefficient of x 2 = 0, the sum of roots = 0
æz - i ö p é
æz ö
æz öù
arg ç
Q arg ç 1 ÷ = - arg ç 2 ÷ ú
÷=
ê
èz2 ø
è z1 ø û
èz + 1ø 4 ë
Þ
= 162
Ex. 8 If w is complex cube root of unity and a, b, c are
such that 1 + 1 + 1 = 2 w 2 and
a +w b+w c +w
1
1
1
+
+
= 2 w, then the value of
2
2
a +w
b+w
c + w2
1
1
1
is equal to
+
+
a +1 b +1 c +1
\ The image of the given curve is
æiz + i ö p
arg ç
÷=
èi z - 1ø 4
Þ
n-4
1ö
æ
= - ç w + ÷ = - ( w + w2 ) = 1
è
wø
\
b=6-1=5
2
Hence, a + b 2 = 12 + 52 = 26
Sol. (c) Q The image of z in the line y = x is iz.
Þ
1
z 2017
l
Þ
æz + 1ö p
arg ç
÷=
èz + i ø 4
n
22 = 24× 2
and
Sol. (d) Q The image of z in the real axis is z.
The image is given by
æz - 3ö p
arg ç
÷=
èz - i ø 6
a = z 2017 +
\
47
z 2017 = ( - w2 )2017 = - w2
| z | 2 = zz = (a + bw + cw2 )(a + b w + c w2 )
= (a + bw + cw2 )(a + bw2 + cw)
= a 2 + b 2 + c 2 - ab - bc - ca
1
= [(a - b )2 + (b - c )2 + (c - a )2 ]
2
ù
éQ a ¹ b ¹ c
1
2
ê
\ | a - b | ³ 1, | b - c | ³ 1ú
Þ |z | ³ ´ 6 = 3
ú
ê
2
úû
êë and | a - c | ³ 2
48
Textbook of Algebra
\
|z | ³ 3
\ | a + bw + cw2 | + | a + bw2 + cw|
2
Sol. (b) Q | 3 - i (z - 1)| = | - i (z - 1 - 3i )| = | -i || z - 1 - 3i |
= | z - 1 - 3i |
2
= | a + bw + cw | + | a + b w + c w|
= | a + bw + cw2 | + | a + bw + cw2 |
= 2| a + bw + cw2 | = 2| z | ³ 2 3
and | z - 1 - 3i | = |(z - 2i ) + ( -1 - i )| £ | z - 2i | + | -1 - i |
[Q | z1 + z 2 | £ | z1 | + | z 2 | ]
Hence, the minimum value of | a + bw + cw2 | + | a + bw2 + cw|
is 2 3.
Ex. 10 If | z - 2i | £ 2, where i = -1, then the maximum
value of | 3 - i ( z - 1) | , is
(b) 2 2
(c) 2 + 2
\
| z - 1 - 3i | £ | z - 2i | + 2
Þ
| z - 1 - 3i | £ 2 + 2
Þ
| z - 1 - 3i | £ 2 2
[Q | z - 2i | £ 2 ]
From Eq. (i), we get
l
(a) 2
…(i)
| 3 - i (z - 1)| £ 2 2
Hence, the maximum value of | 3 - i (z - 1)| is 2 2.
(d) 3 + 2 2
JEE Type Solved Examples :
More than One Correct Option Type Questions
n
This section contains 5 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
more than one may be correct.
Q
(z 2 - z 3 ) = (1 + i )(z1 - z 3 )
B(z2)
Ex. 11 If z 1 = a + ib and z 2 = c + id are complex numbers
such that | z 1 | = | z 2 | = 1 and Re( z 1 z 2 ) = 0, where i = -1,
then the complex numbers w1 = a + ic and w 2 = b + id
satisfy
l
(a) | w1| = 1
(c) Re( w1 w2 ) = 0
π/2
(b) | w2 | = 1
(d) None of these
Þ
Þ
or
Sol. (a, b, c)
Q
Þ
and
\
and
\
and
| z1| = 1, | z 2 | = 1 Þ z1 = CiS q, z 2 = CiS f
[given]
Re(z1z 2 ) = Re(CiS (q - f )) = 0
p
p
cos(q - f ) = 0 Þ q - f =
Þ f=q 2
2
a = cosq, b = sinq, c = cos f, d = sin f
w1 = a + ic = cos q + i cos f = cos q + i sin q
pù
é
êQ f = q - 2 ú
ë
û
w2 = b + id = sin q + i sin f = sin q - i cos q
pù
é
êQ f = q - 2 ú
ë
û
| w1| = 1, | w2 | = 1
Re ( w1 w2 ) = Re (cos q + i sin q )(sin q + i cos q ) = 0
Ex. 12 The complex numbers z 1 , z 2 , z 3 satisfying
( z 2 - z 3 ) = (1 + i )( z 1 - z 3 ), where i = -1, are vertices of a
triangle which is
l
(a) equilateral
(c) right angled
Sol. (b,c)
(b) isosceles
(d) scalene
A(z1)
C(z3)
Þ
( z 2 - z1 ) = i ( z1 - z 3 )
( z 2 - z1 ) = - i ( z 3 - z1 )
( z 3 - z1 ) = i ( z 2 - z1 )
z 3 - z1
= e i p/ 2
z 2 - z1
p
AC is obtained by rotating AB about A through
2
anti-clockwise.
p
\
AB = AC , ÐCAB =
2
Hence, z1, z 2 , z 3 form isosceles right angled triangle.
Ex. 13 If z satisfies the inequality | z - 1 | < | z + 1 |, then
one has
l
(a) | z - 2 - i | < | z + 2 - i |, i = -1
(b) |arg( z + i )| <
p
, i = -1
2
(c) Re( z ) < 0
(d) Im(i z ) > 0, i = -1
Sol. (a, b, d)
On putting z = x + iy in the given relation, we have
( x - 1) 2 + y 2 < ( x + 1) 2 + y 2
Þ
x >0
Chap 01 Complex Numbers
i.e.
Re(z ) > 0
and on putting z = x + iy in alternate (a), then
( x - 2) 2 + ( y - 1) 2 < ( x + 2) 2 + ( y - 1) 2
Þ
x >0
which is true.
\Real part of (z + i ) = x > 0,
p
p
then arg(z + i ) lies between - and
2
2
p
and hence |arg (z + i )| <
2
and
Im(i z ) = Im(i ( x - iy )) = Im(y + ix )
= x >0
which is true.
l
Þ ( x - 1) 2 + 3x
[from Eq. (i)]
2
[from Eq. (i)]
- y 2 + 2ixy - i (( x - 1)2 + y 2 ) = 0
Ex. 15 Let z 1 and z 2 be two complex numbers represented
by points on circles | z | =1 and | z | = 2 respectively, then
0 = - 2x + 1
1
x = =y
2
1+i
z = x + iy =
2
(a) max.| 2z 1 + z 2 | = 4
1½
½
(c)½z 2 + ½ £ 3
z 1½
½
\
2x 2 = ( x - 1) 2 + x 2
Þ
…(ii)
1+i
From Eqs. (i) and (ii), we get z =
2
i.e., no real and no purely imaginary roots
1
and
|z | =
<1
2
Q
\
Þ
\
Þ
and
Þ
\
Case I When y = x , then 2xy = ( x - 1)2 + y 2 reduces to
\
= 0 which is not possible.
(b) min.| z 1 - z 2 | = 1
2½
½
(d)½z 1 + ½ £ 2
z2½
½
Sol. (a, b, c, d)
On comparing real and imaginary parts, we get
x 2 - y 2 = 0 and 2xy = ( x - 1)2 + y 2
Þ
2
l
(a) no real root
(b) no purely imaginary root
(c) all roots inside | z | = 1
(d) atleast two roots
Sol. (a, b, c)
On putting z = x + iy , we have
( x + iy )2 - i | x + iy - 1| 2 = 0
x
Case II When y = - x , then 2xy = ( x - 1)2 + y 2 reduces to
- 2x 2 = ( x - 1) 2 + x 2
…(i)
Ex. 14 The equation z 2 - i| z - 1 | 2 = 0, where i = -1, has
Þ
49
and
…(i)
\
| z1| = 1 and | z 2 | = 2
|2z1 + z 2 | £ | 2z1| + | z 2 | = 2| z1| + | z 2 |
| 2z1 + z 2 | £ 2| z1| + | z 2 | = 2 + 2 = 4
| 2z1 + z 2 | £ 4
[alternate (a)]
max.|2z1 + z 2 | = 4
| z1 - z 2 | ³ || z1| - | z 2 || = |1 - 2| = 1
| z1 - z 2 | ³ 1
[alternate (b)]
min.| z1 - z 2 | = 1
1
1
1
1
½
½
½
½
½z 2 + ½ £ | z 2 | +½ ½ = | z 2 | +
=2+ =3
1
z1½
| z1|
½
½z1½
½
½z 2 + 1½
½£ 3
[alternate (c)]
z1½
½
½
½z1 + 2 ½
½ £ | z1| +½
½ 2½
½ = | z1| + 2 = 1 + 2 = 2
2
z 2½
|z 2 |
½
½z 2½
½
½z1 + 2 ½
½£ 2
z 2½
½
[alternate (d)]
JEE Type Solved Examples :
Passage Based Questions
n
This section contains 2 solved passages based upon each
of the passage 3 multiple choice examples have to be
answered. Each of these examples has four choices (a), (b),
(c) and (d) out of which ONLY ONE is correct.
(b) (ab + ab ) (bc + bc ) + (ca - ca )2 = 0
(c) (ab - ab ) (bc - bc ) + (ca + ca )2 = 0
(d) (ab + ab ) (bc - bc ) + (ca - ca )2 = 0
Sol. (b) Q
az 2 + bz + c = 0
Consider a quadratic equation az + bz + c = 0, where a, b and c
are complex numbers.
\
az 2 + bz + c = 0
Þ
a (z )2 + bz + c = 0
16. The condition that the equation has one purely
imaginary root, is
For purely imaginary root,
z = -z
Then,
az 2 - bz + c = 0
Passage I (Ex. Nos. 16 to 18)
2
(a) (ab - ab ) (bc + bc ) + (ca - ca )2 = 0
…(i)
…(ii)
50
Textbook of Algebra
From Eqs. (i) and (ii), we get
z2
z
1
=
=
ca
ca
bc + bc
- ab - ab
Þ
\
z=
Passage II (Ex. Nos. 19 to 21)
Let P be a point denoting a complex number z on the complex
plane.
i.e.
z = Re (z ) + i Im (z ), where i = -1
bc + bc
ca - ca
=
ca - ca - ab - ab
(ab + ab ) (bc + bc ) + (ca - ca ) = 0
17. The condition that the equation has one purely real
root, is
(b) (ab - ab ) (bc + bc ) = (ca + ca )2
(c) (ab - ab ) (bc - bc ) = (ca - ca )2
(d) (ab - ab ) (bc - bc ) = (ca + ca )2
…(i)
Þ
az 2 + bz + c = 0
Þ
a (z )2 + bz + c = 0
19. If P moves such that
|Re ( z )| + | Im( z )| = a (a Î R + )
The locus of P is
(a) a parallelogram which is not a rhombus
(b) a rhombus which is not a square
(c) a rectangle which is not a square
(d) a square
Sol. (d) Q
|Re( z )| + |Im( z )| = a
(a) (ab + ab ) (bc - bc ) = (ca + ca )2
Sol. (c) Q az 2 + bz + c = 0
Re (z ) = x and Im (z ) = y , then z = x + iy
If
2
Þ
Y
a
For purely real root, z = z
X′
2
az + bz + c = 0
Then,
| x | + |y | = a
–a
Y′
2
z
z
1
=
=
bc - bc ca - ca ab - ab
Þ
z=
\ Locus of P is a square.
20. The area of the circle inscribed in the region denoted
by |Re ( z )| + |Im ( z )| =10 equals to
bc - bc
ca - ca
=
ca - ca ab - ab
(a) 50 p sq units
(c) 55 sq units
Sol. (a) From above, a = 10
(ab - ab ) (bc - bc ) = (ca - ca )2
18. The condition that the equation has two purely
imaginary roots, is
(b) -
Sol. (d) Q
az 2 + bz + c = 0
\
az 2 + bz + c = 0
Þ
a ( z )2 + bz + c = 0
…(i)
z = -z
az - bz + c = 0
2
Hence, Eqs. (i) and (ii) are identical.
a
b c
\
=- =
a
b c
or
2r = 10 2
Þ
r =5 2
\ Area of a circle = pr 2 = 50p sq units
21. Number of integral solutions satisfying the inequality
|Re ( z )| + |Im ( z )| < 21, is
(a) 841
(c) 840
Sol. (c) Q
Since, both roots are purely imaginary.
\
Then,
(b) 100 p sq units
(d) 110 sq units
Diameter of circle = Distance between sides of square
= Length of side of square = a 2 = 10 2
a b c
= =
a b c
a
b c
(d) = - =
a
b c
a b c
= =
a b c
a b
c
(c) = = a b
c
(a)
X
–a
From Eqs. (i) and (ii), we get
Þ
a
O
…(ii)
…(ii)
(b) 839
(d) 842
| x | + | y | < 21 Þ 0 £ | x | + | y | £ 20
x > 0, y > 0, 0 £ x + y £ 20
21 × 20
Number of solutions = 21C 2 =
= 210
2
\ Total integral solutions = 4 ´ 210 = 840
If
Chap 01 Complex Numbers
51
JEE Type Solved Examples :
Single Integer Answer Type Questions
n
This section contains 2 examples. The answer to each
example is a single digit integer ranging from 0 to 9
(both inclusive).
Y
B
Ex. 22 If z 1 , z 2 Î C , z 12 + z 22 Î R , z 1 ( z 12 - 3 z 22 ) = 2 and
z 2 (3 z 12 - z 22 ) = 11, the value of z 12 + z 22 is
X′
l
Sol. (5) We have,
and
z1 (z12 - 3z 22 ) = 2
…(i)
z 2 (3z12 - z 22 ) = 11
…(ii)
multiplying Eq. (ii) by i ( -1 ) and then adding in Eq. (i), we
get
z13 - 3z1z 22 + i (3z12z 2 - z 32 ) = 2 + 11i
Þ
(z1 + iz 2 )3 = 2 + 11i
…(iii)
Again, multiplying Eq. (ii) by (- i) and then adding in Eq. (i),
we get
2
z13 - 3z1z 22 - i (3z1 z 2 - z 23 ) = 2 - 11i
Þ
(z1 - iz 2 )3 = 2 - 11i
…(iv)
Now, on multiplying Eqs. (iii) and (iv), we get
(z12 + z 22 )3 = 4 + 121 = 125 = 53
\
z12 + z 22 = 5
Ex. 23 The number of solutions of the equations
| z - ( 4 + 8i )| = 10 and | z - (3 + 5i )| + | z - ( 5 + 11i )| = 4 5,
where i = - 1.
l
Sol. (2) Here, | z - ( 4 + 8i )| = 10
represents a circle with centre (4, 8) and radius 10.
A′
S1
O
S2
A
X
B′
Y′
Also, | z - ( 3 + 5i )| + | z - ( 5 + 11i )| = 4 5
represents an ellipse.
\ |(3 + 5i ) - (5 + 11i )| = 4 + 36 = 40 < 4 5
with foci S1(3, 5) and S 2 (5, 11).
Distance between foci = S1S 2 = 40 = 2 10 = Diameter of
circle
i.e.,
2ae = 2 10
ae = 10 and 2a = 4 5 Þ a = 2 5
1
ae
\
=
e=
a
2
æ 1ö
2
Now, b = a (1 - e ) = 2 5 ç1 - ÷ = 10 = Radius of circle
è 2ø
Þ
\ Centre of the ellipse = Mid-point of S1 and S 2
3 + 5i + 5 + 11i
= 4 + 8i i.e., ( 4, 8)
=
2
which coincides with the centre of the circle and length of
minor-axis is equal to the radius of the circle. Hence, there
are only (2) two solutions of the given equations.
JEE Type Solved Examples :
Matching Type Questions
n
This section contains 2 examples. Examples 24 and 25 have three statements (A, B and C) given in Column I and four
statements (p, q, r and s) in Column II. Any given statement in Column I can have correct matching with one or more
statement(s) given in Column II.
l
Ex. 24
Column I
Column II
(A)
If l and m are the greatest and least values of | z - 1 | , if | z + 2 + i | £ 1, where i = -1, then
(p) l + m = rational
(B)
If l and m are the greatest and least values of | z - 2 | , if | z + i | £ 1, where i = -1, then
(q) l + m = irrational
(C)
If l and m are the greatest and least values of | z + 2i | , if 1 £ | z - 1| £ 3, where i = -1, then
(r) l - m = rational
(s) l - m = irrational
52
Textbook of Algebra
Sol. (A) ® (q, r); (B) ® (q, r); (C) ® (p, s)
From the figure, the greatest value of | z - 2| = | w |
= | w - 0| = OB
= OP + PB = 5 + 1
|z + 2 + i | £ 1
(A) Q
Þ
|(z - 1) + (3 + i )| £ 1
Þ
| w + (3 + i )| £ 1 where, w = z - 1
From the figure, the greatest value of
| z - 1| = | w| = | w - 0| = OB = OP + PB = 10 + 1
l = 10 + 1
\
P
1
A
X′
O
X
and the least value of | z - 1| = | w|
= | w - 0 | = OP - AP = 10 - 1
\
m = 10 - 1
Þ
l + m = ( 10 + 1) + ( 10 - 1) = 2 10 = irrational
and
l - m = ( 10 + 1) - ( 10 - 1) = 2 = rational
Aliter
Q
|z + 2 + i | £ 1
Þ
|(z - 1) + (3 + i )| £ 1
Þ
| w + (3 + i )| £ 1
where, w = z - 1
\
| w + (3 + i )| ³ || w| - | 3 + i ||
or
| w + (3 + i )| ³ || w| - 10 |
From Eqs. (i) and (ii), we get
|| w| - 10 | £ | w + 3 + i | £ 1
Þ
l - m = ( 5 + 1) - ( 5 - 1) = 2 = rational
…(ii)
|z + i | £ 1
|(z - 2) + (2 + i )| £ 1
| w + (2 + i )| £ 1
w= z - 2
| w + (2 + i )| ³ || w| - | 2 + i ||
| w + (2 + i )| ³ || w| - 5 |
or
10 - 1 £ | w| £ 10 + 1
Þ
|| w| - 5| £ 1
or
- 1 £ | w| - 5 £ 1 or
5 - 1 £ | w| £ 5 + 1
l = 5 + 1 and m = 5 - 1
Þ
l + m = 2 5 = irrational
and
(C) Q
Þ
Þ
where,
l - m = 2 = rational
1 £ | z - 1| £ 3
1 £ |(z + 2i ) - (1 + 2i )| £ 3
1 £ | w - (1 + 2i )| £ 3
w = z + 2i
Y
P
(1, 2)
l + m = 2 10 = irrational
O
and l - m = 2 = rational
(B) Q
|z + i | £ 1
Þ
|(z - 2) + ( 2 + i )| £ 1
Þ
| w + (2 + i )| £ 1
where,
w= z - 2
X
B
From the figure, the greatest value of | z + 2i | = | w|
= | w - 0| = OA = OP + PA = 5 + 3
Y
\
l =3+ 5
and the least value of | z + 2i | = | w|
= | w - 0| = OB = PB - OP = 3 - 5
(–2, –1)
P
1
…(ii)
From Eqs. (i) and (ii), we get
|| w| - 5| £ | w + 2 + i | £ 1
l = 10 + 1 and m = 10 - 1
B
…(i)
A
- 1 £ | w| - 10 £ 1
Þ
and
|| w| - 10 | £ 1
or
\
l + m = ( 5 + 1) + ( 5 - 1) = 2 5 = irrational
\
…(i)
m= 5 -1
Þ
Aliter
Q
Þ
Þ
where,
Q
or
(–3, –1)
l = 5 +1
and the least value of | z - 2| = | w |
= | w - 0| = OA = OP - AP = 5 - 1
\
Y
B
\
\
A
O
X
m=3- 5
Þ
l + m = (3 + 5 ) + (3 - 5 ) = 6 = rational
and
l - m = (3 + 5 ) - (3 - 5 ) = 2 5 = irrational
…(i)
Chap 01 Complex Numbers
l
Ex. 25
Column I
(A) If (3 - 4 i ) +
Column II
(- 3 - 4 i ) = z, the principal
(p)
0
value of arg (z) can be (where i = -1)
(B) If (5 + 12i ) +
(q)
±
p
4
(- 15 - 8i ) = z, the principal (r)
±
p
2
±
3p
4
(- 5 + 12i ) = z, the principal
value of arg (z) can be (where i = -1)
(C) If (-15 + 8i ) +
value of arg (z) can be (where i = -1)
(s)
Sol. (A) ® (q, s); (B) ® (q, s); (C) ® (p, r)
Q
æ | z | + Re (z )
z =±ç
+i
2
è
| z | - Re(z ) ö
÷, Im (z ) > 0
2
ø
æ | z | + Re(z )
=±ç
-i
2
è
| z | - Re (z ) ö
÷, Im (z ) < 0
2
ø
= 3 - 3i, 1 + i, - 1 - i, - 3 + 3i
p p
3p 3p
\ Principal values of arg (z ) = - , , ,
4 4
4 4
æ 13 + 5
13 - 5 ö
(B)
+i
(5 + 12i ) = ± ç
÷
2
2 ø
è
= ± (3 + 2i )
æ 13 - 5
( - 5 + 12i ) = ± ç
+i
2
è
= ± (3 + 2i ) ± (2 + 3i )
= 5 + 5i , 1 - i , - 1 + i , - 5 - 5i
p
p 3p
3p
\ Principal values of arg (z ) = , - , , 4
4 4
4
æ 17 - 15
17 + 15 ö
(C)
- 15 + 8i = ± ç
+i
÷
2
2 ø
è
= ± (1 + 4i )
5 - 3ö
÷ = ± (2 - i )
2 ø
and
æ 5-3
( - 3 - 4i ) = ± ç
-i
2
è
5 + 3ö
÷ = ± (1 - 2i )
2 ø
Q
z = ( 3 - 4i ) + ( - 3 - 4i )
\
z = ± (2 - i ) ± (1 - 2i )
13 + 5 ö
÷
2 ø
= ± (2 + 3i )
z = (5 + 12i ) + ( - 5 + 12i )
Q
æ 5+3
(A) (3 - 4i ) = ± ç
-i
2
è
Q
- 15 - 8i = - 15 + 8i = ± (1 + 4i )
= ± (1 - 4i )
z = ( - 15 + 8i ) + ( - 15 - 8i )
= ± (1 + 4i ) ± (1 - 4i ) = 2, 8i , - 8i , - 2
p
p
\ Principal values of arg (z ) = 0, , - , p.
2
2
JEE Type Solved Examples :
Statement I and II Type Questions
n
Directions Example numbers 26 and 27 are
Assertion-Reason type examples. Each of these examples
contains two statements:
Statement-1 (Assertion) and Statement-2 (Reason)
Each of these examples also has four alternative choices,
only one of which is the correct answer. You have to select
the correct choice as given below.
(a) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
Ex. 26 Consider four complex numbers z 1 = 2 + 2i,
z 2 = 2 - 2i, z 3 = - 2 - 2i and z 4 = - 2 + 2i, where i = -1.
Statement-1 z 1 , z 2 , z 3 and z 4 constitute the vertices of a
square on the complex plane because
Statement-2 The non-zero complex numbers z , z , - z , - z
always constitute the vertices of a square.
53
Sol. (c) Statement-1
AB = | z1 - z 2 | = | 4i | = 4,
BC = | z 2 - z 3 | = | 4 | = 4,
CD = | z 3 - z 4 | = | - 4i | = 4
DA = | z 4 - z1| = | - 4 | = 4
AC = | z1 - z 3 | = | 4 + 4i | = 4 2
BD = | z 2 - z 4 | = | 4 - 4i | = 4 2
and
Y
z4 = –2 + 2i
D
z1 = 2 + 2i
A
l
X′
X
O
B
z2 = 2 – 2i
C
z3 = –2 – 2i
Y′
54
Textbook of Algebra
It is clear that, AB = BC = CD = DA and AC = BD
Hence, z1, z 2 , z 3 and z 4 are the vertices of a square.
\ Statement-1 is true.
Statement-2 If z = a + ib
If a ¹ b
Then, AB = | z - z | = |(a + ib ) - (a - ib )| = 2| b |
BC = | z - ( - z )| = | z + z | = | a - ib + a + ib | = 2 | a |
\
AB ¹ BC
Statement-2 is false.
Þ (| z1 | + | z 2 | )2 = | z1| 2 + | z 2 | 2 + 2 | z1| | z 2 | cos (q 1 - q 2 )
[from Eq. (i)]
Þ
\
z2
O
æz ö
amp ç 1 ÷ = 0
èz2 ø
Þ
…(i)
If amp (z1 ) = q 1 and amp (z 2 ) = q 2 , then
2
cos (q 1 - q 2 ) = 1
Þ
q1 - q 2 = 0
or
amp (z1 ) - amp (z 2 ) = 0
\ Statement-1 is true.
Statement-2 Since, z1, z 2 and O (origin) are collinear,
then
æ O - z1 ö
amp ç
÷ =0
èO - z2 ø
Sol. (b) Statement-1
2
| z1 | + | z 2 | + 2 | z1 | | z 2 |
z1
Ex. 27 Consider z 1 and z 2 are two complex numbers
such that | z 1 + z 2 | = | z 1 | + | z 2 |
Statement-1 amp ( z 1 ) - amp ( z 2 ) = 0
Statement-2 The complex numbers z 1 and z 2 are collinear
with origin.
| z1 + z 2 | = | z1| + | z 2 |
2
= | z1 | 2 + | z 2 | 2 + 2 | z1 | | z 2 | cos (q 1 - q 2 )
l
Q
2
Þ
amp (z1 ) - amp (z 2 ) = 0
\ Statement-2 is true, which is not a correct explanation of
Statement-1.
2
| z1 + z 2 | = | z1 | + | z 2 | + 2 | z1 | | z 2 | cos (q 1 - q 2 )
Subjective Type Examples
n
l
Ex. 28 If z - i Re ( z ) = z - Im ( z ) , then prove that z
lies on the bisectors of the quadrants, where i = -1.
Sol. Let z = x + iy , where x , y Î R and i =
-1
\
Re (z ) = x and Im (z ) = y
Then, z - i Re (z ) = z - Im (z )
Þ
x + iy - ix = x + iy - y
x - i ( x - y ) = ( x - y ) + iy
Þ
Ex. 30 Let S denotes the real part of the complex number
5 + 2i 20 + 5i
z=
+
+ 3i, where i = -1, K denotes the sum
2 - 5i 7 + 6i
of the imaginary parts of the roots of the equation
z 2 - 8 (1 - i ) z + 63 - 16i = 0 and G denotes the value of
l
2012
S i r , where i = -1, find the value of S - K + G.
Þ
x
2
+ ( x - y )2 = ( x - y )2 + y 2
r =4
Þ
x
2
+ ( x - y )2 = ( x - y )2 + y 2
Sol. For S,
x
2
Þ
= y 2 or y = ± x
Hence, z lies on the bisectors of the quadrants.
l
Hence, the least value of z1 + z 2 is 19 and the greatest
value of | z1 + z 2 | is 31.
In this section, there are 24 subjective solved examples.
Ex. 29 Find the greatest and the least values of z 1 + z 2 ,
if z 1 = 24 + 7i and z 2 = 6, where i = -1.
Sol. Q z1 = 24 + 7i
\ z1 = (24 )2 + (7 )2 = 25
z1 - z 2
Þ
or
£ z1 + z 2 £ z1 + z 2
25 - 6 £ z1 + z 2 £ 25 + 6
19 £ z1 + z 2 £ 31
z=
(5 + 2i ) (20 + 5i )
+ 3i
+
(2 - 5i ) (7 + 6i )
(5 + 2i ) (2 + 5i ) (20 + 5i ) (7 - 6i )
+
+ 3i
29
85
0 + 29i 170 - 85i
=
+
+ 3i
29
85
= i + 2 - i + 3i = 2 + 3i
\ Re (z ) = 2
\ S =2
For K,
Put z = x + iy in the given equation, then
( x + iy )2 - 8 (1 - i ) ( x + iy ) + 63 - 16i = 0
=
Chap 01 Complex Numbers
On comparing the real and imaginary parts, we get
x 2 - y 2 - 8 ( x + y ) + 63 = 0
…(i)
and
…(ii)
xy + 4 ( x - y ) = 8
On substituting the value of x from Eq. (ii) in Eq. (i), we get
y 4 + 16y 3 + ... = 0
\
For G,
\
1
Ex. 32 If arg ( z 1/ 3 ) = arg ( z 2 + z z
2
of z .
l
G=
S
r =4
r
i =
2009
S
r =1
i
r +3
=i
1+ 3
Þ
1
arg (z 2 + z z 1 / 3 )
2
2 arg (z 1/ 3 ) = arg (z 2 + z z 1/ 3 )
Þ
arg (z 2 / 3 ) = arg (z 2 + z z 1/ 3 )
+ 0 =1
) - arg (z
2/3
)=0
Þ
æz + z z
arg ç
è z2/ 3
1/ 3
ö
÷ =0
ø
Þ
z ö
æ
arg çz 4 / 3 + 1/ 3 ÷ = 0
è
z ø
Þ
S - K + G = 2 - ( - 16) + 1 = 19
Ex. 31 If z - 1 = 1, where z is a point on the argand
z -2
plane, show that
= i tan (arg z ), where i = -1 .
z
2
arg (z + z z
z -1 =1 Þ z -1
2
Þ
=1
Þ
( z - 1) ( z - 1) = 1 Þ z z - z - z = 0
z
(z + z ) = z z Þ
Þ
+1=z
z
z
…(i)
Þ
_ =z -1
z
æ Im (z ) ö
Now, RHS = i tan (arg z ) = i ç
÷
è Re (z ) ø
ì z -z ü
ïï
ïï
æ z -z ö
= i í 2i ý = i ç
÷
z +z
è i (z + z ) ø
ï
ï
ïî 2 ïþ
z
-1
z -z z
( z - 1) - 1 z - 2
[from Eq. (i)]
=
=
=
=
z
z +z
z
+ 1 ( z - 1) + 1
z
= LHS
Aliter
We have, z - 1 = 1 i.e.(z - 1) is unimodular, so we can take
z - 1 = cos q + i sin q
\
z - 2 = - 1 + cos q + i sin q
q
q
q
= - 2 sin 2 + 2i sin cos
2
2
2
q
q
2
2 q
= 2i sin
+ 2i sin cos
2
2
2
q æ
q
qö
…(i)
or
z - 2 = 2i sin ç cos + i sin ÷
2 è
2
2ø
and
z = 1 + cos q + i sin q
q
q
q
= 2 cos 2 + 2i sin cos
2
2
2
q æ
q
qö
z = 2 cos ç cos + i sin ÷
2 è
2
2ø
[by property]
[by property]
is purely real.
1/ 3
z ö
æ
Im çz 4 / 3 + 1/ 3 ÷ = 0
è
z ø
z ö
z ö æ
æ 4/3
+ 1/ 3 ÷ - çz 4 / 3 + 1/ 3 ÷
çz
è
z ø =0
z ø è
2i
z
Þ
z4/ 3 +
4/3
+
—
z
z 1/ 3
(z ) (z )1 / 3
z
2/3
= (z )
4/3
+
= (z )4 / 3 +
(z )
(z )1 / 3
z (z )1 / 3
z
2/3
[Qz 1 / 3 (z )1 / 3 = (z z )1 / 3 = z
1
Þ z 4 / 3 - (z )4 / 3 z
Þ
2/3
2/3
]
((z )4 / 3 - (z )4 / 3 ) = 0
é
1 ù
ú =0
{z 4 / 3 - (z )4 / 3 } ê1 2 /3
úû
êë
z
\
z
2/3
=1
[Qz ¹ z ]
z =1
Therefore,
Ex. 33 C is the complex numbers f : C ® R is defined by
f ( z ) = z 3 - z + 2 . Find the maximum value of f ( z ), if
l
z =1.
\
From Eqs. (i) and (ii), we get
q
z -2
= i tan
2
z
z -2
Therefore,
= i tan (arg z ) [Q arg (z ) = q /2 from Eq. (ii)]
z
z
Þ
Sol. Q
…(ii)
z
z 4/3 +
Þ
Þ
1/ 3
2
l
Sol. Given,
), find the value
Sol. We have, arg (z 1 / 3 ) =
K = - 16
2012
1/ 3
55
z =1
z = e iq
\ f ( e iq ) = e
= e
3iq
- e iq + 2 = e
2iq
2 iq
( e iq - e - iq ) + 2
× 2i sin q + 2
= (cos 2q + i sin 2q ) × 2i sin q + 2
= (2 - 2 sin 2q sin q ) + 2i sin q cos 2q
= 2 (1 - sin 2q sin q ) + i sin q cos 2q
= 2 (1 - sin 2q sin q )2 + (sin q cos 2q )2
56
Textbook of Algebra
II. Aliter
Here,
OA = OB
From Rotation theorem,
z1 - 0 OA 2 pi / 3
=
e
z 2 - 0 OB
= 2 (1 + sin 2 q - 2 sin q sin 2q )
= 2 1 + sin 2 q (1 - 4 cos q )
= 2 1 + (1 - cos q ) (1 + cos q ) (1 - 4 cos q )
For maximum value, cosq = -
1ù
é
êQ cos q ¹ - 1, 1, 4 ú
ë
û
1
2
æ3ö æ1ö
\ Maximum value of f (z ) = 2× 1 + ç ÷ ç ÷ (3) = 13
è2ø è2ø
Ex. 34 Prove that the complex numbers z 1 and z 2 and
2p
the origin form an isosceles triangle with vertical angle
, if
3
2
2
z 1 + z 2 + z 1 z 2 = 0.
Þ
z1 æ
2p
2p ö
= ç cos
+ i sin
÷
z2 è
3
3 ø
Þ
z1
1 i 3
=- +
z2
2
2
l
Sol. Given,
2
z1 +
2
z 2 + z1z 2 = 0
z2
z1 = wz 2 or z1 = w z 2
2
l
¾®
¾®
¾®
2
z1 + z 2 + z1z 2 = 0
+ z1z 2 + z 2 = 0
2
2
Ex. 35 If a = e 2 pi / 7 , where i = -1 and
20
0
+ å A k x k , then find the value of
k =1
\
a = e 2 pi / 7
a 7 = e 2 pi = cos 2p + i sin 2p = 1 + 0 = 1 or a = (1)1/ 7
\ 1, a , a 2 , a 3 , a 4 , a 5 , a 6 are the seven, 7th roots of unity.
Q f (x ) = A0 +
2 pi ù
é
êQ w = e 3 ú
úû
êë
20
å
Ak x k =
20
å
Ak x k
k =0
Now, f ( x ) + f (ax ) + f (a 2 x ) + ... + f (a 6 x )
=
20
å
Ak x k [( (1)k + (a )k + (a 2 )k + ... + (a 6 )k ]
k =0
¾®
i.e., OA is obtained by rotating OB through angle of
= A 0 x 0 (7 ) + A 7 x 7 (7 ) + A14 x 14 (7 )
2p
.
3
= 7 ( A 0 + A 7 x 7 + A14 x 14 )
é
k
k
2 k
6 k
ê Q(1) + (a ) + (a ) + ... + (a )
ë
ù
ì 7, k is multiple of 7
=í
ú
0,
k
is
not
multiple
of
7
î
û
O
2π/3
B(z2)
z1
+1=0
z2
\
Sol. Q
OA = OB e 2 pi / 3
Þ
+
Þ
k =1
( z 1 - 0) = ( z 2 - 0) e 2 pi / 3
2
f ( x ) + f (ax ) + f (a 2 x ) + ... + f (a 6 x ) independent of a.
2
In the first case, z1 = wz 2
Þ
1 z1
3
+
=4 z2
4
2
z1
f (x ) = A
z1 = wz 2 or z1 = w2z 2
Þ
z2
Þ z1 = z 2
(z1 - wz 2 ) (z1 - w2z 2 ) = 0
Þ
z1
Þ
z1 + z 2 + z1z 2 = 0
Given,
+
2
2
2
Hence, two sides equal
amp (z1 ) = amp (w) z 2 = amp (w) + amp (z 2 )
2p
Þ amp (z1 ) =
+ amp (z 2 )
3
2p
Þ
amp (z1 ) - amp (z 2 ) =
3
2p
So, the angle between two sides is .
3
Similarly, the other case
I. Aliter
A(z1)
2p
3
Thus, triangle formed by z1, z 2 and origin is isosceles with
2p
vertical angle .
3
Þ
z1
Þ
In the first case, z1 = wz 2 Þ z1 = w z 2
…(ii)
On squaring both sides in Eq. (ii), we get
2
( z 1 - wz 2 ) ( z 1 - w z 2 ) = 0
Þ
[from Eq. (i)]
æ z1 1 ö i 3
ç + ÷=
èz2 2ø
2
Þ
2
Þ
…(i)
OA = OB and ÐAOB =
Ex. 36 Show that all the roots of the equation
a 1 z 3 + a 2 z 2 + a 3 z + a 4 = 3, (where a i £ 1, i = 1, 2, 3, 4) lie
l
outside the circle with centre at origin and radius 2 / 3.
Sol. Given that, a1z 3 + a 2 z 2 + a 3z + a 4 = 3
Chap 01 Complex Numbers
We have, 3 = a1z 3 + a 2 z 2 + a 3z + a 4
On squaring both sides, we get
(z 3 - z 2 )2
z 3 + a2
3 £ a1
Þ
3 £ a1 z
Þ
Þ
Þ
3£ z
3
3
+ a2
+ z
2
2
z
z + a4
+ a3
z + a4
+ z +1
3£1+ z + z
2
3<1+ z + z
2
=
z 2 + a3
[Q | ai | £ 1]
+ z
3
< 1+ z + z
+ z
3
+ ...
1
1 - |z |
Þ 3<
2
+ z
3
+ ...
1
1- z
Aliter
Q
1- z <
æp a ö
ÐABC = ç - ÷
è2 2ø
From Coni method, we have
æp
z1 - z 2
AB i çè 2 =
e
z 3 - z 2 BC
aö
÷
2ø
…(i)
æp a ö
ÐACB = ç - ÷
è2 2ø
From Coni method, we have
and
Ex. 37 If A, B and C represent the complex numbers
z 1 , z 2 and z 3 respectively on the complex plane and the
1
angles at B and C are each equal to ( p - a ), then prove
2
a
2
that ( z 2 - z 3 ) = 4 ( z 3 - z 1 ) ( z 1 - z 2 ) sin 2 æç ö÷ .
è2ø
l
æp
aö
z 2 - z 3 BC i çè 2 - 2 ÷ø
=
e
z1 - z 3 AC
On dividing Eq. (ii) by Eq. (i), we get
…(ii)
(z 2 - z 3 )2
( BC )2
æ BC ö
=ç
=
÷
(z 3 - z1 ) (z1 - z 2 ) AB × AC è AB ø
Sol. It is given that,
Ð ABC = ÐACB =
= - 4 sin 2
Therefore, (z 2 - z 3 )2 = 4 sin 2 (a / 2) (z 3 - z1 ) (z1 - z 2 )
2
1
Þ - z <0
3
3
Þ
z > 2 / 3 or z - 0 > 2 / 3
Hence, all the roots lie in the exterior of circle,
z - 0 = 2 / 3.
Þ
2
a æ
a
aö
ç cos + i sin ÷
2 è
2
2ø
( z 2 - z1 ) 2
a
= - 4 sin 2 (cos a + i sin a )
2
[from De-Moivre’s theorem]
æ z - z1 ö
(z 3 - z 2 )2
a
= - 4 sin 2 ç 3
÷ [from Eq. (ii)]
2
2 è z 2 - z1 ø
( z 2 - z1 )
3 £ a1z 3 + a 2 z 2 + a 3z + a 4
Þ
57
p a
2 2
ö
æ
÷
ç
sin a
÷
=ç
ç
æp a ö÷
sin
ç
÷
ç
è 2 2 ø ÷ø
è
Þ
ÐA = a
…(i)
\
AC = AB
So, D ABC is an isosceles triangle.
Considering rotation of AB about A through angle a, we get
2
[Q AB = AC ]
2
[using sine rule]
2
æ
æa ö
æa ö ö
ç 2 sin ç ÷ cos ç ÷ ÷
è
ø
è 2 ø÷
2
æa ö
=ç
= 4 sin 2 ç ÷
è2ø
ç
÷
æa ö
cos ç ÷
ç
÷
è2ø
è
ø
A(z1)
a
Therefore, (z 2 - z 3 )2 = 4 (z 3 - z1 ) (z1 - z 2 ) sin 2 a / 2
p– a
2 2
Ex. 38 If z 1 and z 2 are two complex numbers such that
z1 - z 2
iz
= 1, then prove that 1 = k , where k is a real
z1 + z 2
z2
number. Find the angle between the lines from the origin to
the points z 1 + z 2 and z 1 - z 2 in terms of k.
p– a
2 2
B(z2)
Þ
z 3 - z1
=eia
z 2 - z1
Þ
z 3 - z1
= cos a + i sin a
z 2 - z1
l
C(z3)
…(ii)
z 3 - z1
- 1 = cos a - 1 + i sin a
z 2 - z1
Þ
a
a
a
z3 - z2
= - 2 sin 2 + 2i sin cos
2
2
2
z 2 - z1
or
z3 - z2
a æ
a
aö
= 2i sin ç cos + i sin ÷
è
z 2 - z1
2
2
2ø
Sol. (i) Given,
Þ
…(iii)
Þ
z1 - z 2
=1
z1 + z 2
z1
-1
z2
=1
z1
+1
z2
z1
z
-1 = 1 +1
z2
z2
…(i)
58
Textbook of Algebra
(ii) Now, let the angle between OB and OA be q, then from
Coni method,
z1 + z 2 - 0
OB iq
=
e
z1 - z 2 - 0 OA
On squaring Eq. (i) both sides, we have
z1
z2
Þ
2
z
æz ö
+ 1 - 2 Re ç 1 ÷ = 1
èz2 ø
z2
æz ö
4 Re ç 1 ÷ = 0
èz2 ø
2
æz ö
+ 1 + 2 Re ç 1 ÷
èz2 ø
=
z1
is purely imaginary number
z2
z
z
Þ 1 can be written as i 1 = k , where k is a real number.
z2
z2
Þ
æ z1 + z 2 ö
iq
ç
÷ =e
è z1 - z 2 ø
Þ
(ii) Let q be the angle between z1 - z 2 and z1 + z 2 , then
Þ
Q(z1 + z2)
z 1 + z 2 iq
e
z1 - z 2
[from Eq. (i)]
æ z1
ö
ç + 1÷
- ki + 1
z
ç 2
÷ = e iq Þ
= e iq
z
- ki - 1
ç 1 - 1÷
ç
÷
è z2
ø
[from Eq. (ii)]
B(z1 + z2)
O
θ
P(z1 – z2)
ö
æ z1
ç + 1÷
æ - ik + 1 ö
æ z1 + z 2 ö
z
2
÷ = arg ç
q = arg ç
÷
÷ = arg ç
è - ik - 1 ø
è z1 - z 2 ø
ç z1 - 1 ÷
÷
ç
ø
è z2
O
æ k 2 - 1 + 2ik ö
æ -1 + ik ö
= arg ç
÷
÷ = arg ç
è 1 + ik ø
è k 2 +1 ø
Therefore, q = tan
Aliter (i) Given,
Let
Þ
A(z1 – z2)
Þ
- 1 + ki
= e iq
1 + ki
Þ
( - 1 + ki ) (1 - ki )
= e iq
(1 + ki ) (1 - ki )
( k 2 - 1)
2ki
= e iq
æ 2k ö
ç 2
÷
èk - 1ø
Þ
z1 - z 2
=1
z1 + z 2
\
Re (e iq ) = cos q =
and
Im(e iq ) = sin q =
-1
…(i)
z1 - z 2 cos a + i sin a
=
z1 + z 2
1
+
1 + k2
tan q =
l
k2 - 1
k2 + 1
2k
k2 + 1
2k
2
k -1
æ 2k ö
q = tan - 1 ç 2
÷
èk -1ø
Therefore,
Ex. 39 If z = x + iy is a complex number with rationals x
and y and z =1, then show that z 2 n - 1 is a rational
æa ö æ
æa ö
æa öö
2 cos ç ÷ ç cos ç ÷ + i sin ç ÷ ÷
è2ø è
è2ø
è 2 øø
=
æa ö æ
æa ö
æa öö
-2i sin ç ÷ ç cos ç ÷ + i sin ç ÷ ÷
è2ø è
è2ø
è 2 øø
number for every n Î N .
Sol. Since, z = 1, where z is unimodular
\
z = cos q + i sin q
As x and y are rational, cos q , sinq are rationals
1ö
n
æ
zn - z - n
\ z 2 n - 1 = z n çz n - n ÷ = z
è
z ø
æa ö
cot ç ÷
è2ø
z1
=z2
i
iz1
æa ö
= - cot ç ÷ = k ( say) = real
Þ
è2ø
z2
iz1
Hence,
=k
z2
( k 2 + 1)
\
(z1 - z 2 ) + (z1 + z 2 ) 1 + cos a + i sin a
[ by
Þ
=
(z1 + z 2 ) - (z1 - z 2 ) 1 - cos a - i sina
componendo and dividendo]
æa ö
æa ö
æa ö
2 cos 2 ç ÷ + 2i sin ç ÷ cos ç ÷
è2ø
è2ø
è2ø
z1
Þ
=
z2
æa ö
æa ö
2 æa ö
2 sin ç ÷ - 2i sin ç ÷ cos ç ÷
è2ø
è2ø
è2ø
Þ
θ
= 1 2i sin nq
= 2 sin nq
Since, sinnq is rational, therefore z 2 n - 1 is a rational
...(ii)
number.
Chap 01 Complex Numbers
l
Ex. 40 If a is a complex number such that a =1, then
Therefore, z lies on the right bisector of the segment
connecting the points 0 + i × 0 and - 1 + 0 × i. Thus,
Re (z ) = - 1 / 2. Hence, roots are collinear and will have
their real parts equal to - 1 / 2. Hence, sum of the real parts
æ 1
ö
of roots is ç - (n - 1)÷.
è 2
ø
find the value of a, so that equation az 2 + z + 1 = 0 has one
purely imaginary root.
Sol. We have, az 2 + z + 1 = 0
…(i)
On taking conjugate both sides, we get
Aliter
Q
az 2 + z + 1 = 0
_
Þ
a(z )2 + z + 1 = 0
Þ
a ( -z ) 2 + ( -z ) + 1 = 0
n
Þ
= (cos 2r p + i sin 2r p)1/ n
Þ
On dividing each by 4, we get
æa - a ö
æa + a ö
-ç
÷ +ç
÷ =0
è 2i ø
è 2 ø
or
- ( Im (a ))2 + Re (a ) = 0
or
æ1ö
z =-ç ÷i×
è2ø
-1± 1+ 4
2
5 -1
Only feasible value of cos a =
2
Hence,
a = cos a + i sin a
æ 5 -1ö
where,
a = cos - 1 ç
÷
è 2 ø
cosa =
Ex. 41 If n Î N >1, find the sum of real parts of the roots
of the equation z n = ( z + 1) n .
l
Sol. The equation z n = (z + 1)n will have exactly n - 1 roots.
We have,
n
æz + 1ö
ç
÷ =1 Þ
è z ø
z +1
|z |
=1
z +1 = z
z - ( - 1) = z - 0
æz + 1ö
ç
÷
è z ø
n
= 1
r pi
n
2 r pi / n
æ pr ö
× 2i sin ç ÷
èn ø
1
×e
æ pr ö
sin ç ÷
èn ø
pr i
n
pr
pr ö
æ
- i sin
ç cos
÷
è
n ø
n
æi ö
=-ç ÷×
pr
è2ø
sin
n
1
\
[ here r ¹ 0]
Re (z ) = 2
where, r = 1, 2, 3, ... , n - 1
1 1 1
Sum of real parts of z = - - - - ... - (n - 1) times
2 2 2
1
= - (n - 1).
2
…(iii)
Given,
a =1
Let
a = cos a + i sin a
\
Re (a ) = cos a , Im (a ) = sin a
Then, from Eq. (iii), we get
- sin 2 a + cos a = 0 or cos 2 a + cos a - 1 = 0
\
1
æ 2rp ö
æ 2r p ö
= cos ç
÷ + i sin ç
÷=e
è n ø
è n ø
z
1
= (e 2r pi / n - 1) = e
z
2
Þ
1+
or
2
æa - aö
æa + a ö
ç
÷ +ç
÷ =0
è 2 ø
è 2 ø
Þ
Þ
z +1
æz + 1ö
= (1)1 / n
ç
÷ = 1 or
è z ø
z
= (cos 0 + i sin 0)1 /n
Eliminating z from Eqs. (i) and (ii) by cross-multiplication
rule, we get
( a - a )2 + 2 (a + a ) = 0
Þ
z n = (z + 1)n
[since, z is purely imaginary, z = - z ]
…(ii)
a z2 - z + 1 = 0
or
59
Ex. 42 Prove that the angle between the line a z + a z = 0
and its reflection in the real axis is
ì 2 Re (a ) Im (a ) ü
q = tan -1 í
.
2
2ý
î {Im (a )} - {Re (a )} þ
l
Sol. Let z = x + iy , then equation a z + a z = 0 can be written as
Þ
( a + a) x + i ( a - a) y = 0
æa - a ö
æa + a ö
÷y =0
÷x+ç
ç
è 2i ø
è 2 ø
Þ
{Re (a )} x + {Im (a )} y = 0
{Re (a )}
\ Slope of the given line (m ) = {Im (a )}
Then,
tan (180°- f ) = -
{Re (a )}
{Re (a )}
Þ tan f =
{Im (a )}
{Im (a )}
Hence, angle between the given line and its reflection in
real axis
60
Textbook of Algebra
ide
nt
az+
ra
az= y
0φ
Imaginary axis
Inc
\
\
z P = 12 + 16i
Similarly,
zQ = - 12 + 16i
From the figure, E is the point with least modulus and D is
the point with maximum modulus.
ra y
ed
t
c
fle
Re
φ
φ
O
Real axis
ì 2 tan f ü
= 2f = tan -1 {tan 2f } = tan -1 í
2 ý
î 1 - tan f þ
ì
{Re (a )} ü
ï 2
ï
ì 2 Re (a ) Im (a ) ü
ï
{Im(a )} ï
= tan -1 í
= tan -1 í
2ý
2
2 ý
î {Im (a )} - {Re(a )} þ
ï 1 - {Re (a )} ï
ïî
{Im(a )} 2 ïþ
P
R
P
are represented by the points inside and on the circle of
radius 15 and centre at the point C (0, 25).
The complex numbers having least positive argument and
maximum positive arguments in this region are the points
of contact of tangents drawn from origin to the circle.
Here, q = Least positive argument
and f = Maximum positive argument
\ In DOCP , OP = (OC )2 - (CP )2 = (25)2 - (15)2 = 20
\
|a |2 = |b |2 = |c |2 = |d |2 = r 2
Þ
aa =bb =c c =d d =r2
OP 20 4
=
=
OC 25 5
4
æ4ö
Þ q = tan -1 ç ÷
tan q =
è3ø
3
sin q =
Thus, complex number at P has modulus 20 and argument
æ4ö
q = tan -1 ç ÷
è3ø
Q
r2
r2
r2
r2
and d =
,b = ,c =
c
b
a
d
For line PQ, points a, b and z are collinear, then
z z 1
a a 1 =0
a=
Þ
N
z
c
b b
φ
S
or
\
40i
C 25i
and
a
d
…(i)
θ
¾®
| z | = r in the points a, b and c , d , then
| a | = r , | b | = r , | c | = r and | d | = r
Tangent from
origin
O
¾®
b
| z - 25i | £ 15
E
¾®
z D = OD = OC + CD = 25i + 15i = 40i
Sol. Let two non-parallel straight lines PQ , RS meet the circle
Sol. The complex numbers z satisfying the condition
θ
¾®
and
least positive argument.
maximum positive argument.
least modulus.
maximum modulus.
Q
¾®
z E = OE = OC - EC = 25i - 15i = 10i
Ex. 44 Two different non-parallel lines meet the circle
| z | = r in the points a , b and c , d , respectively. Prove that these
a -1 + b -1 - c -1 - d -1
lines meet in the point z given by z =
,
a -1b -1 - c -1d -1
where a, b, c, d are complex constants.
Ex. 43 Among the complex numbers z which satisfies
| z - 25i | £ 15, find the complex numbers z having
D
¾®
Hence,
l
l
(i)
(ii)
(iii)
(iv)
4ö
æ3
z P = 20 (cos q + i sin q ) = 20 ç + i ÷
è5
5ø
1
z ( a - b ) - z (a - b ) + (ab - ab ) = 0
æ ar 2 br 2 ö
ær
r2ö
Þ z ç - ÷ - z (a - b ) + ç
÷ =0
a ø
bø
è b
èa
2
On dividing both sides by (b - a ), we get
r2
r2
z +z (a + b ) = 0
ab
ab
(a + b )
z
z
Þ
+ 2 =0
ab r
ab
Similarly, for line RS, we get
z
z
(c + d )
+ 2 =0
cd r
cd
On subtracting Eq. (ii) from Eq. (i), we get
1 ö (a + b ) (c + d )
æ1
zç - ÷+
=0
è ab cd ø
ab
cd
Þ
z (a -1b -1 - c -1d -1 ) = a -1 + b -1 - c -1 - d -1
Therefore,
z=
a -1 + b -1 - c -1 - d -1
a -1b -1 - c -1d -1
…(i)
…(ii)
Chap 01 Complex Numbers
l Ex. 45 If n is an odd integer but not a multiple of 3, then
prove that xy ( x + y ) ( x 2 + y 2 + xy ) is a factor of
(x + y )n - x n - y n .
Sol. We have, xy ( x + y ) ( x 2 + y 2 + xy ) = xy ( x + y )
( x - w y ) ( x - w2y )
and let
f ( x , y ) = ( x + y )n - x n - y n
…(i)
On putting x = 0 in Eq. (i), we get
f (0, y ) = y n - 0 - y n = 0
\ x - 0 is a factor of Eq. (i).
On putting y = 0 in Eq. (i), we get
f ( x , 0) = x n - x n = 0
\ y - 0 is a factor of Eq. (i).
On putting x = - y in Eq. (i), we get
f ( - y , y ) = ( - y + y )n - ( - y )n - y n
61
Sol. (i) Since, | z - 1 | + | z + 1 | = 4
i.e., (distance of z from the point 1 + 0 × i ) + (distance
of z from the point - 1 + 0 × i ) = 4 (constant)
i.e., The sum of the distances of z from two fixed
points 1 + 0 ×i and - 1 + 0 × i is constant, which is the
definition of an ellipse.
Therefore, locus of z satisfying the given condition
will be an ellipse with foci at 1 + 0 ×i and - 1 + 0 × i and
centre at origin.
(ii) Given that,
p
arg(z + i ) - arg(z - i ) =
2
æz + i ö p
or
…(i)
arg ç
÷=
èz - i ø 2
Let the points A and B have affixes i and - i and the
point P has affix z. Then, Eq. (i) can be written as
= 0 - ( - y )n - y n = y n - y n = 0 [because n is odd]
\ x + y is a factor of Eq. (i).
On putting x = wy in Eq. (i), we get
f ( wy , y ) = ( wy + y )n - ( wy )n - y n
= y n [( w + 1)n - wn - 1]
= y n [( - w2 )n - wn - 1)]
[Q1 + w + w2 = 0]
= - y n { w2 n + wn + 1}
[because n is odd]
ÐBPA =
Thus, locus of P (z ) is such that the angle subtended at
p
P by the line joining points A and B is . This is the
2
definition of a circle with diameter AB.
Y
A(i)
Since, n is odd but not a multiple of 3, then n = 3k + 1 or
n = 3k + 2 , where k is an integer.
[in both cases] …(ii)
\
w2 n + wn + 1 = 0
\
f ( wy , y ) = 0
\
x - wy is a factor of Eq. (i).
On putting x = w2y in Eq. (i), we get
é
æz + i öù
÷ú
êQ ÐBPA = arg ç
èz - i øû
ë
p
2
π/2
P(z)
X
O
B(–i)
f ( w2y , y ) = ( w2y + y )n - ( w2y )n - y n
= y n {( w2 + 1)n - w2 n - 1}
n
n
2n
= y {( - w) - w - 1}
= - y n { wn + w2 n + 1}
=0
\ x - w2y is a factor of Eq. (i).
[because, n is odd]
[from Eq. (ii)]
Combining all the factors, we get
( x - 0) (y - 0) ( x + y ) ( x - wy ) ( x - w2y )
2
2
Therefore, xy ( x + y ) ( x + xy + y ) is a factor of
f ( x , y ) = ( x + y )n - x n - y n .
Ex. 46 Interpret the following equations geometrically on
the Argand plane.
p
(i) | z - 1 | + | z + 1 | = 4 (ii) arg ( z + i ) - arg ( z - i ) =
2
p
p
(iii) 1 < | z - 2 - 3i | < 4 (iv) < arg ( z ) <
4
3
ì | z - 1| + 4 ü
(v) log cos p/ 3 í
ý>1
î 3 | z - 1 | - 2þ
Therefore, locus of point z is a circle with diameter AB
and centre at origin with radius 1.
(iii) We have, 1 < | z - 2 - 3i | < 4 represents a circle with
centre at ( 2, 3) and radius r Î(1, 4 ).
1
C
(2, 3)
4
l
Since, | z - 2 - 3i | > 1 represents the region in the
plane outside the circle.
…(i)
\
| z - 2 - 3i | = 1
and | z - 2 - 3i | < 4 represents the region inside circle.
…(ii)
\
| z - 2 - 3i | = 4
Hence, 1 < | z - 2 - 3i | < 4 represent the angular space
between the two circles (i) and (ii).
62
Textbook of Algebra
p
p
< arg (z ) <
4
3
æy ö
Let z = x + iy Þ arg (z ) = tan -1 ç ÷
èx ø
On multiplying Eqs. (i) and (ii), we get
( z1 - z 2 ) ( z 2 - z 3 ) = ( z1 - z 3 ) 2
(iv) We have,
2
2
2
z1 + z 2 + z 3 = z1z 2 + z 2 z 3 + z 3z1
or
Hence, the triangle whose vertices are z1, z 2 and z 3 is
equilateral.
Similarly, it can be shown that the triangle whose vertices
are z1 ¢ , z 2 ¢ and z 3 ¢ is also equilateral.
y=
√3x
Y
y=
Ex. 48 Show that the triangle whose vertices are
z 1 , z 2 , z 3 and z 1 ¢ , z 2 ¢ , z 3 ¢ are directly similar, if
z1 z1¢ 1
z 2 z 2 ¢ 1 = 0.
l
x
π/3 π/4
O
X
z3 z3 ¢ 1
The given inequality can be written as
p
æy ö p
< tan -1 ç ÷ <
èx ø 3
4
Þ
Þ
Sol. Let A , B, C be the points of affix z1, z 2 , z 3 and A ¢ , B ¢ , C ¢ be
the points of affix z1 ¢ , z 2 ¢ , z 3 ¢.
Since, the triangles ABC and A ¢ B ¢ C ¢ are similar, if
p y
p
< < tan
4 x
3
y
1< < 3
x
¾®
Þ
¾¾®
BC = l B ¢ C ¢
i.e.,
(z 3 - z 2 ) = l (z 3 ¢ - z 2 ¢ )
tan
¾®
x <y < 3 x
and
CA = l C ¢ A ¢
i.e.,
( z1 - z 3 ) = l ( z1 ¢ - z 3 ¢ )
This inequality represents the region between the lines
or
Þ
A(z1)
ì |z - 1 | + 4 ü
ý>1
í
î 3 | z - 1 | - 2þ
ì |z - 1 | + 4 ü
log1 / 2 í
ý>1
î 3 | z - 1 | - 2þ
1
|z - 1 | + 4
<
3 |z - 1 | - 2 2
α
B(z2)
or
2 |z - 1 | + 8 < 3 |z - 1 | - 2
Þ
| z - 1 | > 10
Hence, the inequality represents exterior of a circle of
radius 10 with centre at (1, 0).
Ex. 47 Show that the triangles whose vertices are
z 1 , z 2 , z 3 and z 1¢, z 2 ¢ , z 3 ¢ are equilateral, if
(z 1 - z 2 ) (z 1 ¢ - z 2 ¢ ) = (z 2 - z 3 ) (z 2 ¢ - z 3 ¢ )
= (z 3 - z 1 ) (z 3 ¢ - z 1 ¢ )
l
Also, from the last two relations
( z 2 - z 3 ) ( z 2 ¢ - z 3 ¢ ) = ( z 3 - z1 ) ( z 3 ¢ - z1 ¢ )
C ′ (z3′ )
On dividing Eq. (i) by Eq. (ii), we get
z3 - z2 z3 ¢ - z2 ¢
=
z1 - z 3 z1 ¢ - z 3 ¢
z 3 ( z1 ¢ - z 3 ¢ ) - z 2 ( z1 ¢ - z 3 ¢ )
= z1 ( z 3 ¢ - z 2 ¢ ) - z 3 ( z 3 ¢ - z 2 ¢ )
Þ z1(z 2 ¢ - z 3 ¢ ) - z 2 (z1 ¢ - z 3 ¢ ) + z 3 (z1 ¢ - z 2 ¢ ) = 0
z1 z1 ¢ 1
Hence,
z2 z2 ¢ 1 = 0
Þ
[say]
Then, from Coni method in D ABC and D A ¢ B ¢C ¢, we have
z1 - z 2
AB ia
…(i)
=
e
z 3 - z 2 BC
( z1 - z 2 ) + ( z 2 - z 3 )
z - z3
= 1
( z 2 ¢ - z 3 ¢ ) + ( z1 ¢ - z 2 ¢ ) z1 ¢ - z 3 ¢
z1 - z 2
z - z3
= 1
z 2 ¢ - z 3 ¢ z1¢ - z 3¢
B ′ (z2′)
Aliter
Since, DABC and DA ¢ B ¢ C ¢ are similar.
AB
BC
If
and ÐABC = ÐA ¢ B ¢C ¢ = a
=
A ¢B ¢ B ¢C ¢
z1 - z 2
z - z3
= 2
z 2 ¢ - z 3 ¢ z1 ¢ - z 2 ¢
\
α
C(z3)
z3 z3 ¢ 1
Sol. From the first two relations, we have
=
…(ii)
A ′ (z1′ )
y = x and y = 3x
(v) We have, log cos p/ 3
…(i)
¾¾®
…(i)
…(ii)
and
z 1 ¢ - z 2 ¢ A ¢ B ¢ ia
=
e
z 3 ¢ - z 2 ¢ B ¢C ¢
…(ii)
63
Chap 01 Complex Numbers
AB
BC
AB A ¢ B ¢
\
=
=
A ¢ B ¢ B ¢C ¢
BC B ¢ C ¢
z - z 2 z1 ¢ - z 2 ¢
'
From Eqs. (i) and (ii), we get 1
=
z3 - z2 z3 ¢ - z2 ¢
Þ
(a1z1 + a 3z 3 ) = - (a 2z 2 + a 4z 4 )
On dividing Eq. (ii) by Eq. (i), we get
a1z1 + a 3z 3 a 2 z 2 + a 4 z 4
=
a1 + a 3
a2 + a4
z1 z1 ¢ 1
On simplifying as in Ist method, we get z 2 z 2 ¢ 1 = 0
z3 z3 ¢ 1
Eq. (iii) implies that point O divides PR in the ratio a 3 : a1
and O divides QS in the ratio a 4 : a 2 .
Let OR = a1k , OP = a 3k , OQ = a 4l , OS = a 2 l
Now, in DOPQ,
( PQ )2 = (OP )2 + (OQ )2 - 2 (OP ) (OQ ) cos q
Since,
Ex. 49 If w is the nth root of unity and z 1 , z 2 are any two
complex numbers, then prove that
l
n -1
å
2
k
z1 + w z 2
2
= n { z1
2
+ z2
}, where n Î N .
\
S
k =0
n -1
S
k
w = 0 and
k =0
k
( w) = 0
n -1
LHS =
S
k =0
2
+ a 3a 4a 22l 2
- 2a1a 2 a 3a 4lk cos q
2
From given condition, a1a 2 | z1 - z 2 | = a 3a 4 | z 3 - z 4 | 2
Similarly, a 3a 4 | z 3 - z 4 | =
…(i)
2
z1 + wk z 2
S
k =0
k
( z1 + w z 2 ) ( z1 + ( w ) z 2 )
Þ
k =0
n -1
S |z1|2 + k S= 0 z1z 2 ( w)k
k =0
n -1
= | z1| 2
n -1
S z1z 2 wk
k =0
+
n -1
n -1
+
n -1
S 1 + z1z 2 k S= 0 ( w)k
k =0
+ z1z 2
S
k =0
n -1
[from Eq. (i)]
= n {| z1 | 2 + | z 2 | 2 } = RHS
l
4
4
i =1
j =1
4
i =1
Þ
Þ
z 1 , z 2 , z 3 and z 4 are concyclic, if
a 1a 2 | z 1 - z 2 | 2 = a 3 a 4 | z 3 - z 4 | 2 .
S
Sol. Since, z +
\
Ex. 50 Let S a i = 0 and S a j z j = 0, then prove that
Sol. Q
Ex. 51 If a and b are the roots of
1
z + = 2 (cos q + i sin q ), where 0 < q < p and i = -1 ,
z
show that | a - i | = | b - i |.
l
| z 2 |2
S ( w)k + |z 2 |2 k S= 0 1
k =0
= n | z1| 2 + 0 + 0 + n | z | 2
(a1k ) (a 3k ) = (a 2 l ) (a 4 l )
Þ
OP × OR = OQ × OS
So, P , Q , R and S are concyclic.
= å {z1z1 + z1z 2 ( w)k + z1z 2 wk + z 2 z 2 ( wk ) ( w)k }
n -1
k 2a 3a1 (a 2 a 3 - a1a 4 ) = l 2a 2 a 4 (a 2 a 3 - a1a 4 )
Þ
k
n -1
=
Þ
z+
1
= 2 (cos q + i sin q )
z
1
= 2e iq Þ z 2 - 2e iqz + 1 = 0
z
2e iq ± ( 4e 2 i q - 4 )
z=
2
iq
2 iq
z = e ± (e
- 1) Þ z = e i q ± e i q × 2i sin q
z - i = e i q - i ± e i q × 2i sin q
= (e i q - e i p/ 2 ) ± e i (q + p/ 2 ) × 2sin q
ai = 0
P(z1)
=e
O
θ
S(z4)
\
R(z3)
a1 + a 2 + a 3 + a 4 = 0
(a1 + a 3 ) = - (a 2 + a 4 )
4
and
\
S
j =1
aj z j = 0
a1z1 + a 2 z 2 + a 3z 3 + a 4 z 4 = 0
æ q pö
iç + ÷
è 2 4ø
æ q pö
ç - ÷
4ø
× 2i sin è 2
± e i(
q /2+ p /4
)
× 2sin q
ü
ì
æq p ö
= e i (q / 2 + p/ 4 )í2i sin ç - ÷ ± 2sin q ý
è
ø
2
4
þ
î
θ
Q(z2)
\
Þ
a 3a 4a12k 2
\ a1a 2 a 32k 2 + a1a 2 a 42l 2 = a 3a 4a12k 2 + a 3a 4a 22 l 2
n -1
=
a1a 2 | z1 - z 2 | 2 = a1a 2 a 23k 2 + a1a 2 a 42l 2
-2a1a 2 a 3a 4lk cos q
Sol. If 1, w, w2 , w3 , ..., wn - 1 are the n, nth roots of unity, then
n -1
…(iii)
| z1 - z 2 | 2 = a 32k 2 + a 42l 2 - 2a 3a 4lk cos q
Þ
k =0
…(ii)
…(i)
æq p ö
| z - i | = 1 × 4 sin 2 ç - ÷ + 2 sin q
è2 4ø
æ
p öö
æ
= 2 ç1 - cos çq - ÷ ÷ + 2 sin q
è
è
2 øø
= 2 - 2sin q + 2sin q = 2
Þ | a - i | = | b - i | = 2 [here, a ,b are two values of z - i ]
#L
Complex Numbers Exercise 1 :
Single Option Correct Type Questions
n
This section contains 30 multiple choice questions.
Each question has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct
1. If cos (1 − i ) = a + ib, where a, b ∈ R and i = − 1, then
1
1
1
1
e −  cos 1, b = e +  sin 1
2
2
e
e
1
1
1
1
(b) a = e +  cos 1, b = e −  sin 1
2
2
e
e
1
1
1
1
(c) a = e +  cos 1, b = e +  sin 1
e
e
2
2
(a) a =
(d) a =
(d) 6
polygon of n sides with centre at origin and if
Im (z )
= 2 − 1, the value of n is equal to
Re (z )
(b) 4
(c) 6
(d) 8
r
4. If
Π e ipθ = 1, where Π denotes the continued product
p =1
and i = − 1, the most general value of θ is
2nπ
(a)
, n ∈I
r (r − 1 )
4nπ
, n ∈I
(c)
r (r − 1 )
2nπ
(b)
, n ∈I
r (r + 1 )
4nπ
(d)
, n ∈I
r (r + 1 )
(where, n is an integer)
5. If (3 + i ) (z + z ) − (2 + i ) (z − z ) + 14 i = 0, where i = − 1,
then z z is equal to
(a) 10
(c) − 9
(b) 8
(d) – 10
6. The centre of a square ABCD is at z = 0, A is z 1 . Then,
the centroid of ∆ABC is
(a) z1 (cos π ± i sin π )
π
π

(c) z1  cos ± i sin 

2
2
z1
3
z1
(d)
3
(b)
(cos π ± i sin π )
π
π

 cos ± i sin 

2
2
(where, i = − 1)
7. If z =
3 −i
, where i = − 1, then (i 101 + z 101 )103 equals
2
to
(a) iz
(c) z
(c) ab − ab = 0
(d) ab + ab = 0
(a)
3. If z and z represent adjacent vertices of a regular
(a) 2
(b) ab − ab = 0
Re (α + α 2 + α 3 + α 4 + α 5 ) is
real parts are negative, is
(c) 5
(a) ab + ab = 0
8π 
 8π 
 + i sin   , where i = − 1, then
 11 
 11 
1
1
1
1
e −  cos 1, b = e −  sin 1
e
e
2
2
(b) 4
z is a variable complex number. If the lines
a z + a z + 1 = 0 and b z + b z − 1 = 0 are mutually
perpendicular, then
9. If α = cos 
2. Number of roots of the equation z 10 − z 5 − 992 = 0, where
(a) 3
8. Let a and b be two fixed non-zero complex numbers and
(b) z
(d) None of these
1
2
(b) −
1
2
(c) 0
(d) None of these
10. The set of points in an Argand diagram which satisfy both
π
| z | ≤ 4 and 0 ≤ arg (z ) ≤ , is
3
(a) a circle and a line
(c) a sector of a circle
(b) a radius of a circle
(d) an infinite part line
11. If f ( x ) = g ( x 3 ) + xh( x 3 ) is divisible by x 2 + x + 1, then
(a)
(b)
(c)
(d)
g( x ) is divisible by ( x − 1 ) but not h( x )
h( x ) is divisible by ( x − 1 ) but not g( x )
both g ( x ) and h ( x ) are divisible by ( x − 1 )
None of the above
12. If the points represented by complex numbers
z 1 = a + ib, z 2 = c + id and z 1 − z 2 are collinear, where
i = − 1, then
(a) ad + bc = 0
(c) ab + cd = 0
(b) ad − bc = 0
(d) ab − cd = 0
13. Let C denotes the set of complex numbers and R is the
set of real numbers. If the function f : C → R is defined
by f (z ) = | z |, then
(a) f
(b) f
(c) f
(d) f
is injective but not surjective
is surjective but not injective
is neither injective nor surjective
is both injective and surjective
14. Let α and β be two distinct complex numbers, such that
| α | = | β |. If real part of α is positive and imaginary part
of β is negative, then the complex number
(α + β ) / (α − β ) may be
(a) zero
(c) real and positive
(b) real and negative
(d) purely imaginary
15. The complex number z, satisfies the condition
 z − 25  = 24. The maximum distance from the origin of
z

coordinates to the point z, is
(a) 25
(c) 32
(b) 30
(d) None of these
Chap 01 Complex Numbers
16. The points A, B and C represent the complex numbers
z 1 , z 2 , (1 − i ) z 1 + iz 2 respectively, on the complex plane
(where, i = − 1 ). The ∆ABC, is
(a) isosceles but not right angled
(b) right angled but not isosceles
(c) isosceles and right angled
(d) None of the above
(b) one solution
(d) None of these
(b) 1 + z
(d) None of these
on the complex plane, is
(b)
1
(c)
3
(d) None of these
− 1, then arg ( x + yz ) is
r =1
equal to
(a) 0
 2
(b) − tan −1  
 3 
 2 
(c) − tan −1  
 3
 2
(d) π − tan −1  
 3 
n −1
the value of
∑
r =1
1
, is
(3 − z r )
n ⋅ 3n − 1 1
(a) n
+
2
3 −1
(b)
n ⋅ 3n − 1
+1
3n − 1
n ⋅ 3n − 1
−1
3n − 1
(d) None of these
26. If z = (3 + 7i ) ( λ + iµ ), when λ , µ ∈ I ~ { 0 } and i = − 1,
(b) 58
(d) 3364
27. Given, z = f ( x ) + ig ( x ), where i = −1 and
f , g : (0, 1) → (0, 1) are real-valued functions, which of the
following hold good?
(a) z =
 1 
 1 
1
1
+i 
+i 
 (b) z =

 1 + ix 
 1 − ix 
1 − ix
1 + ix
(c) z =
 1 
 1 
1
1
+i 
+i 
 (d) z =

 1 + ix 
 1 − ix 
1 + ix
1 − ix
28. If z 3 + (3 + 2i ) z + ( − 1 + ia ) = 0, where i = −1, has one
real root, the value of a lies in the interval (a ∈ R )
21. If centre of a regular hexagon is at origin and one of the
vertices on Argand diagram is 1 + 2i, where i = − 1,
(a) ( − 2, − 1 )
(c) ( 0, 1 )
(b) ( − 1, 0 )
(d) (1, 2 )
29. If m and n are the smallest positive integers satisfying
then its perimeter is

the relation 2 CiS

(b) 4 5
(d) 8 5

n

22. Let | z r − r | ≤ r , ∀ r = 1, 2, 3, …, n, then ∑ z r is less than
r = 1
(c) n(n + 1 )
locus of z, is
(a) 0
3364
(c)
3
20. If x = 9 1 / 3 ⋅ 9 1 / 9 ⋅ 9 1 / 27 … ∞, y = 4 1 / 3 ⋅ 4 − 1 / 9 ⋅ 4 1 / 27 … ∞ and
(a) n

is purely imaginary then minimum value of | z | 2 is
5
3
(a) 0
(a) 2 5
(c) 6 5

(c)
19. The centre of the circle represented by | z + 1 | = 2 | z − 1 |
∞
π
4
(d) 26
25. If 1, z 1 , z 2 , z 3 , …, z n − 1 are the n, nth roots of unity, then
and dividing it by z + i, we get the remainder 1 + i.
Then, the remainder upon the division of f (z ) by z 2 + 1,
is
∑ (1 + i ) − r , where i =
(c) 10
(a) a pair of straight lines (b) circle
(c) parabola
(d) ellipse
18. Dividing f (z ) by z − i, we obtain the remainder 1 − i
z=
(b) 2 2
sin  − arg z 
, where i = − 1, then
24. If | z − 2 − i | = | z |
(where, i = − 1)
(a) i + z
(c) 1 − z
z 

z1 − 
|
z | π

z
23. If arg 
= and − z 1 = 3, then | z 1 | equals to
 z  2

| z |


 |z | 
(a) 3
17. The system of equations | z + 1 − i | = 2 and | z | = 3 has
(a) no solution
(c) two solutions
65
(b) 2n
n(n + 1 )
(d)
2

π

6
m

=  4 CiS

n
π
 , where i = − 1,
4
(m + n ) equals to
(a) 60
(b) 72
(c) 96
(d) 120
30. Number of imaginary complex numbers satisfying the
equation, z 2 = z ⋅ 21 − | z | is
(a) 0
(b) 1
(c) 2
(d) 3
66
#L
Textbook of Algebra
Complex Numbers Exercise 2 :
More than One Option Correct Type Questions
n
This section contains 15 multiple choice questions.
Each question has four choices (a), (b), (c) and (d) out of
which MORE THAN ONE may be correct.
z +1
is a purely imaginary number (where i = − 1),
z +i
then z lies on a
31. If
(a) straight line
(b) circle
i = − 1) satisfies
33. If the complex number is (1 + ri ) = λ (1 + i ), when
3
i = − 1, for some real λ, the value of r can be
(b) cosec
(d) tan
π
12
3π
2
34. If z ∈ C , which of the following relation(s) represents a
circle on an Argand diagram?
(b) | z − 3 | = 2
(d) (z − 3 + i ) ( z − 3 − i ) = 5
be a non-real complex cube root of unity, then
n −1
∏ (ω − z r ) can be equal to
r =1
(b) − 1
(d) 1
36. If z is a complex number which simultaneously satisfies
the equations
3 | z − 12 | = 5 | z − 8i | and | z − 4 | = | z − 8 |, where
i = − 1 , then Im (z ) can be
(a) 8
(c) 7
(c) 3.5
(d) 4
π
2π
39. If arg (z + a ) = and arg (z − a ) = (a ∈ R + ), then
6
3
π
(c) arg (z ) =
3
(b) | z | = 2a
(d) arg (z ) =
π
2
40. If z = x + iy, where i = −1, then the equation

 = m represents a circle, then m can be
 (2z − i )
 (z + i ) 
(a)
1
2
(b) 1
(c) 2
(d) ∈(3, 2 3 )
41. Equation of tangent drawn to circle | z | = r at the point
A (z 0 ), is
z
(a) Re   = 1
z0
z
(b) Im   = 1
z0
z 
(c) Im  0  = 1
z
(d) z z 0 + z 0 z = 2 r 2
42. z 1 and z 2 are the roots of the equation z 2 − az + b = 0,
35. If 1, z 1 , z 2 , z 3 , …, z n − 1 be the n, nth roots of unity and ω
(a) 1 + ω
(c) 0
(b) 2.5
(a) | z | = a
(a) | ω − 5 − i | < | ω + 3 + i | (b) | ω − 5 | < | ω + 3 |
π
(c) Im (iω ) > 1
(d) | arg (ω − 1 )| <
2
(a) | z − 1 | + | z + 1 | = 3
7
(c) | z − 2 + i | =
3
(where, i = −1 )
 z − z4 
z − z4 
(d) amp  1
 ≠ amp  2

z2 − z4 
z3 − z4 
(a) 2
32. If z satisfies | z − 1 | < | z + 3 |, then ω = 2 z + 3 − i (where,
π
5
π
(c) cot
12
(c) | z1 − z 3 | ≠ | z 2 − z 4 |
38. If | z − 3 | = min {| z − 1 |, | z − 5 | }, then Re(z ) is equal to
1
2
(d) circle passing through the origin
(c) circle with radius =
(a) cos
z1 − z 4
is purely real
z2 − z3
z − z3
is purely imaginary
(b) 1
z2 − z4
(a)
(b) 17
(d) 15
37. If P (z 1 ), Q(z 2 ), R(z 3 ) and S(z 4 ) are four complex
numbers representing the vertices of a rhombus taken in
order on the complex plane, which one of the following
is hold good?
where | z 1 | = | z 2 | = 1 and a, b are non-zero complex
numbers, then
(a) | a | ≤ 1
(c) arg (a ) = arg (b 2 )
(b) | a | ≤ 2
(d) arg (a 2 ) = arg (b )
43. If α is a complex constant, such that αz 2 + z + α = 0 has
a real root, then
(a) α + α = 1
(b) α + α = 0
(c) α + α = −1
(d) the absolute value of real root is 1
44. If the equation z 3 + (3 + i ) z 2 − 3z − (m + i ) = 0, where
i = −1 and m ∈ R, has atleast one real root, value of m is
(a) 1
(b) 2
(c) 3
(d) 5
45. If z + (3 + 2i ) z + ( − 1 + ia ) = 0, where i = − 1, has one
3
real root, the value of a lies in the interval (a ∈ R )
(a) ( − 2, 1 )
(b) ( − 1, 0 )
(c) ( 0, 1 )
(d) (− 2, 3)
Chap 01 Complex Numbers
67
Complex Numbers Exercise 3 :
Passage Based Questions
n
This section contains 4 passages. Based upon each of
the passage 3 multiple choice questions have to be
answered. Each of these questions has four choices (a),
(b), (c) and (d) out of which ONLY ONE is correct.
Passage I
(Q. Nos. 46 to 48)
if arg ( z ) < 0
π ,
arg ( z ) + arg ( − z ) = 
, where
−
,
π
if arg ( z ) > 0

− π < arg ( z ) ≤ π.
46. If arg (z ) > 0, then arg ( − z ) − arg (z ) is equal to
(a) − π
(c)
(b) −
π
2
π
2
that | z 1 | = | z 2 | and arg (z 1 z 2 ) = π, then z 1 is equal to
(b) z 2
(d) − z 2
 z1 
is equal to
z2 
48. If arg ( 4 z 1 ) − arg (5 z 2 ) = π, then
(Q. Nos. 49 to 51)
Sum of four consecutive powers of i (iota) is zero.
i.e., i n + i n + 1 + i n + 2 + i n + 3 = 0, ∀ n ∈ I .
25
Σ i n ! = a + ib, where i =
n =1
(a)
(b)
(c)
(d)
Σ
r=−2
ir +
Σ
+b
(a) 2
(b) 4
(c) 6
(d) 8
(a) 12
(b) 18
(c) 24
(d) 30
3
54. If z −  = 6 and sum of greatest and least values of | z | is
z

2λ, then λ 2 , is
(a) 12
(b) 18
(c) 24
(d) 30
Passage IV
(Q. Nos. 55 to 57)
Consider the two complex numbers z and w, such that
z −1
= a + ib, where a, b ∈ R and i = − 1.
w=
z+2
55. If z = C iS θ, which of the following does hold good?
9b
1 − 4a
1 − 5a
(b) cos θ =
1 + 4a
(a) sin θ =
b
a
56. Which of the following is the value of − , whenever it
, is
(b) 3
(d) 6
101
Σ i r ! + rΠ= 1 i r
r=4
(a) 11
(c) 33
z

is λ, then λ 2 , is
(d) All of these
i r ! = a + ib, where i = −1, the unit place
r=0
2011
2012
(a) 2
(c) 5
100
1
52. If z −  = 2 and sum of greatest and least values of | z |
(c) (1 + 5a ) 2 + (3b ) 2 = (1 − 4a ) 2
50
digit of a
51. If
−1, then a − b , is
prime number
even number
composite number
perfect number
95
50. If
and equality holds iff origin z 1 and z 2 are collinear
and z 1 , z 2 lie on the same side of the origin.
(b) 1.25
(d) 2.50
Passage II
49. If
For any two complex numbers z1 and z 2 ,
| z | − | z 2 |
| z1 − z 2 | ≥  1
 | z 2 | − | z1 |
z

is λ, then λ 2 , is
47. Let z 1 and z 2 be two non-zero complex numbers, such
(a) 1
(c) 1.50
(Q. Nos. 52 to 54)
2
53. If z +  = 4 and sum of greatest and least values of | z |
(d) π
(a) z 2
(c) − z 2
Passage III
= a + ib, where i = −1, then a + 75b , is
(b) 22
(d) 44
exists?
 θ
(a) 3 tan  
 2
1
(c) − cot θ
3
1
 θ
tan  
 2
3
θ
(d) 3 cot
2
(b)
57. Which of the following equals to | z | ?
(b) (a + 1 ) 2 + b 2
(a) | w |
(c) a + (b + 2 )
2
2
(d) (a + 1 ) 2 + (b + 1 ) 2
68
#L
Textbook of Algebra
Complex Numbers Exercise 4 :
Single Integer Answer Type Questions
n
This section contains 10 questions. The answer to
each question is a single digit integer, ranging from 0
to 9 (both inclusive).
π
(1 + i ) 4
4
 |z | 
then 

 amp (z )
63. If z =
58. The number of values of z (real or complex)
simultaneously satisfying the system of equations
A n = ( A + 1)n = 1, then the least value of n is
1 + z + z 2 + z 3 + … + z 13 = 0 is
65. Let z r ; r = 1, 2, 3, …, 50 be the roots of the equation
59. Number of complex numbers z satisfying z 3 = z is
50
∑ ( z )r
60. Let z = 9 + ai, where i = − 1 and a be non-zero real.
3
r=0
2
50
∑ (z
r =1
1
= − 5λ, then λ equals to
r − 1)
p
32
61. Number of complex numbers z, such that | z | = 1
z z
and +  = 1 is
z z
67. The least positive integer n for which
62. If x = a + ib, where a, b ∈ R and i = − 1 and x = 3 + 4i,
2
x 3 = 2 + 11i, the value of (a + b ) is
#L
= 0. If
 10  2qπ
2qπ  
66. If P = Σ (3p + 2)  Σ sin
− i cos
 , where
q = 1 
p =1
11
11  

i = − 1 and if (1 + i ) P = n (n !), n ∈ N , then the value of n is
If Im (z ) = Im (z ), sum of the digits of a is
2
equals to
64. Suppose A is a complex number and n ∈ N , such that
1 + z + z 2 + z 3 + … + z 17 = 0
and
1 − π i
π −i 
 , where i = −1,

+
1 + π i
 π +i
n
1 + i 
2
−1
 = sin

π
1 − i 
1 + x 2 
 , where x > 0 and i = −1 is

 2x 
Complex Numbers Exercise 5 :
Matching Type Questions
n
This section contains 4 questions. Questions 68 and 69 have three statements (A, B and C) given in Column I and
four statements (p, q, r and s) in Column II and questions 70 and 71 have four statements (A, B, C and D) given in
Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct
matching with one or more statement(s) given in Column II.
68.
(A)
Column I
1
If z −  = 2 and if greatest and least values of | z | are G and L respectively, then G − L , is
z

Column II
(p)
natural number
(B)
2
Ifz +  = 4 and if greatest and least values of | z | are G and L respectively, then G − L , is
z

(q)
prime number
(C)
3
If z −  = 6 and if greatest and least values of | z | are G and L respectively, then G − L , is
z

(r)
composite number
(s)
perfect number
69.
Column I
Column II
(A)
If (6 + 8 i ) +
(− 6 + 8 i ) = z1 , z2 , z3 , z4 (where i = − 1), then | z1 |2 + | z2 |2 | + | z3 |2 + | z4 |2 is divisible by
(p)
7
(B)
If (5 − 12i ) +
(− 5 − 12i ) = z1 , z2 , z3 , z4 (where i = − 1), then | z1 |2 + | z2 |2 + | z3 |2 + | z4 |2 is divisible by
(q)
8
(C)
If (8 + 15 i ) +
(− 8 − 15 i ) = z1 , z2 , z3 , z4(where i = − 1), then | z1 |2 + | z2 |2 + | z3 |2 + | z4 |2 is divisible by
(r)
13
(s)
17
Chap 01 Complex Numbers
70.
71.
Column I
#L
Column
II
(A) If λ and µ are the unit’s place digits of
(143)861 and (5273)1358 respectively,
then λ + µ is divisible by
(p)
2
(B) If λ and µ are the unit’s place digits
of (212)7820 and (1322)1594
respectively, then λ + µ is divisible
by
(q)
3
(C) If λ and µ are the unit’s place digits of
(136)786 and (7138)13491 respectively,
then λ + µ is divisible by
(r)
4
(s)
5
(t)
6
Column I
69
Column II
(A)
6
If z −  = 5 and maximum and
z

minimum values of | z | are λ and µ
respectively, then
(p)
λµ + µ λ = 8
(B)
7
If z −  = 6 and maximum and
z

minimum values of | z | are λ and µ
respectively, then
(q)
λµ − µ λ = 7
(C)
8
If z −  = 7 and maximum and
z

minimum values of | z | are λ and µ
respectively, then
(r)
λµ + µ λ = 7
(s)
λµ − µ λ = 6
(t)
λµ + µ λ = 9
Complex Numbers Exercise 6 :
Statement I and II Type Questions
n
Directions (Q. Nos. 72 to 78) are Assertion-Reason
type questions. Each of these questions contains two
statements:
Statement-1 (Assertion) and Statement-2 (Reason)
Each of these questions also has four alternative
choices, only one of which is the correct answer. You
have to select the correct choice as given below.
(a) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1
(c) Statement1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
72. Statement-1 3 + 7i > 2 + 4i, where i = − 1.
Statement-2 3 > 2 and 7 > 4
73. Statement-1 (cos θ + i sin φ) 3 = cos 3 θ + i sin 3 φ,
i = −1
2
π
π

Statement-2  cos + i sin  = i

4
4
74. Statement-1 Let z 1 , z 2 and z 3 be three complex
numbers, such that | 3z 1 + 1 | = | 3z 2 + 1 | = | 3z 3 + 1 | and
1 + z 1 + z 2 + z 3 = 0, then z 1 , z 2 , z 3 will represent vertices
of an equilateral triangle on the complex plane.
Statement-2 z 1 , z 2 and z 3 represent vertices of an
equilateral triangle, if
z 12 + z 22 + z 32 + z 1z 2 + z 2 z 3 + z 3 z 1 = 0.
75. Statement-1 Locus of z satisfying the equation
| z − 1 | + | z − 8 | = 5 is an ellipse.
Statement-2 Sum of focal distances of any point on
ellipse is constant for an ellipse.
76. Let z 1 , z 2 and z 3 be three complex numbers in AP.
Statement-1 Points representing z 1 , z 2 and z 3 are
collinear.
Statement-2 Three numbers a, b and c are in AP, if
b − a = c − b.
77. Statement-1 If the principal argument of a complex
number z is θ, the principal argument of z 2 is 2θ.
Statement-2 arg (z 2 ) = 2 arg (z )
78. Consider the curves on the Argand plane as
π
C 1 : arg (z ) = ,
4
3π
C 2 : arg (z ) =
4
and C 3 : arg (z − 5 − 5i ) = π, where i = −1.
Statement-1 Area of the region bounded by the curves
25
C 1 , C 2 and C 3 is .
2
Statement–2 The boundaries of C 1 , C 2 and C 3 constitute
a right isosceles triangle.
70
Textbook of Algebra
Complex Numbers Exercise 7 :
Subjective Type Questions
n
In this section, there are 24 subjective questions.
79. If z 1 , z 2 and z 3 are three complex numbers, then prove
that z 1 Im ( z 2 z 3 ) + z 2 Im ( z 3 z 1 ) + z 3 Im (z 1 z 2 ) = 0.
80. The roots z 1 , z 2 and z 3 of the equation
x 3 + 3ax 2 + 3bx + c = 0 in which a, b and c are complex
numbers, correspond to the points A, B, C on the
Gaussian plane. Find the centroid of the ∆ ABC and
show that it will be equilateral, if a 2 = b.
81. If 1, α 1 , α 2 , α 3 and α 4 are the roots of x
− 1 = 0, then
prove that
ω − α1 ω − α 2 ω − α 3 ω − α 4
⋅
⋅
⋅
= ω, where ω is
ω 2 − α1 ω 2 − α 2 ω 2 − α 3 ω 2 − α 4
a non-real complex root of unity.
5
82. If z 1 and z 2 both satisfy the relation z + z = 2 z − 1 and
π
arg (z 1 − z 2 ) = , find the imaginary part of (z 1 + z 2 ).
4
83. If ax + cy + bz = X , cx + by + az = Y , bx + ay + cz = Z,
show that
(i) (a 2 + b 2 + c 2 − bc − ca − ab ) ( x 2 + y 2
+ z − yz − zx − xy ) = X
2
2
+Y
2
+ Z − YZ − ZX − XY
2
(ii) (a 3 + b 3 + c 3 − 3abc ) ( x 3 + y 3 + z 3 − 3xyz )
= X 3 + Y 3 + Z 3 − 3XYZ .
84. For every real number c ≥ 0, find all complex numbers z
which satisfy the equation | z | 2 − 2iz + 2c (1 + i ) = 0,
where i = −1 .
85. Find the equations of two lines making an angle of 45°
with the line (2 − i ) z + (2 + i ) z + 3 = 0, where i = −1
and passing through ( − 1, 4 ).
86. For n ≥ 2, show that
88. Show that if a and b are real, then the principal value of
arg (a ) is 0 or π, according as a is positive or negative and
π
π
that of b is or − , according as b is positive or negative.
2
2
89. Two different non-parallel lines meet the circle | z | = r .
One of them at points a and b and the other which is
tangent to the circle at c. Show that the point of
2c −1 − a −1 − b −1
intersection of two lines is
.
c − 2 − a −1b −1
90. A, B and C are the points representing the complex
numbers z 1 , z 2 and z 3 respectively, on the complex
plane and the circumcentre of ∆ABC lies at the origin. If
the altitude of the triangle through the vertex A meets
the circumcircle again at P, prove that P represents the
 z z 
complex number  − 2 3  .
z1 

91. If | z | ≤ 1 and | ω | ≤ 1, show that
| z − ω | 2 ≤ (| z | − | ω | ) 2 + {arg (z ) − arg (ω )} 2 .
92. Let z, z 0 be two complex numbers. It is given that | z | = 1
and the numbers z, z 0 , z z 0 , 1 and 0 are represented in an
Argand diagram by the points P, P0 , Q, A and the origin,
respectively. Show that ∆POP0 and ∆AOQ are
congruent. Hence, or otherwise, prove that
| z − z 0 | = | z z 0 − 1 |.
93. Suppose the points z 1 , z 2 , …, z n (z i ≠ 0) all lie on one
side of a line drawn through the origin of the complex
planes. Prove that the same is true of the points
1 1
1
, , …, . Moreover, show that
z1 z 2
zn
1
1
1
+
+…+
≠ 0.
z 1 + z 2 + … + z n ≠ 0 and
z1 z 2
zn
94. If a, b and c are complex numbers and z satisfies
2 
2 
 1 + i   1 + i    1 + i  
+
+
1
1
+
1








  2 

   2     2  


2n 



1
1 + i  
… 1 + 
= (1 + i ) 1 − n  , where i = −1.

  2  
 22 


2
87. Find the point of intersection of the curves
arg (z − 3i ) = 3π / 4 and arg (2z + 1 − 2i ) = π , where
4
i = − 1.
az 2 + bz + c = 0, prove that | a | | b | = a (b ) 2 c and
| a | = | c | ⇔ | z | = 1.
95. Let z 1 , z 2 and z 3 be three non-zero complex numbers
| z1 | | z 2 | | z 3 |
and z 1 ≠ z 2 . If | z 2 | | z 3 | | z 1 | = 0, prove that
| z 3 | | z1 | | z 2 |
(i) z 1 , z 2 , z 3 lie on a circle with the centre at origin.
2
z 
z − z1 
(ii) arg  3  = arg  3
 .
z 2 
z 2 − z1 
Chap 01 Complex Numbers
π
3π
5π 1
=
sin
sin
14
14
14 8
π
3π
5π
1
(iv) tan
tan
tan
=
14
14
14
7
96. Prove that, if z 1 and z 2 are two complex numbers and
(iii) sin
1

c > 0, then | z 1 + z 2 | 2 ≤ (1 + c ) | z 1 | 2 + 1 +  | z 2 | 2 .
 c
97. Find the circumcentre of the triangle whose vertices are
Also, show that
given by the complex numbers z 1 , z 2 and z 3 .
π

(1 + y ) 7 + (1 − y ) 7 = 14 y 2 + tan 2 

14 
 2
2 3π   2
2 5π 
y + tan
 y + tan


14  
14 
and then deduce that
 π
 3π 
 5π 
tan 2   + tan 2   + tan 2   = 5
 14 
 14 
 14 
98. Find the orthocentre of the triangle whose vertices are
given by the complex numbers z 1 , z 2 and z 3 .
99. Prove that the roots of the equation
π
3π
5π
and cos .
, cos
7
7
7
Hence, obtain the equations whose roots are
π
3π
5π
(i) sec 2 , sec 2
, sec 2
7
7
7
2 π
2 3π
2 5π
(ii) tan
, tan
, tan
7
7
7
π
3π
5π
(iii) Evaluate sec + sec
+ sec
7
7
7
8x 3 − 4 x 2 − 4 x + 1 = 0 are cos
100. Solve the equation z 7 + 1 = 0 and deduce that
101. If the complex number z is to satisfy
| z | = 3, | z − {a (1 + i ) − i }| ≤ 3 and| z + 2a − (a + 1) i | > 3, where
i = − 1 simultaneously for atleast one z, then find all
a ∈ R.
102. Write equations whose roots are equal to numbers
π
2π
3π
nπ
.
, sin 2
, sin 2
, ..., sin 2
2n + 1
2n + 1
2n + 1
2n + 1
π
2π
3π
nπ
(ii) cot 2
.
, cot 2
, cot 2
, ..., cot 2
2n + 1
2n + 1
2n + 1
2n + 1
π
3π
5π
1
cos
cos
=−
7
7
7
8
π
3π
5π
7
(ii) cos
=
cos
cos
14
14
14
8
(i) sin 2
(i) cos
#L
71
Complex Numbers Exercise 8 :
Questions Asked in Previous 13 Years’ Exams
n
This section contains questions asked in IIT-JEE,
AIEEE, JEE Main & JEE Advanced from year 2005
to year 2017.
103. If ω is a cube root of unity but not equal to 1, then
minimum value of | a + bω + cω 2 |, (where a, b and c are
integers but not all equal), is
[IIT-JEE 2005, 3M]
3
(b)
2
(a) 0
(c) 1
(d) 2
104. PQ and PR are two infinite rays. QAR is an arc. Point
lying in the shaded region excluding the boundary
satisfies
[IIT-JEE 2005, 3M]
Y
(–1+√2, √2i)
X′
P
(1,0)
A
O
105. If one of the vertices of the square circumscribing the
circle | z − 1 | = 2 is 2 + 3i, where i = − 1. Find the
other vertices of the square.
X
that | z 1 + z 2 | = | z 1 | + | z 2 |, then arg (z 1 ) − arg (z 2 ) is
equal to
[AIEEE 2005, 3M]
(b) − π / 2
(d) 0
107. If 1, ω, ω 2 are the cube roots of unity, then the roots of
the equation ( x − 1) 3 + 8 = 0 are
R
Y′
(–1+√2, –√2 i )
[IIT-JEE 2005, 4M]
106. If z 1 and z 2 are two non-zero complex numbers, such
(a) − π
(c) π / 2
Q
0)
(–1,
π
4
π
(b) | z − 1 | > 2; | arg (z − 1 )| <
2
π
(c) | z + 1 | > 2; | arg (z + 1 )| <
4
π
(d) | z + 1 | > 2; | arg (z + 1 )| <
3
(a) | z − 1 | > 2; | arg (z − 1 )| <
(a) − 1, 1 + 2 ω, 1 + 2 ω
(c) − 1, − 1, − 1
2
[AIEEE 2005, 3M]
(b) − 1, 1 − 2 ω, 1 − 2 ω 2
(d) None of these
72
Textbook of Algebra
108. If ω =
z
1
z− i
3
and | ω | = 1, where i = − 1, then z lies on
[AIEEE 2005, 3M]
(a) a straight line
(c) an ellipse
109. If ω = α + iβ, where β ≠ 0, i = − 1 and z ≠ 1, satisfies the
 ω − ωz 
condition that 
 is purely real, the set of values
 1 −z 
of z is
[IIT-JEE 2006, 3M]
(b) {z : z = z }
(d) {z : | z | = 1, z ≠ 1 }
110. The value of Σ sin

10
k =1
2kπ
2kπ 
+ i cos
 (where i = − 1)
11
11 
is
[AIEEE 2006, 3M]
(a) i
(c) − 1
(a) 0
(c) 2
(b) 1
(d) ∞
116. Let z be any point in A ∩ B ∩ C . Then,
(b) a parabola
(d) a circle
(a) {z : | z | = 1 }
(c) {z : z ≠ 1 }
115. The number of elements in the set A ∩ B ∩ C , is
(b) 1
(d) − i
111. If z 2 + z + 1 = 0, where z is a complex number, the value of
| z + 1 − i | 2 + | z − 5 − i | 2 lies between
(a) 25 and 29
(b) 30 and 34
(c) 35 and 39
(d) 40 and 44
117. Let z be any point in A ∩ B ∩ C and ω be any point
satisfying | ω − 2 − i | < 3. Then, | z | − | ω | + 3 lies between
(a) − 6 and 3
(c) − 6 and 6
118. A particle P starts from the point z 0 = 1 + 2i, i = −1. It
moves first horizontally away from origin by 5 units and
then vertically away from origin by 3 units to reach a
point z 1 . From z 1 , the particle moves 2 units in the
direction of the vector i$ + $j and then it moves through
π
in anti-clockwise direction on a circle with
2
centre at origin, to reach a point z 2 , then the point z 2 is
given by
[IIT-JEE 2008, 3M]
1
1
1
1
 3
 6
 2

z +  + z + 2  + z + 3  + … + z + 6 




z
z 
z 
z 
is
[AIEEE 2006, 6M]
an angle
(a) 18
(c) 6
(a) 6 + 7i
(c) 7 + 6i
2
2
2
2
(b) 54
(d) 12
112. A man walks a distance of 3 units from the origin
towards the North-East (N 45° E) direction. From there,
he walks a distance of 4 units towards the North-West
(N 45° W) direction to reach a point P. Then, the position
of P in the Argand plane, is
[IIT-JEE 2007, 3M]
(a) 3 e iπ / 4 + 4i
(b) (3 − 4i ) e iπ / 4
(c) ( 4 + 3i ) e iπ / 4
(d) (3 + 4i ) e iπ / 4
(where i = −1)
113. If | z | = 1 and z ≠ ± 1, then all the values of
z
1−z2
(b) − 7 + 6i
(d) − 6 + 7i
119. If the conjugate of a complex numbers is
i = − 1. Then, the complex number is
−1
(a)
i −1
−1
(c)
i +1
(b)
(d)
i
i
1
, where
i −1
[AIEEE 2008, 3M]
1
+1
1
−1
120. Let z = x + iy be a complex number, where x and y are
lie on
[IIT-JEE 2007, 3M]
(a) a line not passing through the origin
(b) | z | = 2
integers and i = − 1. Then, the area of the rectangle
whose vertices are the roots of the equation
z z 3 + z z 3 = 350, is
[IIT-JEE 2009, 3M]
(a) 48
(c) 40
(c) the X -axis
(d) the Y -axis
114. If | z + 4 | ≤ 3, the maximum value of | z + 1 | is
[AIEEE 2007, 3M]
(a) 4
(c) 6
(b) − 3 and 6
(d) − 3 and 9
(b) 10
(d) 0
Passage (Q. Nos. 115 to 117)
Let A , B and C be three sets of complex numbers as defined
below:
A = {z : Im ( z ) ≥ 1}
B = {z : | z − 2 − i | = 3}
C = {z : Re ((1 − i ) z ) = 2}, where i = − 1
[IIT-JEE 2008, 4+4+4M]
(b) 32
(d) 80
121. Let z = cos θ + i sin θ, where i = − 1. Then the value of
15
Σ Im (z 2 m − 1 ) at θ = 2° is
m =1
1
(a)
sin 2 °
1
(c)
2 sin 2 °
[IIT-JEE 2009, 3M]
1
(b)
3 sin 2 °
1
(d)
4 sin 2 °
4
122. If  z −  = 2, the maximum value of | z | is equal to

z
[AIEEE 2009, 4M]
(a) 2 + 2
(b) 3 + 1
(c) 5 + 1
(d) 2
Chap 01 Complex Numbers
123. Let z 1 and z 2 be two distinct complex numbers and
73
128. If z is any complex number satisfying | z − 3 − 2i | ≤ 2 ,
z = (1 − t ) z 1 + iz 2 , for some real number t with 0 < t < 1
where i = − 1 , then the minimum value of | 2z − 6 + 5i |,
and i = − 1 . If arg (w ) denotes the principal argument
is
of a non-zero complex number w, then
[IIT-JEE 2010, 3M]
(a) | z − z1| + | z − z 2| = | z1 − z 2 |
(b) arg (z − z1 ) = arg (z − z 2 )
z − z1 z − z1
(c)
=0
z 2 − z1 z 2 − z1
[IIT-JEE 2011, 2M]
2π
2π
+ i sin , where
3
3
i = − 1, then the number of distinct complex numbers z
124. Let ω be the complex number cos
ω
ω
z + ω2
ω2
1
ω2
1
z +ω
satisfying
(a) 0
(c) 2
= 0, is equal to
[IIT-JEE 2010, 3M]
(b) 1
(d) 3
Column II.
(b) ( − ∞, 0 ) ∪ ( 0, ∞ )
(d) [2, ∞ )
 1 
 for | z | = 1, z ≠ 1, is
1 − z 
130. The maximum value of arg 
(a)
π
6
[IIT-JEE 2011, 2M]
(b)
π
3
(c)
π
2
(d)
2π
3
131. Let w = e iπ / 3 , where i = − 1 and a, b, c , x , y and z be
non-zero complex numbers such that
a +b +c = x
a + bw + cw 2 = y
a + bw 2 + cw = z.
[Note Here, z takes values in the complex plane and Im
(z ) and Re (z ) denote respectively, the imaginary part
and the real part of z.]
[IIT- JEE 2010, 8M]
Column I
Column II
(A) The set of points z satisfying
| z − i | z || = | z + i | z ||, where
i = − 1, is contained in or equal to
(p) an ellipse with
eccentricity 4/5
(B) The set of points z satisfying
| z + 4 | + | z − 4 | = 10 is contained in
or equal to
(q) the set of points z
satisfying
(C) If | w | = 2, the set of points z = w −
1
w
is contained in or equal to
(D) If | w | = 1, the set of points z = w +
is contained in or equal to
1
w
Im (z) = 0
(r) the set of points z
satisfying
|Im (z)| ≤ 1
(s) the set of points
satisfying
|Re (z) | ≤ 2
(t)
the set of points z
satisfying | z | ≤ 3
126. If α and β are the roots of the equation x 2 − x + 1 = 0,
α 2009 + β 2009 is equal to
[AIEEE 2010, 4M]
(b) 1
(d) – 2
127. The number of complex numbers z, such that
| z − 1 | = | z + 1 | = | z − i |, where i = − 1, equals to
[AIEEE 2010, 4M]
(a) 1
(c) ∞
(a) ( − ∞, − 1 ] ∩ [1, ∞ )
(c) ( − ∞, − 1 ) ∪ (1, ∞ )
given by
125. Match the statements in Column I with those in
(a) – 1
(c) 2
129. The set

  2iz 
 : z is a complex number | z | = 1, z ≠ ± 1 is
 Re 

  1 − z 2 
(d) arg (z − z1 ) = arg (z 2 − z1 )
z +1
[IIT-JEE 2011, 4M]
(b) 2
(d) 0
The value of
| x |2 + | y |2 + | z |2
| a |2 + | b |2 + | c |2
, is
[IIT-JEE 2011, 4M]
132. Let α and β be real and z be a complex number. If
z 2 + α z + β = 0 has two distinct roots on the line
Re (z ) = 1, then it is necessary that
[AIEEE 2011, 4M]
(a) β ∈ ( − 1, 0 )
(c) β ∈ (1, ∞ )
(b) | β | = 1
(d) β ∈( 0, 1 )
133. If ω ( ≠ 1) is a cube root of unity and (1 + ω ) 7 = A + Bω,
then ( A, B ) equals to
(a) (1, 1)
(c) (−1, 1)
[AIEEE 2011, 4M]
(b) (1, 0)
(d) (0, 1)
134. Let z be a complex number such that the imaginary part
of z is non-zero and a = z 2 + z + 1 is real. Then, a cannot
take the value
[IIT-JEE 2012, 3M]
(a) −1
(b)
1
3
(c)
1
2
(d)
3
4
z2
is real, the point represented by the
z −1
complex number z lies
[AIEEE 2012, 4M]
135. If z ≠ 1 and
(a) on a circle with centre at the origin
(b) either on the real axis or on a circle not passing through
the origin
(c) on the imaginary axis
(d) either on the real axis or on a circle passing through the
origin
74
Textbook of Algebra
136. If z is a complex number of unit modulus and argument
1 + z
θ, then arg 
 equals to
1 + z 
π
−θ
2
(c) π − θ
(a) is strictly greater than
(b) is equal to
[JEE Main 2013, 4M]
(c) is strictly greater than
(d) −θ
1
lie on circles
α
( x − x 0 ) 2 + (y − y 0 ) 2 = r 2 and
137. Let complex numbers α and
 2k π 
 2k π 
 ; k = 1, 2,K, 9. Then,
 + i sin 
 10 
 10 
match the column.
142. Let z k = cos 
( x − x 0 ) 2 + (y − y 0 ) 2 = 4 r 2 , respectively. If
z 0 = x 0 + i y 0 satisfies the equation 2 | z 0 | 2 = r 2 + 2 ,
then | α | equals to
[JEE Advanced 2013, 2M]
(a)
1
2
(b)
1
2
(c)
5
2
3
5
but less than
2
2
(d) lies in the interval (1, 2)
(b) θ
(a)
5
2
1
7
(d)
1
3
3 +i
and P = {w n : n = 1, 2, 3, K }. Further,
2

1

 1 
H 1 = z ∈ C : Re(z ) >  and H 2 = z ∈ C : Re(z ) <  −   ,
 2 
2


where C is the set of all complex numbers. If
z 1 ∈ P ∩ H 1 , z 2 ∈ P ∩ H 2 and O represents the origin,
then ∠z 1O z 2 equals to
Column I
Column II
(A)
For each zk there exists a z j such that
zk ⋅ z j = 1
(1)
True
(B)
There exists a k ∈{1, 2, K , 9} such that
z1 ⋅ z = zk has no solution z in the set of
complex numbers
(2)
False
(C)
|1 − z1 ||1 − z2 | K |1 − z9 |
equals to
10
(3)
1
(D)
1−
(4)
2
138. Let w =
9
 2kπ 
∑ cos  10  equals to
k=1
[JEE Advanced 2014, 3M]
[JEE Advanced 2013, 3M]
π
(a)
2
2π
(c)
3
π
(b)
6
5π
(d)
6
Codes
A B C D
(a) 1 2 4 3
(c) 1 2 3 4
Passage (Q. Nos. 139 to 140)
Let
and
[JEE Advanced 2013, 3+3M]
139. min | 1 − 3i − z | equals to
z ∈S
2− 3
2
3− 3
(c)
2
(a)
140. Area of S equals to
10 π
(a)
3
16 π
(c)
3
(a) circle of radius z
(b) circle of radius 2
(c) straight line parallel to X -axis
(d) straight line parallel to Y -axis
144. Let ω ≠ 1 be a complex cube root of unity.
2+ 3
2
3+ 3
(d)
2
(b)
If (3 − 3 ω + 2 ω 2 ) 4n + 3 + (2 + 3 ω − 3 ω 2 ) 4n + 3
+( −3 + 2 ω + 3 ω 2 ) 4n + 3 = 0, then possible value(s) of n is
(are)
[JEE Advanced 2015, 2M]
(a) 1
(c) 3
20 π
(b)
3
32 π
(d)
3
(b) 2
(d) 4
kπ 
 kπ 
 + i sin   , where
7
7
145. For any integer k, let α k = cos 
12
141. If z is a complex number such that | z | ≥ 2, then the

 1 
minimum value ofz +   , is
 2 

143. A complex number z is said to be unimodular if | z | = 1.
Suppose z 1 and z 2 are complex numbers such that
z 1 − 2z 2
is unimodular and z 2 is not unimodular. Then
2 − z 1z 2
the point z 1 lies on a
[JEE Main 2015, 4M]
S = S 1 ∩ S 2 ∩ S 3 , where
S 1 = {z ∈ C : | z | < 4},


 z − 1 + 3i
S 2 =  z ∈ C : Im 
 > 0
 1 − 3i 


S 3 = {z ∈ C : Re z > 0}.
A B C D
(b) 2 1 3 4
(d) 2 1 4 3
i = −1. The value of the expression
[JEE Main 2014, 4M]
is
∑ | αk +1 − αk |
k =1
3
∑ | α 4k −1 − α 4k − 2 |
k =1
[JEE Advanced 2015, 4M]
Chap 01 Complex Numbers
146. A value of θ for which
2 + 3i sin θ
is purely imaginary, is
1 − 2i sin θ
(a) the circle with radius
[JEE Main 2016, 4M]
π
(a)
6
(b) sin
 1 
(c) sin −1  
 3
(d)
−1
 3
 
 4
75
1
1 
and centre  , 0 for a > 0, b ≠ 0
 2a 
2a
(b) the circle with radius −
1
 1 
and centre  − , 0 for
 2a 
2a
a < 0, b ≠ 0
(c) the X -axis for a ≠ 0, b = 0
(d) the Y -axis for a = 0, b ≠ 0
π
3
147. Let a, b ∈ R and a 2 + b 2 ≠ 0.
148. Let ω be a complex number such that 2ω + 1 = z, when


1
Suppose S = z ∈ C : z =
, t ∈ R, t ≠ 0 , where
a + ibt


i = −1. If z = x + iy and z ∈ S, then ( x , y ) lies on
1
1
1
z = −3 if 1 − ω − 1 ω 2 = 3k, then k is equal to
1
ω2
ω7
[JEE Main 2017, 4M]
2
(b) − z
(a) 1
[JEE Advanced 2016 4M]
(d) − 1
(c) z
Answers
Exercise for Session 1
1. (d)
2. (c)
7. (d)
8. (a)
3. (b)
4. (b)
5. (c)
6. (b)
71. A → (r); B → (p, s); C → (q, t)
72. (d)
73. (d)
74. (c)
75. (d)
78. (d)
84. z = c + i (−1 ±
1. (b)
2. (b)
3. (b)
4. (b)
5. (b)
6. (b)
7. (d)
13. (c)
8. (c)
14. (a)
9. (b)
10. (b)
11. (a)
12. (c)
c>
3. (d)
9. (c)
15. (a)
4. (a)
10. (d)
5. (b)
11. (c)
6. (a)
12. (a)
2. (d)
8. (d)
14. (c)
3. (c)
9. (d)
2. (c)
8. (d)
14. (d)
20. (b)
26. (d)
98.
97.
∑ z12 ( z 2 − z 3 ) + ∑ | z1 | 2 ( z2 − z3 )
∑( z1 z2 − z2 z1 )
∑ | z1 |2 ( z2 − z3 )
∑ z1 ( z2 − z3 )
99. (i) x3 − 24x2 + 80x − 64 = 0
4. (d)
10. (b)
5. (b)
11. (b)
6. (b)
12. (a)
(ii) x3 − 21x2 + 35x − 7 = 0
(iii) 4
–
–3 –5
100. Roots of z 7 + 1 = 0 are −1, α , α3, α 5, α , α , α , where
3. (d)
9. (b)
15. (a)
21. (c)
27. (b)
π
π
+ i sin
7
7
α = cos
Chapter Exercises
1. (b)
7. (b)
13. (c)
19. (b)
25. (d)
2 − 1 and no solution for
2 −1
87. No solution
Exercise for Session 4
1. (a)
7. (a)
13. (b)
(1 − c2 − 2c)) for 0 ≤ c ≤
85. (1 − 3i ) z + (1 + 3i ) z − 22 = 0 and (3 + i ) z + (3 − i ) z + 14 = 0
Exercise for Session 3
2. (b)
8. (b)
14. (b)
77. (d)
82. 2
Exercise for Session 2
1. (a)
7. (c)
13. (b)
76. (a)
4. (d)
10. (c)
16. (c)
22. (c)
28. (b)
31. (b,c,d) 32. (b,c,d) 33. (b,c,d) 34. (b,c,d)
37. (a,b,c) 38. (a,d)
39. (a,c)
40. (a,b,d)
42. (b,d) 43. (a,c,d) 44. (a,d) 45. (a,b,d)
46. (a)
47. (d)
48. (b)
49. (a)
52. (d)
53. (c)
54. (a)
55. (c)
58. (1)
59. (5)
60. (9)
61. (8)
64. (6)
65. (5)
66. (4)
67. (4)
68. A → (p, q); B → (p, r); C → (p, r, s)
69. A → (q); B → (q, r); C → (q, s)
70. A → (p, q, r, t); B → (p, s); C → (p, r)
5. (a)
11. (c)
17. (a)
23. (c)
29. (b)
6. (d)
12. (b)
18. (c)
24. (c)
30. (c)
35. (a,c,d) 36. (a,b)
41. (a,d)
50. (c)
56. (d)
62. (3)
51. (b)
57. (b)
63. (4)
 1 − 71 −1 − 4 11   −1 + 4 11 1 + 71 
,
,

 ∪
5
5
2 
 
 2
101. a ∈ 
102. (i)
(ii)
103. (c)
2n+ 1
C 1 (1 − x)n − 2n+ 1C 3 (1 − x)n− 1 x + ... + (−1)n xn = 0
2n+ 1
C 1xn − 2n+ 1 C 3xn− 1 + 2n+ 1 C 5xn− 2 −... = 0
104. (c)
105. (1 −
3 ) + i, −i 3, ( 3 + 1) − i 106. (d)
107. (b) 108. (a)
109. (d)
110. (d) 111. (d) 112. (d)
113. (d) 114. (c)
115. (b)
116. (c)
117. (d) 118. (d)
119. (c) 120. (a)
121. (d)
122. (c)
123. (a, c, d) 124. (b)
125. A → (q, r); B → (p); C → (p, s); D → (q, r, s, t)
126. (b) 127. (a)
128. (5) 129. (a)
130. (c)
131. (3) 132. (c) 133. (a)
134. (d) 135. (d)
136. (b)
137. (c)
138. (c) 139. (c)
140. (b) 141. (d)
142. (c)
143. (a)
144. (a, b, d)
145. (4) 146. (c)
147. (a,c,d)
148. (b)
Solutions
r
Π e ipθ = 1
4. We have,
p =1
e i θ ⋅ e 2i θ ⋅ e 3i θ … e ri θ = 1
⇒
 r (r + 1) 
iθ 

2 
e iθ(1 + 2 + 3 + … + r ) = 1 ⇒ e 
r (r + 1 ) 
r (r + 1 ) 
or cos 
θ  + i sin 
θ = 1 + i ⋅ 0
2
 2



⇒
1. We have,
a + ib = cos (1 − i ) = cos 1 cos i + sin 1 sin i
= cos1 cosh 1 + sin 1 i sinh 1
[Q cos i = cosh 1, sin i ⋅ 1 = i sinh 1 ]
 e + e −1 
 e − e −1 
= cos 1 
 + i sin 1 

 2 
 2 
1
1
1
1
= e +  cos 1 + i ⋅ e −  sin 1
e
e
2
2
∴
1
1
a = e +  cos1
e
2
and
1
1
b = e −  sin 1
2
e
⇒
⇒
5. Let z = x + iy , then
5
t 2 − t − 992 = 0
t=
r (r + 1 ) 
r (r + 1 ) 
cos 
θ  = 1 and sin 
θ = 0
 2

 2

r (r + 1 )
r (r + 1 )
θ = 2m π and
θ = m1π
2
2
4m π
2m1π
and θ =
θ=
r (r + 1 )
r (r + 1 )
where, m, m1 ∈ I
4nπ
Hence, θ =
, n ∈I.
r (r + 1 )
t =z
⇒
On comparing, we get
⇒
2. Given that, z 10 − z 5 − 992 = 0
Let
=1
1 ± 1 + 3968 1 ± 63
=
= 32, − 31
2
2
∴
z 5 = 32
and
z 5 = − 31
But the real part is negative, therefore z 5 = 32 does not hold.
∴ Number of solutions is 5.
3. From Coni method,
O
(3 + i ) (z + z ) − (2 + i ) (z − z ) + 14i = 0 reduces to
(3 + i ) 2 x − (2 + i ) (2iy ) + 14i = 0
⇒
6 x + 2y + i (2 x − 4y + 14 ) = 0
On comparing real and imaginary parts, we get
6 x + 2y = 0
⇒
3x + y = 0
and
2 x − 4y + 14 = 0
⇒
x − 2y + 7 = 0
On solving Eqs. (i) and (ii), we get
x = − 1 and y = 3
∴
z = − 1 + 3i
zz = | z | 2 = | − 1 + 3i | 2 = ( −1 ) 2 + (3 ) 2 = 10
∴
…(i)
…(ii)
6. Since, affix of A is z1.
2π
n
B
C
π
An(z)
A1(z)
z −0
= e 2π i /n or
z −0
Im(z )
But given
= 2 −1
Re(z )
z −z
z
1 z
i
2
⇒
= 2 −1 ⇒ 
z +z
i z
z
2
π
/
2
i
n
e
− 1
⇒  2 π i /n
 = i ( 2 − 1)
+ 1
e
 π
⇒
i tan   = i ( 2 − 1 )
 n
 π
 π
⇒
tan   = tan  
 8
 n
∴
n =8
O
z
= e 2π i /n
z
… (i)

− 1
 = 2 −1
+ 1

[from Eq. (i)]
A(z1)
D
→
→
π/2
→
→
∴ OA = z1 and OB and OC are obtained by rotating OA
→
→
π
through and π. Therefore, OB = iz1 and OC = -z1.
2
z + iz1 + ( −z1 )
Hence, centroid of ∆ABC = 1
3
i
z1 
π
π
= z1 =  cos + i sin 
3
3 
2
2
If A, B and C are taken in clockwise, then centroid of ∆ABC
1 
π
π
= z1  cos − i sin 
3 
2
2
∴
Centroid of ∆ABC =
z1 
π
π
 cos ± i sin 
3 
2
2
Chap 01 Complex Numbers
 −1 − i 3 
3 −i
= i
 = iω 2
2
2


= (iω 2 )101 = i 101 ω 202 = iω
7. Given that, z =
∴
z 101
z1
z1
1
z2
z2
1 =0
z1 − z 2 z1 − z 2 1
z1
⇒
Now, i 101 + z 101 = i + iω = i ( − ω 2 )
∴
+z
101 103
)
= −i
103
ω
206
3
2
2
Applying R3 → R3 − R1 + R2, then z 2
Expand w.r.t. R3, then
⇒
 8π 
 8π 
 + i sin  
 11 
 11 
Now, Re (α + α 2 + α 3 + α 4 + α 5 )
9. We have, α = cos 
α + α2 + α3 + α4 + α5 + α + α2 + α3 + α4 + α 5
2
− 1 + (1 + α + α 2 + α 3 + α 4 + α 5 + α + α 2 + α 3 + α 4 + α 5
=
2
−1 + 0
[sum
of 11, 11th roots of unity]
=
2
1
=−
2
|z | ≤ 4
…(i)
π
0 ≤ arg(z ) ≤
3
and
0…
O
…(ii)
Real axis
which implies the set of points in an argand plane, is a sector
of a circle.
11. Since, x 2 + x + 1 = ( x − ω ) ( x − ω 2 ), where ω is the cube root
of unity and f ( x ) = g ( x 3 ) + x h ( x 3 ) is divisible by x 2 + x + 1.
Therefore, ω and ω 2 are the roots of f ( x ) = 0.
⇒
f (ω ) = 0 and f (ω 2 ) = 0
⇒
g (ω 3 ) + ω h (ω 3 ) = 0
and
g ((ω 2 ) 3 ) + ω 2h (ω 2 ) 3 = 0
⇒
and
g (1 ) + ω h (1 ) = 0
g (1 ) + ω 2h (1 ) = 0
⇒
g (1 ) = h (1 ) = 0
Hence, g ( x ) and h ( x ) both are divisible by ( x − 1 ).
12. Since, z1, z 2 and z1 − z 2 are collinear.
∴
z2
1 =0
M
0 …1
z1z 2 − z1 z 2 = 0
z1z 2 − (z1z 2 ) = 0
⇒
⇒
Im (z1 z 2 ) = 0
Im ((a + ib ) (c + id )) = 0
⇒
⇒
Im ((a + ib ) (c − id )) = 0
bc − ad = 0 ⇒ ad − bc = 0
∴ f (a + ib ) = (a 2 + b 2 )
⇒
f (z ) = f (z ) = f ( −z ) = f ( −z ) = (a 2 + b 2 )
∴ f is not injective (i.e., it is many-one).
but | z | > 0 i. e. f (z ) > 0 ⇒ f (z ) ∈ R + (Range)
⇒ R+ ⊂ R
∴ f is not surjective (i.e., into).
Hence, f is neither injective nor surjective.
14. Let α = re iθ , β = re iφ
[Q| α | = | β |, given]
 π π
where, θ ∈  − ,  and φ ∈ ( − π , 0 )
 2 2
P(z)
π/3
1
13. Let z = a + ib
=
10.
z1
M
= − i ω = iω = z
a
8. The complex slope of the line az + a z + 1 = 0 is α = −
a
b
and the complex slope of the line bz + b z − 1 = 0 is β = −
b
Since, both lines are mutually perpendicular, then
∴
α+β=0
a b
− − =0
⇒
a b
⇒
ab + a b = 0
(i
101
77
z1
z1
1
z2
z2
1 =0
z1 − z 2 z1 − z 2 1
α + β re iθ + reiφ
∴
=
=
α − β re iθ − reiφ
e
e
θ+ φ
i

 2 
θ+ φ
i

 2 
 θ − φ
⋅ 2 cos 

 2 
 θ − φ
⋅ 2i sin 

 2 
 θ − φ
= − i cot 
 = Purely imaginary
 2 
25
25
25 25
−
≤ z+
+
z
z
z
−z
25
| z | ≤ 24 +
⇒
|z |
⇒ | z | 2 − 24 | z | − 25 ≤ 0 ⇒ (| z | − 25 ) (| z | + 1 ) ≤ 0
15. We have, | z | = z +
∴
| z | − 25 ≤ 0
[Q | z | + 1 > 0 ]
⇒
| z | ≤ 25 or | z − 0 | ≤ 25
Hence, the maximum distance from the origin of coordinates
to the point z is 25.
16. Q A ≡ z1, B ≡ z 2, C ≡ (1 − i ) z1 + iz 2
∴
AB = | z1 − z 2 |
BC = | z 2 − (1 − i ) z1 − iz 2| = | (1 − i ) (z 2 − z1 )|
= 2 | z1 − z 2 |
CA = | (1 − i ) z1 + iz 2 − z1| = | − i (z1 − z 2)|
= | − i || z1 − z 2 | = | z1 − z 2 |
It is clear that, AB = CA and ( AB ) 2 + (CA ) 2 = ( BC ) 2
∴ ∆ ABC is isosceles and right angled.
and
78
Textbook of Algebra
17. Centre and radius of circle | z | = 3
are C1 ≡ 0, r1 = 3
and centre and radius of circle
|z + 1 −i | = 2
1 /3
= 41 + 1 /3 = 41 /4 = 2
…(ii)
C 2 = − 1 + i , r2 = 2
and
Q
| C1C 2 | = | − 1 + i | = 2
and
| C1C 2 | < r1 − r2
and z =
∞
1
1
1
∑ (1 + i )− r = (1 + i ) + (1 + i )2 + (1 + i )3 + …∞
r =1
1
1
(1 + i )
=
= = −i
1
i
1−
(1 + i )
|z|=3
O′
y = 41 /3 ⋅ 4 − 1 /9 ⋅ 41 /27…∞ = 41 /3 − 1 /9 + 1 /27 … ∞
…(i)
Now, x + yz = 3 − i 2
 2
∴ arg ( x + yz ) = arg (3 − i 2 ) = − tan −1  
 3 
|z+1–i|=√2
O
21. Q A1 ≡ 1 + 2i
A2
A1(1+2i )
π/3
A3
Hence, circle (ii) completely inside circle (i)
∴ Number of solutions = 0
18. We have, f (z ) = g(z ) (z 2 + 1) + h(z )
A4
where, degree of h (z ) < degree of (z 2 + 1 )
…(i)
⇒ f (z ) = g (z ) (z − i ) (z + i ) + az + b ; a, b ∈ C
Now, f (i ) = 1 − i
[given]
[from Eq. (i)] …(ii)
⇒ ai + b = 1 − i
and f ( −i ) = 1 + i
[given]
[from Eq. (i)] …(iii)
⇒ a ( −i ) + b = 1 + i
On solving Eqs. (ii) and (iii) for a and b, we get
a = − 1 and b = 1
∴ Required remainder, h (z ) = az + b = − z + 1 = 1 − z
1 i 3 1 i 3
= (1 + 2i )  +
+i − 3
= +
2
2  2
2
 3


1
+ 1
=  − 3 + i 

2
 2

 3

1

+ 1
∴ | A1A2 | = 1 + 2i −  − 3 − i 
2

 2

=
⇒
z = x + iy , we get
( x + 1 ) 2 + y 2 = 4 [( x − 1 ) 2 + y 2 ]
3 x + 3y − 10 x + 3 = 0
10
⇒
x 2 + y2 − x + 1 = 0
3
On comparing Eq. (i) with the standard equation
x 2 + y 2 + 2 gx + 2 fy + c = 0
10
5
g=−
= − and f = 0
⇒
6
3
5 
∴ Required centre of circle ≡ ( − g, − f ) ≡  , 0
3 
2
i.e.
∴ Perimeter = 6 | A1A2 | = 6 5
2
5
5
+ 0⋅i =
3
3
20. Q x = 91 /3 ⋅ 91 /9 ⋅ 91 /27 …∞
1 /3
= 91 /3 + 1 /9 + 1 /27 + … ∞ = 91 − 1/3 = 91 /2 = 3

1
3
+ 3 + i 1 −

2
2 

2

3

1
=  + 3 + 1 −
 = 5

2
2 

19. We have, | z + 1| = 2 | z − 1|
2
A5
∴ A2 = (1 + 2i ) e iπ /3
⇒ h(z ) = az + b ; a, b ∈ C
∴ f (z ) = g (z ) (z 2 + 1 ) + az + b ; a, b ∈ C
Put
A6
O
22. We have,
n
∑ zr
…(i)
=
r =1
r =1
=
∴
n
∑ zr
n
∑ (zr − r ) + r
≤
n
∑ (| zr − r | + | r | )
r =1
n
n
n
n
r =1
r =1
r =1
r =1
∑ | zr − r | + ∑ | r | ≤ ∑ r + ∑ | r |
n(n + 1 ) n(n + 1 )
=
+
= n(n + 1 )
2
2
≤ n (n + 1 )
r =1
z 

 z1 − 
z
|
z| π
23. We have, arg 
= and
− z1 = 3
|z |
 z  2


 |z | 
Chap 01 Complex Numbers
which implies the following diagram
λ 7
=
µ 3
⇒
Y
Q
λ, µ ∈ I − { 0}
For minimum value λ = 7, µ = 3
∴
| z | 2 = | (3 + 7i ) ( λ + iµ ) | 2
z
|z|
3
1
90°
z1
X′
= | 3 + 7i | 2 | λ + iµ| 2 = 58 ( λ2 + µ 2 )
X
O
= 58 (7 2 + 3 2 ) = (58 ) 2 = 3364
27. We have,
Y′
⇒
z
− z1 = 3 ⇒ | z1| = 9 + 1 = 10
|z |
24. Let z = x + iy = r (cos θ + i sin θ )
z = f ( x ) + i g( x )
where, i = −1 and f , g : ( 0,1 ) → ( 0,1 ) are real-valued
functions.
 1 
1
(a) z =
+ i

 1 + ix 
1 − ix
(1 + x )
x+i
1 + ix
1+x
+
+i
=
1 + x2 1 + x2 1 + x2
1 + x2
1+x
1+x
and g( x ) =
⇒ f (x ) =
2
1+x
1 + x2
∴ | z | = r , arg (z ) = θ
=

π
Given, | z − 2 − i | = | z | sin  − arg (z )

4
π

⇒ | x + iy − 2 − i | = r sin  − θ
4

⇒ | ( x − 2 ) + i (y − 1 ) | = r
1
(cos θ − sin θ )
2
1
x −y
2
On squaring both sides, we get
2 ( x 2 + y 2 − 4 x − 2y + 5 ) = x 2 + y 2 − 2 xy
⇒
( x − 2 ) 2 + (y − 1 ) 2 =
⇒
But for x = 0.5, f ( 0.5 ) > 1 and g( 0.5 ) > 1, which is out of
range.
Hence, (a) is not a correct option.
 1 
1
(b) z =
+ i

 1 − ix 
1 + ix
=
( x + y ) 2 = 2( 4 x + 2y − 5 )
which is a parabola.
25. Since, 1, z1, z 2, z 3, …, zn − 1 are the n, nth roots of unity.
∴ (z n − 1 ) = (z − 1 ) (z − z1 ) (z − z 2 ) (z − z 3 ) … (z − zn − 1 )
n −1
= (z − 1 ) ∏ (z − zr )
r =1
Taking log on both sides, we get
n −1
loge (z n − 1 ) = loge (z − 1 ) +
∑
loge (z − zr )
⇒ f (x ) =
Hence, (c) is not a correct option.
 1  1 + ix i (1 + ix )
1
(d) z =
+
+ i
=
 1 − ix  1 + x 2 (1 + x 2 )
1 − ix
=
On differentiating both sides w.r.t.z, we get
n −1
Putting z = 3, we get
n −1
∑
r =1
n ⋅ 3n − 1 1
1
−
= n
(3 − zr ) (3 − 1 ) 2
26. We have,
z = (3 + 7i ) ( λ + iµ )
= (3 λ − 7µ ) + i (7 λ + 3 µ )
Since, z is purely imaginary.
∴
3 λ − 7µ = 0
1−x
1−x
and g( x ) =
1+ x2
1+ x2
Clearly, f ( x ), g( x ) ∈ ( 0, 1 ), if x ∈( 0,1 )
Hence, (b) is the correct option.
1 − ix
i (1 − x )
i (1 − ix ) (1 + x )
(c) z =
+
+
=
1 + x2
1 + x 2 (1 + x 2 ) (1 + x 2 )
r =1
1
1
nz n − 1
−
=∑
n
(
)
(
1
z
z
zr )
−
−
(z − 1 )
r =1
 1−x 
(i − x )  1 − x 
1 − ix
=
+
 + i

1 + x 2
1 + x2 1 + x2 1 + x 2
i (1 + x )
(1 − x )
+
(1 + x 2 ) (1 + x 2 )
Hence, (d) is not a correct option.
28. Let z = α be a real roots of equation.
z 3 + (3 + 2i )z + ( −1 + ia ) = 0
⇒
α 3 + (3 + 2i ) α + ( −1 + ia ) = 0
⇒
(α 3 + 3 α − 1 ) + i (a + 2 α ) = 0
On comparing the real and imaginary parts, we get
α 3 + 3 α − 1 = 0 and a + 2 α = 0
a
α=−
⇒
2
3
3a
a
⇒
− −
−1 = 0
8
2
⇒
a 3 + 12a + 8 = 0
79
80
Textbook of Algebra
f (a ) = a 3 + 12a + 8
Let
which is a circle and passing through the origin
⇒
∴
f ( −1 ) < 0 and f ( 0 ) > 0
a ∈ ( −1, 0 )
π
π
π
29. CiS = cos + i sin
6
6
6
 3 + i  1  −1 + i 3  ω
=
 = 
 = = − iω
2
 2  i 
 i
π

∴ 2 CiS 

6
and radius =
32. Given,
⇒
n
= (2 2 (1 + i )n
= (8(1 + i ) 2 )n /2 = (16i )n /2
Thus, (8i )m / 3 = (16i )n / 2
which is satisfy only when m = 48 and n = 24
∴
m + n = 72
30. We have, z 2 = z ⋅ 21 − | z|
Taking modulus on both sides, we get
| z | 2 = | z | ⋅ 21 − | z|
⇒
| z | (| z | − 21 − | z | ) = 0
and
arg (z ) = arg (z ⋅ 2
1 − | z|
2
…(i)
)
⇒
2 arg (z ) = arg (z ) = − arg (z )
⇒
3 arg (z ) = 0
∴
arg (z ) = 0
Then,
y =0
From Eq. (i), | z | = 0 ⇒ x = 0
One solution is z = 0 + i ⋅ 0 = 0.
Also, from Eq. (i),
y=2
[Q z = x + iy ]
[Q y = 0 ]
–x
Y
/2
(0, 1) y = |x|
X′
o
X
Y′
| z | = 21 − | z| ⇒ | x | = 21 − x
|x|
= 2 − x = y (say)
⇒
2
Hence, total number of solutions = 2
z +1
is a purely imaginary number.
31. Q
z +i
 z + 1
 z + 1
 z + 1
z +1
∴
=−
 =−

 ⇒

z + i
z + i
z −i
z + i
⇒
(z + 1 ) (z + i ) + (z − i ) (z + 1 ) = 0
2zz + z (1 + i ) + z (1 − i ) = 0
1 − i
1 + i
zz + 
z =0
z +
 2 
 2 
1+i
1
=
2
2
| z | 2 + 1 − 2 Re (z ) < | z | 2 + 9 + 2 Re (3z )
2 Re ( 4z ) > − 8
Re ( 4z ) > − 4
4z + 4z
⇒
>−4
2
z + z > −2
∴
and
ω = 2z + 3 − i
∴
ω + ω = 2z + 3 − i + 2z + 3 + i
= 2(z + z ) + 6 > − 4 + 6
ω + ω >2
⇒
Option (a) | ω − 5 − i | < | ω + 3 + i |
⇒ | 2z + 3 − i − 5 − i | < | 2z + 3 − i + 3 + i |
⇒
| 2z − 2 − 2i | < | 2z + 6 |
⇒
|z −1 −i | <|z + 3|
which is false.
Option (b) | ω − 5 | < | ω + 3 |
⇒ | 2z + 3 − i − 5 | < | 2z + 3 − i + 3 |
⇒
| 2z − 2 − i | < | 2z + 6 − i |
i
i
z −1 − < z + 3 −
⇒
2
2
⇒
|z −1| <|z + 3|
which is true.
Option (c)
Im (iω ) > 1
iω − iω
>1
⇒
2i
iω + i ω
⇒
>1
2i
ω + ω >2
⇒
which is true.
Option (d)
⇒
⇒
−0=
⇒
⇒
= ( −2iω )m = (( −2iω ) 3 )m /3 = (8i )m /3
n
2
|z − 1 | < |z + 3 |
|z − 1 |2 < |z + 3 |2
⇒
m
π
π
π
 

and  4 CiS  =  4  cos + i sin  

 
4
4
4
1+i
2
⇒
⇒
⇒
π
2
π
| arg (2z + 3 − i − 1 ) | <
2
π
| arg (2z + 2 − i ) | <
2
π
−1  Im(2z + 2 − i ) 
tan 
 <
 Re(2z + 2 − i ) 
2
| arg (ω − 1 ) | <
∴
Re(2z + 2 − i ) > 0
(2z + 2 − i ) + (2z + 2 + i )
>0
⇒
2
z +z +2>0
⇒
z + z > −2
⇒
which is true.
33. ∴ (1 + ri ) 3 = λ (1 + i )
⇒ 1 + (ri ) 3 + 3(1 ) 2ri + 3(1 ) (ri ) 2 = λ (1 + i )
81
Chap 01 Complex Numbers
⇒
⇒
1 − r 3i + 3ri − 3r 2 = λ + i λ
On comparing real and imaginary parts, we get
and
− r 3 + 3r = λ
Then,
− r + 3r = 1 − 3r
3
…(i)
| z − 4 | = | z − 8 | ⇒| z − 4 | = | z − 8 |
2
⇒
8 Re (z ) = 48
∴
Re (z ) = 6
⇒
x =6
From Eqs. (i) and (ii), we get
2 (36 + y 2 ) + 162 − 50y + 38 = 0
2
⇒
r 3 − 3r 2 − 3r + 1 = 0
⇒
(r 3 + 1 ) − 3r (r + 1 ) = 0
⇒
(r + 1 ) (r 2 − r + 1 − 3r ) = 0
⇒
y 2 − 25y + 136 = 0
⇒
(r + 1 ) (r − 4r + 1 ) = 0
⇒
⇒
∴
(y − 17 ) (y − 8 ) = 0
y = 17,8
Im (z ) = 17, 8
2
r = − 1, 2 ± 3
3π
π
π
r = cosec
, tan , cot
2
12
12
∴
⇒
37.
Q(z2)
P(z1)
34. Option (a) | z − 1 | + | z + 1 | = 3
Here, | 1 − ( −1 ) | < 3
i. e. 2 < 3, which is an ellipse.
Option (b) | z − 3 | = 2
It is a circle with centre 3 and radius 2.
7
Option (c) | z − 2 + i | =
3
7
It is a circle with centre (2 − i ) and radius .
3
Option (d)
(z − 3 + i ) (z − 3 − i ) = 5
⇒
(z − 3 + i ) (z − 3 + i ) = 5
⇒
| z − 3 + i |2 = 5
⇒
|z −3 + i | = 5
It is a circle with centre at (3 − i ) and radius 5.
35. Since, 1, z1, z 2, z 3, …, zn − 1 are the n, nth roots of unity.
Therefore,
z − 1 = (z − 1 ) (z − z1 ) (z − z 2 ) … (z − zn − 1 )
n
⇒
zn − 1
= (z − z1 ) (z − z 2 ) … (z − zn − 1 )
z −1
n −1
= ∏ (z − zr )
r =1
90°
∏ (ω − zr ) =
r =1
ωn − 1
ω −1
if n = 3r, r ∈ Z
 0,

if n = 3r + 1 , r ∈ Z
=  1,
1 + ω, if n = 3r + 2, r ∈ Z

36. Q 3 | z − 12 | = 5 | z − 8i |
∴ 9 | z − 12 | 2 = 25 | z − 8i | 2
9 (z − 12 ) (z − 12 ) = 25(z − 8i ) (z + 8i )
⇒
⇒ 9 (zz − 12(z + z ) + 144 ) = 25(zz + 8i (z − z ) + 64 )
⇒
16zz + 108 (z + z ) + 200 (z − z ) i + 304 = 0
⇒
16( x 2 + y 2 ) + 216 x − 400y + 304 = 0
S(z4)
(z3)R
Option (a)Q PS || QR
∴
z − z4 
arg  1
 =0
z2 − z3
⇒
z1 − z 4
is purely real.
z2 − z3
Option (b) Q Diagonals of rhombus are perpendicular.
 z − z3  π
Then, arg  1
 =
z2 − z4  2
⇒
z1 − z 3
is purely imaginary.
z2 − z4
Option (c) Q
PR ≠ QS
∴
| z1 − z 3 | ≠ | z 2 − z 4 |
Option (d) Q
∠QSP = ∠RSQ
∴
Now, putting z = ω, we get
n −1
2
⇒ | z | 2 + 16 − 2 Re ( 4z ) = | z | 2 + 64 − 2 Re(8z )
1 − 3r 2 = λ
and
2( x 2 + y 2 ) + 27 x − 50y + 38 = 0
⇒
⇒
 z − z4 
z − z4 
amp  2
 = amp  3

 z1 − z4 
z2 − z4 
 z − z4 
− amp  1
 = − amp
z2 − z4 
z2 − z4 


z3 − z4 
 z − z4 
z − z4 
amp  1
 = amp  2

z2 − z4 
z3 − z4 
38. Q | z − 3 | = min {| z − 1 |, | z − 5 | }
Case I If | z − 3 | = | z − 1 |
On squaring both sides, we get
| z − 3 |2 = | z − 1 |2
⇒ | z | 2 + 9 − 2 Re (3z ) = | z | 2 + 1 − 2 Re (z )
⇒
⇒
4 Re (z ) = 8
Re (z ) = 2
…(ii)
82
Textbook of Algebra
Case II If | z − 3 | = | z − 5 |
On squaring both sides, we get
| z − 3 |2 = | z − 5 |2
⇒
⇒
| z | 2 + 9 − 2 Re (3z ) = | z | 2 + 25 − 2 Re (5z )
⇒
4 Re (z ) = 16 ⇒ Re (z ) = 4
39.
π/6 π/3
a
2π/3
π/3 π/3
π/6
a
A(– a,0) a O
B(a,0)
From figure, it is clear that z lies on the point of intersection of
the rays from A and B.
Q ∠ACB = 90 ° and OBC is an equilateral triangle.
Hence,
OC = a
⇒
| z − 0 | = a or | z | = a
π
and arg (z ) = arg (z − 0 ) =
3
2z − i
40. Q
=m
z +i
For circle,
⇒
z − i/2 m
=
z +i
2
m
≠1
2
m ≠ 2 and m > 0
| z 0| = r ⇒ | z 0| 2 = r 2 ⇒ z 0 z 0 = r 2
r
O
P(z)
Let P (z ) be any point on tangent, then
π
∴
∠PAO =
2
Complex slope of AP + Complex slope of OA = 0
z − z0 z0 − 0
⇒
+
=0
z − z 0 z0 − 0
⇒
Also,
⇒
z z 0 + z 0 z = 2z 0 z 0
zz 0 + z 0z = 2r 2
zz 0 = zz 0
zz 0 z 0 z
+ 2 =2
r2
r
zz 0
z 0z
+
=2
z 0z 0 z 0z 0
| a | = | z 1 + z 2 | ≤ | z 1| + | z 2 | = 1 + 1 = 2
|a | ≤2
θ+φ
Also, arg (a ) = arg (z1 + z 2 ) = arg (ei θ + ei φ ) =
2
and arg (b ) = arg (z1z 2 ) = arg (ei θ + φ ) ) = θ + φ
Q
∴
∴ 2 arg (a ) = arg (b ) ⇒ arg (a 2 ) = arg (b )
43. Q
Then,
αz2 + z + α = 0
αz + z + α = 0
⇒
α (z ) 2 + z + α = 0
⇒
αz 2 + z + α = 0
On subtracting Eq. (ii) from Eq. (i), we get
(α − α ) z 2 − (α − α ) = 0
⇒
α − α = 0 and z 2 = 1
|z | =| ± 1| =1
44. Let z = α be a real root of equation
z 3 + (3 + i ) z 2 − 3z − (m + i ) = 0
⇒
α 3 + (3 + i ) α 2 − 3 α − (m + i ) = 0
⇒
(α + 3 α 2 − 3 α − m ) + i (α 2 − 1 ) = 0
3
On comparing real and imaginary parts, we get
α 3 + 3α 2 − 3α − m = 0
α2 − 1 = 0 ⇒ α = ± 1
and
For α = 1, we get
1 + 3 −3 −m = 0 ⇒ m =1
For α = − 1, we get
−1 + 3 + 3 −m = 0 ⇒ m =5
45. Let z = α be a real root of equation
z 3 + (3 + 2i ) z + ( −1 + ia ) = 0
⇒
α 3 + (3 + 2i ) α + ( − 1 + ia ) = 0
⇒
(α 3 + 3 α − 1 ) + i (a + 2α ) = 0
On comparing real and imaginary parts, we get
α 3 + 3α − 1 = 0
and
…(i)
2
i.e.,
A(z0)
⇒
⇒
and given | z1| = | z 2 | = 1
Let
z1 = ei θ and z 2 = ei φ
α = α and z = ± 1
∴
Put z = ± 1 in Eq. (i), we get
α+α =±1
and absolute value of real root = 1
41. ∴ A(z 0 ) lie on | z | = r
⇒
∴
42. Q z1 + z 2 = a, z1z 2 = b
C(z)
⇒
z 
z
+ 
z0 z0
=1
2
z
Re   = 1
z0
a + 2α = 0
[Q z = z ] …(ii)
Chap 01 Complex Numbers
⇒
α=−
⇒
−
Let
a
2
=
a 3 3a
−
− 1 = 0 ⇒ a 3 + 12a + 8 = 0
8
2
f (a ) = a 3 + 12a + 8
f ( −1 ) < 0, f ( 0 ) > 0, f ( −2 ) < 0
f (1 ) > 0 and f (3 ) > 0
⇒
a ∈ ( −2, 1 ) or a ∈ ( −1, 0 ) or a ∈ ( −2,3 )
Sol. (Q. Nos. 46 to 48)
= ( − 1 − i ) + (i + i − 1 − 1
+ (i 0 + i 0 + i 0 + … 47 times))
= ( − 1 − i ) + (2i − 2 + 47 )
= 44 + i = a + ib
46. Q arg (z ) > 0
= ( 44 ) (( 44 ) 2 )1005 = ( 44 ) (1936 )1005
47. Q arg (z1z 2) = π
= (Unit place of 44)
× (Unit place digit of (1936 )1005)
arg (z1 ) + arg (z 2 ) = π
⇒
arg (z1 ) − arg (z 2 ) = π
= Unit place of ( 4 × 6 ) = 4
| z 1| = | z 2 |
and unit place digit of b 2012 = (1 ) 2012 = 1
Given,
∴
| z 1| = | z 2 | = | z 2 |
Hence, the unit place digit of a 2011 + b 2012 = 4 + 1 = 5.
z1 + z 2 = 0
Then,
⇒
100
= (i 0 + i 0 + i 0 + … 97 times) + i 1 + 2 + 3 + … + 101
= 97 + i 5151 = 97 + i 3 = 97 − i
4z1 + 5z 2 = 0
∴
z1
5
= = 1 . 25
z2
4
∴
a = 97 and b = − 1
Hence, a + 75b = 97 − 75 = 22
Sol. (Q. Nos. 52 to 54)
If z ±
Sol. (Q. Nos. 49 to 51)
∴
49. Q n ! is divisible by 4, ∀ n ≥ 4.
25
22
n=4
n =1
= i 0 + i 0 + i 0 + … (22 times) = 22
25
25
n =1
n=4
∑ i n ! = i 1! + i 2! + i 3! + ∑
∴
6
= i − 1 − 1 + 22 = 20 + i
∴
a = 20, b = 1
∴
a − b = 20 − 1 = 19
…(i)
⇒
∴
in!
= i + i + i + 22
2
[from Eq. (i)]
and
Also,
⇒
which is a prime number.
50. ∴
95
∑
r = −2
ir +
50
⇒
r=0
⇒
∑ ir !
a
= b, where a, b > 0
z
z±
⇒
∑ i n ! = ∑ i (n + 3)!
∴
∑ i (r + 3)! + i 1 ⋅ i 2 ⋅ i 3 … i 101
r =1
z1
5
= = 1 . 25
z2
4
z1
5
=−
z2
4
97
=
is possible only when | 4z1| = | 5z 2 |
⇒
r =1
r=4
48. arg ( 4z1 ) − arg (5z 2 ) = π
and also
101
51. Q ∑ i r ! + ∏ i r
z1 = − z 2
⇒
[given]
∴
a = 44, b = 1
Unit place digit of a 2011 = ( 44 ) 2011
arg (z ) + arg ( −z ) = − π
− arg (z ) + arg ( −z ) = − π
arg ( −z ) − arg (z ) = − π
⇒
50


0!
1!
2!
3!
r−3

+
+
+
+
+
i
i
i
i
i
∑
∑ i r ! 



r =1
r=4
98
47


= (i −2 + i −1 + 0 ) + i 1 + i 1 + i 2 + i 6 + ∑ i (r + 3)! 




r =1
∴
∴
⇒
⇒
83
a
a
≤|z | +
z
|z |
a
b ≤|z | +
|z |
| z |2 − b | z | + a ≥ 0
|z | ≤
b − b 2 − 4a
2
b + b 2 − 4a
2
a
a
z±
≥ |z | −
z
|z |
|z | ≥
a
|z |
a
−b ≤|z | −
≤b
|z |
b ≥ |z | −
−b | z | ≤ | z | 2 − a ≤ b | z |
…(i)
84
Textbook of Algebra
Case I −b | z | ≤ | z | 2 − a
On squaring and adding Eqs. (i) and (ii), we get
(1 − a ) 2 + b 2 = (2a + 1 ) 2 + (2b ) 2
⇒ | z |2 + b | z | − a ≥ 0
∴
|z | ≤
− b − b 2 + 4a
2
⇒
3a 2 + 3b 2 + 6a = 0
⇒
a 2 + b 2 + 2a = 0
and
|z | ≥
− b + b 2 − 4a
2
From option (c),
(1 + 5a ) 2 + (3b ) 2 = (1 − 4a ) 2
Case II | z | 2 − a ≤ b | z |
⇒
9a 2 + 9b 2 + 18a = 0
⇒ | z |2 − b | z | − a ≤ 0
∴
a 2 + b 2 + 2a = 0
56. From Eq. (i), we get
b − b 2 + 4a
b + b 2 + 4a
≤|z | ≤
2
2
From Case I and Case II, we get
∴
− b + b 2 + 4a
b + b 2 + 4a
≤|z | ≤
2
2
From Eqs. (i) and (ii), we get
 1 − tan 2 θ / 2 
 2 tan θ / 2 
(1 − a ) 
 +b
 = 2a + 1
 1 + tan 2 θ / 2
 1 + tan 2 θ / 2
…(ii)
and the least value of | z | is
⇒ (2 + a ) tan 2
b + b 2 + 4a
2
∴
− b + b 2 + 4a
.
2
λ = Sum of the greatest and least values of | z |
= b 2 + 4a = 4 + 4 = 8
∴
∴ λ2 = 8
= b 2 + 4a = 16 + 8 = 24
λ2 = 24
54. Here, a = 3 and b = 6
λ = Sum of the greatest and least value of | z |
= b 2 + 4a = 36 + 12 = 48 = 4 3
λ = 12
58. Q
1 + z + z 2 + z 3 + … + z 17 = 0
∴
1 ⋅ (1 − z 18 )
=0
(1 − z )
⇒
1 − z 18 = 0, 1 − z ≠ 0
∴
z 18 = 1, z ≠ 1
∴
Sol. (Q. Nos. 55 to 57)
∴
4b 2 − 12( −b 2 )
and 1 + z + z + z + … + z
2
z −1
= a + ib
z +2
55. Q z = CiS θ = e
2b ±
[Q a 2 + b 2 + 2a = 0 ]
− 2b 2 / a
(2b ± 4b ) a
3a a
6ba
− 2ab
or
or
=
=
=−
2
2
2
b
b
− 2b
− 2b
− 2b
θ
b
θ
b
θ
b
or
or − = 3 cot or − cot
cot = −
2
3a a
a
2
2
2
λ =2 3
Q W =
4b 2 − 12a (2 + a )
2(2 + a )
Now, | z | = 1 = (a + 1 ) 2 + b 2
λ = Sum of the greatest and least value of | z |.
∴
θ 2b ±
=
2
iθ
3
13
e −1
= a + ib
e iθ + 2
=0
1 ⋅ (1 − z )
=0
(1 − z )
⇒
1 − z 14 = 0, 1 − z ≠ 0
∴
z 14 = 1, z ≠ 1
∴
…(i)
…(ii)
…(i)
14
From Eqs. (i) and (ii), we get
z 14 ⋅ z 4 = 1 ⇒ 1 ⋅ z 4 = 1
iθ
⇒(cosθ + i sin θ − 1 ) = (a + ib ) (cosθ + i sin θ + 2 )
On comparing real and imaginary parts, we get
cosθ − 1 = a cosθ + 2a − b sin θ
⇒ (1 − a ) cosθ + b sin θ = 2a + 1
and sin θ = a sin θ + b cosθ + 2b
(1 − a ) sin θ − b cosθ = 2b
θ
2
57. ∴ a 2 + b 2 + 2a = 0 ⇒ (a + 1) 2 + b 2 = 1
53. Here, a = 2 and b = 4
⇒
tan
θ
θ
− 2b tan + 3a = 0
2
2
=
52. Here, a = 1 and b = 2
∴
θ
θ
+ 2b tan
2
2
= (2a + 1 ) + (2a + 1 ) tan 2
− b + b 2 + 4a
b + (b 2 + 4a )
≤|z | ≤
2
2
∴ The greatest value of | z | is
⇒ (1 − a ) − (1 − a ) tan 2
z4 = 1
Then,
z = 1, − 1, i , − i
Q
z ≠1
Q
z = − 1, i , − i
Hence, only z = − 1 satisfy both Eqs. (i) and (ii).
∴ Number of values of z is 1.
…(ii)
Chap 01 Complex Numbers
z3 = z
59. We have,
|z | =|z | =|z |
⇒
⇒
⇒
…(i)
3
| z | (| z | 2 − 1 ) = 0
| z | = 0 and | z | 2 = 1
Now, | z | 2 = 1
1
z
On putting this value in Eq. (i), we get
1
z3 =
z
⇒
z4 = 1
⇒
zz = 1 ⇒ z =
1
3
[from Eq. (i)]
,y = ±
2
2
3
1
For x = −
[from Eq. (i)]
,y = ±
2
2
1 i 3
1 i 3 3 i
3 i
∴ Solutions are ±
,− ±
,
± ,−
±
2
2
2
2
2
2
2
2
Hence, number of solutions is 8.
For x =
⇒
x 2 = (a 2 − b 2 ) + 2iab = 3 + 4i
∴
a − b = 3 and ab = 2
2
⇒
z 2 = (81 − a 2 ) + 18ai
and 3a 2b − b 3 = 11
∴
⇒ a (a − 225 ) = 0
a = 0 or a = 225
a≠0
a 2 = 225
∴ The sum of digits of a 2 = 2 + 2 + 5 = 9
∴
a 2 = 4, b 2 = 1
⇒
and
Finally, a = 2, b
Hence,
a
a = 2, b = 1
a = − 2, b = − 1
= 1 satisfies Eq. (ii).
+ b =2 + 1 =3
63. Q
61. Let z = x + iy
and
z z
+
=1
z z
⇒
x + iy
x − iy
+
=1
x − iy
x + iy
⇒
( x + iy ) 2 + ( x − iy ) 2
=1
x2 + y 2
⇒
2 (x 2 − y 2 )
=1
1
2
[Q ab = 2 ]
(1 + i ) 4 = [(1 + i ) 2 ]2
= (1 + i 2 + 2i ) 2 = (1 − 1 + 2i ) 2
…( i )
= 4i 2 = − 4
and
…(i)
1− πi
π −i
+
1+ πi
π +i
=
(1 − π i ) ( π − i ) ( π − i ) (1 − π i )
+
π+1
1+π
π − i − πi − π + π − πi − i − π
π +1
− 2 πi − 2i
=
= − 2i
π+1
=
[from Eq. (i)]
1
…(ii)
x −y = ±
⇒
2
From Eqs. (i) and (ii), we get
1 1 3
2x 2 = 1 ± = ,
2 2 2
1
3
1
3
⇒
x2 = , ⇒ x = ± , ±
2
2
4 4
1
3
For x = , y = ±
[from Eq. (i)]
2
2
1
3
For x = − , y = ±
[from Eq. (i)]
2
2
2
2a 2 = 8, 2b 2 = 2
Then,
2
|z | =1
x2 + y 2 = 1
…(ii)
a 2 + b 2 = (a 2 − b 2 ) 2 + 4a 2 b 2 = 5
2
Q
∴
[given]
From Eq. (i), we get
Im (z 2 ) = Im (z 3 )
But
2
2
According to the question, we have
⇒
2
∴ a − 3ab = 2
3
18a = 243a − a
…(i)
and x = x ⋅ x = (a + ib ) [(a − b ) + 2iab ]
2
= (a 3 − 3ab 2 ) + i (3a 2b − b 3 ) = 2 + 11i
z 3 = (729 − 27a 2 ) + (243a − a 3 ) i
⇒
[given]
= (a 3 − ab 2 − 2ab 2 ) + i [2a 2b + b(a 2 − b 2 )]
z = 9 + ai
3
2
3
Clearly, Eq. (ii) has 4 solutions.
Therefore, the required number of solutions is 5.
60. We have,
x = a + ib
62. We have,
…(ii)
85
Given, z =
1 − π i
π
π −i 
+
(1 + i ) 4 

4
1 + π i
 π +i
π
( − 4 ) ( −2i ) = 2πi
4
 |z | 
2π
Now, 
=4
 =
 amp (z) π / 2
=
64. Q
[from Eqs. (i) and (ii)]
An = 1
⇒
A = (1 )1/n = e 2π r i /n , r = 0, 1, 2, …, n − 1
∴
A = 1, e 2 π i /n , e 4 π i /n , e 6 πi /n , …, e 2 π (n − 1) i /n
and
…(ii)
( A + 1 )n = 1 ⇒ A + 1 = (1 )1/n = e 2 π pi /n
86
Textbook of Algebra
 πp 
A = e 2 π pi /n − 1 = e p π i /n ⋅ 2i sin   ,
 n
p = 0, 1, 2, …, n − 1
 2π 
 π
A = 0, e π i /n ⋅ 2i sin   , e 2 π i /n 2i sin   ,...,
n
 n
⇒
∴
e
π i (n − 1)/n
 π (n − 1 )
⋅ 2i sin 



n
n = 6,
e 4 π i /n = e 4 π i / 6 = e 2 π i / 3
2π
2π
1 i 3
= cos
+ i sin
=− +
2
2
3
3
 π
 π
π i /n
πi /6
and
⋅ 2i sin  
e
⋅ 2i sin   = e
 6
 n
π
π

=  cos + i sin  ⋅ i

6
6
 3 i
1 i 3
=
+ i =− +
2
2
 2 2
Hence, the least value of n is 6.
For
65. Given, z1, z 2, z 3, … , z 50 are the roots of the equation
2q π
2q π 

 10 
= − i  ∑  cos
+ i sin
 − 1
11
11 

q = 0
= − i {(sum of 11, 11th roots of unity) − 1}
= − i (0 − 1) = i
32
2qπ
2qπ  
 10 
− i cos
P = ∑ (3 p + 2 )  ∑ sin



p =1
q =1
11
11  
Q
32
= ∑ (3 p + 2 ) (i ) p
p =1
32
= 3 ∑ p(i ) p + 0 = 3S (say)
p =1
32
where, S = ∑ p(i ) p
p =1
S = 1 ⋅ i + 2 ⋅ i 2 + 3 ⋅ i 3 + … + 31 ⋅ i 31 + 32 ⋅ i 32
iS = 1 ⋅ i 2 + 2 ⋅ i 3 + … + 31 ⋅ i 32 + 32i 33
= ( 0 ) − 32i
32i ⋅ (1 + i )
S =−
(1 − i ) ⋅ (1 + i )
∑ (z )r = 0, then
∑ (z )
∴
50
= (z − z1 ) (z − z 2 ) (z − z 3 ) … (z − z 50 ) = Π (z − zr )
= − 16 (i − 1 ) = 16 (1 − i )
P = 3S = 48 (1 − i )
r =1
r=0
∴
Taking log on both sides on base e, we get
Given,
50
 50

loge  ∑ (z )r  = ∑ loge (z − zr )


r = 0
 r =1
On differentiating both sides w.r.t. z, we get
⇒
r=0
50
∑ (z )r
50
∑
r =1
1
(z − zr )
n
Given,
r=0
⇒
On putting z = 1 in both sides, we get
r=0
50
∑
1
=
50
∑
r =1
⇒
1
(1 − zr )
⇒
(1 + 2 + 3 + … + 50 )
1
=−∑
51
(zr − 1 )
r =1
50
= − ( − 5λ )
50
× 51
2
⇒
= 5λ
51
⇒
λ =5
10
2q π
2q π 

66. Q ∑ sin
− i cos

q =1
11
11 
2q π
2q π 

= − i ∑  cos
+ i sin

q = 1
11
11 
10
Now,
[given]
⇒
∴
2
sin − 1
π
1 + x2


 2x 
1 + x2 π n
sin − 1 
 = (i )
 2x  2
⇒
r=0
2
1 + i
2
−1 1 + x 


 = sin 
1 − i
π
 2x 
in =
50
∑r
4( 4 !) = n(n !)
n=4
1 + i 2 + 2i
1+i
(1 + i ) 2
=
=i
=
2
1 − i (1 − i ) (1 + i )
67. Q
=
(1 + i ) P = n(n !) ⇒ (1 + i ) ⋅ 48 (1 − i ) = n(n !)
96 = n(n !) ⇒
∴
50
∑ r (z )r − 1
32
p =1
(1 − i ) S = (i + i 2 + i 3 + … + i 32 ) − 32i 33
r=0
r
32
p =1
= 3 ∑ p(i ) p + 2 ∑ (i ) p
50
50
p
1 + x2
π

= sin  (i )n 
2

2x
AM ≥ GM
1
x+
2
x ≥1 ⇒ x + 1 ≥1
2x
2
 π n
sin  (i )  ≥ 1
2


π
sin  (i )n  = 1

2
⇒
n = 4, 8, 12, 16, …
∴ Least positive integer, n = 4
…(i)
[Q −1 ≤ sin θ ≤ 1]
Chap 01 Complex Numbers
68. (A) → ( p,q ), (B) → ( p,r ),(C) → ( p,r ,s )
and
− b + b 2 + 4a
b + b 2 + 4a
≤ |z | ≤
2
2
(A) Here, a = 1 and b = 2
Then, − 1 + 2 ≤ | z | ≤ 1 + 2
∴
G =1 + 2
and
L = −1 + 2
Then, − 2 + 6 ≤ | z | ≤ 2 + 6
G =2 + 6
and
L = −2 + 6
⇒
(C) Here, a = 3 and b = 6
Then, − 3 + 2 3 ≤ | z | ≤ 3 + 2 3
∴
G =3 + 2 3
and
L = −3 + 2 3
= 5 − 5i , − 1 − i , − 5 + 5i , 1 + i
z1 = 5 − 5i , z 2 = − 1 − i ,
z 3 = − 5 + 5i and z 4 = 1 + i
∴ | z1| 2 + | z 2 | 2 + | z 3| 2 + | z 4 | 2 = 50 + 2 + 50 + 2
= 104 = 8 × 13
 17 + 8
8 + 15i = ± 
+i
2

3i 
1
 5
=±
+
(5 + 3i )
=±

 2
2
2
 17 − 8
and − 8 − 15i = ± 
−i
2

∴
z = 8 + 15i +
We know that,
 | z | − Re (z )
z =±
+i
2

| z | − Re (z ) 

2

| z | − Re (z ) 

2

If Im (z ) < 0
10 − 6 

2 
= ± (2 2 + i 2 )
= ± 2 (2 + i )
 10 − 6
and − 6 + 8i = ± 
+i
2

10 + 6 

2 
= ± ( 2 + i 2 2 ) = ± 2 (1 + 2i )
z = 6 + 8i +
− 6 + 8i
= ± 2 (2 + i ) ± 2 (1 + 2i )
= 3 2 (1 + i ), 2 (1 − i ), − 3 2 (1 + i ), 2 ( − 1 + i )
z1 = 3 2 (1 + i ), z 2 = 2 (1 − i ),
z 3 = − 3 2 (1 + i )
and
z 4 = 2 (− 1 + i )
∴ | z 1 | + | z 2 | + | z 3| + | z 4 |
2
− 8 − 15i
∴
z1 = 2 ( 4 − i ), z 2 = 2 ( − 1 − 4i )
z 3 = 2 ( − 4 + i ) and z 4 = 2 (1 + 4i )
∴ | z1| + | z 2 | 2 + | z 3| 2 + | z 4 | 2 = 34 + 34 + 34 + 34
= 136 = 17 × 8
2
 10 + 6
(A) 6 + 8i = ± 
+i
2

∴
17 + 8

2 
1
1
=±
(5 + 3i ) ±
(3 − 5i )
2
2
1
1
(8 − 2i ),
( − 2 − 8i ),
z=
2
2
1
1
( − 8 + 2i ),
(2 + 8i )
2
2
69. (A) → (q), B → (q, r), C → (q, s)
∴
17 − 8 

2 
1
5 
 3
i = ±
(3 − 5i )
=±
−
 2
2 
2
⇒
G − L =6
[natural number, composite number and perfect number]
 | z | + Re (z )
If Im(z ) > 0 = ± 
−i
2

13 + 5 
 = ± (2 − 3i )
2 
z = 5 − 12i + −5 − 12i = ± (3 − 2i ) ± (2 − 3i )
∴
(C)
G − L = 4 [natural number and composite number]
 13 − 5
− 5 − 12i = ± 
−i
2

∴
[natural number and prime number]
⇒
G − L =2
(B) Here, a = 2 and b = 4
∴
 13 + 5
13 − 5 
5 − 12i = ± 
−i
 = ± (3 − 2i )
2
2 

(B)
a
If z ±
= b, where a > 0 and b > 0, then
z
87
2
2
2
= 36 + 4 + 36 + 4 = 80 which is divisible by 8.
70. (A) → (p, q, r, t);(B) → (p,s);(C) → (p, r)
(A) Here, the last digit of 143 is 3. The remainder when 861 is
divided by 4 is 1. Then, press switch number 1 and we get
3. Hence, the digit in the units place of (143 ) 861 is 3.
∴
λ =3
Next, the last digit of 5273 is 3. The remainder when 1358
is divided by 4 is 2. Then, press switch number 2 and we
get 9. Hence, the digit in the units place of (5273 )1358 is 9.
∴
µ =9
Hence,
λ + µ = 3 + 9 = 12
which is divisible by 2, 3, 4 and 6.
(B) Here, the last digit of 212 is 2. The remainder when 7820
is divided by 4 is 0. Then, press switch number 0 and we
get 6. Hence, the digit in the unit’s place of (212 ) 7820 is 6.
∴
λ =6
Next, the last digit of 1322 is 2. The remainder when 1594
is divided by 4 is 2. Then, press switch number 2 and we
get 4.
88
Textbook of Algebra
Hence, the digit in the unit’s place of (1322 )1594 is 4.
∴
µ=4
Hence, λ + µ = 6 + 4 = 10, which is divisible by 2 and 5.
(C) Here, the last digit of 136 is 6. Therefore, the unit’s place
of (136 ) 786 is 6.
∴
λ =6
Next, the last digit of 7138 is 8. The remainder when
13491 is divided by 4 is 3. Then, press switch number 3
and we get 2. Hence, unit’s place of (7138 )13491 is 2.
∴
µ =2
Hence,
λ + µ =6 + 2 =8
which is divisible by 2 and 4.
71. (A) → (r); ( B) → (p,s); ( C) → (q, t)
If z −
a
= b, where a > 0 and b > 0, then
z
b + b 2 + 4a
− b + b 2 + 4a
≤ |z | ≤
2
2
b + b 2 + 4a
− b + b 2 + 4a
and µ =
2
2
(A) Here, a = 6 and b = 5
∴
λ = 6 and µ = 1
⇒
λµ + µ λ = 61 + 1 6 = 7
∴ λ=
and
λµ − µ λ = 61 − 1 6 = 5
(B) Here, a = 7 and b = 6
∴
λ = 7 and µ = 1
µ
λ
∴
λ + µ = 71 + 1 7 = 8
and
λµ − µ λ = 81 − 1 8 = 7
72. Statement-1 is false because 3 + 7i > 2 + 4i is meaningless in
the set of complex number as set of complex number does not
hold ordering. But Statement-2 is true.
73. Statement-1 is false as
(cos θ + i sin φ )n ≠ cos nθ + i sin nφ
2
π
π
π
π

Now,  cos + i sin  = cos + i sin

4
4
2
2
=i
[by De-Moivre’s theorem]
∴ Statement-2 is true.
74. We have,
| 3z1 + 1 | = | 3z 2 + 1 | = | 3z 3 + 1 |
 1 
∴ z1, z 2 and z 3 are equidistant from  − , 0 and circumcentre
 3 
 1 
of triangle is  − , 0 .
 3 
Also,
1 + z1 + z 2 + z 3 = 0
1 + z 1 + z2 + z3
=0
3
1
z 1+ z2 + z3
=−
⇒
3
3
1


