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The Most Accepted
CRASH COURSE
PROGRAMME
JEE Main in
40 DAYS
MATHEMATICS
ARIHANT PRAKASHAN (Series), MEERUT
Arihant Prakashan (Series), Meerut
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PREFACE
It is a fact that nearly 10 lacs students would be in the race with you in JEE Main, the gateway
to some of the prestigious engineering and technology institutions in the country, requires
that you take it seriously and head-on. A slight underestimation or wrong guidance will ruin
all your prospects. You have to earmark the topics in the syllabus and have to master them in
concept-driven-problem-solving ways, considering the thrust of the questions being asked in
JEE Main.
The book 40 Days JEE Main Mathematics serves the above cited purpose in perfect manner.
At whatever level of preparation you are before the exam, this book gives you an accelerated
way to master the whole JEE Main Physics Syllabus. It has been conceived keeping in mind the
latest trend of questions, and the level of different types of students.
The whole syllabus of Physics has been divided into day-wise-learning modules with clear
groundings into concepts and sufficient practice with solved and unsolved questions on that
day. After every few days you get a Unit Test based upon the topics covered before that day.
On last three days you get three full-length Mock Tests, making you ready to face the test. It is
not necessary that you start working with this book in 40 days just before the exam. You may
start and finish your preparation of JEE Main much in advance before the exam date. This will
only keep you in good frame of mind and relaxed, vital for success at this level.
Salient Features
Ÿ Concepts discussed clearly and directly without being superfluous. Only the required
Ÿ
Ÿ
Ÿ
Ÿ
Ÿ
material for JEE Main being described comprehensively to keep the students focussed.
Exercises for each day give you the collection of only the Best Questions of the concept,
giving you the perfect practice in less time.
Each day has two Exercises; Foundation Questions Exercise having Topically Arranged
Questions & Progressive Question Exercise having higher Difficulty Level Questions.
All types of Objective Questions included in Daily Exercises (Single Option Correct,
Assertion & Reason, etc).
Along with Daywise Exercises, there above also the Unit Tests & Full Length Mock Tests.
At the end, there are all Online Solved Papers of JEE Main 2019; January & April attempts.
We are sure that 40 Days Physics for JEE Main will give you a fast way to prepare for Physics
without any other support or guidance.
Publisher
CONTENTS
Preparing JEE Main 2020 Mathematics in 40 Days !
Day 1.
Sets, Relations and Functions
Day 2.
Complex Numbers
10-19
Day 3.
Sequences and Series
20-30
Day 4.
Quadratic Equation and Inequalities
31-44
Day 5.
Matrices
45-54
Day 6.
Determinants
55-67
Day 7.
Binomial Theorem and Mathematical Induction
68-77
Day 8.
Permutations and Combinations
78-86
Day 9.
Unit Test 1 (Algebra)
87-94
Day 10. Real Function
1-9
95-103
Day 11. Limits, Continuity and Differentiability
104-116
Day 12. Differentiation
117-126
Day 13. Applications of Derivatives
127-137
Day 14. Maxima and Minima
138-149
Day 15. Indefinite Integrals
150-162
Day 16. Definite Integrals
163-175
Day 17. Area Bounded by the Curves
176-187
Day 18. Differential Equations
188-198
Day 19. Unit Test 2 (Calculus)
199-208
Day 20. Trigonometric Functions and Equations
209-221
Day 21. Properties of Triangle, Height and Distances
222-232
Day 22. Inverse Trigonometric Function
233-242
Day 23. Unit Test 3 (Trigonometry)
243-250
Day 24. Cartesian System of Rectangular Coordinates
251-262
Day 25. Straight Line
263-274
Day 26. The Circle
275-288
Day 27. Parabola
289-300
Day 28. Ellipse
301-313
Day 29. Hyperbola
314-325
Day 30. Unit Test 4 (Coordinate Geometry)
326-335
Day 31. Vector Algebra
336-350
Day 32. Three Dimensional Geometry
351-366
Day 33. Unit Test 5 (Vector & 3D Geometry)
367-374
Day 34. Statistics
375-383
Day 35. Probability
384-394
Day 36. Mathematical Reasoning
395-405
Day 37. Unit Test 6 (Statistics, Probability & Mathematical Reasoning)
406-411
Day 38. Mock Test 1
412-416
Day 39. Mock Test 2
417-423
Day 40. Mock Test 3
424-430
Online JEE Main Solved Papers 2019
1-32
SYLLABUS
MATHEMATICS
UNIT 1 Sets, Relations and Functions
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.
UNIT 2 Complex Numbers and Quadratic Equations
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,
Quadratic equations in real and complex number system and their solutions. Relation between
roots and
co-efficients, nature of roots, formation of quadratic equations with given roots.
UNIT 3 Matrices and Determinants
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.
UNIT 4 Permutations and Combinations
Fundamental principle of counting, permutation as an arrangement and combination as
selection, Meaning of P (n,r) and C (n,r), simple applications.
UNIT 5 Mathematical Induction
Principle of Mathematical Induction and its simple applications.
UNIT 6 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 7 Sequences and Series
Arithmetic and Geometric progressions, insertion of arithmetic, geometric means between two
given numbers. Relation between AM and GM Sum upto n terms of special series: ∑ n, ∑ n2, ∑ n3.
Arithmetico - Geometric progression.
UNIT 8 Limit, Continuity and Differentiability
Real valued functions, algebra of functions, polynomials, rational, trigonometric, logarithmic and
exponential functions, inverse functions. Graphs of simple functions. Limits, continuity and
differentiability.
Differentiation of the sum, difference, product and quotient of two functions. Differentiation of
trigonometric, inverse trigonometric, logarithmic exponential, composite and implicit functions
derivatives of order upto two. Rolle's and Lagrange's Mean Value Theorems. Applications of
derivatives: Rate of change of quantities, monotonic - increasing and decreasing functions,
Maxima and minima of functions of one variable, tangents and normals.
UNIT 9 Integral Calculus
Integral as an anti - derivative. Fundamental integrals involving algebraic, trigonometric,
exponential and
logarithmic functions. Integration by substitution, by parts and by partial fractions. Integration
using trigonometric identities.
Evaluation of simple integrals of the type
dx
,
x2 ± a2
dx
2
x ± a
dx
,
ax 2 + bx + c
(px + q) dx
ax 2 + bx + c
,
2
,
dx
a2 – x2
dx
ax 2 + bx + c
a 2 ± x 2 dx
,
dx
a 2 – x 2
,
(px + q) dx ,
ax 2 + bx + c
,
x 2 – a 2 dx
and
Integral as limit of a sum. Fundamental Theorem of Calculus. Properties of definite integrals.
Evaluation of definite integrals, determining areas of the regions bounded by simple curves in
standard form.
UNIT 10 Differential Equations
Ordinary differential equations, their order and degree. Formation of differential equations.
Solution of differential equations by the method of separation of variables, solution of
homogeneous and linear differential equations of the type dy +p (x) y = q(x)
dx
UNIT 11 Coordinate Geometry
Cartesian system of rectangular coordinates in a plane, distance formula, section formula, locus
and its equation, translation of axes, slope of a line, parallel and perpendicular lines, intercepts of
a line on the coordinate axes.
Ÿ Straight Lines
Various forms of equations of a line, intersection of lines, angles between two lines,
conditions for concurrence of three lines, distance of a point from a line, equations of
internal and external bisectors of angles between two lines, coordinates of centroid,
orthocentre and circumcentre of a triangle, equation of family of lines passing through the
point of intersection of two lines.
Ÿ Circles, Conic Sections
Standard form of equation of a circle, general form of the equation of a circle, its radius
and centre, equation of a circle when the end points of a diameter are given, points of
intersection of a line and a circle with the centre at the origin and condition for a line to be
tangent to a circle, equation of the tangent. Sections of cones, equations of conic sections
(parabola, ellipse and hyperbola) in standard forms, condition for y = mx + c to be a tangent
and point (s) of tangency.
UNIT 12 Three Dimensional Geometry
Coordinates of a point in space, distance between two points, section formula, direction ratios
and direction cosines, angle between two intersecting lines. Skew lines, the shortest distance
between them and its equation. Equations of a line and a plane in different forms, intersection of
a line and a plane, coplanar lines.
UNIT 13 Vector Algebra
Vectors and scalars, addition of vectors, components of a vector in two dimensions and three
dimensional space, scalar and vector products, scalar and vector triple product.
UNIT 14 Statistics and Probability
Measures of Dispersion Calculation of mean, median, mode of grouped and ungrouped data.
Calculation of standard deviation, variance and mean deviation for grouped and ungrouped data.
Probability Probability of an event, addition and multiplication theorems of probability, Baye's
theorem, probability distribution of a random variate, Bernoulli trials and Binomial distribution.
UNIT 15 Trigonometry
Trigonometrical identities and equations. Trigonometrical functions. Inverse trigonometrical
functions and their properties. Heights and Distances.
UNIT 16 Mathematical Reasoning
Statements, logical operations and implies, implied by, if and only if. Understanding of
tautology, contradiction, converse and contra positive.
HOW THIS BOOK IS
USEFUL FOR YOU ?
As the name suggest, this is the perfect book for your recapitulation of the whole syllabus, as it
provides you a capsule course on the subject covering the syllabi of JEE Main, with the smartest
possible tactics as outlined below:
1.
REVISION PLAN
The book provides you with a practical and sound revision plan.
The chapters of the book have been designed day-wise to guide the students in a planned
manner through day-by-day, during those precious 35-40 days. Every day you complete a
chapter/a topic, also take an exercise on the chapter. So that you can check & correct your
mistakes, answers with hints & solutions also have been provided. By 37th day from the date
you start using this book, entire syllabus gets revisited.
Again, as per your convenience/preparation strategy, you can also divide the available 30-35
days into two time frames, first time slot of 3 weeks and last slot of 1 & 1/2 week. Utilize first
time slot for studies and last one for revising the formulas and important points. Now fill the
time slots with subjects/topics and set key milestones. Keep all the formulas, key points on a
couple of A4 size sheets as ready-reckner on your table and go over them time and again. If you
are done with notes, prepare more detailed inside notes and go over them once again. Study
all the 3 subjects every day. Concentrate on the topics that have more weightage in the exam
that you are targeting.
2.
MOCK TESTS
Once you finish your revision on 37th day, the book provides you with full length mock tests
for day 38th, 39th, & 40th, thereby ensures your total & full proof preparation for the final
show.
The importance of solving previous years' papers and 10-15 mock tests cannot be
overemphasized. Identify your weaknesses and strengths. Work towards your strengths i.e.,
devote more time to your strengths to be 100% sure and confident. In the last time frame of 1
& 1/2 week, don't take-up anything new, just revise what you have studied before. Be examready with quality mock tests in between to implement your winning strategy.
3.
FOCUS TOPICS
Based on past years question paper trends, there are few topics in each subject which have
more questions in exam than other. So far Mathematics is concerned it may be summed up as
below:
Calculus, Trigonometry, Algebra, Coordinate Geometry & Vector 3D.
More than 80% of questions are normally asked from these topics.
However, be prepared to find a completely changed pattern for the exam than noted above as
examiners keep trying to weed out 'learn by rot practice'. One should not panic by witnessing a
new pattern , rather should be tension free as no one will have any upper hand in the exam.
4.
IMPROVES STRIKE RATE AND ACCURACY
The book even helps to improve your strike rate & accuracy. When solving practice tests or mock
tests, try to analyze where you are making mistakes-where are you wasting your time; which
section you are doing best. Whatever mistakes you make in the first mock test, try to improve that
in second. In this way, you can make the optimum use of the book for giving perfection to your
preparation.
What most students do is that they revise whole of the syllabus but never attempt a mock and
thus they always make mistake in main exam and lose the track.
5.
LOG OF LESSONS
During your preparations, make a log of Lesson's Learnt. It is specific to each individual as to where
the person is being most efficient and least efficient. Three things are important - what is working,
what's not working and how would you like to do in your next mock test.
6.
TIME MANAGEMENT
Most candidates who don't make it to good medical colleges are not good in one area- Time
Management. And, probably here lies the most important value addition that's the book
provides in an aspirant's preparation. Once the students go through the content of the book
precisely as given/directed, he/she learns the tactics of time management in the exam.
Realization and strengthening of what you are good at is very helpful, rather than what one
doesn't know. Your greatest motto in the exam should be, how to maximize your scoring with
the given level of preparation. You have to get about 200 plus marks out of a total of about 400
marks for admission to a good NIT (though for a good branch one needs to do much better
than that). Remember that one would be doomed if s/he tries to score 400 in about 3 hours.
7.
ART OF PROBLEM SOLVING
The book also let you to master the art of problem solving. The key to problem solving does not lie
in understanding the solution to the problem but to find out what clues in the problem leads you
to the right solution. And, that's the reason Hints & Solutions are provided with the exercises after
each chapter of the book. Try to find out the reason by analyzing the level of problem & practice
similar kind of problems so that you can master the tricks involved. Remember that directly going
though the solutions is not going to help you at all.
8.
POSITIVE PERCEPTION
The book put forth for its readers a 'Simple and Straightforward' concept of studies, which is the
best possible, time-tested perception for 11th hour revision / preparation.
The content of the book has been presented in such a lucid way so that you can enjoy what you are
reading, keeping a note of your already stressed mind & time span.
Cracking JEE Main is not a matter of life and death. Do not allow panic and pressure to create
confusion. Do some yoga and prayers. Enjoy this time with studies as it will never come back.
DAY ONE
Sets, Relations
and Functions
Learning & Revision for the Day
u
Sets
u
Law of Algebra of Sets
u
Composition of Relations
u
Venn Diagram
u
Cartesian Product of Sets
u
Functions or Mapping
u
Operations on Sets
u
Relations
u
Composition of Functions
Sets
●
●
●
A set is a well-defined class or collection of the objects.
Sets are usually denoted by the symbol A, B, C, ... and its elements are denoted by a, b , c,
… etc.
If a is an element of a set A, then we write a ∈ A and if not then we write a ∉ A.
Representations of Sets
There are two methods of representing a set :
●
●
In roster method, a set is described by listing elements, separated by commas, within
curly braces{≠}. e.g. A set of vowels of English alphabet may be described as {a, e, i, o, u}.
elements x. In such a case the set is written as { x : P ( x) holds} or { x| P ( x) holds}, which
is read as the set of all x such that P( x) holds. e.g. The set P = {0, 1, 4 , 9, 16,...} can be
written as P = { x2 | x ∈ Z }.
Types of Sets
●
●
●
PRED
MIRROR
In set-builder method, a set is described by a property P ( x), which is possessed by all its
The set which contains no element at all is called the null set (empty set or void
set) and it is denoted by the symbol ‘φ ’ or ‘{}’ and if it contains a single element, then it is
called singleton set.
A set in which the process of counting of elements definitely comes to an end, is called a
finite set, otherwise it is an infinite set.
Two sets A and B are said to be equal set iff every element of A is an element of B and
also every element of B is an element of A. i.e. A = B, if x ∈ A ⇔ x ∈ B.
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
2
●
●
●
●
●
●
ONE
40
A ∩ B = { x : x ∈ A and x ∈ B}.
i.e.
Equivalent sets have the same number of elements but not
exactly the same elements.
A
A set that contains all sets in a given context is called
universal set (U).
B
U
Let A and B be two sets. If every element of A is an element
of B, then A is called a subset of B, i.e. A ⊆ B.
If A is a subset of B and A ≠ B, then A is a proper subset of
B. i.e. A ⊂ B.
The null set φ is a subset of every set and every set is a
subset of itself i.e. φ ⊂ A and A ⊆ A for every set A. They
are called improper subsets of A.
A∩B
●
●
If S is any set, then the set of all the subsets of S is called
the power set of S and it is denoted by P(S ). Power set of a
given set is always non-empty. If A has n elements, then
P( A) has 2 n elements.
If A ∩ B = φ, then A and B are called disjoint sets.
Let U be an universal set and A be a set such that A ⊂ U.
Then, complement of A with respect to U is denoted by A′
or Ac or A or U − A. It is defined as the set of all those
elements of U which are not in A.
U
A′
A
NOTE
• The set { φ} is not a null set. It is a set containing one
element φ.
• Whenever we have to show that two sets A and B are equal
show that A ⊆ B and B ⊆ A.
●
• If a set A has m elements, then the number m is called
cardinal number of set A and it is denoted by n( A). Thus,
n( A) = m.
The difference A − B is the set of all those elements of A
which does not belong to B.
i.e.
A − B = { x : x ∈ A and x ∉ B}
and B − A = { x : x ∈ B and x ∉ A}.
U
Venn Diagram
The combination of rectangles and circles is called Venn Euler
diagram or Venn diagram. In Venn diagram, the universal set
is represented by a rectangular region and a set is represented
by circle on some closed geometrical figure. Where, A is the
set and U is the universal set.
U
B–A
The symmetric difference of sets A and B is the set
( A − B) ∪ (B − A) and is denoted by A ∆ B.
i.e.
A ∆ B = ( A − B) ∪ ( B − A)
A
B
Operations on Sets
●
The union of sets A and B is the set of all elements which
are in set A or in B or in both A and B.
i.e. A ∪ B = { x : x ∈ A or x ∈ B}
A
B
U
A∆B
Law of Algebra of Sets
If A, B and C are any three sets, then
1. Idempotent Laws
(i) A ∪ A = A
A∪B
●
The intersection of A and B is the set of all those elements
that belong to both A and B.
B
A
A–B
●
A
U
B
A
(ii) A ∩ A = A
2. Identity Laws
(i) A ∪ φ = A
(ii) A ∩ U = A
3. Distributive Laws
(i) A ∪ (B ∩ C) = ( A ∪ B) ∩ ( A ∪ C)
(ii) A ∩ (B ∪ C) = ( A ∩ B) ∪ ( A ∩ C)
U
3
DAY
4. De-Morgan’s Laws
(i)
(ii)
(iii)
(iv)
( A ∪ B)′ = A′ ∩ B ′
( A ∩ B)′ = A′ ∪ B ′
A − (B ∩ C) = ( A − B) ∪ ( A − C )
A − (B ∪ C) = ( A − B) ∩ ( A − C)
5. Associative Laws
Relations
●
●
●
(i) ( A ∪ B) ∪ C = A ∪ (B ∪ C)
(ii) A ∩ (B ∩ C) = ( A ∩ B) ∩ C
●
Let A and B be two non-empty sets, then relation R from A to B
is a subset of A × B, i.e. R ⊆ A × B.
If (a, b ) ∈ R, then we say a is related to b by the relation R and we
write it as aRb.
Domain of R = {a :(a, b ) ∈ R} and range of R = {b : (a, b ) ∈ R}.
If n( A) = p and n(B) = q , then the total number of relations from A
to B is 2 pq .
6. Commutative Laws
(i) A ∪ B = B ∪ A
(iii) A ∆ B = B ∆ A
(ii) A ∩ B = B ∩ A
Types of Relations
●
Let A be any non-empty set and R be a relation on A. Then,
(i) R is said to be reflexive iff ( a, a) ∈ R, ∀ a ∈ A.
Important Results on Operation of Sets
1. A − B = A ∩ B ′
(ii) R is said to be symmetric iff
2. B − A = B ∩ A ′
3. A − B = A ⇔ A ∩ B = φ
4. ( A − B) ∪ B = A ∪ B
5. ( A − B) ∩ B = φ
6. A ⊆ B ⇔ B ′ ⊆ A ′
7. ( A − B) ∪ ( B − A) = ( A ∪ B) − ( A ∩ B)
⇒
(iii) R is said to be a transitive iff ( a, b) ∈ R and (b, c) ∈ R
⇒
( a, c) ∈ R, ∀ a, b, c ∈ A
i.e. aRb and bRc ⇒ aRc, ∀ a, b, c ∈ A.
●
8. n ( A ∪ B) = n ( A) + n ( B) − n ( A ∩ B)
9. n ( A ∪ B) = n ( A) + n ( B)
⇔ A and B are disjoint sets.
10. n ( A − B) = n ( A) − n ( A ∩ B)
11. n ( A ∆ B) = n ( A) + n ( B) − 2n ( A ∩ B)
●
The relation I A = {(a, a) : a ∈ A} on A is called the identity
relation on A.
R is said to be an equivalence relation iff
(i) it is reflexive i.e. ( a, a) ∈ R, ∀ a ∈ A.
(ii) it is symmetric i.e. ( a, b) ∈ R ⇒ (b, a) ∈ R, ∀ a, b ∈ A
(iii) it is transitive
12. n ( A ∪ B ∪ C ) = n ( A) + n ( B) + n (C ) − n ( A ∩ B)
− n ( B ∩ C) − n ( A ∩ C) + n ( A ∩ B ∩ C)
13. n ( A ′ ∪ B ′ ) = n ( A ∩ B) ′ = n (U ) − n ( A ∩ B)
14. n ( A ′ ∩ B ′ ) = n ( A ∪ B) ′ = n (U ) − n ( A ∪ B)
Cartesian Product of Sets
Let A and B be any two non-empty sets. Then the
cartesian product A × B, is defined as set of all ordered
pairs (a, b ) such that a ∈ A and b ∈ B.
i.e.
A × B = {(a, b ) : a ∈ A and b ∈ B}
B × A = {(b , a) : b ∈ B and a ∈ A}
and A × A = {(a, b ) : a, b ∈ A}.
A × B = φ, if either A or B is an empty set.
If n ( A) = p and n (B) = q , then
n ( A × B) = n( A) ⋅ n(B) = pq .
A × (B ∪ C) = ( A × B) ∪ ( A × C)
A × (B ∩ C) = ( A × B) ∩ ( A × C)
A × (B − C) = ( A × B) − ( A × C)
( A × B) ∩ (C × D) = ( A ∩ C) × (B ∩ D).
●
●
●
●
●
●
●
●
(a, b ) ∈ R
(b , a) ∈ R, ∀ a, b ∈ A
i.e.
( a, b) ∈ R and (b, c) ∈ R
⇒
( a, c) ∈ R, ∀ a, b, c ∈ A
Inverse Relation
Let R be a relation from set A to set B, then the inverse of R, denoted
by R −1 , is defined by
R −1 = {(b , a) : (a, b ) ∈ R}. Clearly, (a, b ) ∈ R ⇔ (b , a) ∈ R −1 .
NOTE
• The intersection of two equivalence relations on a set is an
equivalence relation on the set.
• The union of two equivalence relations on a set is not necessarily
an equivalence relation on the set.
• If R is an equivalence relation on a set A, then R −1 is also an
equivalence relation A.
Composition of Relations
Let R and S be two relations from set A to B and B to C respectively,
then we can define a relation SoR from A to C such that
(a, c) ∈ SoR ⇔ ∃ b ∈ B such that (a, b ) ∈ R and (b , c) ∈ S . This relation
is called the composition of R and S.
RoS ≠ SoR
4
ONE
40
Functions or Mapping
●
●
If A and B are two non-empty sets, then a rule f which
associates each x ∈ A, to a unique member y ∈ B, is called a
function from A to B and it is denoted by f : A → B.
i.e. f : A → B is a many-one function, if it is not a one-one
function.
●
f is said to be onto function or surjective function, if each
element of B has its pre-image in A.
A
The set A is called the domain of f (D f ) and set B is called
the codomain of f (C f ).
●
The set consisting of all the f -images of the elements of the
domain A, called the range of f (R f ).
NOTE
• A relation will be a function, if no two distinct ordered pairs
have the same first element.
• Every function is a relation but every relation is not
necessarily a function.
• The number of functions from a finite set A into finite set
B is { n(B )}n ( A).
●
●
●
●
different elements of A have different images in B.
A
B
b1
f
a2
b2
a3
b3
a4
a2
b2
a3
b3
Find the range of f ( x) and show that range of
f ( x) = codomain of f ( x).
f is said to be one-one function or injective function, if
a1
●
Any polynomial function of odd degree is always onto.
The number of onto functions that can be defined from a
finite set A containing n elements onto a finite set B
containing 2 elements = 2 n − 2 .
If n ( A) ≥ n (B), then number of onto function is 0.
If A has m elements and B has n elements, where m < n,
then number of onto functions from A to B is
nm − nC1 (n − 1)m + nC2 (n − 2)m − ..., m < n.
f is said to be an into function, if there exists atleast one
element in B having no pre-image in A. i.e. f : A → B is an
into function, if it is not an onto function.
A
a1
b4
Methods to Check One-One Function
Method I
If f ( x) = f ( y ) ⇒ x = y , then f is one-one.
Method II A function is one-one iff no line parallel to
X-axis meets the graph of function at more
than one point.
●
The number of one-one function that can be defined from a
n( B )
finite set A into finite set B is
0,
●
A
a1
f
●
B
b1
a2
b2
a3
b3
a4
b4
a5
b5
B
f
b1
a2
b2
a3
b3
a4
b4
a5
b5
f is said to be a bijective function, if it is one-one as well as
onto.
NOTE
• If f : A → B is a bijective, then A and B have the same
number of elements.
Pn( A ) , if n (B) ≥ n ( A)
.
otherwise
f is said to be a many-one function, if two or more
elements of set A have the same image in B.
b1
Method to Check Onto Function
Different Types of Functions
Let f be a function from A to B, i.e. f : A → B. Then,
B
f
a1
• If n ( A) = n (B ) = m, then number of bijective map from A to
B is m!.
Composition of Functions
Let f : A → B and g : B → C are two functions. Then, the
composition of f and g, denoted by
gof : A → C, is defined as,
gof ( x) = g[ f ( x)], ∀ x ∈ A.
NOTE
• gof is defined only if f ( x ) is an element of domain of g.
• Generally, gof ≠ fog.
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 If Q = x : x =
1
, where y ∈ N , then
y
(a) 0 ∈Q
(b) 1∈Q
(c) 2∈Q
(d)
2
∈Q
3
2 If P ( A ) denotes the power set of A and A is the void set,
then what is number of elements in P {P {P {P ( A )}}}?
(a) 0
(b) 1
(c) 4
(d) 16
3 If X = {4n − 3n − 1: n ∈ N} and Y = {9 (n − 1): n ∈ N}; where N
is the set of natural numbers,then X ∪ Y is equal to
j
(a) N
(b) Y-X
(c) X
JEE Mains 2014
(d) Y
4 If A, B and C are three sets such that A ∩ B = A ∩ C and
A ∪ B = A ∪ C, then
(a) A = C
(b) B = C
(c) A ∩ B = φ (d) A = B
5 Suppose A1, A2,… , A 30 are thirty sets each having
5 elements and B1, B2,… , Bn are n sets each having
30
n
i =1
j =1
3 elements. Let ∪ A i = ∪ Bj = S and each element of S
belongs to exactly 10 of Ai ’s and exactly 9 of Bj ’ s. The
j NCERT Exemplar
value of n is equal to
(a) 15
(c) 45
(b) 3
(d) None of these
6 If A and B are two sets and A ∪ B ∪ C = U. Then,
{( A − B ) ∪ (B − C ) ∪ (C − A )}′ is equal to
(a) A ∪ B ∪ C
(c) A ∩ B ∩ C
(b) A ∪ (B ∩ C)
(d) A ∩ (B ∪ C)
7 Let X be the universal set for sets A and B, if
n( A ) = 200, n(B ) = 300 and n( A ∩ B ) = 100, then
n ( A′ ∩ B′) is equal to 300 provided n( X ) is equal to
(a) 600
(b) 700
(c) 800
(d) 900
8 If n( A ) = 1000, n(B ) = 500, n( A ∩ B ) ≥ 1 and n( A ∪ B ) = P ,
then
(a) 500 ≤ P ≤ 1000
(c) 1000 ≤ P ≤ 1498
(b) 1001 ≤ P ≤ 1498
(d) 1000 ≤ P ≤ 1499
9 If n( A ) = 4, n(B ) = 3, n( A × B × C ) = 24, then n(C ) is equal to
(a) 2
(b) 288
(c) 12
(d) 1
10 If R = {( 3 , 3) ,( 6, 6), ( 9, 9), (12 , 12), ( 6 , 12), ( 3 , 9), (3,12),
(3, 6)} is a relation on the set A = {3 , 6 , 9 , 12}.
The relation is
(a)
(b)
(c)
(d)
an equivalence relation
reflexive and symmetric
reflexive and transitive
only reflexive
11 Let R = {( x , y ) : x , y ∈ N and x 2 − 4xy + 3y 2 = 0}, where
N is the set of all natural numbers. Then, the relation R is
j
JEE Mains 2013
(a) reflexive but neither symmetric nor transitive
(b) symmetric and transitive
(c) reflexive and symmetric
(d) reflexive and transitive
12 If g ( x ) = 1 + x and f {g ( x )} = 3 + 2 x + x , then f ( x ) is
equal to
(a) 1 + 2 x 2
(c) 1 + x
(b) 2 + x 2
(d) 2 + x
13 Let f ( x ) = ax + b and g ( x ) = cx + d , a ≠ 0, c ≠ 0. Assume
a = 1, b = 2 , if ( fog ) ( x ) = ( gof ) ( x ) for all x. What can you
say about c and d?
(a) c and d both arbitrary
(c) c is arbitrary and d = 1
(b) c = 1and d is arbitrary
(d) c = 1, d = 1
14 If R is relation from {11, 12 , 13 } to {8 , 10 , 12} defined by
y = x − 3. Then, R −1 is
(a) {(8 , 11), (10, 13)}
(c) {(10, 13), (8 , 11)}
(b) {(11, 18), (13 , 10)}
(d) None of these
15 Let R be a relation defined by R = {(4, 5), (1, 4), (4, 6),
(7, 6), (3, 7)}, then R −1 OR is
(a) {(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)}
(b) {(1, 1), (4, 4), (7, 7), (3, 3)}
(c) {(1, 5), (1, 6), (3, 6)}
(d) None of the above
16 Let A be a non-empty set of real numbers and f : A → A
be such that f (f ( x )) = x , ∀ x ∈ R. Then, f ( x ) is
(a) a bijection
(c) onto but not one-one
(b) one-one but not onto
(d) neither one-one nor onto
17 The function f satisfies the functional equation
x + 59
3f ( x ) + 2f
= 10x + 30 for all real x ≠ 1. The value
x −1
of f (7) is
(a) 8
(b) 4
(c) − 8
(d) 11
18 The number of onto mapping from the set A = {1, 2, ...100}
to set B = {1, 2} is
(a) 2100 − 2
(b) 2100
(c) 2 99 − 2
(d) 2 99
19 Let f : R − {n} → R be a function defined by f ( x ) =
where m ≠ n . Then,
(a) f is one-one onto
(c) f is many-one onto
x −m
,
x −n
(b) f is one-one into
(d) f is many-one into
20 A function f from the set of natural numbers to integers
n − 1
, when n is odd
defined by f (n ) = 2
is
n
− , when n is even
2
(a) one-one but not onto
(c) both one-one and onto
(b) onto but not one-one
(d) neither one-one nor onto
6
ONE
40
21 Let f : N → N defined by f ( x ) = x 2 + x + 1 , x ∈ N, then f is
(a) one-one onto
(c) one-one but not onto
(b) many-one onto
(d) None of these
22 Let R be the real line. Consider the following subsets of
the plane R × R .
and
S = {( x , y ): y = x + 1and 0 < x < 2}
T = {( x , y ): x − y is an integer}
Which one of the following is true?