∴ Centroid of the triangle is  − , 0 .
 3 
So, the circumcentre and centroid of the triangle coincide.
Hence, required triangle is an equilateral triangle.
Therefore, Statement-1 is true. Also, z1, z 2 and z 3 represent
vertices of an equilateral triangle, if
z12 + z 22 + z 32 − (z1z 2 + z 2 z 3 + z 3z1 ) = 0.
Therefore, Statement-2 is false.
75. We have,
…(i)
|z − 1| + |z − 8| = 5
Here,
z1 = 1, z 2 = 8 and 2a = 5
Now,
| z1 − z 2 | = | 1 − 8 | = | − 7 | = 7
∴
2a = 5 < 7
Therefore, locus of Eq. (i) does not represent an ellipse. Hence,
Statement-1 is false. Statement-2 is true by the property of
ellipse.
76. Since, z1, z 2 and z 3 are in AP.
∴
2z 2 = z1 + z 3
z + z3
z2 = 1
⇒
2
It is clear that, z 2 is the mid-point of z1 and z 3.
∴ z1, z 2 and z 3 are collinear.
Statement-1 is true, Statement-2 is true; Statement-2 is a
correct explanation of Statement-1.
77. Principal argument of a complex number depend upon
λµ − µ λ = 71 − 1 7 = 6
(C) Here, a = 8 and b = 7
∴
λ = 8 and µ = 1
⇒
λµ + µ λ = 81 + 1 8 = 9
and
⇒
quadrant and principal argument lies in ( − π , π ].
Hence, Statement-1 is always not true and Statement-2 is
obviously true.
π
78. We have, C1 : arg (z ) =
4
π
−1 y 
[ let z = x
⇒
tan   =
 x 4
y
π
= tan = 1
⇒
x
4
⇒
y =x
∴
C1 : y = x
3π
C 2 : arg(z ) =
4
3π
−1 y 
[ let z = x
⇒
tan   =
 x
4
y
3π
⇒
= tan
= −1
x
4
⇒
y =−x
∴
C2 : y = − x
and C 3 : arg (z − 5 − 5i ) = π
 y − 5
[ let z = x
⇒
tan − 1 
 =π
 x − 5
+ iy ]
…(i)
+ iy ]
…(ii)
+ iy ]
Chap 01 Complex Numbers
y −5
= tan π = 0 ⇒ y = 5
x −5
⇒
∴
C3 : y = 5
We get the following figure.
On putting x = ω in Eq. (i), we get
ω2 − 1
= (ω − α 1 ) (ω − α 2 )(ω − α 3 ) (ω − α 4 )
ω −1
and putting x = ω 2 in Eq. (i), we get
⇒
=
x
:y
1
C
C3 : y=5
B
(–5, 5)
A(5, 5)
C2 : y= – x
O(0, 0)
⇒
X
=
1 5−0 5−0
= 25
2 −5 − 0 5 − 0
=
Therefore, the boundary of C1, C 2 and C 3 constitutes right
isosceles triangle.
Hence, Statement-2 is true.
⇒
⇒
⇒
79. Since, Im ( z 2 z 3 ) =
⇒
and
...(ii)
and
...(iii)
On subtracting Eq. (ii) from Eq. (i), we get
2 ( x1 − x 2 ) = y12 − y 22
...(ii)
∴
y1 − y 2 = x1 − x 2
From Eqs. (iii) and (iv), we get
y1 + y 2 = 2
∴
Im (z1 + z 2 ) = 2
Hence, the imaginary part (z1 + z 2 ) is 2.
83. (i) LHS = (a 2 + b 2 + c 2 − bc − ca − ab )
( x 2 + y 2 + z 2 − yz − zx − xy )
x 3 + 3ax 2 + 3bx + c = 0,
z1 + z 2 + z 3 = − 3a
z1 + z 2 + z 3
= −a
3
z1z 2 + z 2 z 3 + z 3z1 = 3b
...(iii)
...(iv)
[Q y1 − y 2 ≠ 0]
= (a + bω + c ω 2 ) (a + bω 2 + cω )
( x + yω + zω 2 ) ( x + yω 2 + zω )
(z1 + z 2 + z 3 ) 2 = 3(z1z 2 + z 2 z 3 + z 3z1 )
= {(a + bω + cω 2 ) ( x + yω + zω 2 )}
( − 3a ) = 3(3b )
2
{(a + bω 2 + cω ) ( x + yω 2 + zω )}
= {ax + cy + bz + ω (bx + ay + cz )
+ ω 2 (cx + by + az )} × {ax + cy + bz + ω 2
(bx + ay + cz ) + ω (cx + by + az )}
= ( X + ωΖ + ω 2 Y ) ( X + ω 2 Z + ω Y )
Therefore, the condition is a = b.
81. Q x 5 − 1 = 0 has roots 1, α 1, α 2, α 3, α 4 .
∴ (x 5 −1) = (x − 1) (x − α1 ) (x − α 2 ) (x − α 3 ) (x − α 4 )
x 5 −1
= (x − α1 ) (x − α 2 ) (x − α 3 ) (x − α 4 )
x −1
...(i)
2 x 2 = 1 + y 22
2 ( x1 − x 2 ) = (y1 + y 2 ) (y1 − y 2 )
But, given that arg (z1 − z 2 ) = π / 4
y − y 2 π
y − y2
Then, tan − 1  1
=1
⇒ 1
 =
x1 − x 2
x
x
4
−
 1
2
2
⇒
x = ( x − 1 ) + iy
x 2 = (x − 1)2 + y 2 ⇒ 2x = 1 + y 2
If z1 = x1 + iy1 and z 2 = x 2 + iy 2
Then,
2 x1 = 1 + y12
z12 + z 22 + z 32 = z1z 2 + z 2 z 3 + z 3z1
⇒
x = x + iy − 1
...(i)
Hence, the centroid of the ∆ABC is the point of affix ( − a ).
Now, the triangle will be equilateral, if
⇒
ω 4 + 1 − 2ω 2 ω + 1 − 2 ω 2
=
ω 2 + 1 − 2ω ω 2 + 1 − 2 ω
=
(OA ) 2 + (OB ) 2 = ( AB ) 2 and OA = OB
we get
...(iii)
− ω 2 − 2ω 2 − 3ω 2
=
=ω
− ω − 2ω
− 3ω
z +z
82. Let z = x + iy , then
=x
2
∴ From given relation, we get
∴ Statement-1 is false.
Now,
OA = 5 2, OB = 5 2 and AB = 10
z 2 z 3 − (z 2 z 3 )
1
= {z 2 z 3 − z 2 z 3 }
2i
2i
1
z1 Im ( z 2 z 3 ) = {z1z 2 z 3 − z1z 2 z 3}
2i
1
Similarly, z 2 Im ( z 3 z1 ) = {z 2 z 3 z1 − z 2 z1 z 3}
2i
1
and
z 3 Im ( z1 z 2 ) = {z 3 z1 z 2 − z 3z1z 2}
2i
On adding Eqs. (i), (ii) and (iii), we get
z1 Im ( z 2 z 3 ) + z 2 Im ( z 3 z1 ) + z 3 Im ( z1 z 2 ) = 0
Therefore, this is proved.
80. Since, z1, z 2 and z 3 are the roots of
ω10 − 1
= (ω 2 − α 1 ) (ω 2 − α 2 ) (ω 2 − α 3 )(ω 2 − α 4 )
ω2 − 1
ω −1
= (ω 2 − α 1 ) (ω 2 − α 2 ) (ω 2 − α 3 )(ω 2 − α 4 )
ω 2 −1
...(ii)
On dividing Eq. (ii) by Eq. (iii), we get
ω − α1 ω − α 2 ω − α 3 ω − α 4
(ω 2 − 1 ) 2
=
⋅ 2
⋅ 2
⋅ 2
2
ω − α1 ω − α 2 ω − α 3 ω − α 4
(ω − 1 ) 2
∴ Area of the region bounded by C1, C 2 and C 3
Q
ω5 − 1
= (ω − α 1 ) (ω − α 2 ) (ω − α 3 ) (ω − α 4 )
ω −1
…(iii)
Y
89
...(i)
= RHS
90
Textbook of Algebra
(ii) LHS = (a 3+ b 3 + c 3 − 3abc ) ( x 3+ y 3 + z 3 − 3 xyz )
= (a + b + c ) (a 2 + b 2 + c 2 − ab − bc − ca ) ×
( x + y + z )( x 2+ y 2 + z 2 − xy − yz − zx )
= (a + b + c ) ( x + y + z )
(a 2 + b 2 + c 2 − ab − bc − ca ) ×
2
2
2
( x + y + z − xy − yz − zx ) [using (i) part]
= (ax + ay + az + bx + by + bz + cx + cy + cz )
( X 2 + Y 2 + Z 2 − YZ − ZX − XY )
= {(ax + cy + bz ) + (cx + by + az ) + (bx + ay + cz )}
( X 2 + Y 2 + Z 2 − YZ − ZX − XY )
= ( X + Y + Z ) ( X + Y + Z − YZ − ZX − XY )
2
=X
3
2
2
∴
2
∴ x 2 + y 2 − 2i ( x + iy ) + 2c (1 + i ) = 0
( x + y + 2y + 2c ) + i ( − 2 x + 2c ) = 0
2
On comparing the real and imaginary parts, we get
x 2 + y 2 + 2y + 2c = 0
and
− 2 x + 2c = 0
From Eqs. (i) and (ii), we get
y 2 + 2y + c 2 + 2c = 0
⇒
y =
−2 ±
…(i)
…(ii)
4 − 4(c 2 + 2c )
= − 1 ± (1 − c 2 − 2c )
2
Q x and y are real.
∴
1 − c 2 − 2c ≥ 0 or c 2 + 2c + 1 ≤ 2
(c + 1 ) 2 ≤ ( 2 ) 2 ⇒ − 2 − 1 ≤ c ≤ 2 − 1
∴
n
(1 − x ) (1 + x ) (1 + x 2 ) (1 + x 2 ) … (1 + x 2 )
(1 − x )
n
(1 − x 2 ) (1 + x 2 ) (1 + x 2 ) … (1 + x 2 )
=
(1 − x )
|z |2 = x 2 + y 2
2
n
2
3
z = x + iy
84. Let
2
LHS = (1 + x ) (1 + x 2 ) (1 + x 2 ) … (1 + x 2 )
=
+ Y + Z − 3 XYZ = RHS
3
Using Eqs. (i) and (ii), then equations of lines are
z + z 3 (z − z )
+
− 11 = 0
2
2i
3 (z + z ) (z − z )
and
−
+7=0
2
2i
i.e.,
(1 − 3i ) z + (1 + 3i ) z − 22 = 0
and
(3 + i ) z + (3 − i ) z + 14 = 0
1+i
86. Putting
= x in LHS, we get
2
0 ≤c ≤ 2 −1
[Q given c ≥ 0]
Hence, the solution is z = x + iy = c + i ( − 1 ± 1 − c 2 − 2c )
2
2
n
n
n
=
(1 − x 2 ) (1 + x 2 ) … (1 + x 2 )
(1 − x )
=
(1 − x 2 ) (1 + x 2 ) 1 − ( x 2 ) 2
=
(1 − x )
(1 − x )
n
2n
i
1− 
 2
=
1 + i
1−

 2 
1
1 − n (1 )

1 
(1 + i )
22
=
= (1 + i ) 1 − n 
⋅
 1 − i  (1 + i )

22 


 2 
makes an angle 3 π / 4 with positive real axis as shown in the
figure.
x+y=3
Y
and z = x + iy ≡ no solution for c > 2 − 1
85. Let z = x + iy
1
∴ m1 = − , 3
3
∴ Equation of straight lines through ( − 1, 4 ) and having slopes
1
1
− and 3 are y − 4 = − ( x + 1 ) and y − 4 = 3 ( x + 1 )
3
3
⇒ x + 3y − 11 = 0 and 3 x − y + 7 = 0
= RHS
87. Since, arg (z − 3i ) = 3π / 4 is a ray which is start from 3i and
for 0 ≤ c ≤ 2 − 1
z +z
…(i)
∴
Re (z ) = x =
2
z −z
…(ii)
and
Im (z ) = y =
2i
The equation (2 − i ) z + (2 + i ) z + 3 = 0 can be written as
2(z + z ) − i (z − z ) + 3 = 0
or
4 x + 2y + 3 = 0
∴ Slope of the given line, m = − 2
Let slope of the required line be m1, then
m −m 
 ⇒ 1 =
 m1 + 2 
⇒ ± 1 = m1 + 2
tan 45 ° =  1
1 − 2m1
1 + m1m 
 1 − 2m1 
1 + i

Q x = 2 
(0, 3)
2
3
y=x+ –
2
3/2
1
X′
1/2
O
Y′
∴ Equation of ray in cartesian form is
y − 3 = tan (3 π / 4 ) ( x − 0 )
or
y − 3 = − x or x + y = 3
and
arg (2z + 1 − 2i ) = π / 4
1 
 
⇒
arg 2 z + − i   = π / 4
 
2 
1 

or arg(2 ) + arg z + − i  = π / 4

2 
1 

or
0 + arg z + − i  = π / 4

2 
X
Chap 01 Complex Numbers

 1

arg z −  − + i   = π / 4
 2


1
which is a ray that start from point − + i and makes an angle
2
π / 4 with positive real axis as shown in the figure.
∴Equation of ray in cartesian form is
y − 1 = 1 [ x − ( − 1 / 2 )] ⇒ y = x + 3 / 2
From the figure, it is clear that the system of equations has no
solution.
88. Let a = r cos α and 0 = r sin α
…(i)
or
On subtracting Eq. (i) from Eq. (ii), we get
z (c −2 − a −1b −1 ) = 2c −1 − a −1 − b −1
∴
r = |a |
Then,
[from Eq. (i)]
a = | a | cos α
∴
cos α = ± 1
Then, cos α = 1 or − 1 according as a is + ve or – ve and
sin α = 0.
Hence, α = 0 or π according as a is + ve and – ve.
Again, let 0 = r1 cos β or b = r1 sin β
…(ii)
0 2 + b 2 = r12
So that,
∴
r1 = | b |
∴
sin β = ± 1
Then, sin β = 1 or − 1 according as b is + ve or – ve and cosβ = 0.
π
π
Hence, β = or − according as b is + ve or −ve.
2
2
89. Let two non-parallel straight lines PQ, RS meet the circle
| z | = r in the points a, b and c.
P
which is a required point.
∴ AP is also perpendicular to BC.
A(z1)
O
(z2)B
P(z)
Then,
∴
 z −z 
Re  1
 =0
z3 − z2
⇒
⇒
z1 − z
z −z
+ 1
z3 − z2 z 3 − z 2
=0
2
z1 − z
z1 − z
+
=0
z3 − z2 z 3 − z 2
…(i)
But O is the circumcentre of ∆ABC, then
OP = OA = OB = OC
On squaring the above relation, we get
S
z
| z | 2 = | z 1| 2 = | z 2 | 2 = | z 3 | 2
Q
⇒
zz = z1z1 = z 2z 2 = z 3z 3
z
z
From first two relations 1 =
z z1
z
z
From first and third relation 2 =
z
z2
z
z
and from first and fourth relation 3 =
z
z3
z1
−1
z1 − z
From Eq. (i), we get
+ z
=0
z3 − z2 z3 − z2
z
z
From Eqs. (ii), (iii), (iv) and (v), we get
z
−1
z1 − z
z1
+
=0
z
z
z3 − z2
−
z3 z2
c
R
Then, | a | = r , | b | = r and | c | = r or | a | 2 = | b | 2 = | c | 2 = r 2
a a = b b = c c = r 2,
r2
r2
r2
, b = and c =
a
c
b
z
z
Points a, b and z are collinear, then a a
b b
⇒
C(z3)
| z | = | z1 | = | z 2 | = | z 3 |
O
a =
D
 z −z  π
arg  1
 =
z3 − z2 2
a
b
∴
2c −1 − a −1 − b −1
c −2 − a −1b −1
90. Q AD ⊥ BC
From Eq. (ii), we get b = | b | sin β
then
z=
Hence,
a 2 + 02 = r 2
So that,
∴
91
1
1 =0
1
z ( a − b ) − z (a − b ) + ab − a b = 0
r 2 r 2
r 2a r 2b
z  −  − z (a − b ) +
−
=0
b
b
a
a
On dividing both sides by r 2 (b − a ), we get
z
z
+ 2 = a −1 + b −1
ab r
For RS, replace a = b = c in Eq. (i), then
z
z
+
= 2c −1
c2 r2
…(i)
…(ii)
⇒
⇒

z2z3 
=0
1 +
zz1 

z 2z3
z z
1+
=0 ⇒z =− 2 3
z z1
z1
 z1 − z 


z3 − z2
…(ii)
…(iii)
…(iv)
…(v)
 z1 − z

Q z − z ≠ 0 
2
 3

92
Textbook of Algebra
92. Given, OA = 1 and | z | = 1
91. From the figure,
α = ( arg (z ) − arg (ω ))
α α
and for every α, sin
≤ 
2  2
…(i)
Y
2
2
P0(z0)
…(ii)
Imaginary axis
A(z)
P(z)
A(1)
|z – ω |
|z|
B(ω)
α
Q(zz0)
|ω |
O
∴
∴
= |z − 0 | = |z | = 1
= OA
= |z 0 − 0 | = |z 0 |
and
= | zz 0 − 0 | = | zz 0| = | z | | z 0| = 1 | z 0| = | z 0|
∴
= OQ
z z 
 z − 0
z 
Also, ∠P0OP = arg  0
 = arg  0  = arg  0 


−
z
z
0
zz 


Real axis
In ∆OAB, from cosine rule
( AB ) 2 = (OA ) 2 + (OB ) 2 − 2OA ⋅ OB cos α
⇒ | z − ω | 2 = | z | 2 + | ω | 2 − 2 | z | | ω | cos α
⇒ | z − ω | 2 = (| z | − | ω | ) 2 + 2 | z | | ω | (1 − cos α )
⇒ | z − ω | 2 = (| z | − | ω | ) 2 + 4 | z | | ω |sin 2
α
2
α
⇒ | z − ω | 2 ≤ (| z | − | ω | ) 2 + 4 | z | | ω |  
 2
⇒
| z − ω | ≤ (| z | − | ω | ) + α
2
2
2
 1−0 
= arg 
 = ∠AOQ
 z z 0 − 0
[from Eq. (i)]
| z − ω | 2 = r 2 + r12 − 2rr1 cos (θ − θ1 )
= (r − r1 ) 2 + 2rr1 − 2rr1 cos (θ − θ1 )
Thus, the triangles POP0 and AOQ are congruent.
∴
PP0 = AQ
| z − z 0| = | z z 0 − 1|
93. Let the equation of line passing through the origin be
…(i)
a z + az = 0
According to the question, z1, z 2, …, zn all lie on one side of
line (i)
…(ii)
∴ a zi + azi > 0 or < 0 for all i = 1, 2, 3, …, n
n
n
i =1
i =1
= (r − r1 ) + 2rr1 (1 − cos (θ − θ1 ))
⇒ a ∑ zi + a ∑ zi > 0 or < 0
 θ − θ1 
= (r − r1 ) 2 + 4rr1 sin 2 

 2 
⇒
 θ − θ1 
≤ (r − r1 ) + 4rr1 

 2 
2
2
= (r − r1 ) + rr1(θ − θ1 )
2
[Qr , r1 ≤ 1]
II. Aliter
z = r cos θ
and
ω = r1 cos θ1
∴ r 2 + r12 − 2rr1 cos (θ − θ1 ) ≤ r 2 + r12 − 2rr1 + (θ − θ1 ) 2

Q r , r1 ≤ 1

2
2
and
sin
x
x
≤


…(iii)
n
 n
If ∑ zi = 0, then ∑ zi = 0,
i =1
 i =1
n
n

hence a ∑ zi + a ∑ zi = 0 
i =1
i =1

From Eq. (ii), we get
a zi + azi > 0 or < 0 for all i = 1, 2, 3, …, n
⇒
a zi zi azi zi
+
> 0 or < 0
zi
zi
a
a
| zi | 2  +  > 0 or < 0
z
z
i
 i
a
a
⇒
+ > 0 or < 0 for all i = 1, 2, 3, …, n
zi zi
1 1
1
, , …, lie on one side of the line a z + az = 0
⇒
z1 z 2
zn
⇒
2
i =1
[Q|sin θ | ≤ | θ | ]
| z − ω | 2 ≤ (| z | − | ω | ) 2 + ( arg z − arg ω ) 2
 θ − θ1   θ − θ1 
⇒ rr1 sin 2 

 ≤
 2   2 
n
∑ zi ≠ 0
2
≤ (r − r1 ) 2 + (θ − θ1 ) 2
Let
 1 
= − arg (z z 0 ) = arg 

z z0
[Q | z | ≤ 1, | ω | ≤ 1]
2
2
⇒
z z 
z z 
= arg  20  = arg  0  = − arg (z z 0 )
 1 
 |z | 
[from Eq. (ii)]
I. Aliter
Let z = r (cosθ + i sin θ ) and ω = r1 (cos θ1 + i sin θ1 ) ,
then | z | = r and | ω | = r1
Also, arg (z ) = θ and arg (ω ) = θ1
and
[Q given | z | ≤ 1, | ω | ≤ 1 ]
r ≤ 1 and r1 ≤ 1
We have, z − ω = (r cos θ − r1 cos θ1 ) + i (r sin θ − r1 sin θ1 )
∴
| z − ω | 2 = (r cos θ − r1 cos θ1 ) 2 + (r sin θ − r1 sin θ1 ) 2
⇒
OP
OP
OP0
OQ
OP0
2
| z − ω | ≤ (| z | − | ω | ) + ( arg (z ) − arg (ω )) 2
2
X
O
Chap 01 Complex Numbers
n 1
1
+a ∑
> 0 or < 0
i = 1 zi
i = 1 zi
n
z2
a ∑
or
93
r
z1
r
n 1
 n 1
1
≠ 0 If ∑
= 0, then ∑
=0
i = 1 zi
i = 1 zi
 i = 1 zi
n
o
Therefore, ∑
r
n 1

1
+a ∑
= 0
i = 1 zi
i = 1 zi

n
⇒
a ∑
z3
From the given condition,
r1 r2 r3
94. Given, | a | | b | = ab 2c ; | a | = | c |; az 2 + bz + c = 0, then we
have to prove that | z | = 1
On squaring, we get
| a | 2 | b | 2 = a b 2c and | a | 2 = | c | 2
⇒ a a b b = a b 2c
and
r2 r3 r1 = 0
r3 r1 r2
⇒
a a =c c
and a a = c c
a b =b c
⇒
If z1 and z 2 are the roots of az 2 + bz + c = 0
1
(r1 + r2 + r3 ) {(r1 − r2 ) 2 + (r2 − r3 ) 2 + (r3 − r1 ) 2 } = 0
2
Since,
r1 + r2 + r3 ≠ 0,
⇒
…(i)
Then, z1 and z 2 are the roots of a ( z ) 2 + b z + c = 0
then (r1 − r2 ) 2 + (r2 − r3 ) 2 + (r3 − r1 ) 2 = 0
…(A)
b
c
z1 + z 2 = − , z1z 2 =
a
a
…(ii)

c 
b
and
z1 + z 2 = − , z1z 2 =
a
a 
1
1 z1 + z 2 − b / a
b
b
Q
+
=
=
= − = − = z1 + z 2
z1 z 2
z1 z 2
c /a
c
a
[from Eqs. (i) and (ii)]
1
1 z1 + z 2 −b / a
and
+
=
=
z1 z 2
c /a
z1 z 2
b
a bc
b
=− =−
= − = z1 + z 2
c
ca a
a
[from Eqs. (i) and (ii)]
1
1
Now, it is clear that z1 = and z 2 =
z1
z2
∴
It is possible only when
r1 − r2 = r2 − r3 = r3 − r1 = 0
∴
r1 = r2 = r3
(ii) Again, in ∆ oz 2 z 3 by Coni method
z 
 z − 0
arg  3
…(i)
 = ∠z 2 oz 3 ⇒ arg  3  = ∠z 2 oz 3
z
−
0
z2

 2
In ∆ z 2z1z 3 by Coni method
 z − z1 
1
arg  3
 = ∠z 2 z1 z 3 = ∠z 2 oz 3 [property of circle]
2
 z 2 − z1 
z 
1
arg  3  [from Eq. (i)]
2
 z1 
 z 3 − z1 
z3
arg   = 2 arg 

 z 2 − z1 
 z1 
=
∴
| z | = 1 ⇒| a | | b | = a b 2c
 z − z1 
z 
Hence, arg  3  = arg  3

 z 2 − z1 
 z1 
|a | = |c |
1
z
From Eq. (A), we get
2
 1
a   +b
z
 1
2
  + c = 0 or cz + b z + a = 0
z
[say]
Hence, z1, z 2, z 3 lie on a circle with the centre at the origin.
Hence,
|z | = 1
Conversely For az 2 + bz + c = 0, we have to prove
| z | = 1 ⇒ | z | 2 = 1 ⇒ z z = 1 ⇒z =
| z1 | = | z 2 | = | z 3 | = r
and
Then, | z1 | 2 = 1 and | z 2 | 2 = 1
and
r13 + r23 + r33 − 3r1r2 r3 = 0
2
96. We know that,
Re (z1z 2 ) ≤ | z1z 2 |
∴ | z1 | + | z 2 | + 2 Re (z1 z 2 ) ≤ | z1 | 2 + | z 2 | 2 + 2 | z1 z 2|
2
⇒
2
| z1 + z 2 | 2 ≤ | z1 | 2 + | z 2 | 2 + 2 | z1 | | z 2 |
…(i)
Also, az + bz + c = 0, on comparing
Also, AM ≥ GM
c b a
= =
a b c
∴ a a = c c and a b = b cs
 1

( c | z1 | ) 2 + 
| z |
1 /2
 c 2 
1


∴
| z 2 | 2  [Qc > 0]
≥  c ⋅| z1 | 2 ⋅
2
c


1
⇒ c | z 1| 2 + | z 2 | 2 ≥ 2 | z 1 | | z 2 |
c
1
2
∴ | z1 | + | z 2 | 2 + 2 | z1 | | z 2 | ≤ | z1 | 2 + | z 2 | 2 + c | z1 | 2 + | z 2 | 2
c
⇒ | z1| 2 + | z 2 | 2 + 2 | z1 | | z 2 | ≤ (1 + c ) | z1 | 2 + (1 + c −1 ) (| z 2 | 2 )
2
⇒ | a | = | c | and | a | | b | = a b 2c
95. (i) Let z1 = r1 (cos α + i sin α ),
z 2 = r2 (cos β + i sin β ) and z 3 = r3 (cos γ + i sin γ )
∴ | z1 | = r1, | z 2 | = r2, | z 3 | = r3
and arg (z1 ) = α , arg (z 2 ) = β, arg (z 3 ) = γ
2
…(ii)
94
Textbook of Algebra
98. Let z be the complex number corresponding to the orthocentre
From Eqs. (i) and (ii), we get
| z1 + z 2 | 2 ≤ (1 + c ) | z1 | 2 + (1 + c −1) | z 2 | 2
O, since AD ⊥ BC , we get
A(z 1)
Aliter
Here, (1 + c ) | z1 | 2 + (1 + c −1 ) | z 2 | 2 − | z1 + z 2 | 2
F
1

= (1 + c ) z1z1 + 1 +  z 2 z 2 − (z1 + z 2 ) ( z1 + z 2 )

c
O(z)
1

= (1 + c ) z1z1 + 1 +  z 2 z 2 − z1z1 − z1z 2 − z 2 z1 − z 2 z 2

c
1
= c z1 z1 + z 2 z 2 − z1z 2 − z 2 z1
c
1
= {c 2z1z1 + z 2 z 2 − cz1z 2 − cz 2 z1 }
c
1
= {cz1 (cz1 − z 2 ) − z 2 (cz1 − z 2 )}
c
1
1
= (cz1 − z 2 ) (cz1 − z 2 ) = (cz1 − z 2 ) (cz1 − z 2 )
c
c
1
2
= | cz1 − z 2 | ≥ 0 as c > 0
c
1

∴ (1 + c ) | z1| 2 + 1 +  | z 2 | 2 − | z1 + z 2 | 2 ≥ 0

c
D
B(z 2 )
i. e.
z − z1
is purely imaginary.
z2 − z3
 z − z1 
z − z1
z − z1
i. e. Re 
+
=0
 = 0 or
z
−
z
z
z
−
z
 2
2
3
2 − z3
3
z − z2
z − z2
+
=0
z 3 − z1 z 3 − z1
Similarly,
z = z1 −
O, then we have
z = z2 −
A(z 1)
(z − z 2 ) ( z 3 − z1 )
(z 3 − z1 )
...(iv)
= ( z1 − z 2 ) (z 2 − z 3 ) (z 3 − z1 )
or z {( z 2 − z 3 ) (z 3 − z1 ) − (z 3 − z1 ) (z 2 − z 3 )}
| z − z1 | = | z − z 2 | = | z − z 3 |
| z − z1 | 2 = | z − z 2 | 2 = | z − z 3 | 2
= ( z1 − z 2 ) (z 2 − z 3 ) (z 3 − z1 ) + z1 (z 2 − z 3 ) ( z 3 − z1 )
− z 2 ( z 3 − z1 ) (z 2 − z 3 )
(z − z1 ) ( z − z1 ) = (z − z 2 ) ( z − z 2 )
= (z − z 3 ) ( z − z 3 )
…(i)
From first two members of Eq. (i), we get
z (z 2 − z1 ) = z1 (z − z1 ) − z 2 (z − z 2 )
z (z 3 − z 2 ) = z 2 (z − z 2 ) − z 3 (z − z 3 )
⇒ z [ z 2 z 3 − z 2 z1 − z 3z 3 + z 3z1 − z 3z 2 + z 3z 3 + z1z 2 − z1 z 3 ]
= ( z1 − z 2 ) {z 2 z 3 − z 2z1 − z 32 + z 3z1 }
+ ( z 2 − z 3 ) (z 3z1 − z12 ) + ( z 3 − z1 ) (z 2 z 3 − z 22 )
…(ii)
= − {z12 ( z 2 − z 3 ) + z 22 ( z 3 − z1 ) + z 32 ( z1 − z 2 )}
and from last two members of Eq. (i), we get
…(iii)
Eliminating z from Eqs. (ii) and (iii), we get
(z 2 − z1 ) [ z 2 (z − z 2 ) − z 3 (z − z 3 )] = (z 3 − z 2 )
[ z1 (z − z1 ) − z 2 (z − z 2 )]
or z [z 2 (z 2 − z1 ) − z 3 (z 2 − z1 ) − z1 (z 3 − z 2 ) + z 2 (z 3 − z 2 )]
= z 2 z 2 (z 2 − z1 ) − z 3z 3 (z 2 − z1 ) − z1z1(z 3 − z 2 ) + z 2 z 2 (z 3 − z 2 )
∑ | z1 | 2 (z 2 − z 3 )
∑ z1 (z 2 − z 3 )
...(iii)
or (z − z1 ) ( z 2 − z 3 ) ( z 3 − z1 ) − (z − z 2 ) (z 3 − z1 ) (z 2 − z 3 )
C(z 3)
B(z 2)
z=
(z − z 2 ) ( z 2 − z 3 )
(z 2 − z 3 )
Eliminating z from Eqs. (iii) and (iv), we get
( z − z1 )
(z − z 2 ) ( z 3 − z1 )
z1 − z 2 =
(z 2 − z 3 ) −
(z 2 − z 3 )
(z 3 − z1 )
(z)
O
or
[Q BE ⊥ CA ] ...(ii)
From Eq. (ii), we get
OA = OB = OC
or z ∑ z1 (z 2 − z 3 ) = ∑ z1 z1 (z 2 − z 3 )
...(i)
From Eq. (i), we get
97. If z be the complex number corresponding to the circumcentre
⇒
C(z 3 )
 z − z1  π
arg 
 =
z2 − z3 2
1

Hence, | z1 + z 2 | 2 ≤ (1 + c ) | z1| 2 + 1 +  | z 2 | 2

c
⇒
⇒
E
+ {z1z 2 z 3 − z 2 z1z1 + z 3z1z1 + z 2 z1z 3
− z1z 3z 3 + z 2z 3z 3 − z1z 2z 3 } − z ∑ (z1z 2 − z 2 z1 )
= − ∑ z12 ( z 2 − z 3 ) − Σ z1z1 (z 2 − z 3 )
2
Hence, z =
∑ z12 (z 2 − z 3 ) + ∑ z1 (z 2 − z 3 )
∑ (z1z 2 − z 2 z1 )
1
7
∴ 7θ = (2n + 1 ) π or 4 θ = (2n + 1 ) π − 3 θ
or cos 4 θ = − cos3 θ
or 2 cos2 2 θ − 1 = − ( 4 cos3 θ − 3 cos θ )
99. Let θ = (2n + 1) π , where n = 0, 1, 2, 3, ... , 6
Chap 01 Complex Numbers
or 2 (2 cos2 θ − 1 ) 2 − 1 = − ( 4 cos3 θ − 3 cos θ )
⇒
or 8 cos4 θ + 4 cos3 θ − 8 cos2 θ − 3 cos θ + 1 = 0
or ( x + 1 ) (8 x − 4 x − 4 x + 1 ) = 0
2
x+1≠0
8x 3 − 4x 2 − 4x + 1 = 0
∴
[Q θ ≠ π]
...(i)
Hence, the roots of this equation are
π
3π
5π
cos , cos , cos .
7
7
7
9π
5π
11 π
[ since cos
= cos , cos
7
7
7
3π
13π
π
= cos , cos
= cos and Eq. (i) is cubic]
7
7
7
1
1
in Eq. (i), then Eq. (i) becomes
(i) On putting 2 = y or x =
y
x
8
4
4
⇒
− −
+1=0
y y y
y
2
⇒
or
3π 
π 

= z 2 + 1 − 2z cos  z 2 + 1 − 2z cos 



7
7
5π 
 2
z + 1 − 2z cos  ...(A)

7
Dividing by z 3 on both sides, we get
1  2
1 
1
 3
z + 3  − z + 2  + z +  − 1





z
z
z
1
3π  
1
π 
1
5π 

= z + − 2 cos  z + − 2 cos  z + − 2 cos 

7 
7
z
z
7
z
1
On putting z + = 2 x, we get
z
(8 x 3 − 6 x ) − ( 4 x 2 − 2 ) + 2 x − 1
3π  
5π 
π 

= 8  x − cos   x − cos   x − cos 

7 
7 
7
or
2
 4 

4
2 
1 −  
1 −  = 
y
y

 y 
16 8 16 
4
4
1+ 2 − =
1 + 2 − 
y y 
y
y
y
y 3 − 24y 2 + 80y − 64 = 0
1
1
where
y = 2 =
= sec 2 θ
cos2 θ
x
Thus, the roots of x 3 − 24 x 2 + 80 x − 61 = 0
π
3π
5π
are sec 2 , sec 2 , sec 2
7
7
7
(ii) Again, putting y = 1 + z i. e. z = y − 1
or
...(ii)
…(i)
So, 8 x 3 − 4 x 2 − 4 x + 1 = 0 and this equation has roots
π
3π
5π
cos , cos , cos
7
7
7
π
3π
5π
Constant term
cos cos
=−
7
7
7
Coefficient of x 3
π
3π
5π
1
[ proved (i) part]
cos cos cos
=−
7
7
7
8
On putting x = 1 in Eq. (i), we get
π 
3π  
5π 

1 = 8 1 − cos  1 − cos  1 − cos 

7 
7 
7
(1 + z ) 3 − 24 (1 + z ) 2 + 80 (1 + z ) − 64 = 0
z 3 − 21z 2 + 35z − 7 = 0
…(iii)
π
3π
5π
are the roots of
, tan 2
, tan 2
7
7
7
x 3 − 21 x 2 + 35 x − 7 = 0
1
(iii) Putting x = in Eq. (i), then Eq. (i) reduces to
u
u 3 − 4u 2 − 4u + 8 = 0 whose roots are
π
3π
5π
sec , sec
, sec .
7
7
7
Therefore, sum of the roots is
π
3π
5π
sec + sec
+ sec
=4
7
7
7
100. Let roots of z 7 + 1 = 0 are − 1, α , α 3, α 5, α , α 3, α 5,
π
π
where α = cos + i sin
7
7
∴ (z 7 + 1 ) = (z + 1 ) (z − α ) (z − α ) (z − α 3 )
or
π
3π
5π 

sin 2 
1 = 8 8 sin 2 sin 2

14
14
14 
Since, sinθ > 0 for 0 < θ < π / 2 , we get
Hence, tan 2
∴
sin
π
3π
5π 1
sin sin
=
14
14
14 8
...(ii) [ proved (iii) part]
Again, putting x = − 1 in Eq. (i), we get
3π  
5π 
π 

− 7 = − 8 1 + cos  1 + cos  1 + cos 

7 
7
7
π
3π
5π 

7 = 8 8 cos2 cos2 cos2 

14
14
14 
Since, cosθ > 0 for 0 < θ < π / 2 , we get
cos
π
3π
5π
7
cos cos
=
14
14
14
8
...(iii) [ proved (ii) part]
On dividing Eq. (ii) by Eq. (iii), we get
(z − α ) (z − α ) (z − α )
3
π

8 x 3 − 4 x 2 − 4 x + 1 = 8  x − cos 

7
3π  
5π 

 x − cos   x − cos 

7 
7
∴ cos
= sec 2 θ − 1 = tan 2 θ, Eq. (ii) reduces to
or
(z 7 + 1 )
= (z − α ) (z − α ) (z − α 3 ) (z − α 3 ) (z − α 5 ) (z − α 5 )
(z + 1 )
⇒ z6 − z5 + z4 − z3 + z2 − z + 1
Now, if cos θ = x, then we have
8x 4 + 4x 3 − 8x 2 − 3x + 1 = 0
3
95
5
5
tan
π
3π
5π
1
tan tan
=
14
14
14
7
[ proved (iv) part]
96
Textbook of Algebra
(1 + y )
in Eq. (A), we get
(1 − y )
3π
5π
π
2 6 cos2 cos2 cos2
(1 + y ) 7 + (1 − y ) 7
14
14
14
=
(1 − y ) 6
2 (1 − y ) 6
Again, circles (i) and (iii) should not cut or touch, then distance
between their centres > sum of their radii
On putting z =
 2
2 π 2
2 3π 

y + tan  y + tan

14 
14 
( − 2a − 0 ) 2 + (a + 1 − 0 ) 2 > 3 + 3
5a 2 + 2a + 1 > 6
or
 2
2 5π 

y + tan

14 
7  2
2 π
∴ (1 + y ) 7 + (1 − y ) 7 = 2 7 ⋅
y + tan 

64
14
 2
2 3π   2
2 5π 

 y + tan
y + tan



14
14 
Using result (ii), we get
π

(1 + y ) 7 + (1 − y ) 7 = 14 y 2 + tan 2 

14
 2
2 3π   2
2 5π 

 y + tan
y + tan

14  
14 
4
⇒
5a 2 + 2a + 1 > 36
5a 2 + 2a − 35 > 0
2a
⇒
a2 +
−7 > 0
5

− 1 + 4 11 
− 1 − 4 11  
or a −
 >0
 a −
5
5

 

or
∴

 − 1 + 4 11

− 1 − 4 11 
, ∞
a ∈  − ∞,
 ∪
5
5




Hence, the common values of a satisfying Eqs. (iv) and (v) are
 − 1 + 4 11 1 + 71 
 1 − 71 − 1 − 4 11 
,
,
a ∈

 ∪
2
5
5
2 



Equating the coefficient of y on both sides, we get
π
3π
5π 

7
C 4 + 7C 4 = 14 tan 2 + tan 2
+ tan 2 
14
14
14 

3π
5π
π
Therefore, tan 2 + tan 2
+ tan 2
=5
14
14
14
102. (i) From De-moivre’s theorem, we know that
sin (2n + 1 ) α =
2n + 1
C1 (1 − sin 2 α )n
sin α − 2n + 1C 3 (1 − sin 2 α )n − 1 sin 3 α
101. Equation z = 3 represents boundary of a circle and equation
z − {a (1 + i ) − i } ≤ 3 represents the interior and the
boundary of a circle and equation z + 2a − (a + 1 ) i > 3
represents the exterior of a circle. Then, any point which
satisfies all the three conditions will lie on first circle, on or
inside the second circle and outside the third circle.
+ ... + ( − 1 )n sin 2n + 1 α
It follows that the numbers
π
2π
nπ
sin
, sin
, ..., sin
2n + 1
2n + 1
2n + 1
are the roots of the equation.
2n + 1
B
C
II
A
III
I
For the existence of such a point first two circles must cut or
atleast touch each other and first and third circles must not
intersect each other. The arcABC of first circle lying inside the
second but outside the third circle, represents all such possible
points.
Let z = x + iy , then equation of circles are
...(i)
x2 + y 2 = 9
2
2
...(ii)
( x − a ) + (y − a + 1 ) = 9
and ( x + 2a ) + (y − a − 1 ) = 9
2
2
...(iii)
Circles (i) and (ii) should cut or touch, then distance between
their centres ≤ sum of their radii
⇒
(a − 0 ) 2 + (a − 1 − 0 ) 2 ≤ 3 + 3
⇒
a + (a − 1 ) ≤ 36
⇒
2a 2 − 2a − 35 ≤ 0
⇒
∴
2
a2 − a −
C1(1 − x 2 )n x − 2n + 1 C 3(1 − x 2 )n − 1 x 3 + ... + ( − 1 )n x 2n + 1
= 0 of the (2n + 1 ) th degree
Consequently, the numbers
π
2π
nπ
are the roots of the
sin 2
, sin 2
, ... , sin 2
2n + 1
2n + 1
2n + 1
equation
2n + 1
C1 (1 − x )n − 2n + 1C 3 (1 − x )n − 1 x + ... + ( − 1 )n xn = 0 of
the nth degree
(ii) From De-moivre’s theorem, we know that
sin(2n + 1 ) α = 2n + 1C1(cosα ) 2n sin α
−
2n + 1
C 3(cosα ) 2n − 2 sin 3 α + ... + ( −1 )n sin 2n + 1 α
or sin (2n + 1 ) α = sin 2n + 1 α
{ 2n + 1C1 cot 2n α − 2n + 1 C 3 cot 2n − 2 α + 2n +1C 5 cot 2n − 4 α − ...}
It follows that α =
2π
3π
nπ
π
,
,
, ... ,
2n + 1 2n + 1 2n + 1
2n + 1
Therefore, equality holds
2n +1
2

1 + 71  
1 − 71 
35
≤ 0 or a −
 ≤0
 a −
2  
2 
2

1 − 71
1 + 71
≤a≤
2
2
...(v)
...(iv)
C1 cot 2n α − 2n +1C 3 cot 2n − 2 α + 2n +1C 5 cot 2n − 4 α−... = 0
It follows that the numbers
π
2π
nπ
are the roots of the
cot 2
, cot 2
, ... , cot 2
2n + 1
2n + 1
2n + 1
equation
2n + 1
C1xn −
of the nth degree.
2n + 1
C 3x n − 1 +
2n + 1
C 5xn − 2 − ... = 0
Chap 01 Complex Numbers
103. Let y = | a + bω + cω 2 |. For y to be minimum, y 2 must be
minimum.
∴ y 2 = | a + bω + cω 2 | 2 = (a + bω + cω 2 ) (a + bω + cω 2 )
= (a + bω + cω 2 ) (a + b ω + c ω 2 )
2
y = (a + bω + cω 2 ) (a + bω 2 + cω )3
= (a 2 + b 2 + c 2 − ab − bc − ca )
1
= [(a − b ) 2 + (b − c ) 2 + (c − a ) 2 ]
2
Since, a, b and c are not equal at a time, so minimum value of
y 2 occurs when any two are same and third is differ by 1.
⇒ Minimum of y = 1 (as a, b, c are integers)
π
104. Equation of ray PQ is arg (z + 1) =
4
π
Equation of ray PR is arg (z + 1 ) = −
4
π
π
π
Shaded region is − < arg (z + 1 ) <
⇒ | arg (z + 1 )| <
4
4
4
∴
| PQ | = ( 2 ) 2 + ( 2 ) 2 = 2
So, arc QAR is of a circle of radius 2 units with centre at
P ( − 1, 0 ). All the points in the shaded region are exterior to
this circle | z + 1 | = 2.
π
i.e. | z + 1 | > 2 and | arg (z + 1 )| < .
4
zB − 1
105. In ∆AOB from Coni method,
= e iπ / 2 = i
zA − 1
B(zB)
√2
A(2 + √3i ) = zA
π/2 √2
D (zD)
z B − 1 = (z A − 1 ) i
z B = 1 + (2 + 3i − 1 ) i = 1 + (1 + i 3 ) i
=1 + i − 3 =1 − 3 + i
zC = 2 − z A = 2 − (2 + 3i ) = − 3 i
z D = 2 − z B = 2 − (1 − 3 + i ) = 1 + 3 − i
and
Hence, other vertices are (1 − 3 ) + i , − 3i , (1 + 3 ) − i.
106. Let z1 = r1 (cos θ1 + i sin θ1 ) and z 2 = r2 (cos θ 2 + i sin θ 2 )
∴| z1 + z 2 | = [(r1 cos θ1 + r2 cos θ 2 ) 2 + (r1 sin θ1 + r2 sin θ 2 ) 2 ]1/2
= [| r12 + r22 + 2r1r2 cos ( θ1 − θ 2 )]1/2 = [(r1 + r2 ) 2 ]1/2
∴ | z1 + z 2 | = | z1 | + | z 2 |
Therefore, cos ( θ1 − θ 2 ) = 1
⇒
∴
|z |2 − 1 = 0
⇒
| z | = 1 and z ≠ 1
10
z = ω, ω 2
1
1
= ω + = ω + ω2 = − 1
z
ω
1
1
z 2 + 2 = ω2 + 2 = ω2 + ω = − 1
⇒
z
ω
1
1
z 3 + 3 = ω3 + 3 = 1 + 1 = 2
∴
z
ω
1
1
1
z 4 + 4 = ω4 + 4 = ω + = − 1
ω
z
ω
1
1
z 5 + 5 = ω5 + 5 = ω2 + ω = − 1
z
ω
1
1
and
z 6 + 6 = ω6 + 6 = 2
z
ω
∴ Required sum = ( −1 ) 2+ ( −1 ) 2 + (2 ) 2 + ( −1 ) 2 + ( −1 ) 2 + (2 ) 2 = 12
z+
112. Let OA = 3, so that the complex number associated with A is
3e iπ /4 . If z is the complex number associated with P, then
N (North)
z
θ1 − θ 2 = 0
4
π/2
107. ( x − 1) = − 8 ⇒ x − 1 = ( − 8)
⇒
⇒
−2k π i
11

 2kπ  
 2kπ 
= i ∑ cos 
 =i ∑ e
 − i sin 




11
11

k =1
k =1 
 10 −2k π i 
11


= i  ∑e
− 1 = i ( 0 − 1 ) [Qsum of 11, 11th roots of unity = 0]
k = 0



= −i
111. Q z 2+ z + 1 = 0
10
⇒
θ1 = θ 2
Thus, arg ( z1 ) – arg ( z 2 ) = 0
3
[Q β ≠ 0 ]
2kπ 
 2kπ 
110. ∑ sin 

 + i cos 
 11 


11
k =1
10
Q
√2
(zC)C
∴
i
= 1 ⇒ |z | = z −
i
3
z−
3
Clearly, locus of z is perpendicular bisector of line joining
i
points having complex number 0 + i 0 and 0 + .
3
Hence, z lies on a straight line.
 ω − ωz 
109. Given, 
 is purely real ⇒ z ≠ 1
 1 −z 
 ω − ω z   ω − ω z  ω − ωz
∴
=
 =

1 −z
 1 −z   1 −z 
⇒
( ω − ω z ) ( 1 − z ) = (1 − z ) ( ω − ω z )
( zz − 1 ) ( ω − ω) = 0
⇒
⇒
(| z | 2 − 1 ) (2iβ ) = 0
[Q ω = α + iβ ]
z
108.
∴
(1)
√2
1/ 3
x − 1 = − 2, − 2ω, − 2ω 2
x = − 1, 1 − 2ω, 1 − 2ω 2
97
A 3ei π/4
3
W
π/4
O
(East)
S
E
98
Textbook of Algebra
z − 3e iπ /4 4 − π /2
4i
= e
=−
3
0 − 3e iπ /4 3
⇒
3z − 9e
iπ / 4
= 12 ie
iπ / 4
120. Q z z (z 2 + z 2 ) = 350
⇒ z = (3 + 4i ) e
Put
⇒
iπ / 4
⇒
113. Let z = cos θ + i sin θ
⇒
z
cos θ + i sin θ
=
1 − z 2 1 − (cos 2 θ + i sin 2 θ )
Hence,
=
cos θ + i sin θ
2 sin 2 θ − 2 i sin θ cos θ
=
cos θ + i sin θ
i
=
− 2 i sin θ (cos θ + i sin θ ) 2 sin θ
z = x + iy
( x 2 + y 2 ) ⋅ 2( x 2 − y 2 ) = 350
( x 2 + y 2 ) ( x 2 − y 2 ) = 175 = 25 × 7
⇒
x 2 + y 2 = 25, x 2 − y 2 = 7
⇒
x 2 = 16 , y 2 = 9
∴
x = ± 4, y = ± 3; x , y ∈ I
Area of rectangle = 8 × 6 = 48 sq units
15
121.
∑
1
z
z
1
1
=
=
=−
=−
2
2
z
−
z
z
−
z
z
−
z

1 −z
zz − z

 2i
 2i 
i
which is imaginary.
=
2 Im | z |
Let E =
Y
A
(–1,0)
O
(0, 0)
X
115. Let A = set of points on and above the line y = 1 in the argand
plane.
B = set of points on the circle ( x − 2 ) 2 + (y − 1 ) 2 = 3 2
C = Re (1 − i ) z = Re [(1 − i ) ( x + iy )] = x + y
⇒
x+y = 2
Hence, ( A ∩ B ∩ C ) has only one point of intersection.
116. The points ( − 1 + i ) and (5 + i ) are the extremities of diameter
of the given circle.
Hence, | z + 1 − i | 2 + | z − 5 − i | 2 = 36
117. Q | z − w | ≤ | z | − | w |
and | z − w | = distance between z and w
z is fixed, hence distance between z and w would be maximum
for diametrically opposite points.
⇒
| z − w | < 6 ⇒ || z | − | w || < 6
⇒ − 6 < |z | − |w | < 6
⇒ − 3 < |z | − |w | + 3 < 9
∴ z1 = 6 + 5i
⇒ z 2 = − 6 + 7i
119. Put ( − i ) in place of i.
Hence,
−1
i +1
∑ sin (2m − 1) θ
=
2 sin (2m − 1 ) θ sin θ
2 sin θ
m =1
=
∑
15
∑
cos (2m − 2 ) θ − cos 2mθ
2 sin θ
15
cos 0 ° − cos 30 θ 1 − cos 60 °
=
2 sin θ
2 sin 2 °
1
1−
1
2 =
=
2 sin 2 ° 4 sin 2 °
=
(Qθ = 2 ° )
4
4
4
≥ |z | −
⇒ 2 ≥ |z | −
z
|z |
|z |
4
⇒ − 2 ≤ |z | −
≤ 2 ⇒ − 2 |z | ≤ |z |2 − 4 ≤ 2 |z |
|z |
122. z −
Y′
118. Q z 0 = 1 + 2i
15
m =1
m =1
114. | z + 4 | ≤ 3
⇒z lies inside or on the circle of radius 3 and centre at ( − 4, 0 ).
∴ Maximum value of | z + 1 | is 6.
Im [e ( 2m − 1) iθ ] =
m =1
Aliter
(–4, 0)
∑
m =1
z
lies on the imaginary axis i.e. x = 0 or on Y-axis.
1 − z2
X′
15
Im (z 2m − 1 ) =
⇒
and
|z |2 + 2 |z | − 4 ≥ 0
12 − 2 |z | − 4 ≤ 0
(| z | + 1 ) 2 ≥ 5 and (| z | − 1 ) 2 ≤ 5
⇒
− 5 ≤ | z | − 1 ≤ 5 and | z | + 1 ≥ 5
⇒
5 − 1 ≤ |z | ≤ 5 + 1
123. As z = (1 − t ) z1 + tz 2
z
z1
t
1–t
z2
⇒ z1, z and z 2 are collinear.
Thus, options (a) and (d) are correct.
z − z1
z − z1
Also,
=
z 2 − z1 z 2 − z1
Hence, option (c) is correct.
2π
2π
1
3
124. ω = cos
+ i sin
=− +i
3
3
2
2
ω is one of the cube root of unity.
∴
ω
ω2
z +1
ω
1
=0
z + ω2
2
ω
1
z +ω
Applying R 1 → R 1 + R 2 + R 3, we get
z
z
z
ω z + ω2
1
=0
ω2
1
z +ω
[Q 1 + ω + ω 2 = 0 ]
Chap 01 Complex Numbers
Now, applying C 2 → C 2 − C1, C 3 → C 3 − C1, we get
0
0
z
ω z + ω2 − ω
=0
1 −ω
ω2
1 −w2
z + ω − ω2
⇒ z [(z + ω 2 − ω ) (z + ω − ω 2 ) − (1 − ω ) (1 − ω 2 )] = 0
⇒
z [z 2 − (ω 2 − ω ) 2 − (1 − ω 2 − ω + ω 3 )] = 0
2
⇒
z [z − (ω 4 + ω 2 − 2ω 3 ) − 1 + ω 2 + ω − ω 3 ] = 0
⇒
z3 = 0
∴
(D) Let w = cos θ + i sin θ, then
z = x + iy = w +
⇒
∴
(A) Putting z = x + iy , we get y x 2 + y 2 = 0
i. e. Im(z ) = 0
(B) 2ae = 8, 2a = 10 ⇒ 10e = 8 ⇒e =
4
5
x2 − x + 1 = 0
126. Q
∴
1 ± (1 − 4 ) 1 ± i 3
=
2
2
1+i 3
1 −i 3
=
and
2
2
x = − ω 2, − ω
∴
α = − ω 2, β = − ω
∴
x=
(0,0)
⇒
α 2009 + β 2009 = − ω 4018 − ω 2009
= − ω − ω 2 = − (ω + ω 2 )
= − (− 1) = 1
(0, 3)
(–5, 0)
1
w
x + iy = 2 cos θ
x = 2 cos θ and y = 0
z =0
125. | z − i | z || = | z + i | z ||
99
127. | z − 1 | = | z + 1 | = | z − i |
(5, 0)
⇒
|z − 1|2 = |z + 1|2 = |z − i |2
⇒ (z − 1 ) ( z − 1 ) = (z + 1 ) ( z + 1 ) = (z − i ) ( z + i )
⇒ z z − z − z + 1 = z z + z + z + 1 = z z + iz − i z + 1
(0, – 3)
∴
 16
b 2 = 25 1 −  = 9
 25
x2 y 2
+
=1
25
9
1
(C) z = 2 (cos θ + i sin θ ) −
2 (cos θ + i sin θ )
1
= 2 (cos θ + i sin θ ) − (cos θ − i sin θ )
2
3
5
z = cos θ + i sin θ
2
2
⇒
3
– ,0
2
(0, 0)
3, 0
2
Let z = x + iy , then
3
5
x = cos θ and y = sin θ
2
2
2
2
⇒
 2y 
 2x 
  +   =1
5
3
⇒
4 x 2 4y 2
+
=1
9
25
⇒
⇒
∴
x2
y2
+
=1
9 / 4 25 / 4
9 25
= (1 − e 2 )
4 4
9 16
4
e2 = 1 −
=
⇒ e=
25 25
5
⇒
− z − z = z + z = i (z − z )
From first two relations,
…(i)
2 (z + z ) = 0 ⇒ Re(z ) = 0
From last two relations,
z + z = i (z − z ) ⇒ 2 Re (z ) = − 2 Im (z )
From Eq. (i), Im(z ) = 0
∴
z = Re(z ) + i Im(z ) = 0 + i ⋅ 0 = 0
Hence, number of solutions is one.
128. We have,
| z − 3 − 2i | ≤ 2
⇒
Now,
| 2z − 6 − 4i | ≤ 4
| 2z − 6 − 4i | = |(2z − 6 + 5i ) − 9i |
≥ || 2z − 6 + 5i | − 9 |
From Eqs. (i) and (ii), we get
| 2z − 6 + 5i | − 9 | ≤ 4
⇒ − 4 ≤ | 2z − 6 + 5i | − 9 ≤ 4
⇒
5 ≤ | 2z − 6 + 5i | ≤ 13
Hence, the minimum value of | 2z − 6 + 5i | is 5.
129. Q | z | = 1
∴ z = e iθ
 2ie iθ 


2i
 = Re  − iθ

iθ 
2 iθ
−
e
e
e
1
−







1 
2i
= Re 

 = Re  −
 sin θ 
 − 2i sin θ 
1
=−
= − cosec θ
sin θ
cosec θ ≤ − 1 ⇒ cosec θ ≥ 1
− cosec θ ≥ 1 ⇒ − cosec θ ≤ − 1
− cosec θ ∈ ( − ∞, − 1 ] ∩ [1, ∞ )
 2iz 
∈ ( − ∞, −1 ] ∩ [1, ∞ )
Re 
2
1 − z 
 2iz 
∴ Re 
= Re
2
1 − z 
Q
⇒
⇒
∴
…(i)
…(ii)
100
Textbook of Algebra
130. Q | z | = 1. Let z = e i θ
iθ
134. Given, z 2 + z + 1 = a ⇒ z 2 + z + 1 − a = 0
∴
z −1 =e
⇒
1
1
ie − iθ /2
=
=−
iθ / 2
z − 1 2ie ⋅ sin (θ / 2 )
2 sin (θ / 2 )
⇒
−1=e
iθ / 2
i ⋅ e − i θ /2
1
=
1 − z 2 sin (θ / 2 )
⋅ 2i sin (θ / 2 )
Hence,
 1   π θ
∴ arg 
 = − 
 1 − z   2 2
135. Let z = x + iy , then
z2
( x + iy ) 2
( x 2 − y 2 + 2ix )
=
=
z − 1 ( x + iy − 1 )
( x − 1 + iy )
 1 
π θ
arg 
−
 =
2 2
1 − z
⇒
 1 
π
∴ Maximum value of arg 
 =
2
1 − z
131. Q | x | 2 = x x = (a + b + c ) (a + b + c )
= (a + b + c ) ( a + b + c )
= | a | 2 + | b | 2 + | c | 2 + ab + a b + bc + b c + ca + ca
…(i)
= (a + bω + cω 2 ) ( a + b ω + c ω 2 )
= (a + bω + cω ) ( a + b ω + c ω )
2
= | a | 2 + | b | 2 + | c | 2 + ab ω 2 + a bω
+ bc ω 2+ b cω + ca ω 2+ caω …(ii)
( x 2 − y 2 + 2ixy ) ( x − 1 − iy )
( x − 1 + iy ) ( x − 1 − iy )
=
( x − 1 ) ( x 2 − y 2 ) + 2 xy 2 + i [ 2 xy ( x − 1 ) − y ( x 2 − y 2 )]
(x − 1)2 + y 2
⇒
⇒
⇒
⇒
y = 0 or x 2 + y 2 − 2 x = 0
Hence, z lies on the real axis or on a circle passing through the
origin.
and| z | 2 = zz = (a + bω 2 + cω ) (a + bω 2 + cω )
= (a + bω 2+ cω ) (a + b ω 2 + c ω )
136. Given, | z | = 1 and arg (z ) = θ
= (a + bω 2 + cω ) (a + b ω + c ω 2 )
+ bcw + b cω 2 + ca ω + caω 2
…(iii)
On adding Eqs. (i), (ii) and (iii), we get
| x | 2 + | y | 2 + | z | 2 = 3 (| a | 2 + | b | 2 + | c | 2 )
+ 0 + 0 + 0 + 0 + 0 + 0 (Q 1 + ω + ω 2 = 0 )
| x |2 + |y |2 + |z |2
=3
|a |2 + |b |2 + |c |2
z +z
132. Q Re(z ) = 1 ∴
=1 ⇒ z + z =2
2
Since, α , β ∈ R
∴ The complex roots are conjugate to each other, if z1, z 2 are
two distinct roots, then z1 = z 2 or z1 = z 2
∴ Product of the roots = z1z 2 = β
∴
2
Aliter II Given, | z | = 1 and arg (z ) = θ
Let z = ω (cube root of unity)
1+ω
1 + z
 1 + ω
∴ arg 
 = arg 
 = arg 
2
1 + z 
1 + ω
1 + ω 
 − ω2 
= arg 

 −ω 
z1z1 = β
∴
β = | z1 | = [ Re(z1 )] + Im | z1 |
2
2
2
∴
133. Q (1 + ω ) = ( − ω ) = − ω
7
2 7
14
= −ω =1 + ω
Given, (1 + ω ) = A + Bω ⇒ 1 + ω = A + Bω
On comparing, we get A = 1, B = 1
∴
z 0 = 2α −
137.
Q
2
7
( A, B ) = (1, 1 )
(Q ω = ω 2 )
(Q 1 + ω + ω 2 = 0 )
= arg (ω ) = arg (z ) = θ
= 1 + Im | z1 | 2 > 1
[Q roots are distinct∴ Im (z1 ) ≠ 0]
β > 1 or β ∈ (1, ∞ )
…(i)
|z | = 1 ⇒ z z = 1
1
…(ii)
⇒
z =
z
1 + z
 1+z 
[from Eq. (ii)]
∴
arg 
 = arg 

1 + z 
 1 + 1 /z 
[from Eq. (i)]
= arg (z ) = θ
Aliter I
Given,
| z | = 1 and arg (z ) = θ
⇒
z = eiθ
 1 + eiθ 
1 + z
∴ arg 
 = arg (e i θ ) = arg (z ) = θ
 = arg 
− iθ
1 + z 
1 + e 
⇒
= | a | 2 + | b | 2 + | c | 2 + ab ω + a bω 2
⇒
=
 z2 
Im 
 =0
 z − 1
2 xy ( x − 1 ) − y ( x 2 − y 2 ) = 0
y (2 x 2 − 2 x − x 2 + y 2 ) = 0
y (x 2 + y 2 − 2x ) = 0
Now,
| y | 2 = y y = (a + bω + cω 2 ) (a + b ω + c ω 2 )
2
− 1 ± ( 4a − 3 )
2
3
[for a = 3 / 4, z will be purely real]
a≠
4
z=
∴
1
α
2 |z 0 |2 = r 2 + 2
2
2
2
1
1
1
= α−
+2
= r 2 + 2 ⇒ 2 2α −
α
α
α
1
1
1
− 8 = 0 ⇒ | α | 2 = 1 or ⇒ | α | = 1 or
⇒ 7 |α |2 +
2
7
7
|α |
∴
2 2α −
Chap 01 Complex Numbers
101
(B) z1 ⋅ z = zk is clearly incorrect.
138.
e iπ/6
ei5π/6
z 10 − 1
z →1 z − 1
lim
(C) Expression =
–1
1
e–i 5π/6
e– i π/6
10
=1
9
 2kπ 
(D) 1 + Σzk = 0 ⇒ 1 + ∑ cos 
 =0
 10 
k =1
∴ Expression = 2
1/2
1/2
z1 − 2z 2
=1
2 − z1z 2
143. Q
3+i
ω=
= e iπ /6, P = e inπ /6
2
As z1 ∈ P ∩ H1 ⇒ z1 = 1, e iπ /6, e − iπ /6
⇒ | z1 − 2z 2 | 2 = | 2 − z1z 2 | 2
As z 2 ∈ P ∩ H 2 ⇒ z 2 = − 1, ei 5π /6, e − i 5π /6
⇒ (z1 − 2z 2 ) (z1 − 2z 2 ) = (2 − z1z 2 ) (2 − z1z 2 )
∠z1Oz 2 = 2 π / 3, where z1 = e iπ /6, z 2 = e i 5π /6
⇒ (z1 − 2z 2 ) (z1 − 2z 2 ) = (2 − z1z 2 ) (2 − z1z 2 )
⇒ z1z1 − 2z1z 2 − 2z1z 2 + 4z 2z 2 = 4 − 2z1z 2 − 2z1z 2 + z1z1z 2z 2
Sol. (Q. Nos. 139-140)
Let z = x + iy , S1 : x + y < 16
⇒ | z 1| 2 + 4 | z 2 | 2 + 4 + | z 1 | 2 | z 2 | 2
( x − 1 ) + i (y + 3 ) 
Now, Im 
>0
1 − 3i


⇒
2
2
S 2 : 3x + y > 0 ⇒ S 3 : x > 0
⇒
√3x + y = 0
| z2 | ≠ 1
∴
| z1 | 2 = 4 or | z1 | = 2
144. Let a = 3, b = −3, c = 2, then
(a + bω + cω 2 ) 4n + 3 + (c + aω + bω 2 ) 4n + 3 + (b + cω + aω 2 ) 4n + 3
=0
S
60°
Q
⇒ Point z1 lies on circle of radius 2.
(0, 4)
0
⇒(a + bω + cω 2 ) 4n + 3
(4, 0)
P
  c + aω + bω 2 ) 4n + 3  4n + 3  b + cω + aω 2 ) 4n + 3 
+


=0
1 + 
2
2

 a + bω + cω 

  a + bω + cω
(1, –3)
139. min | 1 − 3i − z | = min | z − 1 + 3i |
⇒ (a + bω + cω 2 ) 4n + 3(1 + ω 4n + 3 + (ω 2 ) 4n + 3 ) = 0
= perpendicular distance of the point (1, − 3 ) from the straight
3 −3 3 − 3
line 3 x + y = 0 =
=
2
2
1
1
20 π
140. Area of S =   π × 4 2 +   π × 4 2 =
 6
 4
3
141. Since, | z | ≥ 2 is the region lying on or outside circle centered at
( 0, 0 ) and radius 2. Therefore, | z + (1 / 2 )| is the distance of z
from ( − 1 / 2, 0 ), which lies inside the circle.
Hence, minimum value of | z + (1 / 2 )|
= distance of ( − 1 / 2, 0 ) from ( − 2, 0 )
⇒ 4n + 3 should be an integer other than multiple of 3.
∴ n = 1, 2, 4, 5
kπ
kπ
145. Q α k = cos  + i sin   = e iπk /7
 7
 7
∴ α k + 1 − α k = e iπ (k + 1) /7− e iπk /7 = e iπk /7(e iπ /7− 1 )
 π
= e iπk /7⋅ e iπ /14⋅ 2 i sin  
 14
 π
⇒ | α k + 1 − α k | = 2 sin  
 14
∴
and α 4k − 1 − α 4k − 2 = e iπ ( 4k − 1) /7− e iπ ( 4k − 2 ) /7= e iπ ( 4k − 2) /7(e iπ /7− 1 )
[Q | z | ≥ 2 ]
| z + (1 / 2 )| ≥ 3 / 2
142. Clearly, zk10
= 1, ∀ k, where zk ≠ 1
(A) zk ⋅ z j = e i ( 2 π /10) (k +
j)
= 1, if (k + j ) is multiple of 10
i.e. possible for each k.
 π
k =1
Aliter
1
1
≥ 2−
2
2
 π
12
∑ |αk + 1 − αk | = 12 × 2 sin  14 = 24 sin  14
2

 1
=  − + 2 + | 0 − 0 | 2 = 3 / 2

 2
Q | z + (1 / 2 )| ≥ | z | −
( | z1 | 2 − 4 ) (1 − | z 2 | 2 ) = 0
 π
= e iπ ( 4k − 2) /7⋅e iπ /14 ⋅ 2i sin  
 14
 π
⇒ | α 4k − 1 − α 4k − 2| = 2 sin  
 14
∴
3
 π
 π
∑ | α 4k −1 − α 4k − 2| = 3 × 2 sin  14 = 6 sin  14
k =1
102
Textbook of Algebra
12
∑ |αk + 1 − αk |
Hence,
k =1
=4
3
∑ | α 4k − 1 − α 4k − 2|
k =1
2 + 3i sin θ
146. Let z =
1 − 2i sin θ
Q z is purely imaginary
∴ z = −z
 2 + 3i sin θ 
 2 + 3i sin θ 
⇒
 =−


 1 − 2i sin θ 
 1 − 2i sin θ 
⇒
 2 − 3i sin θ 
 2 + 3i sin θ 
 =−


 1 + 2i sin θ 
 1 − 2i sin θ 
⇒ (2 − 3i sin θ ) (1 − 2i sin θ ) + (1 + 2i sin θ ) (2 + 3i sin θ ) = 0
1
⇒ 4 − 12 sin 2 θ = 0 or sin 2 θ =
3
1