(a)
(b)
(c)
(d)
T is an equivalence relation on R but S is not
Neither S nor T is an equivalence relation on R
Both S and T are equivalence relations on R
S is an equivalence relation on R but T is not
23 Consider the following relations
R = {(x , y) | x and y are real numbers and x = wy for some
rational number w};
m p
S = , m, n , p and q are integers such that n, q ≠ 0
n q
and qm = pn }. Then,
(a) R is an equivalence relation but S is not an equivalence
relation
(b) neither R nor S is an equivalence relation
(c) S is an equivalence relation but R is not an equivalence
relation
(d) R and S both are equivalence relations
2 + x , x ≥ 0
, then f (f ( x )) is given by
4 − x , x < 0
24 If f ( x ) =
4 + x,
(a) f (f (x)) =
6 − x,
4 − x,
(c) f (f (x)) =
x,
4 + x, x ≥ 0
(b) f (f (x)) =
x<0
x,
4 − 2 x, x ≥ 0
(d) f (f (x)) =
4 + 2 x, x < 0
x≥0
x<0
x≥0
x<0
25 Statement I A relation is defined by
x 2, 0 ≤ x ≤ 3
is a function.
f (x ) =
2 x , 3 ≤ x ≤ 9
Statement II In a function, every member must have a
unique image.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1
x
1 If f ( x ) + 2f = 3x , x ≠ 0 and S = {x ∈ R : f ( x ) = f ( − x )};
then S
j
JEE Mains 2016
(a) is an empty set
(b) contains exactly one element
(c) contains exactly two elements
(d) contains more than two elements
(c) R − {0, − 1, − 3 }
(a) 16
(b) R − {0, 1, 3 }
1
(d) R − 0, − 1, − 3,
2
(c) 6
(d) 8
4 The set ( A ∪ B ∪ C ) ∩ ( A ∩ B ′∩ C ′ )′ ∩ C ′ is equal to
j
(a) B ∩ C ′
(c) B ′∩ C ′
NCERT Exemplar
(b) A ∩ C
(d) None of these
5 Let A = {1, 2, 3, 4}, B = {2, 4, 6}. Then the number of sets
C such that A ∩ B ⊆ C ⊆ A ∪ B is
(a) 6
(b) 9
(c) 8
(b) 64
(c) 4
(d) 44
1
1
8 Let f x + = x 2 + 2 , x ≠ 0, then f ( x ) is equal to
x
x
the minimum number of ordered pairs which when
added to R make it an equivalence relation is
(b) 7
(a) reflexive but not symmetric (b) symmetric only
(c) reflexive and transitive
(d) equivalence relation
f (1) = 4. The value of f ( 21) is
3 Given the relation R = {(1, 2) ( 2, 3)} on the set A = {1, 2, 3},
(a) 5
by aRb iff the g.c.d. of a and b is 2, then R is
7 Suppose f is a function satisfying f ( x + f ( x )) = 4f ( x ) and
2x − 1
2 x ∈ R : 3
∈R is equal to
2
x + 4x + 3x
(a) R − {0}
6 On the set N of all natural numbers define the relation R
(d) 10
(b) x 2 − 1
(d) x 2 + 1
(a) x 2
(c) x 2 − 2
9 Let f ( x ) =
(a)
x
1+ x2
, the fofofo K of ( x ) is
1442443
x
n
1+ Σ r x2
r = 1
x
(c)
1+ x2
x times
(b)
x
n
1 + Σ 1 x 2
r = 1
x
(d)
nx
1 + nx 2
10 If two sets A and B are having 99 elements in common,
then the number of elements common to each of the sets
A × B and B × A are
(a) 2 99
(b) 992
(c) 100
(d) 18
7
DAY
ANSWERS
SESSION 1
1. (b)
11. (a)
21. (c)
2. (d)
12. (b)
22. (a)
3. (d)
13. (b)
23. (c)
4. (b)
14. (a)
24. (a)
5. (c)
15. (a)
25. (d)
6. (c)
16. (a)
7. (b)
17. (b)
8. (d)
18. (a)
9. (a)
19. (b)
10. (c)
20. (c)
SESSION 2
1. (c)
2. (c)
3. (b)
4. (a)
5. (c)
6. (b)
7. (b)
8. (c)
9. (b)
10. (b)
Hints and Explanations
SESSION 1
6 From Venn Euler’s diagram,
1 Clearly, 1 ≠ 0, 2 and 2
y
3
∴
1
can be 1.
y
⇒
x = 1 ∈Q
[Q y ∈ N ]
C
C–A
A–B
2 The number of elements in power set of
A is 1.
∴
⇒
⇒
It is clear that,
{( A − B ) ∪ (B − C ) ∪ (C − A )} ′
= A ∩ B ∩C
P {P {P ( A )}} = 22 = 4
P {P {P {P ( A )}}} = 24 = 16
7 Qn ( A ∪ B ) = n ( A ) + n (B ) − n ( A ∩ B )
X = {4n − 3 n − 1: n ∈ N }
∴ n ( A ∪ B ) = 200 + 300 − 100 = 400
X = {0, 9, 54, 243,...}
[put n = 1,2,3,...]
∴ n ( A ′ ∩ B ′ ) = n ( A ∪ B )′ = n ( X )
− n( A ∪ B )
Y = {9(n − 1): n ∈ N }
Y = {0, 9,18,27,...}
[put n = 1,2,3,...]
It is clear that X ⊂ Y .
∴
X ∪Y = Y
4 Clearly, A ∩ B = A ∩ C and
A ∪ B = A ∪ C possible if
B =C
5 Number of elements in
A 1 ∪ A 2 ∪ A 3 ∪…∪ A 30 is 30 × 5 but
each element is used 10 times, so
n(S ) =
30 × 5
= 15
10
…(i)
Similarly, number of elements in
B1 ∪ B2 ... ∪ B n is 3 n but each element
is repeated 9 times, so
n(S ) =
⇒
B–C
B
P {P ( A )} = 2 = 2
15 =
3n
9
3n
9
n = 45
⇒
⇒
300 = n ( X ) − 400
n ( X ) = 700
8 We know,
n ( A ∪ B ) = n( A ) + n(B ) − n( A ∩ B )
∴
P = 1500 − n ( A ∩ B )
⇒
n( A ∩ B ) = 1500 − P
Clearly, 1 ≤ n( A ∩ B ) ≤ 500
[Qmaximum number of elements
common in A and B = 500]
⇒
1 ≤ 1500 − P ≤ 500
⇒
− 1499 ≤ − P ≤ − 1000
⇒
1000 ≤ P ≤ 1499
9 We know,
n ( A × B × C ) = n ( A ) × n(B ) × n(C )
24
∴
n(C ) =
=2
4×3
10 Since for each a ∈ A, (a, a)∈ R. R is
reflexive relation.
[from Eq. (i)]
Now, (6, 12) ∈ R but (12, 6) ∉ R. So, it is
not a symmetric relation.
Also, (3, 6), (6, 12) ∈ R ⇒ (3, 12) ∈ R
⇒ R is transitive.
∴ (a, a) ∈ R, ∀a ∈ N ⇒ R is reflexive.
Now, as a2 − 4 ab + 3b 2 = 0
but b 2 − 4ba + 3 a2 ≠ 0
A
1
3 We have,
⇒
U
A∩B∩C
11 Qa2 − 4 a ⋅ a + 3 a2 = 4 a2 − 4 a2 = 0
∴ R is not symmetric.
Also, (a, b ) ∈ R and (b, c ) ∈ R
⇒
/ (a, c ) ∈ N
So, R is not transitive.
12 Given, g ( x ) = 1 +
x
and f { g ( x )} = 3 + 2
⇒ f (1 +
…(i)
x) = 3 + 2 x + x
Put 1 +
∴
x + x
x = y ⇒ x = ( y − 1)2
f ( y ) = 3 + 2 ( y − 1) + ( y − 1)2
= 2 + y2
∴
f ( x ) = 2 + x2
13 Here, ( fog )( x ) = f {g ( x )} = a(cx + d ) + b
and (gof ) ( x ) = g { f ( x )} = c (ax + b ) + d
Since, cx + d + 2 = cx + 2c + d
[Q a = 1, b = 2]
Hence, c = 1 and d is arbitrary.
14 R is a relation from {11, 12, 13} to
{8, 10, 12} defined by
y =x−3⇒ x− y =3
∴
R = {(11, 8), (13, 10)}
Hence, R −1 = {( 8, 11), (10, 13)}
15 Clearly, R −1 = {(5, 4), (4, 1), (6, 4), (6, 7),
(7, 3)}
Now, as (4, 5) ∈ R and (5, 4) ∈ R −1 ,
therefore (4, 4) ∈ R −1OR
Similarly, (1, 4) ∈ R and (4, 1) ∈ R −1
⇒ (1, 1) ∈ R −1OR
(4, 6) ∈ R and (6, 7) ∈ R −1
⇒ (4, 7) ∈ R −1OR
(7, 6) ∈ R and (6, 7) ∈ R −1
⇒ (7, 7) ∈ R −1OR
(7, 6) ∈ R and (6, 4) ∈ R −1
8
ONE
40
⇒ (7, 4) ∈ R −1OR
and (3, 7) ∈ R and (7, 3) ∈ R −1
⇒ (3, 3) ∈ R −1OR
Hence, R −1OR = {(1, 1), (4, 4), (4, 7),
(7, 7), (7, 4), (3, 3)}
16 Let x, y ∈ A such that f ( x ) = f ( y ), then
f ( f ( x )) = f ( f ( y ))
⇒
x= y
⇒ f is one-one.
Also,for any a∈ A, we have
f ( f (a)) = a
⇒ f (b ) = a , where b = f (a) ∈ A
Thus, for each a ∈ A (codomain) there
exists b = f (a) ∈ A such that f (b ) = a
∴ f is onto.
Hence f is a bijective function.
x + 59
x− 1
17 We have, 3 f ( x ) + 2 f
… (i)
= 10 x + 30
x + 59
, we get
On replacing x by
x−1
x + 59
3f
+ 2 f ( x)
x−1
40 x + 560
=
x−1
…(ii)
On solving Eqs. (i) and (ii), we get
6 x − 4 x − 242
x−1
6 × 49 − 4 × 7 − 242
f (7) =
=4
6
f ( x) =
∴
⇒ mq ⋅ ps = np ⋅ rq ⇒ ms = nr
m r
=
⇒
n
s
m r
S
⇒
n s
So, the relation S is transitive.
Hence, the relation S is
equivalence relation.
integer is an image of odd natural
number. So, f is onto.
21 Let x, y ∈ N such that f ( x ) = f ( y )
⇒
x2 + x + 1 = y 2 + y + 1
⇒
( x2 − y 2 ) = y − x
⇒
( x − y ) ( x + y + 1) = 0
⇒
x = y or x = − y − 1 ∉ N
⇒
x= y
⇒ f is one-one.
But f is not onto, as 1∈N does not have
any pre-image.
∴ f is one-one but not onto.
the number of onto relations from A to
B = 2n − 2
∴ Required number of relations
= 2100 − 2
19 Suppose for any x, y ∈ R,
f ( x) = f ( y )
x−m
y −m
=
x−n
y −n
⇒
x= y
So, f is one-one.
Let α ∈ R be such that f ( x ) = α
x−m
m − nα
=α ⇒ x=
∴
x−n
1−α
⇒
Clearly, x ∉ R for α = 1
So, f is not onto.
20 Let x, y ∈ N and both be even.
y
x
Then, f ( x ) = f ( y ) ⇒ − = −
2
2
⇒
x=y
Again, x, y ∈ N and both are odd.
Then, f ( x ) = f ( y ) ⇒ x = y
So, f is one-one
Since, each negative integer is an image
of even natural number and positive
2 + f ( x ),
f ( f ( x )) =
4 − f ( x ),
Thus S is not symmetric.
Hence, S is not an equivalence relation.
Given,
T = {( x, y ) : ( x − y )∈ I }
Now, x − x = 0∈ I , it is reflexive
relation.
Again, now ( x − y )∈ I
⇒ y − x ∈ I , it is symmetric relation.
Let
x − y = I1
and
y − z = I2
Then,
x − z = ( x − y ) + ( y − z)
= I1 + I2 ∈ I
So, T is also transitive. Hence, T is an
equivalence relation.
23 Since, the relation R is defined as
R = {( x , y )| x , y are real numbers and
x = wy for some rational number w}.
(a) Reflexive xRx as x = 1 x
Here, w = 1 ∈ Rational number
So, the relation R is reflexive.
(b) Symmetric xRy ⇒
/ yRx as 0R1 but
1 R/ 0
So, the relation R is not
symmetric.
Thus, R is not equivalence
relation.
Now, for the relation S, defined as,
m p
S = , m , n, p and q ∈ integers
n q
such that n , q ≠ 0 and qm = pn}
m m
(a) Reflexive S
⇒ mn = mn
n n
[true]
Hence, the relation S is reflexive.
m p
(b) Symmetric
S ⇒ mq = np
n q
⇒ np = mq ⇒
p m
S
q n
Hence, the relation S is
symmetric.
p r
m p
(c) Transitive S
and
S
n q
q s
⇒
mq = np and ps = rq
f ( x) ≥ 0
f ( x) < 0
2 + (2 + x ), x ≥ 0
=
2 + (4 − x ), x < 0
4 + x, x ≥ 0
=
6 − x, x < 0
22 Since, (1, 2) ∈ S but (2, 1) ∉S
2
18 We know that if n( A ) = n and n(B ) = 2,
24 Clearly,
2
25 Statement I f ( x ) = x , 0 ≤ x ≤ 3
2 x, 3 ≤ x ≤ 9
f (3) = 9
Now,
Also,
f (3) = 2 × 3 = 6
Here, we see that for one value of x,
there are two different values of f ( x ).
Hence, it is not a function but
Statement II is true.
SESSION 2
1 We have, f ( x ) + 2 f 1 = 3 x,
x
x ≠ 0 … (i)
∴
1
3
f + 2 f ( x ) =
x
x
… (ii)
1
replacing x by
x
On multiplying Eq. (ii) by 2 and then
subtracting it from Eq. (i), we get
6
− 3 f ( x) = 3 x −
x
2
⇒
f ( x) = − x
x
Now, consider f ( x ) = f (− x )
2
2
4
⇒
− x=− + x ⇒
= 2x
x
x
x
⇒
x2 = 2 ⇒ x = ± 2
Thus, x contains exactly two elements.
2x −1
2 Clearly, 3
∈ R only when
x + 4 x2 + 3 x
x3 + 4 x2 + 3 x ≠ 0
Consider x3 + 4 x2 + 3 x = 0
⇒
x ( x2 + 4 x + 3) = 0
⇒
⇒
x ( x + 1) ( x + 3) = 0
x = 0, − 1, − 3
2x − 1
∈ R
x ∈ R : 3
2
x
+
4
x
+
3
x
∴
= R − {0, − 1, − 3}
9
DAY
3. For R to be an equivalence relation, R
must be reflexive, symmetric and
transitive.
R will be reflexive if it contains (1, 1),
(2, 2) and (3, 3)
R will be symmetric if it contains
(2, 1) and (3, 2)
R will be transitive if it contains (1, 3)
and (3, 1)
Hence, minimum number of ordered
pairs = 7
4 ( A ∪ B ∪ C ) ∩ ( A ∩ B ′∩ C ′ )′∩ C ′
= ( A ∪ B ∪ C ) ∩ ( A ′∪ B ∪ C ) ∩ C ′
= (φ ∪ B ∪ C ) ∩ C ′
= (B ∪ C ) ∩ C ′
= (B ∩ C ′ ) ∪ φ = B ∩ C ′
5 Here, A ∩ B = {2, 4}
and A ∪ B = {1, 2, 3, 4, 6}
Q A ∩ B ⊆ C ⊆ A∪B
∴ C can be {2, 4}, {1, 2, 4}, {3, 2, 4},
{6, 2, 4}, {1, 6, 2, 4}, {6, 3, 2, 4},
{1, 3, 2, 4}, {1, 2, 3, 4, 6}
Thus, number of set C which satisfy the
given condition is 8.
6 Clearly, g.c.d (a, a) = a, ∀ a ∈ N
∴R is not reflexive.
If g.c.d (a, b ) = 2, then g.c.d (b, a) is
also 2.
Thus,
aRb ⇒ bRa
Hence, R is symmetric.
According to given option, R is
symmetric only.
x
=
1+
7 We have,
f ( x + f ( x )) = 4 f ( x ) and f (1) = 4
On putting x = 1, we get
f (1 + f (1)) = 4 f (1)
⇒
f (1 + f (1)) = 16
⇒
f (1 + 4) = 16
⇒
f (5) = 16
On putting, x = 5, we get
f (5 + f (5)) = 4 f (5)
⇒
f (5 + 16) = 4 × 16
⇒
f (21) = 64
1
1
f x + = x2 + 2
x
x
2
1
= x + − 2
x
f ( x ) = x2 − 2
9 We have, f ( x ) =
⇒
x2
1 + x2
x
=
1 + 2 x2
x
Similarly, f ( f ( f ( x ))) =
M
fofo K of of ( x ) =
1442443
n times
1 + 3 x2
M
x
1 + nx2
=
x
n
1 + Σ 1 x2
r =1
10 We know,
8 We have,
⇒
1 + x2
f ( f ( x )) =
x
1 + x2
f ( x)
1 + ( f ( x ))2
( A × B ) ∩ (C × D ) = ( A ∩ C )
× (B ∩ D )
∴ ( A × B ) ∩ (B × A ) = ( A ∩ B )
× (B ∩ A )
Thus, number of elements common to
A × B and B × A
= n (( A × B ) ∩ (B × A ))
= n (( A ∩ B ) × (B ∩ A ))
= n( A ∩ B ) × n (B ∩ A )
= 99 × 99 = 992
DAY TWO
Complex
Numbers
Learning & Revision for the Day
u
u
u
Complex Numbers and Its
Representation
Algebra and Equality of
Complex Numbers
Conjugate and Modulus of a
Complex Number
u
u
u
u
Argument or Amplitude of a
Complex Number
Different forms of a Complex
Number
Concept of Rotation
Square Root of a Complex
Number
u
De-Moivre’s Theorem
u
Cube Roots of Unity
u
nth Roots of Unity
u
Applications of Complex
Numbers in Geometry
Complex Numbers and Its Representation
●
●
A number in the form of z = x + iy, where x, y ∈ R and
i = −1, is called a complex number. The real numbers
x and y are respectively called real and imaginary parts
of complex number z.
i.e. x = Re (z), y = Im (z) and the symbol i is called iota.
A complex number z = x + iy is said to be purely real if
y = 0 and purely imaginary if x = 0.
X′
P (x, y)
O
(ii) If n is an integer, then i 4 n = 1, i 4 n
and i 4 n + 3 = − i
(iii) i + i
n
n +1
+i
n +2
+i
n +3
+1
Your Personal Preparation Indicator
x
Real
axis
Y′
Argand diagram
= i, i 4 n
+2
PRED
MIRROR
y
Integral power of iota (i)
(i) i = −1, i 2 = − 1, i 3 = − i and i 4 = 1
●
Y
Imaginary axis
●
X
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
= −1
=0
The complex number z = x + iy can be represented by a point P in a plane called
argand plane or Gaussian plane or complex plane. The coordinates of P are referred
to the rectangular axes XOX ′ and YOY′ which are called real and imaginary axes,
respectively.
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
11
DAY
Algebra and Equality of
Complex Numbers
Argument or Amplitude of
a Complex Number
If z1 =
(i)
(ii)
(iii)
Let z = x + iy be a complex number, represented by a point
P( x, y) in the argand plane. Then, the angle θ which OP makes
with the positive direction of Real axis (X -axis) is called
the argument or amplitude of z and it is denoted by arg ( z ) or
amp ( z ).
y
The argument of z, is given by θ = tan −1
x
x1 + i y1 and z2 = x2 + i y2 are two complex numbers, then
z1 + z2 = ( x1 + x2 ) + i ( y1 + y2 )
z1 − z2 = ( x1 − x2 ) + i ( y1 − y2 )
z1 z2 = ( x1 x2 − y1 y2 ) + i( x1 y2 + x2 y1 )
z1 ( x1 x2 + y1 y2 ) + i ( x2 y1 − x1 y2 )
(iv)
=
z2
x22 + y22
(v) z1 and z2 are said to be equal if x1 = x2 and y1 = y2 .
●
NOTE
• Complex numbers does not possess any inequality,
e.g. 3 + 2i > 1 + 2i does not make any sense.
●
Conjugate and Modulus
of a Complex Number
●
If z = x + iy is a complex number, then conjugate of z is
denoted by z and is obtained by replacing i by −i.
i.e.
z = x − iy
If z = x + iy, then modulus or magnitude of z is denoted by
| z| and is given by| z| =
x + y
2
The principal value of arg(z) is θ, π − θ, −π + θ or − θ
according as z lies in the 1st, 2nd, 3rd or 4th quadrants
y
respectively, where θ = tan −1
.
x
Imaginary axis
●
The value of argument θ which satisfies the inequality
− π < θ < π, is called principal value of argument.
2
P (x, y)
y
θ
x
Results on Conjugate and Modulus
(i)
(ii)
(iii)
(iv)
(v)
(z) = z
z + z = 2 Re (z), z − z = 2 i Im(z)
z = z ⇔ z is purely real.
z + z = 0 ⇔ z is purely imaginary.
z1 ± z2 = z1 ± z2
●
Argument of z is not unique. General value of argument of z
is 2nπ + θ.
Results on Argument
If z, z1 and z2 are complex numbers, then
(i) arg (z) = − arg (z)
(ii) arg (z1 z2 ) = arg (z1 ) + arg (z2 )
z1
(iii) arg = arg (z1 ) − arg (z2 )
z2
(vi) z1 z2 = z1 z2
z1 z1
(vii) = , if z2 ≠ 0
z2 z2
a1 a2 a3
a1
(viii) If z = b1 b2 b3 , then z = b1
c1 c2 c3
c1
Real axis
a2
b2
c2
(iv) The general value of arg (z) is 2nπ − arg (z).
π
(v) If z is purely imaginary then arg (z) = ± .
2
(vi) If z is purely real then arg (z) = 0 or π.
a3
b3
c3
where ai , b i , c i ; (i = 1, 2 , 3) are complex numbers.
| z| = 0 ⇔ z = 0
| z| = | z| = |−z| = |−z|
− | z| ≤ Re(z), Im(z) ≤ | z|
| z1 z2| = | z1|| z2|
z1 | z1|
(xiii)
=
, if| z2| ≠ 0
z2 | z2|
(ix)
(x)
(xi)
(xii)
(xiv) | z1 ± z2|2 = | z1|2 + | z2|2 ± z1 z2 ± z1 z2
= | z1|2 + | z2|2 ± 2 Re (z1 z2 )
n
n
(xv) | z | = | z| , n ∈ N
(xvi) Reciprocal of a complex number For non-zero complex
number z = x + iy, the reciprocal is given by
1
z
z −1 = = 2 .
z | z|
(xvii) Triangle Inequality
(a)| z1 + z2 | ≤ | z1 | + | z 2 | (b) | z1 + z2 | ≥ || z1 | − | z 2 ||
(c)| z1 − z 2 | ≤ | z1 | + | z 2 | (d) | z1 − z2 | ≥ || z1 | − | z 2 ||
(vii) If| z1 + z2| = | z1 − z2|, then
z1
π
arg or arg (z1 ) − arg (z2 ) =
z
2
2
(viii) If| z1 + z2| = | z1| + | z2|, then arg (z1 ) = arg (z2 )
Different forms of a
Complex Number
●
Polar or Trigonometrical Form of z = x + iy is
z = r (cos θ + i sin θ), where r =| z| and θ = arg (z).
If we use the general value of the argument θ, then the
polar form of z is z = r [cos (2nπ + θ) + i sin (2nπ + θ)], where
n is an integer.
●
Euler’s form of z = x + iy is z = re iθ , where r =| z|,θ = arg (z)
and e iθ = cos θ + i sin θ.
12 40
(i) The nth roots of unity are 1, α , α 2 ,... , α n − 1 , where
Concept of Rotation
i2 π
Let z1 , z2 , z3 be the vertices of ∆ABC as shown in figure, then
z −z
z − z |z − z |
α = arg 3 1 and 3 1 = 3 1 e iα
z2 − z1
z2 − z1 | z2 − z1 |
C(z3)
Applications of Complex
Numbers in Geometry
1. Distance between
AB = | z2 − z1|.
a
A(z1)
NOTE
B(z2)
• Always mark the direction of arrow in anti-clockwise sense
and keep that complex number in the numerator on which
the arrow goes.
Square Root of a Complex Number
●
●
If z = a + ib , then
1
z = a + ib = ±
[ | z| + a + i z − a]
2
1
If z = a − ib , then z = a − ib = ±
[ | z| + a − i | z| − a]
2
De-Moivre’s Theorem
●
●
●
α =e n
(ii) 1 + α + α 2 + α 3 + ...+ α n −1 = 0
(iii) 1 ⋅ α ⋅ α 2 ... α n −1 = [−1]n −1
If n is any integer, then (cos θ + i sin θ)n = cos nθ + i sin nθ
If n is any rational number, then one of the values of
(cos θ + i sin θ)n is cos nθ + i sin nθ.
If n is any positive integer, then
2 kπ + θ
2 kπ + θ
(cos θ + i sin θ)1 / n = cos
+ i sin
n
n
where, k = 0, 1, 2,... n − 1
Cube Root of Unity
Cube roots of unity are 1, ω, ω2
−1 + 3 i
−1 − 3 i
where, ω =
and ω2 =
2
2
Properties of Cube Roots of Unity
(i) 1 + ω + ω2 = 0
(ii) ω3 = 1
0 if n ≠ 3 m, m ∈ N
(iii) 1 + ω n + ω2 n =
3 if n = 3 m, m ∈ N
nth Roots of Unity
By nth root of unity we mean any complex number z which
satisfies the equation zn = 1.
A(z1 ) and
B (z2 ) is
given
by
2. Let point P (z) divides the line segment joining A (z1 ) and
B (z2 ) in the ratio m : n . Then,
mz2 + nz1
(i) for internal division, z =
m+n
(ii) for external division, z =
mz2 − nz1
m−n
3. Let ABC be a triangle with vertices A (z1 ), B (z2 ) and C(z3 ),
then centroid G (z) of the ∆ ABC is given by z
1
= (z1 + z2 + z3 )
3
1
Area of ∆ ABC is given by ∆ =
2
z1
z2
z3
z1
z2
z3
1
1
1
4. For an equilateral triangle ABC with vertices A(z1 ), B(z2 )
and C(z3 ), z12 + z22 + z23 = z 2 z3 + z 3 z1 + z1 z 2
5. The general equation of a straight line is az + az + b = 0,
where a is a complex number and b is a real number.
6. (i) An equation of the circle with centre at z0
and radius r, is | z − z0| = r
(ii) | z − z0| < r represents the interior of circle and
| z − z0| > r represents the exterior of circle.
(iii) General equation of a circle is zz + az + az + b = 0,
where b is real number, with centre is − a and radius
is aa − b .
7. If z1 and z2 are two fixed points and k > 0, k ≠ 1 is a real
| z − z1|
number, then
= k represents a circle.
| z − z2|
For k = 1, it represents perpendicular bisector of the
segment joining A(z1 ) and B (z2 ).
8. If end points of diameter of a circle are A(z1 ) and B(z 2) and
P(z) be any point on the circle, then equation of circle in
diameter form is
(z − z1 ) (z − z2 ) + (z − z 2) (z − z1 ) = 0
13
DAY
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 Real part of
(a) −
1
2
1
is
1 − cos θ + i sin θ
(b)
1
2
(c)
2 A value of θ, for which
π
(a)
3
13
3
∑ (i
n
11 If z −
1
tanθ / 2
2
−1
(c) sin
1
(d) sin
3
3
4
−1
+ i n + 1 ) is equal to
n =1
(b) i − 1
(a) i
(c) −i
(b) 1
(d) −2
(c) 2
5 If z1 ≠ 0 and z 2 are two complex numbers such that
a purely imaginary number, then
(a) 2
6 If f ( z ) =
(b) 5
z2
is
z1
2z1 + 3z 2
is equal to
2z1 − 3z 2
j
(c) 3
JEE Mains 2013
(d) 1
7−z
, where z = 1 + 2i , then | f ( z )| is equal to
1− z 2
|z |
(a)
2
(c) 2| z |
(d) None of these
(a)
(b)
(c)
(d)
(c) 1
(d) 3 / 4
8 If a complex number z satisfies the equation
z + 2 z + 1 + i = 0, then z is equal to
(a) 2
(b)
3
(c)
5
j
JEE Mains 2013
(d) 1
9 If α and β are two different complex numbers such that
β −α
is equal to
| α | = 1, | β | = 1, then the expression
1 − αβ
1
2
(c) 2
(b) 1
(a)
z −1
10 If | z | = 1 and ω =
(where z ≠ −1), then Re(ω ) is
z +1
(b) −
(c)
2
z+1
2
2
1
2
j
JEE Mains 2014
is equal to 5/2
lies in the interval (1, 2)
is strictly greater than 5/2
is strictly greater than 3/2 but less than 5/2
(a) 8
(b) 18
(c) 10
(d) 5
14 If z < 3 − 1, then z + 2z cos α is
2
(b) 3 + 1
(d) None of these
(a) less than 2
(c) 3 − 1
15 The number of complex numbers z such that
z − 1 = z + 1 = z − i , is
(a) 0
(b) 1
(d) ∞
(c) 2
16 Number of solutions of the equation z
(a) 1
(b) 2
2
+ 7z = 0 is/are
(c) 4
(d) 6
17 If z z + ( 3 − 4i )z + ( 3 + 4i )z = 0 represent a circle, the area
of the circle in square units is
(c) 25 π 2
(b) 10π
π
5
(d) 25 π
π
5
18 If z = 1 + cos + i sin , then {sin (arg( z ))} is equal to
(a)
(c)
10 − 2 5
4
5+1
4
5 −1
4
(b)
(d) None of these
19 If z is a complex number of unit modulus and argument
1 + z
θ, then arg
equals to
1 + z
(a) −θ
π
(b)
−θ
2
j
(c) θ
JEE Mains 2013
(d) π − θ
20 Let z and ω are two non-zero complex numbers such that
(d) None of these
(a) 0
(d) 2 +
(c) 2
equal to
7 If 8 iz 3 + 12z 2 − 18z + 27i = 0, then the value of | z | is
(b) 2 / 3
5 +1
minimum value of z +
(a) 5 π
(b) | z |
(a) 3 / 2
(b)
12 If z is a complex number such that z ≥ 2, then the
13 If | z1| = 2, | z 2 | = 3 then z1 + z 2 + 5 + 12i is less than or
(d) 0
z −1
4 If
is a purely imaginary number ( where, z ≠ −1), then
z +1
the value of | z | is
(a) −1
3 +1
(a)
(d) 2
2 + 3i sin θ
is purely imaginary, is
1 − 2i sin θ
π
(b)
6
4
= 2, then the maximum value of | z | is
j AIEEE 2009
z
1
z+1
2
(d) None of these
z = ω and arg z + arg ω = π, then z equals
(a) ω
(c) − ω
(b) ω
(d) − ω
21 If z − 1 = 1, then arg ( z ) is equal to
1
arg (z)
2
1
(c) arg (z − 1)
2
(a)
(b)
1
arg (z + 1)
3
(d) None of these
14 40
15
22 Let z = cos θ + i sin θ. Then the value of ∑ Im ( z 2 m −1 ) at,
1
z
33 If Re = 3 , then z lies on
m =1
θ = 2°, is
1
(a)
sin 2 °
1
(b)
3 sin 2 °
1
(c)
2 sin 2 °
1
(d)
4 sin 2 °
(i )
23 If z = (i )(i ) , where i = −1, then | z | is equal to
(b) e − π / 2
(a) 1
(d) e π / 2
(c) 0
π
π
1 + i sin + cos
8
8 equals to
24
1 − i sin π + cos π
8
8
(a) a circle
(c) a parabola
(d) 1
25 If 1, α 1, α 2 , K , α n − 1 are the nth roots of unity, then
( 2 − α 1 )( 2 − α 2 ) K ( 2 − α n −1 ) is equal to
(c) 2 n + 1
(b) 2 n
(a) n
(d) 2 n − 1
26 If ω( ≠ 1) is a cube root of unity and (1 + ω )7 = A + Bω.
(a)
(b)
(c)
(d)
(c) (−1, 1)
(b) (1, 0)
(d) (0, 1)
27 If α , β ∈C are the distinct roots of the equation
x 2 − x + 1 = 0, then α 101 + β107 is equal to
(a) −1
(b) 0
j
(c) 1
r=1
25
28 If x 2 + x + 1 = 0, then ∑ x r +
JEE Mains 2018
(d) 2
2
1
is equal to
xr
(b) 25 ω
(d) None of these
(a) 25
(c) 25 ω2
29 Let ω be a complex number such that 2ω + 1 = z,
1
1
1
2
where z = −3. If 1 −ω − 1 ω 2 = 3k, then k is equal to
1
ω2
ω7
j
JEE Mains 2017
(a) −z
(c) −1
(b) z
(d) 1
1+ ω
ω2
1 + ω2
30 The value − ω − (1 + ω 2 ) (1 + ω) , where ω is cube
− 1 − (1 + ω 2 ) 1 + ω
root of unity, is equal to
(a) 2 ω
(b) 3 ω2
(c) − 3 ω2
(d) 3ω
31 If a , b and c are integers not all equal and ω is a cube
root of unity (where, ω ≠ 1), then minimum value of
| a + bω + cω 2 | is equal to
(a) 0
32 Let ω = e
(b) 1
(c)
3
2
(d)
1
2
iπ /3
, and a, b, c, x , y , z be non-zero complex
numbers such that:
a + b + c = x ; a + b ω + cω = y ; a + b ω + c ω = z
2
Then the value of
(a) 1
x
2
a
2
(b) 2
+ y
2
+ b
2
2
+ z
2
+ c
2
(c) 3
is:
(d) 4
z
lie on
1− z 2
a line not passing through the origin
|z | = 2
the X-axis
theY-axis
36 If ω =
Then, ( A, B ) is equal to
(a) (11
,)
(b) a straight line
(d) None of these
35 If | z | = 1 and z ≠ ± 1, then all the values of
(c) −1
(b) 0
34 If the imaginary part of ( 2z + 1) / (iz + 1) is −2, then the
locus of the point representing z in the complex plane is
8
(a) 2 8
(a) circle with centre onY-axis
(b) circle with centre on X-axis not passing through origin
(c) circle with centre on X-axis passing through origin
(d) None of the above
z
z−
i
3
and | ω | = 1, then z lies on
(a) a circle
(c) a parabola
(b) an ellipse
(d) a straight line
37 If z 1 and z 2 are two complex numbers such that
z1 z 2
+
= 1, then
z 2 z1
(a)
(b)
(c)
(d)
z 1, z 2 are collinear
z 1, z 2 and the origin form a right angled triangle
z 1, z 2 and the origin form an equilateral triangle
None of the above
38 A complex number z is said to be unimodular, if z = 1.
Suppose z1 and z 2 are complex numbers such that
z1 − 2z 2
is unimodular and z 2 is not unimodular.