∴ θ = sin −1  
 3
147. Q
⇒
⇒
or
x2 + y 2 −
or
1
a + ibt
a − ibt
x + iy = 2
a + b 2t 2
bt
a
, y =− 2
x= 2
(a + b 2t 2 )
(a + b 2t 2 )
1
x
x2 + y 2 = 2
=
a + b 2t 2 a
x
=0
a
1 
∴ Locus of z is a circle with centre  , 0
 2a 
1
, a > 0.
2a
Also for b = 0, a ≠ 0, we get y = 0.
∴ locus is X -axis and for a = 0, b ≠ 0, we get x = 0
∴ locus is Y -axis.
and radius =
1
1
1
1
1
ω
1
ω2
1 ω2
ω
148. Let ∆ = 1 −ω 2 − 1 ω 2 = 1
ω2
1
ω7
(Q1 + ω + ω = 0 and ω = 1)
2
3
Applying C1 → C1 + C 2 + C 3, then we get
3 L
M
∆= 0
ω
ω2
M
0
ω2
ω
x + iy =
1
L
1
(Q 1 + ω + ω 2 = 0)
= 3 (ω 2 − ω 4 )
= 3 (− 1 − ω − ω )
(Q ω 3 = 1 and 1 + ω + ω 2 = 0)
= − 3 (1 + 2ω )
= − 3z = 3k (given)
∴
k = −z
(Q 1 + 2ω = z)
CHAPTER
02
Theory of Equations
Learning Part
Session 1
● Polynomial in One Variable
● Identity
● Linear Equation
● Quadratic Equations
● Standard Quadratic Equation
Session 2
● Transformation of Quadratic Equations
● Condition for Common Roots
Session 3
● Quadratic Expression
● Wavy Curve Method
● Condition for Resolution into Linear Factors
● Location of Roots (Interval in which Roots Lie)
Session 4
● Equations of Higher Degree
● Rational Algebraic Inequalities
● Roots of Equation with the Help of Graphs
Session 5
● Irrational Equations
● Irrational Inequations
● Exponential Equations
● Exponential Inequations
● Logarithmic Equations
● Logarithmic Inequations
Practice Part
●
●
JEE Type Examples
Chapter Exercises
Arihant on Your Mobile !
Exercises with the
#L
symbol can be practised on your mobile. See inside cover page to activate for free.
Session 1
Polynomial in One Variable, Identity, Linear Equation,
Quadratic Equations, Standard Quadratic Equation
Polynomial in One Variable
An algebraic expression containing many terms of the
form cx n , n being a non-negative integer is called a
polynomial,
f (x ) = a 0 × x n + a1 × x n - 1 + a2 × x n - 2
+ ... + a n - 1 × x + a n ,
where x is a variable, a 0 , a 1 , a 2 , ..., a n are constants and
a 0 ¹ 0.
i.e.,
1. Real Polynomial
Let a 0 , a 1 , a 2 , ..., a n be real numbers and x is a real variable.
Then,
f ( x ) = a 0 × x n + a 1 × x n - 1 + a 2 × x n - 2 + ... + a n - 1 × x + a n
is called a real polynomial of real variable (x) with real
coefficients.
For example, 5 x 3 - 3 x 2 + 7 x - 4, x 2 - 3 x + 1, etc., are real
polynomials.
Let a 0 , a 1 , a 2 , ..., a n are complex numbers and x is a
varying complex number.
Then f ( x ) = a 0 × x n + a 1 × x n - 1 + a 2 × x n - 2 +... + a n - 1 × x + a n
is called a complex polynomial or a polynomial of complex
variable with complex coefficients.
2
2
For example, x - 7ix + (3 - 2i ) x + 13, 3 x - (2 + 3i ) x + 5i,
etc. (where i = -1) are complex polynomials.
3. Rational Expression
or Rational Function
P(x )
, where P ( x ) and Q ( x )
Q(x )
are polynomials in x, is called a rational expression. As a
particular case when Q ( x ) is a non-zero constant, P ( x )
Q(x )
reduces to a polynomial.
An expression of the form
4. Degree of Polynomial
The highest power of variable (x ) present in the
polynomial is called the degree of the polynomial.
For example, f ( x ) = a 0 × x n + a 1 × x n - 1 + a 2 × x n - 2
+ ... + a n - 1 × x + a n is a polynomial in x of degree n.
Remark
A polynomial of degree one is generally called a linear
polynomial. Polynomials of degree 2, 3, 4 and 5 are known as
quadratic, cubic, biquadratic and pentic polynomials,
respectively.
5. Polynomial Equation
2. Complex Polynomial
3
Thus, every polynomial is a rational expression but a
rational expression may or may not be a polynomial.
For example,
2
(ii)
(i) x 2 - 7 x + 8
x -3
3
2
x - 6 x + 11x - 6
x2 +3
3
(iv) x + or
(iii)
( x - 4)
x
x
If f (x ) is a polynomial, real or complex, then f (x ) = 0 is
called a polynomial equation.
(i) A polynomial equation has atleast one root.
(ii) A polynomial equation of degree n has n roots.
Remarks
1. A polynomial equation of degree one is called a linear
equation i.e. ax + b = 0, where a, b Î C, set of all complex
numbers and a ¹ 0.
2. A polynomial equation of degree two is called a quadratic
equation i.e., ax 2 + bx + c, where a, b, c Î C and a ¹ 0.
3. A polynomial equation of degree three is called a cubic
equation i.e., ax 3 + bx 2 + cx + d = 0, where a, b, c, d Î C and
a ¹ 0.
4. A polynomial equation of degree four is called a biquadratic
equation i.e., ax4 + bx 3 + cx 2 + dx + e = 0, where
a, b, c, d, e Î C and a ¹ 0.
5. A polynomial equation of degree five is called a pentic
equation i.e., ax5 + bx4 + cx 3 + dx 2 + ex + f = 0, where
a, b, c, d, e, f Î C and a ¹ 0.
105
Chap 02 Theory of Equations
6. Roots of an Equation
The values of the variable for which an equation is
satisfied are called the roots of the equation.
If x = a is a root of the equation f ( x ) = 0, then f (a ) = 0.
Remark
The real roots of an equation f ( x ) = 0 are the values of x, where
the curve y = f ( x ) crosses X-axis.
7. Solution Set
The set of all roots of an equation in a given domain is
called the solution set of the equation.
For example, The roots of the equation
x 3 - 2 x 2 - 5 x + 6 = 0 are 1, - 2, 3, the solution set is
{1, - 2, 3 }.
Remark
Solve or solving an equation means finding its solution set or
obtaining all its roots.
Identity
If two expressions are equal for all values of x, then the
statement of equality between the two expressions is
called an identity.
For example, ( x + 1) 2 = x 2 + 2 x + 1 is an identity in x.
or
If f ( x ) = 0 is satisfied by every value of x in the domain of
f ( x ), then it is called an identity.
For example, f ( x ) = ( x + 1) 2 - ( x 2 + 2 x + 1) = 0 is an
identity in the domain C, as it is satisfied by every
complex number.
or
If f ( x ) = a 0 × x n + a 1 × x n - 1 + a 2 × x n - 2
+ ... + a n - 1 × x + a n = 0 have more than n distinct roots, it
is an identity, then
a 0 = a 1 = a 2 = ... = a n - 1 = a n = 0
For example, If ax 2 + bx + c = 0 is satisfied by more than
two values of x, then a = b = c = 0.
or
In an identity in x coefficients of similar powers of x on
the two sides are equal.
For example, If ax 4 + bx 3 + cx 2 + dx + e
= 5 x 4 - 3 x 3 + 4 x 2 - 7 x - 9 be an identity in x, then
a = 5, b = - 3, c = 4, d = - 7, e = - 9.
Thus, an identity in x satisfied by all values of x, where as
an equation in x is satisfied by some particular values of x .
y Example 1. If equation
( l2 - 5l + 6 ) x 2 + ( l2 - 3l + 2)x + ( l2 - 4 ) = 0 is
satisfied by more than two values of x , find the
parameter l.
Sol. If an equation of degree two is satisfied by more than two
values of unknown, then it must be an identity. Then, we
must have
l2 - 5l + 6 = 0, l2 - 3l + 2 = 0, l2 - 4 = 0
Þ
l = 2 , 3 and l = 2 , 1 and l = 2 , - 2
Common value of l which satisfies each condition is l = 2.
y Example 2. Show that
(x + b ) (x + c ) (x + c ) (x + a) (x + a) (x + b )
+
=1
+
(b - a ) (c - a ) (c - b ) (a - b ) (a - c ) (b - c )
is an identity.
Sol. Given relation is
( x + b) ( x + c ) ( x + c ) ( x + a) ( x + a) ( x + b)
+
=1
+
(b - a ) (c - a )
(c - b ) (a - b )
(a - c ) (b - c )
When x = - a, then LHS of Eq. (i) =
…(i)
(b - a ) (c - a )
=1
(b - a ) (c - a )
= RHS of Eq. (i)
When x = - b, then LHS of Eq. (i)
(c - b ) (a - b )
= 1 = RHS of Eq. (i)
=
(c - b ) (a - b )
and when x = - c , then LHS of Eq. (i) =
(a - c ) (b - c )
=1
(a - c ) (b - c )
= RHS of Eq. (i).
Thus, highest power of x occurring in relation of Eq. (i) is 2
and this relation is satisfied by three distinct values of
x ( = - a, - b, - c ). Therefore, it cannot be an equation and
hence it is an identity.
y Example 3. Show that x 2 - 3| x | + 2 = 0 is an
equation.
Sol. Put x = 0 in x 2 - 3 | x | + 2 = 0
Þ
02 - 3 | 0| + 2 = 2 ¹ 0
Since, the relation x 2 - 3 | x | + 2 = 0 is not satisfied by x = 0.
Hence, it is an equation.
Linear Equation
An equation of the form
ax + b = 0
where a, b Î R and a ¹ 0, is a linear equation.
b
Eq. (i) has an unique root equal to - .
a
…(i)
106
Textbook of Algebra
y Example 4. Solve the equation
Sol. We have,
or
or
or
x ( 3x - 1)
x
+
=12
6
2
x ( 3x - 1)
x
+
=12
6
2
1
x x x
+ + =1+
6
2 2 2
3x 7
=
2
6
7
x=
9
Nature of Roots
y Example 5. Solve the equation (a - 3)x + 5 = a + 2 .
Sol. Case I For a ¹ 3 , this equation is linear, then
( a - 3) x = ( a - 3)
( a - 3)
\
=1
x=
( a - 3)
1. If a, b, c Î R and a ¹ 0, then
(i) If D < 0, then Eq. (i) has non-real complex roots.
(ii) If D > 0, then Eq. (i) has real and distinct roots,
namely
x1 =
Case II For a = 3,
-b + D
-b - D
and then
, x2 =
2a
2a
ax 2 + bx + c = a( x - x 1 ) ( x - x 2 ).
0× x + 5 = 3 + 2
Þ
5=5
Therefore, any real number is its solution.
Quadratic Equations
An equation in which the highest power of the unknown
quantity is 2, is called a quadratic equation.
Quadratic equations are of two types :
1. Purely Quadratic Equation
A quadratic equation in which the term containing the
first degree of the unknown quantity is absent, is called a
purely quadratic equation.
i.e.,
ax 2 + c = 0,
where a, c Î C and a ¹ 0.
2. Adfected Quadratic Equation
A quadratic equation in which it contains the terms of
first as well as second degrees of the unknown quantity, is
called an adfected (or complete) quadratic equation.
i.e.,
ax 2 + bx + c = 0,
where a, b, c Î C and a ¹ 0, b ¹ 0.
Standard Quadratic Equation
An equation of the form
ax 2 + bx + c = 0
A root of the quadratic Eq. (i) is a complex number a, such
that aa 2 + ba + c = 0. Recall that D = b 2 - 4ac is the
discriminant of the Eq. (i) and its roots are given by the
following formula.
-b ± D
[Shridharacharya method]
x=
2a
…(i)
where a, b, c Î C and a ¹ 0, is called a standard quadratic
equation.
The numbers a, b, c are called the coefficients of this
equation.
…(ii)
(iii) If D = 0, then Eq. (i) has real and equal roots, then
b
and then
x1 = x2 = 2a
ax 2 + bx + c = a( x - x 1 ) 2 .
…(iii)
To represent the quadratic ax 2 + bx + c in form
Eqs. (ii) or (iii), is to expand it into linear factors.
(iv) If D ³ 0, then Eq. (i) has real roots.
(v) If D 1 and D 2 be the discriminants of two
quadratic equations, then
(a) If D 1 + D 2 ³ 0, then
atleast one of D 1 and D 2 ³ 0.
if D 1 < 0, then D 2 > 0 and if D 1 > 0, then
D 2 < 0.
(b) If D 1 + D 2 < 0, then
atleast one of D 1 and D 2 < 0.
If D 1 < 0, then D 2 > 0 and if D 1 > 0, then
D 2 < 0.
2. If a, b, c Î Q and D is a perfect square of a rational
number, the roots are rational and in case it is not a
perfect square, the roots are irrational.
3. If a, b, c Î R and p + iq is one root of Eq. (i) (q ¹ 0 ),
then the other must be the conjugate ( p - iq ) and
vice-versa (where, p, q Î R and i = -1).
l
l
l
l
4. If a, b, c Î Q and p + q is one root of Eq. (i), then the
other must be the conjugate p - q and vice-versa
(where, p is a rational and q is a surd).
5. If a = 1 and b, c Î I and the roots of Eq. (i) are rational
numbers, these roots must be integers.
Chap 02 Theory of Equations
6. If a + b + c = 0 and a, b, c are rational, 1 is a root of the
Eq. (i) and roots of the Eq. (i) are rational.
7. a 2 + b 2 + c 2 - ab - bc - ca =
1
2
= - {(a - b ) (b - c ) + (b - c ) (c - a ) + (c - a ) (a - b )}
y Example 6. Find all values of the parameter a for
which the quadratic equation
(a + 1) x 2 + 2 (a + 1) x + a - 2 = 0
(i) has two distinct roots.
(ii) has no roots.
(iii) has two equal roots.
(5 + 2 6 ) (5 - 2 6 ) = 1
(5 - 2 6 ) =
reduces to (5 + 2 6 )x
\
Þ
Þ
Hence,
t =
= (p - r ) ³ 0
[given, pr = 2(q + s )]
[Q p and q are real]
or
D1 + D 2 ³ 0
Hence, atleast one of the equations x 2 + px + q = 0 and
x 2 + rx + s = 0 has real roots.
2
+ (5 - 2 6 )x
-3
2
-3
y Example 10. Find all roots of the equation
x 4 + 2x 3 - 16 x 2 - 22x + 7 = 0, if one root is 2 + 3.
2
-3
= 10
On dividing x 4 + 2x 3 - 16x 2 - 22x + 7 by x 2 - 4 x + 1, then
the other quadratic factor is x 2 + 6x + 7.
= 10
Then, the given equation reduce in the form
( x 2 - 4 x + 1) ( x 2 + 6x + 7 ) = 0
2
x -3
1
= t , then t + = 10
t
10 ± (100 - 4 )
= (5 ± 2 6 )
2
(5 + 2 6 )x
Sol. Since, a , b are the roots of
( x - a) ( x - b) = c
or
( x - a ) ( x - b ) - c = 0,
Then
( x - a) ( x - b) - c = ( x - a ) ( x - b)
Þ
( x - a ) ( x - b) + c = ( x - a) ( x - b)
Hence, roots of ( x - a ) ( x - b ) + c = 0 are a, b.
= ( x - 2)2 - 3 = x 2 - 4 x + 1.
1
(5 + 2 6 )
æ 1 ö
+ç
÷
è5 + 2 6 ø
- 3
equation ( x - a ) ( x - b ) + c = 0.
\ Product of these roots = ( x - 2 - 3 ) ( x - 2 + 3 )
t 2 - 10t + 1 = 0
Þ
Þ
-3
(5 + 2 6 )x
Put
or
2
2
2
Sol. All coefficients are real, irrational roots will occur in
conjugate pairs.
Hence, another root is 2 - 3.
y Example 7. Solve for x,
2
2
( 5 + 2 6 ) x - 3 + ( 5 - 2 6 ) x - 3 = 10.
(5 + 2 6 )x
= p 2 + r 2 - 2pr
( x - a ) ( x - b ) = c , c ¹ 0. Find the roots of the
= 4 ( a + 1) ( a + 1 - a + 2)
= 12(a + 1)
(i) For a > ( -1), then D > 0, this equation has two distinct
roots.
(ii) For a < ( - 1), then D < 0, this equation has no roots.
(iii) This equation cannot have two equal roots. Since,
D = 0 only for a = - 1 and this contradicts the
hypothesis.
\
Now, D1 + D 2 = p 2 - 4q + r 2 - 4s = p 2 + r 2 - 4(q + s )
y Example 9. If a , b are the roots of the equation
Sol. By the hypothesis, this equation is quadratic and therefore
a ¹ - 1 and the discriminant of this equation,
D = 4 ( a + 1) 2 - 4 ( a + 1) ( a - 2)
\
y Example 8. Show that if p , q, r and s are real numbers
and pr = 2(q + s ), then atleast one of the equations
x 2 + px + q = 0 and x 2 + rx + s = 0 has real roots.
Sol. Let D1 and D 2 be the discriminants of the given equations
x 2 + px + q = 0 and x 2 + rx + s = 0, respectively.
{(a - b ) 2 + (b - c ) 2 + (c - a ) 2 }
Sol. Q
107
= (5 ± 2 6 ) = (5 + 2 6 )± 1
x
2
-3= ±1
x
2
- 3 = 1 or x 2 - 3 = - 1
x 2 = 4 or x 2 = 2
x = ± 2, ± 2
x 2 + 6x + 7 = 0
\
- 6 ± 36 - 28
= -3 ± 2
2
Hence, the other roots are 2 - 3, - 3 ± 2.
Then,
x=
Relation between Roots
and Coefficients
1. Relation between roots and coefficients of
quadratic equation If roots of the equation
ax 2 + bx + c = 0 (a ¹ 0 ) be real and distinct and a < b,
-b + D
-b - D
.
then a =
,b =
2a
2a
108
Textbook of Algebra
(i) Sum of roots
b
Coefficient of x
.
=a
Coefficient of x 2
=S =a +b = -
(vii) a 5 + b5 = (a 2 + b2 ) (a 3 + b 3 ) - a 2 b2 (a + b)
æ b 2 - 2ac ö æ (b 3 - 3abc ) ö c 2
=ç
÷ ç÷- 2
a3
è a2 ø è
ø a
(ii) Product of roots
Constant term
c
.
= P = ab = =
a Coefficient of x 2
(iii) Difference of roots
Discriminant
D
.
= D ¢ =a -b =
=
a
Coefficient of x 2
=
3. Symmetric function of roots A function of a and
b is said to be symmetric function, if it remains
unchanged, when a and b are interchanged.
For example, a 3 + 3 a 2 b + 3 ab2 + b 3 is a symmetric
function of a and b, whereas a 3 - b 3 + 5 ab is not a
symmetric function of a and b. In order to find the
value of a symmetric function in terms of a + b, ab
and a - b and also in terms of a, b and c.
(i) a 2 + b2 = (a + b) 2 - 2 ab
2
2
æ bö
æ c ö b - 2ac
.
= ç- ÷ - 2 ç ÷ =
è aø
èa ø
a2
(ii) a 2 - b2 = (a + b) (a - b)
b D
æ bö æ D ö
= ç- ÷ ç
÷ =- 2 .
è aø è a ø
a
(iii) a 3 + b 3 = (a + b) 3 - 3 ab (a + b)
3
æ b 3 - 3abc ö
æ bö
æc ö æ bö
= ç- ÷ - 3 ç ÷ ç- ÷ = - ç
÷.
è aø
èa ø è a ø
a3
ø
è
(iv) a 3 - b 3 = (a - b) 3 + 3 ab (a - b)
3
æ Dö
D ( D + 3ac )
æc ö æ D ö
.
=ç
÷=
÷ +3ç ÷ ç
è
ø
a
a
a
ø
è
ø
è
a3
=
D (b 4 - 3acb 2 + 3a 2 c 2 )
.
a5
y Example 11. If one root of the equation
x 2 - ix - (1 + i ) = 0, (i = -1 ) is 1 + i, find the other root.
Sol. All coefficients of the given equation are not real, then
other root ¹ 1 - i.
Let other root be a, then sum of roots = i
i.e.
1 + i + a = i Þ a = ( - 1)
Hence, the other root is ( -1).
y Example 12. If one root of the equation
9+ 5
, then find the other root.
x 2 - 5 x - 19 = 0 is
2
Sol. All coefficients of the given equation are not rational,
9- 5
.
then other root ¹
2
Let other root be a, sum of roots = 5
-9 + 5
9+ 5
+a = 5 Þ a =
2
2
-9 + 5
.
Hence, other root is
2
Þ
y Example 13. If the difference between the
corresponding roots of the equations x 2 + ax + b = 0
and x 2 + bx + a = 0 (a ¹ b ) is the same, find the value
of a + b .
Sol. Let a , b be the roots of x 2 + ax + b = 0 and g , d be the
roots of x 2 + bx + a = 0, then given
a -b = g - d
(v) a 4 + b 4 = (a 2 + b2 ) 2 - 2 a 2 b2
2
2
æ b 2 - 2ac ö
b 4 + 2a 2 c 2 - 4acb 2
æc ö
.
-2ç ÷ =
=ç
÷
èa ø
a4
è a2 ø
(vi) a 4 - b 4 = (a 2 + b2 ) (a 2 - b2 )
=-
2
b D (b - 2ac )
a4
.
.
æ b 2 - 2ac ö æ D ( D + 3ac ) ö æ c ö 2 æ D ö
=ç
֍
÷ +ç ÷ ç
÷
ø èa ø è a ø
a3
è a2 ø è
i.e. x 2 - (Sum of roots) x + Product of roots = 0
x 2 - Sx + P = 0.
a5
(viii) a 5 - b5 = (a 2 + b2 ) (a 3 - b 3 ) + a 2 b2 (a - b)
2. Formation of an equation with given roots
A quadratic equation whose roots are a and b, is
given by ( x - a ) ( x - b) = 0 or x 2 - (a + b) x + ab = 0
\
- (b 5 - 5ab 3 c + 5a 2 bc 2 )
æ bö
ç- ÷
è aø
2
Þ
Þ
a - 4b
=
1
b 2 - 4a
1
é
Dù
êQa - b = a ú
û
ë
a 2 - 4b = b 2 - 4a
Þ (a 2 - b 2 ) + 4 (a - b ) = 0 Þ (a - b ) (a + b + 4 ) = 0
Q
\
a-b¹0
a + b + 4 = 0 or a + b = - 4.
Chap 02 Theory of Equations
y Example 14. If a + b + c = 0 and a, b , c are rational.
Prove that the roots of the equation
(b + c - a ) x 2 + (c + a - b ) x + (a + b - c ) = 0
2
1
1
= - or b = - - a
4
2
2
and so 4 a 2 + 2a - 1 = 0, because a is a root of
4 x 2 + 2x - 1 = 0.
then a + b = -
…(i)
Q (b + c - a ) + (c + a - b ) + (a + b - c ) = a + b + c = 0
\ x = 1 is a root of Eq. (i), let other root of Eq. (i) is a, then
a+b-c
Product of roots =
b+c -a
Þ
y Example 16. If a is a root of 4 x 2 + 2x - 1 = 0.
Prove that 4 a 3 - 3 a is the other root.
Sol. Let other root is b,
are rational.
Sol. Given equation is
(b + c - a ) x 2 + (c + a - b ) x + (a + b - c ) = 0
1´a =
Now,
[Qa + b + c = 0]
c
[rational]
\
a=
a
Hence, both roots of Eq. (i) are rational.
Aliter
Let b + c - a = A , c + a - b = B, a + b - c = C
Then,
[Qa + b + c = 0] …(ii)
A + B+C =0
Now, Eq. (i) becomes
…(iii)
Ax 2 + Bx + C = 0
Discriminant of Eq. (iii),
D = B 2 - 4 AC
= ( - C - A )2 - 4 AC
[Q A + B + C = 0]
b = 4 a 3 - 3 a = a ( 4 a 2 - 3)
= (C + A ) - 4 AC
= (C - A )2 = (2a - 2c )2
1
= - (4 a 2 ) - 2 a
2
1
= - (1 - 2 a ) - 2 a
2
1
= - -a =b
2
3
Hence, 4 a - 3 a is the other root.
Hence, roots of Eq. (i) are rational.
y Example 15. If the roots of equation
a (b - c ) x 2 + b (c - a ) x + c (a - b ) = 0
be equal, prove that a, b , c are in HP.
a (b - c ) x + b (c - a ) x + c (a - b ) = 0
Þ
…(i)
Here, coefficient of x 2 + coefficient of x + constant term = 0
i.e.,
a (b - c ) + b (c - a ) + c (a - b ) = 0
Then, 1 is a root of Eq. (i).
Since, its roots are equal.
Therefore, its other root will be also equal to 1.
\
Hence, a, b and c are in HP.
c (a - b )
a (b - c )
[from Eq. (i)]
y Example 17. If a , b are the roots of the equation
l ( x 2 - x ) + x + 5 = 0. If l 1 and l 2 are two values of l
a b 4
for which the roots a , b are related by + = , find
b a 5
l1 l 2
the value of
.
+
l 2 l1
a 2 + b2 4
=
ab
5
Sol. Given equation is
2
[Q 4 a 2 + 2 a - 1 = 0]
Qa , b are the roots of this equation.
l -1
5
and ab =
\
a +b =
l
l
a b 4
But, given
+ =
b a 5
= 4 (a - c )2 = A perfect square
ab - ac = ca - bc
2ac
b=
a+c
[Q 4 a 2 + 2 a - 1 = 0]
= a ( 1 - 2 a - 3)
Sol. The given equation can be written as
l x 2 - ( l - 1) x + 5 = 0
2
Þ
…(i)
= -2 a 2 - 2 a
-c - c
-a - a
Then, product of roots = 1 ´ 1 =
109
Þ
Þ
Þ
2
( l - 1) 2
(a + b ) - 2ab 4
Þ
=
ab
5
l2
5
l
-
10
l =4
5
( l - 1)2 - 10l 4
Þ l2 - 12l + 1 = 4 l
=
5l
5
l2 - 16l + 1 = 0
It is a quadratic in l, let roots be l1 and l 2 , then
l1 + l 2 = 16 and l1l 2 = 1
\
l1 l 2 l21 + l22 ( l1 + l 2 )2 - 2l1l 2
+
=
=
l 2 l1
l1l 2
l1l 2
=
(16)2 - 2(1)
= 254
1
110
Textbook of Algebra
y Example 18. If a , b are the roots of the equation
x 2 - px + q = 0, find the quadratic equation the roots
of which are (a 2 - b 2 ) (a 3 - b 3 ) and a 3b 2 + a 2 b 3 .
= ( a + b ) (a - b )2 {(a + b )2 - ab }
= p ( p 2 - 4q ) ( p 2 - q )
and a 3b 2 + a 2 b 3 = a 2 b 2 (a + b )= pq 2
S = Sum of roots = p ( p 2 - 4q ) ( p 2 - q ) + pq 2
Sol. Since, a , b are the roots of x 2 - px + q = 0.
= p ( p 4 - 5p 2q + 5q 2 )
\
a + b = p , ab = q
Þ
a - b = ( p 2 - 4q )
P = Product of roots = p 2q 2 ( p 2 - 4q ) ( p 2 - q )
\ Required equation is x 2 - Sx + P = 0
Now, (a 2 - b 2 ) (a 3 - b 3 )
i.e. x 2 - p ( p 4 - 5p 2q + 5q 2 ) x + p 2q 2 ( p 2 - 4q ) ( p 2 - q ) = 0
= (a + b ) (a - b ) (a - b ) (a 2 + ab + b 2 )
#L
1.
Exercise for Session 1
If (a 2 - 1)x 2 + (a - 1) x + a 2 - 4a + 3 = 0 be an identity in x , then the value of a is/are
(a) -1
2.
(b) 1
(c) 3
The roots of the equation x + 2 3x + 3 = 0 are
(a) real and unequal
(c) irrational and equal
3.
(b) non-real
10.
(d) AGP
(c) 2 + i
(b) i
(d 2 - i
(b) ± 4
(c) ± 6
(d) ± 8
If 3p 2 = 5p + 2 and 3q 2 = 5q + 2 where p ¹ q , pq is equal to
2
3
(b) -
2
3
(c)
3
2
(d) -
3
2
If a, b are the roots of the quadratic equation x 2 + bx - c = 0, the equation whose roots are b and c, is
(a) x 2 + ax - b = 0
(b) x 2 - [(a + b ) + ab ] x - ab (a + b ) = 0
(c) x 2 + [(a + b ) + ab ] x + ab (a + b ) = 0
(d) x 2 + [(a + b ) + ab ] x - ab (a + b ) = 0
Let p, q Î {1, 2, 3, 4}. The number of equations of the form px 2 + qx + 1 = 0 having real roots, is
(a) 15
11.
(c) HP
If the difference of the roots of x 2 - lx + 8 = 0 be 2, the value of l is
(a)
9.
(b) GP
If one root of the quadratic equation ix 2 - 2 (i + 1) x + (2 - i ) = 0,i = -1 is 2 - i , the other root is
(a) ± 2
8.
(b) two real roots
(d) None of these
If roots of the equation (q - r ) x 2 + (r - p ) x + ( p - q ) = 0 are equal, then p, q , r are in
(a) - i
7.
(d) equal
If P ( x ) = ax + bx + c and Q( x ) = - ax + dx + c, where ac ¹ 0, then P ( x ) Q ( x ) = 0 has atleast
(a) AP
6.
(c) irrational
2
2
(a) four real roots
(c) four imaginary roots
5.
(b) rational and equal
(d) irrational and unequal
Ifa, b , c Î Q, then roots of the equation (b + c - 2a ) x 2 + (c + a - 2b ) x + (a + b - 2c ) = 0 are
(a) rational
4.
(d) -1, 1, 3
2
(b) 9
(c) 8
(d) 7
2
If a and b are the roots of the equation ax + bx + c = 0 (a ¹ 0, a, b , c being different), then
(1 + a + a 2 ) (1 + b + b 2 ) is equal to
(a) zero
(b) positive
(c) negative
(d) None of these
Session 2
Transformation of Quadratic Equations, Condition
for Common Roots
Transformation of
Quadratic Equations
Let a, b be the roots of the equation ax 2 + bx + c = 0, then
the equation
(i) whose roots are a + k, b + k, is
[replace x by ( x - k )]
a (x - k )2 + b (x - k ) + c = 0
(ii) whose roots are a - k, b - k, is
a ( x + k ) 2 + b ( x + k ) + c = 0 [replace x by ( x + k )]
(iii) whose roots are ak, bk, is
é
æ x öù
ê replace x by çè k ÷ø ú
ë
û
ax 2 + kbx + k 2 c = 0
a b
(iv) whose roots are , , is
k k
ak 2 x 2 + bkx + c = 0
(v) whose roots are - a, - b , is
ax 2 - bx + c = 0
1 1
(vi) whose roots are , , is
a b
cx 2 + bx + a = 0
(vii) whose roots are -
1
1
, - , is
a b
cx 2 - bx + a = 0
k k
(viii) whose roots are , , is
a b
[replace x by xk]
[replace x by (- x )]
é
æ 1 öù
ê replace x by çè x ÷ø ú
ë
û
é
æ 1 öù
ê replace x by çè - x ÷ø ú
ë
û
é
æ k öù
ê replace x by çè x ÷ø ú
ë
û
(ix) whose roots are pa + q, pb + q, is
2
é
æ x -q ö
æ x - q öù
æx -qö
aç
÷ +b ç
÷ú
÷ + c = 0 ê replace x by ç
è p ø
è p øû
è p ø
ë
(x) whose roots are a n , bn , n Î N , is
[ replace x by ( x 1 /n )]
(xi) whose roots are a 1 /n , b1 /n , n Î N is
a (x n )2 + b (x n ) + c = 0
q
q
=x Þ a=pp -a
x
q
So, we replacing x by p - in the given equation, we get
x
Sol. Let
2
qö
qö
æ
æ
çp - ÷ - p çp - ÷ + q = 0
è
è
xø
xø
2
q
pq
2pq
Þ
p2 + 2 - p2 +
+q =0
x
x
x
pq q 2
+ 2 =0
qÞ
x
x
or
qx 2 - pqx + q 2 = 0 or x 2 - px + q = 0
q
q
is the required equation whose roots are
and
×
p -a
p -b
y Example 20. If a and b are the roots of
ax 2 + bx + c = 0, then find the roots of the equation
ax 2 - bx ( x - 1) + c ( x - 1) 2 = 0.
Sol. Q ax 2 - bx ( x - 1) + c ( x - 1)2 = 0
…(i)
2
cx 2 + kbx + k 2 a = 0
a ( x 1 /n ) 2 + b ( x 1 /n ) + c = 0
y Example 19. If a , b be the roots of the equation
x 2 - px + q = 0, then find the equation whose roots are
q
q
and
×
p -a
p -b
[ replace x by ( x n )]
Þ
æ x ö
æ x ö
aç
÷ +c =0
÷ - bç
è x - 1ø
è x - 1ø
2
æ x ö
æ x ö
aç
÷ +c =0
÷ + bç
è1 - x ø
è1 - x ø
Now, a , b are the roots of ax 2 + bx + c = 0.
x
x
Then,
and b =
a=
1- x
1- x
a
b
and x =
Þ
x=
a +1
b +1
a
b
Hence,
are the roots of the Eq. (i).
,
a +1 b +1
or
y Example 21. If a , b be the roots of the equation
3
3
æ1 - a ö æ1 - b ö
3x 2 + 2x + 1 = 0, then find value of ç
÷ +ç
÷ .
è 1+ a ø è 1+ b ø
Sol. Let
1-a
=x
1+a
Þ a=
1- x
1+ x
112
Textbook of Algebra
So, replacing x by
1- x
in the given equation, we get
1+ x
2
æ1 - x ö
æ1 - x ö
2
3 ç
÷ + 1 = 0 Þ x - 2x + 3 = 0
÷ + 2ç
è1 + x ø
è1 + x ø
…(i)
It is clear that
1-a
1-b
and
are the roots of Eq. (i).
1+a
1+b
\
æ1 - a ö æ1 - b ö
÷ =2
÷+ç
ç
è1 + a ø è1 + b ø
…(ii)
æ1 - a ö æ1 - b ö
ç
÷ =3
֍
è1 + a ø è1 + b ø
…(iii)
and
3
\
3
roots are reciprocals to each other?
roots are opposite in sign?
both roots are positive?
both roots are negative?
atleast one root is positive?
atleast one root is negative?
\
æ1 - a ö æ1 - b ö æ1 - a 1 - b ö
3
+
֍
÷ = 2 - 3 × 3 × 2 = 8 - 18 = - 10
֍
ç
è1 + a ø è1 + b ø è1 + a 1 + b ø
D = b 2 - 4ac = 4 (m - 1)2 - 4 (m + 5)
= 4 (m 2 - 3m - 4 )
\ D = 4 (m - 4 ) (m + 1) and here a = 1 > 0
(i) b = 0 and D > 0
Þ 2(m - 1) = 0 and 4 (m - 4 ) (m + 1) > 0
Roots Under Special Cases
ax 2 + bx + c = 0
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
Sol. Here, a = 1, b = 2(m - 1) and c = m + 5
3
æ1 - b ö æ1 - a
æ1 - a ö
1 - bö
+
÷ -3
÷ =ç
÷ +ç
ç
è1 + b ø è 1 + a 1 + b ø
è1 + a ø
Consider the quadratic equation
y Example 22. For what values of m, the
equation x 2 + 2 (m - 1) x + m + 5 = 0 has (m ÎR )
(i) roots are equal in magnitude but opposite in
sign?
Þ m = 1 and m Î( - ¥, - 1) È ( 4, ¥ )
…(i)
where a, b, c Î R and a ¹ 0. Then, the following hold good :
(i) If roots of Eq. (i) are equal in magnitude but opposite in
sign, then sum of roots is zero as well as D > 0, i.e. b = 0
and D > 0.
(ii) If roots of Eq. (i) are reciprocal to each other, then product
of roots is 1 as well as D ³ 0 i.e., a = c and D ³ 0.
(iii) If roots of Eq. (i) are of opposite signs, then product of
roots < 0 as well as D > 0 i.e., a > 0, c < 0 and D > 0 or
a < 0, c > 0 and D > 0.
(iv) If both roots of Eq. (i) are positive, then sum and product of
roots > 0 as well as D ³ 0 i.e., a > 0, b < 0, c > 0 and D ³ 0 or
a < 0, b > 0, c < 0 and D ³ 0.
(v) If both roots of Eq. (i) are negative, then sum of roots < 0,
product of roots > 0 as well as D ³ 0 i.e., a > 0, b > 0, c > 0
and D ³ 0 or a < 0, b < 0, c < 0 and D ³ 0.
(vi) If atleast one root of Eq. (i) is positive, then either one root
is positive or both roots are positive i.e., point (iii) È (iv).
(vii) If atleast one root of Eq. (i) is negative, then either one root
is negative or both roots are negative i.e., point (iii) È (v).
(viii) If greater root in magnitude of Eq. (i) is positive, then
sign of b = sign of c ¹ sign of a.
(ix) If greater root in magnitude of Eq. (i) is negative, then
sign of a = sign of b ¹ sign of c.
(x) If both roots of Eq. (i) are zero, then b = c = 0.
æ bö
(xi) If roots of Eq. (i) are 0 and ç - ÷ , then c = 0.
è aø
c
(xii) If roots of Eq. (i) are 1 and , then a + b + c = 0.
a
\
m Îf
[null set]
(ii) a = c and D ³ 0
Þ 1 = m + 5 and 4 (m - 4 ) (m + 1) ³ 0
Þ m = - 4 and m Î (-¥, - 1] È [ 4, ¥ )
\ m = -4
(iii) a > 0, c < 0 and D > 0
Þ 1 > 0, m + 5 < 0 and 4 (m - 4 ) (m + 1) > 0
Þ m < - 5 and m Î( -¥, - 1) È ( 4, ¥ )
\ m Î (- ¥, - 5)
(iv) a > 0, b < 0, c > 0 and D ³ 0
Þ 1 > 0, 2(m - 1) < 0, m + 5 > 0
and
4 ( m - 4 ) ( m + 1) ³ 0
Þ
m < 1, m > - 5 and m Î (- ¥, - 1] È [ 4, ¥ )
Þ
m Î (- 5, - 1]
(v) a > 0, b > 0, c > 0 and D ³ 0
Þ 1 > 0, 2(m - 1) > 0, m + 5 > 0
and 4 (m - 4 ) (m + 1) ³ 0
Þ m > 1, m > - 5 and m Î (-¥, - 1] È [ 4, ¥ )
\
m Î[ 4, ¥ )
(vi) Either one root is positive or both roots are
positive
i.e.,
(c) È (d)
Þ m Î (-¥, - 5) È ( - 5, - 1]
(vii) Either one root is negative or both roots are
negative
i.e.,
(c) È (e)
Þ
m Î (-¥, - 5) È [ 4, ¥ )
Chap 02 Theory of Equations
Condition for Common Roots
y Example 23. Find the value of l, so that the
equations x 2 - x - 12 = 0 and lx 2 + 10x + 3 = 0 may
have one root in common. Also, find the common root.
1. Only One Root is Common
Consider two quadratic equations
ax 2 + bx + c = 0 and a ¢ x 2 + b ¢ x + c ¢ = 0
[where a, a ¢ ¹ 0 and ab ¢ - a ¢ b ¹ 0]
Let a be a common root, then
a a 2 + ba + c = 0 and a ¢ a 2 + b ¢ a + c ¢ = 0.
On solving these two equations by cross-multiplication,
we have
a2
a
1
=
=
bc ¢ - b ¢ c ca ¢ - c ¢ a ab ¢ - a ¢ b
From first two relations, we get
bc ¢ - b ¢ c
…(i)
a=
ca ¢ - c ¢ a
…(ii)
or
a¢ b¢
´
b
c
b¢ c¢
=
c
l=-
\
43
16
and if x = - 3 is a common root, then
l( -3)2 + 10( -3) + 3 = 0
\
l =3
43
Hence, for l = - , common root is x = 4
16
and for l = 3, common root is x = - 3.
a, b , c ÎR and a ¹ 0) and x 2 + 2x + 3 = 0 have a
Sol. Given equations are
ax 2 + bx + c = 0
a
c¢ a¢
2
[remember]
This is the required condition for one root of two
quadratic equations to be common.
2. Both Roots are Common
Let a, b be the common roots of the equations
ax 2 + bx + c = 0 and a ¢ x 2 + b ¢ x + c ¢ = 0, then
a b
b
b¢
…(iii)
=
Þ
a +b = - = a¢ b¢
a
a¢
a c
c c¢
…(iv)
and
=
ab = =
Þ
a¢ c¢
a a¢
a b c
From Eqs. (iii) and (iv), we get
= =
a¢ b¢ c¢
This is the required condition for both roots of two
quadratic equations to be identical.
Remark
To find the common root between the two equations, make the
same coefficient of x 2 in both equations and then subtract of the
two equations.
…(i)
2
x + 2x + 3 = 0
and
(ab ¢ - a ¢ b ) (bc ¢ - b ¢ c ) = (ca ¢ - c ¢ a ) 2
b
Þ
( x - 4 ) ( x + 3) = 0
\
x = 4, - 3
If x = 4 is a common root, then
l( 4 )2 + 10( 4 ) + 3 = 0
common root, then show that a :b : c = 1 : 2 : 3.
From Eqs. (i) and (ii), we get
bc ¢ - b ¢ c ca ¢ - c ¢ a
=
ca ¢ - c ¢ a ab ¢ - a ¢ b
a
x 2 - x - 12 = 0
Sol. Q
y Example 24. If equations ax 2 + bx + c = 0, (where
and from last two relations, we get
ca ¢ - c ¢ a
a=
ab ¢ - a ¢ b
Þ
113
…(ii)
Clearly, roots of Eq. (ii) are imaginary, since Eqs. (i) and (ii)
have a common root. Therefore, common root must be
imaginary and hence both roots will be common.
Therefore, Eqs. (i) and (ii) are identical.
\
a b c
= = or a : b : c = 1 : 2 : 3
1 2 3
y Example 25. If a, b , c are in GP, show that the
equations ax 2 + 2bx + c = 0 and dx 2 + 2ex + f = 0
a b c
have a common root, if , , are in HP.
d e f
Sol. Given equations are
ax 2 + 2bx + c = 0
and
…(i)
2
dx + 2ex + f = 0
…(ii)
Since, a, b, c are in GP.
\
b 2 = ac or b = ac
From Eq. (i), ax 2 + 2 ac x + c = 0
or
( a x + c )2 = 0 or x = -
Q Given Eqs. (i) and (ii) have a common root.
Hence, x = -
c
also satisfied Eq. (ii), then
a
c
a
114
Textbook of Algebra
c
æc ö
+ f =0
d ç ÷ - 2e
èa ø
a
Þ
2e
d
f
+ =0
a
c
ac
or
d 2e f
+ =0
a b
c
#L
1.
\
[Qb = ac ]
(b) 4x 2 + 7x + 6 = 0
(d) 4x 2 - 7x + 16 = 0
(b) x 2 + x + 1 = 0
(c) x 2 - x - 1 = 0
(d) None of these
The equation formed by decreasing each root of ax 2 + bx + c = 0 by 1 is 2x 2 + 8x + 2 = 0, then
(b) b = - c
(c) c = - a
(d) b = a + c
2
If the roots of equation
(a)
5.
(c) 4x 2 + 7x + 1 = 0
æ 1
1 ö
If a, b are the roots of x 2 - 3x + 1 = 0, then the equation whose roots are ç
,
÷ , is
è a - 2 b - 2ø
(a) a = - b
4.
a b c
Hence, , , are in HP.
d e f
If a and b are the roots of the equation 2x 2 - 3x + 4 = 0, then the equation whose roots are a 2 and b 2, is
(a) x 2 + x - 1 = 0
3.
d e f
, , are in AP.
a b c
Exercise for Session 2
(a) 4x 2 + 7x + 16 = 0
2.
2e
d f
+ =
a c
b
or
x - bx m - 1
are equal but opposite in sign, then the value of m will be
=
ax - c
m +1
a -b
a+b
(b)
b -a
a+b
(c)
a+b
a -b
(d)
b +a
b -a
If x 2 + px + q = 0 is the quadratic equation whose roots are a - 2 and b - 2, where a and b are the roots of
x 2 - 3x + 1 = 0, then
(a) p = 1, q = 5
6.
(b) p = 1, q = - 5
(d) m Î(1, 3)
1
(b) m Îæç - , ¥ö÷
è 8 ø
(c) m Î æç -1, è
1ö
÷
8ø
(d) m ÎR
If both the roots of l (6x 2 + 3) + rx + 2x 2 - 1 = 0 and 6l (2x 2 + 1) + px + 4x 2 - 2 = 0 are common, then 2r - p is
equal to
(b) 0
(c) 1
(d) 2
(c) 3
(d) None of these
a3 + b 3 + c 3
If ax + bx + c = 0 and bx + cx + a = 0 have a common root a ¹ 0, then
is equal to
abc
2
2
(a) 1
10.
(c) m Î [9, ¥)
If the equation (1 + m ) x - 2 (1 + 3m ) x + (1 + 8m ) = 0, where m Î R ~ {-1}, has atleast one root is negative, then
(a) -1
9.
(b) m Î(- ¥, 1]
2
(a) m Î (- ¥, - 1)
8.
(d) None of these
If both roots of the equation x - (m - 3) x + m = 0 (m Î R ) are positive, then
(a) m Î (3, ¥)
7.
(c) p = - 1, q = 1
2
(b) 2
If a ( p + q )2 + 2bpq + c = 0 and a ( p + r )2 + 2bpr + c = 0, then qr is equal to
(a) p 2 +
c
a
(b) p 2 +
a
c
(c) p 2 +
a
b
(d) p 2 +
b
a
Session 3
Quadratic Expression, Wavy Curve Method, Condition
for Resolution into Linear Factors, Location of Roots
Quadratic Expression
2
An expression of the form ax + bx + c , where a, b, c Î R
and a ¹ 0 is called a quadratic expression in x. So, in
general quadratic expression is represented by
2
2
f ( x ) = ax + bx + c or y = ax + bx + c .
Graph of a Quadratic Expression
y = ax 2 + bx + c = f ( x ),
We have,
or
bö
Dö
æ
æ
çy + ÷ = a ç x + ÷
è
è
4a ø
2a ø
Now, let y +
D
b
= Y and x +
=X
4a
2a
\
Y = aX
-b ± D
2a
For D > 0, parabola cuts X-axis in two real and
distinct points
\
ax 2 + bx + c = 0 Þ x =
a < 0, D > 0
[a ¹ 0 ]
2
éæ
bö
D ù
y = a êç x + ÷ ú
2a ø
4a 2 úû
êë è
Þ
4. Intersection with axes
(i) Intersection with X-axis
For X-axis, y = 0.
2
X-axis
X-axis
a > 0, D > 0
For D = 0, parabola touches X-axis in one point
b
i.e., x = - .
2a
a<0 , D = 0
2
Y
Þ
X =
a
1. The shape of the curve y = f (x ) is parabolic.
b
2. The axis of parabola is X = 0 or x +
=0
2a
b
i.e. parallel to Y-axis.
or x = 2a
1
3. (i) If > 0 Þ a > 0, the parabola open upwards.
a
X-axis
2
X-axis
a >0 , D =0
For D < 0, parabola does not cut X-axis i.e.,
imaginary values of x.
a < 0, D < 0
X-axis
X-axis
a > 0, D < 0
(ii) If
1
< 0 Þ a < 0, the parabola open downwards.
a
(ii) Intersection with Y-axis
For Y-axis, x = 0.
\
y =c
5. Greatest and least values of f(x)
1
Vertex of the parabola X 2 = Y is
a
X = 0, Y = 0
116
Textbook of Algebra
Þ
or
x+
b
D
= 0, y +
=0
2a
4a
b
D
x = - ,y = 2a
4a
3. a > 0 and D = 0. So, f ( x ) ³ 0 for all x Î R,
i.e. f ( x ) is positive for all real values of x except at
vertex, where f ( x ) = 0.
Dö
æ b
Hence, vertex of y = ax 2 + bx + c is ç - , - ÷ .
è 2a
4a ø
a>0
X-axis
Vertex
a>0
a<0
Vertex
b
.
2a
D
æ bö
This least value is given by f ç - ÷ = è 2a ø
4a
D
or
y least = - .
4a
æ D
ö
\ Range of y = ax 2 + bx + c is ç - , ¥÷ .
è 4a ø
b
For a < 0, f ( x ) has greatest value at x = - .
2a
D
æ bö
This greatest value is given by f ç - ÷ = è 2a ø
4a
D
or y greatest = 4a
Dö
æ
\ Range of y = ax 2 + bx + c is ç - ¥, - ÷ .
è
4a ø
4. a < 0 and D = 0. So, f ( x ) £ 0 for all x Î R,
i.e. f ( x ) is negative for all real values of x except at
vertex, where f ( x ) = 0.
X-axis
For a > 0, f ( x ) has least value at x = -
Sign of Quadratic Expression
Let f ( x ) = ax 2 + bx + c or y = ax 2 + bx + c ,
where a, b, c Î R and a ¹ 0, for some values of x , f ( x ) may
be positive, negative or zero. This gives the following
cases :
1. a > 0 and D < 0.
So, f ( x ) > 0 for all x Î R,
i.e. f ( x ) is positive for all real values of x.
5. a > 0 and D > 0.
Let f ( x ) = 0 have two real roots a and b (a < b), then
f ( x ) > 0 for all x Î( - ¥, a ) È (b, ¥) and f ( x ) < 0 for
all x Î(a, b).
a>0
a
b
X-axis
6. a < 0 and D > 0
Let f ( x ) = 0 have two real roots a and b (a < b),
then f ( x ) < 0 for all x Î ( - ¥, a ) È (b, ¥)
and f ( x ) > 0 for all x Î(a, b).
a
b
X-axis
a<0
Wavy Curve Method
(Generalised Method of Intervals)
a>0
X-axis
2. a < 0 and D < 0. So, f ( x ) < 0 for all x Î R,
i.e. f ( x ) is negative for all real values of x.
X-axis
a<0
a<0
Wave Curve Method is used for solving inequalities of the
form
f (x ) =
( x - a 1 ) k 1 ( x - a 2 ) k 2 K( x - a m ) k m
( x - b 1 ) p 1 ( x - b 2 ) p 2 K( x - b n ) p n
>0
( < 0, ³ 0 or £ 0),
where, k 1 , k 2 , K, k m , p 1, p 2 , ..., p n are natural numbers and
such that a i ¹ b j , where i = 1, 2, K , m and j = 1, 2, K, n.
Chap 02 Theory of Equations
Important Results
We use the following methods:
k1
1. Solve ( x - a 1 ) ( x - a 2 )
k2
K(x - am )
km
= 0 and
( x - b1 )p1 ( x - a 2 )p 2 K ( x - bn )pn = 0, then we get
x = a1, a 2 , K , am , b1, b 2 , K , bn
[critical points]
2. Assume a 1 < a 2 < K < a m < b 1 < b 2 < K < b n
Plot them on the real line. Arrange inked (black)
circles (·) and un-inked (white) circles ( ), such
that
a1 a2 K am b1 b2 K bn
If f ( x ) > 0
...
...
...
...
f (x ) < 0
...
f ( x ) ³ 0 · · ... ·
...
f ( x ) £ 0 · · ... ·
3. Obviously, b n is the greatest root. If in all brackets
before x positive sign and expression has also
positive sign, then wave start from right to left,
beginning above the number line, i.e.
o
oo o oo o
oo o oo o
oo o
oo o
+
117
( x - a 1 ) k 1 ( x - a 2 ) k 2 K( x - a m ) k m
( x - b 1 ) p 1 ( x - b 2 ) p 2 K( x - b n ) p n
, then
1. The point where denominator is zero or function approaches
infinity, will never be included in the answer.
2. For x 2 < a2 or|x| < a Û - a < x < a
i.e.,
x Î( - a, a)
3. For 0 < x 2 < a2 or 0 < |x| < a
Û
i.e.,
- a < x < a ~ {0}
x Î( - a, a) ~{0}
4. For x 2 ³ a2 or|x| ³ a Û x £ - a or x ³ a
x Î( - ¥, - a] È [ a, ¥)
i.e.,
5. For x 2 > a2 or|x| > a Û x < - a or x > a
x Î( - ¥, - a) È ( a, ¥)
i.e.,
6. For a2 £ x 2 £ b2 or a £ |x| £ b
Û
a £ x £ b or - b £ x £ - a
i.e.,
x Î[ - b, - a] È [ a, b]
7. For a2 < x 2 £ b2 or a < |x| £ b
Û
a < x £ b or - b £ x < - a
i.e.,
x Î[ - b, - a) È ( a, b]
8. For a2 £ x 2 < b2 or a £ |x| < b
Û
a £ x < b or - b < x £ - a
i.e.,
x Î( - b, - a] È [ a, b)
9. For a2 < x 2 < b2 or a < |x| < b
+
Û
a < x < b or - b < x < - a
i.e.,
x Î( - b, - a) È ( a, b)
10. For ( x - a)( x - b) < 0 and a < b, then a < x < b
bn
x Î( a, b)
i.e.,
and if in all brackets before x positive sign and
expression has negative sign, then wave start from
right to left, beginning below the number line, i.e.
-
( x - a1 )k 1( x - a 2 )k 2 K( x - am )k m
( x - b1 )p 1 ( x - b 2 )p 2 K( x - bn )p n
, then
bn
–
4. If roots occur even times, then sign remain same
from right to left side of the roots and if roots
occur odd times, then sign will change from right to
left through the roots of
x = a 1 , a 2 , K, a m , b 1 , b 2 , K, b n .
5. The solution of f ( x ) > 0 or f ( x ) ³ 0 is the union
of all intervals in which we have put the plus sign
and the solution of f ( x ) < 0 or f ( x ) £ 0 is the
union of all intervals in which we have put the
minus sign.
11. If ( x - a) ( x - b) £ 0 and a < b,
then
a £ x £ b, x Î[ a, b]
12. If ( x - a)( x - b) > 0 and a < b, then x < a or x > b
i.e.,
x Î( - ¥, a) È ( b, ¥)
13. If ( x - a) ( x - b) ³ 0 and a < b,
then
x £ a or x ³ b
i.e.,
x Î( - ¥, a] È [ b, ¥)
y Example 26. Solve the inequality
( x + 3)( 3x - 2) 5 (7 - x ) 3 ( 5x + 8 ) 2 ³ 0.
Sol. We have,
Þ
Þ
( x + 3)(3x - 2)5 (7 - x )3 (5x + 8)2 ³ 0
- ( x + 3)(3x - 2)5 ( x - 7 )3 (5x + 8)2 ³ 0
( x + 3) ( 3x - 2) 5 ( x - 7 ) 3 ( 5x + 8) 2 £ 0
[take before x , + ve sign in all brackets]
–3
–
+
–8
5
+
+
2
3
–
7
æ 8ö 2
The critical points are ( - 3), ç - ÷, , 7.
è 5ø 3
é 2 ù ì 8ü
Hence, x Î ( - ¥, - 3] È ê , 7 ú È í - ý.
ë 3 û î 5þ
118
Textbook of Algebra
y Example 27. Solve the inequality
971
1ö
æ
( x - 2)10000 ( x + 1) 253 ç x - ÷ ( x + 8 ) 4
è
2ø
500
x
( x - 3) 75 ( x + 2) 93
+
–2
³0
971
1ö
æ
( x - 2)10000 ( x + 1)253 ç x - ÷ ( x + 8)4
è
2ø
Sol. We have,
³0
500
75
x ( x - 3) ( x + 2)93
1
The critical points are ( - 8), ( - 2), ( - 1), 0, , 2, 3.
2
[Q x ¹ - 2, 0, 3 ]
+
+
–
–1
–2
3
–
æ 2 1ö
Hence, x Î ( - 2, - 1) È ç - , - ÷.
è 3 2ø
y Example 30. For x ÎR, prove that the given
x 2 + 34 x - 71
expression 2
cannot lie between 5 and 9.
x + 2x - 7
Sol. Let
x 2 + 34 x - 71
x 2 + 2x - 7
=y
+
+
+
+
+
–2
–8
–
–1
+
2
1
2
0
–
3
æ 1ù
Hence, x Î ( - ¥, - 8] È [ - 8, - 2) È [ - 1, 0) È ç0, ú È (3, ¥ )
è 2û
or
æ 1ù
x Î ( - ¥, - 2) È [ - 1, 0) È ç0, ú È ( 3, ¥ )
è 2û
y Example 28. Let f ( x ) =
( x - 3)( x + 2)( x + 6 )
.
( x + 1)( x - 5)
Find intervals, where f ( x ) is positive or negative.
f (x ) =
Sol. We have,
( x - 3)( x + 2)( x + 6)
( x + 1)( x - 5)
The critical points are ( - 6), ( - 2), ( - 1), 3, 5
+
–6
–
–2
+
–
–1
+
3
–
5
+
–
5
Þ
y Example 29. Find the set of all x for which
2x
1
.
>
( 2x 2 + 5x + 2) ( x + 1)
2x
1
>
2
( 2x + 5x + 2) ( x + 1)
Sol. We have,
Þ
2x
1
>0
( x + 2)(2x + 1) ( x + 1)
Þ
( 2x 2 + 2x ) - ( 2x 2 + 5x + 2)
>0
( x + 2)( x + 1)(2x + 1)
Þ
or
-
( 3x + 2)
>0
( x + 2)( x + 1)(2x + 1)
( 3x + 2)
<0
( x + 2)( x + 1)(2x + 1)
æ 2ö æ 1ö
The critical points are ( - 2), ( -1), ç - ÷, ç - ÷.
è 3ø è 2ø
–
9
x 2 (y - 1) + (2y - 34 )x + 71 - 7y = 0
For real values of x , discriminant ³ 0
\
(2y - 34 )2 - 4(y - 1)(71 - 7y ) ³ 0
Þ
8y 2 - 112y + 360 ³ 0
Þ
y 2 - 14y + 45 ³ 0
Þ
(y - 9 )(y - 5) ³ 0
Þ
y Î ( - ¥, 5] È [9, ¥ )
Hence, y can never lie between 5 and 9.
y Example 31. For what values of the parameter k in
½ x 2 + kx + 1 ½
½< 3, satisfied for all real
the inequality ½ 2
½ x + x+1 ½
values of x ?
Sol. We have,
For f ( x ) > 0, " x Î ( - 6, - 2) È ( - 1, 3) È (5, ¥ )
For f ( x ) < 0, " x Î ( - ¥, - 6) È ( - 2, - 1) È (3, 5)
–1
2
Þ
½ x 2 + kx + 1 ½
½< 3
½
2
½ x + x +1½
x 2 + kx + 1
<3
-3< 2
x + x +1
2
Since,
1ö
3
æ
x 2 + x + 1 = çx + ÷ + > 0
è
2ø
4
\
- 3( x 2 + x + 1) < x 2 + kx + 1 < 3( x 2 + x + 1)
\
4 x 2 + ( k + 3) x + 4 > 0
…(i)
…(ii)
and
2x 2 - ( k - 3) x + 2 > 0
Q
4 > 0 and 2 > 0
The inequality (i) will be valid, if
(k + 3)2 - 4 × 4 × 4 < 0 Þ (k + 3)2 < 64
or
-8<k +3<8
or
…(iii)
- 11 < k < 5
and the inequality (ii) will be valid, if
(k - 3)2 - 4 × 2 × 2 < 0 or (k - 3)2 < 16
or
- 4 <k -3< 4
or
...(iv)
-1<k <7
The conditions (iii) and (iv) will hold simultaneously, if
-1<k <5
Chap 02 Theory of Equations
Condition for Resolution
into Linear Factors
y Example 34. Find the linear factors of
x 2 - 5xy + 4 y 2 + x + 2y - 2.
Sol. Given expression is
x 2 - 5xy + 4y 2 + x + 2y - 2
Its corresponding equation is
x 2 - 5xy + 4y 2 + x + 2y - 2 = 0
or
x 2 - x (5y - 1) + 4y 2 + 2y - 2 = 0
The quadratic function
f ( x , y ) = ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c
may be resolved into two linear factors, iff
D = abc + 2 fgh - af 2 - bg 2 - ch 2 = 0
½a
½h
½
½g
i.e.,
h
b
f
resolved into two rational linear factors.
Sol. Comparing the given expression with
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c , we have
æ
(12)(2)(m ) + 2 ç è
5 ö æ 11 ö
æ
÷ ç ÷ ( - 5) - (12) ç è
ø
è
ø
2 2
5ö
÷
2ø
2
…(i)
é æx ö
æ x ö æy öù
æy ö
æy ö
æx ö
= z êa ç ÷ + b ç ÷ + c + 2a ç ÷ + 2b ç ÷ + 2c ç ÷ ç ÷ ú
è
è z ø è z øú
ø
è
ø
è
ø
è
ø
z
z
z
êë z
û
= z 2 [aX 2 + bY 2 + c + 2aY + 2bX + 2cXY ]
x
y
where,
= X and
=Y
z
z
Expression (i) will have two rational linear factors in x , y
and z, if expression
2
2
2
2
aX + bY + 2cXY + 2bX + 2aY + c will have two linear
factors, if
abc + 2abc - aa 2 - bb 2 - cc 2 = 0
or
a 3 + b 3 + c 3 = 3abc
– b ,– D
2 a 4a
a>0
Sol. Given expression is
2
Let f ( x ) = ax 2 + bx + c , a, b, c Î R, a ¹ 0 and a, b be the
roots of f ( x ) = 0. Suppose k, k 1 , k 2 Î R and k 1 < k 2 . Then,
the following hold good :
(If both the roots of f ( x ) = 0 are less than k)
y Example 33. If the expression
ax 2 + by 2 + cz 2 + 2ayz + 2bzx + 2cxy can be resolved
into two rational factors, prove that
a 3 + b 3 + c 3 = 3abc .
ax 2 + by 2 + cz 2 + 2ayz + 2bzx + 2cxy
(Interval in which Roots Lie)
1. Conditions for Number k
2
æ 11 ö
- (2) ç ÷ - (m )( - 5)2 = 0
è2ø
121
275
- 75 - 25 m = 0 or m = 2
24 m +
2
2
Þ
x=
Location of Roots
11
æ 5ö
, f = ç - ÷, c = m
è 2ø
2
The given expression will have two linear factors, if and
only if
abc + 2 fgh - af 2 - bg 2 - ch 2 = 0
or
…(i)
(5y - 1) ± (5y - 1)2 - 4 × 1 × ( 4y 2 + 2y - 2)
2
(5y - 1) ± (9y 2 - 18y + 9 )
=
2
(5y - 1) ± (3y - 3)2
=
2
(5y - 1) ± (3y - 3)
=
= 4y - 2, y + 1
2
\ The required linear factors are ( x - 4y + 2) and ( x - y - 1).
\
g½
f ½= 0
½
c½
y Example 32. Find the value of m for which the
expression 12x 2 - 10xy + 2y 2 + 11x - 5y + m can be
a = 12, h = - 5, b = 2, g =
119
α
f (k)
– b
2a
β k
β k
α
– b
2a
X-axis
– b ,– D
2a 4a
X-axis
f (k )
a<0
(i) D ³ 0 (roots may be equal)
(ii) af (k ) > 0
b
(iii) k > - , where a £ b.
2a
y Example 35. Find the values of m, for which both
roots of equation x 2 - mx + 1 = 0 are less than unity.
Sol. Let f ( x ) = x 2 - mx + 1, as both roots of f ( x ) = 0 are less
b
than 1, we can take D ³ 0, af (1) > 0 and < 1.
2a
+
–2
+
–
2
120
Textbook of Algebra
(i) Consider D ³ 0 ( - m )2 - 4 × 1 × 1 ³ 0
Þ
Þ
(m + 2)(m - 2) ³ 0
m Î ( - ¥, - 2] È [2, ¥ )
…(i)
(ii) Consider af (1) > 0 1(1 - m + 1) > 0
Þ
Þ
m -2<0 Þ m <2
m Î ( - ¥, 2)
æ b
ö
(iii) Consider ç < 1÷
è 2a
ø
…(ii)
m
< 1 Þm < 2
2
…(iii)
Þ
m Î ( - ¥, 2)
Hence, the values of m satisfying Eqs. (i), (ii) and (iii)
at the same time are m Î ( - ¥, - 2].
k α
ö
æ b
(iii) Consider ç > 3÷
ø
è 2a
6m
>3
2
Þ
m >1
…(iii)
Þ
m Î (1, ¥ )
Hence, the values of m satisfying Eqs. (i), (ii) and (iii)
æ 11 ö
at the same time are m Î ç , ¥ ÷.
ø
è9
– b ,– D
2a 4a
– b ,– D
2a 4 a
X-axis
β
f (k)
– b ,– D
2a 4 a
a>0
– b
2a
X-axis
X-axis
α
a<0
between roots of the equation x 2 + 2(p - 3)x + 9 = 0.
Sol. Let f ( x ) = x 2 + 2( p - 3)x + 9, as 6 lies between the roots
of f ( x ) = 0, we can take D > 0 and af (6) < 0
(i) Consider D > 0
(i) Consider D ³ 0
( - 6m )2 - 4 × 1(9m 2 - 2m + 2) ³ 0 Þ 8m - 8 ³ 0
m ³ 1 or m Î [1, ¥ )
(ii) Consider a f (3 ) ³ 0
1 × (9 - 18m + 9m 2 - 2m + 2) > 0
+
11/9
Þ
9m 2 - 20m + 11 > 0
Þ
(9m - 11)(m - 1) > 0
+
+
As both roots of f ( x ) = 0 are greater than 3, we can take
b
D ³ 0, af (3) > 0 and> 3.
2a
–
(ii) af (k ) < 0, where a < b
y Example 37. Find all values of p, so that 6 lies
Sol. Let f ( x ) = x 2 - 6mx + 9m 2 - 2 m + 2
1
a<0
(i) D > 0
y Example 36. For what values of m ÎR , both roots of
the equation x 2 - 6mx + 9m 2 - 2m - 2 = 0 exceed 3?
+
X-axis
β
f (k )
β
– b ,– D
2a 4a
(i) D ³ 0 (roots may be equal)
(ii) af (k ) > 0
b
(iii) k < - , where a £ b.
2a
\
f(k)
α
β
k α
– b
2a
…(ii)
If k lies between the roots of f ( x ) = 0
If both the roots of f ( x ) = 0 are greater than k
f (k )
æ 11 ö
m Î ( - ¥, 1) È ç , ¥ ÷
ø
è9
Þ
3. Conditions for a Number k
2. Conditions for a Number k
a>0
11 ö
æ
çm - ÷ ( m - 1) > 0
è
9ø
Þ
…(i)
0
–
6
2
{2 ( p - 3)} - 4 × 1 × 9 > 0
Þ
( p - 3) 2 - 9 > 0
Þ
Þ
p ( p - 6) > 0
p Î ( - ¥, 0) È (6, ¥ )
…(i)
(ii) Consider a f (6 ) < 0
Þ
Þ
1 × {36 + 12( p - 3) + 9 } < 0
3
12p + 9 < 0 Þ p + < 0
4
3ö
æ
p Î ç - ¥, - ÷
è
4ø
…(ii)
Hence, the values of p satisfying Eqs. (i) and (ii) at the
3ö
æ
same time are p Î ç - ¥, - ÷.
è
4ø
121
Chap 02 Theory of Equations
4. Conditions for Numbers k1 and k2
If exactly one root of f ( x ) = 0 lies in the interval (k 1 , k 2 )
f(k2)
k1 α
a>0
f(k1)
k2
f(k1)
f (k 2 )
k1 α
k2
β
β
X-axis
a<0
X-axis
( - 2) 2 - 4 × 4 × a ³ 0 Þ a £
Sol. Let f ( x ) = x 2 - 2mx + m 2 - 1, as exactly one root of
f ( x ) = 0 lies in the interval ( -2, 4 ), we can take D > 0 and
f ( - 2) f ( 4 ) < 0.
(i) Consider D > 0
( - 2 m )2 - 4 × 1(m 2 - 1) > 0 Þ 4 > 0
m ÎR
…(i)
(ii) Consider f ( - 2 ) f ( 4 ) < 0
4 ( 4 - 2 + a) > 0 Þ a > - 2
…(iii)
Þ
a Î ( - 2, ¥ )
Hence, the values of a satisfying Eqs. (i), (ii) and (iii) at
1ù
æ
the same time are
a Î ç - 2, ú.
è
4û
a>0
+
+
–1
3
–
α
5
…(ii)
\
m Î ( - 3 , - 1) È (3, 5)
Hence, the values of m satisfying Eqs. (i) and (ii) at the
same time are m Î ( - 3, - 1) È (3, 5).
5. Conditions for Numbers k1 and k2
(If both roots f ( x ) = 0 are confined between k 1 and k 2 )
–
a>0
f (k2)
f (k1)
k1 a
b k2
– b ,– D
2 a 4a
f(k1)
k1
k2
f(k2)
f(k1 )
β
X-axis
β
α
k1
f (k2)
k2
X-axis
a<0
(i) D > 0
(ii) af (k 1 ) < 0
(iii) af (k 2 ) < 0, where a < b.
y Example 40. Find the values of a for which one
root of equation (a - 5)x 2 - 2ax + a - 4 = 0 is smaller
b ,– D
2a 4a
b k2
k1 a
…(ii)
(iii) Consider a f (1 ) > 0
(m + 1)(m + 3)(m - 3)(m - 5) < 0
(m + 3)(m + 1)(m - 3)(m - 5) < 0
–
4( 4 + 2 + a ) > 0
a > - 6 Þ a Î ( - 6, ¥ )
Þ
(If k 1 and k 2 lie between the roots of f ( x ) = 0)
(m 2 + 4m + 3) (m 2 - 8m + 15) < 0
+
–3
…(i)
6. Conditions for Numbers k1 and k2
( 4 + 4m + m 2 - 1) (16 - 8m + m 2 - 1) < 0
Þ
Þ
1
4
(ii) Consider a f ( - 1 ) > 0
y Example 38. Find the values of m, for which exactly
one root of the equation x 2 - 2mx + m 2 - 1 = 0 lies in
the interval ( - 2, 4 ).
Þ
Sol. Let f ( x ) = 4 x 2 - 2x + a as both roots of the equation,
f ( x ) = 0 are lie between ( - 1, 1), we can take D ³ 0,
1
af ( - 1) > 0, af (1) > 0 and - 1 < < 1.
4
(i) Consider D ³ 0
(i) D > 0
(ii) f (k 1 ) f (k 2 ) < 0, where a < b.
\
y Example 39. Find all values of a for which the
equation 4 x 2 - 2x + a = 0 has two roots lie in the
interval ( - 1, 1).
X-axis
f (k1)
f(k2 )
a<0
(i) D ³ 0 (roots may be equal)
(ii) af (k 1 ) > 0
(iii) af (k 2 ) > 0
b
(iv) k 1 < < k 2 , where a £ b and k 1 < k 2 .
2a
X-axis
than 1 and the other greater than 2.
Sol. The given equation can be written as
æa - 4 ö
æ 2a ö
x2 - ç
÷ = 0, a ¹ 5.
÷x + ç
èa - 5ø
èa - 5ø
æa - 4 ö
æ 2a ö
Now, let f ( x ) = x 2 - ç
÷
÷x + ç
èa - 5ø
èa - 5ø
As 1 and 2 lie between the roots of f ( x ) = 0, we can take
D > 0, 1 × f (1) < 0 and 1 × f (2) < 0.
122
Textbook of Algebra
i.e., (m - 1) (m - 9 ) ³ 0
\
m Î ( - ¥, 1] È [9, ¥ )
(i) Consider D > 0
2
æ æ 2a ö ö
æa - 4 ö
ç- ç
÷ >0
÷ ÷ - 4 ×1 × ç
èa - 5ø
è èa - 5øø
20 ö
æ
36 ça - ÷
è
9ø
>0
Þ
( a - 5) 2
20
or
a>
9
f (2 ) > 0
[Qa ¹ 5]
…(i)
(ii) Consider 1 × f (1 ) < 0
æ 2a ö æ a - 4 ö
9
> 0 or a > 5 …(ii)
12 - ç
÷ <0Þ
÷+ç
( a - 5)
èa - 5ø è a - 5 ø
(iii) Consider 1 × f (2 ) < 0
æa - 4ö
4a
+ç
÷ <0
(a - 5 ) è a - 5 ø
( 4a - 20 - 4a + a - 4 )
(a - 24 )
<0 Þ
<0
(a - 5 )
(a - 5 )
…(iii)
5 < a < 24
4-
Þ
or
i.e.,
4 - 2( m - 3) + m > 0
Þ
m < 10
\
m Î ( - ¥, 10)
and x-coordinate of vertex < 2
( m - 3)
i.e.,
<2 Þ m <7
2
\
m Î ( - ¥, 7 )
On combining Eqs. (i), (ii) and (iii), we get
m Î ( - ¥, 1]
y Example 41. Let x 2 - (m - 3)x + m = 0 (m ÎR ) be a
quadratic equation. Find the value of m for which
f(2)
2 α
i.e.
\
β
X-axis
(m - 1)(m - 9 ) ³ 0
m Î ( - ¥, 1] Î [9, ¥ )
…(i)
f (2 ) > 0
(ii) both the roots are greater than 2.
i.e.
4 - 2( m - 3) + m > 0
Þ
m < 10
\
m Î ( - ¥, 10)
and x-coordinate of vertex > 2
( m - 3)
i.e.,
>2 Þ m >7
2
\
m Î ( 7, ¥ )
On combining Eqs. (i), (ii) and (iii), we get
m Î[9, 10)
(iv) exactly one root lies in the interval (1, 2).
(v) both the roots lie in the interval (1, 2).
(vi) one root is greater than 2 and the other root is
smaller than 1.
(vii) atleast one root lie in the interval (1, 2).
(viii) atleast one root is greater than 2.
Here,
and
f ( x ) = x 2 - ( m - 3) x + m
a = 1, b = - (m - 3), c = m
D = b 2 - 4ac = (m - 3)2 - 4 m
= m 2 - 10m + 9 = (m - 1)(m - 9 )
b ( m - 3)
and x-coordinate of vertex = =
2a
2
(i) Both the roots are smaller than 2
D³0
a
b 2
X-axis
…(ii)
…(iii)
(iii) One root is smaller than 2 and the other root is
greater than 2
D > 0
a
i.e.,
\
2
f(2)
b
(m - 1)(m - 9 ) > 0
m Î ( - ¥, 1) È (9, ¥ )
f (2 ) < 0
f(2)
…(iii)
D³0
(i) both the roots are smaller than 2.
(iii) one root is smaller than 2 and the other root is
greater than 2.
…(ii)
(ii) Both the roots are greater than 2
Hence, the values of a satisfying Eqs. (i), (ii) and (iii) at the
same time are a Î(5, 24 ).
Sol. Let
…(i)
i.e. 4 - 2(m - 3) + m < 0
\
m > 10
X-axis
…(i)
Chap 02 Theory of Equations
\
m Î (10, ¥ )
On combining Eqs. (i) and (ii), we get
m Î (10, ¥ ).
…(ii)
(iv) Exactly one root lies in the interval (1, 2)
D > 0
f (1)
1a
2
f (2)
b
…(i)
α
f (1 ) f (2 ) < 0
( 1 - ( m - 3) + m ) ( 4 - 2( m - 3) + m ) < 0
Þ
4 ( - m + 10) < 0
Þ
m - 10 > 0 Þ m > 10
\
m Î (10, ¥ )
On combining Eqs. (i) and (ii), we get
m Î (10, ¥ )
…(ii)
i.e., (m - 1)(m - 9 ) > 0
\
m Î ( - ¥, 1) È (9, ¥ )
f (1 ) < 0
…(i)
[from (v) part]
m Îf
Hence, at least one root lie in the interval (1, 2)
m Î (10, ¥ ) È f or m Î (10, ¥ )
(viii) Atleast one root is greater than 2
…(i)
…(ii)
f (2 ) > 0
i.e., 4 - 2(m - 3) + m > 0 Þ m < 10
\
m Î ( - ¥, 10)
X-axis
Case II Both roots lie in the interval (1, 2).
X-axis
i.e.,
(m - 1)(m - 9 ) ³ 0
\
m Î ( - ¥, 1] È [9, ¥ )
f (1 ) > 0
i.e., (1 - (m - 3) + m ) > 0 Þ 4 > 0
\
m ÎR
f(2) β
(vii) At least one root lie in the interval (1, 2)
Case I Exactly one root lies in (1, 2)
[from (iv) part]
m Î (10, ¥ )
f (2)
β2
2
i.e., 4 < 0, which is not possible.
Thus, no such ‘m’ exists.
D³ 0
f (1)
1
f(1)
(v) Both the roots lie in the interval (1, 2)
1 α
1 < x -coordinate of vertex < 2
( m - 3)
i.e.,
<2
1<
2
Þ
2 < m - 3 < 4 or 5 < m < 7
…(iv)
\
m Î(5, 7 )
On combining Eqs. (i), (ii), (iii) and (iv), we get
m Îf
(vi) One root is greater than 2 and the other root is
smaller than 1 D > 0
X-axis
(m - 1)(m - 9 ) > 0
m Î ( - ¥, 1) È (9, ¥ )
i.e.,
\
123
…(iii)
Case I One root is smaller than 2 and the other root
is greater than 2.
Then,
[from (iii) part]
m Î (10, ¥ )
Case II Both the roots are greater than 2, then
m Î[9, 10).
Hence, atleast one root is greater than 2.
\ m Î (10, ¥ ) È [9, 10) or m Î [9, 10) È (10, ¥ )
124
Textbook of Algebra
#L
1.
Exercise for Session 3
If x is real, the maximum and minimum values of expression
(a) 4, - 5
2.
If x is real, the expression
1
(a) æç ,
è 13
3.
1ö
÷
3ø
(c)
x +2
takes all values in the interval
(2x 2 + 3x + 6)
1
(b) é êë 13 ,
1ù
3 úû
(b) 0
1
2
1
4
If the inequality
1
2
(b) - 1 < m < 5
71
(d) m <
24
( x 2 - 1)
> 0, is
( x - 2)( x - 3)
(b) - 3
(d) - 1
(b) 5
(d) 9
If c > 0 and 4a + c < 2b , then ax 2 - bx + c = 0 has a root in the interval
(a) (0, 2)
(c) (0, 1)
(b) (2, 4)
(d) (- 2, 0)
If the roots of the equation x 2 - 2ax + a 2 + a - 3 = 0 are less than 3 then
(a) a < 2
(c) 3 < a £ 4
10.
(d) 2
If the expression 2x 2 + mxy + 3y 2 - 5y - 2 can be resolved into two rational factors, the value of | m | is
(a) 3
(c) 7
9.
(d) None of these
mx 2 + 3x + 4
< 5 is satisfied for all x Î R then
x 2 + 2x + 2
The largest negative integer which satisfies
(a) - 4
(c) - 2
8.
(c) 1
(d)
(c) 1 < m < 6
7.
1 1
(c) æç - , ö÷
è 3 13 ø
(b) 0
(a) 1 < m < 5
6.
(d) - 4, - 5
1ö
æ
If the expression çmx - 1 + ÷ is non-negative for all positive real x , the minimum value of m must be
è
xø
(a) -
5.
(c) - 4, 5
If x be real, then the minimum value of x 2 - 8x + 17, is
(a) - 1
4.
(b) 5, - 4
x 2 + 14x + 9
will be
x 2 + 2x + 3
(b) 2 £ a £ 3
(d) a > 4
The set of values of a for which the inequation x 2 + ax + a 2 + 6a < 0 is satisfied for all x Î(1, 2) lies in the
interval
(a) (1, 2)
(c) [- 7, 4]
(b) [1, 2]
(d) None of these
Session 4
Equations of Higher Degree, Rational Algebraic
Inequalities, Roots of Equation with the Help of Graphs,
Equations of Higher Degree
The equation a 0 x n + a 1 x n - 1 + a 2 x n - 2
l
l
l
l
+ K+ a n - 1 x + a n = 0,
where a 0 , a 1 , a 2 , ... , a n - 1 , a n are constants but a 0 ¹ 0, is a
polynomial equation of degree n. It has n and only n roots.
Let a 1 , a 2 , a 3 , K, a n - 1 , a n be n roots, then
a
S a 1 = a 1 + a 2 + a 3 + K + a n - 1 + a n = (- 1 )1 1
a0
[sum of all roots]
S a1 a 2 = a1a 2 + a1a 3 + K + a1an + a 2 a 3 +
K + a 2 an +K + an - 1 an
2 a2
[sum of products taken two at a time]
= ( - 1)
a0
a
S a 1 a 2 a 3 = (- 1 ) 3 3
a0
[sum of products taken three at a time]
a
a 1 a 2 a 3 K a n = (- 1 )n n
[ product of all roots]
a0
ap
In general, Sa 1 a 2 a 3 K a p = ( - 1) p
a0
Remark
1. A polynomial equation of degree n has n roots (real or
imaginary).
2. If all the coefficients, i.e., a0, a1, a2, K, an are real, then the
imaginary roots occur in pairs, i.e. number of imaginary roots
is always even.
3. If the degree of a polynomial equation is odd, then atleast one
of the roots will be real.
4. ( x - a1 )( x - a2 )( x - a3 ) K ( x - an )
= x n + ( - 1)1 S a1 × x n - 1 + ( - 1) 2 S a1 a2 × x n - 2
+ K + ( - 1) n a1 a2 a3 K an
In Particular
(i) For n = 3, if a, b, g are the roots of the equation
ax 3 + bx 2 + cx + d = 0, where a, b, c , d are constants
b
b
and a ¹ 0, then Sa = a + b + g = ( - 1) 1 = - ,
a
a
c c
Sab = ab + bg + g a = ( - 1) 2 =
a a
abg = ( - 1) 3
and
d
d
=a
a
or ax 3 + bx 2 + cx + d = a( x - a )( x - b)( x - g )
= a( x 3 - S a × x 2 + S ab × x - abg )]
(ii) For n = 4, if a, b, g, d are the roots of the equation
ax 4 + bx 3 + cx 2 + dx + e = 0, where a, b, c , d , e are
constants and a ¹ 0, then
b
b
S a = a + b + g + d = ( - 1) 1 = - ,
a
a
c c
S ab = (a + b)( g + d) + ab + gd = ( - 1) 2 = ,
a a
d
d
S abg = ab( g + d) + g d (a + b) = ( - 1) 3 = a
a
e
e
and abgd = ( - 1) 4 =
a a
4
3
or ax + bx + cx 2 + dx + e = a( x - a )
( x - b)( x - g )( x - d)
= a( x 4 - S a × x 3 + S ab × x 2 - S abg × x + abgd)
y Example 42. Find the conditions, if roots of the
equation x 3 - px 2 + qx - r = 0 are in
(i) AP
(ii) GP
(iii) HP
Sol. (i) Let roots of the given equation are
A - D , A , A + D , then
p
3
Now, A is the roots of the given equation, then it must
be satisfy
A 3 - pA 2 + qA - r = 0
A-D+A+A+D=p Þ A=
3
2
Þ
æp ö
æp ö
æp ö
ç ÷ - pç ÷ + qç ÷ - r = 0
è3ø
è3ø
è3ø
Þ
p 3 - 3p 3 + 9qp - 27r = 0
or
2p 3 - 9 pq + 27r = 0,
which is the required condition.
126
Textbook of Algebra
(ii) Let roots of the given equation are
Product of roots = (a - b ) ×a × (a + b ) = 6
Þ ( 2 - b ) 2( 2 + b ) = 6 Þ 4 - b 2 = 3
A
, A , AR , then
R
\
b = ±1
\ Roots of Eqs. (i) are 1, 2, 3 or 3, 2, 1.
1 1 1 1
Hence, roots of the given equation are 1, , or , , 1.
2 3 3 2
A
æ rö
× A × AR = ( - 1)3 × ç - ÷ = r
è 1ø
R
A3 = r
Þ
1
Þ
A =r3
y Example 44. If a , b, g are the roots of the equation
x 3 - px 2 + qx - r = 0, find
Now, A is the roots of the given equation, then
A 3 - pA 2 + qA - r = 0
Þ
(i) åa 2 .
r - p (r )2 / 3 = q (r )1/ 3 - r = 0
or
p (r )2 / 3 = q (r )1/ 3
or
p 3r 2 = q 3r
or
p 3r = q 3
which is the required condition.
(iii) Given equation is
x 3 - px 2 + qx - r = 0
1
On replacing x by in Eq. (i), then
x
3
Þ a 2 + b 2 + g 2 + 2 (ab + bg + ga ) = p 2
…(i)
Þ a 2 b + abg + a 2 g + b 2 a + b 2 g + abg
Þ (a 2 b + a 2 a + b 2 g + b 2 g + g 2 a + g 2 b )
Þ
2q 3 - 9 pqr + 27r 2 = 0,
(iii) Q å a 2 × å a = ( p 2 - 2q ) × p
2
y
Þ
11
y
2
+
Now, roots of Eq. (i) are in AP.
Let the roots be a - b, a , a + b.
Then, sum of roots = a - b + a + a + b = 6
Þ
3a = 6
\
a =2
3
+ g 2 a + g 2 b ) = p 3 - 2pq
Þ
Þ
6
-1=0
y
y 3 - 6y 2 + 11y - 6 = 0
[from result (i)]
2
Þ a 3 + b 3 + g 3 + (a 2 b + a 2 g + b 2 a + b 2 g
1
Sol. Put x = in the given equation, then
y
-
2
Þ (a + b + g )(a + b + g ) = p - 2pq
y Example 43. Solve 6 x 3 - 11x 2 + 6 x - 1 = 0, if roots of
the equation are in HP.
3
å a 2 b = pq - 3r
or
which is the required condition.
6
+ 3abg = pq
å a 2 b + 3r = pq
or
2
q 3 - 3q 3 + 9 pqr - 27r 2 = 0
+ g 2 b + g 2 a = pq
…(ii)
æq ö
æq ö
æq ö
Þ r ç ÷ - qç ÷ + pç ÷ - 1 = 0
è 3r ø
è 3r ø
è 3r ø
Þ
å a 2 = p 2 - 2q
(ii) Q å a × å ab = p × q
Þ
(a + b + g ) × (ab + bg + ga ) = pq
Now, roots of Eq. (ii) are in AP.
Let roots of Eq. (ii) are A - P , A , A + P , then
q
q
or A =
A-P+A+A+P=
r
3r
Q A is a root of Eq. (ii), then
rA 3 - qA 2 + pA - 1 = 0
3
å a 2 + 2 å ab = p 2
or
or
æ1ö
æ1ö
æ1ö
ç ÷ - pç ÷ + qç ÷ - r = 0
èx ø
èx ø
èx ø
rx 3 - qx 2 + px - 1 = 0
(iii) åa 3 .
Sol. Since, a , b, g are the roots of x 3 - px 2 + qx - r = 0.
\
å a = p , å a b = q and abg = r
(i) Q å a × å a = p × p
Þ
(a + b + g )(a + b + g ) = p 2
2
Þ
(ii) åa 2 b.
or
å a 3 + å a 2 b = p 3 - 2pq
å a 3 + pq - 3r = p 3 - 2pq
3
[from result (ii)]
3
å a = p - 3pq + 3r
y Example 45. If a , b, g are the roots of the cubic
equation x 3 + qx + r = 0 , then find the equation whose
roots are (a - b ) 2 , (b - g ) 2 , ( g - a ) 2 .
Sol. Qa , b, g are the roots of the cubic equation
…(i)
x 3 + qx + r = 0
Then,
å a = 0, å ab = q, abg = - r
If y is a root of the required equation, then
y = (a - b )2 = (a + b )2 - 4 ab
4 abg
= (a + b + g - g )2 g
…(i)
…(ii)
Chap 02 Theory of Equations
= (0 - g )2 +
Þ
\
or
4r
g
[from Eq. (ii)]
4r
g
[replacing g by x which is a root of Eq. (i)]
4r
y = x2 +
x
…(iii)
x 3 - yx + 4r = 0
y = g2 +
The required equation is obtained by eliminating x between
Eqs. (i) and (iii).
Now, subtracting Eq. (iii) from Eq. (i), we get
(q + y ) x - 3r = 0
3r
or
x=
q +y
On substituting the value of x in Eq. (i), we get
æ 3r ö
÷
ç
èq + y ø
Thus,
3
æ 3r ö
+ qç
÷ +r =0
èq + y ø
y 3 + 6qy 2 + 9q 2y + ( 4q 3 + 27r 2 ) = 0
which is the required equation.
Remark
å( a - b ) 2 = - 6q,
Õ( a - b ) 2 = - ( 4 q3 + 27r 2 )
Some Results on Roots of a
Polynomial Equation
1. Remainder Theorem If a polynomial f ( x ) is
divided by a linear function x - l, then the remainder
is f ( l),
i.e. Dividend = Divisor ´ Quotient + Remainder
Let Q ( x ) be the quotient and R be the remainder, thus
f ( x ) = ( x - l) Q ( x ) + R
Þ
f ( l) = ( l - l) Q ( l) + R = 0 + R = R
y Example 46. If the expression 2x 3 + 3px 2 - 4 x + p
has a remainder of 5 when divided by x + 2, find the
value of p.
Sol. Let
f ( x ) = 2x 3 + 3px 2 - 4 x + p
Q
f ( x ) = ( x + 2) Q ( x ) + 5
Þ
f ( - 2) = 5
Þ 2( - 2)3 + 3p ( - 2)2 - 4( - 2) + p = 5 or 13p = 13
\
p =1
2. Factor Theorem Factor theorem is a special case of
Remainder theorem.
Let f (x ) = ( x - l) Q ( x ) + R = ( x - l) Q (x ) + f ( l)
If f ( l) = 0, f (x ) = ( x - l) Q (x ), therefore f (x ) is
exactly divisible by x - l.
127
or
If l is a root of the equation f ( x ) = 0, then f ( x ) is
exactly divisible by ( x - l) and conversely, if f ( x ) is
exactly divisible by ( x - l), then l is a root of the
equation f (x ) = 0 and the remainder obtained is f ( l).
y Example 47. If x 2 + ax + 1 is a factor of
ax 3 + bx + c , find the conditions.
Sol. Qax 3 + bx + c = ( x 2 + ax + 1)Q (x )
Let
then
Q ( x ) = Ax + B,
ax 3 + bx + c = ( x 2 + ax + 1)( Ax + B )
On comparing coefficients of x 3 , x 2 , x and constant on
both sides, we get
…(i)
a = A,
…(ii)
0 = B + aA,
…(iii)
b = aB + A ,
and
…(iv)
c =B
From Eqs. (i) and (iv), we get
A = a and B = c
From Eqs. (ii) and (iii), a 2 + c = 0 and b = ac + a are the
required conditions.
y Example 48. A certain polynomial f ( x ), x ÎR, when
divided by x - a, x - b and x - c leaves remainders a, b
and c, respectively. Then, find the remainder when f ( x )
is divided by ( x - a )( x - b )( x - c ), where a, b , c are
distinct.
Sol. By Remainder theorem f (a ) = a, f (b ) = b and f (c ) = c
Let the quotient be Q ( x ) and remainder is R( x ).
\
f ( x ) = ( x - a )( x - b )( x - c )Q ( x ) + R( x )
\
f (a ) = 0 + R(a ) Þ R(a ) = a
f (b ) = 0 + R(b ) Þ R(b ) = b and f (c ) = 0 + R(c )
Þ
R(c ) = c
So, the equation R( x ) - x = 0 has three roots a, b andc . But
its degree is atmost two. So, R( x ) - x must be zero
polynomial (or identity).
Hence, R( x ) = x .
3. Every equation of an odd degree has atleast one real
root, whose sign is opposite to that of its last term,
provided that the coefficient of the first term is
positive.
4. Every equation of an even degree has atleast two real
roots, one positive and one negative, whose last term
is negative, provided that the coefficient of the first
term is positive.
5. If an equation has no odd powers of x, then all roots
of the equation are complex provided all the
coefficients of the equation have positive sign.
128
Textbook of Algebra
A (λ , f (λ))
6. If x = a is root repeated m times in f ( x ) = 0
( f ( x ) = 0 is an nth degree equation in x), then
f ( x ) = ( x - a ) m g(x )
where,g( x )isa polynomialofdegree(n - m ) and theroot
x = a isrepeated(m - 1) timein f ¢ ( x ) = 0,(m - 2 ) times
in f ¢¢( x ) = 0,…,(m - (m - 1)) timesin f m - 1 (x ) = 0.
7. Let f (x ) = 0 be a polynomial equation and l, m are
two real numbers.
Then, f (x ) = 0 will have atleast one real root or an
odd number of roots between l and m, if f ( l) and
f (m ) are of opposite signs.
But if f ( l) and f (m ) are of same signs, then either
f ( x ) = 0 has no real roots or an even number of roots
between l and m.
Illustration by Graphs
Since, f (x ) be a polynomial in x, then graph of y = f (x )
will be continuous in every interval.
A (λ ,f ( λ))
f (λ)
(µ,0)
(γ, 0)
–
R
P (α, 0) (,β, 0)
–
Q
+ f (µ)
(δ,0)
S (µ ,0)
+
+
(,λ0)
– (α, 0)
f(λ)
(β, 0)
(δ, 0)
(γ ,0)
–
(µ, 0)
X-axis
–
f(µ)
A (λ , f (λ))
B ( µ , f ( µ))
(vi)
(a) In figure (i), (ii) and (iii), f ( l) and f (m ) have
opposite signs and equationf ( x ) = 0, has one,
three, five roots between l and m, respectively.
(b) In figure (iv), (v) and (vi), f ( l) and f (m ) have
same signs and equation f ( x ) = 0, has no, four
and four roots between l and m, respectively.
2
then
a x - bx - c = 0,
2
2
a b - bb - c = 0
2
Let
f ( x ) = a x + 2bx + 2c
\
f (a ) = a 2 a 2 + 2ba + 2c = a 2 a 2 - 2a 2 a 2
= -a a
X-axis
R (γ, 0)
Q
(β,0)
–
2
(µ, 0)
+
Þ
f ( µ)
B [µ, f (µ)]
(ii)
R
S
(δ,0)
– (γ, 0)
A (λ , f (λ))
–
(iii)
B( µ , f (µ))
f (λ) A [λ , f (λ)]
(λ, 0)
(iv)
+
f (µ)
( µ, 0)
f (a ) < 0 and f ( b ) = a 2 b 2 + 2b b + 2c
2
= 3a b
Þ
B (µ, f (µ))
+
+
2
= a 2 b 2 + 2a 2 b 2
–
Q
(β, 0)
…(ii)
2
[from Eq. (i)]
(λ, 0) P (α, 0)
–
f (λ) (α,0)
…(i)
2
+
( λ ,0)
X-axis
(v)
and b is a root of
B (µ , f ( µ))
(i)
A (λ ,f ( λ))
(λ, 0)
+
Sol. Since, a is a root of a 2 x 2 + bx + c = 0.
Then,
a 2 a 2 + ba + c = 0
f (µ)
–
f (λ)
X-axis
P (α, 0)
f (λ)
y Example 49. If a, b , c are real numbers, a ¹ 0. If a is
root of a 2 x 2 + bx + c = 0 , b is a root of
a 2 x 2 - bx - c = 0 and 0 < a < b, show that the
equation a 2 x 2 + 2bx + 2c = 0 has a root g that always
satisfies a < g < b.
+
(λ, 0)
B (µ , f (µ))
+
X-axis
+ f(µ)
T (µ, 0)
(ν, 0)
X-axis
[from Eq. (ii)]
2
f (b) > 0
Since, f (a ) and f ( b ) are of opposite signs, then it is clear
that a root g of the equation f ( x ) = 0 lies between a and b.
Hence,
[Qa < b]
a < g <b
y Example 50. If a < b < c < d , then show that
( x - a )( x - c ) + 3 ( x - b )( x - d ) = 0 has real and distinct
roots.
Sol. Let
f ( x ) = ( x - a )( x - c ) + 3( x - b )( x - d )
129
Chap 02 Theory of Equations
Then, f (a ) = 0 + 3(a - b )(a - d ) > 0 [Qa - b < 0, a - d < 0]
and f (b ) = (b - a )(b - c ) + 0 < 0
[Qb - a > 0, b - c < 0]
Thus, one root will lie between a and b.
and
f (c ) = 0 + 3(c - b )(c - d ) < 0 [Qc - b > 0, c - d < 0]
and f (d ) = (d - a )(d - c ) + 0 > 0
[Qd - a >0, d - c >0]
Thus, one root will lie between c and d. Hence, roots of
equation are real and distinct.
8. Let f ( x ) = 0 be a polynomial equation then
(a) the number of positive roots of a polynomial
equation f (x ) = 0 (arranged in decreasing order
of the degree) cannot exceed the number of
changes of signs in f ( x ) = 0 as we move from left
to right.
For example, Consider the equation
2 x 3 - x 2 - x + 1 = 0.
The number of changes of signs from left to right
is 2 (+ to -, then - to +). Then, number of positive
roots cannot exceed 2.
(b) The number of negative roots of a polynomial
equation f ( x ) = 0 cannot exceed the number of
changes of signs in f (-x ).
For example, Consider the equation
5x 4 + 3x 3 - 2x 2 + 5x - 8 = 0
Let f ( x ) = 4 x 4 + 3 x 3 - 2 x 2 + 5 x - 8
\ f ( - x ) = 5x 4 - 3x 3 - 2x 2 - 5x - 8
The number of changes of signs from left to right
is (+ to -). Then number of negative roots cannot
exceed 1.
(c) If equation f ( x ) = 0 have atmost r positive roots
and atmost t negative roots, then equation
f ( x ) = 0 will have atmost (r + t ) real roots, i.e. it
will have atleast n - (r + t ) imaginary roots,
where n is the degree of polynomial.
For example, Consider the equation
5x 6 - 8x 3 + 3x 5 + 5x 2 + 8 = 0
9. Rolle’s Theorem If f ( x ) is continuous function in
the interval [a, b ] and differentiable in interval (a, b )
and f (a ) = f (b ), then equation f ¢ ( x ) = 0 will have
atleast one root between a and b. Since, every
polynomial f ( x ) is always continuous and
differentiable in every interval. Therefore, Rolle’s
theorem is always applicable to polynomial function
in every interval [a, b ] if f (a ) = f (b ).
y Example 51. If 2a + 3b + 6c = 0 ; a, b , c ÎR, then show
that the equation ax 2 + bx + c = 0 has atleast one root
between 0 and 1.
Sol. Given, 2a + 3b + 6c = 0
a b
+ +c =0
3 2
f ¢ ( x ) = ax 2 + bx + c ,
Þ
Let
ax 3 bx 2
+
+ cx + d
3
2
a b
Now, f (0) = d and f (1) = + + c + d
3 2
[from Eq. (i)]
=0+d
Since, f ( x ) is a polynomial of three degree, then f ( x ) is
continuous and differentiable everywhere and f (0) = f (1),
then by Rolle’s theorem f ¢ ( x ) = 0 i.e., ax 2 + bx + c = 0 has
atleast one real root between 0 and 1.
f (x) =
Then,
Reciprocal Equation of the Standard
Form can be Reduced to an Equation
of Half Its Dimensions
Let the equation be
ax 2m + bx 2m - 1 + cx 2m - 2 + K + kx m + K + cx 2 + bx + a = 0
On dividing by x m , then
ax m + bx m - 1 + cx m - 2 + K + k + K +
The given equation can be written as
c
x
+
5x 6 + 3x 5 - 8x 3 + 5x 2 + 8 = 0
Let f ( x ) = 5 x 6 + 3 x 5 - 8 x 3 + 5 x 2 + 8
Here, f ( x ) has two changes in signs.
So, f ( x ) has atmost two positive real roots
and f ( - x ) = 5 x 6 - 3 x 5 + 8 x 3 + 5 x 2 + 8
Here, f ( - x ) has two changes in signs.
So, f ( x ) has atmost two negative real roots.
and x = 0 cannot be root of f ( x ) = 0.
Hence, f ( x ) = 0 has atmost four real roots,
therefore atleast two imaginary roots.
…(i)
On rearranging the terms, we have
m -2
b
x
m -1
+
a
xm
=0
1 ö
1 ö
æ
æ
+c
+ bçx m -1 +
a çx m +
m ÷
m -1 ÷
ø
è
ø
è
x
x
1 ö
æ m -2
+
÷ +K + k = 0
çx
m
è
x -2 ø
Now, x p + 1 +
1
x
p +1
1 öæ
1ö
æ
= çx p +
÷ çx + ÷
p
è
xø
x øè
1 ö
æ
- çx p -1 +
÷
p
è
x -1 ø
130
Textbook of Algebra
1
and given to p succession the
x
values 1, 2, 3,K, we obtain
1
x2 +
=z2 -2
2
x
1
3
= z (z 2 - 2 ) - z = z 3 - 3z
x +
3
x
1
4
= z (z 3 - 3z ) - (z 2 - 2 ) = z 4 - 4z 2 + 2
x +
4
x
1
is of m dimensions in
and so on and generally x m +
xm
z and therefore the equation in z is of m dimensions.
Hence, writing z for x +
y Example 52. Solve the equation
2x 4 + x 3 - 11x 2 + x + 2 = 0.
Sol. Since, x = 0 is not a solution of the given equation.
On dividing by x 2 in both sides of the given equation, we
get
1 ö æ
1ö
æ
…(i)
2 ç x 2 + 2 ÷ + ç x + ÷ - 11 = 0
è
xø
x ø è
1
Put x + = y in Eq. (i), then Eq. (i) reduce in the form
x
2(y 2 - 2) + y - 11 = 0
2y 2 + y - 15 = 0
5
\
y1 = - 3 and y 2 =
2
Consequently, the original equation is equivalent to the
collection of equations
1
ì
ïx + x = - 3
,
í
ïx + 1 = 5
x 2
î
-3- 5
-3+ 5
1
, x2 =
, x3 = , x4 = 2
we find that, x 1 =
2
2
2
Þ
Equations which can be Reduced
to Linear, Quadratic and Biquadratic
Equations
Type I An equation of the form
( x - a )( x - b )( x - c )( x - d ) = A
where, a < b < c < d , b - a = d - c , can be solved by a
change of variable.
(x - a ) + (x - b ) + (x - c ) + (x - d )
i.e.
y=
4
(a + b + c + d )
y =x 4
y Example 53. Solve the equation
(12x - 1)(6 x - 1)(4 x - 1)( 3x - 1) = 5.
Sol. The given equation can be written as
1 öæ
1öæ
1 öæ
1ö
5
æ
çx - ÷çx - ÷çx - ÷çx - ÷ =
ø
è
ø
è
ø
è
ø
è
12
6
4
3
12 × 6 × 4 × 3
1 1 1 1
1 1 1 1
Since,
< < < and = 12 6 4 3
6 12 3 4
We can introduced a new variable,
1 éæ
1ö æ
1ö æ
1ö æ
y = êçx - ÷ + çx - ÷ + çx - ÷ + çx è
ø
è
ø
è
4ë
12
6
4ø è
y=x-
...(i)
1öù
÷
3 ø úû
5
24
5
in Eq. (i), we get
24
1 öæ
3ö
5
- ÷ çy - ÷ =
24 ø è
24 ø 12 × 6× 4 × 3
On substituting x = y +
3 öæ
1 öæ
æ
çy + ÷ çy + ÷ çy
è
24 ø è
24 ø è
é 2 æ 1 ö2 ù é 2 æ 3 ö2 ù
5
êy - ç ÷ ú êy - ç ÷ ú =
è
ø
è
ø
×
× 4 ×3
24
24
12
6
úû
úû êë
êë
Hence, we find that
49
y2 = 2
24
7
7
i.e.
and y 2 = y1 =
24
24
Hence, the corresponding roots of the original equation are
1
1
and .
2
12
Þ
Type II An equation of the form
( x - a )( x - b )( x - c )( x - d ) = Ax 2
where, ab = cd can be reduced to a collection of two
quadratic equations by a change of variable y = x +
ab
.
x
y Example 54. Solve the equation
( x + 2)( x + 3)( x + 8 )( x + 12) = 4 x 2 .
Sol. Since, ( - 2)( - 12) = ( - 3)( - 8), so we can write given equation as
( x + 2)( x + 12)( x + 3)( x + 8) = 4 x 2
( x 2 + 14 x + 24 )( x 2 + 11x + 24 ) = 4 x 2
Þ
Now, x = 0 is not a root of given equation.
On dividing by x 2 in both sides of Eq. (i), we get
24
24
æ
öæ
ö
+ 14 ÷ ç x +
+ 11÷ = 4
çx +
è
øè
ø
x
x
…(i)
…(ii)
24
= y , then Eq. (ii) can be reduced in the form
x
(y + 14 )(y + 11) = 4 or y 2 + 25y + 150 = 0
Put x +
\
y1 = - 15 and y 2 = - 10
Chap 02 Theory of Equations
Thus, the original equation is equivalent to the collection of
equations
24
é
ê x + x = - 15,
ê
ê x + 24 = - 10,
ë
x
é x 2 + 15x + 24 = 0
i.e.
ê 2
êë x + 10x + 24 = 0
On solving these collection, we get
x1 =
- 15 - 129
- 15 + 129
, x2 =
, x 3 = - 6, x 4 = - 4
2
2
Type III An equation of the form ( x - a ) 4 + ( x - b ) 4 = A
can also be solved by a change of variable, i.e. making a
(x - a ) + (x - b )
substitution y =
.
2
y Example 55. Solve the equation
(6 - x ) 4 + (8 - x ) 4 = 16.
Sol. After a change of variable,
(6 - x ) + (8 - x )
y=
2
\
y = 7 - x or x = 7 - y
Now, put x = 7 - y in given equation, we get
(y - 1)4 + (y + 1)4 = 16
Þ
y 4 + 6y 2 - 7 = 0
Þ
(y 2 + 7 )(y 2 - 1) = 0
y2 + 7 ¹ 0
[y gives imaginary values]
\
y2 - 1 = 0
Then,
y1 = - 1 and y 2 = 1
Thus, x 1 = 8 and x 2 = 6 are the roots of the given equation.
Rational Algebraic Inequalities
Consider the following types of rational algebraic
inequalities
ìP ( x ) > 0, Q ( x ) > 0
P(x )
ï
(1)
> 0 Þ {P ( x ) Q ( x ) > 0 Þ í
or
Q(x )
ïP ( x ) < 0, Q ( x ) < 0
î
ìP ( x ) > 0, Q ( x ) < 0
P(x )
ï
(2)
or
< 0 Þ {P ( x ) Q ( x ) < 0 Þ í
Q(x )
ïP ( x ) < 0, Q ( x ) > 0
î
ìP ( x ) ³ 0, Q ( x ) > 0
ìP ( x ) Q ( x ) ³ 0 ï
P(x )
(3)
³ 0 Þí
or
Þí
Q(x )
î Q(x ) ¹ 0
ïP ( x ) £ 0, Q ( x ) < 0
î
ìP ( x ) ³ 0, Q ( x ) < 0
ìP ( x ) Q ( x ) £ 0 ï
P(x )
(4)
£ 0 Þí
Þí
or
Q(x )
î Q(x ) ¹ 0
ïP ( x ) £ 0, Q ( x ) > 0
î
y Example 56. Find all values of a for which the set
of all solutions of the system
ì x 2 + ax - 2
<2
ï
ï x2 - x +1
í 2
ï x + ax - 2 > - 3
ïî x 2 - x + 1
is the entire number line.
Sol. The system is equivalent to
ì x 2 - ( a + 2) x + 4
>0
ï
ï
x2 - x + 1
í 2
ï 4 x + ( a - 3) x + 1 > 0
ïî
x2 - x + 1
2
1ö
3
æ
Since, x 2 - x + 1 = ç x - ÷ + > 0, this system is
è
2ø
4
2
ì x - ( a + 2) x + 4 > 0
equivalent to í 2
î 4 x + ( a - 3) x + 1 > 0
Hence, the discriminants of the both equations of this
system are negative.
ì(a + 2)2 - 16 < 0
i.e.,
Þ (a + 6)(a - 2) < 0
í
2
î(a - 3) - 16 < 0
P(x )
P(x )
> 0,
< 0,
Q(x )
Q(x )
P(x )
P(x )
³ 0,
£0
Q(x )
Q(x )
If P ( x ) and Q ( x ) can be resolved in linear factors, then use
Wavy curve method, otherwise we use the following
statements for solving inequalities of this kind.
131
+
–6
i.e.,
Þ
–
+
2
x Î( - 6, 2)
(a + 1)(a - 7 ) < 0
+
–1
–
…(i)
+
7
i.e.
x Î ( -1, 7 )
Hence, from Eqs. (i) and (ii), we get
x Î ( -1, 2)
…(ii)
132
Textbook of Algebra
Equations Containing
Absolute Values
By definition, | x | = x , if x ³ 0 | x | = - x , if x < 0
y Example 57. Solve the equation x 2 - 5| x | + 6 = 0.
Sol. The given equation is equivalent to the collection of
systems
ì x 2 - 5x + 6 = 0, if x ³ 0
ì( x - 2)( x - 3) = 0, if x ³ 0
Þ í
í 2
î( x + 2)( x + 3) = 0, if x < 0
î x + 5x + 6 = 0, if x < 0
Hence, the solutions of the given equation are
x 1 = 2, x 2 = 3, x 3 = - 2, x 4 = - 3
½ x 2 - 8 x + 12 ½
x 2 - 8 x + 12
½= - 2
½ 2
.
x - 10x + 21
½ x - 10x + 21½
Sol. This equation has the form | f ( x )| = - f ( x )
x 2 - 8x + 12
x 2 - 10x + 21
such an equation is equivalent to the collection of systems
ì f ( x ) = - f ( x ), if f ( x ) ³ 0
í
î f ( x ) = f ( x ), if f ( x ) < 0
The first system is equivalent to f ( x ) = 0 and the second
system is equivalent to f ( x ) < 0 the combining both
systems, we get
f (x ) £ 0
x 2 - 8x + 12
£0
\
x 2 - 10x + 21
( x - 2)( x - 6)
Þ
£0
( x - 3)( x - 7 )
+
+
2
–
3
Important Forms Containing
Absolute Values
y Example 60. Solve the equation
2
½
½ x ½
½+ | x | = x .
| x - 1|
½ x - 1½
Sol. Let f ( x ) =
Then,
x
and g ( x ) = x ,
x -1
f (x ) + g(x ) =
x2
x
+x=
x -1
x -1
\ The given equation can be reduced in the form
| f ( x )| + | g ( x )| = | f ( x ) + g ( x )|
Hence,
Þ
f (x )× g(x ) ³ 0
x2
³0
x -1
+
+
6
–
–
7
y Example 59. Solve the equation
| x - | 4 - x || - 2x = 4.
Sol. This equation is equivalent to the collection of systems
ì| x - ( 4 - x )| - 2x = 4, if 4 - x ³ 0
í
î| x + ( 4 - x )| - 2x = 4, if 4 - x < 0
if x £ 4
ì| 2x - 4 | - 2x = 4,
í
if
4
2
x
=
4
,
x>4
î
The second system of this collection
gives
x =0
but
x>4
Hence, second system has no solution.
0
–
1
From Wavy curve method, x Î (1, ¥ ) È {0}.
Hence, by Wavy curve method,
x Î [2, 3) È [6, 7 )
Þ
The first system is failed and second system gives x = 0.
Hence, x = 0 is unique solution of the given equation.
Form 1 The equation of the form
| f ( x ) + g ( x )| = | f ( x )| + | g ( x )|
is equivalent of the system
f ( x ) g ( x ) ³ 0.
y Example 58. Solve the equation
when, f ( x ) =
The first system of collection Eq. (i) is equivalent to the
system of collection
ì 2x - 4 - 2x = 4, if 2x ³ 4
í
î - 2x + 4 - 2x = 4, if 2x < 4
ì - 4 = 4, if x ³ 2
Þ
í
î - 4 x = 0, if x < 2
…(i)
Form 2 The equation of the form
…(i)
| f 1 ( x )| + | f 2 ( x )| + K + | f n ( x )| = g ( x )
where, f 1 ( x ), f 2 ( x ), K, f n ( x ), g( x ) are functions of x and
g( x ) may be constant.
Equations of this form solved by the method of
intervals. We first find all critical points of
f 1 ( x ), f 2 ( x ), ..., f n ( x ), if coefficient of x is positive, then
graph start with positive sign (+) and if coefficient of x is
negative, then graph start with negative sign (-). Then,
using the definition of the absolute value, we pass from
Eq. (i) to a collection of systems which do not contain the
absolute value symbols.
Chap 02 Theory of Equations
y Example 61. Solve the equation
| x - 1| + | 7 - x | + 2| x - 2| = 4.
For
Sol. Here, critical points are 1, 2, 7 using the method of intervals, we find intervals when the expressions x - 1, 7 - x
and x - 2 are of constant signs.
i.e.
x < 1, 1 < x < 2, 2 < x < 7, x > 7
+
(x – 1)
\
–
1
0£ x £1
x +1
£ 0, if x < 0
x -1
For
+
–1
+
–
+
2
–
+
Thus, the given equation is equivalent to the collection of
four systems,
é ìx < 1
é ìx < 1
ê íx = 2
ê í - ( x - 1) + (7 - x ) - 2( x - 2) = 4
êî
ê î
ê ì1 £ x < 2
ê ì1 £ x < 2
êíx = 3
ê íî( x - 1) + (7 - x ) - 2( x - 2) = 4
Þ êî
ê
êìí2 £ x < 7
ê ìí2 £ x < 7
êî x = 1
ê î( x - 1) + (7 - x ) + 2( x - 2) = 4
êì x ³ 7
ê ìx ³ 7
êí
ê í
êëî x = 4
êë î( x - 1) - (7 - x ) + 2(x - 2) = 4
From the collection of four systems, the given equation has
no solution.
Inequations Containing
Absolute Values
| x | < a Þ - a < x < a (a > 0 )
|x | £a Þ-a £ x £a
| x | > a Þ x < - a and x > a
and
| x | ³ a Þ x £ - a and x ³ a.
|x| ½ 1
½
½³ .
y Example 62. Solve the inequation ½1 ½ 1 + | x |½ 2
By definition,
Sol. The given inequation is equivalent to the collection of
systems
ì½
x ½
1
ì 1
½
½ ³ 1 , if x ³ 0
ï 1ïï | 1 + x | ³ 2 , if x ³ 0
1 + x½ 2
ï½
Þ í
í
1
1
½ ³ 1 , if x < 0
½1 + x ½
ï½
ï
³ , if x < 0
ï
1 - x½ 2
ïî½
î |1 - x | 2
1
ì1- x
ì 1
ïï 1 + x ³ 0, if x ³ 0
ïï 1 + x ³ 2 , if x ³ 0
Þ í
Þ í
1
1
1+ x
ï
ï
³ 0, if x < 0
³ , if x < 0
ïî 1 - x
ïî 1 - x 2
ìx -1
ïï x + 1 £ 0, if x ³ 0
Þ
íx +1
ï
£ 0, if x < 0
ïî x - 1
…(i)
+
+
+
–
+
–1
+
7
( x – 2)
x -1
£ 0, if x ³ 0
x +1
+
1
–
(7 – x)
+
–
1
\
-1£ x <0
Hence, from Eqs. (i) and (ii), the solution of the given
equation is x Î [ - 1, 1].
Aliter
½
½³ 1 Þ ½
½1 - | x | ½
½³ 1
½ 1 ½
|
x
|
1
+
2
|
x
|
1
+
½
½
½ 2
½
Þ
\
133
…(ii)
1
1
Þ 1 + | x | £ 2 or | x | £ 1
³
1 + |x | 2
- 1 £ x £ 1 or x Þ [ -1, 1]
Equations Involving Greatest Integer,
Least Integer and Fractional Part
1. Greatest Integer
[x] denotes the greatest integer less than or equal to x i.e.,
[ x ] £ x . It is also known as floor of x.
Thus,
[3.5779] = 3, [0.89] = 0, [3 ] = 3
[ - 8.7285] = - 9
[ - 0.6] = - 1
[ - 7] = - 7
In general, if n is an integer and x is any real number
between n and n + 1.
i.e.
n £ x < n + 1, then [ x ] = n
Properties of Greatest Integer
(i) [ x ± n ] = [ x ] ± n, n Î I
(ii) [ - x ] = - [ x ], x Î I
(iii) [ - x ] = - 1 - [ x ], x Ï I
(iv) [ x ] - [ - x ] = 2n, if x = n, n Î I
(v) [ x ] - [ - x ] = 2n + 1, if x = n + { x }, n Î I and 0 < { x } < 1
(vi) [ x ] ³ n Þ x ³ n, n Î I
(vii) [ x ] > n Þ x ³ n + 1, n Î I
(viii)[ x ] £ n Þ x < n + 1, n Î I
(ix) [ x ] < n Þ x < n, n Î I
(x) n2 £ [ x ] £ n1 Þ n2 £ x < n1 + 1, n1 , n2 Î I
(xi) [ x + y ] ³ [ x ] + [y ]
134
Textbook of Algebra
é[ x ]ù é x ù
(xii) ê ú = ê ú , n Î N
ë n û ënû
2. Least Integer
é n + 1ù é n + 2 ù é n + 4 ù é n + 8 ù
(xiii) ê
ú +ê
ú +ê
ú +ê
ú + K = n, n Î N
ë 2 û ë 4 û ë 8 û ë 16 û
n - 1ù
1ù é
2ù
é
é
(xiv)[ x ] + ê x + ú + ê x + ú + K + ê x +
= [nx ],
nû ë
nû
n úû
ë
ë
n ÎN
Graph of y = [x ]
2
1
–3 –2 –1
–1
1
2
3
4
of x.
Thus,
(3.578 ) = 4 , (0.87 ) = 1,
( 4) = 4
é- 8.239ù = - 8, é- 0.7ù = 0
In general, if n is an integer and x is any real number
between n and n + 1
i.e.,
n < x £ n + 1, then ( x ) = n + 1
Y
3
X′
( x ) or éx ù denotes the least integer greater than or equal
to x i.e., ( x ) ³ x or éx ù ³ x . It is also known as ceilling
[x] = n
X
–2
n
–3
Y′
Domain and Range of [ x ] are R and I, respectively.
y Example 63. If [ x ] denotes the integral part of x for
real x, then find the value of
1 ù é1
1 ù é1
3 ù
é 1ù é 1
êë 4 úû + êë 4 + 200úû + êë 4 + 100úû + êë 4 + 200úû
é 1 199 ù
+ K+ ê +
.
ë 4 200úû
Sol. The given expression can be written as
1 ù é1
2 ù é1
3 ù
é1ù é1
êë 4 úû + êë 4 + 200 úû + êë 4 + 200 úû + êë 4 + 200 úû
é 1 199 ù
+K+ ê +
ë 4 200 úû
[remember]
Remark
If ( x ) = n, then ( n - 1) < x £ n
Graph of y = ( x ) = éxù
Y
sin (–1)
2
1
[from property (xiv)]
equations y = 2 [ x ] + 3 and y = 3 [ x - 2]
X′
–2 –1
0
1
2
3
X
Y′
Remark
Domain and Range of ( x ) are R and [ x ] + 1, respectively.
simulaneously, determine [ x + y ].
y = 2[ x ] + 3 = 3 [ x - 2]
i.e. If x Î I , then x = [ x ] = ( x ) .
3
y Example 64. Let [a ] denotes the larger integer not
exceeding the real number a. If x and y satisfy the
Sol. We have,
n+1
x
Relation between Greatest Integer and Least Integer
x ÎI
ì[ x ],
(x ) = í
î [ x ] + 1, x Ï I
Remark
1ù
é
= ê200 × ú = [50] = 50
4û
ë
x=[x] = n+1
…(i)
[from property (i)]
Þ
2[ x ] + 3 = 3([ x ] - 2)
Þ
2[ x ] + 3 = 3[ x ] - 6
Þ
[x ] = 9
From Eq. (i), y = 2 ´ 9 + 3 = 21
\
[ x + y ] = [ x + 21] = [ x ] + 21 = 9 + 21 = 30
Hence, the value of [ x + y ] is 30.
y Example 65. If [ x ] and ( x ) are the integral part of
x and nearest integer to x, then solve ( x )[ x ] = 1.
Sol. Case I If x Î I , then x = [ x ] = ( x )
\ Given equation convert in x 2 = 1.
\
x = ( ± 1)
Case II If x Ï I , then ( x ) = [ x ] + 1
Chap 02 Theory of Equations
\Given equation convert in
([ x ] + 1 )[ x ] = 1 Þ [ x ]2 + [ x ] - 1 = 0
-1± 5
2
Then, final answer is x = ± 1.
[x ] =
or
y Example 67. If { x } and [ x ] represent fractional and
integral part of x respectively, find the value of
[impossible]
[x ] +
{x + r }
.
2000
2000
å
r =1
2000
{x }
{x + r }
= [x ] + å
2000
2000
r =1
= [x ] +
Then, ( x )2 + ( x + 1)2 = 25 reduces to
2
x 2 + x + 1 = 25 Þ 2x 2 + 2x - 24 = 0
x 2 + x - 12 = 0 Þ ( x + 4 )( x - 3) = 0
\
x = - 4, 3
Case II If x Ï I , then ( x ) = [ x ] + 1
Then, ( x )2 + ( x + 1)2 = 25 reduces to
{[ x ] + 1} 2 + {[ x + 1] + 1} 2 = 25
Þ
{[ x ] + 1} 2 + {[ x ] + 2} 2 = 25
…(i)
[from property (i)]
{ x } 2000
{x }
´ 2000 = [ x ] + { x } = x
1 = [x ] +
å
2000
2000 r = 1
y Example 68. If { x } and [ x ] represent fractional
and integral part of x respectively, then solve the
equation
x - 1 = ( x - [ x ])( x - { x }).
Sol. Q
x = [ x ] + { x }, 0 £ { x } < 1
Thus, given equation reduces to
[ x ] + { x } - 1 = { x }[ x ]
2[ x ]2 + 6[ x ] - 20 = 0
2
Þ
å
Sol. [ x ] +
Sol. Case I If x Î I , then x = ( x ) = [ x ]
Þ
2000
r =1
y Example 66. Find the solution set of
( x ) 2 + ( x + 1) 2 = 25, where ( x ) is the least integer
greater than or equal to x .
Þ
135
[ x ] + 3[ x ] - 10 = 0
Þ
{[ x ] + 5}{[ x ] - 2} = 0
\
[ x ] = - 5 and [ x ] = 2
Þ
x Î [ - 5, - 4 ) È [2, 3)
Q
x Ï I,
\
x Î ( - 5, - 4 ) È (2, 3)
On combining Eqs. (i) and (ii), we get
x Î ( - 5, - 4 ] È (2, 3]
Þ
{ x }[ x ] - [ x ] - { x } + 1 = 0
Þ
([ x ] - 1)({ x } - 1) = 0
Now,
{x } - 1 ¹ 0
\
[x ] - 1 = 0
Þ
\
…(ii)
[Q0 £ { x } < 1]
[x ] = 1
x Î[1, 2 )
Problem Solving Cycle
If a problem has x , | x |,[ x ], ( x ), { x }, then first solve | x |,
then problem convert in x ,[ x ], ( x ), { x }.
3. Fractional Part
x, | x |, [x], (x), {x}
{ x } denotes the fractional part of x, i.e.0 £ { x } < 1.
Thus, {2 × 7} = 0.7, {5 } = 0, { - 3.72} = 0.28
If x is a real number, then x = [ x ] + { x }
i.e.,
x = n + f , where n Î I and 0 £ f < 1
Properties of Fractional Part of x
(i) { x ± n } = { x }, n Î I
(ii) If 0 £ x < 1, then { x } = x
Graph of y = {x }
x = [x] + {x}
x, [x], (x), {x}
[x], {x}
x, [x], {x}
Y
1
X′
–3
–2
–1
0
Y′
1
2
3
4
Remark
1. For proper fraction 0 < {x } < 1.
2. Domain and range of { x } are R and [ 0, 1), respectively.
3. {- 5238
. } = {- 5 - 0.238} = {- 5 - 1 + 1 - 0.238}
= { - 6 + 0.762} = {6.762} = 0.762
X
x ÎI
ì[ x ],
Secondly, solve ( x ) = í
î[ x ] + 1 , x Ï I
Then, problem convert in x ,[ x ], { x }.
Now, put x = [ x ] + { x }
Then, problem convert in [ x ] and { x }.
Since, 0 £ { x } < 1, then we get [ x ]
From Eq. (i), we get {x}
Hence, final solution is x = [ x ] + { x }.
…(i)
136
Textbook of Algebra
y Example 69. Let { x } and [ x ] denotes the fractional
and integral parts of a real number x, respectively.
Solve 4{ x } = x + [ x ].
Sol. Q
x = [x ] + {x }
…(i)
Then, given equation reduces to
4 {x } = [x ] + {x } + [x ]
2
…(ii)
{x } = [x ]
Þ
3
2
3
Q
0 £ { x } < 1 Þ 0 £ [ x ] < 1 or 0 £ [ x ] <
3
2
\
[ x ] = 0, 1
2
From Eq. (ii), { x } = 0,
3
2
5
From Eq. (i), x = 0, 1 + i.e., x = 0,
3
3
y Example 70. Let { x } and [ x ] denotes the fractional
and integral part of a real number ( x ), respectively.
Solve | 2x - 1| = 3[ x ] + 2{ x }.
1
2
Then, given equation convert to
2x - 1 = 3 [ x ] + 2{ x }
Q
x = [x ] + {x }
From Eqs. (i) and (ii), we get
2([ x ] + { x }) - 1 = 3[ x ] + 2 { x }
\
[x ] = - 1
\
-1£ x <0
No solution
1
Case II 2x - 1 < 0 or x <
2
Then, given equation reduces to
1 - 2x = 3[ x ] + 2 { x }
Q
x = [x ] + {x }
From Eqs. (iii) and (iv), we get
1 - 2([ x ] + { x }) = 3[ x ] + 2 { x }
Sol. Case I 2x - 1 ³ 0 or x ³
Þ
\
Now,
Þ
Þ
Þ
…(i)
…(ii)
1ù
é
êëQ x ³ 2 úû
…(iii)
…(iv)
1 - 5[ x ] = 4 { x }
1 - 5[ x ]
4
0 £ {x } < 1
1 - 5[ x ]
0£
<1
4
0 £ 1 - 5[ x ] < 4
0 ³ - 1 + 5[ x ] > - 4
{x } =
3
1
Þ
1 ³ 5[ x ] > - 3 or - < [ x ] £
5
5
\
[x ] = 0
1
From Eq. (v), { x } =
4
1 1
\
x =0+ =
4 4
…(v)
y Example 71. Solve the equation
( x ) 2 = [ x ] 2 + 2x
where, [ x ] and ( x ) are integers just less than or equal
to x and just greater than or equal to x, respectively.
Sol. Case I If x Î I then
x = [x ] = ( x )
The given equation reduces to
x 2 = x 2 + 2x
Þ
2x = 0 or x = 0
Case II If x Ï I , then ( x ) = [ x ] + 1
The given equation reduces to
([ x ] + 1)2 = [ x ]2 + 2x
Þ
1 = 2( x - [ x ]) or { x } =
…(i)
1
2
1
1
…(ii)
= n + ,n ÎI
2
2
1
Hence, the solution of the original equation is x = 0, n + ,
2
n Î I.
\
x = [x ] +
y Example 72. Solve the system of equations in x , y
and z satisfying the following equations:
x + [ y ] + {z } = 3 × 1
{ x } + y + [z ] = 4 × 3
[ x ] + {y } + z = 5 × 4
where, [ × ] and { ×} denotes the greatest integer and fractional parts, respectively.
Sol. Q[ x ] + { x } = x , [y ] + {y } = y and [z ] + {z } = z ,
On adding all the three equations, we get
2( x + y + z ) = 128
.
Þ
x + y + z = 6.4
Now, adding first two equations, we get
x + y + z + [y ] + { x } = 7.4
Þ
6.4 + [y ] + { x } = 7.4
Þ
[y ] + { x } = 1
\
[y ] = 1 and { x } = 0
On adding last two equations, we get
…(i)
[from Eq. (i)]
…(ii)
x + y + z + {y } + [z ] = 9.7
{y } + [z ] = 3.3
\
[z ] = 3 and {y } = 0.3
On adding first and last equations, we get
x + y + z + [ x ] + {z } = 8.5
Þ
[ x ] + {z } = 2.1
\
[ x ] = 2, {z } = 0.1
From Eqs. (i), (ii) and (iii), we get
x = [x ] + {x } = 2 + 0 = 2
y = [y ] + {y } = 1 + 0.3 = 1.3
and
z = [z ] + {z } = 3 + 0.1 = 3.1
[from Eq. (ii)]
…(iii)
[from Eq. (i)]
…(iv)
Chap 02 Theory of Equations
Roots of Equation with
the Help of Graphs
\
1< x <2
We have, f ( x ) = x 3 - 3 and g ( x ) = 1
Here, we will discuss some examples to find the roots of
equations with the help of graphs.
\
Important Graphs
Aliter
Q
x = [ x ] + f , 0 £ f < 1,
Then, given equation reduces to
x 3 - (x - f ) = 3 Þ x 3 - x = 3 - f
x3-3=1 Þ x3= 4
or
1. y = ax 3 + bx 2 + cx + d
137
x = ( 4 )1 / 3
Hence, x = 41 / 3 is the solution of the equation x 3 - [ x ] = 3.
Hence, it follows that
2< x3 - x £3
Þ
2 < x ( x + 1) ( x - 1) £ 3
Further for x ³ 2, we have x ( x + 1) ( x - 1) ³ 6 > 3
For x < - 1, we have x ( x + 1) ( x - 1) < 0 < 2
For x = - 1, we have x ( x + 1) ( x - 1) = 0 < 2
For -1 < x £ 0, we have x ( x + 1) ( x - 1) £ - x < 1
and for 0 < x £ 1, we have x ( x + 1) ( x - 1) < x < x 3 £ 1
a<0
a>0
2. x = ay 3 + by 2 + cy + d
Therefore, x must be 1 < x < 2
\
[x ] = 1
Now, the original equation can be written as
x 3- 1 = 3 Þ x 3= 4
a<0
a>0
3. y = ax 4 + bx 3 + c x 2 + dx + e
Hence, x = 41 / 3 is the solution of the given equation.
y Example 74. Solve the equation x 3 - 3x - a = 0 for
different values of a.
Sol. We have, x 3 - 3x - a = 0 Þ x 3 - 3x = a
a>0
a<0
Let f ( x ) = x 3 - 3x and g ( x ) = a
y Example 73. Solve the equation x 3 - [ x ] = 3, where
[ x ] denotes the greatest integer less than or equal to x .
Q
Þ
f ¢( x ) = 0
3x 2 - 3 = 0
Sol. We have, x 3 - [ x ] = 3
Þ
x = - 1, 1
f ¢ ¢ ( x ) = 6x
3
Þ
x - 3 = [x ]
Y
f ( x ) = x 3 - 3 and g ( x ) = [ x ].
Let
2
It is clear from the graphs, the point of intersection of
two curves y = f ( x ) and y = g ( x ) lies between (1, 0) and
(2 , 0).
y=a
y = f (x)
Y
X'
2
y = g(x)
1
– 3
–1
0
3
X
1
X¢
–2 –1
1
–1
–2
–3
Y¢
2
3
X
y = x3 – 3x
–2
Y'
\ f ¢ ¢ ( -1) = - 6 < 0 and f ¢ ¢ (1) = 6 > 0
\ f ( x ) local maximum at x = ( -1) and local minimum at
x = 1 and f ( -1) = 2 and f (1) = - 2 and y = g ( x ) = a is a
straight line parallel to X -axis.
138
Textbook of Algebra
Following cases arise
Case I When a > 2,
In this case y = f ( x ) and y = g ( x ) intersects at only one
point, so x 3 - 3x - a = 0 has only one real root.
Case II When a = 2,
y Example 76. Find all values of the parameter k for
which all the roots of the equation
x 4 + 4 x 3 - 8 x 2 + k = 0 are real.
Sol. We have, x 4 + 4 x 3 - 8x 2 + k = 0
Y
In this case y = f ( x ) and y = g ( x ) intersects at two points,
so x 3 - 3x - a = 0 has three real roots, two are equal and
one different.
–4
X¢
0
1
X
y = g(x)
Case III When -2 < a < 2,
In this case y = f ( x ) and y = g ( x ) intersects at three points,
so x 3 - 3x - a = 0 has three distinct real roots.
–3
Case IV When a = - 2,
In this case y = f ( x ) and y = g ( x ) touch at one point and
intersect at other point, so x 3- 3x - a = 0 has three real
roots, two are equal and one different.
y = f(x)
128
Case V When a < - 2,
In this case y = f ( x ) and y = g ( x ) intersects at only one
point, so x 3 - 3x - a = 0 has only one real root.
Y¢
Þ
Let
y Example 75. Show that the equation
x 3 + 2x 2 + x + 5 = 0 has only one real root, such that
[a ] = - 3, where [ x ] denotes the integral point of x .
3
2
Sol. We have, x + 2x + x + 5 = 0
x 3 + 2x 2 + x = - 5
Þ
f ( x ) = x 3 + 2x 2 + x and g ( x ) = - 5
Q
f ¢ ( x ) = 0 Þ 3x 2 + 4 x + 1 = 0
1
x = - 1, - and f ¢¢( x ) = 6x + 4
Þ
3
æ 1ö
\ f ¢ ¢ ( - 1) = - 2 < 0 and f ¢ ¢ ç - ÷ = - 2 + 4 = 2 > 0
è 3ø
Let
\ f ( x ) local maximum at x = - 1 and local minimum at
1
x=3
4
æ 1ö
and
f ( -1) = 0, f ç - ÷ = è 3ø
27
Y
X¢
–1
1
–
3
3
0
2
y = f(x) = x + 2x + x
X
Q
Þ
and
4
3
2
x + 4 x - 8x = - k
f ( x ) = x 4 + 4 x 3 - 8x 2 and g ( x ) = - k
f ¢( x ) = 0
4 x 3 + 12x 2 - 16x = 0 Þ x = - 4, 0, 1
f ¢ ¢ ( x ) = 12x 2 + 24 x - 16
\ f ¢ ¢ ( -4 ) = 80, f ¢ ¢ (0) = - 16, f ¢ ¢ (1) = 20
\ f ( x ) has local minimum at x = - 4 and x = 1 and local
maximum at x = 0
and f ( -4 ) = - 128, f (0) = 0, f (1) = - 3.
Following cases arise
Case I When – k > 0 i.e., k < 0
In this case y = x 4 + 4 x 3 - 8x 2 and y = ( -k ) intersect at
two points, so x 4 + 4 x 3 - 8x 2 + k = 0 has two real roots.
Case II When -k = 0 and -k = - 3, i.e. k = 0, 3
In this case y = x 4 + 4 x 3 - 8x 2 and y = - k intersect at four
points, so x 4 + 4 x 3 - 8x 2 + k = 0 has two distinct real roots
and two equal roots.
Case III When - 3 < - k < 0, i.e. 0 < k < 3
In this case y = x 4 + 4 x 3 - 8x 2 and y = - k intersect at four
distinct points, so x 4 + 4 x 3 - 8x 2 + k = 0 has four distinct
real roots.
Case IV When -128 < - k < - 3, i.e. 3 < k < 128
In this case y = x 4 + 4 x 3 - 8x 2 and y = - k intersect at two
distinct points, so x 4 + 4 x 3 - 8x 2 + k = 0 has two distinct
real roots.
Case V When -k = - 128 i.e., k = 128
y =g(x) = – 5
Y′
and f ( -2) = - 2 and f ( -3) = - 12
Therefore, x must lie between ( -3) and ( -2).
i.e.
-3 < a < - 2 Þ [a ] = - 3
In this casey = x 4 + 4 x 3 - 8x 2 andy = - k touch at one
point, so x 4 + 4 x 3 - 8x 2 + k = 0 has two real and equal roots.
Case VI When -k < - 128, i.e. k > 128
In this case y = x 4 + 4 x 3 - 8x 2 and y = - k do not
intersect, so there is no real root.
Chap 02 Theory of Equations
y Example 77. Let -1 £ p £ 1, show that the equation
é1 ù
4 x 3 - 3x - p = 0 has a unique root in the interval ê , 1ú
ë2 û
and identify it.
Sol. We have, 4 x 3 - 3x - p = 0
4 x 3 - 3x = p
Þ
f ( x ) = 4 x 3 - 3x and g ( x ) = p
Let
\
Þ
f ¢( x ) = 0
12x 2 - 3 = 0
1 1
x = - , - and f ¢ ¢ ( x ) = 24 x
2 2
æ 1ö
æ1ö
\
f ¢ ¢ ç - ÷ = - 12 < 0 and f ¢ ¢ ç ÷ = 12 > 0
è 2ø
è2ø
1ö
æ
\ f ( x ) has local maximum at ç x = - ÷ and local minimum
è
2ø
1ö
æ
at ç x = ÷.
è
2ø
Þ
4 3
æ1ö 4 3
æ 1ö
Also, f ç - ÷ = - + = 1 and f ç ÷ = - = - 1
è2ø 8 2
è 2ø
8 2
Y
1
y = f(x)
y = g(x)
X′
– 3
2
1
2
–1
2
0
–1
Y′
3
2
X
139
We observe that, the line y = g ( x ) = p , where -1 £ p £ 1
é1 ù
intersect the curve y = f ( x ) exactly at point a Îê , 1ú.
ë2 û
Hence, 4 x 3 - 3x - p = 0 has exactly one root in the interval
é1 ù
êë 2 , 1úû .
Now, we have to find the value of root a.
Let a = cos q , then 4 cos 3 q - 3 cos q - p = 0
1
Þ
cos3q = p Þ 3q = cos -1( p ) or q = cos -1( p )
3
ì1
ü
\
a = cos q = cos í cos -1( p )ý
î3
þ
Aliter
Let
\
f( x ) = 4 x 3 - 3x - p
1ö æ
1ö
æ
f¢ ( x ) = 12x 2 - 3 = 12 ç x + ÷ ç x - ÷
è
ø
è
2
2ø
+
+
–1
2
–
1
2
é1 ù
Clearly, f¢ ( x ) > 0 for x Î ê , 1ú.
ë2 û
é1 ù
Hence, f( x ) can have atmost one root in ê , 1ú.
ë2 û
æ1ö
Also, f ç ÷ = - 1 - p and f (1) = 1 - p
è2ø
æ1ö
\ f ç ÷ f ( 1) = - ( 1 - p 2 ) = ( p 2 - 1) £ 0
è2ø
[Q - 1 £ p £ 1]
é1 ù
Since, f( x ) being a polynomial, continuous on ê , 1ú and
ë2 û
æ1ö
f ç ÷ f (1) £ 0. Therefore, by intermediate value theorem
è2ø
é1 ù
f( x ) has atleast one root in ê , 1ú.
ë2 û
é1 ù
Hence, f( x ) has exactly one root in ê , 1ú .
ë2 û
140
Textbook of Algebra
#L
1.
Exercise for Session 4
æ1+ aö
If a, b, g are the roots of x 3 - x 2 - 1 = 0, the value of å ç
÷ , is equal to
è 1- a ø
(a) - 7
(c) - 5
2.
(b) - 6
(d) - 4
If r , s, t are the roots of the equation 8x 3+ 1001x + 2008 = 0. The value of
(r + s ) 3 + (s + t ) 3 + (t + r ) 3 is
(a) 751
(c) 753
3.
If a, b, g, d are the roots of equation x 4 + 4x 3 - 6x 2 + 7x - 9 = 0, the value of Õ (1 + a 2 ) is
(a) 9
(c) 13
4.
6.
(b) real and equal
(d) real and distinct
If x 2 + px + 1 is a factor of the expression ax 3 + bx + c then
(a) a 2 - c 2 = ab
(b) a 2 + c 2 = - ab
(c) a 2 - c 2 = - ab
(d) None of these
The number of real roots of the equation x 2 - 3 | x | + 2 = 0 is
(a) 1
(c) 3
7.
(b) 11
(d) 15
If a, b , c, d are four consecutive terms of an increasing AP, the roots of the equation
( x - a ) ( x - c ) + 2 ( x - b ) ( x - d ) = 0 are
(a) non-real complex
(c) integers
5.
(b) 752
(d) 754
(b) 2
(d) 4
Let a ¹ 0 and p ( x ) be a polynomial of degree greater than 2, if p ( x ) leaves remainder a and ( -a ) when divided
respectively by x + a and x - a, the remainder when p ( x ) is divided by x 2 - a 2, is
(a) 2x
(c) x
8.
The product of all the solutions of the equation ( x - 2)2 - 3 | x - 2 | + 2 = 0 is
(a) 2
(c) 0
9.
(b) -4
(d) None of these
é x ù é x ù é x ù 31
If 0 < x < 1000 and ê ú + ê ú + ê ú =
x , where [ x ] is the greatest integer less than or equal to x, the
ë 2 û ë 3 û ë 5 û 30
number of possible values of x is
(a) 32
(c) 34
10.
(b) -2x
(d) -x
(b) 33
(d) None of these
If [ x ] is the greatest integer less than or equal to x and ( x ) be the least integer greater than or equal to x and
[ x ]2 + ( x )2 > 25 then x belongs to
(a) [3, 4]
(c) [4, ¥)
(b) (- ¥, - 4]
(d) (- ¥, - 4] È [4, ¥)
Session 5
Irrational Equations, Irrational Inequations, Exponential
Equations, Exponential Inequations, Logarithmic Equations,
Logarithmic Inequations
Irrational Equations
Here, we consider equations of the type which contain the
unknown under the radical sign and the value under the
radical sign is known as radicand.
If roots are all even (i.e. x , 4 x , 6 x ,..., etc) of an equation
are arithmetic. In other words, if the radicand is negative
( i.e. x < 0 ), then the root is imaginary, if the radicand is
zero, then the root is also zero and if the radicand is
positive, then the value of the root is also positive.
If roots are all odd ( i.e. 3 x , 5 x , 7 x ,... etc) of an equation,
then it is defined for all real values of the radicand. If the
radicand is negative, then the root is negative, if the
radicand is zero, then the root is zero and if the radicand
is positive, then the root is positive.
l
This equation is defined for 2x + 7 ³ 0
7
ìï
x³and
x + 4 ³0 Þ í
2
ïî x ³ - 4
7
\
x³2
7
For x ³ - , the left hand side of the original equation
2
is positive, but right hand side is zero. Therefore, the
equation has no roots.
(ii) We have,
(x - 4) = - 5
l
Some Standard Formulae to
Solve Irrational Equations
(iii) We have,
2k
f
2k
g = 2k fg , f ³ 0, g ³ 0
Consequently, there is no x for which both expressions
would have sense. Therefore, the equation has no roots.
2. 2k f / 2k g = 2k ( f /g ), f ³ 0, g > 0
3. | f | 2k g =
2k
(iv) We have,
( f 2k g ), g ³ 0
5. 2k fg = 2k | f | 2k g , fg ³ 0
y Example 78. Prove that the following equations has
no solutions.
(2 x + 7) + (x + 4) = 0 (ii)
(iii)
(6 - x ) - (x - 8) = 2
(v)
x + (x + 16) = 3
(vii)
(x - 4) = - 5
-2 - x = (x - 7)
15
(vi) 7 x + 8 -x + 3 = 98
x
(iv)
(x - 3) - x + 9 = (x - 1)
Sol. (i) We have,
( 2x + 7 ) + ( x + 4 ) = 0
( -2 - x ) = 5 ( x - 7 )
This equation is defined for
-2 - x ³ 0 Þ x £ - 2
For x £ - 2 the left hand side is positive, but right
hand side is negative.
Therefore, the equation has no roots.
4. 2k ( f /g ) = 2k | f | / 2k | g |, fg ³ 0, g ¹ 0
(i)
(6 - x ) - x - 8 = 2
The equation is defined for
6 - x ³ 0 and x - 8 ³ 0
ìx £ 6
\
í
îx ³ 8
If f and g be functions of x , k Î N . Then,
1.
The equation is defined for x - 4 ³ 0
\
x³4
For x ³ 4, the left hand side of the original equation is
positive, but right hand side is negative.
Therefore, the equation has no roots.
5
(v) We have,
x + ( x + 16) = 3
The equation is defined for
ìx ³0
x ³ 0 and x + 16 ³ 0 Þ í
î x ³ - 16
Hence,
x ³0
For x ³ 0 the left hand side ³ 4, but right hand side is
3. Therefore, the equation has no roots.
15
(vi) We have, 7 x + 8 - x + 3 = 98
x
142
Textbook of Algebra
For x < 0, the expression 7 x is meaningless,
Sol. We have, 3 ( x + 3) - x - 2 = 7
For x > 0, the expression 8 -x is meaningless
15
and for x = 0, the expression 3 is meaningless.
x
Consequently, the left hand side of the original
equation is meaningless for any x Î R. Therefore, the
equation has no roots.
(vii) We have,
On squaring both sides of the equation, we obtain
9 x + 27 = 49 + x - 2 + 14 x - 2
Þ
Again, squaring both sides, we obtain
16x 2 + 100 - 80x = 49 x - 98
Þ 16x 2 - 129 x + 198 = 0
33 ö
æ
( x - 6) ç x - ÷ = 0
Þ
è
16 ø
( x - 3) - ( x + 9 ) = x - 1
x ³3
Hence, for x ³ 3, the left hand side of the original
equation is negative and right hand side is positive.
Therefore, the equation has no roots.
does not satisfy the original equation.
33
is the extraneous root.
\ x2 =
16
( x ) = g 2n ( x ), n Î N is equivalent to f ( x ) = g ( x ).
Then, find the roots of this equation. If root of this
equation satisfies the original equation, then its root of the
original equation, otherwise, we say that this root is its
extraneous root.
ì g (x ) ³ 0
is equivalent to the system í
2n
î f (x ) = g (x )
y Example 81. Solve the equation
(6 - 4 x - x 2 ) = x + 4.
This equation is equivalent to the system
ì x + 4 ³0
í
2
2
î6 - 4 x - x = ( x + 4 )
On solving the equation x 2 + 6x + 5 = 0
We find that, x 1 = ( -1) and x 2 = ( -5) only x 1 = ( -1) satisfies
the condition x ³ - 4.
Consequently, the number -1 is the only solution of the
given equation.
x = x -2
x 2 - 5x + 4 = 0 Þ ( x - 1) ( x - 4 ) = 0
\
x 1 = 1 and x 2 = 4
Hence, x 1 = 4 satisfies the original equation, but x 2 = 1 does
not satisfy the original equation.
\
x 2 = 1 is the extraneous root.
y Example 80. Solve the equation
3 ( x + 3) - ( x - 2) = 7.
ìx ³ -4
í 2
î x + 6x + 5 = 0
Þ
y Example 79. Solve the equation x = x - 2 .
On squaring both sides, we obtain
x = ( x - 2) 2
(6 - 4 x - x 2 ) = x + 4
Sol. We have,
Squaring an Equation May Give Extraneous Roots
Squaring should be avoided as for as possible. If squaring is
necessary, then the roots found after squaring must be checked
whether they satisfy the original equation or not. If some values
of x which do not satisfy the original equation. These values of x
are called extraneous roots and are rejected.
Sol. We have,
33
16
Form 2 An equation in the form
2n
f ( x ) = g ( x ), n Î N
Remark
Þ
33
16
Hence, x 1 = 6 satisfies the original equation, but x 2 =
Form 1 An equation of the form
2n
x 1 = 6 and x 2 =
For x ³ 3, x - 3 < x + 9 i.e. ( x - 3) - ( x + 9 ) < 0
Some Standard Forms to
Solve Irrational Equations
f
8x - 20 = 14 ( x - 2)
( 4 x - 10) = 7 x - 2
This equation is defined for
ì x ³3
ìx - 3 ³ 0
ï
ï
íx + 9 ³ 0 Þ íx ³ - 9
ï x ³1
ïx - 1 ³ 0
î
î
Hence,
3 ( x + 3) = 7 + ( x - 2)
Þ
Form 3 An equation in the form
3
f (x ) + 3 g (x ) = h (x )
…(i)
where f ( x ), g( x ) are the functions of x , but h( x ) is a
function of x or constant, can be solved as follows cubing
both sides of the equation, we obtain
f ( x ) + g( x ) + 3 3 f ( x ) g( x ) ( 3 f ( x ) + 3 g( x )) = h 3 ( x )
Þ
f ( x ) + g( x ) + 3 3 f ( x ) g( x ) (h( x )) = h 3 ( x )
[from Eq. (i)]
Chap 02 Theory of Equations
We find its roots and then substituting, then into the
original equation, we choose those which are the roots of
the original equation.
y Example 82. Solve the equation
3
Sol.
(2x - 1) + 3 ( x - 1) = 1.
3
We have,
( 2x - 1) + 3 ( x - 1) = 1
…(i)
Cubing both sides of Eq. (i), we obtain
2x - 1 + x - 1 + 3 × 3 ( 2x - 1) ( x - 1)
( 3 (2x - 1) + 3 ( x - 1)) = 1
Þ
3x - 2 + 3 × 3 ( 2x 2 - 3x + 1) ( 1) = 1
[from Eq. (i)]
3 × 3 ( 2x 2 - 3x + 1) = 3 - 3x
Þ
Þ
3
( 2x 2 - 3x + 1) = ( 1 - x )
Again cubing both sides, we obtain
2x 2 - 3x + 1 = ( 1 - x ) 3
Þ
( 2x - 1) ( x - 1) = ( 1 - x ) 3
Þ
( 2x - 1) ( x - 1) = - ( x - 1) 3
2
2x + 5x - 2 = 4
\
\
2x 2 + 5x - 18 = 0
\
x 1 = 2 and x 2 = - 9 / 2
Both roots satisfies the original equation.
Hence, x 1 = 2 and x 2 = - 9 / 2 are the roots of the original
equation.
Irrational Inequations
We consider, here inequations which contain the
unknown under the radical sign.
Some Standard Forms
to Solve Irrational Inequations
2n
f ( x ) < 2n g ( x ), n Î N
ì f (x ) ³ 0
is equivalent to the system í
îg ( x ) > f ( x )
Þ ( x - 1) {2x - 1 + ( x - 1) } = 0
and inequation of the form 2n + 1 f ( x ) < 2n + 1 g ( x ), n Î N
( x - 1) ( x 2 ) = 0
Þ
u = 4, v = 3
We get,
Form 1 An inequation of the form
2
\
x 1 = 0 and x 2 = 1
Q x 1 = 0 is not satisfies the Eq. (i), then x 1 = 0 is an
extraneous root of the Eq. (i), thus x 2 = 1 is the only root of
the original equation.
is equivalent to the inequation f ( x ) < g ( x ).
y Example 84. Solve the inequation
Form 4 An equation of the form
n
5
a - f ( x ) + n b + f ( x ) = g ( x ).
u = n a - f ( x ), n = n b + f ( x )
Let
Then, the given equation reduces to the solution of the
system of algebraic equations.
ì u + v = g(x )
í n
n
îu + v = a + b
y Example 83. Solve the equation
2
é 3
7 ù
6
ê x + 1 + x + 2ú < 5 x - 1 .
ë
û
Sol. The given inequation is equivalent to
3
7
6
+
<
x +1 x +2 x -1
Þ
4 x 2 - 15x - 25
<0
( x + 1) ( x + 2) ( x - 1)
Þ
( x + 5 / 4 ) ( x - 5)
<0
( x + 1) ( x + 2) ( x - 1)
From Wavy Curve Method :
2
(2x + 5x - 2) - 2x + 5x - 9 = 1 .
+
–2
+
+
–5
4
–1
Sol. Let
u = ( 2x 2 + 5x - 2)
and
v = ( 2x 2 + 5x - 9 )
\
2
u = 2x + 5x - 2
æ 5 ö
x Î ( -¥, - 2) È ç - , 1÷ È (1, 5)
è 4 ø
and
v 2 = 2x 2 + 5x - 9
Form 2 An inequation of the form
–
2
Then, the given equation reduces to the solution of the
system of algebraic equations.
u -v =1
u2 - v 2 = 7
Þ
Þ
(u + v ) (u - v ) = 7
u +v =7
[Qu - v = 1]
143
2n
–
1
–
f ( x ) < g ( x ), n Î N .
ì f (x ) ³ 0
ï
is equivalent to the system í g ( x ) > 0
ï f ( x ) < g 2n ( x ),
î
5
144
Textbook of Algebra
and inequation of the form 2n + 1 f ( x ) < g ( x ), n Î N
is equivalent to the inequation f ( x ) < g
2n + 1
( x ).
y Example 85. Solve the inequation ( x + 14 ) < ( x + 2).
Sol. We have, ( x + 14 ) < ( x + 2)
This inequation is equivalent to the system
x ³ - 14
ì
ì x + 14 ³ 0
ï
ï
Þí
x > -2
x +2>0
í
ï x 2 + 3x - 10 > 0
ï x + 14 < ( x + 2)2
î
î
Þ
x ³ - 14
x ³ - 14
ì
ì
ï
ï
Þí
x > -2
x > -2
í
ï( x + 5) ( x - 2) > 0 ï x < - 5 and x > 2
î
î
On combining all three inequation of the system, we get
x > 2, i.e. x Î (2 , ¥ )
Exponential Equations
If we have an equation of the form a x = b (a > 0 ), then
(i) x Î f, if b £ 0
(ii) x = log a b, if b > 0, a ¹ 1
(iii) x Î f, if a = 1, b ¹ 1
(iv) x Î R, if a = 1, b = 1 (since, 1 x = 1 Þ 1 = 1, x Î R )
y Example 87. Solve the equation
2
(6 - x ) ( 3 x - 7.2 x + 3.9 - 9 3 ) = 0.
Sol. We have,
( 6 - x ) ( 3x
f ( x ) > g( x ), n Î N
is equivalent to the collection of two systems of
inequations
ìg ( x ) < 0
ì g (x ) ³ 0
i.e.
and í
í
2n
î f (x ) ³ 0
îf (x ) > g (x )
and inequation of the form 2 n + 1 f ( x ) > g ( x ), n Î N
is equivalent to the inequation f ( x ) > g 2 n + 1 ( x ).
y Example 86. Solve the inequation
( - x 2 + 4 x - 3) > 6 - 2x .
Sol. We have, ( - x 2 + 4 x - 3) > 6 - 2x
This inequation is equivalent to the collection of two
systems, of inequations
6 - 2x ³ 0
ì
ì 6 - 2x < 0
i.e. í 2
2 and í
2
4
3
6
2
x
x
(
x
)
+
>
î
î- x + 4 x - 3 ³ 0
x >3
x £3
ì
ì
and í
Þ í
î( x - 1) ( x - 3) £ 0
î( x - 3) (5x - 13) < 0
ìï x £ 3
ì x >3
and í
Þ í 13
< x <3
î1 £ x < 3
ïî 5
The second system has no solution and the first system has
ö
æ 13
solution in the interval ç < x < 3÷.
ø
è5
æ 13 ö
Hence, x Îç , 3÷ is the set of solution of the original
è5 ø
inequation.
- 7.2x + 3.9
- 9 3) = 0
This equation is defined for
…(i)
6 - x ³ 0 i.e., x £ 6
This equation is equivalent to the collection of equations
6 - x = 0 and 3x
Form 3 An inequation of the form
2n
2
\
2
- 7.2 x + 3. 9
x 1 = 6 and 3
-9 3 =0
x 2 - 7. 2 x + 3. 9
= 32 . 5
x 2 - 7.2 x + 3.9 = 2.5
then
x 2 - 7.2 x + 1.4 = 0
1
and x 3 = 7
We find that,
x2 =
5
Hence, solution of the original equation are
[which satisfies Eq. (i)]
1
x 1 = 6, x 2 = .
5
Some Standard Forms to
Solve Exponential Equations
Form 1 An equation in the form a f ( x ) = 1 , a > 0 , a ¹ 1
is equivalent to the equation f ( x ) = 0
y Example 88. Solve the equation 5 x
2
+ 3x + 2
= 1.
Sol. This equation is equivalent to
x 2 + 3x + 2 = 0
Þ
( x + 1) ( x + 2) = 0
\x 1 = - 1, x 2 = - 2 consequently, this equation has two
roots x 1 = - 1 and x 2 = - 2 .
Form 2 An equation in the form
f (a x ) = 0
is equivalent to the equation f (t ) = 0, where t = a x .
If t 1 , t 2 , t 3 , ..., t k are the roots of f (t ) = 0, then
a x = t 1 , a x = t 2 , a x = t 3 , ..., a x = t k
Chap 02 Theory of Equations
y Example 89. Solve the equation 5 2x - 24 × 5 x - 25 = 0.
Sol. Let 5x = t , then the given equation can reduce in the form
t 2 - 24t - 25 = 0
Þ
\
then
Hence, x 1 = 2 is only one root of the original equation.
f (x)
+ gc f ( x ) = 0,
where a, b, g Î R and a, b, g ¹ 0 and the bases satisfy the
condition b 2 = ac is equivalent to the equation
at 2 + bt + g = 0, where t = (a / b ) f ( x )
(a / b )
= t 1 and (a / b )
f (x)
= t2
y Example 92. Solve the equation 3 x - 4 + 5 x - 4 = 34.
Sol. Here, 32 + 52 = 34, then given equation has a solution
x - 4 = 2.
\ x 1 = 6 is a root of the original equation.
Sol. Here, 9 ´ 16 = (12)2 .
Then, we divide its both sides by 12x and obtain
x
x
æ4ö
æ3ö
64 × ç ÷ - 84 + 27 × ç ÷ = 0
è3ø
è4ø
…(i)
x
æ3ö
Let ç ÷ = t, then Eq. (i) reduce in the form
è4ø
where f ( x ) > 0.
3
9
and t 2 =
4
16
t1 =
x
1
Form 6 An equation of the form { f ( x )} g ( x ) is
equivalent to the equation
{ f ( x )} g ( x ) = 10 g ( x ) log f ( x ) ,
64t 2 - 84t + 27 = 0
\
(t - 4 ) (6t - 3) = 0
1
Then,
t 1 = 4 and t 2 =
2
Thus, given equation is equivalent to
1
2x = 4 and 2x =
2
Then,
x 1 = 2 and x 2 = - 1
Hence, roots of the original equation are x 1 = 2 and
x 2 = - 1.
where a, b, c Î R and a, b, c satisfies the condition
a 2 + b 2 = c , then solution of this equation is f ( x ) = 2 and
no other solution of this equation.
y Example 90. Solve the equation
64 × 9 x - 84 × 12 x + 27 × 16 x = 0.
Þ
6t 2 - 24t - 3t + 12 = 0
Form 5 An equation of the form a f ( x ) + b f ( x ) = c ,
If roots of this equation are t 1 and t 2 , then
f (x)
2
Þ
Form 3 An equation of the form
aa f ( x ) + bb
t = 2x ,
30t - 135t + 60 = 0
6t 2 - 27t + 12 = 0
Let
Then,
Þ
Þ
(t - 25) (t + 1) = 0 Þ t ¹ - 1,
t = 25,
x
5 = 25 = 52 , then x = 2
y Example 93. Solve the equation 5 x x 8 x -1 = 500.
5x x 8x - 1 = 53 × 22
Sol. We have,
2
x
æ3ö
æ3ö
æ3ö
æ3ö
then, ç ÷ = ç ÷ and ç ÷ = ç ÷
è4ø
è4ø
è4ø
è4ø
\
x 1 = 1 and x 2 = 2
Hence, roots of the original equation are x 1 = 1 and x 2 = 2.
a×a
+b×b
f (x )
5x × 8
Þ
5x × 2
Þ
+ c = 0,
where a, b, c Î R and a, b, c ¹ 0 and ab = 1(a and b are
inverse positive numbers) is equivalent to the equation
2
at + ct + b = 0, where t = a
f (x)
If roots of this equation are t 1 and t 2, then a
af (x) = t2.
y Example 91. Solve the equation
15 × 2 x +1 + 15 × 2 2 - x = 135.
Sol. This equation rewrite in the form
60
30.2x + x = 135
2
Þ
5x - 3 × 2
(52
.
æ x - 3ö
ç
÷
è x ø
1 / x (x - 3 )
10(x
= t 1 and
3x - 3
x
)
= 53 × 22
= 53 × 22
=1
=1
is equivalent to the equation
.
f (x)
æ x - 1ö
ç
÷
è x ø
Þ
Form 4 An equation in the form
f (x )
145
Þ
- 3 ) log ( 5× 21 / x )
1/ x
( x - 3) log (5 × 2
=1
)=0
Thus, original equation is equivalent to the collection of
equations
x - 3 = 0, log (5 × 21/ x ) = 0
æ1ö
\
x 1 = 3 , 5 × 21/ x = 1 Þ 21/ x = ç ÷
è5ø
\
x 2 = - log 5 2
Hence, roots of the original equation are x 1 = 3 and
x 2 = - log 5 2.
146
Textbook of Algebra
Exponential Inequations
where a, b, g Î R and a, b, g ¹ 0 and the bases satisfy the
condition b 2 = ac is equivalent to the inequation
When we solve exponential inequation
at 2 + bt + g ³ 0 or at 2 + bt + g £ 0,
a f ( x ) > b (a > 0 ), we have
where t = (a / b ) f ( x ) .
(i) x Î D f , if b £ 0
(ii) If b > 0, then we have f ( x ) > log a b , if a > 1
and f ( x ) < log a b, if 0 < a < 1 for a = 1, then b < 1.
Form 3 An inequation of the form
Remark
or
aa f ( x ) + bb f ( x ) + g ³ 0
where a, b, g Î R and a, b, g ¹ 0 and ab = 1(a and b are
inverse (+ve) numbers) is equivalent to the inequation
The inequation af ( x ) £ b has no solution for b £ 0, a > 0, a ¹ 1.
y Example 94. Solve the inequation 3
Sol. We have,
3x
+2
> (3-2 )1 / x Þ 3x
+2
x +2
æ 1ö
>ç ÷
è9ø
1/ x
a t 2 + bt + g ³ 0 or a t 2 + bt + g £ 0
.
> 3-2 / x
+
0
–
aa f ( x ) + bb f ( x ) + g £ 0
t = af (x)
where
Form 4 If an inequation of the exponential form reduces
to the solution of homogeneous algebraic inequation, i.e.
a 0 f n ( x ) + a 1 f n - 1 ( x ) g( x ) + a 2 f n - 2 ( x ) g 2 ( x ) + K
+ a n - 1 f ( x ) g n - 1 ( x ) + a n g n ( x ) ³ 0,
Here, base 3 > 1
Þ
Þ
\
x +2> -
2
x
Þ
2
x + 2x + 2
>0
x
( x + 1) 2 + 1
Þ x >0
>0
x
x Î (0, ¥ )
Some Standard Forms to Solve
Exponential Inequations
Form 1 An inequation of the form
y Example 95. Solve the inequation
4 x +1 - 16 x < 2 log 4 8.
Sol. Let 4 x = t , then given inequation reduce in the form
3
4t - t 2 > 2 ×
2
Þ
t 2 - 4t + 3 < 0 Þ ( t - 1) ( t - 3) < 0
Þ
1<t <3
[Qt > 0]
Þ
1 < 4x < 3
Þ
0 < x < log 4 3
\
x Î(0, log 4 3)
Let
2
2
- 5x )
2
x - 5x
+ 6 × 2x
2
- 5x
× 3x
= f ( x ) and 3
2
- 5x
- 27 × 32 (x
2
x - 5x
2
- 5x )
aa f ( x ) + bb f ( x ) + gc f ( x ) ³ 0
f (x)
³0
= g(x ) ,
2
£0
[Q g ( x ) > 0]
2
æ f (x )ö
æ f (x )ö
Then, 8 ç
÷ - 27 ³ 0
÷ + 6ç
è g(x ) ø
è g(x ) ø
f (x )
and let
=t
g(x )
8t 2 + 6t - 27 ³ 0
3ö
æ
Þ
çt - ÷ ( t + 9 / 4 ) ³ 0
è
2ø
Þ
t ³ 3 / 2 and t £ - 9 / 4
The second inequation has no root.
From the first inequation, t > 3 / 2
[Qt > 0]
then
æ2ö
ç ÷
è3ø
Þ
x 2 - 5x
æ2ö
³ç ÷
è3ø
-1
x 2 - 5x £ - 1 Þ x 2 - 5x + 1 £ 0
5 - 21
5 + 21
£x£
2
2
é 5 - 21 5 + 21 ù
Hence, x Î ê
.
,
2 úû
ë 2
\
Form 2 An inequation of the form
+ gc
8 × 22 ( x
On dividing in each by g 2 ( x )
ìt > 0,
where t = a x
í
î f (t ) ³ 0 or f (t ) £ 0
f (x)
Sol. The given inequation is equivalent to
then 8 × f ( x ) + 6 f ( x ) × g ( x ) - 27 g ( x ) ³ 0
is equivalent to the system of collection
aa f ( x ) + bb
y Example 96. Solve the inequation
2
2
2
2 2x -10x + 3 + 6 x - 5x +1 ³ 3 2x -10x + 3 .
2
f (a x ) ³ 0 or f (a x ) £ 0
or
where a 0 , a 1 , a 2 , ,..., a n are constants (a 0 ¹ 0 ) and f ( x )
and g( x ) are functions of x .
[Qt > 0]
ù
é 2
êëQ 3 < 1úû
Chap 02 Theory of Equations
Logarithmic Equations
y Example 99. Solve the equation log (log 5 x ) 5 = 2 .
If we have an equation of the form
log a f ( x ) = b , (a > 0 ), a ¹ 1
is equivalent to the equation
Sol. We have, log (log 5 x ) 5 = 2
Base of logarithm > 0 and ¹ 1 .
f ( x ) = a b ( f ( x ) > 0 ).
y Example 97. Solve the equation
log 3 ( 5 + 4 log 3 ( x - 1)) = 2 .
log 5 x > 0 and log 5 x ¹ 1
Þ
x > 1 and x ¹ 5
log 5 x = 51 / 2 = 5
is equivalent to the equation (here, base ¹ 1, > 0).
\
5 + 4 log 3 ( x - 1) = 32
1
log 3 ( x - 1) = 1 Þ x - 1 = 3
\
x=4
Hence, x 1 = 4 is the solution of the original equation.
Some Standard Formulae to Solve
Logarithmic Equations
f and g are some functions and a > 0, a ¹ 1, then, if
f > 0, g > 0, we have
(i) log a ( fg ) = log a f + log a g
(ii) log a ( f / g ) = log a f - log a g
a
(iii) log a f 2 a = 2 a log a | f | (iv) log ab f a = log a f
b
loga g
loga f
loga f
(v) f
(vi) a
=g
=f
y Example 98. Solve the equation
2 x log 4 3 + 3log 4 x = 27.
Sol. The domain of the admissible values of the equation is
x > 0. The given equation is equivalent to
2.3log 4 x + 3log 4 x = 27
\
\ The original equation is equivalent to
Sol. We have, log 3 (5 + 4 log 3 ( x - 1)) = 2
Þ
147
[from above result (v)]
Þ
3.3log 4 x = 27
Þ
3log 4 x = 9
Þ
3log 4 x = 32
Þ
Þ
log 4 x = 2
x 1 = 4 2 = 16 is its only root.
Some Standard Forms to
Solve Logarithmic Equations
Form 1 An equation of the form log x a = b, a > 0 has
\
x1 = 5
5
Hence, 5
is the only root of the original equation.
Form 2 Equations of the form
(i) f (log a x ) = 0, a > 0, a ¹ 1 and
(ii) g (log x A ) = 0, A > 0
Then, Eq. (i) is equivalent to
f (t ) = 0, where t = log a x
If t 1 , t 2 , t 3 , ..., t k are the roots of f (t ) = 0, then
log a x = t 1 , log a x = t 2 , ..., log a x = t k
and Eq. (ii) is equivalent to f (y ) = 0, where y = log x A.
If y 1 , y 2 , y 3 ,K, y k are the roots of f (y ) = 0, then
log x A = y 1 , log x A = y 2 , ..., log x A = y k
y Example 100. Solve the equation
1 - 2 (log x 2 ) 2
= 1.
log x - 2 (log x ) 2
Sol. The given equation can rewrite in the form
1 - 2 (2 log x )2
=1
log x - 2 (log x )2
1 - 8 (log x )2
Þ
log x - 2 (log x )2
then
Þ
-1=0
log x = t ,
Let
1 - 8t
2
t - 2t
2
-1=0 Þ
1 - t - 6t 2
Þ
(i) Only root x = a 1 /b , if a ¹ 1 and b = 0.
(ii) Any positive root different from unity, ifa = 1andb = 0.
(iii) No roots, if a = 1, b ¹ 0.
(iv) No roots, if a ¹ 1, b = 0.
5
(t - 2t 2 )
t - 2t 2
=0
(1 + 2t ) (1 - 3t )
=0
t (1 - 2t )
1
1
ì
ì
ì x = 10-1 / 2
ït = - 2
ï log x = - 2
Þí
Þ í 1
í
1/ 3
1
1
î x 2 = 10
ït =
ï log x =
î
î
3
3
Hence, x 1 =
equation.
=0 Þ
1 - 8t 2 - t + 2t 2
1
and x 2 = 3 10 are the roots of the original
10
148
Textbook of Algebra
y Example 101. Solve the equation
log 3x 10 - 6 log 2x 10 + 11 log x 10 - 6 = 0.
x <1
ì
x
x
ïé
ù
é
ù
í æ1ö - 3 æ1ö + 1 = 0
ú
ú êç ÷
êç ÷
ïê è 2 ø
úû
úû êë è 2 ø
îë
ì x <1
ì x <1
ï
x
x
Þí
æ1ö
íæ 1 ö
î x = ( - log 2 3)
ï çè 2 ÷ø = 3, çè 2 ÷ø + 1 ¹ 0
î
Þ
Sol. Put log x 10 = t in the given equation, we get
t 3 - 6t 2 + 11t - 6 = 0 Þ (t - 1) (t - 2) (t - 3) = 0,
ìt = 1
ï
ít = 2
ït = 3
î
then
It follows that
ì log x 10 = 1
ï
í log x 10 = 2 Þ
ï log x 10 = 3
î
Hence, x 1 = - log 2 3 is the root of the original equation.
Example 103.
ì x = 10
ì x = 10
ï 2
ï
í x = 10 Þ í x = 10
ï x 3 = 10
ï x = 3 10
î
î
[Q x > 0 and ¹ 1]
[Q x > 0 and ¹ 1]
\ x 1 = 10, x 2 = 10 and x 3 = 10 are the roots of the
original equation.
3
Form 3 Equations of the form
(i) log a f ( x ) = log a g ( x ), a > 0 , a ¹ 1 is equivalent to
two ways.
ì g (x ) > 0
Method I í
îf (x ) = g (x )
ì f (x ) > 0
Method II í
îf (x ) = g (x )
(ii) log f ( x ) A = log g ( x ) A , A > 0 is equivalent to two
ways.
ì g (x ) > 0
ï
Method I í g ( x ) ¹ 1
ïf (x ) = g (x )
î
ì f (x ) > 0
ï
Method II í f ( x ) ¹ 1
ïf (x ) = g (x )
î
y Example 102. Solve the equation
éæ 1 ö x
ù
é æ 1öx ù
log 1 / 3 ê2 ç ÷ - 1ú = log 1 / 3 ê ç ÷ - 4ú .
è ø
è ø
ë 4
û
ë 2
û
Sol. The given equation is equivalent to
x
ì
æ1ö
2ç ÷ - 1 > 0
ï
è2ø
ï
í
x
x
ï 2æ1ö - 1 = æ 1 ö - 4
ç ÷
ïî çè 2 ÷ø
è4ø
Þ
Þ
ìæ 1 ö x 1
ïç ÷ >
ïè 2 ø
2
í
x
2x
ïæ 1 ö - 2 æ 1 ö - 3 = 0
ç ÷
ïî çè 2 ÷ø
è2ø
Solve the equation log æ 2 + x ö 7 = log æ
ç
÷
è 10 ø
2 ö 7.
ç
÷
è x + 1ø
Sol. The given equation is equivalent to
ì
2
ï x +1 >0
ìx + 1 > 0
ï
2
ï
ï
Þ
¹
1
í
í x ¹1
+
x
1
ï x = - 6, 3
ï
î
ï2 + x = 2
ïî 10
x +1
\ x 1 = 3 is root of the original equation.
Form 4 Equations of the form
(i) log f ( x ) g ( x ) = log f ( x ) h ( x ) is equivalent to two
ways.
ì h( x ) > 0
ì g( x ) > 0
ï f (x ) > 0
ï f (x ) > 0
ï
ï
Method II í
Method I í
ï f (x ) ¹ 1
ï f (x ) ¹ 1
ïîg( x ) = h( x )
ïîg( x ) = h( x )
(ii) log g ( x ) f ( x ) = log h ( x ) f ( x ) is equivalent to two
ways.
ì f (x ) > 0
ï g( x ) > 0
ï
Method I í
ï g( x ) ¹ 1
ïîg( x ) = h( x )
ì f (x ) > 0
ï h( x ) > 0
ï
Method II í
ï h( x ) ¹ 1
ïîg( x ) = h( x )
y Example 104. Solve the equation
log ( x 2 - 1) ( x 3 + 6 ) = log ( x 2 - 1) (2x 2 + 5x ).
Sol. This equation is equivalent to the system
5
ì
ì 2x 2 + 5x > 0
ï x < - 2 and x > 0
ï
2
x -1>0
ï
ï
Þ í x < - 1 and x > 1
í
2
x -1¹1
ï
ï
x¹± 2
ïî x 3 + 6 = 2x 2 + 5x
ï x = - 2, 1, 3
î
Hence, x 1 = 3 is only root of the original equation.
Chap 02 Theory of Equations
y Example 105. Solve the equation
log ( x 3 + 6 ) ( x 2 - 1) = log ( 2 x 2 +5x ) ( x 2 - 1).
y Example 107. Solve the equation
2 log 2x = log (7 x - 2 - 2x 2 ).
Sol. This equation is equivalent to
ì
x 2 -1>0
ï
ï 2x 2 + 5x > 0
í
2
ï 2x + 5x ¹ 1
ïî x 3 + 6 = 2x 2 + 5x
Sol. This equation is equivalent to the system
ì x < - 1 and x > 1
ï
5
ïï x < - 2 and x > 0
Þ
í
- 5 ± 33
ï x¹
4
ï
ïî x = - 2, 1, 3
Hence, x 1 = 3 is only root of the original equation.
Form 5 An equation of the form
log h ( x ) (log g ( x ) f ( x )) = 0 is equivalent to the system
ì h( x ) > 0
ï h( x ) ¹ 1
ïï
í g( x ) > 0
ï g( x ) ¹ 1
ï
ïî f ( x ) = g( x )
y Example 106. Solve the equation
log x 2 - 6 x + 8 [log 2x 2 - 2x + 8 ( x 2 + 5x )] = 0.
Sol. This equation is equivalent to the system
ì
2
ï x - 6x + 8 > 0
ï
x 2 - 6x + 8 ¹ 1
ï
í 2x 2 - 2x - 8 > 0
ï
2
ï 2x - 2x - 8 ¹ 1
2
ïî x + 5x = 2x 2 - 2x - 8
Solve the equations of this system
ì
x < 2 and x > 4
ï
x ¹3± 2
ï
ïï
1 - 17
1 + 17
and x >
Þ
íx <
2
2
ï
±
1
19
ï
x¹
ï
2
ïî
x = - 1, 8
x = - 1, does not satisfy the third relation of this system.
Hence, x 1 = 8 is only root of the original equation.
Form 6 An equation of the form
2m log a f ( x ) = log a g ( x ), a > 0 , a ¹ 1 , m Î N is
equivalent to the system
ì f (x ) > 0
í 2m
î f ( x ) = g( x )
149
2x > 0
ì
í
2
(
)
x
=
2
7 x - 2 - 2x 2
î
Þ
x >0
ì
í 2
6
7x + 2 = 0
x
î
Þ
x >0
ì
í
(
1
/
2
)
( x - 2 / 3) = 0
x
î
Þ
ìx = 1 / 2
í
îx = 2 / 3
Hence, x 1 = 1 / 2 and x 2 = 2 / 3 are the roots of the original
equation.
Form 7 An equation of the form
(2m + 1 ) log a f ( x ) = log a g ( x ), a > 0 , a ¹ 1 , m Î N
ì
is equivalent to the system í
îf
g( x ) > 0
2m + 1
.
(x ) = g (x )
y Example 108. Solve the equation
log ( 3x 2 + x - 2) = 3 log ( 3x - 2).
Sol. This equation is equivalent to the system
ì
3x 2 + x - 2 > 0
í 2
3
î3x + x - 2 = (3x - 2)
Þ
( x - 2 / 3) ( x - 2) > 0
ì
í
(
x
2 / 3) (9 x 2 - 13x + 3) = 0
î
Þ
ì x < 2 / 3 and x > 2
ï
2
13 ± 61
í
x = ,x =
3
18
îï
Original equation has the only root x 1 =
13 - 61
×
18
Form 8 An equation of the form
log a f ( x ) + log a g ( x ) = log a m ( x ), a > 0 , a ¹ 1
is equivalent to the system
f (x ) > 0
ì
ï
g( x ) > 0
í
ï f ( x ) g ( x ) = m( x )
î
y Example 109. Solve the equation
2 log 3 x + log 3 ( x 2 - 3) = log 3 0.5 + 5log 5 (log 3 8 )
Sol. This equation can be written as
log 3 x 2 + log 3 ( x 2 - 3) = log 3 0.5 + log 3 8
log 3 x 2 + log 3 ( x 2 - 3) = log 3 ( 4 )
150
Textbook of Algebra
This is equivalent to the system
ì
x2 > 0
ï
2
í x -3>0
2
ï x ( x 2 - 3) = 4
î
Þ
ì x < 0 and x > 0
ï
Þ í x < - 3 and x > 3
ï ( x 2 - 4 ) ( x 2 + 1) = 0
î
x 2 - 4 = 0 \ x = ± 2, but x > 0
Consequently, x 1 = 2 is only root of the original equation.
Form 9 An equation of the form
log a f ( x ) - log a g ( x ) = log a h ( x ) - log a t ( x ), a > 0, a ¹ 1
is equivalent to the equation
log a f ( x ) + log a t ( x ) = log a g( x ) + log a h( x ),
which is equivalent to the system
f (x ) > 0
ì
ï
t(x ) > 0
ïï
g( x ) > 0
í
ï
h( x ) > 0
ï
ïî f ( x ) × t ( x ) = g( x ) × h( x )
y Example 110. Solve the equation
3p ö
æ
÷
ç sin
4 = 1 + log ( x + 7 ).
log 2 ( 3 - x ) - log 2 ç
÷
2
ç 5-x ÷ 2
ø
è
Logarithmic Inequations
When we solve logarithmic inequations
ìlog f ( x ) > log a g ( x )
(i) í a
îa > 1
ì g (x ) > 0
ï
Þ í a >1
ïf (x ) > g (x )
î
ìlog f ( x ) > log a g ( x )
(ii) í a
î 0 <a <1
ì f (x ) > 0
ï
Þ í 0 <a <1
ïf (x ) < g (x )
î
y Example 111. Solve the inequation
log 2x + 3 x 2 < log 2x + 3 (2x + 3).
Sol. This inequation is equivalent to the collection of the
systems
x > -1
éì
éì 2x + 3 > 1
êí( x - 3) ( x + 1) < 0
êï x 2 < 2x + 3
êî
êïí
Þ êì
3
ï - < x < -1
êï0 < 2x + 3 < 1
2
êí
êï x 2 > 2x + 3
ëî
ëêïî( x - 3) ( x + 1) > 0
Sol. This equation is equivalent to
æ 3p ö
ç sin ÷ 1
4 ÷ + log 2 + log ( x + 7 )
log 2 (3 - x ) = log 2 ç
2
2
ç5- x ÷ 2
ø
è
æ
ö
1
Þ log 2 (3 - x ) = log 2 ç
÷ + log 2 2 + log 2 ( x + 7 )
è 2( 5 - x ) ø
which is equivalent to the system
ì
ï
3- x >0
ï
1
ï
>0
ï
2 (5 - x )
í
ï x +7 >0
ï
ï(3 - x ) = 2 ( x + 7 )
ïî
2 (5 - x )
Þ
x <3
ì
ï
x <5
ï
í
x
> -7
ï
ïî( x - 1) ( x - 8) = 0
Hence, x 1 = 1 is only root of the original equation.
Þ
é ì x > -1
ê í -1 < x < 3
ê î
êìï - 3 < x < - 1
êí
2
êëïî x < - 1 and x > 3
Þ -1 < x < 3
Þ-
3
< x < -1
2
Hence, the solution of the original inequation is
ö
æ 3
x Î ç - , - 1÷ È ( -1, 3).
ø
è 2
Canonical Logarithmic Inequalities
ìlog x > 0
ìx > 1
1. í a
Þ í
î a >1
îa > 1
ì0 < x < 1
ìlog x > 0
2. í a
Þ í
î0 < a < 1
î 0 <a <1
ìlog x < 0
ì0 < x < 1
3. í a
Þ í
î a >1
î a >1
ìlog x < 0
ì x >1
4. í a
Þ í
î 0 <a <1
î0 < a < 1
Chap 02 Theory of Equations
Some Standard Forms to Solve
Logarithmic Inequations
Form 1 Inequations of the form
Forms
Collection of systems
(a) logg (x) f (x ) > 0
(b)
logg (x) f (x ) ³ 0
Û
ì f (x ) > 1, ì 0 < f (x ) < 1
í
í
î g (x ) > 1, î 0 < g (x ) < 1
Û
ì f (x ) ³ 1, ì 0 < f (x ) £ 1
í
í
î g (x ) > 1, î 0 < g (x ) < 1
(c) log g ( x ) f ( x ) < 0 Û
ì f ( x ) > 1, ì0 < f ( x ) < 1
í
í
î0 < g( x ) < 1, î g( x ) > 1
(d) log g ( x ) f ( x ) £ 0 Û
ì f ( x ) ³ 1, ì0 < f ( x ) £ 1
í
í
î0 < g( x ) < 1, î g( x ) > 1
y Example 112. Solve the inequation
2x ö
æ
log æ x 2 -12x + 30 ö ç log 2 ÷ > 0.
5ø
ç
֏
ç
è
10
Û
ì x < 6 - 6 and x > 6 + 6 and 2 < x < 10
í
0< x <5
î
Therefore, the system has solution 2 < x < 6 - 6
combining both systems, then solution of the original
inequations is
x Î (2, 6 - 6 ) È (10, ¥ ).
Form 2 Inequations of the form
Forms
Collection of systems
(a) log f ( x ) f ( x ) > log f ( x ) g( x )
Û
(b) log f ( x ) f ( x ) ³ log f ( x ) g( x ) Û
ì f ( x ) ³ g( x ),
ï
í g( x ) > 0,
ï f( x ) > 1,
î
ì f ( x ) £ g( x )
ï
í f (x ) > 0
ï 0 < f( x ) < 1
î
(c) log f ( x ) f ( x ) < log f ( x ) g( x ) Û
ì f ( x ) < g( x ),
ï
í f ( x ) > 0,
ï f( x ) > 1,
î
Sol. This inequation is equivalent to the collection of two
systems
ì
x 2 - 12x + 30
<1
ïï0 <
10
í
2x
ï 0 < log 2 æç ö÷ < 1
è 5 ø
ïî
ì f ( x ) > g( x )
ï
í g( x ) > 0
ï 0 < f( x ) < 1
î
On solving the first system, we have
Þ
ì x 2 - 12x + 20 > 0
ï
2x
í
>2
ïî
5
Û
ì( x - 10) ( x - 2) > 0
í
x >5
î
Û
ì x < 2 and x > 10
í
x >5
î
Therefore, the system has solution x > 10.
On solving the second system, we have
Þ
Û
ì0 < x 2 - 12x + 30 < 10
ï
2x
í
1<
<2
ïî
5
ì x 2 - 12x + 30 > 0 and x 2 - 12x + 20 < 0
í
5/2< x <5
î
ì f ( x ) > g( x ),
ï
í g( x ) > 0,
ï f( x ) > 1,
î
ì f ( x ) < g( x )
ï
í f (x ) > 0
ï 0 < f( x ) < 1
î
÷
ø
ì x 2 - 12x + 30
> 1,
ïï
10
í
ï log 2 æç 2x ö÷ > 1,
ïî
è 5 ø
151
(d) log f ( x ) f ( x ) £ log f ( x ) g( x ) Û
ì f ( x ) £ g( x ),
ï
í f ( x ) > 0,
ï f( x ) > 1,
î
ì f ( x ) ³ g( x )
ï
í g( x ) > 0
ï 0 < f( x ) < 1
î
y Example 113. Solve the inequation
log ( x - 3) (2 ( x 2 - 10x + 24 )) ³ log ( x - 3) ( x 2 - 9 ).
Sol. This inequation is equivalent to the collection of systems
ì2( x 2 - 10x + 24 ) ³ x 2 - 9,
ï
x 2 - 9 > 0,
í
ï
x - 3 > 1,
î
152
Textbook of Algebra
ì2( x 2 - 10x + 24 ) £ x 2 - 9
ï
2
í 2( x - 10x + 24 ) > 0
ï
0< x -3<1
î
On solving the first system, we have
On solving the second system, we have
ì x 2 - 20x + 57 ³ 0,
ï
í( x + 3) ( x - 3) > 0,
ï
x > 4,
î
3
10 – 43
10 + 43
6
4
ì x - 20x + 57 £ 0,
ï
í( x - 6) ( x - 4 ) > 0,
ï
3 < x < 4,
î
ì x Î [10 - 43, 10 + 43 ]
ï
í x Î ( - ¥, 4 ) È (6, ¥ )
ï
x Î (3, 4 )
î
2
ì x Î ( - ¥, 10 - 43 ] È [10 + 43, ¥ )
ï
x Î ( - ¥, - 3) È (3, ¥ )
í
ï
x Î ( 4, ¥ )
î
Therefore, the system has solution
Û
Û
Therefore, the system has solution
10 - 43 £ x < 4,
–3
3
10 – 43
4
On combining the both systems, the solution of the original
inequation is
x Î [10 - 43, 4 ) È [10 + 43, ¥ ).
x ³ 10 + 43
x Î [10 + 43, ¥ )
i.e.
#L
1.
Exercise for Session 5
The equation ( x + 1) - ( x - 1) = (4x - 1) has
(a) no solution
2.
(c) two solutions
(d) more than two solutions
(b) two
(c) three
(d) None of these
The number of real solutions of (3x 2 - 7x - 30) - (2x 2 - 7x - 5) = x - 5 is
(a) one
4.
(b) one solution
The number of real solutions of ( x 2 - 4x + 3) + ( x 2 - 9) = (4x 2 - 14x + 6) is
(a) one
3.
x Î [10 - 43, 4 )
i.e.,
10 + 43
(b) two
(c) three
(d) None of these
2
The number of integral values of x satisfying ( - x + 10x - 16) < x - 2 is
(a) 0
(b) 1
(c) 2
(d) 3
x
5.
æ9ö
The number of real solutions of the equation ç ÷ = - 3 + x - x 2 is
è 10 ø
(a) 2
6.
The set of all x satisfying 3
(a) 0 < x < 1
7.
(d) None of these
(c) x > 3-2
(d) None of these
- 3 - 6 > 0 is given by
(b) x > 1
/ 2
+ ( 2 + 1)x = (3 + 2 2 )x
(b) two
(c) four
(b) 0
/ 2
is
(d) infinite
2
-3
(c) 2
+ 1 = (32 + 8 15 )x
2
-3
is
(d) None of these
The number of real solutions of the equation log0. 5 x = | x | is
(a) 0
10.
(c) 0
x
The sum of the values of x satisfying the equation (31 + 8 15 )x
(a) 3
9.
2x
The number of real solutions of the equation 2x
(a) one
8.
(b) 1
(b) 1
(c) 2
(d) None of these
(c) 0 < x < 1
(d) None of these
The inequality ( x - 1)ln (2 - x ) < 0 holds, if x satisfies
(a) 1 < x < 2
(b) x > 0
Shortcuts and Important Results to Remember
1 ‘0’ is neither positive nor negative even integer, ‘2’ is the
only even prime number and all other prime numbers are
odd, ‘1’ (i.e. unity) is neither a composite nor a prime
number and 1, -1are two units in the set of integers.
2
(i) If a > 0, b > 0 and a < b Þ a2 < b2
(ii) If a < 0, b < 0 and a < b Þ a2 > b2
(iii) If a1, a2 , a3 , ..., an Î R
and
a12
+
a22
+
a32
+K+
an2
=0
2 a2c = ab2 + bc 2 , i.e. ab2 , bc 2 , ca2 are in AP
2a b c
c a b
or
i.e. , , are in AP.
= +
b c a
a b c
13 Given, y = ax 2 + bx + c
(i) If a > 0, ymin =
4ac - b2
4a
(ii) If a < 0, ymax =
4ac - b2
4a
14 If a, b are the roots of ax 2 + bx + c = 0 and S n = a n + b n ,
then aS n + 1 + bS n + c S n - 1 = 0.
Þ a1 = a2 = a3 = K = an = 0
1
3 (i) Max (a, b) = (| a + b| + | a - b|)
2
1
(ii) Min (a, b) = (| a + b| - | a - b|)
2
15 If D1 and D2 are discriminants of two quadratics P( x ) = 0
and Q( x ) = 0, then
4 If the equation f ( x ) = 0 has two real roots a and b , then
f ¢ ( x ) = 0 will have a real root lying between a and b.
5 If two quadratic equations P( x ) = 0 and Q ( x ) = 0 have an
irrational common root, both roots will be common.
6 In the equation ax 2 + bx + c = 0 [a, b, c Î R ], if
c
a + b + c = 0, the roots are 1, and if a - b + c = 0, the
a
c
roots are -1and .
a
7 The condition that the roots of ax 2 + bx + c = 0 may be in
the ratio p : q, is
pq b2 = ac ( p + q )2 (here, a : b = p : q )
2
p
q
b
+
=±
q
p
ac
(i) If one root of ax 2 + bx + c = 0 is n times that of the
other, then nb2 = ac (n + 1)2 , here a : b = n : 1.
i.e.,
(ii) If one root of ax 2 + bx + c = 0 is double of the other
here n = 2, then 2 b2 = 9ac .
(i) If D1D2 < 0, then the equation P( x )× Q( x ) = 0 will have
two real roots.
(ii) If D1D2 > 0, then the equation P( x )× Q( x ) = 0 has either
four real roots or no real root.
(iii) If D1D2 = 0, then the equation P( x )× Q( x ) = 0 will have
(a) two equal roots and two distinct roots such
D1 > 0 and D2 = 0 or D1 = 0 and D2 > 0.
(b) only one real solution such that
D1 < 0 and D2 = 0 or D1 = 0 and D2 < 0.
16 If a > 0 and x = a + a + a + K+ ¥ , then x =
1
is n 2 .
æ 1 1 1ö
(i) Least value of (a + b + c ) ç + + ÷ = 32 = 9
èa b cø
(ii) Least value of
æ 1 1 1 1ö
(a + b + c + d ) ç + + + ÷ = 42 = 16
èa b c d ø
1
9 If one root of ax 2 + bx + c = 0 is square of the other, then
a2c + ac 2 + b3 = 3abc .
10 If the ratio of the roots of the equation ax 2 + bx + c = 0 is
equal to the ratio of the roots of
Ax 2 + Bx + C = 0 and
2
2
b
B
.
a ¹ 0, A ¹ 0, then
=
ac AC
11 If sum of the roots is equal to sum of their squares then
2 ac = ab + b2 .
12 If sum of roots of ax 2 + bx + c = 0 is equal to the sum of
their reciprocals, then
1 + (4a + 1)
.
2
17 If a1, a2 , a3 , ..., an are positive real numbers, then least
æ1
1
1
1ö
value of (a1 + a2 + a3 + K + an ) ç +
+
+K+ ÷
è a1 a2 a3
an ø
8 If one root of ax 2 + bx + c = 0 is nth power of the other,
then (anc )n + 1 + (ac n )n + 1 = - b.
that
a c e
= = = K, then each of
b d
f
these ratios is also equal to
a+c +e +K
(i)
b+d + f +K
18 Law of Proportions If
æ pan + qc n + re n + K ö
(ii) ç n
÷
è pb + qd n + rf n + K ø
(iii)
1/ n
(where, p, q, r , ..., n Î R )
n (ace ... )
ac
=
n (bdf K )
bd
19 Lagrange’s Mean Value Theorem Let f ( x ) be a function
defined on [a, b] such that
contd...
(i) f ( x ) is continuous on [a, b] and
Let g( x ) = ( x - a ) be a linear monic polynomial a Î R.
(ii) f ( x ) is derivable on (a, b), then c Î(a, b) such that
f (b) - f (a)
f ¢ (c ) =
b-a
When g( x )| f ( x ); we can find quotient and remainder as
follows :
a
20 Lagrange’s Identity If a1, a2 , a3 , b1, b2 , b3 Î R, then
(a12 + a22 + a32 ) (b12 + b22 + b32 ) - (a1b1 + a2 b2 + a3 b3 )2
= (a1b2 - a2 b1 )2 + (a2 b3 - a3 b2 )2 + (a3 b1 - a1b3 )2
or (a12
+
a22
+
a32 ) (b12
=
+
a1 a2
b1 b2
b22
+
2
+
b32 ) - (a1b1
a2
b2
2
2
+ a2 b2 + a3 b3 )
a3
a3
+
b3
b3
a1
b1
2
Remark
If (a12 + a22 + a32 ) (b12 + b22 + b32 ) £ (a1b1 + a2 b2 + a3 b3 )2 ,
a1 a2 a3
then
=
= .
b1 b2 b3
21 Horner’s Method of Synthetic, Division When, we
divide a polynomial of degree ³ 1 by a linear monic
polynomial, the quotient and remainder can be found by
this method. Consider
f ( x ) = a0 x n + a1 x n - 1 + a2 x n - 2 + K + an
where a0 ¹ 0 and a0 , a1, a2 , ..., an Î R.
an
a0
a1
a2
0
a a0
b1a
a bn - 1
a1
a2
an + ab n - 1 = 0
a0
+ a a0
+ b1a
= b0
= b1
= b2
\ f ( x ) = ( x - a ) (b0 x n - 1 + b1 x
3
n-2
…
+ b2 x
n-3
+ K + bn - 1 )
2
e.g. Find all roots of x - 6 x + 11x - 6 = 0.
Q( x - 1) is a factor of x 3 - 6 x 2 + 11x - 6, then
x=1
\
1
-6
11
-6
0
1
-5
6
1
-5
6
0
x 3 - 6 x 2 + 11x - 6 = ( x - 1) ( x 2 - 5 x + 6)
= ( x - 1) ( x - 2 ) ( x - 3)
3
Hence, roots of x - 6 x 2 + 11x - 6 = 0 are 1, 2 and 3.
JEE Type Solved Examples :
Single Option Correct Type Questions
n
This section contains 10 multiple choice examples.
Each example has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct.
l
Ex. 3 Let f ( x ) = ò
(a) 0 < a < b(b) a < 0 < b < |a |
(c) a < b < 0(d) a < 0 < |a | < b
Sol. (b) Q
…(i)
a + b = - b, ab = c
Q
c < 0 Þ ab < 0
Let
a < 0, b > 0
\
| a | = - a and a < 0 < b [Q a < b] …(ii)
From Eq. (i), we get - | a | + b < 0
…(iii)
Þ
b < |a |
From Eqs. (ii) and (iii), we get
a < 0 < b < |a |
Ex. 2 Let a, b be the roots of the equation x 2 - x + p = 0
and g, d be the roots of the equation x 2 - 4 x + q = 0. If
a, b, g and d are in GP, the integral values of p and q respectively, are
l
(a) -2, - 32 (b) -2, 3
(c) -6, 3
(d) -6, - 32
Sol. (a) Let r be the common ratio of the GP, then
b = ar , g = ar 2 and d = ar 3
2
or
and
ar (1 + r ) = 4
(ar 2 )(ar 3 ) = q
or
a 2r 5 = q
(d) 0 and 1
Sol. (a) We have, f ( x ) =
(2 - t 2 ) dt
f ¢ ( x ) = (2 - x 2 )
Þ
\
x
ò1
x 2 - f ¢( x ) = 0
Þ x 2 - (2 - x 2 ) = 0 Þ x 4 + x 2 - 2 = 0
Þ
x 2 = 1, - 2
Þ
x = ±1
[only for real value of x ]
Ex. 4 If x 2 + 3 x + 5 = 0 and ax 2 + bx + c = 0 have a
common root and a, b, c Î N , the minimum value of a + b + c
is
l
(a) 3
(c) 6
(b) 9
(d) 12
Sol. (b) Q Roots of the equation x 2 + 3x + 5 = 0 are non-real.
Ex. 5 If x 1 , x 2 , x 3 , K, x n are the roots of the equation
x + ax + b = 0, the value of
( x 1 - x 2 )( x 1 - x 3 )( x 1 - x 4 ) K ( x 1 - x n ) is
l
…(ii)
…(iii)
n
(a) nx 1 + b
(b) n ( x 1 )n - 1
gd = q
Þ
1
2
…(i)
3
2
(c) ±
1
2
Thus, given equations will have two common roots.
a b c
[say]
= = =l
Þ
1 3 5
\
a + b + c = 9l
Thus, minimum value of a + b + c = 9
[Qa, b, c Î N ]
g + d = 4 Þ ar + ar = 4
and
(b) ±
(a) ±1
Ex. 1 If a and b (a < b), are the roots of the equation
2
x + bx + c = 0, where c < 0 < b, then
a + b = 1 Þ a + ar = 1
a(1 + r ) = 1
ab = p Þ a (ar ) = p
a 2r = p
( 2 - t 2 ) dt, the real roots of the
equation x 2 - f ¢ ( x ) = 0 are
l
\
or
and
or
x
1
(c) n ( x 1 )n - 1 + a
…(iv)
On dividing Eq. (iii) by Eq. (i), we get
r 2 = 4 Þ r = - 2, 2
If we take r = 2, then a is not integer, so we take r = - 2.
On substituting r = - 2 in Eq. (i), we get a = - 1
(d) n ( x 1 )n - 1 + b
Sol. (c) Q x n + ax + b = ( x - x 1 )( x - x 2 )( x - x 3 )K( x - x n )
Þ ( x - x 2 )( x - x 3 )K( x - x n ) =
On taking lim both sides, we get
x ®x 1
Now, from Eqs. (ii) and (iv), we get
p = a 2r = ( -1)2 ( -2) = - 2
and
Hence,
q = a 2r 5 = ( -1)2 ( -2)5 = - 32
( p , q ) = ( -2, - 32)
x n + ax + b
x - x1
x n + ax + b
® x1
x - x1
( x 1 - x 2 )( x 1 - x 3 )K( x 1 - x n ) = lim
x
= lim
x ® x1
nx
n -1
1
+a
= n ( x 1 )n - 1 + a
ù
é0
êë 0 form úû
156
Textbook of Algebra
Ex. 6 If a, b are the roots of the equation ax 2 + bx + c = 0
and An = a n + bn , then a An + 2 + bAn + 1 + cAn is equal to
l
1
1
= u and
= v, then
x
(1 - x 2 )
35
and u 2 + v 2 = u 2v 2
u +v =
12
2
æ 35 ö
Þ
(u + v )2 = ç ÷
è 12 ø
2
æ 35 ö
u 2 + v 2 + 2uv = ç ÷
Þ
è 12 ø
2
æ 35 ö
2 2
Þ
u v + 2uv = ç ÷
[Qu 2 + v 2 = u 2v 2 ]
è 12 ø
2
æ 35 ö
Þ
u 2v 2 + 2uv - ç ÷ = 0
è 12 ø
49 ö æ
25 ö
æ
Þ
çuv + ÷ çuv - ÷ = 0
è
ø
è
12
12 ø
49
25
Þ
uv = - , uv =
12
12
49
Case I If uv = - , then
12
1
1
49
[here x < 0]
×
=2
x (1 - x )
12
Sol. (d) Let
(b) 1
(c) a + b + c (d) abc
b
c
Sol. (a) Qa + b = - and ab =
a
a
n+2
n+2
+b
\ An + 2 = a
(a) 0
= (a + b )(a n + 1 + bn + 1 ) - abn + 1 - ba n + 1
= (a + b )(a n + 1 + bn + 1 ) - ab(a n + bn )
c
b
= - An + 1 - An
a
a
Þ a An + 2 + b An + 1 + c An = 0
Ex. 7 If x and y are positive integers such that
xy + x + y = 71, x 2 y + xy 2 = 880, then x 2 + y 2 is equal to
l
(a) 125
(b) 137
(c) 146
(d) 152
Sol. (c) Q xy + x + y = 71 Þ xy + ( x + y ) = 71
and
x 2y + xy 2 = 880 Þ xy ( x + y ) = 880
Þ xy and ( x + y ) are the roots of the quadratic equation.
t 2 - 71t + 880 = 0
Þ
Þ
\
So,
x4 - x2 +
Þ
(t - 55)(t - 16) = 0
t = 55, 16
x + y = 16 and xy = 55
x 2 + y 2 = ( x + y )2 - 2xy = (16)2 - 110 = 146
(c) x 2 - 15x + 158 = 0
(d) x 2 + 15x + 158 = 0
= - 3(a 2 + b 2 ) + 5( g 2 + d 2 )
= - 3[(a + b )2 - 2ab ] + 5[( g + d )2 - 2gd ]
= - 3[9 - 10] + 5[25 + 6] = 158
\ Required equation is x 2 + 15x + 158 = 0.
Ex. 9 The number of roots of the equation
1
1
35
+
= is
2
x
12
(1 - x )
(b) 1
(d) 3
(25)2
=0
On combining both cases,
x=-
(5 + 73 ) 3 4
, ,
14
5 5
Hence, number of roots = 3
Ex. 10 The sum of the roots of the equation
2
+ 2 11x + 2 = 2 22 x + 1 + 1 is
l
33 x - 2
(a)
1
11
(b)
2
11
(c)
3
11
(d)
4
11
Sol. (b) Let 211x = t , given equation reduces to
Þ
t3
+ 4t = 2t 2 + 1
4
t 3 - 8t 2 + 16t - 4 = 0 Þ t 1 × t 2 × t 3 = 4
Þ
211x 1 × 211x 2 × 211x 3 = 4
l
(a) 0
(c) 2
(12)2
[here x > 0]
3 4
æ 2 9 ö æ 2 16 ö
çx - ÷ çx - ÷ = 0 Þ x = ,
è
ø
è
ø
5 5
25
25
Þ
Sum of roots = (ag + bd ) + (ad + bg )
= (a + b ) ( g + d ) = 3 ´ ( -5) = ( -15)
Product of roots = (ag + bd ) (ad + bg )
= a 2 gd + abg 2 + bad 2 + b 2 gd
= gd (a 2 + b 2 ) + ab( g 2 + d 2 )
(5 + 73 )
14
25
, then
12
1
1
25
×
=
x (1 - x 2 ) 12
x4 - x2 +
Þ
Sol. (d) Q a + b = 3, ab = 5, g + d = ( - 5), g d = ( - 3)
=0
x=-
Case II If uv =
Ex. 8 If a, b are the roots of the equation x 2 - 3 x + 5 = 0
and g, d are the roots of the equation x 2 + 5 x - 3 = 0, then
the equation whose roots are a g + b d and ad + bg, is
(b) x 2 + 15x - 158 = 0
( 49 )2
Þ
l
(a) x 2 - 15x - 158 = 0
(12)2
Þ
\
Þ 211(x 1 + x 2 + x 3 ) = 22
11( x 1 + x 2 + x 3 ) = 2
2
x1 + x 2 + x 3 =
11
Chap 02 Theory of Equations
157
JEE Type Solved Examples :
More than One Correct Option Type Questions
n
l
This section contains 5 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
more than one may be correct.
Ex. 11 For the equation 2 x 2 - 6 2 x - 1 = 0
Ex. 14 If cos 4 q + p, sin 4 q + p are the roots of the equation x 2 + a ( 2 x + 1) = 0 and cos 2 q + q, sin 2 q + q are the
roots of the equation x 2 + 4 x + 2 = 0 then a is equal to
l
(a) -2
(a) roots are rational
(b) roots are irrational
(c) if one root is ( p + q ), the other is ( - p + q )
need not appear in conjugate pair.
1
Here, a + b = 3 2 and ab = 2
Let a = p + q , then prove that other root b = - p + q .
Ex. 12 Given that a, g are roots of the equation
Ax 2 - 4 x + 1 = 0 and b, d the roots of the equation
Bx 2 - 6 x + 1 = 0, such that a, b, g and d are in HP then
l
(c) B = 2
(a - 3d ) + (a + d ) = 4 Þ 2(a - d ) = 4
and
(a - d ) + (a + 3d ) = 6 Þ 2(a + d ) = 6
5
1
\
a = and d =
2
2
æ5 3ö æ5 1ö
Now, A = (a - 3d )(a + d ) = ç - ÷ ç + ÷ = 3
è2 2ø è2 2ø
æ5 1ö æ5 3ö
and B = (a - d )(a + 3d ) = ç - ÷ ç + ÷ = 8
è2 2ø è2 2ø
2
Ex. 13 If | ax + bx + c | £ 1 for all x in [0, 1], then
(b) |b | > 8
(d) |a | + |b | + |c | £ 17
1
Sol. (a,c,d) On putting x = 0, 1 and , we get
2
|c | £ 1
(a) |a | £ 8
(c) |c | £ 1
and
…(ii)
| a + 2b + 4c | £ 4
…(iii)
| b | £ 8 and | a | £ 8
Þ
…(i)
|a + b + c | £ 1
From Eqs. (i), (ii) and (iii), we get
| a | + | b | + | c | £ 17
Q
cos 4 q - sin 4 q = cos 2q
Þ
cos 4 q - sin 4 q = cos 2 q - sin 2 q
Þ
Þ
or
é
Dù
êQa - b = a ú
ë
û
2
2
4a - 4a = 8 or a - a - 2 = 0
(a - 2)(a + 1) = 0 or a = 2, - 1
4a 2 - 4a
16 - 8
=
1
1
Ex. 15 If a, b, g are the roots of x 3 - x 2 + ax + b = 0 and
b, g, d are the roots of x 3 - 4 x 2 + mx + n = 0. If a, b, g and d
are in AP with common difference d then
l
(d) B = 8
1 1 1
1
Sol. (a,d) Since, a , b, g and d are in HP, hence , , and
a b g
d
are in AP and they may be taken as a - 3d , a - d , a + d
1
and a + 3d . Replacing x by , we get the equation whose
x
roots are a - 3d , a + d is x 2 - 4 x + A = 0 and equation
whose roots are a - d , a + 3d is x 2 - 6x + B = 0, then
l
(d) 2
Þ (cos 4 q + p ) - (sin 4 q + p ) = (cos 2 q + q ) - (sin 2 q + q )
Sol. (b,c) As the coefficients are not rational, irrational roots
(b) A = 4
(c) 1
Sol. (b,d)
(d) if one root is ( p + q ), the other is ( p - q )
(a) A = 3
(b) -1
(a) a = m
(c) n = b - a - 2
(b) a = m - 5
(d) b = m + n - 3
Sol. (b,c,d)
Qa,b, g , d are in AP with common difference d, then
b = a + d, g = a + 2d and d = a + 3d
Given, a,b, g are the roots of x 3 - x 2 + ax + b = 0, then
...(i)
...(ii)
a + b + g =1
...(iii)
ab + bg + ga = a
...(iv)
abg = -b
Also, b, g , d are the roots of x 3 - 4 x 2 + mx + n = 0, then
b+g+d= 4
bg + gd + db = m
bgd = -n
From Eqs. (i) and (ii), we get
3a + 3d = 1
and from Eqs. (i) and (v), we get
3a + 6d = 4
From Eqs. (viii) and (ix), we get
2
d = 1, a = 3
Now, from Eq. (i), we get
1
4
7
b = , g = and d =
3
3
3
From Eqs. (iii), (iv), (vi) and (vii), we get
2
8
13
28
a = - , b = ,m = ,n = 3
27
3
27
\
a = m - 5, n = b - a - 2 and b = m + n - 3
...(v)
...(vi)
...(vii)
...(viii)
...(ix)
158
Textbook of Algebra
JEE Type Solved Examples :
Passage Based Questions
n
This section contains 2 solved passages based upon each
of the passage 3 multiple choice examples have to be
answered. Each of these examples has four choices (a), (b),
(c) and (d) out of which ONLY ONE is correct.
19. (b - c ) 2 + (c - a ) 2 + (d - b ) 2 is equal to
(a) a - d
Sol. (b) Let b = ar , c = ar 2 and d = ar 3
(Ex. Nos. 16 to 18)
If G and L are the greatest and least values of the expression
x2 - x + 1
, x Î R respectively, then
x2 + x + 1
= (ar - ar 2 )2 + (ar 2 - a )2 + (ar 3 - ar )2
= a 2r 2 (1 - r )2 + a 2 (r 2 - 1)2 + a 2r 2 (r 2 - 1)2
= a 2 (1 - r )2 {r 2 + (r + 1)2 + r 2 (r + 1)2 }
= a 2 (1 - r )2 (r 4 + 2r 3 + 3r 2 + 2r + 1)
5
16. The least value of G + L is
(a) 0
(b) 2
= a 2 (1 - r )2 (1 + r + r 2 )2 = a 2 (1 - r 3 )2
(c) 16
y=
Sol. (b) Let
(d) 32
x - x +1
17. G and L are the roots of the equation
(b) 4 x 2 - 17 x + 4 = 0
(d) x 2 - 5x + 6 = 0
Sol. (a) Equation whose roots are G and L, is
Þ
= (a - ar 3 )2 = (a - d )2
2
x2 + x + 1
Þ
x 2y + xy + y = x 2 - x + 1
[Q x Î R ]
Þ
( y - 1) x 2 + ( y + 1) x + y - 1 = 0
2
2
[Qb - 4ac ³ 0]
\
( y + 1) - 4 × ( y - 1) ( y - 1) ³ 0
Þ
(y + 1)2 - (2y - 2)2 ³ 0
Þ
(3y - 1) (y - 3) £ 0
1
1
\
£ y £ 3 Þ G = 3 and L = \ GL = 1
3
3
G 5 + L5
1/ 5
1/ 5
=1
³ (GL ) = (1)
2
5
5
G +L
³ 1 or G 5 + L5 ³ 2
Þ
2
\ Minimum value of G 5 + L5 is 2.
(a) 3 x 2 - 10x + 3 = 0
(c) x 2 - 7 x + 10 = 0
x 2 - (G + L ) x + GL = 0
10
x 2 - x + 1 = 0 or 3x 2 - 10x + 3 = 0
3
20. (u + v + w ) is equal to
1
(c) 20
2
Sol. (a) Now,
u + 2v + 3w = 6
4 u + 5v + 6w = 12
and
6u + 9v = 4
From Eqs. (i) and (ii), we get
2u + v = 0
Solving Eqs. (iii) and (iv), we get
1
2
u = - ,v =
3
3
5
Now, from Eq. (i), we get w =
3
1 2 5
v +u +w = - + + =2
\
3 3 3
(a) 2
(b) 3
(c) 4
1
< l <3
3
Sum of values of l = 1 + 2 = 3
Sol. (b) Q L < l < G Þ
(d) 5
\ l = 1, 2
Passage II
(Ex. Nos. 19 to 21)
Let a, b, c and d are real numbers in GP. Suppose u, v, w satisfy
the system of equations u + 2v + 3w = 6, 4u + 5v + 6w = 12 and
6u + 9v = 4. Further, consider the expressions
æ1 1 1ö
f ( x ) = ç + + ÷ x 2 + [( b - c ) 2 + ( c - a ) 2 + ( d - b ) 2 ]
è u v wø
x + u + v + w = 0 and g (x ) = 20x 2 + 10 ( a - d ) 2 x - 9 = 0.
(b)
(d)
1
20
…(i)
…(ii)
…(iii)
…(iv)
21. If roots of f ( x ) = 0 be a, b, the roots of g ( x ) = 0 will be
(a) a , b
Sol. (c) Now,
18. If L < l < G and l Î N , the sum of all values of l is
(a) 2
(d) (a + d )2
Now, (b - c )2 + (c - a )2 + (d - b )2
Passage I
5
(b) (a - d )2 (c) a 2 - d 2
(b) - a , - b
(c)
1 1
,
a b
(d) -
1
1
,a
b
æ1 1 1 ö
f ( x ) = ç + + ÷ x 2 + [(b - c )2
èu v w ø
+ (c - a )2 + (d - b )2 ] x + u + v + w = 0
9 2
x + (a - d )2 x + 2 = 0
10
Þ
f (x ) = -
Þ
f ( x ) = - 9 x 2 + 10(a - d )2 x + 20 = 0
Given, roots of f ( x ) = 0 are a and b.
1
Now, replace x by in Eq. (v), then
x
-9 10(a - d )2
+
+ 20 = 0
x
x2
2
2
Þ
20x + 10(a - d ) x - 9 = 0
g(x ) = 0
1 1
Roots of g ( x ) = 0 are , .
\
a b
…(v)
159
Chap 02 Theory of Equations
JEE Type Solved Examples :
Single Integer Answer Type Questions
n
Ex. 23 If a root of the equation
n 2 sin 2 x - 2 sin x - ( 2n + 1) = 0 lies in [0, p / 2 ], the
minimum positive integer value of n is
This section contains 2 examples. The answer to each
example is a single digit integer ranging from 0 to 9
(both inclusive).
l
Ex. 22 If the roots of the equation 10 x 3 - cx 2
- 54 x - 27 = 0 are in harmonic progression, the value of c is
l
Sol. (3) Q n 2 sin 2 x - 2sin x - (2n + 1) = 0
sin x =
Þ
2 ± 4 + 4n 2 (2n + 1)
2n 2
[by Shridharacharya method]
Sol. (9) Given, roots of the equation
10x 3 - cx 2 - 54 x - 27 = 0 are in HP.
1
Now, replacing x by in Eq. (i), we get
x
27 x 3 + 54 x 2 + cx - 10 = 0
Hence, the roots of Eq. (ii) are in AP.
Let a - d , a and a + d are the roots of Eq. (ii).
54
Then,
a -d + a + a +d = 27
2
Þ
a=3
Since, a is a root of Eq. (ii), then
27a 3 + 54a 2 + ca - 10 = 0
æ 8ö
æ4ö
æ 2ö
Þ 27 ç - ÷ + 54 ç ÷ + c ç - ÷ - 10 = 0
è 27 ø
è9 ø
è 3ø
2c
Þ
= 0 or c = 9
63
…(i)
=
…(ii)
Þ
Þ
…(iii)
[from Eq. (iii)]
n2
0 £ sin x £ 1
1 + (2n 3 + n 2 + 1)
0£
£1
n2
0 £ 1 + (2n 3 + n 2 + 1) £ n 2
Q
Þ
1 ± (2n 3 + n 2 + 1)
[Q x Î [0, p/2]]
(2n 3 + n 2 + 1) £ (n 2 - 1)
[Qn > 1]
On squaring both sides, we get
2n 3 + n 2 + 1 £ n 4 - 2n 2 + 1
n 4 - 2n 3 - 3n 2 ³ 0
Þ
Þ
n 2 - 2n - 3 ³ 0 Þ (n - 3) (n + 1) ³ 0
Þ
n ³3
\
n = 3, 4, 5, K
Hence, the minimum positive integer value of n is 3.
JEE Type Solved Examples :
Matching Type Questions
n
This section contains 2 examples. Examples 24 and 25
have three statements (A, B and C) given in Column I and
four statements (p, q, r and s) in Column II. Any given
statement in Column I can have correct matching with
one or more statement(s) given in Column II.
Ex. 24 Column I contains rational algebraic expressions
and Column II contains possible integers which lie in their
range. Match the entries of Column I with one or more
entries of the elements of Column II.
l
Column I
2
Column II
(A)
y=
x - 2x + 9
, x ÎR
x 2 + 2x + 9
(p)
1
(B)
x 2 - 3x - 2
y=
, x ÎR
2x - 3
(q)
3
2
(C)
y=
2x - 2x + 4
, x ÎR
x2 - 4 x + 3
(r)
-4
Sol. (A) ® (p); (B) ® (p, q, r, s); (C) ® (p, q, s)
Þ
-9
x 2 - 2x + 9
x 2 + 2x + 9
Þ x 2y + 2xy + 9y = x 2 - 2x + 9
( y - 1) x 2 + 2x ( y + 1) + 9 ( y - 1) = 0
Q
\
x ÎR
4 ( y + 1) 2 - 4 × 9 × ( y - 1) 2 ³ 0
Þ
(y + 1)2 - (3y - 3)2 ³ 0
Þ
Þ
\
(B) Q
Þ
\
Þ
(s)
y=
(A)
\
( 4y - 2) ( -2y + 4 ) ³ 0
(2y - 1) (y - 2) £ 0
1
£ y £ 2 Þ y = 1, 2 (p)
2
x 2 - 3x - 2
Þ 2xy - 3y = x 2 - 3x - 2
y=
2x - 3
x 2 - x (3 + 2y ) + 3y - 2 = 0 Q x Î R
(3 + 2y )2 - 4 × 1 × (3y - 2) ³ 0
4y 2 + 17 ³ 0
y Î R (p, q, r, s)
160
Textbook of Algebra
y=
(C)Q
Sol. (A) ® ( r, s , t ); ( B) ® ( p , q , r ); ( C) ® ( r , s , t )
2x 2 - 2x + 4
x 2 - 4x + 3
(A) Let
Then,
f ( 1) = a + b + c = - c
and
f ( 0) = c
\
f ( 0) f ( 1) = - c 2 < 0
x 2y - 4 xy + 3y = 2x 2 - 2x + 4
x 2 (y - 2) + 2x (1 - 2y ) + 3y - 4 = 0
x ÎR
4 (1 - 2y )2 - 4 (y - 2) (3y - 4 ) ³ 0
( 4y 2 - 4y + 1) - (3y 2 - 10y + 8) ³ 0
y 2 + 6y - 7 ³ 0
( y + 7 ) ( y - 1) ³ 0
y £ - 7 or y ³ 1(p,q,s)
Þ
Þ
Q
\
Þ
Þ
Þ
\
(B) Let
\
\
Ex. 25 Entries of Column I are to be matched with one
or more entries of Column II.
Column II
(A)
If a + b + 2c = 0 but c ¹ 0, then
ax 2 + bx + c = 0 has
atleast one root in
(p) (-2, 0)
(B)
If a, b, c Î R such that
2a - 3 b + 6 c = 0, then equation has
atleast one root in
(q) (-1, 0)
Let a, b, c be non-zero real numbers
such that
(r)
(C)
1
ò0 (12+ cos x ) (ax + bx + c) dx
= ò (1 + cos8 x ) (ax 2 + bx + c) dx ,
0
2
8
2
[Qa + b + 2c = 0]
[Qc ¹ 0]
\Equation f ( x ) = 0 has a root in (0, 1).
\ f ( x ) has a root in (0, 2) as well as in ( -1, 1) (r)
l
Column I
f ( x ) = ax 2 + bx + c
and
f ¢( x ) = ax 2 + bx + c
ax 3 bx 2
+
+ cx + d
3
2
f ( 0) = d
a b
æ 2a - 3b + 6c ö
f ( - 1) = - + + c + d = - ç
÷ +d
è
ø
3 2
6
f (x ) =
[Q 2a - 3b + 6c = 0]
= 0+d = d
Hence, f (0) = f ( -1)
Hence, f ¢ ( x ) = 0 has atleast one root in ( -1,0) (q)
\
f ( x ) = 0 has a root in ( -2,0) (p) as well as (-1,1) (r)
(C) Let f ( x ) = ò(1 + cos 8 x )(ax 2 + bx + c )dx
atleast one root in
(-1, 1)
Given, f (1) - f (0) = f (2) - f (0)
Þ
f ( 1) = f ( 2)
Þ
f ¢ ( x ) = 0 has atleast one root in (0,1).
Þ (1 + cos 8 x )(ax 2 + bx + c ) = 0 has atleast one root in (0,1).
(s) atleast one root in
(0, 1)
the equation ax + bx + c = 0 has
Þ ax 2 + bx + c = 0 has atleast one root in (0, 1) (s)
\ ax 2 + bx + c = 0 has a root in (0, 2) (t) as well as in
(-1, 1)(r)
(t) atleast one root in
(0, 2)
JEE Type Solved Examples :
Statement I and II Type Questions
n
Directions Example numbers 26 and 27 are
Assertion-Reason type examples. Each of these examples
contains two statements:
Statement-1 (Assertion) and Statement-2 (Reason)
Each of these examples also has four alternative choices,
only one of which is the correct answer. You have to select
the correct choice as given below.
(a) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
Ex. 26 Statement 1 Roots of x 2 - 2 3 x - 46 = 0 are
rational.
Statement 2 Discriminant of x 2 - 2 3 x - 46 = 0 is a
perfect square.
l
2
Sol. (d) In ax + bx + c = 0, a, b, c Î Q
[here Q is the set of rational number]
If D > 0 and is a perfect square, then roots are real, distinct
and rational.
But, here
2 3 ÏQ
\ Roots are not rational.
2 3 ± (12 + 184 )
2
3 ± 7. [irrational]
Here, roots are
i.e.
But
D = 12 + 184 = 196 = (14 )2
\ Statement-1 is false and Statement-2 is true.
Ex. 27 Statement 1 The equation a x + b x + c x - d x = 0
has only one real root, if a > b > c > d .
Statement 2 If f (x ) is either strictly increasing or decreasing function, then f (x ) = 0 has only one real root.
l
Sol. (c) Q a x + b x + c x - d x = 0
Þ
ax + bx + c x = d x
Chap 02 Theory of Equations
x
x
\ f ( x ) is increasing function and
Þ f ( x ) has only one real root.
But Statement-2 is false.
x
æc ö
æb ö
æa ö
Let f ( x ) = ç ÷ + ç ÷ + ç ÷ - 1
èd ø
èd ø
èd ø
x
\
x
x
æb ö æc ö
æc ö
æa ö æb ö
æa ö
f ¢( x ) = ç ÷ ln ç ÷ + ç ÷ ln ç ÷ + ç ÷ ln ç ÷ > 0
èd ø èd ø
èd ø
èd ø èd ø
èd ø
For example, f ( x ) = e
solution.
and f (0) = 2
x
161
lim f ( x ) = - 1
x ®- ¥
is increasing but f ( x ) = 0 has no
Subjective Type Examples
n
On substituting p = 18 in the given equation, we obtain
In this section, there are 24 subjective solved examples.
x
Ex. 28 If a, b are roots of the equation
2
x - p ( x + 1) - c = 0, show that (a + 1) (b + 1) = 1 - c . Hence,
a 2 + 2 a +1 b2 + 2 b +1
prove that
+
= 1.
a 2 + 2 a + c b2 + 2 b + c
l
x
\
a +b = p
and
ab = - p - c
Now, ( a + 1) ( b + 1) = ab + a + b + 1
( a + 1) ( b + 1) = 1 - c
=
=
a + 2a + 1
2
a + 2a + c
+
…(i)
2
b + 2b + 1
b 2 + 2b + c
( a + 1) 2
2
( a + 1) - ( 1 - c )
Ex. 30 If the roots of the equation ax 2 + bx + c = 0 (a ¹ 0 )
be a and b and those of the equation Ax 2 + Bx + C = 0
( A ¹ 0 ) be a + k and b + k. Prove that
2
b 2 - 4ac æ a ö
=ç ÷ .
B 2 - 4 AC è A ø
+
( b + 1) 2
Sol. Qa - b = ( a + k ) - ( b + k )
2
( b + 1) - ( 1 - c )
b 2 - 4ac
( B 2 - 4 AC ) é
Dù
=
Qa -b =
ê
a úû
a
A
ë
Þ
( a + 1) 2
2
( a + 1) - ( a + 1) ( b + 1)
( b + 1)
+
2
- 18x + 45 = 0
l
= - p -c + p + 1 = 1-c
Second Part LHS =
2
( x - 3) ( x - 15) = 0
Þ
x = 3, 15
Hence, the roots of the given equation are ( - 3), ( -15), 3
and 15.
x 2 - px - p - c = 0
2
+ 18x + 45 = 0
Þ
( x + 3) ( x + 15) = 0
Þ
x = - 3, 5
and substituting p = - 18 in the given equation, we obtain
Sol. Since, a and b are the roots of the equation,
Hence,
2
æ b 2 - 4ac ö æ a ö
÷ =ç ÷
ç 2
è B - 4 AC ø è A ø
Þ
2
( b + 1) - ( a + 1) ( b + 1)
a +1 b +1 a -b
+
= 1 = RHS
=
=
a -b b -a a -b
[from Eq. (i)]
On squaring both sides, then we get
b 2 - 4ac
æaö
=ç ÷
2
B - 4 AC è A ø
2
Hence, RHS = LHS
Ex. 31 Let a , b and c be real numbers such that
a + 2b + c = 4. Find the maximum value of (ab + bc + ca ).
l
Ex. 29 Solve the equation x 2 + px + 45 = 0. It is given
that the squared difference of its roots is equal to 144.
l
Sol. Let a , b be the roots of the equation x 2 + px + 45 = 0 and
given that
(a - b )2 = 144
Þ
Þ
\
p 2 - 4 × 1× 45 = 144
2
p = 324
p = ( ± 18)
é
Dù
êQ a - b = a ú
ë
û
Sol. Given,
Þ
Let
Þ
a + 2b + c = 4
a = 4 - 2b - c
y = ab + bc + ca = a (b + c ) + bc
= ( 4 - 2b - c ) (b + c ) + bc
= - 2b 2 + 4b - 2bc + 4c - c 2
2b 2 + 2(c - 2)b - 4c + c 2 + y = 0
Since, b Î R, so
4 ( c - 2) 2 - 4 ´ 2 ´ ( - 4c + c 2 + y ) ³ 0
162
Textbook of Algebra
(c - 2)2 + 8c - 2c 2 - 2y ³ 0
Þ
Sol. Given equations are
x 2 + ax + b = 0
2
Þ
c - 4c + 2y - 4 £ 0
2 ³ (ab + bc + ca + b 2 )
Þ
ab + bc + ca £ 4 - b
Let a , b be the roots of Eq. (i), b, g be the roots of Eq. (ii)
and g , d be the roots of Eq. (iii), then
…(iv)
a + b = - a, ab = b
…(v)
b + g = - c , bg = d
…(vi)
g + a = - e , ga = f
\ LHS = (a + c + e )2 = ( - a - b - b - g - g - a )2
[Qa + 2b + c = 4 ]
2
[from Eqs. (iv), (v) and (vi)]
…(vii)
= 4 (a + b + g )2
Ex. 32 Find a quadratic equation whose roots x 1 and x 2
satisfy the condition
x 12 + x 22 = 5, 3 ( x 15 + x 25 ) = 11 ( x 13 + x 23 ) (assume that x 1 , x 2
are real).
l
Þ
x 13 + x 23
= 4 (a 2 + b 2 + g 2 + 2ab + 2bg + 2ga )
11
=
3
= 4 (a + b + g )2
Þ
( x 12 + x 22 ) ( x 13 + x 23 ) - x 12 x 22 ( x 1 + x 2 )
Þ
( x 12 + x 22 ) -
( x 13 + x 23 )
x 12 x 22 ( x 1 + x 2 )
( x1 +
x 2 ) ( x 12
+
x 22
- x 1x 2 )
=
11
3
=
11
3
2
5-
2
2
Sol. Qa , b are the roots of the equation
x 12 x 22
4
=
3 5 - x 1x 2
x
3x 12 x 22 + 4 x 1x 2 - 20 = 0
Þ
Þ 3x 12 x 22 + 10x 1x 2 - 6x 1x 2 - 20 = 0
Þ
\
( x 1x 2 - 2) (3x 1x 2 + 10) = 0
æ 10 ö
x 1x 2 = 2, ç - ÷
è 3ø
We have,
\
\
( x1 + x 2 ) = 5 + 4 = 9
[if x 1x 2 = 2]
+ px + q = 0
\
a + b = - p , ab = q
and g , d are the roots of the equation x 2 + rx + s = 0
…(i)
\
g + d = - r, g d = s
Now, (a - g ) (a - d ) ( b - g ) ( b - d )
…(ii)
= (a 2 + r a + s ) ( b 2 + r b + s )
2
2
2
10 ù
5é
æ 10 ö
( x 1 + x 2 )2 = 5 + 2 ç - ÷ = - ê if x 1x 2 = - ú
è 3 ø
3û
3ë
which is not possible, since x 1, x 2 are real.
Thus, required quadratic equations are x 2 ± 3x + 2 = 0.
Ex. 33 If each pair of the three equations
x 2 + ax + b = 0, x 2 + cx + d = 0 and x 2 + ex + f = 0 has
exactly one root in common, then show that
(a + c + e ) 2 = 4 (ac + ce + ea - b - d - f ).
[from Eq. (ii)]
= a b + r ab (a + b ) + r ab + s (a + b 2 )
2
+ sr (a + b ) + s 2
x1 + x 2 = ± 3
and
l
2
= [a 2 - a ( g + d ) + g d] [ b 2 - b ( g + d ) + g d ]
( x 1 + x 2 )2 = x 12 + x 22 + 2x 1x 2 = 5 + 2x 1x 2
2
Ex. 34 If a, b are the roots of the equation
x + px + q = 0 and g, d are the roots of the equation
x 2 + rx + s = 0, evaluate (a - g ) (a - d) ( b - g ) ( b - d) in
terms of p, q, r and s . Deduce the condition that the equations have a common root.
l
x 12 x 22
11
=
5 - x 1x 2
3
Þ
…(viii)
From Eqs. (vii) and (viii), then we get
(a + c + e )2 = 4 (ac + ce + ea - b - d - f )
[Q x 1 + x 2 = 5]
Þ
RHS = 4(ac + ce + ea - b - d - f )
= 4 {(a + b ) ( b + g ) + ( b + g ) ( g + a ) + ( g + a )
(a + b ) - ab - bg - ga )}
[from Eqs. (iv), (v) and (vi)]
3( x 15 + x 25 ) = 11( x 13 + x 23 )
x 15 + x 25
…(iii)
x + ex + f = 0
\ Maximum value of (ab + bc + ca) is 4.
Sol. We have,
…(ii)
2
x + cx + d = 0
Since, c Î R, so 16 - 4 (2y - 4 ) ³ 0 Þ y £ 4
Hence, maximum value of ab + bc + ca is 4.
Aliter
AM ³ GM
Q
(a + b ) + (b + c )
Þ
³ (a + b ) (b + c )
2
Þ
…(i)
2
= a 2 b 2 + r ab (a + b ) + r 2 ab + s [(a + b )2 - 2ab ]
+ sr (a + b ) + s 2
= q 2 - pqr + r 2q + s ( p 2 - 2q ) + sr ( - p ) + s 2
= (q - s )2 - rpq + r 2q + sp 2 - prs
= (q - s )2 - rq ( p - r ) + sp ( p - r )
= (q - s )2 + ( p - r ) (sp - rq )
…(iii)
For a common root (let a = g or b = d ),
then (a - g ) (a - d ) ( b - g ) ( b - d ) = 0
…(iv)
Chap 02 Theory of Equations
From Eqs. (iii) and (iv), we get
(q - s )2 + ( p - r ) (sp - rq ) = 0
Ex. 37 Solve for ‘ x ’
1 ! + 2 ! + 3 ! + K + ( x - 1) ! + x ! = k 2 and k Î I .
Þ (q - s )2 = ( p - r ) (rq - sp ), which is the required
condition .
Sol. For x < 4 , the given equation has the only solutions
l
x = 1, k = ± 1 and x = 3, k = ± 3. Now, let us prove that
there are no solutions for x ³ 4. The expressions
1! + 2! + 3! + 4 !
= 33 ü
1! + 2! + 3! + 4 ! + 5!
= 153 ïï
ý ends with the digit
1! + 2! + 3! + 4 ! + 5! + 6!
= 873 ï
1! + 2! + 3! + 4 ! + 5! + 6! + 7 ! = 5913ïþ
3.
Now, for x ³ 4 the last digit of the sum 1! + 2! + K + x ! is
equal to 3 and therefore, this sum cannot be equal to a
square of a whole number k (because a square of a whole
number cannot end with 3).
Ex. 35 Find all integral values of a for which the
quadratic Expresion ( x - a ) ( x - 10 ) + 1 can be factored as a
product ( x + a ) ( x + b) of two factors and a, b ÎI.
l
Sol. We have, ( x - a ) ( x - 10) + 1 = ( x + a ) ( x + b )
On putting x = - a in both sides, we get
( - a - a ) ( - a - 10) + 1 = 0
\
(a + a ) (a + 10) = - 1
a + a and a + 10 are integers.
\
a + a = - 1 and a + 10 = 1
or
a + a = 1 and a + 10 = - 1
(i) If a + 10 = 1
\
a = - 9, then a = 8
Similarly, b = - 9
[Qa, a Î I ]
l
n radical signs
Sol. Rewrite the given equation
(ii) If a + 10 = - 1
\
a = - 11, then a = 12
Similarly, b = 12
x + 2 x + 2 x + K + 2 x + 2 x + 2x = x
a = 8, 12
x = x + 2 x + 2 x + K + x + 2x
14444442444444
3
2 n radical signs
l
Ex. 36 Solve the equation
x + 3 - 4 ( x - 1) + x + 8 - 6 ( x - 1) = 1.
Sol. Let
Further, let us replace the last letter x by the same
expression again and again yields.
\
( x - 1) = t
x = x + 2 x + 2 x + K + 2 x + 2x
144444424444443
3n radical signs
x = t 2 + 1, t ³ 0
We have,
The given equation reduce in the form
2
= x + 2 x + 2 x + K + 2 x + 2x = …
144444424444443
2
(t + 4 - 4t ) + (t + 9 - 6t ) = 1
Þ
4 n radical signs
We can write,
| t - 2| + | t - 3| = 1
x = x +2 x +2 x +K
+
3
–
–
+
–
= lim x + 2 x + 2 x + K + 2 x + 2x
N ®¥
1
44444444244444444
3
+
2
N radical signs
If follows that
\
2£t £3
x = x +2 x +2 x +K
2
Þ
4 £t £9
Þ
4 £ x -1£9
…(i)
On replacing the last letter x on the LHS of Eq. (i) by the
value of x expressed by Eq. (i), we get
Here, ( x - 12) ( x - 10) + 1 = ( x - 11)2
Hence,
Ex. 38 Find the real roots of the equation
x + 2 x + 2 x + K + 2 x + 2 3x = x
1444444424444444
3
( x - 8) ( x - 10) + 1 = ( x - 9 )2
Here,
163
Þ
5 £ x £ 10
\ Solution of the original equation is x Î[ 5, 10].
= x + 2 ( x + 2 x + K ) = ( x + 2x )
x 2 = x + 2x
x 2 - 3x = 0
x = 0, 3
Hence,
Þ
\
164
l
Textbook of Algebra
Ex. 39 Solve the inequation, ( x
2
1
Þ x2 - x - 1 = 0
x
1± 5
x=
\
2
1+ 5
1- 5
, x2 =
x1 =
\
2
2
satisfy the given equation and this equation has no other
roots.
+ x + 1) x < 1.
Þ
Sol. Taking logarithm both sides on base 10,
then
x log ( x
2
+ x + 1) < 0
which is equivalent to the collection of systems
éì x > 0,
éì x > 0,
êí 2
êí
2
êî log ( x + x + 1) < 0, Þ êî x + x + 1 < 1,
êì x < 0,
êì x < 0,
êí
êí
2
êëî x 2 + x + 1 > 1,
êëî log ( x + x + 1) > 0,
Þ
Þ
éì x > 0,
éì x > 0,
êí - 1 < x < 0
êí x ( x + 1) < 0,
î
êî
Þ ê
êì x < 0,
êì x < 0,
êí
êí
ëî x > 0 and x < ( - 1)
ëî x ( x + 1) > 0
ì x Î f,
í
î x < ( - 1)
l
first equation of this system, we get
2 - y + |y - 2| = 1
Now, consider the following cases
If
y ³ 2,
then
2-y +y -2=1 Þ 0=1
No value of y for y ³ 2.
If
y < 2,
3
then
2 - y + 2 - y = 1 Û y = , which is true.
2
From the second equation of this system,
3
= 2 - | x - 1|
2
1
1
Þ
| x - 1| = Þ x - 1 = ±
2
2
1
1 3
Þ
x =1± Þx = ,
2
2 2
Consequently, the set of all solutions of the original system
1 3
3
is the set of pairs ( x , y ), where x = , and y = .
2 2
2
? Remark
When the inequation is in power, then it is better to take log.
Ex. 40 Solve the equation
1
=x
1+
1
1+
1
1+
1+ O
1
1+
x
When in expression on left hand side the sign of a fraction is
repeated n times.
l
Sol. Given equation is
1
1
1+
=x
1
1+ O
1
1+
x
Let us replace x on the LHS of the given equation by the
expression of x . This result in an equation of the same
form, which however involves 2n fraction lines. Continuing
this process on the basis of this transformation, we can
write
1
x = 1 + lim 1 +
1
n ®¥
1+
1
1+
1+ O
1
[n fractions]
1+
x
1+
Ex. 41 Solve the system of equations
ì| x - 1 | + | y - 2 | = 1
.
í
î y = 2 - | x -1|
Sol. On substituting | x - 1 | = 2 - y from second equation in
Consequently, the interval x Î ( - ¥, - 1) is the set of all
solutions of the original inequation.
1+
x =1+
Ex. 42 Let a , b, c be real and ax 2 + bx + c = 0 has two real
roots a and b, where a < -1 and b >1, then show that
c
b
1+ +
< 0.
a
a
l
Sol. Since, a < - 1 and b > 1
a + l = - 1 and b = 1 + m
c
b
Now, 1 + +
= 1 + ab + | a + b |
a
a
and
\
[ l, m > 0 ]
= 1 + ( - 1 - l ) (1 + m ) + | - 1 - l + 1 + m |
= 1 - 1 - m - l - lm + | m - l |
[if m > l]
= - m - l - lm + m - l
[if l > m]
= - m - l - lm + l - m
c
b
1+ +
= - 2l - lm or - 2m - lm
a
a
On both cases, 1 +
c
b
+
<0
a
a
[Q l, m > 0]
Chap 02 Theory of Equations
It follows that
Aliter
Q
ax 2 + bx + c = 0, a ¹ 0
c
b
x2 + x + =0
a
a
c
b
2
Let f ( x ) = x + x +
a
a
f ( -1) < 0 and f (1 ) < 0
b c
b c
Þ 1 - + < 0 and 1 + + < 0
a a
a a
b
c
Then, 1 +
+ <0
a
a
l
æ3 - x ö
Ex. 43 Solve the equation x ç
÷
è x +1ø
Sol. Hence, x + 1 ¹ 0
and let
\
and
85
åbi = - 2 and å å
i =1
85
a
–1
b
l
æ
3-xö
÷ = 2.
çx +
x +1ø
è
Ex. 44 Show that for any real numbers a 3 , a 4 , a 5 , ..., a 85 ,
the roots of the equation
a 85 x 85 + a 84 x 84 + K + a 3 x 3 + 3 x 2 + 2 x + 1 = 0 are not real.
84
x + 1 = 0, 2 - 1 = 0
\
x = - 1, x = 0
Now, consider the following cases :
x < -1
2- (x + 1) - 2x = - (2x - 1) + 1
Þ
2- (x + 1) = 2
\
- ( x + 1) = 1
…(i)
\
x = -2
-1 £ x < 0
2 x + 1 - 2 x = - ( 2 x - 1) + 1
Þ
2 x +1 = 2
\
x +1=1
\
x =0
x ¹0
[Q - 1 £ x < 0]
x³ 0
2 x +1 - 2 x = 2 x - 1 + 1
Þ
2 x + 1 = 2 ×2 x
Þ
2 x +1 = 2 x +1
which is true for
…(ii)
x ³ 0.
Now, combining all cases, we have the final solution as
x Î [0, ¥ ) È { - 2}
Q (y ) = y
+ 2y
84
+ 3y
83
l
Ex. 46 Solve the inequation
- | y | + x - ( x 2 + y 2 - 1) ³ 1.
Sol. We have, - | y | + x - ( x 2 + y 2 - 1) ³ 1
Þ
…(i)
+K + a 3 x 3 + 3 x 2 + 2 x + 1 = 0
Since, P(0) = 1, then 0 is not a root of Eq. (i).
Let a 1, a 2 , a 3 , ..., a 85 be the complex roots of Eq. (i).
1 ö
æ
Then, the bi ç let ÷ the complex roots of the polynomial
è ai ø
+ a 3y
82
+ K + a 85
0
x
…(i)
l
85
Ex. 45 Solve the equation
2 |x +1| - 2 x = | 2 x - 1| + 1.
–1
ìx 2 - x + 2 = 0
ì x 2 - 2x + 1 = 0
or í 2
í 2
îx - x + 2 = 0
î x - 2x + 1 = 0
2
ìæ
1ö
7
ì x2 - x + 2 = 0
ïç x - ÷ + ¹ 0
Þ
Þ íè
ø
í 2
2
4
î x - 2x + 1 = 0
ï ( x - 1) 2 = 0
î
\
( x - 1) 2 = 0
Þ x = 1 is a unique solution of the original equation.
+ a 84 x
ö
Sol. Find the critical points :
Þ
P ( x ) = a 85 x
85
åbi2 = çç åbi ÷÷
i =1
æ3 - x ö
= ( x + 1) ç
÷ + x =3- x + x =3
è x + 1ø
\
u + v = 3 and uv = 2
Then,
u = 2, v = 1 or u = 1, v = 2
Given equation is equivalent to the collection
ì æ3 - x ö
ì æ3 - x ö
÷ =2
÷ =1
ïï x ç
ïï x ç
è x + 1ø
è x + 1ø
or
\
í
í
ïx + 3 - x = 1
ïx + 3 - x = 2
ïî
ïî
x +1
x +1
Sol. Let
æ
bi b j = 3
1 £ i < j £ 85
2
- 2 å å bi b j
èi = 1 ø
1 £ i < j £ 85
= 4 - 6 = -2 < 0
Thus, the bi ’s is not all real and then a i ’s are not all real.
Then,
1
æ3 - x ö
3- x
=v
xç
÷ = u and x +
è x + 1ø
x +1
uv = 2
æ3 - x ö
æ3 - x ö
u +v = xç
÷
÷+x+ç
è x + 1ø
è x + 1ø
85
165
if
x - | y | ³ 1 + ( x 2 + y 2 - 1)
x ³ | y |,
then squaring both sides,
x 2 + y 2 - 2x | y | ³ 1 + x 2 + y 2 - 1 + 2 ( x 2 + y 2 - 1)
Þ
Since,
- x | y | ³ ( x 2 + y 2 - 1)
x ³ |y | ³ 0
…(i)
…(ii)
166
Textbook of Algebra
Then, LHS of Eq. (i) is non-positive and RHS of Eq. (ii) is
non-negative. Therefore, the system is satisfied only, when
both sides are zero.
\ The inequality Eq. (i) is equivalent to the system.
ì x |y | = 0
í 2
2
îx + y - 1 = 0
The Eq.(i) gives x = 0 or y = 0. If x = 0, then we find y = ± 1
from Eq. (ii) but x ³ | y | which is impossible.
If y = 0, then from Eq. (ii), we find
and
-a2 £ t £ 0
But
t >0
\
t ³
- a 2 + ( a 4 + 8)
2
\
ax ³
- a 2 + ( a 4 + 8)
2
For
0 < a < 1,
æ - a 2 + ( a 4 + 8) ö
÷
x £ loga ç
ç
÷
2
è
ø
x2 = 1
\
[from Eq. (ii)]
x = 1, - 1
Taking
x =1
[Q x ³ | y | ]
\ The pair (1, 0) satisfies the given inequation. Hence, (1, 0)
is the solution of the original inequation.
\
é
æ - a 2 + ( a 4 + 8) ö ù
÷ú
x Îê - ¥, loga ç
ç
÷ú
2
êë
è
øû
æ - a 2 + ( a 4 + 8) ö
÷
and for a > 1, x ³ loga ç
ç
÷
2
è
ø
Ex. 47 If a 1 , a 2 , a 3 , ..., a n (n ³ 2 ) are real and
(n - 1) a 12 - 2na 2 < 0, prove that atleast two roots of the
equation x n + a 1 x n - 1 + a 2 x n - 2 + K + a n = 0 are
imaginary.
l
\
æ
æ - a 2 + ( a 4 + 8) ö ö
÷, ¥ ÷
x Îç loga ç
÷ ÷
ç
ç
2
ø ø
è
è
Sol. Let a 1, a 2 , a 3 , ..., a n are the roots of the given equation.
l
Then,
å a 1 = a 1 + a 2 + a 3 + K + a n = - a1
and å a 1 a 2 = a 1 a 2 + a 1 a 3 + K + a n - 1 a n = a 2
Now, (n - 1)a12 - 2na 2 = (n - 1) ( å a 1 )2 - 2n å a 1 a 2
Sol. We rewrite the given inequation in the form,
= n {( å a 1 )2 - 2 å a 1a 2 } - ( å a 1 )2
= n å a 12 - ( å a1 )2
=
2
ïì (9 - x ) - x - 1 ³ | x |, if | x | > 1
í
|, 0 < | x | < 1
ïî (9 - x 2 ) - x - 1 £ | x if
1£i < j £n
But given that (n - 1)a12 - 2na 2 < 0
Þ
2
å × å (a i - a j ) < 0
1£i < j £n
which is true only, when atleast two roots are imaginary.
Ex. 48 Solve the inequation | a 2 x + a x + 2 - 1 | ³ 1 for all
values of a (a > 0, a ¹ 1).
l
Sol. Using a x = t ,
the given inequation can be written in the form
| t 2 + a 2t - 1| ³ 1
Q
…(i)
a > 0 and a ¹ 1, then a x > 0
\
Inequation (i) write in the forms,
t >0
t 2 + a 2t - 1 ³ 1 and t 2 + a 2t - 1 £ - 1
\
t £
-a2 - a4 + 8
- a 2 + ( a 4 + 8)
,t ³
2
2
log| x | ( (9 - x 2 ) - x - 1) ³ log| x | (| x | )
This inequation is equivalent to the collection of systems.
å (a i - a j )2
å
Ex. 49 Solve the inequation
log | x | ( ( 9 - x 2 ) - x - 1) ³ 1.
…(ii)
éì
For x > 1
For x > 1
éì
êï
êï
2
2
êïí (9 - x ) - x - 1 ³ x
êïí (9 - x ) ³ 2x + 1
For x < - 1
For x < - 1
êï
êï
êï
êï
2
(
)
(9 - x 2 ) ³ 1
9
1
x
x
³
x
and êî
Þ êî
For 0 < x < 1
êì
êì For 0 < x < 1
êï ( 9 - x 2 ) - x - 1 £ x
êï ( 9 - x 2 ) £ 2x + 1
êïí
êïí
For - 1 < x < 0
êï
êï For - 1 < x < 0
êï ( 9 - x 2 ) - x - 1 £ - x
êï
(9 - x 2 ) £ 1
ëî
ëî
é ì
For x > 1
ê ï 2
2
ê ï - ( 11 + 1) £ x £ ( 11 - 1)
5
5
í
ê
For x < - 1
ï
ê
-2 2 £ x £ 2 2
ê ïî
Þ
êì
For 0 < x < 1
êï
2
2
êï x £ - ( 11 + 1) and x ³ ( 11 - 1)
5
5
êí
For - 1 < x < 0
êï
êïî
x £ - 2 2 and x ³ 2 2
ë
167
Chap 02 Theory of Equations
é ì
x Îf
ê í- 2 2 £ x < - 1
ê î
Þ
êìï 2 ( 11 - 1) £ x < 1
êí 5
êëïî
x Îf
Hence, the original inequation consists of the intervals
2
-2 2 £ x < - 1 and ( 11 - 1) £ x < 1 .
5
ö
é2
Hence, x Î [ -2 2, - 1) È ê ( 11 - 1), 1÷
ø
ë5
Sol. Putting t = 3x in the original equation, then we obtain
T-axis
0
(ii)
(i)
0
Ex. 50 Find all values of ‘a’ for which the equation
4 - a2 x - a + 3 = 0 has atleast one solution.
T-axis
0
0
T-axis
T-axis
l
Sol. Putting 2x = t > 0, then the original equation reduced in
the form
t 2 - at - a + 3 = 0
…(i)
that the quadratic Eq. (i) should have atleast one positive
root (t > 0), then
Discriminant, D = ( -a )2 - 4 × 1 × ( -a + 3) ³ 0
a 2 + 4a - 12 ³ 0
Þ
Þ
( a + 6) ( a - 2) ³ 0
+
+
–6
\
–
at 2 + 4 (a - 1)t + a > 1
Þ
at 2 + 4 (a - 1)t + (a - 1) > 0
[t > 0, Q 3x > 0]
This is possible in two cases. First the parabola
f (t ) = at 2 + 4(a - 1)t + (a - 1) opens upwards, with its
vertex (turning point) lying in the non-positive part of the
T -axis, as shown in the following four figures.
\ a > 0 and sum of roots £ 0
4( a - 1)
£ 0 and f (0) ³ 0
Þ
2a
\
a > 0, a - 1 ³ 0 and a - 1 ³ 0
Hence,
a³1
2
a Î ( -¥, - 6] È [2, ¥ )
If roots of Eq. (i) are t 1 and t 2 , then
0
ì t1 + t 2 = a
í
ît 1t 2 = 3 - a
For
a Î ( - ¥, - 6]
t 1 + t 2 < 0 and t 1t 2 > 0. Therefore, both roots are negative
and consequently, the original equation has no solutions.
a Î[2, ¥ )
>
t 1 + t 2 > 0 and t 1t 2 < 0, consequently, atleast one of the
roots t 1 or t 2 , is greater than zero.
Thus, for a Î [2, ¥ ), the given equation has atleast one
solution.
For
Ex. 51 Find all the values of the parameter a for which
the inequality a 9 x + 4 (a - 1) 3 x + a > 1, is satisfied for all
real values of x .
l
(iv)
(iii)
x
T-axis
Second the parabola f (t ) opens upward, with its vertex
lying in positive direction of t, then
4( a - 1)
a > 0, > 0 and D £ 0
2a
Þ
a > 0,(a - 1) < 0
and
16(a - 1)2 - 4(a - 1)a £ 0
Þ
and
a > 0,a < 1
4(a - 1)(3a - 4 ) £ 0
4
3
These inequalities cannot have simultaneously.
Hence, a ³ 1 from Eq. (i).
Þ
a > 0,a < 1 and 1 £ a £
#L
Theory of Equations Exercise 1 :
Single Option Correct Type Questions
n
This section contains 30 multiple choice questions.
Each question has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct
1. If a, b, c are real and a ¹ b, the roots of the equation
2
2 (a - b ) x - 11 (a + b + c ) x - 3 (a - b ) = 0 are
(a) real and equal
(c) purely imaginary
(b) real and unequal
(d) None of these
2. The graph of a quadratic polynomial y = ax 2
+ bx + c ; a, b, c Î R is as shown.
Y
X
O
8. If the roots of the quadratic equation
( 4 p - p 2 - 5)x 2 - (2p - 1) x + 3p = 0 lie on either side of
unity, the number of integral values of p is
(a) 1
(b) 2
(c) 3
(d) 4
9. Solution set of the equation
2
32 x - 2×3x
(a) { -3, 2 }
2
+x +6
+ 3 2 (x + 6 ) = 0 is
(b) {6, - 1 }
(c) { -2, 3 }
(d) {1, - 6 }
10. Consider two quadratic expressions f ( x ) = ax 2 + bx + c
and g ( x ) = ax 2 + px + q (a, b, c , p , q Î R, b ¹ p ) such that
their discriminants are equal. If f ( x ) = g ( x ) has a root
x = a, then
(a) a will be AM of the roots of f ( x ) = 0 and g( x ) = 0
(b) a will be AM of the roots of f ( x ) = 0
(c) a will be AM of the roots of f ( x ) = 0 or g( x ) = 0
(d) a will be AM of the roots of g( x ) = 0
Which one of the following is not correct?
c
(a) b 2 - 4ac < 0
(b) < 0
a
(c) c is negative
æ bö
(d) Abscissa corresponding to the vertex is ç - ÷
è 2a ø
3. There is only one real value of ‘a’ for which the
quadratic equation ax 2 + (a + 3) x + a - 3 = 0 has two
positive integral solutions. The product of these two
solutions is
(a) 9
(b) 8
(c) 6
(d) 12
4. If for all real values of a one root of the equation
x 2 - 3ax + f (a ) = 0 is double of the other, f ( x ) is equal to
(b) x 2
(a) 2x
(c) 2 x 2
(d) 2 x
5. A quadratic equation the product of whose roots x 1 and
x 2 is equal to 4 and satisfying the relation
x1
x2
+
= 2, is
x1 - 1 x 2 - 1
(a) x 2 - 2 x + 4 = 0
(b) x 2 - 4 x + 4 = 0
(c) x 2 + 2 x + 4 = 0
(d) x 2 + 4 x + 4 = 0
6. If both roots of the quadratic equation
x 2 - 2ax + a 2 - 1 = 0 lie in ( - 2, 2), which one of the
following can be [a ] ? (where [×] denotes the greatest
integer function)
(a) -1
(b) 1
(c) 2
(d) 3
7. If ( - 2, 7 ) is the highest point on the graph of
2
y = - 2x - 4ax + l, then l equals
(a) 31
(b) 11
(c) -1
1
(d) 3
11. If x 1 and x 2 are the arithmetic and harmonic means of
the roots of the equation ax 2 + bx + c = 0, the quadratic
equation whose roots are x 1 and x 2 , is
(a) abx 2 + (b 2 + ac ) x + bc = 0
(b) 2abx 2 + (b 2 + 4ac ) x + 2bc = 0
(c) 2abx 2 + (b 2 + ac ) x + bc = 0
(d) None of the above
12. f ( x ) is a cubic polynomial x 3 + ax 2 + bx + c such that
f ( x ) = 0 has three distinct integral roots and f ( g ( x )) = 0
does not have real roots, where g ( x ) = x 2 + 2x - 5, the
minimum value of a + b + c is
(a) 504
(b) 532
(c) 719
(d) 764
13. The value of the positive integer n for which the
n
quadratic equation å ( x + k - 1) ( x + k ) = 10n has
k =1
solutions a and a + 1 for some a, is
(a) 7
(b) 11
(c) 17
(d) 25
2
14. If one root of the equation x - lx + 12 = 0 is even
prime, while x 2 + lx + m = 0 has equal roots, then m is
(a) 8
(b) 16
(c) 24
(d) 32
15. Number of real roots of the equation
x + x - (1 - x ) = 1 is
(a) 0
(b) 1
(c) 2
(d) 3
16. The value of 7 + 7 - 7 + 7 - K upto ¥ is
(a) 5
(c) 3
(b) 4
(d) 2
Chap 02 Theory of Equations
17. For any real x, the expression 2 (k - x ) [x + x 2 + k 2 ]
24. The roots of the equation
cannot exceed
(a + b ) x
2
2
(b) 2k
(d) None of these
(a) k
(c) 3k 2
18. Given that, for all x Î R, the expression
x
2
x
2
- 2x + 4
+ 2x + 4
lies
3
(b) and 2
2
(d) 0 and 2
(c) -1 and 1
20. If one root of the equation ix 2 - 2 (1 + i ) x + 2 - i = 0 is
(3 - i ), where i = -1, the other root is
(c) -1 + i
(d) -1 - i
-1
21. The number of solutions of |[x ] - 2x | = 4, where [x ]
denotes the greatest integer £ x is
(a) infinite
22. If x
2
(b) 4
(c) 3
+ x + 1 is a factor of ax
root of ax
3
d
(a) a
+ bx
2
(d) 2
3
+ bx
2
+ cx + d , the real
+ cx + d = 0 is
d
(b)
a
(c)
a
d
(d) None of these
23. The value of x which satisfy the equation
(5x 2 - 8x + 3) - (5x 2 - 9 x + 4 ) = (2x 2 - 2x )
- (2x 2 - 3x + 1), is
(a) 3
(c) 1
#L
2
- 15
(a) ± 2, ± 3
(b) ± 4, ± 14
(c) ± 3, ± 5
(d) ± 6, ± 20
= 2a,
25. The number of pairs ( x , y ) which will satisfy the
equation
x
2
- xy + y
2
= 4 ( x + y - 4 ), is
(a) 1
(c) 4
(b) 2
(d) None of these
(a) 0
(b) b, c, d
(d) a + d , b + d , c + d
(b) 3 +
+ (a - b ) x
x 4 - y 4 = 3789108 is
( x - a ) ( x - b ) ( x - c ) = d , d ¹ 0, the roots of the equation
( x - a ) ( x - b ) ( x - g ) + d = 0 are
(a) 3 + i
- 15
26. The number of positive integral solutions of
19. Let a , b, g be the roots of the equation
(a) a, b, d
(c) a, b, c
2
where a 2 - b = 1, are
1
between and 3, the values between which the
3
9 ×32x + 6×3x + 4
lies, are
expression
9 ×32x - 6×3x + 4
(a) -3 and 1
169
(b) 2
(d) 0
(b) 1
(c) 2
(d) 4
27. The value of ‘a’ for which the equation x
4
and x + ax
2
3
+ ax + 1 = 0
+ 1 = 0, have a common root, is
(a) a = 2
(c) a = 0
(b) a = - 2
(d) None of these
28. The necessary and sufficient condition for the equation
(1 - a 2 ) x 2 + 2ax - 1 = 0 to have roots lying in the
interval (0, 1), is
(a) a > 0
(c) a > 2
(b) a < 0
(d) None of these
29. Solution set of x - 1 - | x | < 0, is
é
-1 + 5 ö
(a) ê -1,
÷
ø
2
ë
é
ù
-1 + 5
(c) ê -1,
ú
2
ë
û
(b) [ -1, 1 ]
æ
-1 + 5 ö
(d) ç -1,
÷
è
ø
2
30. If the quadratic equations ax
2
+ 2cx + b = 0 and
2
ax + 2bx + c = 0 (b ¹ c ) have a common root, a + 4b + 4c ,
is equal to
(a) -2
(c) 0
(b) -1
(d) 1
Theory of Equations Exercise 2 :
More than One Correct Option Type Questions
n
This section contains 15 multiple choice questions.
Each question has four choices (a), (b), (c) and (d) out of
which MORE THAN ONE may be correct.
31. If 0 < a < b < c and the roots a , b of the equation
ax 2 + bx + c = 0 are non-real complex numbers, then
(a) | a | = | b |
(c) | b | < 1
(b) | a | > 1
(d) None of these
32. If A, G and H are the arithmetic mean, geometric mean
and harmonic mean between unequal positive integers.
Then, the equation Ax 2 - | G | x - H = 0 has
(a) both roots are fractions
(b) atleast one root which is negative fraction
(c) exactly one positive root
(d) atleast one root which is an integer
170
Textbook of Algebra
33. The adjoining graph of y = ax 2 + bx + c shows that
40. For which of the following graphs of the quadratic
expression f ( x ) = ax 2 + bx + c , the product ofabc is
negative
Y
Y
Y
(a)
X¢
O
Y¢
(a,0)
(b, 0)
(b)
X
O
(a) a < 0
(b) b 2 < 4ac
X
Y
Y
(c) c > 0
(d) a and b are of opposite signs
X
O
O
X
(d)
(c)
34. If the equation ax 2 + bx + c = 0 (a > 0) has two roots a
X
O
and b such that a < - 2 and b > 2, then
(a) b 2 - 4ac > 0
(b) c < 0
(c) a + | b | + c < 0
(d) 4a + 2 | b | + c < 0
41. If a, b Î R and ax
35. If b 2 ³ 4ac for the equation ax 4 + bx 2 + c = 0, then all
the roots of the equation will be real, if
(a) b > 0, a < 0, c > 0
(c) b > 0, a > 0, c > 0
36. If roots of the equation x
(b) b < 0, a > 0, c > 0
(d) b > 0, a < 0, c < 0
3
+ bx
2
+ cx - 1 = 0 from an
increasing GP, then
(a) b + c = 0
(b) b Î ( -¥, - 3 )
(c) one of the roots is 1
(d) one root is smaller than one and one root is more than one
2
+ bx + c , where a, b, c Î R, a ¹ 0. Suppose
| f ( x )| £ 1, " x Î[0, 1], then
37. Let f ( x ) = ax
(a) | a | £ 8
(c) | c | £ 1
(b) | b | £ 8
(d) |a | + |b | + |c | £ 17
38. cos a is a root of the equation 25x 2 + 5x - 12 = 0,
-1 < x < 0, the value of sin 2a is
24
25
24
(c) 25
(a)
12
25
20
(d)
25
(b) -
39. If a, b, c Î R (a ¹ 0) and a + 2b + 4c = 0, then equation
ax
2
+ bx + c = 0 has
(a) atleast one positive root
(b) atleast one non-integral root
(c) both integral roots
(d) no irrational root
2
+ bx + 6 = 0, a ¹ 0 does not have two
distinct real roots, the
(a) minimum possible value of
(b) minimum possible value of
(c) minimum possible value of
(d) minimum possible value of
3a + b is -2
3a + b is 2
6a + b is -1
6a + b is 1
+ 3x 2 - 9 x + l is of the form ( x - a ) 2 ( x - b ), then
l is equal to
42. If x
3
(b) -27
(d) -5
(a) 27
(c) 5
2
+ (b - c ) x + a - b - c = 0 has unequal real roots
for all c Î R, then
43. If ax
(a) b < 0 < a
(c) b < a < 0
(b) a < 0 < b
(d) b > a > 0
44. If the equation whose roots are the squares of the roots
of the cubic x 3 - ax 2 + bx - 1 = 0 is identical with the
given cubic equation, then
(a) a = b = 0
(b) a = 0, b = 3
(c) a = b = 3
(d) a, b are roots of x 2 + x + 2 = 0
45. If the equation ax 2 + bx + c = 0 (a > 0) has two real roots
a and b such that a < - 2 and b > 2, which of the
following statements is/are true?
(a) 4a - 2 | b | + c < 0
(b) 9a - 3 | b | + c < 0
(c) a - | b | + c < 0
(d) c < 0, b 2 - 4ac > 0
Chap 02 Theory of Equations
#L
171
Theory of Equations Exercise 3 :
Passage Based Questions
n
This section contains 6 passages. Based upon each of
the passage 3 multiple choice questions have to be
answered. Each of these questions has four choices (a),
(b), (c) and (d) out of which ONLY ONE is correct.
(Q. Nos. 46 to 48)
(b) y =
53. Minimum value of y = f ( x ) is
If G and L are the greatest and least values of the
2x 2 - 3x + 2
expression
, x Î R respectively.
2x 2 + 3x + 2
(c) 7100
(a) -4 2
(c) 0
(b) -2 2
(d) 2 2
54. Number of integral value of l for which
the roots of f ( x ) = 0, is
46. The least value of G 100 + L100 is
(b) 3100
x2
-2 2
2 2
x2
(d) y =
- 2
2
(a) y = x 2 - 8
(c) y = x 2 - 4
Passage I
(a) 2100
52. y = f ( x ) is given by
(d) None of these
(a) 9
(b) 10
(a) 5 x - 26 x + 5 = 0
(b) 7 x - 50 x + 7 = 0
(d) 11 x 2 - 122 x + 11 = 0
48. If L2 < l < G 2 , l Î N , the sum of all values of l is
(b) 1081
(c) 1225
(d) 1176
Passage II
(Q. Nos. 49 to 51)
If roots of the equation x 4 - 12x 3 + cx 2 + dx + 81 = 0 are
positive.
49. The value of c is
(a) -27
(b) 27
(b) -54
(d) 54
(c) -81
(d) -108
51. Root of the equation 2cx + d = 0, is
1
(b) 2
(a) -1
1
(d)
2
(c) 1
Let f (x ) = x 2 + bx + c and g (x ) = x 2 + b1 x + c1 .
Let the real roots of f ( x ) = 0 be a, b and real roots of
g ( x ) = 0 be a + k , b + k for same constant k. The least value
1
7
of f ( x ) is - and least value of g (x ) occurs at x = .
4
2
55. The value of b1 is
(a) -8
(b) -7
(c) -6
(d) 5
1
3
(d) -
56. The least value of g ( x ) is
(c) -54
50. The value of d is
(a) -27
(Q. Nos. 55 to 57)
2
(c) 9 x 2 - 82 x + 9 = 0
(a) 1035
(d) 12
Passage III
47. G and L are the roots of the equation
2
(c) 11
l
lies between
2
(a) -1
(b) -
1
2
(c) -
1
4
57. The roots of f ( x ) = 0 are
(b) -3, 4
(d) 3, - 4
(a) 3, 4
(c) - 3, - 4
Passage IV
Passage II
(Q. Nos. 52 to 54)
In the given figure vertices of DABC lie on
y = f ( x ) = ax 2 + bx + c. The DABC is right angled isosceles
triangle whose hypotenuse AC = 4 2 units.
Y
y = f(x) = ax2 + bx + c
O
A
90°
B
C
X
(Q. Nos. 58 to 60)
2
If ax - bx + c = 0 have two distinct roots lying in the
interval (0, 1); a, b, c Î N .
58. The least value of a is
(a) 3
(c) 5
(b) 4
(d) 6
59. The least value of b is
(a) 5
(c) 7
(b) 6
(d) 8
60. The least value of log 5 abc is
(a) 1
(c) 3
(b) 2
(d) 4
172
Textbook of Algebra
3
Passage V
Passage VI
(Q. Nos. 61 to 63)
(Q. Nos. 64 to 66)
2
If 2x + ax + bx + 4 = 0 (a and b are positive real
numbers) has three real roots.
61. The minimum value of a 3 is
(a) 108
(c) 432
(b) 216
(d) 864
62. The minimum value of b 3 is
(a) 432
(c) 1728
(b) 864
(d) None of these
63. The minimum value of (a + b ) 3 is
(a) 1728
(c) 6912
#L
(b) 3456
(d) 864
If a, b, g, d are the roots of the equation
x 4 + A x 3 + B x 2 + Cx + D = 0 such that ab = gd = k and
A, B , C , D are the roots of x 4 - 2 x 3 + 4 x 2 + 6x - 21 = 0
such that A + B = 0.
C
64. The value of is
A
(a) -
k
2
(b) -k
(c)
k
2
(d) k
65. The value of (a + b ) ( g + d ) in terms of B and k is
(a) B - 2k
(b) B - k
(c) B + k
(d) B + 2k
66. The correct statement is
(a) C 2 = AD (b) C 2 = A 2D (c) C 2 = AD 2 (d) C 2 = ( AD ) 2
Theory of Equations Exercise 4 :
Single Integer Answer Type Questions
n
This section contains 10 questions. The answer to
each question is a single digit integer, ranging from 0
to 9 (both inclusive).
67. The sum of all the real roots of the equation
| x - 2 | 2 + | x - 2 | - 2 = 0 is
68. The harmonic mean of the roots of the equation
(5 + 2 ) x 2 - ( 4 + 5 ) x + 8 + 2 5 = 0 is
69. If product of the real roots of the equation,
x 2 - ax + 30 = 2 ( x 2 - ax + 45), a > 0,
is l and minimum value of sum of roots of the equation
is m. The value of (m) (where ( × ) denotes the least integer
6
function) is
1ö
1 ö
æ 6
æ
çx + ÷ - çx + 6 ÷ - 2
è
ø
è
x
x ø
is
70. The minimum value of
3
1ö
1
æ
3
çx + ÷ + x + 3
è
xø
x
(for x > 0)
71. Let a, b, c , d are distinct real numbers and a, b are the
roots of the quadratic equation x 2 - 2cx - 5d = 0. If c and
d are the roots of the quadratic equation
x 2 - 2ax - 5b = 0, the sum of the digits of numerical
values of a + b + c + d is
72. If the maximum and minimum values of y =
x
2
- 3x + c
x
2
+ 3x + c
1
are 7 and respectively, the value of c is
7
73. Number of solutions of the equation
x 2 - ( x - 1) 2 + ( x - 2) 2 = 5 is
74. If a and b are the complex roots of the equation
(1 + i ) x
2
+ (1 - i ) x - 2i = 0, where i = -1, the value of
2
| a - b | is
75. If a , b be the roots of the equation
4 x 2 - 16x + c = 0, c Î R such that 1 < a < 2 and 2 < b < 3,
then the number of integral values of c, are
76. Let r , s and t be the roots of the equation
8x 3 + 1001x + 2008 = 0 and if
99 l = (r + s ) 3 + (s + t ) 3 + (t + r ) 3 , the value of [l ] is
(where [ × ] denotes the greatest integer function)
173
Chap 02 Theory of Equations
#L
Theory of Equations Exercise 5 :
Matching Type Questions
n
This section contains 4 questions. Questions 78 and 80 have three statements (A, B and C) given in Column I and
four statements (p, q, r and s) in Column II and questions 77 and 79 have three statements (A, B and C) given in
Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct
matching with one or more statement(s) given in Column II.
77. Column I contains rational algebraic expressions and
79. Column I contains rational algebraic expressions and
Column II contains possible integers which lie in their
range. Match the entries of Column I with one or more
entries of the elements of Column II.
Column I
(A)
y=
x 2 - 2x + 4
, x ÎR
x 2 + 2x + 4
Column II contains possible integers of a.
Column I
Column II
(p)
-2
(A)
y=
ax 2 + 3x - 4
, x Î R and y Î R
3x - 4 x 2 + a
(B)
y=
ax 2 + x - 2
, x Î R and y Î R
a + x - 2x 2
(q)
(C)
y=
x 2 + 2x + a
, x Î R and y Î R
x 2 + 4 x + 3a
(r)
2
(B)
y=
2x + 4 x + 1
, x ÎR
x2 + 4 x + 2
(q)
-1
2
(C)
y=
x - 3x + 4
, x ÎR
x-3
(r)
2
(s)
3
(t)
8
Column II
(p)
0
1
3
(s)
5
(t)
7
80.
78.
Column I
Column I
(p) a + b + c = 0
(A) If a, b, c, d are four non-zero real
numbers such that
(d + a - b)2 + (d + b - c)2 = 0 and
the roots of the equation
a (b - c)x 2 + b (c - a) x + c (a - b) = 0
are real and equal, then
(B)
(C)
If the equation ax 2 + bx + c = 0
and x 3 - 3x 2 + 3x - 1 = 0 have a
common real root, then
Column II
Column II
(A)
The equation x 3 - 6x 2 + 9x + l = 0 have
exactly one root is (1, 3), then |[ l + 1]| is
(where [ × ] denotes the greatest integer
function)
(p)
0
(B)
x 2 - lx - 2
< 2, " x Î R , then
x2 + x + 1
|[ l ]| is (where [ × ] denotes the greatest
integer function)
(q)
1
(C)
If x 2 + lx + 1 = 0 and
(b - c) x 2 + (c - a) x + (a - b) = 0 have
both the roots common, then |[ l - 1]|,
(where [ × ] denotes the greatest integer
function)
(r)
2
(s)
3
(q) a, b, c are in AP
Let a, b, c be positive real numbers (r) a, b, c are in GP
such that the expression
bx 2 + ( (a + c)2 + 4 b2 ) x + (a + c)
is non-negative, " x Î R, then
(s) a, b, c are in HP
If - 3 <
174
#L
Textbook of Algebra
Theory of Equations Exercise 6 :
Statement I and II Type Questions
n
Directions (Q. Nos. 81 to 87) are Assertion-Reason
type questions. Each of these questions contains two
statements:
Statement-1 (Assertion) and Statement-2 (Reason)
Each of these questions also has four alternative
choices, only one of which is the correct answer. You
have to select the correct choice as given below.
(a) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1
(c) Statement1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
81. Statement-1 If the equation ( 4 p - 3) x
2
+ ( 4q - 3) x + r = 0 is satisfied by x = a, x = b and x = c
3
(where a, b, c are distinct), then p = q = and r = 0.
4
Statement-2 If the quadratic equation
ax 2 + bx + c = 0 has three distinct roots, then a, b and c
are must be zero.
82. Statement-1 The equation
x 2 + (2m + 1) x + (2n + 1) = 0, where m, n Î I , cannot have
any rational roots.
Statement-2 The quantity (2m + 1) 2 - 4 (2n + 1), where
m, n Î I , can never be perfect square.
#L
83. Statement-1 In the equation ax 2 + 3x + 5 = 0, if one
root is reciprocal of the other, then a is equal to 5.
Statement-2 Product of the roots is 1.
+ Bx 2 + Cx + D = 0,
A ¹ 0, is the arithmetic mean of the other two roots, then
the relation 2B 3 + k 1 ABC + k 2 A 2 D = 0 holds good and
then (k 2 - k 1 ) is a perfect square.
84. Statement-1 If one root of Ax
3
Statement-2 If a, b, c are in AP, then b is the arithmetic
mean of a andc .
85. Statement-1 If x , y, z be real variables satisfying
x + y + z = 6 and xy + yz + zx = 8, the range of variables
x , y and z are identical.
Statement-2 x + y + z = 6 and xy + yz + zx = 8 remains
same, if x , y, z interchange their positions.
3
+ bx + c = 0, where a, b, c Î R cannot
have 3 non-negative real roots.
86. Statement-1 ax
Statement-2 Sum of roots is equal to zero.
87. Statement-1 The quadratic polynomial
y = ax 2 + bx + c (a ¹ 0 and a, b, c Î R ) is symmetric about
the line 2ax + b = 0.
Statement-2 Parabola is symmetric about its axis of
symmetry.
Theory of Equations Exercise 7 :
Subjective Type Questions
n
In this section, there are 24 subjective questions.
88. For what values of m, the equation
(1 + m ) x 2 - 2 (1 + 3m ) x + (1 + 8m ) = 0 has (m Î R )
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
both roots are imaginary?
both roots are equal?
both roots are real and distinct?
both roots are positive?
both roots are negative?
roots are opposite in sign?
roots are equal in magnitude but opposite in sign?
atleast one root is positive?
atleast one root is negative?
roots are in the ratio 2 : 3?
89. For what values of m, then equation
2x 2 - 2 (2m + 1) x + m (m + 1) = 0 has (m Î R )
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
both roots are smaller tha 2?
both roots are greater than 2?
both roots lie in the interval (2, 3)?
exactly one root lie in the interval (2, 3)?
one root is smaller than 1 and the other root is
greater than 1?
one root is greater than 3 and the other root is
smaller than 2?
atleast one root lies in the interval (2, 3)?
atleast one root is greater than 2?
atleast one root is smaller than 2?
roots a and b, such that both 2 and 3 lie between a
and b?
Chap 02 Theory of Equations
90. If r is the ratio of the roots of the equation
2
2
(r + 1)
b
= .
r
ac
1
1
1
91. If the roots of the equation
+
= are equal
x + p x +q r
in magnitude but opposite in sign, show that p + q = 2r
æ p2+q2 ö
÷.
and that the product of the roots is equal to ç 2 ø
è
ax 2 + bx + c = 0, show that
92. If one root of the quadratic equation ax 2 + bx + c = 0 is
equal to the nth power of the other, then show that
1
n n +1
(ac )
n
+ (a c )
1
n +1
+ b = 0.
2
+ bx + c = 0 and
2
g , d those of equation lx + mx + n = 0, then find the
equation whose roots are ag + bd and ad + bg .
94. Show that the roots of the equation
2
2
2
(a - bc ) x + 2 (b - ac ) x + c - ab = 0
are equal, if either b = 0 or a 3 + b 3 + c 3 - 3abc = 0.
95. If the equation x 2 - px + q = 0 and x 2 - ax + b = 0 have
a common root and the other root of the second
equation is the reciprocal of the other root of the first,
then prove that (q - b ) 2 = bq ( p - a ) 2 .
96. If the equation x 2 - 2px + q = 0 has two equal roots,
then the equation (1 + y ) x 2 - 2 ( p + y ) x + (q + y ) = 0
will have its roots real and distinct only, when y is
negative and p is not unity.
97. Solve the equation x
log x (x + 3 )2
(2 + 3 )
+ (2 - 3 )
x 2 - 2 x -1
2
2
æ| x + 4 | - | x | ö
÷ > 0.
x -1
ø
è
102. Solve the inequation log x 2 + 2 x - 3 ç
103. Solve the system | x 2 - 2x | + y = 1, x 2 + | y | = 1.
104. If a , b, g are the roots of the cubic x 3 - px 2 + qx - r = 0.
Find the equations whose roots are
1
1
1
(i) bg + , ga + , ab +
a
b
g
Also, find the value of ( b + g - a ) ( g + a - b ) (a + b - g ).
105. If A 1 , A 2 , A 3 , ..., A n , a1 , a 2 , a 3 , ..., an , a, b, c Î R, show
that the roots of the equation
A 12
A 22
A 32
A n2
+
+
+K+
x - a1 x - a 2 x - a 3
x - an
2
2
= ab + c x + ac are real.
106. For what values of the parameter a the equation
x 4 + 2ax 3 + x 2 + 2ax + 1 = 0 has atleast two distinct
negative roots?
107. If [x ] is the integral part of a real number x. Then solve
[2x ] - [x + 1] = 2x .
108. Prove that for any value of a, the inequation (a 2 + 3)
x 2 + (a + 2) x - 6 < 0 is true for atleast one negative x .
6x 2 - 77 [x ] + 147 = 0, where [x ] is the integral part of x ?
101
.
=
10 (2 - 3 )
æ x ö
÷ = 8.
è x - 1ø
99. Solve the equation x 2 + ç
110. If a , b are the roots of the equation x 2 - 2x - a 2 + 1 = 0
and g , d are the roots of the equation
x 2 - 2 (a + 1) x + a (a - 1) = 0, such that a , b Î ( g , d ), find
the value of ‘a’.
111. If the equation x 4 + px 3 + qx 2 + rx + 5 = 0 has four
100. Solve the equation
( x + 8) + 2 ( x + 7 ) + ( x + 1) - ( x + 7 ) = 4.
#L
2
4 x + 2 (2a + 1) 2 x + 4a 2 - 3 > 0 is satisfied for any x .
109. How many real solutions of the equation
= 16.
98. Solve the equation
x 2 - 2 x +1
101. Find all values of a for which the inequation
(ii) ( b + g - a ), ( g + a - b ), (a + b - g )
93. If a , b are the roots of the equation ax
2
175
positive real roots, find the minimum value of pr.
Theory of Equations Exercise 8 :
Questions Asked in Previous 13 Years' Exam
n
This section contains questions asked in IIT-JEE,
AIEEE, JEE Main & JEE Advanced from year 2005
to year 2017.
112. If a , b are the roots of ax 2 + bx + c = 0 and a + b ,
a 2 + b 2 , a 3 + b 3 are in GP, where D = b 2 - 4ac , then
[IIT-JEE 2005, 3M]
(a) D ¹ 0
(b) bD = 0
(c) cb ¹ 0
(d) cD = 0
113. If S is a set of P( x ) is polynomial of degree £ 2 such that
P(0) = 0, P(1) = 1, P ¢( x ) > 0, "x Î(0, 1), then [IIT-JEE 2005, 3M]
(a) S = 0
(b) S = ax + (1 - a ) x 2, " a Î ( 0, ¥ )
(c) S = ax + (1 - a ) x 2, " a Î R
(d) S = ax + (1 - a ) x 2, " a Î ( 0, 2 )
176
Textbook of Algebra
114. If the roots of x 2 - bx + c = 0 are two consecutive
2
integers, then b - 4c is
(a) 1
(c) 3
[AIEEE 2005, 3M]
(b) 2
(d) 4
115. If the equation an x n + an - 1 x n - 1 + ... + a1 x = 0, a1 ¹ 0,
n ³ 2, has a positive root x = a, then the equation
nan x n - 1 + (n - 1) an - 1 x n - 2 + K + a1 = 0 has a positive
root, which is
[AIEEE 2005, 3M]
(a) greater than or equal to a
(b) equal to a
(c) greater than a
(d) smaller than a
(a) ( - ¥, 4 )
(c) (5, 6 )
[AIEEE 2005, 3M]
(b) [ 4, 5 ]
(d) (6, ¥ )
and those of x 2 - 10ax - 11b = 0 are c and d, the value of
IIT-JEE 2006, 6M]
a + b + c + d , when a ¹ b ¹ c ¹ d , is
118. Let a, b, c be the sides of a triangle. No two of them are
equal and l Î R . If the roots of the equation
x 2 + 2 (a + b + c ) x + 3 l (ab + bc + ca ) = 0 are real, then
[IIT-JEE 2006, 3M]
5
(b) l <
3
æ 4 5ö
(d) l Î ç , ÷
è 3 3ø
(b) ( -3, ¥ )
(d) ( -¥, - 3 )
123. Let a, b, c , p , q be real numbers. Suppose a , b are roots of
the equation x 2 + 2px + q = 0 and a ,
124. The quadratic equation x 2 - 6x + a = 0 and
x 2 - cx + 6 = 0 have one root in common. The other
roots of the first and second equations are integers in
the ratio 4 : 3. The common root is
[AIEEE 2008, 3M]
(a) 4
(b) 3
(c) 2
120. If the roots of the quadratic equation x 2 + px + q = 0 are
125. How many real solutions does the equation
x 7 + 14 x 5 + 16x 3 + 30x - 560 = 0 have?
(b) 3
(c) 5
[AIEEE 2006, 3M]
121. Let a, b be the roots of the equation x 2 - px + r = 0 and
a
, 2 b be the roots of the equation x 2 - qx + r = 0. The
2
value of r is
[IIT-JEE 2007, 3M]
2
(q - p ) (2 p - q )
9
2
(d) (2 p - q ) (2q - p )
9
(b)
[AIEEE 2008, 3M]
(d) 7
pö
÷ and maxima at
3ø
æ
(a) The cubic has minima at ç è
(b) The cubic has minima at both
æ
p
and ç è
3
pö
÷
3ø
(c) The cubic has maxima at both
æ
p
and ç è
3
pö
÷
3ø
tan 30° and tan 15°, respectively, the value of 2 + q - p is
2
(a) ( p - q ) (2q - p )
9
2
(c) (q - 2 p ) (2q - p )
9
(d) 1
roots, where p > 0 and q < 0. Which one of the following
holds?
[AIEEE 2008, 3M]
(b) m > 3
(d) 1 < m < 4
(b) 3
(d) 1
[IIT-JEE 2008, 3M]
(a) Statement-1 is true, Statement-2, is true; Statement-2 is a
correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
126. Suppose the cubic x 3 - px + q = 0 has three distinct real
x 2 - 2mx + m 2 - 1 = 0 are greater than - 2 but less than
4, lie in the interval
[AIEEE 2006, 3M]
(a) 2
(c) 0
1
are the roots of
b
the equation ax 2 + 2bx + c = 0, where b 2 Ï {-1, 0, 1}.
(a) 1
119. All the values of m for which both roots of the equation
(a) - 2 < m < 0
(c) - 1 < m < 3
(a) ( - 3, 3 )
(c) (3, ¥ )
Statement-2 b ¹ pa or c ¹ qa
117. Let a and b be the roots of equation x 2 - 10cx - 11d = 0
4
(a) l <
3
æ1 5ö
(c) l Î ç , ÷
è3 3ø
x 2 + ax + 1 = 0 is less than 5, the set of possible values
of a, is
[AIEEE 2007, 3M]
Statement-1 ( p 2 - q ) (b 2 - ac ) ³ 0 and
116. If both the roots of the quadratic equation
x 2 - 2kx + k 2 + k - 5 = 0
are less than 5, k lies in the interval
122. If the difference between the roots of the equation
(d) The cubic has minima at
æ
p
and maxima at ç 3
è
p
3
pö
÷
3ø
127. The smallest value of k , for which both roots of the
equation x 2 - 8kx + 16 (k 2 - k + 1) = 0 are real, distinct
and have value at least 4 , is
[IIT-JEE 2009, 4M]
(a) 6
(b) 4
(c) 2
(d) 0
2
128. If the roots of the equation bx + cx + a = 0 be
imaginary, then for all real values of x, the expression
3b 2 x 2 + 6bcx + 2c 2 , is
[AIEEE 2009, 4M]
(a) less than ( - 4ab )
(c) less than 4ab
(b) greater than 4ab
(d) greater than ( - 4ab )
Chap 02 Theory of Equations
129. Let p and q be real numbers such that p ¹ 0, p 3 ¹ - q . If
a and b are non-zero complex numbers satisfying
a + b = - p and a 3 + b 3 = q, a quadratic equation
b
a
having and as its roots, is
[IIT-JEE 2010, 3M]
a
b
(a) ( p 3 + q ) x 2 - ( p 3 + 2q ) x + ( p 3 + q ) = 0
137. The equation e
sin x
- e - sin x - 4 = 0 has
[AIEEE 2012, 4M]
(a) exactly one real root
(b) exactly four real roots
(c) infinite number of real roots
(d) no real roots
138. If the equations x 2 + 2x + 3 = 0 and ax 2 + bx + c = 0,
(b) ( p 3 + q ) x 2 - ( p 3 - 2q ) x + ( p 3 + q ) = 0
a, b, c Î R have a common root, then a : b : c is
(c) ( p 3 - q ) x 2 - (5 p 3 - 2q ) x + ( p 3 - q ) = 0
(a) 3 : 2 : 1
[JEE Main 2013, 4M]
(d) ( p 3 - q ) x 2 - (5 p 3 + 2q ) x + ( p 3 - q ) = 0
+ 4 x 3 . Let
s be the sum of all distinct real roots of f ( x ) and let
t = | s |, real number s lies in the interval [IIT-JEE 2010, 3M]
2
3ö
æ 1 ö
æ
æ 3 1ö
æ 1ö
(a) ç - , 0 ÷ (b) ç - 11, ÷ (c) ç - , - ÷ (d) ç 0, ÷
è 4 ø
è
è 4 2ø
è 4ø
4ø
131. Let a and b be the roots of x 2 - 6x - 2 = 0, with a > b. If
a10 - 2a 8
is
2a 9
[IIT-JEE 2011, 3 and JEE Main 2015,4M]
(a) 1
(b) 2
(c) 3
(d) 4
2
x + bx - 1 = 0 x + x + b = 0
have one root in common, is
[IIT-JEE 2011, 3M]
(a) - 2
(b) - i 3, i = -1
(c) i 5, i = -1
(d) 2
x - 4x
3
+ 12x
2
+ x - 1 = 0 is
[IIT-JEE 2011, 4M]
g ( x ) = a1 x 2 + b1 x + c 1 and p ( x ) = f ( x ) - g ( x ). If p ( x ) = 0
only for x = ( - 1) and p ( - 2) = 2, the value of p (2) is
[AIEEE 2011, 4M]
(b) 3
(c) 9
(d) 6
135. Sachin and Rahul attempted to solve a quadratic
equation. Sachin made a mistake in writing down the
constant term and ended up in roots ( 4, 3). Rahul made a
mistake in writing down coefficient of x to get roots
(3, 2). The correct roots of equation are [AIEEE 2011, 4M]
(a) - 4, - 3
(b) 6, 1
(c) 4, 3
-3 ( x - [x ]) 2 + 2 ( x - [x ]) + a 2 = 0 (where [ × ] denotes
the greatest integer function) has no integral solution,
then all possible values of a lie in the interval
[JEE Main 2014, 4M]
(a) ( -2, - 1 )
(b) ( -¥, - 2 ) È (2, ¥ )
(c) ( -1, 0 ) È ( 0, 1 )
(d) (1,2)
140. Let a, b be the roots of the equation px 2 + qx + r = 0,
p ¹ 0. If p , q , r are in AP and
34
(a)
9
61
(c)
9
1 1
+ = 4 , the value of
a b
[JEE Main 2014, 4M]
2 13
(b)
9
2 17
(d)
9
f ( x ) = x 5 - 5x + a. Then,
134. Let for a ¹ a1 ¹ 0, f ( x ) = ax 2 + bx + c ,
(a) 18
(d) 1 : 2 : 3
141. Let a Î R and let f : R ® R be given by
133. The number of distinct real roots of
4
(c) 3 : 1 : 2
| a - b |, is
132. A value of b for which the equations
2
(b) 1 : 3 : 2
139. If a Î R and the equation
130. Consider the polynomial f ( x ) = 1 + 2x + 3x
an = a n - b n for n ³ 1, the value of
177
(d) - 6, - 1
[JEE Advanced 2014, 3M]
(a) f ( x ) has three real roots, if a > 4
(b) f ( x ) has only one real root, if a > 4
(c) f ( x ) has three real roots, if a < - 4
(d) f ( x ) has three real roots, if -4 < a < 4
142. The quadratic equation p ( x ) = 0 with real coefficients
has purely imaginary roots. Then, p ( p ( x )) = 0 has
[JEE Advanced 2014, 3M]
(a) only purely imaginary roots
(b) all real roots
(c) two real and two purely imaginary roots
(d) neither real nor purely imaginary roots
143. Let S be the set of all non-zero real numbers a such that
( 3 (1 + a ) - 1)x 2 + ( (1 + a ) - 1)x + ( 6 (1 + a ) - 1) = 0,
the quadratic equation ax 2 - x + a = 0 has two distinct
real roots x 1 and x 2 satisfying the inequality
| x 1 - x 2 | < 1.
where a > - 1, then lim a (a ) and lim b(a ), are
Which of the following intervals is (are) a subset(s) of S ?
136. Let a(a ) and b(a ) be the roots of the equation
a ® 0+
a ® 0+
[JEE Advanced 2015, 4M]
[IIT-JEE 2012, 3M]
æ 5ö
(a) ç - ÷ and 1
è 2ø
æ 1ö
(b) ç - ÷ and ( -1 )
è 2ø
æ 7ö
(c) ç - ÷ and 2
è 2ø
æ 9ö
(d) ç - ÷ and 3
è 2ø
1 ö
æ 1
(a) ç - , ÷
è 2
5ø
æ 1 ö
(c) ç 0, ÷
è
5ø
æ 1 ö
(b) ç - , 0 ÷
è 5 ø
æ 1 1ö
(d) ç , ÷
è 5 2ø
178
Textbook of Algebra
144. The sum of all real values of x satisfying the equation
( x 2 - 5x + 5) x
(a) 6
2
+ 4 x - 60
(b) 5
= 1 is
(c) 3
[JEE Main 2016, 4M]
(d) -4
p
p
< q < - . Suppose a 1 and b 1 are the roots of
6
12
equation x 2 - 2x sec q + 1 = 0 and a 2 and b 2 are the roots
of the equation x 2 + 2x tan q - 1 = 0. If a 1 > b 1 and
[JEE Advanced 2016, 3M]
a 2 > b 2 , then a 1 + b 2 equals
145. Let -
(a) 2 (sec q - tan q)
(c) - 2 tan q
(b) 2 sec q
(d) 0
146. If for a positive integer n, the quadratic equation
x ( x + 1) + ( x + 1) ( x + 2) K + ( x + n - 1) ( x + n ) = 10n has
two consecutive integral solutions, then n is equal to
[JEE Main 2017, 4M]
(a) 11
(c) 9
(b) 12
(d) 10
Answers
Exercise for Session 1
1.(b)
7.(c)
2. (c)
8. (b)
3. (a)
9. (c)
4. (b)
10. (d)
5. (a)
11. (b)
6. (a)
æ 7 - 33 11 - 73 ö æ 7 + 33 11 + 73 ö
(iv) m Îç
,
,
÷
÷ Èç
ø
2
2
2
2
ø è
è
Exercise for Session 2
1.(a)
7. (c)
2. (c)
8. (b)
3. (b)
9. (c)
4. (a) 5.(d)
10. (a)
6. (c)
4. (c) 5.(d)
10. (d)
6. (c)
4. (d) 5.(a)
10. (d)
6. (d)
4. (c) 5.(c)
10. (d)
6. (b)
Exercise for Session 3
1.(a)
7.(c)
2. (b)
8. (a)
3. (c)
9. (a)
Exercise for Session 4
1.(c)
7.(d)
2. (c)
8. (c)
3. (c)
9. (b)
Exercise for Session 5
1.(a)
7.(a)
2. (a)
8. (b)
3. (b)
9. (b)
æ
æ 7 + 33 ö
7 - 33 ö
89. (i) m Î ç -¥ ,
, ¥ ÷ (iii) m Î f
÷ (ii) m Î ç
è
è
ø
2 ø
2
Chapter Exercises
1.(b)
2. (b)
3. (b)
4. (c)
5. (a)
6. (a)
7.(c)
8. (b)
9. (c)
10. (a)
11. (b)
12. (c)
13.(b)
14. (b) 15. (b)
16. (c)
17. (b)
18. (b)
19.(c)
20. (d) 21. (b)
22. (a)
23. (c)
24. (b)
25.(a)
26. (a) 27. (b)
28. (c)
29. (a)
30. (c)
31.(a,b) 32.(b,c)
33.(a,d)
34.(a,b,c,d)
35. (b,d) 36. (a,b,c,d)
37.(a, b,c,d) 38. (a,c) 39. (a,b) 40. (a,b,c,d) 41. (a,c) 42. (b,c)
43.(c,d) 44.(a,c,d) 45.(a,c,d)
46.(d)
47. (b)
48. (d)
49.(d)
50.(d)
51. (c)
52.(b)
53. (b)
54. (c)
55. (b)
56. (d)
57. (a)
58.(c)
59. (a)
60. (b)
61. (c)
62. (b)
63. (c)
64.(d)
65. (a)
66. (b)
67. (4)
68. (4)
69. (9)
70. (6) 71. (3)
72. (4)
73. (2)
74. (5)
75.(3)
76.(7)
77. (A) ® (r,s), (B) ® (p,q,r,s,t), (C) ® (p,q,t)
78. (A) ® (q,r,s), (B) ® (p), (C) ® (q)
79. (A) ® (q,r,s,t), (B) ® (q,r), (C) ® (p,q)
80.(A) ® (p,q,r,s), (B) ® (p,q), (C) ® (s)
81.(d)
82. (a)
83. (a)
84. (a)
85. (a)
86. (d)
87.(a)
88. (i) m Î (0, 3) (ii) m = 0, 3
(iii) m Î (-¥ , 0) È (3, ¥ )
(iv) m Î (-¥ , - 1) È [ 3, ¥ )
(v) m Î f
(vi) m Î (-1, - 1/ 8)
(vii) m = - 1/ 3
(viii) m Î (-¥ , - 1) È (-1, - 1/ 8) È [ 3, ¥ )
81 ± 6625
(ix) m Î (-1, - 1/ 8)
(x) m =
32
æ 7 - 33 7 + 33 ö
(vi) m Î ç
,
÷
ø
2
2
è
æ 7 - 33 11 - 73 ö æ 7 + 33 11 + 73 ö
(vii) m Î ç
,
,
÷Èç
÷
ø è
ø
2
2
2
2
è
æ 7 - 33 7 + 33 ö æ 7 + 33 ö
(viii) m Î ç
,
, ¥÷
÷ Èç
ø
2
2
2
ø è
è
æ
7 - 33 ö æ 7 - 33 7 + 33 ö
(ix) m Î ç -¥ ,
,
÷
÷ Èç
ø
2 ø è
2
2
è
æ 11 - 73 7 + 33 ö
(x) m Î ç
,
÷
è
ø
2
2
(v) m Î (0, 3)
93. a2l 2x 2 - ablmx + (b2 - 2ac) ln + (m2 - 2ln) ac = 0
97. x Î f
98. x1 = 1 + 1 + log2 +
99. x1 = 2, x2 = - 1 +
3 10 , x2
= 1 - 1 + log2 +
3 10
3 and x3 = - 1 - 3
100. x1 = 2
æ 3 ö
101. a Î (-¥ , - 1) È ç
, ¥÷
ø
è 2
102. x Î (-1 - 5 , - 3) È ( 5 - 1, 5)
æ1 - 5 1 - 5 ö
103. The pairs (0, 1), (1, 0), ç
,
÷ are solutions of the
è 2
2 ø
original system of equations.
104. (i) ry3 - q (r + 1) y2 + p (r + 1)2 y - (r + 1)3 = 0
(ii) y3 - py2 + (4 q - p2 ) y + (8r - 4 pq + p3 ) = 0 and
4 pq - p3 - 8r
æ3 ö
106. a Î ç , ¥ ÷ 107. x1 = - 1, x2 = - 1/ 2
109. Four
è4 ø
1
110. a Î æç - , 1ö÷
è 4 ø
111. 80
112. (d)
113. (d)
115. (d)
121. (d)
127. (c)
133.(2)
139. (c)
144. (c)
117. 1210
123. (b)
129. (b)
135. (b)
141.(b,d)
146. (a)
118.(a)
124. (c)
130. (c)
136. (b)
142. (d)
119. (c) 120. (b)
125. (a) 126. (d)
131. (c) 132. (b)
137. (d) 138. (d)
143. (a, d)
116. (a)
122. (a)
128. (d)
134. (a)
140. (b)
145. (c)
114. (a)
5. Q
Solutions
and
2 (a − b ) x 2 − 11 (a + b + c ) x − 3 (a − b ) = 0
∴ D = { − 11 (a + b + c )} 2 − 4 ⋅ 2 (a − b ) ⋅ ( −3 ) (a − b )
6. Let f ( x ) = x 2 − 2ax + a 2 − 1
Therefore, the roots are real and unequal.
2. Here, a < 0
a −3
=I+ +2
a
D = (a + 3 ) 2 − 4a (a + 3 )
9
{( I + − 2 ) 2 − 12 }
= +
(I + 1)2
and
D must be perfect square, then I
From Eq. (ii),
Product of the roots = I
+
+
Now, four cases arise:
Case I D ≥ 0
[from graph]
–2
[Q a < 0 ]
[Qc < 0 , a < 0 ]
[let]
…(i)
…(ii)
[from Eq. (i)]
=6
2
⇒
( − 2a ) 2 − 4 ⋅ 1 (a 2 − 1 ) ≥ 0
⇒
∴
4≥0
a ∈R
⇒
4 + 4a + a 2 − 1 > 0
⇒
a 2 + 4a + 3 > 0
X
⇒
(a + 1 ) (a + 3 ) > 0
∴
a ∈ ( − ∞, − 3 ) ∪ ( − 1, ∞ )
Case III f (2 ) > 0
⇒
4 − 4a + a 2 − 1 > 0
⇒
a 2 − 4a + 3 > 0
⇒
∴
(a − 1 ) (a − 3 ) > 0
a ∈ ( − ∞, 1 ) ∪ (3, ∞ )
Case IV − 2 < x-coordinate of vertex < 2
⇒
− 2 < 2a < 2
∴
a ∈ ( − 1, 1 )
Combining all cases, we get a ∈ ( − 1, 1 )
Hence,
[a ] = − 1, 0
 − 4a 
 = −2
 2 (− 2)
⇒
∴
+ 2 =6 + 2 =8
a =2
y = − 2x 2 − 8x + λ
Since, Eq. (i) passes through points ( − 2, 7 )
∴
7 = − 2 (− 2)2 − 8 (− 2) + λ
α + 2 α = 3a ⇒ 3 α = 3a
α =a
α ⋅ 2 α = f (a )
f (a ) = 2 α 2= 2a 2
f (x ) = 2x
β
7. We have, − 
x 2 − 3ax + f (a ) = 0
⇒
α
Case II f ( − 2 ) > 0
4. Let α be one root of
⇒
⇒
and
⇒
x 2 − 2x + 4 = 0
or
= 121 (a + b + c ) 2 + 24 (a − b ) 2 > 0
Product of the roots = αβ =
...(i)
⇒
2 x1x 2 − x1 − x 2 = 2 ( x1x 2 − x1 − x 2 + 1 )
[from Eq. (i)]
⇒
8 − x1 − x 2 = 2 ( 4 − x1 − x 2 + 1 )
or
…(ii)
x1 + x 2 = 2
From Eqs. (i) and (ii), required equation is
x 2 − ( x1 + x 2 ) x + x1x 2 = 0
1. We have,
Cut-off Y -axis, x = 0
⇒
y =c < 0
∴
c<0
x -coordinate of vertex > 0
b
⇒
−
>0
2a
b
<0
⇒
a
But
a<0
∴
b>0
and y-coordinate of vertex < 0
D
D
−
<0 ⇒
⇒
>0
4a
4a
∴
D<0
i.e.
b 2 − 4ac < 0
c
∴
>0
a
(a + 3 )
3. Sum of the roots = −
=I+
a

3 
∴
a = − +

 I + 1
x1x 2 = 4
x1
x2
+
=2
x1 − 1 x 2 − 1
2
…(i)
[using Eq. (i)]
⇒
7 = − 8 + 16 + λ
∴
λ = −1
8. Since, the coefficient of n 2 = ( 4p − p 2 − 5) < 0
Therefore, the graph is open downward.
According to the question, 1 must lie between the roots.
…(i)
180
Textbook of Algebra
f (1) > 0
4p − p − 5 − 2p + 1 + 3p > 0
Hence,
⇒
⇒
− p + 5p − 4 > 0
⇒
p 2 − 5p + 4 < 0
∴
α 3 ≤ − 7, α 2 ≤ − 8, α 1 ≤ − 9
∴
a + b + c ≥ 719
∴Minimum value of a + b + c is 719.
Q
α1 + α 2 + α 3 = − a
⇒
− a ≤ − 24
⇒
a ≥ 24
α 1α 2 + α 2 α 3 + α 3α 1 = b
⇒
b ≥ 191
and
α 1α 2 α 3 = − c
⇒
− c ≤ − 504
⇒
c ≥ 504
∴
a + b + c ≥ 719
Hence, minimum value of a + b + c is 719.
2
⇒
( p − 4) ( p − 1) < 0
⇒
1<p<4
∴
p = 2, 3
Hence, number of integral values of p is 2.
2
9. We have, 3 2 x − 2 ⋅ 3 x
2
+ x+ 6
+ 3 2 ( x + 6) = 0
2
⇒
(3 x − 3 x + 6 ) 2 = 0
⇒
3x − 3x + 6 = 0
⇒
2
2
3x = 3x + 6 ⇒ x 2 = x + 6
⇒
x2 − x − 6 = 0
n
13. Q ∑ ( x + k − 1) ( x + k ) = 10n
⇒
(x − 3) (x + 2) = 0
∴
x = { − 2, 3 }
10. Given, b 2 − 4ac = p 2 − 4aq
and
⇒
⇒
∴
k =1
…(i)
⇒
[given] …(ii)
From Eq. (i), we get
(b + p ) (b − p ) + 4a (q − c ) = 0
⇒
(b + p ) (b − p ) + 4aα (b − p ) = 0 [from Eq. (ii)]
(b + p )
or
[Qb ≠ p]
α=−
4a
 b  p
−  + − 
 a  a 
=
4
Sum of the roots of ( f (x ) = 0) 
+ Sum of the roots of (g (x ) = 0)

=
4
= AM of the roots of f ( x ) = 0
and
g( x ) = 0
11. Let α and β be the roots of ax 2 + bx + c = 0.
b
α+β
∴
x1 =
=−
2
2a
c
2⋅
2c
2αβ
a
and
x2 =
=
=−
b
α +β −b
a
∴The required equation is
2bc
 b   2c  
x 2 −  −  +  −   x +
=0
2ab
 2a   b  
i.e.
2abx + (b + 4ac ) x + 2bc = 0
2
2
12. Let α 1, α 2 and α 3 be the roots of f ( x ) = 0, such that
α1 < α 2 < α 3
and g( x ) can take all values from [ − 6, ∞ ).
⇒
n
∑ x 2 + x (2k − 1) + (k − 1)k = 10n
k =1
f ( x ) = g( x )
ax 2 + bx + c = ax 2 + px + q
(b − p ) x = q − c
q −c
x=
=α
b−p
g( x ) = ( x + 1 ) 2 − 6 ≥ − 6
Q
2
nx 2 + x (1 + 3 + 5 +…+ (2n − 1 ))
+ ( 0 + 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 +…+ (n − 1 )n ) = 10n
n
⇒ nx 2 + x ⋅ (1 + 2n − 1 )
2
 n (n + 1 ) (2n + 1 ) n(n + 1 )
+
−
 = 10n

6
2 
⇒
⇒
Q
⇒
⇒
⇒
n (n 2 − 1 )
= 10n
3
(n 2 − 31 )
x 2 + nx +
=0
3
D
(α + 1 ) − α =
1
1= D
nx 2 + n 2x +
[dividing by n]
D =1
(n − 31 )
n2 − 4 ⋅1 ⋅
=1
3
3n 2 − 4n 2 + 124 = 3
2
⇒
n 2 = 121
∴
n = 11
14. Since, 2 is only even prime.
Therefore, we have
2 2 + λ ⋅ 2 + 12 = 0
⇒
∴
λ =8
x 2 + λx + µ = 0
⇒
x 2 + 8x + µ = 0
But Eq. (i) has equal roots.
∴
D=0
2
⇒
8 − 4 ⋅1 ⋅µ = 0
⇒
µ = 16
…(i)
Chap 02 Theory of Equations
x +
15. We have,
x − (1 − x ) = 1
Let y =
x − 1 − x =1 − x
⇒
=
On squaring both sides, we get
x − 1 − x =1 + x −2 x
1 − x = 1 + 4x − 4 x
4 x = 5x
⇒
⇒
[on squaring both sides]
Hence, the number of real solutions is 1.
16. Let x = 7 + 7 − 7 + 7 −…∞
⇒
x= 7+ 7−x
⇒
[on squaring both sides]
x2 − 7 = 7 − x
⇒
(x 2 − 7)2 = 7 − x
⇒
[again, squaring on both sides]
x − 14 x + x + 42 = 0
4
2
⇒
( x − 3 ) ( x + 3 x 2 − 5 x − 14 ) = 0
⇒
(x − 3) (x + 2) (x 2 + x − 7) = 0
3
⇒
x = 3, − 2,
∴
x =3
− 1 ± 29
2
[Q x > 7]
17. Let y = 2 (k − x ) ( x + ( x + k )
2
⇒
2
y − 2 (k − x ) x = 2 (k − x ) ( x 2 + k 2 )
On squaring both sides, we get
⇒ y 2 + 4 (k − x ) 2 x 2 − 4 xy (k − x ) = 4 (k − x ) 2 ( x 2 + k 2 )
⇒
⇒ 4 (k − y ) x − 4(2k − ky ) x − y + 4k = 0
2
Since, x is real.
∴
⇒
3
2
4
D≥0
16 (2k 3 − ky ) 2 − 4 ⋅ 4 (k 2 − y ) ( 4k 4 − y 2 ) ≥ 0
[using, b 2 − 4ac ≥ 0]
⇒ 4k 6 + k 2y 2 − 4k 4y − ( − k 2y 2 + 4k 6 + y 3 − 4yk 4 ) ≥ 0
⇒
2k 2y 2 − y 3 ≥ 0
⇒
y 2 (y − 2k 2 ) ≤ 0
∴
18. We have,
y ≤ 2k 2
1 x 2 − 2x + 4
< 3, ∀ x ∈ R
<
3 x 2 + 2x + 4
1 x 2 + 2x + 4
< 3, ∀ x ∈ R
<
3 x 2 − 2x + 4
…(i)
We have, product of the roots = 4 > 0, which is true.
2 (y + 1 )
And sum of the roots =
>0
(y − 1 )
y +1
⇒
>0
y −1
∴
y ∈ ( − ∞, − 1 ) ∪ (1, ∞ )
From Eqs. (i) and (ii), we get
1 <y <3
19. Since α , β and γ are the roots of
…(ii)
(x − a ) (x − b ) (x − c ) = d
⇒ (x − a ) (x − b ) (x − c ) − d = (x − α ) (x − β) (x − γ )
⇒ (x − α ) (x − β) (x − γ ) + d = (x − a ) (x − b ) (x − c )
⇒ a, b and c are the roots of
(x − α ) (x − β) (x − γ ) + d = 0
20. Since, all the coefficients of given equation are not real.
Therefore, other root ≠ 3 + i.
Let other root be α.
2 (1 + i )
Then, sum of the roots =
i
2 (1 + i )
⇒
α + 3 −i =
i
⇒
α + 3 − i = 2 − 2i
∴
α = −1 −i
21. We have,
|[ x ] − 2 x | = 4
⇒ |[ x ] − 2 ([ x ] + { x })| = 4
⇒
|[ x ] + 2 { x }| = 4
which is possible only when
2 { x } = 0, 1
y 2 − 4 xy (k − x ) = 4 (k − x ) 2k 2
2
t 2 + 2t + 4
, where t = 3 x + 1
t 2 − 2t + 4
By the given condition, for every t ∈ R,
1
<y <3
3
But
t = 3x + 1 > 0
Again, squaring on both sides, we get
4
x =
5
16
x=
25
9 ⋅ 3 2 x + 6 ⋅ 3 x + 4 (3 x + 1 ) 2 + 2 ⋅ 3 x + 1 + 4
=
9 ⋅ 3 2 x − 6 ⋅ 3 x + 4 (3 x + 1 ) 2 − 2 ⋅ 3 x + 1 + 4
⇒ (y − 1 ) t 2 − 2 (y + 1 ) t + 4 (y − 1 ) = 0
− 1 − x =1−2 x
⇒
181
1
If { x } = 0, then [ x ] = ± 4 and then x = − 4, 4 and if { x } = ,
2
then
[x ] + 1 = ± 4
⇒
[ x ] = 3, − 5
1
1
∴
x = 3 + and − 5 +
2
2
7 9
9 7
⇒
x = , − ⇒ x = − 4, − , , 4
2 2
2 2
22. We know that, x 2 + x + 1 is a factor of ax 3 + bx 2 + cx + d .
Hence, roots of x 2 + x + 1 = 0 are also roots of
ax 3 + bx 2 + cx + d = 0. Since, ω and ω 2
182
Textbook of Algebra
1 3i 

2
 where ω = − +  are two complex roots of x + x + 1 = 0.

2 2
Therefore, ω and ω 2 are two complex roots of
ax 3 + bx 2 + cx + d = 0.
We know that, a cubic equation has atleast one real root. Let
real root be α. Then,
d
d
α ⋅ ω ⋅ ω2 = −
⇒α=−
a
a
23. We have,
(5 x 2 − 8 x + 3 ) − (5 x 2 − 9 x + 4 )
= (2 x 2 − 2 x ) − (2 x 2 − 3 x + 1 )
⇒
(5 x − 3 ) ( x − 1 ) − (5 x − 4 ) ( x − 1 )
= 2 x ( x − 1 ) − (2 x − 1 ) ( x − 1 )
⇒
x − 1 ( 5x − 3 − 5x − 4 ) = x − 1 ( 2x − 2x − 1 )
x −1 = 0
⇒
⇒
x =1
(a + b ) (a − b ) = a 2 − b = 1
24. We have,
∴
(a + b )
x 2 − 15
(a + b )
⇒
[given]
x 2 − 15
+ (a − b )
= 2a
1
+
= 2a
2
(a + b ) x − 15
26. Since, 3789108 is an even integer. Therefore, x 4 − y 4 is also an
even integer. So, either both x and y are even integers or both
of them are odd integers.
Now,
x 4 − y 4 = (x − y ) (x + y ) (x 2 + y 2 )
⇒ x − y , x + y , x 2 + y 2 must be even integers.
Therefore, ( x − y ) ( x + y ) ( x 2 + y 2 ) must be divisible by 8. But
3789108 is not divisible by 8. Hence, the given equation has no
solution.
∴ Number of solutions = 0
27. We have,
x 3 + ax + 1 = 0
or
x 4 + ax 2 + x = 0
…(i)
and
x + ax + 1 = 0
…(ii)
1 − a2 ≠ 0
x 2 − 15
 2a 
 1 
x2 + 
x − 
 =0
2
1 − a 
1 − a2
y = (a + b ) x − 15
1
y + = 2a ⇒ y 2 − 2ay + 1 = 0
y
⇒
2
From Eqs. (i) and (ii), we get
x −1 = 0
⇒
x =1
which is a common root.
∴
1 +a + 1 = 0
⇒
a = −2
28. Q(1 − a 2 ) x 2 + 2ax − 1 = 0
2
Let
4
Let
 2a 
 1 
f (x ) = x 2 + 
x − 

1 − a2
 (1 − a 2 )
4a 2 − 4
= a ± a2 − 1
2
[Qa 2 − b = 1]
y = a ± b = (a + b ) ± 1
y =
⇒
∴
⇒ (a + b ) x
∴
2
− 15
2a ±
= (a + b ) ± 1
x 2 − 15 = ± 1
⇒
x 2 = 15 ± 1 ⇒ x 2 = 16, 14
⇒
x = ± 4, ± 14
25. We have,
x 2 − x (y + 4 ) + y 2 − 4y + 16 = 0
Q
∴
x ∈R
( − (y + 4 )) 2 − 4 ⋅ 1 ⋅ (y 2 − 4y + 16 ) ≥ 0
⇒
2
⇒
∴
∴
Then,
y 2 − 8y + 16 ≤ 0
⇒ (y − 4 ) 2 ≤ 0
⇒
Case II f ( 0 ) > 0
⇒
y =4
x 2 − 4 x + 16 = 4( x + 4 − 4 )
⇒
(x − 4)2 = 0
x=4
Number of pairs is 1 i.e., ( 4, 4 ).
4a 2 + 4 − 4a 2
≥0
(1 − a 2 ) 2
4
≥0
(1 − a 2 ) 2
⇒
(y − 4 ) 2 = 0
x 2 − 8 x + 16 = 0
4a 2
4
+
≥0
2 2
(1 − a )
(1 − a 2 )
⇒
[using, b − 4ac ≥ 0 ]
y 2 + 8y + 16 − 4y 2 + 16y − 64 ≥ 0
3y 2 − 24y + 48 ≤ 0
X
 2a 
 −1 

 − 4 ⋅1 ⋅ 
 ≥0
1 − a2
1 − a2
2
⇒
β1
The following cases arise:
Case I D ≥ 0
x 2 − xy + y 2 = 4 ( x + y − 4 )
⇒
α
0
−1
>0 ⇒
(1 − a 2 )
1 − a2 < 0
∴
Case III f (1 ) > 0
⇒
1
<0
1 − a2
1+
a ∈ ( − ∞, − 1 ) ∪ (1, ∞ )
2a
1
−
>0
(1 − a 2 ) (1 − a 2 )
[always true]
Chap 02 Theory of Equations
 − 1 + 5
x ∈ 0,

2


Case II If x < 0, i.e., −1 ≤ x < 0
x − (1 + x ) < 0
1 − a 2 + 2a − 1
a 2 − 2a
<0
>0 ⇒
2
(1 − a )
1 − a2
⇒
+
+
–1
–
0
∴
+
1
–
2
⇒
a (a − 2 )
>0
(a + 1 ) (a − 1 )
⇒
∴
Case IV 0 < x -coordinate of vertex < 1
2a
a
<1 ⇒ 0 < 2
0<−
<1
⇒
2 (1 − a 2 )
a −1
a
a
and 1 − 2
0<
>0
⇒
(a + 1 ) (a − 1 )
a −1
a
⇒
>0
(a + 1 ) (a − 1 )
⇒
0
–
⇒
+
–1
–
1– √5
2
–
a + 4b + 4c = 0
c
 b
31. 0 < a < b < c, α + β =  −  and αβ =
 a
a
For non-real complex roots,
b 2 − 4ac < 0
which is defined only when
1 − |x | ≥ 0
⇒
|x | ≤ 1
⇒
x ∈ [ − 1, 1 ]
Now, from Eq. (i), we get
x < 1 − |x|
Case I If x ≥ 0, i.e., 0 ≤ x ≤ 1
x − (1 − | x | ) < 0
⇒
x < (1 − x )
On squaring both sides, we get
x2 + x − 1 < 0
⇒
But
−1 − 5
−1 + 5
<x<
2
2
x≥0
⇒
(α − β ) 2 < 0
Q
0 <a <b <c
Roots
are
conjugate,
then
∴
| α | = | β|
c
But
αβ =
a
c
| αβ | =
>1
a
1+√5
2
1 − 5 
1 + 5 
a ∈
, 0 ∪ 
, ∞
 2

 2

Combining all cases, we get
a >2
29. We have,
x − 1 − |x| < 0
⇒
b 2 4c
−
<0
a2 a
(α + β ) 2 − 4 αβ < 0
⇒
1 − 5 
1 + 5 
and a ∈ ( − ∞, − 1 ) ∪ 
, 1 ∪ 
, ∞
 2

 2

∴
4a (c + b ) = − a
[Qb ≠ c ]
2
⇒
+
1
4a (c − b ) (c + b ) = a 2(b − c )
⇒
1
+
5


2a (b − c ) ⋅ 2 (c 2 − b 2 ) = a 2(b − c ) 2
⇒
a ∈ ( − 1, 0 ) ∪ (1, ∞ )

1 + 5 
1 − 5
a −
 a −

2  
2 

>0
(a + 1 ) (a − 1 )
and
[always true]
30. We have, (a ⋅ 2b − 2c ⋅ a ) (2c ⋅ c − b ⋅ 2b ) = (ba − ca ) 2
+
+
–1
x < 1+ x
x ∈ [ − 1, 0 )
Combining both cases, we get

−1 +
x ∈ − 1,
2

a ∈ ( − ∞, − 1 ) ∪ ( 0, 1 ) ∪ (2, ∞ )
–
183
⇒
⇒
c


Q a < c, ∴ > 1


a
| α| | β| > 1
| α | 2 > 1 or | α | > 1
32. Given equation is
…(i)
Ax 2 − | G | x − H = 0
…(i)
∴ Discriminant = ( − | G | ) − 4 A ( − H )
2
= G 2 + 4 AH
= G 2 + 4G 2
[QG 2 = AH ]
= 5G 2 > 0
∴ Roots of Eq. (i) are real and distinct.
a+b
2ab
> 0, G = ab > 0, H =
>0
Q A=
2
a+b
[Qa and b are two unequal positive integers]
Let α and β be the roots of Eq. (i). Then,
| G|
α+β=
>0
A
H
and
αβ = −
<0
A
D G 5
and
α −β =
=
>0
A
A
184
Textbook of Algebra
|G | + G 5
>0
2A
|G | − G 5
and
β=
<0
2A
Exactly one positive root and atleast one root which is
negative fraction.
33. It is clear from graph that the equation y = ax 2 + bx + c = 0
has two real and distinct roots. Therefore,
…(i)
b 2 − 4ac > 0
∴
α=
Q Parabola open downwards.
∴
a<0
andy = ax 2 + bx + c cuts-off Y -axis at, x = 0.
∴
y =c < 0
⇒
c<0
and x-coordinate of vertex > 0
b
b
<0
⇒
−
>0 ⇒
a
2a
⇒
b>0
It is clear that a and b are of opposite signs.
34. Let y = ax 2 + bx + c
[Qa < 0]
a>0
α
–2 2
β
The given equation will have four real roots, i.e. Eq. (i) has two
non-negative roots.
b
Then,
− ≥0
a
af ( 0 ) ≥ 0
and
[given]
b 2 − 4ac ≥ 0
b
⇒
≤0
a
ac ≥ 0
⇒
a > 0, b < 0, c > 0
or
a < 0, b > 0, c < 0
a
36. Let the roots be , a and ar , where a > 0, r > 1
r
∴ Product of the roots = 1
a
⋅ a ⋅ ar = 1
⇒
r
⇒
a3 = 1
∴
a =1
1
Now, roots are , 1 and r . Then,
r
1
+ 1 + r = −b
r
1
⇒
+ r = −b −1
r
1
Q
r + >2
r
⇒
or
Also,
Case II af ( − 2 ) < 0
⇒
a ( 4a − 2b + c ) < 0
⇒
4a − 2b + c < 0
Case III af (2 ) > 0
⇒
a ( 4a + 2b + c ) > 0
⇒
4a + 2 b + c > 0
Combining Case II and Case III, we get
4a + 2| b| + c < 0
Also, at x = 0,
⇒
b ∈ ( − ∞, − 3 )
1
1
⋅1 + 1 ⋅r + r ⋅ = c
r
r
1
+ r + 1 =c = −b
r
[from Eq. (i)]
[from Eq. (i)]
b +c = 0
1
Now, first root = < 1
r
[Q one root is smaller than one]
Second root = 1
Third root = r > 1
[Q one root is greater than one]
f ( x ) = ax 2 + bx + c
y <0 ⇒c<0
37. We have,
a, b , c ∈ R
1
On putting x = 0, 1, , we get
2
y <0
ax 2 + bx + c < 0
For x = 1,
a+b+c<0
…(i)
and for x = − 1,
a −b + c < 0
…(ii)
Combining Eqs. (i) and (ii), we get
and
a + | b| + c < 0
⇒
35. Put x 2 = y .
Then, the given equation can be written as
f (y ) = ay 2 + by + c = 0
b < −3
∴
Also, since for − 2 < x < 2,
⇒
…(i)
−b −1 >2
⇒
Consider the following cases:
Case I D > 0
⇒
b 2 − 4ac > 0
[one root is 1]
…(i)
and
⇒
and
|c | ≤ 1
|a + b + c | ≤ 1
1
1
a + b + c ≤1
4
2
−1 ≤ c ≤ 1,
−1 ≤a + b + c ≤1
− 4 ≤ a + 2b + 4c ≤ 4
− 4 ≤ 4a + 4b + 4c ≤ 4
− 4 ≤ − a − 2b − 4c ≤ 4
[Q a ≠ 0 ]
Chap 02 Theory of Equations
On adding, we get
− 8 ≤ 3a + 2b ≤ 8
Also,
− 8 ≤ a + 2b ≤ 8
∴
− 16 ≤ 2a ≤ 16
⇒
| a| ≤ 8
⇒
|b | ≤ 8
∴
38. Q x =
or
or
∴
41. Here, D ≤ 0
− 16 ≤ 2b ≤ 16
We get,
| a| + | b| + | c | ≤ 17
− 5 ± 25 + 1200 −5 ± 35 30 − 40
=
= ,
50
50
50 50
3 −4
cos α = ,
5 5
But
∴
∴
∴
∴
−1 < x < 0
4
cos α = − [ lies in II and III quadrants]
5
3
sin α =
[ lies in II quadrant]
5
3
sin α = −
[lies in III quadrant]
5
24
sin 2α = 2 ⋅ sin α ⋅ cos α = −
25
[ lies in II quadrant]
24
∴ sin 2α = 2 ⋅ sin α ⋅ cosα =
25
39. Qa + 2b + 4c = 0
2
∴
 1
 1
a   + b  + c = 0
 2
 2
1
It is clear that one root is .
2
Let other root be α. Then, α +
⇒
which depends upon a and b.
40. Q Cut-off Y -axis, put x = 0, i.e. f ( 0) = c
b
<0
2a
a < 0, c < 0, b < 0
abc < 0
b
a < 0, c > 0, −
>0
2a
Option (a) a < 0, c < 0, −
or
∴
Option (b)
a < 0, c > 0, b > 0
abc < 0
b
Option (c) a > 0, c > 0, −
>0
2a
or
∴
or
∴
a > 0, c > 0, b < 0
abc < 0
[lies in III quadrant]
and
f ( x ) ≥ 0, ∀ x ∈ R
∴
f (3 ) ≥ 0
⇒
9a + 3b + 6 ≥ 0
or
3a + b ≥ − 2
⇒ Minimum value of 3a + b is − 2.
and
f (6 ) ≥ 0
⇒
36a + 6b + 6 ≥ 0
⇒
6a + b ≥ − 1
⇒ Minimum value of 6a + b is −1.
42. Since, f ( x ) = x 3 + 3x 2 − 9x + λ = ( x − α ) 2( x − β)
∴ α is a double root.
∴ f ′( x ) = 0 has also one root α.
i.e. 3 x 2 + 6 x − 9 = 0 has one root α.
∴
x 2 + 2 x − 3 = 0 or ( x + 3 ) ( x − 1 ) = 0
has the root α which can either −3 or 1.
If α = 1, then f (1 ) = 0 gives λ − 5 = 0 ⇒ λ = 5.
If α = − 3, then f ( − 3 ) = 0 gives
− 27 + 27 + 27 + λ = 0
⇒
λ = − 27
43. We have, D = (b − c ) 2 − 4a (a − b − c ) > 0
⇒
b 2 + c 2 − 2bc − 4a 2 + 4ab + 4ac > 0
⇒
c 2 + ( 4a − 2b ) c − 4a 2 + 4ab + b 2 > 0, ∀c ∈ R
Since, c ∈ R, so we have
( 4a − 2b ) 2 − 4 ( − 4a 2 + 4ab + b 2 ) < 0
⇒
1
b
=−
2
a
1 b
α=− −
2 a
b
<0
2a
a < 0, c < 0, b < 0
abc < 0
Option (d) a < 0, c < 0, −
−1 ≤ − c ≤ 1, − 8 ≤ − a ≤ 8
Q
185
4a 2 − 4ab + b 2 + 4a 2 − 4ab − b 2 < 0
⇒
a (a − b ) < 0
If a > 0, then a − b < 0
i.e.
0 <a <b
or
b >a > 0
If a < 0, then a − b > 0
i.e.
0 >a >b
or
b <a < 0
44. We have,
x 3 − ax 2 + bx − 1 = 0
Then,
…(i)
α 2 + β 2 + γ 2 = (α + β + γ ) 2 − 2 (αβ + βγ + γα )
= a 2 − 2b
α 2 β 2 + β 2 γ 2 + γ 2 α 2 = (αβ + βγ + γα ) 2
− 2 αβγ (α + β + γ ) = b 2 − 2a
and
α 2 β2 γ 2 = 1
Therefore, the equation whose roots are α 2, β 2 and γ 2, is
x 3 − (a 2 − 2b ) x 2 + (b 2 − 2a ) x − 1 = 0
Since, Eqs. (i) and (ii) are indentical, therefore
a 2 − 2b = a and b 2 − 2a = b
…(ii)
186
Textbook of Algebra
⇒
Eliminating b, we have
(a 2 − a ) 2
a2 − a
− 2a =
4
2
a {a (a − 1 ) 2 − 8 − 2 (a − 1 )} = 0
⇒
⇒
a (a − 2a − a − 6 ) = 0
⇒
a (a − 3 ) (a 2 + a + 2 ) = 0
⇒
a = 0 or a = 3 or a 2 + a + 2 = 0
⇒
or
b = 0 or b = 3
b2 + b + 2 = 0
3
∴
∴
2
∴
Now,
⇒
46. Least value of G100 + L100 is 2.
∴
a =b = 0
or
a =b =3
or a and b are roots of x 2 + x + 2 = 0.
47. The quadratic equation having roots G and L, is
⇒
45. Here, D > 0
⇒
a > 0
α
–2
2
48. We have,
β
X
b 2 > 4ac
f (0) < 0
c<0
f (1 ) < 0
⇒
a+b+c<0
f (− 1) < 0
⇒
a −b + c < 0
f (2 ) < 0
⇒
4a + 2b + c < 0
f (− 2) < 0
⇒
4a − 2b + c < 0
From Eqs. (i) and (ii), we get
c < 0, b2 − 4ac > 0
and
⇒
From Eqs. (iii) and (iv), we get
a − |b | + c < 0
and from Eqs. (v) and (vi), we get
4a − 2 | b | + c < 0
Solutions (Q. Nos. 46 to 48)
Let
⇒
2x 2 − 3x + 2
y = 2
2x + 3x + 2
2 x 2y + 3 xy + 2y = 2 x 2 − 3 x + 2
⇒ 2 (y − 1 ) x 2 + 3 (y + 1 ) x + 2 (y − 1 ) = 0
As x ∈ R
∴
D≥0
2
⇒
9 (y + 1 ) − 4 ⋅ 2 (y − 1 ) ⋅ 2 (y − 1 ) ≥ 0
⇒
9 (y + 1 ) 2 − 16 (y − 1 ) 2 ≥ 0
⇒
(3y + 3 ) 2 − ( 4y − 4 ) 2 ≥ 0
⇒
(7y − 1 ) (7 − y ) ≥ 0
x 2 − (G + L ) x + GL = 0
50
x2 − x + 1 = 0
7
7 x 2 − 50 x + 7 = 0
L2 < λ < G 2
2
⇒
⇒
b 2 − 4ac > 0
or
(7y − 1 ) (y − 7 ) ≤ 0
1
≤y ≤7
7
1
G = 7 and L =
7
GL = 1
G100 + L100
G100 + L100
≥ (GL )100 ⇒
≥1
2
2
G100 + L100 ≥ 2
…(i)
⇒
 1
2
  < λ <7
 7
1
< λ < 49
49
λ = 1, 2, 3,…, 48 as λ ∈ N
∴ Sum of all values of λ = 1 + 2 + 3 +…+ 48 =
…(ii)
48 × 49
= 1176
2
Solutions (Q. Nos. 49 to 51)
…(iii)
…(iv)
…(v)
…(vi)
Let roots be α , β, γ, δ > 0.
∴
α + β + γ + δ = 12
(α + β ) ( γ + δ ) + αβ + γδ = c
αβ ( γ + δ ) + γδ (α + β ) = − d
α β γδ = 81
α+β+γ+δ
AM =
Q
=3
4
and
GM = (α β γ δ )1 /4 = (81 )1 /4 = 3
AM = GM
∴
⇒
α =β = γ =δ =3
49. c = (α + β) ( γ + δ ) + αβ + γδ
= (3 + 3 ) (3 + 3 ) + 3 ⋅ 3 + 3 ⋅ 3 = 36 + 18 = 54
50. Qαβ ( γ + δ ) + γδ (α + β) = − d
∴
d = − {3 ⋅ 3 ⋅ (3 + 3 ) + 3 ⋅ 3 ⋅ (3 + 3 )} = − 108
d
( − 108 )
51. Required root = − = −
=1
2c
2 × 54
Solutions (Q. Nos. 52 to 54)
Given that, AC = 4 2 units
AC
∴
AB = BC =
= 4 units
2
and
OB = ( BC ) 2 − (OC ) 2
= ( 4 ) 2 − (2 2 ) 2
= 2 2 units
AC 

Q OC =

2 
Chap 02 Theory of Equations
∴ Vertices are A ≡ ( − 2 2, 0 ),
Solutions (Q. Nos. 58 to 60)
B ≡ ( 0, −2 2 )
and
C ≡ (2 2, 0)
52. Since, y = f ( x ) = ax 2 + bx + c passes through A, B and C, then
0 = 8a − 2 2b + c − 2 2 = c
0 = 8a + 2 2b + c
1
We get, b = 0, a =
and c = − 2 2
2 2
x2
∴
y = f (x ) =
−2 2
2 2
Let f ( x ) = ax 2 − bx + c has two distinct roots α and β. Then,
f ( x ) = a ( x − α ) ( x − β ). Since, f ( 0 ) and f (1 ) are of same sign.
Therefore,
c (a − b + c ) > 0
⇒
c (a − b + c ) ≥ 1
2
∴
a αβ (1 − α ) (1 − β ) ≥ 1
and
2
But
∴
⇒
x2
53. Minimum value of y =
− 2 2 is at x = 0.
2 2
∴
(y ) min = − 2 2
⇒
Also,
⇒
54. f ( x ) = 0
⇒
Given,
or
⇒
(α − β ) = (α + k ) − ( β + k )
b 2 − 4c
b 2 − 4c1
= 1
1
1
2
2
b − 4c = b1 − 4c1
…(i)
1 (b 2 − 4c )
1
Given, least value of f ( x ) = − −
=−
4 ×1
4
4
⇒
∴
b 2 − 4c = 1
b 2 − 4c = 1 = b12 − 4c1
[from Eq. (i)] …(ii)
7
Also, given least value of g( x ) occurs at x = .
2
b1
7
−
=
∴
2 ×1 2
∴
55. b 1 = − 7
b1 = − 7
56. Least value of g ( x ) = −
b12 − 4c1
1
=−
4 ×1
4
57. Q g( x ) = 0
∴
x 2 + b1x + c1 = 0
− b1 ± b12 − 4c1
2
7±1
=
= 3, 4
2
∴ Roots of g( x ) = 0 are 3, 4.
⇒
x=
1 1
1

−  − α ≤

4 2
4
a 2 αβ (1 − α ) (1 − β ) <
a2
16
a2
>1 ⇒ a > 4
16
a ≥ 5 as a ∈ I
b 2 − 4ac ≥ 0
[Q α ≠ β]
b 2 ≥ 4ac ≥ 20
Solutions. (Q. Nos. 61 to 63)
Solutions. (Q. Nos. 55 to 57)
⇒
α (1 − α ) =
⇒
b ≥5
Next, a ≥ 5, b ≥ 5, we get c ≥ 1
∴
abc ≥ 25
∴
log 5 abc ≥ log 5 25 = 2
58. Least value of a is 5.
59. Least value of b is 5.
60. Least value of logb abc is 2.
x2
−2 2 = 0 ⇒ x = ± 2 2
2 2
λ
−2 2 < <2 2
2
−4 2<λ<4 2
∴Initial values of λ are
−5, − 4, − 3, − 2, − 1, 0, 1, 2, 3, 4, 5.
∴Number of integral values is 11.
We have,
187
[from Eq. (ii)]
Let α , β and γ be the roots of 2 x 3 + ax 2 + bx + 4 = 0.
a
α+β+γ=−
∴
2
b
αβ + βγ + γα = and αβγ = − 2
2
61. Q AM ≥ GM
(− α ) + (− β) + (− γ )
≥ {( − α ) ( − β ) ( − γ )}1 /3
∴
3
a
2 ≥ (2 )1/ 3
⇒
3
∴
a ≥ 6 (2 )1/ 3
or
…(i)
a 3 ≥ 432
Hence, minimum value of a 3 is 432.
62. Q AM ≥ GM
(− α ) (− β) + (−β) (− γ ) + (− γ ) (− α )
∴
3
≥ {( − α ) ( − β ) ( − β ) ( − γ ) ( − γ ) ( − α )}1 / 3
b/ 2
≥ ( 4 )1/ 3
⇒
3
…(ii)
⇒
b ≥ 6 ( 4 )1/ 3
or
b 3 ≥ 864
Hence, minimum value of b 3 is 864.
63. From Eqs. (i) and (ii), we get
ab ≥ 6 (2 )1 / 3 ⋅ 6( 4 )1 / 3
188
Textbook of Algebra
⇒
Q
∴
or
ab ≥ 36 × 2
a+b
a+b
≥ ab ≥ 6 2 ⇒
≥6 2
2
2
a + b ≥ 12 2
68. We have,
(5 + 2 ) x 2 − ( 4 + 5 ) x + 8 + 2 5 = 0
∴Sum of the roots =
(a + b ) 3 ≥ 3456 2
Hence, minimum value of (a + b ) 3 is 3456 2.
and product of the roots =
Solutions (Q. Nos. 64 to 66)
α+β+γ+δ=−A
(α + β ) ( γ + δ ) + αβ + γδ = B
αβ ( γ + δ ) + γδ (α + β ) = − C
and
αβγδ = D
C αβ ( γ + δ ) + γδ (α + β )
64. Q
=
A
α+β+γ+δ
k ( γ + δ ) + k (α + β )
=
α+β+γ+δ
∴
…(i)
…(ii)
…(iii)
…(iv)
=
…(v)
[Qαβ = γδ = k]
66. From Eq. (iv), we get
⇒
αβγδ = D
k ⋅k = D
[Qαβ = γδ = k]
2
⇒
∴
C
  =D
 A
x − 3x = 0
2
⇒
x (x − 3) = 0
⇒
x = 0, 3
Here, 0 is not possible.
∴
x =3
Case II If x < 2, then
(x − 2)2 − x + 2 − 2 = 0
⇒
y − 4y − 60 = 0
⇒
(y − 10 ) (y + 6 ) = 0
∴
y = 10, − 6
⇒
y = 10, y ≠ − 6
Now,
x − ax + 30 = 10
⇒
x 2 − ax + 20 = 0
Given,
αβ = λ = 20
α+β
≥ αβ = 20
2
α + β ≥ 2 20
µ=4 5
µ = 4 5 = 8.9 ⇒ (µ ) = 9
i.e.,


1
x
6


70. Q N r =  x +  −  x 6 +
1
 −2
x6
2
3
6

1 
1
1
1  


=  x +  −  x 3 + 3  =   x +  +  x 3 + 3  





x
x

x 
x  
3

1
1 
  x +  −  x 3 + 3 

x
x 

1
 
= Dr ⋅ 3  x +  
 
x
∴
1
Nr

= 3 x +  ≥ 6
r


x
D
Nr
is 6.
Dr
a + b = 2c
Hence, minimum value of
71.
⇒
(y − 1 ) (y + 2 ) = 0 ⇒ y = 1, − 2
But − 2 is not possible.
Hence,
| x − 2| = 1 ⇒ x = 1, 3
∴ Sum of the roots = 1 + 3 = 4
[Qy > 0]
2
∴ Minimum value of µ is 4 5.
x 2 − 5x + 4 = 0
y2 + y −2 = 0
…(i)
2
or
⇒
(x − 1) (x − 4) = 0
⇒
x = 1, 4
Here, 4 is not possible.
∴
x =1
∴The sum of roots = 1 + 3 = 4
Aliter
Let | x − 2 | = y .
Then, we get
y = 2 y + 15
⇒
C 2 = A 2D
There are two cases:
Case I If x ≥ 2, then ( x − 2 ) 2 + x − 2 − 2 = 0
⇒
∴
∴
[from Eq. (v)]
67. The given equation is | x − 2 | 2 + | x − 2 | − 2 = 0.
⇒
2 × Product of the roots 2 × (8 + 2 5 )
=
=4
Sum of the roots
(4 + 5 )
69. Let x 2 − ax + 30 = y
[Qαβ = γδ =k]
=k
8+2 5
5+ 2
∴The harmonic mean of the roots
65. From Eq. (ii), we get
(α + β ) ( γ + δ ) = B − (αβ + γδ ) = B − 2k
4+ 5
5+ 2
ab = − 5d
c + d = 2a
cd = − 5b …(iv)
From Eqs. (i) and (iii), we get
a + b + c + d = 2 (a + c )
∴
a + c =b +d
…(i)
…(ii)
…(iii)
…(v)
Chap 02 Theory of Equations
…(vi)
⇒
∴
a 2 − 2ac − 5d = 0
…(vii)
⇒
c 2 − 2ac − 5b = 0
…(viii)
∴
From Eqs. (vii) and (viii), we get
a 2 − c 2 − 5 (d − b ) = 0
⇒
⇒
⇒
(a + c ) (a − c ) + 5 (b − d ) = 0
(a + c ) (a − c ) + 15 (c − a ) = 0
(a − c ) (a + c − 15 ) = 0
∴
a + c = 15, a − c ≠ 0
∴
[from Eq. (vi)]
75. Q 4x − 16x + c = 0
∴
9 (y + 1 ) 2 − 4c (y − 1 ) 2 ≥ 0
c
4
Then, the following cases arises:
1
( 2 cy − 2 c ) 2 − (3y + 3 ) 2 ≤ 0
⇒ {( 2 c + 3 ) y − (2 c − 3 )} {(2 c − 3 )y − (2 c + 3 )} ≤ 0
Case I
⇒
∴
Case II
2 c −3
2 c +3
≤y ≤
2 c +3
2 c −3
⇒
2 c +3
=7
2 c −3
⇒
But given,
⇒
2 c + 3 = 14 c − 21
or
12 c = 24 or
or
∴
73. We have,
c =2
c=4
x − (x − 1) + (x − 2)2 = 5
2
2
| x | − | x − 1| + | x − 2| = 5
Case I If x < 0, then
− x + (x − 1) − (x − 2) = 5
x =1 − 5
Case II If 0 ≤ x < 1, then
x + (x − 1) − (x − 2) = 5
x = 5 − 1, which is not possible.
Case III If 1 ≤ x < 2, then
x − (x − 1) − (x − 2) = 5
⇒
c
=0
4
f (x ) = x 2 − 4x +
x 2(y − 1 ) + 3 x (y + 1 ) + c (y − 1 ) = 0
x ∈R
⇒
x 2 − 4x +
x 2 − 3x + c
x 2 + 3x + c
Q
⇒
| α − β| = 5
2
Let
a + b + c + d = a + c + b + d = 15 + 15 = 30
9 + 16 = 5
2
⇒
⇒ Sum of digits of a + b + c + d = 3 + 0 = 3
⇒
α + β = i , and αβ = − (1 + i )
α − β = (α + β ) 2 − 4 αβ = i 2 + 4 (1 + i ) = (3 + 4i )
| α − β| =
From Eq. (v), we get b + d = 15
72. Q y =
x 2 − ix − (1 + i ) = 0
∴
And c is a root of
∴
(1 + i ) x 2 + (1 − i ) x − 2i = 0
2i
(1 − i )
x2 +
x−
=0
(1 + i )
(1 + i )
74. Q
From Eqs. (i) and (iii), we get
b − d = 3 (c − a )
Also, a is a root of x 2 − 2cx − 5d = 0
189
x = 3 − 5, which is not possible.
Case IV If x > 2, then
x − (x − 1) + (x − 2) = 5
⇒
Hence, number of solutions is 2.
x =1 + 5
∴
Case III
⇒
⇒
∴
Case IV
⇒
⇒
⇒
α
2
β
3
D>0
16 − c > 0
c < 16
f (1 ) > 0
c
1−4+ >0
4
c
>3
4
c > 12
f (2 ) < 0
c
4 −8 + < 0
4
c
<4
4
c < 16
f (3 ) > 0
c
9 − 12 + > 0
4
c
>3
4
c > 12
Combining all cases, we get
12 < c < 16
Thus, integral values of c are 13, 14 and 15.
Hence, number of integral values of c is 3.
76. We have,
r +s +t=0
1001
rs + st + tr =
8
2008
and
rst = −
= − 251
8
…(i)
…(ii)
…(iii)
190
Textbook of Algebra
Now, (r + s ) 3 + (s + t ) 3 + (t + r ) 3 = ( − t ) 3 + ( − r ) 3 + ( − s ) 3
= − (t 3 + r 3 + s 3 ) = − 3 rst
[Qr + s + t = 0]
[Qr + s + t = 0]
⇒
∴ a, b and c are in HP.
From Eqs. (i) and (ii), we get
a =b =c
∴ a,b and c are in AP, GP and HP.
(B)Q
x 3 − 3x 2 + 3x − 1 = 0
⇒ x 2 (y − 1 ) + 2 (y + 1 ) x + 4 (y − 1 ) = 0
As x ∈ R , we get
D≥0
4 (y + 1 ) 2 − 16 (y − 1 ) 2 ≥ 0
⇒
3y 2 − 10y + 3 ≤ 0
⇒
(y − 3 ) (3y − 1 ) ≤ 0
1
≤y ≤3
⇒
3
2x 2 + 4x + 1
(B) We have, y = 2
x + 4x + 2
As x ∈ R , we get
⇒
4 (y − 1 ) − (y − 2 ) (2y − 1 ) ≥ 0
2y 2 − 3y + 2 ≥ 0
3
y2 − y + 1 ≥ 0
2
⇒
⇒
y ∈R
x − 3x + 4
y =
x −3
D≤0
(a + c ) 2 + 4b 2 − 4b (a + c ) ≤ 0
⇒
(a + c − 2b ) 2 ≤ 0
or
(a + c − 2b ) 2 = 0
⇒
ax 2 + 3 x − 4
3x − 4x 2 + a
x 2 (a + 4y ) + 3 (1 − y ) x − (ay + 4 ) = 0
D≥0
9 (1 − y ) 2 + 4 (a + 4y ) (ay + 4 ) ≥ 0
⇒ ( 9 + 16a ) y 2 + ( 4a 2 + 46 )y + (9 + 16a ) ≥ 0, ∀ y ∈ R
⇒ If 9 + 16a > 0, then D ≤ 0
Now,
D≤0
⇒
( 4a 2 + 46 ) 2 − 4 ( 9 + 16a ) 2 ≤ 0
x 2 − (3 + y ) x + 3y + 4 = 0
As x ∈ R , we get
D ≥ 0 ⇒ (3 + y ) 2 − 4 (3y + 4 ) ≥ 0
⇒
∴
⇒
As x ∈R , we get
2
(C) We have,
bx + ( (a + c ) 2 + 4b 2 ) x + (a + c ) ≥ 0
⇒
2
∴
(C) Given,
(A) We have, y =
3
7

≥0
y −  +

4
16
⇒
a+b+c=0
2
∴
a + c = 2b
Hence a, b and c are in AP.
79. A → (q,r,s,t); B → (q,r); C → (p,q)
2
⇒
(x − 1)3 = 0
⇒
⇒ x 2(y − 2 ) + 4(y − 1 ) x + 2y − 1 = 0
⇒
…(iii)
∴
x = 1, 1, 1
⇒ Common root, x = 1
∴
a (1 ) 2 + b (1 ) + c = 0
⇒
D≥0
16 (y − 1 ) 2 − 4 (y − 2 ) (2y − 1 ) ≥ 0
a (b − c ) = c (a − b )
2ac
b=
a+c
or
x 2 − 2x + 4
x 2 + 2x + 4
⇒
…(ii)
Given, roots [Eq. (ii)] are equal.
c (a − b )
1 ×1 =
∴
a (b − c )
= − 3 ( − 251 ) = 753
Now,
99 λ = (r + s ) 3 + (s + t ) 3 + (t + r ) 3 = 753
753
λ=
= 7.6
∴
99
∴
[λ ] = 7
77. A → (r,s); B → (p, q, r,s, t); C → (p, q, t)
(A) We have, y =
Q
a (b − c ) + b (c − a ) + c (a − b ) = 0
∴ x = 1 is a root of
a (b − c ) x 2 + b (c − a ) x + c (a − b ) = 0
⇒
y 2 − 6y − 7 ≥ 0 ⇒ (y + 1 ) (y − 7 ) ≥ 0
4 [( 2a 2 + 23 ) 2 − ( 9 + 16a ) 2 ] ≤ 0
⇒ [(2 a 2 + 23 ) + ( 9 + 16a )] [( 2a 2 + 23 ) − ( 9 + 16a )] ≤ 0
⇒
y ∈ ( − ∞, − 1 ] ∪ [7, ∞ )
78. A → (q,r,s); B → (p); C → (q)
⇒
( 2a 2 + 16a + 32 ) ( 2a 2 − 16a + 14 ) ≤ 0
(A)Q(d + a − b ) 2 + (d + b − c ) 2 = 0
⇒
4 (a + 4 ) 2 (a 2 − 8a + 7 ) ≤ 0
which is possible only when
d + a − b = 0, d + b − c = 0
⇒
b −a =c −b
⇒
2b = a + c
∴ a, b and c are in AP.
⇒
a 2 − 8a + 7 ≤ 0
⇒
⇒
∴
⇒
(a − 1 ) (a − 7 ) ≤ 0
1 ≤a ≤7
9 + 16a > 0 and 1 ≤ a ≤ 7
1 ≤a ≤7
…(i)
Chap 02 Theory of Equations
y =
(B) We have,
⇒
f (1 ) > 0
⇒
1 −6 + 9 + λ > 0
⇒
λ>−4
and
f (3 ) < 0
⇒
27 − 54 + 27 + λ < 0
⇒
λ<0
From Eqs. (i), (ii) and (iii), we get
−4<λ<0
⇒
−3 < λ + 1 < 1
∴
[ λ + 1 ] = − 3, − 2, − 1, 0
∴
|[ λ + 1 ]| = 3, 2, 1, 0
ax 2 + x − 2
a + x − 2x 2
x 2 (a + 2y ) + x (1 − y ) − (2 + ay ) = 0
As x ∈R , we get
⇒
D≥0
( 1 − y ) 2 + 4 ( 2 + ay ) (a + 2y ) ≥ 0
⇒
( 1 + 8a ) y 2 + ( 4a 2 + 14 ) y + (1 + 8a ) ≥ 0
⇒ If 1 + 8a > 0, then D ≤ 0
⇒
( 4a 2 + 14 ) 2 − 4 (1 + 8a ) 2 ≤ 0
⇒
4 [(2a 2 + 7 ) 2 − (1 + 8a ) 2 ] ≤ 0
⇒ [(2a 2 + 7 ) + (1 + 8a )] [(2a 2 + 7 ) − (1 + 8a )] ≤ 0
(B)Q
⇒
(2a 2 + 8a + 8 ) (2a 2 − 8a + 6 ) ≤ 0
Given,
⇒
4 (a + 2 ) 2 (a 2 − 4a + 3 ) ≤ 0
⇒
a 2 − 4a + 3 ≤ 0
⇒
⇒
x 2(y − 1 ) + 2 (2y − 1 ) x + a (3y − 1 ) = 0
⇒
D≥0
4 (2y − 1 ) 2 − 4 (y − 1 ) a (3y − 1 ) ≥ 0
⇒
( 4 − 3a ) y 2 − ( 4 − 4a )y + (1 − a ) ≥ 0
⇒ If 4 − 3a > 0, then D ≤ 0
⇒
( 4 − 4a ) 2 − 4 ( 4 − 3a ) (1 − a ) ≤ 0
⇒
4 (2 − 2a ) 2 − 4 ( 4 − 3a ) (1 − a ) ≤ 0
⇒
4 + 4a − 8a − ( 4 − 7a + 3a ) ≤ 0
⇒
a2 − a ≤ 0
2
Also,
x = 1, 3
f ′′ ( x ) = 6 x − 12
f ′′(1 ) < 0 and f ′′ (3 ) > 0
3
1
f (0) < 0 ⇒ λ < 0
4x 2 − (λ − 3) x + 1 > 0
x 2 + (λ + 2)x + 4 > 0
∴
(λ − 3)2 − 4 ⋅ 4 ⋅ 1 < 0
and
(λ + 2)2 − 4 ⋅ 1 ⋅ 4 < 0
⇒
(λ − 3)2 − 42 < 0
and
(λ + 2)2 − 42 < 0
⇒
∴
Now,
f ′ ( x ) = 3 x 2 − 12 x + 9 = 0
λ
x 2 − λx − 2
<2
x2 + x + 1
− 4 < λ −3 < 4
−4<λ+2<4
− 1 < λ <7
−6 < λ <2
−1 < λ <2
[ λ ] = − 1, 0, 1
|[ λ ]| = 0, 1
Also, x = 1 satisfies
x2 + λ x + 1 = 0
(A) Let y = f ( x ) = x 3 − 6 x 2 + 9 x + λ
0
−3 <
(C)Q
(b − c ) + (c − a ) + (a − b ) = 0
∴ x = 1 is a root of
(b − c ) x 2 + (c − a ) x + (a − b ) = 0
2
⇒
a (a − 1 ) ≤ 0
⇒
0 ≤a ≤1
80. A → (p,q,r,s);B → (p,q); C → (s)
∴
…(iii)
x 2 + x + 1 > 0, ∀ x ∈ R
and
⇒
and
or
and
We get,
∴
⇒
As x ∈ R , we get
…(ii)
− 3 x 2 − 3 x − 3 < x 2 − λx − 2 < 2 x 2 + 2 x + 2
⇒
⇒
(a − 1 ) (a − 3 ) ≤ 0
⇒
1 ≤a ≤3
Thus, 1 + 8a > 0 and 1 ≤ a ≤ 3
⇒
1 ≤a ≤3
x 2 + 2x + a
(C) We have,
y = 2
x + 4 x + 3a
191
…(i)
1+λ+1=0
λ = −2
λ −1 = −3
[ λ − 1] = − 3
⇒
|[ λ − 1 ]| = 3
81. If quadratic equation ax 2 + bx + c = 0 is satisfied by more than
two values of x, then it must be an identity.
Therefore, a = b = c = 0
∴ Statement-2 is true.
But in Statement-1,
4 p − 3 = 4q − 3 = r = 0
3
Then,
p =q = ,r = 0
4
which is false.
Since, at one value of p or q or r, all coefficients at a time ≠ 0.
∴ Statement-1 is false.
192
Textbook of Algebra
82. We have,
x 2 + (2m + 1 ) x + (2n + 1 ) = 0
…(i)
m, n ∈ I
D = b 2 − 4ac
∴
= ( 2m + 1 ) 2 − 4 ( 2n + 1 )
is never be a perfect square.
Therefore, the roots of Eq. (i) can never be integers. Hence, the
roots of Eq. (i) cannot have any rational root as a = 1, b, c ∈ I .
Hence, both statements are true and Statements –2 is a correct
explanation of Statement-1.
83. Let α be one root of equation ax 2 + 3x + 5 = 0. Therefore,
1 5
α⋅ =
α a
5
⇒
1=
a
⇒
a =5
Hence, both the statements are true and Statement-2 is the
correct explanation of Statement-1.
84. Let roots of
…(i)
Ax 3 + Bx 2 + Cx + D = 0
are α − β, α , α + β (in AP).
B
Then,
(α − β) + α + (α + β) = −
A
B
, which is a root of Eq. (i).
⇒
α=−
3A
Then, Aα 3 + Bα 2 + Cα + D = 0
3
2
 B
 B
 B
⇒ A −
 +D=0
 + C −
 + B −
 3A 
 3A 
 3A 
−
⇒
⇒
B3
B3
BC
+
−
+D=0
2
27 A
9A 2 3A
4
4 

,2 +
z ∈ 2−

3
3 
Since, Eqs. (i) and (ii) remains same, if x, y , z interchange their
positions.
Hence, both statements are true and Statement-2 is a correct
explanation of Statement-1.
86. Let y = ax 3 + bx + c
dy
∴
= 3ax 2 + b
dx
dy
For maximum or minimum
= 0, we get
dx
b
x=± −
3a
dy
Case I If a > 0, b > 0, then
>0
dx
In this case, function is increasing, so it has exactly one root
dy
Case II If a < 0, b < 0, then
<0
dx
In this case, function is decreasing, so it has exactly one root.
and
Case III a > 0, b < 0 or a < 0, b > 0, then y = ax 3 + bx + c is
maximum at one point and minimum at other point.
Hence, all roots can never be non-negative.
∴Statement-1 is false. But
Coefficient of x 2
Sum of roots = −
=0
Coefficient of x 3
i.e., Statement-2 is true.
87. Statement-2 is obviously true.
But
y = ax 2 + bx + c
b
c

y = a x2 + x + 

a
a
2 B 3 − 9 ABC + 27 A 2D = 0
2

b
D 
= a  x +  − 2 


2a
4a 

Now, comparing with 2 B 3 + k1ABC + k2 A 2D = 0, we get
k1 = − 9, k2 = 27
k2 − k1 = 27 − ( − 9 ) = 36 = 6 2
∴
2
Hence, both statements are true and Statement-2 is a correct
explanation of Statement-1.
85. Q x, y , z ∈ R
x + y + z =6
and
xy + yz + zx = 8
⇒
xy + ( x + y ) {6 − ( x + y )} = 8
⇒ xy + 6 x + 6y − ( x 2 + 2 xy + y 2 ) = 8
or
∴
…(i)
…(ii)
[from Eq. (i)]
y 2 + (x − 6) y + x 2 − 6x + 8 = 0
( x − 6 ) 2 − 4 ⋅ 1 ⋅ ( x 2 − 6 x + 8 ) ≥ 0, ∀ y ∈ R
⇒
or
or
Similarly,
[where, D = b 2 − 4ac]
− 3 x 2 + 12 x + 4 ≥ 0 or 3 x 2 − 12 x − 4 ≤ 0
4
4
≤ x ≤2 +
2−
3
3
4
4 

x∈ 2−
,2 +

3
3 
4 
4

,2 +
y ∈ 2−

3 
3
b
1 
D

 x +  = y + 



a
2a
4a 
b
D
Let
x+
= X and y +
= Y.
2a
4a
1
∴
X2 = Y
a
b
Equation of axis, X = 0 i.e. x +
=0
2a
or
2ax + b = 0
2
Hence, y = ax + bx + c is symmetric about the line
2ax + b = 0.
∴ Both statements are true and Statement-2 is a correct
explanation of Statement-1.
88. Q(1 + m ) x 2 − 2 (1 + 3m ) x + (1 + 8m ) = 0
⇒
∴ D = 4 (1 + 3m ) 2 − 4 (1 + m ) (1 + 8m ) = 4m (m − 3 )
(i) Both roots are imaginary.
∴
D<0
⇒
4m (m − 3 ) < 0
Chap 02 Theory of Equations
⇒
0 <m <3
or
m ∈( 0, 3 )
(ii) Both roots are equal.
∴
D=0
⇒
4m (m − 3 ) = 0
⇒
m = 0, 3
(iii) Both roots are real and distinct.
∴
D>0
⇒
4m (m − 3 ) > 0
⇒
m < 0 or m > 3
∴
m ∈ ( − ∞, 0 ) ∪ (3, ∞ )
(iv) Both roots are positive.
Case I Sum of the roots > 0
2 (1 + 3m )
>0
⇒
(1 + m )
⇒
 1 
m ∈ ( − ∞, − 1 ) ∪  − , ∞
 3 
Case II Product of the roots > 0
(1 + 8m )
⇒
>0
(1 + m )
 1 
m ∈ ( − ∞, − 1 ) ∪  − , ∞
 8 
Case III
D≥0
⇒
4m (m − 3 ) ≥ 0
m ∈ ( − ∞, 0 ] ∪ [ 3, ∞ )
Combining all Cases, we get
m ∈ ( − ∞, − 1 ) ∪ [ 3, ∞ )
(v) Both roots are negative.
Consider the following cases:
2 (1 + 3m )
Case I Sum of the roots < 0 ⇒
<0
(1 + m )
⇒
Combining all cases, we get
1

m ∈  − 1, − 

8
(vii) Roots are equal in magnitude but opposite in sign, then
Consider the following cases:
Case I Sum of the roots = 0
2 (1 + 3m )
=0
⇒
(1 + m )
1
m = − ,m ≠1
⇒
3
Case II D > 0 ⇒ 4m(m − 3 ) > 0
⇒
m ∈ ( − ∞, 0 ) ∪ ( 3, ∞ )
Combining all cases, we get
1
m=−
3
(viii) Atleast one root is positive, then either one root is positive
or both roots are positive.
i.e.
(d ) ∪ ( f )
1

or
m ∈ ( − ∞, − 1 ) ∪  − 1, −  ∪ [3, ∞ )

8
(ix) Atleast one root is negative, then either one root is
negative or both roots are negative.
1

i.e.
(e ) ∪ ( f ) or m ∈  − 1, − 

8
(x) Let roots are 2α are 3α. Then,
Consider the following cases:
Case I Sum of the roots = 2 α + 3 α =
⇒
1

m ∈  − 1, − 

3
⇒
(1 + 8m )
>0
(1 + m )
 1 
m ∈ ( − ∞, 1 ) ∪  − , ∞
 8 
Case III D ≥ 0
4m (m − 3 ) ≥ 0 ⇒ m ∈ ( − ∞, 0 ] ∪ [3, ∞ )
Combining all cases, we get
m ∈φ
(vi) Roots are opposite in sign, then
Case I Consider the following cases:
Product of the roots < 0
(1 + 8m )
⇒
<0
(1 + m )
1

m ∈  − 1, − 

8
Case II D > 0 ⇒ 4m (m − 3 ) > 0
⇒
m ∈ ( − ∞, 0 ) ∪ (3, ∞ )
α=
2 (1 + 3m )
(1 + m )
2 (1 + 3m )
5 (1 + m )
Case II Product of the roots = 2 α ⋅ 3 α =
Case II Product of the roots > 0 ⇒
⇒
193
6α 2 =
(1 + 8m )
(1 + m )
(1 + 8m )
(1 + m )
From Eqs. (i) and (ii), we get
2
2 (1 + 3m ) 
(1 + 8m )
6
 =
5
1
(
+
)
(1 + m )
m


⇒
24 (1 + 3m ) 2 = 25 (1 + 8m ) (1 + m )
⇒ 24 ( 9m 2 + 6m + 1 ) = 25 ( 8m 2 + 9m + 1 )
16m 2 − 81m − 1 = 0
or
⇒
81 ± ( − 81 ) 2 + 64
32
81 ± 6625
m=
32
m=
89. Q2x 2 − 2 (2m + 1) x + m (m + 1) = 0
∴
D = [ − 2 (2m + 1 )] − 8m (m + 1 )
2
= 4 {(2m + 1 ) 2 − 2m (m + 1 )}
= 4 (2m 2 + 2m + 1 )
[Q m ∈ R ]
[ D = b 2 − 4ac ]
194
Textbook of Algebra
2

1
1
1

= 8  m 2 + m +  = 8  m +  +  > 0



2
2
4

D > 0, ∀ m ∈ R
or
…(i)
1
b 2 (2m + 1 ) 
=
= m + 
x -coordinate of vertex = −

2a
4
2
…(ii)
Combining all cases, we get
 7 + 33 
m ∈
, ∞
2


(iii) Both roots lie in the interval (2, 3).
Consider the following cases:
and let
1
f ( x ) = x 2 − (2m + 1 ) x + m (m + 1 )
2
(i) Both roots are smaller than 2.
…(iii)
f(3)
f(2)
α
2
α
β
2
∴
[from Eq. (i)]
[from Eq. (ii)]
Combining all cases, we get

7 − 33 
m ∈  − ∞,

2 

(ii) Both roots are greater than 2.
Consider the following cases:
⇒
or
1
9 − 3 (2m + 1 ) + m (m + 1 ) > 0
2
m 2 − 11m + 12 > 0

 11 + 73 
11 − 73 
m ∈  − ∞,
, ∞
 ∪
2
2




Combining all cases, we get
m ∈φ
(iv) Exactly one root lie in the interval (2,3) .
Consider the following cases:
Case I D > 0
∴
m ∈R
2
β
Case I D ≥ 0
∴
m ∈R
Case II x -coordinate of vertex > 2
1
m + >2
⇒
2
3
∴
m>
2
Case III f (2 ) > 0


7 − 33 
33 
m ∈  − ∞,
, ∞
 ∪ 7 +
2 
2



[from part (a)]
Case IV 2 < x -coordinate of vertex < 3
1
2 <m + <3
⇒
2
3
5
 3 5
or
< m < or m ∈  , 
 2 2
2
2
f(2)
α
[from Eq. (i)]
Case III f (3 ) > 0
∴

 7 + 33 
7 − 33 
m ∈  − ∞,
, ∞
 ∪
2 
2



2
X
3
Case I D ≥ 0
∴
m ∈R
Case II f (2 ) > 0

 7 + 33 
7 − 33 
, ∞
∴ m ∈  − ∞,
 ∪
2 
2



X
Consider the following cases:
Case I D ≥ 0
∴
m ∈R
Case II x -coordinate of vertex < 2.
1
m + <2
⇒
2
3
or
m<
2
Case III f (2 ) > 0
1
4 − (2m + 1 ) 2 + m (m + 1 ) > 0
⇒
2
⇒
m 2 − 7m + 4 > 0
β
α
3
β
[from Eq. (i)]
X
X
[from Eq. (i)]
Case II f (2 ) f (3 ) < 0
1


 4 − 2 ( 2m + 1 ) + m ( m + 1 )


2
1


9 − 3 ( 2m + 1 ) + m ( m + 1 ) < 0


2
[from Eq. (ii)]
⇒
⇒
[from part (a)]
( m 2 − 7m + 4 ) ( m 2 − 11m + 12 ) < 0

7 − 33  
7 + 33 
m −
 m −

2  
2 


11 + 73 
11 − 73  
m −
 m −
 <0
2
2

 

Chap 02 Theory of Equations
+
7 – √33
2
+
7 – √73
2
–
7 + √33
2
–
11 + √73
2
 7 − 33 11 − 73 
 7 + 33 11 + 73 
∴ m ∈
,
,
 ∪

2
2
2
 2



Combining all cases, we get
 7 − 33 11 − 73 
 7 + 33 11 + 73 
m ∈
,
,
 ∪

2
2
2
 2



(v) One root is smaller than 1 and the other root is greater
than 1.
Consider the following cases:
α
Case I D > 0
∴
Case II f (1 ) < 0
⇒
⇒
1
β
[from Eq. (iii)]
⇒
m (m − 3 ) < 0
∴
m ∈( 0, 3 )
Combining both cases, we get
m ∈( 0, 3 )
(vi) One root is greater than 3 and the other root is smaller
than 2.
Consider the following cases:
2
3
∴
(d ) ∪ (c )
 7 − 33 11 − 73 
 7 + 33 11 + 73 
∴ m ∈
,
,
 ∪

2
2
2
 2



i.e.
(viii) Atleast one root is greater than 2.
i.e. (Exactly one root is greater than 2) ∪ (Both roots are
greater than 2)
2
β
X
or(Exactly one root is greater than 2) ∪ (b )
Consider the following cases:
Case I D > 0
∴
m ∈R
[from Eq. (i)]
Case II f (2 ) < 0
⇒
m 2 − 7m + 4 < 0
 7 − 33 7 + 33 
m ∈
,

2
2 

∴
Combining both cases, we get
 7 − 33 7 + 33 
m ∈
,

2 
 2
…(II)
β
X
7 − 33
7 + 33
<m <
2
2
 7 − 33 7 + 33 
m ∈
,

2 
 2
[from Eq. (i)]
(ix) Atleast one root is smaller than 2.
i.e. (Exactly one root is smaller than 2) ∪(Both roots are
smaller than 2)
or
(h) (II) ∪ (a)

 7 − 33 7 + 33 
7 − 33 
We get, m ∈  − ∞,
,
 ∪

2 
2 

 2
(x) Both 2 and 3 lie between α and β.
Consider the following cases:
Case I D > 0
∴
m ∈R
[from Eq. (i)]
Case III f (3 ) < 0
⇒
m 2 − 11m + 12 < 0
∴
…(I)
Finally from Eqs. (I) and (II), we get
 7 − 33 7 + 33 
 7 + 33 
m ∈
,
, ∞
 ∪
2 
 2
 2

Case I D > 0
∴
m ∈R
Case II f (2 ) < 0
⇒
m 2 − 7m + 4 < 0
∴
(vii) Atleast one root lies in the interval (2, 3).
α
[from Eq. (i)]
1
1 − (2m + 1 ) + m (m + 1 ) < 0
2
m 2 − 3m < 0
α
Combining all cases, we get
 7 − 33 7 + 33 
m ∈
,

2 
 2
X
m ∈R
 11 − 73 11 + 73 
m ∈
,

2
2


∴
+
195
11 − 73
11 + 73
<m <
2
2
α
2
3
β
X
196
Textbook of Algebra
Case II f (2 ) < 0
1
⇒
 7 − 33 7 + 33 
m ∈
,

2 
 2
∴
Case III f (3 ) < 0
⇒
∴ From Eq. (i), we get
1
1
 11 − 73 11 + 73 
m ∈
,

2
2


⇒ (c )n + 1 ⋅ a
⇒
or
91. We have,
⇒
α+β r +1
=
α −β r −1
[using componendo and dividendo method]
− b /a r + 1
=
⇒ b (1 − r ) = (1 + r ) D
r −1
D
a
(1 + r ) 2 b 2
=
r
ac
1
1
1
+
=
x+p x+q r
1
(x + q ) + (x + p )
=
x 2 + ( p + q ) x + pq r
(1 + r ) 2 ⋅ 4ac = b 2( 4r ) or
2 pq − p 2 − q 2 − 2 pq
=
2
p2 + q2
2
92. Let α be one root of the equation ax 2 + bx + c = 0.
=−
c
a
c
=
a
α ⋅ αn =
⇒
αn + 1
1
c n + 1 ⋅ a n + 1 + (cn )n + 1 ⋅ a n + 1 + b = 0
1
 b  2 2c  n c  m  2 2n 
=  −  −  +  −  − 


a  l a  l 
l 
 a
2
2
b − 2ac  n c m − 2nl  (b 2 − 2ac ) ln + ( m 2 − 2nl ) ac
=
 + 
=
2
2
a 2l 2
 a
l a  l

∴ Required equation is
(b 2 − 2ac ) ln + (m 2 − 2nl ) ac
 mb 
=0
x2 −   x +
 al 
a 2l 2
⇒ a 2l 2x 2 − mbalx + (b 2 − 2ac ) ln + (m 2 − 2nl ) ac = 0
94. Since, the roots are equal.
∴
⇒
D=0
4 (b 2 − ac ) 2 − 4 (a 2 − bc ) (c 2 − ab ) = 0
⇒
(b 2 − ac ) 2 − (a 2 − bc ) (c 2 − ab ) = 0
⇒
b (a 3 + b 3 + c 3 − 3abc ) = 0
⇒
b = 0 or a 3 + b 3 + c 3 − 3abc = 0
95. Let α and β be the roots of x 2 − px + q = 0. Then,
α+β=p
αβ = q
Then, other root be α .
and
1
n
+b=0
(anc )n + 1 + (c na )n + 1 + b = 0
b
93. We have,
α+β=−
a
c
m
n
and γδ =
αβ =
⇒ γ+δ=−
a
l
l
Now, sum of the roots
= (αγ + βδ ) + (αδ + βγ ) = (α + β ) γ + (α + β ) δ
= (α + β ) ( γ + δ )
 b   m  mb
= −  −  =
 a  l 
al
n
b
a
n
+1
n+1
= {(α + β ) 2 − 2αβ } γδ + αβ {( γ + δ ) 2 − 2 γδ }
Now, since the roots are equal in magnitudes, but opposite in
sign. Therefore,
Sum of the roots = 0
⇒
p + q − 2r = 0
…(i)
⇒
p + q = 2r
and product of the roots = pq − ( p + q ) r
 p + q
[from Eq. (i)]
= pq − ( p + q ) 

 2 
α + αn = −
−
and product of the roots
= (αγ + βδ ) (αδ + βγ )
= (α 2 + β 2 ) γδ + αβ ( γ 2 + δ 2 )
b 2(1 − r ) 2 = (1 + r ) 2 (b 2 − 4ac )
⇒ x 2 + ( p + q − 2r ) x + pq − ( p + q ) r = 0
∴
1
+ (c n )n + 1 ⋅ a
1
On squaring both sides, we get
⇒
1
+1
n+1
⇒
α
=r
β
⇒
−
1
⇒
Combining all cases, we get
 11 − 73 7 + 33 
m ∈
,

2
2 

90. Q
n
b
cn + 1 cn + 1
=−
+ 
 
a
a
a
m 2 − 11m + 12 < 0
∴
cn + 1
α= 
a
⇒
m − 7m + 4 < 0
2
…(i)
And α and
1
be the roots of x 2 − ax + b = 0. Then,
β
1
α + =a
β
α
=b
β
…(i)
…(ii)
…(iii)
... (iv)
197
Chap 02 Theory of Equations
LHS = (q − b ) 2
Now,
∴
α

= αβ − 

β
2
[from Eqs. (ii) and (iv)]
2

1 
1

= α β −  = α 2 (α + β ) − α +  


β 
β

2
2
= α ( p − a ) [from Eqs. (i) and (iii)]
α
= αβ ⋅ ( p − a ) 2
β
2
λ = 10,
2
= bq ( p − a ) 2
⇒
(2 + 3 ) x
= RHS
…(i)
Now, (1 + y ) x 2 − 2 ( p + y ) x + (q + y ) = 0
x 2 − 2 x = log 2 +
⇒
( x − 1 ) 2 = 1 + log 2 +
∴
( x − 1 ) = 1 + log 2 +
= 4 ( p 2 + 2 py + y 2 − q − y − qy − y 2 )
[from Eq. (i)]
= − 4 ( p − 1 ) 2y
[Qy < 0 and p ≠ 1]
>0
Hence, roots of (1 + y ) x 2 − 2 ( p + y ) x + (q + y ) = 0 are real
and distinct.
2
97. x log x ( x + 3) = 16
…(i)
⇒
∴
But
i.e. no solution.
∴
98. Q(2 + 3 ) x
⇒
2
x ∈φ
− 2x + 1
(2 + 3 )
[by property]
x+3=±4
x = 1 and x = − 7
x ≠ 1, x ≠ − 7
+ (2 − 3 ) x
x2 − 2x
2
− 2x − 1
x = 1 ± (1 + log 2 +
3
10 )
⇒
x 1 = 1 + (1 + log 2 +
3
10 )
x 2 = 1 − (1 + log 2 +
3
⇒
or
2
− 2x
= λ, then Eq. (i) reduces to
1 101
λ+ =
λ 10
10 λ2 − 101 λ + 10 = 0
( λ − 10 ) (10 λ − 1 ) = 0
10
3
10
10
3
10 ]
10 )
2
99. We have,
 x 
x2+ 
 =8
 x − 1
⇒

x 
x
=8
x +
 − 2⋅ x ⋅

x − 1
(x − 1)
⇒
 x2 
 x2 

 −2 
 −8 = 0
 x − 1
 x − 1
Let y =
y 2 − 2y − 8 = 0
⇒
∴
(y − 4 ) (y + 2 ) = 0
y = 4, − 2
x2
If y = 4, then
4=
x −1
or
x 2 − 4x + 4 = 0
or
(x − 2)2 = 0
or
∴
x =2
x1 = 2
−2 =
− 2 ± (4 + 8)
2
x = −1 ± 3
x=
∴
x 2 = − 1 + 3, x 3 = − 1 − 3
100. We have,
Let
x2
x −1
x 2 + 2x − 2 = 0
⇒
…(i)
…(i)
x2
. Then, Eq. (i) reduces to
x −1
∴

1 
Q 2 − 3 = 2 + 3 


Let (2 + 3 ) x
3
10, 1 − log 2 +
3
⇒
or
⋅ (2 + 3 ) (2 − 3 )
101
+ (2 − 3 )
⋅ (2 − 3 ) =
10
101
x2 − 2x
x2 − 2x
(2 + 3 )
+ (2 − 3 )
=
10
1
101
x2 − 2x
(2 + 3 )
+
=
2
10
(2 + 3 ) x − 2 x
or
3
2
and if y = − 2, then
101
=
10 (2 − 3 )
x2 − 2x − 1
⇒
10, − log 2 +
2
= 4 [(2 p − q − 1 ) y + p 2 − q ]
Equation is defined, when
x > 0, x ≠ 1, x ≠ − 3,
Then,
(x + 3)2 = 42
3
2
∴ Discriminant = 4 ( p + y ) 2 − 4 (1 + y ) (q + y )
= 4 [(2 p − p 2 − 1 ) y + 0 ]
= 10, 10 − 1
[Q ( x − 1 ) 2 ≠ 1 − log 2 +
[from Eqs. (ii) and (iv)]
D=0
( −2 p ) 2 − 4q = 0 or p 2 = q
− 2x
⇒
96. Since, roots of x 2 − 2px + q = 0 are equal.
∴
i.e.,
2
1
10
x + 8 + 2 ( x + 7 ) + ( x + 1 ) − ( x + 7 ) = 4 …(i)
(x + 7) = λ
x = λ2 − 7
or
Then, Eq. (i) reduces to
( λ2 − 7 + 8 + 2 λ ) + ( λ2 − 7 + 1 − λ ) = 4
⇒
or
( λ + 1 ) + ( λ2 − λ − 6 ) = 4
( λ2 − λ − 6 ) = 3 − λ
…(ii)
198
Textbook of Algebra
Case I If 0 < x 2 + 2 x − 3 < 1
On squaring both sides, we get
λ2 − λ − 6 = 9 + λ2 − 6 λ
⇒
∴
⇒
5 λ = 15
λ =3
(x + 7) = 3
[from Eq. (ii)]
or
x + 7 =9
∴
x =2
and x = 2 satisfies Eq. (i).
Hence, x 1 = 2
101. We have,
Putting t = 2
2
2
…(i)
in the Eq. (i), we get
t 2 + 2 (2a + 1 ) t + 4a 2 − 3 > 0
Let
Q
4 < x 2 + 2x + 1 < 5
⇒
4 < (x + 1)2 < 5
2
[Q t > 0,∴2 x > 0]
f (t ) = t 2 + 2 (2a + 1 ) t + 4a 2 − 3
f (t ) > 0
⇒
⇒
Consider the following cases:
Case I Sum of the roots > 0
( 2a + 1 )
−2
>0
1
1

∴
a ∈  − ∞, − 

2
∴
⇒
Case II Product of the roots > 0
4a 2 − 3
⇒
>0
1
3
or
a2 >
4

 3 
3
or
a ∈  − ∞, −
 ∪  , ∞
2 

 2

⇒
⇒
x ∈ ( − 5, 1 )
x ∈ [ − 4, 0 )
[Q − 4 ≤ x < 0] …(iv)
(x + 4) − x
<1
x ≥ 0, then
(x − 1)
4
>0
x −1
(x − 5)
>0
(x − 1)
1−
x 2 + 2x − 3 > 1
⇒
x 2 + 2x + 1 > 5 ⇒ (x + 1)2 > 5
⇒
x+1<− 5
or
x+1> 5
Then,
 | x + 4| − | x |
 >0

x −1 
102. We have, log x 2 + 2 x − 3 
Now,
The given inequation is valid for
| x + 4| − | x |
>0
(x − 1)
⇒
…(i)
[Q x ≥ 0] …(v)
Case II If
∴
⇒
∴
…(ii)
[Q x < − 4] …(iii)
∴
x ∈ ( − ∞, 1 ) ∪ ( 5, ∞ )
⇒
x ∈ [ 0, 1 ) ∪ ( 5, ∞ )
From Eqs. (iii), (iv) and (v), we get
x ∈ ( − ∞, 1 ) ∪ ( 5, ∞ )
Now, common values in Eqs. (ii) and (iv) is
x ∈ ( − 5 − 1, − 3 )
⇒
4a + 4 < 0
∴
a < −1
or
a ∈ ( − ∞, − 1 )
Combining all cases, we get
 3 
a ∈ ( − ∞, − 1 ) ∪  , ∞
 2

x 2 + 2 x − 3 > 0, ≠ 1
x ∈ ( − ∞, − 3 ) ∪ (1, ∞ )
x ∈ ( − ∞, − 4 )
x+4+x
− 4 ≤ x < 0, then
−1 < 0
(x − 1)
(x + 5)
<0
(x − 1)
and
Case III
D<0
⇒
4 ( 2a + 1 ) 2 − 4 ⋅ 1 ⋅ ( 4a 2 − 3 ) < 0
Now, consider the following cases:
− 5 − 1 < x < − 3 or 1 < x < 5 − 1
x ∈ ( − 5 − 1, − 3 ) ∪ (1, 5 − 1 )
| x + 4| − | x |
Then,
<1
(x − 1)
− (x + 4) + x
Now, x < − 4, then
<1
(x − 1)
4
1+
⇒
>0
x −1
(x + 3)
>0
⇒
(x − 1)
∴
⇒
T-axis
and
− 5 < ( x + 1 ) < − 2 or 2 < x + 1 < 5
⇒
∴
4 x + 2 (2a + 1 ) 2 x + 4a 2 − 3 > 0
x2
⇒
x ∈ ( − ∞, − 1 − 5 ) ∪ ( 5 − 1, ∞ )
| x + 4| − | x |
>1
(x − 1)
−4
x < − 4, then
>1
x −1
4
1+
<0
x −1
x+3
<0
x −1
x ∈ ( − 3, 1 )
…(vi)
…(vii)
…(viii)
199
Chap 02 Theory of Equations
[Qx < − 4]
which is false.
⇒
2x + 4
− 4 ≤ x < 0, then
−1 > 0
(x − 1)
(x + 5)
>0
(x − 1)
∴
x ∈ ( − ∞, − 5 ) ∪ (1, ∞ )
which is false.
4
and
>1
x ≥ 0, then
x −1
4
1−
<0
⇒
x −1
x −5
⇒
<0
x −1
x2 − x − 1 = 0
∴
[Q− 4 ≤ x < 0 ]
∴
x ∈(1, 5 )
which is false.
Now, common values in Eq. (viii) and (ix) is
x ∈ ( 5 − 1, 5 )
∴
…(ix)
[Qx ≥ 0]
103. Let y ≥ 0, then | y | = y
and then given system reduces to
| x 2 − 2x | + y = 1
…(i)
x + y =1
…(ii)
2
x =1
1± 5
∴
x=
2
1± 5
fail
∴
x=
2
1− 5
⇒
x=
2
1− 5
1− 5
x=
, 1, then y =
,0
⇒
2
2
1 − 5 1 − 5
,
∴ Solutions are 
 and (1, 0).
2 
 2
From Eqs. (i) and (ii), we get
x 2 = | x 2 − 2x |
[Q x < 0]
1 − 5 1 − 5
Hence, all pairs ( 0, 1 ), (1, 0 ) and 
,
 are solutions
2 
 2
of the original system of equations.
104. Given, α , β and γ are the roots of the cubic equation
…(x)
Combining Eqs. (viii) and (x), we get
x ∈ ( − 5 − 1, − 3 ) ∪ ( 5 − 1, 5 )
and
⇒
x 3 − px 2 + qx − r = 0
…(i)
∴ α + β + γ = p, αβ + βγ + γα = q , αβγ = r
1
(i) Let
y = βγ +
α
αβγ + 1 r + 1
⇒
y =
=
α
α
r +1
α=
∴
y
From Eq. (i), we get
α 3 − pα 2 + qα − r = 0
⇒
x 2 = | x | | x − 2|
⇒
Now,
x < 0, 0 ≤ x < 2, x ≥ 2
x 2 = x ( x − 2 ), x 2 = − x ( x − 2 )
(r + 1 ) 3 p (r + 1 ) 2 q (r + 1 )
−
+
−r = 0
y
y3
y2
or
ry 3 − q(r + 1 )y 2 + p (r + 1 ) 2y − (r + 1 ) 3 = 0
(ii) Let y = β + γ − α = (α + β + γ ) − 2α = p − 2 α
p −y
∴
α=
2
From Eq. (i), we get
α 3 − pα 2 + qα − r = 0
x 2 = x (x − 2)
∴
x=0
⇒
x (x + x − 2) = 0
∴
x=0
fail ∴ x = 0, 1 fail
⇒
x = 0, 1, then y = 1, 0
∴Solutions are (0, 1) and (1, 0).
If y < 0 then | y | = − y and then given system reduces to
| x 2 − 2x | + y = 1
and
x2 − y = 1
From Eqs. (iii) and (iv), we get
| x 2 − 2x | + x 2 = 2
⇒
Now,
| x | | x − 2| + x 2 = 2
x < 0, 0 ≤ x < 2, x ≥ 2
x (x − 2) + x 2 = 2
⇒
(p − y ) 3 p (p − y )2 q (p − y )
−
+
−r = 0
8
4
2
or y 3 − py 2 + ( 4q − p 2 )y + ( 8r − 4 pq + p 3 ) = 0
…(iii)
…(iv)
Also product of roots = − (8r − 4 pq + p 3 )
105. Assume α + iβ is a complex root of the given equation, then
conjugate of this root, i.e. α − iβ is also root of this equation.
On putting x = α + iβ and x = α − iβ in the given equation, we
get
2
A3
An2
A12
A22
+
+
+… +
α + iβ − an
α + iβ − a1 α + iβ − a 2 α + iβ − a 3
= ab 2 + c 2(α + iβ ) + ac
− x (x − 2) + x 2 = 2
x (x −2 ) + x = 2
2
⇒
⇒
2x 2 − 2x − 2 = 0 ⇒ 2x = 2
x2 − x − 1 = 0
and
A12
α − iβ − a1
+
A22
α − iβ − a 2
+
A32
α − iβ − a 3
+… +
= ab + c (α − iβ ) + ac
2
2
…(i)
An2
α − iβ − an
…(ii)
200
Textbook of Algebra
On subtracting Eq. (i) from Eq. (ii), we get

A12
A22
A32
2iβ 
+
+
2
2
2
2
(α − a2 ) + β
(α − a 3 )2 + β2
( α − a1 ) + β

An2
+…+
+ c2 = 0
2
2
(α − an ) + β

The expression in bracket ≠ 0
∴
2iβ = 0 ⇒ β = 0
Hence, all roots of the given equation are real.
106. Given equation is
…(i)
x 4 + 2ax 3 + x 2 + 2ax + 1 = 0
On dividing by x 2, we get
x 2 + 2ax + 1 +
1
 2
 x + 2  + 2a

x 
⇒

x

2a
1
+ 2 =0
x
x
1
+  +1=0
x
2
1
1


 x +  − 2 + 2a  x +  + 1 = 0


x
x
⇒
or
1

 x +  + 2a

x
1

x +  − 1 = 0

x
1
y 2 + 2ay − 1 = 0, where y = x +
x
− 2a ± ( 4a 2 + 4 )
= − a ± (a 2 + 1 )
2
Taking ‘+’ sign, we get
y = − a + (a 2 + 1 )
⇒
or
x+
1
= − a + (a 2 + 1 )
x
y = − a − (a 2 + 1 )
⇒
1
= − a − (a 2 + 1 )
x
or
x+
x 2 + (a + (a 2 + 1 ) ) x + 1 = 0
(a + (a 2 + 1 ) + 2 ) (a + (a 2 + 1 ) − 2 ) > 0
Q
a + (a 2 + 1 ) + 2 > 0
∴
a + (a 2 + 1 ) − 2 > 0
⇒
(a 2 + 1 ) > 2 − a
a ≥ 2

2
2
or a + 1 > (2 − a ) , if a < 2
⇒
a ≥ 2

or a > 3 , if a < 2

4
⇒
a ≥ 2

or 3 < a < 2
 4
Hence,
3
< a < ∞ or
4
or
x 2 + (a − (a 2 + 1 ) ) x + 1 = 0
Taking ‘−’ sign, we get
⇒
Since,
∴
⇒
Now,
⇒
or
or
or
∴
y =
∴
(a + (a 2 + 1 ) ) 2 − 4 > 0
107. We have,
2
or
∴
…(ii)
or
108. We have,
3 
a ∈  , ∞
4 
[2 x ] − [ x + 1 ] = 2 x
LHS = Integer
RHS = 2x = Integer
[2 x ] = 2 x
− [ x + 1] = 0
[ x + 1] = 0
0 ≤ x + 1 <1
−1 ≤ x < 0
− 2 ≤ 2x < 0
2 x = − 2, − 1
1
x = − 1, −
2
1
x 1 = − 1, x 2 = −
2
(a 2 + 3 ) x 2 + (a + 2 ) x − 6 < 0
…(iii)
Let α , β be the roots of Eq. (ii) and γ, δ be the roots of Eq. (iii).
Then,
α + β = (a 2 + 1 ) − a
and
γ + δ = − (a 2 + 1 ) − a
and
γδ =1
Clearly, α + β > 0 and αβ > 0
∴Either α , β will be imaginary or both real and positive
according to the Eq. (i) has atleast two distinct negative roots.
Therefore, both γ and δ must be negative. Therefore,
(i) γδ > 0, which is true as γ δ = 1.
(ii)
γ+δ<0
⇒
⇒
∴
(iii)
X
αβ = 1
and
− (a + (a 2 + 1 ) ) < 0
a + (a + 1 ) > 0, which is true for all a.
2
a ∈R
D>0
Let
Q
f ( x ) = (a 2 + 3 ) x 2 + (a + 2 ) x − 6
(a 2 + 3 ) > 0 and f ( x ) < 0
∴
D>0
2
2
⇒(a + 2 ) + 24 (a + 3 ) > 0 is true for all a ∈ R .
109. We have,
⇒
⇒
Q
∴
or
6 x 2 − 77[ x ] + 147 = 0
6 x 2 + 147
= [x ]
77
(0.078) x 2 = [x ] − 1.9
(0.078 ) x 2 > 0 ⇒ x 2 > 0
[ x ] − 1.9 > 0
[ x ] > 1.9
Chap 02 Theory of Equations
∴
If
Then,
∴
If
Then,
∴
If
Then,
∴
If
Then,
∴
If
Then,
∴
If
Then,
∴
If
Then,
[ x ] = 2, 3, 4, 5,…
[ x ] = 2, i. e. 2 ≤ x < 3
2 − 1 .9
x2 =
= 1. 28
0.078
x = 1.13
[ x ] = 3, i. e. 3 ≤ x < 4
3 − 1.9
x2 =
= 14.1
0.078
x = 3.75
[ x ] = 4, i.e. 4 ≤ x < 5
4 − 1.9
x2 =
= 26.9
0.078
x = 5.18
[ x ] = 5, i.e. 5 ≤ x < 6
5 − 1.9
x2 =
= 39.7
0.078
x = 6.3
[ x ] = 6, i. e. 6 ≤ x < 7
6 − 1.9
4.1
x2 =
=
= 52.56
0.078 0.078
x = 7.25
[ x ] = 7, i. e. 7 ≤ x < 8
7 − 1.9
5.1
x2 =
=
= 65.38
0.078 0.078
x = 8.08
[ x ] = 8, i. e. 8 ≤ x < 9
8 − 1.9
6.1
x2 =
=
= 78.2
0.078 0.078
∴
If
x = 8.8
[ x ] = 9, i. e. 9 ≤ x < 10
9 − 1.9
7.1
Then,
x2 =
=
= 91.03
0.078 0.078
∴
x = 9.5
If
[ x ] = 10, i. e.10 ≤ x < 11
10 − 1.9
8.1
Then,
x2 =
=
= 103.8
0.078
0.078
∴
x = 10.2
If
[ x ] = 11, i.e. 11 ≤ x < 12
11 − 1.9
Then,
x2 =
0.078
9.1
=
= 116.7
0.078
∴
x = 10.8
Other values are fail.
Hence, number of solutions is four.
110. Since, the given equation is
x 2 − 2x − a 2 + 1 = 0
⇒
∴
⇒
γ
[fail]
[true]
β
X
δ
Let f ( x ) = x 2 − 2(a + 1 ) x + a (a − 1 ), thus the following
conditions hold good:
Consider the following cases:
Case I
D>0
⇒
4 (a + 1 ) 2 − 4a (a − 1 ) > 0
⇒
3a + 1 > 0
[fail]
∴
a>−
[fail]
Case II
f (α ) < 0
⇒
f (1 + a ) < 0
⇒ (1 + a ) 2 − 2 (1 + a ) (1 + a ) + a (a − 1 ) < 0
[fail]
[fail]
[true]
1
3
⇒
− (1 + a ) 2 + a (a − 1 ) < 0
⇒
− 3a − 1 < 0
⇒
a>−
1
3
Case III
f (s ) = 0
⇒
f (1 − a ) < 0
⇒ (1 − a ) 2 − 2 (a + 1 ) (1 − a ) + a (a − 1 ) < 0
⇒
( 4a + 1 ) (a − 1 ) < 0
1
∴
− <a <1
4
Combining all cases we get
 1 
a ∈  − , 1
 4 
111. pr = ( − p ) ( − r )
= ( α + β + γ + δ ) ( αβγ + αβδ + γδα + γδβ )
= α 2 βγ + α 2 βδ + α 2 γδ + αβγδ + β 2 γα
[true]
+ β 2 αδ + αβγδ + β 2 γδ + γ 2 αβ + αβγδ
+ γ 2 δα + γ 2δβ + αβγδ + αβδ 2 + γαδ 2 + γβδ 2
[true]
Q
⇒
⇒
[fail]
or
∴
112. (α 2
AM ≥ GM
pr
≥ (α 16 β16 γ16 δ16 )1/6 = α β γδ = 5
16
pr
≥5
16
pr ≥ 80
Minimum value of pr is 80.
+ β 2 ) 2 = (α + β ) (α 3 + β 3 )
⇒ {(α + β ) 2 − 2 αβ } 2 = (α + β ) {(α + β ) 3 − 3 αβ (α + β )}
2
⇒
3
 b 2 2c 
3bc 
 b  − b
 2 −  = −   3 + 2 


a
a  a
a 
a
⇒
3
 b 2 − 2ac 
 − b   − b + 3abc 
=






 a  
a3

 a2 
(x − 1)2 = a 2
x − 1 ≠ a or x = 1 ± a
α = 1 + a and β = 1 − a
α
201
2
202
Textbook of Algebra
⇒
4a 2c 2 = acb 2
⇒
Combining all cases, we get
ac (b − 4ac ) = 0
2
117. We have,
a≠0
c∆ = 0
As
⇒
113. Let P ( x ) = bx 2 + ax + c
P (0) = 0
c=0
P (1 ) = 1
a + b =1
P ( x ) = ax + (1 − a ) x 2
As
⇒
As
⇒
Now,
P ′ ( x ) = a + 2 (1 − a ) x
As
P ′ ( x ) > 0 for x ∈( 0, 1 )
Only option (d) satisfies above condition.
114. Let the roots are α and α + 1, where α ∈ I .
Then, sum of the roots = 2 α + 1 = b
Product of the roots = α (α + 1 ) = c
and
∴
and
⇒
and
Next,
c + d = 10a, cd = − 11b
a + b + c + d = 10 (a + c )
abcd = 121 bd
b + d = 9 (a + c )
ac = 121
2
a − 10 ac − 11d = 0
and
c 2 − 10ac − 11b = 0
⇒ a 2 + c 2 − 20 ac − 11 (b + d ) = 0
⇒ (a + c ) 2 − 22 × 121 − 99 (a + c ) = 0
⇒
a + c = 121 or − 22
If a + c = − 22 ⇒a = c , rejecting these values, we have
a + c = 121
∴
a + b + c + d = 10 (a + c ) = 1210
118.
D≥0
b 2 − 4c = ( 2α + 1 ) 2 − 4α (α + 1 )
Now,
4 (a + b + c ) 2 − 12 λ (ab + bc + ca ) ≥ 0
= 4α + 1 + 4α − 4α − 4α = 1
2
∴
k ∈ ( −∞,4 )
a + b = 10c, ab = − 11d
2
b − 4c = 1
(a 2 + b 2 + c 2 ) − (3 λ − 2 ) (ab + bc + ca ) ≥ 0
2
115. Let f ( x ) = an x + an − 1 x
n
n− 1
∴
+…+ a1 x,
f ( 0 ) = 0;f (α ) = 0
⇒ f ′( x ) = 0 has atleast one root between ( 0, α ).
i.e. Equation
nan xn − 1 + (n − 1 ) an − 1xn − 2 +…+ a1 = 0
(i)
⇒
|b − c | < a
b 2 + c 2 − 2bc < a 2
…(ii)
⇒
| c − a| < b
c 2 + a 2 − 2ca < b 2
…(iii)
From Eqs. (i), (ii) and (iii), we get
a2 + b2 + c2
<2
ab + bc + ca
Consider the following cases:
f(5)
α
(a 2 + b 2 + c 2 )
(ab + bc + ca )
|a − b | < c
a 2 + b 2 − 2ab < c 2
Since,
⇒
has a positive root smaller than α.
116. Let f ( x ) = x 2 − 2kx + k 2 + k − 5
(3 λ − 2 ) ≤
P
S
…(iv)
From Eqs. (i) and (iv), we get
3λ − 2 < 2 ⇒ λ <
X
4
3
119. Q x 2 − 2mx + m 2 − 1 = 0
D≥0
Case I
⇒
4k − 4 .1(k + k − 5 ) ≥ 0
⇒
−4(k − 5 ) ≥ 0
2
2
⇒
k −5 ≤ 0
⇒
k ≤ 5 or k ∈ ( −∞, 5 ]
Case II x-Coordinate of vertex x < 5
2k
⇒
<5
2
⇒
Case III
k < 5 or k ∈ ( −∞,5 )
f (5 ) > 0
⇒
25 − 10k + k 2 + k − 5 > 0
⇒
k − 9k + 20 > 0
⇒
2
(k − 4 )(k − 5 ) > 0 or k ∈ ( −∞,4 ) ∪ (5, ∞ )
⇒
(x − m )2 = 1
∴
x − m = ± 1 or x = m − 1, m + 1
According to the question,
m − 1 > − 2, m + 1 > − 2
⇒
m > − 1, m > − 3
Then,
m > −1
and
m − 1 < 4, m + 1 < 4
⇒
m < 5, m < 3 and m < 3
From Eqs. (i) and (ii), we get − 1 < m < 3
120. x 2 + px + q = 0
…(i)
…(ii)
Sum of the roots = tan 30 ° + tan 15 ° = − p
Product of the roots = tan 30 °⋅ tan 15 ° = q
tan 30 ° + tan 15 °
tan 45 ° = tan (30 ° + 15 ° ) =
1 − tan 30 °⋅ tan 15 °
Chap 02 Theory of Equations
⇒
1=
−p
⇒ − p =1 −q
1 −q
203
125. Let f ( x ) = x 7 + 14x 5 + 16x 3 + 30x − 560
⇒
q− p =1
∴
2 + q − p =3
121. The equation x 2 − px + r = 0 has roots (α , β) and the equation
α

x 2 − qx + r has roots  , 2β .
2

α
⇒
r = αβ and α + β = p and + 2β = q
2
2q − p
2 (2 p − q )
and α =
⇒
β=
3
3
2
⇒
αβ = r = (2q − p ) (2 p − q )
9
∴ f ′( x ) = 7 x 6 + 70 x 4 + 48 x 2 + 30 > 0, ∀ x ∈ R
⇒ f ( x ) is an increasing function, for all x ∈ R
Hence, number of real solutions is 1.
126. Let f ( x ) = x 3 − px + q
∴
f ′( x ) = 3 x 2 − p
⇒
f ′′( x ) = 6 x
p
√3
p
–
√3
α + β = −a
122.
⇒
| α − β | < 5 ⇒ (α − β ) 2 < 5
For maxima or minima, f ′( x ) = 0
a − 4 < 5 ⇒ a ∈ ( − 3, 3 )
∴
2
x=±
123. Suppose roots are imaginary, then β = α
and
1
=α
β
⇒
β=−
[not possible]
⇒ Roots are real ⇒( p 2 − q ) (b 2 − ac ) ≥ 0
and
2b
1
=α +
a
β
x=−
α c
= , α + β = − 2 p, αβ = q
β a
α + 4 β = 6 and 4αβ = a
α + 3 β = c and 3α β = 6
From Eqs. (i) and (ii), we get
…(i)
…(ii)
x 2 − 6x + 8 = 0
⇒
x = 2, 4
α = 2, 4 β = 4, then 3 β = 3
3
α = 4, 4 β = 2, then 3 β =
2
If
∴Common root is x = 2.
p
and maxima at
3
p
.
3
[non-integer]
X
4
∴
⇒
D>0
64k 2 − 4 ⋅ 16 (k 2 − k + 1 ) > 0
⇒
k >1
−b
8k
>4 ⇒
>4
2a
2
k >1
f (4) ≥ 0
16 − 32k + 16 (k 2 − k + 1 ) ≥ 0
⇒
⇒
and
⇒
⇒
a = 8, αβ = 2
Now, first equation becomes
If
p
p
<0
 = −6
3
3
127. Let f ( x ) = x 2 − 8kx + 16 (k 2 − k + 1)
If β = 1, then α = q
[not possible]
⇒
c = qa
− 2b
Also,
α+1=
a
− 2b
⇒
− 2p =
a
[not possible]
⇒
b = ap
⇒ Statement −2 is true but it is not the correct explanation of
Statement-1.
124. Let α,4β be roots of x 2 − 6x + a = 0 and α , 3 β be the roots of
x 2 − cx + 6 = 0.
Then,

f ′′  −

and
Hence, given cubic minima at x =
⇒ Statement −1 is true.
−
 p
 p
f ′′ 
 =6   > 0
 3
 3
⇒
1
β
p
3
⇒
…(ii)
k 2 − 3k + 2 ≥ 0
⇒
(k − 1 ) (k − 2 ) ≥ 0
⇒
k ≤ 1 or k ≥ 2
From Eqs. (i), (ii) and (iii), we get
k ≥2
kmin = 2
128. Since, roots of bx 2 + cx + a = 0 are imaginary.
∴
…(i)
…(iii)
c 2− 4ab < 0
− c 2 > − 4ab
…(i)
204
Textbook of Algebra
132. Let α be the common root.
f ( x ) = 3b 2x 2 + 6bcx + 2c 2
Let
Then, α 2 + bα − 1 = 0 and α 2 + α + b = 0
3b > 0
2
Since,
D = (6bc ) 2 − 4 (3b 2 ) (2c 2 ) = 12b 2c 2
and
12b 2c 2
D
= −c 2 > −4ab
f (x ) = −
=−
4a
4 (3b 2 )
∴ Minimum value of
α β α 2 + β 2 (α + β ) 2 − 2 αβ
129. + =
=
β α
αβ
αβ
…(i)
1 1
×
−1
b
1
b
=
−1 1
2
b 1
⇒
(1 − b ) (b + 1 ) = ( − 1 − b ) 2
⇒
b 3 + 3b = 0
∴
2
b = 0, i 3, − i 3 , where i = − 1.
133. Let f ( x ) = x 4 − 4 x 3 + 12x 2 + x − 1
α 3 + β 3 = q, α + β = − p
and given,
1 b
⇒
⇒
(α + β ) 3 − 3 αβ (α + β ) = q
∴
f ′( x ) = 4 x 3 − 12 x 2 + 24 x + 1
⇒
− p 3 + 3 pαβ = q
⇒
f ′′( x ) = 12 x 2 − 24 x + 24
αβ =
or
∴ From Eq. (i), we get
= 12 ( x 2 − 2 x + 2 )
q + p3
3p
= 12 [( x − 1 ) 2 + 1 ] > 0
i.e. f ′′( x ) has no real roots.
Hence, f ( x ) has maximum two distinct real roots, where
f ( 0 ) = − 1.
134. Given, p ( x ) = f ( x ) − g ( x )
2 (q + p 3 )
p 3 − 2q
α β
3p
=
+ =
3
β α
(q + p 3 )
(q + p )
3p
α β
and product of the roots = ⋅ = 1
β α
p2 −
⇒
It is clear that p( x ) = 0 has both equal roots − 1, then
(b − b1 )
−1 −1 = −
(a − a1 )
c − c1
and
− 1 × −1 =
a − a1
 p 3 − 2q 
∴ Required equation is x 2 − 
 x+1=0
 q + p3 
or (q + p 3 ) x 2 − ( p 3 − 2q ) x + (q + p 3 ) = 0
130. Since, f ′ ( x ) = 12x 2 + 6x + 2
…(i)
⇒
b − b1 = 2 (a − a1 ) and c − c1 = (a − a1 )
Also given, p( −2 ) = 2
…(ii)
⇒ 4 (a − a1 ) − 2 (b − b1 ) + (c − c1 ) = 2
From Eqs. (i) and (ii), we get
4 (a − a1 ) − 4 (a − a1 ) + (a − a1 ) = 2
…(iii)
∴
(a − a1 ) = 2
[from Eq. (i)] …(iv)
⇒
b − b1 = 4 and c − c1 = 2
Now, p(2 ) = 4 (a − a1 ) + 2 (b − b1 ) + (c − c1 )
[from Eqs. (iii) and (iv)]
= 8 + 8 + 2 = 18
135. Let the quadratic equation be
ax 2 + bx + c = 0
Here, D = 6 − 4 ⋅ 12 ⋅ 2 = 36 − 96 = − 60 < 0
2
∴
f ′( x ) > 0, ∀ x ∈ R
⇒ Only one real root for f ( x ) = 0
Also,
f ( 0 ) = 1, f ( − 1 ) = − 2
⇒ Root must lie in ( − 1, 0 ).
 1 1
Taking average of 0 and ( − 1 ), f  −  =
 2 4
1

⇒ Root must lie in  − 1, −  .

2
1
 3
Similarly, f  −  = −
 4
2
 3 1
⇒ Root must lie in  − , −  .
 4 2
131. Qα 2 − 6 α − 2 = 0 ⇒α 2 − 2 = 6 α
and
∴
…(i)
β − 6β − 2 = 0 ⇒ β − 2 = 6β
2
2
a10 − 2a 8 ( α
=
2a 9
10
…(ii)
− β ) − 2 (α − β )
2 (α 9 − β9 )
10
8
8
=
α 8( α 2 − 2 ) − β 8 ( β 2 − 2 )
2 (α 9 − β9 )
=
α ⋅ 6α − β ⋅ 6β
2 (α 9 − β9 )
=
6 (α 9 − β9 )
=3
2 (α 9 − β9 )
8
p( x ) = (a − a1 ) x 2 + (b − b1 ) x + (c − c1 )
8
[from Eqs. (i) and (ii)]
Sachin made a mistake in writing down constant term.
∴ Sum of the roots is correct.
i.e.
α + β =7
Rahul made a mistake in writing down coefficient of x .
∴ Product of the roots is correct.
i.e.
αβ = 6
⇒ Correct quadratic equation is
x 2 − (α + β ) x + αβ = 0
⇒
x 2 − 7x + 6 = 0
⇒ ( x − 6 )( x − 1 ) = 0 ⇒ x = 6,1
Hence, correct roots are 1 and 6.
136. Let a + 1 = h 6
∴
(h 2 − 1 ) x 2 + (h 3 − 1 ) x + (h − 1 ) = 0
Chap 02 Theory of Equations
 h 2 − 1 2  h 3 − 1
 x+1=0
x + 

 h −1
 h −1
⇒
q
p
−
⇒
As a → 0, then h → 1
 h 2 − 1 2
 h 3 − 1
lim 
 x + lim 
 x+1=0
h → 1 h − 1 
h → 1 h − 1 
r
p
=4
⇒
q = − 4r
Also, given p, q, r are in AP.
⇒
2x 2 + 3x + 1 = 0
∴
2q = p + r
⇒
2x 2 + 2x + x + 1 = 0
⇒
p = − 9r
⇒
(2 x + 1 ) ( x + 1 ) = 0
Now, | α − β | =
1
x = − 1 and x = −
2
∴
137. Let e sin x = t
…(i)
Then, the given equation can be written as
1
t − − 4 = 0 ⇒ t 2 − 4t − 1 = 0
t
t=
∴
4 ± (16 + 4 )
2
⇒
e sin x = (2 + 5 )
⇒
sin x = loge ( 2 + 5 )
Q
…(ii)
(2 + 5 ) > e
[Qe = 2.71828… ]
loge ( 2 + 5 ) > 1
⇒
From Eqs. (ii) and (iii), we get
sin x > 1
Hence, no real root exists.
138. Given equations are
ax 2 + bx + c = 0
⇒
⇒
(16r 2 + 36r 2
52 | r |
[from Eqs. (i) and (ii)]
=
9 |r |
9 |r |
2 13
9
141. f ( x ) = x 5 − 5x and g( x ) = − a
f ′( x ) = 5 x 4 − 5
∴
4
…(ii)
[fractional part of x ]
…(i)
2 ± ( 4 + 12a ) 1 + (1 + 3a )
=
6
3
2
2
[Q 0 < { x } < 1 ]
1 + (1 + 3a 2 )
< 1 ⇒ (1 + 3a 2 ) < 2
3
…(ii)
a2 < 1 ⇒ − 1 < a < 1
0<
From Eqs. (i) and (ii), we get
a ∈ ( − 1, 0 ) ∪ ( 0, 1 )
1 1
α+β
140. Q + = 4 ⇒
=4
α β
αβ
1
–4
…(i)
3 { x } 2 − 2{ x } − a 2 = 0
{x } =
=
–1
For no integral solution, { x } ≠ 0
∴
a≠ 0
The given equation can be written as
⇒
(q 2 − 4 pr )
| p|
[which is impossible]
Clearly, roots of Eq. (ii) are imaginary, since Eqs. (i) and (ii)
have a common root, therefore common root must be
imaginary and hence both roots will be common. Therefore,
Eqs. (i) and (ii) are identical.
a b c
∴
= = or a : b : c = 1 : 2 : 3
1 2 3
139. Q x − [ x ] = { x }
[from Eq. (i)] …(ii)
Q for ax 2 + bx + c = 0, α − β = D 

a 
D
| a|
…(iii)
x 2 + 2x + 3 = 0
and
... (i)
=
=
[Qe sin x > 0,∴taking + ve sign]
205
= 5 (x 2 + 1) (x − 1) (x + 1)
Clearly, f ( x ) = g( x ) has one real root, if a > 4 and three real
roots, if | a| < 4.
142. Since, b = 0 for p ( x ) = ax 2 + bx + c, as roots are pure
imaginary.
(− c ± i c )
, which are clearly neither pure real nor
a
pure imaginary, as c ≠ 0.
⇒x = ±
143. Qαx 2 − x + α = 0 has distinct real roots.
∴
D>0
⇒
 1 1
1 − 4α 2 > 0 ⇒ α ∈  − , 
 2 2
Also,
⇒
⇒
...(i)
| x1 − x 2 | < 1 ⇒ | x1 − x 2 | 2 < 1
1
D
1 − 4α 2
< 1 ⇒α 2 >
<1 ⇒
2
5
α2
a
1   1 

α ∈  −∞,−  ∪  , ∞

5  5 
From Eqs. (i) and (ii), we get
 1 1   1 1
S =  − ,−  ∪  , 
 2
5   5 2
...(ii)
206
Textbook of Algebra
144. ( x 2 − 5x + 5) x
2
+ 4 x − 60
Q α 1, β1 are roots of x 2 − 2 x sec θ + 1 = 0 and α 1 > β1
=1
Case I
x 2 − 5 x + 5 = 1 and x 2 + 4 x − 60 can be any real number
⇒
x = 1, 4
Case II
x 2 − 5 x + 5 = − 1 and x 2 + 4 x − 60 has to be an even number
⇒
x = 2, 3
For x = 3, x 2 + 4 x − 60 is odd, ∴ x ≠ 3
Hence,
x =2
Case III x 2 − 5 x + 5 can be any real number and
x 2 + 4 x − 60 = 0
⇒
x = − 10, 6
⇒ Sum of all values of x = 1 + 4 + 2 − 10 + 6 = 3
145. Q x 2 − 2x sec θ a + 1 = 0 ⇒ x = sec θ ± tan θ
π
π
and
− <θ < −
6
12
 π
 π
⇒
sec  −  > sec θ > sec  − 
 12
 6
or
 π
sec   > sec θ > sec
 6
 π
 