2 − z 1z 2
Then, the point z1 lies on a
(a)
(b)
(c)
(d)
j
JEE Mains 2015
straight line parallel to X −axis
straight line parallel toY −axis
circle of radius 2
circle of radius 2
39 If | z 2 − 1 | = | z |2 +1, then z lies on
(a) a real axis
(c) a circle
(b) an ellipse
(d) imaginary axis
40 Let z satisfy z = 1 and z = 1 − z
j
JEE Mains 2013
Statement I z is a real number.
Statement II Principal argument of z is π /3.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 For positive integers n1 and n 2 the value of the expression
8 If a complex number z lies in the interior or on the
(1 + i )n1 + (1 + i 3 )n1 + (1 + i 5 )n 2 + (1 + i 7 )n 2
where i = −1, is a real number iff
(a) n1 = n2
boundary of a circle of radius 3 and centre at ( − 4 , 0),
then the greatest and least value of | z + 1| are
(b) n2 = n2 − 1 (c) n1 = n2 + 1 (d) ∀n1 and n2
(a) 5, 0
z
is real, then the point represented by the
z −1
complex number z lies
2 If z ≠ 1 and
(a) on the imaginary axis
(b) either on the real axis or on a circle passing through the
origin
(c) on a circle with centre at the origin
(d) either on the real axis or on a circle not passing through
the origin
2π
2π
3 Let ω be the complex number cos
. Then the
+ i sin
3
3
number of distinct complex numbers z satisfying
z +1
ω
ω2
1 = 0 is equal to
ω
z + ω2
ω2
1
(a) 0
(a) 2
(d) 4
(c) x − y < 0
(a) 3eiπ / 4 + 4i
(c) (4 + 3i )eiπ / 4
11 If 1, a1, a 2 ... a n −1 are n th roots of unity, then
1
1
1
+
+ ... +
equals to
1 − a1 1 − a 2
1 − a n −1
(c) 3 / 2
(d) None of these
6 Let z = x + iy be a complex number where x and y are
(b) 32
(c) 40
+
+
+
+
y2
y2
y2
y2
(c)
n
n −1
(d) None of these
2
(b) ω or ω /|ω | 2
(d) None of these
k =1
2kπ
2kπ
+ i cos
is
11
11
(b) −1
(a) 1
(c) − i
(d) i
14 Let z1 and z 2 be roots of the equation z + pz + q = 0,
2
p, q ∈ C. Let A and B represents z1 and z 2 in the complex
plane. If ∠AOB = α ≠ 0 and OA = OB; O is the origin, then
p 2 / 4q is equal to
(a) sin2 (α / 2) (b) tan2 (α / 2) (c) cos2 (α / 2) (d) None of these
15 If 1, ω and ω 2 are the three cube roots of unity α , β, γ are
the cube roots of p, q < 0, then for any x , y , z the
x α + y β + z γ
expression
is equal to
x β + y γ + z α
then the locus of z is
x2
x2
x2
x2
n −1
2
10
(d) 80
7 If α + i β = cot −1( z ), where z = x + iy and α is a constant,
(a)
(b)
(c)
(d)
(b)
13 The value of ∑ sin
integers. Then the area of the rectangle whose vertices
are the roots of the equation zz 3 + zz 3 = 350 is
(a) 48
2n − 1
n
(a) ω or ω
(c) ω or ω /|ω | 2
(d) x + y > 0
such that arg ( z − z1 ) / ( z − z 2 ) = π / 4, then z − 7 − 9i is
equal to
(b) 3 2
(d) 6
(b) (3 − 4i )eiπ / 4
(d) (3 + 4i )eiπ / 4
2
5 Let z1 = 10 + 6i , z 2 = 4 + 6i . If z is any complex number
(a) 18
(c) 5
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
log1/ 2 z − 1 > log1/ 2 z − i is given by
(b) x − y > 0
(b) 3
12 For z ,ω ∈ C, if z ω − ω z = z − ω, then z is equal to
4 The locus of z = x + iy which satisfying the inequality
(a) x + y < 0
(d) None of these
10 A man walks a distance of 3 units from the origin towards
(a)
(c) 2
(c) 6, 0
then the minimum value of 2z − 6 + 5i is
z +ω
(b) 1
(b) 6, 1
9 If z is any complex number satisfying z − 3 − 2i ≤ 2,
2
− x cot 2 α − 1 = 0
− 2 x cot α − 1 = 0
− 2 x cot 2 α + 1 = 0
− 2 x cot 2 α − 1 = 0
(b) ω
(a) 1
(c) ω 2
(d) None of these
ANSWERS
SESSION 1
1.
11.
21.
31.
SESSION 2
1. (d)
11. (b)
(b)
(b)
(c)
(b)
2.
12.
22.
32.
(d)
(b)
(d)
(c)
2. (b)
12. (b)
3.
13.
23.
33.
(b)
(b)
(a)
(c)
3. (b)
13. (c)
4.
14.
24.
34.
(b)
(a)
(c)
(b)
4. (b)
14. (c)
5.
15.
25.
35.
(d)
(b)
(d)
(d)
5. (b)
15. (c)
6.
16.
26.
36.
(a)
(b)
(a)
(d)
6. (a)
7.
17.
27.
37.
(a)
(d)
(c)
(c)
7. (d)
8.
18.
28.
38.
(c)
(b)
(d)
(c)
8. (c)
9.
19.
29.
39.
(b)
(c)
(a)
(d)
9. (c)
10.
20.
30.
40.
(a)
(c)
(c)
(d)
10. (d)
16 40
Hints and Explanations
SESSION 1
3
13
∑ (i
n
+ i n +1 ) = (1 + i )
n =1
13
∑i
n
n =1
= (1 + i )
i (1 − i 13 )
1− i
= i − 1 [Q i 13 = i , i 2 = −1]
4 Let z = x + iy
x + iy − 1 ( x − 1) + iy
z−1
=
=
z + 1 x + iy + 1 ( x + 1) + iy
( x + 1) − iy
×
( x + 1) − iy
( x − 1)( x + 1) − iy ( x − 1) + iy
( x + 1) − i 2 y 2
( x + 1)2 − i 2 y 2
=
=
x2 − 1 + iy ( x + 1 − x + 1) + y 2
( x + 1)2 + y 2
i (2 y )
z − 1 ( x2 + y 2 − 1)
=
+
z + 1 ( x + 1)2 + y 2 ( x + 1)2 + y 2
z−1
is purely imaginary.
Since,
z+1
z − 1
∴
Re
=0
z + 1
2
2
x + y −1
⇒
=0
( x + 1)2 + y 2
⇒
⇒
x + y −1 = 0
2
2
x2 + y 2 = 1
|z |2 = 1 ⇒ |z |= 1
10 Given,|z | = 1
⇒
5 Given, z2 is a purely imaginary
zz = 1
Let z = ni. Then,
2z1 + 3z2
2z1 − 3z2
6
=
z2
2 + 3ni
z1
=
=
z2
2 − 3ni
2 − 3⋅
z1
2+ 3⋅
4 + 9n2
=
4 + 9n2
=
∴
=1
z+1
2
=0
[Q zz = 1]
z
z
4
4
|z | ≤ z − +
z |z |
4
|z | ≤ 2 +
|z|
⇒
⇒
|z |2 − 2 |z | − 4
≤0
|z |
⇒
Since,|z | > 0
⇒
|z |2 − 2 |z | − 4 ≤ 0
⇒ [|z | − ( 5 + 1)] [|z |− (1 − 5)] ≤ 0
⇒
Now,| f (z )| =
5]
7 Given, 8 iz 3 + 12z 2 − 18z + 27i = 0
4z 2 (2iz + 3) + 9 i (2iz + 3) = 0
(2iz + 3) (4z2 + 9 i ) = 0
2iz + 3 = 0 or 4z2 + 9 i = 0
3
|z | =
2
∴
2zz − 2
2
Re(ω ) = 0.
7− z
and z = 1 + 2i
1 − z2
7 − (1 + 2i )
∴ f (z ) =
1 − (1 + 2i )2
6 − 2i
6 − 2i
=
=
1 − (1 − 4 + 4i ) 4 − 4i
6 − 2i
1+ i
6 + 4i + 2
=
×
=
4(1 − i ) (1 + i )
4(12 − i 2 )
8 + 4i 1
=
= (2 + i )
4(2)
2
⇒
⇒
⇒
z+1
z − 4 + 4
11 |z | =
Given, f (z ) =
4+ 1
5 |z |
=
=
2
2
2
[Qz = 1 + 2i , given ⇒|z |=
z−1 z−1
+
z+1 z+1
(z − 1)(z + 1) + (z − 1)(z + 1)
Now, 2Re(ω ) = ω + ω =
z1
1−
5 ≤ |z | ≤
5+ 1
12 z ≥ 2 is the region on or outside circle
whose centre is (0,0) and radius is 2.
1
Minimum z + is distance of z, which
2
1
lie on circle z = 2 from − , 0 .
2
Y
8 We have, ( x + iy ) + 2 x + iy + 1 + i = 0
[put z = x + iy ]
⇒ ( x + iy ) + 2 ( x + 1)2 + y 2 + i = 0
⇒
(– 1
2
2
,0
(0, 0)
(2, 0)
X
y+1= 0
x + 2 ( x + 1) + (−1)2 = 0
2
and
y = −1
⇒
x2 = 2[( x + 1)2 + 1]
⇒
x2 = 2 x2 + 4 x + 4
2
⇒ x + 4 x + 4 = 0 ⇒ ( x + 2)2 = 0
⇒
x = −2
∴ z = −2 − i ⇒ z = 4 + 1 = 5
β −α
β −α
9
=
1 − αβ
β ⋅ β − αβ
β −α
β (β − α )
=
Y′
1
∴ Minimum z +
2
1
= Distance of − , 0 from (−2, 0)
2
2
=
Q |β | = 1
and |β|2 = ββ = 1
=
A
(–2, 0)
x + 2 ( x + 1) + y = 0
2
and
⇒
X′
(
1
1 − cos θ + i sin θ
1
=
2sin2 (θ / 2) + 2i sin(θ / 2)cos(θ / 2)
1
1
=
2i sin(θ / 2) [cos(θ / 2) − i sin(θ / 2)]
cos(θ / 2) + i sin(θ / 2) 1 1
=
= + cot(θ / 2)
2i sin(θ / 2)
2 2i
1
1
= − i ⋅ cot θ / 2
2
2
2 Let z = 2 + 3i sin θ is purely imaginary
1 − 2i sin θ
then we have
Re (z ) = 0
2 + 3i sin θ
Consider, z =
1 − 2i sin θ
(2 + 3i sin θ)(1 + 2i sin θ)
=
(1 − 2i sin θ)(1 + 2i sin θ)
(2 − 6sin2 θ) + (4sin θ + 3sin θ )i
=
1 + 4sin2 θ
Q Re(z ) = 0
2 − 6sin2 θ
∴
=0
1 + 4sin2 θ
1
1
sin2 θ = ⇒ sinθ = ±
⇒
3
3
−1 1
⇒
θ = ± sin
3
1 Let z =
⇒
⇒
1 |β − α| |β − α|
=
=1
|β| | β − α| |β − α|
[Q|z|=|z|]
−2 + 1 + 0 = 3
2
2
Alternate Method
We know,|z1 + z2 |≥||z1 |−|z2||
∴
z+
1
1
1
≥ |z| −
= |z|−
2
2
2
≥ z−
1
3
=
2
2
17
DAY
∴
z+
1
3
≥
2
2
20 Let z = ω = r and let arg ω = θ
1
3
∴ Minimum value of z + is ⋅
2
2
13 Fact: z1 + z2 + ... + z n
∴
≤ z1 + z2 + ... + z n
z1 + z2 + (5 + 12i )
≤ z1 + z2 + 5 + 12i
= 2 + 3 + 13 = 18
2
14 Consider z2 + 2z cos α ≤ z + 2 z
2
cos α ≤ z + 2 z
< ( 3 − 1)2 + 2( 3 − 1)
= 3+ 1−2 3 + 2 3 −2= 2
∴ z + 2z cos α < 2
2
|z − 1| = |z + 1|
Re z = 0 ⇒ x = 0
|z − 1| = |z − i| ⇒ x = y
|z + 1| = |z − i| ⇒ y = − x
Since, only (0, 0) will satisfy all
conditions.
∴ Number of complex number z = 1.
2
16 Given z + 7z = 0
⇒
z z + 7z = 0 ⇒ z (z + 7) = 0
Case (i) : z = 0,
∴ z = 0 = 0 + i0
Case (ii) : z = −7 ∴ z = −7 + 0i
Hence, there is only two solutions.
z = 0 and z = −7
17 Given zz + (3 − 4i )z + (3 + 4i )z = 0
Let z = x + iy
Then, zz = x2 + y 2
∴ x2 + y 2 + (3 − 4i )( x + iy )
+ (3 + 4i )( x − iy ) = 0
⇒ x2 + y 2 + 6 x + 8 y = 0
⇒ ( x + 6 x) + ( y + 8 y ) = 0
2
⇒ ( x + 3)2 + ( y + 4)2 = 32 + 42
⇒ [ x − (−3)]2 + [ y − (−4)]2 = 52
So, area of circle be π R = 25π
[QR = radius = 5]
18 If z = 1 + cos θ + i sin θ, then arg(z ) = θ
2
π/5
π
∴
arg(z ) =
=
2
10
⇒ sin(arg z )
5−1
π
= sin = sin18° =
10
4
2
19 Given, z = 1 and arg z = θ
∴ z = e iθ and z =
21 Given,|z − 1| = 1 ⇒ z − 1 = e i θ ,
where arg(z − 1) = θ
…(i)
⇒
z = eiθ + 1
⇒
z = 1 + cos θ + i sin θ
[Qe i θ = cos θ + i sin θ]
θ
θ
2θ
= 2 cos + 2i sin ⋅ cos
2
2
2
θ 1
⇒ arg (z ) = = arg(z − 1) [from Eq. (i)]
2 2
22 Given that z = cos θ + i sin θ = e iθ
15 Let z = x + iy
2
Then, ω = r (cos θ + i sin θ) = re i θ
and arg z = π − θ
Hence, z = r (cos( π − θ) + i sin( π − θ))
= r (− cos θ + i sin θ)
= − r (cos θ − i sin θ)
z = −ω
1
z
1+ z
1+ z
Now, arg
= arg
1 + z
1 + 1
z
= arg (z ) = θ
∴
15
∑ lm(z
2 m −1
)=
m=1
15
∑ lm(e
iθ 2 m −1
)
=
∑ lme
i(2 m −1 )θ
m=1
= sin θ + sin 3θ + sin 5θ + ... + sin 29θ
14 ⋅ 2θ
15⋅ 2θ
sin θ +
sin
2
2
=
2θ
sin
2
sin(15θ)sin(15θ)
1
=
=
[Qθ = 2° ]
sin θ
4sin 2°
23 Clearly, i = cos π + i sin π = e iπ /2
2
2
∴
(i )i = (e iπ /2 )i = e
( i )(i )
Now, (i )
e − π /2
= (i )
i2 ⋅
π
= e − π /2
2
⇒ z = (i )e
− π /2
e − π /2
⇒
|z|=|i |
=1
π
24 Let z = cos + i sin π
8
8
1
π
π
Then, = cos − i sin
z
8
8
8
1 + cos π + i sin π
8
8
8 = 1 + z
Now,
−1
π
π
1 + z
1 + cos − i sin
8
8
8
8
π
π
(1 + z )z
8
=
= z = cos + i sin
(1 + z )
8
8
π
π
= cos 8 ⋅ + i sin 8 ⋅
8
8
[using De-moivre’s theorem]
= cos π = −1(Q sin π = 0)
25 Clearly, ( x − 1)( x − α1 )( x − α2 ) …
( x − α n −1 ) = x – 1
Putting x = 2, we get
( 2 − α1 )( 2 − α2 )K( 2 − α n − 1 ) = 2n − 1
n
26 We have, (1 + ω )7 = A + Bω
We know that 1 + ω + ω = 0
∴
1 + ω = −ω2
⇒
(− ω2 ) 7 = A + Bω
⇒
− ω14 = A + Bω
2
27 α,β are the roots of x2 − x + 1 = 0
Q Roots of x2 − x + 1 = 0 are − ω, − ω 2
∴ Let α = −ω and β = − ω 2
⇒
α101 + β107 = (− ω )101 + (− ω2 )107
= − (ω101 + ω214 ) = − (ω2 + ω )
[Qω3 n + 2 = ω2 and ω3 n+1 = ω]
= −(−1) = 1
[1 + ω + ω2 = 0]
28 x2 + x + 1 = 0
⇒
x = ω , ω2
1
1
So, x + r = ω r + r = − 1
x
ω
or 2 according as r is not divisible by 3
or divisible by 3.
r
∴ Required sum
= 17(−1) 2 + 8 ⋅ 2 2 = 49
m=1
15
⇒ − ω2 = A + Bω ⇒ 1 + ω = Α + Βω
[Qω14 = ω12 ⋅ ω2 = ω2 ]
On comparing both sides, we get
A = 1, B = 1
29 Given, z = 2ω + 1
⇒ω =
−1 + z
−1 + 3i
⇒ω =
2
2
[Qz =
−3]
⇒ ω is complex cube root of unity
1
1 1
Now,
1 −ω2 − 1 ω2 = 3k
1
ω2 ω7
⇒
1 1 1
1 ω ω2 = 3k
1 ω2 ω
Q1 + ω + ω2 = 0
ω7 = ω
Applying R1 → R1 + R2 + R3 , we get
3 1 + ω + ω2 1 + ω + ω2
1
ω
ω2 = 3k
2
1
ω
ω
3 0 0
⇒ 1 ω ω2 = 3k
1 ω2 ω
⇒ 3(ω2 − ω 4 ) = 3k
⇒
k = ω2 − ω ⇒ k = −1 − 2ω
⇒
k = −(1 + 2ω ) ⇒ k = − z
30 Using 1 + ω + ω2 = 0, we get
1 + ω ω2
∆ = 1 + ω2 ω
ω2 + ω ω
−ω
− ω2
− ω2
Applying C1 → C1 + C2 ,
0
ω2
∆ =
0
ω
ω2 + 2ω ω
−ω
− ω2
− ω2
= (ω2 + 2ω )(− ω + ω2 ) = − 3ω2
18 40
31 |a + bω + c ω2|2 = (a + bω + c ω2 )
38 Given, z2 is not unimodular i.e. z2 ≠ 1
(a + b ω + c ω )
= (a + bω + c ω2 )(a + bω2 + c ω )
[Q ω = ω2 and ω 2 = ω]
2
2
= a + b + c 2 − ab – bc − ca
1
= [(a − b ) 2 + (b − c ) 2 + (c − a) 2 ]
2
So, it has minimum value 1 for a = b = 1
and c = 2.
2
2
2
z − 2z2
and 1
is unimodular.
2 − z1 z2
⇒
⇒
32 Clearly, x + y + z = x x + y y + zz
z1
2
2
2
+ (a + bω + cω )(a + bω + c ω )
2
2
2
2
= 3( a + b + c )
⇒
2
2
2
2
x + y
+ z
a + b + c
2
=3
2
|z|
1
z
Q z = |z |2
x
⇒ 2
= 3 ⇒ 3 x2 + 3 y 2 − x = 0
x + y2
So, it is a circle whose centre is on
X-axis and passes through the origin.
2z + 1 (2 x + 1) + 2iy
34
=
iz + 1
(1 − y ) + ix
[(2 x + 1) + 2iy ] ⋅ [(1 − y ) − ix]
=
(1 − y )2 − i 2 x2
(2 x − y + 1) − (2 x2 + 2 y 2 + x − 2 y )i
=
1 + x2 + y 2 − 2 y
∴ Imaginary part
− (2 x2 + 2 y 2 + x − 2 y )
=
= −2
1 + x2 + y 2 − 2 y
⇒ x + 2 y − 2 = 0, which is a straight line.
35 Clearly,
z
z
1
, which
=
=
1 − z2
zz − z2 z − z
is always imaginary.
36 |ω | = 1 ⇒|z | = z − i
3
It is the perpendicular bisector of the
1
line segment joining (0, 0) to 0, i.e.
3
the line y =
1
⋅
6
37 Given, z1 + z2 = 1 ⇒ z12 + z22 = z1 z2
z2
2
= 4 + z1
2
2
z2
2
− 2z1 z2 − 2z1 z2
2
⇒ ( z2 − 1)( z1 − 4) = 0
Q
z2
∴
z1
Let z1 =
Point z1
Either z = z ⇒ real axis
or
z1
⇒ z12 + z22 + z32 = z1 z2 + z1 z3 + z2 z3 ,
where z3 = 0
So, z1 , z2 and the origin form an
equilateral triangle.
and
y =±
3 ω = e 2 πi / 3 = imaginary cube root of unity
∴
1 + ω + ω2 = 0
z+1
Now, ∆ =
ω
ω2
z
= ω
ω2
ω
z + ω2
1
z
z + ω2
1
ω2
1
z+ω
z
1
z+ω
(applying R1 → R1 + R2 + R3 )
1
1
1
= z ω z + ω2
1
ω2
1
z+ω
1
2
3
2
= z {[(z + ω2 )(z + ω ) − 1]
+ [ω2 − ω( z + ω)] + [ω − ω2 (z + ω2 )]}
= z {z2 + z(ω + ω2 ) + ω3
−1 − ωz − ω2 z} = z3
3
∴ ∆ = 0 ⇒ z = 0 ⇒ z = 0 is the only
solution.
1
3
±
i
2
2
1
3
Now, take, z = +
i
2
2
3 / 2
π
∴
θ = tan −1
=
3
1/2
∴
2
z = z + z ⇒ zz − z − z = 0
i.e. ( x2 + y 2 = 2 x )
represents a circle passing through origin.
40 Let z = x + iy
x2 + y 2 = 1 and 2 x = 1 ⇒ x =
z−1 z −1
zzz − z2 = zzz − z2
2
z (z − z ) − (z − z )(z + z ) = 0
2
Then, |r 2e 2 iθ − 1| = r 2 + 1
⇒ (r 2 cos 2θ − 1) 2 + (r 2 sin 2θ) 2
= (r 2 + 1) 2
⇒ r 4 − 2r 2 cos 2θ + 1 = r 4 + 2 r 2 + 1
π
⇒ cos 2θ = − 1 ⇒ θ =
2
π
π
⇒ z = r cos + i sin = ir
2
2
⇒
⇒
⇒
⇒ (z − z ) ( z − (z + z )) = 0
≠1
=2
x + iy ⇒ x2 + y 2 = (2)2
lies on a circle of radius 2.
Then,
x2 + y 2 = 1
and x + iy = 1 − ( x − iy )
which is purely real ∀ n1 , n2 .
2
2
2 Clearly, z = z
2
39 Let z = re i θ
33 Given, Re 1 = 3 ⇒ Re z 2 = 3
z
= 2 − z1 z2
+ 4 z2 − 2z1 z2 − 2z1 z2
= (a + b + c )(a + b + c )
+ (a + bω + cω2 )(a + bω + c ω )
2
z1 − 2z2
⇒ (z1 − 2z2 )(z1 − 2z2 ) = (2 − z1 z2 )
2
(2 − z1 z2 ) [Qzz = z ]
⇒
2
z1 − 2z2
=1
2 − z1 z2
n π
n π
+ ( 2 )n1 cos 1 − i sin i 1
4
4
n π
π
+ ( 2 )n2 cos 2 − i sin n2
4
4
n
π
n
π
+ ( 2 )n2 cos 2 + i sin 2
4
4
n1 π
n π
n1
= ( 2 ) 2cos
+ ( 2 )n2 2cos 2
4
4
z=
4 In the problem, base = 1 / 2∈(0,1)
SESSION 2
∴ z−1 < z− i ⇒
n1
1 Clearly, (1 + i )
+ (1 + i 3 )n1
n1
π
π
+ 2 cos − + i sin −
4
4
n2
π
π
2 cos + i sin
4
4
n
π
n
π
= ( 2 )n1 cos 1 + i sin 1
4
4
n2
+
2
⇒ (z − 1)(z − 1) < (z − i )( z + i )
+ (1 + i 5 )n2 + (1 + i 7 )n2
n1
n1
= (1 + i ) + (1 − i ) + (1 + i )n2 + (1 − i )n2
n1
π
π
= 2 cos + i sin
4
4
π
π
+ 2 cos − + i sin −
4
4
2
z−1 < z− i
[Q z
2
= z z]
⇒ (1 + i )z + (1 − i )z > 0
⇒ (z + z ) + i (z − z ) > 0
z − z
z + z
⇒
>0
+ i
2
2
z + z − z − z > 0
⇒
2 2i
⇒ Re(z ) − Im(z ) > 0 ⇒ x − y > 0
5 arg z − (10 + 6i ) = π
z − (4 + 6i )
4
y−6 π
−1 y − 6
− tan −1
=
⇒ tan
x − 10
x−4
4
[take z = x + iy ]
19
DAY
y−6 y−6
−
x − 10 x − 4 = 1
⇒
( y − 6)( y − 6)
1+
( x − 10)( x − 4)
⇒
z − (7 + 9i ) = 3 2
iπ / 4
6 We have, zz = (z + z ) = 350
2
k =1
5
= 2 Distance of z from 3, −
2
where z lies on circle (i).
9
∴ min 2z − 6 + 5i = 2PA = 2 − 2 = 5
2
⇒ x2 + y 2 − 14 x − 18 y + 112 = 0
⇒ ( x − 7)2 + ( y − 9)2 = 18 = (3 2 )2
2
⇒ 2( x2 + y 2 )( x2 − y 2 ) = 350
⇒ ( x2 + y 2 )( x2 − y 2 ) = 175
Since x, y ∈ I , the only possible case
which gives integral solutions, is
…(i)
x2 + y 2 = 25
…(ii)
x2 − y 2 = 7
From Eqs. (i) and (ii) x2 = 16; y 2 = 9
⇒ x = ±4; y = ±3 ⇒ Area = 48
10 Clearly, 0 − 3e iπ / 4 = 3 e iπ /2
P(z)
N
( x + y − 1)
2x
x2 + y 2 − 2 x cot 2α − 1 = 0
=
∴
2
8 Given,|z + 4 | ≤ 3
Now,|z + 1 | = |z + 4 − 3 |
≤ |z + 4 | + | 3 | ≤ 3 + 3 = 6
Hence, greatest value of|z + 1 | = 6
Since, least value of the modulus of a
complex number is 0.
Consider,|z + 1 | = 0 ⇒ z = − 1
Now,
|z + 4 | = |− 1 + 4 | = 3
⇒| z + 4 | ≤ 3 is satisfied by z = − 1.
∴ Least value of|z + 1| = 0
9 z − 3 − 2i ≤ 2
∴
z − 3e iπ / 4 = 4ie iπ / 4
z = (3 + 4i )e iπ / 4
…(i)
z = 1, a1 , a2 ,... an −1
1
1
Let α =
, then z = 1 −
1− z
α
n
1
[by (i)]
∴ 1 − = 1
α
n
n
⇒ (α − 1) − α = 0
⇒ − C1 α n −1 + C2α n −2 + ... + (−1)n = 0
1
1
1
where, α =
,
...,
1 − a1 1 − a2 1 − an −1
1
1
1
+
+ ...+
1 − a1 1 − a2
1 − an −1
⇒
=
2
C2
n(n − 1) (n − 1)
=
=
2/ n
2
C1
2
12 z ω − ω z = z − ω
⇒
( x − 3)2 + ( y − 2)2 = 22 = 4
... (i)
2
+ 1)z
z
zω
=z ⇒
= zω
ω
ω
B(z2)
A(z1)
α
⇒
⇒
⇒
∴
... (ii)
⇒ zz ω − ωz ω − z + ω = 0
⇒ (z ω − 1)(z − ω ) = 0 ⇒ z = ω or zω = 1
i.e.
zω = 1
2
1
= ω/ ω
⇒
z = ω or z =
ω
X
O
⇒
Also, from Eq. (i), zz ω − ωω z = z − ω
... (i)
z2
= e iα = cos α + i sin α
z1
11 Given z n = 1, where
2
⇒ z lies on or inside the circle
z1 + z2 = − p and z1 z2 = q
OA = OB
z1 = z2
Y
z +1
z
⇒
=
= real
2
ω
ω +1
P(3, – 5/2)
k
k =1
where α = e − i 2 π /11
= i [α + α2 + α3 + ... + α10 ]
⇒
Q
⇒
E
O
2
A
k =1
10
∑α
14 Given, z2 + pz + q = 0
3eiπ/4
⇒ ( z + 1)ω = ( ω
(3, 2)
=i
11
= −i
7 We have, α + iβ = cot (z )
2
− i 2 kπ
=i
−1
⇒
cot(α + iβ ) = x + iy
and cot(α − iβ ) = x − iy
Now, consider
cot 2α = cot [(α + iβ ) + (α − iβ )]
cot(α + iβ ) ⋅ cot(α − iβ ) − 1
=
cot(α + iβ ) + cot(α − iβ )
10
= i ∑e
α(1 − α10 )
(α − α11 )
=i
1−α
1−α
(α − 1)
11
=i
[Qα = cos 2 π − sin 2 π = 1]
(1 − α )
z − 3e
4
−3e − iπ / 4
3
= i
z − 3e iπ / 4
4
∴
2k π
2k π
+ i cos
11
11
10
2kπ
2kπ
= i ∑ cos
− i sin
11
11
k =1
10
13 We have, ∑ sin
5
2z − 6 + 5i = 2 z − 3 − i
2
z1 + z2
= 1 + cos α + i sin α
z1
α
α
α
= 2cos cos + i sin
2
2
2
(z1 + z2 )2
2 α iα
= 4cos e
z12
2
z
= 4cos 2 α / 2 ⋅ 2
z1
α
(z1 + z2 )2 = 4cos 2 z1 z2
2
p2 = 4 q cos 2 α / 2
p2
α
= cos 2
4q
2
15 Q p < 0, take
p = − q 3 (q > 0)
∴ p1 /3 = q(−1)1 /3 = − q , − qω, − qω2
Now, take α = −q , β = − qω, γ = − qω2
Then, given expression
=
x + yω + zω2
= ω2
xω + yω2 + z
DAY THREE
Sequence
and Series
Learning & Revision for the Day
u
u
u
Definition
Arithmetic Progression (AP)
Arithmetic Mean (AM)
u
u
u
Geometric Progression (GP)
Geometric Mean (GM)
Arithmetico-Geometric
Progression (AGP)
u
u
Sum of Special Series
Summation of Series by the
Difference Method
Definition
●
By a sequence we mean a list of numbers, arranged according to some definite rule.
or
We define a sequence as a function whose domain is the set of natural numbers or some
subsets of type {1, 2, 3, ... k}.
●
●
If a1, a2 , a3 ,..., an,.... is a sequence, then the expression a1 + a2 + a3 + ... + an+... is called the
series.
If the terms of a sequence follow a certain pattern, then it is called a progression.
PRED
MIRROR
Arithmetic Progression (AP)
●
Your Personal Preparation Indicator
It is a sequence in which the difference between any two consecutive terms is always
same.
An AP can be represented as a , a + d, a + 2 d, a + 3 d, … where, a is the first term, d is the
common difference.
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
●
The nth term, t n = a + (n − 1)d
u
●
Common difference d = t n − t n−1
●
●
●
●
The nth term from end, t n = l − (n − 1)d, where l is the last term.
n
n
Sum of first n terms, S n = [2 a + (n − 1)d ] = [a + l ], where l is the last term.
2
2
1
If sum of n terms is S n, then nth term is t n = S n − S n−1 , t n = [t n− k + t n+ k ], where k < n
2
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
21
DAY
NOTE
• Three non-zero numbers a, b, c are in GP iff b2 = ac.
• If a, b and c are in AP as well as GP, then a = b = c.
• If a > 0 and r > 1 or a < 0 and 0 < r < 1, then the GP will be
• Any three numbers in AP can be taken as a − d , a, a + d.
• Any four numbers in AP can be taken as
a − 3d , a − d , a + d , a + 3d.
an increasing GP.
• Any five numbers in AP can be taken as
• If a > 0 and 0 < r < 1 or a < 0 and r > 1, then the GP will be a
a − 2d , a − d , a, a + d , a + 2d.
decreasing GP.
• Three numbers a, b, c are in AP iff 2b = a + c.
Important Results on GP
An Important Result of AP
●
●
In a finite AP, a1 , ..., an, the sum of the terms equidistant
from the beginning and end is always same and equal to
the sum of first and last term
i.e. a1 + an = ak + an –( k − 1 ), ∀ k = 1, 2, 3,..., n − 1.
●
Arithmetic Mean (AM)
●
●
●
●
If a, A and b are in AP, then A =
mean of a and b.
a+b
is the arithmetic
2
If a, A1 , A2 , …, An , b are in AP, then A1, A2,..., An are
the n arithmetic means between a and b.
The n arithmetic means, A1, A2 , ..., An, between a and b are
r (b − a)
given by the formula, Ar = a +
∀ r = 1, 2, ... n
n+1
Sum of n AM’s inserted between a and b is n A i.e.
a + b
A1 + A2 + A3 + K + An = n
2
NOTE
●
●
●
●
●
It is a sequence in which the ratio of any two consecutive
terms is always same.
A GP can be represented as a, ar , ar 2 , …
where, a is the first term and r is the common ratio.
The nth term, t n = ar n –1
l
The nth term from end, t n′ = n − 1 , where l is the last term.
r
1 – r
, r ≠1
a
Sum of first n terms, S n = 1 – r
na,
r =1
If| r | < 1, then the sum of infinite GP is S ∞ =
NOTE
●
●
a
r
a a
• Any four numbers in GP can be taken as 3 , , ar, ar 3 .
r r
a a
• Any five numbers in GP can be taken as 2 , , a, ar, ar 2 .
r r
• Any three numbers in GP can be taken as , a, ar.
If a, G1, G2,K , Gn, b are in GP, then G1, G2 ,K , Gn are the n
geometric means between a and b.
The n GM’s, G1, G2 , ..., Gn, inserted between a and b, are
r
Product of n GM’s, inserted between a and b, is the nth
power of the single GM between a and b,
i.e. G1 ⋅ G2 ⋅ ... ⋅ Gn = G n = (ab )n/2 .
NOTE
• If a and b are of opposite signs, then their GM can not exist.
• If A and G are respectively the AM and GM between two
numbers a and b, then a, b are given by
[ A ± ( A + G)( A − G) ] .
• If a1 , a2 , a3 ,... , an are positive numbers, then their GM
= ( a1 a2 a3 ... an ) 1 / n .
Arithmetico-Geometric
Progression (AGP)
●
a
1−r
If a, G and b are in GP, then G = ab is the geometric mean
of a and b.
b n + 1
given by the formula, Gr = a
.
a
( a1 + a2 + a3 + K + an )
n
n
●
●
●
Geometric Progression (GP)
In a finite GP, a1 , a2 , ..., an, the product of the terms
equidistant from the beginning and the end is always
same and is equal to the product of the first and the last
term.
i.e. a1 an = ak ⋅ an − ( k − 1 ), ∀ k = 1, 2, 3, ..., n − 1.
Geometric Mean (GM)
• The AM of n numbers a1 , a2 , ... , an is given by
AM =
If a1 , a2 , a3 ,K , an is a GP of positive terms, then
log a1 , log a2 ,K , log an is an AP and vice-versa.
●
A progression in which every term is a product of a term
of AP and corresponding term of GP, is known as
arithmetico-geometric progression.
If the series of AGP be a + (a + d)r + (a + 2 d)r 2 + ...
+ {a + (n − 1)d}r n−1 + ..., then
a
dr (1 − r n−1 ) {a + (n − 1) d }r n
+
−
,r ≠1
1−r
1−r
(1 − r )2
a
dr
=
+
,| r | < 1
1 − r (1 − r )2
(i) S n =
(ii) S ∞
22
40
Method to find the Sum of n-terms of
Arithmetic Geometric Progression
●
Usually, we do not use the above formula to find the sum of n
terms.
Infact we use the mechanism by which we derived the
formula, shown below:
Let, S n = a + (a + d) r + (a + 2d) r 2
+ K + (a + (n − 1) d) r n − 1 …(i)
Step I Multiply each term by r (Common ratio of GP) and
obtain a new series
⇒
r S n = ar + (a + d) r 2 + K +
(a + (n − 2) d) r
n −1
+ (a + (n − 1) d) r …(ii)
n
Step II Subtract the new series from the original series by
shifting the terms of new series by one term
⇒ (1 − r ) S n= a + [dr + dr 2 +K + dr n−1 ] − (a + (n − 1) d) r n
1 − rn −1
⇒ S n(1 − r ) = a + dr
− (a + (n − 1) d) r n
1−r
⇒
1 − r n − 1 (a + (n − 1) d) n
a
Sn =
+ dr
r
−
1−r
1−r
(1 − r )2
●
Sum of squares of first n natural numbers,
n (n + 1)(2 n + 1)
12 + 22 + K + n2 = Σ n 2 =
6
Sum of cubes of first n natural numbers,
2
n (n + 1)
13 + 23 + 33 + ... + n3 = Σ n3 =
2
(i) Sum of first n even natural numbers
2 + 4 + 6 + K+ 2 n = n (n + 1)
(ii) Sum of first n odd natural numbers
1 + 3 + 5 + K + (2 n − 1) = n2
Summation of Series by
the Difference Method
If nth term of a series cannot be determined by the methods
discussed so far. Then, nth term can be determined by the
method of difference, if the difference between successive
terms of series are either in AP or in GP, as shown below:
Let T1 + T2 + T3 + ... be a given infinite series.
If T2 − T1 , T3 − T2 ,K are in AP or GP, then Tn can be found by
following procedure.
Clearly, S n = T1 + T2 + T3 + K + Tn
…(ii)
∴ S n − S n = T1 + (T2 − T1 ) + (T3 − T2 ) + ... + (Tn − Tn−1 ) − Tn
Sum of Special Series
●
…(i)
S n = T1 + T2 + K + Tn−1 + Tn
Again,
Sum of first n natural numbers,
n (n + 1)
1 + 2 + ... + n = Σ n =
2
⇒
Tn = T1 + (T2 − T1 ) + (T3 − T2 ) + K + (Tn − Tn−1 )
⇒
Tn = T1 + t 1 + t 2 + t 3 + t n−1
n
where, t 1, t 2 , t 3 ,K are terms of the new series ⇒ S n = ∑ Tr
r =1
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
7
2
1 If log3 2, log3 ( 2x − 5) and log3 2 x − are in AP, then x
is equal to
(a) 2
n −1
a1 an + 1
(b)
1
a1 an + 1
(c)
n+1
a1 an + 1
(d)
n
a1 an
+1
5 A man arranges to pay off a debt of ` 3600 by 40 annual
(b) 3
(c) 4
(d) 2, 3
2 The number of numbers lying between 100 and 500 that
are divisible by 7 but not by 21 is
(a) 57
(a)
(b) 19
(c) 38
(d) None of these
3 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
(a) − 150
(c) 150
(b) 150 times its 50th term
(d) zero
4 If a1, a 2, ..., an + 1 are in AP, then
1
1
1
is
+
+K+
a1a 2 a 2a 3
anan + 1
instalments which are in AP. When 30 of the instalments
are paid, he dies leaving one-third of the debt unpaid.
The value of the 8th instalment is
(a) ` 35
(c) ` 65
(b) ` 50
(d) None of these
6 Let a1, a 2, a 3, ... be an AP, such that
a1 + a 2 + ... + a p
a1 + a 2 + a 3 + ... + aq
=
p3
a
; p ≠ q , then 6 is equal to
a 21
q3
j
41
(a)
11
11
(c)
41
121
(b)
1681
121
(d)
1861
JEE Mains 2013
23
DAY
19 If | a | < 1 and | b | < 1, then the sum of the series
7 A person is to count 4500 currency notes.
Let an denotes the number of notes he counts in the n th
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
(a) 24 min
(b) 34 min
8 If log
3
(c) 125 min
(d) 135 min
a + log(3)1/ 3 a + log31/ 4 a + K upto 8th term
2
2
2
(a) ±
3
(b) 2 2
(d) None of these
9 n arithmetic means are inserted between 7 and 49 and
(b) 12
(c) 13
(d) 14
(a) 1
(b) 2
(a) g 2
(d) 3
(a)
8
5
(b)
4
3
j
(c) 1
JEE Mains 2016
7
(d)
4
12 Three positive numbers form an increasing GP. If the
middle term in this GP is doubled,then new numbers are
in AP. Then, the common ratio of the GP is j JEE Mains 2014
(a)
2+
3
(b) 3 +
2
(c) 2 − 3
(d) 2 +
3
13 A GP consists of an even number of terms. If the sum of
all the terms is 5 times the sum of terms occupying odd
places,then its common ratio is
(a) 2
(b) 3
(c) 4
(d) 5
14 The sum of first 20 terms of the sequence 0.7, 0.77,
0.777, …, is
j
JEE Mains 2013
7
− 20
(b) [99 − 10 ]
9
7
(d) [99 + 10 − 20 ]
9
7
(a)
[179 − 1020 ]
81
7
(c)
[179 + 10 − 20 ]
81
15 If x , y and z are distinct prime numbers, then
(a) x, y and z may be in AP but not in GP
(b) x, y and z may be in GP but not in AP
(c) x, y and z can neither be in AP nor in GP
(d) None of the above
(c) 64
(b) −10 < x < 0 (c) 0 < x < 10 (d) x > 10
(a) a m
2
(b) 2a m
2
2
(c) 3 a m
2
(b) 6
(d) − 6
(c) 12
23 The sum to 50 terms of the series
2
1
1
1 + 2 1 +
+ 3 1 +
+ K , is given by
50
50
(a) 2500
(c) 2450
(b) 2550
(d) None of these
24 If (10)9 + 2(11)1(10)8 + 3(11)2(10)7 + ... + 10 (11)9 = k (10)9 ,
then k is equal to
121
(a)
10
j
441
(b)
100
(c) 100
JEE Mains 2014
(d) 110
25 The sum of the infinity of the series
2
6
10 14
1 + + 2 + 3 + 4 + ... is
3 3
3
3
(a) 3
(b) 4
(c) 6
(d) 2
26 The sum of the series1 + 3 + 5 + K upto 20 terms is
3
3
(a) 319600
(c) 306000
3
(b) 321760
(d) 347500
∑ a 4k + 1 = 416
2
and a 9 + a 43 = 66. If a12 + a 22 + K + a17
= 140 m, then m is
j
JEE Mains 2018
equal to
(b) 68
(c) 34
(d) 33
28 Let a, b, c ∈ R . if f ( x ) = ax + bx + c be such that
2
2
(d) 4 a m
a + b + c = 3 and f ( x + y ) = f ( x ) + f ( y ) + xy , ∀ x , y ∈ R ,
10
square is formed by joining the mid-points of these
squares. Then, a third square is formed by joining the
mid-points of the second square and so on. Then, sum of
the area of the squares which carried upto infinity is
2
(a) 4
(a) 66
(d) 16
18 The length of a side of a square is a metre. A second
2
GM will be
k =0
17 An infinite GP has first term x and sum 5, then x belongs
(a) x < − 10
2
3
12
such that (nm + 1) divides (1 + n + n 2 + K + n127 ), is
(b) 8
(d) 3 g 2
(c) 2g
27 Let a1, a 2, a 3, ..., a 49 be in AP such that
16 Let n ( > 1) be a positive integer, then the largest integer m
(a) 32
(b) −g 2
22 If five GM’s are inserted between 486 and , then fourth
11 If the 2nd, 5th and 9th terms of a non-constant AP are in
GP, then the common ratio of this GP is
(b) 20 months
(d) 18 months
two numbers a and b, then ( 2p − q )( p − 2 q ) is equal to
x −y
is equal to
z
2
(c) 1 / 2
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
j AIEEE 2011
will be ` 11040 after
21 If one GM, g and two AM’s, p and q are inserted between
10 If x = 111K1 (20 digits), y = 333K3 (10 digits) and
z = 222K 2 (10 digits), then
1
(1 − a) (1 − ab)
1
(d)
(1 − a) (1 − b) (1 − ab)
(b)
(a) 19 months
(c) 21 months
their sum is found to be 364, then n is
(a) 11
1
(1 − a) (1 − b)
1
(c)
(1 − b) (1 − ab)
(a)
20 A man saves ` 200 in each of the first three months of his
= 44, then the value of a is
1
(c)
2
1 + (1 + a ) b + (1 + a + a 2 ) b 2 + (1 + a + a 2 + a 3 ) b 3 + K is
2
then
∑ f (n ) is equal to
n =1
(a) 330
j
(b) 165
(c) 190
JEE Mains 2017
(d) 255
29 The sum of the series ( 2)2 + 2( 4)2 + 3 ( 6)2 + ... upto 10
terms is
(a) 11300
j
(b) 11200
(c) 12100
JEE Mains 2013
(d) 12300
24
40
30 Let A be the sum of the first 20 terms and B be the sum
of the first 40 terms of the series
12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + K
(a)
If B − 2A = 100λ, then λ is equal to
(a) 232
(b) 248
j
(c) 464
7
2
(b)
11
4
(c)
33 The sum of the series1 +
11
2
(d)
j
(c) 142
JEE Mains 2015
18
(a)
11
60
11
1
1
+
+ ...
1+ 2 1+ 2 + 3
upto 10 terms is
13 13 + 23 13 + 23 + 33
+
+... is
+
1
1+ 3
1+ 3 + 5
(b) 96
3
5
7
+
+
+ ... upto 11 terms is
12 12 + 22 12 + 22 + 32
j JEE Mains 2013
JEE Mains 2018
(d) 496
31 The sum of first 9 terms of the series
(a) 71
32 The sum
j
22
(b)
13
20
(c)
11
JEE Mains 2013
16
(d)
9
(d) 192
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 The value of12 + 32 + 52 + ... + 252 is
(a) 2925
(b) 1469
j
(c) 1728
JEE Mains 2013
(d) 1456
2 If the function f satisfies the relation f ( x + y ) = f ( x ) ⋅ f ( y )
for all natural numbers x , y , f (1) = 2 and
8 If Sn is the sum of first n terms of a GP : {an } and Sn′ is the
sum of another GP : {1 / an }, then Sn equals
(a)
n
∑ f (a + r ) = 16 ( 2n −1), then the natural number a is
(b) 3
3 If the sum of an infinite GP is
(c) 4
(d) 5
7
and sum of the squares
2
147
of its terms is
, then the sum of the cubes of its terms
16
is
(a)
315
19
(b)
700
39
4 The sum of the infinite series
(a)
31
18
(b)
65
32
(c)
985
13
(d)
1029
38
5
55
555
+ 2 +
+ K is
13 13
133
(c)
65
36
(d)
75
36
5 Given sum of the first n terms of an AP is 2n + 3 n 2.
Another AP is formed with the same first term and double
of the common difference, the sum of n terms of the new
AP is
(a) n + 4n 2
6 For 0 < θ <
z =
∞
(b) 6n 2 − n
(c) n 2 + 4n
(d) 3 n + 2n 2
∞
∞
π
, if x = ∑ cos 2n θ , y = ∑ sin2n θ and
2
n =0
n =0
∑ cos 2n θ sin2n θ, then xyz is equal to
n =0
(a) xz + y
(b) x + y + z (c) yz + x
(d) x + y − z
1
1
1
π4
1
1
1
7 If 4 + 4 + 4 + K to ∞ = , then 4 + 4 + 4 + K to
90
1
2
3
1
3
5
∞ is equal to
(a)
π4
96
(b)
π4
45
(c)
89 π 4
90
(d)
π4
90
(b) a1an Sn′
(c)
a1 ′
Sn
an
(d)
an ′
Sn
a1
9 If the sum of the first ten terms of the series
r =1
(a) 2
Sn′
a1an
2
2
2
2
16
3
2
1
4
2
m,
1 + 2 + 3 + 4 + 4 + ..., is
5
5
5
5
5
then m is equal to
(a) 102
j
(b) 101
(c) 100
JEE Mains 2016
(d) 99
10 If m is the AM of two distinct real numbers I and n (l, n > 1)
and G1, G 2 and G 3 are three geometric means between I
j
and n, then G14 + 2G 24 + G 34 equals
JEE Mains 2015
(a) 4l 2mn
(c) 4 lmn 2
(b) 4 lm 2n
(d) 4l 2m 2n 2
11 The sum of the series ( 2 + 1) + 1 + ( 2 − 1) + K ∞ is
(a)
2
(b) 2 + 3 2
(c) 2 − 3 2
(d)
4+ 3 2
2
12 The largest term common to the sequences 1, 11, 21, 31,
... to 100 terms and 31, 36, 41, 46, ... to 100 terms is
(a) 531
(b) 471
(c) 281
(d) 521
13 If a, b, c are in GP and x is the AM between a and b, y the
AM between b and c, then
a c
+ =1
x
y
a c
(c)
+ =3
x
y
(a)
(b)
a c
+ =2
x
y
(d) None of these
14 Suppose a, b and c are in AP and a 2, b 2 and c 2 are in
3
GP. If a < b < c and a + b + c = , then the value of a is
2
(a)
1
2 2
(b)
1
2 3
(c)
1
1
−
2
3
(d)
1
1
−
2
2
25
DAY
15 For any three positive real numbers a, b and c, if
17 Statement I The sum of the series
1 + (1 + 2 + 4) + ( 4 + 6 + 9) + ( 9 + 12 + 16) + K
9 ( 25a + b ) + 25 (c − 3ac ) = 15b ( 3a + c ), then
2
(a)
(b)
(c)
(d)
2
2
b, c and a are in GP
b, c and a are in AP
a, b and c are in AP
a, b and c are in GP
j
+ ( 361 + 380 + 400) is 8000.
JEE Mains 2017
n
Statement II
k =1
number n.
16 If S1, S 2, S 3,K are the sum of infinite geometric series
whose first terms are 1, 2 , 3,K and whose common ratios
1 1 1
, , ,... respectively, then
2 3 4
2
is equal to
S12 + S 22 + S 32 + K + S10
(a) 485
(c) 500
∑ [k 3 − (k − 1)3 ] = n 3, for any natural
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is ture; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
(b) 495
(d) 505
ANSWERS
SESSION 1
SESSION 2
1 (b)
2 (c)
3 (d)
4 (d)
5 (c)
6 (b)
7 (b)
8 (a)
9 (c)
10 (a)
11 (b)
12 (d)
13 (c)
14 (c)
15 (a)
16 (c)
17 (c)
18 (b)
19 (c)
20 (c)
21 (b)
22 (b)
23 (a)
24 (c)
25 (a)
26 (a)
27 (c)
28 (a)
29 (c)
30 (b)
31 (b)
32 (c)
33 (c)
1 (a)
11 (d)
2 (b)
12 (d)
3 (d)
13 (b)
4 (c)
14 (d)
5 (b)
15 (b)
6 (b)
16 (d)
7 (a)
17 (a)
8 (b)
9 (b)
10 (b)
Hints and Explanations
SESSION 1
1 Q2 log 3 (2 − 5) = log 3 2 + log 3 2x − 7
2
7
x
2
x
⇒
(2 − 5) = 2 2 −
2
⇒
t 2 + 25 − 10 t = 2t − 7
x
[put 2x = t ]
⇒
t − 12t + 32 = 0
⇒
⇒
(t − 8) (t − 4) = 0
2x = 8 or 2x = 4
2
∴
x = 3 or x = 2
At, x = 2, log 3 (2x − 5) is not defined.
Hence, x = 3 is the only solution.
2 The numbers between 100 and 500
that are divisible by 7 are 105, 112,
119, 126, …, 490, 497.
Let such numbers be n.
∴
t n = an + (n − 1) d
⇒
497 = 105 + (n − 1) × 7
⇒
n − 1 = 56
⇒
n = 57
The numbers between 100 and 500
that are divisible by 21 are 105,
126, 147, …, 483.
Let such numbers be m.
∴
483 = 105 + (m − 1) × 21
⇒
18 = m − 1 ⇒ m = 19
∴ Required number
= n − m = 57 − 19 = 38
3 Let a be the first term and d (d ≠ 0)
be the common difference of a given
AP, then
T100 = a + (100 − 1) d = a + 99 d
T 50 = a + (50 − 1) d = a + 49 d
T150 = a + (150 − 1) d = a + 149 d
Now, according to the given condition,
100 × T100 = 50 × T 50
⇒
100(a + 99 d ) = 50(a + 49 d )
⇒
2(a + 99 d ) = (a + 49 d )
⇒
2a + 198 d = a + 49 d
⇒
∴
a + 149 d = 0
T150 = 0
4 Let d be the common difference of
given AP and let
1
1
1
+
+K+
a1 a2
a2 a3
an an
S =
. Then,
+1
d
d
d
+
+K+
a
a
a
a
a
a
2 3
n n +1
1 2
a2 − a1 + a3 − a2
a1 a2
a2 a3
1
=
a
a
−
d
n +1
n
+K+
an an + 1
1
1 1
1
a − a + a − a
1
2
2
3
1
=
1
d
1
+K+
−
a
an + 1
n
a
a
−
1 1
1
1
n +1
1
=
−
=
d a1
an + 1 d a1 an + 1
1 a + nd − a1
n
= 1
=
d a1 an + 1
a
a
1
n +1
S =
1
d
26
40
5 Given, 3600 = 40 [ 2a + (40 − 1) d ]
⇒
⇒
180 = 2a + 39d
…(i)
After 30 instalments one-third of the
debt is unpaid.
3600
Hence,
= 1200 is unpaid and 2400
3
is paid.
30
Now, 2400 =
{2a + (30 − 1) d }
2
…(ii)
∴
160 = 2a + 29d
On solving Eqs. (i) and (ii), we get
a = 51, d = 2
Now, the value of 8th instalment
= a + (8 − 1) d
= 51 + 7 ⋅ 2 = ` 65
6 Given that,
a1 + a2 + K + a p
a1 + a2 + K + a q
=
1
1
+
1
1
log
log
3
3
8 S n = log a
2
4
3
+ K upto 8th term
log a2
⇒
[2 + 3 + 4 + K + 9] = 44
log 3
p3
q3
p
[ 2a1 + ( p − 1 )d ]
p3
2
⇒
= 3
q
q
[ 2a2 + (q − 1 )d ]
2
where, d is a common difference of an
AP.
2a 1 + ( p − 1)d
p2
= 2
⇒
2a 2 + (q − 1)d
q
d
2
2 = p
⇒
2
d
q
a2 + (q − 1)
2
On putting p = 11 and q = 41 , we get
d
a1 + (11 − 1)
2
2 = (11)
d
(41)2
a2 + (41 − 1)
2
a1 + 5d
121
⇒
=
a2 + 20d 1681
a6
a21
=
121
1681
7 Number of notes that the person counts
in 10 min
= 10 × 150 = 1500
Since, a10, a11 , a12 ,... are in AP with
common difference − 2.
Let n be the time taken to count
remaining 3000 notes.
n
Then, [2 × 148 + (n − 1) × −2] = 3000
2
⇒
n2 − 149 n + 3000 = 0
⇒
(n − 24)(n − 125) = 0
∴
n = 24 and 125
Then, the total time taken by the person
to count all notes
= 10 + 24
= 34 min
+
Given, sum of all terms = 5 × sum of
terms occupying odd places, i.e.
a + ar + ar 2 + ... + a r 2 n −1
= 5 × (a + a r 2 + a r 4 + ... + a r 2 n −2 )
⇒
[given]
⇒
44 log a2 = 44 log 3
a= ±
∴
3
9 We know that,
A1 + A2 + K + A n = nA, where
a+ b
A=
2
7 + 49
∴ 364 =
n
2
364 × 2
⇒
n=
= 13
56
10 Given, x = 1 (999K 9) = 1 (1020 − 1)
9
9
1
1
y = (999K 9) = (1010 − 1)
3
3
2
2
and
z = (999K 9) = (1010 − 1)
9
9
∴
x − y 2 1020 − 1 − (1010 − 1)2
=
z
2(1010 − 1)
1010 + 1 − (1010 − 1)
=
=1
2
a1 + ( p − 1)
⇒
1
2 1 log 3
2
3600 = 20 (2a + 39d )
11 Let a be the first term and d be the
common difference.
Then, we have a + d , a + 4d , a + 8 d in
GP,
i.e.
(a + 4d )2 = (a + d ) (a + 8d )
⇒ a2 + 16d 2 + 8ad = a2 + 8ad
+ ad + 8d 2
⇒
8d 2 = ad
⇒
8d = a
[Qd ≠ 0]
Now, common ratio,
a + 4d
8d + 4d 12d
4
r =
=
=
=
a+ d
8d + d
9d
3
12 Let a, ar , ar 2 be in GP ( where, r > 1).
On multiplying middle term by 2, we
get that a, 2ar , ar 2 are in AP.
⇒
⇒
⇒
∴
r = 2 + 3 [QAP is increasing]
2
3
13 Let the GP be a, a r , a r , a r ,
ar 2 n −2 , ar 2 n −1 .
where, a, ar 2 , ar 4 , ar 6, ... occupy odd
places and ar , ar 3 , ar 5, ar 7 ,... occupy
even places.
5(r 2 n − 1)
r2n − 1
=
r −1
(r − 1)(r + 1)
1=
5
⇒ r + 1 = 5⇒ r = 4
r+1
14 Let S = 0.7 + 0.77 + 0.777 + … upto 20
terms
7
77 777
=
+
+
+ … upto 20 terms
10 102 103
1 + 11 + 111
= 7 10 102 103
+ … upto 20 terms
=
9 + 99 + 999
7
10 100 1000
9
+ … upto 20 terms
1
1
1
7 1 − + 1 − 2 + 1 − 3
10
10
10
9
+ … upto 20 terms
7
=
(1 + 1 + … upto 20 terms)
9
=
1 + 1 + 1
− 10 102 103
+ … upto 20 terms
20
1
1
1 −
10
10
7
=
20 −
1
9
1−
10
Qsum of n terms of GP,
a(1 − r n )
, where r < 1
Sn =
1− r
=
20
7
1
1
20 − 1 −
10
9
9
=
20
7 179 1 1
7
+ =
[179 + 10−20]
9 9
9 10 81
⇒ 4 ar = a + ar 2 ⇒ r 2 − 4r + 1 = 0
4 ± 16 − 4
r =
= 2± 3
2
a (r 2 n − 1) 5a[(r 2 )n − 1]
=
r −1
r2 − 1
a(r n − 1)
QS n = r − 1
15 x, y , z are in GP
⇔
y 2 = xz
⇔ x is factor of y. Which is not
possible, as y is a prime number.
If x = 3, y = 5and z = 7, then they are in
AP.
Thus, x, y and z may be in AP but not
in GP.
27
DAY
16 Clearly,
1+ n+ n + K+ n
2
=
(n
64
127
− 1) (n
n−1
64
n128 − 1
=
n−1
a (r n − 1)
QS n = r − 1
+ 1)
= (1 + n + n2 + K + n 63 ) (n 64 + 1)
Thus, the largest value of m for which
n m + 1 divides
1 + n + n2 + K + n127 is 64.
5− x
17 Since, S ∞ = x = 5 ⇒ r =
1− r
5
a + a2 + K + an − 1 ) b n − 1
n =1
n =1
⇒
n + 11 n − 522 = 0
⇒
n2 + 29 n − 18 n − 522 = 0
⇒
n (n + 29) − 18(n + 29) = 0
1 − an n − 1
b
1− a
∞
1 ∞
n −1
=
− ∑ an b n − 1
∑ b
1 − a n = 1
n =1
∞
∞
1
n −1
n −1
=
− a ∑ (ab )
∑ b
1 − a n = 1
n =1
a
1
2
=
[1 + b + b + K ∞] −
1− a
1− a
[1 + ab + (ab )2 + K ]
1
1
a
1
=
⋅
−
⋅
1 − a 1 − b 1 − a 1 − ab
[Q|b | < 1 and |ab | = |a||b | < 1]
1 − ab − a (1 − b )
=
(1 − a) (1 − b ) (1 − ab )
1 − ab − a + ab
=
(1 − a) (1 − b ) (1 − ab )
1
=
(1 − b ) (1 − ab )
20 Let the time taken to save
` 11040 be (n+3) months.
For first 3 months he saves ` 200 each
month.
n
In (n+3) months, 3 × 200 + {2(240)
2
+ (n − 1) × 40} = 11040
S 50 = 2500.
24 Given, k ⋅ 109 = 109 + 2(11)1 (10)8
+ 3(11)2 (10)7 + ... + 10(11)9
2
⇒
∴ n = 18
11
11
k = 1 + 2 + 3
10
10
[neglecting n = − 29 ]
∴ Total time = (n + 3) = 21 months
b −a
.