 12
and
 π
 π
tan  −  < tan θ < tan  − 
 6
 12
⇒
 π
 π
− tan   < tan θ < − tan  
 12
 6
or
 π
 π
tan   > − tan θ > tan  
 6
 12
∴ α 1 = sec θ − tan θ and β1 = sec θ + tan θ
⇒ α 2, β 2 are roots of x 2 + 2 x tan θ − 1 = 0
α 2 > β2
α 2 = − tan θ + sec θ
and
∴
and
β 2 = − tan θ − sec θ
α 1 + β 2 = − 2 tan θ
Hence,
146. Q x ( x + 1) + ( x + 1) ( x + 2) + .... + ( x + n − 1) ( x + n ) = 10n
⇒ nx 2 + x (1 + 3 + 5 + K + (2n − 1 )) + (1 . 2 + 2 . 3
+ ...+ (n − 1 ) . n ) = 10n
1
or nx + n x + (n − 1 ) n (n + 1 ) = 10n
3
2
or
2
3 x 2 + 3nx + (n 2 − 1 ) = 30
or
3 x + 3nx + (n − 31 ) = 0
Q
|α −β| =1
or
(α − β ) 2 = 1
or
D
=1
a2
or
D = a2
or
2
2
9n 2 − 12 . (n 2 − 31 ) = 9
or
n 2 = 121
∴
n = 11
(Q n ≠ 0)
CHAPTER
03
Sequences
and Series
Learning Part
Session 1
● Sequence
● Series
● Progression
Session 2
● Arithmetic Progression
Session 3
● Geometric Sequence or Geometric Progression
Session 4
● Harmonic Sequence or Harmonic Progression
Session 5
● Mean
Session 6
● Arithmetico-Geometric Series (AGS)
● Sigma ( S) Notation
● Natural Numbers
Session 7
● Application to Problems of Maxima and Minima
Practice Part
●
●
JEE Type Examples
Chapter Exercises
Arihant on Your Mobile !
Exercises with the
#L
symbol can be practised on your mobile. See inside cover page to activate for free.
208
Textbook of Algebra
The word “Sequence” in Mathematics has same meaning as in ordinary English. A collection of objects listed in a
sequence means it has identified first member, second member, third member and so on. The most common examples
are depreciate values of certain commodity like car, machinery and amount deposits in the bank for a number of years.
Session 1
Sequence, Series, Progression
Sequence
Recursive Formula
A succession of numbers arranged in a definite order or
arrangement according to some well-defined law is called
a sequence.
Or
A sequence is a function of natural numbers (N) with
codomain is the set of real numbers (R) [complex numbers
(C )]. If range is subset of real numbers (complex numbers),
it is called a real sequence (complex sequence).
Or
A mapping f : N ® C, then f (n ) = t n , n Î N is called a
sequence to be denoted it by
{ f (1), f (2 ), f (3 ), ... } = {t 1 , t 2 , t 3 , ... } = {t n }.
The nth term of a sequence is denoted by
Tn , t n , a n , a(n ), u n , etc.
A formula to determine the other terms of the sequence in
terms of its preceding terms is known as recursive
formula.
For example,
If
T1 = 1 and Tn + 1 = 6 Tn , n Î N .
Then, T2 = 6 T1 = 6 × 1 = 6
T 3 = 6 T2 = 6 × 6 = 36
T 4 = 6 T 3 = 6 × 36 = 216 ...
Then, sequence is 1, 6, 36, 216,...
Remark
The sequence a1, a2, a3, ... is generally written as {an}.
For example ,
(i) 1, 3, 5, 7, ... is a sequence, because each term (except
first) is obtained by adding 2 to the previous term and
Tn = 2n - 1, n Î N .
Or
If T1 = 1, Tn + 1 = Tn + 2 , n ³ 1
(ii) 1, 2, 3, 5, 8, 13, ... is a sequence, because each term
(except first two) is obtained by taking the sum of
preceding two terms.
Or
If T1 = 1, T2 = 2, Tn + 2 = Tn + Tn + 1 , n ³ 1
(iii) 2, 3, 5, 7, 11, 13, 17, 19, ... is a sequence.
Here, we cannot express Tn , n Î N by an algebraic
formula.
Types of Sequences
There are two types of sequences
1. Finite Sequence
A sequence is said to be finite sequence, if it has finite
number of terms. A finite sequence is described by
a 1 , a 2 , a 3 , ... , a n or T1 , T2 , T 3 , ..., Tn , where n Î N.
For example
(i) 3, 5, 7, 9, …, 37
(ii) 2, 6, 18, 54, …, 4374
2. Infinite Sequence
A sequence is said to be an infinite sequence, if it has
infinite number of terms. An infinite sequence is described
by a 1 , a 2 , a 3 , ... or T1 , T2 , T 3 ,K
For example,
1 1 1
(i) 1, , , , …
3 9 27
1 1 1 1
(ii) 1, , , , , ...
2 4 8 16
209
Chap 03 Sequences and Series
Series
In a sequence, the sum of the directed terms is called a
series.
For example, If 1, 4, 7, 10, 13, 16,... is a sequence, then its
sum i.e., 1 + 4 + 7 + 10 + 13 + 16 +K is a series.
In general, if T1 , T2 , T 3 , ..., Tn ,... denote a sequence, then the
symbolic expression T1 + T2 + T 3 + ... + Tn + ... is called a
series associated with the given sequence.
Each member of the series is called its term.
In a seriesT1 + T2 + T 3 + ... + Tr + ..., the sum of first n terms
is denoted by S n . Thus,
Or
A sequence is said to be progression, if its terms increases
(respectively decreases) numerically.
For example, The following sequences are progression :
1 1 1 1
(i) 1, 3, 5, 7, ...
(ii) , , , , ...
2 6 18 54
1 1
1
(iii) 1, - , , - , K
(iv) 1, 8, 27, 256, ...
3 9 27
1
(v) 8, - 4, 2, - 1, , K
2
The sequences (iii) and (v) are progressions, because
n
S n = T1 + T2 + T 3 + ... + Tn = å Tr = å Tn
| 1| > -
r =1
If S n denotes the sum of n terms of a sequence.
Then,
S n - S n - 1 = (T1 + T2 + T 3 + ... + Tn )
- (T1 + T2 + ... + Tn - 1 ) = Tn
Thus,
Tn = S n - S n - 1
Types of Series
1
1
1
>
> > ...
3
9
27
1>
i.e.
1 1 1
> >
>K
3 9 27
| 8 | > | - 4 | > | 2 | > | - 1| >
and
8 > 4 >2 >1>
i.e.
There are two types of series
1. Finite Series
A series having finite number of terms is called a finite
series.
For example,
(i) 3 + 5 + 7 + 9 + ... + 21
(ii) 2 + 6 + 18 + 54 + ... + 4374
2. Infinite Series
A series having an infinite number of terms is called an
infinite series.
For example,
1 1 1
(i) 1 + + +
+ ...
3 9 27
1 1 1
(ii) 1 + + + + ...
2 4 8
Progression
If the terms of a sequence can be described by an explicit
formula, then the sequence is called a progression.
1
> ...
2
1
> ...
2
Remark
All the definitions and formulae are valid for complex numbers
in the theory of progressions but it should be assumed (if not
otherwise stated) that the terms of the progressions are real
numbers.
y Example 1. If f : N ® R, where f (n ) = an =
n
(2n + 1) 2
write the sequence in ordered pair form.
Sol. Here, an =
n
( 2n + 1) 2
On putting n = 1, 2, 3, 4, ... successively, we get
1
1
2
2
a1 =
= , a2 =
=
( 2 × 1 + 1) 2 9
(2 × 2 + 1)2 25
a3 =
3
( 2 × 3 + 1) 2
M
=
3
4
4
, a4 =
=
49
(2 × 4 + 1)2 81
M
M
1 2 3 4
Hence, we obtain the sequence , , , , ...
9 25 49 81
Now, the sequence in ordered pair form is
ìæ 1 ö æ 2 ö æ 3 ö æ 4 ö ü
í ç1, ÷, ç2, ÷, ç3, ÷, ç 4, ÷, ...ý
î è 9 ø è 25 ø è 49 ø è 81 ø þ
,
210
Textbook of Algebra
y Example 2. The Fibonacci sequence is defined by
an + 1
for
a1 = 1 = a 2 , an = an - 1 + an - 2 , n > 2. Find
an
n = 1, 2, 3, 4, 5.
Sol. Q
\
and
\
n
= a 2 + a1 = 1 + 1 = 2,
= a3 + a2 = 2 + 1 = 3
= a4 + a3 = 3 + 2 = 5
= a5 + a4 = 5 + 3 = 8
a
2
a
3 a
5
a
8
= 1, 3 = = 2 , 4 = , 5 = and 6 =
a2 1
a3 2 a4 3
a5 5
y Example 3. If the sum of n terms of a series is
2n 2 + 5n for all values of n, find its 7th term.
Sol. (i) On putting r = 1, 2, 3, 4, K, n in (r 2 + 2),
we get 3, 6, 11, 18, ... , (n 2 + 2)
n
Hence,
(ii) The r th term of series =
Hence, the given series can be written as
n
æ r ö
n
1 2 3 4
+ + + + ... +
= å ç
÷
n + 2 r = 1 èr + 2ø
3 4 5 6
First term of a sequence is 1 and the (n + 1) th term is obtained by adding (n + 1) to the nth term for all natural
numbers n, the 6th term of the sequence is
(b) 13
(d) 27
The first three terms of a sequence are 3, 3, 6 and each term after the second is the sum of two terms
preceding it, the 8th term of the sequence is
(a) 15
(c) 39
3.
6
(b) 3
(d) 7
If for a sequence {a n}, Sn = 2n 2 + 9n, where Sn is the sum of n terms, the value of a 20 is
(a) 65
(c) 87
5.
(b) 24
(d) 63
æ np ö
If a n = sin ç ÷ , the value of å a n2 is
è 6 ø
n =1
(a) 2
(c) 4
4.
r
.
r +2
Exercise for Session 1
(a) 7
(c) 21
2.
å(r 2 + 2) = 3 + 6 + 11 + 18 + K + (n 2 + 2)
r =1
Hence, T 7 = 4 ´ 7 + 3 = 31
1
n
1 2 3 4
in sigma
+ + + +K +
n +2
3 4 5 6
form.
Tn = Sn - Sn - 1 = (2n 2 + 5n ) - (2n 2 + n - 3) = 4n + 3
#L
in expanded form.
r =1
Sol. Given, Sn = 2n 2 + 5n
Þ
Sn - 1 = 2 (n - 1)2 + 5 (n - 1) = 2n 2 + n - 3
\
å(r 2 + 2)
(i) Write
(ii) Write the series
a1 = 1 = a 2
a3
a4
a5
a6
a2
a1
y Example 4.
(b) 75
(d) 97
5
If a1 = 2, a 2 = 3 + a1 and a n = 2 a n - 1 + 5 for n > 1, the value of
å a r is
r =2
(a) 130
(c) 190
(b) 160
(d) 220
Session 2
Arithmetic Progression (AP)
Types of Progression
Progressions are various types but in this chapter we will
studying only three special types of progressions which
are following :
1. Arithmetic Progression (AP)
2. Geometric Progression (GP)
3. Harmonic Progression (HP)
Arithmetic Progression (AP)
An arithmetic progression is a sequence in which the
difference between any term and its just preceding term
(i.e., term before it) is constant throughout. This constant
is called the common difference (abbreviated as CD) and is
generally denoted by ‘d’.
Or
An arithmetic progression is a sequence whose terms
increase or decrease by a fixed number. This fixed number
is called the common difference of the AP.
A finite or infinite sequence {t 1 , t 2 , t 3 , K, t n }
or {t 1 , t 2 , t 3 , ... } is said to be an arithmetic progression
(AP), if t k - t k - 1 = d , a constant independent of k, for
k = 2, 3, 4, ..., n or k = 2, 3, 4,K as the case may be :
The constant d is called the common difference of the AP.
d = t 2 - t 1 = t 3 - t 2 = ... = t n - t n - 1
i.e.
Remarks
1. If a be the first term and d be the common difference, then
AP can be written as
a, a + d, a + 2d,... , a + (n - 1) d,... ," n Î N.
Algorithm to determine whether a sequence is
an AP or not
Step I Obtain t n (the nth term of the sequence).
Step II Replace n by n - 1 in t n to get t n - 1 .
Step III Calculate t n - t n - 1 .
If t n - t n - 1 is independent of n, the given sequence is an
AP otherwise it is not an AP.
y Example 5.
(i) 1 , 3, 5, 7, ...
(ii) p , p + e p , p + 2e p , K
(iii) a , a - b , a - 2b , a - 3b , K
Sol. (i) Here, 2nd term – 1st term = 3rd term – 2nd term = ...
Þ 3 - 1 = 5 - 3 = ... = 2, which is a common
difference.
(ii) Here, 2nd term – 1st term = 3rd term – 2nd term = ...
Þ ( p + e p ) - p = ( p + 2e p ) - ( p + e p ) = ...
= e p, which is a common difference.
(iii) Here, 2nd term – 1st term = 3rd term – 2nd term = ...
Þ (a - b ) - a = (a - 2b ) - (a - b ) = ...
= - b, which is a common difference.
y Example 6. Show that the sequence < t n > defined by
t n = 5n + 4 is an AP, also find its common difference.
Sol. We have, t n = 5n + 4
On replacing n by (n - 1), we get
t n - 1 = 5 ( n - 1) + 4
Þ
t n - 1 = 5n - 1
\
t n - t n - 1 = (5n + 4 ) - (5n - 1) = 5
Clearly, t n - t n - 1 is independent of n and is equal to 5. So,
the given sequence is an AP with common difference 5.
2. If we add the common difference to any term of AP, we get
the next following term and if we subtract it from any term,
we get the preceding term.
y Example 7. Show that the sequence < t n > defined
by t n = 3n 2 + 2 is not an AP.
3. The common difference of an AP may be positive, zero,
negative or imaginary.
Sol. We have, t n = 3n 2 + 2
4. Constant AP common difference of an AP is equal to zero.
5. Increasing AP common difference of an AP is greater than
zero.
6. Decreasing AP common difference of an AP is less than
zero.
7. Imaginary AP common difference of an AP is imaginary.
On replacing n by (n - 1), we get
t n - 1 = 3 ( n - 1) 2 + 2
Þ
t n - 1 = 3n 2 - 6n + 5
t n - t n - 1 = (3n 2 + 2) - (3n 2 - 6n + 5)
= 6n - 3
Clearly, t n - t n - 1 is not independent of n and therefore it is
not constant. So, the given sequence is not an AP.
\
212
Textbook of Algebra
Remark
y Example 8. Find first negative term of the sequence
If the nth term of a sequence is an expression of first degree in n.
For example, tn = An + B, where A, B are constants, then that
sequence will be in AP for tn - tn - 1 = ( An + B) - [ A ( n - 1) + B]
= A [ n - ( n - 1)] = A = constant = Common difference or
coefficient of n in tn Students are advised to consider the above
point as a behaviour of standard result.
3
1
1
20, 19 , 18 , 17 , ...
4
2
4
Sol. The given sequence is an AP in which first term, a = 20
3
and common difference, d = - . Let the nth term of the
4
given AP be the first negative term. Then,
t n < 0 Þ a + ( n - 1) d < 0
æ 3ö
Þ
20 + (n - 1) ç - ÷ < 0
è 4ø
General Term of an AP
Let ‘a’ be the first term, ‘d’ be the common difference and ‘
l ’ be the last term of an AP having ‘n’ terms, where n Î N .
Then, AP can be written as a, a + d , a + 2d , ..., l - 2d , l - d , l
(i) nth Term of an AP from Beginning
1st term from beginning = t 1 = a = a + (1 - 1) d
2nd term from beginning = t 2 = a + d = a + (2 - 1) d
3rd term from beginning = t 3 = a + 2d = a + (3 - 1) d
M
M
M
M
n th term from beginning = t n = a + (n - 1) d , " n Î N
Hence, n th term of an AP from beginning
= t n = a + (n - 1) d = l [last term]
(ii) nth Term of an AP from End
1st term from end = t ¢1 = l = l - (1 - 1) d
2nd term from end = t ¢2 = l - d = l - (2 - 1) d
3rd term from end = t ¢ 3 = l - 2d = l - (3 - 1) d
M
M
M
M
¢
nth term from end = t n = l - (n - 1) d , " n Î N
Hence, n th term of an AP from end
[first term]
t ¢n = l - (n - 1) d = a
Now, it is clear that
t n + t ¢n = a + (n - 1) d + l - (n - 1) d = a + l
or
t n + t ¢n = a + l
i.e. In a finite AP, the sum of the terms equidistant from
the beginning and end is always same and is equal to the
sum of first and last term.
Remark
1. nth term is also called the general term.
2. If last term of AP is tn and common difference be d, then
terms of AP from end are tn, tn - d, tn - 2d, ...
3. If in a sequence, the terms an alternatively positive and
negative, then it cannot be an AP.
l-a
4. Common difference of AP =
, where, a = first term of AP,
n+ 1
l = last term of AP and n = number of terms of AP.
5. If tn, tn + 1, tn + 2 are three consecutive terms of an AP, then
2 tn + 1 = tn + tn + 2 . In particular, if a, b and c are in AP, then
2b = a + c .
Þ
80 - 3n + 3 < 0
83
2
or n > 27
n>
Þ
3
3
Þ
n = 28
Thus, 28th term of the given sequence is the first negative
term.
y Example 9. If the mth term of an AP is
nth term is
1
and the
n
1
, then find mnth term of an AP.
m
Sol. If A and B are constants, then r th term of AP is
t r = Ar + B
1
1
Given,
tm =
Þ Am + B =
n
n
1
1
and
Þ An + B =
tn =
m
m
1
From Eqs. (i) and (ii), we get A =
and B = 0
mn
1
× mn + 0 = 1
mn th term = t mn = Amn + B =
mn
Hence, mn th term of the given AP is 1.
…(i)
…(ii)
y Example 10. If | x - 1 |, 3 and | x - 3 | are first three
terms of an increasing AP, then find the 6th term of on AP.
Sol. Case I For x < 1,
| x - 1 | = - ( x - 1)
and
| x - 3 | = - ( x - 3)
\1 - x, 3 and 3 - x are in AP.
Þ
6=1- x +3- x
Þ
x = -1
Then, first three terms are 2, 3, 4,
which is an increasing AP.
\ 6th term is 7.
Case II For 1 < x < 3,
| x - 1| = x -1
and
| x - 3 | = - ( x - 3) = 3 - x
\x - 1, 3 and 3 - x are in AP.
Þ
6= x -1+3- x
Þ
6=2
[Qd = 1]
[impossible]
Chap 03 Sequences and Series
Case III For x > 3, | x - 1 | = x - 1 and | x - 3 | = x - 3
\ x - 1, 3 and x - 3 are in AP.
Þ
6= x -1+ x -3 Þ x =5
Then, first three terms are 4, 3, 2, which is a decreasing AP.
y Example 11. In the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4,
4, ..., where n consecutive terms have the value n, find
the 150th term of the sequence.
Sol. Let the 150th term = n
Then, 1 + 2 + 3 + ... + (n - 1) < 150 < 1 + 2 + 3 + ... + n
( n - 1) n
n ( n + 1)
Þ
< 150 <
2
2
Þ
n (n - 1) < 300 < n (n + 1)
Taking first two members
n (n - 1) < 300 Þ n 2 - n - 300 < 0
announced the answer almost at once. The teacher
overawed at this asked Gauss to explain how he got this
answer. Gauss explained that he had added these numbers
in pairs as follows
(1 + 100 ), (2 + 99 ), (3 + 98 ) , K
100
There are
= 50 pairs. The answer can be obtained by
2
multiplying 101 by 50 to get 5050.
Sum of n Terms of an AP
2
1
1ö
æ
çn - ÷ < 300 +
è
4
2ø
1
1201
Þ
0<n < +
2
2
Þ
0 < n < 17.8
and taking last two members,
n (n + 1) > 300
2
1
1ö
æ
Þ
n
+
ç
÷ > 300 +
è
4
2ø
1
1201
\
n>- +
2
2
Þ
n > 16. 8
From Eqs. (i) and (ii), we get
16.8 < n < 17.8
Þ
n = 17
Þ
…(i)
…(ii)
y Example 12. If a1 , a 2 , a 3 , a 4 and a 5 are in AP with
5
common difference ¹ 0, find the value of å ai when
i =1
a 3 = 2.
Let ‘a’ be the first term, ‘d’ be the common difference, ‘l ’
be the last term of an AP having n terms and S n be the
sum of n terms, then
S n = a + (a + d ) + (a + 2d ) + K + (l - 2d ) + (l - d ) + l …(i)
Reversing the right hand terms
S n = l + (l - d ) + (l - 2d ) + ... + (a + 2d ) + (a + d ) + a …(ii)
On adding Eqs. (i) and (ii), we get
2 S n = (a + l ) + (a + l ) + (a + l ) + ...
+ (a + l ) + (a + l ) + (a + l )
= (a + l ) + (a + l ) + ... upto n terms = n (a + l )
n
…(iii)
S n = (a + l )
\
2
Now, if we substitute the value of l viz., l = a + (n - 1) d , in
this formula, we get
n
n
S n = [a + a + (n - 1) d ] = [2a + (n - 1)d ]
2
2
n
S n = [2a + (n - 1) d ]
\
2
If we substitute the value of a viz.,
l = a + (n - 1) d
or
Sol. Qa1, a 2 , a 3 , a 4 and a 5 are in AP, we have
a1 + a 5 = a 2 + a 4 = a 3 + a 3
a1 + a 5 = a 2 + a 4 = 4
a1 + a 2 + a 3 + a 4 + a 5 = 4 + 2 + 4 = 10
Þ
213
[Qt n + t ¢n = a + l ]
[Qa 3 = 2]
If we substitute the value of a + l viz.,
t n + t ¢ n = a + l in Eq. (iii), then
n
S n = (t n + t ¢ n )
2
5
åai = 10
i =1
Sum of a Stated Number of
Terms of an Arithmetic Series
More than 200 yr ago, a class of German School Children
was asked to find the sum of all integers from 1 to 100
inclusive. One boy in the class, an eight year old named
Carl Fredrick Gauss (1777-1855) who later established
his reputation as one of the greatest Mathematicians
a = l - (n - 1) d in Eq. (iii), then
n
S n = [ 2l - (n - 1) d ]
2
Corollary I Sum of first n natural numbers
i.e. 1 + 2 + 3 + 4 + ... + n
Here,
\
a = 1 and d = 1
n
S = [2 × 1 + (n - 1) × 1]
2
n (n + 1)
=
2
214
Textbook of Algebra
Corollary II Sum of first n odd natural numbers
i.e., 1 + 3 + 5 + ...
Here,
a =1
and
d =2
n
S = [2 × 1 + (n - 1) × 2 ] = n 2
\
2
Corollary III If sum of first n terms is S n , then sum of next
m terms is S m + n - S n .
Important Results with Proof
1. If S n , t n and d are sum of n terms, nth term
and common difference of an AP respectively,
then
d = tn - tn -1
[n ³ 2]
[n ³ 2]
t n = Sn - Sn - 1
d = Sn - 2 Sn - 1 + Sn - 2
[n ³ 3]
Proof
Q
Sn = t1 + t2 + t 3 +K + tn - 1 + tn
Þ
Sn = Sn - 1 + tn
\
tn = Sn - Sn - 1
but
d = tn - tn -1
= (S n - S n - 1 ) - (S n - 1 - S n - 2 )
\
d = Sn - 2 Sn - 1 + Sn - 2
2. A sequence is an AP if and only if the sum of
its n terms is of the form An 2 + Bn, where A
and B are constants independent of n.
In this case, the nth term and common difference of
the AP are A (2n - 1) + B and 2A, respectively.
n
Sum of first n terms = [2a + (n - 1) d ]
2
=
where, A =
\
Corollary Q
S n = An 2 + Bn
\
t n = A (2n - 1) + B
t n = A (replacing n 2 by 2n - 1) + coefficient of n
and
d = 2A
i.e.
d =2
[coefficient of n 2 ]
Sn
tn
d
1.
5n2 + 3n
5 (2n - 1) + 3 = 10n - 2
2.
- 7n2 + 2n
- 7 (2n - 1) + 2
= - 14 n + 9
- 14
3.
- 9n2 - 4 n
- 9 (2n - 1) - 4
= - 18n + 5
- 18
4.
4 n2 - n
4 (2n - 1) - 1 = 8n - 5
10
8
3. If S n = an 2 + bn + c , where S n denotes the sum
of n terms of a series, then whole series is not
an AP. It is AP from the second term onwards.
Proof As
S n = an 2 + bn + c for n ³ 1, we get
S n - 1 = a (n - 1) 2 + b (n - 1) + c for n ³ 2
S n - 1 = A (n - 1) 2 + B (n - 1)
tn = Sn - Sn - 1
2
d
d
,B =a 2
2
Hence, S n = An 2 + Bn, where A and B are constants
independent of n.
Hence, the converse is true.
Proof As S n = An 2 + Bn
\
dn 2 æ
dö
+ ç a - ÷ n = An 2 + Bn,
è
2
2ø
2
= ( An + Bn ) - [ A (n - 1) + B (n - 1)]
= A [n 2 - (n - 1) 2 ] + B
t n = A (2n - 1) + B
Þ
t n - 1 = A [2 (n - 1) - 1] + B
= A (2n - 3 ) + B
Now, t n - t n - 1 = [ A (2n - 1) + B ] - [ A (2n - 3 ) + B ]
[a constant]
= 2A
Hence, the sequence is an AP.
Conversely, consider an AP with first term a and
common difference d.
Now,
tn = Sn - Sn - 1
Þ
t n = a (2n - 1) + b, n ³ 2
\
t n - 1 = a [2 (n - 1) - 1] + b, n ³ 3
Þ
t n - 1 = a (2n - 3 ) + b, n ³ 3
\
t n - t n - 1 = 2a = constant, n ³ 3
\
t 3 - t 2 = t 4 - t 3 = t 5 - t 4 = ...
But
t 2 - t 1 = (S 2 - S 1 ) - S 1 = S 2 - 2 S 1
= ( 4a + 2b + c ) - 2 (a + b + c )
= (2a - c )
[QS 1 = t 1 ]
\
t2 - t1 ¹ t 3 - t2
Hence, the whole series is not an AP. It is AP from
the second term onwards.
Chap 03 Sequences and Series
Ratio of Sums is Given
1. If ratio of the sums of m and n terms of an AP
is given by
Sm
Am 2 + Bm
=
Sn
An 2 + Bn
\
S m = ( Am + Bm ) k ,
t m = S m - S m - 1 = [ A (2m - 1) + B ]k
t n = S n - S n - 1 = [ A (2n - 1) + B ] k
t m A (2m - 1) + B
=
tn
A (2n - 1) + B
y Example 13. The ratio of sums of m and n terms of
an AP is m 2 : n 2 . The ratio of the m th and nth terms is
(a) (2m + 1) : (2n - 1)
(c) (2m - 1) : (2n - 1)
Sol. (c) Here,
\
Þ
(b) m : n
(d) None of these
Sm m 2
= 2
Sn
n
t m ( 2m - 1)
=
tn
( 2n - 1)
t m : t n = ( 2m - 1) : ( 2n - 1)
[QA = 1, B = 0]
2. If ratio of the sums of n terms of two AP’s is
given by
Sn
An + B
=
S ¢n Cn + D
where, A, B, C , D are constants and A, C ¹ 0
\ S n = n ( An + B ) k, S ¢n = n (Cn + D ) k
Þ t n = [ A (2n - 1) + B ] k , t n¢ = [C (2n - 1) + D ] k
Þ d = t n - t n - 1 = 2 A , d ¢ = t n¢ - t n¢ -1 = 2C
t n A (2n - 1) + B
d A
and
\
=
=
t n¢ C (2n - 1) + D
d¢ C
Note If A = 0, C = 0
S
B
t
B
d 0
and
Then, n = Þ n =
= = not defined
S n¢ D
t ¢n
D
d¢ 0
Remark
an + b
tn
=
¢
tn
cn + d
where, a, b, c, d are constants and a, c ¹ 0, then
n + 1ö
a æç
÷+ b
è 2 ø
Sn
=
n + 1ö
S ¢n
c æç
÷+ d
è 2 ø
If
Sn : Sn¢ = (7n + 1) : ( 4n + 17 )
A = 7, B = 1, C = 4 and D = 17
7 (2n - 1) + 1 14n - 6
tn
=
=
t ¢n 4 (2n - 1) + 17 8n + 13
Here,
d
A 7
=
=
d¢ C 4
and
S n = ( An + Bn ) k
\
Sol. Given,
2
2
Þ
y Example 14. The sums of n terms of two arithmetic
progressions are in the ratio (7n + 1) :(4n + 17 ). Find the
ratio of their nth terms and also common differences.
\
where A, B are constants and A ¹ 0.
215
Hence,
t n : t n¢ = (14n - 6) : (8n + 13) and d : d ¢ = 7 : 4
y Example 15. The sums of n terms of two AP’s are in
the ratio ( 3n - 13) :( 5n + 21). Find the ratio of their 24th
terms.
Sol. Given, Sn : Sn¢ = (3n - 13) : (5n + 21)
Here,
A = 3, B = - 13, C = 5 and D = 21
\
\
t 24
t 24 3 (2 ´ 24 - 1) - 13 128 1
=
=
=
t ¢24 5 (2 ´ 24 - 1) + 21 256 2
: t ¢24 = 1 : 2
y Example 16. How many terms of the series
2
1
20 + 19 + 18 + ... must be taken to make 300?
3
3
Explain the double answer.
Sol. Here, given series is an AP with first term a = 20 and the
2
common difference, d = - .
3
Let the sum of n terms of the series be 300.
n
Then,
Sn = {2a + (n - 1) d }
2
nì
æ 2 öü
Þ
300 = í2 ´ 20 + (n - 1) ç - ÷ý
è 3 øþ
2î
n
Þ
300 = {60 - n + 1}
3
2
Þ
n - 61n + 900 = 0
Þ
(n - 25) (n - 36) = 0
Þ
n = 25 or n = 36
\ Sum of 25 terms = Sum of 36 terms = 300
Explanation of double answer
Here, the common difference is negative, therefore terms
æ -2 ö
go on diminishing and t 31 = 20 + (31 - 1) ç ÷ = 0 i.e., 31st
è 3 ø
term becomes zero. All terms after 31st term are negative.
These negative terms (t 32 , t 33 , t 34 , t 35 , t 36 ) when added to
positive terms (t 26 , t 27 , t 28 , t 29 , t 30 ), they cancel out each
other i.e., sum of terms from 26th to 36th terms is zero.
Hence, the sum of 25 terms as well as that of 36 terms is
300.
216
Textbook of Algebra
but given
y Example 17. Find the arithmetic progression
consisting of 10 terms, if the sum of the terms
occupying the even places is equal to 15 and the sum
1
of those occupying the odd places is equal to 12 .
2
Þ
Þ
Þ
Sol. Let the successive terms of an AP be t 1, t 2 , t 3 , ..., t 9 , t 10 .
\
By hypothesis,
t 2 + t 4 + t 6 + t 8 + t 10 = 15
5
(t 2 + t 10 ) = 15
Þ
2
Þ
t 2 + t 10 = 6
Þ
( a + d ) + ( a + 9d ) = 6
Þ
2a + 10d = 6
1
and
t 1 + t 3 + t 5 + t 7 + t 9 = 12
2
5
25
(t 1 + t 9 ) =
Þ
2
2
Þ
t1 + t 9 = 5
Þ
a + a + 8d = 5
Þ
2a + 8d = 5
1
1
From Eqs. (i) and (ii), we get d = and a =
2
2
1
1
1
Hence, the AP is , 1, 1 , 2, 2 , K
2
2
2
…(i)
t 1 + t 5 + t 10 + t 15 + t 20 + t 24 = 225
(t 1 + t 24 ) + (t 5 + t 20 ) + (t 10 + t 15 ) = 225
3 (t 1 + t 24 ) = 225
t 1 + t 24 = 75
24
S 24 =
(t 1 + t 24 ) = 12 ´ 75 = 900
2
y Example 20. If (1 + 3 + 5 + ... + p ) + (1 + 3 + 5 + ... + q )
= (1 + 3 + 5 + ... + r ), where each set of parentheses
contains the sum of consecutive odd integers as
shown, then find the smallest possible value of
p + q + r (where, p > 6).
Sol. We know that, 1 + 3 + 5 + ... + (2n - 1) = n 2
Thus, the given equation can be written as
2
2
æ p + 1ö
æq + 1ö
ær + 1ö
ç
÷ +ç
÷ =ç
÷
è 2 ø
è 2 ø
è 2 ø
Þ
…(ii)
y Example 18. If N, the set of natural numbers is
partitioned into groups S 1 = {1}, S 2 = {2, 3},
S 3 = {4, 5, 6}, ..., find the sum of the numbers in S 50 .
Sol. The number of terms in the groups are 1, 2, 3, ...
Q The number of terms in the 50th group = 50
The last term of 1st group = 1
\
The last term of 2nd group = 3 = 1 + 2
The last term of 3rd group = 6 = 1 + 2 + 3
M
M
M
M
M
M
The last term of 49th group = 1 + 2 + 3 + ... + 49
\ First term of 50th group = 1 + (1 + 2 + 3 + ... + 49 )
49
(1 + 49 ) = 1226
=1+
2
50
{2 ´ 1226 + (50 - 1) ´ 1}
S 50 =
\
2
= 25 ´ 2501 = 62525
y Example 19. Find the sum of first 24 terms of on AP
t 1 , t 2 , t 3 , ... , if it is known that
t 1 + t 5 + t 10 + t 15 + t 20 + t 24 = 225.
Sol. We know that, in an AP the sums of the terms equidistant
from the beginning and end is always same and is equal
to the sum of first and last term.
Then, t 1 + t 24 = t 5 + t 20 = t 10 + t 15
2
( p + 1) 2 + ( q + 1) 2 = ( r + 1) 2
Therefore, ( p + 1, q + 1, r + 1) form a Pythagorean triplet as
p > 6Þp + 1 > 7
The first Pythagorean triplet containing a number > 7 is
(6, 8, 10).
Þ
p + 1 = 8, q + 1 = 6, r + 1 = 10
Þ
p + q + r = 21
Properties of Arithmetic Progression
1. If a 1 , a 2 , a 3 , ... are in AP with common difference d,
then a 1 ± k, a 2 ± k, a 3 ± k, ... are also in AP with
common difference d.
2. If a 1 , a 2 , a 3 , ... are in AP with common difference d,
a a a
then a 1 k, a 2 k, a 3 k, ... and 1 , 2 , 3 , ... are also in AP
k k k
d
(k ¹ 0 ) with common differences kd and ,
k
respectively.
3. If a 1 , a 2 , a 3 , .. and b 1 , b 2 , b 3 , .. . are two AP’s with
common differences d 1 and d 2 , respectively. Then,
a 1 ± b 1 , a 2 ± b 2 , a 3 ± b 3 , ... are also in AP with
common difference (d 1 ± d 2 ).
4. If a 1 , a 2 , a 3 , ... and b 1 , b 2 , b 3 , ... are two AP’s with
common differences d 1 and d 2 respectively, then
a a a
a 1 b 1 , a 2 b 2 , a 3 b 3 , ... and 1 , 2 , 3 ,K are not in AP.
b1 b2 b 3
5. If a 1 , a 2 , a 3 , ..., a n are in AP, then
ar - k + ar + k
ar =
, " k, 0 £ k £ n - r
2
Chap 03 Sequences and Series
6. If three numbers in AP whose sum is given are to be
taken as a - b, a, a + b and if five numbers in AP
whose sum is given, are to be taken as a - 2b, a - b,
a, a + b, a + 2b, etc.
In general, If (2 r + 1) numbers in AP whose sum is
given, are to be taken as (r Î N ).
a - r b, a - (r - 1) b, ..., a - b, a , a + b, ... ,
a + (r - 1) b, a + rb
Remark
1. Sum of three numbers = 3a
Sum of five numbers = 5a
M
M
M
M
M
Sum of ( 2r + 1) numbers = ( 2r + 1) a
2. From given conditions, find two equations in a and b and
then solve them. Now, the numbers in AP can be obtained.
7. If four numbers in AP whose sum is given, are to be
taken as
a - 3b, a - b, a + b, a + 3b and if six numbers in AP,
whose sum is given are to be taken as a - 5b, a - 3b,
a - b, a + b, a + 3b, a + 5 b, etc.
In general If 2 r numbers in AP whose sum is given,
are to be taken as (r Î N ).
a - (2 r - 1) b, a - (2 r - 3 ) b,..., a - 3 b, a - b,
a + b, a + 3 b, ..., a + (2 r - 3 ) b, a + (2 r - 1) b
Remark
1. Sum of four numbers = 4 a
Sum of six numbers = 6 a
M
M
M
M
M
Sum of 2r numbers = 2ra
2. From given conditions, find two conditions in a and b and
then solve them. Now, the numbers in AP can be obtained.
y Example 21. If S 1 , S 2 , S 3 , ..., S p are the sums of
n terms of p AP’s whose first terms are 1, 2, 3, ..., p and
common differences are 1, 2, 3, ..., (2p - 1) respectively,
1
show that
S 1 + S 2 + S 3 + .... + S p = np (np + 1).
2
Sol. Q1, 2, 3, ..., p are in AP.
Then, 2×1, 2×2, 2×3, K, 2p are also in AP.
…(i)
[multiplying 2 to each term]
and 1, 3, 5, ..., (2p - 1) are in AP.
217
From Eq. (iii),
n
n
{2 × 1 + (n - 1) × 1}, {2 × 2 + (n - 1) × 3},
2
2
n
{2 × 3 + (n - 1) × 5}, …,
2
n
{2p + (n - 1) (2p - 1)} are also in AP
2
n
[multiplying to each term]
2
i.e. S1, S 2 , S 3 , ..., S p are in AP.
p
\ S1 + S 2 + S 3 + ... + S p = {S1 + S p }
2
p ìn
n
ü
= í [2 × 1 + (n - 1) × 1] + [2 × p + (n - 1) (2p - 1)]ý
2 î2
2
þ
np
=
{2 + (n - 1) + 2p + (n - 1) (2p - 1)}
4
np
1
=
(2np + 2) = np (np + 1)
2
4
Aliter
n ( n + 1)
Here, S1 = 1 + 2 + 3 + ... upto n terms =
2
n
S 2 = 2 + 5 + 8 + ... upto n terms = [ 2 × 2 + (n - 1)3]
2
n (3n + 1)
=
2
n (5n + 1)
Similarly, S 3 = 3 + 8 + 13 + ... upto n terms =
, etc.
2
Now, S1 + S 2 + S 3 + ... + S p
n ( n + 1) n ( 3n + 1) n ( 5n + 1)
=
+
+
+ ... upto p terms
2
2
2
n
= [(n + 3n + 5n + ... upto p terms)
2
+ (1 + 1 + 1 + ... upto p terms)]
n ép
ù
= ê (2n + ( p - 1) 2n ) + p ú
2 ë2
û
np
1
=
[n + n ( p - 1) + 1] = np (np + 1)
2
2
y Example 22. Let a and b be roots of the equation
x 2 - 2x + A = 0 and let g and d be the roots of the
equation x 2 - 18 x + B = 0. If a < b < g < d are in
arithmetic progression, then find the values of A and B.
Sol. Qa , b, g , d are in AP.
Then, (n - 1) × 1, (n - 1) ×3, (n - 1)×5, ..., (n - 1) (2p - 1) are
also in AP.
…(ii)
[multiplying (n - 1) to each term]
From Eqs. (i) and (ii), we get
2×1 + (n - 1) ×1, 2×2 + (n - 1) ×3, 2×3 + (n - 1) ×5, K,
Let
…(iii)
2p + (n - 1) (2p - 1) are also in AP.
[adding corresponding terms of Eqs. (i) and (ii)]
and
Given,
Þ
Þ
b = a + d, g = a + 2d, d = a + 3d, d > 0
[here, sum of a , b, g , d is not given]
a + b = 2, ab = A
2a + d = 2, ab = A
…(i)
g + d = 18, gd = B
2a + 5d = 18, gd = B
…(ii)
218
Textbook of Algebra
From Eqs. (i) and (ii), we get
d = 4, a = - 1
\
b = 3, g = 7, d = 11
Þ
A = ab = ( - 1) (3) = - 3
and
B = gd = (7 )(11) = 77
y Example 23. The digits of a positive integer having
three digits are in AP and their sum is 15. The number
obtained by reversing the digits is 594 less than the
original number. Find the number.
Sol. Let the digit in the unit’s place be a - d , digit in the ten’s
place be a and the digit in the hundred’s place be a + d .
Sum of digits = a - d + a + a + d = 15
[given]
Þ
3a = 15
…(i)
\
a=5
\ Original number = (a - d ) + 10a + 100 (a + d )
= 111a + 99d = 555 + 99d
and number formed by reversing the digits
= (a + d ) + 10a + 100 (a - d )
= 111a - 99d = 555 - 99d
Given, (555 + 99d ) - (555 - 99d ) = 594 Þ 198d = 594
\
d =3
Hence, original number = 555 + 99 ´ 3 = 852
y Example 24. If three positive real numbers are in AP
such that abc = 4, then find the minimum value of b.
Sol. Qa, b, c are in AP.
Let a = A - D , b = A , c = A + D
Then,
a = b - D, c = b + D
Now,
abc = 4
(b - D ) b (b + D ) = 4
Þ
b (b 2 - D 2 ) = 4
b2 - D 2 < b2
Þ
b (b 2 - D 2 ) < b 3 Þ 4 < b 3
Þ
b > ( 4 )1/ 3 or b > (2)2 / 3
\
Hence, the minimum value of b is (2)2 / 3 .
y Example 25. If a, b , c , d are distinct integers form an
increasing AP such that d = a 2 + b 2 + c 2 , then find the
value of a + b + c + d .
Sol. Here, sum of numbers i.e., a + b + c + d is not given.
Let b = a + D , c = a + 2D , d = a + 3D , " D Î N
According to hypothesis,
a + 3D = a 2 + ( a + D ) 2 + ( a + 2D ) 2
Þ
5D 2 + 3 (2a - 1) D + 3a 2 - a = 0
2
\
D=
…(i)
2
- 3 (2a - 1) ± 9 (2a - 1) - 20 (3a - a )
10
- 3 (2a - 1) ± ( - 24a 2 - 16a + 9 )
10
Now,
- 24a 2 - 16a + 9 ³ 0
Þ
24a 2 + 16a - 9 £ 0
1
70
1
70
Þ
- £a£ - +
3
3
3
12
Þ
a = - 1, 0
[ Qa Î I ]
3
When a = 0 from Eq. (i), D = 0, (not possible Q D Î N ) and
5
for a = - 1
4
From Eq. (i),
D = 1,
5
\
D =1
[Q D Î N ]
\
a = - 1, b = 0, c = 1, d = 2
Then, a + b + c + d = - 1 + 0 + 1 + 2 = 2
=
Chap 03 Sequences and Series
#L
1.
Exercise for Session 2
If nth term of the series 25 + 29 + 33 + 37 + ... and 3 + 4 + 6 + 9 + 13 + K are equal, then n equals
(a) 11
2.
20
5r + 3
8
5
20
5r 2 + 3
(d) -13
(c) 1 : 3
(d) 3 : 1
(c) 0
(d)
1
2
5
4
(c)
2
3
(d)
1
3
The sum of first 2n terms of an AP is a and the sum of next n terms is b, its common difference is
a - 2b
3n 2
(b)
2b - a
3n 2
(c)
a - 2b
3n
(d)
2b - a
3n
The sum of three numbers in AP is - 3 and their product is 8, then sum of squares of the numbers is
(b) 10
(c) 12
Let Sn denote the sum of n terms of an AP, if S2n
(b) 6
(d) 21
S
= 3Sn, then the ratio 3n is equal to
Sn
(c) 16
(d) 12
The sum of the products of the ten numbers ± 1, ± 2, ± 3, ± 4, ± 5 taking two at a time, is
(a) - 65
11.
(b) 2 : 1
(b)
(a) 9
10.
(d)
(c) - 12
(b) -1
(a) 9
9.
(c) 20 (5r + 3)
The 6th term of an AP is equal to 2, the value of the common difference of the AP which makes the product
a1a4a 5 least is given by
(a)
8.
20
5r - 3
If the p th, q th and rth terms of an AP are a, b and c respectively, the value of a (q - r ) + b (r - p ) + c ( p - q ) is
(a)
7.
(d) 14
If the 9th term of an AP is zero, the ratio of its 29th and 19th terms is
(a) 1
6.
(b)
(b) -1
(a) 1 : 2
5.
(c) 13
In a certain AP, 5 times the 5th term is equal to 8 times the 8th term, its 13th term is
(a) 0
4.
(b) 12
1
7
1 20
The rth term of the series 2 + 1 + 1 +
+ ... is
2
13
9 23
(a)
3.
219
(c) - 55
(b) 165
(d) 95
If a1, a 2, a 3, ..., a n are in AP, where a i > 0 for all i , the value of
1
+
a1 + a 2
(a)
1
a1 + an
1
+ ... +
a2 + a3
(b)
1
is
an - 1 + an
1
a1 - an
(c)
n
a1 - an
(d)
(n - 1)
a1 + an
Session 3
Geometric Sequence or Geometric Progression (GP)
Geometric Sequence or
Geometric Progression (GP)
A geometric progression is a sequence, if the ratio of any
term and its just preceding term is constant throughout.
This constant quantity is called the common ratio and is
generally denoted by ‘ r’.
Or
A geometric progression (GP) is a sequence of numbers,
whose first term is non-zero and each of the term is
obtained by multiplying its just preceding term by a
constant quantity. This constant quantity is called
common ratio of the GP.
Let t 1 , t 2 , t 3 , ..., t n ; t 1 , t 2 , t 3 , ... be respectively a finite or an
infinite sequence. Assume that none of t n¢ ’s is 0 and that
tk
= r , a constant (i.e., independent of k).
tk -1
For k = 2, 3, 4, ..., n or k = 2, 3, 4, ... as the case may be. We
then call {t k } nk = 1 or {t k } k¥= 1 as the case may be a
geometric progression (GP). The constant ratio r is called
the common ratio (CR) of the GP.
t
t
t
i.e., r = 2 = 3 = ... = n
t1 t2
tn -1
If
t 1 = a is the first term of a GP, then
t 2 = ar , t 3 = t 2 r = ar 2 , t 4 = t 3 r = ar 3 ,K ,
t n = t n - 1 r = ar n - 1
It follows that, given that first term a and the common ratio
r, the GP can be rewritten as
a, ar , ar 2 , ..., ar n - 1 (standard GP) or a, ar , ar 2 ,..., ar n - 1 ,…
(standard GP)
according as it is finite or infinite.
Important Results
1. In a GP, neither a = 0 nor r = 0.
2. In a GP, no term can be equal to ‘0’.
3. If in a GP, the terms are alternatively positive and negative,
then its common ratio is always negative.
4. If we multiply the common ratio with any term of GP, we get
the next following term and if we divide any term by the
common ratio, we get the preceding term.
5. The common ratio of GP may be positive, negative or
imaginary.
6. If common ratio of GP is equal to unity, then GP is known as
Constant GP.
7. If common ratio of GP is imaginary or real, then GP is known
as Imaginary GP.
8. Increasing and Decreasing GP
For a GP to be increasing or decreasing r > 0. If r < 0, terms
of GP are alternatively positive and negative and so neither
increasing nor decreasing.
a
a>0
a>0
a<0
a<0
r
0 < r <1
r >1
r >1
0 < r <1
Result
Decreasing
Increasing Decreasing Increasing
y Example 26.
(i) 1, 2, 4, 8, 16, …
1 1
(ii) 9, 3, 1, , , …
3 9
1
(iv) - 8, - 4, - 2 , - 1, - , ...
2
(v) 5 , - 10, 20, K
(vi) 5, 5, 5, 5,...
(vii) 1, 1 + i , 2i, - 2 + 2i , ...; i = -1
(iii) - 2 , - 6 , - 18,K
Sol. (i) Here, a = 1
2 4 8 16
= = =
= ... = 2 i.e. a = 1 and r = 2
1 2 4
8
Increasing GP (a > 0, r > 1)
(ii) Here, a = 9
1 1
3 1 3 9
1
1
i.e. a = 9, r =
and r = = = = = ... =
3
9 3 1 1
3
3
Decreasing GP (a > 0, 0 < r < 1)
(iii) Here, a = - 2
-6 -18
and r =
= ... = 3
=
-6
-2
i.e. a = - 2, r = 3
Decreasing GP (a < 0, r > 1)
(iv) Here, a = - 8
1
1
-4 -2 -1
and r =
=
=
= 2 = .... =
2
-8 -4 -2 -1
1
i.e. a = - 8, r =
2
Increasing GP (a < 0, 0 < r < 1)
and r =
Chap 03 Sequences and Series
(v) Here, a = 5
-10
20
and r =
=
= ... = - 2 i.e., a = 5, r = - 2
5
-10
Neither increasing nor decreasing (r < 0)
(vi) Here, a = 5
5 5 5
and r = = = = ... = 1 i.e., a = 5, r = 1
5 5 5
Constant GP (r = 1)
(vii) Here, a = 1
and r =
1+i
2i
- 2 + 2i
=
=
= ...
1
1+i
2i
2i (1 - i )
(- 1 + i) i
= (1 + i ) =
=
= ...
(1 + i ) (1 - i )
i2
= ( 1 + i ) = ( i + 1) = ( 1 + i ) = K
i.e., a = 1, r = 1 + i
Imaginary GP (r = imaginary)
y Example 27. Show that the sequence < t n > defined
2 2n - 1
for all values of n Î N is a GP. Also, find
by t n =
3
its common ratio.
22n - 1
3
On replacing n by n - 1, we get
Sol. We have, t n =
tn - 1 =
Clearly,
2
2n - 3
3
tn
tn - 1
Þ
22n - 1
= 2n3- 3 = 22 = 4
2
3
tn
is independent of n and is equal to 4. So, the
tn - 1
given sequence is a GP with common ratio 4.
y Example 28. Show that the sequence < t n >
defined by t n = 2 × 3n + 1 is not a GP.
Sol. We have, t n = 2 × 3n + 1
On replacing n by (n - 1) in t n , we get
t n - 1 = 2 × 3n - 1 + 1
tn - 1 =
Þ
tn
\
Clearly,
tn - 1
tn
tn - 1
=
(2 × 3n + 3)
3
( 2 × 3n + 1)
n
( 2 × 3 + 3)
3
=
3 ( 2 × 3n + 1)
( 2 × 3n + 3)
is not independent of n and is therefore not
constant. So, the given sequence is not a GP.
221
General Term of a GP
Let ‘a’ be the first term, ‘r’ be the common ratio and ‘l ’ be
the last term of a GP having ‘n’ terms. Then, GP can be
l l
written as a, ar , ar 2 , ..., , , l
r2 r
(i) nth Term of a GP from Beginning
1st term from beginning = t 1 = a = ar 1 - 1
2nd term from beginning = t 2 = ar = ar 2 - 1
3rd term from beginning = t 3 = ar 2 = ar 3 - 1
M
M
M
M
M
nth term from beginning = t n = ar
n -1
,"nÎN
Hence, n th term of a GP from beginning
t n = ar n -1 = l
[last term]
(ii) n th Term of a GP from End
1st term from end = t ¢ 1 = l =
l
1 -1
r
l
l
2nd term from end = t ¢ 2 = =
2
r r -1
l
l
3rd term from end = t ¢ 3 =
=
2
3 -1
r
r
M
M
M
M
l
,"n ÎN
nth term from end = t ¢ n =
n -1
r
Hence, nth term of a GP from end
l
= t¢n =
=a
n -1
r
Now, it is clear that t k ´ t ¢k = ar k - 1 ´
[first term]
l
r
k -1
=a ´l
or
t k ´ t ¢k = a ´ l, " 1 £ k £ n
i.e. in a finite GP of n terms, the product of the k th
term from the beginning and the k th term form the
end is independent of k and equals the product of the
first and last terms.
Remark
1. nth term is also called the general term.
2. If last term of GP be tn and CR is r, then terms of GP from
t t
end are tn, n , n2 , K
r r
3. If in a GP, the terms are alternatively positive and negative,
then its common ratio is always negative.
222
Textbook of Algebra
4. If a and l represent first and last term of a GP respectively,
1
ön - 1
We get,
l
then common ratio of GP = r = æç ÷
è aø
\
5. If tn , tn + 1, tn + 2 are three consecutive terms of a GP, then
tn + 1 tn + 2
=
Þ tn2 + 1 = tn tn + 2
tn
tn + 1
b c
In particular, if a, b, c are in GP, then = Þ b2 = ac
a b
b2 c 2
On squaring, 2 = 2
a
b
Hence, a2, b2, c 2 are also in GP.
y Example 29. If first term of a GP is a, third term is b
and (n + 1)th term is c . The (2n + 1)th term of a GP is
b
a
(a) c
(b)
bc
a
(c) abc
(d)
c2
a
b
a
c
n
Þ r =
a
b = ar 2 Þ r =
Also,
c = ar n
2
(c)
y Example 32. The 1025th term in the sequence 1, 22,
4444, 88888888, ... is
(b) 2 10
(d) 2 12
(2 - 1) (1 + 2 + 22 + 23 + K + 2n - 1 )
< 1025
( 2 - 1)
£
Þ
(2 - 1) (1 + 2 + 22 + 23 + ... + 2n )
( 2 - 1)
2n - 1 < 1025 £ 2n + 1 - 1 Þ 2n < 1026 £ 2n + 1
or
n +1
³ 1026 > 1024
n +1
> 210 Þ n + 1 > 10
2
Þ
(b) pq
2
…(i)
\
p
q
n > 9 \ n = 10
[which is always satisfy Eq. (i)]
Hence, (b) is the correct answer.
(d) None of these
Sol. Let a be the first term and r be the common ratio, then
…(i)
t m + n = p Þ ar m + n - 1 = p
t m - n = q Þ ar m - n - 1 = q
…(ii)
From Eqs. (i) and (ii), we get
ar m + n - 1 ´ ar m - n - 1 = p ´ q
Þ
a 2 r 2 m - 2 = pq Þ ar m - 1 = pq
Þ
t m = pq
y Example 33. If a, b , c are real numbers such that
3 (a 2 + b 2 + c 2 + 1) = 2 (a + b + c + ab + bc + ca ), then
a, b , c are in
(a) AP only
(c) GP and AP
y Example 31. If sin q , 2 (sin q + 1), 6 sin q + 6 are in
GP, then the fifth term is
(b) 81 2
(c) 162 (d) 162 2
Sol. [ 2 (sin q + 1)]2 = sin q (6 sin q + 6)
Þ [(sin q + 1) 2 (sin q + 1) - 6sin q ] = 0
(b) GP only
(d) None of these
Sol. Given, 3 (a 2 + b 2 + c 2 + 1) = 2 (a + b + c + ab + bc + ca )
Þ
Hence, (b) is the correct answer.
(a) 81
[sinq = - 1 is not possible]
1
and common ratio
2
æ1
ö
2 ç + 1÷
è2
ø
=3 2
=r =
1
æ ö
ç ÷
è2ø
1
4
\
t 5 = ar = (3 2 )4 = 162
2
Hence, (c) is the correct answer.
Þ
y Example 30. The (m + n )th and (m - n )th terms of a
GP are p and q, respectively. Then, the m th term of
the GP is
(a) p ç ÷
èpø
1
2
then first term = a = sinq =
Hence, (d) is the correct answer.
m
æ q ö 2n
1
2
Sol. The number of digits in each term of the sequence are 1,
2, 4, 8, .... which are in GP. Let 1025th term is 2n .
Then,
1 + 2 + 4 + 8 + ... + 2n - 1 < 1025 £ 1 + 2 + 4 + 8 + ... + 2n
c2
æc ö
t 2 n + 1 = ar 2n = a (r n )2 = a ç ÷ =
èa ø
a
\
sinq =
(a) 2 9
(c) 2 11
Sol. Let common ratio = r
\
sin q = - 1,
Þ
2 (a 2 + b 2 + c 2 - ab - bc - ca ) +
(a 2 + b 2 + c 2 - 2a - 2b - 2c + 3)
{(a - b )2 + (b - c )2 + (c - a )2 } +
{(a - 1)2 + (b - 1)2 + (c - 1)2 } = 0
Þ a - b = b - c = c - a = 0 and a - 1 = b - 1 = c - 1 = 0
Þ
a=b=c =1
Þ a, b, c are in GP and AP.
Hence, (c) is the correct answer.
Chap 03 Sequences and Series
Sum of a Stated Number of
Terms of a Geometric Series
The game of chess was invented by Grand Vizier Sissa
Ben Dhair for the Indian king Shirham. Pleased with the
game, the king asked the Vizier what he would like as
reward. The Vizier asked for one grain of wheat to be
placed on the first square of the chess, two grains on the
second, four grains on the third and so on (each time
doubling the number of grains). The king was surprised of
the request and told the vizier that he was fool to ask for
so little.
The inventor of chess was no fool. He told the king ‘‘What
I have asked for is more wheat that you have in the entire
kingdom, in fact it is more than there is in the whole
world’’ He was right. There are 64 squares on a chess
board and on the nth square he was asking for 2 n - 1 grains,
if you add the numbers
S = 1 + 2 + 2 2 + 2 3 + .... + 2 62 + 2 63
i.e.,
…(i)
On multiplying both sides by 2, then
2S = 2 + 2 2 + 2 3 + 2 4 + .... + 2 63 + 2 64
…(ii)
223
If r = 1, the above formulae cannot be used. But, then the
GP reduces to a, a, a, ...
\ S n = a + a + a +K n times = na
Sum to Infinity of a GP, when the Numerical Value
of the Common Ratio is Less than Unity, i.e. It is a
Proper Fraction
If a be the first term, r be the common ratio of a GP, then
Sn =
a (1 - r n )
a
ar n
=
(1 - r )
(1 - r ) (1 - r )
Let - 1 < r < 1 i.e. | r | < 1, then lim r n ® 0
n®¥
Let S ¥ denote the sum to infinity of the GP, then
a
,
S¥ =
(1 - r )
where - 1 < r < 1
Recurring Decimal
Recurring decimal is a very good example of an infinite
geometric series and its value can be obtained by means of
infinite geometric series as follows
··
0 . 3 2 7 = 0.327272727... to infinity
On subtracting Eq. (i) from Eq. (ii), we get
= 0.3 + 0.027 + 0.00027 + 0.0000027 + K upto infinity
3
27
27
27
=
+
+
+
+ ... upto infinity
3
5
10 10
10
10 7
3
27 æ
1
1
ö
=
+
ç 1 + 2 + 4 + K upto infinity ÷
3
ø
10 10 è 10
10
64
S = 2 - 1 = 18, 446, 744, 073, 709, 551, 615 grains i.e.,
represent more wheat that has been produced on the Earth.
Sum of n Terms of a GP
Let a be the first term, r be the common ratio, l be the last
term of a GP having n terms and S n the sum of n terms,
then
l
l
…(i)
S n = a + ar + ar 2 + ... +
+ +l
2
r
r
On multiplying both sides by r (the common ratio)
l
r S n = ar + ar 2 + ar 3 + K + + l + lr
r
On subtracting Eq. (ii) from Eq. (i), we have
S n - r S n = a - lr or S n (1 - r ) = a - lr
a - lr
, when r < 1
\
Sn =
1 -r
lr - a
, when r > 1
Sn =
r -1
Now,
l = t n = ar n - 1
Then, above formula can be written as
a (1 - r n )
a (r n - 1 )
when r < 1, S n =
,
Sn =
(1 - r )
(r - 1 )
when r > 1
…(ii)
æ
ö
ç
÷
3
27 ç 1 ÷
=
+
1 ÷
10 10 3 ç
ç1 - 2 ÷
è
10 ø
3
27 297 + 27
=
+
=
990
10 990
324
[rational number]
=
990
Aliter (Best method)
Let P denotes the figure which do not recur and suppose
them p in number, Q denotes the recurring period
consisting of q figures. Let R denotes the value of the
recurring decimal.
Then,
R = 0 × PQQQ ...
\
10 p ´ R = P × QQQ ...
10 p + q ´ R = PQ × QQQ ...
PQ - P
.
\ Therefore, by subtraction R =
p +q
- 10 p )
(10
and
224
Textbook of Algebra
Corollary I If R = 0 × QQQ ...
Q
(when Q denote the recurring period
Then, R =
q
10 - 1
consisting of q figures)
·
3
1
For example, If R = 0. 3, then R =
=
1
10 - 1 3
Corollary II The value of recurring decimal is always
rational number.
In Particular
1 ì10
ü
(i) For a = 1, 1 + 11 + 111 + ... = í (10 n - 1) - ný
9î9
þ
· ·
y Example 34. Find the value of 0. 32 58.
··
R = 0.3258
Sol. Let
…(i)
Þ
R = 0.3258585858 ...
Here, number of figures which are not recurring is 2 and
number of figures which are recurring is also 2.
Then,
and
100R = 32.58585858 ...
…(ii)
10000R = 3258.58585858...
…(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
\
Hence,
9900R = 3226
3226
R=
9900
1613
R=
4950
Shortcut Methods for
Recurring Decimals
1. The numerator of the vulgar fraction is obtained by
subtracting the non-recurring figure from the given
figure.
2. The denominator consists of as many 9’s as there are
recurring figure and as many zero as there are
non-recurring figure.
For example,
· ·
3654 - 36 3618
(i) 0. 36 5 4 =
=
9900
9900
··
··
327 - 3 1314
(ii) 1 . 3 2 7 = 1 + 0. 3 2 7 = 1 +
=
990
990
·
3 -0 1
(iii) 0. 3 =
=
9
3
y Example 35. Find the sum upto n terms of the
series a + aa + aaa + aaaa + K, " a Î N and 1 £ a £ 9.
Sol. Let S = a + aa + aaa + aaaa + ... upto n terms
= a (1 + 11 + 111 + 1111 + ... upto n terms)
a
= (9 + 99 + 999 + 9999 + K upto n terms)
9
a
{(101 - 1) + (102 - 1) + (103 - 1) + (104 - 1) + ...
9
upto n terms}
a
= {(10 + 102 + 103 + K upto n terms)
9
- (1 + 1 + 1 + ...n times)}
ü a ì 10 n
a ì 10 (10n - 1)
ü
- n ý = í (10 - 1) - n ý
= í
9 î 10 - 1
9
9
î
þ
þ
[Remember]
=
2 ì10
ü
(ii) For a = 2, 2 + 22 + 222 + ... = í (10 n - 1) - ný
9î9
þ
3 ì10
ü
(iii) For a = 3, 3 + 33 + 333 + ... = í (10 n - 1) - ný
9î9
þ
(iv) For a = 4, 4 + 44 + 444 + ... =
4 ì10
ü
n
í (10 - 1) - ný
9î9
þ
5 ì10
ü
(v) For a = 5, 5 + 55 + 555 + ... = í (10 n - 1) - ný
9î9
þ
6 ì10
ü
(vi) For a = 6, 6 + 66 + 666 + ... = í (10 n - 1) - ný
9î9
þ
7 ì10
ü
(vii) For a = 7, 7 + 77 + 777 + ... = í (10 n - 1) - ný
9î9
þ
8 ì10
ü
(viii) For a = 8, 8 + 88 + 888 + ... = í (10 n - 1) - ný
9î9
þ
9 ì10
ü
(ix) For a = 9, 9 + 99 + 999 + ... = í (10 n - 1) - ný
9î9
þ
y Example 36. Find the sum upto n terms of the series
0.b + 0.bb + 0.bbb + 0.bbbb + ..., " b Î N and 1 £ b £ 9.
Sol. Let S = 0.b + 0.bb + 0.bbb + 0.bbbb + ... upto n terms
= b (01
. + 011
. + 0111
.
+ 01111
.
+ K upto n terms)
b
. + 099
. + 0999
.
.
= (09
+ 09999
+ .... upto n terms)
9
b
. ) + (1 - 001
. ) + (1 - 0001
. ) + (1 - 00001
.
) + ... upto
= {(1 - 01
9
n terms}
b
= {(1 + 1 + 1 + 1 + ... upto n times)
9
- (01
. + 001
. + 0001
.
+ 00001
.
+ ... upto n terms)}
bì
1
1
1
æ1
öü
= ín - ç + 2 + 3 + 4 + ... upto n terms ÷ý
è 10 10
øþ
9î
10
10
n
ì
1 æ
1 öü
ç1 - æç ö÷ ÷ ï
ï
n
è 10 ø ÷ø ï b ìï
10 çè
bï
1é
æ 1 ö ùüï
1
n
= ín =
úý
ê
ç
÷
ý
í
1
è 10 ø úï
9 ê
9ï
ï 9 îï
1ûþ
ë
10
ï
ï
þ
î
[Remember]
Chap 03 Sequences and Series
In Particular
(i) For b = 1,
249
{2 ´ 249 + ( 249 - 1) ´ 1}
2
= 248 ( 250 + 249 - 1)
\ S 50 =
n
1 ìï
1 é æ 1 ö ù üï
0.1 + 0.11 + 0.111 + ... = ín - ê1 - ç ÷ ú ý
9ï
9 ê è 10 ø ú ï
ë
ûþ
î
(ii) For b = 2,
n
2 ìï
1 é æ 1 ö ù üï
0.2 + 0.22 + 0.222 + ... = ín - ê1 - ç ÷ ú ý
9ï
9 ê è 10 ø ú ï
ë
ûþ
î
(iii) For b = 3,
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
225
n
3 ìï
1 é æ 1 ö ù üï
0.3 + 0.33 + 0.333 + ... = ín - ê1 - ç ÷ ú ý
9ï
9 ê è 10 ø ú ï
ë
ûþ
î
For b = 4,
n
4 ìï
1 é æ 1 ö ù üï
0.4 + 0.44 + 0.444 + ... = ín - ê1 - ç ÷ ú ý
9ï
9 ê è 10 ø ú ï
ë
ûþ
î
For b = 5,
n
5 ìï
1 é æ 1 ö ù üï
0.5 + 0.55 + 0.555 + ... = ín - ê1 - ç ÷ ú ý
9ï
9 ê è 10 ø ú ï
ë
ûþ
î
For b = 6,
n
6 ìï
1 é æ 1 ö ù üï
0.6 + 0.66 + 0.666 + ... = ín - ê1 - ç ÷ ú ý
9ï
9 ê è 10 ø ú ï
ë
ûþ
î
For b = 7,
n
7 ìï
1 é æ 1 ö ù üï
0.7 + 0.77 + 0.777 + ... = ín - ê1 - ç ÷ ú ý
9ï
9 ê è 10 ø ú ï
ë
ûþ
î
For b = 8,
n
8 ìï
1 é æ 1 ö ù üï
0.8 + 0.88 + 0.888 + ... = ín - ê1 - ç ÷ ú ý
9ï
9 ê è 10 ø ú ï
ë
ûþ
î
For b = 9,
n
9 ïì
1 é æ 1 ö ù ïü
0.9 + 0.99 + 0.999 + ... = ín - ê1 - ç ÷ ú ý
9ï
9 ê è 10 ø ú ï
ë
ûþ
î
y Example 37. If N, the set of natural numbers is
partitioned into groups S 1 = {1}, S 2 = {2, 3},
S 3 = {4, 5, 6, 7}, S 4 = {8, 9, 10, 11, 12, 13, 14, 15}, ..., then find
the sum of the numbers in S 50 .
Sol. The number of terms in the groups are 1, 2, 22 , 23 , ...
\ The number of terms in the 50th group = 250 - 1 = 249
0
1-1
Q The first term of 1st group = 1 = 2 = 2
The first term of 2nd group = 2 = 21 = 22 - 1
The first term of 3rd group = 4 = 22 = 23 - 1
M
M
M
M
M
The first term of 50th group = 250 - 1 = 249
= 248 [249 ( 2 + 1) - 1] = 248 (3 × 249 - 1)
1 1
1
y Example 38. If S n = 1 + + 2 + ... + n - 1 , then
2 2
2
calculate the least value of n such that
1
.
2 - Sn <
100
Sol. Given, Sn = 1 +
1 1
1
+
+ ... + n - 1
2 22
2
Þ
Sn = 2 -
Þ
1
2 - Sn =
n
é
æ1ö ù
1 × ê1 - ç ÷ ú
è2ø ú
ê
û
= ë
1ö
æ
ç1 - ÷
è
2ø
1
n -1
2
n- 1
2
<
1 ù
é
êëQ2 - Sn < 100 úû
1
100
Þ
2n - 1 > 100 > 26
Þ
2n - 1 > 26
\
n -1>6 Þ n >7
Hence, the least value of n is 8.
y Example 39. If x = 1 + a + a 2 + a 3 + ... + ¥ and
y = 1 + b + b 2 + b 3 + ... + ¥ show that
xy
, where
1 + ab + a 2b 2 + a 3 b 3 + ... + ¥ =
x+ y -1
0 < a < 1 and 0 < b < 1.
Sol. Given, x = 1 + a + a 2 + a 3 + ... + ¥ =
Þ
1
1-a
x - ax = 1
æ x - 1ö
a=ç
÷
è x ø
\
…(i)
and
y = 1 + b + b 2 + b 3 + ... + ¥
Similarly,
æy - 1ö
b=ç
÷
è y ø
…(ii)
0 < a < 1, 0 < b < 1
0 < ab < 1
Since,
\
1
1 - ab
Now, 1 + ab + a 2b 2 + a 3b 3 + ... + ¥ =
1
æ x - 1ö æy - 1ö
1- ç
֍
÷
è x øè y ø
xy
=
xy - xy + x + y - 1
=
[from Eqs. (i) and (ii)]
Hence, 1 + ab + a 2b 2 + a 3b 3 + ... + ¥ =
xy
x +y -1
226
Textbook of Algebra
Properties of Geometric Progression
1. If a 1 , a 2 , a 3 , ... are in GP with common ratio r, then
a a a
a 1 k, a 2 k, a 3 k, ... and 1 , 2 , 3 , ... are also in GP
k k k
(k ¹ 0 ) with common ratio r.
2. If a 1 , a 2 , a 3 , ... are in GP with common ratio r, then
a 1 ± k, a 2 ± k, a 3 ± k, .... are not in GP (k ¹ 0 ).
3. If a 1 , a 2 , a 3 , ... are in GP with common ratio r, then
1 1 1
(i)
, , , ... are also in GP with common
a1 a2 a 3
1
ratio .
r
(ii) a 1n , a 2n , a n3 , ... are also in GP with common ratio
r n and n Î Q .
(iii) log a 1 , log a 2 , log a 3 , ... are in AP (a i > 0, " i )
In this case, the converse also holds good.
4. If a 1 , a 2 , a 3 , ... and b 1 , b 2 , b 3 , ... are two GP’s with
common ratios r1 and r2 , respectively. Then,
a a a
(i) a 1 b 1 , a 2 b 2 , a 3 b 3 , ... and 1 , 2 , 3 ,K are also in
b1 b2 b 3
r
GP with common ratios r1 r2 and 1 ,
r2
respectively.
(ii) a 1 ± b 1 , a 2 ± b 2 , a 3 ± b 3 , ... are not in GP.
5. If a 1 , a 2 , a 3 , ..., a n - 2 , a n - 1 , a n are in GP. Then,
(i) a 1 a n = a 2 a n - 1 = a 3 a n - 2 = K
(ii) a r = a r - k a r + k , " k, 0 £ k £ n - r
(iii)
a2 a 3 a 4
a
=
=
= ... = n
a1 a2 a 3
an - 1
Þ
Also,
a 22 = a 3 a 1 , a 23 = a 2 a 4 , ...
Remark
1. Product of three numbers = a3
Product of five numbers = a5
M
M
M
M
M
Product of ( 2m + 1) numbers = a2 m + 1
2. From given conditions, find two equations in a and r and
then solve them. Now, the numbers in GP can be obtained.
7. If four numbers in GP whose product is given are to
a a
be taken as , , ar , ar 3 and if six numbers in GP
r3 r
whose product is given are to be taken as
a a a
, , , ar , ar 3 , ar 5 , etc.
r5 r 3 r
In general If (2m ) numbers in GP whose product is
given are to be taken as (m Î N )
a
a
a a
,
,..., , , ar , ar 3 , ... , ar 2m - 3 , ar 2m - 1
2m - 1
2m - 3
r
r
r3 r
Remark
1. Product of four numbers = a4
Product of six numbers = a6
M
M
M
M
M
Product of ( 2m) numbers = a2 m
2. From given conditions, find two equations in a and r and
then solve them. Now, the numbers in GP can be obtained.
y Example 40. If S 1 , S 2 , S 3 , ..., S p are the sum of
infinite geometric series whose first terms are 1, 2, 3,
1 1 1
1
..., p and whose common ratios are , , , ...,
2 3 4
p+1
respectively, prove that
p (p + 3)
.
S 1 + S 2 + S 3 + ... + S p =
2
Sol. Q
2
a 2 = a 1r , a 3 = a 1r ,
a 4 = a 1 r 3 , ..., a n = a 1 r n - 1
where, r is the common ratio of GP.
6. If three numbers in GP whose product is given are
a
to be taken as , a, ar and if five numbers in GP
r
whose product is given are to be taken as
a a
, , a, ar , ar 2 , etc.
2 r
r
In general If (2m + 1) numbers in GP whose
product is given are to be taken as (m Î N )
a
a
a
,
, ..., , a, ar , ..., ar m - 1 , ar m
m
m -1
r
r
r
p
Sp =
1-
\
\
1
p +1
= ( p + 1)
S1 = 2, S 2 = 3, S 3 = 4, ...
LHS = S1 + S 2 + S 3 + ... + S p
= 2 + 3 + 4 + ... + ( p + 1) =
=
p ( p + 3)
= RHS
2
p
( 2 + p + 1)
2
y Example 41. Let x 1 and x 2 be the roots of the
equation x 2 - 3x + A = 0 and let x 3 and x 4 be the
roots of the equation x 2 - 12x + B = 0. It is known
that the numbers x 1 , x 2 , x 3 , x 4 (in that order) form
an increasing GP. Find A and B.
Chap 03 Sequences and Series
Sol. Q x 1, x 2 , x 3 , x 4 are in GP.
Let
[here, product of x 1, x 2 , x 3 , x 4 are not given]
Given, x 1 + x 2 = 3, x 1x 2 = A
…(i)
Þ
x 1(1 + r ) = 3, x 12 r = A
and x 3 + x 4 = 12, x 3 x 4 = B
Þ x 1r 2 (1 + r ) = 12, x 12 r 5 = B
and
B=
x 12 r 5
2
æ1 + r 2 + r ö
æ1
ö
a 2 ç + r + 1÷ = 156 Þ 36 ç
÷ = 156
èr
ø
r
è
ø
æ1 + r + r 2 ö
2
Þ
3ç
÷ = 13 Þ 3r - 10r + 3 = 0
r
è
ø
1
Þ
( 3r - 1) (r - 3) = 0 Þ r = or r = 3
3
Putting the values of a and r , the required numbers are 18,
6, 2 or 2, 6, 18. Hence, the sum of numbers is 26.
Þ
x 2 = x 1r , x 3 = x 1r 2 , x 4 = x 1r 3
From Eqs. (i) and (ii),
r2 = 4 Þ r =2
From Eq. (i), x 1 = 1
Now,
A = x 12 r = 12 × 2 = 2
227
…(ii)
[for increasing GP]
5
= 1 × 2 = 32
[from Eq. (i)]
[from Eq. (ii)]
y Example 42. Suppose a, b , c are in AP and a 2 , b 2 , c 2
3
are in GP, if a > b > c and a + b + c = , then find the
2
values of a and c.
Sol. Since, a, b, c are in AP and sum of a, b, c is given.
a = b - D, c = b + D
[ D < 0] [Qa > b > c ]
3
and given a + b + c =
2
3
Þ b- D +b+b+ D =
2
1
\
b=
2
1
1
Then, a = - D and c = + D
2
2
Also, given a 2 , b 2 , c 2 are in GP, then (b 2 )2 = a 2c 2
1 1
Þ
± b 2 = ac Þ ± = - D 2
4 4
1 1 1
2
Þ
D = ± =
[Q D ¹ 0]
4 4 2
1
1
\ D=±
Þ D=[Q D < 0]
2
2
1
1
1
1
Hence, a = +
and c = 2
2
2
2
Let
y Example 43. If the continued product of three
numbers in GP is 216 and the sum of their products in
pairs is 156, then find the sum of three numbers.
Sol. Here, product of numbers in GP is given.
a
\ Let the three numbers be , a, ar .
r
a
Then,
× a × ar = 216
r
Þ
a 3 = 216
\
a=6
Sum of the products in pairs = 156
a
a
Þ
× a + a × ar + ar × = 156
r
r
y Example 44. Find a three-digit number whose
consecutive digits form a GP. If we subtract 792 from
this number, we get a number consisting of the same
digits written in the reverse order. Now, if we increase
the second digit of the required number by 2, then the
resulting digits will form an AP.
Sol. Let the three digits be a, ar , ar 2 , then according to hypothesis
100a + 10ar + ar 2 - 792 = 100ar 2 + 10ar + a
Þ
99a (1 - r 2 ) = 792
Þ
a (1 + r ) (1 - r ) = 8
and a, ar + 2, ar 2 are in AP.
2 (ar + 2) = a + ar 2
Then,
Þ
…(i)
a (r 2 - 2r + 1) = 4 Þ a (r - 1)2 = 4
…(ii)
On dividing Eq. (i) by Eq. (ii), we get
r +1
1
= -2 Þ r =
r -1
3
From Eq. (ii), a = 9
Thus, digits are 9, 3, 1 and so the required number is 931.
Examples on Application of
Progression in Geometrical Figures
y Example 45. A square is drawn by joining the
mid-points of the sides of a given square. A third
square is drawn inside the second square in the
same way and this process continues indefinitely. If
a side of the first square is 16 cm, then determine
the sum of the areas of all the squares.
Sol. Let a be the side length of square, then
G
D
O
K
H
J
T
S
Q
R
N
P
L
A
C
M
E
F
I
B
AB = BC = CD = DA = a
228
Textbook of Algebra
Q E, F , G , H are the mid-points of AB, BC , CD and DA,
respectively.
a
\
EF = FG = GH = HE =
2
and I , J , K , L are the mid-points of EF , FG , GH and HE,
respectively.
a
IJ = JK = KL = LI =
\
2
a
Similarly, MN = NO = OP = PM =
and
2 2
a
QR = RS = ST = TQ = , ...
4
S = Sum of areas
= ABCD + EFGH + IJKL + MNOP + QRST + ...
2
2
2
æ a ö
æa ö
æ a ö
= a2 + ç ÷ + ç ÷ + ç
÷ + ...
è 2ø
è2ø
è2 2 ø
1 1 1
æ
ö
= a 2 ç1 + + + + ...÷
è
ø
2 4 8
ö
æ
ç 1 ÷
2
2
2
=a ç
÷ = 2a = 2 (16)
1
ç1 - ÷
è
2ø
a a a
æ
ö
P = Sum of perimeters = 3 ça + + + + K÷
è
ø
2 4 8
æ
ö
ç a ÷
=3ç
[Qa = 24 cm ]
÷ = 6a = 6 ´ 24 = 144 cm
1
ç1 - ÷
è
2ø
y Example 47. Let S 1 , S 2 , ... be squares such that for
each n ³ 1, the length of a side of S n equals the
length of a diagonal of S n + 1 . If the length of a side
of S 1 is 10 cm and the area of S n less than 1 sq cm.
Then, find the value of n.
Sol. We have, length of a side of
Sn = length of diagonal of Sn + 1
Length of a side of Sn = 2 (length of a side of Sn + 1)
Þ
Þ
Þ
[Qa = 16 cm]
Length of a side of Sn + 1
Length of a side of Sn
æ 1 ö
Side of Sn = 10 ç ÷
è 2ø
= 512 sq cm
Sol. Let a be the side length of equilateral triangle, then
AB = BC = CA = a
A
H
F
M
I
B
E
L
K
D
1
, for all n ³ 1
2
Sides of S1, S 2 , S 3 , ... form a GP with common ratio
1
and first term 10.
2
\
y Example 46. One side of an equilateral triangle is 24
cm. The mid-points of its sides are joined to form
another triangle whose mid-points, in turn, are joined
to form still another triangle. This process continues,
indefinitely. Find the sum of the perimeters of all the
triangles.
=
Þ Area of Sn = ( Side)2 =
n -1
=
10
(n - 1)
2 2
100
2n - 1
Now, given area of Sn < 1
100
Þ
< 1 Þ 2n - 1 > 100 > 26
n -1
2
Þ
2n - 1 > 26 Þ n - 1 > 6
\
n > 7 or n ³ 8
y Example 48. The line x + y = 1 meets X-axis at A and
Y-axis at B, P is the mid-point of AB, P1 is the foot of
perpendicular from P to OA, M1 is that of P1 from OP;
P2 is that of M1 from OA, M 2 is that of P2 from OP; P3
is that of M 2 from OA and so on. If Pn denotes the nth
foot of the perpendicular on OA, then find OPn .
J
Y
B
C
P
Q D , E, F are the mid-points of BC , CA and AB, respectively.
a
\
EF = FD = DE =
2
and H , I , J are the mid-points of EF , FD and DE,
respectively.
a
\
IJ = JH = HI =
4
a
Similarly,
KL = ML = KM = , ...
8
M1
M2
O P3 P2 P 1
A
X
Sol. We have,
(OMn - 1 )2 = (OPn )2 + ( Pn Mn - 1 )2
= (OPn )2 + (OPn )2 = 2 (OPn )2 = 2 a n2 [say]
Also, (OPn - 1 )2 = (OMn - 1 )2 + ( Pn - 1Mn - 1 )2
Chap 03 Sequences and Series
Þ
Þ
Þ
\
1
1
a n2 - 1 = 2 a n2 + a n2 - 1 Þ a n2 = a n2 - 1
4
2
1
an = an - 1
2
1
1
1
OPn = a n = a n - 1 = 2 a n - 2 = ... = n
2
2
2
n
æ1ö
OPn = ç ÷
è2ø
Use of GP in Solving
Practical Problems
1
in GP with a = 1, r = . Let time taken by the insect in
2
covering 3 mm be n seconds.
1 1
\1 + + + ... + n terms = 3
2 4
n
é
æ1ö ù
1 × ê1 - ç ÷ ú
è2ø ú
êë
û =3
Þ
1
12
n
Þ
In this part, we will see how the formulae relating to GP
can be made use of in solving practical problems.
y Example 49. Dipesh writes letters to four of his
friends. He asks each of them to copy the letter and
mail to four different persons with the request that
they continue the chain similarly. Assuming that the
chain is not broken and that it costs 25 paise to mail
one letter, find the total money spent on postage till
the 8th set of letters is mailed.
Sol. Number of letters in the 1st set = 4 (These are letters sent
by Dipesh)
Number of letters in the 2nd set = 4 + 4 + 4 + 4 = 16
Number of letters in the 3rd set
= 4 + 4 + 4 + ... + 16 terms = 64
M
M
M
M
The number of letters sent in the 1st set, 2nd set, 3rd set, ...
are respectively 4, 16, 64, ... which is a GP with a = 4,
16 64
r =
=
=4
4 16
\Total number of letters in all the first 8 sets
4 ( 4 8 - 1)
= 87380
=
4 -1
\ Total money spent on letters = 87380 ´
229
25
= ` 21845
100
y Example 50. An insect starts from a point and
travels in a straight path 1 mm in the first second and
half of the distance covered in the previous second in
the succeeding second. In how much time would it
reach a point 3 mm away from its starting point.
Sol. Distance covered by the insect in the 1st second = 1 mm
1 1
Distance covered by it in the 2nd second = 1 ´ = mm
2 2
1 1 1
Distance covered by it in the 3rd second = ´ = mm
2 2 4
M
M
M
M
The distance covered by the insect in 1st second, 2nd
1 1
second, 3rd second, ... are respectively 1, , , ..., which are
2 4
3
æ1ö
1- ç ÷ =
è2ø
2
n
Þ
1
æ1ö
ç ÷ =è2ø
2
Þ
2n = - 2
which is impossible because 2n > 0
\Our supposition is wrong.
\ There is no n Î N , for which the insect could never 3 mm
in n seconds.
Hence, it will never to able to cover 3 mm.
Remark
The maximum distance that the insect could cover is 2 mm.
1 1
1
i.e.,
1 + + + ... =
=2
1
2 4
12
y Example 51. The pollution in a normal atmosphere is
less than 0.01%. Due to leakage of a gas from a
factory, the pollution is increased to 20%. If every day
80% of the pollution is neutralised, in how many days
the atmosphere will be normal?
Sol. Let the pollution on 1st day = 20
The pollution on 2nd day = 20 ´ 20% = 20 (0.20)
The pollution on 3rd day = 20 (0.20)2
M
M
M
M
Let in n days the atmosphere will be normal
\
Þ
20 (0.20)n - 1 < 001
.
æ2ö
ç ÷
è 10 ø
n -1
<
1
2000
Taking logarithm on base 10, we get
(n - 1) (log 2 - log 10) < log 1 - log 2000
Þ
(n - 1) (0.3010 - 1) < 0 - (0.3010 + 3)
Þ
n -1>
3.3010
0.6990
Þ
n > 5722
.
Hence, the atmosphere will be normal in 6 days.
230
Textbook of Algebra
#L
1.
Exercise for Session 3
The fourth, seventh and the last term of a GP are 10, 80 and 2560, respectively. The first term and number of
terms in GP are
4
(a) , 12
5
2.
4
(b) , 10
5
(b) (ab )n
(b) - 27
10
31
(b)
1 n
(10 - 1) + n 2
9
(d)
32
31
(d) 18
(b)
1 n
(10 - 1) + 2n
9
(c)
10
(10n - 1) + n 2
9
(d)
10
(10n - 1) + 2n
9
(c) 10
(d) 12
(c) 3
(d) 4
If | a | < 1and | b | < 1, then the sum of the series 1 + (1 + a )b + (1 + a + a 2 ) b 2 + (1 + a + a 2 + a 3 ) b 3 + ... is
1
(1 - a ) (1 - b )
(b)
1
(1 - a ) (1 - ab )
(c)
1
(1 - b ) (1 - ab )
(d)
1
(1 - a ) (1 - b ) (1 - ab )
If the sides of a triangle are in GP and its larger angle is twice the smallest, then the common ratio r satisfies
the inequality
3
2
(b) 1 < r <
(c) 1 < r < 2
2
2
(d) r >
2
2
If ax + bx + cx + d is divisible by ax + c, then a, b , c, d are in
(b) GP
(c) HP
(d) None of these
If (r )n denotes the number r r r ... (n digits), where r = 1, 2, 3, ..., 9 and a = (6)n, b = (8)n, c = (4)2n, then
(b) a 2 + b - c = 0
(c) a 2 + b - 2c = 0
(d) a 2 + b - 9c = 0
· ·
0.4 2 7 represents the rational number
(a)
15.
(c) 10
(b) 2
(a) a 2 + b + c = 0
14.
32
31
S (S - S2 )
If S1, S2, S3 be respectively the sum of n, 2n and 3n terms of a GP, then 1 3
is equal to
(S2 - S1)2
(a) AP
13.
(c) -
(b) 8
(a) 0 < r <
12.
(d) - 13. 5
In an increasing GP, the sum of the first and last term is 66, the product of the second and the last but one is
128 and the sum of the sum of the terms is 126, then the number of terms in the series is
(a)
11.
(c) 13.5
The sum to n terms of the series 11 + 103 + 1005 + ... is
(a) 1
10.
10
31
(b) -2
(a) 6
9.
(d) None of these
Three distinct numbers x , y , z form a GP in that order and the numbers 7x + 5y , 7y + 5z , 7z + 5x form an AP
in that order. The common ratio of GP is
(a)
8.
(c) r < 1 or r > 3
In a sequence of 21 terms the first 11 terms are in AP with common difference 2 and the last 11 terms are in
GP with common ratio 2, if the middle term of the AP is equal to the middle term of GP, the middle term of the
entire sequence is
(a) - 4
7.
(d) None of these
If x , 2x + 2, 3x + 3 are in GP, the fourth term is
(a) -
6.
(c) (ab )n
(b) - 3 < r < - 1
(a) 27
5.
/2
If a1, a 2, a 3, (a1 > 0) are three successive terms of a GP with common ratio r, the value of r for which
a 3 > 4a 2 - 3a1 holds is given by
(a) 1 < r < 3
4.
5
(d) , 10
4
If the first and the nth terms of a GP are a and b respectively and if P is the product of the first n terms, then P 2
is equal to
(a) ab
3.
5
(c) , 12
4
47
99
(b)
47
110
(c)
47
999
(d)
49
99
If the product of three numbers in GP be 216 and their sum is 19, then the numbers are
(a) 4, 6, 9
(b) 4, 7, 8
(c) 3, 7, 9
(d) None of these
Session 4
Harmonic Sequence or Harmonic Progression (HP)
Harmonic Sequence or
Harmonic Progression (HP)
A Harmonic Progression (HP) is a sequence, if the
reciprocals of its terms are in Arithmetic Progression (AP)
1 1 1
i.e., t 1 , t 2 , t 3 , ... is HP if and only if , , , ... is an AP.
t1 t2 t 3
For example, The sequence
1 1 1
5 10
(ii) 2, , ,K
(i) , , , ...
2 5 8
2 3
1 1
1
(iii) ,
,
, ... are HP’s.
a a + d a + 2d
Remark
1. No term of HP can be zero.
2. The most general or standard HP is
1
1
1
1
,
,
,
, ....
a a + d a + 2d a + 3d
y Example 52. If a, b , c are in HP, then show that
a -b a
= .
b -c c
Sol. Since, a, b, c are in HP, therefore
i.e.
or
1 1 1
, , are in AP
a b c
1 1 1 1
- = b a c b
a -b b -c
a-b a
or
=
=
ab
bc
b-c c
Remark
A HP may also be defined as a series in which every three
I - II
I
consecutive terms (say I, II, III) satisfy
= this relation.
II - III III
y Example 53. Find the first term of a HP whose
5
1
second term is and the third term is .
4
2
5 1
Sol. Let a be the first term. Then, a, , are in HP.
4 2
5
a4 =a
Then,
[from above note]
5 1 1
4 2 2
Þ
Þ
\
4a - 5
= 2a
5-2
4a - 5 = 6a or 2a = - 5
5
a=2
(i) n th Term of HP from Beginning
Let a be the first term, d be the common difference of an
AP. Then, nth term of an AP from beginning = a + (n - 1) d
Hence, the nth term of HP from beginning
1
, " n ÎN
=
a + (n - 1) d
(ii) n th Term of HP from End
Let l be the last term, d be the common difference of an
AP. Then,
nth term of an AP from end = l - (n - 1) d
1
Hence, the nth term of HP from end =
,"n ÎN
l - (n - 1) d
Remark
1.
1
1
+
nth term of HP from beginning nth term of HP from end
1
1
= a+ l =
+
first term of HP last term of HP
2. There is no general formula for the sum of any number of
quantities in HP are generally solved by inverting the terms
and making use of the corresponding AP.
1 1
1
1
+ +
+
= 0, then prove
a c a -b c -b
that a, b , c are in HP, unless b = a + c .
y Example 54. If
Sol. We have,
1 1
1
1
+ +
+
=0
a c a-b c -b
Þ
æ1
1 ö æ1
1 ö
÷ =0
ç +
÷+ç +
èa c - b ø èc a - b ø
Þ
(c - b + a ) (a - b + c )
=0
+
a (c - b )
c (a - b )
Þ
ù
é
1
1
(a + c - b ) ê
+
ú =0
ë a (c - b ) c (a - b ) û
Þ
(a + c - b ) [ 2ac - b (a + c )] = 0
232
Textbook of Algebra
a + c - b ¹ 0, then 2ac - b (a + c ) = 0
If
b=
or
Þ
2ac
a+c
Þ
Therefore, a, b,c are in HP and if 2ac - b (a + c ) ¹ 0, then
a + c - b = 0 i.e., b = a + c .
Sol. Given, a1, a 2 , a 3 , ..., an are in HP.
1 1 1
1
are in AP.
, , , ...,
a1 a 2 a 3
an
an - 1 - an
a1 - a 2 a 2 - a 3 a 3 - a 4
=
=
= ... =
=D
a1a 2
a2 a3
a 3a 4
an - 1 an
a1a 2 =
1ù
é1
ê a + (n - 1) D - a ú = (n - 1) a1an
1
1û
ë
Hence, a1a 2 + a 2 a 3 + a 3a 4 + ... + an - 1an = (n - 1) a1an
=
a1an
D
Remark
In particular case,
1. when n = 4 a1a2 + a2 a3 + a3a4 = 3a1a4
2. when n = 6 a1a2 + a2 a3 + a3a4 + a4 a5 + a5 a6 = 5 a1a6
y Example 56. The sum of three numbers in HP is 37
1
and the sum of their reciprocals is . Find the
4
numbers.
Sol. Three numbers in HP can be taken as
Then,
and
\
From Eq. (i),
1 1 1
, ,
.
a-d a a+d
1
1
1
+ +
= 37
a-d a a+d
1
4
1
a=
12
a-d +a+a+d =
12
12
+ 12 +
= 37
1 + 12d
1 - 12d
d=±
respectively a, b and c, then prove that
a - a4
a - a3
a1 - a 2
, a2 a3 = 2
, a 3a 4 = 3
,
D
D
D
an - 1 - an
..., an - 1an =
D
On adding all such expressions, we get
a - an a1an æ 1
1ö
a1a 2 + a 2 a 3 + a 3a 4 + ... + an - 1an = 1
=
ç - ÷
D
D è an a1 ø
Þ
1
24
or d 2 =
25 ´ 144
25
y Example 57. If p th, qth and r th terms of a HP be
Let D be the common difference of the AP, then
1
1
1
1
1
1
1
1
- =
=
= ... =
=D
a 2 a1 a 3 a 2 a 4 a 3
an an - 1
Þ
1 - 144d 2 =
1
60
1 1 1
1 1 1
\ a - d , a, a + d are , , or , , ×
15 12 10 10 12 15
Hence, three numbers in HP are 15, 12, 10 or 10, 12, 15.
\
y Example 55. If a1 , a 2 , a 3 , ..., an are in HP, then prove
that a1a 2 + a 2 a 3 + a 3a 4 + ... + an - 1an = (n - 1) a1an
\
12
12
24
+
= 25 Þ
= 25
1 - 12d 1 + 12d
1 - 144d 2
(q - r ) bc + (r - p ) ca + (p - q ) ab = 0.
Sol. Let A and D be the first term and common difference of
the corresponding AP. Now, a, b, c are respectively the p
th, q th and r th terms of HP.
1 1 1
\ , , will be respectively the p th, q th and r th terms of
a b c
the corresponding AP.
1
…(i)
Þ
= A + ( p - 1) D
a
1
…(ii)
= A + ( q - 1) D
b
1
…(iii)
= A + ( r - 1) D
c
On subtracting Eq. (iii) from Eq. (ii), we get
1 1
(c - b )
(b - c )
- = (q - r ) D Þ bc (q - r ) =
=b c
D
D
So, LHS = (q - r ) bc + (r - p ) ca + ( p - q ) ab
1
{b - c + c - a + a - b } = 0 = RHS
=D
Theorem Relating to the Three Series
If a, b, c are three consecutive terms of a series, then
a -b a
if
= , then a, b, c are in AP.
b -c a
a -b a
a -b a
= , then a, b, c are in GP and if
= , then
b -c b
b -c c
a, b, c are in HP.
if
…(i)
Mixed Examples on AP, GP and HP
y Example 58. If a, b , c are in AP and a 2 , b 2 , c 2 be in
a
HP. Then, prove that - , b , c are in GP or else
2
a =b =c.
Sol. Given, a, b, c are in AP.
233
Chap 03 Sequences and Series
\
b=
2
2
a+c
2
…(i)
2
and a , b , c are in HP.
b2 =
\
2a c
…(ii)
a2 + c 2
b 2 {(2b )2 - 2ac } = 2a 2c 2
4
2
[from Eq. (i)]
2 2
Þ
2b - acb - a c = 0
Þ
(2b 2 + ac ) (b 2 - ac ) = 0
Þ
2b 2 + ac = 0 or b 2 - ac = 0
Sol. Given, a, b, c are in HP.
2ac
ab
or c =
a+c
2a - b
…(i)
=
Sol. Since a, b, c are in AP, GP and HP as well
or
{2a - (2a - b )}
2ac
b=
a+c
…(i)
…(ii)
…(iii)
(a + c )2 = 4ac
or
(a + c )2 - 4ac = 0
or
(a - c )2 = 0
\
a=c
…(iv)
a+a
…(v)
On putting c = a in Eq. (i), we get b =
=a
2
From Eqs. (iv) and (v), a = b = c , thus the three numbers will
be equal.
ab 2
(2a - b )2
Sol. Given, a, b, c are in AP,
a+c
2
a+c
2
2
b = ac
b=
æa + c ö
ç
÷ = ac
è 2 ø
[from Eq. (ii)] …(iii)
y Example 60. If a, b , c , d and e be five real numbers
such that a, b , c are in AP; b , c , d are in GP; c , d , e are in
HP. If a = 2 and e = 18, then find all possible values of
b , c and d.
b=
a = 2, e = 18
c 2 = (2) (18) = 36
2
æ ab ö
2 æ ab ö
ç
÷ -ç
÷
b è 2a - b ø
è 2a - b ø
(2a - b )2
…(iv)
From Eqs. (i) and (ii), we have
2
=
c (c + e ) = e (a + c )
c 2 = ae
y Example 61. If three positive numbers a, b and c are
in AP, GP and HP as well, then find their values.
and
æ 2c 2
ö
- c÷
e=ç
è b
ø
ab
…(iii)
c = ±6
2±6
From Eq. (i), b =
= 4, - 2
2
c 2 36 36 36
and from Eq. (ii),
or
=
d=
=
-2
4
b
b
\
d = 9 or - 18
Hence,
c = 6, b = 4, d = 9 or c = - 6, b = - 2, d = - 18
…(ii)
and given, c , d , e are in AP.
c +e
\
d=
2
Þ
e = 2d - c
From Eqs. (i) and (iii), e =
2ce
c +e
Now, substituting the values of b and d in Eq. (ii), then
æ a + c ö æ 2ce ö
c2 = ç
֍
÷
è 2 ø èc + e ø
\
Given, b, c , d are in GP.
\
c 2 = bd
\
…(ii)
\
y Example 59. If a, b , c are in HP, b , c , d are in GP
ab 2
and c , d , e are in AP, then show that e =
.
(2a - b ) 2
b=
d=
Given,
From Eq. (iv),
2
But given, a, b, c are in AP.
Which is possible only when a = b = c
\
\
Þ
Þ
a
1
If 2b + ac = 0, then b = - ac or - , b, c are in GP
2
2
2
and if b - ac = 0 Þ a, b, c are in GP.
2
c 2 = bd
and c , d , e are in HP.
2 2
From Eq. (ii) b 2 {(a + c )2 - 2ac } = 2a 2c 2
Þ
b, c , d are in GP,
\
…(i)
Remark
1. If three positive numbers are in any two of AP, GP and HP,
then it will be also in third.
2. Thus, if three positive numbers are in any two of AP, GP and
HP, then they will be in the third progression and the
numbers will be equal.
234
Textbook of Algebra
#L
1.
Exercise for Session 4
If a, b , c are in AP and b , c, d be in HP, then
(a) ab = cd
2.
(a) AP
3.
2
n
If a, b , c are in HP, then
If
(c) HP
(d) None of these
(b) GP
(c) HP
(d) None of these
(c) - 1
(d) None of these
(b)
2
n
(c) -
n
2
(d)
n
2
(b) 2b = 3a + c
(c) b 2 =
æ ac ö
ç ÷
è 8ø
(d) None of these
a
b
c
are in
,
,
b +c c+a a+b
(b) GP
(c) HP
(d) None of these
(c) HP
(d) None of these
(c) HP
(d) None of these
x +y
y +z
are in HP, then x , y , z are in
, y,
2
2
(a) AP
10.
(d) None of these
If a, b , c are in AP and a 2, b 2, c 2 are in HP, then
(a) AP
9.
(b) GP
(b) 1
(a) a = b = c
8.
(c) HP
If (m + 1)th, (n + 1)th and (r + 1)th terms of an AP are in GP and m , n, r are in HP, then the value of the ratio of
the common difference to the first term of the AP is
(a) -
7.
(b) GP
If a, b , c are in GP, a - b , c - a, b - c are in HP, then a + 4b + c is equal to
(a) 0
6.
(d) abcd = 1
If x , 1, z are in AP and x , 2, z are in GP, then x , 4, z will be in
(a) AP
5.
(c) ac = bd
If a, b , c are in AP and a, b , d are in GP, then a, a - b , d - c will be in
(a) AP
4.
(b) ad = bc
a 1 2
If a, b , c are in AP, then
, , are in
bc c b
(b) GP
a+b
b +c
1
are in AP, then a, , c are in
If
,b,
1 - ab
1 - bc
b
(a) AP
(b) GP
Session 5
Mean
Mean
Arithmetic Mean
If three terms are in AP, then the middle term is called the
Arithmetic Mean (or shortly written as AM) between the
other two, so if a, b, c are in AP, then b is the AM of a
and c.
(i) Single AM of n Positive Numbers
Let n positive numbers be a 1 , a 2 , a 3 , ..., a n and A be the
AM of these numbers, then
a + a 2 + a 3 + ... + a n
A= 1
n
In particular Let a and b be two given numbers and A be
the AM between them, then a, A, b are in AP.
a +b
A=
\
2
Remark
2a + 3b + 5c
.
3
a + a2 + a3 + ... + an - 1 + 2an
2. AM of a1, a2, a3, ..., an - 1, 2an is 1
.
n
1. AM of 2a, 3b, 5c is
(ii) Insert n-Arithmetic Mean Between
Two Numbers
Let a and b be two given numbers and A 1 , A 2 , A 3 , ..., A n
are AM’s between them.
Then, a, A 1 , A 2 , A 3 , ..., A n , b will be in AP.
Now, b = (n + 2 ) th term = a + (n + 2 - 1) d
\
\
Þ
æb -a ö
d =ç
÷
è n + 1ø
`
[Remember] [where, d = common difference] …(i)
A1 = a + d , A2 = a + 2d , ...., An = a + nd
æb -a ö
æb -a ö
A1 = a + ç
÷ , ... , An
÷ , A2 = a + 2 ç
è n + 1ø
è n + 1ø
æb -a ö
=a +n ç
÷
è n + 1ø
Corollary I The sum of n AM’s between two given
quantities is equal to n times the AM between them.
Let two numbers be a and b and A 1 , A 2 , A 3 , K, A n are n
AM’s between them.
Then, a, A 1 , A 2 , A 3 , ..., A n , b will be in AP.
\ Sum of n AM’s between a and b
= A1 + A 2 + A 3 + ... + An
n
= ( A1 + An )
[Q A1, A 2 , A 3 , ..., An are in AP]
2
n
n
= (a + d + a + nd ) = [ 2a + (n + 1) d ]
2
2
n
[from Eq. (i)]
= (2a + b - a )
2
æa + b ö
=n ç
[AM between a and b]
÷=n
è 2 ø
[Remember]
Aliter A1 + A2 + A 3 + ... + An
= (a + A 1 + A2 + A 3 + ... + A n + b ) - (a + b )
(n + 2 )
æa +b ö
=
(a + b ) - (a + b ) = n ç
÷
è 2 ø
2
[AM of a and b]
=n
Aliter
[This method is applicable only when n is even]
A1 + A2 + A 3 + ... + An - 2 + An - 1 + An
= ( A1 + An ) + ( A2 + An - 1 ) + ( A 3 + An - 2 ) + ...
n
upto terms
2
n
= (a + b ) + (a + b ) + (a + b ) + ... upto times
2
[QTn + T ¢n = a + l ]
n
æa +b ö
[AM of a and b]
= (a + b ) = n ç
÷ =n
è 2 ø
2
Corollary II The sum ofmAM’s between any two
numbers is to the sum of n AM’s between them asm : n .
Let two numbers be a and b.
\ Sum of m AM’s between a and b =m [AM of a and b]
…(i)
Similarly, sum of n AM’s between a and b = n
[AM of a and b] …(ii)
Sum of m AM’s m ( AM of a and b ) m
=
\
=
n
Sum of n AM’s n ( AM of a and b )
236
Textbook of Algebra
y Example 62. If a, b , c are in AP and p is the AM
between a and b and q is the AM between b and c ,
then show that b is the AM between p and q.
According to the example,
A8
3
=
An - 2 5
Sol. Qa, b, c are in AP.
Þ
…(i)
\
2b = a + c
Q p is the AM between a and b.
a+b
…(ii)
\
p=
2
Qq is the AM between b and c .
b+c
…(iii)
q=
\
2
On adding Eqs. (ii) and (iii), then
a + b b + c a + c + 2b 2b + 2b
[using Eq. (i)]
p +q =
+
=
=
2
2
2
2
p +q
\
p + q = 2b or b =
2
Hence, b is the AM between p and q.
y Example 63. Find n, so that
an + 1 + b n + 1
an + b n
(a ¹ b ) be
the AM between a and b.
an + 1 + bn + 1
Sol. Q
Þ
n
a +b
\
Þ
Þ
a+b
2
2l
l
n +1
n +1
+2= l
n
n +1
- l - l +1=0 Þ
y Example 65. If 11 AM’s are inserted between 28 and
10, then find the three middle terms in the series.
Sol. Let A1, A 2 , A 3 , ..., A11 be 11 AM’s between 28 and 10.
If d be the common difference, then
10 - 28
3
d=
=12
2
Total means = 11 (odd)
æ 11 + 1 ö
Middle mean = ç
\
÷th = 6th = A 6
è 2 ø
Then, three middle terms are A 5 , A 6 and A 7 .
15 41
\
=
A 5 = 28 + 5d = 28 2
2
A 6 = 28 + 6d = 28 - 9 = 19
21 35
and
=
A 7 = 28 + 7d = 28 2
2
\
a+c
i.e., 2b = a + c
2
LHS = a 2 (b + c ) + b 2 (c + a ) + c 2 (a + b )
b=
…(i)
= (a 2b + a 2c ) + b 2 ( 2b ) + (c 2a + c 2b )
= b (a 2 + c 2 ) + ac (a + c ) + 2b 3
+ l + l +1
= b [(a + c )2 - 2ac ] + ac ( 2b ) + 2b 3
n
( l - 1) ( l - 1) = 0
[Qa ¹ b ]
n =0
y Example 64. There are n AM’s between 3 and 54
such that 8th mean is to (n - 2) th mean as 3 to 5.
Find n.
= b (a + c )2 + 2b 3 = b ( 2b )2 + 2b 3 = 6b 3
2
2
(a + b + c )3 = (2b + b )3
9
9
2
= ´ 27b 3 = 6b 3
9
Hence, LHS = RHS
RHS =
Geometric Mean
Sol. Let A1, A 2 , A 3 , ..., An be n AM’s between 3 and 54.
If d be the common difference, then
54 - 3
51
d=
=
n +1 n +1
6n + 6 = 153n - 2346 Þ 147n = 2352
n = 16
Sol. Qa, b, c are in AP.
n
l -1¹0
ln - 1 = 0 Þ ln = 1 = l0
Þ
\
y Example 66. If a, b , c are in AP, then show that
2
a 2 (b + c ) + b 2 (c + a ) + c 2 (a + b ) = (a + b + c ) 3 .
9
é æ a ön + 1
ù é æ a ön
ù æa
ö
+ 1ú = ê ç ÷ + 1ú ç + 1÷
2 êç ÷
è
ø
ø
è
ø
è
b
b
b
êë
úû êë
úû
a
=l
b
2 ln + 1 + 2 = ( ln + 1 ) ( l + 1 )
Let
\
=
é æ a ön + 1
ù
bn + 1 ê ç ÷
+ 1ú
ù
êë è b ø
úû b é æ a ö
= ê ç ÷ + 1ú
n
è
ø
2
b
é
ù
ë
û
æa ö
bn ê ç ÷ + 1ú
è
ø
b
êë
úû
Þ
Þ
n
Þ
5 (3 + 8d ) = 3 [3 + (n - 2)d ] Þ 6 = d (3n - 46)
51
[from Eq. (i)]
6 = (3n - 46)
( n + 1)
…(i)
If three terms are in GP, then the middle term is called the
Geometric Mean (or shortly written as GM) between the
other two, so if a, b, c are in GP, then b is the GM of a
and c.
Chap 03 Sequences and Series
(i) Single GM of n Positive Numbers
n (n + 1 )
= an × r 2
Let n positive numbers be a 1 , a 2 , a 3 , ..., a n and G be the
GM of these numbers, then G = (a 1 a 2 a 3 ... a n ) 1 /n
n
In particular Let a and b be two numbers and G be the
GM between them, then a, G , b are in GP.
Hence, G = ab ; a > 0, b > 0
Remark
1. If a < 0, b < 0, then G = - ab
2. If a < 0, b > 0 or a > 0, b < 0, then GM between a and b does
not exist.
Example
(i) The GM between 4 and 9 is given by
G = 4 ´9 =6
(ii) The GM between - 4 and - 9 is given by
G = -4 ´ - 9 = - 6
(iii) The GM between - 4 and 9 or 4 and - 9 does not
exist.
(- 4) ´ 9 = -1 36 = 6i
i.e.
Two Numbers
Let a and b be two given numbers and G 1 , G 2 , G 3 , K, G n
are n GM’s between them.
Then, a, G 1 , G 2 , G 3 , ..., G n , b will be in GP.
Now, b = (n + 2 ) th term = ar n + 2 - 1
\
r =ç ÷
èa ø
[where r = common ratio] [Remember]
G 1 = ar , G 2 = ar 2 , ..., G n = ar n
1
æb ö n +1
Þ G1 = a ç ÷
èa ø
2
æb ö n +1
, G2 = a ç ÷
èa ø
n
æb ö n +1
, ..., G n = a ç ÷
èa ø
Corollary The product of n geometric means between
a and b is equal to the nth power of the geometric
mean between a and b.
Let two numbers be a and b and G 1 , G 2 , G 3 , ..., G n are n
GM’s between them.
Then, a, G 1 , G 2 , G 3 , ..., G n , b will be in GP.
\ Product of n GM’s between a and b
= G 1 G 2 G 3 K G n = (ar ) (ar 2 ) (ar 3 ) ...(ar n )
=a
æb ö 2
= a ç ÷ = a n /2 b n /2 = ( ab ) n
èa ø
= [GM of a and b] n
1 + 1 + 1 + ... + 1
×r
1 + 2 + 3 + ... + n
[Remember]
Aliter [This method is applicable only when n is even]
G 1 G 2 G 3 KG n - 2 G n -1 G n = (G 1 G n ) (G 2 G n - 1 )
n
(G 3 G n - 2 ) ... factors
2
n
= (ab ) (ab ) (ab ) ... factors
[QTn ´ Tn¢ = a ´ l ]
2
= (ab ) n /2 = ( ab ) n = [GM of a and b] n
y Example 67. If a be one AM and G1 and G 2 be two
geometric means between b and c , then prove that
G13 + G 23 = 2abc .
Sol. Given, a = AM between b and c
a=
and
G1G 2 = bc
From Eqs. (i) and (ii),
G 2 G 2 G 3 + G 23 G13 + G 23
2a = 1 + 2 = 1
=
G 2 G1
G1G 2
bc
Þ
…(i)
\
[from Eq. (i)]
b+c
Þ 2a = b + c
2
Again, b, G1, G 2 , c are in GP.
G1 G 2
c
G2
G2
\
=
=
Þ b = 1 ,c = 2
b
G1 G 2
G1
G2
(ii) Insert n-Geometric Mean Between
1
æb ö n +1
n (n + 1 )
2
n
Þ
4 ´ (-9) = -1 36 = 6i
and
1 ù
é
n êæ b ö n + 1 ú
=a × ç ÷
êè a ø
ú
êë
úû
237
…(i)
…(ii)
[QG1G 2 = bc ]
G13 + G 23 = 2abc
y Example 68. If one geometric mean G and two
arithmetic means p and q be inserted between two
quantities, then show that G 2 = (2p - q ) (2q - p ).
Sol. Let the two quantities be a and b, then
G 2 = ab
Again, a, p , q , b are in AP.
\
p -a=q - p =b-q
Þ
a = 2p - q
b = 2q - p
From Eqs. (i) and (ii), we get
G 2 = (2p - q ) (2q - p )
y Example 69. Find n, so that
the GM between a and b.
Sol. Q
an + 1 + bn + 1
an + bn
= ab
an + 1 + b n + 1
an + b n
…(i)
…(ii)
(a ¹ b ) be
238
Textbook of Algebra
é æ a ön + 1
ù
n +1
æa ö
bn + 1 ê ç ÷
+ 1ú
1
+1
ç
÷
èb ø
a
êë è b ø
úû
æa ö 2
Þ
Þ
=ç ÷
=b
n
n
èb ø
b
é
ù
æa ö
n æa ö
b ê ç ÷ + 1ú
ç ÷ +1
èb ø
êë è b ø
úû
a
Let
=l
b
Þ
Þ
ln + 1 + 1
ln + 1
n+
l
=
1
2
1
l2
n +1
Þ l
1
+1= l
+
1
l2
1
1
2
n+
( l2 - 1) ( l
- 1) = 0
1
l2
-1¹0
\
1
n+
l 2
-1=0
Þ
1
n+
l 2
= 1 = l0
Þ
Þ
1
2
( l2 - 1) - ( l2 -1) = 0
1
Þ
n+
[Qa ¹ b ]
x 2 - 2 Ax + G 2 = 0
[Remember]
(ii) a and b are given by A ± ( A + G ) ( A - G )
[Remember]
(iii)
[Remember]
A >G
Proof Q A is the AM between a and b, then
a +b
…(i)
A=
Þ a + b = 2A
2
and G is the GM between a and b, then
…(ii)
G = ab Þ ab = G 2
x 2 - (a + b ) x + ab = 0
Þ
i.e. x 2 - 2 Ax + G 2 = 0 is the required equation.
1
3
and 9 and verify that their product is the fifth power of
1
the geometric mean between and 9.
3
y Example 70. Insert five geometric means between
1
and 9.
3
1
Then, , G1, G 2 , G 3 , G 4 , G 5 , 9 are in GP.
3
Þ
\
x=
2A ± ( - 2A) 2 - 4 × 1 × G 2
2×1
x = A ± (A + G ) (A - G )
= A ± ( A2 - G 2 )
Now, for real, positive and unequal numbers of a and b,
(A + G ) (A - G ) > 0 Þ (A - G ) > 0
\
A >G
Remark
1. If a and b are real and positive numbers, then A ³ G
2. If a1, a2, a3, ..., an are n positive numbers, then AM ³ GM i.e.,
a1 + a2 + a3 + ... + an
³ ( a1a2 a3 ... an )1/ n
n
1/ 6
æ ö
1
ç9 ÷
Here, r = common ratio = ç ÷ = 3 2 = 3
1
ç ÷
è3ø
1
1
\
G1 = ar = × 3 =
3
3
1
2
G 2 = ar = × 3 = 1
3
1
3
G 3 = ar = × 3 3 = 3
3
1
4
G 4 = ar = × 9 = 3
3
1
5
G 5 = ar = × 9 3 = 3 3
3
Now, Product = G1 ´ G 2 ´ G 3 ´ G 4 ´ G 5
5
æ 1
ö
1
=
´ 1 ´ 3 ´ 3 ´ 3 3 = 9 3 = ( 3) 2 = ç
´ 9÷
3
è 3
ø
1
ù
é
= ê GM of and 9 5ú
3
û
ë
Let a and b be two real, positive and unequal numbers and
A, G are arithmetic and geometric means between them,
then
(i) a and b are the roots of the equation
\a and b are the roots of the equation, then
x 2 - (sum of roots) x + product of roots = 0
1
1
n + = 0 or n = 2
2
Sol. Let G1, G 2 , G 3 , G 4 , G 5 be 5 GM’s between
An Important Theorem
3. (i) If a > 0, b > 0 or a < 0, b < 0 and l1 > 0, l 2 > 0, then
a
b
l1 + l 2 ³ 2 l1l 2
b
a
1
a
if = x > 0 and l1 = l 2 = 1, then x + ³ 2
x
b
(ii) If a > 0, b < 0 or a < 0, b > 0 and l1 > 0, l 2 > 0, then
a
b
l1 + l 2 £ - 2 l1l 2
b
a
1
a
if = x < 0 and l1 > 0, l 2 > 0 then, x + £ - 2
x
b
5
y Example 71. AM between two numbers whose sum is
100 is to the GM as 5 : 4, find the numbers.
Sol. Let the numbers be a and b.
Then,
a + b = 100
or
2A = 100
Chap 03 Sequences and Series
Þ
…(i)
A = 50
A 5
50 5
and given,
[from Eq. (i)]
= Þ
=
G 4
G 4
…(ii)
\
G = 40
From important theorem a, b = A ± ( A + G ) ( A - G )
= 50 ± (50 + 40) (50 - 40)
= 50 ± 30 = 80, 20
a = 80, b = 20
a = 20, b = 80
\
or
y Example 72. If a1 , a 2 , K, an are positive real numbers
whose product is a fixed number c , then find the
minimum value of a1 + a 2 + ... + an - 1 + 3an .
Sol. Q AM ³ GM
a1 + a 2 + ... + an - 1 + 3an
³ (a1a 2 ...an - 13an )1/n = (3c )1/n
n
Þ a1 + a 2 + ... + an - 1 + 3an ³ n (3c )1/n
\
Hence, the minimum value of a1 + a 2 + ... + an - 1 + 3an is
n (3c )1/n .
Harmonic Mean
If three terms are in HP, then the middle term is called the
Harmonic Mean (or shortly written as HM) between the
other two, so if a, b, c are in HP, then b is the HM of a and c.
(i) Single HM of n Positive Numbers
Let n positive numbers be a 1 , a 2 , a 3 , ..., a n and H be the
HM of these numbers, then
n
H=
1
1
1ö
æ 1
+
+ ... + ÷
ç +
an ø
è a1 a2 a 3
In particular Let a and b be two given numbers and H be
the HM between them a, H , b are in HP.
2ab
2
i.e., H =
Hence, H =
1 1
(a + b )
+
a b
Remark
HM of a, b, c is
3
3abc
or
.
1 1 1
ab + bc + ca
+ +
a b c
a +b
.
2
It does not follow that HM between the same numbers is
1 1
+
2
2ab
a b
. The HM is the reciprocals of
.
i.e.,
a +b
(a + b )
2
Caution The AM between two numbers a and b is
239
(ii) Insert n-Harmonic Mean Between
Two Numbers
Let a and b be two given numbers and H 1 , H 2 , H 3 ,K, H n
are n HM’s between them.
Then, a, H 1 , H 2 , H 3 , ..., H n , b will be in HP, if D be the
common difference of the corresponding AP.
\b = (n + 2 ) th term of HP.
1
=
( n + 2 )th term of corresponding AP
1
=
1
+ (n + 2 - 1) D
a
1 1
[Remember]
Þ
D= b a
(n + 1)
1
1
1
1
1
1
= + D,
= + 2 D , …,
= + nD
\
Hn a
H1 a
H2 a
(a - b )
1
1
1
1 2 (a - b )
1
,
= +
Þ
= +
, ...,
H 1 a ab (n + 1) H 2 a ab (n + 1)
Hn
1 n (a - b )
= +
a ab (n + 1)
Corollary The sum of reciprocals of n harmonic
means between two given numbers is n times the
reciprocal of single HM between them.
Let two numbers be a and b and H 1 , H 2 , H 3 , ..., H n are n
HM’s between them. Then, a, H 1 , H 2 , H 3 , ..., H n , b will be
in HP.
1
1
1
1
1 ö
næ 1
\
+
+
+ ... +
= ç
+
÷
H1 H2 H 3
Hn 2 è H1 Hn ø
n
é
ù
êëQS n = 2 (a + l )úû
n æ1
1
ö n æ 1 1ö
ç + D + - D÷ = ç + ÷
ø 2 èa b ø
2 èa
b
n
n
=
=
æ
ö [HM of a and b ]
ç 2 ÷
ç
÷
ç 1 + 1÷
èa bø
=
Aliter [This method is applicable only when n is even]
1
1
1
1
1
1
+
+
+ ... +
+
+
H1 H2 H 3
Hn -2 Hn -1 Hn
1 ö æ 1
1 ö
æ 1
=ç
+
+
÷
÷ +ç
è H1 Hn ø è H2 Hn -1 ø
æ 1
1 ö
n
+ç
+
÷ + ... upto terms
2
è H 3 Hn -2 ø
240
Textbook of Algebra
1
1
ö
ö æ1
æ1
= ç + D + - D ÷ + ç + 2D + - 2D ÷
ø
ø èa
èa
b
b
n
1
ö
æ1
+ ç + 3 D + - 3 D ÷ + ...upto terms
ø
èa
2
b
n
æ 1 1ö æ 1 1ö æ 1 1ö
= ç + ÷ + ç + ÷ + ç + ÷ + ... upto terms
èa b ø èa b ø èa b ø
2
=
n æ 1 1ö
n
n
=
ç + ÷=
2 èa b ø æ
ö ( HM of a and b )
ç 2 ÷
ç
÷
ç 1 + 1÷
èa bø
y Example 75. Find n, so that
\
2xy
x +y
Þ
H + y 2x + x + y 3x + y
=
=
H - y 2x - x - y
x -y
Let
Then,
2 (y - x )
x + 3y - 3x - y
=2
=
=
(y - x )
y-x
H + x H +y
Aliter
=2
+
H - x H -y
æH + x
ö æ
-2y
H +yö
2x
Þç
- 1÷ = ç1 =
÷ Þ
H -yø
H - x H -y
èH - x
ø è
i.e.
an + 1 + bn + 1
H
2y
H
2x
and
=
=
x
x +y
y
x +y
H + x H + y x + 3y 3x + y
\
+
=
+
H - x H -y
y-x
x -y
i.e.
Sol. Q
Þ
By componendo and dividendo, we have
H + x 2y + x + y x + 3y
=
=
H - x 2y - x - y
y-x
and
Hx - xy = - Hy + xy Þ H ( x + y ) = 2xy
2xy
H =
(x + y )
an + b n
n
a +b
=
(a ¹ b ) be
2ab
a+b
é æ a ön + 1
ù
bn + 1 ê ç ÷
+ 1ú b 2 é2 æç a ö÷ ù
ê èb øú
êë è b ø
úû
ë
û
=
n
a
æ
ö
é
ù
æa ö
b ç + 1÷
bn ê ç ÷ + 1ú
èb
ø
è
ø
b
êë
úû
æa ö
ç ÷
èb ø
n +1
æa ö
2ç ÷
èb ø
=
n
a
æ
ö
æa ö
ç ÷ +1
ç ÷ +1
è
ø
b
èb ø
a
=l
b
ln +1 + 1
2l
=
n
l
+1
l +1
+1
( l + 1) ( ln + 1 + 1) = 2 l ( ln + 1)
Þ
ln + 2 + l + ln + 1 + 1 = 2 ln + 1 + 2l
Þ
ln + 2 - ln + 1 - l + 1 = 0
Þ
ln + 1 ( l - 1) - 1 ( l - 1) = 0
Þ
( l - 1)( ln + 1 - 1) = 0
Þ
\
l -1¹0
ln + 1 - 1 = 0
Þ
y Example 74. If a1 , a 2 , a 3 , ..., a10 be in AP and
h1 , h 2 , h 3 , ..., h10 be in HP. If a1 = h1 = 2 and
a10 = h10 = 3, then find the value of a 4 h 7 .
n
Þ
Þ
which is true as, x , H , y are in HP. Hence, the required
result.
[Qa ¹ b ]
ln + 1 = 1 = l 0
n + 1 = 0 or n = - 1
y Example 76. Insert 6 harmonic means
6
between 3 and .
23
Sol. Q a1, a 2 , a 3 , ..., a10 are in AP.
If d be the common difference, then
a - a1 3 - 2 1
d = 10
=
=
9
9
9
3
1 7
\
a 4 = a1 + 3d = 2 + = 2 + =
9
3 3
an + 1 + b n + 1
HM between a and b.
y Example 73. If H be the harmonic mean between x
H+x H+y
and y, then show that
+
=2
H-x H-y
Sol. We have, H =
and given h1, h 2 , h 3 , ..., h10 are in HP.
If D be common difference of corresponding AP.
1
1
1 1
1
h10 h1 3 2
Then,
D=
=
=9
54
9
1
1
18
1 6 1 1 7
Þ h7 =
\
= - =
=
+ 6D = 7
2 54 2 9 18
h 7 h1
7 18
Hence, a 4 × h 7 = ´
=6
3 7
Sol. Let H 1, H 2 , H 3 , H 4 , H 5 , H 6 be 6 HM’s between 3 and
6
are in HP.
23
1 1 1 1 1 1 1 23
,
,
,
,
,
,
, are in AP.
3 H1 H 2 H 3 H 4 H 5 H 6 6
Then, 3, H 1, H 2 , H 3 , H 4 , H 5 , H 6 ,
…(i)
Þ
6
.
23
Chap 03 Sequences and Series
Let common difference of this AP be D.
23 1
21
1
(23 - 2)
D= 6 3 =
=
=
\
7
7 ´6
7 ´6 2
\
1
1
1 1 5
= +D= + =
3 2 6
H1 3
Þ
1
6
=1
5
5
4
1
1
1
3
= + 2D = + 1 =
Þ H2 =
H2 3
3
3
4
Remark
If a1, a2, a3, K, an are n positive numbers, then AM ³ GM ³ HM i.e.,
a1 + a2 + ... + an
n
³ (a1 a2 ...an )1 / n ³
æ1
n
1
1ö
+ ... + ÷
ç +
è a1 a2
an ø
Sign of equality (AH = GM = HM) holds when numbers are equal
i.e., a1 = a2 = ... = an .
H1 =
Important Theorem 2
If A, G , H are arithmetic, geometric and harmonic means
of three given numbers a, b and c, then the equation
having a, b, c as its roots is
1 3 11
1
1
6
= + 3D = + =
Þ H3 =
H3 3
3 2 6
11
3G 3
x - G3 = 0
H
a +b +c
Proof QA = AM of a, b, c =
3
i.e.,
a + b + c = 3 A…(i)
G = GM of a, b, c = (abc ) 1 / 3
x 3 - 3Ax 2 +
7
1
1
1
3
= + 4D = + 2 = Þ H 4 =
H4 3
3
3
7
1 5 17
1
1
6
= + 5D = + =
Þ H5 =
3 2 6
17
H5 3
10
1
1
1
3
= + 6D = + 3 =
Þ H6 =
3
3
10
H6 3
and
Let a and b be two real, positive and unequal numbers and
A, G and H are arithmetic, geometric and harmonic means
respectively between them, then
Proof
(i) Q
Now,
[Remember]
[Remember]
A=
…(ii)
H = HM of a, b, c
3abc
3G 3
3
=
=
=
1 1 1 ab + bc + ca ab + bc + ca
+ +
a b c
[from Eq. (ii)]
3
3G
…(iii)
i.e.
ab + bc + ca =
H
\a, b, c are the roots of the equation
x 3 - (a + b + c ) x 2 + (ab + bc + ca ) x - abc = 0
and
Important Theorem 1
(ii) A > G > H
[Remember]
abc = G 3
i.e.
1 3 6 3 6 3
\ HM’s are 1 , , , , , ×
5 4 11 7 17 10
(i) A, G , H form a GP i.e., G 2 = AH
241
a +b
2ab
, G = ab and H =
2
a +b
3G 3
x -G3 =0
H
[from Eqs. (i), (ii) and (iii)]
Geometrical Proof of A > G > H
Let OA = a unit and OB = b unit and AB be a diameter of
semi-circle. Draw tangent OT to the circle and TM
perpendicular to AB.
x 3 - 3 Ax 2 +
i.e.,
æ a + b ö æ 2ab ö
2
AH = ç
÷ = ab = G
֍
è 2 ø èa +b ø
Therefore, G 2 = AH i.e. A, G , H are in GP.
Remark
The result AH = G2 will be true for n numbers, if they are in GP.
(ii) Q
or
A > G [from important theorem of GM]
A
>1
G
Þ
G
>1
H
Þ
G >H
é A G
ù
2
êëQ G = H Þ G = AH úû
From Eqs. (i) and (ii), we get
A >G > H
…(i)
…(ii)
T
Y
O
A
M
C
B
X
Let C be the centre of the semi-circle.
OA + OB (OC - AC ) + (OC + CB )
=
Q
2
2
2 OC
=
= OC
[Q AC = CB = radius of circle]
2
242
Textbook of Algebra
a +b
[i.e. OC = arithmetic mean]
2
a +b
Þ
A=
2
Now, from geometry
(OT ) 2 = OA ´ OB = ab = G 2
\
OC =
Sol. Let the two numbers be a and b.
1
Given,
G= H
n
Now,
G 2 = AH
\
OT = G , the geometric mean
Now, from similar DOCT and DOMT, we have
(OT ) 2
2ab
ab
OM OT
or OM =
=
=
=
a +b a +b
OC
OT OC
2
\
OM = H , the harmonic mean
Also, it is clear from the figure, that
OC > OT > OM i.e. A > G > H
\
y Example 77. If A x = G y = H z , where A, G, H are AM,
GM and HM between two given quantities, then prove
that x , y , z are in HP.
\
x
y
Q
Þ
=
G 2 = AH Þ (k 1/ y )2 = k 1/ x × k 1/ z
Þ
2 1 1
1 1 1
= + Þ , , are in AP.
y x z
x y z
…(i)
2A + G = 27
Þ
[from Eq. (i)]
2A + 4 A = 27
9
A=
\
2
9
From Eq. (i), G 2 = 4 ´ = 18
2
Now, from important theorem of GM
9
ö
æ 81
a, b = A ± ( A 2 - G 2 ) = ± ç - 18÷
ø
è4
2
9 3
= ± = 6, 3 or 3, 6
2 2
1
times the
n
harmonic mean between two numbers, then show that
the ratio of the two numbers is
1 + (1 - n 2 ) : 1 - (1 - n 2 ).
æH2 H2 ö
±
ç 4 - 2÷
n2
n ø
èn
[1 ± (1 - n 2 )]
[1 + (1 - n 2 )]
a n2
=
b H [1 - (1 - n 2 )]
n2
a : b = 1 + (1 - n 2 ) : 1 - (1 - n 2 )
Sol. Q G > H
ac > b
\
n
n
(ac ) 2 > bn or a 2 c 2 > bn
n
Also,
n
n
…(i)
n
(a 2 - c 2 )2 > 0 Þ an + c n - 2a 2 c 2 > 0
n
n
an + c n > 2 a 2 c 2 > 2bn
Þ
n
\
2
y Example 79. If the geometric mean is
\
H
n2
H
Þ
y Example 78. The harmonic mean of two numbers is
4, their arithmetic mean A and geometric mean G
satisfy the relation 2A + G 2 = 27. Find the numbers.
and given
…(ii)
H
n
Hence, x , y , z are in HP.
Sol. Let the numbers be a and b.
Given,
H =4
Q
G 2 = AH = 4 A
H
y Example 80. If three positive unequal quantities
a, b , c be in HP, then prove that a n + c n > 2b n , n Î N
A = k 1/ x , G = k 1/ y , H = k 1/ z
+ 1/ z
A=
[from Eq. (i)]
a, b = A ± ( A 2 - G 2 ) =
z
k 2 / y = k 1/ x
= AH
n2
n2
Now, from important theorem of GM
Sol. Let A = G = H = k
Then,
H2
Þ
…(i)
n
[from Eq. (i)]
n
a + c > 2b
y Example 81.
(i) If a , b , c , d be four distinct positive quantities in AP,
then
(a) bc > ad
(b) c -1d -1 + a -1b -1 > 2 (b -1d -1 + a -1c -1 - a -1d -1 )
(ii) If a , b , c , d be four distinct positive quantities in GP,
then
(a) a + d > b + c
(b) c -1d -1 + a -1b -1 > 2 (b -1d -1 + a -1c -1 - a -1d -1 )
(iii) If a , b , c , d be four distinct positive quantities in HP,
then
(a) a + d > b + c
(b) ad > bc
Sol. (i) Q a, b, c , d are in AP.
(a) Applying AM > GM
For first three members, b > ac
Þ
b 2 > ac
…(i)
Chap 03 Sequences and Series
and for last three members, c > bd
2
Þ
c > bd
…(ii)
From Eqs. (i) and (ii), we get
b 2c 2 > (ac ) (bd )
Hence,
bc > ad
(b) Applying AM > HM
For first three members,
2ac
b>
a+c
Þ
ab + bc > 2ac
For last three members, c >
2bd
b+d
(ii) Q a, b, c , d are in GP.
(a) Applying AM > GM
For first three members,
a+c
>b
2
Þ
a + c > 2b
b+d
For last three members,
>c
2
Þ
b + d > 2c
From Eqs. (v) and (vi), we get
a + c + b + d > 2b + 2c or a + d > b + c
(b) Applying GM > HM
2ac
For first three members, b >
a+c
Þ
ab + bc > 2ac
2bd
For last three members, c >
b+d
…(vii)
Þ
bc + cd > 2bd
From Eqs. (vii) and (viii), we get
ab + bc + bc + cd > 2ac + 2bd
or
ab + cd > 2 (ac + bd - bc )
Dividing in each term by abcd, we get
…(viii)
c -1d -1 + a -1b -1 > 2 (b -1d -1 + a -1c -1 - a -1d -1 )
…(iii)
bc + cd > 2bd
From Eqs. (iii) and (iv), we get
ab + bc + bc + cd > 2ac + 2bd
or
ab + cd > 2 (ac + bd - bc )
Dividing in each term by abcd, we get
c -1d -1 + a -1b -1 > 2 (b -1d -1 + a -1c -1 - a -1d -1 )
243
…(iv)
(iii) Qa, b, c , d are in HP.
(a) Applying AM > HM
For first three members,
a+c
>b
2
Þ
a + c > 2b
b+d
For last three members,
>c
2
…(ix)
Þ
…(x)
b + d > 2c
From Eqs. (ix) and (x), we get
a + c + b + d > 2b + 2c
…(v)
or
a+d >b+c
(b) Applying GM > HM
For first three members, ac > b
ac > b 2
Þ
…(xi)
For last three members,
bd > c
…(vi)
Þ
bd > c 2
From Eqs. (xi) and (xii), we get
(ac ) (bd ) > b 2c 2
or
ad > bc
…(xii)
244
Textbook of Algebra
#L
1.
Exercise for Session 5
If the AM of two positive numbers a and b (a > b ) is twice of their GM, then a : b is
(a) 2 +
3 :2 -
3
(c) 2 : 7 + 4 3
2.
If A1, A2; G1, G2 and H1, H2 are two arithmetic, geometric and harmonic means, respectively between two
quantities a and b , then which of the following is not the value of ab is?
(a) AH
1 2
(c)G1G2
3.
(b) - 12
(d) None of these
Let n Î N, n > 25. If A, G and H denote the arithmetic mean, geometric mean and harmonic mean of 25 and n.
Then, the least value of n for which A, G, H Î {25, 26, ..., n}, is
(a) 49
(c) 169
5.
(b) A2 H1
(d) None of these
The GM between - 9 and - 16, is
(a) 12
(c) - 13
4.
(b) 7 + 4 3 : 7 - 4 3
(d) 2 : 3
(b) 81
(d) 225
If 9 harmonic means be inserted between 2 and 3, then the value of A +
6
+ 5 (where A is any of the AM’s and
H
H is the corresponding HM), is
(a) 8
(c) 10
6.
If H1, H2, ..., Hn be n harmonic means between a and b , then
(a) n
(c) 2n
7.
H1 + a Hn + b
is
+
H1 - a Hn - b
(b) n + 1
(d) 2n - 2
The AM of two given positive numbers is 2. If the larger number is increased by 1, the GM of the numbers
becomes equal to the AM to the given numbers. Then, the HM of the given numbers is
3
2
1
(c)
2
(a)
8.
(b) 9
(d) None of these
(b)
2
3
(d) 2
If a, a1, a 2, a 3, ..., a 2n, b are in AP and a, b1, b 2, b 3, K , b 2n, b are in GP and h is the HM of a and b , then
an + an + 1
a1 + a 2n a 2 + a 2n - 1
is equal to
+
+ ... +
b1b 2n
b 2 b 2n - 1
bn bn + 1
(a)
2n
h
(c) nh
(b) 2nh
(d)
n
h
Session 6
Arithmetico-Geometric Series (AGS), Sigma (Σ)
Notation, Natural Numbers
Arithmetico-Geometric
Series (AGS)
=a +
Definition
\
A series formed by multiplying the corresponding terms
of an AP and a GP is called Arithmetico - Geometric
Series (or shortly written as AGS)
For example, 1 + 4 + 7 + 10 + ... is an AP and
1 + x + x 2 + x 3 + ... is a GP.
Multiplying together the corresponding terms of these
series, we get
1 + 4 x + 7 x 2 + 10 x 3 + ... which is an
Arithmetico-Geometric Series.
Again, a + (a + d ) + (a + 2d ) + ... + [a + (n - 1)d ] is a typical
AP
and 1 + r + r 2 + ... + r n - 1 is a typical GP.
Sn =
dr (1 - r n - 1 )
- [a + (n - 1) d ]r n
(1 - r )
dr (1 - r n - 1 ) [a + (n - 1 ) d]r n
a
+
(1 - r )
(1 - r )
(1 - r ) 2
…(iii)
Remark
The above result (iii) is not used as standard formula in any
examination. You should follow all steps as shown above.
To Deduce the Sum upto Infinity from the Sum upto
n Terms of an Arithmetico - Geometric Series, when
|r | < 1
From Eq. (iii), we have
[a + (n - 1) d ]r n
a
dr
dr n
Sn =
+
1 - r (1 - r ) 2 (1 - r ) 2
(1 - r )
If | r | < 1, when n ® ¥, r n ® 0
Multiplying together the corresponding terms of these
series, we get
a + (a + d ) r + (a + 2d ) r 2 + ... + [a + (n - 1)d ]r n - 1
and
which is called a standard Arithmetico-Geometric series.
\
Sum of n Terms of an
Arithmetico-Geometric Series
Independent method Let S ¥ denotes the sum to
infinity, then
S ¥ = a + (a + d ) r + (a + 2d ) r 2 + (a + 3d ) r 3
Let the series be a + (a + d ) r + (a + 2d ) r 2 + ...
dr n
(1 - r ) 2
and
[a + (n - 1) d ]r n
both ® 0
(1 - r )
S¥ =
a
dr
+
(1 - r ) (1 - r ) 2
+ ... upto ¥ …(iv)
+ [a + (n - 1)d ]r n - 1
Multiplying both sides of Eq. (iv) by r, we get
rS ¥ = ar + (a + d ) r 2 + (a + 2d ) r 3 + ... upto ¥
Let S n denotes the sum to n terms, then
S n = a + (a + d ) r + (a + 2d ) r 2 + ... + [a + (n - 2 ) d ] r n - 2
+ [a + (n - 1) d ]r
n -1
…(i)
Subtracting Eq. (v) from Eq. (iv), we get
Multiplying both sides of Eq. (i) by r, we get
rS n = ar + (a + d ) r 2 + (a + 2d ) r 3 + ...
(1 - r ) S ¥ = a + (dr + dr 2 + dr 3 + ... upto ¥)
=a +
+ [a + (n - 2 ) d ]r n - 1 + [a + (n - 1) d ]r n …(ii)
Subtracting Eq. (ii) from Eq. (i), we get
(1 - r ) S n = a + (dr + dr 2 + K + dr n - 1 ) - [a + (n - 1) d ]r n
\
S¥ =
dr
(1 - r )
a
dr
+
(1 - r ) (1 - r ) 2
…(v)
246
Textbook of Algebra
y Example 82. Find the sum of the series
4 7 10
1+ + 2 + 3 +K
5 5
5
(i) to n terms.
(ii) to infinity.
=1+
n - 1ù
n -1
5 15 é
( 3n - 2) æ 1 ö
æ1ö
+
ê1 - ç ÷
úç ÷
è5ø
è5ø
4
4 16 ê
úû
ë
n -1
35 (12n + 7 ) æ 1 ö
= ç ÷
è5ø
16
16
3
æ1ö
æ1ö
æ1ö
1 + 4 ç ÷ + 7 ç ÷ + 10 ç ÷ + ...
è5ø
è5ø
è5ø
The series is an Arithmetico-Geometric series, since each
term is formed by multiplying corresponding terms of the
series 1, 4, 7, ... which are in AP and
1 1
1, , 2 , ... which are in GP.
5 5
\
æ1ö
(ii) S ¥ = 1 + 4 ç ÷ + 7
è5ø
æ1ö
\ Tn -1 = (3n - 5) ç ÷
è5ø
n -1
æ1ö
= ( 3n - 2) ç ÷
è5ø
2
2
æ1ö æ1ö
Then, Sn = 1 + 4 ç ÷ + 7 ç ÷ + ...
è5ø è5ø
æ1ö
+ ( 3n - 2) ç ÷
è5ø
Þ
n -1
…(i)
1
Multiplying both the sides of Eq. (i) by , we get
5
2
\
Sol. Let S ¥ = 1 + 4 x + 7 x 2 + 10x 3 + ... upto ¥
n - 1ù
ú
úû
n
æ1ö
- ( 3n - 2) ç ÷
è5ø
or
\
4
7
S¥ =
5
4
35
S¥ =
16
n
æ1ö
+ (3n - 2) ç ÷ …(ii)
è5ø
æ1ö
ç ÷
è5ø
…(iv)
y Example 83. If the sum to infinity of the series
35
1 + 4 x + 7 x 2 + 10x 3 + ... is , find x .
16
1
1
æ1ö
æ1ö
æ1ö
Sn = + 4 ç ÷ + 7 ç ÷ + ... + (3n - 5) ç ÷
è5ø
è5ø
è5ø
5
5
Subtracting Eq. (ii) from Eq. (i), we get
é1 æ1ö2 æ1ö3
1ö
æ
ç1 - ÷ Sn = 1 + 3 ê + ç ÷ + ç ÷ + ... +
è
è5ø
5ø
êë 5 è 5 ø
3
n -1
3
ù
éæ1ö æ1ö2 æ1ö3
4
Sn = 1 + 3 ê ç ÷ + ç ÷ + ç ÷ +... + (n - 1) terms ú
è5ø è5ø
è5ø
5
úû
ëê
n
æ1ö
- ( 3n - 2) ç ÷
è5ø
n - 1ùü
ì1 é
æ1ö
ï ê1 - ç ÷
úï
n
è5ø
ï 5 êë
úû ï
æ1ö
3
)
= 1 + 3í
(
n
2
ç ÷
ý
1
è5ø
ï
ï
15
ï
ï
þ
î
…(iii)
Subtracting Eq. (iv) from Eq. (iii), we get
éæ1ö æ1ö2 æ1ö3
ù
æ 1ö
1
3
1
S
=
+
ê ç ÷ + ç ÷ + ç ÷ + ...upto ¥ ú
ç
÷ ¥
è 5ø
êë è 5 ø è 5 ø è 5 ø
úû
æ 1 ö
÷
ç
3
= 1+3ç 5 ÷ = 1 +
1
4
ç1 - ÷
è
5ø
n -1
(i) Let sum of n terms of the series is denoted by Sn .
æ1ö
+ ( 3n - 5) ç ÷
è5ø
3
1
æ1ö
æ1ö
æ1ö
S ¥ = ç ÷ + 4 ç ÷ + 7 ç ÷ + ... upto ¥
è5ø
è5ø
è5ø
5
n-2
n-2
2
æ1ö
æ1ö
ç ÷ + 10 ç ÷ + ... upto ¥
è5ø
è5ø
1
Multiplying both sides of Eq. (i) by , we get
5
2
ù
é
1 æ1ö
Tn = [n th term of 1, 4, 7, ...] ên th term of 1, , ç ÷ , ...ú
5 è5ø
ë
úû
æ1ö
= [1 + (n - 1) 3] ´ 1 × ç ÷
è5ø
n - 1ü
n
ìï
ï
æ1ö
æ1ö
3
2
1
(
n
)
ç
÷
ç
÷
ý
í
è5ø
è5ø
ïþ
ïî
\ Sn =
Sol. The given series can be written as
2
3
4
…(i)
Multiplying both sides of Eq. (i) by x we get
x S ¥ = x + 4 x 2 + 7 x 3 + 10x 4 + ... upto ¥
…(ii)
Subtracting Eq. (ii) from Eq. (i), we get
(1 - x ) S ¥ = 1 + 3x + 3x 2 + 3x 3 + ... upto ¥
æ x ö ( 1 + 2x )
= 1 + 3 ( x + x 2 + x 3 + ... upto ¥ ) = 1 + 3 ç
÷=
è 1 - x ø (1 - x )
\
Þ
S¥ =
( 1 + 2x )
(1 - x )2
=
35
16
35 ù
é
êëQS ¥ = 16 úû
16 + 32x = 35 - 70x + 35x 2
Þ
35x 2 - 102x + 19 = 0
Þ
(7 x - 19 ) (5x - 1) = 0
19
x¹
7
[Q for infinity series common ratio - 1 < x < 1]
1
x=
5
Hence,
Chap 03 Sequences and Series
n
y Example 84. Find the sum of the series
1 + 2 2 x + 3 2 x 2 + 4 2 x 3 + ... up to ¥, | x | < 1.
2
2
2
= 1 + 4 x + 9 x 2 + 16x 3 + ... upto ¥
r =1
xS ¥ = x + 4 x 2 + 9 x 3 + 16x 4 + ... upto ¥
(1 - x ) S ¥ = 1 + 3x + 5x 2 + 7 x 3 + ... upto ¥
…(ii)
n
¢ ¹ æç T ö÷
T
T
r
r
å
çå r÷
èr = 1 ø
r =1
x (1 - x ) S ¥ = x + 3x 2 + 5x 3 + 7 x 4 + ... upto ¥
æ n ö
ç åT r ÷
ç
÷
èr = 1 ø
æT ö
r ÷
¹
¢÷
æ n ¢ö
r = 1 è Tr ø
ç åT r ÷
÷
ç
èr = 1 ø
n
…(iii)
å çç
4.
…(iv)
[sigma operator is not distributive over division]
Subtracting Eq. (iv) from Eq. (iii), we get
2
3
(1 - x ) (1 - x ) S ¥ = 1 + 2x + 2x + 2x + ... upto ¥
n
å aTr = a
5.
= 1 + 2 ( x + x 2 + x 3 + ... upto ¥)
r =1
æ x ö (1 + x )
=1+2ç
÷=
è 1 - x ø (1 - x )
n
k
n
j =1 i =1
åu n .
æ n öæ n
ö
= ç åT i ÷ ç åT j ÷
ç
֍
÷
èi = 1 ø è j = 1 ø
Examples on Sigma Notation
m
(i)
n=9
9
n =1
1
å a = a + a + a + K upto m times = am
i =1
å a = a + a + a + ... upto n times = an
i.e. å 5 = 5n, å 3 = 3n
5
5
5
(iii) å (i 2 - 3i ) = å i 2 - 3 å i
(ii)
i =1
2
9
(iv)
n
Shortly S is written in place of å.
1
Properties of Sigma Notation
i =1
2
æ 1+1 ö
æ r +1 ö
æ 2 +1 ö
æ 3 +1 ö
å çè 2 r + 4 ÷ø = çè 2 × 1 + 4 ÷ø + çè 2 × 2 + 4 ÷ø + çè 2 × 3 + 4 ÷ø
r =1
=
1
Remark
2
= 55 - 45 = 10
3
Thus, å n = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
i =1
2
2
= (1 + 2 + 3 + 4 + 5 ) - 3 (1 + 2 + 3 + 4 + 5 )
å n, i.e. å n only means the sum of n similar
terms when n varies from 1 to 9.
n
[where a is constant]
[where i and j are independent]
n =1
åT r
åT r
r =1
(1 - x )3
S is a letter of greek alphabets and it is called ‘sigma’. The
symbol sigma ( S ) represents the sum of similar terms.
Usually sum of n terms of any series is represented by
placing S the nth term of the series. But if we have to find
the sum of k terms of a series whose nth term is u n , this
1.
n
å åT i T j
6.
(1 + x )
Sigma (S) Notation
æ n ¢ö
ç åT r ÷
ç
÷
èr = 1 ø
[sigma operator is not distributive over multiplication]
Again, multiplying both sides of Eq. (iii) by x, we get
For example,
r =1
n
3.
Subtracting Eq. (ii) from Eq. (i), we get
will be represented by
r =1
[sigma operator is distributive over
addition and subtraction]
…(i)
Multiplying both sides of Eq. (i) by x, we get
S¥ =
n
åTr ± åTr¢
2
Sol. Here, the numbers 1 , 2 , 3 , 4 , ... i.e. 1, 4, 9, 16, K are not
in AP but 1, 4 - 1 = 3, 9 - 4 = 5, 16 - 9 = 7, K are in AP.
Let S ¥ = 1 + 22 x + 32 x 2 + 4 2 x 3 + ... upto ¥
\
n
± Tr¢ ) =
å (Tr
2.
247
2 3 4 40 + 45 + 48 133
13
+ + =
=
=1
6 8 10
120
120
120
Important Theorems on
å (Sigma) Operator
n
Theorem 1
å f ( r + 1) -
f ( r ) = f (n + 1) - f (1)
r =1
= T1 + T2 + T 3 + ... + Tn , when Tn is the
r =1
general term of the series.
Theorem 2
n
å f (r + 2) - f (r ) = f (n + 2) + f (n + 1) - f (2) - f (1)
r =1
248
Textbook of Algebra
n
Proof (Theorem 1)
n
å f (r + 1) - f (r )
Taking
r =1
= [ f (2 ) - f (1)] + [ f (3 ) - f (2 )]
+ [ f ( 4 ) - f (3 )] + ... + [ f (n + 1) - f (n )]
= f (n + 1) - f (1)
Proof (Theorem 2)
n
n
r =1
r =1
å f (r + 2) - f (r ) = å[ f (r + 2) - f (r + 1)]
+ [ f (r + 1) - f (r )]
=
n
n
r =1
r =1
å f (r + 2) - f (r + 1) + å f (r + 1) - f (r )
= [ f (n + 2 ) - f (2 )] + [ f (n + 1) - f (1)]
= f (n + 2 ) + f (n + 1) - f (2 ) - f (1)
[from Theorem 1]
Remark
1.
n
k
k
r =1
m=1
m=1
å f ( r + k ) - f ( r ) = å f ( n + m) - å f ( m), " k Î N
n
2.
åf (2r + 1) - f (2r - 1) = f (2n + 1) - f (1)
r =1
n
3.
å
on both sides, we get
r=1
n
n
n
n
r =1
r =1
r =1
r =1
å r 3 - (r - 1) 3 = 3 å r 2 - 3 å r + å 1
n 3 - 0 3 = 3 å n2 - 3 å n + n
Þ
…(i)
[from important Theorem 1]
Substituting the value of å n in Eq. (i), we get
3 × n (n + 1)
Þ
n 3 = 3 å n2 +n
2
3n (n + 1)
n
3å n 2 = n 3 +
- n = (2n 2 + 3n + 1)
Þ
2
2
n (n + 1) (2n + 1)
=
2
n (n + 1) (2n + 1)
[Remember]
Þ
å n2 =
6
Independent Proof We know that,
(2 r + 1) 3 - (2 r - 1) 3 = 24 r 2 + 2
n
Taking
å f (2r ) - f (2r - 2) = f (2n) - f (0 )
å on both sides, we get
r =1
r =1
n
Natural Numbers
r =1
The positive integers 1, 2, 3, ... are called natural numbers.
These form an AP with first term and common difference,
each equal to unity.
(i) Sum of the First n Natural Numbers
n (n + 1)
= Sn
2
n (n + 1)
Sn =
2
1 + 2 + 3 + ... + n =
Þ
[Remember]
(ii) Sum of the First n Odd Natural
Numbers
n
1 + 3 + 5 + ... upto n terms = [2 × 1 + (n - 1) × 2 ] = n 2
2
2
[Remember]
Þ
(
2
n
1
)
=
n
å
(iii) Sum of the Squares of the First n
Natural Numbers
n (n + 1) (2n + 1)
6
3
3
2
Proof We know that, r - (r - 1) = 3r - 3r + 1
12 + 2 2 + 3 2 + ... + n 2 = å n 2 =
n
å(2 r + 1) 3 - (2 r - 1) 3 = å(24 r 2 + 2)
r =1
n
n
r =1
r =1
Þ (2n + 1) 3 - 1 3 = 24 å r 2 + 2 å 1
[from points to consider-2]
3
(2n + 1) - 1 = 24 å n 2 + 2n
Þ
Þ
(2n + 1) 3 - (2n + 1) = 24 å n 2
Þ
(2n + 1) [(2n + 1) 2 - 1] = 24 å n 2
Þ
(2n + 1) (2n + 1 + 1) (2n + 1 - 1) = 24 å n 2
å n2 =
Þ
n (n + 1) (2n + 1)
6
[Remember]
(iv) Sum of the Cubes of the First n
Natural Numbers
2
ìn (n + 1) ü
1 3 + 2 3 + 3 3 + ... + n 3 = å n 3 = (å n) = í
ý
î 2 þ
Proof We know that,
r 4 - (r - 1) 4 = 4 r 3 - 6r 2 + 4 r - 1
n
Taking
å
r =1
on both sides, we get
2
Chap 03 Sequences and Series
n
n
n
n
r =1
r =1
r =1
r =1
å r 4 - (r - 1) 4 = 4 å r 3 - 6 å r 2 + 4 å r -
Substituting the values of å n, å n 2 , å n 3 in Eq. (i), we get
n
å1
r =1
n 4 - 0 4 = 4 å n 3 - 6 å n2 + 4 å n - n
Þ
n5 = 5 å n 4 -
Þ
…(i)
[from important theorem 1]
2
Substituting the values of ån and ån in Eq. (i), we get
6 n (n + 1) (2n + 1) 4n (n + 1)
n 4 = 4 ån 3 Þ
+
-n
6
2
Þ 4 å n 3 = n 4 + n (n + 1) (2n + 1) - 2n (n + 1) + n
= n [n 3 + (n + 1) (2n + 1) - 2 (n + 1) + 1]
= n 2 (n + 1) 2
2
ìn (n + 1) ü
2
ån = í
ý = (ån)
î 2 þ
3
[Remember]
Independent Proof We know that,
r 2 (r + 1) 2 - r 2 (r - 1) 2 = 4 r 3
å
r =1
n
n
å r 2 (r + 1) 2 - r 2 (r - 1) 2 = 4 å r 3
r =1
This can be evaluated using the above results.
r =1
2
y Example 85. Find the sum to n terms of the series
1 2 + 3 2 + 5 2 + K upto n terms.
n (n + 1) - 1 × 0 = 4 å n 3
Þ
2
2
[from important Theorem 1]
2
ìn (n + 1) ü
2
ån 3 = í
ý = (ån)
î 2 þ
Þ
3
3
3
2
4n (n + 1) (2n + 1) 4n (n + 1)
+n
6
2
n
= ( 4n 2 + 6n + 2 - 6n - 6 + 3)
3
n ( 4n 2 - 1)
=
3
=
(v) Sum of the Powers Four of the
First n Natural Numbers
1 4 + 2 4 + 3 4 + ... + n 4 = å n 4
n (n + 1) (2n + 1) (3n 2 + 3n - 1)
30
y Example 86. Find the sum to n terms of the series
1 × 2 2 + 2 × 3 2 + 3 × 4 2 + ... .
Proof We know that,
r 5 - (r - 1) 5 = 5r 4 - 10r 3 + 10r 2 - 5r + 1
Sol. Let Tn be the nth term of this series, then
Tn = (n th term of 1, 2, 3, …) (nth term of 22 , 32 , 4 2 , ... )
n
å
on both sides, we get
= n (n + 1)2 = n 3 + 2n 2 + n
r =1
n
år
r =1
Þ
5
n
- (r - 1) = 5 å r - 10
5
4
r =1
5
5
4
n
år
r =1
3
Tn = [1 + (n - 1)2]2 = (2n - 1)2 = 4n 2 - 4n + 1
\ Sum of n terms Sn = STn = 4 Sn 2 - 4 Sn + S1
Corollary 1 + 2 + 3 + K + n = (1 + 2 + 3 + ... + n )
Taking
Sol. Let Tn be the nth term of this series, then
[Remember]
3
=
n
{6n 4 + 15n (n 2 + 2n + 1) - 10 (2n 2 + 3n + 1)
6
+ 15n + 15 - 6 }
n
Þ å n 4 = (6n 4 + 15n 3 + 10n 2 - 1)
30
n (n + 1) (2n + 1) (3n 2 + 3n - 1)
=
30
If nth term of a sequence is given by Tn = an3 + bn2 + cn + d,
where a, b, c, d are constants.
Then, sum of n terms, Sn = STn = a Sn3 + b Sn2 + c Sn + d S1
on both sides, we get
2
ì
5n (n + 1) 2 5 (n + 1) (2n + 1)
\ 5 å n 4 = n ín 4 +
2
3
î
5 (n + 1) ü
+
- 1ý
2
þ
Remark
n
Taking
10 n 2 (n + 1) 2
4
10 n (n + 1) (2n + 1) 5n (n + 1)
+
+n
6
2
=
= n (n 3 + 2n 2 + n )
\
249
3
n
n
+ 10 å r - 5 å r +
r =1
2
r =1
n
å1
r =1
2
n - 0 = 5 å n - 10 å n + 10 å n - 5 å n + n …(i)
[from important Theorem 1]
\ Sum of n terms Sn = STn
2 = S n 3 + 2 S n 2 + Sn
2
ì n (n + 1) (2n + 1)ü n (n + 1)
ì n ( n + 1) ü
=í
ý+
ý + 2í
6
2
2
î
þ
î
þ
250
Textbook of Algebra
n (n + 1) ì n (n + 1) 2(2n + 1)
ü
+
+ 1ý
í
2
2
3
î
þ
n ( n + 1)
=
(3n 2 + 3n + 8n + 4 + 6)
12
n (n + 1) (3n 2 + 11n + 10) n (n + 1) (n + 2) (3n + 5)
=
=
12
12
=
y Example 87. Find the sum of n terms of the series
whose nth terms is (i) n (n - 1) (n + 1) (ii) n 2 + 3n .
y Example 89. Show that
1 × 2 2 + 2 × 3 2 + ... + n × (n + 1) 2
2
2
2
1 × 2 + 2 × 3 + ... + n × (n + 1)
\
\
Tn = n (n + 1)2 and Tn ¢ = n 2 (n + 1)
LHS =
=
\ Sum of n terms Sn = STn = Sn 3 - Sn
ì n ( n + 1) ü
ì n ( n + 1) ü
=í
ý -í
ý
2
2
î
î
þ
þ
n ( n + 1) ì n ( n + 1)
ü
=
- 1ý
í
2
2
î
þ
=
n ( n + 1) ( n - 1) ( n + 2)
4
(ii) We have, Tn = n 2 + 3n
\ Sum of n terms Sn = STn = Sn 2 + S3n
= Sn 2 + (31 + 32 + 33 + ... + 3n )
=
n (n + 1) (2n + 1) 3 (3n - 1)
+
6
( 3 - 1)
=
n (n + 1) (2n + 1) 3 n
+ ( 3 - 1)
6
2
y Example 88. Find the sum of the series
3
3
13 + 2 3 + 3 3
13 1 + 2
+
+
+ ... upto n terms.
1
1+ 3
1+ 3+ 5
ì n ( n + 1) ü
í
ý
(13 + 23 + 33 + ... + n 3 )
2
þ
Tn =
=î
(1 + 3 + 5 + ... + (2n - 1)) n (1 + 2n - 1)
2
( n + 1) 2 1 2
=
= (n + 2n + 1)
4
4
Let Sn denotes the sum of n terms of the given series. Then,
1
Sn = STn = S(n 2 + 2n + 1)
4
1
2
= ( Sn + 2Sn + S1)
4
1 ì n (n + 1) (2n + 1) 2n (n + 1)
ü
= í
+
+ ný
4î
6
2
þ
n
=
{2n 2 + 3n + 1 + 6n + 6 + 6}
24
n (2n 2 + 9n + 13)
Hence, Sn =
24
STn Sn (n + 1)2 S(n 3 + 2n 2 + n )
=
=
2
S(n 3 + n 2 )
STn¢ Sn (n + 1)
Sn 3 + 2 S n 2 + S n
Sn 3 + Sn 2
2
ì n (n + 1) (2n + 1)ü ì n (n + 1)ü
ì n ( n + 1) ü
ý
ý+í
ý + 2í
í
6
2 þ
2 þ
þ î
î
=î
2
ì n (n + 1) (2n + 1)ü
ì n ( n + 1) ü
ý
í
ý +í
2 þ
6
î
þ
î
n (n + 1) ì n (n + 1) 2(2n + 1)
ü
+
+ 1ý
í
2
2
3
î
þ
=
n (n + 1) ì n (n + 1) (2n + 1)ü
+
í
ý
3 þ
2
2
î
1
(3n 2 + 3n + 8n + 4 + 6)
=6
1
(3n 2 + 3n + 4n + 2)
6
(3n 2 + 11n + 10) (3n + 5) (n + 2) (3n + 5)
= RHS
=
=
=
(3n + 1) (n + 2) (3n + 1)
(3n 2 + 7n + 2)
y Example 90. Find the sum of the series
1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + ... upto n terms.
Sol. Here, Tn = {nth term of 1, 2, 3, ...}
Sol. Let Tn be the nth term of the given series. Then,
2
3n + 5
.
3n + 1
Sol. Let Tn and Tn ¢ be the nth terms of the series in numerator
and denominator of LHS. Then,
Sol. (i) We have, Tn = n (n - 1) (n + 1) = n 3 - n
2
=
\
´ {nth term of 2, 3, 4, ...} ´ {nth term of 3, 4, 5, ...}
Tn = n (n + 1) (n + 2) = n 3 + 3n 2 + 2n
\ Sn = Sum of n terms of the series
= STn = Sn 3 + 3Sn 2 + 2Sn
2
ì n ( n + 1) ü
ì n (n + 1) (2n + 1)ü
ì n ( n + 1) ü
=í
ý
ý + 2í
ý + 3í
2 þ
6
2
î
î
þ
î
þ
n ( n + 1) ì n ( n + 1)
ü
=
+ (2n + 1) + 2ý
í
2
2
î
þ
n ( n + 1) 2
=
( n + n + 4n + 2 + 4 )
4
n ( n + 1) ( n + 2) ( n + 3)
=
4
y Example 91. Find sum to n terms of the series
1 + (2 + 3) + (4 + 5 + 6 ) + ... .
Sol. Now, number of terms in first bracket is 1, in the second
bracket is 2, in the third bracket is 3, etc. Therefore, the
number of terms in the nth bracket will be n.
Chap 03 Sequences and Series
Let the sum of the given series of n terms = S
n (n +1)
\Number of terms in S = 1 + 2 + 3 + ... + n =
2
Also, the first term of S is 1 and common difference is also 1.
ì n ( n + 1) ü
í
ý
2 þ
S=î
2
\
é
ö ù
æ n ( n + 1)
- 1÷ × 1ú
ê2 × 1 + çè
ø û
2
ë
n ( n + 1)
( 4 + n 2 + n - 2)
8
n ( n + 1) ( n 2 + n + 2)
=
8
=
y Example 92. Find the sum of the series
1 × n + 2 × (n - 1) + 3 × (n - 2 ) + 4 × (n - 3 ) + ... + (n - 1) × 2 + n × 1
also, find the coefficient of x n - 1 in the expansion of
(1 + 2 x + 3 x 2 + ... + nx n - 1 ) 2 .
Sol. The r th term of the given series is
Tr = r × (n - r + 1) = (n + 1) r - r 2
\Sum of the series
Sn =
n
n
n
r =1
r =1
r =1
å T r = ( n + 1 ) å r - å r 2 = ( n + 1 ) Sn - Sn 2
n (n + 1) n (n + 1) (2n + 1)
= ( n + 1)
2
6
n ( n + 1)
n ( n + 1) ( n + 2)
=
(3n + 3 - 2n - 1) =
6
6
Now,
(1 + 2x + 3x 2 + ... + nx n - 1 )2 = (1 + 2x + 3x 2 + ... + nx n - 1 )
´ (1 + 2x + 3x 2 + ... + nx n - 1 )
\Coefficient of x n - 1 in (1 + 2x + 3x 2 + ... + nx n - 1 )2
= 1 × n + 2 × (n - 1) + 3 × (n - 2) + ... + n × 1
= Sn =
n ( n + 1) ( n + 2)
6
Method of Differences
If the differences of the successive terms of a series are in
AP or GP, we can find the nth term of the series by the
following steps.
Step I Denote the nth term and the sum of the series
upto n terms of the series by Tn and S n ,
respectively.
Step II Rewrite the given series with each term shifted
by one place to the right.
Step III Then, subtract the second expression of S n from
the first expression to obtain Tn .
251
y Example 93. Find the nth term and sum of n terms
of the series, 1 + 5 + 12 + 22 + 35 + ... .
Sol. The sequence of differences between successive terms is 4,
7, 10, 13,... . Clearly, it is an AP with common difference
3. So, let the nth term of the given series be Tn and sum
of n terms be Sn .
Then, Sn = 1 + 5 + 12 + 22 + 35 + ... + Tn - 1 + Tn
…(i)
…(ii)
Sn = 1 + 5 + 12 + 22 + ... + Tn - 1 + Tn
Subtracting Eq. (ii) from Eq. (i), we get
0 = 1 + 4 + 7 + 10 + 13 + ... + (Tn - Tn - 1 ) - Tn
Þ
Tn = 1 + 4 + 7 + 10 + 13 + ... n terms
n
1
= {2 × 1 + (n - 1) 3} = (3n 2 - n )
2
2
3 2 1
Hence, Tn = n - n
2
2
3
1
\ Sum of n terms Sn = STn = Sn 2 - Sn
2
2
3 æ n (n + 1) (2n + 1) ö 1 æ n (n + 1) ö
= ç
÷- ç
÷
ø 2è
ø
2è
6
2
n ( n + 1)
=
(2n + 1 - 1)
4
1
1
= n 2 ( n + 1) = ( n 3 + n 2 )
2
2
y Example 94. Find the nth term and sum of n terms
of the series, 1 + 3 + 7 + 15 + 31 + ... .
Sol. The sequence of differences between successive terms is 2,
4, 8, 16, ... . Clearly, it is a GP with common ratio 2. So, let
the nth term and sum of the series upto n terms of the
series be Tn and Sn , respectively. Then,
…(i)
Sn = 1 + 3 + 7 + 15 + 31 + ... + Tn - 1 + Tn
…(ii)
Sn = 1 + 3 + 7 + 15 + ... + Tn - 1 + Tn
Subtracting Eq. (ii) from Eq. (i), we get
0 = 1 + 2 + 4 + 8 + 16 + ... + (Tn - Tn - 1 ) - Tn
Þ
Tn = 1 + 2 + 4 + 8 + 16 + ... upto n terms
1 × (2n - 1)
=
2-1
Hence,
Tn = (2n - 1)
\ Sum of n terms Sn = STn = S(2n - 1) = S2n - S1
= (2 + 22 + 23 + ... + 2n ) - n
=
2 × (2n - 1)
- n = 2n + 1 - 2 - n
( 2 - 1)
y Example 95. Find the nth term of the series
1 + 4 + 10 + 20 + 35 + ...
Sol. The sequence of first consecutive differences is 3, 6, 10,
15, ... and second consecutive differences is 3, 4, 5, ... .
Clearly, it is an AP with common difference 1. So, let the
nth term and sum of the series upto n terms of the series
be Tn and Sn , respectively.
252
Textbook of Algebra
Then,
…(i)
Sn = 1 + 4 + 10 + 20 + 35 + ... + Tn - 1 + Tn
…(ii)
Sn = 1 + 4 + 10 + 20 + ... + Tn - 1 + Tn
Subtracting Eq. (ii) from Eq. (i), we get
0 = 1 + 3 + 6 + 10 + 15 + ... + (Tn - Tn - 1 ) - Tn
Þ
Tn = 1 + 3 + 6 + 10 + 15 + ... upto n terms
or
…(iii)
Tn = 1 + 3 + 6 + 10 + 15 + ... + t n - 1 + t n
…(iv)
Tn = 1 + 3 + 6 + 10 + ... + t n - 1 + t n
Now, subtracting Eq. (iv) from Eq. (iii), we get
0 = 1 + 2 + 3 + 4 + 5 + ... + (t n - t n - 1 ) - t n
or
t n = 1 + 2 + 3 + 4 + 5 + ... upto n terms
n ( n + 1)
= Sn =
2
1
\
T n = St n = ( Sn 2 + Sn )
2
1 æ n (n + 1) (2n + 1) n (n + 1) ö
= ç
+
÷
ø
2è
6
2
1 n ( n + 1)
1
(2n + 1 + 3) = n (n + 1) (n + 2)
= ×
2
6
6
y Example 96. Find the nth term of the series
1 + 5 + 18 + 58 + 179 + ...
Sol. The sequence of first consecutive differences is 4, 13, 40,
121, ... and second consecutive differences is 9, 27, 81, ... .
Clearly, it is a GP with common ratio 3. So, let the nth
term and sum of the series upto n terms of the series be
Tn and Sn , respectively. Then,
…(i)
Sn = 1 + 5 + 18 + 58 + 179 + ... + Tn - 1 + Tn
…(ii)
Sn = 1 + 5 + 18 + 58 + ... + Tn - 1 + Tn
Subtracting Eq. (ii) from Eq. (i), we get
0 = 1 + 4 + 13 + 40 + 121 + ... + (Tn - Tn - 1 ) - Tn
Þ
Tn = 1 + 4 + 13 + 40 + 121 + ... upto n terms
or
…(iii)
Tn = 1 + 4 + 13 + 40 + 121 + ... + t n - 1 + t n
…(iv)
Tn = 1 + 4 + 13 + 40 + ... + t n - 1 + t n
Now, subtracting Eq. (iv) from Eq. (iii), we get
0 = 1 + 3 + 9 + 27 + 81 + ... + (t n - t n - 1 ) - t n
or
t n = 1 + 3 + 9 + 27 + 81 + ... upto n terms
=
\
1 × (3n - 1) 1 n
= ( 3 - 1)
2
( 3 - 1)
T n = St n =
=
1
( S3n - S1)
2
1
{(3 + 32 + 33 + ... + 3n ) - n }
2
ü
1 ì 3 (3n - 1)
- ný
í
2 î ( 3 - 1)
þ
1
3 n
= ( 3 - 1) - n
2
4
=
Method of Differences (Shortcut)
to find nth term of a Series
The nth term of the series can be written directly on the
basis of successively differences, we use the following steps
to find the nth termTn of the given sequence.
Step I If the first consecutive differences of the given
sequence are in AP, then take
Tn = a (n - 1) (n - 2 ) + b (n - 1) + c , where a, b, c
are constants. Determine a, b, c by putting
n = 1, 2, 3 and putting the values of T1 , T2 , T 3 .
Step II If the first consecutive differences of the given
sequence are in GP, then take
Tn = ar n - 1 + bn + c , where a, b, c are constants
and r is the common ratio of GP. Determine
a, b, c by putting n = 1, 2, 3 and putting the values
of T1 , T2 , T 3 .
Step III If the differences of the differences computed in
Step I are in AP, then take
Tn = a (n - 1) (n - 2 ) (n - 3 ) + b (n - 1) (n - 2 )
+ c (n - 1) + d , where a, b, c , d are
constants.Determine by putting n = 1, 2, 3, 4 and
putting the values of T1 , T2 , T 3 , T 4 .
Step IV If the differences of the differences computed in
Step I are in GP with common ratio r, then take
Tn = ar n - 1 + bn 2 + cn + d , where a, b, c , d are
constants. Determine by putting n = 1, 2, 3, 4 and
putting the values of T1 , T2 , T 3 , T 4 .
y Example 97. Find the nth term and sum of n terms of
the series 2 + 4 + 7 + 11 + 16 + ... .
Sol. The sequence of first consecutive differences is 2, 3, 4, 5,
... . Clearly, it is an AP.
Then, nth term of the given series be
…(i)
Tn = a ( n - 1 ) ( n - 2 ) + b ( n - 1 ) + c
Putting n = 1, 2, 3, we get
2 = c Þ 4 = b + c Þ 7 = 2a + 2b + c
1
After solving, we get a = , b = 2, c = 2
2
Putting the values of a, b, c in Eq. (i), we get
1
1
Tn = ( n - 1 ) ( n - 2 ) + 2 ( n - 1 ) + 2 = ( n 2 + n + 2 )
2
2
1
2
Hence, sum of series Sn = STn = ( Sn + Sn + 2S1)
2
1 æ n (n + 1) (2n + 1) n (n + 1)
ö
= ç
+
+ 2n ÷
ø
2è
6
2
1
= n (n 2 + 3n + 8)
6
Chap 03 Sequences and Series
Putting n = 1, 2, 3, 4, we get
2=a+b+c +d
5 = 3a + 4b + 2c + d
12 = 9a + 9b + 3c + d
31 = 27a + 16b + 4c + d
After, solving these equations, we get
a = 1, b = 0, c = 1, d = 0
Putting the values of a, b, c , d in Eq. (i), we get
y Example 98. Find the nth term and sum of n terms
of the series 5 + 7 + 13 + 31 + 85 + ... .
Sol. The sequence of first consecutive differences is 2, 6, 18,
54, ... . Clearly, it is a GP with common ratio 3. Then, nth
term of the given series be
…(i)
Tn = a (3)n - 1 + bn + c
Putting n = 1, 2, 3, we get
5=a+b+c
7 = 3a + 2b + c
13 = 9a + 3b + c
Solving these equations, we get
a = 1, b = 0, c = 4
Putting the values of a, b, c in Eq. (i), we get
Tn = 3n - 1 + 4
…(ii)
…(iii)
…(iv)
Tn = 3n - 1 + n
(Maha Shortcut)
To find t 1 + t 2 + t 3 + ... + t n - 1 + t n
Let S n = t 1 + t 2 + t 3 + ... + t n - 1 + t n
[1st order differences]
Then,
Dt 1 , Dt 2 , Dt 3 ,..., Dt n - 1
2
2
2
2
[2nd order differences]
D t 1 , D t 2 , D t 3 ,..., D t n - 1
= (1 + 3 + 32 + ... + 3n - 1 ) + 4n
1
(3n - 1)
+ 4n = (3n + 8n - 1)
2
( 3 - 1)
M
\ tn =
y Example 99. Find the nth term of the series
1 + 2 + 5 + 12 + 25 + 46 + ... .
M
n -1
C 0 t1 +
M
n -1
C 1 Dt 1 +
n -1
C 2 D2 t 1 +...
+
Sol. The sequence of first consecutive differences is 1, 3, 7, 13,
21, ... . The sequence of the second consecutive differences
is 2, 4, 6, 8, ... . Clearly, it is an AP. Then, nth term of the
given series be
Tn = a ( n - 1 ) ( n - 2 ) ( n - 3 ) + b ( n - 1 ) ( n - 2 )
+ c (n - 1) + d …(i)
Putting n = 1, 2, 3, 4, we get
…(ii)
1=d
…(iii)
2=c +d
…(iv)
5 = 2b + 2c + d
…(v)
12 = 6a + 6b + 3c + d
After, solving these equations, we get
1
a = , b = 1, c = 1, d = 1
3
Putting the values of a, b, c , d in Eq. (i), we get
1
Tn = (n 3 - 6n 2 + 11n - 6) + (n 2 - 3n + 2) + (n - 1) + 1
3
1
n
= (n 3 - 3n 2 + 5n ) = (n 2 - 3n + 5)
3
3
Sol. The sequence of first consecutive differences is 3, 7, 19,
55, ... . The sequence of the second consecutive differences
is 4, 12, 36, ... . Clearly, it is a GP with common ratio 3.
Then, nth term of the given series be
…(i)
n -1
C r - 1 Dr - 1 t 1
S n = n C 1 t 1 + n C 2 Dt 1 + n C 3 D2 t 1 + ... + n C r Dr t 1
and
where,
Dt 1 = t 2 - t 1 , Dt 2 = t 3 - t 2 , etc.
D2t 1 = Dt 2 - Dt 1 , D3 t 1 = D2t 2 - D2t 1 , etc.
y Example 101. Find the n th term and sum to n terms
of the series 12 + 40 + 90 + 168 + 280 + 432 + ... .
Sol. Let Sn = 12 + 40 + 90 + 168 + 280 + 432 + ..., then
1st order differences are 28, 50, 78, 112, 152, ...
(i.e. Dt 1, Dt 2 , Dt 3 , ...)
and 2nd order differences are
22, 28, 34, 40, ... (i.e. D2t 1, D2t 2 , D2t 3 , ...)
and 3rd order differences are
6, 6, 6, 6,... (i.e. D3t 1, D3t 2 , D3t 3 , ...)
and 4th order differences are
0, 0, 0, 0, ... (i.e. D4t 1, D4t 2 , D4t 3 , ...)
\
t n = 12 × n - 1C 0 + 28 × n - 1C 1 + 22 . n - 1C 2 + 6 × n - 1C 3
= 12 + 28 (n - 1) +
y Example 100. Find the nth term of the series
2 + 5 + 12 + 31 + 86 + ... .
Tn = a (3)n - 1 + bn 2 + cn + d
…(ii)
…(iii)
…(iv)
…(v)
Method of Differences
Hence, sum of the series
Sn = STn = S(3n - 1 + 4 ) = S(3n - 1 ) + 4 S1
= 1×
253
22(n - 1) (n - 2)
2
6 ( n - 1) ( n - 2) ( n - 3)
+
1 ×2 ×3
= n 3 + 5n 2 + 6n
and Sn = 12 × n C 1 + 28 × n C 2 + 22 × n C 3 + 6 × n C 4
254
Textbook of Algebra
28n (n - 1) 22n (n - 1) (n - 2)
+
2
1 ×2 ×3
6× n ( n - 1) ( n - 2) ( n - 3)
+
1 ×2 ×3 × 4
n
2
= (n + 1) (3n + 23n + 46)
12
= 12n +
Vn Method
To find the sum of the series of the forms
I. a 1 a 2 ... a r + a 2 a 3 ... a r + 1 + ... + a n a n + 1 ... a n + r - 1
1
1
1
II.
+
+ ... +
a 1 a 2 ... a r a 2 a 3 ... a r + 1
a n a n + 1 ... a n + r - 1
where, a 1 , a 2 , a 3 , ..., a n , ... are in AP.
Solution of form I Let S n be the sum and Tn be the nth
term of the series, then
S n = a 1 a 2 ... a r + a 2 a 3 ... a r + 1 + ...
+ a n a n + 1 + ... + a n + r - 1
…(i)
\
Tn = a n a n + 1 a n + 2 ... a n + r - 2 a n + r - 1
Let Vn = a n a n + 1 a n + 2 ... a n + r - 2 a n + r - 1 a n + r
[taking one extra factor in Tn for Vn ]
\
Vn - 1 = a n - 1 a n a n + 1 ... a n + r - 3 a n + r - 2 a n + r - 1
Þ Vn - Vn - 1 = a n a n + 1 a n + 2 ... a n + r - 1 (a n + r - a n - 1 )
[from Eq. (i)] …(ii)
= Tn (a n + r - a n - 1 )
Let d be the common difference of AP, then
a n = a 1 + (n - 1) d
Then, from Eq. (ii)
Vn - Vn - 1 = Tn [{a 1 + (n + r - 1) d }
- {a 1 + (n - 2 ) d }] = (r + 1) d Tn
1
Þ
(Vn - Vn - 1 )
Tn =
(r + 1) d
\
n
1
S n = ST n = å T n =
(r + 1) d
n =1
1
=
(Vn - V 0 )
(r + 1) d
Corollary II
1
{n (n + 1)
3
n (n + 1) (n + 2 )
(n + 2 ) - 0 × 1 × 2 } =
3
(ii) 1 × 3 × 5 × 7 + 3 × 5 × 7 × 9 + ... + (2n - 1) × (2n + 1) (2n + 5 )
×(2n + 3 ) ×
1
= {(2n - 1) (2n + 1) (2n + 3 ) (2n + 5 ) (2n + 7 )
11
- ( -1) × 1 × 3 × 5 × 7 }
1
= {(2n - 1) (2n + 1) (2n + 3 ) (2n + 5 ) (2n + 7 ) + 105 }
11
Solution of form II Let d be the common difference of
AP, then a n = a 1 + (n - 1) d
Let sum of the series and nth term are denoted by S n and
Tn , respectively. Then,
1
1
1
Sn =
+
+ ... +
a n a n + 1 ... a n + r - 1
a 1 a 2 ... a r a 2 a 3 ... a r + 1
(i) 1 × 2 + 2 × 3 + ... + n (n + 1) =
\
Tn =
Let Vn =
So,
a n a n + 1a n + 2
an + 1 an + 2
Þ Vn - Vn - 1 =
=
n
å ( Vn - Vn - 1 )
n =1
1
(a n a n + 1 ... a n + r - a 0 a 1 ... a r )
(r + 1) (a 2 - a 1 )
Corollary I If a 1 , a 2 , a 3 , ..., a n , ... are in AP, then
1
3 (a 2 - a 1 )
(a n a n + 1 a n + 2 - a 0 a 1 a 2 )
1
4 (a 2 - a 1 )
(a n a n + 1 a n + 2 a n + 3 - a 0 a 1 a 2 a 3 )
an + 1 an + 2
1
... a n + r - 2 a n + r - 1
1
a n a n + 1 ... a n + r - 3 a n + r - 2
an - an + r - 1
anan + 1 an + 2 - an + r - 2 an + r - 1
S n = ST n =
n
å
n =1
(i) For r = 2, a 1 a 2 + a 2 a 3 + ... + a n a n + 1 =
(ii) For r = 3,
a 1 a 2 a 3 + a 2 a 3 a 4 + ... + a n a n + 1 a n + 2 =
…(ii)
[from Eq. (i)]
= Tn (a n - a n + r - 1 )
= Tn [{a 1 + (n - 1) d } - {a 1 + (n + r - 2 ) d }]
= d (1 - r ) Tn
(Vn - Vn - 1 )
Tn =
d (1 - r )
\
\
1
... a n + r - 2 a n + r - 1
…(i)
[leaving first factor from denominator of Tn ]
1
Vn - 1 =
a n a n + 1 ... a n + r - 3 a n + r - 2
[from important Theorem 1 of S]
=
1
... a n + r - 2 a n + r - 1
(Vn - Vn - 1 )
d (1 - r )
=
1
(Vn - V 0 )
d (1 - r )
[from important Theorem 1 of S]
=
ì
1
1
í
(a 2 - a 1 ) (1 - r ) îa n + 1 a n + 2 ... a n + r - 2 a n + r - 1
-
ü
1
ý
a 1 a 2 ... a r - 2 a r - 1 þ
Chap 03 Sequences and Series
1
(r - 1) (a 2 - a 1 )
ì
ü
1
1
í
ý
îa 1 a 2 ... a r - 1 a n + 1 a n + 2 ... a n + r - 1 þ
Corollary I If a 1 , a 2 , a 3 , ..., a n , ... are in AP, then
(i) For r = 2,
1
1
1
1
+
+ ... +
=
a 1a 2 a 2a 3
a n a n + 1 (a 2 - a 1 )
ì1
ü
æ
a
1
1 n + 1 - a1 ö
÷
í ý= ç
îa 1 a n + 1 þ d è a 1 a n + 1 ø
Hence, the sum of n terms is S n =
=
(ii) For r = 3,
=
1 æ a 1 + nd - a 1 ö
n
÷=
ç
d è a 1a n + 1 ø a 1 a n + 1
1
1
1
+
+ ... +
a 1a 2a 3 a 2a 3a 4
an an + 1 an + 2
ì 1
ü
1
1
í
ý
2 (a 2 - a 1 ) îa 1 a 2 a n + 1 a n + 2 þ
(iii) For r = 4,
1
1
1
+
+ ... +
a 1a 2a 3a 4 a 2a 3a 4a 5
a n a n + 1a n + 2a n + 3
=
ì 1
ü
1
1
í
ý
3 (a 2 - a 1 ) îa 1 a 2 a 3 a n + 1 a n + 2 a n + 3 þ
Corollary II
1
1
1
1
n
(i)
+
+
+ ... +
=
1× 2 2 × 3 3 × 4
n (n + 1) n + 1
1
1
1
1
(ii)
+
+
+ ... +
1× 2 × 3 2 × 3 × 4 3 × 4 × 5
n (n + 1) (n + 2 )
ü 1
1ì 1
1
1
= í
ý= 2 î1 × 2 (n + 1) (n + 2 ) þ 4 2 (n + 1) (n + 2 )
1
1
(iii)
+
1× 3 × 5 × 7 3 × 5 × 7 × 9
1
+ ... +
(2n - 1) (2n + 1) (2n + 3 ) (2n + 5 )
ü
1ì 1
1
= í
ý
6 î1 × 3 × 5 (2n + 1) (2n + 3 ) (2n + 5 ) þ
1
1
=
90 6 (2n + 1) (2n + 3 ) (2n + 5 )
y Example 102. Find the sum upto n terms of the
series 1 × 4 × 7 × 10 + 4 × 7 × 10 × 13 + 7 × 10 × 13 × 16 + ...
Tn = (3n - 2) (3n + 1) (3n + 4 ) (3n + 7 )
(nth term of 7, 10, 13, ...) (nth term of 10, 13, 16, ...)
…(i)
\Vn = (3n - 2) (3n + 1) (3n + 4 ) (3n + 7 ) (3n + 10)
Vn - 1 = (3n - 5) (3n - 2) (3n + 1) (3n + 4 ) (3n + 7 )
Þ
Vn = ( 3n + 10) Tn
[from Eq. (i)]
Vn - 1 = ( 3n - 5) Tn
and
\
Vn - Vn - 1 = 15 Tn
1
Tn = ( Vn - Vn - 1 ))
15
\
\
Sn = STn =
n
1
å 15 (Vn - Vn - 1 )
n =1
=
1
(Vn - V0 )
15
[from important Theorem 1 of S]
1
{(3n - 2) (3n + 1) (3n + 4 ) (3n + 7 ) (3n + 10)
=
15
- ( -2)(1)( 4 )(7 )(10)}
1
{(3n - 2)(3n + 1)(3n + 4 )(3n + 7 )(3n + 10) + 560}
=
15
Shortcut Method
1
Sn =
(last factor of III term - first factor of I term)
=
1
(16 - 1)
(Taking one extra factor in Tn in last
- Taking one extra factor in I term in start)
{(3n - 2)(3n + 1)(3n + 4 )(3n + 7 )(3n + 10)
- ( -2) × 1 × 4 × 7 × 10}
1
{(3n - 2)(3n + 1)(3n + 4 )(3n + 7 )(3n + 10) + 560}
=
15
y Example 103. Find the sum to n terms of the series
1
1
1
+
+
+ ... .
1 × 3 × 5 × 7 × 9 3 × 5 × 7 × 9 × 11 5 × 7 × 9 × 11 × 13
Also, find the sum to infinity terms.
Sol. Let Tn be the nth term of the given series.
1
Then, Tn =
…(i)
(2n - 1)(2n + 1)(2n + 3)(2n + 5)(2n + 7 )
\
Vn =
Vn - 1
1
(2n + 1)(2n + 3)(2n + 5)(2n + 7 )
[leaving first factor from denominator of Tn ]
1
=
(2n - 1)(2n + 1)(2n + 3)(2n + 5)
Þ Vn - Vn - 1 =
Sol. Let Tn be the nth term of the given series.
\Tn = (n th term of 1, 4, 7, ...(nth term of 4, 7, 10, ...)
255
=
1
(2n + 1)(2n + 3)(2n + 5)(2n + 7 )
1
(2n - 1)(2n + 1)(2n + 3)(2n + 5)
(2n - 1) - (2n + 7 )
(2n - 1)(2n + 1)(2n + 3)(2n + 5)(2n + 7 )
= - 8 Tn
[from Eq. (i)]
256
Textbook of Algebra
1
( Vn - Vn - 1 )
8
n
1 n
1
\ Sn = STn = å Tn = - å ( Vn - Vn - 1 ) = - ( Vn - V0 )
8n = 1
8
n =1
\
Tn = -
[from Important Theorem 1 of S]
1
= (V0 - Vn )
8
ü
1ì 1
1
= í
ý
8 î 1 × 3 × 5 × 7 (2n + 1)(2n + 3)(2n + 5)(2n + 7 )þ
=
1
1
840 8 (2n + 1) (2n + 3) (2n + 5) (2n + 7 )
1
1
1
1
- =
- 0=
840
840 ¥ 840
Shortcut Method
1
1
1
+
+
+ ...
1 × 3 × 5 × 7 × 9 3 × 5 × 7 × 9 × 11 5 × 7 × 9 × 11 × 13
1
+
(2n - 1) (2n + 1) (2n + 3) (2n + 5) (2n + 7 )
and S ¥ =
Maha Shortcut Method
1
outside the bracket
8
ö
æ
1
1
1
=
=
= ...÷ and in bracket leaving last
ç i.e.
9 - 1 11 - 3 13 - 5
ø
è
factor of denominator of first term - leaving first factor of
denominator of last term
Taking
i.e., S n =
ö
1æ
1
1
÷
ç
8 è 1 × 3 × 5 × 7 (2n + 1) (2n + 3 ) (2n + 5 ) (2n + 7 ) ø
\ S¥ =
1
1æ
1
ö
- 0÷ =
ç
ø 840
8 è1× 3 × 5 × 7
n
y Example 104. If
…(i)
1ì 1
1
1
1
+
í
8 î 1 × 3 × 5 × 7 3 × 5 × 7 × 9 3 × 5 × 7 × 9 5 × 7 × 9 × 11
1
1
+
+ ...
5 × 7 × 9 × 11 7 × 9 × 11 × 13
1
+
(2n - 1) (2n + 1) (2n + 3) (2n + 5)
ü
1
ý
(2n + 1) (2n + 3) (2n + 5) (2n + 7 )þ
ü
1ì 1
1
ý
í
8 î 1 × 3 × 5 × 7 (2n + 1) (2n + 3) (2n + 5) (2n + 7 )þ
[middle terms are cancelled out]
=
1
1
= Sn
840 8 (2n + 1) (2n + 3) (2n + 5) (2n + 7 )
\ Sum to infinity terms = S ¥ =
1
1
-0=
840
840
[say]
n (n + 1) (n + 2) (n + 3)
,
12
where Tr denotes the rth term of the series. Find
n
1
lim å .
n ®¥
r = 1 Tr
Sol. We have, Tn =
n
n -1
r =1
r =1
å Tr - å Tr
n ( n + 1) ( n + 2) ( n + 3) ( n - 1) n ( n + 1) ( n + 2)
12
12
n ( n + 1) ( n + 2)
=
[(n + 3) - (n - 1)]
12
n ( n + 1) ( n + 2) 1
3
=
=
3
Tn n ( n + 1 ) ( n + 2 )
=
ü
(2n + 7 ) - (2n - 1)
+
ý
(2n - 1) (2n + 1) (2n + 3) (2n + 5) (2n + 7 )þ
=
=
r =1
Now, in each term in denominator
9 - 1 = 11 - 3 = 13 - 5 = ... = (2n + 7 ) - (2n - 1) = 8
Then, Eq. (i) can be written as
1ì 9 - 1
11 - 3
13 - 5
+
+
= í
+ ...
8 î 1 × 3 × 5 × 7 × 9 3 × 5 × 7 × 9 × 11 5 × 7 × 9 × 11 × 13
=
å Tr
\
n
1
åT
n ®¥
lim
n
3
å r ( r + 1) ( r + 2)
n ®¥
= lim
r =1 r
r =1
n
1
å r ( r + 1) ( r + 2)
n ®¥
= 3 lim
r =1
æ 1
ö
1
1
1
= 3 lim ç
+
+
+ ... +
÷
n ® ¥ è 1 ×2 ×3
2 ×3 × 4 3 × 4 ×5
n ( n + 1) ( n + 2) ø
Maha Shortcut Method
ö
1æ 1
1
ç
÷
n ® ¥ 2 è 1 ×2
( n + 1) ( n + 2) ø
= 3 lim
=
3 æ1
ö 3
ç - 0÷ =
ø 4
2 è2
Chap 03 Sequences and Series
#L
1.
Exercise for Session 6
The sum of the first n terms of the series
(b)
3
2
(b) 2n + 1 - n - 2
(b) 9999
1
n (n + 1)
2n
(c)
n+ 1
(c) 10000
(d) 100000
(b) (n + 1) (n + 2) (n + 3)
1
(d) (n + 1) (n + 2) (n + 3)
4
n
n+ 1
2
(d)
n (n + 1)
Sum of the n terms of the series
2n
n+ 1
6n
(c)
n+ 1
If t n =
3
5
7
+
+ 2
+ ... is
12 12 + 22
1 + 22 + 33
4n
n+ 1
9n
(d)
n+ 1
(b)
1
1 1
1
1
equals
(n + 2) (n + 3) for n = 1, 2, 3, ... , then +
+
+ ... +
4
t1 t 2 t 3
t 2003
4006
3006
4006
(c)
3008
4003
3007
4006
(d)
3009
(a)
9.
(d) None of these
(b)
(a)
8.
(c) 2n - n - 2
5
2
1
1
1
1
equals
+
+
+ ... +
1× 2 2 × 3 3 × 4
n (n + 1)
(a)
7.
(d)
The sum of the series 1× 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + ... upto n terms is
(a) n (n + 1) (n + 2)
1
(c) n (n + 1) (n + 2) (n + 3)
4
6.
(c) 2
99th term of the series 2 + 7 + 14 + 23 + 34 + ... is
(a) 9998
5.
(d) 2n - 1
1 + 3 + 7 + 15 + 31 + ... upto n terms equals
(a) 2n + 1 - n
4.
(c) n + 2- n - 1
21/ 4 × 41/ 8 × 81/ 16 × 161/ 32 ... is equal to
(a) 1
3.
1 3 7 15
+ + +
+ ... is
2 4 8 16
(b) 1 - 2-n
(a) 2n - n - 1
2.
257
The value of
(b)
1
1
1
+
+
+ ... upto ¥ is
(1 + a ) (2 + a ) (2 + a ) (3 + a ) (3 + a ) (4 + a )
(where, a is constant)
(a)
1
1+ a
(b)
(c) ¥
10.
2
1+ a
(d) None of these
n
If f ( x ) is a function satisfying f ( x + y ) = f ( x ) f ( y ) for all x , y ÎN such that f (1) = 3 and å f ( x ) = 120. Then, the
x =1
value of n is
(a) 4
(b) 5
(c) 6
(d) None of these
Session 7
Application to Problems of Maxima
and Minima (Without Calculus)
Application to Problems
of Maxima and Minima
(Without Calculus)
Suppose that a 1 , a 2 , a 3 , ..., a n are n positive variables and k
is constant, then
(i) If a 1 + a 2 + a 3 + . . . . . + a n = k (constant), the
value of a 1 a 2 a 3 . . . a n is greatest when
a 1 = a 2 = a 3 = . . . . = a n , so that the greatest
n
ækö
value of a 1 a 2 a 3 . . . a n is ç ÷ .
èn ø
Proof Q AM ³ GM
a 1 + a 2 + a 3 + ... + a n
³ (a 1 a 2 a 3 ... a n ) 1 /n
\
n
k
Þ
³ (a 1 a 2 a 3 ... a n ) 1 /n
n
n
æk ö
or
(a 1 a 2 a 3 ... a n ) £ ç ÷
ènø
Here,
a 1 = a 2 = a 3 = ... = a n
n
æk ö
\ Greatest value of a 1 a 2 a 3 ... a n is ç ÷ .
ènø
y Example 105. Find the greatest value of xyz for
positive values of x , y , z subject to the condition
yz + zx + xy = 12.
Sol. Given, yz + zx + xy = 12 (constant), the value of
(yz ) (zx ) ( xy ) is greatest when yz = zx = xy
Here,
n = 3 and k = 12
3
æ 12 ö
Hence, greatest value of (yz )(zx )( xy ) is ç ÷ i.e. 64.
è3ø
\ Greatest value of x 2y 2z 2 is 64.
Thus, greatest value of xyz is 8.
Aliter
Given yz + zx + xy = 12, the greatest value of (yz )(zx )( xy )
is greatest when
[say]
yz = zx = xy = c
Since, yz + zx + xy = 12
\
c + c + c = 12
Þ
3c = 12 or c = 4
\
yz = zx = xy = 4
Hence, greatest value of (yz ) (zx ) ( xy ) is 4 × 4 × 4
i.e., greatest value of x 2y 2z 2 is 64.
Hence, greatest value of xyz is 8.
y Example 106. Find the greatest value of x 3y 4 , if
2x + 3y = 7 and x ³ 0, y ³ 0.
Sol. To find the greatest value of x 3y 4 or
( x )( x )( x )(y )(y )(y )(y )
Here, x repeats 3 times and y repeats 4 times.
Given,
2x + 3y = 7,
then multiplying and dividing coefficients of x and y by 3
and 4, respectively.
æ 2x ö
æ 3y ö
Rewrite
3ç ÷ + 4 ç ÷ = 7
è 3 ø
è4ø
æ 2x ö æ 2x ö æ 2x ö æ 3y ö æ 3y ö æ 3y ö æ 3y ö
or ç ÷ + ç ÷ + ç ÷ + ç ÷ + ç ÷ + ç ÷ + ç ÷ = 7
è 3 ø è 3 ø è 3 ø è4ø è4ø è4ø è4ø
1
2
3
4
5
6
7
Here, k = 7 and n = 7
Hence, greatest value of
7
æ 2x ö æ 2x ö æ 2x ö æ 3y ö æ 3y ö æ 3y ö æ 3y ö æ 7 ö
ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ is ç ÷ .
è 3 ø è 3 ø è 3 ø è 4 ø è 4 ø è 4 ø è 4 ø è7 ø
or greatest value of
23 × 34
33 × 4 4
x 3y 4 is 1.
Thus, greatest value of x 3y 4 is
32
.
3
(ii) If a 1 a 2 a 3 . . . a n = k (constant), the value of
a 1 + a 2 + a 3 + . . . + a n is least when
a 1 = a 2 = a 3 = . . . = a n , so that the least of
a 1 + a 2 + a 3 + . . . + a n is n (k ) 1 /n .
Proof Q AM ³ GM
a + a 2 + a 3 + ... + a n
³ (a 1 a 2 a 3 ... a n ) 1 /n = (k )1 /n
\ 1
n
a 1 + a 2 + a 3 + ... + a n
Þ
³ (k ) 1 /n
n
or a 1 + a 2 + a 3 + ... + a n ³ n (k ) 1 /n
Here, a 1 = a 2 = a 3 = ... = a n
\ Least value of a 1 + a 2 + a 3 + ... + a n is n (k ) 1 /n
Chap 03 Sequences and Series
y Example 107. Find the least value of 3x + 4 y for
positive values of x and y, subject to the condition
x 2 y 3 = 6.
Sol. Given, x 2y 3 = 6
y Example 109. If a, b , c be positive real numbers,
3
a
b
c
prove that
+
+
³ .
b + c c + a a+b 2
Sol. Arithmetic mean of ( -1) th powers
³ ( - 1) th power of arithmetic mean
or
( x ) ( x ) (y ) (y ) (y ) = 6
Here, x repeats 2 times and y repeats 3 times
æ 3x ö
æ 4y ö
\
3x + 4y = 2 ç ÷ + 3 ç ÷
è 2 ø
è 3 ø
æ b+c ö
ç
÷
èa + b + c ø
æ 3x ö æ 3x ö æ 4y ö æ 4y ö æ 4y ö
=ç ÷+ç ÷+ç ÷+ç ÷+ç ÷
è 2 ø è 2 ø è 3 ø è 3 ø è 3 ø
1
2
3
4
5
multiplying and dividing coefficient of x and y by 2 and 3
respectively and write x 2y 3 = 6
æ 3x ö
Þç ÷
è 2 ø
æ 3x ö
ç ÷
è 2 ø
æ 4y ö
ç ÷
è 3 ø
æ 4y ö
ç ÷
è 3 ø
2
43
æ 4y ö 3
ç ÷ = 2 ´ 3 ´ 6 = 32
è 3 ø 2
3
Hence, least value of
Þ
Þ
Here, n = 5 and k = 32
3x 3x 4y 4y 4y
+
+
+
+
2
2
3
3
3
= 5 (32)1/ 5 = 10
Þ
or
i.e. least value of 3x + 4y = 10
y Example 108. Find the minimum value of
bcx + cay + abz , when xyz = abc .
[constant]
Here,
n =3
Hence, minimum value of bcx + cay + abz = n (k )1/n
= 3 (a 3b 3c 3 )1/ 3 = 3abc
An Important Result
If a i > 0, i = 1, 2, 3, ..., n which are not identical, then
(ii)
+ a 2m
+ ... + a nm
n
If 0 < m < 1
m
æ a + a 2 + ... + a n ö
<ç 1
÷ ;
ø
è
n
Remark
If a1 = a2 = .... = an , then use equal sign in inequalities.
-1
-1
æ c +a ö
æ a+b ö
+ç
÷ +ç
÷
èa + b + c ø
èa + b + c ø
3
c +a
a+b
æ b+c
+
+
ç
+b+c a+b+c a+b+c
a
³ç
3
ç
ç
è
a +b +c a +b + c a + b + c
+
+
-1
b+c
c +a
a+b
æ2ö
³ç ÷
è3ø
3
9
a
b
c
+1+
+1+
+1³
2
b+c
c +a
a+b
9
a
b
c
+
+
³ -3
b+c c +a a+b 2
3
a
b
c
+
+
³
b+c c +a a+b 2
ö
÷
÷
÷
÷
ø
-1
Sol. Since, AM of 2nd powers > 2nd power of AM
2
2
2
1ö
1ö
æ
æ
1
1ö
æ
ça + ÷ + çb + ÷
+
+
+
a
b
ç
÷
è
è
aø
bø
a
b
>ç
\
÷
2
2
ç
÷
è
ø
1
1
-1
-1 2
-1
-1 2
= (a + b + a + b ) = (1 + a + b )
[Qa + b = 1]
4
4
2
2
1ö
1ö
1
æ
æ
-1
-1 2
…(i)
\
ça + ÷ + çb + ÷ > ( 1 + a + b )
è
è
2
aø
bø
Again,
m
a 1m + a 2m + ... + a nm æ a 1 + a 2 + ... + a n ö
>ç
÷ ; If m < 0
è
ø
n
n
or m > 1
a 1m
-1
y Example 110. If a and b are positive and a + b = 1,
2
2
1ö
1ö
25
æ
æ
show that ç a + ÷ + çb + ÷ > .
è
è bø
aø
2
Sol. To find the minimum value of
bcx + cay + abz ,
write,
xyz = abc
or
(bcx ) (cay ) (abz ) = a 3b 3c 3 = k
(i)
259
or
Þ
a -1 + b -1 æ a + b ö
>ç
÷
è 2 ø
2
-1
-1
æ1ö
= ç ÷ =2
è2ø
a -1 + b -1
>2
2
a -1 + b -1 > 4
(1 + a -1 + b -1 ) > 5 or (1 + a -1 + b -1 )2 > 25
25
1
…(ii)
( 1 + a -1 + b -1 ) 2 >
Þ
2
2
From Eqs. (i) and (ii), we get
\
2
2
1ö
1ö
25
æ
æ
ça + ÷ + çb + ÷ >
è
è
2
aø
bø
260
Textbook of Algebra
#L
1.
Exercise for Session 7
The minimum value of 4x + 42 - x , x Î R is
(a) 0
(c) 4
2.
(b) 2
(d) 8
If 0 < q < p, then the minimum value of sin3 q + cosec3 q + 2, is
(a) 0
(c) 4
3.
(b) 2
(d) 8
If a, b , c and d are four real numbers of the same sign, then the value of
(a) [2, ¥)
(c) (4, ¥)
4.
If 0 < x <
(b) [3, ¥)
(d) [4, ¥)
p
, then the minimum value of 2 (sin x + cos x + cosec 2x )3 is
2
(a) 27
(c) 6.75
5.
(b) 13.5
(d) 0
If a + b + c = 3 and a > 0, b > 0, c > 0, then the greatest value of a 2b 3c 2 is
(a)
34 × 210
7
7
3 × 212
(b)
2
(c)
6.
77
310 × 24
77
3 × 22
12
(d)
77
a a a
If x + y + z = a and the minimum value of + + is 81l , then the value of l is
x y z
1
(a)
2
1
(c)
4
7.
a b
c d
+ + + lies in the interval
b
c d a
(b) 1
(d) 2
a, b , c are three positive numbers and abc 2 has the greatest value
1
1
,c =
2
4
1
1
(c) a = b = , c =
4
2
(a) a = b =
1
, then
64
1
3
1
(d) a = b = c =
4
(b) a = b = c =
Shortcuts and Important Results to Remember
18 If a, b, c are in GP, then a + b, 2 b, b + c are in HP.
1 If Tn = An + B, i.e. nth term of an AP is a linear expression
in n, where A, B are constants, then coefficient of n i.e., A
is the common difference.
19 If a, b, c are in AP, then l a , l b, lc are in GP, where
l > 0, l ¹ 1.
2 If S n = Cn 2 + Dn is the sum of n terms of an AP, where C
and D are constants, then common difference of AP is 2C
i.e., 2 times the coefficient of n 2 .
20 If - 1 < r < 1, then GP is said to be convergent, if r < - 1or
r > 1, then GP is said to be divergent and if r = -1, then
series is oscillating.
3
(i) d = Tn - Tn - 1 [n ³ 2 ]
(ii) Tn = S n - S n - 1 [n ³ 2 ]
(a ± b)n , (b ± c )n , (c ± d )n are in GP, " n Î I
(iii) d = S n - 2 S n - 1 + S n - 2 [n ³ 3]
22 If a, b, c are in AP as well as in GP, then a = b = c .
4 If for two different AP’s
Sn
¢
Sn
Then,
Tn
¢
Tn
2
=
An + Bn
Cn 2 + Cn
=
A (2 n - 1) + B
C (2 n - 1) + D
or
An + B
Cn + D
23 The equations a1 x + a2 y = a3 , a4 x + a5 y = a6 has a unique
solution, if a1, a2 , a3 , a4 , a5 , a6 are in AP and common
difference ¹ 0.
24 For n positive quantities a1, a2 , a3 , ..., an
AM ³ GM ³ HM
5 If for two different AP’s
æ n + 1ö
Aç
÷+B
è 2 ø
Tn An + B
Sn
=
=
, then
Tn¢ Cn + D
S n¢ C æç n + 1ö÷ + D
è 2 ø
sign of equality (AM = GM = HM) holds when quantities
are equal
7 If pTp = qTq of an AP, then Tp + q = 0
8 If S p = Sq for an AP, then S p + q = 0
9 If S p = q and Sq = p of an AP, then S p + q = - ( p + q )
é Pn - q ù
10 If Tp = P and Tq = Q for a GP, then Tn = ê n - p ú
ëQ
û
11 If Tm + n = p, Tm - n = q for a GP, then
æq ö
pq , Tn = p ç ÷
è pø
a1 = a2 = a3 = ... = an .
i.e.
6 If Tp = q and Tq = p, then Tp + q = 0, Tr = p + q - r
Tm =
21 If a, b, c , d are in GP, then
1/( p - q )
25 For two positive numbers a and b (AM) (HM)
= (GM) 2 , the result will be true for n numbers, if they are in
GP.
26 If odd numbers of (say 2 n + 1) AM’s, GM’s and HM’s be
inserted between two numbers, then their middle means
[i.e., (n + 1) th mean] are in GP.
27 If a2 , b2 , c 2 are in AP.
1
1
1
are in AP.
,
,
b+c c + a a+ b
Þ
m / 2n
12 If Tm = n, Tn = m for a HP, then
mn
mn
Tm + n =
, Tmn = 1,Tp =
(m + n )
p
13 If Tp = qr , Tq = pr for a HP, then Tr = pq
14 No term of HP can be zero and there is no formula to find
S n for HP.
a-b a
a
a
or .
15 a, b, c are in AP, GP or HP as
= or
b-c a
b
c
16 If A, G, H be AM, GM and HM between a and b, then
ì A, when n = 0
1
an + 1 + bn + 1 ï
= íG, when n = 2
an + bn
ï
î H, when n = - 1
17 If A and G are the AM and GM between two numbers
a, b, then a, b are given by A ± ( A + G ) ( A - G )
28 Coefficient of x n - 1 and x n - 2 in
( x - a1 ) ( x - a2 ) ( x - a3 ) ... ( x - an )
are - (a1 + a2 + a3 + ... + an ) and S a1a2 , respectively
where, S a1a2 =
(S a1 )2 - S a12
.
2
29 1 + 3 + 5 + ... upto n terms = n 2
30 2 + 6 + 12 + 20 + K upto n terms =
31 1 + 3 + 7 + 13 + ... upto n terms =
n (n + 1) (n + 2 )
3
n (n 2 + 2 )
3
32 1 + 5 + 14 + 30 + K upto n terms =
n (n + 1)2 (n + 2 )
12
33 If a1, a2 , a3 , ..., an are the non-zero terms of a non-constant
AP, then
(n - 1)
1
1
1
1
+
+
+ ... +
=
a1a2 a2 a3 a3 a4
an - 1 an
a1an
JEE Type Solved Examples :
Single Option Correct Type Questions
n
l
This section contains 10 multiple choice examples.
Each example has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct.
Ex. 1 If b - c , 2b - l, b - a are in HP, then a -
b-
l
l
, c - are is
2
2
l
,
2
=
=
(b) GP
(d) None of these
2(b - c )(b - a )
Sol. (b) (2b - l ) =
(b - c ) + (b - a )
Þ
(2b - l ) = (2b - (a + c )) = 2 [b - (a + c )b + ac ]
Þ
2
1
2
Sol. (b) Q x =
2b - 2bl + l(a + c ) - 2ac = 0
b 2 - bl +
l
(a + c ) - ac = 0
2
and
2
lö
l
l
æ
Þ çb - ÷ + (a + c ) - ac = 0
è
ø
2
4 2
\
2
Þ
lö
l2 l
æ
- (a + c ) + ac
çb - ÷ =
è
2ø
4 2
Þ
lö
löæ
lö
æ
æ
çb - ÷ = ça - ÷ çc - ÷
è
ø
è
ø
è
2
2
2ø
125
50 2
(a1r )
(b) 1
Ex. 2 Let a 1 , a 2 , a 3 , ..., a 10 are in GP with a 51 = 25 and
101
101
æ1ö
å a i = 125, then the value of å çè a i ÷ø equals
i =1
i =1
a1(1 - r 101 )
= 125
(1 - r )
1
i =1 i
=
=
125
a1
125
(25)
2
=
[from Eq. (i)]
1
5
1
1
1
1
+
+
+...+ 100 =
ai ai r ai r 2
ai r
(d) 4
æ n ( n + 1) ö
occurs as ç
÷
è 2 ø
é æ 1 ö101 ù
ê ç ÷ - 1ú
êë è r ø
úû
æ1 ö
ç - 1÷
èr
ø
1
ù
é
ê here r > 1ú
û
ë
æ 62 ´ 63 ö
ç
÷
è 2 ø
th
æ 63 ´ 64 ö
ç
÷
è 2 ø
th
th
term, the last 62 occurs as
= 1953 rd term and the last 63 occurs as
= 2016 th term.
\ 63 occurs from 1954th term to 2016th term.
Hence, (2011)th term is 63.
[let 0 < r < 1] …(i)
1
ai
(b) 62
(d) 64
Sol. (c) The last 4 occurs as 1 + 2 + 3 + 4 = 10th term. The last n
Þ (a1 + a1r + a1r 2 + ...+ a1r 100 ) = 125
åa
(a 51 )
2
´
1010 + 1 - (1010 - 1)
=1
2
(a) 61
(c) 63
1
1
1
(c)
(d)
(a)5
(b)
5
25
125
Sol. (b) Let 1st term be a and common ratio be r , then
101
1
å a = 125
i =1 i
\
125
(c) 2
=
l
101
a1r
100
l Ex. 4 Consider the sequence 1, 2, 2, 3, 3, 3, …, where n
occurs n times. The number that occurs as 2011th terms is
l
l
l
, b - , c - are in GP.
2
2
2
Þ
=
1
1
1
(999...9 ) = (1020 - 1),
9
9
1
1
y = (999...9 ) = (1010 - 1)
3
3
2
2
z = (999...9 ) = (1010 - 1) ]
9
9
1
1 20
(10 - 1) - (1010 - 1)2
x -y2 9
9
=
2 10
z
(10 - 1)
9
2
Hence, a -
(1 - r )
=
Ex. 3 If x =111 ...1 (20 digits), y = 333 ... 3 (10 digits) and
x -y 2
equals
z = 222 ... 2 (10 digits), then
z
(a)
2
2
a1r
100
l
(a) AP
(c) HP
Þ
(1 - r 101 )
117
Ex. 5 Let S = å
1
, when[ × ] denotes the greatest
2[ r ] + 1
p
integer function and if S = , when p and q are co-primes,
q
the value of p + q is
l
r =1
(a) 20
(b) 76
(c) 19
(d) 69
263
Chap 3 Sequences and Series
Sol. (b) Q
S=
117
1
r ]+ 1
å 2[
r =1
3
5
7
19
18
+
+
+ ... +
+
2 ×1 + 1 2 ×2 + 1 2 ×3 + 1
2 × 9 + 1 2 × 10 + 1
18
6 69
=9+ =9+ =
21
7
7
=
\
Ex. 9 Let l be the greatest integer for which
5 p 2 - 16, 2 pl, l2 are distinct consecutive terms of an AP,
where p ÎR . If the common difference of the AP is
æmö
ç ÷ , m, n Î N and m, n are relative prime, the value of m + n
ènø
is
l
p = 69 and q = 7 Þ p + q = 69 + 7 = 76
Ex. 6 If a, b, c are non-zero real numbers, then the minimum value of the expression
(a 8 + 4a 4 + 1)(b 4 + 3b 2 + 1)(c 2 + 2c + 2 )
equals
a 4b 2
l
(a) 12
(b) 24
Sol. (c) Let P =
(c) 30
a
4
³ 6, b 2 + 3 +
1
b2
2
(d) 60
…(i)
\
B - 4 AC ³ 0
[Q p Î R ]
Þ
16l2 - 4 × 5 × ( l2 - 16) ³ 0
Þ
- l2 + 80 ³ 0 or l2 ³ 80
Þ
- 80 £ l £ 80
\
³ 5 and (c + 1)2 + 1 ³ 1
1
ù
é
êQx + x ³ 2 for x > 0ú
û
ë
l =8
From Eq. (i),
5p 2 - 32p + 48 = 0
Þ
( p - 4 )(5p - 12) = 0
\
p = 4, p =
Þ
Ex. 7 If the sum of m consecutive odd integers is m 4 , then
the first integer is
l
\
Hence,
Sol. (d) Let 2a + 1, 2a + 3, 2a + 5, ... be the AP, then
4
m = (2a + 1) + (2a + 3) + (2a + 5) + ... upto m terms
m
{2(2a + 1) + (m - 1) ×2} = m(2a + 1 + m - 1)
2
m 3 = (2a + 1) + m - 1
Þ
2a + 1 = m 3 - m + 1
\
( 4r + 5 ) 5 -r
is
r = 1 r (5r + 5 )
(a) 5104
(c) 5504
1
2
(c)
(d)
25
125
¥
n
æ (5r + 5) - r ö 1
( 4r + 5)5-r
Sol. (a) å
= lim å ç
÷× r
n ®¥
r = 1 è r (5r + 5) ø 5
r =1 r (5r + 5)
n
æ1
n ®¥
n
æ 1
Hence,
r =1
1
r +1
(r + 1)5
ö
÷
ø
æ1
ö 1
1
1
= lim å ç ÷ = -0 =
n +1
n ®¥
5
ø 5
r = 1 è 5 (n + 1)5
n
(b) 5304
(d) 5704
l
= [ l2 - 14 ]
2
\ l must be an even integer
1 ö1
-
m = 128 and n = 5
m + n = 143
Þ
r =1
å çè r ×5-r
[given]
l2 = 2l [ l2 - 14 ]
å çè r - 5r + 5 ÷ø 5r
n ®¥
= lim
12
,p ¹4
5
[for p = 4 all terms are equal]
Sol. (b) Q2l, l, [ l2 - 14 ] are in GP, then
2
(b)
5
= lim
12
5
Ex. 10 If 2l, l and [ l2 - 14 ], l Î R - {0 } and [.] denotes
the greatest integer function are the first three terms of a GP
in order, then the 51th term of the sequence,
1, 3 l, 6 l, 10 l, . . . is
Ex. 8 The value of å
1
(a)
5
[greatest integer]
l
¥
l
p=
Now, common difference = l2 - 2pl
12
æ 3 ö 128 m
= 64 - 16 ´ = 64 ç1 - ÷ =
=
è 5ø
5
5
n
(b) m 3 + m - 1
(d) m 3 - m + 1
=
(d) 148
5p 2 - 4 pl + l2 - 16 = 0
Þ
\
P ³ 6 × 5 × 1 = 30 Þ P ³ 30
Hence, the required minimum value is 30.
(a) m 3 + m + 1
(c) m 3 - m - 1
(c) 143
4 pl = 5p 2 - 16 + l2
a 4b 2
1 öæ
1ö
æ
= ça 4 + 4 + 4 ÷ çb 2 + 3 + 2 ÷ {(c + 1)2 + 1)}
è
ø
è
a
b ø
1
(b) 138
2
Sol. (c) Q5p - 16, 2pl, l are in AP, then
(a 8 + 4a 4 + 1)(b 4 + 3b 2 + 1)(c 2 + 2c + 2)
Q = a4 + 4 +
(a) 133
l=4
Now, required sequence 1,12, 24, 40, ...
or 1, 4(1 + 2), 4(1 + 2 + 3), 4(1 + 2 + 3 + 4 ), ...
\
51th term = 4(1 + 2 + 3 +... + 51)
51
= 4 × (1 + 51) = 4 × 51 × 26 = 5304
2
264
Textbook of Algebra
JEE Type Solved Examples :
More than One Correct Option Type Questions
n
3 1
1
1
l
l
+ Sn + Sn - - n +n 2 - n +n 1
4 4
2
4 2
2
1
1
ln
ln
l
l
Þ Sn = - n + 2 - n + 1 Þ Sn = 2 - n n+ 1 - n n- 1 < 2
4
2 2
2
2
2
This section contains 5 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
more than one may be correct.
Ex. 11 The first three terms of a sequence are 3, - 1, -1.
The next terms are
Þ Sn =
l
5
5
(d) 27
9
Sol. (b, d) The given sequence is not an AP or GP or HP. It is
an AGP, 3, (3 + d )r ,(3 + 2d )r 2 ,...
(a) 2
(c) -
(b) 3
Ex. 14 If Sr = r + r + r + ... ¥ , r > 0 then which the
following is/are correct.
l
(a) Sr ,S 6 , S12 , S 20 are in AP
(b) S 4 ,S 9 , S16 are irrational
(3 + d )r = -1, (3 + 2d )r 2 = -1
Þ
(c) ( 2S 4 - 1 )2 ,( 2S 5 - 1 )2 ( 2S 6 - 1 )2 are in AP
(3 + d )2 = - (3 + 2d )
Eliminating r, we get
(d) S 2 ,S12 , S 56 are in GP
Sol. (a, b, c, d)
2
d + 8d + 12 = 0 Þ d = -2, - 6,
Þ
r = -1,
then
QSr = r + r + + ... ¥ = r + Sr
1
3
\ Next term is (3 + 3d )r 3 = 3, -
5
9
1 + ( 1 + 4r )
[Qr > 0]
2
Alternate (a) S 2 ,S 6 , S12 , S 20 i.e., 2, 3, 4, 5 are in AP.
1 + 17 1 + 37 1 + 65
Alternate (b) S 4 ,S 9 , S16 i.e.,
are
,
,
2
2
2
irrationals.
Alternate (c)(2S 4 - 1 )2 , (2S 5 - 1 )2 , (2S 6 - 1 )2 i.e., 17, 21, 25 are in AP
\
Ex. 12 There are two numbers a and b whose product is
192 and the quotient of AM by HM of their greatest common
divisor and least common multiple is 169 . The smaller of a
48
and b is
l
(a) 2
(b) 4
(c) 6
have GL = ab = 192
…(i)
AM
æ G + L ö æ G + L ö 169
of G and L is ç
÷=
֍
è 2 ø è 2GL ø
HM
48
169
Þ (G + L ) =
GL =
12
Þ
G + L = 52 but GL
Þ
G = 4, L = 48
169
´ 192 = 132 × 4 2
12
= 192
Þ a = 4, b = 48 or a = 12, b = 16
ln
1 1
2
3
5
Ex. 13 Consider a series + 2 + 3 + 4 + 5 +...+ n .
2 2
2
2
2
2
If Sn denotes its sum to n terms, then Sn cannot be
l
(a) 2
Sol. (a, b, c, d)
Q
(b) 3
(c) 4
Sr =
Alternate (d) S 2 , S12 , S 56 i.e., 2, 4, 8 are in GP.
(d) 12
Sol. (b, d) If G = GED of a and b, L = LCM of a and b, we
2
2
Sr - Sr - r = 0
Þ
(d) 5
1 1
2
3
5
l
Sn = + 2 + 3 + 4 + 5 +... + nn
2 2
2
2
2
2
l ö
3 1 æ1 1
2
3
5
= + ç + 2 + 3 + 4 + 5 +... + nn ÷
4 4 è2 2
2
2
2
2 ø
l
l
l ö 1
1 æ1 1
2
+ ç + 2 + 3 + ... + nn ÷ - - n +n 2 - n n+ 1
2 è2 2
2
2
2 ø 4 2
1 1 1
Ex. 15 If , , are in AP and a, b, -2c are in GP, where
a b c
a, b, c are non-zero, then
l
(a) a 3 + b 3 + c 3 = 3abc
(b) -2a,b , - 2c are in AP
(c) -2a,b , - 2c are in GP
(d) a 2 ,b 2 , 4c 2 are in GP
Sol. (a, b, d)
Q
\
1 1 1
, , are in AP Þ a, b, c are in HP
a b c
2ab
b=
a +c
and a, b, - 2c are in GP, then b 2 = -2ac
…(i)
…(ii)
From Eqs. (i) and (ii), we get
b=
-b 2
Þ a +b +c = 0
a +c
\ a 3 + b 3 + c 3 = 3abc and a, b, -2c are in GP
Þ
a 2 , b 2 , 4c 2 are also in GP and a + b + c = 0
Þ
2b = -2a - 2c
\ -2a,b, - 2c are in AP.
[Qb ¹ 0]
Chap 3 Sequences and Series
265
JEE Type Solved Examples :
Passage Based Questions
n
This section contains 3 solved passages based upon each
of the passage 3 multiple choice examples have to be
answered. Each of these examples has four choices (a), (b),
(c) and (d) out of which ONLY ONE is correct.
Passage I
(Ex. Nos. 16 to 18)
Consider a sequence whose sum to n terms is given by the
quadratic function S n = 3n 2 + 5n.
Sol. (a) Q
\
(b) GP
(c) HP
(d) AGP
2
Sn = 3n + 5n
Tn = Sn - Sn - 1
= (3n 2 + 5n ) - [3(n - 1)2 + 5(n - 1)]
= 3(2n - 1) + 5 = 6n + 2
The nth term is a linear function in n. Hence, sequence
must be an AP.
17. For the given sequence, the number 5456 is the
(a) 153 th term
(c) 707 th term
(b) 932 th term
(d) 909 th term
Tn = 5456
Sol. (d) Given,
Þ
6n + 2 = 5456 Þ 6n = 5454
\
n = 909
\ The number 5456 is the 909 th term.
18. Sum of the squares of the first 3 terms of the given
series is
(a) 1100
(b) 660
(c) 799
(d) 1000
Sol. (b) T12 + T 22 + T 32 = 82 + 14 2 + 202 = 64 + 196 + 400 = 660
(Ex. Nos. 19 to 21)
Let r be the number of identical terms in the two AP’s.
Form the sequence of identical terms, it will be an AP, then
the rth term of this AP make t r £ the smaller of the last
term of the two AP’s.
19. The number of terms common to two AP’s 3, 7, 11, …,
407 and 2, 9, 16, …, 709 is
(b) 21
(c) 28
(b) 191
(c) 211
(d) 213
Sol. (b) Series 3 + 7 + 11 + ... has common difference = 4 and
series 1 + 6 + 11 + ... has common difference = 5
Hence, the series with common terms has common
difference LCM of 4 and 5 which is 20.
The first common terms is 11.
Hence, the series is 11 + 31 + 51 + 71 + ...
\
t 10 = 11 + (10 - 1) (20) = 191
Aliter t n for 3 + 7 + 11 + ... = 3 + (n - 1) ( 4 ) = 4n - 1
and t m for 1 + 6 + 11 + ... = 1 + (m - 1) (5) = 5m - 4
For a common term, 4n - 1 = 5m - 4 i.e., 4n = 5m - 3
For m = 3, n = 3 gives the first common term i.e., 11.
For m = 7, n = 8 gives the second common term i.e., 31.
For m = 11, n = 13 gives the third common term i.e., 51.
Hence, the common term series is 11 + 31 + 51 + ...
\
t 10 = 11 + (10 - 1) 20 = 191
21. The value of largest term common to the sequences 1,
11, 21, 31, ... upto 100 terms and 31, 36, 41, 46, ... upto
100 terms, is
(a) 281
(b) 381
(c) 471
(d) 521
Sol. (d) Sequence 1, 11, 21, 31, ... has common difference = 10
Passage II
(a) 14
20. The 10th common term between the series 3 + 7 + 11 + ...
and 1 + 6 + 11 + ... is
(a) 189
16. The nature of the given series is
(a) AP
Aliter By inspection, first common term to both the series
is 23, second common term is 51, third common term is 79
and so on. These numbers form an AP 23, 51, 79, ...
Since,
T14 = 23 + 13 (28) = 387 < 407
and
T15 = 23 + 14 (28) = 415 > 407
Hence, number of common terms = 14
(d) 35
Sol. (a) Sequence 3, 7, 11, ... , 407 has common difference = 4
and sequence 2, 9, 16, ... , 709 has common difference = 7.
Hence, the sequence with common terms has common
difference LCM of 4 and 7 which is 28.
The first common term is 23.
Hence, the sequence is 23, 51, 79, ... , 387 which has 14 terms.
and sequence 31, 36, 41, 46, ... has common difference = 5.
Hence, the sequence with common terms has common
difference LCM of 10 and 5 which is 10.
The first common term is 31.
Hence, the sequence is 31, 41, 51, 61, 71, ...
…(i)
Now, t 100 of first sequence = 1 + (100 - 1) 10 = 991
and t 100 of second sequence = 31 + (100 - 1) 5 = 526
Value of largest common term < 526
\t n of Eq. (i) is 31 + (n - 1) 10 = 10n + 21
t 50 = 10 ´ 50 + 21 = 521
is the value of largest common term.
Aliter Let mth term of the first sequence be equal to the
nth term of the second sequence, then
1 + (m - 1) 10 = 31 + (n - 1) 5
266
Textbook of Algebra
Þ
Þ
10m - 9 = 5n + 26 Þ 10m - 35 = 5n
2m - 7 = n £ 100 Þ 2m £ 107
1
m £ 53
Þ
2
\ Largest value of m = 53
\ Value of largest term = 1 + (53 - 1) 10 = 521
x
y
z
+
+
³ 06
.
2- x 2-y 2-z
Þ
Thus, minimum value of
n
23. If å a i2 = l, " a i ³ 0 and if greatest and least values of
i =1
2
Passage III
(Ex. Nos. 22 to 24)
We are giving the concept of arithmetic mean of mth
power. Let a1 , a 2 , a 3 , ..., a n be n positive real numbers (not
all equal) and m be a real number. Then,
m
a1m + am2 + am3 + ... + anm æ a1 + a 2 + a 3 + ... + an ö
>ç
÷ ,
è
ø
n
n
æ n ö
ç å a i ÷ are l1 and l 2 respectively, then ( l1 - l 2 ) is
÷
ç
èi = 1 ø
(a) n l
(b) (n - 1) l
(c) (n + 2) l
(d) (n + 1) l
Sol. (b) Q AM of 2nd powers ³ 2nd power of AM
\
if m Î R ~ [0, 1]
However, if m Î(0, 1), then
a1m + am2 + am3 + ... + anm æ a1 + a 2 + a 3 + ... + an ö
<ç
÷
ø
è
n
n
m
a12 + a 22 + a 32 + ... + an2 æ a1 + a 2 + a 3 + ... + an ö
³ç
÷
è
ø
n
n
Þ
Obviously, if m = {0, 1}, then
22. If x > 0, y > 0, z > 0 and x + y + z =1, the minimum
y
x
z
value of
+
+
, is
2 - x 2 -y 2 - z
é6 - (x + y + z )ù
=ê
ú
3
û
ë
-1
æ6 - 1ö
=ç
÷
è 3 ø
-1
=
3
5
or
Þ
1
1
1
9
+
+
³
2- x 2-y 2-z 5
or
2
2
2
18
+
+
³
2- x 2-y 2-z
5
or
1+
2
æ n ö
\ çç å ai ÷÷ £ n l
èi = 1 ø
…(i)
= l + 2 å a1a 2 ³ l
æ n ö
ç å ai ÷ ³ l
ç
÷
èi = 1 ø
\
…(ii)
From Eqs. (i) and (ii), we get
2
æ n ö
l £ çç å ai ÷÷ £ n l
èi = 1 ø
-1
[Q x + y + z = 1]
( 2 - x ) -1 + ( 2 - y ) -1 + ( 2 - z ) -1 3
³
3
5
1é 1
1
1 ù 3
+
+
ú³
ê
3 ë2 - x 2 - y 2 - z û 5
Þ
2
2
(a) 0.2
(b) 0.4
(c) 0.6
(d) 0.8
Sol. (c) Since, AM of ( -1) th powers ³ ( -1) th powers of AM
( 2 - x ) -1 + ( 2 - y ) -1 + ( 2 - z ) -1 æ 2 - x + 2 - y + 2 - z ö
³ç
÷
è
ø
3
3
æ n ö
ç å ai ÷
l çi = 1 ÷
³ç
n
n ÷
ç
÷
ç
÷
è
ø
2
Also, (a1 + a 2 + a 3 + ... + an )2 = a12 + a 22 + a 32 +
... + an2 + 2 å a1 a 2
m
a1m + am2 + am3 + ... + anm æ a1 + a 2 + a 3 + ... + an ö
=ç
÷ .
è
ø
n
n
\
x
y
z
is 0.6.
+
+
2- x 2-y 2-z
\
Then,
l1 = n l and l 2 = l
l1 - l 2 = ( n - 1 ) l
24. If sum of the mth powers of first n odd numbers is l,
" m > 1, then
(a) l < n m
(b) l > n m (c) l < n m + 1 (d) l > n m + 1
Sol. (d) Qm > 1
\
1m + 3m + 5m + ... + (2n - 1)m
n
m
æ 1 + 3 + 5 + ... + (2n - 1) ö
>ç
÷
è
ø
n
m
æn
ö
ç (1 + 2n - 1) ÷
2
=ç
÷ = nm
n
ç
÷
è
ø
x
y
z
18
+1+
+1+
³
2- x
2-y
2-z
5
or
x
y
z
18
+
+
³
-3
2- x 2-y 2-z
5
\ 1m + 3m + 5m + ... + (2n - 1)m > n m + 1
x
y
z
3
+
+
³ = 06
.
2- x 2-y 2-z 5
Hence,
Hence,
l > nm + 1
267
Chap 3 Sequences and Series
JEE Type Solved Examples :
Single Integer Answer Type Questions
n
This section contains 2 examples. The answer to each
example is a single digit integer ranging from 0 to 9
(both inclusive).
Ex. 25 A sequence of positive terms A1 , A 2 , A3 , ..., An
3 (1 + An )
satisfies the relation An + 1 =
. Least integral value
(3 + An )
of A1 for which the sequence is decreasing can be
A1 >
l
Sol. (2) Q An + 1 =
3 ( 1 + An )
3 ( 1 + A1 )
. For n = 1, A 2 =
( 3 + A1 )
( 3 + An )
3 (1 + A 2 )
(3 + A 2 )
æ
3 ( 1 + A1 ) ö
3 ç1 +
÷
( 3 + A1 ) ø 6 + 4 A1 3 + 2 A1
è
=
=
=
3 ( 1 + A1 )
4 + 2 A1
2 + A1
3+
( 3 + A1 )
Q Given, sequence can be written as
3 ( 1 + A1 ) ( 3 + 2 A1 )
, ...
,
A1,
( 3 + A1 ) ( 2 + A1 )
Given, A1 > 0 and sequence is decreasing, then
For n = 2, A 3 =
3 ( 1 + A1 ) 3 ( 1 + A1 ) ( 3 + 2 A1 )
,
>
( 2 + A1 )
( 3 + A1 ) ( 3 + A1 )
Þ
A12 > 3 or A1 > 3
\
A1 = 2
[least integral value of A1]
Ex. 26 When the ninth term of an AP is divided by its
second term we get 5 as the quotient, when the thirteenth
term is divided by sixth term the quotient is 2 and the
remainder is 5, then the second term is
l
Sol. (7) Let a be the first term and d be the common difference,
then T 9 = 5 T 2
Þ
(a + 8d ) = 5 (a + d )
\
4a = 3d
and
T13 = T 6 ´ 2 + 5
Þ
a + 12d = 2 (a + 5d ) + 5
Þ
2d = a + 5
From Eqs. (i) and (ii), we get
a = 3 and d = 4
\
T2 = a + d = 7
…(i)
…(ii)
JEE Type Solved Examples :
Matching Type Questions
n
l
This section contains 2 examples. Examples 27 has three
statements (A, B and C) given in Column I and four
statements (p, q, r and s) in Column II example 28 has
four statements (A, B, C and D) given in Column I and
five statements (p, q, r, s and t) in Column II. Any given
statement in Column I can have correct matching with
one or more statement(s) given in Column II.
åai =
i =1
Column I
Column II
If a1 , a2 , a3 , ... are in AP and a1 + a6 + a10 + a21
30
+ a25 + a30 = 120, then
(p)
å ai is
400
i=1
(B)
30
Then,
Ex. 27
(A)
(A)Qa1, a 2 , a 3 , ... are in AP.
\
a1 + a 30 = a 6 + a 25 = a10 + a 21 = l
Q
a1 + a 6 + a10 + a 21 + a 25 + a 30 = 120
\
3l = 120
Þ
l = 40
25
+ a13 + a17 + a21 + a25 = 112, then
å ai is
(B)Qa1, a 2 , a 3 , ... are in AP.
\ a1 + a 25 = a 5 + a 21
= a 9 + a17 = a13 + a13 = l
(q)
600
\
3l +
i=1
(C)
If a1 , a2 , a3 , ... are in AP and
a1 + a4 + a7 + a10 + a13
(r)
800
å ai is
Þ
i=1
25
(s)
Sol. (A) ® (q); (B) ® (p); (C) ® (s)
1000
Then,
l
= 112
2
7l
= 112
2
l = 32
Þ
16
+ a16 = 375, then
30
(a1 + a 30 ) = 15 ´ l = 15 ´ 40 = 600
2
Q a1 + a 5 + a 9 + a13 + a17 + a 21 + a 25 = 112
If a1 , a2 , a3 , ... are in AP and a1 + a5 + a9
[say]
åai =
i =1
25
25
(a1 + a 25 ) =
´ 32 = 400
2
2
[say]
268
Textbook of Algebra
For c , a, we get
(C)Qa1, a 2 , a 3 , .... are in AP.
\ a1 + a16 = a 4 + a13 = a 7 + a10 = l
Q a1 + a 4 + a 7 + a10 + a13 + a16 = 375
\
3l = 375 \ l = 125
16
åai =
Then,
i =1
Þ
Þ
Ex. 28
Column I
(A) If a > 0, b > 0, c > 0 and the minimum value
of a (b2 + c2 ) + b (c2 + a2 ) + c (a2 + b2 ) is
l abc, then l is
Column II
(p)
2
(B) If a, b, c are positive, a + b + c = 1and the
(q)
æ1 ö æ1 ö æ1 ö
minimum value of ç - 1÷ ç - 1÷ ç - 1÷ is
èa ø èb ø èc ø
l, then l is
4
(C) If a > 0, b > 0, c > 0, s = a + b + c and the
2s
2s
2s
is
minimum value of
+
+
s-a s-b s-c
(l - 1), then l is
(r)
6
(D) If a > 0, b > 0, c > 0, a, b, c are in GP and the
l
l
æ aö
æ cö
the minimum value of ç ÷ + ç ÷ is 2,
è bø
è bø
then l is
(s)
(a + b ) ³ 2 ab
æ1
ö
ç - 1÷
èa
ø
Þ
(A)Q AM ³ GM
ab 2 + ac 2 + bc 2 + ba 2 + ca 2 + cb 2
6
³ (ab 2 × ac 2 × bc 2 × ba 2 × ca 2 × cb 2 )1/ 6 = abc
Here,
l -1=9
\
l = 10
(D) If a, b, c are in GP.
Then, a l , b l , c l are also in GP.
AM ³ GM
al + c l
³ bl
2
a l + c l ³ 2b l
Then,
Þ
l
(B)Q AM ³ GM
For b, c , we get
(b + c )
³ bc
2
Þ
(b + c ) ³ 2 bc
l
æa ö
æc ö
ç ÷ + ç ÷ ³2
èb ø
èb ø
\
l ÎR
Hence, l = 2, 4, 6, 8, 10
Þ
…(i)
3
1
1
1
+
+
s -a s -b s -c
2s
2s
2s
+
+
³9
s -a s -b s -c
\ a (b 2 + c 2 ) + b (c 2 + a 2 ) + c (a 2 + b 2 ) ³ 6 abc
l =6
l =8
3
1
1
1
+
+
s -a s -b s -c
3s - s
3
³
æ 1
3
1
1 ö
+
+
ç
÷
ès - a s - b s - c ø
Þ
Þ
æ1
ö
ç - 1÷ ³ 8
èc
ø
3s - (a + b + c )
³
3
Þ
Sol. (A) ® (r); (B) ® (s); (C) ® (t); (D) ® (p, q, r, s, t)
Þ
æ1
ö
ç - 1÷
èb
ø
\
(C)Q AM ³ HM
(s - a ) + (s - b ) + (s - c )
³
3
10
…(iii)
On multiplying Eqs. (i), (ii) and (iii), we get
(b + c ) (c + a ) (a + b ) ³ 8abc
Þ
(1 - a ) (1 - b ) (1 - c ) ³ 8abc [Qa + b + c = 1]
8
(t)
\
…(ii)
and for a, b, we get
(a + b )
³ ab
2
16
(a1 + a16 )
2
= 8 ´ l = 8 ´ 125 = 1000
l
(c + a )
³ ca
2
(c + a ) ³ 2 ca
[say]
Chap 3 Sequences and Series
269
JEE Type Solved Examples :
Statement I and II Type Questions
n
l
Directions Example numbers 29 to 32 are
Assertion-Reason type examples. Each of these examples
contains two statements:
Statement-1 (Assertion) and Statement-2 (Reason)
Each of these examples also has four alternative choices,
only one of which is the correct answer. You have to select
the correct choice as given below.
(a) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
Ex. 31 Consider an AP with a as the first term and d is
the common difference such that Sn denotes the sum to
n terms and a n denotes the nth term of the AP. Given that for
S
m2
(m ¹ n ).
some m, n Î N , m =
Sn
n2
Statement 1 d = 2a because
2m + 1
a
Statement 2 m =
an
2n + 1
l
Sol. (c) Q
Ex. 29. Statement 1 The sum of first n terms of the series
12 - 22 + 32 - 4 2 + 52 -... can be = ±
n ( n + 1)
.
2
Statement 2 Sum of first n natural numbers is
n (n +1)
.
2
Sol. (a) Clearly, nth term of the given series is negative or
positive according as n is even or odd, respectively.
Case I When n is even, in this case the given series is
12 - 22 + 32 - 4 2 + 52 - 62 + ... + (n - 1)2 - n 2
= (12 - 22 ) + (32 - 4 2 ) + (52 - 62 ) + ...+ [(n - 1)2 - n 2 ]
= (1 - 2) (1 + 2) + (3 - 4 ) (3 + 4 ) + (5 - 6) (5 + 6) + ...
+ (n - 1 - n ) (n - 1 + n )
n ( n + 1)
= - (1 + 2 + 3 + 4 + 5 + 6 + ... + (n - 1) + n ) =
2
Case II When n is odd, in this case the given series is
12 - 22 + 32 - 4 2 + 52 - 62 + ... + (n - 2)2 - (n - 1)2 + n 2
= (12 - 22 ) + (32 - 4 2 ) + (52 - 62 )+... + [(n - 2)2 - (n - 1)2 ] + n 2
= (1 - 2) (1 + 2) + (3 - 4 ) (3 + 4 ) + (5 - 6) (5 + 6) + ...
+ [(n - 2) - (n - 1)] [(n - 2) + (n - 1)] + n 2
= - [1 + 2 + 3 + 4 + 5 + 6 + ... + (n - 2) + (n - 1)] + n 2
( n - 1) ( n - 1 + 1)
n ( n + 1)
=+ n2 =
2
2
It is clear that Statement-1 is true, Statement-2 is true and
Statement-2 is correct explanation for Statement-1.
Let
Sm = m 2k , Sn = n 2k
\
am = Sm - Sm - 1 = m 2k - (m - 1)2 k
Þ
am = (2m - 1) k
Similarly,
an = (2n - 1) k \
Ex. 32 Statement-1 1, 2, 4, 8, … is a GP, 4, 8, 16, 32, …
is a GP and 1 + 4, 2 + 8, 4 + 16, 8 + 32, ... is also a GP.
Statement-2 Let general term of a GP with common ratio r
be Tk + 1 and general term of another GP with common ratio r
be Tk¢ + 1 , then the series whose general term
l
Tk¢¢+ 1 = Tk + 1 + Tk¢ + 1 is also a GP with common ratio r.
Sol. (a) 1, 2, 4, 8, …
Common ratio r = 2
\
\
Statement-2 (AM) (HM) =(GM) 2 is true for positive numbers.
Then,
Sol. (c) If a, b be two real, positive and unequal numbers, then
a+b
2ab
, GM = ab and HM =
2
a+b
AM) (HM) = (GM) 2
\
This result will be true for n numbers, if they are in GP.
Hence, Statement-1 is true and Statement-2 is false.
AM =
am 2m - 1
=
2n - 1
an
Statement-2 is false.
Also, Q
a1 = k , a 2 = 3k , a 3 = 5k , …
Given,
a1 = a = k
\
a1 = a, a 2 = 3a, a 3 = 5a, …
Common difference d = a 2 - a1 = a 3 - a 2 = ...
\
Þ
d = 2a
\ Statement-1 is true.
Ex. 30 Statement 1 If a , b, c are three positive numbers
ö
3abc
æa +b +c ö æ
2 /3
in GP, then ç
.
֍
÷ = (abc )
ø è ab + bc + ca ø
è
3
l
Sm m 2
= 2
Sn
n
Tk + 1 = 1 × (2)k + 1 - 1 = 2k
and 4, 8, 16, 32, ...
Common ratio, r = 2
Tk¢ + 1 = 4 × (2)k + 1 - 1 = 4 × 2k
Tk + 1 + Tk¢ + 1 = 5 × 2k = Tk¢¢+ 1
Common ratio of Tk¢¢+ 1 =
5 × 2k
5 × 2k - 1
= 2, which is true.
Hence, Statement-1 and Statement-2 both are true and
Statement-2 is the correct explanation of Statement-1.
270
Textbook of Algebra
Subjective Type Examples
n
In this section, there are 24 subjective solved examples.
l Ex. 33 In a set of four numbers, the first three are in GP
and the last three are in AP with a common difference of 6. If
the first number is same as the fourth, then find the four
numbers.
Sol. Let the last three numbers in AP, be a, a + 6, a + 12.
[Q 6 is the common difference]
If first number is b, then four numbers are
b, a, a + 6, a + 12
But given, b = a + 12
…(i)
\ Four numbers are a + 12, a, a + 6, a + 12
Since, first three numbers are in GP.
Then,
a 2 = (a + 12) (a + 6)
Þ
a 2 = a 2 + 18a + 72
Þ
2a + 1 = 16 = 24
Þ
a+1= 4
\
a=3
Ex. 35 If n is a root of x 2 (1 - ac ) - x (a 2 + c 2 )
- (1 + ac ) = 0 and if n harmonic means are inserted between
a and c , find the difference between the first and the last
means.
l
Sol. Let H 1, H 2 , H 3 , ..., H n , are n harmonic means, then
a, H 1, H 2 , H 3 , ..., H n ,b are in HP.
If d be the common difference, then
[from Eq. (i)]
n
Ex. 34 Find the natural number a for which
å f (a + k )
k =1
= 16 ( 2 n - 1), where the function f satisfies
f ( x + y ) = f ( x ) f (y ) for all natural numbers x , y and
further f (1) = 2.
f ( x + y ) = f ( x ) f (y )
and
f ( 1) = 2
On putting x = y = 1 in Eq. (i), we get
f ( 1 + 1) = f ( 1) f ( 1) = 2 × 2
\
f ( 2) = 22
Sol. Given,
Þ
1 1
1
1
= -d
= + d and
h1 a
hn c
…(i)
…(ii)
=
M
Similarly,
=
or
M M
f ( l ) = 2l , l Î N
n
n
k =1
k =1
2a
Þ
n
å2k
= 16 (2n - 1)
k =1
Þ 2 (2 + 22 + 23 + ... + 2n ) = 16 (2n - 1)
a
1
æ 1
ac (n + 1) ac (n + 1)
1 ö
= ac (n + 1) ç
÷
cn + a
an + c
è cn + a an + c ø
ac (a - c ) (n 2 - 1)
…(ii)
acn 2 + (a 2 + c 2 ) n + ac
x 2 (1 - ac ) - x (a 2 + c 2 ) - (1 + ac ) = 0.
[from Eq. (iii)]
Then,
å f (a + k ) = 16 (2n - 1) Þ å2a + k
Q
a
c
a
c
=
1 + ad 1 - cd 1 + (a - c ) 1 - (a - c )
c ( n + 1)
a ( n + 1)
But given n is a root of
n 2 (1 - ac ) - n (a 2 + c 2 ) - (1 + ac ) = 0
acn 2 + (a 2 + c 2 )n + ac = n 2 - 1,
then from Eq. (ii), h1 - hn =
f (a + k ) = 2a + k , a + k Î N
\
…(i)
æ
ö
an + c - cn - a
= ac (n + 1) ç
÷
2
2
2
è acn + (a + c )n + ac ø
…(iii)
f ( 4 ) = 24
\
(a - c )
ac (n + 1)
d=
\ h1 - hn =
f ( 3) = 23
On putting x = y = 2 in Eq. (i), we get
f ( 2 + 2) = f ( 2) f ( 2) = 22 × 22
1 1
= + ( n + 2 - 1) d
c a
\
Now, on putting x = 1, y = 2 in Eq. (i), we get
f (1 + 2) = f (1) f (2) = 2 × 22 [from Eqs. (ii) and (iii)]
\
2 (2n - 1)
= 16 (2n - 1)
( 2 - 1)
1 1 1 1
1 1
\ ,
,
,
, ...,
, are in AP.
a H1 H 2 H 3
Hn b
Þ
18a + 72 = 0
\
a=-4
Hence, four numbers are 8, - 4, 2, 8.
l
2a ×
Þ
= 16 (2n - 1)
ac (a - c ) (n 2 - 1)
( n 2 - 1)
= ac (a - c )
Ex. 36 A number consists of three-digits which are in GP
the sum of the right hand and left hand digits exceeds twice
the middle digit by 1 and the sum of the left hand and
middle digits is two third of the sum of the middle and right
hand digits. Find the number.
l
Chap 3 Sequences and Series
Sol. Let the three digits be a, ar and ar 2 , then number is
100a + 10ar + ar
2
20
+ 20r = 50
r
or
2 + 2r 2 = 5r or 2r 2 - 5r + 2 = 0
1
or
(r - 2) (2r - 1) = 0 \
r = 2 or
2
Hence, the three numbers are 10, 20, 40 or 40, 20, 10.
or
…(i)
a + ar 2 = 2ar + 1
Given,
a(r 2 - 2r + 1) = 1
or
…(ii)
a ( r - 1) 2 = 1
2
2
Also, given a + ar = (ar + ar ) Þ 3 + 3r = 2r + 2r 2
3
or
2r 2 - r - 3 = 0 or (r + 1) (2r - 3) = 0
3
\
r = - 1,
2
1
1
For
= ÏI
r = - 1, a =
( r - 1) 2 4
or
Ex. 39 If the sum of m terms of an AP is equal to the sum
of either the next n terms or the next p terms, then prove that
æ 1 1ö
æ 1 1ö
(m + n ) ç - ÷ = (m + p ) ç - ÷ .
èm nø
èm pø
l
Sol. Let the AP be a, a + d , a + 2d , ...
Given, T1 + T 2 + ... + Tm = Tm + 1 + Tm + 2 + ... + Tm + n …(i)
On adding T1 + T 2 + ... + Tm both sides in Eq. (i), we get
2 (T1 + T 2 + ... + Tm ) = T1 + T 2 + ... + Tm + Tm + 1
+ ... + Tm + n
Þ
2Sm = Sm + n
m
m +n
\
2 × [ 2a + (m - 1) d ] =
[ 2a + (m + n - 1) d ]
2
2
Let
2a + (m - 1) d = x
m +n
mx =
{ x + nd }
Þ
2
…(ii)
Þ
(m - n ) x = ( m + n ) nd
Again, T1 + T 2 + ... + Tm = Tm + 1 + Tm + 2 + ... + Tm + p
Similarly,
…(iii)
(m - p ) x = (m + p ) pd
On dividing Eq. (ii) by Eq. (iii), we get
m - n (m + n ) n
=
m - p (m + p ) p
\
r ¹ -1
3
1
[from Eq. (ii)]
For
r = ,a =
=4
2
2
æ3
ö
ç - 1÷
è2
ø
9
3
From Eq. (i), number is 400 + 10 × 4 × + 4 × = 469
4
2
n
l
Ex. 37 Find the value of the expression å
j
i
å å1.
i =1 j =1 k =1
n
Sol.
i
j
n
i
i ( i + 1)
2
i =1
n
å å å1 = å å j = å
i =1 j =1 k =1
i =1 j =1
n ù
é n 2
1
ê åi + åi ú =
ån 2 + ån
2
êë i = 1
i =1 ú
û
1 é n (n + 1) (2n + 1) n (n + 1) ù
= ê
+
ú
2ë
6
2
û
[
]
=
1
2
=
n ( n + 1)
n ( n + 1) ( n + 2)
[ 2n + 1 + 3] =
12
6
Þ
(m - n ) (m + p ) p = (m - p ) (m + n ) n
On dividing both sides by mnp, we get
æ1
1ö
æ1 1 ö
(m + p ) ç - ÷ = (m + n ) ç - ÷
èn m ø
èp mø
l Ex. 38 Three numbers are in GP whose sum is 70. If the
extremes be each multiplied by 4 and the mean by 5, then
they will be in AP. Find the numbers.
Sol. Let the three numbers in GP be
Given,
a
+ a + ar = 70
r
æ1
1ö
æ 1 1ö
Hence, (m + n ) ç - ÷ = (m + p ) ç - ÷
èm n ø
èm p ø
a
, a, ar .
r
…(i)
4a
, 5a, 4ar are in AP.
r
10a a
4a
\
= + ar
+ 4ar or
10a =
4
r
r
5a
[from Eq. (i)]
or
= 70 - a
2
or
5a = 140 - 2a or 7a = 140
\
a = 20
From Eq. (i), we get
20
+ 20 + 20r = 70
r
and
271
Ex. 40 Find the sum of the products of every pair of the
first n natural numbers.
l
Sol. We find that
S = 1 × 2 + 1 × 3 + 1 × 4 + ... + 2 × 3 + 2 × 4 + ... + 3 × 4
+ 3 × 5 + ... + (n - 1) × n …(i)
Q
[1 + 2 + 3 + ... + (n - 1) + n ]2 = 12 + 22 + 32 +
... + (n - 1)2 + n 2
+ 2 [1 × 2 + 1 × 3 + 1 × 4 + ... + 2 × 3 + 2 × 4 +... + 3 × 4 + 3 × 5
+ ... + (n - 1) × n ]
( å n )2 = å n 2 + 2S
[from Eq. (i)]
272
Textbook of Algebra
( å n )2 - å n 2
2
2
n (n + 1) (2n + 1)
ì n ( n + 1) ü
í
ý 2
6
î
þ
=
2
n 2 (n + 1)2 n (n + 1) (2n + 1)
4
6
=
2
n ( n + 1)
=
[3n (n + 1) - 2 (2n + 1)]
24
n (n + 1) (3n 2 - n - 2)
=
24
(n - 1) n (n + 1) (3n + 2)
S=
24
Þ
Hence,
l
Let a, ar , ar 2 , K be a GP with common ratio | r | < 1 [Q given
infinitely GP]
and also given
S ¥ = 27
a
…(i)
so,
= 27
1-r
and
a - ar = f ¢(0)
Þ
a (1 - r ) = f ¢(0) = 3
[Q f ¢ (0) = 3]
…(ii)
\
a (1 - r ) = 3
From Eqs. (i) and (ii), we get
1
1
(1 - r )2 =
Þ 1-r = ±
9
3
1
r =1±
\
3
4 2
4
So,
[Q | r | < 1]
r = , Þ r ¹
3 3
3
2
Hence,
r =
3
S=
p /4
Ex. 41 If I n = ò
0
tan n x dx , show that
1
1
1
1
,
,
,
,Kform an AP. Find its
I 2 + I 4 I3 + I 5 I 4 + I 6 I 5 + I7
common difference.
l
Ex. 43 Solve the following equations for x and y
1
1
log10 x + log10 x + log10 x + ... = y
2
4
Sol. We have,
p/ 4
and
n+2
ò0 (tan x + tan x ) dx
p/ 4
= ò tann x (1 + tan 2 x ) dx
0
In + In + 2 =
=
Hence,
p/ 4
ò0
n
Sol. From the first equation
p/ 4
é tann + 1 x ù
tann x × sec 2 x dx = ê
ú
êë n + 1 úû 0
1
=n +1
In + In + 2
=
1
n +1
On putting n = 2, 3, 4, 5, ...
1
1
1
1
\
= 3,
= 4,
= 5,
= 6, ...
I2 + I4
I3 + I5
I4 + I6
I5 + I7
1
1
1
1
Hence,
,
,
,
, K are in AP with
I2 + I4 I3 + I5 I4 + I6 I5 + I7
common difference 1.
Ex. 42 If the sum of the terms of an infinitely decreasing
GP is equal to the greatest value of the function
f ( x ) = x 3 + 3 x - 9 on the interval [ -5, 3 ] and the difference
between the first and second terms is f ¢ (0 ), then show that
2
the common ratio of the progression is .
3
l
Sol. Given,
\
f ( x ) = x 3 + 3x - 9
f ¢( x ) = 3x 2 + 3
Hence, f ¢( x ) > 0 in [ - 5, 3], then f ( x ) is an increasing
function on [ -5, 3] and therefore, f ( x ) will have greatest
value at x = 3.
Thus, greatest value of f ( x ) is
f ( x ) = 33 + 3 × 3 - 9 = 27
1 + 3 + 5 + ... + ( 2y - 1)
20
=
.
4 + 7 + 10 + ... + (3y + 1) 7 log 10 x
1 1
ì
ü
log10 x í1 + + + ... + ¥ý = y
2
4
î
þ
ü
ì
ï 1 ï
log10 x í
Þ
ý=y
1
ï1 - ï
2þ
î
…(i)
Þ
2 log10 x = y
From the second equation
1 + 3 + 5 + ... + (2y - 1)
20
=
4 + 7 + 10 + ... + (3y + 1) 7 log10 x
y
(1 + 2y - 1)
20
2
Þ
=
y
7
log
10 x
( 4 + 3y + 1)
2
2y
20
Þ
=
3y + 5 7 log10 x
Þ
7y (2 log10 x ) = 60y + 100
[from Eq. (i)]
Þ
7y (y ) = 60y + 100
Þ
7y 2 - 60y - 100 = 0
\
(y - 10) (7y + 10) = 0
-10
\
y = 10, y ¹
7
[because y being the number of terms in series Þ y Î N ]
From Eq. (i), we have
2 log10 x = 10 Þ log10 x = 5
\
x = 10 5
Hence, required solution is x = 10 5, y = 10
273
Chap 3 Sequences and Series
p
Ex. 44 If 0 < x < ,
2
exp [(sin 2 x + sin 4 x + sin 6 x + ... + ¥) log e 2 ] satisfies the
quadratic equation x 2 - 9 x + 8 = 0, find the value of
sin x - cos x
.
sin x + cos x
l
successive rows are 1, 2, 4, 8, ..., respectively.
=
4
=2
2
x
)
2tan x = 20
tan 2 x = 0
\
x =0
Þ
and let
y =8=2
[impossible] [Q x > 0]
{2 + 2
-1
]
- 1}
tan x = 3
\
sin x - cos x
=2- 3
sin x + cos x
[from Eq. (i)]
a + 3c 3a + c
3 æc a ö
+
=1+ ç + ÷
2a
2c
2 èa c ø
…(ii)
[Qa ¹ c ]
3 æc a ö
ç + ÷ >3
2 èa c ø
Þ
( 3 - 1) 2 3 + 1 - 2 3
=
3-1
2
3 æc a ö
ç + ÷ > 1 + 3 or P > 4
2 èa c ø
or
1+
Hence,
a+b
c +b
>4
+
2a - b 2c - b
Ex. 47 Find the sum of n terms of the series
1
2
3
+
+
+ ... .
2
4
2
4
2
1+ 1 + 1
1+ 2 + 2
1 + 3 + 34
l
Ex. 45 The natural numbers are arranged in the form
given below
1
l
2
3
4
5
6
7
9
10
11
12
13
14
..................................................................................
..................................................................................
…(i)
AM > GM
æc a ö
ç + ÷ >2
èa c ø
Q
2
3 -1
3 -1
sin x - cos x tan x - 1
=
´
=
sin x + cos x tan x + 1
3 +1
3 -1
Hence,
8
=
tan 2 x
= 23
=
r -1
r
2ac
2ac
c +
a+c
a+c
+
=
2ac
2ac
2a 2c a+c
a+c
tan x = 3
\
- 1) × 1}
a+
2
Þ
Þ
\
-1
2 1 1
= +
b a c
a+b
c +b
+
P=
2a - b 2c - b
\
Let
y =2
Because y satisfies the quadratic equation.
Then,
y 2 - 9y + 8 = 0
So,
y = 1, 8
2
if
y = 1 = 2tan x
x
+ (2r
Sol. Since, a, b, c are in HP.
tan 2 x
tan 2 x
2
-1
Ex. 46 If a , b, c are in HP, then prove that
a +b
c +b
+
> 4.
2a - b 2c - b
6
2
loge 2 tan x
2tan
{2 × 2r
l
= exp (tan 2 x × loge 2) = exp (loge 2tan
Þ
2
Hence, sum of the numbers in the nth group is
2n - 2 [2n + 2n - 1 - 1].
\ exp [(sin x + sin x + sin x + ... + ¥ ) loge 2]
Now, if
-1
=2
= tan 2 x
1 - sin 2 x
=e
2r
r -2
sin 2 x
2
-1
[Q number of terms in r th group is 2r
Then, sin 2 x + sin 4 x + sin 6 x + K + ¥
=
= 2r
Hence, the sum of the numbers in the r th group is
0 < sin 2 x < 1
\
-1
Tr = 1 × 2r
p
2
0< x <
Sol.
Sol. Let 1st term of the r th group be Tr and the 1st terms of
15
The rth group containing 2r - 1 numbers. Prove that sum
of the numbers in the nth group is 2n - 2 [2n + 2n - 1 - 1].
Sol. The n th term of the given series is Tn =
\Sum of n terms = Sn =
=
n
(1 + n 2 + n 4 )
n
å Tn = å ( 1 + n 2 + n 4 )
n
å (1 + n + n 2 ) (1 - n + n 2 )
274
Textbook of Algebra
æ
1
2
=
ö
1æ
1
1
ç
÷
2
2 è1 - 1 + 1 1 + n + n ø
=
1
ö
÷
1+n +n ø
=
å çè 1 - n + n 2
2
(n + n )
2
2 (1 + n + n )
=
-
1
Ex. 49 Find the sum of the first n terms of the series
1 + 3 × 2 2 + 3 3 + 3 × 4 2 + 5 3 + 3 × 6 2 + ...
l
2
3
[ by property]
If (i) n is even, (ii) n is odd.
Sol. Case I If n is even.
n ( n + 1)
2
2 ( n + n + 1)
Let
n = 2m
\
S = 13 + 3 × 22 + 33 + 3 × 4 2 + 53 + 3 × 62 +
... + (2m - 1)3 + 3 (2m )2
343
according as the
55
series a , x , y , z , b is an AP or HP. Find the values of a and b
given that they are positive integers.
l
Ex. 48 The value of xyz is 55 or
= {13 + 33 + 53 + ... + (2m - 1)3 } + 3 {22 + 4 2 + 62
+ ... + (2m )2 }
=
Sol. If a, x , y , z , b are in AP.
Then, b = Fifth term = a + (5 - 1) d
where, d is common difference]
b-a
\
d=
4
[given]
\
x × y × z = (a + d ) (a + 2d ) (a + 3d ) = 55
Þ (a + 3b ) (a + b ) (3a + b ) = 55 ´ 32
If they are in HP.
The common difference of the associated AP is
\
Þ
(a - b )
4ab
1 1 (a - b )
= +
x a
4ab
4ab
x=
a + 3b
\
1 1 2 (a - b )
= +
y a
4ab
Þ
y=
and
Þ
32 a 3b 3 343
=
55 ´ 32 55
or
Þ
Hence,
or
æ1 1ö
ç - ÷.
èb a ø
3 3
a b = 343
ab = 7
a = 7, b = 1
a = 1, b = 7
r =1
m
m
r =1
r =1
m
m
m
r =1
r =1
r =1
år 3 - 12 år 2 + 6 år - å1 + 12 år 2
m
m
m
r =1
2
r =1
2
r =1
år 3 + 6 år - å1
m ( m + 1)
m ( m + 1)
+6
-m
4
2
= 2m 2 (m + 1)2 + 3m (m + 1) - m
= m [ 2m 3 + 4m 2 + 5m + 2]
3
2
ù é
n é æn ö
nù
æn ö
æn ö
ê2 ç ÷ + 4 ç ÷ + 5 ç ÷ + 2ú êQm = ú
è
ø
è
ø
è
ø
2û
2 ê 2
2
2
úû ë
ë
n
Hence, S = (n 3 + 4n 2 + 10n + 8) …(i)
8
Case II If n is odd.
Then, (n + 1) is even in the case
=
Sum of first n terms = Sum of first (n +1) terms - (n + 1) th
term
=
=
4ab
2ab
4ab
×
×
= 343
(a + 3b ) (a + b ) (3a + b )
Þ
m
å {8r 3 - 12r 2 + 6r - 1} + 12 år 2
= 8×
1 1 3 (a - b )
= +
z a
4ab
4ab
z=
3a + b
xyz =
r =1
=8
4ab
2ab
=
2a + 2b a + b
\
r =1
m
=8
…(i)
1
4
m
r =1
æ b + 3a ö æ 2a + 2b ö æ a + 3b ö
ç
֍
֍
÷ = 55
è 4 øè 4 ø è 4 ø
Þ
i.e.
=
m
å(2r - 1)3 + 3 × 4 år 2
[given]
[from Eq. (i)]
( n + 1)
[(n + 1)3 + 4 (n + 1)2 + 10 (n + 1) + 8] - 3 (n + 1)2
8
1
(n + 1) [n 3 + 3n 2 + 3n + 1 + 4n 2 + 8n + 4 + 10n
8
+ 10 + 8 - 24n - 24 ]
Hence, S =
l
1
(n + 1) [n 3 + 7n 2 - 3n - 1]
8
Ex. 50 Find out the largest term of the sequence
1
4 9 16
,
,
,
, ... .
503 524 581 692
Sol. General term can be written as Tn =
n2
500 + 3n 3
Chap 3 Sequences and Series
1
500
= 2 + 3n
Tn
n
dU n
1000
=+3
dn
n
Un =
Let
Then,
æ 1000 ö
n=ç
÷
è 3 ø
1/ 3
(n 2 + 3n + 2)
2
Hence, this is the required result.
æ 1000 ö
(not an integer) and 6< ç
÷
è 3 ø
1/ 3
<7
Ex. 52 If the equation x 4 - 4 x 3 + ax 2 + bx + 1 = 0 has
four positive roots, find the values of a and b.
l
Sol. Let x 1, x 2 , x 3 , x 4 are the roots of the equation
x 4 - 4 x 3 + ax 2 + bx + 1 = 0
l
is more close to 7 than to 6. Thus, we take
d 2U n
i.e., AM = GM
which is true only when x 1 = x 2 = x 3 = x 4 = 1
Hence, given equation has all roots identical, equal to 1 i.e.,
equation have form
( x - 1) 4 = 0
å ( 2r + 1) f (r ).
¥
r =1
l
n
n
å(r 2 + 2r + 1 - r 2 ) f (r ) = å {(r + 1)2 - r 2 } f (r )
r =1
n
å {(r + 1)
2
f ( r ) - ( r + 1) f ( r + 1) + ( r + 1)
r =1
=
¥
¥
å å 3m (n × 3m + m × 3n )
=
am =
Now, let
r =1
f (r + 1) - r 2 f (r )}
n -1
( r + 1)
=- å
+ å (r + 1)2 f (r + 1) + (n + 1)2
r = 1 ( r + 1)
r =1
n
é
1 ù
f ( n + 1) - å r 2 f ( r )
ú
êQ f (r + 1) - f (r ) =
r + 1û
ë
r =1
S=
Then,
¥
å(r + 1) + {22 f (2) + 32 f (3) + ... + n 2 f (n )}
2
2
1
æ 3m ö æ 3m 3n ö
ç ÷ç
+ ÷
nø
èm øèm
3m
3n
and an =
m
n
¥
¥
å åa
m =1 n =1 m
1
(am + an )
…(i)
1
(an + am )
…(ii)
By interchanging m and n, then
S=
¥
¥
å åa
m =1 n =1 n
n
r =1
¥
å å
m =1 n =1
n
2
On adding Eqs. (i) and (ii), we get
2
+ (n + 1) f (n + 1) - {1 f (1) + 2 f (2)
+ 32 f (3) + ... + n 2 f (n )}
.
m 2n
f (r + 1) - r f (r )}
r =1
+ m × 3n )
m =1 n =1
2
å( r + 1)2 { f (r ) - f (r + 1)} + å {(r + 1)2
n
m 2n
2
n
=-
Sol. Let S =
r =1
2
¥
å å 3m (n × 3m
m =1 n =1
r =1
=
Ex. 53 Evaluate
å(2r + 1) f (r )
n
…(ii)
On comparing Eqs. (i) and (ii), we get
a = 6, b = - 4
n
=
x 4 - 4 x 3 + 6x 2 - 4 x + 1 = 0
Þ
1 1
1
Ex. 51 If f (r ) = 1 + + + ... + and f (0 ) = 0, find
2 3
r
Sol. Since,
…(i)
\ x 1 + x 2 + x 3 + x 4 = 4 and x 1 x 2 x 3 x 4 = 1
x + x2 + x3 + x4 4
Q AM = 1
= =1
4
4
and GM = ( x 1 x 2 x 3 x 4 )1/ 4 = (1)1/ 4 = 1
1/ 3
= + ve , then U n will be minimum and
dn 2
therefore, Tn will be maximum for n = 7.
Hence, T 7 is largest term. So, largest term in the given
49
sequence is
.
1529
Further
r =1
= ( n + 1) 2 f ( n + 1) -
But n is an integer, therefore for the maxima or minima of
1/ 3
æ 1000 ö
U n we will take n as the nearest integer to ç
÷ .
è 3 ø
æ 1000 ö
Since, ç
÷
è 3 ø
n = 7.
r =1
f ( n + 1) - 12 f ( 1)
n ( n + 1)
- n + ( n + 1) 2 f ( n + 1) - f ( 1)
2
n ( n + 3)
[Q f (1) = 1]
= ( n + 1) 2 f ( n + 1) -1
2
3000
= 4
dn 2
n
For maxima or minima of U n , we have
1000
dU n
= 0 Þ n3 =
3
dn
Þ
n
å r - å 1 + ( n + 1) 2
=-
d 2U n
and
n
=-
275
2S =
¥
¥
¥
¥
mn
1
= å å m n
a
m n
m =1 n =1 3 3
å åa
m =1 n =1
276
Textbook of Algebra
2
2
3
é æ1ö
ù
æ ¥ nö
æ1ö
æ1ö
= çç å n ÷÷ = ê1 ç ÷ + 2 ç ÷ + 3 ç ÷ + ...ú
è3ø
è3ø
êë è 3 ø
úû
èn = 13 ø
2
=
2
3
æ1ö
æ1ö
æ1ö
where, S ¢ = 1 ç ÷ + 2 ç ÷ + 3 ç ÷ + ... + ¥
è3ø
è3ø
è3ø
æ1ö
æ1ö
1ç ÷ +2ç ÷
è3ø
è3ø
–
–
1
S¢ =
3
–
2
1 æ1ö
æ1ö
+ç ÷ +ç ÷
è3ø
3 è3ø
2 ¢
S =
3
1
3
=
\
1
13
3
S¢=
4
3
+ ... + ¥
=
¥
¥
i=0 j=0 k =0
=
1
Sol. Let S =
¥
¥
å å å
i =0 j =0 k =0
3i 3 j 3k
1
[i ¹ j ¹ k ]
3i 3 j 3k
Let
S1 =
¥
¥
å å å
i =0 j =0 k =0
æ ¥ 1ö
çå ÷
=
3i 3 j 3k çèi = 0 3i ÷ø
3
1
3
Case I If i = j = k
S2 =
¥
¥
¥
1
å å å
i =0 j =0 k =0
=
¥
1
å 33i
=1+
i =0
3 i 3 j 3k
1
33
+
1
36
+ ... =
1
1-
1
=
27
26
33
Case II If i = j ¹ k
Let
S3 =
¥
¥
¥
å å å
i =0 j =0 k =0
i =0
Þ Sn = n + 1
1
n +1
n
n
r =1
r =1
n
å [(n + 1) r - r 2 + (n + 2)]
n
n
n
r =1
r =1
å r - å r 2 + ( n + 2) å 1
= ( n + 1) å n -
ån
2
r =1
+ ( n + 2) × n
(n + 1) n (n + 1) n (n + 1) (2n + 1)
+ ( n + 2) n
2
6
n
= (n 2 + 9n + 14 )
6
=
and S12 + S 22 + ... + Sn2 =
27
æ3ö
=ç ÷ =
è2ø
8
Let
1
åSr Sn - r + 1 = å(r + 1) (n - r + 2)
= ( n + 1)
We will first of all find the sum without any restriction on
i, j , k .
¥
¥
å 33i
r =1
(i ¹ j ¹ k )
¥
-
\ S1 Sn + S 2 Sn - 1 + S 3 Sn - 2 + ... + Sn S1
å å å
Ex. 54 Find the value of
i =0
n
Sn =
2
9
S=
32
¥
1
Ex. 55 Let Sn , n =1, 2, 3, K be the sum of infinite geomet1
ric series, whose first term is n and the common ratio is
.
n +1
Evaluate
S1Sn + S 2 Sn - 1 + S3 Sn - 2 + ... + Sn S1
.
lim
n ®¥
S12 + S 22 + ... + Sn2
1-
\
3
l
Sol. Q
æ3ö
From Eq. (iii), we get 2S = ç ÷
è4ø
l
+ ... + ¥
1
2
=
¥
1ö
3 9 27 135
= × =
2 8 26 208
Hence required sum, S = S1 - S 2 - 3 S 3
27 27
æ 135 ö 27 ´ 26 - 27 ´ 8 - 3 ´ 135 81
=
-3ç
=
÷=
è 208 ø
208
208
8
26
…(iii)
3
1 æ3
i =0
= (S ¢ )2
2
¥
å 32 i çè 2 - 3i ÷ø = å 2 × 32 i
æ ¥ 1 öæ ¥ 1ö
çå
÷çå
÷
=
3i 3 j 3k çèi = 0 3 2 i ÷ø çèk = 0 3k ÷ø
1
[Qk ¹ i ]
n
n
n
r =1
r =1
r =0
…(i)
åSr2 = å( r + 1)2 = å( r + 1)2 - 12
(n + 1) (n + 2)( 2n + 3)
=
-1
6
n
= (2n 2 + 9n + 13)
6
From Eqs. (i) and (ii), we get
S1 Sn + S 2 Sn - 1 + S 3 Sn - 2 + ... + Sn S1
lim
n ®¥
S12 + S 22 + ... + Sn2
n 2
(n + 9n + 14 )
= lim 6
= lim
n ®¥ n
n ®¥
(2n 2 + 9n + 13)
6
1+0+0 1
=
=
2+0+0 2
æ
ç1 +
è
æ
ç2 +
è
9 14 ö
+
÷
n n2 ø
9 13 ö
+
÷
n n2 ø
…(ii)
Chap 3 Sequences and Series
Ex. 56 The nth term of a series is given by
n 5 + n3
and if sum of its n terms can be expressed as
tn =
n 4 + n 2 +1
1
, where a n and bn are the nth terms of
Sn = a n2 + a +
2
bn + b
some arithmetic progressions and a , b are some constants,
b
prove that n is a constant.
an
l
Sol. Since,
tn =
5
n +n
n2 n 1
1
+ - + 2
2
2 2 2n + 2n + 2
2
æn
1 ö
5
1
=ç
+
÷ - +
2
8 æ
è 2 2 2ø
1 ö
3
2
n
+
ç
÷ +
è
2
2ø
n + n2 + 1
=n +
=
but given, Sn = an2 + a +
n4 + n2 + 1
1
2 ( n 2 + n + 1)
-
1
2 ( n 2 - n + 1)
åt n
1ì æ
1
1
öü
= ån + í å ç 2
- 2
÷ý
è
2
n + n + 1 n - n + 1ø
Sum of n terms Sn =
î
þ
[ by property]
2
3
n
ö
n ( n + 1) 1 æ
1
+ ç 2
- 1÷
2
2 èn + n + 1
ø
1 ö
1 1
1
æn
=ç
+
÷ - - +
2
è 2 2 2ø
8 2 æ
1 ö
3
çn 2 +
÷ +
è
2
2ø
4
=n-
=
1
bn2 + b
On comparing, we get
1
5
n
1 ö
3
æ
, a = - , bn = çn 2 +
an =
+
÷, b =
è
8
2
2 2 2
2ø
\
bn
= 2 , which is constant.
an
277
#L
Sequences and Series Exercise 1 :
Single Option Correct Type Questions
n
This section contains 30 multiple choice questions.
Each question has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct
1. If the numbers x , y, z are in HP, then
xy
x + y
yz
y + z
,
xz
,
x + z
(a) AP
(c) HP
(b) 1 - 2 -n
(c) n + 2 -n - 1
(d) 2n - 1
(a) the altitudes are in AP (b) the altitudes are in HP
(c) the medians are in GP (d) the medians are in AP
(b) GP
(d) None of these
8. Let a1 , a 2 , ..., a10 be in AP and h 1 , h 2 , ..., h 10 be in HP. If
n
2. If a1 , a 2 , ... are in HP and f k = å ar - ak , then
r =1
2 a1 , 2 a2 , 2 a3 , 2 a4 , ... are in
ü
ì
a1
a2
a
, a 3 = 3 , Ký
í wherea 1 = , a 2 =
f
f
f
3
2
1
þ
î
(a) AP
(c) HP
a1 = h 1 = 2 and a10 = h 10 = 3, then a 4 h 7 is
(a) 2
(c) 5
(b) 3
(d) 6
9. If I n = ò
- sin 2nx
dx , then I 1 , I 2 , I 3 ,... are in
1 - cos 2x
p1
0
(a) AP
(c) HP
(b) GP
(d) None of these
3. ABC is a right angled triangle in which ÐB = 90° and
BC = a. If n points L 1 , L 2 , ..., L n on AB are such that AB
is divided in n + 1 equal parts and
L 1 M 1 , L 2 M 2 , K , L n M n are line segments parallel to BC
and M 1 , M 2 , K , M n are on AC, the sum of the lengths of
L 1 M 1 , L 2 M 2 , ..., L n M n is
a (n + 1 )
a (n - 1 )
(b)
2
2
an
(c)
2
(d) impossible to find from the given data
(a)
(b) GP
(d) None of these
10. If a (b - c ) x 2 + b (c - a ) xy + c (a - b ) y 2 is a perfect
square, the quantities a, b, c are in
(a) AP
(c) HP
(b) GP
(d) None of these
11. The sum to infinity of the series,
2
1ö
1ö
æ
æ
1 + 2 ç1 - ÷ + 3 ç1 - ÷ + ... is
è nø
è nø
(a) n 2
1ö
æ
(c) n ç1 + ÷
è
nø
4. Let S n (1 £ n £ 9 ) denotes the sum of n terms of the series
1 + 22 + 333 + ... + 999... 9, then for 2 £ n £ 9
123
(b) n (n + 1 )
2
(d) None of these
7
12. If log 3 2, log 3 (2 x - 5) and log 3 æç2 x - ö÷ are in AP, x is
è
2ø
equal to
9 times
(a) 2
(c) 4
1 n
(10 - n 2 + n )
9
(b) 3
(d) 2, 3
13. Let a, b, c be three positive prime numbers. The
1 n
(10 - n 2 + 2n - 2 )
9
(c) 9 (Sn - Sn - 1 ) = n (10n - 1 )
(b) Sn =
progression in which a , b , c can be three terms (not
necessarily consecutive), is
(d) None of the above
2
5. If a, b, c are in GP, then the equations ax + 2bx + c = 0
and dx 2 + 2ex + f = 0 have a common root, if
are in
(a) AP
(c) HP
is equal to
(a) 2n - n - 1
3 7 15
+ +
+ ...
4 8 16
7. If in a DPQR, sin P, sin Q, sin R are in AP, then
are in
(a) Sn - Sn - 1 =
1
2
6. Sum of the first n terms of the series +
(b) GP
(d) None of these
d e f
, ,
a b c
(a) AP
(c) HP
(b) GP
(d) None of these
14. If n is an odd integer greater than or equal to 1, the value
of n 3 - (n - 1) 3 + (n - 2) 3 - ... + ( - 1)n - 1 1 3 is
(n + 1 ) 2 (2n - 1 )
4
(n + 1 ) 2 (2n + 1 )
(c)
4
(a)
(b)
(n - 1 ) 2 (2n - 1 )
4
(d) None of these
Chap 03 Sequences and Series
15. If the sides of a right angled triangle form an AP, the
sines of the acute angles are
3 4
(a) ,
5 5
5 -1
,
2
(c)
(b)
5+1
2
3,
(a) 1
(c) 16
3 1
(d)
,
2 2
(b)
5
4
(d) None of these
is non-zero, the sum of first 3n terms is equal to the sum
of the next n terms. The ratio of the sum of the first 2n
terms to the next 2n terms is
1
5
3
(c)
4
(b)
2
3
( x - 1) ( x - 2) ( x - 3) ... ( x - n ), is
n (n 2 + 2 ) (3n + 1 )
24
ratio r, then
an - 1 an
a a
a1a 2
a a
is equal
+ 2 2 3 2 + 2 3 4 2 + ... + 2
2
2
a1 - a 2 a 2 - a 3 a 3 - a 4
an - 1 - an2
to
(a)
nr
1 - r2
(b)
(n - 1 ) r
1 - r2
(c)
nr
1 -r
(d)
sum of the first five terms, the ratio of the first term to
the common difference is
(a)
1
2
(b) 2
(c)
1
4
(d) 4
æy ö
cos x sec ç ÷ is equal to
è2ø
(a) ± 2
(c) -
å a 2r
(d) None of these
(b) 6
(d) 8
28. If x , y, z are in GP ( x , y, z > 1), then
1
1
1
are in
,
,
2x + ln x 4 x + ln y 6x + ln z
(a) AP
(c) HP
100
1
2
of integral AM’s is
(a) 5
(c) 7
(b) 3519
(d) 3731
20. Let an be the nth term of an AP. If
1
2
(b)
27. If 11 AM’s are inserted between 28 and 10, the number
19. Consider the pattern shown below:
Row 1 1
Row 2 3 5
Row 3 7 9 11
Row 4 13 15 17, 19, etc.
The number at the end of row 60 is
= a and
(b) GP
(d) None of these
29. The minimum value of the quantity
å a 2r - 1 = b, the common difference of the AP is
(a 2 + 3a + 1) (b 2 + 3b + 1) (c 2 + 3c + 1)
,
abc
where a, b, c Î R +, is
(a) a - b
a -b
(c)
2
11 3
23
(c) 25
r =1
100
r =1
(b) b - a
(d) None of these
21. If a1 , a 2 , a 3 , a 4 , a 5 are in HP, then
a1a 2 + a 2 a 3 + a 3 a 4 + a 4 a 5 is equal to
(a) 2 a1a 5
(c) 4a1a 5
(n - 1 ) r
1 -r
26. If cos ( x - y ), cos x and cos ( x + y ) are in HP, the
n (n 2 - 1 ) (3n + 2 )
(b)
24
n (n 2 + 1 ) (3n + 4 )
(c)
24
(d) None of the above
(a) 3659
(c) 3681
(b) 2
(d) None of these
25. The sum of the first ten terms of an AP is four times the
(d) None of these
18. The coefficient of x n - 2 in the polynomial
(a)
= l abc , then l is
(a) 1
(c) 4
24. If a1 , a 2 , a 3 , ... are in GP with first term a and common
17. If the arithmetic progression whose common difference
(a)
(b) 4
(d) 64
23. If a, b, c are in AP and (a + 2b - c ) (2b + c - a ) (c + a - b )
common difference of the AP which makes the product
a1 a 4 a 5 least, is given by
8
5
2
(c)
3
22. If a, b, c and d are four positive real numbers such that
abcd = 1, the minimum value of
(1 + a ) (1 + b ) (1 + c ) (1 + d ) is
1
3
16. The sixth term of an AP is equal to 2. The value of the
(a)
279
(b) 3a1a 5
(d) 6a1a 5
(a)
(b) 125
(d) 27
30. Let a1 , a 2 , ... be in AP and q 1 , q 2 , ... be in GP. If
a1 = q 1 = 2 and a10 = q 10 = 3, then
(a) a 7 q19 is not an integer (b) a19 q 7 is an integer
(c) a 7 q19 = a19 q10
(d) None of these
280
#L
Textbook of Algebra
Sequences and Series Exercise 2 :
More than One Correct Option Type Questions
n
This section contains 15 multiple choice questions.
Each question has four choices (a), (b), (c) and (d) out of
which MORE THAN ONE may be correct.
1
2
1
3
31. If a(n ) = 1 + + +
40. Let E =
(b) a (100 ) > 100
(d) a (200 ) < 100
(a) E < 3
32. If the first and (2n - 1) th term of an AP, GP and HP are
equal and their nth terms are a, b and c respectively, then
(a) a = b = c
(c) a + c = b
(b) a ³ b ³ c
(d) ac - b 2 = 0
p
2
¥
z=
å cos 2n
¥
å cos 2n f, y =
n=0
¥
å sin 2n
f and
n=0
n=0
(b) xyz = xy + z
(d) xyz = yz + x
34. If a, b, c are in AP and a 2 , b 2 , c 2 are in HP, then which of
the following could hold true?
a
(a) - , b, c are in GP
2
(c) a 3, b 3, c 3 are in GP
(d) None of these
35. The next term of the GP x , x + 2, x + 10 is
(b) 6
2
36. If the sum of n consecutive odd numbers is 25 - 11 , then
(b) n = 16
(d) last odd number is 49
37. The GM of two positive numbers is 6. Their AM is A and
HM is H satisfy the equation 90A + 5H = 918, then A
may be equal to
1
(a)
5
(b) 5
5
(c)
2
+
1
32
+ ..., then
3
2
(c) E < 2
(d) E > 2
41. Let S n (n ³ 1) be a sequence of sets defined by
ì 8 11 14 ü
ì 3 5ü
S 1 = {0}, S 2 = í , ý, S 3 = í , , ý,
2
2
þ
î3 3 3 þ
î
ì 15 19 23 27 ü
S 4 = í , , , ý, ..., then
î4 4 4 4þ
439
20
431
(b) third element in S 20 is
20
(c) sum of the elements in S 20 is 589
(d) sum of the elements in S 20 is 609
42. Which of the following sequences are unbounded?
an =
(d) 54
2
(a) n = 14
(c) first odd number is 23
22
n
2
n
æ 2n + 1 ö
1ö
æ
(b) ç
÷ (c) ç1 + ÷ (d) tann
è
èn+2ø
nø
43. Let a sequence {an } be defined by
3
729
(c)
16
1
+
(b) E >
1ö
æ
(a) ç1 + ÷
è
nø
(b) a = b = c
2
1
12
(a) third element in S 20 is
f sin 2n f, then
(a) xyz = xz + y
(c) xyz = x + y + z
(a) 0
(b) 7th term is 18
505
405
(d) sum of first 10 terms is
(c) sum of first 10 terms is
4
4
1
1
, then
+ ... + n
4
2 -1
(a) a (100 ) < 100
(c) a (200 ) > 100
33. For 0 < f < , if x =
(a) 7th term is 16
(d) 10
38. If the sum to n terms of the series
1
1
1
1
+
+
+ ... + , then
n +1 n +2 n +3
3n
(a) a 2 =
11
12
(9n + 5 )
(3n + 1 ) (3n + 2 ) (3n + 3 )
-2
- an =
3 (n + 1 )
(d) an + 1
æ
è
44. Let S n ( x ) = ç x n - 1 +
(a) f ( 0 ) = 15
640
(c) f ( l ) =
27
(c) S100( x ) =
1
1
(1 + 2) 2 +
(1 + 2 + 3) 2
S =1+
(1 + 3)
(1 + 3 + 5)
1
+
(1 + 2 + 3 + 4 ) 2 + ...
(1 + 3 + 5 + 7 )
1 ö
n -1 ÷
ø
x
1ö
æ
+ (n - 1) ç x + ÷ + n, then
è
xø
(a) S1( x ) = 1
39. For the series,
19
20
(c) an + 1 - an =
l
1
1
1
1
, then
+
+
+ ... is 1 × 3 × 5 × 7 3 × 5 × 7 × 9 5 × 7 × 9 × 11
90 f (n )
(b) f (1 ) = 105
1
(d) l =
3
(b) a 2 =
1 ö
æ
+ 2 ç x n - 2 + n - 2 ÷ + ...
ø
è
x
(b) S1( x ) = x +
1
x
2
1 æ x100 - 1 ö
1 æ x100 - 1 ö
ç
÷ (d) S100( x ) = 100 ç
÷
99
x è x -1 ø
x è x -1 ø
2
45. All the terms of an AP are natural numbers and the sum
of the first 20 terms is greater than 1072 and less than
1162. If the sixth term is 32, then
(a) first term is 7
(b) first term is 12
(c) common difference is 4 (d) common difference is 5
Chap 03 Sequences and Series
#L
281
Sequences and Series Exercise 3 :
Passage Based Questions
n
This section contains 8 passages. Based upon each of the
passage 3 multiple choice questions have to be
answered. Each of these questions has four choices (a), (b),
(c) and (d) out of which ONLY ONE is correct.
Passage I
(Q. Nos. 46 to 48)
8 16 24
+
+ ...
S n be the sum of n terms of the series +
5 65 325
46. The value of lim S n is
n ®¥
(a) 0
1
(b)
2
(c) 2
(d) 4
56
2505
(b)
56
6505
(c)
56
5185
(d)
81
41
(d)
56
9605
48. The value of S 8 , is
(a)
288
145
(b)
1088
545
(c)
107
245
Passage II
(Q. Nos. 49 to 51)
Two consecutive numbers from 1, 2, 3, ..., n are removed.
105
.
The arithmetic mean of the remaining numbers is
4
49. The value of n lies in
(a) (41, 51)
(b) (52, 62)
(d) (74, 84)
(b) lies between 1200 to 1500
(d) None of these
(Q. Nos. 52 to 54)
There are two sets A and B each of which consists of three
numbers in AP whose sum is 15 and where D and d are the
common differences such that D = 1 + d, d > 0. If
p = 7 ( q - p), where p and q are the product of the
numbers respectively in the two series.
(c) 175
(b) 160
(b) 22
(d) 84
(b) 56
R
(c) 58
(d) 60
(c) 32
(d) 4
r
57. The value of r + R is
(a) 5392
(b) 368
Passage V
(Q. Nos. 58 to 60)
The numbers 1, 3, 6, 10, 15, 21, 28, ... are called triangular
numbers. Let t n denotes the nth triangular number such
that t n = t n - 1 + n, " n ³ 2.
58. The value of t 50 is
(a) 1075
(b) 1175
(c) 1275
(d) 1375
t 101 are
(c) 67
(c) 101
(d) 102
60. If (m + 1) is the nth triangular number, then (n - m) is
(a) 1 + (m 2 + 2m )
(b) 1 + (m 2 + 2 )
(c) 1 + (m 2 + m )
(d) None of these
(Q. Nos. 61 to 63)
Let A1 , A2 , A3 , ..., Am be arithmetic means between - 3
and 828 and G1 , G2 , G3 , ..., Gn be geometric means
between 1 and 2187. Product of geometric means is 3 35
and sum of arithmetic means is 14025.
61. The value of n is
(b) 30
(c) 25
(d) 10
(c) 51
(d) 68
(d) 80
63. The value of G 1 + G 2 + G 3 + ... + G n is
(a) 17
(c) 120
(b) 100
62. The value of m is
54. The value of 7 D + 8d is
(a) 37
(c) 78
(d) 210
53. The value of q is
(a) 200
(a) 54
(a) 45
52. The value of p is
(b) 140
(b) 72
Passage VI
Passage III
(a) 105
(a) 66
(a) 99
(b) lies between 10 to 30
(d) greater than 70
51. Sum of all numbers is
(a) less than 1000
(c) greater than 1500
55. The value of p is
59. The number of positive integers lying between t 100 and
(c) (63, 73)
50. The removed numbers
(a) are less than 10
(c) lies between 30 to 70
(Q. Nos. 55 to 57)
There are two sets A and B each of which consists of three
numbers in GP whose product is 64 and R and r are the
p 3
common ratios such that R = r + 2. If = , where p and q
q 2
are sum of numbers taken two at a time respectively in the
two sets.
56. The value of q is
47. The seventh term of the series is
(a)
Passage IV
(d) 52
(a) 2044
(c) 511
(b) 34
(b) 1022
(d) None of these
282
Textbook of Algebra
Passage VII
(Q. Nos. 64 to 66)
Suppose a, b are roots of ax 2 + bx + c = 0 and g, d are roots
of Ax 2 + Bx + C = 0.
64. If a , b, g , d are in AP, then common difference of AP is
1 æb B ö
(a) ç - ÷
4 èa A ø
(c)
1 æc B ö
ç - ÷
2 èa A ø
1 æb B ö
(b) ç - ÷
3 èa A ø
(d)
67. The value of p is
(a)
na + b
n+1
(b)
nb + a
n+1
(c)
na - b
n+1
(d)
nb - a
n+1
68. The value of q is
1 æc C ö
ç - ÷
3 èa A ø
(a)
65. If a, b, c are in GP as well as a , b, g , d are in GP, then
A, B, C are in
(n - 1 ) ab
(n + 1 ) ab
(n + 1 ) ab
(n - 1 ) ab
(b)
(c)
(d)
nb + a
nb + a
na + b
na + b
69. Final conclusion is
2
æn + 1ö
(a) q lies between p and ç
÷ p
è n -1 ø
(a) AP only
(b) GP only
(c) AP and GP
(d) None of these
æ bA ö
(a) ç ÷
è aB ø
æ aB ö
(b) ç ÷
è bA ø
æn + 1ö
(b) q lies between p and ç
÷p
è n -1 ø
2
æn + 1ö
(c) q does not lie between p and ç
÷ p
è n -1 ø
æ bC ö
(c) ç ÷
è cB ø
æ cB ö
(d) ç ÷
è bC ø
æn + 1ö
(d) q does not lie between p and ç
÷p
è n -1 ø
66. If a , b, g , d are in GP, then common ratio of GP is
#L
Passage VIII (Q. Nos. 67 to 69)
Suppose p is the first of n ( n >1) arithmetic means between
two positive numbers a and b and q the first of n harmonic
means between the same two numbers.
Sequences and Series Exercise 4 :
Single Integer Answer Type Questions
n
This section contains 10 questions. The answer to each
question is a single digit integer, ranging from 0 to 9
(both inclusive).
70. Let a, b, c , d be positive real numbers with a < b < c < d .
Given that a, b, c , d are the first four terms of an AP and
ad p
is , where p and q are
a, b, d are in GP. The value of
bc q
prime numbers, then the value of q is
110
71. If the coefficient of x in the expansion of Õ (1 + rx ) is
r =1
l (1 + 10) (1 + 10 + 10 2 ), then the value of l is
72. A 3-digit palindrome is a 3-digit number (not starting
with zero) which reads the same backwards as forwards
For example, 242. The sum of all even 3-digit palindromes
is 2n 1 × 3n 2 × 5n 3 × 7 n 4 × 11n 5 , alue of n 1 + n 2 + n 3 + n 4 + n 5 is
73. If n is a positive integer satisfying the equation
2 + (6 × 2 2 - 4 × 2) + (6 × 3 2 - 4 × 3) + ... + (6 × n 2 - 4 × n ) = 140,
then the value of n is
74. Let S( x ) = 1 + x - x 2 - x 3 + x 4 + x 5 - x 6 - x 7
+ ... + ¥, where 0 < x < 1. If S( x ) =
of ( x + 1) 2 is
2 +1
, then the value
2
75. The sequence a1 , a 2 , a 3 , ... is a geometric sequence with
common ratio r. The sequence b1 , b 2 , b 3 , ... is also a
geometric sequence. If b1 = 1, b 2 = 4 7 - 4 28 + 1, a1 = 4 28
¥
¥
1
and å
= å bn , then the value of (1 + r 2 + r 4 ) is
n = 1 an
n =1
76. Let (a1 , b1 ) and (a 2 , b 2 ) are the pair of real numbers such
that 10, a, b, ab constitute an arithmetic progression.
æ 2a a + b1b 2 ö
Then, the value of ç 1 2
÷ is
ø
è
10
77. If one root of Ax 3 + Bx 2 + Cx + D = 0, A ¹ 0, is the
arithmetic mean of the other two roots, then the relation
2B 3 + lABC + mA 2 D = 0 holds good. Then, the value of
2l + m is
1
2
22
78. If | x | > 1, then sum of the series
+
+
1+ x 1+ x2 1+ x4
1
23
, then the value of l is
+
+ ... upto ¥ is
x-l
1+ x8
79. Three non-zero real numbers form an AP and the
squares of these numbers taken in same order form a
GP. If the possible common ratios are (3 ± k ) where
ék 8 ù
k Î N , then the value of ê - ú is (where [ ] denotes
ë8 k û
the greatest integer function).
Chap 03 Sequences and Series
#L
283
Sequences and Series Exercise 5 :
Matching Type Questions
n
This section contains 4 questions. Questions 80, 81 and 82 have three statements (A, B and C) and question 83 has four
statements (A, B, C and D) given in Column I and questions 80 and 81 have four statements (p, q, r and s), question 82
has five statements (p, q, r, s and t) and question 83 has three statements (p, q and r) in Column II, respectively. Any
given statement in Column I can have correct matching with one or more statement(s) given in Column II.
80.
Column I
Column II
(A)
a, b, c, d are in AP, then
(p)
a+ d > b+ c
(B)
a, b, c, d are in GP, then
(q)
ad > bc
(C)
a, b, c, d are in HP, then
(r)
1 1 1 1
+ > +
a d b c
(s)
ad < bc
81.
Column I
(A)
(B)
(C)
Column II
For an AP a1 , a2 , a3 , ..., an , K;
5
a1 = ; a10 = 16. If a1 + a2
2
+ ... + an = 110, then ‘n’ equals
(p)
The interior angles of a convex
non-equiangular polygon of 9 sides
are in AP. The least positive integer
that limits the upper value of the
common difference between the
measures of the angles in degrees is
(q)
For an increasing GP,
a1 , a2 , a3 , ..., an , K;
a6 = 4 a4; a9 - a7 = 192,
if a4 + a5 + a6 + ... + an = 1016, then
n equals
(r)
Column I
(A) If a1 , a2 , a3 , ... are in AP and
a1 + a4 + a7 + a14 + a17 +
a20 = 165,
a = a2 + a6 + a15 + a19 and
b = 2 (a9 + a12 ) - (a3 + a18 ),
then
9
11
12
Column II
(p)
Column I
Column II
(B)
(q)
If a1 , a2 , a3 , ... are in AP and
a1 + a5 + a10 + a15 + a20 + a24
= 195,
a = a2 + a7 + a18 + a23 and
b = 2 (a3 + a22 ) - (a8 + a17 ),
then
a + 2b = 260
(C)
If a1 , a2 , a3 , ... are in AP and
a1 + a7 + a10 + a21 +
a24 + a30 = 225,
a = a2 + a7 + a24 + a29 and
b = 2 (a10 + a21 ) - (a3 + a28 ),
then
(r)
a + 2b = 220
(s)
a - b = 5l , l Î I
(t)
a + b =15m , m ÎI
10
83.
(s)
82.
82.
a = 2b
Column I
Column II
(A) If 4 a2 + 9b2 + 16 c2
= 2 (3ab + 6bc + 4 ca), where a, b, c
are non-zero numbers, then a, b, c are
in
(p)
AP
(B) If 17a2 + 13b2 + 5 c2
= (3ab + 15bc + 5 ca), where
a, b, c are non-zero numbers, then
a, b, c are in
(q)
GP
2
2
2
(C) If a + 9b + 25 c
æ 15 5 3 ö
= abc ç + + ÷ ,where a, b, c are
è a
b cø
non-zero numbers, then a, b, c are in
(r)
HP
(D) If (a2 + b2 + c2 ) p2 - 2p (ab + bc + ca)
+ (a2 + b2 + c2 ) £ 0, where a, b, c, p
are non-zero numbers, then a, b, c are
in
284
#L
Textbook of Algebra
Sequences and Series Exercise 6 :
Statement I and II Type Questions
n
Directions (Q. Nos. 84 to 90) are Assertion-Reason type
questions. Each of these questions contains two
statements:
Statement-1 (Assertion) and Statement-2 (Reason)
Each of these questions also has four alternative choices,
only one of which is the correct answer. You have to select
the correct choice as given below.
(a) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1
(c) Statement1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
84. Statement 1 4, 8, 16 are in GP and 12, 16, 24 are in HP.
Statement 2 If middle term is added in three
consecutive terms of a GP, resultant will be in HP.
85. Statement 1 If the nth term of a series is 2n 3 + 3n 2 - 4,
then the second order differences must be an AP.
Statement 2 Ifnth term of a series is a polynomial of
degreem, thenmth order differences of series are constant.
86. Statement 1 The sum of the products of numbers
n
± a1 , ± a 2 , ± a 3 , K , ± an taken two at a time is - å ai2 .
i =1
#L
Statement 2 The sum of products of numbers a1 , a 2 , a 3 ,
..., an taken two at a time is denoted by å
å ai a j .
1£i < j £n
87. Statement 1 a + b + c = 18 (a, b, c > 0), then the
maximum value of abc is 216.
Statement 2 Maximum value occurs when a = b = c
88. Statement 1 If 4a 2 + 9b 2 + 16c 2 = 2 (3ab + 6bc + 4ca ),
where a, b, c are non-zero real numbers, then a, b, c are in
GP.
Statement-2 If (a1 - a 2 ) 2 + (a 2 - a 3 ) 2 + (a 3 - a1 ) 2 = 0,
then a1 = a 2 = a 3 , " a1 , a 2 , a 3 Î R .
89. Statement 1 If a and b be two positive numbers, where
a > b and 4 ´ GM = 5 ´ HM for the numbers. Then, a = 4b.
Statement 2 (AM) (HM) = (GM) 2 is true for positive
numbers.
90. Statement1 The difference between the sum of the first
100 even natural numbers and the sum of the first 100
odd natural numbers is 100.
Statement 2 The difference between the sum of the
first n even natural numbers and sum of the first n odd
natural numbers is n.
Sequences and Series Exercise 7 :
Subjective Type Questions
n
In this section, there are 24 subjective questions.
91. The p th, (2p ) th and ( 4p ) th terms of an AP, are in GP,
then find the common ratio of GP.
92. Find the sum of n terms of the series
93.
94.
95.
96.
(a + b ) + (a 2 + ab + b 2 ) + (a 3 + a 2 b + ab 2 + b 3 ) + ...,
where a ¹ 1, b ¹ 1 and a ¹ b.
The sequence of odd natural numbers is divided into
groups 1; 3, 5; 7, 9, 11; ... and so on. Show that the sum of
the numbers in nth group is n 3 .
Let a, b, c are respectively the sums of the first n terms,
the next n terms and the next n terms of a GP. Show that
a, b, c are in GP.
If the first four terms of an arithmetic sequence are
a, 2a, b and (a - 6 - b ) for some numbers a and b, find the
sum of the first 100 terms of the sequence.
1
1
1
p2
, find
If 2 + 2 + 2 + ... upto ¥ =
6
1
2
3
1
1
1
(i) 2 + 2 + 2 + ... upto ¥
1
3
5
1
1
1
+
+ K upto ¥
22 32 4 2
97. If the arithmetic mean of a1 , a 2 , a 3 , ..., an is a and
b1 , b 2 , b 3 , ..., bn have the arithmetic mean b and
ai + bi = 1 for i = 1, 2, 3, ..., n, prove that
(ii) 1 -
n
å (ai - a )2 +
i =1
n
å ai bi = nab
i =1
98. If a1 , a 2 , a 3 , ... is an arithmetic progression with common
difference 1 and a1 + a 2 + a 3 + ... + a 98 = 137, then find
the value of a 2 + a 4 + a 6 + ... + a 98 .
99. If t 1 = 1, t r - t r - 1 = 2r - 1 , r ³ 2, find
n
åt r .
r =1
100. Prove that I 1 , I 2 , I 3 , ... form an AP, if
(i) I n = ò
p
0
sin 2nx
dx
sin x
(ii) I n = ò
p
0
2
æ sin nx ö
ç
÷ dx
è sin x ø
Chap 03 Sequences and Series
101. Consider the sequence S = 7 + 13 + 21 + 31 + ... + T n , find
the value of T 70 .
3
3
3
1
1 ö
1 ö
æ
æ
102. Find value of æç x + ö÷ + ç x 2 + 2 ÷ +...+ ç x n + n ÷ .
è
ø
ø
è
ø
è
x
x
x
a12 - a 22 + a 32 - a 42 + ... + a 22n - 1 - a 22n =
108. Show that,
n
(1 + 5 -1 ) (1 + 5 -2 ) (1 + 5 -4 ) (1 + 5 -8 ) ... (1 + 5 -2 )
n+1
5
= (1 - 5 - 2
)
4
109. Evaluate S =
103. If am be the mth term of an AP, show that
n
(a12 - a 22n ).
(2n - 1)
n=0
(a 2 + 1)
n
(where a > 1).
111. Find the sum to n terms, whose nth term is
105. If a1 , a 2 , a 3 , ..., an are in AP with a1 = 0, prove that
æ1
1
1 ö
÷
+ ... +
- a 2 çç +
an - 2 ÷ø
èa2 a3
an - 1
a
=
+ 2
a2
an - 1
106. Balls are arranged in rows to form an equilateral triangle.
The first row consists of one ball, the second row of two
balls and so on. If 669 more balls are added, then all the
balls can be arranged in the shape of a square and each of
the sides, then contains 8 balls less than each side of the
triangle. Determine the initial number of balls.
107. If q 1 , q 2 , q 3 , ..., q n are in AP whose common difference
is d, then show that
sin d {sec q 1 sec q 2 + sec q 2 sec q 3 + ...
+ sec q n - 1 sec q n } = tan q n - tan q 1 .
#L
2n
æ 2n - 1 ö
æ1ö
æ2ö
÷ + ...
tan -1 ç ÷ + tan -1 ç ÷ + ... + tan -1 ç
è3ø
è9 ø
è 1 + 2 2n - 1 ø
in AP, show that they are in the ratio
a3 a4
a
+
+ ... + n
a2 a3
an - 1
n
å
110. Find the sum to infinite terms of the series
104. If three unequal numbers are in HP and their squares are
1 + 3 : - 2 : 1 - 3 or 1 - 3 : - 2 : 1 + 3.
285
tan [ a + (n - 1) b ] tan (a + nb ).
n
112. If
åTr =
r =1
n
n
1
.
(n + 1) (n + 2) (n + 3), find å
8
r = 1 Tr
113. If S 1 , S 2 , S 3 denote the sum of n terms of 3 arithmetic
series whose first terms are unity and their common
difference are in HP, prove that
n=
2S 3 S 1 - S 1S 2 - S 2 S 3
.
S1 - 2 S 2 + S 3
114. Three friends whose ages form a GP divide a certain
sum of money in proportion to their ages. If they do that
three years later, when the youngest is half the age of
the oldest, then he will receive ` 105 more than he gets
now and the middle friend will get ` 15 more than he
gets now. Find the ages of the friends.
Sequences and Series Exercise 8 :
Questions Asked in Previous 13 Year’s Exam
n
This section contains questions asked in IIT-JEE,
AIEEE, JEE Main & JEE Advanced from year 2005 to
year 2017.
115. If a, b, c are in AP and | a |, | b |, | c | < 1 and
z = 1 + c + c 2 + ... + ¥
2
[AIEEE 2005, 3M]
(b) GP
3
(c) HP
n
3 æ3ö
æ3ö
æ3ö
- ç ÷ + ç ÷ - ... + ( -1)n - 1 ç ÷ and
è
ø
è
ø
è4ø
4
4
4
bn = 1 - an , then find the least natural number n 0 such
that bn > an , " n ³ n 0 .
[IIT-JEE 2006, 6M]
116. If an =
a1 + a 2 + ... + aq
41
(a)
11
2
(c)
7
y = 1 + b + b 2 + ... + ¥
Then, x , y, z will be in
a1 + a 2 + ... + a p
=
p2
q
2
, p ¹ q , then
a6
equals
a 21
[AIEEE 2006, 4.5M]
x = 1 + a + a 2 + ... + ¥
(a) AP
(d) None of these
117. Let a1 , a 2 , a 3 , ... be terms are in AP, if
7
(b)
2
11
(d)
41
118. If a1 , a 2 , ..., an are in HP, then the expression
a1a 2 + a 2 a 3 + ... + an - 1an is equal to
(a) n (a1 - an )
(b) (n - 1 ) (a1 - an )
(c) na1an
(d) (n - 1 ) a1an
[AIEEE 2006, 6M]
286
Textbook of Algebra
119. Let Vr denotes the sum of the first r terms of an
arithmetic progression whose first term is r and the
common difference is (2 r - 1). Let T r = Vr + 1 - Vr - 2 and
[IIT-JEE 2007, 4+4+4M]
Q r = T r + 1 - T r for r = 1, 2, K
(i) The sum V1 + V2 + ... + Vn is
1
(a) n (n + 1 ) (3n 2 - n + 1 )
12
1
(b) n (n + 1 ) (3n 2 + n + 2 )
12
1
(c) n (2n 2 - n + 1 )
2
1
(d) (2n 3 - 2n + 3 )
3
(ii) Tr is always
(a) an odd number
(c) a prime number
(b) an even number
(d) a composite number
harmonic means respectively, of two distinct positive
numbers. For n ³ 2, let A n - 1 , G n - 1 and H n - 1 has
arithmetic, geometric and harmonic means as
[IIT-JEE 2007, 4+4+4M]
A n , G n , H n , respectively.
(i) Which one of the following statement is correct?
(b) G1 < G2 < G3 < K
(c) G1 = G2 = G3 = ...
(d) G1 < G3 < G5 < ... and G2 > G4 > G6 > ...
(ii) Which of the following statement is correct?
(a) A1 > A2 > A3 > K
(b) A1 < A2 < A3 < K
(c) A1 > A3 > A5 > ... and A2 < A4 < A6 < ...
(d) A1 < A3 < A5 < K and A2 > A4 > A6 > K
(iii) Which of the following statement is correct?
(a) H1 > H 2 > H 3 > ...
(b) H1 < H 2 < H 3 < ...
(c) H1 > H 3 > H 5 > ... and H 2 < H 4 < H 6 < K
(d) H1 < H 3 < H 5 < K and H 2 > H 4 > H 6 > ...
121. If a geometric progression consisting of positive terms,
each term equals the sum of the next two terms, then
the common ratio of this progression equals
[AIEEE 2007, 3M]
(c)
1
(1 - 5 )
2
5
1
5
2
1
(d) ( 5 - 1 )
2
(b)
Statement 1 The numbers b1 , b 2 , b 3 , b 4 are neither in
AP nor in GP.
Statement 2 The numbers b1 , b 2 , b 3 , b 4 are in HP.
[IIT-JEE 2008, 3M]
120. Let A 1 , G 1 , H 1 denote the arithmetic, geometric and
(a)
in GP. Let b1 = a1 , b 2 = b1 + a 2 , b 3 = b 2 + a 3
and b 4 = b 3 + a 4 .
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a
correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
(iii) Which one of the following is a correct statement?
(a) Q1, Q2, Q3, K are in AP with common difference 5
(b) Q1, Q2, Q3, K are in AP with common difference 6
(c) Q1, Q2, Q3, K are in AP with common difference 11
(d) Q1 = Q2 = Q3 = ...
(a) G1 > G2 > G3 > ...
122. Suppose four distinct positive numbers a1 , a 2 , a 3 , a 4 are
123. The first two terms of a geometric progression add upto
12 the sum of the third and the fourth terms is 48, if the
terms of the geometric progression are alternately
positive and negative, then the first term is
[AIEEE 2008, 3M]
(a) - 12
(b) 12
(d) -4
(c) 4
124. If the sum of first n terms of an AP is cn 2 , then the sum
of squares of these n terms is
2
2
(a)
n ( 4n - 1 )c
6
(c)
n ( 4n 2 - 1 )c 2
3
[IIT-JEE 2009, 3M]
2
2
(b)
n ( 4n + 1 ) c
3
(d)
n ( 4n 2 + 1 ) c 2
6
125. The sum to infinity of the series
1+
2 6 10 14
+
+
+
+ ... is
3 32 33 34
(a) 6
(b) 2
(c) 3
[AIEEE 2009, 4M]
(d) 4
126. Let S k , k = 1, 2, ..., 100, denote the sum of the infinite
geometric series whose first term is
k -1
and common
k!
100
1
100 2
ratio is . Then, the value of
+ å | (k 2 - 3k + 1) S k | is
100 ! k = 2
k
[IIT-JEE 2010, 3M]
127. Let a1 , a 2 , a 3 , ..., a11 be real numbers satisfying
a1 = 15, 27 - 2a 2 > 0 and ak = 2ak - 1 - ak - 2 for
2
a 2 + a 22 + ... + a11
k = 3, 4, K , 11. If 1
= 90 , then the value
11
a + a 2 + ... + a11
is equal to
of 1
11
[IIT-JEE 2010, 3M]
128. A person is to count 4500 currency notes. Let an denotes
the number of notes he counts in the nth minute. If
a1 = a 2 = ... = a10 = 150 and a10 , a11 , ... are in AP with
common difference - 2, then the time taken by him to
count all notes is
[AIEEE 2010, 8M]
(a) 34 min
(c) 135 min
(b) 125 min
(d) 24 min
Chap 03 Sequences and Series
129. The minimum value of the sum of real numbers
-5
a ,a
-4
, 3a
-3
8
, 1, a and a
10
with a > 0 is
[IIT-JEE 2011, 4M]
130. A man saves ` 200 in each of the first three months of
his service. In each of the subsequent months his saving
increases by ` 40 more than the saving of immediately
previous month. His total saving from the start of
service will be ` 11040 after
[AIEEE 2011, 4M (Paper I)]
(a) 19 months
(c) 21 months
(b) 20 months
(d) 18 months
100
131. Let an be the nth term of an AP, if
å a 2r
= a and
r =1
100
å a 2r - 1 = b, then the common difference of the AP is
r =1
[AIEEE 2011, 4M (Paper II)]
a -b
(a)
200
a -b
(c)
100
(b) a - b
a 20 = 25 . The least positive integer n for which an < 0 is
(b) 23
(d) 25
[IIT-JEE 2012, 3M]
133. Statement 1 The sum of the series
1 + (1 + 2 + 4 ) + ( 4 + 6 + 9 ) + (9 + 12 + 16)
+ K + (361 + 380 + 400) is 8000.
Statement 2
å (k
3
3
3
- (k - 1) ) = n for any natural
k =1
number n.
[AIEEE 2012, 4M]
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a
correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2 is
not a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
134. If 100 times the 100th term of an AP with non-zero
common difference equals the 50 times its 50th term,
then the 150th term of this AP is
[AIEEE 2012, 4M]
(a) 150 times its 50th term (b) 150
(c) zero
(d) -150
135. If x, y, z are in AP and tan -1 x , tan -1 y , tan -1 z are also
in AP, then
[JEE Main 2013, 4M]
(a) 2 x = 3y = 6z
(b) 6 x = 3y = 2z
(c) 6 x = 4y = 3z
(d) x = y = z
7
(a) (99 - 10 -20 )
9
7
(c) (99 + 10 -20 )
9
[JEE Main 2013, 4M]
7
(179 + 10 -20 )
81
7
(d) (179 - 10 -20 )
81
(b)
k (k + 1)
2
× k 2 , then S n can take value(s)
k =1
(a) 1056
[JEE Advanced 2013, 4M]
(b) 1088
(c) 1120
(d) 1332
138. A pack contains n cards numbered from 1 to n. Two
consecutive numbered cards are removed from the pack
and the sum of the numbers on the remaining cards is
1224. If the smaller of the numbers on the removed cards
is k, then k - 20 is equal to
[JEE Advanced 2013, 4M]
139. If (10) 9 + 2 (11)1 (10) 8 + 3 (11) 2 (10) 7 + K+ (10)(11) 9
= k (10) 9 , then k is equal to
(a) 100
[JEE Main 2014, 4M]
121
(c)
10
(b) 110
(d)
441
100
140. Three positive numbers form an increasing GP. If the
(a) 2 - 3
(b) 2 + 3
(c) 2 + 3
(d) 3 + 2
b
a
a, b, c are in geometric progression and the arithmetic
a 2 + a - 14
is
mean of a, b, c is b + 2, the value of
a +1
141. Let a, b, c be positive integers such that is an integer. If
[JEE Advanced 2014, 3M]
142. The sum of first 9 terms of the series
13 13 + 23 13 + 23 + 33
+ ... is
+
+
1
1+3
1+3+5
[JEE Main 2015, 4M]
(a) 192
(d) 142
(b) 71
(c) 96
143. If m is the AM of two distinct real numbers l and n
(l , n > 1) and G 1 , G 2 and G 3 are three geometric means
between l and n, then G 14 + 2G 24 + G 34 equals
[JEE Main 2015, 4M]
(a) 4 l 2m 2n 2
(c) 4 lm 2n
(b) 4 l 2mn
(d) 4 lmn 2
144. Suppose that all the terms of an arithmetic progression
(AP) are natural numbers. If the ratio of the sum of the first
seven terms to the sum of the first eleven terms is6 : 11 and
the seventh term lies between 130 and 140, then the
common difference of this AP is
[JEE Main 2015, 4M]
145. If the 2nd, 5th and 9th terms of a non-eustant AP are in
GP, then the common ratio of this GP is
[JEE Main 2016, 4M]
(a) 1
136. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, …,
is
å ( -1)
[JEE Main 2014, 4M]
132. If a1 , a 2 , a 3 ,…be in harmonic progression with a1 = 5 and
n
4n
middle terms in this GP is doubled, the new numbers are
in AP. Then, the common ratio of the GP is
(d) b - a
(a) 22
(c) 24
137. Let S n =
287
(b)
7
4
(c)
8
5
(d)
4
3
146. If the sum of the first ten terms of the series
2
2
2
2
16
æ 3ö
æ 2ö
æ 1ö
æ 4ö
2
ç1 ÷ + ç2 ÷ + ç3 ÷ + 4 + ç 4 ÷ + K is m, then
è 5ø
è 5ø
è 5ø
è 5ø
5
[JEE Main 2016, 4M]
m equal to
(a) 100
(b) 99
(c) 102
(d) 101
288
Textbook of Algebra
147. Let bi > 1 for i = 1, 2, ..., 101. Suppose loge b1 , loge b 2 ,
148. For any three positive real numbers a, b and c ,
loge b 3 , ..., loge b101 are in Arithmetic Progression (AP)
with the common difference loge 2. Suppose
a1 , a 2 , a 3 , K , a101 are in AP. Such that, a1 = b1 and
a 51 = b 51 . If t = b1 + b 2 + K + b 51 and
s = a1 + a 2 + K + a 51 , then
[JEE Advanced 2016, 3M]
(a) s > t and a101 > b101
(c) s < t and a101 > b101
9 (25a 2 + b 2 ) + 25 (c 2 - 3ac ) = 15b (3a + c ). Then
[JEE Main 2017, 4M]
(a) a, b and c are in GP
(b) b, c and a are in GP
(c) b, c and a are in AP
(d) a, b and c are in AP
(b) s > t and a101 < b101
(d) s < t and a101 < b101
Answers
Exercise for Session 1
1. (c)
2. (d)
3. (b)
4. (c)
5. (a)
4. (b)
5. (c)
4. (c)
5. (d)
4. (d)
5. (a)
Exercise for Session 2
1. (b)
2. (a)
3. (a)
6. (c)
Exercise for Session 3
1. (b)
2. (d)
3. (b)
Exercise for Session 4
1. (c)
2. (c)
3. (c)
6. (a)
Exercise for Session 5
1. (c)
7. (b)
2. (a)
8. (b)
3. (a)
9. (a)
4. (c) 5. (b)
10. (b)
6. (b)
Exercise for Session 6
1. (b)
7. (a)
2. (d)
8. (c)
3. (b)
9. (b)
4. (c) 5. (a)
10. (c)
6. (a)
4. (d)
6. (c)
2. (d)
8. (a)
3. (b)
5. (c)
2. (c)
8. (d)
3. (b)
9. (a)
4. (a)
5. (c)
6. (b)
5. (b)
6. (a)
10. (a)
2. (c)
3. (d)
4. (a)
Chapter Exercises
1. (a)
2. (d)
99. 2 n +
102.
Exercise for Session 9
1. (d)
7. (c)
92.
1
-n-2
3
Exercise for Session 8
1. (c)
7. (c)
91. 2.
95. - 5050
Exercise for Session 7
1. (a)
7. (a)
31. (a,c) 32. (b,d)
33. (b,c) 34. (a,b) 35. (c,d) 36. (a,c,d)
37. (a,d) 38. (a,b,c) 39. (a,c)
40. (b,c) 41. (a,c) 42. (c,d)
43. (b,c) 44. (a,c)
45. (a,d)
46. (c)
47. (d)
48. (a)
49. (a)
50. (a)
51. (b)
52. (a)
53. (c)
54. (b)
55. (d)
56. (b)
57. (c)
58. (c)
59. (b)
60. (d)
61. (d)
62. (b)
63. (d)
64. (a)
65. (b)
66. (b)
67. (a)
68. (b)
69. (c)
70. (3)
71. (5)
72. (8)
73. (4)
74. (2)
75. (7)
76. (3)
77. (9)
78. (1)
79. (0)
80. (A) ® (r, s); (B) ® (p, r);
(C) ® (p, q)
81. (A) ® (r);
(B) ® (p);
(C) ® (q)
82. (A) ® (p,r,s,t); (B) ® (p,q,s,t); (C) ® (p,s,t)
83. (A) ® (r);
(B) ® (p);
(C) ® (r); (D) ® (q)
84. (a)
85. (a)
86. (b)
87. (a)
88. (d)
89. (c)
90. (a)
3. (c)
4. (c)
5. (a)
6. (c)
7. (b)
8. (d)
9. (a)
10. (c)
11. (a)
12. (b)
13. (d)
14. (a)
15. (a)
16. (c)
17. (a)
18. (b)
19. (a)
20. (d)
21. (c)
22. (c)
23. (c)
24. (b)
25. (a)
26. (a)
27. (a)
28. (c)
29. (b)
30. (c)
x (1 - x 3n )
3
1 ïì a2 (1 - an ) b2 (1 - bn ) ïü
í
ý
(a - b) ïî (1 - a)
(1 - b) ïþ
p2
p2
(ii)
96. (i)
98. 93
8
12
101. 5113
(1 - x 3n )
+
3n
3
+
3x (1 - x n ) 3 (1 - x n )
+ n
(1 - x)
x (1 - x)
(1 - x )
1
106. 1540
109.
a-1
sin nb
- n tan b
cos (a + nb ) cos a
p
110.
111.
tan b
4
n (n + 3 )
112.
114.
12, 18, 27
2 (n + 1) (n + 2)
1- x
x
115. (c)
116. (7) 117. (d)
118. (d)
119. (i) (b), (ii) (d), (iii) (b)
120. (i) (c), (ii) (a), (iii) (b) 121. (d) 122. (c) 123. (a) 124. (c)
125. (c) 126. (3)
127. (0)
128. (a)
129. (8) 130. (c)
131. (c) 132. (d)
133. (a)
134. (c)
135. (d) 136. (b)
137. (a,d) 138. (5)
139. (a)
140. (b) 141. (4) 142. (c)
143. (c) 144. (a)
145. (d)
146. (d) 147. (b) 148. (c)
3. Q
Solutions
2a
n+1
3a
L3M3 =
n+1
Similarly, L2M2 =
1. Q x, y , z are in HP.
\
\
M
1 1 1
, , are in AP.
x y z
1 1 1 1
- = x y y z
Q
…(i)
[say]
zx
1
=
=b
1
1
x + z
+
z
x
xy
1
=
=c
1
1
x + y
+
x
y
[say]
n times
Then,
b 2 = ac
…(i)
anddx 2 + 2ex + f = 0 have a common root.
Þ
Þ
…(i)
…(A)
2
ax + 2bx + c = 0
Now,
2
ax + 2 ac x + c = 0
[ by Eq. (i)]
2
( ax + c ) = 0 Þ
x=-
Þ
ax + c = 0
c
a
[repeated]
æ
cö
2
By the condition (A), ç ÷ be the root of dx - 2ex + f = 0
aø
è
[say]
So, it satisfy the equation
2
r =1
æ
æ
cö
cö
d ç÷ + 2e ç ÷+ f =0
aø
è
è
aø
Þ
Subtracting from each term by 1, we get
f1 f 2 f 3
, , , ¼are also in AP.
a1 a 2 a 3
1 1 1
,
,
, ¼are in AP.
Þ
a1 a 2 a 3
\
a1, a 2, a 3, ¼are in HP.
\
2 a1 , 2 a 2 , 2 a 3 , ¼are not in AP/GP/HP.
n (10n - 1 )
10 - 1
and equations ax 2 + 2bx + c = 0
r =1
n
a1 + f1 = a 2 + f 2 = a 3 + f 3 = ¼ = l
l l l
From Eq. (i), , , , ¼are also in AP.
a1 a 2 a 3
a + f1 a 2 + f 2 a 3 + f 3
Þ 1
,
,
, ¼ are also in AP.
a1
a2
a3
n times
5. Given that a, b, c are in GP.
fk = å ar - ak
ak + fk = å ar = l
Sn - Sn - 1 = nnnn ¼ n = n (111 ¼ 1 )
1424
3 14243
\ 9 (Sn - Sn - 1 ) = n (10n - 1 )
n
Þ
C
a
= n (10n - 1 + 10n - 2 + ¼ + 10 + 1 ) =
2. Qa1, a 2, a 3, ¼are in HP.
Q
Mn
Ln
4. Q Sn = 1 + 22 + 333 + ¼ + nnnn
1
42¼
4
3n
\
xy
are in AP.
x + y
1 1 1
, , , ¼are in AP.
Þ
a1 a 2 a 3
M
L1M1 + L2M2 + ¼ + Ln Mn
B
a
=
(1 + 2 + 3 + ¼ + n )
(n + 1 )
a
n (n + 1 ) na
=
×
=
(n + 1 )
2
2
[say]
1
1
1
1
1
1
1 1
+
+
y
z
z
x
a -b
x y a
=
=
=
1
1
1 1 a
b -c
1
1
1
1
y z
+
+
z
x
x
y
yz
zx
,
,
y + z z + x
M1
M2
M3
L1
L2
L3
n terms
[from Eq. (i)]
Hence,
M
A
na
Ln Mn =
n+1
yz
1
=
=a
1
1
y + z
+
y
z
and
a
AL1 L1M1
=
Þ L1M1 =
n+1
AB
BC
dc
d
f
c
2e
+ =0
- 2e
+ f =0 Þ
a
a
ac
c
a
d 2e
f
d
f
æe ö
+ = 0 Þ + =2 ç ÷
Þ
èb ø
a
c
a b
c
d e f
So, , , are in AP.
a b c
1 3 7 15
6. Q Sn = + + +
+ ¼ n up to terms
2 4 8 16
1ö æ
1ö æ
1ö
æ
= ç1 - ÷ + ç1 - ÷ + ç1 - ÷ + ¼ n up to terms
è
ø
è
ø
è
2
4
8ø
Þ
1ö
1ö æ
1ö æ
æ
= ç1 - ÷ + ç1 - 2 ÷ + ç1 - 3 ÷ + ¼ +
è
2ø è
2 ø
2 ø è
1ö
æ
ç1 - n ÷
è
2 ø
290
Textbook of Algebra
=n -
1æ
1 1
1 ö
ç1 + + 2 + ¼ + n - 1 ÷
ø
è
2
2 2
2
1
1 1 1
-1
-1
h
h1 3 2
D = 10
=
=
9
9 ´ 6 54
10 - 1
n
é
æ1ö ù
1 ê1 - ç ÷ ú
è2ø ú
1
ê
û
=n - + ë
1ö
æ
2
ç1 - ÷
è
2ø
1
1
1 1
æ -1 ö 1 1
= + 6D Þ
= + 6ç ÷ = è 54 ø 2 9
l7 2
h7 h1
1
7
18
=
Þ h7 =
h7 18
7
7 18
So,
a 4h7 = ´
=6
3 7
p 1 - sin 2nx
p 1 - sin 2nx
9. Q I n = ò
dx
dx Þ I n = ò
0 1 - cos 2 x
0 2 sin 2 x
So,
é
ù
a (1 - r n )
, if 0 < r < 1 ú
ê by sum GP ,Sn =
1 -r
ë
û
1
= n - 1 + 2 -n
2n
7. Let triangle be the area of DPQR.
=n -1 +
Þ I n + 1 + I n - 1 - 2I n
P
r
=
h1
q
h2
p
R
1
[h1, h2, h3 are altitudes]
´ p ´ h1
2
2D
…(i)
h1 =
Þ
p
2D
Similarly,
…(ii)
h2 =
q
2D
and
…(iii)
h3 =
r
According to the question, sin P , sin Q, sin R are in AP.
Then, kp, kq, kr are in AP
[by sine rule]
Þ p, q, r are in AP.
2D 2D 2D
are in AP.
[ by Eqs. (i), (ii) and (iii)]
,
,
Þ
h1 h2 h3
1 1 1
Þ
, , are in AP.
h1 h2 h3
\
D=
Þ h1, h2, h3 are in HP.
Þ Altitudes are in HP.
8. Given that, a1, a 2, ¼, a10 be in AP.
Let d be the common difference of AP.
a - a1
\
d = 10
10 - 1
3 -2
[given that, a1 = h1 = 2 and a10 = h10 = 3]
d =
9
1
d =
9
3
1 7
\
a 4 = a1 + 3d = 2 + = 2 + =
9
3 3
Now, h1, h2, ¼, h10 be in HP.
So, common difference of respective AP.
ò0
[1 - sin 2 (n + 1 ) x + 1 - sin 2 (n - 1 ) x - 2
+ 2 sin 2nx ]
dx
sin 2 x
- sin 2 (n + 1 ) x ] + [sin 2nx - sin 2 (n - 1 ) x ]
dx
sin 2 x
p - 2 cos (2n + 1 ) x sin ( x ) + 2 cos (2n - 1 ) x sin x
dx
ò0
sin 2 x
p sin x [cos (2n - 1 ) x - cos (2n + 1 ) x ]
dx
=ò
0
sin 2 x
p 2 sin 2nx sin x
p
2
dx = 2 ò sin 2nx dx =
=ò
[ - cos 2nx ]p0
0
0
sin x
2n
1
= - (1 - 1 ) = 0
n
\
I n + 1 + I n - 1 = 2I n
Þ I n - 1 + I n , I n + 1 are in AP.
\ I 1, I 2, I 3,¼are in AP.
10. Given that,
a (b - c ) x 2 + b (c - a ) xy + c (a - b ) y 2 is perfect square.
1
2
1
=
2
=
h3
Q
1
2
p
p [sin 2nx
ò0
b 2 (c - a ) 2 = 4a (b - c ) × c (a - b )
\
Þ b 2 (c - a ) 2 = 4ac (a - b ) (b - c )
Þ
[a (b - c ) + c (a - b )]2 = 4ac (a - b ) (b - c )
Þ
[Q a (b - c ) + b (c - a ) + c (a - b ) = 0 ]
[a (b - c ) - c (a - b )]2 = 0
Þ
Þ
a (b - c ) - c (a - b ) = 0
ab - ac - ca + bc = 0 Þ b (a + c ) = 2ac
2ac
b=
a+b
Þ
Þ a, b, c are in HP.
2
1ö
1ö
æ
æ
S = 1 + 2 ç1 - ÷ + 3 ç1 - ÷ + ¼ + ¥
è
ø
è
n
nø
11. Let
1ö
1ö
1ö
æ
æ
æ
ç1 - ÷ S = ç1 - ÷ + 2 ç1 - ÷ + ¼ + ¥
è
è
è
nø
nø
nø
-
-
2
1ö
1ö æ
1ö
æ
æ
S ç1 - 1 + ÷ = 1 + ç1 - ÷ + ç1 - ÷ + ¼ + ¥
è
ø
è
ø
è
n
n
nø
Chap 03 Sequences and Series
S
=
n
Þ
n
1
n
S = n2
Þ
é
ù
a
by GP ú
êS ¥ =
r
1
ë
û
1
Þ
1ö
æ
1 - ç1 - ÷
è
nø
S=
æ
è
7ö
2ø
12. Q log 3 2, log 3 (2 x - 5) and log 3 ç2 x - ÷ are in AP.
For defined, 2 x - 5 > 0 and 2 x -
7
>0
2
2
é n - 1 æn - 1
öù
ê 2 çè 2 + 1 ÷ø ú
é n (n + 1 ) ù
=ê
ú
úû - 16 × ê
2
2
ë
ú
ê
úû
êë
n 2 (n + 1 ) 2 4 (n - 1 ) 2 (n + 1 ) 2 (n + 1 ) 2 2
=
=
[n - (n - 1 ) 2 ]
4
16
4
(n - 1 ) 2
(2n - 1 ) (n + 1 ) 2
=
× (2n - 1 ) (1 ) =
4
4
15. Let the sides of right angled triangle be
(a - d ), a, (a + d ) (a > d ).
2
A
x
\
…(i)
2 >5
…(ii)
7
From Eq. (i), 2, 2 x - 5, 2 x - are in GP.
2
7ö
æ
x
\
(2 - 5 ) 2 = 2 × ç2 x - ÷
è
2ø
2x
2
Þ
(2 x - 8 ) (2 x - 4 ) = 0
Þ
\
B
- 12 × 2 + 32 = 0
3
x
C
2 =8 =2 ,2 ¹ 4
a 2 + d 2 + 2ad = a 2 + a 2 + d 2 - 2ad
[fromEq. (ii)]
x =3
Let a , b , c are 3 terms of AP. [not necessarily consecutive]
a = A + (p - 1) D
…(i)
b = A + (q - 1 ) D
…(ii)
c = A + (r - 1 ) D
…(iii)
[A and D be the first term and common difference of AP]
a - b = (p - q ) D
…(iv)
b - c = (q - r ) D
…(v)
c - a = (r - p ) D
…(vi)
On dividing Eq. (iv) by Eq. (v), we get
a - b p -q
=
b - c q -r
a 2 = 4ad
a = 4d
13. Qa, b, c are positive prime numbers.
Then,
a
By Pythagoras theorem,
(a + d ) 2 = a 2 + (a - d ) 2
2 x = 8, 4
x
(a + b )
(a – d )
x
Þ
\
291
…(vii)
Since, p, q, r are natural numbers and a, b, c are positive prime
numbers, so
Eq. (vii) does not hold.
So, a , b and c cannot be the 3 terms of AP.
[not necessarily consecutive]
Similarly, we can show that a , b , c cannot be any 3 terms
of GP and HP.
[not necessarily, consecutive]
14. Given that n is an odd integer greater than or equal to 1.
Sn = n 3 - (n - 1 ) 3 + (n - 2 ) 3 - ¼ + ( - 1 )n - 11 3
= 1 3 - 2 3 + ¼ + (n - 2 ) 3 - (n - 1 ) 3 + n 3
[Qn is odd integer, so (n - 1 ) is even integer]
n -1
æ
ö
3
3
= (1 + 2 + ¼ + n 3 ) - 2 × 2 3 ç1 3 + 2 3 + ¼ +
terms÷
è
ø
2
[since a ¹ 0] …(i)
a
4d 4
According to the question, sin A =
=
=
a + d 5d 5
a - d 3d 3
sin C =
=
=
a + d 5d 5
16. T6 = 2
Let d be common difference of AP and a be the first term of
AP.
T6 = 2
…(i)
Þ
a + 5d = 2
Let
A = a1a 4a 5
A = a (a + 3d ) (a + 4d )
[using Tn = a + (n - 1 ) d and from Eq. (i) a = 2 - 5d ]
A = (2 - 5d ) (2 - 2d ) (2 - d )
A = 8 - 32d + 34d 2 - 10d 3
dA
For max and min values of A,
=0
dd
- 30d 2 + 68d - 32 = 0 Þ 15d 2 - 34d + 16 = 0
15d 2 - ( 24d + 10d ) + 16 = 0
15d 2 - 24d - 10d + 16 = 0
For
So,
3d ( 5d - 8 ) - 2 ( 5d - 8 ) = 0
( 5d - 8 ) ( 3d - 2 ) = 0
8
2
or d =
d =
5
3
2
2 d A
d = ,
>0
3 dd 2
2
A is least for d = .
3
292
Textbook of Algebra
17. Given, common difference ¹ 0
S 3n = S 4n - S 3n
2 × S 3n = S 4n
Þ
[ let Sn = Pn 2 + Qn]
Þ 2 × [ P (3n ) 2 + Q(3n )] = P ( 4n ) 2 + Q( 4n )
2 Pn 2 + 2 Qn = 0
Þ
Q = - nP
or
…(i)
2
S 2n
P (2n ) + Q(2n )
=
S 4n - S 2n [ P ( 4n ) 2 + Q( 4n )] - [ P (2n ) 2 + Q(2n )]
\
=
2n (2nP + Q ) 2nP + Q
=
12 Pn 2 + 2nQ 6nP + Q
=
2nP - nP 1
=
6nP - nP 5
[from Eq. (i)]
18. Let f ( x ) = ( x - 1) ( x - 2) ( x - 3) ¼ ( x - n )
= xn - S1xn - 1 + S 2 xn - 2 - ¼ + ( - 1 )n (1 × 2 × 3 ¼ n )
So, coefficient of xn - 2 in f ( x ) = S 2 = (1 × 2 + 1 × 3 + ¼)
= Sum of product of first n natural number taken 2 at time
=
1
[(1 + 2 + ¼ + n ) 2 - (1 2 + 2 2 + ¼ + n 2 )]
2
22. Q(1 + a ) (1 + b ) (1 + c ) (1 + d )
= 1 + a + b + c + d + ab + ac + ad + bc + bd + cd
[16 terms]
+ abc + abd + cda + cdb + abcd
\ AM ³ GM
(1 + a ) (1 + b ) (1 + c ) (1 + d )
³ (a 8b 8c 8d 8 )1/16
16
= (abcd )1/2 = (1 )1/2 = 1
[Q abcd = 1 ]
(1 + a ) (1 + b ) (1 + c ) (1 + d )
Þ
³1
16
Þ
(1 + a ) (1 + b ) (1 + c ) (1 + d ) ³ 16
\Minimum value of (1 + a ) (1 + b ) (1 + c ) (1 + d ) is 16.
23. Qa, b, c are in AP.
…(i)
\
2b = a + c
Now, (a + 2b - c ) (2b + c - a ) (c + a - b )
[from Eq. (i)]
= (a + a + c - c ) (a + c + c - a ) (2b - b )
= (2a ) (2c ) (b ) = 4abc
\ l=4
24. a1, a 2,¼, an are in GP with first term a and common ratio r.
an - 1 an
aa
aa
…(i)
Sn = 2 1 2 2 + 2 2 3 2 + ¼ + 2
a1 - a 2 a 2 - a 3
an - 1 - an2
144444424444443
2
n (n + 1 ) (2n + 1 ) ù
1 éì n (n + 1 )ü
= êí
ú
ý 2 êëî
2
6
þ
úû
1 n (n + 1 ) é n (n + 1 ) 2n + 1 ù
= ×
êë
2
2
2
3 úû
1 n (n + 1 ) é 3n 2 + 3n - 4n - 2 ù
= ×
ú
ê
2
2
6
û
ë
n (n + 1 ) (3n 2 - n - 2 ) n (n + 1 ) (3n + 2 )(n - 1 )
=
24
24
n (n 2 - 1 ) (3n + 2 )
=
24
19. If last term of nth row is Tn , then
Let S = 1 + 5 + 11 + 19 + ¼ + Tn
S = 1 + 5 + 11 + ¼ + Tn - 1 + Tn
- - - - 0 = 1 + 4 + 6 + 8 + ¼ + n terms - Tn
=
Tn = 1 + 2 ( 2 + 3 + 4 + ¼+ (n - 1 ) terms)
(n - 1 )
[2 × 2 + (n - 2 ) × 1]
=1 + 2
2
= 1 + (n - 1 ) (n + 2 )
= 1 + n2 + n - 2
(n - 1) times
Tn =
Tn = n 2 + n - 1
Þ
T60 = (60 ) + 60 - 1 = 3600 + 59 = 3659
100
20. Given that,
å
a 2r = a
r =1
Þ
a 2 + a 4 + ¼ + a 200 = a
and
å
100
a 2r - 1 = b
r =1
Þ
a1 + a 3 + ¼ + a199 = b
an - 1 an
an2 - 1
- an2
=
an - 1 an
(an - 1 - an ) (an - 1 + an )
1
ö
æ
an - 1 ö
æ
a
ç1 - n ÷ ç1 +
÷
an - 1 ø è
an ø
è
r
1
=
=
1
ö (r + 1 ) (1 - r )
æ
(1 - r ) ç1 + ÷
è
rø
=
2
\
On subtracting Eq. (ii) from Eq. (i), we get
(a 2 - a1 ) + (a 4 - a 3 ) + ¼ + (a 200 - a199 ) = a - b
d + d + ¼ up to 100 terms = a - b
[beacause an be the nth term of AP with common difference d ]
100 d = a - b
a -b
d =
100
21. Given that, a1, a 2, a 3, a 4 , a 5 are in HP.
1 1 1 1 1
\
, , , , are in AP.
a1 a 2 a 3 a 4 a 5
1
1
1
1
1
1
1
1
[say]
- =
- =
Þ
=
=d
a 2 a1 a 3 a 2 a 4 a 3 a 5 a 4
\
a1 - a 2 = a1a 2 d Þ a 2 - a 3 = a 2 a 3d
a 3 - a 4 = a 3a 4d Þ a 4 - a 5 = a 4 a 5d
On adding all, we get
1ö
æ 1
ç - ÷
a1 - a 5
a
a1 ÷
= 4 a1a 5
a1a 2 + a 2 a 3 + a 3a 4 + a4 a 5 =
= a1a 5 ç 5
d
ç d ÷
÷
ç
ø
è
…(i)
n
n
r
(n - 1 ) r
=
2
(1 - r 2 )
n = 2 (1 - r )
\ Sn = åTn = å
n=2
[ by GP]
Chap 03 Sequences and Series
25. According to the question, for AP
S10 = 4 S 5
n
10
5
é
ù
(2a + 9d ) = 4 × (2a + 4d ) ê by Sn = [2a + (n - 1 ) d ]ú
2
2
2
ë
û
10a + 45d = 20a + 40d
a 1
=
Þ
10a = 5d Þ
d 2
26. Q cos ( x - y ), cos x, cos ( x + y ) are in HP.
cos x =
Þ
2 (cos2 x - sin 2 y )
cos x =
2 cos x cos y
30. a1, a 2,¼are
, in AP and q1, q 2, ¼ are in GP.
a1 = q1 = 2 and a10 = q10 = 3
Let d be the common diference of AP
3 -2 1
i.e.,
d =
=
9
9
Þ cos2 x cos y = cos2 x - sin 2 y
Þ
cos2 x (1 - cos y ) = sin 2 y
cos2 x = (1 + cos y )
Þ
cos2 x = 2 cos2
[Q1 - cos y ¹ 0]
æ3ö
Let r be the common ratio of GP i.e., r = ç ÷
è2ø
æy ö
cos x sec ç ÷ = ± 2
è2ø
\
a19 = a1 + 18d = 2 + 18d
1 36
= 2 + 18 ´ =
=4
9 9
y
2
æy ö
cos2 x sec 2 ç ÷ = 2
è2ø
Þ
æ3ö
= 2× ç ÷
è2ø
Given, 28, A1, A2, A3, ¼, A11, 10 are in AP.
10 - 28
3
\
d =
=12
2
3
\
A i = 28 + id = 28 - i
2
It is clear that A2, A4 , A6, A8, A10 are integral AM’s.
Hence, number of integral AM’s are 5.
28. Q x, y , z are in GP
[ x, y , z > 1 ]
2 x + ln x, 4 x + ln y , 6 x + ln z are also in AP.
1
1
1
are in HP.
\
,
,
2 x + ln x 4 x + ln y 6 x + ln z
[ x > 1]
1/ 9
q 7 = q1r 6 = 2r 6
Then,
27. Let 11 AM’s are A1, A2, A 3,¼, A 11.
\ ln x, ln y , ln z are in AP
and 2 x, 4 x, 6 x are also in AP.
By property,
1 8
=
9 3
a 7 = a1 + 6d = 2 + 6d = 2 + 6 ´
Then,
= (1 + cos y ) (1 - cos y )
Þ
1
1
³2 Þ a + + 3 ³5
a
b
1
1
Similarly, b + ³ 2 Þ b + + 3 ³ 5
b
b
1
1
and
c + ³2 Þ c + + 3 ³5
c
c
1
1
1
æ
öæ
öæ
ö
\ ça + + 3 ÷ çb + + 3 ÷ çc + + 3 ÷ ³ 125
è
øè
øè
ø
a
b
c
a+
So,
A ³ 5 × 5 × 5 Þ A ³ 125
Minimum value of A is 125.
2 cos ( x - y ) cos ( x + y )
cos ( x - y ) + cos ( x + y )
\
293
6´
1
9
æ3ö
q10 = q1r = 2r = 2 × ç ÷
è2ø
9
2 /3
æ3ö
=2 ç ÷
è2ø
9´
9
1
9
=3
q19 = q1 × r 18 = 2 × r 18
æ3ö
= 2× ç ÷
è2ø
18 ´
1
9
æ3ö
= 2ç ÷
è2ø
18 / 9
=
9
2
8 9
´ = 12, which is an integer.
3 2
2 /3
2 /3
æ3ö
æ3ö
(b) a19q 7 = 4 ´ 2 ´ ç ÷
=8 ç ÷ ,
è2ø
è2ø
(a) a 7q19 =
which is not an integer.
8 9
(c) a 7q19 = ´ = 12; a19q10 = 4 ´ 3 = 12
3 2
1 1
1
31. Qa(n ) = 1 + + + ¼ + n
2 3
2 -1
(a 2 + 3a + 1 ) (b 2 + 3b + 1 ) (c 2 + 3c + 1 )
abc
æ a 2 + 3a + 1 ö æ b 2 + 3b + 1 ö æ c 2 + 3c + 1 ö
=ç
֍
֍
÷
a
b
c
è
øè
øè
ø
æ1 1ö æ 1 1 1 1ö
=1 + ç + ÷ + ç + + + ÷
è2 3ø è 4 5 6 7ø
1ö
1
æ1
+ ç + ¼+ ÷ + ¼+ n
è8
15 ø
2 -1
1ö æ
1ö æ
1ö
æ
= ça + 3 + ÷ çb + 3 + ÷ çc + 3 + ÷,
è
aø è
bø è
cø
1
1 ö æ1 1 1
1 ö
æ1
=1+ ç + 2
÷ + ¼+ n
÷+ç + + +
è 2 2 - 1 ø è 22 5 6 23 - 1 ø
2 -1
29. Let A =
where a, b, c Î R + .
1
Applying AM ³ GM on a and ,
a
\
Þ
Þ
a(n ) < 1 + 1 + ¼+ n terms
a(n ) < n
a (100 ) < 100
294
Textbook of Algebra
1 æ1 1 ö æ1 1 1 1ö
1
+ ç + ÷ + ç + + + ÷+ ¼ + n
2 è3 4ø è5 6 7 8ø
2 -1
Also, a(n ) = 1 +
1 æ 1
1ö æ 1
1 1 1ö
=1 + + ç 1
+ 2÷ + ç 2
+ + + 3÷
2 è2 + 1 2 ø è2 + 1 6 7 2 ø
æ
1
1ö 1
+ ¼ + ç n -1
+ ¼+ n ÷ - n
è2
+1
2 ø 2
1 2 4
2
1
+ + + ¼+ n - n
2 4 8
2
2
1ö n
n
æ
Þ a(n ) >
a(n ) > ç1 - n ÷ +
è
2
2 ø 2
\ a (200 ) > 100
32.
n
middle term
1
2n – 1
x=
¥
å
cos2n f = 1 + cos2 f + cos4 f + ¼ + ¥
1
1
=
1 - cos2 f sin 2 f
1
sin 2 f =
x
…(i)
n =0
or
…(ii)
and
1 1
+
x y
xy = x + y
z=
Þ
b 2 {(a + c ) 2 - 2ac } = 2a 2c 2
Þ
b 2 ( 4b 2 - 2ac ) = 2a 2c 2
Þ
2b 4 - ac (b 2 ) - a 2c 2 = 0
Þ
(b 2 - ac ) (2b 2 + ac ) = 0
…(ii)
[from Eq. (i)]
b 2 - ac = 0
( x 2 + 2 ) 2 = x ( x 3 + 10 )
Þ
x 4 + 4 + 4 x 2 - x 4 - 10 x = 0
Þ
4 x 2 -10 x + 4 = 0
2x 2 - 5x + 2 = 0
2
2x - 4x - x + 2 = 0
Þ
Þ
2x (x - 2) - 1 (x - 2) = 0
( x - 2 ) (2 x - 1 ) = 0
1
Þ
x = 2 or x =
2
For x = 2, first 3 terms are 2, 6, 18.
So, 4th term of GP = 2 × (3 ) 3 = 54
1 9 81
1
For x = , first 3 terms are , , .
2 4 8
2
So,
T4 =
1 æ9ö
1 729 729
=
ç ÷ = ´
2 è2ø
2
8
16
36. Let n consecutive odd numbers be
1 1
1= +
x y
\
2a 2c 2
a2 + c2
3
From Eqs. (i) and (ii),
sin 2 f + cos2 f =
…(i)
b 2 {a 2 + c 2 } = 2a 2c 2
Þ
¥
1
1
=
1 - sin 2 f cos2 f
1
cos2 f =
y
[from Eq. (iii)]
Þ
Þ
y = å sin 2n f = 1 + sin 2 f + sin 4 f + ... + ¥
=
b2 =
So,
=
and
xyz = z + xy
xyz = z + x + y
a+c
34. Qa, b, c are in AP Þ b =
2
2 2 2
anda , b , c are in HP.
a, b, c are in GP.
But given a, b, c are in AP.
\
a =b =c
and if
2b 2 + ac = 0
-a
then
, b, c are in GP.
2
35. According to the question, x, x 2 + 2 and x 3 + 10 are in GP.
n=0
or
Þ
and
If
In a AP of (2n - 1 ) terms, nth term = a
In a GP of (2n - 1 ) terms, nth term = b
In a HP of (2n - 1 ) terms, nth term = c
a, b, c will be arithmetic mean, geometric mean, harmonic
mean, respectively.
So, a ³ b ³ c and b 2 = ac
p
33. Q 0 < f <
2
\
0 < sin f < 1 and 0 < cos f < 1
\
xy
xy - 1
Þ
n -1
a(n ) > 1 +
z=
Þ
…(iii)
¥
å cos2n f sin2n f
n=0
= 1 + cos2 f sin 2 f + cos4 f sin 4 f + ...
1
1
[from Eqs. (i) and (ii)]
=
=
2
2
1 - sin f cos f 1 - 1
xy
2k + 1, 2k + 3, 2k + 5, ..., 2k + 2n - 1
According to question, sum of these n numbers
n
= [ 2k + 1 + 2k + 2n - 1 ] = n (2k + n )
2
= n 2 +2kn = (n + k ) 2 - k 2
Given that, (n + k ) 2 - k 2 = 25 2 - 11 2
Þ
n + k = 25 and k = 11 Þ n = 14 and k = 11
So, first term = 2k + 1 = 23
Last term = 2k + 2n - 1 = 22 + 28 - 1 = 22 + 27 = 49
Chap 03 Sequences and Series
37. QG = 6 and G 2 = AH
40.
36
A
Given, 90 A + 5 H = 918
36
10
Þ
90 A + 5 ´
= 918 Þ 5 A +
= 51
A
A
Þ 5 A 2 - 51 A + 10 = 0 Þ( A - 10 ) ( 5 A - 1 ) = 0
1
A = 10,
\
5
1
38. Q
Tn =
(2n -1 ) (2n + 1 ) (2n + 3 ) (2n + 5 )
H =
Þ
n
Sn = å Tn a
\
n =1
Sn =
=
1 n
(2n + 5 ) - (2n - 1 )
å
6 n = 1 (2n - 1 ) (2n + 1 ) (2n + 3 ) (2n + 5 )
ö
1
÷
(2n + 1 ) (2n + 3 ) (2n + 5 ) ø
ö
1æ 1
1
ç
÷
6 è 1 × 3 × 5 (2n +1 ) (2n + 3 ) (2n + 5 ) ø
1
1
6
=
90 (2n + 1 ) (2n + 3 ) (2n + 5 )
1
l=
\
6
and
f (n ) = (2n + 1 ) (2n + 3 ) (2n + 5 )
\
f ( 0 ) = 15
f (1 ) = 105
ö 640
ö æ1
ö æ1
æ1ö æ1
and f ( l ) = f ç ÷ = ç + 1 ÷ ç + 3 ÷ ç + 5 ÷ =
ø 27
ø è3
ø è3
è6ø è3
1
1
39. QS = 1 +
(1 + 2 ) 2 +
(1 + 2 + 3 ) 2 + ...
(1 + 3 )
(1 + 3 + 5 )
1
Tn =
× (1 + 2 + 3 + 4 + ... n terms) 2
(1 + 3 + 5 + 7 + ...n terms)
=
=
2
(n + 1 ) 2
æ n (n + 1 ) ö
×ç
÷ =
ø
2
ù è
4
én
êë 2 [2 × 1 + (n - 1 ) × 2 ]úû
(a) T7 =
1
2
10
10
1
æn + 1ö
2
(b) S10 = å ç
÷ = å (n + 2n + 1 )
è
ø
2
4
n =1
n =1
1
4
10
10
ö
æ 10
ç å n 2 + 2 å n + å 1÷
÷
ç
èn = 1
n =1
n =1 ø
1 æ 10 ´ 11 ´ 21 2 ´ 10 ´ 11
ö
+
+ 10 ÷
ç
è
ø
4
6
2
1
505
= ( 385 + 110 + 10 ) =
4
4
=
1ö æ1 1ö
æ
E < 1 + ç1 - ÷ + ç - ÷ + ...
è
2ø è2 3ø
E <2
…(i)
1
1
E >1 +
+
+ ...
(2 ) (3 ) (3 ) ( 4 )
æ1 1ö æ1 1 ö
E > 1 + ç - ÷ + ç - ÷ + ...
è2 3ø è3 4ø
1
3
E >1 + ; E >
2
2
41. QS1 = { 0}
ì 3 5ü
S2 = í , ý
î 2 2þ
ì 8 11 14ü
S3 = í , , ý
î3 3 3 þ
ì 15 19 23 27ü
S4 = í , , , ý
î 4 4 4 4þ
M M M
Let S = 3 + 8 + 15 + ... + T19
S = 3 + 8 + ... + T18 + T19
- 0 = 3 + 5 + 7 + ... + 19 terms - T19
T19 = 3 + 5 + 7 + ... + 19 terms
19
19
T19 = (6 + 18 ´ 2 ) =
´ 42 = 399
\
2
2
ì 399 419 439 ü
,
,
, ...ý
S 20 = í
î 20 20 20 þ
439
\ Third element of S 20 =
20
20 1
Sum of elements of S 20 =
´ [ 2 ´ 399 + 19 ´ 20 ]
2 2
= 399 + 190 = 589
æ
è
1ö
nø
n
42. (a) QS = ç1 + ÷
2
3
æ1ö
æ1ö
æ1ö
æ1ö
S = 1 + nC1 ç ÷ + nC 2 ç ÷ + nC 3 ç ÷ + ... + nCn ç ÷
èn ø
èn ø
èn ø
ènø
S = 1 + n×
(7 +1 ) 2 64
=
= 16
4
4
=
1
1
1
+
+
+ ...
12 22 32
1
1
E <1 +
+
+ ...
(1 ) (2 ) (2 ) (3 )
E=
1 n æ
1
åç
6 n = 1 è (2n - 1 ) (2n + 1 ) (2n + 3 )
-
295
1 n (n - 1 )
+
2!
n
æ 1 ö n (n - 1 ) (n - 2 )
ç 2÷ +
èn ø
3!
n
æ1ö
ç 3÷
èn ø
n (n - 1 ) ... 1 æ 1 ö
ç n÷
èn ø
n!
1 æ
1ö 1 æ
1ö æ
2ö
S = 1 + 1 + ç1 - ÷ + ç1 - ÷ ç1 - ÷ + ...
n ø 3! è
nø è
nø
2! è
+ ... +
1 æ
1ö
ç1 - ÷
n! è
nø
1
1
1
S < 1 + 1 + + + ... +
n!
2! 3!
+
2ö æ
n - 1ö
æ
÷
ç1 - ÷ ... ç1 è
nø è
n ø
296
Textbook of Algebra
1
1
1
+
+ ... +
1 ×2 1 ×2 ×3
1 × 2 × 3 ... n
1
1
1
S <1 + 1 + +
+ ... +
2 2 ×2
2 × 2 ... 2
1
1- n
2 Þ S < 1 + 2 æç1 - 1 ö÷
S <1 +
1
è
2n ø
12
1
S < 3 - n - 1 \S < 3, " n
2
S <1 + 1 +
=
1
1
1
1
+
+
3n + 1 3n + 2 3n + 3 n + 1
1
1
2
=
+
3n + 1 3n + 2 3 (n + 1 )
=
1 æ
1ö æ
2ö æ
n - 1ö
÷
ç1 - ÷ ç1 - ÷ ... ç1 n! è
nø è
nø è
n ø
9n 2 + 15n + 6 + 9n 2 + 12n + 3 - 18n 2 - 18n - 4
(3n + 1 ) (3n + 2 ) (3n + 3 )
9n + 5
=
(3n + 1 ) (3n + 2 ) (3n + 3 )
=
æ
è
44. Sn ( x ) = ç xn - 1 +
1ö
æ
S ¢ ç1 - ÷ = xn - 1 + xn - 2 + xn - 3 + ... + x - (n - 1 )
è
xø
n2
Þ
For n = 1, a1 = 2,
4
4
1ö
3 4 81
æ
æ3ö
n = 2, a 2 = ç1 + ÷ = ç ÷ = 4 =
= 5.06
è
è2ø
2ø
16
2
[approximate]
for
1ö
æ
lim ç1 + ÷
n®¥è
nø
n2
lim
1
= en ®¥ n
´ n2
an =
2n
å
a =1
1
n+a
Þ
lim n
= en ® ¥ = e¥ = ¥
\ {an } represents unbounded sequence.
(d) Qan = tan n
2
n3
an = n +
+ n 5 + ... + ¥
3
15
and we know that - ¥ < tan n < ¥
So, {an } is unbounded sequence.
1
1
1
1
43. Q an =
+
+
+ ... +
n+1 n+2 n+3
3n
1
1
1
1
an =
+
+
+ ... +
n+1 n+2 n+3
n + 2n
1 ö
1 ö
æ n-2
+ n -2÷
÷ + 2 çx
è
ø
xn - 1 ø
x
1ö
æ
+ ... + (n - 1 ) ç x + ÷ + n
è
xø
Let S ¢ = xn - 1 + 2 xn - 2 + 3 xn - 3 + ... + (n - 1 ) x
S¢
=
xn - 2 + 2 xn - 3 + ... + (n - 2 ) x + (n - 1 )
x
-
S¢
\ {an } is bounded sequence.
1ö
æ
an = ç1 + ÷
è
nø
57 19
=
60 20
æ 1
1
1ö
-ç
+
+ ... + ÷
èn + 1 n + 2
3n ø
S > 1 + 1; S > 2
\ S is bounded.
2n + 1
(b) Q an =
n+2
3
For n1 = 1, a1 = = 1,
3
5
for n = 2 , a 2 = = 1.25
4
M
M M
Now,
an + 1 - an > 0 Þ an + 1 > an
\ an represents the increasing sequence
1ö
æ
n ç2 + ÷
è
2n + 1
nø 2
= =2
lim an = lim
= lim
n®¥
n®¥
n®¥ n + 2
2ö 1
æ
n ç1 + ÷
è
nø
(c) Q
1
1 1 1 1 20 + 15 + 12 + 10
= + + + =
a
2
+
3 4 5 6
60
a =1
æ 1
1
1 ö
Now, an + 1 - an = ç
+
+ ... +
÷
èn + 2 n + 3
3n + 3 ø
1 æ
1ö 1 æ
1ö æ
2ö
Also, S = 1 + 1 + ç1 - ÷ + ç1 - ÷ ç1 - ÷ + ...
n ø 3! è
nø è
nø
2! è
+
4
å
a2 =
(x - 1) x × (xn - 1 - 1)
=
- (n - 1 )
x
(x - 1)
x2
(n - 1 ) x
(xn - 1 - 1) 2
(x - 1)
(x - 1)
1
2
(n - 1 )
S ¢¢ = n - 1 + n - 2 + ... +
x
x
x
1
S ¢¢ = n [ x + 2 x 2 + ... + (n - 1 ) xn - 1 ]
x
1 [(n - 1 ) xn - nxn - 1 + 1 ]
[similarly as above]
= n
(x - 1)2
x
S¢ =
\
Sn ( x ) = S ¢ + S ¢¢ + n
Þ
Sn ( x ) =
For
1
x (n - 1)
æ xn - 1 ö
÷
ç
è x -1 ø
2
…(i)
n = 1, S1 ( x ) = 1
S100 ( x ) =
1 æ x100 - 1 ö
÷
ç
x 99 è x - 1 ø
2
45. Let the AP start with n and common difference d, then
according to question,
n + 5d = 32
n = 32 - 5d
and1072 < n + (n + d ) + ... + (n + 19d ) < 1162
…(i)
Chap 03 Sequences and Series
19 ´ 20
d < 1162
2
1072 < 640 - 100d + 190d < 1162
432 < 90d < 522
4.8 < d < 5.8
Let d is natural number, so d = 5
\
n = 32 - 5 ´ 5 = 7
First term is 7 and common difference is 5.
1072 < 20n +
Sol. (Q. Nos. 46 to 48)
8 16
24
+
+
+ ...
5 65 325
8r
8r
Tr = 4
=
4r + 1 (2r 2 + 2r + 1 ) (2r 2 - 2r + 1 )
Let Sn =
é (2r 2 + 2r + 1 ) - (2r 2 - 2r + 1 ) ù
=2 ê
ú
2
2
ë (2r + 2r + 1 ) (2r - 2r + 1 ) û
é
ù
1
1
=2 ê 2
- 2
ú
ë 2r - 2r + 1 2r + 2r + 1 û
46. lim Sn = lim
n®¥
n®¥
å
Tr
æ
ö
1
= 2 lim ç1 - 2
÷ = 2 (1 - 0 ) = 2
n®¥è
2n + 2n + 1 ø
8
8
r =1
r =1
Sol. (Q. Nos. 4-6)
Let p and ( p + 1 ) be removed numbers from 1, 2, 3, ... n, then
Sum of the remaining numbers
n (n + 1 )
=
- (2 p + 1 )
2
From given condition,
n (n + 1 )
- (2 p + 1 )
105
2
=
(n - 2 )
4
2n 2 - 103n - 8 p + 206 = 0
Since, n and p are integers, so n must be even.
Let n = 2r, we get
4r 2 + 103 (1 - r )
p=
4
Since, p is an integer, then (1 - r ) must be divisible by 4.
Let r = 1 + 4t , we get
n = 2 + 8t and p = 16t 2 - 95t + 1
Now,
51. Sum of all numbers =
1 £ p <n
50 (50 + 1 )
= 1275
2
Sol. (Q. Nos. 52 to 54)
Let A = { A - D, A, A + D }; B = {a - d , a, a + d }
According to the question,
A - D + A + A + D = 15
Þ
3 A = 15
Þ
A =5
and
a - d + a + a + d = 15
Þ
a =5
and
D =1 + d
p = (A - D ) A (A + D )
p = A (A 2 - D 2 )
2
p = 5 (25 - D )
...(i)
…(ii)
…(iii)
…(iv)
…(v)
2
q = 5 (25 - d )
p = 7 (q - p )
8 p = 7q
From Eqs. (iv) and (v), we get
8 ´ 5 (25 - D 2 ) = 7 ´ 5 (25 - d 2 )
Given that,
200 - 8 D 2 = 175 - 7d 2
25 = 8 D 2 - 7d 2
25 = 8 (1 + d ) 2 - 7d 2
æ
ö
1
1
- 2
ç 2
÷
è2 r - 2 r + 1 2 r + 2 r + 1ø
æ
ö
1 ö 288
1
æ
= 2 ç1 ÷ = 2 ç1 ÷=
2
è
145 ø 145
è
2 (8 ) + 2 (8 ) + 1 ø
Þ
50. Hence, removed numbers are 7 and 8.
8 ´7
56
=
4
4 ´ 7 + 1 9605
S 8 = å Tr = 2 å
48.
Þ
t = 6 Þn = 50 and p = 7
49. Hence, the value of n lies in ( 41,51).
Similarly,
r =1
æ
ö
1
1
= lim å 2 ç 2
- 2
÷
n®¥
2r + 2r + 1 ø
r = 1 è 2r - 2r + 1
T7 =
1 £ 16t 2 - 95t + 1 < 8t + 2
n
n
47.
Þ
297
[from Eq. (iii)]
2
25 = 8 + 8d + 16d - 7d 2
17 - d 2 - 16d = 0
d 2 + 16d -17 = 0
(d + 17 ) (d - 1 ) = 0
d = - 17 or d = 1
Þ
d =1
Þ
D =2
52. p = 5 (25 - D 2 ) = 5 (25 - 4) = 5 (21) = 105
[Q d > 0]
53. q = 5 (25 - d 2 ) = 5 (25 - 1) = 120
54. 7D + 8d = 14 + 8 = 22
Sol. (Q. Nos. 55 to 57)
ìA
ü
A = í , A, ARý
îR
þ
ü
ìa
B = í , a, ar ý
þ
îr
A
According to the question, × A × AR = 64
R
3
Þ
A = 64 Þ A = 4
a
× a × ar = 64 Þ a 3 = 64 Þa = 4
r
and
R =r + 2
Let
…(i)
…(ii)
…(iii)
298
Textbook of Algebra
60. According to the question, (m + 1) is the nth triangular number,
A
A
× A + A × AR + AR ×
R
R
16
A2
2
2
+ 16 R + 16
=
+A R+A =
R
R
a
a
q = × a + a × ar + ar ×
r
r
a2
16
=
+ a 2r + a 2 =
+ 16r + 16
r
r
p 3
=
q 2
p=
Given that,
then
n (n + 1 )
=m + 1
2
n 2 + n - 2 (m + 1 ) = 0
\
(16 + 16 R 2 + 16 R ) r 3
=
2
(16 + 16r 2 +16r ) R
So,
Sol. (Q. Nos. 61 to 63)
A1, A2, A3, ..., Am are arithmetic means between - 3 and 828.
(a + b )
So,
A1 + A2 + ... + Am = m
2
æ - 3 + 288 ö
Þ
A1 + A2 + ... + Am = m ç
÷
è
ø
2
2
(1 + R + R ) r 3
=
(1 + r 2 + r ) R 2
From Eq. (iii),
R =r + 2
(1 + r 2 + 4 + 4r + r + 2 ) r 3
Þ
=
2
(1 + r + r 2 ) (r + 2 )
Þ
3
r 3 + 5r 2 + 7r
=
3
r + 3r 2 + 3r + 2 2
Þ
r 3 - r 2 - 5r + 6 = 0
Þ
(r - 2 ) (r 2 + r - 3 ) = 0
Þ
- 1 ± 13
2
So,
R=4
ö 16
æ1
ö
æ1
55. p = 16 ç + R + 1÷ = 16 ç + 4 + 1÷ = (21) = 84
ø 4
è4
ø
èR
æ1
èr
ö
ø
æ1
è2
ö
ø
56. q = 16 ç + r + 1÷ = 16 ç + 2 + 1÷ =
Þ
So,
Þ
61. n = 10
16
´ 7 = 8 ´ 7 = 56
2
3 35 = (1 ´ 2187 )n / 2 Þ 3 35 = 3 7n / 2
7n
35 =
2
…(ii)
n = 10
[by Eq. (ii)]
62. m = 34
57. r R + Rr = ( 4) 2 + (2) 4 = 16 + 16 = 32
[by Eq. (i)]
2
3
63. G1 + G2 + ... + Gn = r + r + r + ... + r
Sol. (Q. Nos. 58 to 60)
Given sequence, 1, 3, 6, 10, 15, 21, 28, ...
where
tn = tn - 1 + n, " n ³ 2
So,
tn = [tn - 2 + (n - 1 )] + n
= tn - 3 + (n - 2 ) + (n - 1 ) + n
M
M
M
tn = t1 + 2 + 3 + ... + (n - 1 ) + n
tn = 1 + 2 + 3 + ... + n
n (n + 1 )
tn =
2
50 ´ 51
58. t 50 =
= 25 ´ 51 = 1275
2
100 ´ 101
59. t100 =
= 5050
2
101 ´ 102
t101 =
= 101 ´ 51 = 5151
2
Number of positive integers lying between t100 and t101
= 5151 - 5050 - 1
= 101 - 1 = 100
æ 825 ö
14025 = m ç
÷
è 2 ø
[given that sum of AM’s = 14025]
Þ
m = 17 ´ 2
…(i)
\
m = 34
Now, G1, G2, ..., Gn be the GM’s between 1 and 2187.
\
G1G2 G3 ... Gn = (ab )n / 2
r = 2 or r =
Þ
- 1 ± 1 + 8 (m + 1 )
2
- 1 + (8m + 9 )
=
2
- 1 + 8m + 9 - 2m
n -m =
2
n=
n
= r + r 2 + r 3 + ... + r 10 = r
(1 - r 10 )
1 -r
1/n + 1
1/11
é
ù
æl ö
æ 2187 ö
=ç
= 3 7/11 ú
÷
êQ r = ç ÷
è
ø
è
ø
a
1
êë
úû
= 3 7 /11
…(i)
(1 - 3 70 /11 )
(1 - 3 7/11 )
Solution (Q. Nos. 64 to 66)
Q
b 2 - 4ac
b
c
a + b = - , ab = , a - b =
a
a
a
and
g + d=-
B
C
, gd = , g - d =
A
A
B 2 - 4 AC
A
64. Since, a, b, g are in AP.
Let
Q
or
b = a + D , g = a + 2D and d = a + 3D
-b
b
a+b=
Þa+ a+ D=a
a
b
…(i)
2a + D = a
Chap 03 Sequences and Series
B
B
Þ 2 a + 5D = A
A
From Eqs. (i) and (ii), we get
1 æb B ö
æ B bö
4D = ç - + ÷ or D = ç - ÷
è A aø
4 èa A ø
and
g + d=-
…(ii)
69. Again,
Þ
a+b
=
g+d
So,
g + d = ar 2 + ar 3 = -
and
B
Þ
ar (1 + r ) = A
From Eqs. (i) and (ii), we get
Ba
r2 =
bA
aB
r=
\
bA
2
…(iii)
From Eqs. (i) and (ii), we get
2
æn + 1ö
q<p<ç
÷ p
èn - 1ø
70. a, b, c, d are positive real numbers with
a <b <c <d
According to the question, a, b, c, d are in AP.
Þ b = a + a, c = a + 2 a and d = a + 3 a
a be the common difference
and a, b, d are in GP.
Þ
b 2 = ad
…(i)
…(i)
…(ii)
a 2 + a 2 + 2aa = a 2 + 3aa
a 2 = aa
Þ
…(ii)
…(A)
From Eqs. (i) and (ii), we get
(a + a ) 2 = a (a + 3 a )
Þ
B
A
2
Þ
a (a - a ) = 0
Þ
a = 0 or a = a
a ¹ 0 by (A), so a = a
From Eq. (i), b = 2a, c = 3a and d = 4a
ad
a × 4a 2 æ p ö
=
= =ç ÷
bc 2a × 3a 3 è q ø
where, p and q are prime numbers.
So,
q =3
Sol. (Q. Nos. 67 to 69)
For n > 1, we have n + 1 > n - 1
110
2
Þ
æ a
bö
æa b
ö
÷
n ç + - 2÷ n ç
aø
èb a
ø
è b
p
-1 =
=
q
(n + 1 ) 2
(n + 1 ) 2
p
p
-1 > 0 Þ
>1 Þ p >q
q
q
Þ
ab
gd
c
b
b 2A 2 c A
a
a
Þ
=
Þ
=
C
B
a 2B 2 aC
A
A
ac A 2 c A
=
Þ B 2 = AC
Þ
C
aB 2
Hence, A, B, C are in GP.
66. Since, a, b, g, d, ... are in GP.
b g d
r= = =
\
a b g
b
Þ
a + b = a + ar = a
b
Þ
a (1 + r ) = a
a 2n + abn 2 + b 2n + ab
ab (n + 1 ) 2
æa b ö
n ç + ÷ + (n 2 + 1 )
èb a ø
=
(n + 1 ) 2
b g d
= =
a b g
b d
a b
= Þ =
a g
g d
Þ
p æ an + b ö é a + bn ù
=ç
÷´
q è n + 1 ø êë ab (n + 1 ) úû
=
65. Since, a, b, g, d, ... are in GP.
\
299
æn + 1ö
n+1
>1 Þ p ç
÷ >p
èn - 1ø
n -1
Now,
p =a + d
Since, a, p, b, are in AP.
b -a
And
d =
n+1
(b - a ) na + b
67.
p =a +
=
n+1
n+1
1 1
1 1
1
68.
= +D= + b a
q a
a
n+1
ab (n + 1 )
Þ
q=
a + bn
[Q p > 0] …(i)
71. Q å (1 + rx ) = (1 + x ) (1 + 2x ) (1 + 3x )...(1 + 110x )
r =1
= 1110 + ( x + 2 x + 3 x + ... + 110 x ) 1109 + ...
So, coefficient of x in
110
å
r =1
(1 + rx ) = (1 + 2 + 3 + ... + 110 ) =
110 ´ 111
= 55 ´ 111
2
= 6105
Now, l (1 + 10 ) (1 + 10 + 10 2 ) = l (11 ) (111 )
Þ
l (111 ) (11 ) = 6105 Þ l = 5
72. Let number of the form palindrome be aba.
Now, If aba is even, then a may be 2, 4, 6, 8 and b take values
0, 1, 2, ..., 9 .
So, total number of palindrome (even) = 10 ´ 4 = 40
300
Textbook of Algebra
To find the sum of all even 3 digit plaindrome
So, sum of number start with 2
= (200 + 2 ) ´ 10 + ( 0 + 1 + 2 + 3 + ... + 9 ) ´ 10
= 2020 + 450 = 2470
Sum of number start with 4 = ( 404 ) ´ 10 + 450
Similarly, sum of number start with 6 = (606 ) ´ 10 + 450
Similarly, sum of number start with 8 = (808 ) ´ 10 + 450
\Total sum = (202 + 404 + 606 + 808 ) ´ 10 + 450 ´ 4
= 20200 + 1800 = 22000
= 2 4 ´ 5 3 ´ 11
4
3
1
n1
n2
n3
On comparing 2 ´ 5 ´ 11 with
2
´3
´5
n4
´7
n5
´ 11 ,
n1 = 4, n2 = 3, n3 = 0, n4 = 0, n5 = 1
Now,
n1 + n2 + n3 + n4 + n5 = 8
2
73. Q2 + ( 6 × 2 - 4 × 2) + ( 6 × 3 2 - 4 × 3)
+ ... + (6 × n 2 - 4 × n ) = 140
Þ 2 + 6 (2 2 + 3 2 + ... + n 2 ) - 4 × (2 + 3 + ... + n ) = 140
æ n (n + 1 ) (2n - 1 )
ö
Þ 2+6ç
- 1÷ - 4
è
ø
6
ö
æ n (n + 1 )
- 1 ÷ = 140
ç
ø
è
2
Þ 2 + n (n + 1 ) (2n + 1 ) - 6 - 2n (n + 1 ) + 4 = 140
Þ
n (n + 1 ) (2n + 1 ) - 2n (n + 1 ) - 140 = 0
Þ
2n 3 + 3n 2 + n - 2n 2 - 2n - 140 = 0
Þ
2n 3 + n 2 - n - 140 = 0
Þ
(n - 4 ) (2n 2 + 9n + 35 ) = 0
Þ
n = 4 or 2n 2 + 9n + 35 = 0
Þ
2n 2 + 9n + 35 = 0
- 9 ± 81 - 280
Þ
n=
4
9 ± - 199
[complex values]
n=
\
4
Only positive integer value of n is 4.
74. S ( x ) = 1 + x - x 2 - x 3 + x 4 + x 5 - x 6 - x 7+ ... + ¥
where x Î( 0, 1 )
S ( x ) = (1 + x ) - x 2 (1 + x ) + x 4 (1 + x ) - x 6 (1 + x ) + ... + ¥
Þ S ( x ) = (1 + x ) [1 - x 2 + x 4 - x 6 + ... + ¥ ]
æ 1 ö
Þ S ( x ) = (1 + x ) ç
÷
è1 + x2 ø
é
ù
a
êQ S ¥ = 1 - r for GP ú
ë
û
2+1
2
1+x
2+1
=
2
1 + x2
According to the question, S ( x ) =
So,
Þ
2 + 2x = ( 2 + 1)x 2 + 2 + 1
Þ
( 2 + 1) x 2 - 2x - 2 + 2 + 1 = 0
Þ
( 2 + 1) x 2 - 2x + 2 - 1 = 0
1
=0
2+1
( 2 + 1) x 2 - 2x +
Þ
Þ
[( 2 + 1 ) x ]2 - 2 ( 2 + 1 ) x + 1 = 0
Þ
[( 2 + 1 ) x - 1 ]2 = 0
Þ
x=
1
2+1
[repeated]
x = 2 -1
So,
(x + 1)2 = 2
\
75. a1, a 2, a 3, ... are in GP with common ratio r
and b1, b2, b3, ... is also a GP i.e. b1 = 1
b2 = 4 7 - 4 28 + 1, a1 = 4 28
¥
¥
1
å an = åbn
and
n =1
n =1
1
1
1
+
+
+ ... + ¥ = b1 + b2 + b3 + ... + ¥
a1 a 2 a 3
1
1
1
+
+
+ ... + ¥
Þ 4
28 4 28 r 4 28 r 2
= 1 + ( 4 7 - 4 28 + 1 ) + ( 4 7 - 4 28 + 1 ) 2 + ... + ¥
Þ
Þ
Þ
Let
Þ
Þ
Þ
Now,
1
1
28 =
1 1 - 4 7 + 4 28 - 1
1r
r
1
=4 4
4
(r - 1 ) 28
7 ( 4 - 1)
4
r
1
1
=
(r - 1 ) 4 4 ( 4 4 - 1 )
4
4 = a, we get
r
1
=
(r - 1 ) a a - 1
ra - r = r a - a Þ r = a
r =44
1 + r 2 + r 4 = 1 + (4 4 )2 + (4 4 )4
= 1 + 41/2 + 4 = 1 + 2 + 4 = 7
76. Let
a = 10 + D
b = 10 + 2 D
ab = 10 + 3 D
On substituting the values of a and b in Eq. (iii), we get
(10 + D ) (10 + 2 D ) = (10 + 3 D )
Þ
2 D 2 + 27 D + 90 = 0
15
\
D = -6,D = 2
\
a1 = 10 - 6 = 4 ,
15 5
a 2 = 10 =
2 2
and b1 = 10 - 12 = - 2 , b2 = 10 - 15 = - 5
Now,
æ 2a1a 2 + b1b2 ö æ 2 ´ 10 + 10 ö
÷ =3
÷=ç
ç
ø
ø è
è
10
10
…(i)
…(ii)
…(iii)
Chap 03 Sequences and Series
77. Given equation, Ax 3 + Bx 2 + Cx + D = 0
…(i)
A¹0
a+g
Let roots are a, b, g, then b =
2
3
Given relation, 2 B + l ABC + m A 2D = 0
b
From Eq. (i), a + b + g = A
B
Þ
3b = A
B
b= Þ
3A
Now, b satisfy Eq. (i), so
where,
3
…(ii)
- B3
B3
BC
+ 2 +D=0
2
3A
27 A
9A
2 B 3 BC
+D=0
27 A 2 3 A
2 B 3 - 9 ABC + 27 DA 2 = 0
Þ
æ
n
2r
å çç
n®¥
r =0 è 1
+ x2
r
+
2r
1 - x2
r
æ 2r + 1
2r
= lim å ç
r
+
1
r
ç
2
n®¥
1 - x2
r =0 è 1 - x
n
1 - x2
r
ö
÷
÷
ø
n+1
n®¥
2n + 1
-1
[given]
l =1
According to the question, (a - d ) 2, a 2, (a + d ) 2 are in GP.
(a 2 ) 2 = (a - d ) 2 (a + d ) 2
Þ
a 4 = (a 2 - d 2 ) 2
Þ
a 4 = a 4 + d 4 - 2a 2d 2
Þ
Þ
Þ
a 2 (a 2 - 2d 2 ) = 0
a ¹ 0, so a 2 = 2d 2
a = ± 2d
Let common ratio of GP is r.
Þ
r2 =
(3 + 2 2 ) d 2
(3 - 2 2 ) d 2
Þ
r2 =
(3 + 2 2 ) (3 + 2 2 )
9 -8
Þ
r 2 = (3 + 2 2 ) 2
Þ
r 2 = (3 + 8 ) 2
2d 2 + d 2 + 2 2d 2
2d 2 + d 2 - 2 2 d 2
[from Eq. (i) for a = 2 d ]
\
r = ± (3 + 8 )
Þ
r =3 + 8
[Q r is positive]
Þ
r = (3 - 8 )
[Q r is positive]
Compare r with 3 ± k, we get
k =8
ék 8 ù é8 8ù
\
êë 8 - k úû = êë 8 - 8 úû
= [1 - 1 ] = [ 0 ] = 0
80. (A) a, b, c, d are in AP
[a, b, c, d are positive real numbers]
b 2 > ac
…(i)
…(ii)
From Eqs. (i) and (ii), we get
b 2c 2 > (ac ) (bd ) Þ bc > ad
79. Let number of AP are (a - d ), a, (a + d ) .
\
r2 =
Now, applying for b, c, d
c > bd Þ c 2 > bd
1
1
=01-x
1 -x
x
1
1
=
=
x -1 x - l
\
Þ
Þ
2n +1
x2
1
a 2 + d 2 + 2ad
a 2 + d 2 - 2ad
By AM > GM, for a, b, c
b > ac
ö
÷
÷
ø
æ 2n + 1
1 ö÷
= lim ç
n
+
1
n®¥ç
1 - x ÷ø
è1 - x2
= lim
r2 =
r = ± (3 - 8 )
2r
-
Þ
Similarly, for a = - 2 d , we get
Compare with Eq. (iii), we get
l = - 9 , m = 27
2 l + m = - 18 + 27 = 9
æ 1
ö
2
22
78. Let P = lim ç
+
+
+ ... upto n terms÷
2
4
n ® ¥ è1 + x
1+x
1+x
ø
= lim
(a + d ) 2
(a - d ) 2
[from Eq. (ii)]
2
Þ
r2 =
…(iii)
æ- Bö
æ- Bö
æ- Bö
Aç
÷+D=0
÷ +C ç
÷ +Bç
è 3A ø
è 3A ø
è 3A ø
Þ
\
301
…(i)
Again, applying AM > HM for a,b, c
1 1 2
2
b>
Þ + >
1 1
a
c b
+
a c
For last 3 terms b, c, d
1 1 2
2
c>
Þ + >
1 1
b d c
+
b d
From Eqs. (iii) and (iv), we get
1 1 1 1 2 2
+ + + > +
a c b d b c
1 1 1 1
Þ
+ > +
a d b c
(B) a, b, c, d are in GP.
For a, b, c applying AM > GM,
a+c
> b Þ a + c > 2b
Þ
2
…(iii)
…(iv)
…(i)
302
Textbook of Algebra
Similarly, for b, c, d
b + d > 2c
From Eqs. (i) and (ii), we get
a + b + c + d > 2 b + 2c Þ a + d > b + c
Now, applying GM > HM for a, b c
2 ac
b>
a+c
1 1 2
Þ
+ >
c a c
Similarly, for b, c, d , we get
1 1 2
+ >
d b c
On adding Eqs. (iii) and (iv), we get
1 1 1 1
æ 1 1ö
+ + + >2 ç + ÷
èb c ø
a b c d
1 1 1 1
+ > +
a d b c
(C) a, b, c, d are in HP.
Applying AM > HM for a, b, c
a+c
>b
2
Þ
a + c > 2b
...(ii)
Þ
Þ
ac > b 2
Þ
Similarly, for b, c, d
Þ
bd > c 2
…(iii)
…(iv)
…(i)
…(ii)
Þ
and
…(iii)
…(iv)
ad > bc
81. (A) a1, a 2, a 3,..., an , ... are in AP
and
From Eqs. (i) and (ii), we get
né
3ù
5 + (n - 1 ) ú = 110
2 êë
2û
n (3n + 40 ) - 11 (3n + 40 ) = 0
(3n + 40 ) (n - 11 ) = 0
40
or n = 11
So,
n=3
\
n = 11
[n Î N ]
(B) Let first angle = a
[in degrees]
Common difference = d
[in degrees]
Number of sides n = 9
\Sum of interior angles = (n - 2 ) ´ 180°
n
[ 2a + (n - 1 ) d ] = (n - 2 ) ´ 180°
Þ
2
9
Þ
( 2 a + 8d ) = 7 ´ 180°
2
Þ
a + 4d = 140°
and largest angle T9 = a + 8d < 180°
Þ
4d < 40
Þ
d < 10
\
d =9
(C) Given increasing GP,
a1, a 2, ..., an , ...
where
a 6 = 4a 4
[r is the common ratio]
a1r 5 = 4a1r 3
Þ
On multiplying Eqs. (iii) and (iv), we get
abcd > b 2c 2
5
a1 = , a10 = 16
2
\
a1 + a 2 + ... + an = 110
n
Þ
(a1 + an ) = 110
2
n é5 5
ù
Þ
+ + (n - 1 ) d ú = 110
2 êë 2 2
û
5
16 a - a1
2 = 27 = 3
Now, d = 10
=
10 - 1
9
9 ´2 2
3n 2 + 40n - 33n - 440 = 0
Þ
Þ
Þ
Similarly, for last 3 terms b, c, d
Þ
b + d > 2c
On adding Eqs. (i) and (ii), we get
a + b + c + d > 2b + 2c
Þ
a + d >b + c
Again, applying GM > HM for a, b, c
ac > b
3
= 220
2
3n 2 + 7n - 440 = 0
5n + (n 2 - n )
Þ
r2 = 4
r =2
a 9 - a 7 = 192
a1 (r 8 - r 6 ) = 192
[Q increasing GP]
a1 (256 - 64 ) = 192
192
a1 =
192
a1 = 1
Then, a 2 = 2 , a 3 = 4 and a 4 + a 5 + ... + an = 1016
(a1 + a 2 + ... + an ) - (a1 + a 2 + a 3 ) = 1016
1 (2n - 1 )
= 1016 + 7
2 -1
2n = 1023 + 1 = 1024 = 210
\
n = 10
82. (A) a1, a 2, ... are in AP.
…(i)
…(ii)
a1 + a 4 + a 7 + a14 + a17 + a 20 = 165 [In an AP, sum of the
terms equidistant from the 1st and last is equal to sum of
1st and last terms]
Þ
3 (a1 + a 20 ) = 165
Þ
a1 + a1 +19d = 55
d is the common difference of AP.
…(i)
2a1 + 19d = 55
Now,
a = a 2 + a 6 + a15 + a19
a = 2 (a 2 + a19 )
Chap 03 Sequences and Series
a = 2 (a1 + d + a1 + 18d )
a = 2 (2a1 + 19d ) …(ii)
and
b = 2 (a 9 + a12 ) - (a 3 + a18 )
b = 2 (a1 + 8d + a1 + 11d ) - (a1 + 2d + a1 + 17d )
b = 2 (2a1 + 19d ) - (2a1 + 19d )
…(iii)
b = 2a1 + 19d
From Eqs. (i) and (iii), we get
a = 2b
From Eqs. (i), (ii) and (iii), we get
[from Eq.(i)]
a + 2 b = 4 (2a1 +19d ) = 4(55 ) = 220
a + b = 3 (2a1 + 19d )
= 3 ´ 55 = 165 = 15 ´ 11 = 15m, where m Î I
a - b = 2a1 + 19d
= 55 = 5 ´ 11 = 5l, where l Î I
(B) a1, a 2, ... are in AP.
a1 + a 5+ a10 + a15 + a 20 + a 24 = 195
3 (a1 + a 24 ) = 195
...(i)
Þ
a1 + a 24 = 65
Þ
2a1 + 23d = 65
Now,
a = a 2 + a 7 + a18 + a 23
= 2 (a 2 + a 23 ) = 2 (2a1 + 23d )
[from Eq. (i)]
a = 130
b = 2 (a 2 + a 22 ) - (a 8 + a17 )
= 2 (2a1 + 23d ) - (2a1 + 23d )
= 130 - 65 = 65
Then, a = 2 b
a + 2 b = 130 + 130 = 260
a + b = 195 = 15 ´ 13 = 15 m, where m = 13
and a - b = 130 - 65 = 65
= 5 ´ 13 = 5l, where l = 13
(C) a1, a 2,... are in AP.
a1 + a 7 + a10 + a 21 + a 24 + a 30 = 225
3 (a1 + a 30 ) = 225
…(i)
2a1 + 29d = 75
Now,
a = a 2 + a 7 + a 24 + a 29
a = 4a1 + 58d = 2 (2a1 + 29d )
= 2 ´ 75 = 150
…(ii)
a = 150
and
b = 2 (a10 + a 21 ) - (a 3 + a 28 )
= 2 (2a1 + 29d ) - (2a1 + 29d ) = 150 - 75
…(iii)
b = 75
Then, a = 2b
a + 2b = 150 + 150 = 300 and a - b = 150 - 75 = 75
= 5 ´ 15 = 5l, where l = 15
and a + b = 150 + 75 = 225 = 15 ´ 15 = 15m, where m = 15
83. (A) 4a 2 + 9b 2 + 16c 2 = 2 (3ab + 6bc + 4ca )
(2a ) 2 + (3b ) 2 + ( 4c ) 2 - (2a ) (3b ) ( 4c ) - (2a ) ( 4c ) = 0
1
{(2a - 3b ) 2 + (3b - 4c ) 2 + ( 4c - 2a ) 2 } = 0
2
Þ
2a - 3b = 0 and 3b - 4c = 0
and
4c - 2a = 0 Þ2a = 3b
303
3b = 4c and 4c = 2a
3
4
1
a = b and b = c and c = a
Þ
2
3
2
3
4
3
Þ
a = b and b = c and c = b
2
3
4
3
3
So,
a, b, c are b, b, b
2
4
2 1 4
Reciprocal of the terms , , ,
3b b 3b
which is in AP.
So, these a, b, c are in HP.
(B) 17a 2 + 13b 2 + 5c 2 = 3ab + 15bc + 5ca
and
Þ 34a 2 + 26a 2 + 10c 2 - 6ab - 30bc - 10ca = 0
Þ
(3a - b ) 2 + (5b - 3c ) 2 + (c - 5a ) 2 = 0
Þ
3a - b = 0 and 5b - 3c = 0 and c - 5a = 0
a b c
= = =l
Þ
1 3 5
\
a = l, b = 3 l, c = 5 l
Hence, a, b, c are in AP.
æ 15 5 3 ö
(C) a 2 + 9b 2 + 25c 2 = abc ç + + ÷
èa
b cø
[say]
Þ (a ) 2 + (3b ) 2 + (5c ) 2 -15bc - 5ac - 3ab = 0
1
Þ
{(a - 3b ) 2 + (3b - 5c ) 2 + (5c - a ) 2 } = 0
2
Þ a - 3b = 0 and 3b - 5c = 0 and 5c - a = 0
Þ
a = 3b and 3b = 5c and 5c = a
5
a
Þ
a = 3b and b = c and c =
5
3
5
3
Þ
a = 3b and b = c and c = b
3
5
3b
So, a,b, c are of the form 3b, b, .
5
3b
1 1 5
are , , , which are in AP.
Reciprocal of 3b, b,
5
3b b 3b
2
5 1 2ù
é 1 1
êëQ b - 3b = 3b and 3b - b = 3b úû
(D) (a 2 + b 2 + c 2 ) p 2 - 2 p (ab + bc + ca ) + a 2 + b 2 + c 2 £ 0
Þ (a 2p 2 + b 2 - 2abp ) + (b 2p 2 + c 2 - 2 pbc )
+ (c 2p 2 + a 2 - 2acp ) £ 0
Þ
(ap - b ) 2 + (bp - c ) 2 + (cp - a ) 2 £ 0
Þ
(ap - b ) 2 + (bp - c ) 2 + (cp - a ) 2 = 0
Þ
ap - b = 0 and bp - c = 0 and cp - a = 0
b
c
a
Þ
p = and p = and p =
a
b
c
b c a
= =
Þ
a b c
Þa, b, c are in GP.
84. If a, b, c are in GP.
Then,
b 2 = ac
If middle term is added, then a + b, 2b and c + b are in GP.
304
Textbook of Algebra
I - II
a + b - 2b
=
II - III 2b - (c + b )
[here, I = a + b, II = 2b, III = c + b]
a - b ab - b 2 ab - ac
=
=
b - c b 2 - bc ac - bc
a (b - c ) (a + b ) (b + c )
=
c (a - b ) (a + b ) (b + c )
[Q b 2 = ac ]
=
=
a (b 2 - c 2 ) (a + b )
c (a 2 - b 2 ) (b + c )
=
I
a (ac - c 2 ) (a + b ) a + b
;
=
2
c (a - ac ) (b + c ) b + c III
Hence, a + b, 2b, b + c are in HP.
Hence, both statements are true and Statement-2 is correct
explanation for Statement-1.
85. QTn = 2n 3 + 3n 2 - 4
Sequence is 1, 24, 77, 172, 321, ...
First order difference 23, 53, 95, 149, ...
Second order difference 30, 42, 54, ...
which are in AP.
\Statenemt-1 is true.
QTn is of three degree and third order difference will be constant.
Statement-2 is true, which is correct explanation for
Statement-1.
86. Statement-1 Let S be the required sum of product of numbers.
æ n
ç å xi
ç
è i =1
2
n
ö
÷ = å xi2 + 2 å å xi x j
÷
ø
i =1
1 £i < j £n
n
\ (a1 - a1 + a 2 - a 2 + ... + an - an ) 2 = 2 åai2 + 2S
i =1
\
n
S = - åai2
i =1
\Statement-1 is true.
Statement-2 is true but not correct explanation for
Statement-1.
87. Statement-1 a + b + c = 18, a, b, c > 0
Applying
AM ³ GM for a, b, c
a+b+c 3
³ abc Þ 3 abc £ 6 Þ abc £ 216
3
Maximum value of abc is 216 which occurs at a = b = c .
Statement-2 is the correct explanation for Statement-1.
88. Statement-1
4a 2 + 9b 2 + 16c 2 - 2 (3ab + 6bc + 4ca ) = 0
Þ (2a ) 2 + (3b ) 2 + ( 4c ) 2 - (2a ) (3b ) - (3b ) ( 4c ) - (2a ) ( 4c ) = 0
1
Þ
{(2a - 3b ) 2 + (3b - 4c ) 2 + ( 4c - 2a ) 2 } = 0
2
Þ
2a - 3b = 0 and 3b - 4c = 0 and 4c - 2a = 0
4c
a
3b
4c
3b
and c = Þ a =
and b =
and c =
Þ and b =
2
3
2
3
4
3b
3b
, which are in HP.
Then, a, b, c are of the form , b,
2
4
So, Statement-1 is false.
Statement-2 If
(a1 - a 2 ) 2 + (a 2 - a 3 ) 2 + (a 3 - a1 ) 2 = 0
Þ
a1 - a 2 = 0 and a 2 - a 3 = 0 and a 3 - a1 = 0
Þ
a1 = a 2 = a 3, " a1, a 2, a 3 Î R
So, Statement-2 is true.
a+b
2ab
89. Q A =
, G = ab and H =
2
a+b
4G = 5 H
G 2 = AH
Given,
and
\
H =
G2
A
…(i)
…(ii)
From Eqs. (i) and (ii), we get
5G 2
4G =
Þ 4 A = 5G
A
2 (a + b ) = 5 ab
Þ
Þ
4 (a 2 + b 2 + 2ab ) = 25ab
Þ
4a 2 - 17ab + 4b 2 = 0
Þ
(a - 4b ) ( 4a - b ) = 0
[Q a > b]
a = 4b, 4a - b ¹ 0
\Statement-1 is true.
Statement-2 is true only for two numbers, if numbers more
than two, then this formula (AM) (HM) = (GM) 2 is true, if
numbers are in GP.
Statement-2 is false for positive numbers.
90. Statement-1 Sum of first 100 even natural numbers
2 (100 ´ 101 )
= 10100
2
Sum of 100 odd natural numbers = 1 + 3 + ... + 199
100
O=
(1 + 199 ) = 10000
2
\
E - O = 100
So, Statement-1 is true.
Statement-2 Sum of first n natural even numbers
2n (n + 1 )
= n2 + n
E = 2 + 4 + ... + 2n =
2
Sum of first n odd natural numbers
O = 1 + 3 + ... + (2n - 1 )
n
= [1 + 2n - 1 ] = n 2
2
So,
E - O = n2 + n - n2 = n
E1 = 2 + 4 + ... + 200 =
Statement-2 is true and correct explanation for Statement-1.
91. Let
Tn = An + B
\
Tp = Ap + B,
T2p = 2 Ap + B, T4 p = 4 Ap + B
\ Tp , T2p , T4 p are in G P.
\
(2 Ap + B ) 2 = ( Ap + B ) ( 4 Ap + B )
Þ
\
ABp = 0
B = 0, A ¹ 0, p ¹ 0
T2p 2 Ap + 0
Þ Common ratio, r =
=
=2
Tp
Ap + 0
Chap 03 Sequences and Series
92. a ¹ 1, b ¹ 0 and a ¹ b
2
2
3
2
2
3
Let S = (a + b ) + (a + ab + b ) + (a + a b + ab + b ) + ¼+ n
terms
1
=
[(a 2 - b 2 ) + (a 3 - b 3 ) + (a 4 - b 4 ) + ¼+ n terms]
(a - b )
1
[a 2(1 + a + ¼ + n terms)
=
(a - b )
- b 2 (1 + b + b 2 + ¼ + n terms)]
1 é 2 1 × (an - 1 )
1 × (bn - 1 ) ù
- b2 ×
ú
êa ×
(a - b ) ë
(a - 1 )
(b - 1 ) û
1 é 2 (1 - an )
(1 - bn ) ù
=
- b2
êa
ú
(a - b ) ë (1 - a )
(1 - b ) û
=
p2 1 p2 3 p2 p2
= ´
=
6
4 6
4
6
8
1
1
1
1
ö
æ1
(ii) 1 - 2 + 2 - 2 + ¼ + ¥ = ç 2 _ 3 + ¼÷
ø
è1
2
3
4
3
1 æ1
1
ö
- 2 ç 2 - 2 + ¼ + ¥÷
ø
2 è1
2
=
p2 1 p2 p2
- ´
=
8
4
6
12
2
97. Saibi = Sai (1 - ai ) = na - Sai
=
= na - S[(ai - a ) 2 + a 2 + 2a (ai - a )]
= na - S[(ai - a ) 2 - Sa 2 - 2aS(ai - a )
\ Saibi + S(ai - a ) 2 = na - na 2 - 2a (na - na )
1, 3, 5, 7, 9, 11, …
\nth row contains n elements
1st element of nth row = n 2 - (n - 1 )
= na (1 - a ) = nab
Least element of nth row = n 2 + (n - 1 )
b = S 2n - Sn =
c = S 3n
=
a(r 2n - 1 ) a (r n - 1 ) a (r n - 1 ) n
=
(r )
(r - 1 )
(r - 1 )
(r - 1 )
[by part (i)]
= na - S(ai - a + a ) 2
93. Sequence of natural number is divided into group.
\Sum of the element in the nth row
n
n
= (a + l ) = [n 2 - (n - 1 ) + n 2 + (n - 1 )]
2
2
n 2
n
= [n - n + 1 + n 2 + n - 1 ] = [ 2n 2 ] = n 3
2
2
a (r n - 1 )
94.
a = Sn =
r -1
305
éQ Sbi = S1 - Sai ù
ê \ nb = n - na ú
ê
ú
êë or a + b = 1 úû
98. a1 + a 2 + ¼ + a 98 = 137
98
(a1 + a 98 ) = 137
2
137
137
; 2a1 + 97 =
a1 + a 2 + 97 =
49
49
137
1 (137 - 4753
2a1 =
- 97; a1 =
49
2
49
4616
2308
; a1 =
a1 = 2 ´ 49
49
…(i)
…(ii)
a (r 3n - 1 ) a (r 2n - 1 )
- S 2n =
(r - 1 )
(r - 1 )
a (r n - 1 ) 2n
a (r n - 1 ) n 2
× (r )
(r + r n + 1 - r n - 1 ) =
(r - 1 )
(r - 1 )
Now, a 2 + a 4 +¼ + a 98 = (a1 + 1 ) + (a1 + 3 ) + ¼ + (a1 + 97 )
[Q d = 1 ]
= 49a1 + (1 + 3 + ¼+ 97 )
2308 49
= - 49 ´
+
(1 + 97 )
49
2
= - 2308 + 49 2
…(iii)
From Eqs. (i), (ii) and (iii), b 2 = ac , so a, b, c are in GP.
95. First four terms of an AP are a, 2a, b and (a - 6 - b ).
So,
2a - a = a - 6 - b - b
Þ
a = a - 6 - 2b
Þ
- 2b = 6 Þb = - 3
and
2a - a = b - 2a
Þ
b = 3a Þ a = - 1
\First terms a = - 1 and d = a = - 1
100
S100 =
[ 2a + (100 - 1 ) d ]
2
= 50 [ - 2 + 99 ( - 1 )]
= 50 ( - 2 - 99 ) = 50 ( - 101 ) = - 5050
1
1
1
p2
96. (i) 2 + 2 + 2 + ¼+ ¥ =
...(i)
6
1
2
3
1
1
1
\
+
+
+ ¼+ ¥
12 32 52
1
1
1
1
1
ö æ1
ö
æ1
= ç 2 + 2 + 2 + 2 + ¼+ ¥ ÷ - ç 2 + 2 + 2 + ¼ + ¥ ÷
ø è2
ø
è1
2
3
4
4
6
…(i)
= - 2308 + 2401 = 93
99. t1 = 1 and tr - tr - 1 = 2r - 1, r ³ 2
t 2 - t1 = 2
t3 - t2 = 22
t4 - t3 = 23
M M M
tn - tn - 1 = 2n - 1
Addiing columnwise, we get
tn - t1 = 2 + 2 2 + ¼ + 2n - 1
tn = 1 + 2 + 2 2 + ¼ + 2n - 1
tn =
n
So,
1 × (2n - 1 )
Þ tn = 2n - 1
2 -1
åtr = t1 + t2 + ¼ + tn = (2 - 1) + (22 - 1) + ¼ + (2n - 1)
r =1
= (2 + 2 2 + ¼ + 2n ) - n =
= 2n + 1 - n - 2
2 × (2n - 1 )
- n = 2n + 1 - 2 - n
(2 - 1 )
306
Textbook of Algebra
p sin 2n ( p - x )
dx = ò
dx
0
sin x
sin x
p sin (2np - 2nx )
dx
=ò
0
sin x
p sin 2nx
In = - ò
dx
0 sin x
100. (i) I n = ò
p sin 2nx
=
0
(1 - x 3n )
x 3 (1 - x 3n )
3 x (1 - xn ) 3 (1 - xn )
+ 3n
+
+ n
3
3
(1 - x )
(1 - x )
x (1 - x )
x (1 - x )
103. Let d be the common difference of AP.
LHS = a12 - a 22 + a 32 - a 42 + ¼ + a 22n - 1 - a 22n
(a1 - a 2 ) (a1 + a 2 ) + (a 3 - a 4 ) (a 3 + a 4 )
+ ¼ + (a 2n - 1 - a 2n ) (a 2n - 1 + a 2n )
= - d (a1 + a 2 + ¼ + a 2n - 1 + a 2n )
= - d [(a1 + a 2n ) + (a 2 + a 2n - 1 ) + ¼ + (an + an + 1 )]
= - dn (a1 + a 2n )
2
2
(a 2 - a 22n ) -dn (a1 - a 2n )
[Q a 2n = a1 + (2n - 1 ) d ]
=
= - dn 1
(1 - 2n ) d
(a1 - a 2n )
n
=
(a12 - a 22n )
2n - 1
I n = - I n Þ 2I n = 0 Þ I n = 0
\
I1 = I 2 = I 3 = ¼ = 0
which is a constant series.
\ This series is AP with common difference 0 and first
term o.
p sin 2 nx
(ii) I n = ò
dx
0 sin 2 x
sin 2 nx
Let
f (x ) =
sin 2 x
Hence, f ( p - x ) = f ( x )
p/ 2 sin 2 nx
So,
dx
In = 2 ò
0
sin 2 x
Now, I n + 1 + I n - 1 - 2 I n
p/ 2 ì (sin 2(n + 1 ) x - sin 2 nx )
= 2ò í
0
sin 2 x
î
(sin 2(n - 1 ) x - sin 2 nx )ü
+
ý dx
sin 2 x
þ
Þ
b = 2l and ac = l (a + c )
2 2 2
Now, a , b , c are in AP
Þ
Þ
(a + c ) 2 - 2 l (a + c ) - 8 l2 = 0
3
3
3
n
1ö
1 ö
1 ö
1 ö
æ n
æ 2
æ n
÷ + çx + 2 ÷ + ¼ + çx + n ÷ = å çx + n ÷
ø
è
ø
è
ø
ø
è
x
x
x
x
n =1
n
1
1 öö
æ
æ
å çè x 3n + x 3n + 3 çè xn + xn ÷ø ÷ø
n =1
n
n
n =1
n =1
1
å x 3n + å x 3n
+3
n
n
n =1
n =1
1
å xn + 3 å xn
So,
…(i)
b2 =
Þ
(a + c - 4 l ) (a + c + 2 l ) = 0
Þ
a + c = 4l or a + c = - 2l
Case I If a + c = 4l
\
ac = 4 l2
2
[from Eq. (i)]
2
Þ
(a - c ) = (a + c ) - 4ac
Þ
(a - c ) 2 = 16 l2 - 16 l2
Þ
(a - c ) 2 = 0 Þa = c
Let given that a, b, c are distinct, so a + c = 4l is not valid.
Case II If a + c = - 2l
[from Eq. (i)]
Þ
ac = - 2 l2
Tn = 7 + 6 + 8 + 10 + ¼+ n terms
Tn = 7 + {6 + 8 + 10 + ¼ + (n - 1 ) terms}
(n - 1 )
Tn = 7 +
(12 + (n - 2 ) 2 )
2
(n - 1 )
Tn = 7 +
(8 + 2n )
2
Tn = 7 + (n - 1 ) ( 4 + n )
T70 = 7 + 69 ´ 74 = 7 + 5106 = 5113
=
[say]
p/ 2
S=
7 + 13 + 21 + ¼ + Tn -1 + Tn
- - - - 0 = 7 + 6 + 8 + 10 + ¼+ n terms - Tn
=
Þ
2ac
a+c
b
ac
=
=l
2 a+c
b=
0
sin (2n + 1 ) x sin x - sin (2n - 1 ) x sin x
dx
sin 2 x
p/ 2 sin (2n + 1 ) x - sin (2n - 1 ) x
dx
=2 ò
0
sin x
p/ 2 2 cos 2nx sin x
dx
=2 ò
0
sin x
p /2
4
2
[sin 2nx ]p0 /2 = × 0 = 0
= 4 ò cos 2nx dx =
0
n
2n
\ I n + 1 + I n - 1 = 2 I n \I 1, I 2, I 3, ¼are in A P.
101. S = 7 + 13 + 21 + 31 + ¼ + Tn
æ
è
\
a2 + c2
Þ2b 2 = a 2 + c 2
2
2(2 l ) 2 = (a + c ) 2 - 2ac
=2 ò
102. ç x +
104. Let a, b, c (unequal number) are in HP
\
(a - c ) 2 = (a + c ) 2 - 4ac
Þ
(a - c ) 2 = 4 l2 + 8 l2 Þ (a - c ) = ± 2 3l
If
a - c = 2 3l,
...(ii)
...(iii)
then
a + c = 2l
From Eqs. (ii) and (iii), we get
a = ( 3 - 1 ) l and c = - (1 + 3 ) l
3
\
a : b : c = ( 3 - 1)l : 2l : - ( 3 + 1)l
a : b : c = ( 3 - 1) : 2 : - ( 3 + 1)
Þ
a : b : c = (1 - 3 ) : - 2 : ( 3 + 1 )
If
a - c = - 2 3l,
…(iv)
then
a + c = - 2l
…(v)
Chap 03 Sequences and Series
n
108. LHS = (1 + 5 - 1 ) (1 + 5 -2 ) (1 + 5 - 4 ) ¼ (1 + 5 - 2 )
From Eqs. (iv) and (v), we get
a = - ( 3 + 1 ) l and c = ( 3 - 1 ) l
\
a : b : c = - ( 3 + 1) l : 2l : ( 3 - 1)l
Þ
a : b : c = (1 + 3 ) : - 2 : (1 - 3 )
æ
= ç1 +
è
1ö
÷
5ø
1ö
æ
ç1 - ÷
è
5ø
=
1ö
æ
ç1 - ÷
è
5ø
105. a1, a 2, a,¼, an are in AP with a1 = 0 and common difference d
[d ¹ 0 ]
\ a 2 = d , a 3 = 2d , ¼, an = (n - 1 ) d
1
1
[(n - 2 ) d - d ] (n - 1 ) d
+
(2d - d ) +
(3d - d ) + ... +
d
2d
(n - 2 ) d
(n - 3 ) d
n-1
= [1 + 1 + ¼ + (n - 3 ) times ] +
n-2
(n - 1 )
(n - 2 ) + 1
= (n - 3 ) +
= (n - 3 ) +
n -2
(n - 2 )
1
1
= (n - 3 ) + 1 +
=n -2 +
n -2
n -2
an - 1
a (n - 2 ) d
d
a
=
+
=
+ 2 = RHS
d
(n - 2 ) d
a2
an - 1
106. Let one side of equilateral triangle contains n balls. Then
Number of balls (initially) = 1 + 2 + 3 + ¼ + n =
n (n + 1 )
2
n (n + 1 )
+ 669 = (n - 8 ) 2
2
n 2 + n + 1338 = 2n 2 - 32n + 128
According to the question,
Þ
n 2 - 33n - 1210 = 0
Þ
Þ
(n - 55 ) (n + 22 ) = 0 Þ n = 55 or n = - 22
which is not possible
\
n = 55
n (n + 1 ) 55 ´ 56
So,
=
= 55 ´ 28 = 1540
2
2
107. q1, q2, q3, ¼, qn are in AP.
So,
q2 - q1 = q3 - q2 = ¼ = qn - qn - 1 = d
\LHS = sin d [sec q1 sec q2 + sec q2 sec q3 + ¼
+ sec qn - 1 sec qn ]
é
1
1
= sin d ê
+
q
q
q
cos
cos
cos
cos q3
1
2
2
ë
ù
ú
cos qn - 1 cos qn û
sin d
sin d
sin d
=
+
+ ¼+
cos q1 cos q2 cos q2 cos q3
cos qn - 1 cos qn
+ ¼+
=
1
sin ( qn - qn - 1 )
sin ( q2 - q1 )
sin ( q3 - q2 )
+
+ ¼+
cos q1 cos q2 cos q2 cos q3
cos qn -1 cos qn
= (tan q2 - tan q1 ) + (tan q3 - tan q2 )
+ ¼ + (tan qn - tan qn - 1 )
= tan qn - tan q1 = RHS
1 ö
1ö æ
1ö æ
1öæ
æ
ç1 + ÷ ç1 + 2 ÷ ç1 + 4 ÷ ¼ ç1 + n ÷
è
5ø è
5 ø è
5 øè
52 ø
5
4
=
(an - 1 - a 2 )
1
1
a
(a 3 - a 2 ) +
(a 4 - a 2 ) + ¼ +
+ n
a2
a3
an - 2
an - 1
=
1 ö
1ö æ
1öæ
æ
ç1 + 2 ÷ ç1 + 4 ÷ ¼ ç1 + n ÷
è
ø
è
ø
è
5
5
52 ø
éæ
1 öù
1ö æ
1öæ
1öæ
ê çè1 + 2 ÷ø çè1 + 2 ÷ø çè1 + 4 ÷ø ¼ ç1 + 2 n ÷ ú
è
5
5
5
5 øû
ë
M
M
M
5 æ
1 ö 5
- 2n + 1
= ç1 - n + 1 ÷ = (1 - 5
) = RHS
4è
ø 4
52
æ1
1
1 ö
a
a
a
a
÷
LHS = 3 + 4 + 5 + ¼ + n - a 2 ç +
+¼+
an - 2 ø
a2 a3 a4
an - 1
è a2 a3
=
307
¥
109. S = å
n=0
Sn =
2n
n
a2 + 1
n
, (a > 1 )
2n
å
n
a2 + 1
1
2
4
8
2n
=
+ 2
+ 4
+ 8
+ ¼+ n
a+1 a +1 a +1 a +1
a2 + 1
1
2
4
8
2n
=
+
+
+
+ ¼+
n
2
4
8
1+a 1+a
1+a
1+a
1 + a2
n =0
æ
2n
1
1 ö
1
2
4
= ç+
+ ¼+
+
+
÷+
n
2
4
è 1 -a 1 -aø 1 + a 1 + a
1+a
1 + a2
æ 1
4
2n
1
1 ö
2
=
+ç
+
+
+
¼
+
÷+
n
a - 1 è1 -a 1 + a ø 1 + a2 1 + a4
1 + a2
æ 2
1
2 ö
4
2n
+ç
+
+
+
¼
+
÷
n
a - 1 è1 - a2 1 + a2 ø 1 + a4
1 + a2
M
M
M
M
1
2n + 1
Sn =
+
a - 1 1 - a2n + 1
=
æ 1
2n + 1 ö÷
S = lim Sn = lim ç
+
n+1 ÷
n®¥
n ® ¥ ça - 1
è
ø
1 - a2
2n + 1
æ
ö
n+1
ç
÷
a2
1
÷= 1 + 0 = 1
= lim ç
+
1
n ® ¥ ça - 1
- 1 ÷÷ a - 1 0 - 1 a - 1
ç
2n + 1
è
ø
a
æ
ö
2n - 1 ö
2n -1
÷ = tan - 1 ç
÷
n
n -1
2n - 1
è1 + 2
ø
è 1 + 2 ×2
ø
æ
110. Tn = tan - 1 ç
æ 2n - 2n -1 ö
= tan - 1 ç
÷ = tan - 1 2 n - tan - 1 2 n - 1
è 1 + 2 n ×2 n - 1 ø
Sn = T1 + T2 + ¼ + Tn
= (tan - 1 21 - tan - 1 2 0 ) + (tan - 1 2 2 - tan - 1 21 ) + ¼
+ (tan - 1 2n - tan - 1 2n - 1 )
= (tan - 1 2n - tan - 1 1 )
p
Sn = tan - 1 2n 4
pö p p p
æ
S = lim Sn = lim ç tan - 1 2n - ÷ = - =
n®¥
n®¥è
4ø 2 4 4
308
Textbook of Algebra
111. Tn = tan [ a + (n - 1) b} tan ( a + nb)
tan b = tan [( a + nb ) - { a + (n - 1 ) b }]
tan ( a + nb ) - tan ( a + (n - 1 ) b }]
tan b =
1 + tan ( a + nb ) tan { a + (n - 1) b }]
\1 + Tn = cot b [tan ( a + nb ) - tan { a + (n - 1 ) b }]
Tn = cot b [tan ( a + nb ) - tan { a + (n - 1 ) b }] - 1
For n = 1,
T1 = cot b [tan ( a + b ) - tan a ] - 1
For n = 2,
T2 = cot b [tan ( a + 2b ) - tan ( a + b )] - 1
For n = 3,
T3 = cot b [tan ( a + 3b ) - tan ( a + 2b )] - 1
M
M
M
M
For n = n,
Tn = cot b [tan ( a + nb ) - tan ( a + (n - 1 ) b }] - 1
Sum columnwise,
Sn = T1 + T2 + T3 + ...+ Tn = cot b [tan ( a + nb ) - tan a ] - n
sin ( a + nb ) sin a
sin ( a + nb - a )
+
cos ( a + nb ) cos a
cos a cos ( a + nb )
=
-n=
-n
tan b
tan b
sin nb
=
- n tan b
cos ( a + nb ) cos a
tan b
n
112. Sn = å Tr =
r =1
n (n + 1 ) (n + 2 ) (n + 3 )
8
r (r + 1 ) (r + 2 ) (r + 3 ) (r - 1 ) r (r + 1 ) (r + 2 )
8
8
r (r + 1 ) (r + 2 )
=
2
æ 1
ö
1
1
2
(r + 2 ) - r
=
=
=ç
÷
Tr r (r + 1 )(r + 2 ) r (r + 1 ) (r + 2 ) è r (r + 1 ) (r + 1 )(r + 2 ) ø
Tr = Sr - Sr - 1 =
n
æ
ö
1
1
1
å Tr = å çè r (r + 1) - (r + 1) (r + 2) ÷ø
r =1
r =1
n
n ì
æ1
1 ö æ 1
1 öü
= å íç ÷-ç
÷ý
è
ø
è
r
r
1
r
1
r
2 øþ
+
+
+
r =1 î
æ1
1 ö æ1
1 ö
=ç ÷-ç ÷
è1 n + 1ø è2 n + 2ø
=
1
1
1
n (n + 3 )
+
=
2 n + 2 n + 1 2 (n + 1 ) (n + 2 )
\
Þ
Þ
Þ
2
1
1
=
+
d 2 d1 d 3
2
1
1
=
+
2 (S 2 - n ) 2 (S1 - n ) 2 (S 3 - n )
n (n - 1 )
n (n - 1 )
n (n - 1 )
2
1
1
=
+
S 2 - n S1 - n S 3 - n
S + S1 - 2n
2
= 3
S 2 - n (S1 - n ) (S 3 - n )
Þ2 [S1S 3 - (S1 + S 3 ) n + n 2 ] = (S 2 - n ) (S1 + S 3 - 2n )
Þ2S1S 3 - 2(S1 + S 3 ) n + 2n 2
= S1S 2 + S 2 S 3 - 2nS 2 - n (S1 + S 3 ) + 2n 2
Þ 2S1S 3 - S 2 S 3 - S1S 2 = n (S1 + S 3 - 2S 2 )
(2S S - S 2 S 3 - S1S 2 )
Þ
n= 1 3
(S1 - 2S 2 + S 3 )
114. Let their ages be a, ar , ar 2.
After 3 yr, their ages will be a + 3, ar + 3, ar 2 + 3.
Given, 2 (a + 3 ) = ar 2 + 3
Let x rupees be the sum of the money divided.
And let y = a + ar + ar 2
n
\
Si = [2 ´ a + (n - 1 ) di ], " i = 1, 2, 3
2
n
Þ
Si = [2 + (n - 1 ) di ]
2
n (n - 1 )
2 (Si - n )
Þ
Si = n +
di Þ di =
2
n (n - 1 )
Given that d1, d 2, d 3 are in HP.
1 1 1
are in AP.
, ,
\
d1 d 2 d 3
…(ii)
2
Then, y + 9 = a + 3 + (ar + 3 ) + (ar + 3 )
x (a + 3 ) x a
We have,
=
+ 105
(y + 9 )
y
æa + 3 a ö
Þ
xç
- ÷ = 105
èy + 9 y ø
Also,
Þ
…(iii)
x (ar + 3 ) xar
=
+ 15
(y + 9 )
y
é ar + 3 ar ù
xê
- ú = 15
y û
ëy +9
…(iv)
On dividing Eq. (iii) by Eq. (iv), we get
y (a + 3 ) - a (y + 9 )
y - 3a
=7 Þ
=7
y (ar + 3 ) - ar (y + 9 )
y - 3ar
a (7r - 1 )
Þ
6y = 21ar - 3a Þ y =
2
From Eq. (ii),
a(7r - 1 )
= a + ar + ar 2
2
Þ
5ar = 3a + 2ar 2
…(v)
From Eqs. (i) and (ii),
a = 12, r =
113. Let d1, d 2 and d 3 be the common differences of the 3 arithmetic
progressions.
…(i)
3
2
2
3
æ3ö
Let ages of these friends are 12, 12 ´ , 12 ´ ç ÷ i.e. 12, 18, 27.
è2ø
2
1
1
1
.
and z =
115. Clearly, x =
,y =
1 -a
1 -b
1 -c
Since, a, b, c are in AP.
Þ
1 - a, 1 - b, 1 - c are also in AP.
1
1
1
are in HP.
Þ
,
,
1 -a 1 -b 1 -c
\x, y , z are in HP.
Chap 03 Sequences and Series
1
2
nö
3 æ
3
ç1 - æç - ö÷ ÷÷
n
è 4ø ø 1
4 çè
1
æ 3ö
Now, An =
<
Þ ç- ÷ > 3
è 4ø
2
6
1+
4
Obviously, it is true for all even values of n.
3
1
But for
n = 1, - < 4
6
3
27
1
æ 3ö
n = 3 ç- ÷ = <è 4ø
24
6
116. Q Bn = 1 - An > An Þ An <
5
243
1
æ 3ö
n = 5, ç - ÷ = >è 4ø
1024
6
which is true for n = 7 obviously, n0 = 7
Aliter
Bn = 1 - An > An
n
æ
3 ö
çç1 - æç - ö÷ ÷÷
n
è 4ø ø 1
1
1
3 è
æ 3ö
An <
Þ
Þ
< Þ ç- ÷ > 3
è 4ø
6
2
4
2
1+
4
Obviously, it is true for all even values of n.
3
1
But for
n = 1, - < 4
6
3
27
1
æ 3ö
n = 3, ç - ÷ = <è 4ø
64
6
5
243
1
æ 3ö
n = 5, ç - ÷ = <è 4ø
1024
6
7
2187
1
æ 3ö
n = 7 Þ ç- ÷ = >è 4ø
12288
6
Hence, minimum natural number n0 = 7.
p
[2a1 + ( p - 1 ) d ]
p2
2
117. Q
= 2
q
[2a1 + (q - 1 ) d ] q
2
æ p - 1ö
a1 + ç
÷d
è 2 ø
2a1 + ( p - 1 ) d p
p
=
Þ
=
Þ
2a1 + (q - 1 ) d q
æq - 1ö
q
a1 + ç
÷d
è 2 ø
a
a 6 11
For 6 , p = 11 and q = 41 Þ
=
a 21
a 21 41
1
1
1
1
1
1
[say]
118.
- =
=¼= =d
a 2 a1 a 3 a 2
an an - 1
an - 1 - an
a - a2
a - a3
Then, a1a 2 = 1
, a 2a 3 = 2
, ¼, an - 1an =
d
d
d
a1 - an
\ a1a 2 + a 2 a 3 + ¼ + an - 1an =
d
1
1
a1 - an
Also,
= + (n - 1 ) d Þ
= (n - 1 ) a1an
an a1
d
\ a1a 2 + a 2 a 3 + ¼ + an - 1an = (n - 1 ) a1an
r
1
119. (i)
Vr = [(2 r + (r - 1 ) (2 r - 1 )] = (2r 3 - r 2 + r )
2
2
n
1é n 3 n 3 n ù
\ åVr = ê2 år - år + år ú
2 êë r = 1
r =1
r =1
r =1 ú
û
and for
309
2
1 é æ n (n + 1 ) ö
n (n + 1 ) (2n + 1 ) n (n + 1 ) ù
+
÷ ê2 ç
úû
ø
2 êë è
2
6
2
1
= n (n + 1 ) (3n 2 + n + 2 )
12
1
1
(ii) Vr + 1 - Vr = (r + 1 ) 3 - r 3 - [(r + 1 ) 2 - r 2 ] + ( 1 )
2
2
= 3r 2 + 2r + 1
\
Tr = 3r 2 + 2r - 1
= (r + 1 )(3r - 1 ), which is a composite number.
(iii) Since,
Tr = 3r 2 + 2r - 1
\
Tr + 1 = 3 (r + 1 ) 2 + 2 (r + 1 ) - 1
=
\
Qr = Tr + 1 - Tr = 3 [2r + 1 ] + 2[1 ]
Þ
Qr = 6r + 5
Þ
Qr + 1 = 6(r + 1 ) + 5
Common difference = Qr + 1 - Qr = 6
a+b
2ab
120. (i) A 1 =
; G 1 = ab ; H1 =
2
a+b
2 An - 1Hn - 1
An - 1 + Hn - 1
, Gn = An - 1Hn - 1 and Hn =
An =
2
An - 1 + Hn - 1
Clearly, G1 = G2 = G3 = ¼ = ab
(ii) A2 is AM of A1 and H1 and A1 > H1
Þ
A1 > A2 > H1
A3 is AM of A2 and H 2 and A2 > H 2
Þ
A2 > A3 > H 2
\
A1 > A2 > A3 > ¼
(iii) As above A1 > H 2 > H1, A1 > H 3 > H 2
\
H1 < H 2 < H 3 < ¼
121. Let geometric progression is a, ar , ar 2,¼,
Q
Þ
a = ar + ar
r2 + r - 1 = 0
[a, r > 0 ]
2
Þ r=
-1 ± 5
2
5 -1
2
122. b1 = a1, b2 = b1 + a 2 = a1 + a 2, b3 = b2 + a 3 = a1 + a 2 + a 3
and b4 = b3 + a 4 = a1 + a 2 + a 3 + a 4
Hence, b1, b2, b3, b4 are neither in AP nor in GP and nor in HP
123. Let a, ar , ar 2, ¼
\
and
r=
a + ar = 12
ar 2 + ar 3 = 48
…(i)
…(ii)
On dividing Eq. (ii) by Eq. (i), we get
r 2 = 4, if r ¹ - 1
\
r = -2
[Q terms are alternatively positive and negative]
Now, from Eq. (i), a = - 12
124. Q Sn = cn 2
\ tn = Sn - Sn - 1 = c (2n - 1 )
Stn2 = c 2 S(2n - 1 ) 2
= c 2 S ( 4n 2 - 4n + 1 ) = c 2 { 4 Sn 2 - 4 Sn + S1}
ì 4n (n + 1 ) (2n + 1 ) 4n (n + 1 )
ü
= c2 í
+ ný
6
2
î
þ
310
Textbook of Algebra
Þ
ü
ì2
= c 2n í ( 2n 2 + 3n + 1 ) - 2n - 2 + 1ý
þ
î3
2
2
…(i)
…(ii)
ì 1 ü
ï 2 ï 4 2
4
1
1
ö 4
æ1
= + 4 ç 2 + 3 + 4 + ¼÷ = + 4 í 3 ý= + = 2
1
ø
è
3
3
3
3
3
ï1 - ï 3 3
3þ
î
\ S =3
k -1
1
a
k
126. Sk =
= k! = =
1 - r 1 - 1 k ! (k - 1 )!
k
100
100
k=2
k=2
Now, å |(k 2 - 3k + 1 ) Sk | = å | (k 2 - 3k + 1 ) ×
100
k=2
127. Q
æ 99 100 ö
÷
ç
è 98 ! 99 ! ø
(100 ) 2
100
=3 100 !
99!
(100 ) 2 100
+ å |(k 2 - 3k + 1 ) Sk | = 3
100 ! k = 2
ak = 2ak - 1 - ak - 2 or ak - 1 =
ak - 2 + ak
\ a1, a 2, a 3, ¼are in AP.
2
a 2 + a 22 + a 32 + ¼ + a11
Q 1
= 90 Þ
11
11
å
Þ
2
11
å ai2 = 11 ´ 90
k =1
2
(a1 + (i - 1 ) d ) = 11 ´ 90
k =1
Þ
11
å {a12 + 2a1d (i - 1) + d 2 (i - 1)2 } = 11 ´ 90
k =1
Þ
11 ´ a12 + 2a1d ( 0 + 1 + 2 + 3 + ¼ + 10 )
+ d 2 ( 0 2 + 1 2 + 2 2 + ¼ + 10 2 ) = 11 ´ 90
æ 10 × 11 ö
2 æ 10 × 11 × 21 ö
Þ 11 ´ 15 2 + 2 ´ 15 ´ d × ç
÷
÷ +d ×ç
è
ø
è 2 ø
6
= 11 ´ 90
Þ 385d 2 + 1650d + 1485 = 0
[Q a1 = 15 ]
Þ
7d 2 + 30d + 27 = 0
Þ
\
(n - 125 ) (n - 24 ) = 0
n = 125, 24
n = 125 is not possible.
\
n = 24
\Total time = 10 + 24 = 34 min
129. Q AM ³ GM
\
Þ
\
a - 5 + a - 4 + a - 3 + a - 3 + a - 3 + 1 + a 8 + a10
8
³ (a - 5 × a - 4 × a - 3 × a - 3 × a - 3 × 1 × a 8 × a10 )1/8 = (1 )1/8 = 1
a - 5 + a - 4 + 3a - 3 + 1 + a 8 + a10
³1
8
-5
-4
-3
8
10
a + a + 3a + 1 + a + a ³ 8
Þ Required minimum value = 8
æ2 1 ö æ2 3ö æ3 4ö
= ç - ÷ + ç - ÷ + ç - ÷ + ¼+
è 1! 0!ø è 1! 2!ø è 2! 3!ø
=3 -
d = - 3, d ¹ -
130. Let the time taken to save ` 11040 be (n + 3) months.
1 2
2 3
3 4
99 100
+
+
+ ¼+
0! 1!
1! 2!
2! 3!
98 ! 99 !
=
\
1
|
(k - 1 )!
k
(k - 1 )
=
(k - 2 ) (k - 1 )!
=å
9
[Q 27 - 2a 2 > 0 ]
7
11
{2a1 + (11 - 1 ) d }
a1 + a 2 + a 3 + ¼ + a11
= 2
\
11
11
= a1 + 5d = 15 - 15 = 0
128. Till 10th minute, number of counted notes = 1500
n
\3000 = {2 ´ 148 + (n - 1 ) ´ - 2 } = n (148 - n + 1 )
2
Þ
n 2 - 149n + 3000 = 0
\
2
c n
n ( 4n - 1 ) c
( 4n 2 - 1 ) =
3
3
2 6 10 14
125. Let S = 1 + + 2 + 3 + 4 + ¼
3 3
3
3
1
1
2
6
10
\ S=
+
+ 3 + 4 +¼
3
3 32
3
3
On subtracting Eq. (ii) from Eq. (i), we get
2
1
4
4
4
S =1 + + 2 + 3 + 4 + ¼
3
3 3
3
3
=
(7d + 9 ) (d + 3 ) = 0
For first 3 months, he saves ` 200 each month.
\In (n + 3 ) month,
n
3 ´ 200 + [2 (240 ) + (n - 1 ) ´ 40 ] = 11040
2
n
600 + [ 40 (12 + n - 1 )] = 11040
Þ
2
Þ
600 + 20n (n + 11 ) = 11040
Þ
n 2 + 11n - 522 = 0
Þ
(n - 18 ) (n + 29 ) = 0
\
n = 18, neglecting n = - 29
\Total time = n + 3 = 21 months
131. Given,
a 2 + a 4 + a 6 + ¼ + a 200 = a
and
a1 + a 3 + a 5 + ¼ + a199 = b
On subtracting Eq. (ii) from Eq. (i), we get
(a 2 - a1 ) + (a 4 - a 3 ) + (a 6 - a 5 ) + ¼+
(a 200 - a199 ) = a - b
Þ
d + d + d +¼+ d = a - b Þ 100d = a - b
( a - b)
d =
\
100
132. Qa1, a 2, a 3, ¼are in HP.
1 1 1
\
, , , ¼are in AP.
a1 a 2 a 3
Let D be the common difference of this AP, then
1
1
= + (20 - 1 ) D
a 20 a1
1 1
4
25
5 =Þ
D=
19
25 ´ 19
…(i)
…(ii)
311
Chap 03 Sequences and Series
and
1 4 (n - 1 ) æ 95 - 4n + 4 ö
1
1
= + (n - 1 ) D = =ç
÷
5 25 ´ 19 è 25 ´ 19 ø
an a1
æ 99 - 4n ö
=ç
÷<0
è 25 ´ 19 ø
[Q an < 0 ]
Þ
99 - 4n < 0 Þ n > 24 .75
Hence, the least positive integer is n = 25.
133. Q
(1 ) = (1 - 0 ) (1 2 + 1 × 0 + 0 2 ) = 1 3 - 0 3
(1 + 2 + 4 ) = (2 - 1 ) (2 2 + 2 × 1 + 1 2 ) = 2 3 - 1 3
( 4 + 6 + 9 ) = (3 - 2 ) (3 2 + 3 × 2 + 2 2 ) = 3 3 - 2 3
M
M
M
(361 + 380 + 400 ) = (20 - 19 ) (20 2 + 20 × 19 + 19 2 ) = 20 3 - 19 3
Required sum
= (1 3 - 0 3 ) + (2 3 - 1 3 ) + (3 3 - 2 3 ) +¼ + (20 2 - 19 3 ) = 20 3 = 8000
n
Also, åk 3 - (k - 1 ) 3 =
k =1
=
n
å(3k
2
n
å {k - (k - 1)} {k 2 + k(k - 1) + (k - 1)2 }
k =1
- 3k + 1 ) = 3 Sn 2 - 3 Sn + S1
...(iii)
\x, y , z are in GP.
From Eqs. (i) and (ii) x, y , z are in AP and also in GP, then
x = y = z.
136. S = 0.7 + 0.77 + 0.777 + K upto 20 terms
7
= (0.9 + 0.99 + 0.999 + K upto 20 terms)
9
7
= [(1 - 0.1 ) + (1 - 0.01 ) + (1 - 0.001 ) + K upto 20 terms]
9
7
= [(1 + 1 + 1 + ¼ upto 20 times)
9
1
1
æ1
öù
- ç + 2 + 3 + ¼ upto 20 terms÷ ú
è 10 10
øû
10
20 ö ù
é
æ
1
1
çç 1 - æç ö÷ ÷÷ ú
ê
è
ø ø ú 7 é 180 - 1 + 10 -20 ù
10
10
è
7ê
= ê 20 ú
ú = 9ê
1
9
9
û
ë
1ú
ê
10
ú
ê
û
ë
7
=
(179 + 10 - 20 )
81
137. Sn = - 1 2 - 2 2 + 3 2 + 4 2 - 5 2 - 6 2 + 7 2 + 8 2 -¼
k =1
3n(n + 1 ) (2n + 1 ) 3n (n + 1 )
=
+n
6
2
n
= (2n 2 + 3n + 1 - 3n - 3 + 2 ) = n 3
2
Both statements are correct and Statement-2 is the correct
explanation of Statement-1.
134. Let a be the first term and d be the common difference. Then,
100 T100 = 50 T50
Þ
100 (a + 99d ) = 50 (a + 49d )
Þ
2 (a + 99d ) = (a + 49d ) Þ a + 149 d = 0
\
T150 = 0
135. Q x, y , z are in AP.
Let x = y - d , z = y + d
Also, given tan - 1 x, tan - 1 y , tan - 1 z are in AP.
\
Þ
Þ
2 tan - 1 y = tan - 1 x + tan - 1 z
æ 2y ö
æx+z ö
tan - 1 ç
÷ = tan - 1 ç
÷
2
è 1 - xz ø
è1 - y ø
x+z
2y
2y
2y
=
=
Þ
1 - y 2 1 - xz
1 - y 2 1 - (y 2 - d 2 )
2
2
2
Þ
y =y -d
\
d =0
From Eq. (i), x = y and z = y
\
x =y =z
Aliter
Q
x, y , z are in AP.
\
2y = x + z
Also, tan - 1 x, tan - 1 y , tan - 1 z are in AP.
\
Þ
Þ
Þ
...(i)
2 tan - 1 y = tan - 1 x + tan - 1 z
æ 2y ö
æx+z ö
tan - 1 ç
÷ = tan - 1 ç
÷
2
è 1 - xz ø
è1 - y ø
x+z
2y
2y
=
=
1 - y 2 1 - xz 1 - xz
y 2 = zx
[from Eq. (i)]
+ ( 4n - 1 ) 2 + ( 4n ) 2
= (3 2 - 1 2 ) + ( 4 2 - 2 2 ) + (7 2 - 5 2 ) + (8 2 - 6 2 ) + ¼
+ [{( 4n - 1 ) 2 - ( 4n - 3 ) 2 } + {( 4n ) 2 - ( 4n - 2 ) 2 }]
= 4 [2 + 3 + 6 + 7 + 10 + 11 + ¼ + ( 4n - 2 ) + ( 4n - 1 )]
= 8 {(1 + 3 + 5 + ¼ + (2n - 1 )} + 4 {3 + 7 + 11
+ ¼ + ( 4n - 1 )}
= 16n 2 + 4n = 4n ( 4n + 1 ), n Î N
Satisfied by (a) and (d), where n = 8, 9, respectively.
138. Let two consecutive numbers are k and k + 1 such that
1 £ k £ n - 1, then
(1 + 2 + 3 + ¼ + n ) - (k + k + 1 ) = 1224
n (n + 1 )
n 2 + n - 2450
- (2k + 1 ) = 1224 or k =
Þ
2
4
n 2 + n - 2450
Now,
1£
£ n - 1 Þ 49 < n < 51
4
\
n = 50 Þ k = 25
Hence,
k - 20 = 25 - 20 = 5
139. The given series can be written as
2
8
æ 11 ö
æ 11 ö
æ 11 ö
æ 11 ö
k = 1 + 2 ç ÷ + 3 ç ÷ + ¼ + 9 ç ÷ + 10 ç ÷
è 10 ø
è 10 ø
è 10 ø
è 10 ø
æ 11 ö
On multiplying both sides by ç ÷ , then
è 10 ø
2
3
9
9
…(i)
10
11k æ 11 ö
æ 11 ö
æ 11 ö
æ 11 ö
æ 11 ö
= ç ÷ + 2 ç ÷ + 3 ç ÷ + ¼ + 9 ç ÷ + 10 ç ÷ …(ii)
è 10 ø
è 10 ø
è 10 ø
è 10 ø
10 è 10 ø
...(i)
...(ii)
Now, on subtracting Eq. (ii) from Eq. (i), then
2
-
10 times
[from Eq. (ii)]
2
k
æ 11 ö
æ 11 ö æ 11 ö
æ 11 ö
= 1 + ç ÷ + ç ÷ + ¼ + ç ÷ - 10 ç ÷
è 10 ø
è 10 ø è 10 ø
è 10 ø
10
1444442444443
üï
ìï æ 11 ö 10
1 × í ç ÷ - 1ý
10
ïþ
ï è 10 ø
æ 11 ö
- 10 ç ÷
= î
è 10 ø
æ 11
ö
ç - 1÷
è 10
ø
10
312
Textbook of Algebra
Þ
ïì æ 11 ö
k = - 100 × í ç ÷
è ø
îï 10
10
10
ïü
æ 11 ö
- 1ý + 100 ç ÷ = 100
è 10 ø
þï
2
4 ± (16 - 4 )
[Qr > 1]
=2 + 3
2
b c
141. Let
= =r
a b
\
b = ar , c = ar 2
a+b+c
b c
æb ö 6
Given,
=b + 2 Þ 1 + + =3 ç ÷ +
èa ø a
3
a a
6
Þ
1 + r + r 2 = 3r +
a
Now, for a = 6, only we get r = 0, 2
[rational]
So,
r =2
Þ
(a, b, c ) = (6, 12, 24 )
2
a + a - 14 36 + 6 - 14
=
=4
\
a+1
6+1
2
æ n (n + 1 ) ö
ç
÷
è
ø
(n + 1 ) 2
13 + 23 + 33 + ¼ + n 3
2
142. Tn =
=
=
4
1 + 3 + 5 + ¼ + (2n - 1 ) n (1 + 2n - 1 )
2
1 2
= (n + 2n + 1 )
4
1
1 é n (n + 1 )(2n + 1 ) 2n(n + 1 )
ù
+
+ nú
\ Sn = ( Sn 2 + 2 Sn + S1 ) = ê
4ë
6
2
4
û
r=
1
[285 + 90 + 9 ] = 96
4
l +n
143. Given, m =
Þl + n = 2m
2
and l , G1, G2, G3, n are in GP.
G1 G2 G3
n
\
=
=
=
l
G1 G2 G3
S9 =
…(i)
(l +
2G22
2
+n )
[from Eq. (ii)]
[from Eq. (ii)]
[from Eq. (i)] = 4lm 2n
144. Let first term = a and common difference = d
Þ
a = 8d
a + 4d 8d + 4d 4
\Common ratio =
=
=
a+d
8d + d
3
(Qa = 8d )
Aliter
Let the GP be a, ar , ar 2 and terms of AP and A + d , A + 4d ,
A + 8d , then
ar 2 - ar ( A + 8d ) - ( A + 4d ) 4
r=
=
=
ar - a
( A + 4d ) - ( A + d ) 3
æ8ö
è5ø
2
2
2
2
æ 12 ö
æ 16 ö
æ 20 ö
æ 44 ö
÷ + ç ÷ + ç ÷ +K+ ç ÷
è5ø
è5ø
è5ø
è5ø
2
146. ç ÷ + ç
16 2
(2 + 3 2 + 4 2 + 5 2 + K 11 2 )
25
16 æ 11 . (11 + 1 ) . (22 + 1 )
ö
=
- 1÷
ç
ø
25 è
6
=
=
16
16
16
´ 505 =
´ 101 =
m (given)
25
5
5
\m = 101
147. Q loge b1, loge b2, loge b3, ..., loge b101 are in AP.
Þ b1, b2, b3, K, b101 are in GP with common ratio 2.
(Q common difference = loge 2)
Also, a1, a 2, a 3, K a101 are in AP.
where, a1 = b1 and a 51 = b51
Þ
GM < AM
b2 < a 2, b3 < a 3, K, b50 < a 50
Þ t <s
Also, a1, a 2, a 3, K, a101 are in AP and b1, b2, b3, .... b101 are in GP.
Q
= ln (l 2 + 2 ln + n 2 )
Q
Þ
\ b1 + b2 + b3 + K + b51 < a1 + a 2 + a 3 + K + a 51
= G3G1 (l 2 + 2G3G1 + n 2 ) [from Eq. (ii)]
= ln (l + n ) 2 = ln (2m ) 2
(Qa, d Î N )
\ (a + 4d ) 2 = (a + d ) (a + 8d )
Q
…(ii)
Now, G14 + 2 G24 + G34 = l 2 G22 + 2 G24 + n 2 G22
2
130 < a1 + 6d < 140 Þ 130 < 15d < 140
\ b2, b3, K, b50 are GM’s and a 2, a 3, ...., a 50 are AM’s between b1
and b51.
G1G3 = ln, G12 = lG2, G22 = G3G1, G32 = nG2
= G22
Þ
\
\
r >1
New numbers a, 2ar , ar 2 are in AP.
\
4ar = a + ar 2 Þ r 2 - 4r + 1 = 0
Þ
a = 9d and 130 < a 7 < 140
2
1
8 <d <9
Þ d =9
3
3
145. Q a + d , a + 4d , a + 8d are in GP (d ¹ 0)
140. Let a, ar , ar are in GP. Q GP is increasing.
Hence,
or
sum of seven terms
6
=
sum of eleven terms 11
7
7
(2a + 6d )
(a1 + a 7 )
6
6
2
=
=
Þ 2
11
11
11
11
(2a + 10d )
(a1 + a11 )
2
2
\
a1 = b1 and a 51 = b51
b101 > a101
148. (15a ) 2 + (3b ) 2 + (5c ) 2 - 45ab - 15bc - 75ac = 0
1
{(15a - 3b ) 2 + (3b - 5c ) 2 + (5c - 15a ) 2 } = 0
2
Þ
(15a - 3b ) 2 + (3b - 5c ) 2 + (5c - 15a ) 2 = 0
Þ
or
15a - 3b = 0, 3b - 5c = 0, 5c - 15a = 0
Q
b = 5a, c = 3a
Þ 5a, 3a, a are in AP i.e. b, c, a are in AP.
CHAPTER
04
Logarithms and
Their Properties
Learning Part
Session 1
● Definition
● Characteristic and Mantissa
Session 2
● Principle Properties of Logarithm
Session 3
● Properties of Monotonocity of Logarithm
● Graphs of Logarithmic Functions
Practice Part
●
●
JEE Type Examples
Chapter Exercises
Arihant on Your Mobile !
Exercises with the
#L
symbol can be practised on your mobile. See inside cover page to activate for free.
314
Textbook of Algebra
The technique of logarithms was introduced by John Napier (1550-1617). The logarithm is a form of indices which is
used to simplify the algebraic calculations. The operations of multiplication, division of a very large number becomes
quite easy and get converted into simple operations of addition and subtraction, respectively. The results obtained are
correct upto some decimal places.
Session 1
Definition, Characteristic and Mantissa
Definition
The logarithm of any positive number, whose base is a
number ( > 0 ) different from 1, is the index or the power to
which the base must be raised in order to obtain the given
number.
x
i.e. if a = b (where a > 0, ¹ 1), then x is called the
logarithm of b to the base a and we write log a b = x ,
clearly b > 0. Thus, log a b = x Û a x = b, a > 0, a ¹ 1 and
b > 0.
If a = 10, then we write log b rather than log 10 b. If a = e ,
we write ln b rather than log e b. Here, ‘e’ is called as
Napier’s base and has numerical value equal to 2.7182.
Also, log 10 e is known as Napierian constant.
i.e.
log 10 e = 0 . 4343
\
ln b = 2.303 log 10 b
é
1
êsince, ln b = log 10 b ´ log e 10 = log e ´ log 10 b
10
ë
1
ù
=
log 10 b = 2.303 log 10 bú
0.4343
û
Remember
Corollary II The function defined by
f ( x ) = log a x , a > 0, a ¹ 1 is called logarithmic function. Its
domain is (0, ¥) and range is R (set of all real numbers).
Corollary III a x > 0, " x Î R
(i) If a > 1 , then a x is monotonically increasing.
For example, 5 2.7 > 5 2.5 , 3 222 > 3 111
(ii) If 0 < a < 1 , then a x is monotonically decreasing.
æ 1ö
For example, ç ÷
è5ø
2.7
æ 1ö
<ç ÷
è5ø
2.5
, (0.7 ) 222 < (0.7 ) 212
Corollary IV
(i) If a > 1, then a - ¥ = 0
\
log a 0 = - ¥ (if a > 1)
(ii) If 0 < a < 1 , then a ¥ = 0
\ log a 0 = + ¥ (if 0 < a < 1)
Corollary V (i) log a b ® ¥, if a > 1, b ® ¥
(ii) log a b ® - ¥, if 0 < a < 1 , b ® ¥
Remark
(i) log 2 = log 10 2 = 0.3010
(ii) log 3 = log 10 3 = 0.4771
(iii) ln 2 = 2.303log2 = 0.693
(iv) ln 10 = 2.303
Corollary I From the definition of the logarithm of the
number b to the base a, we have an identity
a loga b = b, a > 0, a ¹ 1 and b > 0
which is known as the Fundamental Logarithmic
Identity.
1. ‘log’ is the abbreviation of the word ‘logarithm’.
2. Common logarithm (Brigg’s logarithms) The base is 10.
3. If x < 0, a > 0 and a ¹ 1, then log a x is an imaginary.
x >1
ì+ ve,
ï
4. If a > 1, log a x = í 0,
x =1
ï - ve, 0 < x < 1
î
ì+ ve, 0 < x < 1
ï
And if 0 < a < 1,log a x = í 0,
x =1
ï - ve,
x >1
î
5. log a 1 = 0 ( a > 0, a ¹ 1)
log a a = 1 ( a > 0, a ¹ 1) and log ( 1/ a) a = -1 ( a > 0, a ¹ 1)
Chap 04 Logarithms and Their Properties
y Example 1. Find the value of the following :
(ii) log 3 2 324
(i) log 9 27
(iii) log 1/ 9 (27 3 )
(iv) log ( 5 + 2 6 ) (5 - 2 6 )
(vi) 2 2log 4 5
(v) log 0 . 2 0008
.
(vii)
ì1
ü
1
1
-log 2.5 í +
+
+ Ký
3 32 33
î
þ
ï
(0.4)
y Example 2. Find the value of the following:
(ii) log (sec 2 60 ° - tan 2 60 °)cos60°
(i) log tan 45 °cot 30°
(iii) log (sin 2 30 ° + cos 2 30 °) 1
(iv) log 30 1
Sol. (i) Here, base = tan45° = 1 tan
\ log is not defined.
log 20 ( 0. 3 )
(viii) (0.05)
(ii) Here, base = sec2 60° - tan 2 60° = 1
x = log 9 27
Þ
9 x = 27 Þ 32 x = 33 Þ 2x = 3
\ log is not defined.
(iii) Q log (sin 2 30 ° + cos 2 30 ° ) 1 = log1 1 ¹ 1
\
3
2
x = log 3
\ log is not defined.
(iv) log 30 1 = 0
Sol. (i) Let
Q Here, base = 1
x=
(ii) Let
315
2 324
2 4
Þ (3 2 ) = 324 = 2 ×3 Þ (3 2 )x = (3 2 )4
x
\
Characteristic and Mantissa
x=4
x = log1 / 9 (27 3 )
(iii) Let
x
æ1ö
-2 x
= 37 / 2 Þ - 2x = 7 / 2
ç ÷ = 27 3 Þ 3
è9 ø
Þ
\
x=-
7
4
(iv) Q (5 + 2 6 ) (5 - 2 6 ) = 1
or 5 + 2 6 =
1
5 -2 6
Now, let x = log (5 + 2
6 ) (5
= log1/ (5 - 2
- 2 6)
6)5 - 2
6 = -1
[from Eq. (i)]
x = log 0.2 0.008
(v) Let
Þ
…(i)
(0.2)x = 0.008 Þ (0.2)x = (0.2)3 Þ x = 3
(vi) Let x = 22 log 4 5 = 4 log 4 5 = 5
(vii) Let x =
ì1
ü
1
1
- log 2.5 í +
+
+ Ký
îï 3 3 2 3 3
þ
(0.4)
ì 1 ü
ï
ï
- log 2.5 í 3 ý
ö
ï 1- 1 ï
÷
3þ
ïî
æ4
=ç
è 10 ø
(viii) Let x =
æ2ö
=ç ÷
è5ø
log
( 0. 3 )
(0.05) 20
=
æ 1ö
- log 2.5 ç ÷
è 2ø
æ5ö
=ç ÷
è2ø
æ 1ö
log 5 / 2 ç ÷
è 2ø
log
(l )
(0.05) 20
=
1
2
…(i)
where, l = 0.3
Then, l = 0.33333 ...
Þ 10l = 3.33333 ...
On subtracting Eq. (ii) from Eq. (iii), we get
1
9l = 3 Þ l =
3
Now, from Eq. (i), x = (0.05)
æ1ö
=ç ÷
è 20 ø
…(ii)
…(iii)
æ 1ö
log 20 ç ÷
è 3ø
log
(3 )
(20)1/2
-1
æ1ö
=ç ÷
è 20 ø
2
-
1
log 20 3
1/ 2
= 20(2 log 20 3 ) = 20log 20 3 = 32 = 9
The integral part of a logarithm is called the
characteristic and the fractional part (decimal part) is
called mantissa.
i.e., log N = Integer + Fractional or decimal part ( +ve)
¯
¯
Characteristic
Mantissa
The mantissa of log of a number is always kept positive.
i.e., if log564 = 2.751279, then 2 is the characteristic and
0.751279 is the mantissa of the given number 564.
And if log 0.00895 = - 2.0481769
= - 2 - 0.0481769
= ( -2 - 1) + (1 - 0.0481769 )
= - 3 + 0.9518231
Hence, -3 is the characteristic and 0.9518231
(not 0.0481769) is mantissa of log 0.00895.
In short, -3 + 0.9518231 is written as 3.9518231 .
Remark
1. If N > 1, the characteristic of log N will be one less than the
number of digits in the integral part of N.
For example, If log 23568
. = 2.3723227
Here,
N = 23568
.
\ Number of digits in the integral part of N = 3
Þ Characteristic of log 23568
. = N -1= 3-1=2
2. If 0 < N < 1, the characteristic of log N is negative and
numerically it is one greater than the number of zeroes
immediately after the decimal part in N.
For example, If log0 .0000279 = 5 .4456042
Here, four zeroes immediately after the decimal point in the
number 0.0000279 is ( 4 + 1), i.e. 5.
3. If the characteristics of log N be n, then number of digits in N
is ( n + 1) (Here, N > 1).
4. If the characteristics of log N be -n, then there exists ( n - 1)
number of zeroes after decimal part of N (here, 0 < N < 1).
316
Textbook of Algebra
y Example 3. If log 2 = 0.301 and log 3 = 0.477, find
Sol. Let
P = (0.036)16 Þ log P = 16log (0.036)
the number of digits in 6 20 .
Sol. Let P = 6
\
20
= ( 2 ´ 3)
æ 22 × 32 ö
æ 36 ö
= 16 log ç
÷
÷ = 16 log ç
è 1000 ø
è 1000 ø
20
log P = 20 log(2 ´ 3) = 20 {log 2 + log 3}
= 20 {0.301 + 0.477}
= 16 {log 22 + log 32 - log 103 }
= 16 {2 log 2 + 2 log 3 - 3}
= 20 ´ 0.778 = 15.560
= 16 {2 ´ 0.301 + 2 ´ 0.477 - 3}
Since, the characteristic of log P is 15, therefore the number
of digits in P will be 15 + 1, i.e. 16.
y Example 4. Find the number of zeroes between the
decimal point and first significant digit of (0.036)16 ,
where log 2 = 0.301 and log 3 = 0.477.
#L
1.
The value of log2
3
(b) 8
(d) 5
The value of log( 8 - 3
The value
(a) 2
(c) 6
4.
7 )(8
+ 3 7 ) is
(b) -1
(d) Not defined
ü
ì1
1
+ Ký
log 2.5í +
þ
î 3 32
of (0.16)
is
(b) 4
(d) 8
If log 2 = 0.301, the number of integers in the expansion of 417 is
(a) 9
(c) 13
5.
= -24 + 0.896 = 24 + 0.896
\ The required number of zeroes = 24 - 1 = 23.
1728 is
(a) -2
(c) 0
3.
= -48 + 24 + 0.896
Exercise for Session 1
(a) 6
(c) 3
2.
= 16 {1.556 - 3} = 24.896 - 48
(b) 11
(d) 15
If log 2 = 0.301, then the number of zeroes between the decimal point and the first significant figure of 2-34 is
(a) 9
(c) 11
(b) 10
(d) 12
Session 2
Principle Properties of Logarithm
Principle Properties
of Logarithm
Let m and n be arbitrary positive numbers,
a > 0, a ¹ 1, b > 0, b ¹ 1 and a, b be any real numbers, then
(i) log a (m n ) = log a m + log a n
In general, log a ( x 1 x 2 x 3 K x n ) = log a x 1
+ log a x 2 + log a x 3 + K + log a x n
(where, x 1 , x 2 , x 3 , ..., x n > 0)
Or
æ n
ö n
log a ç Õ x i ÷ = å log a x i , "x i > 0
èi =1 ø i = 1
where, i = 1 , 2, 3, ..., n.
æm ö
(ii) log a ç ÷ = log a m - log a n
ènø
(iii) log a m a = a log a m
log a m
(v) log b m =
log a b
1
(iv) log ab m = log a m
b
(vii) log a x 2 ¹ 2 log a x , a > 0, a ¹ 1
Since, domain of log a ( x 2 ) is R ~ {0 } and domain of
log a x is (0, ¥) are not same.
(viii)a logb a = a , if b = a 2 , a > 0, b > 0, b ¹ 1
(ix) a logb a = a 2 , if b = a , a > 0, b > 0, b ¹ 1
y Example 5. Solve the equation 3 × x log 5 2 + 2log 5 x = 64.
Sol. Q
3 × x log 5 2 + 2log 5 x = 64
Þ
3 × 2log 5 x + 2log 5 x = 64
4 ×2
Þ
log 5 x
[by extra property (i)]
= 64
Þ
2log 5 x = 4 2 = 24
\
log 5 x = 4
x = 54 = 625
Þ
y Example 6. If 4 log16 4 + 9log 3 9 = 10log x 83 , find x .
4 log16 4 = 4 = 2
Sol. Q
and
9
log 3 9
[by extra property (ix)]
2
= 9 = 81
[by extra property (viii)]
\ 4 log16 4 + 9 log 3 9 = 2 + 81 = 83 = 10log x 83
Remark
1. log b a × log a b = 1 Û log b a =
1
log a b
2. log b a × log c b × log a c = 1
3. log y x × log z y × log a z = log a x
x
Þ
\
y Example 7. Prove that a
Sol. Q
4. eln a = ax
log10 83 = log x 83
x = 10
a
(loga b )
=a
Extra Properties of Logarithm
(i) a
logb x
=x
(ii) a
loga x
= x , a > 0, a ¹ 1, x > 0
(iii) log ak
logb a
, b ¹ 1, a, b, x are positive numbers.
1
x = log a x , a > 0, a ¹ 1, x > 0
k
(iv) log a x 2k = 2k log a | x |, a > 0, a ¹ 1, k Î I
1
(v) loga 2k x = log|a | x , x > 0, a > 0, a ¹ ± 1 and k Î I ~ {0}
2k
b
(vi) log a a x b = log a x , x > 0, a > 0, a ¹ 1, a ¹ 0
a
=a
=b
Hence, a
(loga b )
loga b ´
loga b ´
-b
log b a
= 0.
logb a
loga b × (logb a )
logb a
-b
(logb a )
y Example 8. Prove that
Sol. LHS =
log a b
[by extra property (ii)]
=0
log 2 24 log 2 192
= 3.
log 96 2
log 12 2
log 2 24 log 2 192
log 96 2
log12 2
= log 2 24 ´ log 2 96 - log 2 192 ´ log 2 12
Now, let 12 = l, then
LHS = log 2 2l ´ log 2 8l - log 2 16l ´ log 2 l
318
Textbook of Algebra
= (log 2 2 + log 2 l ) (log 2 8 + log 2 l )
- (log 2 16 + log 2 l )log 2 l
3
= (log 2 2 + log 2 l ) (log 2 2 + log 2 l )
- (log 2 24 + log 2 l )log 2 l
= (1 + log 2 l ) (3 log 2 2 + log 2 l )
- ( 4 log 2 2 + log 2 l )log 2 l
= (1 + log 2 l ) (3 + log 2 l ) - log 2 l ( 4 + log 2 l )
=3
= RHS
#L
1
We have,
log l a × {log 5 l × log l 25} = 2
Þ
(log l a ){log 5 25} = 2
Þ
(log l a ){log 5 52 } = 2
Þ
(log l a ){2 log 5 5} = 2
Þ
(log l a ){ 2 } = 2
\
log l (a ) = 1 or a = l
If a = log24 12, b = log48 36 and c = log36 24, 1 + abc is equal to
(b) 2bc
(c) 2ca
(d) ba + bc
(c) 1
(d) 2
(c) n
(d) -n
The value of log4[log2 {log2(log3 81)}] is equal to
(a) -1
3
Sol. Here, l > 0, l ¹ 1
Exercise for Session 2
(a) 2ab
2
y Example 9. Solve for a, l, if
log l a× log 5 l × log l 25 = 2.
(b) 0
log2log2 ( ... 2 ) is equal to
14243
n times
(a) 0
4
If a = log3 5, b = log1725, which one of the following is correct?
(a) a < b
5
(b) 1
The value of log0.75 log2
(a) -1
(b) a = b
-2
(c) a > b
(d) None of these
(c) 1
(d) None of these
(0.125) is equal to
(b) 0
Session 3
Properties of Monotonocity of Logarithm,
Graphs of Logarithmic Functions
Properties of Monotonocity
of Logarithm
1. Constant Base
Graphs of Logarithmic
Functions
1. Graph of y = log a x , if a > 1 and x > 0
Y
ì x > y > 0, if a > 1
(i) log a x > log a y Û í
î0 < x < y , if 0 < a < 1
X′
O
ì 0 < x < y , if a > 1
(ii) log a x < log a y Û í
îx > y > 0, if 0 < a < 1
ìx > a p , if a > 1
(iii) log a x > p Û í
p
î0 < x < a , if 0 < a < 1
Y′
2. Graph of y = log a x , if 0 < a < 1 and x > 0
Y
ì 0 < x < a p , if a > 1
(iv) log a x < p Û í
p
îx > a , if 0 < a < 1
(1, 0)
X′
(ii) If a > 1 , then log x a is monotonically decreasing in
(0, 1) È (1, ¥).
Y′
Remark
1. If the number x and the base ‘a’ are on the same side of the
unity, then the logarithm is positive.
Case I y = log ax , a > 1, x > 1 Case II y = log a x , 0 < a < 1, 0 < x < 1
X′
Very Important Concepts
O (1, 0)
(i) If a > 1 , p > 1 , then log a p > 0
(iii) If a > 1, 0 < p < 1, then log a p < 0
(iv) If p > a > 1, then log a p > 1
(vii) If 0 < p < a < 1, then log a p > 1
X
X′
O
Y′
X
(1, 0)
Y′
2. If the number x and the base a are on the opposite sides of
the unity, then the logarithm is negative.
Case I y = log a x, a > 1, 0 < x < 1
Case II y = log a x, 0 < a < 1, x > 1
Y
Y
(v) If a > p > 1, then 0 < log a p < 1
(vi) If 0 < a < p < 1, then 0 < log a p < 1
Y
Y
(iii) If 0 < a < 1 , then log x a is monotonically increasing in
(0, 1) È (1, ¥).
(ii) If 0 < a < 1, p > 1, then log a p < 0
X
O
2. Variable Base
(i) log x a is defined, if a > 0, x > 0, x ¹ 1.
X
(1, 0)
X′
(1, 0)
O
Y′
X X′
(1, 0)
O
Y′
X
320
Textbook of Algebra
3. Graph of y = log a | x |
Y
Y
0<a<1
a>1
X′
O
(–1, 0)
y Example 10. Arrange in ascending order
log 2 ( x ), log 3 ( x ), log e ( x ), log 10 ( x ), if
(i) x > 1
(ii) 0 < x < 1.
(1, 0)
(–1, 0)
X′
X
(1, 0)
O
Sol. Q2 < e < 3 < 10
X
(i) For x > 1, log x 2 < log x e < log x 3 < log x 10
Þ
1
1
1
1
<
<
<
log 2 ( x ) loge ( x ) log 3 ( x ) log10 ( x )
Remark
Þ
log 2 ( x ) > loge ( x ) > log 3 ( x ) > log10 ( x )
Graphs are symmetrical about Y-axis.
Hence, ascending order is
Y′
Y′
4. Graph of y = | log a | x ||
log10 ( x ) < log 3 ( x ) < loge ( x ) < log 2 ( x )
Y
0<a<1
a>1
X′
(ii) For 0 < x < 1, log x 2 > log x e > log x 3 > log x 10
1
1
1
1
>
>
>
Þ
log 2 ( x ) loge ( x ) log 3 ( x ) log10 ( x )
Y
(–1, 0) O
(1, 0)
X X′
(–1, 0) O
Y′
\ log 2 ( x ) < loge ( x ) < log 3 ( x ) < log10 ( x )
which is in ascending order.
X
(1, 0)
Y′
y Example 11. If log 11 = 1.0414, prove that 1011 > 1110 .
Remark
Graphs are same in both cases i.e., a > 1and 0 < a < 1.
5. Graph of | y | = log a | x ||
Sol. Q
log 1011 = 11 log 10 = 11
and log 1110 = 10 log 11 = 10 ´ 1.0414 = 10.414
It is clear that,
a > 0 and a ≠ 1
X′
(–1, 0)
(1, 0)
O
11 > 10.414
log 1011 > log 1110
Þ
Y
11
Þ
[Q here, base = 10]
10
10 > 11
y Example 12. If log 2 ( x - 2) < log 4 ( x - 2), find the
X
interval in which x lies.
Sol. Here, x - 2 > 0
Þ
Y′
6. Graph of y = log a [ x ], a > 1 and x ³ 1
(where [ × ] denotes the greatest integer function)
Since, when 1 £ x < 2,[ x ] = 1 Þ log a [ x ] = 0
when 2 £ x < 3,[ x ] = 2 Þ log a [ x ] = log a 2
when 3 £ x < 4,[ x ] = 3 Þ log a [ x ] = log a 3 and so on.
Y
and
Þ
x >2
1
log 2 ( x - 2) < log 2 2 ( x - 2) = log 2 ( x - 2)
2
1
log 2 ( x - 2) < log 2 ( x - 2)
2
1
log 2 ( x - 2) < 0 Þ log 2 ( x - 2) < 0
2
Þ
x - 2 < 20 Þ x - 2 < 1
Þ
x <3
From Eqs. (i) and (ii), we get
2 < x < 3 or x Î(2, 3)
…(i)
Þ
…(ii)
y Example 13. Prove that log n (n + 1) > log (n +1) (n + 2)
loga 2
X′
0
Y′
loga 3
for any natural number n > 1.
loga 4
Sol. Since,
X
1
2
3
4
5
For
æn + 2ö
n +1
1
1
=ç
=1+ >1+
÷
n
n
n + 1 èn + 1ø
n > 1,
æn + 2ö
æn + 1ö
æn + 1ö
logn ç
÷ > logn + 1 ç
÷ > logn + 1 ç
÷
è n ø
è n ø
èn + 1ø
Chap 04 Logarithms and Their Properties
Þ
Þ
\
logn (n + 1) - logn n > log (n + 1)(n + 2) - log (n + 1)(n + 1)
logn (n + 1) - 1 > log (n + 1)(n + 2) - 1
logn (n + 1) > log (n + 1)(n + 2) Hence proved.
y Example 14. Find the least value of the
expression 2 log 10 x - log x 0.01, where x > 0, x ¹ 1.
Sol. Let P = 2 log10 x - log x 001
. = 2 log10 x - log x (10-2 )
= 2(log10 x + log x 10)
How to Find Minimum Value of
³2 ×2 = 4
l1 log a x + l2 log x a, a > 0, x > 0,
a ¹ 1, x ¹ 1 and l1, l2 Î R+
Q
Þ
Þ
\
P³4
y Example 15. Which is smaller 2 or (log p 2 + log 2 p )?
Sol. Let
P = log p 2 + log 2 p > 2 [by above article] [Q p ¹ 2]
\
P >2
Þ
(log p 2 + log 2 p ) > 2
Hence, the smaller number is 2.
l1 loga x + l 2 log x a ³ 2 l1l 2
Hence, the minimum value of l1 loga x + l 2 log x a is 2 l1l 2 .
1
[by above article]
Hence, the least value of P is 4.
AM ³ GM
l1 loga x + l 2 log x a
³ ( l1 loga x ) ( l 2 log x a ) = l1l 2
2
#L
321
Exercise for Session 3
If log0.16(a + 1) < log0.4(a + 1), then a satisfies
(a) a > 0
(b) 0 < a < 1
(c) -1 < a < 0
(d) None of these
1
2
The value of x satisfying the inequation x log10 x × log10 x < 1, is
(a) 0 < x < 10
3
(c) 0 < x < 101/ 10
(d) None of these
(b) x < 0
(c) -1 < x < 1
(d) None of these
2
(c) æç 0, ö÷
è 5ø
(d) None of these
If logcosec x sin x > 0, then
(a) x > 0
4
(b) 0 < x < 1010
The value of log10 3 lies in the interval
2
(a) æç ,
è5
1ö
÷
2ø
1
(b) æç 0, ö÷
è 2ø
2
5
3
3 æ3ö
æ3ö
+ ç ÷ + ç ÷ + ..., should
è4ø
4 è4ø
differ from the sum of the series by less than 10-6, is (given, log 2 = 0.30103, log 3 = 0.47712)
The least value of n in order that the sum of first n terms of the infinite series 1 +
(a) 14
(b) 27
(c) 53
(d) 57
Shortcuts and Important Results to Remember
1 For a non-negative number ‘a’ and n ³ 2, n Î N,
n
a = a1/ n .
5 log 2 log 2
... 2 = - n
14
4244
3
n times
2 The number of positive integers having base a and
characteristic n is an + 1 - an .
3 Logarithm of zero and negative real number is not
defined.
4 |log b a + log a b| ³ 2, " a > 0, a ¹ 1, b > 0, b ¹ 1.
6 a
log a b
=b
log b a
7 Logarithms to the base 10 are called common logarithms
(Brigg’s logarithms).
8 If x = logc b + log b c , y = log a c + logc a,
z = log a b + log b a, then x 2 + y 2 + z 2 - 4 = xyz.
322
Textbook of Algebra
JEE Type Solved Examples :
Single Option Correct Type Questions
n
The pairs (a, b) are
This section contains 8 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
ONLY ONE is correct.
4
æ æ kp ö ö
Ex. 1 The expression log 2 5 - å log 2 ç sin ç ÷ ÷ reduces
è è 5 øø
k =1
p
to , where p and q are co-prime, the value of p 2 + q 2 is
q
(3, 32 ), ( 4, 4 2 ), (5, 52 ), (6, 62 ),…, (44, 44 2 ) and (3, 33 ), (4, 4 3 ),
(5, 53 ),…, (12, 123 ).
Hence, there are 42 + 10 = 52 pairs.
l
(a) 13
(b) 17
(c) 26
æ æ kp ö ö
Sol. (b) Let p = log 2 5 - å log 2 çsin ç ÷ ÷
è è 5 øø
k =1
(d) 29
4
ì
æ æ 2p ö ö
æ æ p öö
= log 2 5 - í log 2 çsin ç ÷ ÷ + log 2 çsin ç ÷ ÷
ø
è
è è 5 øø
ø
è
5
î
æ æ 4 p ö öü
æ æ 3p ö ö
+ log 2 çsin ç ÷ ÷ + log 2 çsin ç ÷ ÷ý
è è 5 ø øþ
è è 5 øø
2p
3p
4pü
ì p
= log 2 5 - log 2 ísin × sin × sin × sin ý
5
5
5
5 þ
î
ì
æpö
æ 2p öü
= log 2 5 - log 2 ísin 2 ç ÷ × sin 2 ç ÷ý
è
ø
è 5 øþ
5
î
ì (1 - cos 72° )(1 - cos 144° )ü
= log 2 5 - log 2 í
ý
4
þ
î
1
18
°
1
+
36
°
(
sin
)(
cos
)
ì
ü
= log 2 5 - log 2 í
ý
4
î
þ
ìæ
5 + 1öü
5 - 1öæ
÷ ç1 +
֕
ï ç1 4 øï
4
ø
è
è
ï
= log 2 5 - log 2 í
ý
4
ï
ï
ïþ
ïî
ì (5 - 5 )(5 + 5 )ü
æ5ö
= log 2 5 - log 2 í
ý = log 2 5 - log 2 ç ÷
è 16 ø
64
î
þ
4 p
æ 16 ö
[given]
= log 2 ç5 ´ ÷ = log 2 24 = =
è
1 q
5ø
\ p = 4, q = 1
Hence, p 2 + q 2 = 4 2 + 12 = 17
(b) 50
(c) 52
(d) 54
x = loga b
6
Þ x + = 5 Þ x 2 - 5x + 6 = 0 Þ x = 2, 3
x
From Eq. (i), we get loga b = 2, 3
(a) 48
Þ
b = a 2 or a 3
(a) 221
(b) 222
(c) 223
(d) 224
Sol. (c) In a triangle,
log10 12 + log10 75 > log10 n Þ n < 12 ´ 75 = 900
…(i)
\
n < 900
and
log10 12 + log10 n > log10 75
75 25
Þ
n>
=
12 4
25
…(ii)
\
n>
4
25
From Eqs. (i) and (ii), we get
< n < 900
4
\
n = 7, 8, 9, 10, …, 899
Hence,
a = 7, b = 899
\
b - a = 892 = 4 ´ 223
Hence, b - a is divisible by 223.
æ 1 + log 3 (abc ) ö
Ex. 4 If 5 log abc (a 3 + b 3 + c 3 ) = 3 lç
÷ and
è log 3 (abc ) ø
m
(abc ) a + b + c =1 and l = , where m and n are relative primes,
n
the value of | m + n| + | m - n| is
l
(a) 8
Sol. (b) Q
\
(b) 10
a +b +c
(abc )
(c) 12
= 1 = (abc )
(d) 14
0
a + b + c = 0 Þ a 3 + b 3 + c 3 = 3abc
Now, LHS = 5 logabc (a 3 + b 3 + c 3 ) = 5 logabc (3abc )
…(i)
æ 1 + log 3 (abc ) ö
æ log 3 (3abc ) ö
and RHS = 3l ç
÷ = 3l ç
÷
è log 3 (abc ) ø
è log 3 (abc ) ø
Ex. 2 If 3 £ a £ 2015, 3 £ b £ 2015 such that
log a b + 6 log b a = 5, the number of ordered pairs (a, b) of
integers is
l
Sol. (c) Let
Ex. 3 The lengths of the sides of a triangle are log 10 12 ,
log 10 75 and log 10 n, where n Î N . If a and b are the least
and greatest values of n respectively, the value of b - a is
divisible by
l
…(i)
= 3l logabc (3abc )
From Eqs. (i) and (ii), we get
5 logabc (3abc ) = 3l logabc (3abc )
5 m
l= =
\
3 n
Þ
m = 5, n = 3
Hence, | m + n | + | m - n | = 8 + 2 = 10
…(ii)
[given]
323
Chap 04 Logarithms and Their Properties
log 4 3
Ex. 5 If a logb c = 3 × 3 log 4 3 × 3 log 4 3
where a, b, c ÎQ, the value of abc is
l
(a) 9
Sol. (c)
(b) 12
a
logb c
(c) 16
log 4 3 log 4 3
× 3 log 4 3
Now,
(d) 20
\
1+ log 4 3 + (log 4 3 ) 2 + (log 4 3 ) 3 + ... ¥
=3
æ bcd ö
æ acd ö
Ex. 8 If x = log 2a ç
÷ , y = log 3b ç
÷,
è 2 ø
è 3 ø
æ abd ö
æ abc ö
z = log 4c ç
÷ and w = log 5d ç
÷ and
è 4 ø
è 5 ø
1
1
1
1
+
+
+
= log abcd N + 1, the value of N is
x +1 y +1 z +1 w +1
4
a = 3, b = , c = 4
3
4
Hence, abc = 3 × × 4 = 16
3
l
\
Ex. 6 Number of real roots of equation
3 log 3 ( x
(a) 0
Sol. (a) Q
2 - 4x +3)
(b) 1
3log 3 (x
= ( x - 3 ) is
(c) 2
2 - 4x + 3)
(a) 40
(c) 120
(d) infinite
= ( x - 3)
…(i)
Sol. (c) Q
2
Eq. (i) is defined, if x - 4 x + 3 > 0
Þ
( x - 1)( x - 3) > 0
Þ
x < 1 or x > 3
Þ
…(ii)
Eq. (i) reduces to x 2 - 4 x + 3 = x - 3 Þ x 2 - 5x + 6 = 0
\
x = 2, 3
From Eqs. (ii) and (iii), use get x Î f
\ Number of real roots = 0
Þ
abc = 66
Þ
b 3 = 66
Þ
b = 36
æ 2abcd ö
x + 1 = log 2a ç
÷ = log 2a (abcd )
è 2 ø
1
= logabcd 2a
x +1
1
1
Similarly,
= logabcd 3b,
= logabcd 4c
y +1
z +1
Ex. 7 If log 6 a + log 6 b + log 6 c = 6, where a, b, c Î N and
a, b, c are in GP and b - a is a square of an integer, then the
value of a + b - c is
(b) 15
(c) 9
log 6 a + log 6 b + log 6 c = 6
log 6 (abc ) = 6
(b) 80
(d) 160
æ bcd ö
x = log 2a ç
÷
è 2 ø
\
…(iii)
l
(a) 21
Sol. (b) Q
Þ
b 2 362
is an integer for a = 27
=
a
a
a = 27, b = 36, c = 48
c =
Hence, a + b - c = 27 + 36 - 48 = 15
= 31/ (1 - log 4 3 ) = 31/ log 4 (4 / 3 ) = 3log 4 /3 4
l
Also, b - a = 36 - a is a square for a = 35, 32, 27, 20, 11
... ¥,
1
= logabcd 5d
w +1
1
1
1
1
+
+
+
= logabcd (2a × 3b × 4c × 5d )
\
x +1 y +1 z +1 w +1
and
(d) 3
= logabcd (120abcd )
= logabcd 120 + 1
[Qb 2 = ac ]
= logabcd N + 1
N = 120
Hence,
[given]
JEE Type Solved Examples :
More than One Correct Option Type Questions
n
Þ
Þ
or
This section contains 4 multiple choice examples. Each
example has four choices (a), (b), (c) and (d) out of which
more than one may be correct.
Þ
Ex. 9 The equation
(log 10 x + 2 ) 3 + (log 10 x - 1) 3 = ( 2 log 10 x + 1) 3 has
l
(a) no natural solution (b) two rational solutions
(c) no prime solution
(d) one irrational solution
Sol. (b, c, d) Let log10 x + 2 = a and log10 x - 1 = b
\ a + b = 2 log10 x + 1, then given equation reduces to
a 3 + b 3 = (a + b )3
Hence,
l
3ab(a + b ) = 0 Þ a = 0 or b = 0 or a + b = 0
log10 x + 2 = 0 or log10 x - 1 = 0
2 log10 x + 1 = 0
x = 10-2 or x = 10 or x = 10-1/ 2
1
1
or x = 10 or x =
x=
100
10
Ex. 10 The value of
log 5 9 × log 7 5 × log 3 7
log 3 6
co-prime with
(a)1
(b) 3
(c) 4
(d) 5
+
1
log 4 6
is
324
Textbook of Algebra
Sol. (a, b, d) Let P =
=
1
log 5 9 × log 7 5 × log 3 7
+
log 3 6
log 4 6
log 3 9
+ log
log 3 6
6
4 = log
69
+ log
6
4
= log 6 (36) = log 6 ( 6 )4 = 4 Þ P = 4
which is co-prime with 1, 3, 4 and 5.
Ex. 11 Which of the following quantities are irrational
for the quadratic equation
(log 10 8 ) x 2 - (log 10 5 ) x = 2(log 2 10 ) -1 - x ?
l
(a) Sum of roots
(c) Sum of coefficients
(b) Product of roots
(d) Discriminant
Sol. (c, d) Q (log10 8)x 2 - (log10 5)x = 2(log 2 10)-1 - x
Þ (3 log10 2)x 2 + (1 - log10 5)x - 2 log10 2 = 0
Þ
(3 log10 2)x 2 + (log10 2)x - 2 log10 2 = 0
1
Now, Sum of roots = - = Rational
3
2
Product of roots = - = Rational
3
Sum of coefficients = 3 log10 2 + log10 2 - 2 log10 2
= 2 log10 2 = Irrational
Discriminant = (log10 2)2 + 24(log10 2)2
= 25 (log10 2)2 = Irrational
Ex. 12 The system of equations
log 10 ( 2000 xy ) - log 10 x × log 10 y = 4
log 10 ( 2yz ) - log 10 y × log 10 z = 1
and log 10 ( zx ) - log 10 z × log 10 x = 0
has two solutions (x 1 , y 1 , z 1 ) and (x 2 , y 2 , z 2 ), then
l
(b) y 1 + y 2 = 25
(a) x 1 + x 2 = 101
(c) x 1x 2 = 100
(d) z 1z 2 = 100
Sol. (a, b, c, d) Let log10 x = a, log10 y = b and log10 z = c
Then, given equations reduces to
a + b - ab = 4 - log10 2000 = log10 5
b + c - bc = 1 - log10 2 = log10 5
and
c + a - ca = 0
From Eqs. (i) and (ii), we get
a + b - ab = b + c - bc
Þ
(c - a ) - b(c - a ) = 0
Þ
(c - a )(1 - b ) = 0
1 - b ¹ 0, c - a = 0 Þ c = a
From Eq. (iii), we get
2a - a 2 = 0 Þ a = 0, 2
…(i)
…(ii)
…(iii)
c = a Þ c = 0, 2
b = log10 5, 2 - log10 5
Then,
and
\
log10 x = 0, 2 Þ x = 100 , 102
Þ
Þ
and
Þ
Þ
x = 1, 100
x 1 = 1, x 2 = 100
log10 y = log10 5, 2 - log10 5
= log10 5, log10 20
y = 5, 20
y1 = 5, y 2 = 20
and
log10 z = 0, 2 Þ z = 100 , 102
Þ
z = 1, 100
Þ
z1 = 1, z 2 = 100
Finally, x 1 + x 2 = 1 + 100 = 101, y1 + y 2 = 5 + 20 = 25,
x 1x 2 = 1 ´ 100 = 100 and z1z 2 = 1 ´ 100 = 100
JEE Type Solved Examples :
Passage Based Questions
n
This section contains 2 solved passages based upon each
of the passage 3 multiple choice examples have to be
answered. Each of these examples has four choices (a), (b),
(c) and (d) out of which ONLY ONE is correct.
15. If x 2 t
(a) 1
2 -6
2
+ y 6 - 2t = 6, the value of t 1 t 2 t 3 t 4 is
(b) 2
(c) 4
Passage I
Q
log10 ( x - 2) + log10 y = 0
(Ex. Nos. 13 to 15)
\
x - 2 > 0, y > 0
Suppose that log 10 ( x - 2 ) + log 10 y = 0 and
x + (y - 2 ) = ( x + y ) .
Þ
x > 2, y > 0
and
log10 {( x - 2)y } = 0
13. The value of x is
Þ
( x - 2)y = 100 = 1
\
( x - 2)y = 1
(a) 2 + 2 (b) 1 + 2
(c) 2 2
(d) 4 - 2
(c) 1 + 2
(d) 2 + 2 2
Also, given that x + (y - 2) = ( x + y )
14. The value of y is
(a) 2
(b) 2 2
(d) 8
Sol. (Ex. Nos. 13-15)
\
x ³ 0, y - 2 ³ 0, x + y ³ 0
…(i)
…(ii)
Chap 04 Logarithms and Their Properties
Þ
x ³ 0, y ³ 2
On squaring both sides, we get
…(iii)
Passage II
(Ex. Nos. 16 to 18)
x + y - 2 + 2 x ( y - 2) = x + y
Þ
If 10
x y -2 =1
Þ
x ( y - 2) = 1
From Eqs. (i) and (iii),we get
x > 2, y ³ 2
and from Eqs. (ii) and (iv), we get y = x
From Eq. (ii), ( x - 2)x = 1
…(iv)
\
x=
2± 4 + 4
2
(a) q r
2-6
[neglect -ve sign, since x > 2]
Þ
2-6
2 t2 - 3
Þ
(x )
(3 + 2 2 )t
2-3
+ ( x -1 )2t
+ (x
2-6
x = rq
and
Þ
logq {logr (log p x )} = 0
logr (log p x ) = 1 Þ log p x = r
-2 t 2 - 3
…(i)
x = pr
=6
2-3
=6
)
+ (3 - 2 2 )t
(b) rq
q
…(ii)
x =r = p
r
(d) r r /q
(c) 1
r q = pr
Sol. (a) Q
Þ
p =r
…(iii)
q /r
18. The value of q is
(a) r p / r
(b) p log p r
(c) r logr p
(d) r r
/p
Sol. (c) From Eq. (iii),
t = ± 2, ± 2
\
log p {logq (logr x )} = 0
logq (logr x ) = 1 Þ logr x = q
(a) r q / r
=6
t = 4, 2
\
= 1 = 100
Þ
2
Þ
(d) rq
17. The value of p is
t2 - 3 = ± 1
Now, we get
log p [logq (logr x )]
From Eqs. (i) and (ii), we get
2
x 2t
(c) r p
\
+ y 6 - 2t = 6
Þ
10
Þ
Þ
14. (c) y = x = 1 + 1
x 2t
= 1 and log q {log r (log p x )} = 0.
(b) r q
Sol. (b) Q
13. (b) x = ( 2 + 1).
15. (d) Q
log p {logq (logr x )}
16. The value of x is
x 2 - 2x - 1 = 0
Þ
æ log p ö
q log r = r log p Þ q = r ç
÷ = r logr p
è log r ø
t 1 t 2 t 3 t 4 = ( 2) ( - 2) ( 2 ) ( - 2 ) = 8
JEE Type Solved Examples :
Single Integer Answer Type Questions
n
This section contains 2 examples. The answer to each
example is a single digit integer ranging from 0 to 9
(both inclusive).
Ex. 19 If x 1 and x 2 are the solutions of the equation
x x
x
= 100 x such that x 1 >1 and x 2 < 1, the value of 1 2 is
2
log x
l
log10 x
Sol. (5) Q
x
10
= 100x
Taking logarithm on both sides on base 10, then we get
log10 x × log10 x = log10 100 + log10 x
Þ
(log10 x )2 - log10 x - 2 = 0
Þ
(log10 x - 2) (log10 x + 1) = 0
\
\
\
325
log10 x = 2, - 1 Þ x = 102 , 10-1
x 1 = 100, x 2 =
x 1x 2
=5
2
1
10
Ex. 20 If (31.6) a = (0.0000316) b = 100, the value of
1 1
- is
a b
l
Sol. (3) Q
(31.6)a = (0.0000316)b = 100
Þ
a lo g10 (31.6) = b l og10 (0.0000316) = log10 100
Þ
a log10 (31.6) = b log10 (31.6 ´ 10-6 ) = 2
Þ
a log10 (31.6) = b log10 (31.6) - 6b = 2
2
= log10 (31.6)
a
2
= log10 (31.6) - 6
b
2 2
- =6
a b
1 1
- =3
a b
Þ
and
\
Þ
326
Textbook of Algebra
JEE Type Solved Examples :
Matching Type Questions
n
l
This section contains 2 examples. Examples 24 and 25
have three statements (A, B and C) given in Column I and
four statements (p, q, r and s) in Column II. Any given
statement in Column I can have correct matching with
one or more statement(s) given in Column II.
1
, 25
5
1
\ Product of the values of x =
´ 25 = 5 5
5
(C) Q
logb a = - 3 and logb c = 4
3
\
logc a = 4
and
a3x = c x - 1
(A)
(B)
(C)
Column II
If x1 and x 2 satisfy the equation
( x + 1 ) log10 ( x + 1) = 100 ( x + 1 ), then
the value of( x1 + 1 ) ( x 2 + 1 ) + 5 is
(p) irrational
Þ
Þ
The product of all values of x
which make the following
statement true
(log 3 x ) (log 5 9 ) - log x 25 + log 3 2
= log 3 54, is
(q) rational
Þ
l
(A)
= 100 ( x + 1)
( x + 1) = 102 , 10-1
( x 1 + 1) ( x 2 + 1) = 102 ´ 10-1 = 10
( x 1 + 1) ( x 2 + 1) + 5 = 10 + 5
= 15 = 3 ´ 5
(B) Q(log 3 x ) (log 5 9 ) - log x 25 + log 3 2 = log 3 54
Þ
2 log 5 x - 2 log x 5 = log 3 54 - log 3 2
= log 3 (27 ) = 3
Let log 5 x = l, then
2
2l - = 3
l
Þ
2l2 - 3l - 2 = 0
Þ
2l2 - 4 l + l - 2 = 0
1
2l ( l - 2) + 1( l - 2) = 0 Þ l = - , 2
2
log 2 3
-3
(p) divisible by 2
log 3 2
and c = log 2 log 2
2 ,
(q) divisible by 4
then HM of a and b is
{log10 ( x + 1)} 2 = 2 + log10 ( x + 1)
Þ {log10 ( x + 1) - 2} {log10 ( x + 1) + 1} = 0
\
log10 ( x + 1) = 2, - 1
Þ
4
13
[prime and rational]
Column II
If a and b are the roots of
ax 2 + bx + c = 0, where
a = 2 log 2 3 - 3 log 3 2,
b =1 + 2
Þ {log10 ( x + 1)} 2 - log10 ( x + 1) - 2 = 0
\
[from Eq. (i)]
Column I
Taking logarithm on both sides on base 10, then we get
log10 ( x + 1) × log10 ( x + 1) = log10 100 + log10 ( x + 1)
Þ
q = 13
…(i)
Ex. 22
Sol. A ® (q, s, t), B ®(p), C ® (q, r)
Þ
- 9 x = 4 x - 4 or x =
\
(s) composite
(t) twin prime
+ 1)
3x log a = ( x - 1)log c
3x logc a = x - 1
3
3x ´ - = x - 1
4
Þ
(r) prime
If logb a = - 3, logb c = 4 and if the
value of x satisfying the equation
a 3 x = c x - 1 is expressed in the form
p / q, where p and q are relatively
prime, then q is
(A) ( x + 1)log10 (x
x = 5-1 / 2 , 52 or x =
Þ
Ex. 21
Column I
1
log 5 x = - , 2
2
\
(B)
The sum of the solutions of the
equation
| x - 1 | log 2 x
(C)
2 - 2 log 4
x
= ( x - 1 ) 7 is
If 5 (log y x + log x y ) = 26, xy = 64,
then the value of | x - y | is
(r)
divisible by 6
(s) divisible by 8
(t)
divisible by 10
Sol. A ® (p, q, r), B ® (p, r), C ® (p, r, t)
(A) Q a = 3 - 2 = 1, b = 1, c = log 2 log 2 22
-6
= log 2 (2-6 ) = - 6
The equation reduces to x 2 + x - 6 = 0
\
\
a + b = - 1, ab = - 6
2ab
2( - 6)
HM =
= 12
=
a +b
( - 1)
(B) Obviously, x = 2 is a solution. Since, LHS is positive,
x - 1 > 0. The equation reduces to
log 2 x 2 - 2 log x 4 = 7
Þ
2l -
4
= 7, where l = log 2 x
l
327
Chap 04 Logarithms and Their Properties
Þ
2l2 - 7 l - 4 = 0 Þ l = 4, -
1
2
Let
1
Þ x = 2 4 , 2-1 / 2
2
1
Þ
x = 16,
2
1
x = 16, x ¹
[Q x > 1]
Þ
2
\ Solutions are x = 2, 16
\ Sum of solutions = 2 + 16 = 18
(C) If a = log x , b = log y
a b
\
log y x + log x y = +
b a
\
\
Þ
log 2 x = 4, -
a
1 26
= l, then l + =
b
5
l
Þ
5l2 - 26l + 5 = 0
Þ
5l2 - 25l - l + 5 = 0
Þ
( l - 5) ( 5l - 1) = 0
1
5
1
a
a
= 5, Þ
=5
5
b
b
l = 5,
Þ
\
Þ
a = 5b
and a + b = log x + log y = log( xy ) = log(64 )
\
a + b = 6 log 2
From Eqs. (i) and (ii), we get
b = log2 and a = 5 log 2
Þ
y = 2, x = 32 or y = 32, x = 2
\
| x - y | = 30
5(log y x + log x y ) = 26
a b 26
+ =
5
b a
…(i)
…(ii)
JEE Type Solved Examples :
Statement I and II Type Questions
n
Directions Example numbers 23 to 24 are
Assertion-Reason type examples. Each of these examples
contains two statements:
Statement-1 (Assertion) and Statement-2 (Reason)
Each of these examples also has four alternative choices,
only one of which is the correct answer. You have to select
the correct choice as given below.
(a) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
Ex. 24 Statement-1 If p, q Î N satisfy the equation
x x = ( x ) x and q > p, then q is a perfect number.
Statement-2 If a number is equal to the sum of its factor,
then number is known as perfect number.
l
Sol. (d) Q
ln ( x )
Þ
Þ
l
Sol. (d) Q
æ 1 ö
N =ç ÷
è 0.4 ø
20
æ 10 ö
= ç 2÷
è2 ø
20
Þ log10 N = 20(1 - 2 log10 2) = 20 (1 - 2 ´ 0.3010)
= 20 ´ 0.3980 = 7.9660
Since, characteristic of log10 N is 7, therefore the number of
digits in N will be 7 + 1, i.e. 8.
Hence, Statement-1 is false and Statement-2 is true.
x
= ( x )x
Taking logarithm on both sides on base e, then
20
æ 1 ö
Ex. 23 Statement-1 If N = ç ÷ , then N contains
è 0.4 ø
7 digits before decimal.
Statement-2 Characteristic of the logarithm of N to the
base 10 is 7.
x
Þ
Þ
x
= ln ( x )x
x ln x = x ln x Þ
x ln x =
x
ln x
2
xö
æ
ln x ç x - ÷ = 0
è
2ø
æ
xö
ln x × x × ç1 ÷ =0
2 ø
è
ln x = 0, x = 0, 1 -
x
=0
2
\
x = 1, 0, 4
Q
x ÎN
\
x = 1, 4 Þ p = 1 and q = 4
Q
4 =1´2´2 Þ 4 ¹1+2+2
\ q is not a perfect number.
Hence, Statement-1 is false and Statement-2 is true.
328
Textbook of Algebra
Subjective Type Questions
n
In this section, there are 21 subjective solved examples.
l
Ex. 25 Prove that log 3 5 is an irrational.
and
8>7
Þ
log 8 > log 7
From Eqs. (i) and (ii), we get
log 11 × log 8 > log 7 × log 5
log 11 log 5
Þ
>
Þ log 7 11 > log 8 5
log 7
log 8
Sol. Let log 3 5 is rational.
\
Þ
log 3 5 =
p
, where p and q are co-prime numbers.
q
5 = 3p /q Þ 3p = 5q
which is not possible, hence our assumption is wrong.
Hence, log 3 5 is an irrational.
l
…(i)
3
Þ
Ex. 30 Given, a 2 + b 2 = c 2 . Prove that
log b + c a + log c - b a = 2 log c + b a × log c - b a, " a > 0, a ¹ 1
c - b > 0, c + b > 0
c - b ¹ 1, c + b ¹ 1.
l
Ex. 26 Find the value of the expression
(log 2 ) 3 + log 8 × log 5 + (log 5 ) 3 .
Sol. Q log 2 + log 5 = log(2 × 5) = log10 = 1
(log 2 + log 5) = 1
Sol. LHS = logb + c a + logc - b a
Þ (log 2)3 + (log 5)3 + 3 log 2 log 5(log 2 + log 5) = 13
Þ (log 2)3 + (log 5)3 + log 23 log 5(1) = 1
Þ
l
Ex. 27 If l
Sol. Q
[from Eq. (i)]
(log 2)3 + log 8 log 5 + (log 5)3 = 1
log 3 5
(log 3 5 ) 2
= 81, find the value of l
.
llog 3 5 = 81
=
1
1
+
loga (c + b ) loga (c - b )
=
loga (c + b ) + loga (c - b )
loga (c + b )loga (c - b )
=
loga (c 2 - b 2 )
loga (c + b ) × loga (c - b )
=
loga a 2
loga (c + b ) × loga (c - b )
=
2loga a
loga (c + b ) × loga (c - b )
=
2
loga (c + b ) × loga (c - b )
\ ( llog 3 5 )log 3 5 = (81)log 3 5
Þ
4
2
l(log 3 5 ) = 34 log 3 5 = 3log 3 5 = 54 = 625
Sol. Given, (2009 )( x )
=x
Ex. 31 Let a > 0, c > 0, b = ac , a , c and ac ¹1, N > 0.
log a N log a N - log b N
.
Prove that
=
log c N log b N - log c N
l
2
Taking logarithm both sides on base 2009, then
log 2009 (2009 ) + log 2009 x × log 2009 x = log 2009 x 2
1
+ (log 2009 x )2 = 2 log 2009 x
[ for x >0]
2
1
Þ (log 2009 x )2 - 2 log 2009 x + = 0
2
If roots are x 1 and x 2 , then log 2009 x 1 + log 2009 x 2 = 2
Þ
Þ
l
log 2009 ( x 1x 2 ) = 2 or x 1x 2 = (2009 )2
Þ
11 > 5
log 11 > log 5
1
1
loga N - logb N
log N a log N b
Sol. RHS =
=
1
1
logb N - logc N
log N b log N c
=
(log N b - log N a ) log N c
×
(log N c - log N b ) log N a
æb ö
log N ç ÷
è a ø loga N loga N
×
=
= LHS
=
æ c ö logc N logc N
log N ç ÷
èb ø
b cù
é
2
êëQb = ac Þ b = ac Þ a = b úû
Ex. 29 Prove that log 7 11 is greater than log 8 5.
Sol. Q
[Qc 2 - b 2 = a 2 ]
= 2logc + b a × log c - b a = RHS
Ex. 28 Find the product of the positive roots of the
equation ( 2009 )( x ) log 2009 x = x 2 .
l
log 2009 x
…(ii)
…(i)
Chap 04 Logarithms and Their Properties
\
(b + c ) ln a + (c + a ) ln b + (a + b ) ln c = 0
Þ
ln ab + c + ln bc + a + ln c a + b = 0
[Qc = b y ]
Þ
ln {ab + c × bc + a × c a + b } = 0
[Qb = a x ]
Þ
ab + c × bc + a × c a + b = e 0 = 1
Ex. 32 If a x = b, b y = c , c z = a , x = log b a k1 , y = log c b k 2 ,
z = log a c k 3 , find the minimum value of 3k 1 + 6k 2 + 12k 3 .
l
Sol. Q
a = c z = (b y )z
= b yz = (a x )yz = a xyz
\
k2
Also, xyz = logb a × logc b × loga c
\
k3
= k1 × k 2 × k 3 × logb a × logc b × loga c
1 = k1k 2k 3
AM ³ GM
3k1 + 6k 2 + 12k 3
³ (3k1 × 6k 2 × 12k 3 )1 / 3
3
= (3 × 6 × 12 × k1k 2k 3 )1 / 3
= (3 × 6 × 12)1 / 3
3
3 1/ 3
= (2 × 3 )
= (1)1 / 3 = 1
l
Sol. Q 5
æ 1ö
log1 / 5 ç ÷
è 2ø
log
2
ö
æ
4
÷
ç
è 7 + 3ø
= 5log 5 (2 ) = 2
æ
ö
4
÷ = log
è 7 + 3ø
2ç
2
ö
æ
4( 7 - 3)
÷
ç
è( 7 + 3) ( 7 - 3)ø
= log 2 ( 7 - 3 )
= log 21 / 2 ( 7 - 3 )1
=
…(i)
1
log b
Similarly,
=
y log a + log b + log c
…(ii)
1
log c
=
z log a + log b + log c
…(iii)
1
log 2 ( 7 - 3 )
1/2
= log 2 ( 7 - 3 )2 = log 2 (10 - 2 21 )
æ
ö
1
and log1 / 2 ç
÷ = log 2 (10 + 2 21 )
è 10 + 2 21 ø
Hence,
5
ln a
ln b
ln c
=
=
, prove that
(b - c ) (c - a ) (a - b )
a b + c × b c + a × c a + b =1
a b + c + b c + a + c a + b ³ 3.
(b + c ) ln a + (c + a ) ln b + (a + b ) ln c
=
0
[using ratio and proportion]
+ log
ö
ö
æ
æ
4
1
÷
÷ + log1/ 2 ç
è 10 + 2 21 ø
è 7 + 3)ø
2ç
= 2 + log 2 {(10 - 2 21 ) (10 + 2 21 )}
= 2 + log 2 (100 - 84 ) = 2 + log 2 (2)4 = 2 + 4 = 6
Ex. 36 Find the square of the sum of the roots of the
equation log 2 x × log 3 x × log 5 x = log 2 x × log 3 x
+ log 3 x × log 5 x + log 5 x × log 2 x.
l
Sol. Let log 2 x = A , log 3 x = B and log 5 x = C , then the given
equation can be written as
Sol. Since, a > 0, b > 0, c > 0
ln a
ln b
ln c
=
=
(b - c ) (c - a ) (a - b )
log1 / 5(1 / 2 )
= 2 + log 2 (10 - 2 21 ) + log 2 (10 + 2 21 )
xyz = xy + yz + zx
Also, prove that
+ log
æ
ö
1
+ log 1 / 2 ç
÷.
è 10 + 2 21 ø
log a + log b + log c
log a
or
[from Eq. (i)]
³3
log 1 / 5 (1 / 2)
log bc
log b + log c
=1+
log a
log a
1
log a
=
x log a + log b + log c
Ex. 34 If
+c
[Qk1k 2k 3 = 1]
On adding Eqs. (i), (ii) and (iii), we get
1 1 1
+ + =1
x y z
l
+b
a +b
=6
x = 1 + loga bc = 1 +
and
a
c +a
Ex. 35 Simplify 5
Ex. 33 If x = 1 + log a bc , y = 1 + log b ca , z = 1 + log c ab,
prove that xyz = xy + yz + zx .
or
b +c
or
l
=
ab + c + bc + a + c a + b
³ (ab + c × bc + a × c a + b )1 / 3
3
Þ
or 3k1 + 6k 2 + 12k 3 ³ 18
\ Minimum value of 3k1 + 6k 2 + 12k 3 is 18.
Sol. Q
…(i)
Again, AM ³ GM
xyz = 1
k1
Q
329
1
1ö
æ1
+ ÷
ABC = AB + BC + CA = ABC ç +
èC A B ø
Þ
1 1
æ1
ö
ABC ç + + - 1÷ = 0
èA B C
ø
or A = 0, B = 0, C = 0,
1
1 1
+ + -1=0
A B C
330
Textbook of Algebra
1
2a +
2 log 2 3 + log 2 7
c
=
=
2 + log 2 5 + log 2 7 2 + ab + 1
c
[from Eqs. (i), (ii) and (iii)]
log 2 x = 0, log 3 x = 0, log 5 x = 0, log x 2 + log x 3 + log x 5 = 0
1444442444443 14444244443
x >0
or
x > 0, x ¹ 1
x = 20 , x = 30 , x = 50 , log x (2 × 3 × 5) = 0
x = 1, x = 1, x = 1, x = 30
or
æ 2ac + 1 ö
=ç
÷
è 2c + abc + 1 ø
\ Roots are 1 and 30.
Hence, the required value
= (1 + 30)2 = (31)2 = 961
Ex. 39 Show that the sum of the roots of the equation
x + 1 = 2 log 2 ( 2 x + 3 ) - 2 log 4 (1980 - 2 - x ) is log 2 11.
l
Ex. 37 Given that log 2 a = l, log 4 b = l2 and
æ a 2b 5 ö
2
log c 2 (8 ) =
, write log 2 ç
÷ as a function of ‘l’
è c4 ø
l3 + 1
(a , b, c > 0, c ¹ 1).
l
Sol. Given,
x + 1 = 2 log 2 (2x + 3) - 2 log 4 (1980 - 2- x )
= 2 log 2 (2x + 3) - 2 log 2 2 (1980 - 2- x )1
1
= 2 log 2 (2x + 3) - 2 × log 2 (1980 - 2- x )
2
Sol. Q log 2 a = l Þ a = 2l
log 4 b = l2
Þ
2
b = 4l = 2 2l
Þ
2
ì ( 2x + 3) 2 ü
= log 2 í
-x ý
î 1980 - 2 þ
2
logc 2 (8) =
and
3
l +1
or 2x + 1 =
2
3
logc 2 = 3
2
l +1
Þ
Þ
logc 2 =
Þ
4
Þ
3( l3 + 1)
log 2 c =
or
\
= log 2 (2x + 3)2 - log 2 (1980 - 2- x )
3
3( l + 1)
or c = 2
4
2
3 + 1)
2 l + 10 l2 - 3 ( l3 + 1)
Ex. 40 Solve the following equations for x and y
1
log 100 | x + y | = , log 10 y - log 10 | x | = log 100 4 .
2
}
}
Sol. Q
Sol. Q log 2 3 = a
…(i)
log 2 5 log 2 5
=
log 2 3
a
log 2 5 = ab
and
log 7 2 = c
Þ
Now,
log100 | x + y | =
3
Ex. 38 Given that log 2 3 = a , log 3 5 = b, log 7 2 = c ,
express the logarithm of the number 63 to the base 140 in
terms of a , b and c.
\
…(i)
x 1 + x 2 = log 2 11
Þ
l
b = log 3 5 =
22 x - 3954 × 2x + 11 = 0
2x 1 × 2x 2 = 11 or 2x 1 + x 2 = 11
= 2l + 10l - 3( l + 1)
Þ
1980(2x + 1 ) - 2 = 22 x + 9 + 6 × 2x
l
= log 2 {2 2 l × 2 10 l × 2-3 (l
2
1980 - 2- x
If x 1, x 2 are the roots of Eq. (i), then
ìï 3 ( l3 + 1) üï
ý
í
4
ïþ
ïî
æ a 2b 5 ö
log 2 ç 4 ÷ = log 2 (a 2b 5c -4 )
è c ø
= log 2 {2
( 2x + 3) 2
[from Eq. (i)]
1
1
= c or log 2 7 =
c
log 2 7
log 2 63
log 2 (32 ´ 7 )
=
log140 63 =
log 2 140 log 2 (22 ´ 5 ´ 7 )
...(ii)
1
2
Þ
| x + y | = (100)1/ 2 = 10
Þ
| x + y | = 10
and
log10 y - log10 | x | = log100 4, y > 0
Þ
æy ö
2
log10 ç ÷ = log10 2 22 = log10 2
è|x |ø
2
Þ
æy ö
y
=2
log10 ç ÷ = log10 2 Þ
|x |
è|x |ø
Þ
y = 2| x |
…(i)
…(ii)
From Eqs. (i) and (ii),we get
| x + 2| x || = 10
…(iii)
Case I If x > 0, then | x | = x
From Eq. (iii),
| x + 2x | = 10
…(iii)
Chap 04 Logarithms and Their Properties
Þ
3| x | = 10 Þ | x | =
10
3
10
20
,y =
3
3
Case II If x < 0, then | x | = - x
From Eq. (iii),
x=
\
[from Eq. (ii)]
Þ
| - x | = 10 Þ | x | = 10
- x = 10
Þ
Þ
log( x - 1) (log 3 x - 4 ) (2 log 3 x + 1) = 0
Þ
| x - 2x | = 10
\
Þ
ì
ü
4
log( x - 1)í2 log 3 x - 7ý = 0
log
x
3
î
þ
log( x - 1){2(log 3 x )2 - 7 log 3 x - 4 } = 0
Þ
log( x - 1) = 0, log 3 x = 4, log 3 x = -
Þ
Þ
x = - 10
\
y = 20
ì 10 20ü
Hence, solutions are í , ý, { -10, 20}.
î3 3þ
x = 2, 81
From Eq. (ii),
6 log x × log 10 a× loga 5
log
x
Sol. Q × a a
- 3log 10 (x / 10 ) = 9 100
5
Sol. Given,
+
2 log 2 log 2 x + log1/ 2 log 2 (2 2x ) = 1
Þ
1ö
÷
2ø
Þ
Þ
6 log10 x 3log10 x
×5
= 3log10 x + 1 [by property]
5
3
2 log 2 log 2 x - log 2 log 2 (2 2x ) = 1
Þ 2 log 2 log 2 x - log 2 {log 2 (2 2 ) + log 2 x } = 1
ì3
ü
Þ
2 log 2 log 2 x - log 2 í + log 2 x ý = 1
î2
þ
ö
æ3
Let log 2 x = l, then 2 log 2 l - log 2 ç + l ÷ = 1
ø
è2
+ log 4 2
2 ç log10 x
6 log10 5
è
×x
- 3(log10 x - 1) = 3 2
5
[by property]
ö
æ3
log 2 l2 - log 2 ç + l ÷ = 1
ø
è2
Þ
Let log10 x = l, then
Þ
Þ
Þ
\
l
6 l 3l
×5 = 3 × 3l
3
5
5
=3
l-2
ü
ì
ï l2 ï
l2
=1 Þ
= 21
log 2 í
ý
3
3
ï + lï
+l
þ
î2
2
l2 = 3 + 2l Þ l2 - 2l - 3 = 0
Þ
6 l
æ1
ö 10
× 5 = 3l ç + 3÷ = × 3l
è3
ø 3
5
l-2
Þ
Þ
\
or
Þ
which is possible only, where l = 2.
log10 x = 2
Ex. 42 Find the value of x satisfying the equation
2 - 2 log 9
x
( l - 3) ( l + 1) = 0
l = 3, - 1
log 2 x = 3, - 1
x = 23 , 2-1
1
x = 8,
Þ
2
But the given equation is valid only when,
x = 102 = 100
| x - 1 |log 3 x
= ( x - 1) 7 .
x > 0, 2 2x > 0, log 2 x > 0, log 2 (2 2x ) > 0
Sol. The given equation is,
|x -
2
1| log 3 x - 2 log x 9
= ( x - 1) 7
…(i)
Þ
This equation is defined for
2 - 2 log 9
x
= ( x - 1) 7
Taking log on both sides, then
(log 3 x 2 - 2 log x 9 )log( x - 1) = 7 log( x - 1)
Þ
2
log( x - 1){log 3 x - 2 log x 9 - 7 } = 0
1
2 2
From Eq. (i), the solution of the given equation is x = 8.
x ³ 2, then Eq. (i) reduces to
( x - 1)log 3 x
x > 0, x > 0, x > 1, x >
Hence,x > 1
x 2 > 0, x > 0, x ¹ 1 and x - 1 ³ 1
Þ
x - 1 = (10)0 , x = 34 , x = 3-1 / 2
1
x - 1 = 1, x = 81, x =
3
1 ù
é
êëQ x ³ 2, \ x ¹ 3 úû
Ex. 43 Find all real numbers x which satisfy the
equation 2 log 2 log 2 x + log 1/ 2 log 2 ( 2 2 x ) = 1.
Ex. 41 Solve the following equation for x
6 loga x ×log10 a×loga 5
a
- 3 log10 ( x / 10 ) = 9 log100 x + log 4 2 .
5
Þ
1
2
l
l
æ1
331
Ex. 44 Solve for x ,
log 3 / 4 log 8 ( x 2 + 7 ) + log 1 / 2 log 1 / 4 ( x 2 + 7 ) -1 = - 2.
l
Sol. Given,
log 3 / 4 log 8 ( x 2 + 7 ) + log1/ 2 log1/ 4 ( x 2 + 7 )-1 = - 2
…(i)
332
Textbook of Algebra
Þ log 3 / 4 log 2 3 ( x 2 + 7 ) + log 2 -1 log 2 -2 ( x 2 + 7 )-1 = - 2
1
ö
æ
1
ö
æ
+ 2÷
÷
ç | loga b | +
ç loga b +
|
log
|
b
b
log
a
a
÷
÷ = ç
= ç
4
2
÷
ç
÷
ç
÷
ç
÷
ç
ø
è
ø
è
ì1
ì1
ü
ü
Þ log 3 / 4 í log 2 ( x 2 + 7 )ý- log 2 í log 2 ( x 2 + 7 )ý = - 2
î3
î2
þ
þ
log 2 ( x 2 + 7 ) = 6l
Let
...(i)
log 3 / 4 (2l ) - log 2 (3l ) = - 2
log 2 (2l )
- log 2 (3l ) = - 2
log 2 (3 / 4 )
Then,
Þ
Þ
1 + log 2 l
- (log 2 3 + log 2 l ) = - 2
log 2 3 - log 2 4
Þ
1 + log 2 l
- (log 2 3 + log 2 l ) = - 2
log 2 3 - 2
ö
1æ
1
= ç | loga b | +
÷
2è
| loga b | ø
loga 4 (b / a ) + logb 4 (a / b )
and
Again, let log 2 l = A and log 2 3 = B, then
1+ A
- (B + A) = - 2
B-2
Þ
Þ
Þ
Þ
Þ
Þ
=
1
æa ö
æb ö 1
loga ç ÷ + logb ç ÷
èb ø
èa ø 4
4
=
1
(loga b - 1 + logb a - 1)
4
loga b +
=
1 + A - B 2 - AB + 2B + 2A = - 2B + 4
A ( 3 - B ) = B 2 - 4 B + 3 = ( B - 1) ( B - 3)
Þ
Þ
log 2 ( x 2 + 7 ) = 4
Þ
x 2 + 7 = 24 = 16 or x 2 = 9
\
=
\ loga 4 ab + logb 4 ab - loga 4 b / a + logb 4 (a / b )
P (say)
1 ìï
1
= í | loga b | +
+
2 ïî
| loga b |
ì 2, b ³ a > 1.
.
= í log b
a ,1 < b < a
î2
Sol. Since,
loga 4 ab + logb 4 ab =
=
\ 2P
üï
ý
ïþ
1
loga b
loga b
= 21 = 2
Case II If 1 < b < a, then
ü
1ì
1
1
P = í | loga b | +
+ | loga b | ý
2î
| loga b |
| loga b | þ
1
1
loga (ab ) + logb (ab )
4
4
1
(1 + loga b + logb a + 1)
4
1
| loga b |
ü
1ì
1
1
P = í | loga b | +
- | loga b | +
ý
2î
| loga b |
| loga b | þ
x = ±3
( loga 4 ab + logb 4 ab - loga 4 b /a + logb 4 a /b ) loga b
| loga b | -
Case I If b ³ a > 1 , then
Ex. 45 Prove that
2
1
| loga b |
2
=
l
1
-2
loga b
4
| loga b | -
A = - ( B - 1)
[Q B - 3 ¹ 0, i.e. log 2 3 ¹ 3]
A + B = 1 Þ log 2 l + log 2 3 = 1
log 2 (3l ) = 1
3l = 2
1
3 × log 2 ( x 2 + 7 ) = 2 [from Eq. (i)]
6
2
= | loga b |
\ 2P
loga b
= 2loga b
#L
Logarithms and Their Properties Exercise 1 :
Single Option Correct Type Questions
n
This section contains 20 multiple choice questions.
Each question has four choices (a), (b), (c) and (d) out of
which ONLY ONE is correct
1. If log10 2 = 0.3010..., the number of digits in the number
2000 2000 is
(a) 6601
(b) 6602
(c) 6603
(d) 6604
2. There exist a positive number λ, such that
log 2 x + log 4 x + log 8 x = log λ x , for all positive real
numbers x .
If λ = b a , where a, b ∈ N , the smallest possible value of
(a + b ) is equal to
(a) 12
(b) 63
(c) 65
(d) 75
3. If a, b and c are the three real solutions of the equation
x log10 x + log10 x
2
3
+3
2
=
1
−
x +1 −1
where, a > b > c , then a, b, c are in
(a) AP
(c) HP
1
x + 1 +1
(b) GP
(d) a −1 + b −1 = c −1
n −1
4. If f (n ) = ∏ logi (i + 1), the value of
i=2
(a) 5010
100
∑ f (2k ) equals
k =1
(b) 5050
(c) 5100
(d) 5049
5. If log 3 27 ⋅ log x 7 = log 27 x ⋅ log 7 3, the least value of x, is
(a) 7 −3
(b) 3 −7
(c) 7 3
(d) 3 7
6. If x = log 5 (1000) and y = log 7 (2058), then
(a) x > y
(c) x = y
8. If x n > x n −1 > ... > x 2 > x 1 > 1, the value of
..
log x1 log x 2 log x 3 K log xn x n
(a) 0
(c) 2
..x 1
is
(b) 1
(d) undefined
x (y + z − x ) y (z + x − y ) z ( x + y − z )
9. If
=
=
,
log x
log y
log z
then x y = z y is equal to
y
x
(a) z x x z
10. If y = a
(a) a
y
z
(b) x zy x
1
1 − loga x
1
1 + log a z
and z = a
(b) a
1
2 + log a z
(c) x yy z
1
1 − loga y
(c) a
(a) 1
(c) 0
(b) 2
(d) −1
13. If x = 1 + loga bc , y = 1 + logb ca, z = 1 + logc ab, then
xyz
is equal to
xy + yz + zx
(a) 0
(c) −1
(b) 1
(d) 2
14. The value of a
(a) loga N
(c) log N a
logb ( logb N)
logb a
is
(b) logb N
(d) log N b
15. The value of 49 A + 5 B , where A = 1 − log 7 2 and
B = − log 5 4 is
(a) 10.5
(c) 12.5
(b) 11.5
(d) 13.5
16. The number of real values of the parameter λ for which
(log16 x ) 2 − log16 x + log16 λ = 0 with real coefficients
will have exactly one solution is
(a) 1
(c) 2
(d) 4
xn. − 1
y = logb a , a > 0, b > 0 and a, b ≠ 1)
(b) 2
(d) 4
2
= − log 5 (0.2 − 5 x − 4 ), then x is
(c) 3
12. The value of a x − b y is (where x = loga b and
17. The number of roots of the equation x logx (x + 3 ) = 16 is
7. If log 5 120 + ( x − 3) − 2 log 5 (1 − 5 x − 3 )
(b) 2
(a) ( − ∞, 1 )
(b) (1, 2 )
(c) (2, ∞ )
(d) None of the above
(a) 1
(c) 3
(b) x < y
(d) None of these
(a) 1
11. If log 0. 3 ( x − 1) < log 0.09 ( x − 1), then x lies in the interval
(d) x xy y
(b) 0
(d) 4
18. The point on the graph y = log 2 log 6 {2
1
1 − log a z
(d) a
1
2 − log a z
+ 4 },
whose y-coordinate is 1 is
(a) (1, 1 )
(c) ( 8, 1 )
(b) (6, 1 )
(d) (12, 1 )
19. Given, log2 = 0.301 and log 3 = 0.477, then the number of
digits before decimal in 312 × 2 8 is
(a) 7
(c) 9
(b) 8
(d) 11
20. The number of solution(s) for the equation
2 log x a + logax a + 3 loga 2 x a = 0, is
, then x is equal to
(2 x + 1)
(a) one
(c) three
(b) two
(d) four
334
#L
Textbook of Algebra
Logarithms and Their Properties Exercise 2 :
More than One Correct Option Type Questions
n
This section contains 9 multiple choice questions. Each
question has four choices (a), (b), (c) and (d) out of which
MORE THAN ONE may be correct.
21. If x
(log 2 x )2 − 6 log 2 x + 11
(a) 2
(b) 4
= 64, then x is equal to
(c) 6
(d) 8
22. If log λ x ⋅ log 5 λ = log x 5, λ ≠ 1, λ > 0, then x is equal to
(a) λ
(b) 5
(c)
1
5
(d) None of these
23. If S = {x : log x 3x , where log 3 x > − 1}, then
(b) S ∈ φ
(a) S is a finite set
1 
(d) S properly contains  , ∞
3 
(c) S ⊂ ( 0, ∞ )
24. If x satisfies log 2 (9 x −1 + 7 ) = 2 + log 2 (3 x −1 + 1), then
(a) x ∈ Q
(b) x ∈ N
(c) x ∈{ x ∈ Q : x < 0 }
(d) x ∈ Ne (set of even natural numbers)
α+β+γ+δ
16
1
(c) −1
α + β −1 + γ −1 + δ −1
(a) ≤
α+β+γ+δ
16
1
(d)
αβγδ
(b) ≥
27. If log10 5 = a and log10 3 = b, then
(a + b )
(3 − 2a )
(d) All of these
(a) log10 8 = 3 (1 − a )
(b) log 40 15 =
1 − a
(c) log 243 32 = 

 b 
28. If x is a positive number different from 1, such that
loga x , logb x and logc x are in AP, then
(a) log b =
2 (log a ) (log c )
(log a + log c )
(b) b =
a+c
2
(d) c 2 = (ac ) log a b
29. If | a | < | b |, b − a < 1 and a, b are the real roots of the
p p
... p p , p > 0 and p ≠ 1 is equal to
14243
n times
#L
and a, b, c , d ≠ 0, > 1, then logabcd x equals
(c) b = ac
25. log p log p p
(b) −n
(a) n
1
(c)
n
26. If loga x = α, logb x = β, logc x = γ and logd x = δ, x ≠ 1
(d) log 1 /p( pn )
equation x 2 − | α | x − | β | = 0, the equation
x
log| b |
− 1 = 0 has
a
(a) one root lying in interval ( − ∞, a )
(b) one root lying in interval (b, ∞ )
(c) one positive root
(d) one negative root
Logarithms and Their Properties Exercise 3 :
Passage Based Questions
n
This section contains 4 passages. Based upon each of the
passage 3 multiple choice questions have to be
answered. Each of these questions has four choices (a), (b),
(c) and (d) out of which ONLY ONE is correct.
(b) 8
(d) 20
Passage I
Passage II
(Q. Nos. 33 to 35)
30. If a1 = 5 and a 2 = 3, the number of integral values of N is
(b) 32
(c) 48
(d) 64
31. If a1 = 6, a 2 = 4 and a 3 = 3, the largest integral value of
N is
(a) 124
(c) 624
smallest integral values of N, is
(a) 2
(c) 14
(Q. Nos. 30 to 32)
Let log 2 N = a1 + b1 , log 3 N = a 2 + b2 and
log 5 N = a 3 + b3 , where a1 , a 2 , a 3 ∈ I and
b1 , b2 , b3 ∈[ 0, 1).
(a) 16
32. If a1 = 6, a 2 = 4 and a 3 = 3, the difference of largest and
(b) 63
(d) 127
Let ‘S ’ denotes the antilog of 0.5 to the base 256 and ‘K ’
denotes the number of digits in 610 (given
log 10 2 = 0.301, log 10 3 = 0.477) and G denotes the number of
positive integers, which have the characteristic 2, when the
base of logarithm is 3.
33. The value of G is
(a) 18
(b) 24
(c) 30
(d) 36
(c) 216
(d) 288
34. The value of KG is
(a) 72
(b) 144
Chap 04 Logarithms and Their Properties
35. The value of SKG is
(a) 1440
(c) 2016
Passage IV (Q. Nos. 39 to 41)
(b) 17280
(d) 2304
Let G , O , E and L be positive real numbers such that
log (G ⋅ L ) + log (G ⋅ E ) = 3, log ( E ⋅ L ) + log ( E ⋅ O ) = 4,
log (O ⋅ G ) + log (O ⋅ L ) = 5 (base of the log is 10).
Passage III
(Q. Nos. 36 to 38)
39. If the value of the product (GOEL ) is λ, the value of
Suppose ‘U’ denotes the number of digits in the number
( 60)100 and ‘M’ denotes the number of cyphers after decimal,
−296
. If the fraction
before a significant figure comes in ( 8)
U/M is expressed as rational number in the lowest term as p /q
(given log 10 2 = 0.301and log 10 3 = 0.477 ).
36. The value of p is
(a) 1
(b) 2
(c) 3
(d) 4
37. The value of q is
(a) 5
(c) 3
(b) 2
(d) 4
38. The equation whose roots are p and q, is
#L
335
(a) x 2 − 3 x + 2 = 0
(b) x 2 − 5 x + 6 = 0
(c) x 2 − 7 x + 12 = 0
(d) x 2 − 9 x + 20 = 0
log λ log λ log λK is
(a) 3
(c) 5
(b) 4
(d) 7
40. If the minimum value of 3G + 2L + 2O + E is 2 λ 3 µ 5 ν ,
where λ , µ and ν are whole numbers, the value of
∑ (λ µ + µ λ ) is
(a) 7
(c) 19
(b) 13
(d) None of these
G
O
41. If log   and log   are the roots of the equation
O
 E
(a) x + x = 0
(b) x 2 − x = 0
(c) x 2 − 2 x + 3 = 0
(d) x 2 − 1 = 0
2
Logarithms and Their Properties Exercise 4 :
Single Integer Answer Type Questions
n
This section contains 10 questions. The answer to each
question is a single digit integer, ranging from 0 to 9
(both inclusive).
42. If x , y ∈ R + and log10 (2x ) + log10 y = 2 and
m
, where m and n
n
are relative prime, the value of m − 3n 6 is
log10 x 2 − log10 (2y ) = 4 and x + y =
43. A line x = λ intersects the graph of y = log 5 x and
y = log 5 ( x + 4 ). The distance between the points of
intersection is 0.5. Given λ = a + b, where a and b are
integers, the value of (a + b ) is
46. If x > 2 is a solution of the equation
| log
3
x − 2 | + | log 3 x − 2 | = 2, then the value of x is
47. Number of integers satisfying the inequality
log 2
48. The value of b (> 0) for which the equation
2 log1 / 25 (bx + 28) = − log 5 (12 − 4 x − x 2 ) has coincident
roots, is
49. The value of
44. If the left hand side of the equation
a (b − c ) x 2 + b (c − a ) xy + c (a − b )y 2 = 0 is a perfect
square, the value of
2
 log(a + c ) + log(a − 2b + c )
+