3
b − a 2a + b
∴ p = a+ d = a+
=
3
3
b − a a + 2b
q =b −d =b −
=
3
3
Now, (2 p − q )( p − 2q )
(4a + 2b − a − 2b ) (2a + b − 2a − 4b )
⋅
3
3
22 Here, a = 486 and b = 2
2 1
∴ G 4 = 486 ⋅
3 486
1
= 486
3 ⋅ 243
1
= 486
729
r
n +1
4/6
[Q here, n = 5]
4/6
4/6
= 486 ⋅
1
34
11 k = 1 11 + 2 11
10
10
10
=6
23 Let x = 1 + 1 and S 50 be the sum of
50
first 50 terms of the given series.
Then,
S 50 = 1 + 2 x + 3 x2
...(i)
2
9
11
11
+ ... + 9 + 10
10
10
10
...(ii)
On subtracting Eq.(ii) from Eq.(i),
we get
11
11 11
k 1 − = 1 +
+
10
10 10
2
9
11
11
+ ... + 9 − 10
10
10
10
11 10
1 − 1
10
10 − 11
10
−10 11
⇒ k
=
10
10
11 − 1
10
10
10
11
11
⇒ − k = 10 10 − 10 − 10
10
10
∴ k = 100
14
25 Let S = 1 + 2 + 62 + 10
+ 4 +K
3 3
33
3
2
3
5
7
=1+
1 + + 2 + 3 + K
3
3 3
3
2
1
2⋅1 / 3
+
3 1 − 1 / 3 (1 − 1 / 3)2
Qsum of infinite AGP, is
a
dr
S∞ =
+
1− r
(1 − r )2
=1+
+ ... + 50 x 49 …(i)
⇒
9
a( r n − 1)
Q in GP,sum of n terms = r − 1 ,
when r > 1
3
b
We know that, G r = a
a
2
11
+ ... + 10
10
(n − 18)(n + 29) = 0
2
+ (1 + a + a2 + a3 ) b 3 + K ∞
∞
⇒
30 + n2 + 11 n = 552
= − ab = − g 2
19 Clearly, 1 + (1 + a)b + (1 + a + a )b
∑
600 + 20 n (n + 11) = 11040
⇒
=
2
=
⇒
So, common difference d is
carried upto infinity
a2
a2
= a2 +
+
+ ...
2
4
2
a
=
= 2a2 m2
1
1−
2
∑ (1 +
− 1
50
50
⇒ S 50
= − 50 + 50 x − 50 x
50
in AP.
18 Sum of the area of the squares which
=
n
{40(12 + n − 1)} = 11040
2
21 Since, g = ab . Also, a, p , q and b are
For infinite GP,|r | < 1
5− x
⇒ −1 <
< 1 ⇒ −10 < − x < 0
5
∴ 0 < x < 10
∞
⇒ 600 +
x S 50 = x + 2 x2
+ K + 49 x 49 + 50 x 50 …(ii)
⇒ (1 − x )S 50 = 1 + x + x2 + x3
+ K + x 49 − 50 x 50
[subtracting Eq. (ii) from Eq. (i)]
1 − x 50
⇒ S 50 (1 − x ) =
− 50 x 50
1− x
50
− 1 1 − x
⇒ S 50
− 50 x 50
=
50
− 1
50
Q x = 1 + 1
50
=1+
2 3 2 9
2
3
+ ⋅
= 1 + ⋅2⋅ = 3
3 2 3 4
3
2
26 13 + 33 + K + 393 = 13 + 23 + 33
+ K + 403 – (23 + 43 + 63 + ... + 403 )
2
40 × 41
3
3
=
− 8(1 + 2
2
+ 33 + K + 203 )
= (20 × 41)2 − 8
20 × 21
2
= 202 [412 − 2(21)2 ]
= 319600
2
28
40
27 Let a1 = a and d = common difference
Q a1 + a5 + a9 + L + a49 = 416
∴ a + (a + 4d ) + (a + 8d )
+ … (a + 48d ) = 416
13
(2a + 48d ) = 416
⇒
2
⇒
a + 24d = 32
…(i)
Also, we have a9 + a43 = 66
∴ a + 8d + a + 42d = 66
⇒
2a + 50d = 66
⇒
a + 25d = 33
…(ii)
Solving Eqs. (i) and (ii), we get
a = 8 and d = 1
Now,
a12
+
a22
+
+L+
a32
= 140m
2
a17
8 + 9 + 10 + … + 24 = 140m
2
2
2
2
⇒ (12 + 22 + 32 + … + 242 ) − (12 + 22
+ 32 + … + 72 ) = 140m
⇒
⇒
⇒
⇒
⇒
24 × 25 × 49 7 × 8 × 15
−
= 140m
6
6
3×7× 8× 5
(7 × 5 − 1) = 140m
6
7 × 4 × 5 × 34 = 140m
140 × 34 = 140m
m = 34
28 We have, f ( x ) = ax2 + bx + c
Now, f ( x + y ) = f ( x ) + f ( y ) + xy
Put y = 0 ⇒ f ( x ) = f ( x ) + f (0) + 0
⇒
f (0) = 0
⇒
c =0
Again, put y = − x
∴
f (0) = f ( x ) + f (− x ) − x2
⇒
0 = ax2 + bx + ax2 − bx − x2
1
⇒
2ax2 − x2 = 0 ⇒ a =
2
Also, a + b + c = 3
1
5
⇒
+ b + 0 = 3⇒ b =
2
2
x2 + 5x
f ( x) =
∴
2
n2 + 5n 1 2
5
Now, f (n ) =
= n + n
2
2
2
10
1 10 2
5 10
∴
∑ f (n ) = 2 ∑ n + 2 ∑ n
n =1
n =1
n =1
12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + …
A = sum of first 20 terms
B = sum of first 40 terms
∴ A = 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52
+ 2 ⋅ 62 + … + 2 ⋅ 202
A = (1 + 2 + 3 + … + 202 ) + (22 + 42
2
2
2
+ 62 + … + 202 )
A = (12 + 22 + 32 + … + 202 )
+ 4(12 + 22 + 32 + …+102 )
20 × 21 × 41 4 × 10 × 11 × 21
+
6
6
20 × 21
20 × 21 × 63
A=
(41 + 22) =
6
6
Similarly,
A=
B = (12 + 22 + 32 + … + 402 ) + 4(12
+ 22 +… + 202 )
40 × 41 × 81 4 × 20 × 21 × 41
+
6
6
40 × 41 × 123
40 × 41
B =
(81 + 42) =
6
6
Now, B − 2 A = 100λ
40 × 41 × 123
∴
6
2 × 20 × 21 × 63
−
= 100λ
6
40
⇒
(5043 − 1323) = 100λ
6
40
⇒
× 3720 = 100λ
6
⇒
40 × 620 = 100λ
40 × 620
⇒
λ=
= 248
100
B =
31 Write the nth term of the given series
and simplify it to get its lowest
form.Then, apply, S n = ΣT n .
Given series is
13 13 + 23 13 + 23 + 33
+
+
+ K∞
1
1+ 3
1+ 3+ 5
Let T n be the nth term of the given
series.
∴ Tn =
1 10 × 11 × 21 5 10 × 11
⋅
+ ×
2
6
2
2
385 275 660
=
+
=
= 330
2
2
2
=
29 Series (2)2 + 2(4)2 + 3(6)2 + K
13 + 23 + 33 + ... + n3
1 + 3 + 5 + K upto n terms
2
n(n + 1)
2
2
= (n + 1)
=
2
4
n
(n + 1)2 1
=
4
4
n =1
9
= 4 {1 ⋅ 12 + 2 ⋅ 22 + 3 ⋅ 32 + K }
∴
30 We have,
Now, S 9 = ∑
[(22 + 32 + ... + 102 ) + 12 − 12 ]
Tn = 4n ⋅ n
2
and S n = ΣT n = 4Σ n3 = 4
Now, S 10 = [10 ⋅ (10 + 1)]
2
= (110)2 = 12100
2
n(n + 1)
2
=
1 10(10 + 1)(20 + 1)
− 1
4
6
384
=
= 96
4
2n + 1
32 T n =
(12 + 22 + ... + n2 )
2n + 1
6
=
=
n (n + 1 ) (2n + 1 ) n (n + 1 )
6
1
1
= 6 −
n ( n + 1 )
1 1
1 1
T1 = 6 − , T2 = 6 − , …
2 3
1 2
1
1
T11 = 6
−
11 12
1
1 6 × 11 11
∴S = 6 −
= 12 = 2
1
12
33 n th term of the series is
1
2
=
n (n + 1 ) n (n + 1 )
2
1
1
⇒ Tn = 2 −
n n + 1
Tn =
1 1
1 1
⇒ T1 = 2 − , T2 = 2 − ,
2 3
1 2
1 1
1
1
T3 = 2 − , ..., T10 = 2
−
3 4
10 11
∴ S 10 = T1 + T2 + K + T20
1 − 1 + 1 − 1 + 1 − 1
2 2 3 3 4
= 2
1
1
+K+
−
10 11
1
= 2 1 −
11
10 20
= 2⋅
=
11 11
SESSION 2
1 Let S = 12 + 32 + 52 + K + 252
= (12 + 22 + 32 + 42 + K + 252 )
− (22 + 42 + 62 + K + 242 )
= (1 + 2 + 3 + 42 + K + 252 )
2
2
2
− 22 (1 + 22 + 32 + K + 122 )
25(25 + 1) (2 × 25 + 1)
=
6
12 (12 + 1) (2 × 12 + 1)
− 4×
6
25 × 26 × 51 4 × 12 × 13 × 25
=
−
6
6
= 25 × 13 × 17 − 4 × 2 × 13 × 25
= 5525 − 2600 = 2925
2 Now, f (2) = f (1 + 1)
= f (1) ⋅ f (1) = 22 and f (3) = 23
Similarly, f (n ) = 2n
∴ 16(2n − 1) =
n
∑ f (a +
r =1
a
r )=
n
∑2
a+r
r =1
= 2 (2 + 22 + K + 2n )
29
DAY
2n − 1
= 2a ⋅ 2
2−1
[GP series]
1
= 1 − cos 2 θ sin2 θ
z
xy − 1
1
=1−
=
xy
xy
and
= 2a + 1 (2n − 1)
2a + 1 = 16 = 24
⇒
∴
⇒
∴
a= 3
3 Let GP be a, ar , ar ,K,|r | < 1.
2
According to the question,
a
7 a2
147
= ,
=
1 − r 2 1 − r2
16
On eliminating a, we get
2
147
7
(1 − r 2 ) = (1 − r ) 2
2
16
⇒ 3(1 + r ) = 4 (1 − r ) ⇒ r =
∴ Sum of cubes
a3
=
=
1 − r3
(3)3
1
1 −
7
3
1
,a = 3
7
1029
=
38
4 Let S = 5 + 552 + 555
…(i)
+…
13 13
133
S
5
55
and
...(ii)
=
+ 3 + …
13 132
13
On subtracting Eq. (ii) from Eq. (i), we
get
12
5
50
500
+…
S =
+
+
13
13 132
133
10
which is a GP with common ratio .
13
13 5
10 65
∴
S =
×
÷ 1 −
=
12 13
13 36
a
QS ∞ = 1 − r
5 Here, T1 = S1 = 2(1 ) + 3(1 )2 = 5
T2 = S 2 − S 1 = 16 − 5 = 11
[QS 2 = 2(2) + 3(2) = 16]
2
T3 = S 3 − S 2 = 33 − 16 = 17
[Q S 3 = 2(3) + 3 (3)2 = 33 ]
x y = x yz − z
x yz = x y + z = x + y + z
7 Let S = 14 + 14 + 14 + K + to ∞
1
3
5
1
1
1
π4
Since, 4 + 4 + 4 + K to ∞ =
90
1
2
3
1
1
1
∴ 4 + 4 + 4 + K to ∞
1
3
5
1
1
1
π4
+ 4 + 4 + 4 + K to ∞ =
2
90
4
6
1
1
1
⇒ 4 + 4 + 4 + K to ∞
1
3
5
1 1
1
1
π4
+ 4 + 4 + K + to ∞ =
4
4
90
2 1
2
3
1 π4
π4
⇒S +
⋅
=
16 90
90
1
1
1
π4
Q 4 + 4 + 4 + K to ∞ =
90
2
3
1
+
15 π 4
π4
1
π4
⇒ S =
=
1 −
=
90
16 16 × 90 96
8 Let an = a r
x=
1
1 − cos 2 θ
=
1
sin2 θ
y =
1
cos 2 θ
1
1
Now, +
=1
x
y
⇒
and z =
x + y = xy
.
1 1 − 1
a
r
and S n′ =
1
1−
r
1
1
Q first term of a is a
n
1
and common ratio is
r
1 (r n − 1)
a
=
⋅r
r n (r − 1)
=
1
1 − cos 2 θ sin2 θ
1− r
1
⋅
1 − r a⋅ r n − 1
n
1 − rn 1
a (1 − r n ) 1
⋅
=
⋅
1 − r an
1− r
a an
1
= Sn ⋅
a1 an
=
S n = a1 an S n′
[Q sin2 θ + cos 2 θ = 1]
+ ... to 10 terms
=
1
(8 + 12 + 16 + 202 + 242
2
52
2
2
+ ... to 10 terms)
=
=
42
52
42
(2 + 3 + 4 + 5
2
2
2
2
+ ... to 10 terms)
(22 + 32 + 42 + 52 + ... + 112 )
52
16 2
=
((1 + 22 + ... + 112 ) − 12 )
25
16 11 ⋅ (11 + 1) (2 ⋅ 11 + 1)
=
− 1
25
6
16
16
=
(506 − 1) =
× 505
25
25
16
16
m=
× 505
⇒
5
25
⇒
m = 101
10 Given, m is the AM of l and n.
∴
l + n = 2m
and G1 , G2 , G3 are geometric means
between l and n.
So, l ,G1 ,G2 ,G3 , n are in GP.
n
n = lr 4 ⇒ r =
l
1/4
Now, G14 + 2G24 + G34 = (lr )4
+ 2(lr 2 )4 + (lr 3 )4
= l 4 × r 4 (1 + 2r 4 + r 8 ) = l 4 × r 4 ( r 4 + 1)2
2
= l4 ×
nn+ l
2
2
= ln × 4 m = 4 l m n
l l
[Q n + l = 2m]
11 Given series is a geometric series with
a = 2 + 1 and r = 2 − 1.
∴ Required sum
2+1
2+1
a
=
=
=
1 − r 1 − ( 2 − 1) 2 − 2
=
=
( 2 + 1) (2 +
2)
(2 − 2 ) (2 +
2)
2 2+ 2+ 2+
4−2
2
=
4+ 3 2
2
12 Clearly, the common terms of the given
the series.
Then, we have
2
2
∴ G1 = lr , G2 = lr 2 , G3 = lr 3 ;
9 Let S10 be the sum of first ten terms of
Similarly,
2
n
⇒
,
2
Let r be the common ratio of this GP.
a (1 − r n )
Then, S n =
1− r
Hence, sequence is 5, 11, 17.
∴ a = 5and d = 6
For new AP, A = 5, D = 2 × 6 = 12
n
S ′ n = [2 × 5 + (n − 1 )12]
∴
2
= 6 n2 − n
6 Sum of three infinite GP’s are
n −1
2
8
12
16
24
= + + + 42 +
5
5
5
5
2
3
2
1
S 10 = 1 + 2 + 3
5
5
5
2
2
4
+ 42 + 4 + ... to 10 terms
5
sequences are
31, 41, 51, ...
Now, 100th term of 1, 11, 21, 31, ... is
1 + 99 × 10 = 991
and 100th term of 31, 36, 41, 46, ... is
31 + 99 × 5 = 526 .
Let the largest common term be 526.
Then, 526 = 31 + (n − 1) 10
⇒ (n − 1) 10 = 495
30
40
⇒
n − 1 = 49.5
⇒
n = 50.5
But n is an integer, n = 50.
Hence, the largest common term is
31 + (50 − 1) 10 = 521 .
In this case,
(a − c )2 = (a + c )2 − 4ac = 0
b 2 = ac
Case II When a + c = 1 and ac = −
…(i)
Also, as x is A between a and b
a+ b
…(ii)
x=
∴
2
b +c
Similarly, y =
…(iii)
2
a c
2a
2c
Now, consider +
=
+
x y
a+ b b + c
[using Eqs. (ii) and (iii)]
ab + ac + ac + bc
=2
2
ab + ac + b + bc
ab + bc + 2ac
=2
ab + bc + 2ac
[using Eq. (i)]
=2
14 Since, a, b, c are in AP
∴
2b = a + c
…(i)
Also, as a2 , b 2 and c 2 are in GP
∴
b 4 = a2c 2
Q
a+ b + c =
3
2
1
b =
⇒
2
⇒
a+ c =1
1
1
and ac =
or −
4
4
∴
3b =
Let S=(1)+(1+2+4)+(4+6+9)
⇒
a=c
But a ≠ c , as a < c .
13 Since, a, b, c are in GP.
∴
17 Statement I
…(ii)
+(9+12+16)+... +(361+380+400)
= (0+0+1)+(1+2+4)+(4+6+9)
1
4
In this case, (a − c )2 = 1 + 1 = 2
⇒
a−c = ± 2
But a < c , a − c = − 2
On solving a + c = 1
and a − c = − 2, we get
1
1
.
a= −
2
2
[using Eq. (i)]
[using Eq. (i)]
Case I When a + c = 1 and ac =
1
4
=
⇒
10
∑ Sr
r =1
2
= 22 + 32 + K + 112
= 12 + 22 + 32 + K + 112 − 1
=
11 × 12 × 23
− 1 = 505
6
+ (r − 1)r + (r )2 ]
r 3 − (r − 1)3
r =1 r − ( r − 1)
n
∑
[Q(a3 − b 3 ) = (a − b )(a2 + ab + b 2 )]
=
1
⇒ [(15a − 3b )2 + (3b − 5 c )2
2
+ (5c − 15a)2 ] = 0
1
is common ratio
r +1
r
Sr =
=r +1
1
1−
r +1
2
r =1
− (15a)(5c ) − (15a)(3b ) − (3b )(5c ) = 0
⇒ 15a = 3b , 3b = 5 c and 5 c = 15a
∴ 15a = 3b = 5c
a b c
= = = λ (say)
⇒
1 5 3
⇒
a = λ, b = 5λ, c = 3λ
Hence, b, c and a are in AP.
n
∑ [(r − 1)
Sn =
− 75ac − 45ab − 15bc = 0
⇒ (15a)2 + (3b )2 + (5c )2
first term and
[using Eq. (ii)]
Now, the sum to n terms of the series is
15 We have, 225a2 + 9b2 + 25c 2
16 Here, S r is sum of an infinite GP, r is
3
2
+(9+12+16) +... +(361+380+400)
Now, we can clearly observe the
elements in each bracket.
The general term of the series is
T r = (r − 1)2 + (r − 1) r + (r 2 )
n
∑ [r
3
− (r − 1)3 ]
r =1
= (13 − 03 ) + (23 − 13 ) + (33 − 23 )
+ ... + [n3 − (n − 1)3 ]
Rearranging the terms, we get
S n = − 03 + (13 − 13 ) + (23 − 23 )
+ (33 − 33 ) + ... + [(n − 1)3 − ( n − 1)3 ] + n3
= n3
⇒ S 20 = 8000
Hence, Statement I is correct.
Statement II We have, already proved
in the Statement I that
Sn =
n
∑ (r
3
− (r − 1)3 ) = n3
r =1
Hence, Statement II is also correct and
it is a correct explanation for
Statement I.
DAY FOUR
Quadratic
Equation and
Inequalities
Learning & Revision for the Day
u
u
u
u
Quadratic Equation
u
Relation between Roots and
Coefficients
Formation of an Equation
u
Transformation of Equations
u
u
Maximum and Minimum
Value of ax 2 + bx + c
Sign of Quadratic Expression
Position of Roots
Inequalities
u
u
Arithmetic-GeometricHarmonic Mean
Inequality
Logarithm Inequality
Quadratic Equation
●
●
●
An equation of the form ax2 + bx + c = 0, where a ≠ 0, a, b and c, x ∈ R, is called a real
quadratic equation. Here a, b and c are called the coefficients of the equation.
PRED
MIRROR
The quantity D = b – 4ac is known as the discriminant of the equation ax + bx + c = 0
−b ± D
and its roots are given by x =
2a
2
2
An equation of the form az2 + bz + c = 0, where a ≠ 0, a, b and c, z ∈C (complex) is called a
−b ± D
complex quadratic equation and its roots are given by z =
.
2a
Nature of Roots of Quadratic Equation
Let a, b , c ∈ R and a ≠ 0, then the equation ax + bx + c = 0
2
(i) has real and distinct roots if and only if D > 0.
(ii) has real and equal roots if and only if D = 0.
(iii) has complex roots with non-zero imaginary parts if and only if D < 0.
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
32
40
Some Important Results
(i) If p + iq (where, p, q ∈ R, q ≠ 0) is one root of
ax2 + bx + c = 0, then second root will be p − iq
(ii) If a, b , c ∈ Q and p + q is an irrational root of
ax + bx + c = 0, then other root will be p − q .
2
(iii) If a, b , c ∈Q and D is a perfect square,
then ax2 + bx + c = 0 has rational roots.
(iv) If a = 1, b , c ∈ I and roots of ax2 + bx + c = 0 are
rational numbers, then these roots must be integers.
(v) If the roots of ax2 + bx + c = 0 are both positive, then
the signs of a and c should be like and opposite to the
sign of b.
(vi) If the roots of ax2 + bx + c = 0 are both negative, then
signs of a, b and c should be like.
(vii) If the roots of ax2 + bx + c = 0 are reciprocal to each
other, then c = a.
(viii) In the equation ax2 + bx + c = 0 (a, b , c ∈ R),
if a + b + c = 0, then the roots are 1, c and if
a
c
a − b + c = 0, then the roots are −1 and − .
a
Relation between Roots
and Coefficients
Quadratic Roots
If α and β are the roots of quadratic equation ax2 + bx + c = 0;
b
a ≠ 0, then sum of roots = α + β = −
a
c
and product of roots = αβ = .
a
And, also ax2 + bx + c = a( x − α )( x − β)
Formation of an Equation
Quadratic Equation
If the roots of a quadratic equation are α and β, then the
equation will be of the form x2 − (α + β) x + αβ = 0.
Cubic Equation
If α , β and γ are the roots of the cubic equation, then the
equation will be form of
x3 − (α + β + γ) x2 + (αβ + βγ + γα ) x − αβγ = 0.
Transformation of Equations
Let the given equation be
a0 x n + a1 x n −1 + ... + an −1 x + an = 0
…(A)
Then, the equation
(i) whose roots are k (≠ 0) times roots of the Eq. (A),
x
is obtained by replacing x by in Eq. (A).
k
(ii) whose roots are the negatives of the roots of Eq. (A),
is obtained by replacing x by − x in Eq. (A).
(iii) whose roots are k more than the roots of Eq. (A),
is obtained by replacing x by ( x − k ) in Eq. (A).
(iv) whose roots are reciprocals of the roots of Eq. (A),
is obtained by replacing x by 1/x in Eq. (A) and then
multiply both the sides by x n .
Maximum and Minimum
Value of ax2 + bx + c
(i) When a > 0, then minimum value of ax2 + bx + c is
−D
4ac − b 2
−b
or
at x =
4a
4a
2a
(ii) When a < 0, then maximum value of ax2 + bx + c is
−D
−b
4ac − b 2
or
at x =
4a
2a
4a
Cubic Roots
If α , β and γ are the roots of cubic equation
b
ax3 + bx2 + cx + d = 0; a ≠ 0, then α + β + γ = −
a
c
βγ + γα + αβ =
a
d
and
αβγ = −
a
Common Roots (Conditions)
Suppose that the quadratic equations are ax + bx + c = 0 and
a′ x2 + b ′ x + c′ = 0.
2
(i) When one root is common, then the condition is
(a′ c − ac′ )2 = (bc′ − b ′ c)(ab ′ − a′ b ).
(ii) When both roots are common, then the condition is
a b
c
=
= .
a′ b ′ c′
Sign of Quadratic Expression
Let f ( x) = ax2 + bx + c, where a, b and c ∈ R and a ≠ 0.
(i) If a > 0 and D < 0, then f ( x) > 0, ∀x ∈ R.
(ii) If a < 0 and D < 0, then f ( x) < 0, ∀x ∈ R.
(iii) If a > 0 and D = 0, then f ( x) ≥ 0, ∀x ∈ R.
(iv) If a < 0 and D = 0, then f ( x) ≤ 0, ∀x ∈ R.
(v) If a > 0, D > 0 and f ( x) = 0 have two real roots α and
β, where (α < β), then f ( x) > 0, ∀ x ∈(− ∞, α ) ∪ (β, ∞) and
f ( x) < 0, ∀ x ∈(α , β).
(vi) If a < 0, D > 0 and f ( x) = 0 have two real roots α and β,
where (α < β), then f ( x) < 0, ∀ x ∈ (−∞, α ) ∪ (β, ∞) and
f ( x) > 0, ∀ x ∈(α , β).
DAY
33
Position of Roots
Let ax + bx + c = 0 has roots α and β. Then, we have the
following conditions:
2
(i) with respect to one real number (k ).
Situation
Required conditions
α<β< k
D ≥ 0, af (k ) > 0, k > − b / 2a
k<α<β
D ≥ 0, af (k ) > 0, k < − b /2a
α< k<β
D > 0, af (k ) < 0
(ii) with respect to two real numbers k1 and k2 .
Situation
Required conditions
k1 < α < β < k2
D ≥ 0, af (k1 ) > 0,
af (k2 ) > 0, k1 < − b /2a < k2
α < k1 < k2 < β
D > 0, af (k1 ) < 0, af (k2 ) < 0
k1 < α < k2 < β
D > 0, f (k1 ) f (k2 ) < 0
Inequalities
Let a and b be two real numbers. If a − b is negative, we say
that a is less than b ( a < b ) and if a − b is positive, then a is
greater than b (a > b ). This shows the inequalities concept.
Important Results on Inequalities
(i) If a > b , then a ± c > b ± c, ∀ c ∈ R.
a
(a) for m > 0, am > bm, >
m
a
(b) for m < 0, am < bm, <
m
(iii) (a) If a > b > 0 , then
•a >b
2
b
m
b
m
•| a| > |b |
(b) If a < b < 0, then
• a2 > b 2
The Arithmetic-Geometric-Harmonic Mean of positive real
numbers is defined as follows
Arithmetic Mean ≥ Geometric Mean ≥ Harmonic Mean
(i) If a, b > 0 then
• | a| > |b |
(iv) If a < 0 < b , then
Logarithm Inequality
If a is a positive real number other than 1 and ax = m, then x
is called the logarithm of m to the base a, written as log a m. In
log a m, m should always be positive.
(i) If m ≤ 0, then log a m will be meaningless.
(ii) log a m exists, if m , a > 0 and a ≠ 1.
Important Results on Logarithm
(i) alog a x = x ; a > 0, ≠ 1, x > 0
log b x
=x
log b a
; a , b > 0, ≠ 1, x > 0
(iii) log a a = 1, a > 0, ≠ 1
1
(iv) log a x =
; x , a > 0, ≠ 1
log x a
1 1
• <
a b
1 1
• >
a b
(a) a2 > b 2 , if| a| > |b | (b) a2 < b 2 , if| a| < |b |
(v) If a < x < b and a, b are positive real numbers, then
a2 < x2 < b 2 .
(vi) If a < x < b and a is negative number and b is positive
number, then
(a) 0 ≤ x2 < b 2 , if |b | >| a | (b) 0 ≤ x2 < a2 , if | a | >|b |
(vii) If ai > b i > 0 , where i = 1, 2, 3, ..., n, then
a1 a2 a3 ... an > b1b2b3 ... b n .
(viii) If ai > b i , where i = 1, 2, 3, ..., n, then
a1 + a2 + a3 + ... + an > b1 + b2 + ... + b n .
(ix) If| x | < a, then
(a) for a > 0, − a < x < a.
(b) for a < 0, x ∈ φ .
a+b
2
≥ ab ≥
1 1
2
+
a b
(ii) If ai > 0 , where i = 1, 2, 3, K , n, then
a 1 + a 2 + ... + an
n
≥ (a1 ⋅ a2 ... an )1 / n ≥
1
1
1
n
+
+ ... +
a1
a2
an
(ii) a
(ii) If a > b , then
2
Arithmetic-GeometricHarmonic Mean Inequality
(v) log a x = log a b logb x =
logb x
; a, b > 0, ≠ 1, x > 0
logb a
(vi) For x > 0; a > 0, ≠ 1
1
(a) log an ( x) = log a x
n
m
(b) log an x m = log a x
n
(vii) For x > y > 0
(a) log a x > log a y, if a > 1
(b) log a x < log a y, if 0 < a < 1
(viii) If a > 1 and x > 0, then
(a) log a x > p ⇒ x > a p
(b) 0 < log a x < p ⇒ 0 < x < a p
(ix) If 0 < a < 1 , then
(a) log a x > p ⇒ 0 < x < a p
(b) 0 < log a x < p ⇒ a p < x < 1
34
40
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 If 3x 2 − 7x − 30 + 2x 2 − 7x − 5 = x + 5, then x is equal to
(a) 2
(b) 3
(c) 6
(d) 5
2 The number of solutions for equation x − 5 x + 6 = 0 is
2
(a) 4
(b) 3
(c) 2
(d) 1
2
3 The roots of the equation 2x − 1 − 3 2x − 1 + 2 = 0 are
1 1
1 3
3 1
1
3
(a) − , 0, (b) − , 0, (c) − , , 0,1 (d) − , 0,1,
2
2 2
2 2
2 2
2
4 The product of all the values of x satisfying the equation
(5 + 2 6)
x2 − 3
(a) 4
+ (5 − 2 6)
x2 − 3
(b) 6
(c) 8
(d) 19
5 The root of the equation 2(1 + i )x − 4( 2 − i )x − 5 − 3i = 0,
where i = −1, which has greater modulus, is
5 − 3i
(b)
2
3+i
(c)
2
3i + 1
(d)
2
6 x 2 + x + 1 + 2k ( x 2 − x − 1) = 0 is perfect square for how
(b) 0
(c) 1
(d) 3
7 If the roots of (a + b )x − 2(bc + ad )x + c + d 2 = 0 are
2
2
2
2
equal, then
(a)
a c
=
b d
(b)
a b
a b
+ = 0 (c) =
c d
d c
(d) a + b = c + d
8 The least value of α for which tan θ and cot θ are roots of
the equation x 2 + ax + 1 = 0, is
(a) 2
(a) four real roots
(b) exactly two real roots
(c) either two or four real roots (d) atmost two real roots
15 The rational values of a in ax 2 + bx + 1 = 0 if
root, are
(a) a = 13, b = − 8
(c) a = 13, b = 8
(b) 1
(c) 1/2
(d) 0
9 If one root of the equation x 2 − λx + 12 = 0 is even prime
16 If 1 − i , is a root of the equation x 2 + ax + b = 0, where
(a) 1,−1
(c) 3 ,−3
(b) 16
(c) 24
(d) 32
10 If a + b + c = 0, then the roots of the equation
4ax 2 + 3bx + 2c = 0, where a , b, c ∈ R are
(a) real and distinct
(c) real and equal
x 2 − 30x + p = 0 is the square of the other, is/are
(b) 125 and − 216
(d) Only 216
(a) Only 125
(c) 125 and 215
variable x has real roots, then β lies in the interval
(b) (− π, 0)
π π
(c) − ,
2 2
12 If ax 2 + 2bx − 3 c = 0 has no real root and
then the range of c is
(a) (−1, 1 )
(c) (0, ∞)
(d) (0, π)
3c
< a + b,
4
(b) (0, 1)
(d) (− ∞, 0)
13 If a , b and c are real numbers in AP, then the roots of
ax 2 + bx + c = 0 are real for
(a) all a and c
c
(c)
−7 ≥ 4 3
a
x −m
x +n
are
=
mx + 1 nx + 1
reciprocal to each other, then
(a) n = 0
(b) m = n
(c) m + n = 1
(d) m 2 + n 2 = 1
19 Let α and α 2 be the roots of x 2 + x + 1 = 0, then the
equation whose roots are α 31 and α 62 , is
(a) x 2 − x + 1 = 0
(c) x 2 + x + 1 = 0
(b) x 2 + x − 1 = 0
(d) x 60 + x 30 + 1 = 0
20 If α and β are the roots of x 2 − a( x − 1) + b = 0, then the
value of
(a)
1
1
2
is
+ 2
+
α − aα β − aβ a + b
4
a+b
2
(b)
1
a+b
(c) 0
(d) −1
21 The value of a for which the sum of the squares of the
roots of the equation x 2 − (a − 2)x − a − 1 = 0 assume the
j
least value is
AIEEE 2005
(b) imaginary
(d) infinite
11 The equation (cos β − 1) x 2 + (cos β)x + sin β = 0 in the
(a) (0, 2 π)
(b) 2,− 2
(d) None of these
17 The values of p for which one root of the equation
while x 2 + λx + µ = 0 has equal roots, then µ is equal to
(a) 8
1
is a
4+ 3
(b) a = − 13, b = 8
(d) a = − 13, b = − 8
18 If the roots of the quadratic equation
many value of k
(a) 2
ac ≠ 0, then the equation P ( x ) ⋅ Q ( x ) = 0 has
a, b ∈ R , then the values of a and b are
= 10 is
2
3 − 5i
(a)
2
14 If P ( x ) = ax 2 + bx + c and Q ( x ) = − ax 2 + dx + c = 0 ;
(b) no a and c
a
(d)
+7 ≥2 3
c
(a) 2
(b) 3
(c) 0
(d) 1
22 If α and β be the roots of the equation
2x 2 + 2(a + b )x + a 2 + b 2 = 0, then the equation whose
roots are (α + β )2 and (α − β )2 , is
(a) x 2 − 2abx − (a 2 − b 2 )2 = 0 (b) x 2 − 4abx − (a 2 − b 2 )2 = 0
(c) x 2 − 4abx + (a 2 − b 2 )2 = 0(d) None of these
23 Let α , β be the roots of x 2 − 2x cos φ + 1 = 0, then the
equation whose roots are α n and β n , is
(a) x 2 − 2 x cosnφ − 1 = 0
(c) x 2 − 2 x sinnφ + 1 = 0
(b) x 2 − 2 x cosnφ + 1 = 0
(d) x 2 + 2 x sinnφ − 1 = 0
24 The harmonic mean of the roots of the equation
( 5 + 2 )x 2 − ( 4 + 5 )x + 8 + 2 5 = 0 is
(a) 2
(b) 4
(c) 6
(d) 8
DAY
35
25 If the ratio of the roots of λx 2 + µx + ν = 0 is equal to the
ratio of the roots of x + x + 1 = 0, then λ , µ and ν are in
2
(a) AP
(c) HP
(a) 1 : 2 : 3
1
1
1
26 If the roots of the equation
+
= are equal in
x +p x +q r
magnitude but opposite in sign, then the product of the
roots is
(b) − (p 2 + q 2 )
k +1
k+2
and
, then (a + b + c )2 is equal to
k
k +1
(c) b 2 − 4ac
(d) b 2 − 2ac
such that β < α < 0, then the quadratic equation whose
roots are α , β , is given by
(a) a x + b x + c = 0
(c) a x − b x + c = 0
2
(b) ax − b x + c = 0
2
(d) a x
2
+ b| x + c = 0
(ωα + ω 2β)(ω 2α + ωβ)
is equal to
α 2 β2
+
β
α
q
p
(b) αβ
(c) −
(d) ω
(c) {− 2, 5 }
j
p
= 0,
4
(c) 4, 3
(d) −6, − 1
(c) 24
(d) 44
π
P
Q
and tan are the roots of the
2
2
2
equation ax 2 + bx + c = 0, then
(b) b = c + a
(d) b = c
34 If the equation k ( 6x 2 + 3) + rx + 2x 2 − 1 = 0 and
6 k ( 2x 2 + 1) + px + 4x 2 − 2 = 0 have both roots common,
then 2r − p is equal to
(a) 2
(b) 1
(c) 0
(b) 0 < c < b 2
(d) | c | > | b | 2
(a) a1 + a2 + … + an
(c) n (a1 + a2 + … + an )
(b) 2 (a1 + a2 + a3 + ... + an )
(d) None of these
39 If the roots of the equation bx 2 + cx + a = 0 is imaginary,
then for all real values of x, the expression
3b 2 x 2 + 6bcx + 2c 2 is
j
AIEEE 2009
(b) less than 4ab
(d) less than −4ab
(d) k
(b) a < − 5
(d) 2 < a < 5
1
is non-negative for all
x
positive real x, then the minimum value of a must be
(c)
33 In ∆PQR , R = . If tan
(a) a = b + c
(c) c = a + b
(d) b = a + c
2
( x − a1 )2 + ( x − a 2 )2 +… + ( x − a n )2 assumes its least
value at x =
(d) {3, − 5 }
roots of x 2 + 3x + 4 = 0, then (α + γ )(α + δ )(β + γ )
(β + δ ) is equal to
(b) 18
2
(a) | c| < | b| 2
(c) | c| < | b| 2
(a) 0
32 If α and β are the roots of x 2 + x + 2 = 0 and γ,δ are the
(a) −18
(c) c = − a
37 If f ( x ) = x + 2bx + 2c and g ( x ) = − x − 2cx + b 2 such
JEE Mains 2013
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).
j
The correct roots of equation are
AIEEE 2011
(b) 6, 1
(b) b = − c
2
41 If the expression ax − 1 +
31 Sachin and Rahul attemped to solve a quadratic
(a) −4, − 3
(a) a = − b
(a) − 5 < a < 2
(c) a > 5
p
q
such that | α − β | = 10, then p belongs
(b) {− 3, 2 }
ax 2 + bx + c = 0 by 1 is 2x 2 + 8x + 2 = 0, then
40 If x 2 + 2ax + 10 − 3a > 0 for all x ∈ R , then
30 If α and β are roots of the equation x 2 + px + 3.
(a) {2, − 5 }
(d) 3 : 1 : 2
36 The equation formed by decreasing each root of
(a) greater than 4ab
(c) greater than −4ab
29 If α and β be the roots of x 2 + px + q = 0, then
(a) −
(c) 1 : 3 : 2
38 If a ∈ R and a1, a 2 , a 3 … , a n ∈ R , then
28 If α and β are the roots of the equation ax 2 + bx + c = 0
2
(b) 3 : 2 : 1
AIEEE 2012
that min f ( x ) > max g ( x ), then the relation between b and
c is
(d) −pq
27 If the roots of the equation ax 2 + bx + c = 0 of the form
(a) 2b 2 − ac (b) ∑a 2
a, b, c ∈ R , have a common root, then a : b : c is
j
(b) GP
(d) None of these
(a) −2 (p 2 + q 2 )
− (p 2 + q 2 )
(c)
2
35 If the equations x 2 + 2x + 3 = 0 and ax 2 + bx + c = 0,
1
2
1
4
(d) None of these
42 The number of real solutions of the equation
x
9
2
= − 3 + x − x is
10
(a) 0
(c)2
(b) 1
(d) None of these
43 If α and β be the roots of the quadratic equation
ax 2 + bx + c = 0 and k be a real number, then the
condition, so that α < k < β is given by
(b) ak 2 + bk + c = 0
(d) a 2 k 2 + abk + ac < 0
(a) ac > 0
(c) ac < 0
44 2x − 3 < x + 5 , then x belongs to
(a) (−3,5)
(b) (5, 9)
2
(c) − , 8
3
45 The least integral value α of x such that
satisfies
(a) α + 3 α − 4 = 0
(c) α 2 − 7α + 6 = 0
2
2
(d) −8,
3
x −5
> 0,
x 2 + 5x − 14
j
JEE Mains 2013
(b) α − 5 α + 4 = 0
(d) α 2 + 5 α − 6 = 0
2
36
40
46 The solution set of
| x − 2 | −1
≤ 0 is
|x −2| −2
51 If a, b, c are positive real numbers such that a + b + c = 1,
then the greatest value of (1 − a ) (1 − b ) (1 − c ), is
(a) [0, 1] ∪ (3, 4)
(b) [0, 1] ∪ [3, 4]
(c) [−1, 1) ∪ (3, 4]
(d) None of these
47 Number of integral solutions of
(a) 0
(b) 1
x +2 1
> is
x2 +1 2
(c) 2
(d) 3
48 If the product of n positive numbers is 1, then their sum is
(a) a positive integer
1
(c) equation to n +
n
(b) divisible by n
(d) greater than or equal to n
49 The minimum value of P = bcx + cay + abz , when
xyz = abc, is
(a) 3abc
(c) abc
(a)
1
27
(b)
(c)
4
27
(d) 9
52 If a, b, c, d are positive real numbers such that
a + b + c + d = 2, then M = (a + b ) (c + d ) satisfies the
relation
(a) 0 ≤ M ≤ 1 (b) 1 ≤ M ≤ 2
(c) 2 ≤ M ≤ 3
(d) 3 ≤ M ≤ 4
53 log2 ( x − 3x + 18) < 4, then x belongs to
2
(a) (1, 2)
(c) (1, 16)
(b) (2, 16)
(d) None of these
54 If log0. 3 ( x − 1) > log0. 09 ( x − 1), then x lies in
(b) (− ∞, 1)
(d) None of these
(a) (1, 2)
(c) (2, ∞)
(b) 6abc
(d) 4abc
55 What is the solution set of the following inequality?
x + 5
logx
>0
1 − 3x
50 If a, b and c are distinct three positive real numbers, then
1 1 1
(a + b + c ) + + is
a b c
(a) > 1
(c) < 9
8
27
(a) 0 < x <
(b) > 9
(d) None of these
(c)
1
3
(b) x ≥ 3
1
< x<1
3
(d) None of these
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 Let S = {x ∈ R : x ≥ 0 and 2
x − 3 + x ( x − 6) + 6 =.0
Then, S
(a)
(b)
(c)
(d)
j
JEE Mains 2018
2 The roots of the equation 2
(a) 1 − log2 3, 2
(c) 2 , − 2
⋅3
3 x / ( x − 1)
= 9 are given by
2
(b) log2 , 1
3
(log 3)
(d) − 2 , 1 −
(log 2)
a − 2a 8
is equal
a n = α − β for n ≥ 1, then the value of 10
2a 9
n
to
(a) 6
j
(b) − 6
4 If x 2 − 5x + 1 = 0, then
(a) 2424
(c) 2525
(c) 3
(b) real and equal
(d) None of these
6 Let α and β be the roots of equation px 2 + qx + r = 0,
3 Let α and β be the roots equation x 2 − 6x − 2 = 0. If
n
( x − a )( x − c ) + 2 ( x − b )( x − d ) = 0 are
(a) real and distinct
(c) imaginary
is an empty set
contains exactly one element
contains exactly two elements
contains exactly four elements
x + 2
5 If a < b < c < d , then the roots of the equation
JEE Mains 2015
(d) − 3
x 10 + 1
is equal to
x5
(b) 3232
(d) None of these
p ≠ 0. If p, q and r are in AP and
1 1
+ = 4, then the value
α β
of α − β is
(a)
61
9
j
(b)
2 17
9
(c)
34
9
JEE Mains 2014
(d)
2 13
9
7 If α and β are the roots of the equation ax 2 + bx + c = 0
(a ≠ 0, a, b, c being different), then (1 + α + α 2 ) (1 + β + β 2 ) is
(a) zero
(c) negative
(b) positive
(d) None of these
8 The minimum value of the sum of real numbers
a −5 , a −4 , 3a −3 , 1, a 8 and a10 a > 0 is
(a) 9
(b) 8
(c) 2
(d) 1
9 For a > 0, the roots of the equation
logax a + logx a 2 + loga 2 x a 3 = 0 is given by
(a) a −4 / 3
(b) a −3 / 4
(c) a1 / 2
(d) a −1
DAY
37
19 If α and β are the roots of the equation ax 2 + bx + c = 0,
10 If a , b and c are in AP and if the equations
(b − c ) x + (c − a )x + (a − b ) = 0 and 2 (c + a ) x
+ (b + c ) x = 0 have a common root, then
2
(a) a 2 , b 2 and c 2 are in AP
(c) c 2 , a 2 and b 2 are in AP
2
(b) a 2 , c 2 and b 2 are in AP
(d) None of these
11 If the equations x 2 + ax + 12 = 0, x 2 + bx + 15 = 0 and
x 2 + (a + b )x + 36 = 0 have a common positive root, then
the ordered pair (a , b ) is
(a) (− 6, − 7)
(c) (− 6, − 8)
(b) (− 7, − 8)
(d) (− 8, − 7)
x 2 − 3x + 4
will be
x 2 + 3x + 4
(a) 2, 1
(c) 7,
13 If a ∈ R and the equation
−3( x − [ x ])2 + 2( x − [ x ]) + a 2 = 0 (where, [ x ] denotes the
greatest integer ≤ x) has no integral solution, then all
j JEE Mains 2014
possible values of a lie in the interval
(b) (1, 2)
(d) (−∞, − 2) ∪ (2, ∞)
2π
2π
, then the quadratic equation
14 If a = cos
+ i sin
7
7
whose roots are α = a + a 2 + a 4 and β = a 3 + a 5 + a 6 , is
(a) x 2 − x + 2 = 0
(c) x 2 + x + 2 = 0
(b) x 2 + 2 x + 2 = 0
(d) x 2 + x − 2 = 0
15 If α and β are roots of 375 x 2 − 25x − 2 = 0 and
S n = α n + β n , then lim
n→ ∞
n
∑S
r
is equal to
7
116
29
(c)
358
(b)
1
12
16 If S = {a ∈ N, 1 ≤ a ≤ 100} and [tan x ] − tan x − a = 0 has
2
real roots, where [. ] denotes the greatest integer
function, then number of elements in set S equals
(b) 5
(c) 6
(d) 9
17 The sum of all real values of x satisfying the equation
( x 2 − 5 x + 5 )x
(a) 3
2
+ 4 x − 60
= 1 is
(b) −4
j
(c) 6
JEE Mains 2016
(d) 5
18 If λ is an integer and α , β are the roots of
λ
= 0 such that 1 < α < 2 and 2 < β < 3, then
4
the possible values of λ are
4x 2 − 16x +
(a) {60, 64, 68}
(c) {49, 50,…, 62, 63}
x 2 − 2kx + k 2 + k − 5 = 0 are less than 5, then k lies in the
j AIEEE 2005
interval
(b) (−∞, 4)
(c) (6, ∞)
(d) (5, 6)
x 2 − 2mx + m 2 − 1 = 0 are greater than −2 but less than 4
lie in the interval
(a) m > 3
(c) 1 < m < 4
(b) −1 < m < 3
(d) −2 < m < 0
22 Let α and β be real and z be a complex number. If
z 2 + αz + β = 0 has two distinct roots on the line
j AIEEE 2011
Re ( z ) = 1, then it is necessary that
(a) β ∈ (− 1, 0) (b) | β | = 1
23 The equation e
(a)
(b)
(c)
(d)
sin x
−e
(c) β ∈ [1, ∞)
− sin x
− 4 = 0 has
(d) β ∈ (0, 1)
j
AIEEE 2012
infinite number of real roots
no real root
exactly one real root
exactly four real roots
24 Ifa, b, c, d are positive real numbers such that
a+
1
1
1
1
= 4, b + = 1, c + = 4 and d + = 1, then
b
c
d
a
(b) b = d but a ≠ c
(d) cd = 1and ab ≠ 1
(ω and ω2 are complex cube roots of unity)
(d) None of these
(a) 2
− b (1 − x) + c (1 − x)2 = 0
− b (x − 1) + c (x − 1)2 = 0
+ b (1 − x) + c (1 − x)2 = 0
+ b (x + 1) + c (1 + x)2 = 0
(a) a = c and b = d
(c) ab = 1and cd ≠ 1
r =1
(a)
ax 2
ax 2
ax 2
ax 2
21 All the values of m for which both roots of the equation
1
5
(d) None of these
(a) (−1, 0) ∪ (0,1)
(c) (−2, − 1)
(a)
(b)
(c)
(d)
(a) [4,5]
(b) 5,
1
7
β
is
1+ β
α
and
1+ α
20 If both the roots of the quadratic equation
12 If x is real, then the maximum and minimum value of the
expression
then the quadratic equation whose roots are
(b) {61, 62, 63}
(d) {62, 65, 68, 71, 75}
25 Let f : R → R be a continuous function defined by
f (x ) =
1
e x + 2e − x
1
Statement I f (c ) = , for some c ∈ R .
3
Statement II 0 < f ( x ) ≤
1
2 2
, ∀ x ∈R .
j
AIEEE 2010
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true.
Statement II is a correct explanation of Statement I
(c) Statement I is true, Statement II is true;
Statement II is not a correct explanation for Statement I
(d) Statement I is true, Statement II is false
38
40
ANSWERS
SESSION 1
SESSION 2
1.
11.
21.
31.
41.
51.
(c)
(d)
(d)
(b)
(c)
(b)
2.
12.
22.
32.
42.
52.
1. (c)
11. (b)
21. (b)
(a)
(d)
(b)
(d)
(a)
(a)
3.
13.
23.
33.
43.
53.
2. (d)
12. (c)
22. (c)
(d)
(c)
(b)
(c)
(d)
(a)
(c)
(c)
(b)
(c)
(c)
(a)
4.
14.
24.
34.
44.
54.
3. (c)
13. (a)
23. (b)
5.
15.
25.
35.
45.
55.
4. (c)
14. (c)
24. (a)
(a)
(a)
(b)
(a)
(d)
(d)
6.
16.
26.
36.
46.
5. (a)
15. (b)
25. (b)
(a)
(d)
(c)
(b)
(b)
7.
17.
27.
37.
47.
6. (d)
16. (d)
(c)
(b)
(c)
(d)
(d)
8.
18.
28.
38.
48.
7. (b)
17. (a)
(a)
(a)
(c)
(d)
(d)
8. (b)
18. (c)
9.
19.
29.
39.
49.
(b)
(c)
(a)
(c)
(a)
10.
20.
30.
40.
50.
9. (a)
19. (c)
(a)
(c)
(c)
(a)
(b)
10. (b)
20. (b)
Hints and Explanations
SESSION 1
4 Q 5− 2 6 =
1 We have,
3 x2 − 7 x − 30 + 2 x2 − 7 x − 5 = x + 5
⇒ 3 x2 − 7 x − 30 = ( x + 5) − 2 x2 − 7 x − 5
On squaring both sides, we get
∴
t +
1
5+ 2 6
1
= 10,
t
2
where t = (5 + 2 6 )x
−3
⇒ t − 10t + 1 = 0
7 Since, roots are real.
3 x2 − 7 x − 30
= x2 + 25 + 10 x + (2 x2 − 7 x − 5)
−2( x + 5) 2 x2 − 7 x − 5
⇒
t = 5± 2 6
or
t = (5 + 2 6 )± 1
⇒ −10 x − 50 = −2( x + 5) 2 x2 − 7 x − 5
From Eqs. (i) and (ii),
x + 5 ≠ 0, 2 x2 − 7 x − 5 = 5
[Q x = −5does not satisfy the given
equation]
⇒ 2 x2 − 7 x − 30 = 0
∴
x=6
2 Given equation is x2 − 5 x + 6 = 0
and
⇒
x2 + 3 x + 2 x + 6 = 0; x < 0
( x − 3)( x − 2) = 0, x ≥ 0
and
( x + 3)⋅ ( x + 2) = 0, x < 0
∴ x = 3, x = 2 and x = −3, x = −2
There are four solutions of this equation.
3 Given equation is
2
2 x − 1 − 32 x − 1 + 2 = 0
Let 2 x − 1 = t , then
t 2 − 3t + 2 = 0
⇒
(t − 1)(t − 2) = 0 ⇒ t = 1,2
⇒
2 x − 1 = 1 and 2 x − 1 = 2
⇒
2 x − 1 = ±1 and 2 x − 1 = ±2
3 1
⇒
x = 1, 0 and x = , −
2 2
∴ {2(bc + ad )}2 = 4(a2 + b 2 )(c 2 + d 2 )
…(ii)
⇒
x = − 2 , 2 , − 2, 2
2(1 + i )x2 − 4(2 − i )x − 5 − 3i = 0
=−
16(2 − i )2
+ 8(1 + i )(5 + 3i )
4(1 + i )
Now,
−1 − i
=
2
1 1
+ =
4 4
1
2
and
3 − 5i
=
2
9 25
+
=
4 4
17
2
Also,
1
2
Hence, required root is
2
2
Since, roots are real
∴ a2 − 4 ≥ 0 ⇒ a ≥ 2
Thus, the least value of a is 2.
9 We know that only even prime is 2,
∴ (2)2 − λ(2) + 12 = 0.
i
4− i
−1 − i
3 − 5i
or
or
=
1+ i
1+ i
2
2
17
>
2
4a c + 4b d − 8abcd = 0
4(ac − bd )2 = 0
ac = bd
a b
=
d c
2 2
8 Given equation is x2 + ax + 1 = 0
5 The given equation is
⇒x=
+4b 2c 2 + 4b 2d 2
⇒
∴ Required product = 8
4(2 − i ) ±
⇒ 4b 2c 2 + 4a2d 2 + 8abcd = 4a2c 2 + 4a2d 2
⇒
⇒
⇒
x2 − 3 = ± 1
x2 = 2, 4
⇒
When x ≥ 0, x2 − 5x + 6 = 0
and when x < 0, x2 + 5x + 6 = 0
⇒
x2 − 3 x − 2 x + 6 = 0; x ≥ 0
…(i)
2
If equation is a perfect square then root
are equal
i.e. (1 − 2k )2 − 4(1 + 2k ) (1 − 2k ) = 0
1 −3
i.e.
k = , .
2 10
Hence, total number of values = 2.
3 − 5i
.
2
6 Given equation
(1 + 2k ) x2 + (1 − 2k )x + (1 − 2k ) = 0
⇒
λ=8
...(i)
Qx2 + λx + µ = 0 has equal roots.
[QD = 0]
∴
λ2 − 4µ = 0
⇒ (8)2 − 4µ = 0 ⇒ µ = 16
10 Here, D = (3b )2 − 4 (4 a)(2c )
= 9b 2 − 32ac = 9 (− a − c )2 − 32ac
= 9 a2 − 14 ac + 9c 2
a 2 14 a
= 9c 2 −
⋅ + 1
c
9
c
2
a
7
49
= 9c 2 − −
+ 1 > 0
c
9
81
Hence, the roots are real and distinct.
DAY
39
11 For real roots, discriminant,
D = b − 4 ac ≥ 0
= cos 2 β − 4(cos β − 1)sin β ≥ 0
= cos 2 β + 4(1 − cos β )sin β ≥ 0
So, sin β should be > 0.
[Q cos 2 β ≥ 0,1 − cos β ≥ 0]
⇒
β ∈ (0, π )
2
12 Here, D = 4b + 12ca < 0
2
⇒
b 2 + 3ca < 0
⇒
ca < 0
If c > 0, then a < 0
3c
Also,
< a+ b
4
⇒
…(i)
3ca > 4 a2 + 4 ab
⇒ b + 3ca > 4 a2 + 4 ab + b 2
…(ii)
= (2a + b )2 ≥ 0
From (i) and (ii), c > 0, is not true.
∴ c<0
2
13 Since, D ≥ 0
b 2 − 4 ac ≥ 0
2
c + a
⇒
− 4 ac ≥ 0 [Q 2b = a + c ]
2
⇒
c 2 − 14ca + a2 ≥ 0
∴
2
c
c
⇒ − 14 + 1≥ 0
a
a
2
c − 7 ≥ 48
a
⇒
c
−7 ≥ 4 3
a
⇒
14 Let D1 and D2 be the discriminants of
given equation, respectively. Then
D1 + D2 = b 2 − 4 ac + d 2 + 4 ac
= b2 + d 2 > 0
So, either D1 and D2 are positive or
atleast one D’s is positive.
Therefore, P ( x )⋅ Q ( x ) = 0 has either two
or four real roots.
15 One root =
4− 3 4− 3
1
×
=
13
4+ 3 4− 3
4+ 3
13
∴The quadratic equation is
4+ 3
4 − 3
x2 −
+
x
13
13
∴ Other root =
4+ 3 4− 3
+
⋅
=0
13
13
or
13 x2 − 8 x + 1 = 0
This equation must be identical with
ax2 + bx + 1 = 0;
∴
a = 13 and b = − 8.
16 Sum of roots
−a
= (1 − i ) + (1 + i ) ⇒ a = − 2.
1
[since, non-real complex roots occur in
conjugate pairs]
Product of roots,
b
= (1 − i )(1 + i ) ⇒ b = 2
1
17 Let roots be α and α2 .
Then, α + α2 = 30 and α3 = p
⇒
α2 + α − 30 = 0
⇒
(α + 6)(α − 5) = 0
∴
⇒
and
∴
α
p
p
p
= −6, 5
= α3 = (−6)3 = −216
= (5)3 = 125
= 125and −216
x−m
x+ n
18 Given,
=
mx + 1 nx + 1
⇒ x2 (m − n ) + 2mnx + (m + n ) = 0
1
Roots are α, respectively, then
α
1 m+ n
α⋅ =
α
m−n
⇒ m − n = m + n ⇒ n = 0.
19 Since, α,α2 be the roots of the equation
x2 + x + 1 = 0
∴
α + α 2 = −1
... (i)
3
and α = 1
... (ii)
Now,
α 31 + α 62 = α31 (1 + α31 )
31
⇒
α + α 62 = α30 ⋅ α(1 + α 30 ⋅ α)
⇒ α31 + α 62 = (α3 )10 ⋅ α{1 + (α3 )10 ⋅ α}
⇒ α31 + α 62 = α(1 + α) [from Eq. (ii)]
[from Eq. (i)]
⇒ α31 + α 62 = −1
Again, α31 ⋅ α 62 = α 93
⇒
α31 ⋅ α 62 = [α3 ]31 = 1
∴ Required equation is
x2 − (α31 + α 62 )x + α31 ⋅ α 62 = 0
⇒
x2 + x + 1 = 0
20 Since, α and β are the roots of
x2 − ax + a + b = 0, then
a+ β = a
and
αβ = a + b
⇒
a2 + αβ = aα
⇒
α2 − aα = − (a + b )
and
αβ + β2 = aβ
⇒
β2 − aβ = − (a + b )
1
1
2
+
+
∴
α2 − aα β2 − aβ
a+ b
1
1
2
=
+
+
=0
− (a + b ) − (a + b ) (a + b )
21 Let α and β be the roots of equation
x2 − (a − 2)x − a − 1 = 0
Then, α + β = a − 2 and αβ = − a − 1
Now, α2 + β2 = (α + β )2 − 2αβ
⇒
α2 + β2 = (a − 2)2 + 2(a + 1)
⇒
α2 + β2 = a2 − 2a + 6
⇒
α2 + β2 = (a − 1)2 + 5
The value of α2 + β2 will be least, if
a−1 = 0
a=1
⇒
22 Since, α and β are the roots of the
equation
2 x2 + 2(a + b )x + a2 + b 2 = 0
∴
(α + β )2 = (a + b )2 and αβ =
a2 + b 2
2
Now, (α − β )2 = (α + β )2 − 4αβ
a2 + b 2
= (a + b )2 − 4
2
= − (a − b )2
Now, the required equation whose
roots are (α + β )2 and (α − β )2 is
x2 − {(α + β )2 + (α − β )2 } x
+ (α + β )2 (α − β )2 = 0
⇒ x2 − {(a + b )2 − (a − b )2 } x
− (a + b )2 (a − b )2 = 0
2
⇒
x − 4abx − (a2 − b 2 )2 = 0
23 The given equation is
x2 − 2 x cos φ + 1 = 0
4cos 2 φ − 4
2
= cos φ ± i sin φ
Let α = cos φ + i sin φ, then
β = cos φ − i sin φ
∴ α n + β n = (cos φ + i sin φ)n
+ (cos φ − i sin φ)n
= 2cos nφ
and α nβ n = (cos nφ + i sin nφ)
⋅ (cos nφ − i sin nφ)
= cos 2 nφ + sin2 nφ = 1
∴ Required equation is
x2 − 2 x cos nφ + 1 = 0
24 Given equation is
(5 + 2 )x2 − (4 + 5)x + 8 + 2 5 = 0
Let x1 and x2 are the roots of the
equation, then
∴
x=
2cos φ ±
x1 + x2 =
4+ 5
5+ 2
... (i)
8 + 2 5 2(4 + 5)
=
5+ 2
5+ 2
... (ii)
= 2( x1 + x2 )
Clearly, harmonic mean
2 x1 x2
4( x1 + x2 )
=
=
=4
x1 + x2
( x1 + x2 )
and
x1 x2 =
[from Eq. (ii)]
25 Let α, β and α ′, β ′ be the roots of the
given equations, respectively.
µ
ν
∴ α + β = − , αβ =
λ
λ
and α ′ = ω and β ′ = ω2
α α′
[given]
=
Q
β β′
α
ω
⇒
=
⇒ β = αω
β ω2
From Eq. (i),
…(i)
40
40
µ 2
ν
,α ω =
λ
λ
µ
ν
− αω2 = − , α2ω =
λ
λ
[Q− ω2 = 1 + ω]
µ2
ν
=
⇒ µ 2 = λν
λ2
λ
α + αω = −
⇒
⇒
26 Simplified form of given equation is
(2x + p + q ) r = ( x + p )( x + q )
⇒ x + ( p + q − 2r )x
− ( p + q ) r + pq = 0
Since, sum of roots = 0
⇒
− ( p + q − 2r ) = 0
p+q
r =
⇒
2
and product of roots
= − ( p + q ) r + pq
( p + q )2
=−
+ pq
2
1
= − ( p2 + q 2 )
2
2
27 Clearly, sum of roots,
k+1 k+2
b
... (i)
+
=−
k
k+1
a
and product of roots,
k+1 k+2 c
×
=
k
k+1 a
k+2 c
=
⇒
k
a
2 c
c−a
2a
⇒
= −1 =
⇒k =
k
a
a
c−a
On putting the value of k in Eq. (i), we
get
c+a
2c
b
+
=−
2a
c+a
a
⇒
(c + a)2 + 4ac = −2b(a + c )
⇒ (a + c )2 + 2b(a + c ) = −4ac
⇒ (a + c )2 + 2b(a + c ) + b 2 = b 2 − 4ac
⇒ (a + b + c )2 = b 2 − 4ac
(Qω + ω2 = −1)
= α2 + β2 − αβ
2
2
= (α + β ) − 3αβ = p − 3q
α 2 β2 α 3 + β3
Also,
+
=
β
α
αβ
p(3q − p2 )
(α + β )3 − 3αβ(α + β)
=
=
αβ
q
∴ The given expression
( p2 − 3q )
q
=
=−
p(3q − p2 )
p
q
30 Clearly, α + β = − p and αβ =
Also,
(α − β )2 = 10
⇒ (α + β )2 − 4αβ = 10
⇒
p2 − 3 p = 10
⇒ ( p + 2)( p − 5) = 0
∴ p = −2, 5
ax2 + bx + c = 0 and its roots are α and
β.
Sachin made a mistake in writing down
constant term.
∴ Sum of roots is correct i.e. α + β = 7
Rahul made mistake in writing down
coefficient of x.
∴ Product of roots is correct.
i.e. α ⋅ β = 6
⇒ Correct quadratic equation is
x2 − (α + β )x + αβ = 0.