 , (a, b, c ∈ R , a > c ) is
a
−
c
log(
)


45. Number of integers satisfying the inequality
 1
 
 3
| x + 2|
2 − |x |
> 9 is
x − 2 log12/ 4 x + 1 > 0, is
2
log
21 / 4
2
− 3 log27 125 − 4
7 4 log49 2 − 3
is
50. If x 1 and x 2 ( x 2 > x 1 ) are the integral solutions of the
equation
 5
(log 5 x ) 2 + log 5 x   = 1, the value of | x 2 − 4 x 1 | is
x
1
2
log λ c = nx n +1 , the value of n is
51. If x = log λ a = loga b = log b c and
336
#L
Textbook of Algebra
Logarithms and Their Properties Exercise 5 :
Matching Type Questions
This section contains 3 questions. Questions 52 to 54 have four statements (A, B, C and D) given in Column I and four
statements (p, q, r and s) in Column II. Any given statement in Column I can have correct matching with one or more
statement(s) given in Column II.
n
52.
Column I
Column I
Column II
(A)
log 3 243
log 2 32
(p)
positive integer
(B)
2 log 6
(log 12 + log 3 )
(q)
negative integer
(C)
 1
log1 / 3  
 9
(r)
rational but not
integer
(D)
−2
log 5 16 − log 5 4
log 5 128
(A)
prime
Column I
Column II
(A) The expression log 02.5 8 has the value
equal to
(p)
(q)
(B) The value of the expression
(log10 2 ) 3 + log10 8 ⋅ log10 5 + (log10 5 ) 3 + 3, is
1
(C) Let N = log 2 15 ⋅ log1 / 6 2 ⋅ log 3   . The
 6
value of [ N ] is (where [ ⋅ ] denotes the
greatest integer function)
#L
Column I
(B)
53.
b
54.
(s)
Column II
(D) If (52.6 ) = ( 0.00526 ) = 100, the value of
1 1
− is
a b
a
(r)
1
(s)
4
Column II
 2 (x − 2) 
If log1 /x 
 ≥ 1, then x can
( x + 1 ) ( x − 5 ) 
belongs to
(p)  1 
 0, 
 3
3
If log 3 x − log 23 x ≤ log (1/2
2
can belongs to
(q) (1, 2 ]
2 ) 4, then
x
(C)
If log1 / 2( 4 − x ) ≥ log1/2 2 − log1 / 2( x − 1 ),
then x belongs to
(r)
[3, 4)
(D)
Let α and β are the roots of the quadratic
equation
( λ2 − 3 λ + 4 ) x 2 − 4 (2 λ − 1 ) x + 16 = 0,
if α and β satisfy the condition β > 1 > α ,
then p can lie in
(s)
(3, 8)
2
3
Logarithms and Their Properties Exercise 6 :
Statement I and II Type Questions
n
Directions Question numbers 55 to 60 are AssertionReason type questions. Each of these questions contains
two statements:
Statement-1 (Assertion) and Statement-2 (Reason)
Each of these questions also has four alternative choices,
only one of which is the correct answer. You have to select
the correct choice as given below.
(a) Statement-1 is true, Statement-2 is true; Statement-2
is a correct explanation for Statement-1
(b) Statement-1 is true, Statement-2 is true; Statement-2
is not a correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is false
(d) Statement-1 is false, Statement-2 is true
55. Statement-1 log10 x < log 3 x < loge x < log 2 x
( x > 0, x ≠ 1).
Statement-2 If 0 < x < 1, then log x a > log x b ⇒0 < a < b .
56. Statement-1 The equation 7 log7 (x
distinct real roots.
3
+1)
− x 2 = 1 has two
Statement-2 a loga N = N , where a > 0, a ≠ 1 and N > 0.
7
57. Statement-1   <  
 3
 3
1
⇒
1
4
 1
 1
7 log   < 4 log   ⇒ 7 < 4
 3
 3
Statement-2 If ax < ay, where a < 0, x , y > 0, then x > y.
58. Statement-1 The equation x logx (1− x ) = 9 has two
2
distinct real solutions.
Statement-2 a loga b = b, when a > 0, a ≠ 1, b > 0.
59. Statement-1 The equation (log x ) 2 + log x 2 − 3 = 0 has
two distinct solutions.
Statement-2 log x 2 = 2 log x .
60. Statement-1 log x 3 ⋅ log x
3 = log 81 (3) has a solution.
Statement-2 Change of base in logarithms is possible.
/9
Chap 04 Logarithms and Their Properties
Logarithms and Their Properties Exercise 7 :
#L
Subjective Type Questions
n
In this section, there are 27 subjective questions.
61.
(i) If log 7 12 = a, log12 24 = b, then find value of log 54 168 in
terms of a and b.
(ii) If log 3 4 = a, log 5 3 = b, then find the value of log 3 10 in
terms of a and b.
62. If
ln a
ln b
ln c
=
=
, prove the following.
b −c c −a a −b
(i) abc = 1
+ bc + c 2
2
⋅ bc
2
+ ca + a 2
⋅ ca
2
+ ab + b 2
2
+ bc + c 2
log 4 (2000)
6
+
+ bc
2
+ ca + a 2
3
log 5 (2000) 6
+ ca
2
+ ab + b 2
≥3
1
4
64. If log2 = 0.301 and log 3 = 0.477, find the number of
integers in
(ii) 6 20
(i) 5 200
(iii) the number of zeroes after the decimal is 3 −500.
65. If log2 = 0.301 and log3 = 0.477, find the value of
log (3.375).
66. Find the least value of log 2 x − log x (0.125) for x > 1.
67. Without using the tables, prove that
1
1
+
> 2.
log 3 π log 4 π
68. Solve the following equations.
71. Find the value of x satisfying
loga {1 + logb {1 + logc (1 + log p x )}} = 0.
2
(3 −
6 ) − 6 log 8 ( 3 −
(v) 3 log 3
( x − 1)
(vi) log1 /2( 3 x − 1 ) 2 < log1 /2( x + 5 ) 2
2
(vii) log10 x + 2 ≤ log10
x
(viii) log10( x 2 − 2 x − 2 ) ≤ 0
3

(ix) log x 2 x −  > 2

4
(x) log1 / 3 x < log1 / 2 x
(xi) log 2x + 3 x 2 < log 2x + 3(2 x + 3 )
5
(xii) log 22 x + 3 log 2 x ≥ log 4
2
2
x
(xiii) ( x + x + 1 ) < 1
1
(xiv) log ( 3x 2 + 1) 2 <
2
= 10 x
2
− 3 log10 x + 1
2 16
> 1000
(xvi) log 4 {14 + log 6( x − 64 )} ≤ 2
2
(xvii) log 2(9 − 2 x ) ≤ 10 log10 ( 3 − x )
(iii) 2 ⋅ x log 4 3 + 3 log 4 x = 27
 2x + 3
(xviii) loga 
 ≥ 0 for
 x 
(a) a > 1, (b) 0 < a < 1
(xix) 1 + log 2( x − 1 ) ≤ log x − 1 4
(xx) log 5x + 4 ( x 2 ) ≤ log 5x + 4 (2 x + 3 )
(iv) log 4 log 3log 2 x = 0
= 10 5 + log10 x
1


(vi) log 3  log 9 x + + 9 x  = 2 x


2
(vii) 4 log10 x + 1 − 6 log10 x − 2 ⋅ 3 log10 x
1
log10( x − 3 )
(viii)
=
2
log10( x − 21 ) 2
(ix) x log 2 x + 4 = 32
(x) loga x = x, where a = x log 4 x
(xi) log
2 sin x (1
.
< 3 log 3 ( x − 6) + 3
(ii) log 2( 9 + 2 ) = 3
log10 x + 5
3
2)
(iii) log 2(2 − x ) < log1 / 2( x + 1 )
(iv) log x 2 ( x + 2 ) < 1
(xv) x (log10 x )
x
(v) x
.
(ii) log 2x ( x 2 − 5 x + 6 ) < 1
1
3
(i) x
2
(i) log ( 2x + 3) x 2 < 1
63. Prove that log10 2 lies between and ⋅
1 + log10 x
70. Find the value of the expression
73. Solve the following inequations.
=1
(iv) a + b + c ≥ 3
(v) aa + bb + cc ≥ 3
(vi) ab
logarithm to the base 10.
72. Find the value of 4 5 log4
(ii) aa ⋅ bb ⋅ cc = 1
(iii) ab
69. Find a rational number, which is 50 times its own
+ cos x ) = 2
2
+2
74. Solve log x (ax )1 / 5 + loga (ax )1 / 5
=0
x
+ loga  
a
1/ 5
a
+ log x  
x
1/ 5
= a.
75. It is known that x = 9 is root of the equation,
log π ( x 2 + 15a 2 ) − log π (a − 2) = log π
find the other roots of this equation.
8ax
a −2
337
338
Textbook of Algebra
76. Solve log 4 (log 3 x ) + log 1/ 4 (log1 / 3 y ) = 0 and
x2 +y2 =
83. Find the solution set of the inequality
17
⋅
4
2 log1/ 4 ( x + 5) >
77. Find the real value(s) of x satisfying the equation
log 2 x ( 4 x ) + log 4 x (16x ) = 4.
78. Find the sum and product of all possible values of x
which makes the following statement true
 4
log 6 54 + log x 16 = log 2 x − log 36   .
9
79. Solve the equation
3
log 4 ( x + 2) 3 + 3 = log 4 ( 4 − x ) 3 + log 4 ( x + 6) 3 .
2
80. Solve log 2 ( 4 x + 1 + 4 ) ⋅ log 2 ( 4 x + 1) = log1/
81. Solve the system of equations 2
log10
x+ y
2
 1
 .
 8
(x + 5 )
(2).
84. Solve log 3 ( x + | x − 1 | ) = log 9 ( 4 x − 3 + 4 | x − 1 | ).
85. In the inequality
2

x 5
(log 2 x ) 4 −  log1 /2
 − 20 log 2 x + 148 < 0
4 

holds true in (a, b ), where a, b ∈ N . Find the value of
ab (a + b ).
86. Find the value of x satisfying the equation
(log 3 3 3x + log x
3
3x ) ⋅ log 3 x

+  log 3

= 256 and
3
3
x
  + log x
 3
3
 3 
   log 3 x
x
3
= 2.
87. If P is the number of natural numbers whose logarithm
 3
xy − log10   = 1.
 2
82. Solve the system of equations
log 2 y = log 4 ( xy − 2), log 9 x 2 + log 3 ( x − y ) = 1.
#L
9
log 1 (9 ) + log
4
3 3
to the base 10 have the characteristic P and Q is the
number of natural numbers reciprocals of whose 3
logarithms to the base 10 have the characteristic − q,
show that log10 P − log10 Q = p − q + 1.
Logarithms and Their Properties Exercise 8 :
Questions Asked in Previous 13 Year’s Exam
n
This section contains questions asked in IIT-JEE,
AIEEE, JEE Main & JEE Advanced from year 2005 to
year 2017.
90. The value of

 1
1
1
1
K  is
4−
4−
4−
6 + log 3 / 2 
3 2
3 2 
3 2
3 2

88. Let a = log 3 log 3 2 and an integer k satisfying
1 < 2 (− k + 3
−a
)
< 2, then k equals to
(a) 0
(c) 2
[IIT-JEE 2012, 4M]
[IIT-JEE 2008, 1.5M]
(b) 1
(d) 3
89. Let ( x 0 , y 0 ) be solution of the following equations
(2x ) ln2 = (3y ) ln 3 and 3 ln x = 2 ln y , then x 0 is
[IIT-JEE 2011, 3M]
(a)
1
6
(b)
1
3
(c)
1
2
(d) 6
91. If 3 = 4
x
x −1
, then x equals
[JEE Advanced 2013, 3M]
(a)
2 log 3 2
2 log 3 2 − 1
(b)
2
2 − log 2 3
(c)
1
1 − log 4 3
(d)
2 log 2 3
2 log 2 3 − 1
Answers
Exercise for Session 1
1. (a)
2. (b)
1
71. 1
72. 9
6
3
1
73. (i) x ∈  − , 3 ∪ {− 1, 0 } (ii) x ∈  0,  ∪ (1, 2) ∪ (3, 6)
 2 
 2

1 − 5  1 + 5 
(iii) x ∈  − 1,
, 2
 ∪
2   2


69. 100
3. (b)
4. (b)
5. (b)
4. (c)
5. (c)
4. (a)
5. (c)
Exercise for Session 2
1. (b)
2. (b)
3. (d)
Exercise for Session 3
1. (c)
2. (c)
3. (d)
(iv) x ∈ (− 2, 1) ∪ (2, ∞ ) ~ {−1, 0}
(v) x > 6
(vi) x ∈ ( − ∞ , − 5) ∪ (− 5, − 1) ∪ (3, ∞ )
(vii) x ∈ (0, 10−1 ] ∪ [102 , ∞ )
Chapter Exercises
1. (c)
7. (a)
13. (b)
19. (c)
2. (d)
8. (b)
14. (b)
20. (b)
3. (b)
9. (a)
15. (c)
4. (b)
10. (c)
16. (b)
5. (a)
11. (c)
17. (b)
21. (a, b, d) 22. (b, c) 23. (c, d)
24. (a, b) 25. (b, d)
26. (a, c) 27. (a, b, c, d)
28. (a, d) 29. (c, d)
6. (a)
12. (c)
18. (d)
30. (b)
31. (d)
32. (a)
33. (a)
34. (b)
35. (d)
36. (b)
37. (c)
38. (b)
39. (b)
40. (a)
41. (d)
42. (9)
43. (6)
44. (4)
45. (3)
46. (9)
47. (3)
48. (4)
49. (7)
50. (1)
51. (2)
52. (A) → (p, s), (B) → (p), (C) → (q), (D) → (r)
53. (A) → (r), (B) → (s), (C) → (q), (D) → (q)
54. (A) → (q), (B) → (p), (C) → (q, r), (D) → (s)
55. (d)
56. (d)
57. (d)
58. (d)
59. (c)
60. (d)
ab + 1
ab +
(ii)
a (8 − 5 b )
2b
1
66. 2 3
68. (i) 10,
10
(iii) x = 16
(iv) x = 8
1
1
(vii) x =
(vi) x =
3
100
π
(x) x = 2
(xi) x =
3
61. (i)
2
64. (i) 140 (ii)16
(iii) 238 65. 0.528
(ii) x ∈ φ
(v) {10−5 , 103}
(viii) x = 5 (ix) x = 2 or
70.
1
32
(viii) x ∈ [ −1, 1 − 3 ) ∪ (1 +
3
3 1
(ix) x ∈  ,  ∪  1, 
 8 2  2
3 , 3]
(x) x ∈ (0, 1)
3
(xi) x ∈  − , − 1 ∪ (−1, 3)
 2

(xiii) x ∈ (− ∞ , − 1)
(xv) x ∈ (1000, ∞ )
(xvii) x ∈ ( − ∞ , 0]
1
(xii) x ∈  0,  ∪ [ 2, ∞ )
 16 
(xiv) x ∈ (− ∞ , − 1) ∪ (1, ∞ )
(xvi) x ∈ [ − 10, − 8) ∪ (8, 10]
3
(xviii) (a) x ∈ (− ∞ , − 3] ∪ (0, ∞ ) (b) x ∈  −3, − 