⇒ x2 − 7 x + 6 = 0 having roots 1
and 6.
32 Since, α + β = − 1,αβ = 2,
γ + δ = − 3, and γδ = 4
∴ (α + γ )(α + δ )(β + γ )(β + δ )
= (α2 − 3α + 4)(β2 − 3β + 4)
= 4 − 3αβ2 + 4β2 − 3α2β + 9αβ
− 12β + 4α2 − 12α + 16
= 4 − 3(2)β + 4β2 + 4α2
−3( 2)α + 9( 2) − 12 ( β + α ) + 16
2
= 4 − 6β + 4(α2 + β2 )
∴
Hence, required eqution is
b
c
x2 − x +
=0
a
a
⇒
a x2 − b x + c = 0
− 6α + 18 + 12 + 16
= 50 + 6 + 4[(α + β )2 − 2αβ]
= 56 − 12 = 44
33 Given, R = π ⇒ P + Q = π
2
⇒
⇒
29 Since, α and β are the roots of the
equation x2 + px + q = 0 therefore
α + β = − p and αβ = q
Now, (ωα + ω2β )(ω2α + ωβ )
= α2 + β2 + (ω 4 + ω2 )αβ
(Qω3 = 1)
⇒
and (12k + 4)x2 + px + 6 k − 2 = 0
For both common roots,
6k + 2
r 3k − 1
= =
12 k + 4 p 6 k − 2
r 1
⇒
= ⇒ 2r − p = 0
p 2
35 Given equations are
... (i)
x2 + 2 x + 3 = 0
... (ii)
and
ax2 + bx + c = 0
Since, Eq. (i) has imaginary roots.
So, Eq. (ii) will also have both roots
same as Eq. (i)
a b c
Thus,
= =
1 2 3
Hence, a:b :c is 1:2:3.
36 α, β be the roots of ax2 + bx + c = 0
31 Let the quadratic equation be
28 Since, α and β are the roots of the
equation ax + bx + c = 0
b
c
α + β = − and αβ =
a
a
Now, sum of roots = α + β
= − α −β
(Qβ < α < 0)
b b
= − − =
(Qα + β > 0)
a
a
c
and product of roots = α β =
a
3p
4
34 Given, (6 k + 2)x2 + rx + 3 k − 1 = 0
2
P
Q
π
+
=
2
2
4
Q
P
tan + tan
π
2
2
1 = tan =
4 1 − tan P tan Q
2
2
b
−
a
b
1=
=
⇒a+ b =c
c
c
−a
1−
a
∴ α + β = − b / a, αβ = c / a
Now roots are α − 1, β − 1
Their sum, α + β − 2 = (− b / a) − 2
= − 8 /2 = − 4
Their product,
(α − 1) (β − 1) = αβ − (α + β ) + 1
= c /a + b/a + 1 = 1
∴ b / a = 2 i.e. b = 2a
also c + b = 0 ⇒ b = − c .
4b 2 − 8c 2
37 min f ( x ) = − D = −
4a
4
= − (b 2 − 2c 2 )
(upward parabola)
4c 2 + 4b 2
D
max g ( x ) = −
=
4a
4
= b2 + c 2
(downward parabola)
Now, 2c 2 − b 2 > b 2 + c 2
⇒
c 2 > 2b 2 ⇒ |c| > 2 |b|
38 We have,
( x − a1 )2 + ( x − a2 )2 + … + ( x − an )2
= n x2 − 2 x(a1 + a2 + … + an )
+ (a12 + a22 + …+ a2n )
So, it attains its minimum value at
2(a1 + a2 + … + an )
x=
2n
a + a2 + … + an
using : x = −b
= 1
n
2a
39 Given bx2 + cx + a = 0 has imaginary
roots.
…(i)
⇒ c 2 − 4ab < 0 ⇒ c 2 < 4ab
Let f ( x ) = 3b 2 x2 + 6bcx + 2c 2
Here, 3b 2 > 0
So, the given expression has a
minimum value.
−D
∴ Minimum value =
4a
4(3b 2 )(2c 2 ) − 36b 2c 2
4ac − b 2
=
=
4a
4(3b 2 )
DAY
41
=−
12b 2c 2
= − c 2 > −4ab
12b 2
[from Eq. (i)]
40 According to given condition,
4a2 − 4(10 − 3a) < 0
⇒
a2 + 3a − 10 < 0
⇒
(a + 5) (a − 2) < 0
⇒
− 5 < a < 2.
41 We have, ax − 1 + 1 ≥ 0
x
ax2 − x + 1
≥0
⇒
x
2
⇒ ax − x + 1 ≤ 0 as x > 0
It will hold if a > 0 and D ≤ 0
1
a > 0 and 1 − 4a ≤ 0 ⇒ a ≥
4
1
Therefore, the minimum value of a is .
4
42 Let f ( x ) = − 3 + x − x2 .
Then, f ( x ) < 0 for all x, because coeff. of
x2 < 0 and disc. < 0.
Thus, LHS of the given equation is
always positive whereas the RHS is
always less than zero. Hence, the given
equation has no solution.
43 Let f ( x ) = ax2 + bx + c .
Then, k lies betweenα andβ, if a f (k ) < 0
⇒
a (ak 2 + bk + c ) < 0
2
⇒
a k 2 + abk + ac < 0.
44 We have, 2 x − 3 < x + 5
⇒
⇒
⇒
⇒
⇒
2x − 3 − x + 5 < 0
3 − 2 x + x + 5 < 0, x ≤ − 5
3
3 − 2 x − x − 5 < 0, − 5 < x ≤
2
3
2 x − 3 − x − 5 < 0, x >
2
x > 8, x ≤ −5
2
3
x
>
− ,− 5< x ≤
3
2
3
x < 8, x >
2
2 3 3
x ∈ − , ∪ , 8
3 2 2
2
x ∈ − , 8
3
x−5
>0
x + 5x − 14
45
2
⇒
( x + 7 )( x − 2) ( x − 5)
>0
( x − 2)2 ( x + 7 )2
⇒ x ∈ (−7, 2) ∪ (5, ∞ )
So, the least integral value α of x is − 6,
which satisfy the equation
α2 + 5α − 6 = 0
46 Given,
| x − 2| − 1
≤0
| x − 2| − 2
51 Using GM ≤ AM, we have
{(1 − a) (1 − b ) (1 − c )1 /3 }
1− a+ 1−b + 1−c
≤
3
8
⇒ (1 − a) (1 − b ) (1 − c ) ≤
27
8
Hence, the greatest value is .
27
Let
| x − 2 |= k
Then, given equation,
(k − 1) (k − 2)
k −1
≤0 ⇒
≤0
k −2
(k − 2)2
⇒ (k − 1) (k − 2) ≤ 0 ⇒ 1 ≤ k ≤ 2
⇒
1 ≤ | x − 2|≤ 2
52 Using AM ≥ GM, we have
Case I When 1 ≤ | x − 2|
⇒
⇒
⇒
| x − 2|≥ 1
x − 2 ≥ 1 or x − 2 ≤ −1
x ≥ 3 and x ≤ 1
…(i)
Case II When | x − 2|≤ 2
⇒
−2 ≤ x − 2 ≤ 2
⇒
−2 + 2 ≤ x ≤ 2 + 2
⇒
0≤ x ≤ 4
From (i) and (ii), x ∈ [0, 1] ∪ [3, 4]
x+2 1
47
− >0
x2 + 1 2
...(ii)
54. Clearly, x − 1 > 0 ⇒ x > 1
3 + 2 x − x2
>0
2( x2 + 1)
and log 0 . 3 ( x − 1) > log ( 0 . 3 )2 ( x − 1)
⇒
⇒
such that a1 a2 … an = 1.
Using AM ≥ GM, we have
a + a2 + … + an
⇒ 1
≥ (a1 a2 … an )1 / n
n
⇒ a1 + a2 + … an ≥ n.
49 Since, AM ≥ GM
bcx + cay + abz
≥ (a2b 2c 2 ⋅ xyz )1 /3
3
⇒ bcx + cay + abz ≥ 3abc
[Qxyz = abc ]
∴
⇒
3
1 + 1 + 1 >
a b c (abc )1 /3
x ∈ (1, 2)
55 By definition of log x, x > 0 and
x + 5
>0
1 − 3x
⇒
( x + 5)(1 − 3 x )
>0
(1 − 3 x )2
⇒
( x + 5)(1 − 3 x ) > 0
⇒
( x + 5)(3 x − 1) < 0
⇒
... (i)
− 5 < x < 1/3
x > 0, 0 < x < 1 / 3
As
∴
50 We know that, AM > GM
and
1
log 0 . 3 ( x − 1)
2
log 0 . 3 ( x − 1) > 0 ⇒ x < 2
log 0 . 3 ( x − 1) >
Hence,
48 Let a1 , a2 , … , an be n positive integers
a+ b + c
> (abc )1 /3
3
1 1 1
1 /3
+ +
a b c > 1
abc
3
x2 − 3 x + 18 < 16
(Qlog a b < c ⇔ b < ac , if a > 1)
x2 − 3 x + 2 < 0
( x − 1)( x − 2) < 0 ⇒ x ∈ (1,2)
⇒
⇒
Since, denominator is positive
∴3 + 2 x − x2 > 0
⇒
− 1< x < 3
⇒
x = 0, 1, 2
∴
53 log 2 ( x2 − 3 x + 18) < 4
⇒
− x2 − 1 + 2 x + 4
⇒
>0
2( x2 + 1)
⇒
(a + b ) + (c + d )
≥ {(a + b ) (c + d )}1 /2
2
2
⇒
≥ M 1 /2 ⇒ M ≤ 1
2
As a, b, c , d > 0.
Therefore, M = (a + b ) (c + d ) > 0.
Hence, 0 ≤ M ≤ 1.
x+ 5
< 1⇒ x < − 1
1 − 3x
This does not satisfy 0 < x < 1 / 3.
Hence, there is no solution.
SESSION 2
1 We have
... (ii)
From (i) and (ii), we get
a + b + c 1 + 1 + 1
a b c
3
3
>
⋅ (abc )1 /3
(abc )1 /3
1 1 1
⇒ (a + b + c ) + + > 9
a b c
2 x −3 +
Let
⇒
x ( x − 6) + 6 = 0
x −3 = y
x = y+3
∴ 2 y + ( y + 3)( y − 3) + 6 = 0
⇒
2 y + y2 − 3 = 0
2
⇒
y + 2 y −3 = 0
⇒
⇒
( y + 3)( y − 1) = 0
y ≠ −3 ⇒ y = 1
42
40
⇒
⇒
⇒
y = ±1
x − 3 = ±1
x = 4,2
⇒
5 Given equation can be rewritten as
x = 16, 4
2 We have, 2
x +2
⋅ 33 x / ( x − 1 ) = 9
Taking log on both sides, we get
3x
( x + 2)log 2 +
log 3 = 2 log 3
( x − 1)
⇒ ( x2 + x − 2)log 2 + 3 x log 3
= 2( x − 1)log 3
⇒ x2 log 2 + (log 2 + log 3)x
− 2 log 2 + 2 log 3 = 0
− (log 2 + log 3) ±
{(log 2 + log 3)2 − 4 log 2
(− 2 log 2 + 2 log 3)}
∴ x=
2 log 2
− (log 2 + log 3)
±
{(3 log 2)2 − 6 log 2 log 3
+ (log 3)2 }
=
2 log 2
=
− (log 2 + log 3) ± (3 log 2 − log 3)
2 log 2
∴ x = − 2, 1 −
log 3
log 2
equation x2 − 6 x − 2 = 0
Qan = α n − β n , n ≥ 1 ⇒ a10 = α10 − β10
a8 = α 8 − β 8 and a9 = α 9 − β 9
Now, consider
[Qα and β are the
roots of
α10 − β10 − 2(α 8 − β 8 ) x2 − 6 x − 2 = 0
=
2(α 9 − β 9 )
or
x2 = 6 x + 2
=
=
=
α 8(α2 − 2) − β 8(β2 − 2) ⇒ α2 = 6α + 2
2(α 9 − β 9 )
⇒ α2 − 2 = 6α
and β2 = 6β + 2
α 8 ⋅ 6α − β 8 ⋅ 6β
⇒
2(α 9 − β 9 )
β2 − 2 = 6β]
6α 9 − 6β 9 6
= =3
2(α 9 − β 9 ) 2
4 We have, x2 − 5x + 1 = 0
1
1
= 5 ⇒ x2 + 2 = 52 − 2 = 23
x
x
1
and x3 + 3 = 53 − 3 × 5 = 110
x
1
1
Now, x2 + 2 x3 + 3
x
x
x+
= 23 × 110 = 2530
⇒ x5 +
Since, p,q and r are in AP.
∴
2q = p + r
1 1
α+β
Also,
+ =4 ⇒
=4
α β
αβ
−q
4r
⇒
α + β = 4αβ ⇒
=
p
p
2q = p + r
2(−4r ) = p + r ⇒ p = −9r
−q
4r
4r
4
α+β =
=
=
=−
p
p
−9r
9
r
r
1
αβ =
=
=
p −9r
−9
Q
⇒
Q
and
3 Given, α and β are the roots of the
a10 − 2a8
2a9
3 x2 − (a + c + 2b + 2d )x + ac + 2bd = 0
Now, discriminant, D
= (a + c + 2b + 2d )2 − 4⋅ 3(ac + 2bd )
= {(a + 2d ) + (c + 2b )}2 − 12(ac + 2bd )
= {(a + 2d ) − (c + 2b )}2 + 4(a + 2d )
(c + 2b ) − 12(ac + 2bd )
= {(a + 2d ) − (c + 2b )}2
−8ac + 8ab + 8dc − 8bd
= {(a + 2d ) − (c + 2b )}2 + 8(c − b )(d − a)
Which is + ve, since a < b < c < d .
Hence, roots are real and distinct.
6 Since, α and β are roots of
px2 + qx + r = 0, p ≠ 0
−q
r
, αβ =
∴
α+β =
p
p
1
1
= 2530 − x + = 2525
x5
x
Now, consider
(α − β )2 = (α + β )2 − 4αβ
16 4 16 + 36
=
+ =
81 9
81
52
2
⇒ (α − β ) =
81
2
⇒ α −β =
13
9
7 α + β = −b / a and αβ = c / a
Now, (1 + α + α2 )(1 + β + β2 )
= 1 + (α + β ) + (α2 + β2 )
+ αβ + αβ(α + β) + (αβ) 2
= 1 + (α + β ) + (α + β )2
− αβ + αβ(α + β) + (αβ) 2
b b2 c
c
b
c2
= 1 − + 2 − + − + 2
a a
a a a a
(a2 + b 2 + c 2 − ab − bc − ca)
=
a2
2
= [(a − b ) + (b − c )2 + (c − a)2 ] / 2a2
which is positive.
8 Using AM ≥ GM
−5
−4
a + a
−3
+ a
−3
+ a
−3
+ a
+ 1 + a8 + a10
∴
8
≥ (a−5 ⋅ a−4 ⋅ a−3 ⋅ a−3 ⋅ a−3 ⋅ 1 ⋅ a8 ⋅ a10 )1 / 8
⇒ a−5 + a−4 + 3a−3 + 1 + a8 + a10
≥ 8⋅1
Hence, minimum value is 8.
9 We have,
log a a
2 log a a
+
log a a + log a x
log a x
3 log a a
+
=0
2 log a a + log a x
1
2
3
+ +
=0
⇒
1+ t
t
2+ t
⇒
( let log a x = t )
2t + t 2 + 2t 2 + 6t + 4 + 3t 2 + 3t
=0
t (1 + t ) (2 + t )
⇒
⇒
⇒
⇒
⇒
∴
6t 2 + 11t + 4 = 0
6t + 8t + 3t + 4 = 0
(2t + 1) (3t + 4) = 0
1
4
t = − ,−
2
3
1
4
log a x = − , −
2
3
x = a−1 /2 , a−4 /3
2
10 Since, x = 1 is a root of first equation. If
α is another root of first equation, then
a−b
α × 1= α =
b −c
(Product of roots)
2a − 2b 2a − (a + c )
=
=
=1
2b − 2c
a + c − 2c
So, the roots of first equation are 1 and
1.
Since, the equations have a common root,
1 is also a root of second equation.
⇒
2(c + a) + b + c = 0
⇒
2(2b ) + b + c = 0
[since, a, b and c are in AP]
⇒
c = − 5b
Also,
a + c = 2b ⇒ a = 2b − c
= 2b + 5 b = 7b
∴ a2 = 49b 2 , c 2 = 25 b 2
Hence, a2 , c 2 and b 2 are in AP.
11 On adding first and second equations,
we get
2 x2 + (a + b )x + 27 = 0
On subtracting above equation from
given third equation, we get
x2 − 9 = 0 ⇒ x = 3, − 3
Thus, common positive root is 3.
Now, (3)2 + 3 a + 12 = 0 ⇒ a = − 7
and 9 + 3b + 15 = 0 ⇒ b = − 8
Hence, the order pair (a, b ) is (− 7, − 8).
2
12 Let y = x2 − 3 x + 4
x + 3x + 4
⇒ ( y − 1)x2 + 3( y + 1)x + 4( y − 1) = 0
Q x is real.
∴
D≥0
⇒
9( y + 1)2 − 16( y − 1)2 ≥ 0
⇒
−7 y 2 + 50 y − 7 ≥ 0
⇒
7 y 2 − 50 y + 7 ≤ 0
⇒
( y − 7)(7 y − 1) ≤ 0
DAY
43
⇒
y ≤ 7 and y ≥
1
7
1
≤ y≤7
7
Hence, maximum value is 7 and
1
minimum value is .
7
Now, consider
n
lim
⇒
13 Here, a ∈ R and equation is
−3{ x − [ x]}2 + 2{ x − [ x]} + a2 = 0
Let t = x − [ x], then
−3t 2 + 2t + a2 = 0
1 ± 1 + 3a2
⇒ t =
3
Q t = x − [ x] = { x}
[fractional part]
∴
0≤ t ≤ 1
1 ± 1 + 3a2
⇒
0≤
<1
3
[Q 0 ≤ { x} < 1]
But 1 − 1 + 3a2 < 0 therefore
1 + 1 + 3a
0≤
<1
3
2
1 + 3a < 2
1 + 3a2 < 4 ⇒ a2 − 1 < 0
(a + 1)(a − 1) < 0
2
⇒
⇒
⇒
⇒
a ∈ (−1,1)
For no integral solution we consider the
interval (−1, 0) ∪ (0,1).
14 Given, a = cos 2 π + i sin 2 π
7
7
∴
a7 = cos 2 π + i sin 2 π = 1
[Qe iθ = cos θ + i sin θ]
Also, α = a + a2 + a4 ,
β = a3 + a5 + a6
Then, the sum of roots,
S = α + β = a + a2 + a3 + a4
+ a5 + a6
6
7
a(1 − a ) a − a
⇒ S =
=
1− a
1− a
a−1
=
= −1
[Q a7 = 1]
1− a
and product of the roots,
P = αβ = (a + a2 + a4 ) (a3 + a5 + a6 )
= a4 + a5 + 1 + a6 + 1 + a2 + 1
+ a + a3
[Q a7 = 1]
= 3 + (a + a2 + a3 + a4 + a5 + a6 )
=3−1=2
Hence, the required quadratic equation
is x2 + x + 2 = 0
15 Since, α and β are the roots of
375 x − 25x − 2 = 0.
2
∴
and
α +β=
25
1
=
375 15
αβ = −
2
375
ΣS
n→ ∞ r = 1
n
r
= lim
Σ (α
r
n→ ∞ r = 1
+β )
r
= (α + α2 + α3 + K ∞ )
+ (β + β2 + β3 + K ∞ )
α − αβ + β − αβ
α
β
=
+
=
1−α 1−β
(1 − α )(1 − β )
=
α + β − 2αβ
1 − (α + β ) + αβ
1
4
+
15 375
1
2
1−
−
15 375
25 + 4
=
375 − 25 − 2
=
Here, 64 − λ > 0
∴ λ < 64
29
1
=
=
348 12
Also, 1 < α < 2 and 2 < β < 3
16 Here, [tan2 x] = integer
and a = integer
So, tan x is also an integer.
Then, tan2 x − tan x − a = 0
⇒ a = tan x (tan x − 1) = I (I − 1)
= Product of two
consecutive integers
∴ a = 2, 6, 12, 20, 30, 42, 56, 72, 90
Hence, set S has 9 elements.
x2 + 4 x − 60
17 Given, ( x − 5x + 5)
2
When x = 3,
x2 + 4 x − 60 = 9 + 12 − 60 = −39, which
is not an even integer.
Thus, in this case, we get x = 2
Hence, the sum of all real values of
x = −10 + 6 + 4 + 1 + 2 = 3
18 Given, 4 x2 − 16 x + λ = 0
4
16 ± (256 − 4λ )
x=
∴
8
8 ± (64 − λ )
=
4
(64 − λ )
⇒ α, β = 2 ±
4
=1
Clearly, this is possible when
I. x2 + 4 x − 60 = 0 and
x2 − 5 x + 5 ≠ 0
or
II. x2 − 5 x + 5 = 1
or
III. x − 5 x + 5 = −1 and
x 2 + 4 x − 60 = Even integer
2
Case I When x2 + 4 x − 60 = 0
⇒
x2 + 10 x − 6 x − 60 = 0
⇒
x( x + 10) − 6( x + 10) = 0
⇒
( x + 10)( x − 6) = 0
⇒
x = −10 or x = 6
Note that, for these two values of
x, x2 − 5x + 5 ≠ 0
Case II When x2 − 5x + 5 = 1
⇒
x2 − 5x + 4 = 0
⇒
x2 − 4 x − x + 4 = 0
⇒
( x − 4) ( x − 1) = 0
⇒
x = 4 or x = 1
Case III When x2 − 5x + 5 = −1
⇒
x2 − 5x + 6 = 0
⇒
x2 − 2 x − 3 x + 6 = 0
⇒
x( x − 2) − 3( x − 2) = 0
⇒
( x − 2)( x − 3) = 0
⇒
x = 2 or x = 3
Now, when x = 2,
x2 + 4 x − 60 = 4 + 8 − 60 = −48, which
is an even integer.
64 − λ
∴
1< 2 −
and
2< 2 +
⇒
− 1< −
and
0<
⇒
1>
and
0<
<2
4
64 − λ
<3
4
64 − λ
4
(64 − λ )
4
(64 − λ )
4
(64 − λ )
4
(64 − λ )
<0
<1
>0
<1
i.e.
0<
⇒
0 < (64 − λ ) < 4
4
<1
⇒
0 < 64 − λ < 16 ⇒ λ > 48
or
48 < λ < 64
∴ λ = {49, 50, 51, 52, ... , 63}
19 Since, roots of ax2 + bx + c = 0 are α and
β. Hence, roots of cx2 + bx + a = 0, will
1
1
be and . Now, if we replace x by
α
β
x − 1, then roots of
1
c ( x − 1)2 + b( x − 1) + a = 0 will be 1 +
α
1
1
and 1 + . Now, again replace x by ,
β
x
we will get c (1 − x )2 + b(1 − x ) + ax2 = 0,
α
β
whose roots are
and
.
1+ α
1+ β
20 Let f ( x ) = x2 − 2kx + k 2 + k − 5
Since, both roots are less than 5.
b
< 5and f (5) > 0
∴ D ≥ 0, −
2a
2
Here, D = 4k − 4(k 2 + k − 5)
= −4k + 20 ≥ 0
⇒
k≤5
... (i)
44
40
b
<5⇒ k<5
... (ii)
2a
and
f (5) > 0
⇒
25 − 10k + k 2 + k − 5 > 0
⇒
k 2 − 9k + 20 > 0
⇒
(k − 5)(k − 4) > 0
⇒
k < 4 and k > 5
…(iii)
From (i), (ii) and (iii), we get
k<4
−
21 Since, both roots of equation
x2 − 2mx + m2 − 1 = 0 are greater than −2
but less than 4.
b
∴
D ≥ 0, −2 < −
< 4,
2a
f (4) > 0 and f (−2) > 0
Now,
D≥0
⇒
4m2 − 4m2 + 4 ≥ 0
... (i)
⇒
4> 0⇒m ∈R
(m + 3) (m + 1) > 0
−∞ < m < −3 and
... (iv)
−1 < m < ∞
From (i), (ii), (iii) and (iv), we get
m lie between −1 and 3.
⇒
22 Let z = x + iy , given Re(z ) = 1
∴ x = 1 ⇒ z = 1 + iy
Since, the complex roots are conjugate
to each other.
So, z = 1 + iy and 1 − iy are two roots of
z2 + α z + β = 0.
Q Product of roots = β
⇒ (1 + iy )(1 − iy ) = β
∴ β = 1 + y 2 ≥ 1 ⇒ β ∈ [1, ∞ )
23 Given equation is
e sin x − e − sin x = 4 ⇒ e sin x −
1
=4
e sin x
1
=4
t
2
2
⇒ t − 1 − 4t = 0 ⇒ t − 4t − 1 = 0
4 ± 16 + 4
⇒ t =
2
t = 2 ± 5 ⇒ e sin x = 2 ± 5
But −1 ≤ sin x ≤ 1 ⇒ e −1 ≤ e sin x ≤ e 1
1
⇒ e sin x ∈ ,e
e
Also, 0 < e < 2 + 5
Hence, given equation has no solution.
Let e sin x = t , then t −
b
<4
2a
2m
−2 <
<4
2⋅ 1
−2 < −
⇒
⇒
... (ii)
−2 < m < 4
f (4) > 0
⇒
16 − 8m + m2 − 1 > 0
⇒
m2 − 8m + 15 > 0
⇒
(m − 3)(m − 5) > 0
⇒ −∞ < m < 3 and 5 < m < ∞ ... (iii)
and
f (−2) > 0
⇒
4 + 4m + m2 − 1 > 0
⇒
m2 + 4m + 3 > 0
24 Using AM > GM, we have
1
a
1
b
>2
, b + >2
b
b
c
c
1
c
1
d
c + >2
and d + > 2
d
d
a
a
a + 1 b + 1 c + 1 d + 1 > 16
b
c
d
a
a+
1
1
1
But, a + b + c +
b
c
d
d + 1 = 4 × 1 × 4 × 1 = 16
a
1
1
1
1
and d =
∴ a = ,b = ,c
b
c
d
a
1
1 1
1
⇒ a = = 2, b = = , c = = 2
b
c 2
d
1 1
and
d = =
a 2
1
1
⇒ a = 2, b = ,c = 2 and d =
2
2
⇒ a = c and b = d
1
25 We have, f ( x ) =
2
x
e + x
e
Using AM ≥ GM , we get
2
1 /2
ex + x
e ≥ e x ⋅ 2 , as e x > 0
x
2
e
2
x
⇒ e + x ≥2 2
e
1
1
0<
≤
⇒
2
ex + x 2 2
e
1
∴
0 < f ( x) ≤
,∀x∈R
2 2
Statement II is true and Statement I is
also true as for some ‘c’.
1
[for c = 0]
⇒
f (c ) =
3
1
which lies betwen 0 and
.
2 2
So, Statement II is correct explanation
of Statement I.
DAY FIVE
Matrices
Learning & Revision for the Day
u
u
u
Matrix
Types of Matrices
Equality of Matrices
u
u
u
Algebra of Matrices
Transpose of a Matrix
Some Special Matrices
u
u
u
Trace of a Matrix
Equivalent Matrices
Invertible Matrices
Matrix
●
●
●
A matrix is an arrangement of numbers in rows and columns.
A matrix having m rows and n columns is called a matrix of order m × n and the
number of elements in this matrix will be mn.
a11 a12 a13 ... a1 n
a
a22 a23 ... a2 n
A matrix of order m × n is of the form A = 21
... ... ... ... ...
a
m1 am2 am3 ... amn
Some important terms related to matrices
●
●
●
The element in the ith row and jth column is denoted by aij.
The elements a11 , a22 , a33 , ...... are called diagonal elements.
The line along which the diagonal elements lie is called the principal diagonal or
simply the diagonal of the matrix.
PRED
MIRROR
Your Personal Preparation Indicator
Types of Matrices
●
●
●
●
●
If all elements of a matrix are zero, then it is called a null or zero matrix and it is
denoted by O.
A matrix which has only one row and any number of columns is called a row matrix
and if it has only one column and any number of rows, then it is called a column
matrix.
If in a matrix, the number of rows and columns are equal, then it is called a square
matrix. If A = [aij]n × n , then it is known as square matrix of order n.
If in a matrix, the number of rows is less/greater than the number of columns, then it
is called rectangular matrix.
If in a square matrix, all the non-diagonal elements are zero, it is called a diagonal
matrix.
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
46
●
●
●
FIVE
40
If in a square matrix, all non-diagonal elements are zero
and diagonal elements are equal, then it is called a
scalar matrix.
If in a square matrix, all non-diagonal elements are zero
and diagonal elements are unity, then it is called an unit
(identity) matrix. We denote the identity matrix of order
n by I n and when order is clear from context then we
simply write it as I.
In a square matrix, if aij = 0, ∀ i > j, then it is called an
upper triangular matrix and if aij = 0, ∀i < j , then it is
called a lower triangular matrix.
NOTE
Transpose of a Matrix
Let A be m × n matrix, then the matrix obtained by interchanging
the rows and columns of A is called the transpose of A and is
denoted by A′ or AC or A T .
If A be m × n matrix, A′ will be n × m matrix.
Important Results
(i) If A and B are two matrices of order m × n, then
( A ± B)′ = A′ ± B ′
(ii) If k is a scalar, then (k A)′ = k A′
(iii) ( A′ )′ = A
(iv) ( AB)′ = B′ A′
(v) ( A n )′ = ( A′ )n
• The diagonal elements of diagonal matrix may or may
not be zero.
Equality of Matrices
Two matrices A and B are said to be equal, if they are of
same order and all the corresponding elements are equal.
Algebra of Matrices
●
●
●
If A = [aij]m × n and B = [b ij] m × n be two matrices of same
order, then A + B = [aij + b ij]m × n and A − B = [aij − b ij]m × n ,
where i = 1, 2, ..., m, j = 1, 2, ..., n.
If A = [aij] be an m × n matrix and k be any scalar, then,
kA = [kaij]m × n .
If A = [aij]m × n and B = [b ij]n × p be any two matrices such
that number of columns of A is equal to the number of
rows of B, then the product matrix AB = [c ij], of order
m × p, where c ij =
ik
b kj .
Some Important Properties
●
●
●
●
●
●
●
A + B = B + A (Commutativity of addition)
( A + B) + C = A + (B + C) (Associativity of addition)
α ( A + B) = αA + αB, where α is any scalar.
(α + β) A = αA + βA, where α and β are any scalars.
α (βA) = (αβ) A, where α and β are any scalars.
( AB) C = A (BC) (Associativity of multiplication)
AI = A = IA
A (B + C) = AB + AC (Distributive property)
NOTE
●
●
●
●
●
●
n
∑a
k =1
●
Some Special Matrices
• A2 = A ⋅ A, A3 = A ⋅ A ⋅ A = A2 ⋅ A1 , K
• If the product AB is possible, then it is not necessary that
the product BA is also possible. Also, it is not necessary
that AB = BA.
• The product of two non-zero matrices can be a zero
matrix.
●
A square matrix A is called an idempotent matrix,
if it satisfies the relation A2 = A.
A square matrix A is called nilpotent matrix of order k,
if it satisfies the relation A k = O, for some k ∈ N .
The least value of k is called the index of the nilpotent
matrix A.
A square matrix A is called an involutary matrix,
if it satisfies the relation A2 = I .
A square matrix A is called an orthogonal matrix,
if it satisfies the relation AA′ = I or A′ A = I .
A square matrix A is called symmetric matrix,
if it satisfies the relation A′ = A.