2
3
3

(xix) x ∈ (2, 3]
(xx) x ∈  − , −  ∪ [ −1, 0) ∪ (0, 3]
 5
2
74. x = a 4/ 5a 2
75. x =15 for a = 3
1
1
76. x = 2 or , y = or 2
2
2
9
2
77. x = 1, 2 −3/ 2
78. Sum = , Product = 2
79. x = 2
80. x = 0
82. x = 3, y = 2
83. x ∈ (−5,−4) ∪ (−3,−1)
25
84. x =
64
89. (c)
81. (9, 25) and (25, 9)
85. 3456
86. x ∈ (1, 3]
90. (4)
91. (a, b, c)
88. (b)
340
Textbook of Algebra
Solutions
n -1
log(i + 1 ) log(n )
=
= log 2 n
log(i )
log(2 )
f (n ) = Õ
4.
i=2
\
f (2k ) = k
100
Then,
k =1
1. log10 2 = 0.3010
2. Q l > 0 and l ¹ 1 and x > 0
Þ
Þ
log 2 x + log 4 x + log 8 x = log l x
1
1
Þ log 2 x + log 2 x + log 2 x = log l x
2
3
11
log 2 x = log l x
Þ
6
11
11
Þ
=
6 log x 2 log x l
Þ
l11 = 2 6 Þ
Þ
l = (2 6 )1/11
log 7 x = ± 3
x = 7 3 or x = 7 -3
1
Then, the least value of x is 3 i.e., 7 -3.
7
6. Q x = log 5(5 3 ´ 8) = 3 + log 5 8
Þ
and
l = 2 6 /11
…(i)
Given that, l = b a and a, b Î N
1
l = ab
Þ
…(ii)
From Eqs. (i) and (ii), we get
a = 2 6 and b = 11
Þ
3. x
2
log10
x + log10 x 3 + 3
=
a + b = 64 + 11 = 75
2
1
1
x + 1 -1
x + 1 +1
Given, a, b and c are real solution Eq. (i) and a > b > c and for
Eq. (i) to be defined x > 0, x > - 1 Þ x > 0 from Eq. (i),
2x
log 2 x + 3 log10 x + 3
x 10
=
2
On taking logarithm both sides on base 10, then
2
(log10
x + 3 log10 x + 3 ) log10 x = log10 x
Þ
2
(log10
x
1
log 3(3 3 ) × log x 7 = log 3 x × log 7 3
3
9 × log x 7 = log 7 x
9 = (log 7 x ) 2
Þ
Þ
11 log x l - 6 log x 2 = 0
æ l11 ö
l11
log x ç 6 ÷ = 0 Þ 6 = 1
2
è2 ø
Þ
... (i)
Eq. (i) valid for x > 0, x ¹ 1
On solving Eq. (i),
log10 y = 2000 log10 2000 = 2000 ´ (log10 2 + 3 )
= 2000 ´ 3.3010 = 6602
So, the number of digits in 2000 2000 = 6602 + 1 = 6603.
Þ
100 × (100 + 1 )
= 5050
2
5. log 3 27 × log x 7 = log 27 x × log 7 3
y = 2000 2000
Let
100
å f (2k ) = kå= 1k =
+ 3 log10 x + 2 ) log10 x = 0
Þ(log10 x + 1 )(log10 x + 2 ) log10 x = 0
\
log10 x = - 2, -1, 0
\
x = 10 -2, 10 -1, 10 0
1 1
x=
, ,1
100 10
1
1
So, a, b, c can take values a = 1, b = , c =
(Q a > b > c )
10
100
\
a, b, c ÎGP
x - 3 = log 5 8
y = log 7 (7 3 ´ 6 ) = 3 + log 7 6
Þ
Q
y - 3 = log 7 6
8 > 6 and 7 > 5
Þ
log 8 > log 6 and log 7 > log 5
or
(log 8 )(log 7 ) > (log 6 ) (log 5 )
…(i)
…(ii)
Þ
log 5 8 > log 7 6
[from Eqs. (i) and (ii)]
Þ
x -3 >y -3
\
x >y
7. Q log 5 120 + ( x - 3) - 2 log 5 (1 - 5 x - 3 ) = - log 5( 0 .2 - 5 x - 4 )
Þ
log 5(5 ´ 24 ) + ( x - 3 )
æ 1 - 5x - 3 ö
= log 5(1 - 5 x - 3 ) 2 - log 5 ç
÷
5
è
ø
Þ
1 + log 5 24 + ( x - 3 ) = log 5 {5 × (1 - 5 x - 3 )}
Þ
1 + log 5(24 × 5 x - 3 ) = 1 + log 5(1 - 5 x - 3 )
Þ
24 × 5 x - 3 = 1 - 5 x - 3
Þ
25 × 5 x - 3 = 1
5x - 1 = 50
Þ
\
8. Given, xn > xn
-1
x -1 = 0 Þ x =1
> ××× > x 2 > x1 > 1
x
x
N 1
\ log x1 log x 2 log x 3 ××× log xn xn n - 1
= log x1 log x 2 log x 3 ××× log xn -1xn -1
= log x1 x1 = 1
x1
x n - 2N
(Q loga a = 1 )
x (y + z - x ) y (z + x - y ) z ( x + y - z ) 1
=
=
=
log z
n
log x
log y
Then,
log x = nx (y + z - x )
log y = ny (z + x - y )
and
log z = nz ( x + y - z )
9. Let
…(i)
…(ii)
…(iii)
341
Chap 04 Logarithms and Their Properties
\
logb (logb N )
y log x + x log y = y log z + z log y
14.
= z log x + x log z
Þ
Þ
10. Q y = a
x y ×y x = y z ×z y = z x × x z
1
1 - loga x
1
1 - loga x
loga y =
and
z = a 1 - loga y
1
loga z =
1 - loga y
…(i)
1
or
…(ii)
A = 1 - log 7 2
A = log 7 7 - log 7 2
7
7
49
Þ 49 A =
A = log 7 Þ 7 A =
2
2
4
1
æ1ö
B
and
B = - log 5 4 = log 5 ç ÷ Þ 5 =
è 4ø
4
49 1
50
A
B
\
49 + 5 =
+
=
= 12.5
4
4
4
16. (log16 x ) 2 - log16 x + log16 l = 0
Þ
1
æ
ö
1
1-ç
÷
è 1 - loga x ø
=1-
1ö
1
æ
Eq. (i) defined for x > 0, l > 0 ç log16 x - ÷ - + log16 l = 0
è
2ø
4
1
loga x
For exactly one solution,
1
log16 x - = 0
2
1
1
\
- + log16 l = 0 Þ log16 l =
4
4
1/ 4
or
l = (16 ) = 2
1
1
= (1 - loga z ) Þ loga x =
(1 - loga z )
loga x
1
x = a 1 - loga z
\
11. log 0.3( x - 1) < log 0.09( x - 1)
…(i)
Eq. (i) defined for x > 1,
Þ log 0.3 ( x - 1 ) - log ( 0.3)2 ( x - 1 ) < 0
1
Þ log 0.3( x - 1 ) - log 0.3 ( x - 1 ) < 0
2
1
log 0.3( x - 1 ) < 0
Þ
2
Þ
log 0.3 ( x - 1 ) < 0
Þ
( x - 1 ) > ( 0.3 )0
…(ii)
Þ
From Eqs. (ii) and (iii), we get
x > 2 Þ x Î(2, ¥ )
log b × loga b logb a
…(i)
From Eq. (i), x > 0 and x ¹ 1
By Eq. (i),
( x + 3 ) 2 = 16
... (ii)
18. Given, y = log 2 log 6(2
2x + 1
+ 4)
…(i)
From Eq. (i) to be defined,
2x + 1 > 0 Þ x > -
1
2
We find value of x for which y = 1
loga b
=a a
x
\ a - by = 0
2
x log x ( x + 3) = 16
17.
Þ
x+3=±4
Þ
x = 1 or x = - 7
From Eq. (ii), no values of x satisfy Eq. (i).
\ Number of values of x satisfy Eq. (i)
\ Number of roots = 0
[Q base of log is lie in (0, 1)]
…(iii)
x >2
12. Q a x = a
...(i)
2
From Eqs. (i) and (ii), we get
loga z =
= a loga (logb N ) = logb N
logb a
15. 49 A + 5 B = ?
log( x y × y x ) = log(y z × z y ) = log( x z × z x )
Þ
a
=a
loga b logb a = b logb a
\
= by
Þ
13. Q x = 1 + loga bc = loga a + loga bc = loga (abc )
1
= logabc a
\
x
1
Similarly,
= logabc b
y
1
and
= logabc c
z
On adding Eqs. (i), (ii) and (iii), we get
1 1 1
+ + = logabc abc = 1
x y z
xy + yz + zx
xyz
= 1 or
=1
Þ
xyz
xy + yz + zx
Þ
…(i)
…(ii)
…(iii)
Þ
1 = log 2 log 6 (2
log 6(2
2
2x + 1
+ 4)
+ 4) = 2
2x + 1
2
2x + 1
+ 4 = 36
2x + 1
= 32 = 2 5 Þ 2 x + 1 = 5
Þ
2 x + 1 = 25 Þ x = 12
So, required point is (12, 1).
19. Given that, log2 = 0.301
log3 = 0.477
Let
y = 312 ´ 2 8
log y = 12 log 3 + 8 log 2
= 12 ´ ( 0.477 ) + 8 ( 0.301 ) = 8.132
So, number of digits before decimal in 312 ´ 2 8 = 8 + 1 = 9
... (ii)
342
Textbook of Algebra
20. Given, equation 2 log x a + logax a + 3 loga 2 x a = 0
...(i)
Þ
(3 x - 1 - 3 ) (3 x - 1 - 1 ) = 0
2
1
3
+
+
=0
loga x 1 + loga x 2 + loga x
…(ii)
Þ
Þ
x - 1 = 1 or x - 1 = 0
x = 2 or x = 1
Þ
loga x = t
Let
Then, Eq. (ii),
y = log p log p ( p p ×××p p )
1424
3
25.
2
1
3
+
+
= 0 Þ 6t 2 + 11t + 4 = 0
t 1+t 2+t
4
1
t = - or 3
2
x = a -4 / 3 or x = a -1 / 2
Þ
So,
1ü
ü
ì
ì
ï
ï
ï1
ï
= log p í log p ( p ×××p p ) p ý = log p í log p ( p p ×××p p )ý
14243ï
p
14243ï
ï
ï
(n - 1) times þ
(n - 1) times þ
î
î
ì
ü
ï1 1
ï
= log p í × log p ( p p ×××p p )ý
p
p
1
4
2
4
3
ï
ï
(n - 2 ) times þ
î
æ 1 ö
= log p ç n ÷ = - n, log1/p pn = -n
èp ø
Two value of x possible for which Eq. (i) is defined and satisfy.
21. Decimal on x > 0 and x ¹ 1.
Taking logarithm on both sides on base 2, we get
{(log 2 x ) 2 - 6 log 2 x + 11 } log 2 x = 6
26. loga x = a, logb x = b, logc x = g, logd x = d
log 2 x = t
t - 6t + 11t - 6 = 0
Let
\
3
Þ
Þ
Þ
2
(t - 1 ) (t - 2 ) (t - 3 ) = 0 Þ t = 1, 2, 3
log 2 x = 1, 2, 3
x = 2, 2 2, 2 3
22. log l x × log 5 l = log x 5
…(i)
Þ
l ¹ 1, l > 0 and x > 0, x ¹ 1
log 5 x = log x 5
Þ (log 5 x ) 2 = 1
Þ
log 5 x = ± 1
\
x = 5 and
x = 51 and 5 -1
Þ
1
5
…(i)
y = log x 3 x , x ¹ 1
Þ
3x £ 1
Þ
…(ii)
…(iii)
3x £ 1
x£
x >1
24. Given equation,
log 2(9 x - 1 + 7 ) = 2 + log 2(3 x - 1 + 1 )
Þ
Þ
Þ
log x b = b
log x c = g -1
…(iii)
Þ
log x d = d-1
…(iv)
…(ii)
On adding Eqs. (i), (ii), (iii) and (iv) , we get
1 1 1 1
log x (abcd ) = + + +
a b g d
1
\ logabcd x = -1
-1
a + b + g - 1 + d- 1
…(v)
{3 2 ( x - 1) + 7 }
=2
log 2 ( x - 1)
+1
3
3 2 ( x - 1) + 7 = 4 × {3 ( x - 1) + 1 }
{3 ( x - 1) } 2 - 4 × 3 ( x - 1) + 3 = 0
4
a+b+g+d
³ -1
4
a + b - 1 + g - 1 + d- 1
1
a+b+g+d
£
16
+ b - 1 + g - 1 + d- 1
or
27. Q
a -1
a+b+g+d
16
log10 5 = a and log10 3 = b
logabcd x £
æ 10 ö
log10 2 = log10 ç ÷ = 1 - a
è5ø
[from Eq. (v)]
…(i)
…(ii)
Option (a)
\
log10 8 = 3 log10 2 = 3 (1 - a )
log10 15
log10(5 ´ 3 )
Option (b) log 40 15 =
=
log10 40 log10 (2 3 ´ 5 )
1
3
Now, for x > 1, from Eq. (iii), 3 x ³ 1 Þ x ³
Þ
…(i)
Þ
\
No solution for this case.
\
-1
or
To be defined y, 3 x > 0 Þ x > 0
and
log x 3 x ³ 0
From Eqs. (i) and (iii),
æ1 ö
for
x Î ç , 1÷ Þ
è3 ø
log x a = a -1
AM ³ HM Þ
log 3 x > - 1
1
x>
3
Let
Þ
For a, b, g, d
23. S = { x : log x 3x : log 3 x > - 1}
Þ
[ p > 0, p ¹ 1 ]
n times
1
3
=
log10 5 + log10 3
log10 2 3 + log10 5
=
a+b
(a + b )
=
3 (1 - a ) + a
(3 - 2a )
[from Eq. (ii)]
log10 2
5
Option (c) log 243 32 = log 35 2 5 = log 5 2 =
log10 2
5
1 -a
[from Eqs. (i) and (ii)]
=
a
Hence, all options are correct.
Chap 04 Logarithms and Their Properties
28. Q x > 0 and x ¹ 1
Sol. (Q. Nos. 33 to 35)
Given, loga x , logb x and logc x are in AP.
Þ
2 logb x = loga x + logc x
2 log x log x log x
=
+
Þ
log b
log a
log c
Þ
2
1
1
=
+
log b log a log c
Þ
2 (log a ) (log c )
(log a + log c )
log b
2 log c
=
log a log a + log c
Þ
Þ
loga b =
é log x ¹ 0 ù
ê\ x ¹ 1 ú
ë
û
So,
Given equation, log| b|
…(i)
…(ii)
\
G = 18
33. The value of G = 18
35. The value of SKG = 16 ´ 8 ´ 18 = 16 ´ 144 = 2304
Sol. (Q. Nos. 36 to 38)
x
= | b |1
a
U = Number of digits in (60 )100
Þ
| x | = | ab |
[from Eq. (ii)]
Þ
| x | = | b|
\
x=±b
Sol. (Q. Nos. 30 to 32)
Q
log 2 N = a 1 + b 1
Þ
b1 = log 2 N - a1
Given,
0 £ b1 < 1 Þ 0 £ log 2 N - a1 < 1
Þ
a1 £ log 2 N < 1 + a1
…(i)
Þ
2a 1 £ N < 21 + a 1
1+a2
So,
x = 7 + 1, x = 8
Number of positive integers which have characteristic 2, when the
base of logarithm is 3
= 3 2 + 1 - 3 2 = 18
34. The value of KG = 8 ´ 18 = 144
x
x
- 1 = 0, log| b|
=1
a
a
Þ
Similarly,
3a 2 £ N 3
…(ii)
and
5a 3 £ N < 51 + a 3
…(iii)
Let
2 5 £ N < 2 6 and 3 3 £ N < 3 4
\ Common values of N are 32, 33, 34, K, 63
Number of integral values of N are 32.
31. Here, a1 = 6, a 2 = 4 and a 3 = 3, then from Eqs. (i), (ii) and (iii),
Þ 64, 65, 66, ..., 127, 81, 82, 83,..., 242 and 125, 126, ..., 624
\ Largest common value = 127
32. Here, a1 = 6, a 2 = 4 and a 3 = 3
From question number 31, we get
64, 65, 66,..., 127; 81, 82, 83, ..., 242 and 125, 126, ...,624
\ Largest common value = 127
and smallest common value = 125
\ Difference = 127 - 125 = 2
a = (60 )100
log10 a = 100 log10 60 = 100 (1 + log10 2 + log10 3 )
= 100 (1.778 )
log10 a = 177.8
So,
…(i)
U = 177 + 1 Þ U = 178
M = Number of cyphers after decimal, before a significant
figure comes in (8 ) -296
Let
b = (8 ) -296
log10 b = ( -296 ) log10 8 = ( -296 ) ´ 3 log10 2
log10 b = ( -296 ) ´ 3 ´ ( 0.301 )
= - 267.288 = - 267 - 0.288
= - 267 - 1 + (1 - 0.288 ) = - 268 + 0.712
log10 b = 268.712
30. Here, a1 = 5 and a 2 = 3, then from Eqs. (i) and (ii),
2 6 £ N < 2 7, 3 4 £ N < 3 5 and 5 3 £ N < 5 4
a = 610
log a = 10 log10 6 = 10 ( 0.301 + 0.477 )
= 10 ( 0.778 )
log(610 ) = 7.78
29. | a | < |b |, b - a < 1
a + b = | a| ü
ý
ab = - |b|þ
…(i)
[Q log10 2 = 0.301, log10 3 = 0.477 ]
log c 2
= log (ac ) c 2
log(ac )
a, b Î x 2 - | a | x - | b | = 0
S = 16
K = Number of digits in 610
Let
c 2 = (ac ) loga b
\
S = Antilog of (0.5) to the base 256
log 256 S = 0.5
S = (256 ) 0.5 = (2 8 )1/2
S = 24
log b =
Also,
343
\
M = 268 - 1 = 267
U 178
Now,
=
M 267
According to the question,
U 2
=
M 3
U
p
=
Þ
M q
So,
p =2
and
q =3
36. The value of p = 2.
37. The value of q = 3.
38. The equation whose roots are p and q is x 2 - 5x + 6 = 0.
344
Textbook of Algebra
Sol. (Q. Nos. 39 to 41)
According to question, G,O, E , L > 0 and are real numbers.
Such that,
log10(G × L ) + log10(G × E ) = 3 Þ log10 G 2LE = 3
Þ
and
Þ
G 2LE = 10 3
log10 E × L + log10 E × O = 4
log10 E 2 × L × O = 4
...(i)
Þ
E 2 × L × O = 10 4
and log10(O × G ) + log10(O × L ) = 5
Þ log10 O 2 GL = 5 Þ O 2 GL = 10 5
...(ii)
...(iii)
...(iv)
39. Now, let
1
= (log l )
+
1 1
+ + ×××
4 8
1/ 2
1 - 1/ 2
40. Minimum of 3G + 2L + 2O + E = 2 l3m5 u
8 ´ 8 G 3L 2O 2E = 2 l3m5 u
…(v)
8 ´ (10 8 )1/8 = 2 l3m5 u
8 ´ 10 = 2 l3m5 u
2 4 ´ 51 = 2 l3m5 u
l = 4, u = 1, m = 0
l
S( l + m ) = ( 4 0 + 0 4 ) + ( 01 + 1 0 ) + (1 4 + 41 )
= (1 + 0 ) + ( 0 + 1 ) + 1 + 4 = 7
æG ö
æO ö
æG ö
41. log10 ç ÷ + log10 ç ÷ = log10 ç ÷ = log10 1 = 0
èEø
èEø
èO ø
[divide Eq. (iv) and Eq. (ii) of Q. 39]
G
O
æ1ö
P = log10 × log10 = log ç ÷ log(10 ) = - 1
è 10 ø
O
E
[by dividing Eq. (i) by Eq. (ii) and dividing Eq. (iii) by Eq. (iv) in
Q. 10]
2
2
= x - 0 × x + ( -1 ) = 0 = x - 1
42. log10(2x ) + log10 y = 2 Þ 2xy = 10
and
Þ
…(i)
log10 x 2 - log10 2y = 4
x2
= 10 4
2y
Þ
…(i)
a (b - c ) X 2 + b (c - a ) X + c (a - b ) = 0
Q
a (b - c ) + b (c - a ) + c (a - b ) = 0
\
X =1
Q Eq. (i) is perfect square.
\ Roots are equal.
c (a - b )
1 ´1 =
\
a (b - c )
2ac
b=
Þ
a+c
G 3L2O 2E = 10 8
2
[given]
æxö
æxö
a (b - c ) ç ÷ + b (c - a ) ç ÷ + c (a - b ) = 0
èy ø
èy ø
x
Let
=X
y
(equality hold, if G = L = O = E )
From Eqs. (i) and (iii) of Q. 10, we get
m
Given, AB = 0.5
Þ
log 5 ( l + 4 ) - log 5 l = 0.5
l+4
= (5 )1/2 = 5
Þ
l
4
( 5 + 1)
l=
=4
Þ
5 -1
4
2
where l, m, u ÎW
Apply AM ³ GM for 3G, 2 L, 2O, E
3G + 2 L + 2O + E 8 3
³ G ´ L2 ´ O 2 ´ E
8
\
A º ( l, log 5 l ), l > 0
and solving x = l and y = log 5( x + 4 ), we get
\
a = 1 and b = 5
Then,
a + b =1 + 5 =6
44. \ a (b - c ) x 2 + b (c - a ) xy + c (a - b ) y 2 = b, y ¹ 0
= (log l )
= log 10 = 4 log 10 = 4
From Eq. (v),
43. Solving, x = l and y = log 5 x, we get
=1 + 5 =a + b
4
So,
(given)
B º { l, log 5( l + 4 )}, l > - 4
From Eqs. (i), (ii) and (iii), we get
G 3O 3E 3L3 = 1012
GOEL = 10 4
Þ
l = 10 4
y = log l log l log l ××× = (log l ) 2
From Eqs. (i) and (ii), x 3 = 10 6 Þ x = 100
1
From Eq. (i),
y =
2
1 201 m
x + y = 100 + =
=
\
n
2
2
\
m = 201 and n = 2
Þ m - 3n 6 = 201 - 3 (2 ) 6 = 201 - 192 = 9
Now, log(a + c ) + log(a - 2b + c )
= log {(a + c ) 2 - 2b (a + c )}
= log {(a + c ) 2 - 4ac }
[from Eq. (ii)]
2
Þ
= log(a - c ) = 2 log(a - c )
log(a + c ) + log(a - 2b + c )
=2
log(a - c )
2
ì log(a + c ) + log(a - 2b + c )ü
\ í
ý =4
log(a - c )
þ
î
45. According to the question, x Î I
| x + 2|
æ 1 ö 2 - | x|
Given equation, ç ÷
> 9 [ x ¹ ± 2]
è3ø
Þ
…(ii)
…(ii)
3
| x + 2|
2 - | x|
> 32 Þ -
| x + 2|
>2
2 -| x |
…(i)
Chap 04 Logarithms and Their Properties
| x + 2|
-2 > 0
|x | - 2
| x + 2| - 2 | x | + 4
Þ
>0
| x| - 2
x - 2 + 2x + 4
Case I If x < - 2, >0
-x - 2
Þ
Þ
\
Þ
x
x+2
…(ii)
++
0
– – – –
++++
which is not possible.
Case II -2 < x < 0, then Eq. (ii)
x + 2 + 2x + 4
3x + 6
>0 Þ
>0
Þ
-x - 2
-(x + 2)
-3 ( x + 2 )
> 0-3 > 0
(x + 2)
(log 2 x - 2 ) (log 2 x + 1 ) < 0
- 1 < log 2 x < 2
2 -1 < x < 2 2
1
<x<4
Þ
2
Þ
x Î I , so x = 1, 2, 3
So, number of integer value of x is 3.
48. Given that, b > 0
2 log1/25 (bx + 28 ) = - log 5 (12 - 4 x - x 2 )
2
log 5 (bx + 28 ) = - log 5 (12 - 4 x - x 2 )
(- 2)
2 < x <6
So, the integer values of x = 3, 4, 5
So, the number of integer values of x is 3.
46. x > 2
y–1
Þ
and
and
…(i)
…(ii)
+
3 3
49.
Case I
y < 1, then x < 3
Eq. (ii) becomes - 2y + 2 - y + 2 = 2
2 log1 / 4 2 - 3 log 27 125 - 4 2 4 log 2 2 - 3 log 3
=
4 log 2 21
7 4 log 49 2 - 3
7 7
=
2
- 3y = - 2, y =
3
2
3
x = 32 / 3
log 3 x =
Þ
bx + 28 > 0
12 - 4 x - x 2 > 0
b+4=±8
b = 4 or b = - 12
Since,
b > 0 so b = 4
for this value x > - 7 and - 6 < x < 2
2
+
bx + 28 = 12 - 4 x - x 2
…(ii)
x 2 + ( 4 + b ) x + 16 = 0
- 28
and
and - 6 < x < 2
x>
b
Since, Eq. (i) has coincident roots, so discriminant Eq. (ii) is zero.
( 4 + b ) 2 - 64 = 0
1
–
which is less than 2, so not acceptable.
Case II
1 < y < 2, then 3 < x < 9
From Eq. (ii), 2 (y - 1 ) - (y - 2 ) = 2
…(i)
Þ
+
–
…(i)
Þ
Þ
Þ
which is not possible.
Case III when x > 0
From Eq. (ii),
x + 2 - 2x + 4
-x + 6
>0 Þ
>0
x -2
x -2
x -6
<0
x -2
–
[acceptable]
From Eq. (i),
x>0
1
2
Eq. (i) Þ log 2 x log 22 x + 1 > 0
2
(- 2)2
1
1
Þ
log 2 x - log 22 x + 1 > 0
2
2
Þ
(log 2 x ) 2 - (log 2 x ) - 2 < 0
x+2
> 0 Þ -1 > 0
-(x + 2)
y–2
[impossible]
Case III
y ³ 2, then x ³ 9
From Eq. (ii), 2 (y - 1 ) + (y - 2 ) = 2
\
y = 2, log 3 x = 2
\
x =9
47. Given equation is
log 2 x - 2 log12/4 x + 1 > 0
– – –2
| log 3 x - 2 | + | log 3 x - 2 | = 2
| 2 log 3 x - 2 | + | log 3 x - 2 | = 2
2 | log 3 x - 1 | + | log 3 x - 2 | = 2
Let
log 3 x = y
Then, Eq. (i) Þ2 | y - 1 | + | y - 2 | = 2
y =2
log 3 x = 2
x = 32 = 9
345
24 - 5 - 4
7
[from Eq. (i)]
æ5ö
èxø
2 log 7 2
-3
=
50. (log 5 x ) 2 + log 5x ç ÷ = 1, x > 0, x ¹
Þ
5
-4
16 - 9
=7
22 - 3
1
5
æ5ö
log 5 ç ÷
èxø
1 - log 5 x
= 1 Þ (log 5 x ) 2 +
(log 5 x ) 2 +
=1
log 5 (5 x )
1 + log 5 x
Let log 5 x = t, then
346
Textbook of Algebra
t2 +
1 -t
=1
1+t
9 < 15 < 27
2 < log 3 15 < 3
So, [ N ] = 2 (q)
(D) (52.6 )a = ( 0.00526 )b = 100
t 3 + t 2 - 2t = 0
Þ
Þ
Þ
t (t + 2 ) (t - 1 ) = 0 Þ t = - 2, 0, 1
x = 5 -2, 5 0, 51
1
x = , 1, 5
Þ
25
x1, x 2 Î I
\
x1 = 1, x 2 = 5
\
| x 2 - 4 x1 | = | 5 - 4 | = 1
1
51. Given, x = log l a = loga b = logb c and log l c = nxn + 1
2
x = log l a = loga b = logb c and log l c = nxn +1
(52.6 )a = 100 and ( 0.00526 )b = 100
2
52.6 = 10a
b
(52.6 ) ´ 10
(52.6 ) = 10 2 + 4 b
...(i)
log 3 243
log 3 3 5
5 ´2
=
=
= 2 (p,s)
log 2 32 - 1 log 2 5
5
2
2
2 log 6
2 log 6 2 log 6
(B)
=
=
= 1 (p)
log 12 + log 3 log 36 2 log 6
+
= - log 3 3 4 = - 4 (q)
log 5 16 - log 5 4
log 5 (2 ) 2 2
=
= (r)
=
7
log 5 128
log 5 (2 )
log 5 (2 ) 7 7
53. (A) log (20× 5) 2 8 = log12/2 8 = (log 2-1 2 3 ) 2
\
2
æ 3
ö
= ç
log 2 2 ÷ = ( - 3 ) 2 = 9 = 3 (r)
è-1
ø
(B) (log10 2 ) 3 + log10 8 × log10 5 + (log10 5 ) 3
Þ
Þ
= (log10 2 ) 3 + 3 log10 2 log10 5 + (log10 5 ) 3
= (log10 2 ) 3 + 3 × log10 2 × log10 5 × (log10 2 + log10 5 )
Þ
+ (log10 5 ) 3
[Q log10 2 + log10 5 = log10 10 = 1 ]
= (log10 2 + log10 5 ) 3 = (log10 10 ) 3 = (1 ) 3 = 1
3 + (log10 2 ) 3 + log10 8 × log10 5 + (log10 5 ) 3
1
6
= log 2 15 ( - log 6 2 ) ( - log 3 6 )
log 15 log 2 log 6
=
´
´
= log 3 15
log 2
log 6 log 3
+
2
–1
–
5
x > 0 and x ¹ 1
x Î ( 0, 1 ) È (1, 2 ) È (5, ¥ )
x Î( 0, 1 )
1
>1
x
2 (x - 2)
By Eq. (i), log 1
³1
(
x
+ 1) (x - 5)
x
Let
So,
Case I
æ 16 ö
log 5 ç ÷
è4ø
…(i)
x Î ( - 1, 2 ) È (5, ¥ )
then
–
(C) N = log 2 15 × log1/6 2 × log 3
2 + 4b
2
-2
= 3 + 1 = 4 (s)
…(ii)
10a = 10 b
2 2
= +4
Þ
a b
1 1
Þ
- = 2 (q)
a b
2 (x - 2)
54. (A) Given that, log1/x
³1
(x + 1) (x - 5)
(x - 2)
for log to be defined
> 0,
(x + 1) (x - 5)
52. (A)
(D)
æ2 + 4b ö
ç
÷
è b ø
From Eqs. (i) and (ii), we get
n =2
æ1ö
(C) log1/3 ç ÷
è9ø
52.6 = 10
Þ
log l c = nxn + 1
Þ
= 10
…(i)
2
b
From Eq. (i), log l a ´ loga b + logb c = x 3
1
log l c = x 3, log l c = x 3
2
log l c = 2 x 3
Compare with
- 4b
…(ii)
2 (x - 2)
1
³
(x + 1) (x - 5) x
2 (x - 2)
1
- ³0
(x + 1) (x - 5) x
2x (x - 2) - (x + 1) (x + 5)
³0
x (x + 1) (x - 5)
2x 2 - 4x - x 2 + 4x + 5
³0
x (x + 1) (x - 5)
Þ
Þ
x2 + 5
³0
x (x + 1) (x - 5)
Þ
x (x + 1) (x - 5) > 0
–1
–
+
+
0
–
Þ
x Î ( - 1, 0 ) È (5, ¥ )
But by Eq. (ii), x Î( 0, 1 )
So, no solution for this case.
5
347
Chap 04 Logarithms and Their Properties
Case II Let x Î (1, 2 ) È (5, ¥ )
…(iii)
1
<1
x
Eq. (i) Þ log 1
x
x Î (1, 2 ] È [3, 4 ) (q, r)
(D) Given equation is
( l2 - 3 l + 4 ) x 2 - 4 (2 l - 1 ) x + 16 = 0
2 (x - 2)
³1
(x + 1) (x - 5)
2
So,
+
…(iv)
–
0
Þ
log 23 x - log 3 x - 2 ³ 0
Þ
2
log 1
2
Þ
-
Þ
Þ
…(i)
2
(x - 4) (x - 1)
£1
2
(x - 4) (x - 1) ³ - 2
x 2 - 5x + 4 + 2 ³ 0
–
3
( x - 3 )( x - 2 ) ³ 0
x £ 2 or x ³ 3
From Eq. (i) to be defined, 4 - x > 0 and x - 1 > 0
x < 4 and x > 1
7 log 7
( x 3 + 1)
- x2 = 1
x + 1 - x = 1 ì for this x + 1 > 0
ï
x3 - x2 = 0 íÞ x3 > - 1
x 2 ( x - 1 ) = 0 ïî Þ x > - 1
3
+
2
Þ
log10 x < log 3 x < loge x < log 2 x
and for
0 < x <1
We get,
log10 x > log 3 x > loge x > log 2 x
It is clear that for
x > 0, x ¹ 1
Statement-1 is false.
56. Statement-1
x 2 - 5x + 6 ³ 0
+
…(iii)
Statement-1 If x > 1
Þ
log x a < log x b
\Statement-2 If 0 < x < 1
Þ
log x a > log x b
\ Statement-2 is true, also
10 > 3 > e > 2
If
x > 1,
then log x 10 > log x 3 > log x e > log x 2
1
1
1
1
Þ
<
<
<
log x 10 log x 3 log x e log x 2
(4 - x ) (x - 1)
³0
2
Þ
(l - 3) (l - 8) < 0
3 < l <8
From Eqs. (ii) and (iii), we get
3 < l < 8 Þ l Î(3, 8 ) (s)
55. If 0 < a < b
(log 3 x - 2 ) (log 3 x + 1 ) ³ 0
log 3 x £ - 1
log 3 x ³ 2
1
Þ
x £ or x ³ 9
3
From Eq. (i), x > 0
æ 1ù
So,
x Î ç 0, ú È [9, ¥ ) (p)
è 3û
2
f (1 ) < 0 by graph of f ( x )
…(i)
Þ
Þ
or
(C) log 1 ( 4 - x ) ³ log 1 2 - log 1 ( x - 1 )
…(ii)
l2 - 11 l + 24 < 0
3 æ - 2ö
´ç
÷ ´2 ´1
2 è 3 ø
log 3 x - log 32 x + 2 £ 0
D>0
15
l>
8
and
\
defined, when x > 0
Þ
β
Let f ( x ) = ( l2 - 3 l + 4 ) x 2 = 4 (2 l - 1 ) x + 16
5
From Eqs. (iii) and (iv), x Î(1, 2 ] (q)
3
(B) log 3 x - log 23 x £ log 1 4
2
2 2
log 3 x - log 32 x ³
α
Þ We get
+
–1
l2 - 3 l + 4 > 0, " l Î R
[by case I]
Þ
x (x + 1) (x - 5) < 0
Þ
x Î ( - ¥, - 1 ) È ( 0, 5 )
Eq. (iii), x Î (1, 2 ) È (5, ¥ )
–
…(i)
2
1
x2 + 5
£0
x (x + 1) (x - 5)
Þ
2
3ö
7
æ3ö
æ
æ3ö
l2 - 3 l + 4 = l2 - 3 l + ç ÷ - ç ÷ + 4 = ç l - ÷ +
è2ø
ø
è
è2ø
2
4
2 (x - 2)
1
£
(x + 1) (x - 5) x
2 (x - 2)
1
- £0
(x + 1) (x - 5) x
Þ
From Eqs. (ii) and (iii),
…(ii)
…(iii)
2
3
x = 0 (repeated) or x = 1
Thus, Eq. (i) has 2 repeated roots.
\Statement-1 is false.
Statement-2 a loga N = N , a > 0, a ¹ 1 and N > 0
which is true.
…(i)
348
Textbook of Algebra
æ1ö
è3ø
7
æ1ö
è3ø
4
7
æ1ö
æ1ö
loge ç ÷ < loge ç ÷
è3ø
è3ø
1
1
7 loge < 4 loge
3
3
1
Now,
loge < 0
3
So,
7>4
Statement-1 is false.
Statement-2
ax < ay
and
a < 0, x > 0, y > 0
Eq. (i) divide by a, we get x > y
Statement-2 is true.
58. Statement-1
x log x
(1 - x ) 2
4
and
[Q2 < e < 3]
=9
ì Eq. ( i) is defined, if
í
x ¹ 1, x > 0
î
1-x=±3
\
x = - 2 or 4
x=4
\ Eq. (i) has only one solution.
Statement-1 is false.
Statement-2 a loga b = b, where a > 0, a ¹ 1, b > 0
\
log 168 log (2 3 ´ 3 ´ 7 )
=
log 54
log ( 3 3 ´ 2 )
=
3 log 2 + log 3 + log 7
3 log 3 + log 2
=
3 + log 2 3 + log 2 7 3 + l + m
=
3 log 2 3 + 1
3l + 1
3+
=
…(i)
2
Þ
(log x ) + 2 log x - 3 = 0
Þ
Þ
Þ
(log x + 3 ) (log x - 1 ) = 0
log x = - 3 or log x = 1
x = 10 -3 or x = 10
=
\
Now,
Q LHS has domain x Î R and RHS has domain x Î ( 0, ¥ )
\ Statement-2 is false.
60. Statement-1
log x 3 × log x / 9 3 = log 81 3
Eq. (i) holds, if x > 0, x ¹ 1, x ¹ 9
1
1
1
By Eq. (i),
×
=
log 3 x (log 3 x + 2 ) 4
(ab + 1 )
a (8 - 5b )
ab = log 5 4
log 5 10 2 log 5 10
log 3 10 =
=
log 5 3
2 log 5 3
…(i)
62. Q
ln a
ln b
ln c
=
=
b -c c -a a -b
(i) Q
(log 3 x + 2 ) 2 = 8
[by using law of proportion]
ln a
ln b
ln c
=
=
b -c c -a a -b
=
ln a + ln b + ln c
ln (abc )
=
b -c + c -a + a -b
0
Þ ln (abc ) = 0 Þ abc = 1
log 3 x = 2 ( - 1 ± 2 )
2)
Two values of x satisfying Eq. (i)
So, Statement-1 is false.
Statement-2 Change of bases in logarithm is possible.
\ Statement-2 is true.
…(i)
log 5 (100 ) log 5 ( 4 ´ 25 )
=
2b
2b
log 5 4 + 2 ab + 2
[from Eq. (i)]
=
=
2b
2b
(log 3 x ) 2 + 2 log 3 x + 4 = 8
x = 32 (- 1 ±
3 - 2b
1
+
b - 1 a (b - 1 )
3 (3 - 2b )
+1
b -1
=
(log 3 x ) 2 + 2 log 3 x - 4 = 0
log 3 x + 2 = ± 2 2
…(ii)
(ii) Q a = log 3 4 and b = log 5 3
Eq. (i) is defined for x > 0.
So, Eq. (i) has 2 distinct solutions.
Statement-2 log x 2 ¹ 2 log x
\
log 54 168 =
[acceptable]
which is true.
59. Statement-1 (log x ) 2 + log x 2 - 3 = 0
…(i)
Let log 2 3 = l and log 2 7 = m
2+l
From Eq. (i), a =
m
3+l
and from Eq. (ii), b =
, we get
2+l
1
3 - 2b
and m =
l=
a (b - 1 )
b -1
...(i)
(1 - x ) 2 = 9
log 12 2 log 2 + log 3
=
log 7
log 7
2 + log 2 3
a=
log 2 7
log 24 3 log 2 + log 3
b = log12 24 =
=
log 12 2 log 2 + log 3
3 + log 2 3
=
2 + log 2 3
61. (i) Qa = log 7 12 =
57. Statement-1 ç ÷ < ç ÷ . Taking log on both sides,
(ii)
a ln a + b ln b + c ln c
ln a
ln b
ln c
=
+
+
b - c c - a a - b a (b - c ) + b (c - a ) + c (a - b )
ln aa + ln bb + ln cc
ln (aa × bb × cc )
=
0
0
a b c
Þ ln (a b c ) = 0
=
Þ
aabbcc = 1
Chap 04 Logarithms and Their Properties
(iii)
(ii) a = 6 20
ln a
ln b
ln c
=
=
b -c c -a a -b
\ log a = 20 log 6 = 20 (log 2 + log 3 )
= 20 (0.310 + 0.477)
= 20 ´ 0.778 = 15.560
So, number of integers in 6 20 = 15 + 1 = 16
[(b 2 + bc + c 2 ) ln a + (c 2 + ca + a 2 ) ln b
=
+ (a 2 + ab + b 2 ) ln c ]
[(b 2 + bc + c 2 ) (b - c ) + (c 2 + ca + a 2 ) (c - a )
+ (a 2 + ab + b 2 ) (a - b )]
=
=
ln ab
ln (ab
+ bc + c 2
2
2
2
+ ln bc + ca + a + ln ca + ab + b
3
3
(b - c ) + (c 3 - a 3 ) + (a 3 - b 3 )
2
+ bc + c 2
× bc
2
+ ca + a 2
× ca
2
+ ab + b 2
2
+ bc + c 2
ab
2
× bc
+ bc + c 2
2
+ ca + a 2
× bc
2
× ca
+ ca + a 2
2
+ ab + b 2
2
× ca
Þ
b
65. Given that, log10 2 = 0.301
)=0
and
=1
(iv)Q AM ³ GM
a+b+c
³ (abc )1 / 3 = (1 )1 / 3 = 1
\
3
a+b+c
\
³1 Þ a + b + c ³3
3
(v) Q AM ³ GM
aa + b b + c c
³ (aa × bb × cc )1 / 3
Þ
3
= (1 )1/3 = 1
a
[from Eq. (i)]
ab
2
[from Eq. (ii)]
c
+ bc + c 2
+ bc
2
+ ca + a 2
+ ca
2
+ ab + b 2
3
³ (ab
2
+ bc + c
2
× bc
2
+ ca + a
2
× ca
2
+ ab + b
1/ 3
= (1 )
2
)1 / 3
[from Eq. (iii)]
=1
Þ
Þ
ab
2
+ bc + c 2
+ bc
2
+ ca + a 2
+ ca
2
+ ab + b 2
3
a
b 2 + bc + c 2
+b
c 2 + ca + a 2
1
3
63. To prove log10 2 lies between and
+ ca
2
log103 = 0.477
log 3.375 = log(3375 ) - log 10 3 = log 5 3 ´ 3 3 - 3 log 5 ´ 2
= 3 log 5 + 3 log 3 - 3 log 5 - 3 log 2
= 3 (0.477) - 3 (0.301) = 3 (0.176)
= 0.528
æ1ö
66. Let P = log 2 x - log x (0.125) = log 2 x - log x ç ÷
è8ø
= log 2 x + 3 log x 2
AM ³ GM
log 2 x + 3 log x 2
Þ
³ (log 2 x ) (3 log x 2 ) = 3
2
P
³ 3
\
2
P ³2 3
Þ
\
a +b +c
³ 1 Þ aa + b b + c c ³ 3
3
(vi)Q AM ³ GM
log a = - 500 log 3 = - 500 ´ (0.477) = - 238.5
= - 239 + 0.5 = 239.5
So, number of zeroes after the decimal in
3 -500 = 239 - 1 = 238
)
+ ab + b 2
a = 3 - 500
(iii) Let
2
0
Þ ln (ab
\
2
349
+ ab + b 2
\ Least value of log 2 x - log x (0.125) is 2 3.
1
1
67. Let y =
= log p 3 + log p 4
+
log 3 p log 4 p
= log p 12
Now,
log p 12 > log p p 2 \ y > 2
³1
³3
68. (i) \ x
Þ
212 = 4096
= 10 x
…(i)
x×x
x [x
log x
10
x ¹ 0, so x
1000 < 4096 < 10000
10 3 < 212 < 10 4
log x
10
Þ
Taking logarithm to the base 10,
log10 10 3 < log10 212 < log10 10 4
1
1
3 < 12 log10 2 < 4 Þ < log10 2 <
4
3
64. log 2 = 0.301
log a = 200 log 5 = 200 (log 10 - log 2 ) = 200 (1 - 0.301)
= 200 ´ 0.699 = 139.8
200
So, number of integers in 5 = 139 + 1 = 140.
1 + log x
10
Þ
1
4
log 3 = 0.477
(i) Let a = 5 200
12 > p
2
x
log x
10
= 10 x
- 10 ] = 0
- 10 = 0
log
10 x
= 10
Þ
Þ
log10 x = log x 10
(log10 x ) 2 = 1
Þ
Þ
log10 x = ± 1
x = 10 ± 1
Þ
x = 10 or
log 2 (9 + 2 x ) = 3
(ii)
Þ
Þ
9 + 2x = 8
2x = - 1
which is not possible, so x Î f.
1
10
[Q x > 0]
350
Textbook of Algebra
Þ
2 × x log 4 3 + 3 log 4 x = 27
(iii)
2 ×3
log 4 x
log 4 x
= 27
( x + 1)
= 33
+3
3 log 4
éQ a
ëê
log c
b
x
Let
Þ
log10 x + 5
3
10
Þ
x
log10 ( x - 3 ) 2 = log10 ( x 2 - 21 )
( x - 3 ) 2 = x 2 - 21
Defined for x > 0,
log 2 x + 4 = log x 2 5
5
log 2 x + 4 =
log 2 x
(y + 5 ) (y - 3 ) = 0
y = - 5 or y = 3
1
x = 5 or x = 10 3
10
(log 2 x ) 2 + 4 log 2 x - 5 = 0
x = {10 -5, 10 3}
(log 2 x + 5 ) (log 2 x - 1 ) = 0
log 2 x = - 5 or log 2 x = 1
x = 2 -5 or x = 21
1
x = or x = 2
\
32
which satisfy Eq. (i).
(x)
loga x = x
and
a = x log 4 x
Þ
Þ
1
Þ x =9 2
2
x = 3 -1
1
x=
3
(vii) 4 log10 x + 1 - 6 log10 x - 2 × 3 log10 x
Þ
2
+ 2
a x = x , a = x 1/x
=0
...(i)
From Eq. (ii),
Let log10 x = l, then
2 2 l + 2 - (2 ´ 3 ) l - 2 × 3 2 l + 2 = 0
Þ
l
æ3ö
æ3ö
22 - ç ÷ - 2 × 32 × ç ÷
è2ø
è2ø
2l
=0
l
Let
...(i)
...(ii)
Defined for x > 0
From Eq. (i),
x = ax
Þ 2 2 log10 x + 2 - (2 ´ 3 ) log10 x - 2 × 3 2 log10 x + 2 = 0
Þ
…(i)
\
x =5
satisfy Eq. (i), hence x = 5.
(ix) x log 2 x + 4 = 32
= 10 5+ y
1
\
x > 21
2 log10 ( x - 3 ) = log10 ( x 2 - 21 )
x 2 - 6 x + 9 = x 2 - 21
Defined for x > 0,
1
log 9 x + + 9 x = 9 x
2
Þ
-2
Þ
Þ
y + 2y - 15 = 0
log 9 x = -
4
9
æ3ö
æ3ö
ç ÷ = ç ÷ Þ l = -2
è2ø
è2ø
1
Hence, x = 10 l = 10 -2
100
1
log10 ( x - 3 )
(viii)
=
log10 ( x 2 - 21 ) 2
\
…(i)
1
æ
ö
(vi) log 3 ç log 9 x + + 9 x ÷ = 2 x
è
ø
2
Þ
1
2
is defined for x > 1 and x 2 > 21.
2
\
m=
l
...(i)
y 2 + 5y = 15 + 3y
Þ
Þ
Þ
æ y + 5ö
y× ç
÷
è 3 ø
\
\
Defined for x > 0
log10 x = y
x = 10 y
By Eq. (i),
Þ
= 10 5 + log10
ù
ûú
9 m (2 m + 1 ) - 4 (2 m + 1 ) = 0
1
4
m = - ,m =
2
9
m¹-
log 4 ( x + 1 ) = 3
log 4 x = 2
x = 16
(iv) log 4 log 3 log 2 x = 0
Defined for x > 0, log 2 x > 0 and log 3 log 2 x > 0
Þ
x > 0, x > 1, x > 3
\
x >3
log 3 log 2 x = 1
log 2 x = 3, x = 8
which satisfy Eq. (i).
(v)
=c
log a
b
æ3ö
ç ÷ =m
è2ø
\
18 m 2 + m - 4 = 0
Þ
18 m 2 + 9 m - 8 m - 4 = 0
Þ
1
xx
= x log 4 x
1
= log 4 x
x
x = log x 4 Þ x x = 4
\
x =2
(xi) log 2 sin x (1 + cos x ) = 2
Defined for 1 + cos x > 0, 2 sin x > 0
and 2 sin x ¹ 1, then
1 + cos x = 2 sin 2 x
1 + cos x = 2 - 2 cos2 x
…(i)
Chap 04 Logarithms and Their Properties
x 2 - 2x - 3 > 0
2 cos2 x + cos x - 1 = 0
(2 cos x - 1 ) (cos x + 1 ) = 0
1 + cos x ¹ 0
1
cos x =
2
So,
\
p
, Eq. (i) is defined for that value of x .
3
69. Let rational number be x , then
x = 50 log10 x Þ 2 x = 100 × log10 x
Taking logarithm to the base 10, then
log10 2 + log10 x = 2 + log10 (log10 x )
Let
log10 x = l
\
log10 2 + l = 2 + log10 ( l )
æ lö
log10 ç ÷ = l - 2
Þ
è2ø
(x - 3) (x + 1) < 0 - 1 < x < 3
x Î ( - 1, 3 ]
x¹0
æ 3 ö
Eqs. (ii), (iii) and (iv), x Î ç - , 3 ÷ È { - 1, 0 }
è 2 ø
(ii) log 2 x ( x 2 - 5 x + 6 ) < 1
(x - 6) (x - 1) > 0
1
1
log 2000 2000 =
6
6
[1 + logb {1 + logc (1 + log p x )}] = 0
1 + logb {1 + logc (1 + log p x )} = 1
logb {1 + logc (1 + log p x )} = 0
1 + logc (1 + log p x ) = 1
logc (1 + log p x ) = 0
1 + log p x = 1
log p x = 0
x = p0
x < 1 or x > 6
...(i)
Case II 2 x > 1 Þ x >
1
1
=5 ´
log 2 (3 - 6 ) - 6 ´ log 2 ( 3 - 2 )
5/ 2
3
= log 2 (3 - 6 ) 2 - log 2 ( 3 - 2 ) 2
2
æ 3- 6 ö
æ 3 ( 3 - 2)ö
= log 2 ç
÷ = log 2 ç
÷ = log 2 3
è 3 - 2ø
è ( 3 - 2) ø
5 log 4 2 ( 3 - 6 ) - 6 log 8 ( 3 - 2 )
2 log 2 3
=2
log 2 9
=9
2
73. (i) log 2 x + 3 x < 1
Case I 0 < 2 x + 3 < 1, i.e. Eq. (i),
x 2 > 2x + 3
…(i)
3
< x < -1
2
1
2
…(A)
…(v)
From Eq. (i), log 2 x ( x 2 - 5 x + 6 ) < 1
= 5 log 2 5 / 2 (3 - 6 ) - 6 log 2 3 ( 3 - 2 )
2
…(iv)
From Eqs. (iii), (iv) and (ii)
æ 1ö
x Î ç 0, ÷
è 2ø
Þ
x =1
Eq. (i) is satisfied for this value of x .
72. Q5 log 4 2 (3 - 6 ) - 6 log 8 ( 3 - 2 )
=2
…(iii)
x 2 - 7x + 5 > 0
=
log 2 3
…(ii)
x 2 - 5x + 6 < 2x
= log ( 2000) 6 ( 4 2 ´ 5 3 )
=4
1
2
(x 2 - 5x + 6) < 1
Case I 0 < 2 x < 1 Þ 0 < x <
From Eq. (i), log 2 x
…(iii)
…(iv)
…(i)
For Eq. (i) to be defined 2 x > 0 and 2 x ¹ 1
1ü
So,
x > 0 and x ¹ ï
2ý
and x 2 - 5 x + 6 > 0 Þ x < 2 or x > 3 ïþ
= 2 log ( 2000) 6 4 + 3 log ( 2000) 6 5
= log ( 2000) 6 4 2 + log ( 2000) 6 5 3
\ =4
…(ii)
Þ
Eq. (i),
which is true for l = 2.
\
log10 x = 2
Þ
x = 10 2 = 100
2
3
70. Let y =
+
log 4 (2000 ) 6 log 5 (2000 ) 6
Þ
Þ
Þ
Þ
Þ
Þ
Þ
(x - 3) (x + 1) > 0
x < - 1 or x > 3
ö
æ 3
x Î ç - , - 1÷
ø
è 2
Case II 2 x + 3 > 1 Þ x > - 1
Eq. (i), x 2 < 2 x + 3
x=
71. loga
351
Þ
x 2 - 5x + 6 < 2x
Þ
x 2 - 7x + 6 < 0
Þ
1 < x <6
From Eqs. (ii), (v) and (vi),
x Î (1, 2 ) È (3, 6 )
æ 1ö
From Eqs. (A) and (B), x Î ç 0, ÷ È (1, 2 ) È (3, 6 )
è 2ø
…(vi)
(iii) log 2 (2 - x ) < log1 / 2 ( x + 1 )
From Eq. (i) to be defined 2 - x > 0 Þ x < 2
and
x + 1 > 0Þx > -1
So,
x Î ( - 1, 2 )
Now, from Eq. (i), log 2 (2 - x ) + log 2 ( x + 1 ) < 0
(2 - x ) ( x + 1 ) < 1
(x - 2) (x + 1) + 1 > 0
x2 - x -2 + 1 > 0
…(i)
x2 - x -1 > 0
…(B)
…(ii)
352
Textbook of Algebra
1- 5
2
1+ 5
or
x>
2
From Eqs. (ii) and (iii),
æ
1 - 5 ö æ1 + 5 ö
x Î ç - 1,
, 2÷
÷ Èç
ø
è
2 ø è 2
Þ
(iv) log x 2 ( x + 2 ) < 1
From Eq. (i) to be defined, x + 2 > 0 Þ x > - 2 ü
ý
and
x Î R, x ¹ 0 and x ¹ 1
þ
Case I x Î ( - 1, 1 ) ~ { 0 }
Eq. (i),
(x + 2) > x 2
…(iii)
...(i)
…(A)
…(ii)
log 3
x -1
< 3 log 3 ( x - 6) + 3
From Eq. (i) to be defined
x -1 > 0 Þ x >1
and x - 6 > 0 Þ x > 6
From Eqs. (ii) and (iii), x > 6
Eq. (i), x - 1 - ( x - 6 ) - 3 < 0
(1010 x - 2 ) (log10 x + 1 ) ³ 0
log10 x £ - 1 or log10 x ³ 2
1
x £ or x ³ 100
10
From Eqs. (ii) and (iii),
æ 1ù
x Î ç 0, ú È [100, ¥ )
è 10 û
…(B)
…(iv)
…(v)
…(C)
…(i)
…(ii)
…(iii)
…(iv)
x Î ( - ¥, 1 - 3 ) È (1 + 3, ¥ )
x 2 - 2x - 2 £ 1
x 2 - 2x - 3 £ 0
(x - 3) (x + 1) £ 0
-1 £ x £3
From Eqs. (ii) and (iii), we get
x Î [ - 1, 1 - 3 ) È (1 + 3, 3 ]
3ö
æ
(ix)
log x ç2 x - ÷ > 2
è
4ø
\
From Eq. (i) to be defined x > 0, x ¹ 1, 2 x x > 0, x ¹ 1, x >
0 < x <1
3
2x - < x 2
4
8x - 3 - 4x 2 < 0
x 2 + 9 - 6x - x + 1 > 0
x 2 - 7 x + 10 > 0
1
From Eq. (i) to be defined x ¹ , x ¹ - 5
3
2
2
Eq. (i), (3 x - 1 ) > ( x + 5 )
(3 x - 1 - x - 5 ) (3 x - 1 + x + 5 ) > 0
(2 x - 6 ) ( 4 x + 4 ) > 0
(x - 3) (x + 1) > 0
x < - 1 or x > 3
From Eqs. (ii) and (iii),
x Î ( - ¥, - 5 ) È ( - 5, - 1 ) È (3, ¥ )
3
8
…(ii)
…(iii)
…(i)
3
>0
4
…(ii)
3ö
æ
From Eq. (i), log x ç2 x - ÷ > 2
è
4ø
x - 1 < (x - 3) x - 1 < (x - 3)2
(x - 5) (x - 2) > 0
x < 2 or x > 5
From Eqs. (iv) and (v), x > 6
(vi) log1/2 (3 x - 1 ) 2 < log1/2 ( x + 5 ) 2
…(i)
(x - 1)2 - ( 3 )2 > 0
[ x - (1 + 3 )] [ x - (1 - 3 )] > 0
Case I
x -1 - x + 3 < 0
…(iii)
x Î ( 0, 10 -1 ] È [10 2, ¥ )
or
…(iii)
x2 - x - 2 > 0
(v) 3
…(ii)
(viii) log10 ( x - 2 x - 2 ) £ 0
From Eq. (i),
x 2 - 2x - 2 > 0
2
x - 2x + 1 - 3 > 0
x - x -2 < 0
x < - 1 or x > 2
From Eqs. (iv), (v) and (A),
x Î ( - 2, - 1 ) È (2, ¥ )
From Eqs. (B) and (C),
x Î ( - 2, 1 ) È (2, ¥ ) ~ { - 1, 0 }
…(i)
From Eq. (i),
x>0
2
log10 x + log10 x - 2 ³ 0
2
2
(x - 2) (x + 1) < 0
x -1 < x <2
From Eqs. (ii), (iii) and (A),
x Î ( - 1, 0 ) È ( 0, 1 )
Case II x Î ( - ¥, - 1 ) È (1, ¥ )
Eq. (i),
x + 2 < x2
2
log10 x + 2 £ log10
x
(vii)
x<
…(iii)
4x 2 - 8x + 3 > 0
2
4x - 6x - 2x + 3 > 0
…(v)
…(i)
…(ii)
(2 x - 1 ) (2 x - 3 ) > 0
1
3
x < or x >
2
2
From Eqs. (ii), (iii) and (iv),
æ3 1ö
x Îç , ÷
è8 2ø
Case II
Eq. (i)
…(iii)
x >1
Þ
3
2x - > x 2
4
8x - 3 > 4x 2
4x 2 - 8x + 3 < 0
…(iv)
…(v)
...(vi)
Chap 04 Logarithms and Their Properties
1
3
<x<
2
2
From Eqs. (ii), (vi) and (vii), we get
æ 3ö
x Î ç1, ÷
è 2ø
…(viii)
From Eqs. (v) and (viii), we get
æ3 1ö æ 3ö
x Î ç , ÷ È ç1, ÷
è8 2ø è 2ø
Þ
log x < 0 Þ x < 1
So,
x Î( 0, 1 )
2
(xi) log 2x + 3 x < log 2x + 3 (2 x + 3 )
x>-
Case I
Then,
3
< x < -1
2
If
x>0
log ( x 2 + x + 1 ) < 0
…(i)
Þ
x (x + 1) < 0
Þ
-1 < x < 0
From Eqs. (i) and (ii), x Î f
Case II If x < 0
Then,
log ( x 2 + x + 1 ) > 0
Þ
…(A)
…(ii)
…(iii)
From Eq. (ii), log 2x + 3 x 2 < 1
(x - 3) (x + 1) > 0
x < - 1 or x > 3
æ 3
ö
From Eqs. (A), (iii) and (iv), x Î ç - , - 1 ÷
è 2
ø
Þ
x (x + 1) > 0
\
x Î ( - ¥, - 1 ) È ( 0, ¥ )
From Eqs. (iii) and (iv), we get
x Î ( - ¥, - 1 )
1
(xiv) log ( 3x 2 + 1) 2 <
2
2 < (3 x 2 + 1 )1/2
(3 x 2 + 1 > 1, " x Î R )
Case II If 2 x + 3 > 1 Þ x > - 1
log 2x + 3 x 2 < 1
…(v)
x 2 < 2x + 3
x 2 - 2x - 3 < 0
(x - 3) (x + 1) < 0
Þ
-1 < x <3
So, Eqs. (A), (v) and (vi), x Î ( - 1, 3 )
From Eqs. (B) and (C),
æ 3
ö
x Î ç - , - 1 ÷ È ( - 1, 3 )
è 2
ø
5
log 4 2 16
2
5 2
log 22 x + 3 log 2 x - ´ log 2 16 ³ 0
2 5
2
log 2 x + 3 log 2 x - 4 ³ 0
(xii) log 22 x + 3 log 2 x ³
(log 2 x + 4 ) (log 2 x - 1 ) ³ 0
Þ
…(i)
…(iv)
(log
x < - 1 or x > 1
x Î ( - ¥, - 1 ) È (1, ¥ )
x ) 2 - 3 log
x+1
10
(xv) x 10
> 1000
From Eq. (i) to be defined, x > 0 and x ¹ 1
Let log10 x = y Þ x = 10 y
From Eq. (i), 10 y ( y
…(vi)
…(C)
…(iii)
4 < 3x 2 + 1
3x 2 > 3
x2 > 1
…(iv)
…(B)
…(ii)
x2 + x + 1 > 1
x 2 > 2x + 3 Þ x 2 - 2x - 3 > 0
Þ
Þ
…(i)
x2 + x + 1 < 1
\
3
2
2x + 3 ¹ 1
x ¹ -1
x Î R - { 0}
2
From Eq. (i), log 2x + 3 x < 1
x 2 + x + 1 > 0, " x Î R
Q
From Eq. (i) to be defined,
2x + 3 > 0
Þ
…(ii)
…(iii)
Taking logarithm on both sides, then
x log ( x 2 + x + 1 ) < 0
æ log 3 - log 2 ö
log x ç
÷<0
è log 2 log 3 ø
0 < 2x + 3 < 1 Þ -
x ³2
x>0
æ 1ù
From Eqs. (ii) and (iii), x Î ç 0, ú Î [2, ¥ )
è 16 û
or
From Eq. (i),
1
16
(xiii)Q ( x 2 + x + 1 ) x < 1
(x) log1/3 x < log1/2 x ( x > 0 )
Þ
log 3 x > log 2 x
log x log x
Þ
>0
log 3 log 2
Case I
log 2 x £ - 4 or log 2 x ³ 1 Þ x £
…(vii)
353
2
- 3y + 1)
> 10 3
Þ
y 3 - 3y 2 + y - 3 > 0
Þ
y 2 (y - 3 ) + 1 (y - 3 ) > 0
Þ
(y - 3 ) (y 2 + 1 ) > 0
Þ
Þ
Þ
Þ
(xvi)
…(i)
y >3
log10 x > 3
x > 1000
x Î (1000, ¥ )
log 4 {14 + log 6( x 2 - 64 )} £ 2
2
14 + log 6 ( x - 64 ) £ 16
log 6 ( x 2 - 64 ) £ 2
…(i)
354
Textbook of Algebra
x 2 - 64 £ 36
x 2 £ 100
From Eq. (i),
- 10 £ x £ 10
x - 64 > 0
…(ii)
2
Þ
x < - 8 or x > 8
From Eqs. (ii) and (iii),
x Î [ - 10, - 8 ) È (8, 10 ]
(xvii)
log 2 (9 - 2 x ) £ 10
log
10
(3 - x )
…(iii)
...(i)
From Eq. (i) to be defined,
9 - 2x > 0 Þ 9 > 2x
Þ
2 x < 9 Þ x < log 2 9
3 - x > 0 Þ x <3
Then,
x <3
From Eq. (i), log 2 (9 - 2 x ) £ 3 - x
Þ
9 - 2x - 8 × 2-x £ 0
x 2
+
–
x
Þ
(2 ) - 92 + 8 ³ 0
Þ
(2 x - 8 ) (2 x - 1 ) ³ 0
Þ
…(ii)
9 - 2x £ 23 - x
Þ
x+3
£0
x
Þ
-3 £ x £ 0
Þ
x Î [ - 3, 0 ]
3ö
é
From Eqs. (ii) and (v), we get x Î ê - 3, - ÷
2ø
ë
(xix)
1 + log 2 ( x - 1 ) £ log ( x - 1) 4
From Eq. (i) to be defined, x - 1 > 0 Þ x > 1
and
x -1 ¹1 Þ x ¹2
By Eq. (i), 1 + log 2 ( x - 1 ) £ 2 log ( x - 1) 2
Let log 2 ( x - 1 ) = l, then
2
1+l£
l
l2 + l - 2
£0
Þ
l
(l + 2) (l - 1)
Þ
£0
l
Þ
x
x
2 £ 1 or 2 ³ 8
Þ
x £ 0 or x ³ 3
From Eqs. (ii) and (iii), x £ 0 Þ x Î ( - ¥, 0 ]
æ 2x + 3 ö
(xviii) loga ç
÷³0
è x ø
From inequation (a), a > 1
2x + 3
By Eq. (i),
>0
x
é
æ 3ö ù
ê x - çè - 2 ÷ø ú
Þ
ú>0
ê
ê x-0 ú
úû
êë
3
Þ
x < - or x > 0
2
3ö
æ
From Eq. (i), loga ç2 + ÷ ³ 0
è
xø
…(iii)
…(i)
…(ii)
3
2 + ³1
x
3+x
³0
x
x - (- 3)
³0
x-0
x £ - 3 or x ³ 0
From Eqs. (ii) and (iii),
x £ - 3 or x > 0
Þ
x Î ( - ¥, - 3 ] È ( 0, ¥ )
From inequation in (b), 0 < a < 1
2x + 3
From Eq. (i),
£1
x
…(i)
…(ii)
+
0
–2
…(v)
–
1
Þ
Þ
or
Þ
l £ - 2 or 0 < l £ 1
log 2 ( x - 1 ) £ - 2
0 < log 2 ( x - 1 ) £ 1
x - 1 £ 2 -2 or 2 0 < x - 1 £ 2 1
5
Þ
x £ or 2 < x £ 3
4
From Eqs. (ii) and (iii), we get
x Î(2, 3 ]
2
(xx) log 5x + 4 x £ log 5x + 4 (2 x + 3 )
4
From Eq. (i) to be defined, 5 x + 4 > 0 Þ x > 5
3
5x + 4 ¹ 1 Þ x ¹ 5
3
2x + 3 > 0 Þ x > 2
and
x Î ( - ¥, ¥ ) - { 0 }
æ 4 3ö æ 3 ö
x Î ç - , - ÷ È ç - , 0 ÷ È ( 0, ¥ )
Þ
è 5 5ø è 5 ø
…(iii)
…(i)
…(ii)
From Eq. (i), log 5x + 4 x 2 £ log 5x + 4 (2 x + 3 )
log 5x + 4
Case I
…(iii)
Þ
…(iv)
From Eq. (iii),
x2
£0
2x + 3
0 < 5x + 4 < 1
4
3
- <x<5
5
x2
³1
2x + 3
x 2 - 2x - 3
³0
2x + 3
…(iii)
…(iv)
Chap 04 Logarithms and Their Properties
(x - 3) (x + 1)
³0
é
æ 3öù
x
ç
÷
ê
è 2 ø úû
ë
–
3
2
5
loga x = a 2
4
5 2
a
x = a4
+
–
+
–
–1
Case II 1 < x < a
3
By Eq. (iii),
æ 3
ù
x Î ç - , - 1 ú È [3, ¥ )
è 2
û
…(v)
From Eqs. (ii), (iv) and (v), x Î f
…(vi)
3
Case II
5x + 4 > 1 Þ x > 5
x2
From Eq. (iii),
£1
2x + 3
…(vii)
Þ
ù
é
ú
ê
3
1
(
x
)
(
x
+
)
ú£0
ê
ê ì
æ 3 öü ú
ê í x - çè - 2 ÷øý ú
þ û
ë î
…(viii)
…(ix)
log x (ax )1/5 + loga (ax )1/5
loga x +
Let
1
5
æa ö
+ log x ç ÷
èxø
1
+
| loga x |
loga x +
| loga x | -
1
- 2 = 5a
loga x
1
= 5a …(ii)
| loga x |
| loga x | = y
[y ³ 0 ]
1
1
y +
+ y = 5a
y
y
Case I x ³ a > 1 Eq. (iii) Þy +
Þ
= a …(i)
loga x - 1 + log x a - 1 = a
1
+2+
loga x
8ax
a -2
Eq. (i) is defined, if a - 2 > 0 Þa > 2
8ax
>0
a -2
x 2 + 15a 2 = 8ax
1/ 5
1 + log x a + 1 + loga x
+
| loga x | +
1/ 5
log p ( x 2 + 15a 2 ) - log p (a - 2 ) = log p
…(iii)
1
1
+ y - = 5a
y
y
…(ii)
2 | loga x | = 5a
5
a
2
…(iii)
( x - 3a )( x - 5a ) = 0
\
x = 3a and x = 5a
For
a = 3, x = 9 and x = 15
Þ
x = 15 for a = 3
76. Given that,
log 4 (log 3 x ) + log1/4 (log1/3 y ) = 0
1
1
Þ
log 2 log 3 x - log 2( - log 3 y ) = 0
2
2
æ log 3 x ö ù
1é
Þ
÷ú = 0
ê log 2 ç
è - log 3 y ø û
2ë
Þ
Þ
Þ
Þ
2y = 5a
| loga x | =
…(i)
By Eq. (ii),
a >2
So, ax > 0, then x > 0
Eq. (i) for
x = 9, a > 0
( x 2 + 15a 2 )
8ax
log p
= log p
(a - 2 )
a -2
74. Given equation is
1
5
2
75. Given equation,
From Eqs. (vi) and (ix), we get
æ 3 3ö
x Î ç - , - ÷ È [ - 1, 0 ) È ( 0, 3 ]
è 5 2ø
æxö
loga ç ÷
èa ø
…(v)
1
1
y + - y + = 5a
y
y
2
= 5a
y
2
y =
5a
2
| loga x | =
5a
4
loga x = 2
5a
x = a 4 / 5a
3
Þ x < - or x Î [ - 1, 3 ]
2
From Eqs. (ii), (vii) and (viii),
æ 3 3ö
x Î ç - , - ÷ È [ - 1, 0 ) È ( 0, 3 ]
è 5 2ø
+
355
Also, given that,
-
…(i)
log 3 x
=1
log 3 y
log 3 x = - log 3 y
æ1ö
log 3 x = log 3 ç ÷
èy ø
1
y
17
x2 + y 2 =
4
1
17
x2 + 2 =
4
x
x=
…(ii)
356
Textbook of Algebra
2
1ö
17
æ
+2
çx + ÷ =
è
4
xø
1 5
x+ =
x 2
1
1
x + =2 +
x
2
1
\
x = 2 or
2
1
For these values of x, y = or 2
2
77. log 2x 4x + log 4 x 16x = 4
[by Eq. (i) x > 0, y > 0]
[by Eq. (ii)]
…(ii)
Þ
log 2 { 4 ( x + 2 )} = log 2 {( 4 - x ) ( x + 6 )}
Þ
4 (x + 2) = (4 - x ) (x + 6)
4 x + 8 = - x 2 - 2 x + 24
x 2 + 6 x - 16 = 0
(x + 8) (x - 2) = 0
\
x = - 8, x = 2
From Eqs. (ii) and (iii), we get x = 2
æ 1 ö
80. log 2 ( 4 x + 1 + 4) × log 2( 4 x + 1) = log1/ 2 ç ÷
è 8ø
log 2 [ 4( 4 x + 1 )] × log 2( 4 x + 1 ) = log
x
2 l2 + 9 l + 8 = 4 l2 + 12 l + 8
\
3
l = 0, l = 2
3
log 2 x = 0, log 2 x = 2
x = 2 0, x = 2 - 3 / 2
or
x = 1, x = 2 -3/2
\
Þ
78. Given equation,
log 6 54 + log x 16 = log
2
x - log 36
…(i)
\
(y + 2 ) y = 3
y + 2y - 3 = 0
2
(y + 3 ) (y - 1 ) = 0
y = 1 or y = - 3
log 2 ( 4 x + 1 ) = - 3
1
4x + 1 =
8
1
x
4 = -1
8
7
x
4 = - which is not possible.
8
4 x + 1 = 2 or
4 x = 1 or
x = 0 or
…(i)
Eq. (i) holds, if x > 0, x ¹ 1
From Eq. (i),
\
81. 2
2
1 + log 6 9 + 4 log x 2 = 2 log 2 x - log 6
3
2
1 + log 6 9 + log 6 + 4 log x 2 - 2 log 2 x = 0
3
2 + 4 log x 2 - 2 log 2 x = 0
(log 2 x ) 2 - log 2 x - 2 = 0
log 2 x = 2 or log 2 x = - 1
1
Þ
x = 4 or x =
2
1 9
Sum of the values of x satisfy Eq. (i) = 4 + =
2 2
1
Product of the values of x satisfy Eq. (i) = 4 ´ = 2
2
3
79. Let log 4 ( x + 2) 3 + 3 = log 4 ( 4 - x ) 3 + log 4 ( x + 6) 3
2
Eq. (i) holds, if 4 - x > 0 and x + 0 > 0, x + 2 > 0
8 =3
log 2( 4 x + 1 ) = y
\ log 2 ( 4 x + 1 ) = 1 or
4
9
2
x
Let
2 l2 + 3 l = 0
Þ
…(iii)
[2 + log 2 ( 4 + 1 )] log 2 ( 4 + 1 ) = 3
Let log 2 x = l, then
(2 + l ) 2 + (1 + l ) ( 4 + l ) = 4 (1 + l ) (2 + l )
Þ
Þ
log 2 ( x + 2 ) + 2 = log 2 ( 4 - x ) + log 2 ( x + 6 )
Eq. (i) defined, for 4 x + 1 > 0 which is true for all x Î R.
On dividing by log 2, then
2 + log 2 x 4 + log 2 x
+
=4
1 + log 2 x 2 + log 2 x
Þ
Þ
…(i)
1
1
From Eq. (i) is defined for x > 0, x ¹ , x ¹
2
4
log 4 x log 16 x
Þ
+
=4
log 2 x
log 4 x
2 log 2 + log x 4 log 2 + log x
+
=4
Þ
2 log 2 + log x
log 2 + log x
Þ
i.e.,
…(ii)
-2 < x < 4
From Eq. (i),
3
1
1
´ 2 ´ log 2 |( x + 2 )| + 3 = ´ 3 log 2( 4 - x )
2
2
2
1
+ ´ 3 log 2 ( x + 6 )
2
x=0
x +
y
= 256
x +
y
= 28
Þ
2
Þ
x + y =8
Also, given that, log10
xy - log10
…(i)
3
=1
2
…(ii)
which is defined, xy > 0
3ö
æ
So, Eq. (ii) Þ log10 xy = log10 ç10 ´ ÷
è
2ø
Þ
Þ
xy = 15
Þ
xy = 225
From Eq. (i), x + y + 2 xy = 64
…(i)
x + y = 64 - 30
x + y = 34
From Eq. (iii),
xy = 225
After solving, we get x = 9 or x = 25, then y = 25 or y = 9
Hence, solutions are (9, 25) and (25, 9).
…(iii)
Chap 04 Logarithms and Their Properties
82. Given, log 2 y = log 4 ( xy - 2)
…(i)
Eq. (i) defined for y > 0 and xy - 2 > 0
xy > 0
From Eqs. (ii) and (iii) Þ y > 0, x > 0
By Eq. (i),
y = xy - 2
…(ii)
…(iii)
From Eqs. (ii), (iv) and (v),
x Î ( - 5, - 4 ) È ( - 3, - 1 )
84. log 3 ( x + | x - 1 | ) = log 9 ( 4 x - 3 + 4 | x - 1 | )
then log 3 ( x + | x - 1 | ) = log 32 ( 4 x - 3 + 4 | x - 1 | )
…(iv)
y (x - y ) = 2
…(v)
Þ
2
log 9 x + log 3 ( x - y ) = 1
which is defined for x Î R - { 0 } and x - y > 0
Þ
x >y
By Eq. (vi), x( x - y ) = 3 Þ x 2 - xy = 3
and
x (x - y ) = 3
Form Eqs. (iv) and (vii), y 3 + 2 = x 2 - 3
…(vi)
9 + 4 x - 12 x = 4 x + 4 - 4 x
…(vii)
8 x =5
\
x=
…(ix)
From Eq. (i), x > 0
Þ (log 2 x ) 4 - (5 log 2 x - 2 ) 2 - 20 log 2 x + 148 < 0
(log 2 x ) 4 - 25 log 22 x + 144 < 0
{(log 2 x ) 2 - 16 } {(log 2 x ) 2 - 9 } < 0
…(i)
9 < (log 2 x ) 2 < 16
3 < log 2 x < 4 or - 4 < log 2 x < - 3
8 < x < 16
…(ii)
1
1
or
<x<
16
8
According to the question in Eq. (i) holds, for x Î(a, b )
where a, b Î N
So, from Eq. (ii), a = 8, b = 16
\
ab (a + b ) = 8 ´ 16 (8 + 16 )
= 144 ´ 24 = 3456
86.
+
y 2 - 3y + 2
<0
y
(y - 2 ) (y - 1 )
<0
y
–
æ
ç log 3
è
3
x
3ö
+ log x 3 ÷ log 3 x 3 = 2
3
xø
æ
ö
1
= 2 ç log 3 x +
+ 2÷ +
è
ø
log 3 x
2
y < 0 or 1 < y < 2
log 2 ( x + 5 ) < 0 or 1 < log 2 ( x + 5 ) < 2
x + 5 < 1 or 2 < x + 5 < 4
x<-4
-3 < x < -1
…(iii)
…(i)
Eq. (i) is defined for x > 0, x ¹ 1
1
From Eq. (i),
(1 + log 3 x + 1 + log x 3 ) 3 log 3 x
3
1
+
(log 3 x - 1 + log x 3 - 1 ) 3 log 3 x
3
+
1
…(ii)
(log 3 3 3 x ) + (log x 3 3 x ) log 3 x 3
- y + 3y - 2
>0
y
+
...(i)
(log 2 x ) 4 - 25 log 22 x - 4 + 20 log x x - 20 log 2 x + 148 < 0
2
0
2
è
…(x)
Now, let log 2 ( x + 5 ) = y , then Eq. (iii) becomes
2
-y - + 3 > 0
y
–
25
64
x 5ö
÷ - 20 log 2 x + 148 < 0
4 ø
85. (log 2 x ) 4 - ç log1/2
- log 2 ( x + 5 ) + 3 - 2 log x + 5 2 > 0
2
…(iii)
- log 2 ( x + 5 ) +3>0
log 2 ( x + 5 )
Þ
Þ
Þ
Þ
or
Þ
æ
2
x -y =5
Þ
3 - 2 x = 2 | x - 1|
On squaring both sides, then
...(viii)
On dividing Eq. (v) by Eq. (viii),
y 2
2x
=
Þ y =
x 3
3
From Eqs. (ix) and (x),
x = 3 and y = 2
83. Given that,
9
2 log1/4 ( x + 5 ) > log1/3 3 9 + log x + 5 2
4
By Eq. (i), x + 5 > 0 Þ x > - 5
x + 5 ¹1 Þ x ¹ - 4
So,
x Î ( - 5, - 4 ) È ( - 4, ¥ )
Now, by Eq. (i)
2
9 æ - 2ö
log 2 ( x + 5 ) - ´ ç
÷ log 3 9 - 2 log x + 5 2 > 0
-2
4 è 3 ø
Þ
2( x + | x - 1 | ) = 4 x - 3 + 4 | x - 1 |
Þ
Also given that,
Þ
…(i)
From Eq. (i) is defined, if x ³ 0
y 2 - xy + 2 = 0
2
357
æ
ö
1
- 2÷
ç log 3 x +
è
ø
log 3 x
= 2 | log x 3 |
…(iv)
…(v)
Þ
| log 3 x | +
1
+
| log 3 x |
| log 3 x | -
1
= 2 | log x 3 |
| log 3 x |
…(ii)
358
Textbook of Algebra
Case I If x ³ 3, log 3 x -
1
>0
log 3 x
Taking log with base e on both sides, then
ln x × ln 3 = ln y × ln 2
From Eqs. (i) and (ii), we get
æ
ln x × ln 3 ö
ln 2 (ln 2 + ln x ) = ln 3 ç ln 3 +
÷
è
ln 2 ø
2 log 3 x = 2 log x 3
1
(log 3 x ) 2 = 1 Þ x = 3 or , so x = 3
3
2
Case II If 1 < x < 3,
= 2 log x 3
log 3 x
æ (ln 3 ) 2
ö
ln x ç
- ln 2 ÷ = - ((ln 3 ) 2 - (ln 2 ) 2 )
è ln 2
ø
Þ
Þ
log 3 x × log x 3 = 1
Þ
1 =1
which is true, for all x Î(1, 3 ].
So, x Î(1, 3 ]
87. P = Number of natural numbers, whose logarithms to the base
10 have characteristic p.
Let ‘x ’ represent the natural number, i.e.
x = l ´ 10 p [ l = a1 × a 2 × a 3 ¼]
æ1ö
ln x = - ln 2 = ln ç ÷
è2ø
\
1
2
1
x0 =
2
x=
Þ
\
90. Let S =
1
3 2
4-
1
1
1
¼¥
443 2
3 2
3 2
So, P = Number of natural numbers which have ( p + 1 ) digits
= 9 × 10 p - 1 + 1 = 9 × 10 p
Þ
S=
Q = Number of natural numbers which have (q ) digits.
Q = 9 × 10q - 1 - 1 + 1 = 9 × 10q - 1
or
(3 2 S ) 2 = 4 - S
p
q -1
So, log10 P - log10 Q = log10(9 × 10 ) - log10(9 × 10
88. Q
Þ
3a = log 3 2
\
3 -a = log 2 3
Now,
)
= (log10 9 + p ) - (log10 9 + q - 1 )
= p -q + 1
a = log 3 log 3 2
1 < 2 -k + 3
-a
-a
Þ
2 0 < 2 -k + 3
\
0 < -k + 3 -a < 1
and
3 ln x = 2 ln y
Þ
\
(9 S - 4 ) (2 S + 1 ) = 0
9S - 4 = 0
S=
æ 1
6 + log 3/2 ç
ç3 2
è
4 -S
[Q2S + 1 ¹ 0]
4 æ3ö
=ç ÷
9 è2ø
-2
log 3/2 S = - 2 Þ6 + log 3/2 S = 6 - 2 = 4
4-
1
3 2
4-
1
3 2
4-
ö
¼÷ = 4
÷
3 2
ø
1
91. (3 / 4) x = 1 / 4
Þ
0 < - k + log 2 3 < 1
Þ
0 > k - log 2 3 > - 1
Þ
log 2 3 - 1 < k < log 2 3
\
k =1
89. Q(2x ) ln 2 = (3y ) ln 3
Taking log with base e on both sides, then
ln 2 ( ln 2 + ln x ) = ln 3 (ln 3 + ln y )
18 S 2 + S - 4 = 0
Þ
Hence,
< 21
1
3 2
Þ
or
< 21
…(ii)
Taking log with base 2
Þ
x (log 2 3 - 2 ) = -2
2
1
\
x=
=
Þ (b, c)
2 - log 2 3 1 - log 4 3
…(i)
and taking log with base 3
Þ
x (1 - log 3 4 ) = -2 log 3 2
2 log 3 2
\
x=
2 log 3 2 - 1
CHAPTER
05
Permutations and
Combinations
Learning Part
Session 1
● Fundamental Principle of Counting
● Factorial Notation
Session 2
● Divisibility Test
● Principle of Inclusion and Exclusion
● Permutation
Session 3
● Number of Permutations Under Certain Conditions
● Circular Permutations
● Restricted Circular Permutations
Session 4
● Combination
● Restricted Combinations
Session 5
● Combinations from Identical Objects
Session 6
● Arrangement in Groups
● Multinomial Theorem
● Multiplying Synthetically
Session 7
● Rank in a Dictionary
● Gap Method [when particular objects are never together]
Practice Part
●
●
JEE Type Examples
Chapter Exercises
Arihant on Your Mobile !
Exercises with the
#L
symbol can be practised on your mobile. See inside cover page to activate for free.
360
Textbook of Algebra
In everyday life, we need to know about the number of ways of doing certain work from given number of available
options. For example, Three persons A,B and C are applying for a job in which only one post is vacant. Clearly, vacant
post can be filled either by A or B or C i.e., total number of ways doing this work is three.
Again, let two persons A and B are to be seated in a row, then only two possible ways of arrangement is AB or BA.In
two arrangements, persons are same but their order is different. Thus, in arranging things, order of things is important.
Session 1
Fundamental Principle of Counting, Factorial Notation
Fundamental Principle
of Counting
(ii) Addition Principle
(i) Multiplication Principle
If an operation can be performed in ‘m’ different ways,
following which a second operation can be performed in ‘n’
different ways, then the two operations in succession can be
performed inm × n ways. This can be extended to any finite
number of operations.
Note For AND → ‘×’ (multiply)
y Example 1. A hall has 12 gates. In how many ways
can a man enter the hall through one gate and come
out through a different gate?
Sol. Since, there are 12 ways of entering into the hall. After
entering into the hall, the man come out through a different gate in 11 ways.
Hence, by the fundamental principle of multiplication, total
number of ways is 12 × 11 = 132 ways.
y Example 2. There are three stations A, B and C, five
routes for going from station A to station B and four
routes for going from station B to station C. Find the
number of different ways through which a person can
go from A to C via B.
Sol. Since, there are five routes for going from A to B. So,
there are four routes for going from B to C.
A
C
B
Hence, by the fundamental principle of multiplication, total
number of different ways
[i.e., A to B and then B to C ]
=5× 4
= 20 ways
If an operation can be performed in ‘m’ different ways and
another operation, which is independent of the first
operation, can be performed in ‘n’ different ways. Then,
either of the two operations can be performed in (m + n )
ways. This can be extended to any finite number of
mutually exclusive operation.
Note For OR → ‘+’ (Addition)
y Example 3. There are 25 students in a class in which
15 boys and 10 girls. The class teacher select either a
boy or a girl for monitor of the class. In how many
ways the class teacher can make this selection?
Sol. Since, there are 15 ways to select a boy, so there are 10
ways to select a girl.
Hence, by the fundamental principle of addition, either a
boy or a girl can be performed in 15 + 10 = 25 ways.
y Example 4. There are 4 students for Physics, 6
students for Chemistry and 7 students for Mathematics
gold medal. In how many ways one of these gold
medals be awarded?
Sol. The Physics, Chemistry and Mathematics student’s gold
medal can be awarded in 4, 6 and 7 ways, respectively.
Hence, by the fundamental principle of addition, number
ways of awarding one of the three gold medals.
= 4 + 6 + 7 = 17 ways.
Factorial Notation
Let n be a positive integer. Then, the continued product of
first ‘n’ natural numbers is called factorial n, to be denoted
by n ! or n i.e., n ! = n(n − 1)(n − 2 ) K 3 ⋅ 2 ⋅ 1
Note When n is negative or a fraction, n! is not defined.
Chap 05 Permutations and Combinations
Some Important Properties
Exponent of prime p in n!
(i) n ! = n (n − 1) ! = n (n − 1)(n − 2 ) !
(ii) 0 ! = 1 ! = 1
(iii) (2n ) ! = 2 n n ![1 ⋅ 3 ⋅ 5 K(2n − 1)]
n!
= n(n − 1)(n − 2 ) K (r + 1)
r!
n!
(v)
= n(n − 1) (n − 2 ) K (n − r + 1)
(n − r ) !
[r < n ]
(iv)
(vi)
[r < n ]
1
1
λ
, then λ = (n + 2 ) 2
+
=
n ! (n + 1) ! (n + 2 ) !
y Example 5. Find n, if (n + 2)! = 60 × (n − 1)!.
Sol. Q(n + 2)! = (n + 2)(n + 1)n (n − 1)!
(n + 2)!
= (n + 2)(n + 1)n
(n − 1)!
⇒
⇒
∴
60 = (n + 2)(n + 1)n
5 × 4 × 3 = ( n + 2) × ( n + 1) × n
n =3
[given]
n
y Example 6. Evaluate
∑r × r! .
r =1
Sol. We have,
n
n
n
r =1
r =1
r =1
∑r × r ! = ∑ {(r + 1) − 1}r ! = ∑(r + 1)! − r !
= (n + 1)! − 1!
[ put r = n in (r + 1)! and r = 1 is r !]
= (n + 1)! − 1
n
y Example 7. Find the remainder when
∑ r ! is divided
r =1
by 15, if n ≥ 5.
n 
Now, the last integer amongs 1, 2, 3, …,   which is
p
n 
p  n 
divisible by p is   =  2 . Now, from Eq. (i), we get
 p   p 
 

n  
n 
E p (n !) =   + E p  p, 2 p, 3 p, K,  2  p 
 p  

p
n 
[because the remaining natural numbers from 1 to   are
p
not divisible by p]

 n 
n   n 
E p (n !) =   +  2  + E p  1 ⋅ 2 ⋅ 3 K  2  
∴

 p  
 p   p 
Similarly, we get
n 
n   n   n 
E p (n !) =   +  2  +  3  + K +  s 
 p 
 p   p   p 
where, s is the largest natural number such that
Sol. Let
N =
Exponent of prime p in n ! is denoted by E p (n !), where p is
prime number and n is a natural number. The last integer
n 
amongst 1, 2, 3, …, (n − 1), n which is divisible by p is   p,
p
where [ ⋅ ] denotes the greatest integer function.
∴
E p (n !) = E p (1 ⋅ 2 ⋅ 3 K(n − 1) ⋅ n )

n  
= E p  p ⋅ 2 p ⋅ 3 pK(n − 1) p ⋅   p 

 p  

[because the remaining natural numbers from 1 to n are
not divisible by p]

 n 
n 
…(i)
=   + E p 1 ⋅ 2 ⋅ 3 K  

 p 
p
(vii) If x ! = y ! ⇒ x = y or x = 0, y = 1
or x = 1, y = 0
⇒
361
n
∑ r ! = 1! + 2! + 3! + 4 ! + 5! + 6! + 7 ! + K + n !
r =1
= (1! + 2! + 3! + 4 !) + (5! + 6! + 7 ! +K + n !)
= 33 + (5! + 6! + 7 ! + K+n !)
N 33 (5! + 6! + 7 ! + K+ n !)
⇒
=
+
15 15
15
3
=2+
+ Integer [as 5!, 6!, K are divisible by 15]
15
3
=
+ Integer
15
Hence, remainder is 3.
ps ≤ n < ps + 1
Note Number of zeroes at the end of n! = E5 ( n!).
y Example 8. Find the exponent of 3 in 100! .
Sol. In terms of prime factors 100! can be written as 2a ⋅ 3b ⋅ 5c ⋅ 7d K
Now, b = E 3 (100!)
100 100 100 100
=   +  4  +  3  +  4  +K
 3  3  3  3 
= 33 + 11 + 3 + 1 + 0 + ... = 48
Hence, exponent of 3 is 48.
362
Textbook of Algebra
∴
Aliter
Q
100! = 1 × 2 × 3 × 4 × 5 × K × 98 × 99 × 100
= (1 × 2 × 4 × 5 × 7 × K × 98 × 100)
(3 × 6 × 9 × K × 96 × 99 )
33
= k × 3 (1 × 2 × 3 × K × 32 × 33)
= [k (1 × 2 × 4 × 5 × K × 31 × 32)]
(3 × 6 × 9 × K × 30 × 33)
= 333 k1 × 311(1 × 2 × 3 × K × 10 × 11)
= 344 [k1 (1 × 2 × 4 × 5 × K × 10 × 11)](3 × 6 × 9 )
= k 2 × 344 × 34 × 2 = k 3 × 348
Hence, exponent of 3 in 100! is 48.
y Example 9. Prove that 33! is divisible by 219 and
what is the largest integer n such that 33! is divisible
by 2n ?
Sol. In terms of prime factors, 33! can be written as
2a ⋅ 3b ⋅ 5c ⋅ 7d ⋅ K
33 33 33 33 33
Now, E 2 (33!) =   +  2  +  3  +  4  +  5  + K
 2  2  2  2  2 
= 16 + 8 + 4 + 2 + 1 + 0 + K
= 31
Hence, the exponent of 2 in 33! is 31. Now, 33! is divisible
by 231 which is also divisible by 219 .
∴ Largest value of n is 31.
y Example 10. Find the number of zeroes at the end
of 100!.
Sol. In terms of prime factors, 100! can be written as
2a ⋅ 3b ⋅ 5c ⋅ 7d K
100  100  100  100  100  100 
Now, E 2(100 !) = 
+
+
+
+
+
 2   2 2   2 3   2 4   2 5   2 6 
= 50 + 25 + 12 + 6 + 3 + 1 = 97
100 100
and E 5 (100!) =   +  2 
 5  5 
= 20 + 4 = 24
100! = 297 ⋅ 3b ⋅ 524 ⋅ 7d K = 273 ⋅ 3b ⋅ (2 × 5)24 ⋅ 7d K
= 273 ⋅ 3b ⋅ (10)24 ⋅ 7d K
Hence, number of zeroes at the end of 100! is 24.
or Exponent of 10 in 100! = min (97, 24) = 24.
Aliter
Number of zeroes at the end of 100!
100 100
= E 5 (100!) =   +  2  + K
 5  5 
= 20 + 4 + 0 + K = 24
y Example 11. For how many positive integral values of
n does n! end with precisely 25 zeroes?
Sol. Q Number of zeroes at the end of n ! = 25
⇒
E 5 ( x !) = 25
⇒
n   n   n 
 5  + 25 + 125 + K = 25
[given]
It’s easy to see that n = 105 is the smallest satisfactory value
of n. The next four values of n will also work (i.e., n = 106,
107, 108, 109). Hence, the answer is 5.
y Example 12. Find the exponent of 80 in 180!.
Sol. Q 80 = 24 × 5
∴
180 180 180 180
E 2 (180!) =   +  2  +  3  +  4 
 2  2  2  2 
180 180 180
+  5  +  6  +  7  +K
2  2  2 
= 90 + 45 + 22 + 11 + 5 + 2 + 1 = 176
180 180 180
and E 5 (180!) =   +  2  +  3  + K
 5  5  5 
= 36 + 7 + 1 + 0 + K
= 44
176 
Now, exponent of 16 in 180! is 
= 44, where [ ⋅ ] denotes the
 4 
greatest integer function. Hence, the exponent of 80 in 180! is 44.
Chap 05 Permutations and Combinations
#L
1.
Exercise for Session 1
There are three routes: air, rail and road for going from Chennai to Hyderabad. But from Hyderabad to
Vikarabad, there are two routes, rail and road. The number of routes from Chennai to Vikarabad via
Hyderabad is
(a) 4
2.
(b) 5
(b) 11
(c) 9
If a, b and c are three consecutive positive integers such that a < b < c and
(a) a
4.
(d) 7
There are 6 books on Mathematics, 4 books on Physics and 5 books on Chemistry in a book shop. The
number of ways can a student purchase either a book on Mathematics or a book on Chemistry, is
(a) 10
3.
(c) 6
(b) b
(c) c
(d) 15
1
1
λ
, the value of λ is
+
=
a! b ! c!
(d) a + b + c
If n!, 3 × n ! and (n + 1)! are in GP, then n ! , 5 × n ! and (n + 1)! are in
(a) AP
(b) GP
(c) HP
(d) AGP
(c) n ⋅ (n + 1) !
(d) n ⋅ (n + 2) !
n
5.
Sum of the series
∑ (r 2 + 1) r! is
r =1
(a) (n + 1) !
6.
α
(b) 6
(c) 8
(d) 10
(b) 30
(c) 31
(d) 32
(b) 25
(c) 47
(d) 48
(b) 24 6
(c) 1212
(d) 48 5
(c) 6
(d) 8
The last non-zero digit in 20! is
(a) 2
11.
φ
The number 24! is divisible by
(a) 6 24
10.
θ
The exponent of 12 in 100! is
(a) 24
9.
δ
The number of naughts standing at the end of 125! is
(a) 29
8.
γ
If 15 ! = 2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 , the value of α − β + γ − δ + θ − φ is
(a) 4
7.
(b) (n + 2) ! − 1
β
(b) 4
The number of prime numbers among the numbers 105 ! + 2,105 ! + 3,105 ! + 4, …, 105 ! + 104
and 105 ! + 105 is
(a) 31
(b) 32
(c) 33
(d) None of these
363
Session 2
Divisibility Test, Principle of Inclusion and Exclusion,
Permutation
Divisibility Test
In decimal system all numbers are formed by the digits 0,
1, 2, 3, …, 9. If a b c d e is a five-digit number in decimal
system, then we can write that.
a b c d e = 10 4 ⋅ a + 10 3 ⋅ b + 10 2 ⋅ c + 10 ⋅ d + e.
Number a b c d e will be divisible
(1) by 2, if e is divisible by 2.
(2) by 4, if 2d + e is divisible by 4.
(3) by 8, if 4c + 2d + e is divisible by 8.
(4) by 2t , if number formed by last t digits is divisible by 2 t .
For example, Number 820101280 is divisible by 2 5
because 01280 is divisible by 2 5 .
(5) by 5, if e = 0 or 5.
(6) by 5t , if number formed by last t digits is
divisible by 5 t .
For example, Number 1128375 is divisible by 5 3
because 375 is divisible by 5 3 .
(7) by 3, if a + b + c + d + e (sum of digits) is divisible
by 3.
(8) by 9, if a + b + c + d + e is divisible by 9.
(9) by 6, if e = even and a + b + c + d + e is divisible by 3.
(10) by 18, if e = even and a + b + c + d + e is divisible
by 9.
(11) by 7, if abcd − 2e is divisible by 7.
For example, Number 6552 is divisible by 7 because
655 − 2 × 2 = 651 = 93 × 7 is divisible by 7.
(12) by 11, if
a1
+2
c4
+3e
4
Sum of digits at
odd places
−
b{
+d
Sum of digits
at even places
is divisible by 11.
For example, Number 15222163 is divisible by 11
because
(1 + 2 + 2 + 6 ) − (5 + 2 + 1 + 3 ) = 0 is divisible by 11.
(13) by 13, if abcd + 4e is divisible by 13.
For example, Number 1638 is divisible by 13 because
163 + 4 × 8 = 195 = 15 × 13 is divisible by 13.
Principle of Inclusion
and Exclusion
1. If A and B are finite sets, from the Venn diagram (i), it
is clear that
U
A
B
(i)
n (A ∪ B ) = n (A) + n (B ) − n (A ∩ B )
and n ( A ′ ∩ B ′ ) = n (U ) − n ( A ∪ B )
2. If A , B and C are three finite sets, then from the Venn
diagram (ii), it is clear that
U
B
A
C
(ii)
n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n (C ) − n ( A ∩ B )
− n ( B ∩ C ) −n (C ∩ A ) + n ( A ∩ B ∩ C )
and n ( A ′ ∩ B ′ ∩ C ′ ) = n (U ) − n ( A ∪ B ∪ C )
Note If A1,A2,A 3, …, An are finite sets, then
n ( A1 ∪ A2 ∪ ... ∪ An ) = Σn( Ai ) − Σn( Ai ∩ Aj ) + Σn( Ai ∩ Aj ∩ Ak ) −
... + ( −1) n Σn( A1 ∩ A2 ∩ ... ∩ An )
and n( A′1 ∩ A′ 2 ∩ ... ∩ A′ n ) = n( U ) − n( A1 ∪ A2∪ ... ∪ An ).
y Example 13. Find the number of positive integers
from 1 to 1000, which are divisible by atleast 2, 3 or 5.
Sol. Let Ak be the set of positive integers from 1 to 1000,
which is divisible by k. Obviously, we have to find
n ( A 2 ∪ A 3 ∪ A 5 ). If [ ⋅ ] denotes the greatest integer
function, then
1000
= [500] = 500
n( A 2 ) = 
 2 
1000
= [333.33] = 333
n( A 3 ) = 
 3 
Chap 05 Permutations and Combinations
1000
= [200] = 200
n( A 5 ) = 
 5 
1000
n( A 2 ∩ A 3 ) = 
. ] = 166
= [16667
 6 
1000
n( A 3 ∩ A 5 ) = 
. ] = 66
= [6667
 15 
1000
= [100] = 100
n( A 2 ∩ A 5 ) = 
 10 
1000
and n ( A 2 ∩ A 3 ∩ A 5 ) = 
= [33.33] = 33
 30 
From Principle of Inclusion and Exclusion
n( A 2 ∪ A 3 ∪ A 5 ) = n( A 2 ) + n( A 3 ) + n( A 5 )
− n (A2 ∩ A3 )
−n ( A 3 ∩ A 5 ) − n ( A 2 ∩ A 5 ) + n ( A 2 ∩ A 3 ∩ A 5 )
= 500 + 333 + 200 − 166 − 66 − 100 + 33 = 734
Hence, the number of positive integers from 1 to 1000,
which are divisible by atleast 2, 3 or 5 is 734.
Note
(i) The number of permutations of n different things taken all at
a time = n Pn = n!
(ii)
n
P0 = 1 , n P1 = n and n Pn − 1 = n Pn = n!
(iii)
n
Pr = n( n − 1 Pr
∴n( A′ 2 ∩ A′ 3 ∩ A′ 5 ) = n( U ) − n( A2 ∪ A3 ∪ A5 ) [here , n( U ) = 1000]
= 1000 − 734 = 266
(iv)
n −1
Pr = ( n − r )
Each of the different arrangements which can be made by
taking some or all of a number of things is called a
permutation. In permutation, order of the arrangement is
important.
Important Results
(v)
n
where, n ∈ N , r ∈W and 0 ≤ r ≤ n.
n ways
2
3
(n–1) ways (n–2) ways
r
(n–r+1) ways
= n (n − 1) (n − 2 )...(n − r + 1)
n (n − 1)(n − 2 )...(n − r + 1) × (n − r )
=
(n − r ) !
n!
=
= RHS
(n − r ) !
K
−1
56
y Example 14. If
Sol. We have,
56
Pr
+6
54
Pr
+3
54
Pr
+4
Pr
+3
54
Pr
+4
54
Pr
+3
⇒ (56)(55) 54
⇒
54
Pr + 6 :
=
30800
1
=
30800
1
Pr + 3 = 30800 : 1, find r P2 .
[from note (iii)]
= 10
⇒ 54 − (r + 4 ) + 1 = 10
r = 41
r
y Example 15. If
P2 =
n+5
P2 = 41 ⋅ 40 = 1640
41
Pn + 1 =
n+5
Sol. We have,
[from note (v)]
11 (n − 1) n + 3
⋅
Pn , find n.
2
Pn + 1
n+3
Pn
(n + 5)(n + 4 ) ⋅n + 3 Pn − 1
=
11(n − 1)
2
11(n − 1)
2
[from note (iii)]
(n + 5)(n + 4 ) 11(n − 1)
=
2
( n + 3 − n + 1)
[from note (v)]
n+3
Pn
=
⇒
(n + 5)(n + 4 ) = 22(n − 1)
⇒
n 2 − 13n + 42 = 0
⇒
(n − 6)(n − 7 ) = 0
∴
Proof LHS = n Pr = Number of ways of filling up r
vacant places simultaneously from n different things
1
Pr
− 3) =
Pr
= ( n − r + 1)
Pr − 1
⇒
Pr = n (n − 1 )(n − 2) K (n − r + 1 )
n!
=
(n − r ) !
− 2)
n
1. The number of permutations of n different
things, taking r at a time is denoted by n Pr or
P (n , r ) or A (n , r ), then
n
n −1
∴
Permutation
= n( n − 1)( n − 2 Pr
− 1)
= n( n − 1)( n − 2)( n − 3 Pr
Note
The number of positive integers from 1 to 1000, which are not
divisible by 2, 3 or 5 is n ( A′2 ∩ A′3 ∩ A′3 ).
365
n = 6, 7
y Example 16. If m + n P2 = 90 and
values of m and n.
Sol. Q
∴
and
∴
m +n
P2 = 90 = 10 × 9 =
10
m −n
P2 = 30, find the
P2
m + n = 10
m −n
...(i)
P2 = 30 = 6 × 5 = P2
6
m −n =6
From Eqs. (i) and (ii), we get
m = 8 and n = 2.
...(ii)
366
Textbook of Algebra
Note Total number of letters in English alphabet = 26
y Example 17. Find the value of r, if
(i) 11 Pr = 990
P5 + 5 ⋅ 8 P4 = 9 Pr
(ii)
8
(iii)
22
20
Pr + 1 :
8
⇒
8
⇒
8
P5 + 5 ⋅ 8 P4 = 9 Pr
P4 (8 − 5 + 1 + 5) = 9 Pr
⇒
[from note (v)]
9 ⋅ P4 = Pr
⇒
9
9
P5 = 9 Pr
∴
[from note (iii)]
r =5
22
(iii) Q
Pr
+1 :
⇒
⇒
20
Pr
+2
22
Pr
+1
20
Pr
+2
Pr
−1
(19 − r ) ⋅ (20 − r ) ⋅ (21 − r ) ⋅
20
Pr
−1
11
=
52
[from note (iii) and (iv)]
(21 − r ) ⋅ (20 − r ) ⋅ (19 − r ) = 52 × 2 × 21
= 14 × 13 × 12
r =7
⇒
∴
n +1
Pn + 1 − 1.
Sol. LHS = 1P1 + 2 ⋅ 2 P2 + 3 ⋅ 3 P3 + ... + n ⋅ n Pn
=
∑r ⋅
r =1
=
r
Pr =
n
∑ {(r + 1) − 1)} ⋅
r
10
P9
Out of these, the number of numbers having zero at the
first place = 9 P8
Hence, required number of numbers =
10
P9 − 9 P8
=9 ×
9!
= 9 × 9!
1!
Pn − 9 Pn − 1
10
r =1
n
∑ {(r + 1) ⋅ r Pr − r Pr )}
n
∑(r + 1 Pr + 1 − r Pr )
n +1
Total number of 9-digit numbers =
which have all different digits =
Pr
[from note (iii)]
r =1
=
Sol. Number of digits are 10 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
Note Total number of n digit numbers ( 1 ≤ n ≤ 10 ) ,
r =1
=
y Example 21. Find the total number of 9-digit
numbers which have all different digits.
= 10 × 9 P8 − 9 P8 = 9 × 9 P8
y Example 18. Prove that
1
P1 + 2 ⋅ 2P2 + 3 ⋅ 3P3 + ... + n ⋅ nPn =
n
Sol. The signals can be made by using one or more flags at a
time. Hence, by the fundamental principle of addition, the
total number of signals
= 4 + ( 4 × 3) + ( 4 × 3 × 2) + ( 4 × 3 × 2 × 1)
= 4 + 12 + 24 + 24 = 64
11
=
52
20
y Example 20. How many different signals can be
given using any number of flags from 4 flags of
different colours?
= 4 P1 + 4 P2 + 4 P3 + 4 P4
= 11 : 52
22 ⋅ 21 ⋅
[except vowels]
(iv) Words which do not contains any vowels are
SKY, FLY, TRY, ...
 8P

P4  8 5 + 5 = 9 Pr
 P4

8
[weak vowels]
(iv) Words which contains all vowels are
EDUCATION, EQUATION, ...
r =3
(ii) Q
[strong vowels]
(ii) W and Y an half vowels.
(iii) Number of consonants = 21
i.e., B, C, D, F, G, ..., Y, Z
Pr + 2 = 11 : 52
Sol. (i) Q 11 Pr = 990 = 11 × 10 × 9 = 11P3
∴
(i) Number of vowels = 5
i.e., A, E, I, O, U
Pn + 1 − 1P1 =
n +1
Pn + 1 − 1
= RHS
y Example 19. Determine the number of permutations
of the letters of the word ‘SIMPLETON’ taken all at
a time.
Sol. There are 9 letters in the word ‘SIMPLETON’ and all the
9 letters are different. Hence, the number of permutations
taking all the letters at a time
= 9 P9 = 9 ! = 362880
y Example 22. A 5-digit number divisible by 3 is to be
formed using the numbers 0, 1, 2, 3, 4 and 5 without
repetition. Find total number of ways in which this can
be done.
Sol. A number will be divisible by 3, if sum of the digits in
number be divisible by 3.
Here, 0 + 1 + 2 + 3 + 4 + 5 = 15 , which is divisible by 3.
Therefore, the digit that can be left out, while the sum still
is multiple of 3, is either 0 or 3.
If 0 left out
Then, possible numbers = 5 P5 = 5! = 120
If 3 left out
Then, possible numbers = 5 P5 − 4 P4 = 5! − 4 ! = 120 − 24 = 96
Hence, required total numbers = 120 + 96 = 216
Chap 05 Permutations and Combinations
y Example 23. A 5-digit number is formed by the digits
1, 2, 3, 4, 5 without repetition. Find the number of the
numbers, thus formed divisible by 4.
Sol. Let a 5-digit number be abcde.
It will be divisible by 4, if 2d + e is divisible by 4.
⇒
2d + e is divisible by 4 ∴ e must be even.
123
Even
⇒
e

2 d +  is divisible by 4

2
1424
3
Should be even
Then, e = 2, d = 1, 3, 5
 Total four cases
and e = 4, d = 2

3! = 24
∴ Required number of ways = 4 × {
Number of ways
filling abc after filling de.
Aliter A number will be divisible by 4, if the last two digits
of the number is divisible by 4, then for divisible by 4, last
two digits 12 or 24 or 32 or 52
3! ways
4 ways
Hence, the number formed is divisible by 4 = 3! × 4 = 24.
y Example 24. Find the number of permutations of
letters ab c d e f g taken all together if neither ‘beg’
nor ‘cad’ pattern appear.
Sol. The total number of permutations without any restriction is 7!
n (U ) = 7 ! = 5040
Let n ( A ) be the number of permutations in which ‘beg’
pattern always appears
b e gac d f
i.e.,
n ( A ) = 5! = 120
and let n ( B ) be the number of permutations in which ‘cad’
pattern always appears
c a db e f g
i.e.,
n ( B ) = 5! = 120
Now, n ( A ∩ B ) = Number of permutations in which ‘beg’
and ‘cad’ pattern appear
begcad f
i.e.,
n ( A ∩ B ) = 3! = 6
Hence, the number of permutations in which ‘beg’ and ‘cad’
patterns do not appear is n ( A ′ ∩ B ′ ).
or
n ( A ′ ∩ B ′ ) = n (U ) − n ( A ∪ B )
= n (U ) − [n ( A ) + n ( B ) − n ( A ∩ B )]
= 5040 − 120 − 120 + 6 = 4806
2. The number of permutations of n things taken all
at a time, p are alike of one kind, q are alike of
second kind and r are alike of a third kind and
n!
the rest n − ( p + q + r ) are all different is
p!q !r !
367
Proof Let the required number of permutations be x .
Since, p different things can be arranged among
themselves in p ! ways, therefore if we replace p
identical things by p different things, which are also
different from the rest of things, the number of
permutations will become x × p !
Again, if we replace q identical things by q different
things, the number of permutations will become
( x × p !) × q !
Again, if we replace r identical things by r different
things, the number of permutations will become
( x × p ! × q !) × r ! . Now, all the n things are different
and therefore, number of permutations should be n ! .
Thus, x × p ! × q ! × r ! = n !
n!
x=
∴
p!q!r !
Remark
The above theorem can be extended further i.e. if there are
n things taken all at a time, p1 are alike of one kind, p2 are alike of
second kind,p3 are alike of third kind, ..., pr are alike of rth kind
such that p1 + p2 + p3 + K + pr = n, the number of permutations
n!
of these n things is
.
p1 ! p2 ! p3 ! ... pr !
y Example 25. How many words can be formed with
the letters of the word ‘ARIHANT’ by rearranging them?
Sol. Here, total letters 7, in which 2A’s but the rest are
7!
different. Hence, the number of words formed =
= 2520
2!
y Example 26. Find the number of permutations of the
letters of the words ‘DADDY DID A DEADLY DEED’.
Sol. Here, total letters 19, in which 9D ’s, 3A’s, 2Y’s, 3E’s and
rest occur only once.
19 !
∴ Required number of permutations =
9 ! × 3! × 2! × 3!
y Example 27. How many words can be formed with
the letters of the words
(i) HIGH SCHOOL and
(ii) INTERMEDIATE by rearranging them?
Sol. (i) Here, total letters are 10, in which 3H’s and 2O’s, but
the rest are different. Hence, the number of words
10!
formed =
3! 2!
(ii) Here, total letters are 12, in which 2I’s, 2T’s and 3E’s
but the rest are different. Hence, the number of words
12!
Note [For Remember]
formed =
2! 2! 3!
High School = 10 th class = Total number of letters are 10
Intermediate = 12 th class = Total number of letters
are 12
368
Textbook of Algebra
3. The number of permutations of n different
things taken r at a time when each thing may be
repeated any number of times is n r .
Proof Since, the number of permutations of n
different things taken r at a time = Number of ways in
which r blank places can be filled by n different
things.
Clearly, the first place can be filled in n ways. Since,
each thing may be repeated, the second place can be
filled in n ways. Similarly, each of the 3rd, 4th, K, r th
place can be filled in n ways.
By multiplication principle, the number of
permutations of n different things taken r at a time
when each thing may be repeated any number of
times
= n × n × n × K × r factors
= nr
Corollary When r = n
i.e., the number of permutations of n different things,
taken all at a time, when each thing may be repeated
any number of times in each arrangements is n n .
y Example 28. A child has four pockets and three
marbles. In how many ways can the child put the
marbles in its pockets?
Sol. The first marble can be put into the pocket in 4 ways, so
can the second. Thus, the number of ways in which the
child can put the marbles = 4 × 4 × 4 = 4 3 = 64 ways
y Example 29. There are m men and n monkeys
(n > m ). If a man have any number of monkeys. In how
many ways may every monkey have a master?
Sol. The first monkey can select his master by m ways and
after that the second monkey can select his master again
by m ways, so can the third and so on.
All monkeys can select master = m × m × m K upto n
factors = (m )n ways
y Example 30. How many four digit numbers can be
formed by using the digits 1, 2, 3, 4, 5, 6, 7, if atleast
one digit is repeated ?
Sol. The numbers that can be formed when repetition of digits
is allowed are 7 4 = 2401.
The numbers that can be formed when all the digits are
distinct when repetition is not allowed are 7 P4 = 840.
Therefore, the numbers that can be formed when atleast
one digit is repeated = 7 4 − 7 P4
= 2401 − 840 = 1561
y Example 31. In how many ways can 4 prizes be distributed
among 5 students, if no student gets all the prizes?
Sol. The number of ways in which the 4 prizes can be given
away to the 5 students, if a student can get any number
of prizes = 54 = 625.
Again, the number of ways in which a student gets all the 4
prizes = 5, since there are 5 students and any one of them
may get all the four prizes.
Therefore, the required number of ways in which a student
does not get all the prizes = 625 − 5 = 620.
y Example 32. Find the number of n-digit numbers, which
contain the digits 2 and 7, but not the digits 0, 1, 8, 9.
Sol. The total number without any restrictions containing
digits 2, 3, 4, 5, 6, 7 is n (U ) = 6n .
The total number of numbers that contain 3, 4, 5, 6, 7 is
n ( A ) = 5n .
The total number of numbers that contain 2, 3, 4, 5, 6 is
n ( B ) = 5n .
The total number of numbers that contain 3, 4, 5, 6 is
n ( A ∩ B ) = 4n .
The total number of numbers that do not contain digits 2
and 7 is n ( A ∪ B )
i.e., n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B )
= 5n + 5n − 4 n = 2 (5n ) − 4 n
Hence, the total number of numbers that contain 2 and 7 is
n( A ′ ∩ B ′ )
∴ n ( A ′ ∩ B ′ ) = n (U ) − n ( A ∪ B ) = 6n − 2 ⋅ (5n ) + 4n
y Example 33. Show that the total number of
permutations of n different things taken not more than
r at a time, when each thing may be repeated any
n(n r − 1)
number of times is
.
(n − 1)
Sol. Given, total different things = n
The number of permutations of n things taken one at a time
= n P1 = n , now if we taken two at a time (repetition is
allowed), then first place can be filled by n ways and second
place can again be filled in n ways.
∴The number of permutations of n things taken two at a time
= n P1 × n P1 = n × n = n 2
Similarly, the number of permutations of n things taken
three at a time = n 3
M
M
M
M
M
The number of permutations of n things taken r at a
time = n r . Hence, the total number of permutations
= n + n 2 + n 3 + K + nr
n ( n r − 1)
[sum of r terms of a GP]
=
( n − 1)
Chap 05 Permutations and Combinations
#L
1.
Exercise for Session 2
If nP5 = 20 ⋅ nP3, then n equals
(a) 4
2.
(b) 8
9
If
m+ n
(b) 14
P2 = 56 and
If
Pn − 1 :
11!
2! 2!
(b) 20
(c) 30
(d) 60
(b) 12
(c) 15
(d) 18
(b) 7560
(c) 10080
(d) 576
(b) 480
(c) 624
(d) 696
(b) 30
(c) 34
(d) None of these
(b)
11!
2!
(c)
11!
2! 2! 2!
(d) 11!
(b) 20
(c) 40
(d) 120
(b) 210
(c) 133
(d) 72
(b) 75
(c) 35
(d) 2520
(b) 671
(c) 1256
(d) 1296
The number of 4-digit numbers that can be made with the digits 1, 2, 3, 4 and 5 in which atleast two digits are
identical, is
(a) 45 − 5 !
16.
(d) 6
Four die are rolled. The number of ways in which atleast one die shows 3, is
(a) 625
15.
(b) 24
Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose each of them can leave
the cabin independently at any floor beginning with the first. The total number of ways in which each of the five
persons can leave the cabin at any one of the floor, is
(a) 57
14.
(c) 120
Pn = 7 : 10, then P3 equals
A train time table must be compiled for various days of the week so that two trains twice a day depart for three
days, one train daily for two days and three trains once a day for two days. How many different time tables can
be compiled?
(a) 140
13.
(d) 80
n
Six identical coins are arranged in a row. The number of ways in which the number of tails is equal to the
number of heads, is
(a) 9
12.
2n − 1
The number of words can be formed with the letters of the word ‘MATHEMATICS’ by rearranging them, is
(a)
11.
(c) 60
P3
equals
P2
The number of all five digit numbers which are divisible by 4 that can be formed from the digits 0, 1, 2, 3, 4
(without repetition), is
(a) 36
10.
n
Total number of words that can be formed using all letters of the word ‘DIPESH’ that neither begins with ‘I’ nor
ends with ‘D’ is equal to
(a) 504
9.
P3 = 24, then
The number of nine non-zero digits such that all the digits in the first four places are less than the digit in the
middle and all the digits in the last four places are greater than that in the middle, is
(a) 48
8.
m−n
If a denotes the number of permutations of x + 2 things taken all at a time, b the number of permutations of x
things taken 11 at a time and c the number of permutations of ( x − 11) things taken all at a time such that
a = 182 bc, the value of x is
(a) 10
7.
(d) 16
In a train, five seats are vacant, the number of ways three passengers can sit, is
(a) 10
6.
(c) 15
m
(b) 40
2n + 1
(a) 60
5.
(d) 7
n
(a) 20
4.
(c) 6
If P5 + 5 ⋅ P4 = Pr , then n + r equals
9
(a) 13
3.
369
(b) 505
(c) 600
(d) 120
There are unlimited number of identical balls of three different colours. How many arrangements of atmost 7
balls in a row can be made by using them?
(a) 2187
(b) 343
(c) 399
(d) 3279
Session 3
Number of Permutations Under Certain Conditions,
Circular Permutations, Restricted Circular Permutations
Number of Permutations
Under Certain Conditions
(i) Number of permutations of n different things, taken r
at a time, when a particular thing is to be always
included in each arrangement, is r ⋅ n − 1 Pr − 1
Corollary Number of permutations of n different
things, taken r at a time, when p particular things is
to be always included in each arrangement, is
p ! (r − ( p − 1)) n − p Pr − p .
(ii) Number of permutations of n different things, taken r
at a time, when a particular thing is never taken in
each arrangement, is
n −1
Pr .
(iii) Number of permutations of n different things, taken
all at a time, when m specified things always come
together, is
m ! × (n − m + 1) !
(iv) Number of permutations of n different things, taken
all at a time, when m specified things never come
together, is
n ! − m ! × (n − m + 1) !
y Example 34. How many permutations can be made
out of the letters of the word ‘TRIANGLE’ ? How many
of these will begin with T and end with E ?
Sol. The word ‘TRIANGLE’ has eight different letters, which
can be arranged themselves in 8! ways.
∴ Total number of permutations = 8 ! = 40320
Again, when T is fixed at the first place and E at the last
place, the remaining six can be arranged themselves in
6 ! ways.
∴The number of permutations which begin with T and end with
E = 6! = 720.
y Example 35. In how many ways can the letters of
the word ‘INSURANCE’ be arranged, so that the vowels
are never separate?
Sol. The word ‘INSURANCE’ has nine different letters,
combine the vowels into one bracket as (IUAE) and treating them as one letter we have six letters viz.
(IUAE) N S R N C and these can be arranged among
6!
themselves in
ways and four vowels within the bracket
2!
can be arranged themselves in 4 ! ways.
6!
∴ Required number of words =
× 4 ! = 8640
2!
y Example 36. How many words can be formed with
the letters of the word ‘PATALIPUTRA’ without
changing the relative positions of vowels and
consonants?
Sol. The word ‘PATALIPUTRA’ has eleven letters, in which
2P’s, 3A’s, 2T’s, 1L, 1U, 1R and 1I. Vowels are AAIUA
5!
These vowels can be arranged themselves in
= 20 ways.
3!
The consonants are PTLPTR these consonants can be
6!
arranged themselves in
= 180 ways
2!2!
∴ Required number of words
= 20 × 180 = 3600 ways.
y Example 37. Find the number of permutations
that can be had from the letters of the
word ‘OMEGA’
(i) O and A occuping end places.
(ii) E being always in the middle.
(iii) Vowels occuping odd places.
(iv) Vowels being never together.
Sol. There are five letters in the word ‘OMEGA’.
(i) When O and A occuping end places
i.e.,
M E G (OA)
the first three letters (M, E, G) can be arranged
themselves by 3 ! = 6 ways and last two letters (O, A)
can be arranged themselves by 2 ! = 2 ways.
∴ Total number of such words
= 6 × 2 = 12 ways.
Chap 05 Permutations and Combinations
(ii) When E is the fixed in the middle, then there are four
places left to be filled by four remaining letters O, M,
G and A and this can be done in 4 ! ways.
∴ Total number of such words = 4 ! = 24 ways.
(iii) Three vowels (O, E, A) can be arranged in the odd
places in 3 ! ways (1st, 3rd and 5th) and the two
consonants (M, G) can be arranged in the even places
in 2 ! ways (2nd and 4th)
∴ Total number of such words
= 3 ! × 2 ! = 12 ways.
(iv) Total number of words = 5 ! = 120
Combine the vowels into one bracket as (OEA) and
treating them as one letter, we have
(OEA), M, G and these can be arranged themselves in
3 ! ways and three vowels with in the bracket can be
arranged themselves in 3 ! ways.
∴ Number of ways when vowels come together
= 3 ! × 3 ! = 36 ways.
Hence, number of ways when vowels being
never together = 120 − 36 = 84 ways.
Circular Permutations
D
C
A
D
E
C
D
B
B
C
A
B
E
A
(i)
(ii)
B
A
C
D
(iv)
A
B
E
C
(ii) Arrangements of beads or flowers
(all different) around a circular
necklace or garland
Consider five beads A, B, C , D and E in a necklace or five
flowers A, B, C , D and E in a garland, etc. If the necklace or
garland on the left is turned over, we obtain the
arrangement on the right i.e. anti-clockwise and clockwise
order of arrangement is not different we will get
arrangements as follows:
We see that arrangements in figures are not different.
Flip to right
D
C
(iii)
E
D
(v)
We see that, if 5 persons are sitting at a round table, they
can be shifted five times and five different arrangements.
Thus, obtained will be the same, because anti-clockwise
order of A, B, C , D and E does not change.
But if A, B, C , D and E are sitting in a row and they are
shifted in such an order that the last occupies the place of
first, then the five arrangements will be different. Thus, if
there are 5 things, then for each circular arrangement
number of linear arrangements is 5.
C
B
A
Consider five persons A, B, C , D and E on the
circumference of a circular table in order which has no
head now, shifting A, B, C , D and E one position in
anti-clockwise direction we will get arragements as
follows
E
Similarly, if n different things are arranged along a circle
for each circular arrangement number of linear
arrangements is n.
Therefore, the number of linear arrangements of n
different things = n × number of circular arrangements of
n different things
E
(i) Arrangements round a circular table
371
D
B
E
A
Then, the number of circular permutations of n different
1
things taken all at a time is (n − 1) !, if clockwise and
2
anti-clockwise orders are taken as not different.
y Example 38. Find the number of ways in which 12
different beads can be arranged to form a necklace.
Sol. 12 different beads can be arranged among themselves in a
circular order in (12 − 1) ! = 11 ! ways. Now, in the case of
necklace, there is no distinction between clockwise and
anti-clockwise arrangements. So, the required number of
1
arrangements = (11 !).
2
y Example 39. Consider 21 different pearls on a
necklace. How many ways can the pearls be placed in
on this necklace such that 3 specific pearls always
remain together?
Sol. After fixing the places of three pearls, treating 3 specific
pearls = 1 unit. So, we have now
18 pearls + 1 unit = 19 and the number of arrangement will
be (19 − 1) ! = 18 !
Also, the number of ways of 3 pearls can be arranged
between themselves is 3 ! = 6.
Since, there is no distinction between the clockwise and
anti-clockwise arrangements.
1
So, the required number of arrangements = 18 !⋅ 6 = 3 (18 !).
2
372
Textbook of Algebra
Restricted Circular
Permutations
y Example 42. In how many different ways can five
boys and five girls form a circle such that the boys and
girls alternate?
Case I If clockwise and anti-clockwise orders are taken as
different, then the number of circular permutations of n
different things taken r at a time.
Sol. After fixing up one boy on the table, the remaining can
be arranged in 4 ! ways but boys and girls are to alternate.
There will be 5 places, one place each between two boys
these five places can be filled by 5 girls in 5 ! ways.
=
n
B1
Pr 1
n!
= ⋅
r
r (n − r ) !
( n − 1) !
[result (5) (i)]
y Example 40. In how many ways can 24 persons be
seated round a table, if there are 13 sets ?
Sol. In case of circular table, the clockwise and anti-clockwise
orders are different, the required number of circular
24
P13
24 !
.
permutations =
=
13
13 × 11 !
⇒ n ! = n × number of circular arrangements of n
different things
⇒ Number of circular arrangements of n different things
n!
=
= ( n − 1) !
n
Hence, the number of circular permutations of n different
things taken all at a time is (n − 1) !, if clockwise and
anti-clockwise orders are taken as different.
y Example 41. Find the number of ways in which three
Americans, two British, one Chinese, one Dutch and
one Egyptian can sit on a round table so that persons
of the same nationality are separated.
B4
B3
Hence, by the principle of multiplication, the required
number of ways = 4 ! × 5 ! = 2880.
y Example 43. 20 persons were invited to a party. In
how many ways can they and the host be seated at a
circular table ? In how many of these ways will two
particular persons be seated on either side of the host?
Sol. I Part Total persons on the circular table
= 20 guest + 1 host = 21
They can be seated in (21 − 1) ! = 20 ! ways.
II Part After fixing the places of three persons
(1 host + 2 persons).
Treating (1 host + 2 persons) = 1 unit, so we have now
{(remaining 18 persons + 1 unit) = 19} and the number of
arrangement will be (19 − 1) ! = 18 ! also these two particular
persons can be seated on either side of the host in 2 ! ways.
P20
Sol. The total number of persons without any restrictions is
n (U ) = (8 − 1)!
P19
n ( A ∪ B ) = n ( A ) + nB ) − n ( A ∩ B )
= 720 + 1440 − 288= 1872
Hence,
n ( A ∩ B ′ ) = n (U ) − n ( A ∪ B )
= 5040 − 1872
= 3168
P2
P3
P4
P5
P16
P15
P14
P13
= 720
When, two British ( B1, B 2 ) are sit together
= 1440
When, three Americans ( A1, A 2 , A 3 ) and two British ( B1, B 2 )
are sit together n ( A ∩ B ) = 4 ! × 3! × 2! = 288
P1
P17
n ( A ) = 5! × 3!
n ( B ) = 6! × 2!
H
P18
= 7! = 5040
When, three Americans ( A1, A 2 , A 3 ) are sit together,
∴
B5
B2
Note For checking correctness of formula, put r = n, then we get
P6
P7
P12 P P P9
11 10
P8
Hence, the number of ways of seating 21 persons on the
circular table such that two particular persons be seated on
either side of the host = 18 ! × 2 ! = 2 × 18 !
Case II If clockwise and anti-clockwise orders are taken as
not different, then the number of circular permutations of n
n
P
1
n!
different things taken r at a time = r = ⋅
2r
2r (n − r ) !
Note
For checking correctness of formula put r = n, then we get
( n − 1)!
[result (5) (ii)]
2
Chap 05 Permutations and Combinations
373
y Example 44. How many necklace of 12 beads each can be made from 18 beads of various colours?
Sol. In the case of necklace, there is no distinction between the clockwise and anti-clockwise arrangements, the required
number of circular permutations.
=
#L
1.
Exercise for Session 3
How many words can be formed from the letters of the word ‘COURTESY’ whose first letter is C and the last
letter is Y ?
(a) 6!
2.
(b) 8!
3
(7) !
2
(b) 2 (7) !
(d) 60
(b) 4 P3 × 5 !
(c) 6P3 × 5 !
(d) 5P3 × 3 !
(b) n Pn 1
(c) nC n1 − 1
(d) n Pn1 − 1
(b) 7 ! × 6 !
(c) (6 !)2
(d) (7 !)2
The number of ways that 8 beads of different colours be string as a necklace, is
(a) 2520
(b) 2880
(c) 4320
(d) 5040
If 11 members of a committee sit at a round table so that the President and secretary always sit together, then
the number of arrangements, is
(a) 9 ! × 2
11.
(c) 24
In how many ways can 7 men and 7 women can be seated around a round table such that no two women can
sit together?
(a) 7!
10.
(b) 12
There are n numbered seats around a round table. Total number of ways in whichn1 (n1 < n ) persons can sit
around the round table, is equal to
(a) nCn 1
9.
(b) 72
(d) 360
In how many ways can 5 boys and 3 girls sit in a row so that no two girls are sit together?
(a) 5 ! × 3 !
8.
(d) 3 (7) !
How many words can be made from the letters of the word ‘DELHI’, if L comes in the middle in every word?
(a) 6
7.
5
(7) !
2
All the letters of the word ‘EAMCET’ are arranged in all possible ways. The number of such arrangements in
which two vowels are not adjacent to each other, is
(a) 54
(c) 114
6.
(c)
(b) 3 ! × 4 !
4!
(d)
3!
(c) 3!
5.
(d) 2 (7) !
The number of words can be formed from the letters of the word ‘MAXIMUM’, if two consonants cannot occur
together, is
(a) 4!
4.
(c) 2 (6) !
The number of words that can be made by writing down the letters of the word ‘CALCULATE’ such that each
word starts and ends with a consonant, is
(a)
3.
18 × 17 × 16 × 15 × 14 × 13 ! 119 × 13 !
18 !
P12
=
=
=
6 × 5 × 4 × 3 × 2 × 1 × 24
2
2 × 12 6 ! × 24
18
(b) 10!
(c) 10 ! × 2
(d) 11!
In how many ways can 15 members of a council sit along a circular table, when the secretary is to sit on one
side of the Chairman and the deputy secretary on the other side?
(a) 12 ! × 2
(b) 24
(c) 15 ! × 2
(d) 30
Session 4
Combination, Restricted Combinations
Combination
(ii) n C r = 0, if r > n
Each of the different groups or selections which can be
made by some or all of a number of given things without
reference to the order of the things in each group is called
a combination.
(iii) n C 0 = n C n = 1 , n C 1 = n
Important Result
(vi) If n C x = n C y ⇒ x = y or x + y = n
(1) The number of combinations of n different
things taken r at a time is denoted by n C r or
n 
C (n , r ) or   .
r 
Cr =
n!
r !( n − r )!
n
2
(v) n C r = n C n − r , if r >
(vii) n C r + n C r − 1 =
(viii) n C r =
n
⋅
r
n −1
n +1
[Pascal’s rule]
Cr
Cr − 1
(ix) n ⋅ n − 1 C r − 1 = (n − r + 1) ⋅ n C r − 1
Then,
n
(iv) n Pr = n C r , if r = 0 or 1
[ 0 ≤ r ≤ n]
n
Pr
=
r!
n ( n − 1) ( n − 2) K ( n − r + 1)
=
, n ∈ N and r ∈W
r ( r − 1) ( r − 2) K 2 ⋅ 1
Proof Let the number of combinations of n different
things taken r at a time be n C r .
n
(x)
n
Cr
Cr − 1
n −r +1
r
=
n
2
n −1 n +1
or
(b) If n is odd , n C r is greatest for r =
2
2
(xi) (a) If n is even , n C r is greatest for r =
(xii) n C 0 + n C 1 + n C 2 + K + n C n = 2 n
(xiii) n C 0 + n C 2 + n C 4 +K
Now, each combination consists of r different things and
these r things can be arranged among themselves in r !
ways.
Thus, for one combination of r different things, the
number of arrangements is r ! .
= n C 1 + n C 3 + n C 5 + K = 2n − 1
(xiv) 2n + 1 C 0 +
(xv) n C n +
2n + 1
n +1
C1 +
2n + 1
n +2
C2 +
Cn +
C 2 +K +
n+3
r ! × Cr
…(i)
+
But number of permutations of n different things taken r at
…(ii)
a time is n Pr .
From Eqs. (i) and (ii), we get
r ! × n C r = n Pr =
∴
n
Cr =
n!
(n − r ) !
n!
, r ∈W and n ∈ N
r ! (n − r ) !
Note the following facts:
(i) n C r is a natural number
y Example 45. If
Sol. We have,
15
15
C 3r = 15C r
+3
3r = r + 3
3r + r + 3 = 15
2r = 3 or 4r = 12
3
⇒
r = or r = 3
2
3
but r ∈W , so that r ≠
2
∴
r =3
r
2n − 1
Cn =
C 3r = 15C r + 3 , find r C 2 .
⇒
or
⇒
Then,
C n = 2 2n
Cn + K
Hence, for n C r combinations, number of arrangements is
n
2n + 1
C 2 = 3C 2 = 3C 1 = 3
2n
Cn + 1
Chap 05 Permutations and Combinations
y Example 46. If n C 9 = n C 7 , find n.
n
Sol. We have,
⇒
C 9 = nC 7 ⇒ n = 9 + 7
∴
n
=2
r
⇒
n = 2r
On solving Eqs. (i) and (ii), we get n = 10 and r = 5.
[Q9 ≠ 7]
n = 16
y Example 47. Prove that
 n
 n   n   n + 2
  +2
+
 =

r 
 r − 1  r − 2  r 
y Example 50. If n C r − 1 = 36, n C r = 84 and
n 
Sol. ∴   = n C r
r 
Sol. Here,
n
= (n C r + n C r
=
n +1
Cr +
−1
+ n Cr
+ (n C r
− 1)
n
n +1
−1 =
Cr
−2
−1
n+2
+ n Cr
2n
2n
⇒
n
C3
C3
=
n +1
C r + 1 : nC r :
n +1
Cr
⇒
⇒
n
n −1
=
6n + 6 = 11r + 11
6n − 11r = 5
and
Cr
n −1
Cr
n
⋅
r
n −1
Cr
n −1
Cr
n
+1
Cr
=
n − (r + 1) + 1 126
=
( r + 1)
84
−1
−1
n −r 3
=
r +1 2
−1
⇒
2n − 2r = 3r + 3
⇒
5r − 2n = − 3
or
10r − 4 n = − 6
∴
…(ii)
r =3
y Example 51. Prove that product of r consecutive
positive integers is divisible by r ! .
⇒ 3n = 18
Sol. Let r consecutive positive integers be (m),
(m + 1), (m + 2), K , (m + r − 1), where m ∈ N .
∴ Product = m (m + 1) (m + 2) K (m + r − 1)
(m − 1)! m (m + 1) (m + 2) K (m + r − 1)
=
(m − 1)!
C r − 1 = 11 : 6 : 3 ,
=


n n −1
 n
Cr
Q C r = r ⋅
− 1
(m + r − 1)!
r ! ⋅ (m + r − 1)!
=
(m − 1)!
r ! (m − 1)!
[Q m + r − 1C r is natural number]
= r!⋅ m +r
−1
Cr ,
which is divisible by r ! .
…(i)
y Example 52. Evaluate
47
6
=
3
6
=
3
 n Cr
n − r + 1
=

Q n
r

 C r − 1
10r − 27 = 3 ⇒ 10r = 30
11
6
n + 1 C r 11
⋅
=
6
r + 1 n Cr
n + 1 11
=
r +1 6
n
⇒
+1
Cr
n
⇒
⇒
Cr
…(i)
From Eq. (i), we get
find the values of n and r.
Sol. Here,
10r − 3n = 3
n =9
8 n − 4 = 11n − 22
n =6
y Example 49. If
 n Cr
n − r + 1
=

Q n
r

 C r − 1
On subtracting Eq. (ii) from Eq. (i), we get
11
1
2n ( 2n − 1) (2n − 2)
4 ( 2n − 1)
1⋅2⋅3
= 11 ⇒
= 11
n ( n − 1) ( n − 2)
( n − 2)
1⋅2⋅3
⇒
∴
3n − 3r + 3 = 7 r
⇒
C 3 : n C 3 = 11 : 1
84
36
⇒
and
Sol. We have,
⇒
Cr − 1
=
n −r +1 7
=
r
3
n
Cr
Cr
⇒
− 2)
C 3 : n C 3 = 11 : 1, find the value of n .
2n
n
⇒
n + 2
=
 = RHS
 r 
y Example 48. If
…(ii)
C r + 1 = 126, find r.
n 
 n   n 
∴ LHS =   + 2 
 +

r 
r − 1 r − 2
= n Cr + 2 n Cr
375
n n −1
 n
Cr
Q C r = r ⋅

− 1

C4 +
3
5
∑ 50 − j C 3 + ∑ 56 − k C 53 − k .
j =0
Sol. We have,
k =0
47
C4 +
3
∑ 50 − j C 3
j =0
+
5
∑ 56 − k C 53 − k
k =0
376
Textbook of Algebra
=
C4 +
47
3
∑ 50 − j C 3
5
∑ 56 − k C 3 [Q nC r
+
j =0
=
C 4 + ( 50C 3 +
+ ( 56C 3 +
=
C4 +
47
= ( C4 +
47
=
48
=
49
49
M
C3 +
49
+
52
C3) +
48
+
52
C3 +
49
C3 +
50
M
=
54
48
47
C4 +
C4 +
C3 +
50
C3 +
54
C3 +
50
C3 +
54
C3 +
49
C3 +
53
C 3 + 51C 3 )
52
C3 +
C3 +
55
56
C3
C3 + C3
51
C3 +
C3 +
55
C 3 + 51C 3 + K +
50
56
C3
56
C3
56
C3
M
2n
⇒
∴
Cr
Cr
=
C4
2n − r + 1
r
>1
2n
Cr
−1
2n
Cr
−1
<
2n
[for 1 ≤ r ≤ n]
Cr
2n
C0 <
2n
C 1,
2n
C1 <
2n
C 2 , K,
2n
Cn − 1 <
2n
but
2n
2n
C0 <
Cr =
2n
C 2n <
2n
2n
C1 <
2n
C2 <
2n
C3 < K <
2n
Cn − 1 <
2n
C 2n − 1 <
C 2n − 2 <
C 2n − 3 < K <
2n
Hence, the greatest value of
2n
2n
C r is
2n
Cn + 1 <
2n
Cn .
y Example 54. Thirty six games were played in a
football tournament with each team playing once
against each other. How many teams were there?
Sol. Let the number of teams be n.
Then number of matches to be played is n C 2 = 36
⇒
⇒
n
9 ×8 9
= C2
C2 =
1×2
n =9
(iii) The number of combinations of n different objects
taking r at a time when, p particular objects are
always included and q particular objects are always
excluded = n − p −q C r − p
(i) The number of selections of r consecutive things out of
n things in a row = n − r + 1.
(ii) The number of selections of r consecutive things out of
n, if r < n
.
n things along a circle = 
1, if r = n
y Example 55. In how many ways can a cricket, eleven
players by chosen out of a batch 15 players, if
(i) a particular is always chosen.
(ii) a particular player is never chosen?
Sol. (i) Since, particular player is always chosen. It means
that 11 − 1 = 10 players are selected out of the
remaining 15 − 1 = 14 players.
14 ⋅ 13 ⋅ 12 ⋅ 11
= 1001
1⋅2⋅3⋅ 4
(ii) Since, particular player is never chosen. It means that
11 players are selected out of the remaining 15 − 1 = 14
players.
Cn
C 2n − r , it follows that
2n
(ii) The number of combinations of r things out of n
different things, such that k particular things are not
together in any selection = n C r − n − k C r − k
=
Cn
On combining all inequalities, we get
⇒
Cr .
∴ Required number of ways = 14C 10 = 14C 4
On putting r =1, 2, 3, …, n,
then
n −k
Note
57
 n Cr
n − r + 1
=
Q n

2n
r
C r −1
 C r − 1

2 ( n − r ) + ( r + 1) 1 + 2( n − r ) + 1
=
=
>1
r
r
2n
(i) The number of selections (combinations) of r objects
out of n different objects, when
(a) k particular things are always included= n − k C r − k .
(b) k particular things are never included =
y Example 53. Prove that the greatest value of
2n
C r ( 0 ≤ r ≤ 2n ) is 2n C n (for 1 ≤ r ≤ n).
Sol. We have,
Restricted Combinations
C 3 + 51C 3
C3 +
53
M
C3 =
C3)
53
C3 + K +
56
47
C3 +
C3 +
C3 +
M
56
48
C3 +
55
C3 +
48
C3 +
49
47
C4 +
k =0
C3 +
47
= nC n − r ]
2n
Cn
∴ Required number of ways = 14C 11 = 14C 3
=
14 ⋅ 13 ⋅ 12
= 364
1 ⋅ 2⋅ 3
y Example 56. How many different selections of 6
books can be made from 11 different books, if
(i) two particular books are always selected.
(ii) two particular books are never selected?
Sol. (i) Since, two particular books are always selected. It
means that 6 − 2 = 4 books are selected out of the
remaining 11 − 2 = 9 books.
9 ⋅8⋅7 ⋅6
∴ Required number of ways = 9C 4 =
= 126.
1⋅2⋅3⋅ 4
377
Chap 05 Permutations and Combinations
(ii) Since, two particular books are never selected. It
means that 6 books are selected out of the remaining
11 − 2 = 9 books.
∴ Required number of ways = 9C 6
= 9C 3 =
9 ⋅8⋅7
= 84.
1⋅2⋅3
y Example 57. A person tries to form as many different
parties as he can, out of his 20 friends. Each party
should consist of the same number. How many friends
should be invited at a time? In how many of these
parties would the same friends be found?
Sol. Let the person invite r number of friends at a time. Then,
the number of parties are 20 C r , which is maximum,
when r = 10.
If a particular friend will be found in p parties, then p is the
number of combinations out of 20 in which this particular
friend must be included. Therefore, we have to select 9
more from 19 remaining friends.
Hence, p = 19C 9
(2) The number of ways (or combinations) of n
different things selecting atleast one of them is
2n − 1 . This can also be stated as the total
number of combinations of n different things.
Proof For each things, there are two possibilities, whether
it is selected or not selected.
Hence, the total number of ways is given by total
possibilities of all the things which is equal to
2 × 2 × 2 × ... × n factors = 2 n
But, this includes one case in which nothing is selected.
Hence, the total number of ways of selecting one or more
of n different things = 2 n − 1
Aliter Number of ways of selecting one, two, three, …, n
things from n different things are
n
n
n
n
C 1 , C 2 , C 3 , ..., C n , respectively.
Hence, the total number of ways or selecting atleast one
thing is
n
C1 + nC2 + nC 3 + K + nCn
= ( n C 0 + n C 1 + n C 2 + K + n C n ) − n C 0 = 2n − 1
y Example 58. Mohan has 8 friends, in how many
ways he invite one or more of them to dinner?
Sol. Mohan select one or more than one of his 8 friends. So,
required number of ways
= 8C 1 + 8C 2 + 8C 3 + K + 8C 8
= 2 − 1 = 255.
8
y Example 59. A question paper consists of two
sections having respectively, 3 and 5 questions. The
following note is given on the paper ‘‘It is not
necessory to attempt all the questions one question
from each section is compulsory’’. In how many
ways can a candidate select the questions?
Sol. Here, we have two sections A and B (say), the section A
has 3 questions and section B has 5 questions and one
question from each section is compulsory, according to
the given direction.
∴ Number of ways selecting one or more than one question
from section A is 2 3 − 1 = 7
and number of ways selecting one or more than one
question from section B is 2 5 − 1 = 31
Hence, by the principle of multiplication, the required
number of ways in which a candidate can select the
questions
= 7 × 31 = 217.
y Example 60. A student is allowed to select atleast
one and atmost n books from a collection of (2n + 1)
books. If the total number of ways in which he can
select books is 63, find the value of n.
Sol. Given, student select atleast one and atmost n books from
a collection of (2n + 1) books. It means that he select one
book or two books or three books or … or n books.
Hence, by the given hypothesis.
2n + 1
C1
+
2n + 1
C2
+
2n + 1
C 3 + ... + 2 n + 1 C n = 63
…(i)
Also, the sum of binomial coefficients, is
2n + 1
C0 +
2n + 1
C1 + K +
2n + 1
Cn +
= ( 1 + 1)
⇒
2n + 1
C 0 + 2 ( 2 n + 1C 1 +
+
Cn + 1
+K+
=2
2n + 1
C2 + K +
2n + 1
⇒ 128 = 2
2n + 1
2n + 1
⇒ 7 = 2n + 1
C 2n + 1 = 2
⇒
∴
2 =2
n =3
Cn + 1
2n + 1
2n + 1
1 + 2 × 63 + 1 = 2
2n + 1
2n + 1
2n + 1
⇒
7
2n + 1
2n + 1
Cn )
[Q C r = C n − r ]
n
y Example 61. There are three books of Physics, four
of Chemistry and five of Mathematics. How many
different collections can be made such that each
collection consists of
(i) one book of each subject,
(ii) atleast one book of each subject,
(iii) atleast one book of Mathematics.
Sol. (i) 3 C 1 × 4C 1 × 5C 1 = 3 × 4 × 5 = 60
(ii) (23 − 1) × (24 − 1) × (25 − 1) = 7 × 15 × 31 = 3255
(iii) (25 − 1) × 27 = 31 × 128 = 3968
378
Textbook of Algebra
#L
1.
Exercise for Session 4
If 43Cr − 6 = 43C3r + 1, the value of r is
(a) 6
2.
If
18
(b) 8
C15 + 2 ( C16 ) +
18
If
20
(b) 20
(c) 22
(d) 24
(c) 13
(d) None of these
Cn + 2 = nC16 , the value of n is
(a) 7
4.
(d) 12
n
(a) 18
3.
(c) 10
C16 + 1 = C3, the value of n is
17
If 47C4 +
(b) 10
5
∑ 52 − r C3 is equal to
r =1
(a) 47C6
5.
(b)
If C3 + C4 >
n
n
(b) n < 6
10
Cx
(a) {1, 2, 3}
7.
If
If
2n
(b) 3
(b) r = 4
n
(c) r = 5
(d) r = 6
(c) 4
(d) 5
(c) 5
(d) 7
(c) 8
(d) 10
n
(b) 3
(b) 3
n
(b) 6
(b) 99
(c) 108
(d) 165
(b) 2008
(c) 2020
(d) 8002
(b) 210
(c) 10 ! − 1
(d) 210 − 1
In an examination, there are three multiple choice questions and each question has four choices. Number of
ways in which a student can fail to get all answers correct, is
(b) 12
(c) 27
(d) 63
In an election, the number of candidates is 1 greater than the persons to be elected . If a voter can vote in
254 ways, the number of candidates is
(b) 7
(c) 8
(d) 10
The number of groups that can be made from 5 different green balls, 4 different blue balls and 3 different red
balls, if atleast one green and one blue ball is to be included
(a) 3700
18.
n
A man has 10 friends. In how many ways he can invite one or more of them to a party?
(a) 6
17.
(d) 5
In how many ways a team of 11 players can be formed out of 25 players, if 6 out of them are always to be
included and 5 are always to be excluded
(a) 11
16.
(c) 4
There are 12 volleyball players in all in a college, out of which a team of 9 players is to be formed. If the captain
always remains the same, in how many ways can the team be formed ?
(a) 10!
15.
(d) {9, 10, 11}
If P3 + Cn − 2 = 14n, the value of n is
n
(a) 2002
14.
(c) {8, 9, 10}
> 2 . Cx is
If nPr = 840, nCr = 35, the value of n is
(a) 36
13.
(d) n < 7
If Cr = Cr − 1 and Pr = Pr + 1, the value of n is
n
(a) 5
12.
−1
(c) n > 7
C3 : nC2 = 44 : 3 , for which of the following value of r, the value of nCr will be 15.
(a) 1
11.
(d) None of these
C2 : C2 = 9 : 2 and Cr = 10, then r is equal to
(a) 2
10.
C4
n
(a) r = 3
9.
52
10
(b) {4, 5, 6}
n
(a) 2
8.
(c)
C3 then
The Solution set of
2n
C5
n+1
(a) n > 6
6.
52
(b) 3720
(c) 4340
(d) None of these
A person is permitted to select atleast one and atmost n coins from a collection of (2n + 1) distinct coins. If the
total number of ways in which he can select coins is 255, thenn equals
(a) 4
(b) 8
(c) 16
(d) 32
Session 5
Combinations from Identical Objects
Combinations from
Identical Objects
(i) The number of combinations of n identical objects
taking r objects (r ≤ n ) at a time = 1.
(ii) The number of combinations of zero or more objects
from n identical objects = n + 1.
(iii) The total number of combinations of atleast one out
of a 1 + a 2 + a 3 + K + a n objects, where a 1 are alike of
one kind, a 2 are alike of second kind, a 3 are alike of
third kind, …, a n are alike of nth kind
= (a 1 + 1) (a 2 + 1) (a 3 + 1) K (a n + 1) − 1
y Example 62. How many selections of atleast one red
ball from a bag containing 4 red balls and 5 black
balls, balls of the same colour being identical?
Sol. Number of selections of atleast one red ball from 4 identical red balls = 4
Number of selections of any number of black balls from 5
identical black balls
=5+1=6
∴ Required number of selections of balls
= 4 × 6 = 24
y Example 63. There are p copies each of n different
books. Find the number of ways in which a non-empty
selection can be made from them.
Sol. Since, copies of the same book are identical.
∴ Number of selections of any number of copies of a book
is p + 1. Similarly, in the case for each book.
Therefore, total number of selections is ( p + 1)n .
But this includes a selection, which is empty i.e., zero copy
of each book. Excluding this, the required number of
non-empty selections is ( p + 1)n − 1.
y Example 64. There are 4 oranges, 5 apples and
6 mangoes in a fruit basket and all fruits of the same
kind are identical. In how many ways can a person make
a selection of fruits from among the fruits in the basket?
Sol. Zero or more oranges can be selected out of 4 identical
oranges = 4 + 1 = 5 ways.
Similarly, for apples number of selection = 5 + 1 = 6 ways
and mangoes can be selected in 6 + 1 = 7 ways.
∴ The total number of selections when all the three kinds of
fruits are selected = 5 × 6 × 7 = 210
But, in one of these selection number of each kind of fruit is
zero and hence this selection must be excluded.
∴ Required number = 210 − 1 = 209
Combinations when both Identical
and Distinct Objects are Present
The number of combinations (selections) of one or more
objects out of a 1 + a 2 + a 3 + K + a n objects, where a 1
are alike of one kind, a 2 are alike of second kind, a 3 are
alike of third kind, …, a n are alike of nth kind and k
are distinct.
= {(a 1 + 1) (a 2 + 1) (a 3 + 1) L (a n + 1)}
(k C 0 + k C1 + k C2 + K + k Ck ) − 1
= (a 1 + 1) (a 2 + 1) (a 3 + 1) + K + (a n + 1) 2 k − 1
y Example 65. Find the number of ways in which one
or more letters can be selected from the letters
AAAAA BBBB CCC DD EFG.
Sol. Here, 5A’s are alike, 4 B ’s are alike, 3 C ’s are alike, 2D ’s
are alike and E, F, G are different.
∴ Total number of combinations
= ( 5 + 1) ( 4 + 1) ( 3 + 1) ( 2 + 1) 2 3 − 1
= 6 ⋅5 ⋅ 4 ⋅3 ⋅8 − 1
= 2879
[excluding the case, when no letter is selected]
Explanation Selection from ( AAAAA ) can be made by 6
ways such include no A, include one A, include two A,
include three A, include four A, include five A. Similarly,
selections from ( BBBB ) can be made in 5 ways, selections
from (CCC ) can be made in 4 ways, selections from ( DD )
can be made in 3 ways and from E, F, G can be made in
2 × 2 × 2 ways.
Number of Divisors of N
Every natural number N can always be put in the form
N = p 1α 1 ⋅ p 2α 2 ⋅ p 3α 3 ... p kα k , where p 1 , p 2 , p 3 , ..., p k are
distinct primes and α 1 , α 2 , α 3 , K, α k ∈W .
380
Textbook of Algebra
(i) The total number of divisors of N including 1 and N
= (α 1 + 1) (α 2 + 1) (α 3 + 1) K (α k + 1)
(ii) The total number of divisors of N excluding 1 and N
= (α 1 + 1) (α 2 + 1) (α 3 + 1) K (α k + 1) − 2
(iii) The total number of divisors of N excluding either 1
or N = (α 1 + 1) (α 2 + 1) (α 3 + 1) K (α k + 1) − 1
(iv) Sum of all
divisors = ( p 10 + p 11 + p 12 + p 13 + K + p 1∞1 )
( p 20 + p 21 + p 22 + p 23 + K + p 2α 2 ) ...
( p k0 + p k1 + p k2 + p k3 + K + p kα k )
 1 − p α1 + 1   1 − p α 2 + 1   1 − p αk + 1 
1
k
2

 ... 
 ⋅
= 




−
1
1
p
p
p k 
−
1
−
1
2
 

 
(v) Sum of proper divisors (excluding 1 and the
expression itself)
= Sum of all divisors − ( N + 1)
(vi) The number of even divisors of N are possible only if
p 1 = 2 , otherwise there is no even divisor.
∴ Required number of even divisors
= α 1 (α 2 + 1) (α 3 + 1) + K + (α k + 1)
(vii) The number of odd divisors of N
Case I If p 1 = 2 , the number of odd divisors
= (α 2 + 1) (α 3 + 1) K (α k + 1)
Case II If p 1 ≠ 2 , the number of odd divisors
= (α 1 + 1) (α 2 + 1) (α 3 + 1) K (α k + 1)
(viii) The number of ways in which N can be resolved as a
product of two factors

 1 (α + 1) (α + 1) K (α + 1) ,
2
k
 2 1
=
1
 {(α 1 + 1) (α 2 + 1) K (α k + 1) + 1} ,
2
if N is not a
perfect square
if N is a
perfect square
(ix) The number of ways in which a composite number N
can be resolved into two factors which are relatively
prime ( or coprime) to each other is equal to 2 n − 1 ,
where n is the number of different factors (or different
primes) in N .
y Example 66. Find the number of proper factors of
the number 38808. Also, find sum of all these divisors.
Sol. The number 38808 = 23 ⋅ 32 ⋅ 7 2 ⋅ 11
Hence, the total number of proper factors (excluding 1 and
itself i.e., 38808)
= ( 3 + 1) ( 2 + 1) ( 2 + 1) ( 1 + 1) − 2
= 72 − 2 = 70
And sum of all these divisors (proper)
= ( 20 + 21 + 22 + 23 ) ( 30 + 31 + 32 )
(7 0 + 71 + 7 2 ) (110 + 111 ) − 1 − 38808
= (15) (13) (57 ) (12) − 38809
= 133380 − 38809
= 94571
y Example 67. Find the number of even proper divisors
of the number 1008.
Sol. Q1008 = 24 × 32 × 71
∴ Required number of even proper divisors
= Total number of selections of atleast one 2 and
any number of 3’s or 7’s.
= 4 × (2 + 1) × (1 + 1) − 1 = 23
y Example 68. Find the number of odd proper divisors
of the number 35700. Also, find sum of the odd
proper divisors.
Sol. Q35700 = 22 × 31 × 52 × 71 × 171
∴ Required number of odd proper divisors
= Total number of selections of zero 2 and any
number of 3’s or 5’s or
7’s or 17’s
= (1 + 1) (2 + 1) (1 + 1) (1 + 1 ) − 1 = 23
∴ The sum of odd proper divisors
= (30 + 31 ) (50 + 51 + 52 ) (7 0 + 71 ) (17 0 + 71 ) − 1
= 4 × 31 × 8 × 18 − 1
= 17856 − 1 = 17855
y Example 69. If N = 10800, find the
(i) the number of divisors of the form
4m + 2 , ∀ m ∈W .
(ii) the number of divisors which are multiple of 10.
(iii) the number of divisors which are multiple of 15.
Sol. We have, N = 10800 = 24 × 33 × 52
(i) Q( 4 m + 2) = 2(2m + 1), in any divisor of the form
4 m + 2, 2 should be exactly 1.
So, the number of divisors of the form
( 4 m + 2) = 1 × (3 + 1) × (2 + 1) = 1 × 4 × 3 = 12
(ii) ∴ The required number of proper divisors
= Total number of selections of atleast one 2 and one
5 from 2, 2, 2, 2, 3, 3, 3, 5, 5
= 4 × (3 + 1) × 2 = 32
Chap 05 Permutations and Combinations
(iii) ∴ The required number of proper divisors
= Total number of selections of atleast one 3 and one
5 from 2, 2, 2, 2, 3, 3, 3, 5, 5
= ( 4 + 1) × 3 × 2 = 30
y Example 70. Find the number of divisors of the
number N = 2 3⋅ 3 5 ⋅ 5 7 ⋅7 9 ⋅9 11 , which are
381
Proof The number of ways in which (m + n ) distinct
objects are divided into two groups of the sizem and n
= The number of ways m objects are selected out of
(m + n ) objects to form one of the groups, which can
be done in m + n C m ways. The other group of n objects
is formed by the remaining n objects.
perfect square.
m+n
Sol. QN = 2 3 ⋅ 3 5⋅ 5 7 ⋅ 7 9⋅ 9 11
= 2 3 ⋅ 3 5 ⋅ 5 7 ⋅ 7 9 ⋅ 3 22
= 2 3⋅ 3 27⋅ 5 7 ⋅ 7 9
(Tree diagram)
2 can be taken in 2 ways (i.e., 20 or 22 )
3 can be taken in 14 ways (i.e., 30 or 32 or 34 or 36 or 38 or
310 or 312 or 314 or 316 or 318 or 320 or 322 or 324 or 326 )
5 can taken in 4 ways (i.e., 50 or 52 or 54 or 56 )
and 7 can taken in 5 ways
(i.e., 7 0 or 7 2 or 7 4 or 7 4 or 7 6 or 7 8 )
Hence, total divisors which are perfect squares
= 2 × 14 × 4 × 5 = 560
y Example 71. In how many ways the number 10800
can be resolved as a product of two factors?
Sol. Let N = 10800 = 2 4 × 3 3 × 5 2
Here, N is not a perfect square
[Q power of 3 is odd]
1
Hence, the number of ways = ( 4 + 1) (3 + 1) (2 + 1) = 30
2
=
m +n
Sol. Let N = 18900 = 2 ⋅ 3 ⋅ 5 ⋅ 7
3
2
=
Division of Objects Into Groups
(a) Division of Objects Into Groups of Unequal Size
Theorem Number of ways in which (m + n )
distinct objects can be divided into two unequal
(m + n ) !
groups containing m and n objects is
.
m! n !
(m + n )
m!n!
(m + n )
× 2!
m!n!
Corollary II The number of ways in which (m + n + p )
distinct objects can be divided into three unequal
groups containingm, n and p objects, is
m+n+p
m
p
n
1
Relatively prime or coprime Two numbers not necessarily prime are said to be relatively prime or coprime, if
their HCF (highest common factor) is one as 2, 3, 5, 7 are
relatively prime numbers.
[number of different primes in N ]
∴
n=4
Hence, number of ways in which a composite number N
can be resolved into two factors which are relatively
prime or coprime = 24 − 1 = 23 = 8
C m ⋅n C n =
Corollary I The number of ways to distribute (m + n )
distinct objects among 2 persons in the groups
containing m and n objects
= (Number of ways to divide) × (Number of groups)
y Example 72. In how many ways the number 18900
can be split in two factors which are relatively prime
(or coprime)?
2
n
m
For perfect square of N , each prime factor must occur even
number of times.
(Tree diagram)
m +n +p
C m ⋅n + p C n ⋅p C p =
(m + n + p ) !
m!n! p!
Corollary III The number of ways to distribute
(m + n + p ) distinct objects among 3 persons in the
groups containing m, n and p objects
= (Number of ways to divide) × (Number of groups)
(m + n + p ) !
=
× 3!
m!n! p!
Corollary IV The number of ways in which
( x 1 + x 2 + x 3 + L + x n ) distinct objects can be
divided into n unequal groups containing x 1 ,
x 2 , x 3 , K, x n objects, is
382
Textbook of Algebra
52
x1 + x2 + x3 +...+xn
13
13
13
x1
x2
.
x3 ..
xn
(Tree diagram)
52!
(Tree diagram)
(x 1 + x 2 + x 3 + L + x n ) !
.
x 1 ! x 2 ! x 3 !L x n !
Corollary V The number of ways to distribute
( x 1 + x 2 + x 3 + K + x n ) distinct objects among n
persons in the groups containing x 1 , x 2 , K , x n
objects
= (Number of ways to divide) × (Number of groups)
( x + x 2 + x 3 + ... + x n ) !
= 1
× n!
x 1 ! x 2 ! x 3 !K x n !
(b) Division of Objects Into Groups of Equal Size
The number of ways in which mn distinct objects can
be divided equally into m groups, each containing n
objects and
(i) If order of groups is not important is.
 (mn ) !  1
× .
=
 (n !) m  m !
(ii) If order of groups is important is.
 (mn ) ! 1 
(mn ) !

.
×  ×m ! =
m
m !
 (n !)
(n !) m
Note Division of 14n objects into 6 groups of 2n, 2n, 2n, 2n, 3n, 3n,


( 14 n)!


 ( 2n)! ( 2n)! ( 2n)! ( 2n)! ( 3n)! ( 3n)! 
( 14 n)
1
size is
=
×
4
2
4
!
2!
4 ! 2!
(( 2n)!) (( 3n)!)
Now, the distribution ways of these 6 groups among 6 persons is
( 14 n)!
4
[( 2n)!)] [( 3n)!]
2
×
1
( 14 n)!
× 6!=
× 15
4 ! 2!
[( 2n)!]4 [( 3n)!] 2
y Example 73. In how many ways can a pack of 52
cards be
(i) distributed equally among four players in order?
(ii) divided into four groups of 13 cards each?
(iii) divided into four sets, three of them having 17 cards
each and fourth just one card?
Sol. (i) Here, order of group is important, then the numbers
of ways in which 52 different cards can be divided
equally into 4 players is
13
4 ! (13!)
4
× 4! =
52!
(13!)4
Aliter Each player will get 13 cards. Now, first player can
be given 13 cards out of 52 cards in 52 C 13 ways. Second
player can be given 13 cards out of remaining 39 cards (i.e.,
52 − 13 = 39) in
39
C 13 ways. Third player can be given 13
cards out of remaining 26 cards (i.e., 39 − 13 = 26 ) in
ways and fourth player can be given 13 cards out of
remaining 13 cards (i.e., 26 − 13 = 13) in 13 C 13 ways.
26
C 13
Hence, required number of ways
=
52
C 13 ×
C 13 ×
=
52!
39 !
26!
52!
×
×
× 1=
13! 39 ! 13! 26! 13! 13!
(13!)4
39
C 13 × 13C 13
26
(ii) Here, order of group is not important, then the
number of ways in which 52 different cards can be
divided equally into 4 groups is
52
13
13
13
13
(Tree diagram)
52!
4 ! (13!)4
Aliter Each group will get 13 cards. Now, first group can be
given 13 cards out of 52 cards in 52 C 13 ways. Second group
can be given 13 cards out of remaining 39 cards (i.e.,
52 − 13 = 39) in 39 C 13 ways. Third group can be given 13
cards out of remaining 26 cards
(i.e., 39 − 13 = 26) in 26 C 13 ways and fourth group can be
given 13 cards out of remaining 13 cards (i.e., 26 − 13 = 13)
in 13 C 13 ways. But the all (four) groups can be interchanged
in 4! ways. Hence, the required number of ways
1
= 52 C 13 × 39C 13 × 26C 13 × 13C 13 ×
4!
1
52!
39 !
26!
52!
=
×
×
×1×
=
4 ! (13!)4 4 !
13! 39 ! 13! 26! 13! 13!
(iii) First, we divide 52 cards into two sets which contains
1 and 51 cards respectively, is
52!
1! 51!
Chap 05 Permutations and Combinations
=
52
1
17
17
17
Now, 51 cards can be divided equally in three sets each
contains 17 cards (here order of sets is not important)
51!
in
ways.
3! (17 !)3
383
12! × 2 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 !⋅ 2
= 1584.
=
5! × 7 !
5 ⋅ 4 ⋅3 ⋅2 ⋅1 ⋅7 !
II Part Here, order is not important, then the number of
ways in which 12 different balls can be divided into three
groups of 5,4 and 3 balls respectively, is
12!
12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5!
= 27720
=
=
5! 4 ! 3!
5! ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 3 ⋅ 2 ⋅ 1
12
Hence, the required number of ways
52!
51!
=
×
1! 51!
3 !(17 !)3
=
52!
1! 3!(17 )3
=
5
52!
(17 !) 3 3!
Aliter First set can be given 17 cards out of 52 cards in
52
C 17 . Second set can be given 17 cards out of remaining 35
cards (i.e., 52 − 17 = 35) in 35 C 17 . Third set can be given 17
cards out of remaining 18 cards (i.e., 35 − 17 = 18) in 18 C 17
and fourth set can be given 1 card out of 1 card in 1C 1. But
the first three sets can interchanged in 3! ways. Hence, the
total number of ways for the required distribution
1
= 52 C 17 × 35C 17 × 18C 17 × 1C 1 × !
3
1
52!
35!
18!
(52)!
=
×
×
×1× =
3! (17 !)3 3!
17 ! 35! 17 ! 1! 17 ! 18!
y Example 74. In how many ways can 12 different balls
be divided between 2 boys, one receiving 5 and the
other 7 balls? Also, in how many ways can these 12 balls
be divided into groups of 5, 4 and 3 balls, respectively?
Sol. I Part Here, order is important, then the number of ways
in which 12 different balls can be divided between two
boys which contains
5 and 7 balls respectively, is
4
2
(Tree diagram)
Aliter First group can be given 5 balls out of 12 balls in
C 5 ways. Second group can be given 4 balls out of
remaining 7 balls (12 − 5 = 7 ) in 7 C 4 and 3 balls can be
given out of remaining 3 balls in 3 C 3 .
12
Hence, the required number of ways (here order of groups
are not important)
=
12
C 5 × 7C 4 × 3C 3
=
12!
7!
×
×1
5! 7 ! 4 ! 3!
=
12!
5! 4 ! 3!
y Example 75. In how many ways can 16 different
books be distributed among three students A, B, C so
that B gets 1 more than A and C gets 2 more than B ?
Sol. Let A gets n books, then B gets n + 1 and C gets n + 3.
Now, n + (n + 1) + (n + 3) = 16
⇒
3n = 12
∴
n=4
12
16
5
7
(Tree diagram)
12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 !
12!
⋅ 2 = 1584
=
× 2! =
(5⋅ 4 ⋅ 3 ⋅ 2⋅ 1)7 !
5! 7 !
Aliter First boy can be given 5 balls out of 12 balls in 12 C 5 .
Second boy can be given 7 balls out of 7 balls (i.e.,
12 − 5 = 7) but there order is important boys interchange by
(2 types), then required number of ways
12!
= 12 C 5 × 7C 7 × 2! =
× 1 × 2!
5! 7 !
4
5
7
(Tree diagram)
⇒ A , B, C gets 4, 5 and 7 books, respectively.
Hence, the total number of ways for the required
distribution
16!
=
4 ! 5! 7 !
384
Textbook of Algebra
The number of division ways for tree diagrams (i), (ii) and
(iii) are
9!
1
9!
9!
1
and
× ,
× , respectively.
(2!)2 (5!) 2! 2! 3! 4 !
(3!)3 3!
y Example 76. In how many ways can 9 different
books be distributed among three students if each
receives atleast 2 books?
Sol. If each receives atleast 2 books, then the division as
shown by tree diagrams
9
2
9
2
5
2
(i)
3
1.
4
3
(b) 125
(c) 167
(b) (32)2 − 1
(c) 232 − 1
(b) (m + 1) 2 n
(c) m ⋅ 2n + 1
7.
(b) 35
(d) 37
The sum of the divisors of 2 5 ⋅ 34 ⋅ 52 is
(b) 32 ⋅ 71 ⋅ 112 ⋅ 31
(d) None of these
The number of proper divisors of 2p ⋅ 6q ⋅ 21r , ∀ p, q , r ∈ N, is
(a) (p + q + 1) (q + r + 1) (r + 1)
(b) (p + q + 1) (q + r + 1) (r + 1) − 2
(c) (p + q ) (q + r ) r − 2
(d) (p + q ) (q + r )r
The number of odd proper divisors of 3p ⋅ 6q ⋅ 15r , ∀ p, q , r ∈ N, is
(a) (p + 1) (q + 1) (r + 1) − 2
(c) (p + q + r + 1) (r + 1) − 2
8.
(b) 27
(c) 34
(d) 43
Total number of divisors of 480 that are of the form 4n + 2, n ≥ 0, is equal to
(a) 2
10.
(b) (p + 1) (q + 1) (r + 1) − 1
(d) (p + q + r + 1) (r + 1) − 1
The number of proper divisors of 1800, which are also divisible by 10, is
(a) 18
9.
(d) None of these
Number of proper factors of 2400 is equal to
(a) 32 ⋅ 71 ⋅ 112
(c) 3 ⋅ 7 ⋅ 11⋅ 31
6.
(d) 231
If a1, a 2, a 3, L , a n + 1 be (n + 1) different prime numbers, then the number of different factors (other than 1) of
a1m ⋅ a 2 ⋅ a 3, K , a n + 1, is
(a) 34
(c) 36
5.
(d) 168
In a city no two persons have identical set of teeth and there is no person without a tooth. Also, no person has
more than 32 teeth. If we disguard the shape and size of tooth and consider only the positioning of the teeth,
the maximum population of the city is
(a) m + 1
4.
(iii)
There are 3 oranges, 5 apples and 6 mangoes in a fruit basket (all fruits of same kind are identical). Number of
ways in which fruits can be selected from the basket, is
(a) 2 32
3.
= [378 + 1260 + 280] × 6
= 11508
3
3
Exercise for Session 5
(a) 124
2.
 9!
1
9!
9!
1
× +
+
×  × 3!
 2
3
2
2
3
4
3
!
!
!
!
!
(3!)
(2!) (5!)
9
(ii)
#L
Hence, the total number of ways of distribution of these
groups among 3 students is
(b) 3
(c) 4
Total number of divisors of N = 2 ⋅ 3 ⋅ 5
5
(a) 54
(b) 55
4
10
(d) 5
⋅ 7 that are of the form 4n + 2, n ≥ 1, is equal to
6
(c) 384
(d) 385
Chap 05 Permutations and Combinations
11.
Total number of divisors of N = 3 5 ⋅ 5 7⋅ 79 that are of the form 4n + 1, n ≥ 0 is equal to
(a) 15
12.
12 !
(b)
(4 !)3
12 !
2
(3 !) (2 !)
3
12 !
(3 !)2 (2 !)3
2n !
(n !)2
12 !
12 !
(c)
(3 !)4
(d)
(4 !)4
12 !
(3 !)3
12 !
12 !
(c)
3 ! (4 !)3
(d)
4 ! (3 !)3
12 !
(3 !)4
(b)
12 ! 5 !
2
(3 !) (2 !)
3
(c)
12 !
3
4
(3 !) (2 !)
(d)
12 ! 5 !
(3 !)2 (2 !)4
(b)
12 ! 5 !
(3 !)2 (2 !)3
(c)
12 !
(3 !)2 (2 !)4
(d)
12 ! 5 !
(3 !)2 (2 !)4
(b)
2n !
(2n !)n
(c)
2n !
n ! (2n !)2
(d) None of these
n different toys have to be distributed among n children. Total number of ways in which these toys can be
distributed so that exactly one child gets no toy, is equal to
(a) n !
18.
(b)
The total number of ways in which 2n persons can be divided into n couples, is
(a)
17.
12 !
(4 !)3
Number of ways in which 12 different things can be divided among five persons so that they can get 2, 2, 2, 3,
3 things respectively, is
(a)
16.
(d) 240
Number of ways in which 12 different things can be distributed in 5 sets of 2, 2, 2, 3, 3, things is
(a)
15.
(c) 120
Number of ways in which 12 different things can be distributed in 3 groups, is
(a)
14.
(b) 30
Number of ways in which 12 different books can be distributed equally among 3 persons, is
(a)
13.
385
(b) n ! nC2
(c) (n − 1) ! nC2
(d) n ! n −1C2
In how many ways can 8 different books be distributed among 3 students if each receives atleast 2 books?
(a) 490
(b) 980
(c) 2940
(d) 5880
Session 6
Arrangement in Groups, Multinomial Theorem,
Multiplying Synthetically
Arrangement in Groups
(a) The number of ways in which n different things
can be arranged into r different groups is
r (r + 1 )(r + 2) K (r + n − 1 ) or n !⋅ n − 1 C r − 1
according as blank groups are or are not admissible.
Proof
(i) Let n letters a 1 , a 2 , a 3 , K, a n be written in a row
in any order. All the arrangements of the letters
in r, groups, blank groups being admissible, can
be obtained thus, place among the letters (r − 1)
marks of partition and arrange the (n + r − 1)
things (consisting of letters and marks) in all
possible orders. Since, (r − 1) of the things are
alike, the number of different arrangements is
(n + r − 1) !
= r (r + 1)(r − 2 ) K(r + n − 1).
(r − 1) !
(ii) All the arrangements of the letters in r groups,
none of the groups being blank, can be obtained
as follows:
(I) Arrange the letters in all possible orders. This can be
done in n ! ways.
(II) In every such arrangement, place (r − 1) marks of
partition in (r − 1) out of the (n − 1) spaces between
the letters. This can be done in n − 1 C r − 1 ways.
Hence, the required number is n !⋅ n − 1 C r − 1 .
y Example 77. In how many ways 5 different balls can
be arranged into 3 different boxes so that no box
remains empty?
Sol. The required number of ways = 5!⋅ 5 − 1 C 3 − 1 = 5!⋅ 4C 2
Aliter
Each box must contain atleast one ball, since no box
remains empty. Boxes can have balls in the following
systems
Box
I
II
III
1
1
3
I
II
III
Number of 1
balls
2
2
Box
Or
(b) The number of ways in which n different things
can be distributed into r different groups is
r n −r C 1 (r − 1 ) n + r C 2 (r − 2) n − K + ( −1 ) r −1 ⋅ r C r −1
Or
r
∑ ( −1 ) p ⋅ r C p ⋅ (r − p )n
p=0
Or
Coefficient of x in n !(e x − 1 ) r .
n
Here, blank groups are not allowed.
Proof In any distribution, denote the groups by
g 1 , g 2 , g 3 , K, g r and consider the distributions in
which blanks are allowed.
The total number of these is r n .
The number in which g 1 is blank, is (r − 1) n .
Therefore, the number in which g 1 is not blank, is
r n − (r − 1) n
of these last, the number in which g 2 is blank, is
(r − 1) n − (r − 2 ) n
Therefore, the number in which g 1 , g 2 are not blank, is
r n − 2 (r − 1) n + (r − 2 ) n
of these last, the number in which g 3 is blank, is
(r − 1) n − 2 (r − 2 ) n + (r − 3 ) n
Therefore, the number in which g 1 , g 2 , g 3 are not
blank, is
r n − 3 (r − 1) n + 3 (r − 2 ) n − (r − 3 ) n
 4 ⋅ 3
= (120) ⋅ 
 = 720
 1 ⋅2 
Number of
balls
All 5 balls can be arranged by 5! ways and boxes can be
3!
arranged in each system by .
2!
3!
3!
Hence, required number of ways = 5! × + 5! ×
2!
2!
= 120 × 3 + 120 × 3 = 720
This process can be continued as far as we like and it
is obvious that the coefficients are formed as in a
binomial expansion.
Hence, the number of distributions in which no one
of x assigned groups is blank, is
r n − x C 1(r − 1)n + x C 2 (r − 2)n − K + ( −1)x (r − x )n
Chap 05 Permutations and Combinations
when x = r , then
r n − r C 1 (r − 1) n + r C 2 (r − 2 ) n −K + ( −1) r − 1 ⋅ r C r − 1
(r − (r − 1)) n + ( −1) r ⋅ r C r (r − r ) n
Or
n
r
n
r
r − C 1 (r − 1) + C 2 (r − 2)n − K + ( −1)r − 1 ⋅ r C r − 1
Aliter
By Principle of Inclusion and Exclusion
Let Ai denotes the set of distribution of things, if ith
group gets nothing. Then, n( Ai ) = (r − 1) n
[as n things can be distributed among (r − 1) groups
in (r − 1) n ways]
Then, n ( Ai ∩ A j ) represents number of distribution
ways in which groups i and j get no object. Then,
n ( Ai ∩ A j ) = (r − 2 ) n
Also, n ( Ai ∩ A j ∩ Ak ) = (r − 3 ) n
This process can be continued, then the required
number is
n ( A1 ′ A2 ′ ∩ K ∩ Ar ′ )
= n (U ) − n( A1 ∪ A2 ∪ K ∪ Ar )
= r n − { ∑ n( Ai ) − ∑ n ( Ai ∩ A j )
+ ∑ n ( Ai ∩ A j ∩ Ak ) K
+ ( −1) n ∑ n ( A1 ∩ A2 ∩ K ∩ Ar )}
= r n − { r C 1 (r − 1) n − r C 2 (r − 2 ) n
+ r C 3 (r − 3 ) n − K+ ( −1) r ⋅ r C r − 1 }
= r n − r C 1 (r − 1) n + r C 2 (r − 2 ) n
− r C 3 (r − 3 ) n + K + ( −1) r − 1 ⋅ r C r − 1 ⋅ 1
Note Coefficient of x r in epx =
Box
I
Number of
balls
1
y Example 78. In how many ways 5 different balls can be
distributed into 3 boxes so that no box remains empty?
Sol. The required number of ways
= 35 − 3C 1(3 − 1)5 + 3C 2 (3 − 2)5 − 3C 3 (3 − 3)5
= 243 − 96 + 3 − 0 = 150
Or
Coefficient of x 5 in 5!(e x − 1)3
= Coefficient of x 5 in 5!(e 3 x − 3e 2 x + 3e x − 1)
 35
1
25
= 5!  − 3 × + 3 ×  = 35 − 3⋅25 + 3 = 243 − 96 + 3 = 150
5! 
5!
 5!
Aliter
Each box must contain atleast one ball, since number box
remains empty. Boxes can have balls in the following
systems
1
III
3
I
Box
Or Number of 1
II
III
2
2
balls
The number of ways to distribute the balls in I system
= 5C 1 × 4C 1 × 3C 3
∴Thetotalnumberofwaystodistribute 1,1,3ballstotheboxes
3!
= 5C 1 × 4C 1 × 3C 3 × = 5 × 4 × 1 × 3 = 60
2!
and the number of ways to distribute the balls in II system
= 5C 1 × 4C 2 × 2C 2
∴ The total number of ways to distribute 1, 2, 2 balls to the
boxes
3!
= 5C 1 × 4C 2 × 2C 2 ×
2!
= 5 × 6 × 1 × 3 = 90
∴ The required number of ways = 60 + 90 = 150
y Example 79. In how many ways can 5 different
books be tied up in three bundles?
1 5 3
( 3 − C 1 ⋅ 25 + 3C 2 ⋅ 15 )
3!
150
=
= 25
6
Sol. The required number of ways =
y Example 80. If n ( A ) = 5 and n (B ) = 3, find number
of onto mappings from A to B.
Sol. We know that in onto mapping, each image must be
assigned atleast one pre-image.
This is equivalent to number of ways in which 5 different
balls (pre-images) can be distributed in 3 different boxes
(images), if no box remains empty. The total number of
onto mappings from A to B
= 35 − 3C 1(3 − 1)5 + 3C 2 (3 − 2)5
r
p
.
r!
II
387
= 243 − 96 + 3 = 150
(c) The number of ways in which n identical things
can be distributed into r different groups is
n +r −1
C r − 1 or
n −1
Cr − 1
According, as blank groups are or are not
admissible.
Proof
If blank groups are not allowed Any such
distribution can be effected as follows: place the n
things in a row and put marks of partition in a
selection of (r − 1) out of the (n − 1) spaces between
them. This can be done in n − 1 C r − 1 .
If blank groups are allowed The number of
distribution is the same as that of (n + r ) things of the
same sort into r groups with no blank groups. For
such a distribution can be effected thus, put one of the
388
Textbook of Algebra
(n + r ) things into each of the r groups and distribute
the remaining n things into r groups, blank lots being
allowed. Hence, the required number is n + r − 1 C r − 1 .
Aliter The number of distribution of n identical
things into r different groups is the coefficient of x n
in (1 + x + x 2 + K + ∞ ) r or in ( x + x 2 + x 3 + K + ∞ ) r
according as blank groups are or are not allowed.
These expressions are respectively equal to
(1 − x ) −r and x r (1 − x ) −r
Hence, coefficient of x n in two expressions are
n +r −1
C r − 1 and
n −1
C r − 1 , respectively.
y Example 81. In how many ways 5 identical balls can
be distributed into 3 different boxes so that no box
remains empty?
5 −1
C 3 − 1 = 4C 2 =
Sol. The required number of ways =
4 ⋅3
=6
1 ⋅2
Aliter Each box must contain atleast one ball, since no
box remains empty. Boxes can have balls in the following
systems.
Box
Number
of balls
I
1
II
1
III
3
I
II
III
Number of 1
balls
2
2
Box
Or
Here, balls are identical but boxes are different the number
of combinations will be 1 in each systems.
3!
3!
∴ Required number of ways = 1 × + 1 × = 3 + 3 = 6
2!
2!
y Example 82. Four boys picked up 30 mangoes. In
how many ways can they divide them, if all mangoes
be identical?
Sol. Clearly, 30 mangoes can be distributed among 4 boys such
that each boy can receive any number of mangoes.
Hence, total number of ways =
=
30 + 4 − 1
C4 −1
C3 =
33
33 ⋅ 32 ⋅ 31
= 5456
1 ⋅2 ⋅3
y Example 83. Find the positive number of solutions of
x + y + z + w = 20 under the following conditions
(i) Zero value of x , y, z and w are included.
(ii) Zero values are excluded.
Sol. (i) Since, x + y + z + w = 20
Here, x ≥ 0, y ≥ 0, z ≥ 0, w ≥ 0
The number of Sols of the given equation in this case
is same as the number of ways of distributing 20
things among 4 different groups.
Hence, total number of Sols =
=
20 + 4 − 1
C4 −1
C3 =
23
23 ⋅ 22 ⋅ 21
= 1771
1 ⋅2 ⋅3
…(i)
x + y + z + w = 20
x ≥ 1, y ≥ 1, z ≥ 1, w ≥ 1
− 1 ≥ 0, y − 1 ≥ 0, z − 1 ≥ 0, w − 1 ≥ 0
x1 = x − 1 ⇒ x = x1 + 1
y1 = y − 1 ⇒ y = y1 + 1
z1 = z − 1 ⇒ z = z1 + 1
w1 = w − 1 ⇒ w = w1 + 1
Then, from Eq. (i), we get
x 1 + 1 + y1 + 1 + z1 + 1 + w 1 + 1 = 20
⇒
x 1 + y1 + z1 + w 1 = 16
and
x 1 ≥ 0, y1 ≥ 0, z1 ≥ 0, w 1 ≥ 0
Hence, total number of Solutions = 16 + 4 − 1C 4 − 1
(ii) Since,
Here,
or
x
Let
= 19C 3 =
19 ⋅ 18 ⋅ 17
= 57 ⋅ 17 = 969
1 ⋅2 ⋅3
Aliter
Part (ii) Q x + y + z + w = 20
x ≥ 1, y ≥ 1, z ≥ 1, w ≥ 1
Hence, total number of solutions
20 − 1
C 4 − 1 = 19C 3 = 969
=
y Example 84. How many integral solutions are there
to x + y + z + t = 29, when x ≥ 1, y > 1, z ≥ 3 and t ≥ 0 ?
Sol. Since, x + y + z + t = 29
…(i)
and x, y, z, t are integers
∴
x ≥ 1, y ≥ 2, z ≥ 3, t ≥ 0
⇒
x − 1 ≥ 0, y − 2 ≥ 0, z − 3 ≥ 0, t ≥ 0
Let
x 1 = x − 1, x 2 = y − 2, x 3 = z − 3
or x = x 1 + 1, y = x 2 + 2, z = x 3 + 3 and then x 1 ≥ 0, x 2 ≥ 0,
x 3 ≥ 0, t ≥ 0
From Eq. (i), we get
x 1 + 1 + x 2 + 2 + x 3 + 3 + t = 29
⇒
x 1 + x 2 + x 3 + t = 23
Hence, total number of solutions = 23 + 4 − 1C 4 − 1
=
Aliter
Q
and
Let
or
C3 =
26
26 ⋅ 25 ⋅ 24
= 2600
1 ⋅2 ⋅3
x + y + z + t = 29
x ≥ 1, y − 1 ≥ 1, z − 2 ≥ 1, t + 1 ≥ 1
x 1 = x , y1 = y − 1, z1 = z − 2, t 1 = t + 1
x = x 1, y = y1 + 1, z = z1 + 2, t = t 1 − 1
and then
x 1 ≥ 1, y1 ≥ 1, z1 ≥ 1, t 1 ≥ 1
From Eq. (i), x 1 + y1 + 1 + z1 + 2 + t 1 − 1 = 29
⇒
x 1 + y1 + z1 + t 1 = 27
Hence, total number of solutions =
=
27 − 1
C4 −1 =
26
C3
26 ⋅ 25 ⋅ 24
= 2600
1 ⋅2 ⋅3
…(i)
Chap 05 Permutations and Combinations
y Example 85. How many integral Solutions are there
to the system of equations x 1 + x 2 + x 3 + x 4 + x 5 = 20
and x 1 + x 2 = 15, when x k ≥ 0?
Sol. We have, x 1 + x 2 + x 3 + x 4 + x 5 = 20
…(i)
and
x 1 + x 2 = 15
Then, from Eqs. (i) and (ii), we get two equations
x3 + x4 + x5 = 5
x 1 + x 2 = 15
and given x 1 ≥ 0, x 2 ≥ 0, x 3 ≥ 0, x 4 ≥ 0 and x 5 ≥ 0
Then, number of solutions of Eq. (iii)
=
…(ii)
…(iii)
…(iv)
5 + 3 −1
C 3 − 1 = 7C 2
7 ⋅6
= 21
=
1 ⋅2
and number of solutions of Eq. (iv)
=
15 + 2 − 1
C 2 − 1 = 16C 1 = 16
Hence, total number of solutions of the given system of
equations
= 21 × 16 = 336
y Example 86. Find the number of non-negative
integral solutions of 3x + y + z = 24.
Sol. We have,
3x + y + z = 24 and given x ≥ 0, y ≥ 0, z ≥ 0
Let
x =k
…(i)
∴
y + z = 24 − 3k
Here,
24 ≥ 24 − 3k ≥ 0[Q x ≥ 0]
Hence, 0 ≤ k ≤ 8
The total number of integral solutions of Eq. (i) is
24 − 3k + 2 − 1
C2 −1 =
25 − 3k
C 1 = 25 − 3k
Hence, the total number of Sols of the original equation
=
8
8
8
k =0
k =0
k =0
∑(25 − 3k ) = 25 ∑1 − 3 ∑k
8 ⋅9
= 25 ⋅ 9 − 3 ⋅
= 225 − 108 = 117
2
(d) The number of ways in which n identical things
can be distributed into r groups so that no group
contains less than l things and more than m
things (l < m ) is coefficient of x n − lr in the
expansion of (1 − x m − l + 1 ) r (1 − x ) −r .
Proof Required number of ways
= Coefficient of x n in the expansion of
(x l + x l + 1 + x l + 2 + K + x m )r
[Q no group contains less than l things and
more than m things, here r groups]
389
= Coefficient of x n in the expansion of
x lr (1 + x + x 2 + K + x m − l ) r
= Coefficient of x n − lr in the expansion of
(1 + x + x 2 + K + x m − l ) r
= Coefficient of x n − lr in the expansion of
r
 1 ⋅ (1 − x m − l + 1 ) 


(1 − x )


[sum of m − l + 1 terms of GP]
= Coefficient of x
n − lr
in the expansion of
(1 − x m − l + 1 ) r (1 − x ) −r
y Example 87. In how many ways can three persons,
each throwing a single dice once, make a sum of 15 ?
Sol. Number on the faces of the dice are 1, 2, 3, 4, 5, 6 (least
number 1, greatest number 6)
Here, l = 1, m = 6, r = 3 and n = 15
∴ Required number of ways = Coefficient of x 15 − 1 × 3 in the
expansion of (1 − x 6 )3 (1 − x )−3
= Coefficient of x 12 in the expansion of
(1 − 3x 6 + 3x 12 )(1 + 3C 1 x + 4C 2 x 2 + K + 8C 6 x 6 + K
+ 14C 12 x 12 + K )
= 14C 12 − 3 × 8C 6 + 3 = 14C 2 − 3 × 8C 2 + 3
= 91 − 84 + 3 = 10
y Example 88. In how many ways in which an
examiner can assign 30 marks to 8 questions, giving
not less than 2 marks to any question.
Sol. If examiner given marks any seven question 2 (each)
marks, then marks on remaining questions given by
examiner = − 7 × 2 + 30 = 16
If x i are the marks assigned to ith question, then
x 1 + x 2 + x 3 + K + x 8 = 30 and 2 ≤ x i ≤ 16
for i = 1, 2, 3, K, 8.
Here, l = 2,m = 16, r = 8 and n = 30
∴ Required number of ways
= Coefficient of x
30 − 2 × 8
in the expansion of
(1 − x 16 − 2 + 1 )8 (1 − x )−8
= Coefficient of x 14 in the expansion of
(1 − x 15 )8 (1 + 8C 1x + 9C 2 x 2 + K+ 21C 14 x 14 + K )
= Coefficient of x 14 in the expansion of
(1 + 8C 1x + 9C 2 x 2 + K + 21C 14 x 14 + K )
= 21C 14 = 21C 7
Note Coefficient of x r in the expansion of ( 1 − x ) − n is
n + r −1
Cr .
390
Textbook of Algebra
(e) If a group has n things in which p are identical,
then the number of ways of selecting r things
from a group is
r
∑
n−p
r =0
r
∑ n − p C r ,according asr ≤ p or
C r or
r =r − p
r ≥ p.
y Example 89. A bag has contains 23 balls in which 7
are identical. Then, find the number of ways of
selecting 12 balls from bag.
Sol. Here, n = 23, p = 7, r = 12 (r > p )
∴ Required number of selections =
12
∑
16
[Q the remaining (n − 1) letters can be placed in (n − 1)
envelopes is (n − 1) ! ]
and n ( Ai ∩ A j ) = 1 × 1 × (n − 2 ) ! [Qi and j can be placed in
their corresponding envelopes and remaining (n − 2 )
letters can be placed in (n − 2 ) envelopes in (n − 2 ) ! way]
Also, n ( Ai ∩ A j ∩ Ak ) = 1 × 1 × 1 × (n − 3 ) !
Hence, the required number is
n ( A1 ′ ∩ A2 ′ ∩ A 3 ′ ∩ K ∩ An ′ )
= n (U ) − n ( A1 ∪ A2 ∪ A 3 ∪ K ∪ An )
= n ! − {∑ n ( Ai ) − ∑ n( Ai ∩ A j )
+ ∑ n ( Ai ∩ A j ∩ Ak ) −K + ( −1) n
Cr
∑ n ( A1 ∩ A2 ∩ A 3 ∩ K ∩ An )
r =5
= C 5 + C 6 + C 7 + C 8 + C 9 + C 10 + C 11 + C 12
16
16
16
16
16
16
16
16
= n ! − { n C 1 × (n − 1) ! − n C 2 × (n − 2 ) !
= ( 16C 5 + 16C 6 ) + ( 16C 7 + 16C 8 ) + ( 16C 9 + 16C 10 )
+ n C 3 × (n − 3 ) ! −K + ( −1) n − 1 × n C n × 1 ! }
+ ( 16C 11 + 16C 12 )
= 17C 6 + 17C 8 + 17C 10 + 17C 12
= C 11 + C 9 + C 10 + C 12
17
17
17
17
[Q n C r + n C r
−1
=
n × (n − 1) ! n(n − 1)
= n! − 
−
× (n − 2 ) !
1!
2!

n(n − 1)(n − 2 )

+
× (n − 3 ) ! − K + ( −1) n − 1 × 1
3!


n ! n ! n !
= n ! −  − + − K + ( −1) n ⋅ 1

 1! 2 ! 3 !
n 

( −1)
1 1 1
= n ! 1 − + − + K +

n! 
 1! 2 ! 3 !
n +1
Cr ]
[Q C r = n C n − r ]
n
= ( 17C 11 + 17C 12 ) + ( 17C 9 + 17C 10 )
= 18C 12 + 18C 10 = 18C 6 + 18C 8
Derangements Any change in the order of the things in
a group is called a derangement.
Or
When ‘n’ things are to be placed at ‘n’ specific places but
none of them is placed on its specified position, then we
say that the ‘n’ things are deranged.
Or
Assume a 1 , a 2 , a 3 , K, a n be n distinct things such that their
positions are fixed in a row. If we now rearrange a 1 , a 2 ,
a 3 , K, a n in such a way that no one occupy its original
position, then such an arrangement is called a
derangement.
Consider ‘n’ letters and ‘n’ corresponding envelops. The
number of ways in which letters can be placed in the
envelopes (one letter in each envelope) so that no letter is
placed in correct envelope is
1
1 1 1

n !  1 − + − + K + ( −1) n 
 1! 2 ! 3 !
n !
Proof n letters are denoted by 1, 2, 3, K, n. Let Ai denote
the set of distribution of letters in envelopes (one letter in
each envelop) so that the
i th letter is placed in the corresponding envelope, then
n ( Ai ) = 1 × (n − 1) !
Maha Short Cut Method

( −1) n 
1 1 1
D n = n ! 1 − + − + K +

n! 
 1! 2 ! 3 !
Then, D n + 1 = (n + 1 ) D n + ( −1 ) n + 1 , ∀ x ∈ N
If
and D n + 1 = n ( D n + D n − 1 ), ∀ x ∈ N − {1 }
where D 1 = 0
For n = 1, from result I
D 2 = 2 D 1 + ( −1) 2 = 0 + 1 = 1
For n = 2, from result I
D 3 = 3 D 2 + ( −1) 3 = 3 × 1 − 1 = 2
For n = 3, from result I
D 4 = 4 D 3 + ( −1) 4 = 4 × 2 + 1 = 9
For n = 4, from result I
D 5 = 5 D 4 + ( −1) 5 = 5 × 9 − 1 = 44
For n = 5, from result I
D 6 = 6 D 5 + ( −1) 6 = 6 × 44 + 1 = 265
Note D1 = 0, D2 = 1, D3 = 2, D4 = 9, D5 = 44, D6 = 265 [Remember]
Chap 05 Permutations and Combinations
Remark
If r things goes to wrong place out of n things, then ( n − r ) things
goes to original place (here r < n).
If Dn = Number of ways, if all n things goes to wrong places.
and Dr = Number of ways, if r things goes to wrong places.
If r goes to wrong places out of n, then ( n − r ) goes to correct
places.
Then,
Dn = n Cn − r Dr
1
1
1
1
where,
Dr = r !  1 − + ! − + K + ( −1) r 

1! 2!
3!
r !
If atleast p things goes to wrong places, then Dn =
n
∑
n
Cn − r ⋅ Dr
r =p
y Example 90. A person writes letters to six friends
and addresses the corresponding envelopes. In how
many ways can the letters be placed in the envelopes
so that (i) atleast two of them are in the wrong
envelopes. (ii) all the letters are in the wrong
envelopes.
Sol. (i) The number of ways in which atleast two of them in
the wrong envelopes
=
=
6
∑
6
r =2
6
C4
C 6 − r ⋅ Dr
× D 2 + 6C 3 × D 3 + 6C 2 × D 4 + 6C 1
× D 5 + 6C 0 × D 6
= 15D 2 + 20D 3 + 15D 4 + 6D 5 + D 6
[from note]
= 15 × 1 + 20 × 2 + 15 × 9 + 6 × 44 + 265
= 719
(ii) The number of ways in which all letters be placed in
[from note]
wrong envelopes = D 6 = 265
Aliter
(i) The number of all the possible ways of putting 6
letters into 6 envelopes is 6 ! .
Number of ways to place all letters correctly into
corresponding envelopes = 1
and number of ways to place one letter in the wrong
envelope and other 5 letters in the write envelope = 0
[Q It is not possible that only one letter goes in the
wrong envelope, when if 5 letters goes in the right
envelope, then remaining one letter also goes in the
write envelope]
Hence, number of ways to place atleast two letters
goes in the wrong envelopes
= 6 ! − 0 − 1 = 6 ! − 1 = 720 − 1 = 719
(ii) The number of ways 1 letter in 1 address envelope, so
that one letter is in wrong envelope = 0
…(i)
[because it is not possible that only one letter goes in
the wrong envelope]
391
The number of ways to put 2 letters in 2
addressed envelopes so that all are in wrong
envelopes
= The number of ways without restriction − The
number of ways in which all are in correct envelopes
− The number of ways in which 1 letter is in the
correct envelope
= 2! − 1 − 0 = 2 − 1
…(ii) [from Eq. (i)]
=1
The number of ways to put 3 letters in 3
addressed envelopes so that all are in wrong
envelopes
= The number of ways without restriction − The
number of ways in which all are in correct envelopes
− The number of ways in which 1 letter is in the
correct envelope − The number of ways in which 2
letter are in correct envelope
= 3 ! − 1 − 3 C 1 × 1 − 0 [from Eqs. (i) and (ii)]
=2
[ 3C 1 means that select one envelope to put the letter
correctly]
The number of ways to put 4 letters in 4
addressed envelopes so that all are in wrong
envelopes
= The number of ways without restriction − The
number of ways in which all are in correct envelopes
− The number of ways in which 1 letter is in the
correct envelope − The number of ways in which 2
letters are in correct envelopes − The number of ways
in which 3 letters are in correct envelopes
= 4 ! − 1 − 4C 1 × 2 − 4C 2 × 1 − 4C 3 × 0
[from Eqs. (i), (ii) and (iii)]
…(iv)
= 24 − 1 − 8 − 6 − 0 = 9
The number of ways to put 5 letters in 5
addressed envelopes so that all are in wrong
envelopes
= The number of ways without restriction − The
number of ways in which all are in correct envelopes −
The number of ways in which 1 letter is in the correct
envelopes − The number of ways in which 2 letters are
in correct envelopes − The number of ways in which 3
letters are in correct envelopes − The number of ways
in which 4 letters are in correct envelopes
= 5 ! − 1 − 5C 1 × 9 − 5C 2 × 2 − 5C 3 × 1 − 5C 4 × 0
[from Eqs. (i), (ii), (iii) and (iv)]
= 120 − 1 − 45 − 20 − 10 − 0 = 44
392
Textbook of Algebra
The number of ways to put 6 letters in 6 addressed
envelopes so that all are in wrong envelopes
= The number of ways without restriction − The
number of ways in which all are in correct envelopes
− The number of ways in which 1 letter is in the
correct envelope − The number of ways in which 2
letters are in correct envelopes − The number of ways
in which 3 letters are in correct envelopes − The
number of ways in which 4 letters are in correct
envelopes − The number of ways in which 5 letters
are in correct envelopes.
= 6 ! − 1 − 6 C 1 × 44 − 6 C 2 × 9 − 6 C 3 × 2
− 6C 4 × 1 − 6C 5 × 0
[from Eqs. (i), (ii), (iii), (iv) and (v)]
= 720 − 1 − 264 − 135 − 40 − 15 = 720 − 455 = 265
Multinomial Theorem
(i) If there are l objects of one kind, m objects of second
kind, n objects of third kind and so on, then the
number of ways of choosing r objects out of these
objects (i.e., l + m + n +K) is the coefficient of x r in
the expansion of
(1 + x + x 2 + x 3 + K + x l )(1 + x + x 2 + K + x m )
(1 + x + x 2 + K + x n )
Further, if one object of each kind is to be included,
then the number of ways of choosing r objects out of
these objects (i.e., l + m + n +K) is the coefficient of
x r in the expansion of
( x + x 2 + x 3 + K + x l )( x + x 2 + x 3 + K + x m )
(x + x 2 + x 3 + K + x n ) K
(ii) If there are l objects of one kind, m objects of second
kind, n objects of third kind and so on, then the
number of possible arrangements/permutations of r
objects out of these objects (i.e., l + m + n +K) is the
coefficient of x r in the expansion of

x x2
xl  
x x2
xm 

r ! 1 + +
+ K +  1 + +
+K +
l !   1! 2 !
m! 
 1! 2 !

x x2
xn 
1 + +
 K.
+K +
n! 
 1! 2 !
Different Cases of
Multinomial Theorem
Case I If upper limit of a variable is more than or
equal to the sum required, then the upper limit of
that variable can be taken as infinite.
y Example 91. In how many ways the sum of upper
faces of four distinct die can be five?
Sol. Here, the number of required ways will be equal to the
number of solutions of x 1 + x 2 + x 3 + x 4 = 5 i.e., 1 ≤ x i ≤ 6
for i = 1, 2, 3, 4.
Since, upper limit is 6, which is greater than required
sum, so upper limit taken as infiite. So, number of Sols is
equal to coefficient of α 5 in the expansion of
(1 + α + α 2 + K + ∞ )4
= Coefficient of α 5 in the expansion of (1 − α )−4
= Coefficient of α 5 in the expansion of
(1 + 4C 1α + 5C 2 α 2 + K )
= 8C 5 = 8C 3 =
8 ⋅7 ⋅6
= 56
1 ⋅2 ⋅3
Case II If the upper limit of a variable is less than
the sum required and the lower limit of all variables
is non-negative, then the upper limit of that variable
is that given in the problem.
y Example 92. In an examination, the maximum
marks each of three papers is 50 and the maximum
mark for the fourth paper is 100. Find the number
of ways in which the candidate can score 60%
marks in aggregate.
Sol. Aggregate of marks = 50 × 3 + 100 = 250
60
× 250 = 150
100
∴ 60% of the aggregate =
Let the marks scored by the candidate in four papers be x 1,
x 2 , x 3 and x 4 . Here, the number of required ways will be
equal to the number of Sols of x 1 + x 2 + x 3 + x 4 = 150 i.e.,
0 ≤ x 1, x 2 , x 3 ≤ 50 and 0 ≤ x 4 ≤ 100.
Since, the upper limit is 100 < required sum (150).
The number of solutions of the equation is equal to
coefficient of α150 in the expansion of
(α 0 + α1 + α 2 + K + α 50 )3 (α 0 + α1 + α 2 + K + α100 )
= Coefficient of α150 in the expansion of
(1 − α 51 )3 (1 − α10 )(1 − α )−4
= Coefficient of α150 in the expansion of
(1 − 3α 51 + 3α102 )(1 − α101 )(1 + 4C 1α + 5C 2 α 2 + K + ∞ )
= Coefficient of α150 in the expansion of
(1 − 3α 51 − α101 + 3α102 )(1 + 4C 1α + 5C 2 α 2 + K + ∞ )
=
153
C 150 − 3 ×
=
153
C3 − 3 ×
= 110556
C 99 −
C 49 + 3 × 51C 48
102
C3 −
102
52
C 3 + 3 × 51C 3
52
Chap 05 Permutations and Combinations
Very Important Trick
… on multiplying by 1 + α + α 2 + α 3 → To each coefficient
add 3 preceding coefficients
On multiplying p 0 + p 1 α + p 2 α 2 + p 3 α 3 + K + p n α n
by (1 + α ), we get
p 0 + (p 0 + p1 ) α + (p1 + p2 ) α 2 + (p2 + p 3 ) α 3 + K
+ (pn − 2 + pn − 1 )α
n −1
393
1
4
7
8
7
4
1
...
…on multiplying by 1 + α + α → To each coefficient add 2
preceding coefficients.
2
+ (pn − 1 + pn )α + pnα
n
n +1
i.e., we just add coefficient of α r with coefficient of α r − 1
(i.e., previous term) to get coefficient α r in product.
Now, coefficient of α = p r − 1 + p r
r
On multiplying p 0 + p 1 α + p 2 α 2 + p 3 α 3 + K + p n α n by
(1 + α + α 2 )
we get, p 0 + ( p 0 + p 1 ) α + ( p 0 + p 1 + p 2 ) α 2
+ (p1 + p2 + p 3 ) α 3 + (p2 + p 3 + p 4 ) α 4 + K
i.e., to find coefficient of α r in product and add this with 2
preceding coefficients.
1
5
12
19
22
19
12
...
…on multiplying by 1 + α + α 2 → To each coefficient add 2
preceding coefficients.
...
...
...
...
...
53
...
...
Hence, required coefficient is 53.
y Example 94. Find the number of different
selections of 5 letters which can be made from
5A’s, 4B ’s, 3C’s, 2D’s and 1E
Sol. All selections of 5 letters are given by 5th degree terms in
Now, coefficient of α r = p r − 2 + p r − 1 + p r
(1 + A + A 2 + A 3 + A 4 + A 5 )(1 + B + B 2 + B 3 + B 4 )
Similarly, in product of p 0 + p 1 α + p 2 α 2 + K with
(1 + C + C 2 + C 3 )(1 + D + D 2 )(1 + E )
(1 + α + α 2 + α 3 ), the coefficient of α r in product will be
∴ Number of 5 letter selections
= Coefficient of α 5 in (1 + α + α 2 + α 3 + α 4 + α 5 )
p − 3 + pr − 2 + pr − 1 + pr
1r 444
24443
(1 + α + α 2 + α 3 + α 4 )(1 + α + α 2 + α 3 )
(1 + α + α 2 )(1 + α )
3 preceding coefficients
and in product of p 0 + p 1 α + p 2 α 2 + K with
( 1 + α + α 2 + α 3 + α 4 ), the coefficient of α r in product
will be p r − 4 + p r − 3 + p r − 2 + p r − 1 + p r
14444
4244444
3
4 preceding coefficients
Multiplying synthetically
1
α
α
1
1
1
2
α3
α4
1
1
α 5 ...
1
Finally, in product of p 0 + p 1 α + p 2 α + K with
–––––––––––––––––––––––––––––– × 1 + α + α + α 2 + α 3+ α 4
(1 + α + α 2 + α 3 + K + upto ∞ ), the coefficient of α r in
1
2
product will be p 0 + p 1 + p 2 + K + p r − 1 + p r
1444424444
3
y Example 93. Find the coefficient of α in the
product (1 + α + α 2 )(1 + α + α 2 )(1 + α + α 2 + α 3 )
6
(1 + α )(1 + α )(1 + α ).
or (1 + α + α 2 )(1 + α + α 2 )(1 + α + α 2 + α 3 )
( 1 + 3α + 3α 2 + α 3 )
α
1
3
3
2
α
3
1
α
4
0
5 ...
3
6
10
14
17 ...
1
4
10
19
30
41 ...
–––––––––––––––––––––––––––––– × 1 + α
5
14
29
49
71
α
5
0
y Example 95. Find the number of combinations and
permutations of 4 letters taken from the word
EXAMINATION.
Sol. There are 11 letters
A, A, N, N, X, M, T, O.
Then, number of combinations
Multiplying Synthetically
α
5
Hence, required coefficient is 71.
(1 + α + α 2 )(1 + α + α 2 )(1 + α + α 2 + α 3 )(1 + α )3
1
4
–––––––––––––––––––––––––––––– ×1 + α + α 2
1
Sol. The given product can be written as
3
–––––––––––––––––––––––––––––– × 1 + α + α 2 + α 3
1
all preceding coefficients
2
α
6
0
...
= coefficient of x 4 in (1 + x + x 2 )3 (1 + x )5
[Q2A’s, 2I ’s, 2N’s, 1E, 1X, 1M, 1T and 1O]
= Coefficient of x 4 in {(1 + x )3 + x 6 + 3(1 + x )2 x 2
394
Textbook of Algebra
+ 3(1 + x )x 4 }(1 + x )5
X
A
A
O
7 ×6
∴ Number of selections = 3 C 1 × 7 C 2 = 3 ×
= 63
1×2
{(1 + x ) + x (1 + x ) + 3x (1 + x ) + 3x (1 + x ) }
4!
and number of permutations = 63 ⋅ = 756
7 ⋅6
8 ⋅7 ⋅6 ⋅5
2!
= 8C 4 + 0 + 3 ⋅ 7C 2 + 3 =
+ 3 = 70 + 63 + 3
+ 3⋅
1 ⋅2
1 ⋅2 ⋅3 ⋅ 4
From Case I, II and III
= 136
The required number of combinations = 70 + 3 + 63 = 136
and number of permutations
and
number of permutations = 1680 + 18 + 756 = 2454
3
5
= Coefficient of x in
4
8
6
5
2
7
4
6

x x2 
x
= Coefficient of x 4 in 4 ! 1 + +
 1 + 
1! 2!  
1! 

2
3

x
5
= Coefficient of x 4 in 4 ! 1 + x +
 (1 + x )
2

= Coefficient of x 4 in


3
x6 3
4 ! (1 + x )3 +
+ (1 + x )2 x 2 + x 4 (1 + x ) (1 + x )5
4
8
2


= Coefficient of x 4 in


3
3
x6
4 ! (1 + x )8 +
(1 + x )5 + x 2 (1 + x )7 + x 4 (1 + x )6 
4
2
8


3
3
 8 ⋅7 ⋅6 ⋅5 3 7 ⋅6 3 

+ 
+ ⋅
= 4 !  8 C 4 + 0 + ⋅ 7C 2 +  = 24 
2
4
1 ⋅2 ⋅3 ⋅ 4 2 1 ⋅2 4 

= 8 ⋅ 7 ⋅ 6 ⋅ 5 + 6(3 ⋅ 7 ⋅ 6) + 6 ⋅ 3 = 1680 + 756 + 18 = 2454
Aliter There are 11 letters:
A, A, I, I, N, N, E, X, M, T, O
The following cases arise:
Case I All letters different The required number of
choosing 4 different letters from 8 different (A, I, N, E, X,
M, T, O) types of the letters
8 ⋅7 ⋅6 ⋅5
= 70
= 8C 4 =
1⋅2 ⋅3 ⋅ 4
and number of permutations = 8 P 4 = 8 ⋅ 7 ⋅ 6 ⋅ 5 = 1680
Case II Two alike of one type and two alike of
another type This must be 2A’s, 2I’s or 2I’s, 2N’s, or
2N’s, 2A’s.
∴ Number of selections = 3 C 2 = 3
For example, [for arrangements]
A
A
I
I
4!
and number of permutations = 3 ⋅
= 18
2! 2!
Case III Two alike and two different This must be
2A’s or 2I’s or 2N’s
and for each case 7 different letters.
For example, for 2A’s, 7 differents’s are I, N, E, X, M, T, O
For example, [for arrangements]
Note Number of combinations and permutations of 4 letters
taken from the word MATHEMATICS are 136 and 2454
respectively, as like of EXAMINATION.
Number of Solutions with the
Help of Multinomial Theorem
Case I If the equation
…(i)
α + 2 β + 3γ + K + q θ = n
(a) If zero included, the number of solution of Eq. (i)
= Coefficient of x n in (1 + x + x 2 + K )
(1 + x 2 + x 4 + K )(1 + x 3 + x 6 + K ) K
(1 + x q + x 2q + K )
= Coefficient of x n in
(1 − x ) −1 (1 − x 2 ) −1 (1 − x 3 ) −1 K (1 − x q ) −1
(b) If zero excluded, then the number of solutions of Eq. (i)
= Coefficient of x n in ( x + x 2 + x 3 + K )
( x 2 + x 4 + x 6 + K )( x 3 + x 6 + x 9 + K )
K ( x q + x 2q + K )
= Coefficient of x n in x 1 + 2 + 3 + K + q (1 − x ) −1
(1 − x 2 ) −1 (1 − x 3 ) −1 K (1 − x q ) −1
= Coefficient of x
−1
n−
q (q + 1 )
2
in
2 −1
(1 − x ) (1 − x ) (1 − x 3 ) −1 K(1 − x q ) −1
y Example 96. Find the number of non-negative
integral solutions of x 1 + x 2 + x 3 + 4 x 4 = 20.
Sol. Number of non-negative integral solutions of the
given equation
= Coefficient of x 20 in (1 − x )−1(1 − x )−1(1 − x )−1(1 − x 4 )−1
= Coefficient of x 20 in (1 − x )−3 (1 − x 4 )−1
= Coefficient of x 20 in (1 + 3C 1 + 4C 2 x 2 + 5C 3 x 3 + 6C 4 x 4
+ K + 10C 8 x 8 + K + 14C 12 x 12 + K + 18C 16 x 16 + K
+
C 20 x 20 + K )(1 + x 4 + x 8 + x 12 + x 16 + x 20 + K )
22
= 1 + 6C 4 + 10C 8 + 14C 12 + 18C 16 +
22
= 1 + 6C 2 + 10C 2 + 14C 2 + 18C 2 +
C2
C 20
22
Chap 05 Permutations and Combinations
 6 ⋅ 5  10 ⋅ 9   14 ⋅ 13  18 ⋅ 17   22 ⋅ 21
=1+   + 

 +
 +
 +
 1 ⋅ 2  1 ⋅ 2   1 ⋅ 2   1 ⋅ 2   1 ⋅ 2 
Sol. The number of ways of distributing blankets is equal to
the number of solutions of the equation 3x + 2y + z = 15,
where x ≥ 1, y ≥ 1, z ≥ 1 which is equal to coefficient
of α15 in
= 1 + 15 + 45 + 91 + 153 + 231 = 536
y Example 97. Find the number of positive unequal
integral solutions of the equation x + y + z + w = 20.
x + y + z + w = 20
Sol. We have,
395
(α 3 + α 6 + α 9 + α12 + α15 + ...)
× (α 2 + α 4 + α 6 + α 8 + α10 + α12 + α14 +...)
× (α + α 2 + α 3 + ... + α15 + ...)
…(i)
Assume x < y < z < w . Here, x, y, z , w ≥ 1
Now, let x = x 1, y − x = x 2 , z − y = x 3 and w − z = x 4
∴
x = x 1, y = x 1 + x 2 , z = x 1 + x 2 + x 3 and
w = x1 + x 2 + x 3 + x 4
From Eq. (i), 4 x 1 + 3x 2 + 2x 3 + x 4 = 20
Then, x 1, x 2 , x 3 , x 4 ≥ 1
…(ii)
Q
4 x 1 + 3x 2 + 2x 3 + x 4 = 20
∴ Number of positive integral solutions of Eq. (ii)
= Coefficient of x 20 − 10 in
(1 − x 4 )−1(1 − x 3 )−1(1 − x 2 )−1 (1 − x )−1
= Coefficient of x 10 in
(1 − x 4 )−1(1 − x 3 )−1 (1 − x 2 )−1 (1 − x )−1
= Coefficient of x 10 in (1 + x 4 + x 8 + x 12 + ...)
× (1 + x 3 + x 6 + x 9 + x 12 + K ) ×
(1 + x 2 + x 4 + x 6 + x 8 + x 10 + ...) × (1 + x + x 2 + x 3 + x 4
+ x 5 + x 6 + x 7 + x 8 + x 9 + x 10 + K )
= Coefficient of x 10 in
(1 + x 3 + x 6 + x 9 + x 4 + x 7 + x 10 + x 8 )
× (1 + x 2 + x 4 + x 6 + x 8 + x 10 )(1 + x + x 2 + x 3
+ x 4 + x 5 + x 6 + x 7 + x 8 + x 9 + x 10 )
[neglecting higher powers]
= Coefficient of x 10 in
(1 + x 2 + x 4 + x 6 + x 8 + x 10 + x 3 + x 5 + x 7 + x 9 + x 6
+ x 8 + x 10 + x 9 + x 4 + x 6 + x 8 + x 10 + x 7 + x 9 + x 10
+ x 8 + x 10 )(1 + x + x 2 + x 3 + x 4 + x 5 + x 6 + x 7
+ x 8 + x 9 + x 10 ) [neglecting higher powers]
= Coefficient of α 9 in (1 + α 3 + α 6 + α 9 )
× (1 + α 2 + α 4 + α 6 + α 8 )
× (1 + α + α 2 + α 3 + α 4 + α 5 + α 6 + α 7 + α 8 + α 9 )
[neglecting higher powers]
= Coefficient of α 9 in (1 + α 2 + α 4 + α 6 + α 8 + α 3
+ α 5 + α 7 + α 9 + α 6 + α 8 + α 9 ) × (1 + α + α 2
α3 + α4 + α5 + α6 + α7 + α8 + α9 )
[neglecting higher powers]
= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12
Case II If the inequation
…(i)
x 1 + x 2 + x 3 +K + x m ≤ n
[when the required sum is not fixed]
In this case, we introduce a dummy variable x m + 1 .
So that,
x 1 + x 2 + x 3 + K + x m + x m + 1 = n,
…(ii)
xm +1 ≥ 0
Here, the number of Sols of Eqs. (i) and (ii) will be same.
y Example 99. Find the number of positive integral
solutions of the inequation 3x + y + z ≤ 30.
Sol. Let dummy variable w , then
…(i)
3x + y + z + w = 30, w ≥ 0
Now, let a = x − 1, b = y − 1, c = z − 1, d = w , then
…(ii)
3a + b + c + d = 25, where a, b, c, d ≥ 0
∴ Number of positive integral solutions of Eq. (i)
= Number of non-negative integral solutions of Eq. (ii)
= Coefficient of α 25 in (1 + α 3 + α 6 + K )
=1+1+1+1+1+1+1+1+1+1+1+1+1
(1 + α + α 2 + K )3
+ 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 23
But x, y, z and w can be arranged in 4 P4 = 4 ! = 24
Hence, required number of Sols = (23)(24 ) = 552
y Example 98. In how many ways can 15 identical
blankets be distribted among six beggars such that
everyone gets atleast one blanket and two particular
beggars get equal blankets and another three
particular beggars get equal blankets.
= Coefficient of α
25
in (1 + α + α + K )(1 − α )−3
3
6
= Coefficient of α 25 in
(1 + α 3 + α 6 + K )(1 + 3C 1α + 4C 2 α 2 + K )
=
C 25 +
27
C 22 + 21C 19 + 18C 16 + 15C 13 + 12C 10 + 9C 7
24
+ 6C 4 + 3C 1
=
C2 +
27
C 2 + 21C 2 + 18C 2 + 15C 2 + 12C 2 + 9C 2
24
+ 6C 2 + 3C 1
= 351 + 276 + 210 + 153 + 105 + 66 + 36 + 15 + 3 = 1215
Aliter
396
Textbook of Algebra
From Eq. (ii), 3a + b + c + d = 25, where a, b, c, d ≥ 0
Clearly, 0 ≤ a ≤ 8, if a = k , then
…(iii)
b + c + d = 25 − 3k
Hence, number of non-negative integral solutions of
Eq. (iii) is
(27 − 3k )(26 − 3k )
25 − 3k + 3 − 1
C 3 − 1 = 27 − 3k C 2 =
2
3 2
= (3k − 53k + 23