A square matrix A is called skew-symmetric matrix,
if it satisfies the relation A′ = − A.
NOTE
• If A and B are idempotent matrices, then A + B is idempotent
iff AB = − BA.
a1 a2
• If A = b1 b2
c1 c 2
a3
b3 is orthogonal, then
c 3
Σ ai2 = Σ bi2 = Σ c i2 = 1 and Σ ai bi = Σ bi c i = Σ ai c i = 0
• If A
B are symmetric matrices of the same order, then
(i) AB is symmetric if and only if AB = BA .
(ii) A ± B , AB + BA are also symmetric matrices.
• If A and B are two skew-symmetric matrices, then
(i) A ± B, AB − BA are skew-symmetric matrices.
(ii) AB + BA is a symmetric matrix.
• Every square matrix can be uniquely expressed as the sum of
symmetric and skew-symmetric matrices.
1
1
1
1
i.e. A = ( A + A′ ) + ( A − A′ ), where ( A + A′ ) and ( A − A′ )
2
2
2
2
are symmetric and skew-symmetric respectively.
47
DAY
(iii) Addition of constant multiple of the elements of any row
(column) to the corresponding elements of any other row
(column), indicated as
Ri → Ri + kR j (Ci → Ci + kC j).
Trace of a Matrix
The sum of the diagonal elements of a square matrix A is
called the trace of A and is denoted by tr( A).
(i) tr( λA) = λ tr( A)
(iii) tr( AB) = tr( BA)
(ii) tr( A ) = tr( A ′ )
Invertible Matrices
●
Equivalent Matrices
Two matrices A and B are said to be equivalent, if one is
obtained from the other by one or more elementary operations
and we write A ~ B.
Following types
operations.
of
operations
are
called
elementary
●
Some Important Results
●
(i) Interchanging any two rows (columns).
This transformation is indicated by
Ri ↔ R j (Ci ↔ C j)
(ii) Multiplication of the elements of any row (column) by a
non-zero scalar quantity, indicated as
Ri → kRi (Ci → kCi )
A square matrix A of order n is said to be invertible if there
exists another square matrix B of order n such that
AB = BA = I .
The matrix B is called the inverse of matrix A and it is
denoted by A −1 .
●
●
●
●
●
Inverse of a square matrix, if it exists, is unique.
AA −1 = I = A −1 A
If A and B are invertible, then ( AB)−1 = B −1 A −1
( A − 1 )T = ( A T )− 1
If A is symmetric, then A −1 will also be symmetric matrix.
Every orthogonal matrix is invertible.
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
0 1
0 −1
, then which of the following is
0
1 If A = and B =
1 1
1
correct?
j
(a) (A + B) ⋅ (A − B) = A + B
(c) (A + B) ⋅ (A − B) = I
2
2
NCERT Exemplar
(b) (A + B) ⋅ (A − B) = A 2 − B 2
(d) None of these
2 If p, q , r are 3 real numbers satisfying the matrix
3 4 1
equation, [ p q r ] 3 2 3 = [ 3 0 1 ], then 2p + q − r is
2 0 2
j
equal to
JEE Mains 2013
(a) − 3
(b) − 1
(c) 4
(d) 2
3 In a upper triangular matrix n × n, minimum number of
zeroes is
n (n − 1)
2
2n (n − 1)
(c)
2
(a)
(b)
n (n + 1)
2
(d) None of these
1 2
a 0
and B =
4 Let A =
; a , b ∈ N. Then,
3 4
0 b
(a) there exists more than one but finite number of B’s such
that AB = BA
(b) there exists exactly one B such that AB = BA
(c) there exist infinitely many B’s such that AB = BA
(d) there cannot exist any B such that AB = BA
2 3
1 −2 3
5 If A =
and B = 4 5, then
−4 2 5
2 1
(a) AB, BA exist and are equal
(b) AB, BA exist and are not equal
(c) AB exists and BA does not exist
(d) AB does not exist and BA exists
6 If ω ≠ 1 is the complex cube root of unity and matrix
ω 0
70
H=
, then H is equal to
0 ω
(a) H
(b) 0
(c) −H
(d) H 2
7 If A and B are 3 × 3 matrices such that AB = A and
BA = B, then
(a) A 2 = A and B 2 ≠ B
(c) A 2 = A and B 2 = B
(b) A 2 ≠ A and B 2 = B
(d) A 2 ≠ A and B 2 ≠ B
8 For each real number x such that − 1 < x < 1, let
1
1 − x
A( x ) =
−x
1 − x
(a) A (z) =
(b) A(z) =
(c) A (z) =
(d) A(z) =
−x
x +y
1− x
and z =
. Then,
1
1
+ xy
1 − x
A(x) + A(y)
A(x) [A(y)]−1
A(x) ⋅ A(y)
A(x) − A(y)
48
FIVE
40
cos α
9 If A(α ) = sin α
0
(a) A(αβ)
− sin α 0
cos α 0, then A(α ) A(β ) is equal to
0
1
(b) A(α + β)
(c) A(α − β)
(d) None
10 If A is 3 × 4 matrix and B is a matrix such that A′ B and
(b) 3 × 4
(c) 3 × 3
(d) 4 × 4
2
1 2
11 If A = 2 1 −2 is a matrix satisfying the equation
a 2 b
AAT = 9I , where I is 3 × 3 identity matrix, then the ordered
j JEE Mains 2015
pair (a,b) is equal to
(a) (2, − 1)
(b) (−2, 1)
(c) (2, 1)
0 0 1
1 0 0
0 0 0
0 0 1
(d) (−2, − 1)
12 If E = 0 0 1 and F = 0 1 0, then E 2 F + F 2E
(a) F
(b) E
(c) 0
(d) None
13 If A and B are two invertible matrices and both are
symmetric and commute each other, then
−1
−1
14 If neither α nor β are multiples of π /2 and the product AB
x
equal to
2
y
(a) −1
(b) 1
cos 2 α
sin α cos α
A=
cos
sin
sin2 α
α
α
cos 2 β
cos β sin β
B=
β
β
cos
sin
sin2 β
(a) XY
(c) − YX
1 0 0
1 0 0
20 Let A = 0 1 1 , I = 0 1 0 and
0 −2 4
0 0 1
1 2
( A + cA + dI ) . The values of c and d are
A −1 =
6
(a) (− 6, − 11)
(c) (− 6, 11)
21 Elements of a matrix A of order 9 × 9 are defined as
aij = ωi + j (where ω is cube root of unity), then trace ( A ) of
the matrix is
(b) 1
(b) multiple of π
(d) None of these
1
2
1 −1
4
1
1 − 3 and A −1 =
0
−5
10
1
1
1
1
2
−
α is equal to
(b) 5
2
α , then
3
(d) − 1
(c) 2
symmetric
skew-symmetric
orthogonal
None of the above
24 Let A be a square matrix satisfying A 2 + 5A + 5I = O .
The inverse of A + 2I is equal to
(a) A − 2I
(c) A − 3I
1
(b) nilpotent
(d) orthogonal
1
(a)
48
(1 / 3)
1 0
(c)
16 1
(b) symmetric
(d) orthogonal
(b) A + 3I
(d) does not exist
0
0
1
1
0
(b) 3 1 − 1 1
2
3 48
(d) None of these
26 If X is any matrix of order n × p and I is an identity matrix
a
a −1
−2
2
17 If A = a + 1
1
a + 4 is symmetric, then a is
4a
5
−2
2
(c) −1
(d) ω2
48
25 Let A =
. Then A is
1 / 3 1
cos θ − sin θ
, then
cos θ
(b) 2
(c) ω
22 If A = 2
(a)
(b)
(c)
(d)
16 If A =
sin θ
(a) A is skew-symmetric
(c) idempotent
(b) (6, 11)
(d) (6, − 11)
23 If A is skew-symmetric and B = (I − A ) (I + A ) , then B is
3
1 2
15 The matrix 1 2 3 is
−1 −2 −3
(a) idempotent
(c) involutary
(b)YX
(d) None of these
−1
is null matrix, then α − β is
(a) 0
(c) an odd multiple of π/ 2
(d) − 2
(c) 2
X = AB + BA and Y = AB − BA, then ( XY )T is equal to
(a) − 2
of matrices
(a) −2
2
(a) 0
−1
(a) both A B and A B are symmetric
(b) neither A −1B nor A −1B −1 are symmetric
(c) A −1B is symmetric but A −1B −1 is not symmetric
(d) A −1B −1 is symmetric but A −1B is not symmetric
and
2
19 If A and B are symmetric matrices of the same order and
BA′ are both defined, then B is of the type
(a) 4 × 3
1
18 If A = 2 1 − 2 and AT A = AAT = I , then xy is
(d) None
of order n × n, then the matrix M = I − X ( X ′ X )−1 X ′ is
I. Idempotent matrix
II. MX = O
(a) Only I is correct
(c) Both I and II are correct
(b) Only II is correct
(d) None of them is correct
49
DAY
27 Let A and B be two symmetric matrices of order 3.
Statement I A (BA) and ( AB ) A are symmetric
matrices.
Statement II AB is symmetric matrix, if matrix
multiplication of A with B is commutative.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
28 Consider the following relation R on the set of real square
matrices of order 3.
R = {( A, B ) : A = P −1BP for some invertible matrix P}
Statement I R is an equivalence relation.
Statement II For any two invertible 3 × 3 matrices M
and N,(MN )−1 = N −1M −1.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true; Statement II is
correct explanation of Statement I
(c) Statement I is true, Statement II is true; Statement II is
not a correct explanation of Statement I
(d) Statement I is true, Statement II is false
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
2
1
2
1 If A =
, then I + 2A + 3A + ... ∞ is equal to
− 4 − 2
4 1
(a)
− 4 0
2
1
2
3
5
5
(b)
(c) − 8 − 3 (d) − 3 – 8
− 4 − 1
1 2
is
3 4
2 The matrix A that commute with the matrix
1
2
1
(c) A =
3
(a) A =
2b
2a
2a
1 2b
(b) A =
2 3a 2a + 3b
3b 2a + 3b
2a
2a + 3b
(d) None of these
3
2
+ 3b
a
a
3 The total number of matrices that can be formed using 5
different letters such that no letter is repeated in any
matrix, is
(a) 5!
(c) 2 × (5!)
(b) 2 × 5 5
(d) None of these
4 If A is symmetric and B is a skew-symmetric matrix, then
for n ∈ N, which of the following is not correct?
(a) A n
(b) B n
(c) A n
(d) B n
is symmetric
is symmetric if n is even
is symmetric if n is odd only
is skew-symmetric if n is odd
2 1
5 4
5 − 4
Z =
. Then, the value of the sum
− 6 5
(a) 6
(c) 12
1
1
I and A + I are orthogonal matrices, then
2
2
(a) A is orthogonal
(b) A is skew-symmetric matrix
(c) A is symmetric matrix
(d) None of the above
−1 + i 3
2i
7 If A =
+
i 3
1
i
2
−1 − i 3
2i
, i = −1 and f ( x ) = x 2 + 2,
1−i 3
2i
then f ( A ) is equal to
5
(a)
1
(c)
0
− i 3 1 0
2 0 1
0
1
X (YZ )3
+ K to ∞ is
8
(b) 9
(d) None of these
3 − i 3 1 0
(b)
2
0 1
1 0
(d) (2 + i 3 )
0 1
1 0
n
8 If A =
, then A is equal to
1
1
(a) 2 n − 1 A − (n − 1) I
(c) 2 n − 1 A + (n − 1) I
cos θ
5 Consider three matrices X =
, Y = 6 5 and
4 1
X (YZ )2
XYZ
tr ( X ) + tr
+ tr
+ tr
2
4
6 If both A −
(b) nA − (n − 1) I
(d) nA + (n − 1) I
sin θ
n
9 Let A =
. Let A = [bij ]2 × 2 . Define
− sin θ cos θ
An
lim A n = lim [bij ]2 × 2 . Then lim is
n→ ∞
n→ ∞ n
n→ ∞
(a) zero matrix
0 1
(c)
−1 0
(b) unit matrix
(d) limit does not exist
10 If B is skew-symmetric matrix of order n and A is n × 1
column matrix and AT BA = [ p ], then
(a) p < 0
(c) p > 0
(b) p = 0
(d) Nothing can be said
50
FIVE
40
11 If A, B and A + B are idempotent matrices, then AB is
14 If A1, A3 , ..., A2 n − 1 are n skew-symmetric matrices of
equal to
(b) − BA
(a) BA
3
12 If P = 2
− 1
2
is equal to
(d) O
1 2 + 3
4 −1
1 2019
(d)
4 2 + 3
(a)
(c)
1
7
1
7
2
3
−6
−2
3
6
2r − 1
)2 r
−1
will be
(a) symmetric
(b) skew-symmetric
(c) neither symmetric nor skew-symmetric
(d) data not adequate
a b c
15 Let matrix A = b c a , where a, b, c are real positive
c a b
numbers with abc = 1. If AT A = I , then a 3 + b 3 + c 3 is
4 − 2019 3
6057
(a) 3
(c) 2
1
2 − 3
and B = A −1A′, then BB ′ equals
2 − 3
2019
−3
6
2
−3
6
2
(b) 4
(d) None of these
16 If A is an 3 × 3 non-singular matrix such that AA′ = A′ A
−1
(a) (B )′
(c) I
(b)
(d)
1
7
1
7
6
2
3
6
2
−6
2
−3
6
−2
2
2
j
JEE Mains 2014
(b) I + B
(d) B −1
17 A is a 3 × 3 matrix with entries from the set {−1, 0, 1} . The
13 Which of the following is an orthogonal matrix?
6
2
3
−6
2
−3
n
∑ ( 2r − 1) ( A
r =1
1
2 , A = 1 1 and Q = PAP T , then P T Q 2019P
0 1
3
2
1 2019
(a)
1
0
4 + 2019 3
(b)
2019
(c)
(c) I
same order, then B =
3
6
− 2
3
−3
3
probability that A is neither symmetric nor
skew-symmetric is
(a)
39 − 36 − 33 + 1
39
(b)
39 − 36 − 33
39
(c)
39 − 36 + 1
39
(d)
39 − 33 + 1
39
ANSWERS
SESSION 1
1. (d)
11. (d)
21. (a)
2. (a)
12. (b)
22. (b)
3. (a)
13. (a)
23. (c)
4. (c)
14. (c)
24. (b)
5. (b)
15. (b)
25. (c)
6. (a)
16. (d)
26. (c)
7. (c)
17. (b)
27. (b)
8. (c)
18. (c)
28. (c)
9. (b)
19. (c)
10. (b)
20. (c)
SESSION 2
1. (c)
11. (b)
2. (a)
12. (a)
3. (c)
13. (a)
4. (c)
14. (b)
5. (a)
15. (d)
6. (b)
16. (c)
7. (d)
17. (a)
8. (b)
9. (a)
10. (b)
51
DAY
Hints and Explanations
SESSION 1
5 Here, A is 2 × 3 matrix and B is 3 × 2
1 Here,
1 0 −1 0
+
=
1 1 0 2
1 0 −1 0
−
=
1 1 0 0
0 1 0 1
A2 = A ⋅ A =
1 1 1 1
0
1
0
A−B =
1
A+ B =
0 + 1 0 + 1 1
=
0 + 1 1 + 1 1
0 − 1 0
and B 2 = B ⋅ B =
⋅
1 0 1
−1 0
=
0 −1
=
∴ A2 + B 2 =
0
1
2
1
1
2
−1
0
1 1 −1 0 0 1
+
=
1 2 0 −1 1 1
1 1 −1 0 2 1
−
=
A2 − B 2 =
1 2 0 −1 1 3
0 0 0 2
and ( A + B )( A − B ) =
2 1 0 1
0 + 0 0 + 0 0 0
=
=
0 + 0 4 + 1 0 5
Clearly, ( A + B )( A − B ) ≠ A2 − B 2
≠ A + B ≠ I.
2
2
matrix.
∴Both AB and BA exist, and AB is a 2 × 2
matrix, while BA is 3 × 3 matrix.
∴
AB ≠ BA.
ω 0 ω2 0
=
0 ω 0 ω2
2
3
0
ω
0
0
ω
ω
H3 =
=
2 0 ω
3
0
0
ω
ω
ω70 0 ω 69 ⋅ ω
0
70
=
∴ H =
70
ω 69 ⋅ ω
0 ω 0
(ω3 )23 ⋅ ω
0
=
3 23
(ω ) ⋅ ω
0
ω 0
=
=H
[Qω3 = 1]
0 ω
H2 =
3(3 − 1)
Here, number of zeroes = 3 =
2
∴ Minimum number of zeroes
n (n − 1)
=
2
1 2 a 0 a 2b
4 Clearly, AB =
=
3 4 0 b 3a 4 b
a 0 1 2 a 2 a
and
BA =
=
0 b 3 4 3b 4 b
If AB = BA, then a = b .
Hence, AB = BA is possible for
infinitely many values of B’s.
ω 0
0 ω
B =I
⇒
A=I
⇒
A2 = A
cos α
9 A(α ) A(β ) = sin α
0
∴
and A (z ) =
1−
=
=
1 + xy
1 + xy − x
1 + xy
(1 − x ) (1 −
…(i)
…(ii)
1
(x + y)
1 + xy
( x + y )
1
−
1 + xy
(x + y)
1
−
xy
1
+
( x + y )
1
−
1 + xy
− y − ( x + y )
1
xy
1
+
( x + y )
1
−
1 + xy
y ) − ( x + y )
1
xy
1
+
1
=
(1 − x ) (1 − y )
− ( x + y )
1 + xy
…(iii)
− ( x + y ) 1 + xy
Now, consider
− sin α
cos α
0
cos β
× sin β
0
0
0
1
− sin β
cos β
0
0
0
1
cos(α + β ) − sin (α + β ) 0
= sin(α + β ) cos(α + β ) 0
0
0
1
= A(α + β )
Now, for A ′ B to be defined, order of B
should be 3 × m and for BA′ to be
defined, order of B should be n × 4.
Thus, for both A ′ B and BA′ to be
defined, order of B should be 3 × 4.
8 We have,
1 1 − x
1 − x − x 1
1 1 −y
A( y ) =
1 − y − y 1
1
⋅
(1 − x ) (1 − y )
− ( x + y )
1 + xy
…(iv)
− ( x + y ) 1 + xy
10 Clearly, order of A′ is 4 × 3.
Hence, A2 = A and B 2 = B
A ( x) =
−y
1
From Eqs. (iii) and (iv), we get
A(z ) = A( x ) ⋅ A( y ).
⇒ B2 = B
Similarly, BA = B
3 We know that, a square matrix A = [aij]
is said to be an upper triangular matrix
if aij = 0, ∀ i > j.
Consider, an upper triangular matrix
1 2 3
A = 0 5 6
0 0 7 3 × 3
=
7 Since, AB = A
∴
1
⋅
(1 − x ) (1 − y )
1 − x 1
− x 1 − y
6 Clearly,
2 [3 p + 3q + 2r , 4 p + 2q + 0,
p + 3q + 2r ] = [3 0 1]
⇒ 3 p + 3q + 2r = 3, 4 p + 2q = 0,
p + 3q + 2r = 1
⇒ p = 1,q = −2, r = 3
∴ 2 p + q − r = 2 − 2 − 3 = −3
A ( x ) ⋅ A( y ) =
1
2
2
11 Given, A = 2 1 −2
a 2
1 2
A T = 2 1
2 −2
b
a
2
b
1 2 2
Now, AA T = 2 1 −2
a 2 b
⇒
1 2 a
2 1 2
2 −2 b
9
0
a + 4 + 2b
=
0
9
2a + 2 − 2b
2
2
a + 4 + 2b 2a + 2 − 2b a + 4 + b
It is given that, AA T = 9I
9
0
a + 4 + 2b
0
9
2a + 2 − 2b
2
2
a + 4 + 2b 2a + 2 − 2b a + 4 + b
⇒
9 0 0
= 0 9 0
0 0 9
On comparing, we get
a + 4 + 2b = 0
⇒
a + 2b = −4
2a + 2 − 2b = 0
…(i)
52
FIVE
40
⇒
a − b = −1
…(ii)
and
…(iii)
a2 + 4 + b 2 = 9
On solving Eqs. (i) and (ii), we get
a = −2,b = −1
This satisfies Eq. (iii) also.
Hence, (a,b ) ≡ (−2,−1)
12 F is unit matrix ⇒ F 2 = F
= A −1 B
−1
= A −1 (BA ) A −1 ⇒ BA −1 = A −1 B]
⇒ A −1 B is symmetric.
Now, consider
( A −1 B −1 )T = ((BA )−1 )T
A
T −1
= (A )
[Q AB = BA]
) = (A
T −1
(B )
−1 T
) (B
= A
−1
−1 T
B
)
−1
⇒ A −1 B −1 is also symmetric.
2
14 AB = cos α cos α2sin α
cos α sin α
sin α
cos 2 β
cos β sin β
×
sin2 β
cos β sin β
=
21 Clearly, tr ( A ) = a11 + a22 + a33 + a44
∴
X = AB + BA
will be a symmetric matrix and
Y = AB − BA will be a skew-symmetric
matrix.
Thus, we get X T = X and Y T = − Y
Now, consider ( XY )T = Y T X T
= (− Y )( X ) = − YX
⇒ (6 A −1 )A = ( A2 + cA + dI ) A
= (( AB )−1 )
cos α cos β cos(α − β)
sin α cos β cos(α − β )
cos α sin β cos(α − β )
sin α sin β cos(α − β )
0 0
=
0 0
⇒
cos(α − β ) = 0
⇒
α − β = (2n + 1) π / 2
3
1 2
15 Let A = 1 2 3
− 1 −2 − 3
0 0 0
Then, A2 = 0 0 0
0 0 0
Hence, A is nilpotent matrix of index 2.
cos θ sin θ
16 A ′ =
≠ A or − A.
− sin θ cos θ
cos θ − sin θ
A A′ =
sin θ cos θ
[Q Post multiply both sides by A]
⇒ 6( A −1 A ) = A3 + cA2 + dIA
+ a55 + a66 + a77 + a88 + a99
= ω2 + ω 4 + ω 6 + ω 8 + ω10 + ω12
+ ω14 + ω16 + ω18
= (ω + ω + 1) + (ω + ω + 1)
2
+ (ω2 + ω + 1) [Qω3 n = 1, n ∈ N ]
= 0+ 0+ 0
2
⇒
A + cA + dA − 6I = O
2
22 Clearly, AA −1 = I
Now, if R1 of A is multiplied by C3 of
A −1 , we get 2 − α + 3 = 0 ⇒ α = 5
23 Consider,
= (I − A )−1 (I + A ) (I − A ) (I + A )−1
= (I − A )−1 (I − A ) (I + A ) (I + A )−1
…(i)
1
Here, A2 = A ⋅ A = 0
0
1 0
0 1
0 −2
0 0
1 1 ×
−2 4
0 1
0
0
1 = 0 − 1 5
4 0 −10 14
0
0
1
and A3 = A2 ⋅ A = 0 −1 5 ×
0 −10 14
0
0
1 0 0 1
0 1 1 = 0 −11 19
0 −2 4 0 −38 46
Now, from Eq. (i), we get
0
0
0
0
1
1
0 −11 19 + c 0 −1 5
0 −38 46
0 −10 14
1
+ d 0
0
1 0 0 0
− 6 0 1 0 = 0
0 0 1 0
[Q1 + ω + ω2 = 0]
=0
[Q A −1 A = I and IA = A ]
3
2
BB T = (I − A )−1 (I + A ) (I + A )T [(I − A )−1 ] T
⇒ 6I = A + cA + dA
3
6
0 0 0
= 0 0 0
0 0 0
⇒ 1 + c + d − 6 = 0;
− 11 − c + d − 6 = 0
⇒ c + d = 5; − c + d = 17
On solving, we get c = − 6, d = 11.
These value also satisfy other equations.
a = 2.
20 Clearly, 6 A −1 = A2 + cA + dI
T
= (B
⇒ a2 − a − 2 = 0, a2 − 4a + 4 = 0
19 Since, A and B are symmetric matrices
[Q A T = A and B T = B ]
−1 T
2
orthogonal to the other rows.
⇒
R1 ⋅ R3 = 0
⇒
x + 4 + 2y = 0
Also,
R2 ⋅ R3 = 0
⇒
2x + 2 − 2y = 0
On solving, we get x = − 2, y = − 1
∴
xy = 2
= B T ( A T )−1 = B A −1
−1
⇒ a − 1 = a + 1, a + 4 = 4a
2
18 Since, A is orthogonal, each row is
13 Consider, ( A −1 B )T = B T ( A −1 )T
[Q AB = BA ⇒ A ( AB )A
0
1 + c + d − 6
⇒
0
− 11 − c + d − 6
0
− 38 − 10c − 2d
0
19 + 5c + d
46 + 14c + 4 d −
17 A is symmetric
⇒
+ F 2 E = E2 + E
0 1 0 0 1
0 1 × 0 0 1
0 0 0 0 0
0 0 0
= 0 0 0
0 0 0
∴ E 2 + E = E.
E2 F
0
Also, E 2 = 0
0
and
−1
cos θ sin θ 1 0
=
=I
− sin θ cos θ 0 1
∴ A is orthogonal.
0 0
1 1
−2 4
0 0
0 0
0 0
= I⋅I = I
Hence, B is an orthogonal matrix.
24 We have, A2 + 5A + 5I = O
⇒
A2 + 5A + 6I = I
⇒
⇒
( A + 2I ) ( A + 3I ) = I
A + 2I and A + 3I are inverse of
each other.
25 If A =
, then A2 =
2a 1
a
1
1 0
1
0
1 0
1 0
, …, A n =
3a 1
na 1
Here, a = 1 / 3,
1 0
∴ A 48 =
16 1
A3 =
26 We have, M = I − X ( X ′ X )−1 X ′
= I − X ( X −1 ( X ′ )−1 )X ′
[Q( AB )−1 = B −1 A −1 ]
= I − ( XX −1 ) (( X ′ )−1 X ′ )
=I −I ×I
=I −I
=O
[by associative property]
[Q AA −1 = I = A −1 A]
[Q I 2 = I ]
53
DAY
Clearly, M 2 = O = M
So, M is an idempotent matrix. Also,
MX = O .
27 Given, A = A and B = B
T
T
Statement I [ A (BA )] T = (BA )T ⋅ A T
T
T
= (A B )A
T
= (AB) A = A (BA)
So, A(BA ) is symmetric matrix.
Similarly, ( AB ) A is symmetric matrix.
Hence, Statement I is true. Also,
Statement II is true but not a correct
explanation of Statement I.
28 Given, R = {( A, B ) : A = P −1 BP for
some invertible matrix P}
For Statement I
(i) Reflexive ARA
⇒
A = P −1 AP
which is true only, if P = I .
Thus, A = P −1 AP for some
invertible matrix P.
So, R is Reflexive.
(ii) Symmetric
ARB ⇒ A = P −1 BP
⇒ PAP −1 = P (P −1 BP ) P −1
⇒ PAP −1 = ( PP −1 ) B( PP −1 )
B = PAP −1
∴
Now, let Q = P
Then,
−1
B = Q −1 AQ ⇒ BRA
⇒ R is symmetric.
(iii) Transitive ARB and BRC
⇒ A = P −1 BP
= (P −1Q −1 )C (QP )
= (QP )−1 C (QP )
So, ARC.
⇒ R is transitive
So, R is an equivalence relation.
For Statement II It is always true
that (MN )−1 = N −1 M −1
Hence, both statements are true but
second is not the correct
explanation of first.
SESSION 2
1 Clearly, A2 =
− 4 − 2 − 4 − 2
1
2
a b 1 2 1 2 a b
=
c d 3 4 3 4 c d
a + 3b 2a + 4 b
⇒
c + 3d 2c + 4 d
b + 2d
a + 2c
=
3a + 4c 3b + 4d
On equating the corresponding
elements, we get
…(i)
a + 3b = a + 2c ⇒ 3b = 2c
…(ii)
2a + 4b = b + 2d ⇒ 2a + 3b = 2d
…(iii)
c + 3d = 3a + 4c ⇒ a + c = d
…(iv)
2c + 4d = 3b + 4d ⇒ 3b = 2c
Thus, A can be taken as
b 1 2a
a
2b
3b
3 =
a + b 2 3b 2a + 3b
2
2
3 Clearly, matrix having five elements is
of order 5 × 1 or 1 × 5.
∴Total number of such matrices = 2 × 5!.
4 ( A n )′ = ( A AL A )′ = ( A ′ A ′L A ′ )
= ( A ′ )n = A n for all n
1
0 0
=O
0 0
∴ I + 2 A + 3 A2 + ... = I + 2 A
=
2 5
2
1 0 4
=
+
=
0 1 − 8 − 4 − 8 − 3
1
1
⇒ A ′ − I A − I = I
2
2
1 ′
and A + I
2
⇒
From Eq. (i), we get
1
1
1
A ′ A − IA ′ − IA + I = I
2
2
4
1
1
1
⇒ A ′ A − A ′− A + I = I
2
2
4
…(ii)
…(iii)
Similarly, from Eq. (ii), we get
1
1
1
A ′ A + A ′ + A + I = I …(iv)
2
2
4
On subtracting Eq. (iii) from Eq. (iv), we
get
A + A′ = O
A′ = − A
or
Hence, A is a skew-symmetric matrix.
ω2
ω
i =ω 1
− ω − 1
−ω
i
i
1 − ω2
0
A2 = − ω2
0
1
−
ω2
ω
7 We have, A = i 2
− ω
i
∴
∴ A is symmetric for all n ∈ N .
Q
− ω2 + ω 4
0
=
2
4
0
−
+
ω
ω
− ω2 + ω
0
=
0
− ω2 + ω
2
f ( x) = x + 2
[given]
∴
f ( A ) = A2 + 2I
Also, B is skew-symmetric
⇒
B ′ = − B.
∴ (B n )′ = (B B L B )′ = (B ′ B ′ L B ′ )
= (B ′ )n
− ω2 + ω
2 0
0
=
+ 0 2
2
−
+
0
ω
ω
1 0
2
= (− ω + ω + 2)
0 1
1
0
= (3 + 2ω )
0 1
1 0
= (2 + i 3 )
0 1
5 4 5 − 4 1 0
5 Here, YZ =
=
6
5 − 6 5 0 1
X (YZ )2
XYZ
∴ tr ( X ) + tr
+ tr
2
4
X (YZ )3
+ tr
+K
8
X
X
= tr ( X ) + tr + tr + K
2
4
1
1
= tr ( X ) + tr ( X ) + tr ( X ) + K
2
4
1
1
= tr ( X ) 1 + + 2 + K
2 2
1
= tr ( X )
1
1−
2
= 2 tr ( X ) = 2 (2 + 1) = 6
6 Since, both A − 1 I and A + 1 I are
2
2
orthogonal, therefore, we have
A − 1 I ′ A − 1 I = I
2
2
…(i)
A + 1 I = I
2
A′ + 1 I A + 1 I = I
2
2
n
⇒ B n is symmetric if n is even and is
skew-symmetric if n is odd.
A = P −1 (Q −1CQ ) P
2
a b
be a matrix that
c d
1 2
commute with
. Then,
3 4
= (− B )n = (−1)n B n .
and B = Q −1CQ
⇒
2 Let A =
8 A2 =
=
1 1 1 1 2 1
1 0 1 0
1 0
1 0 1 0 1 0
=
2 1 1 1 3 1
....................
....................
1 0
An =
n 1
A3 =
0
n 0 n − 1
−
n n 0
n − 1
= nA − (n − 1) I
=
54
FIVE
40
cos θ
sin θ
cos θ
sin θ
9 A2 =
− sin θ cos θ − sin θ cos θ
13 We know that a matrix
cos 2θ sin 2θ
=
− sin 2θ cos 2θ
cos 3θ sin 3θ
3
Similarly, A =
etc
− sin 3θ cos 3θ
cos nθ sin nθ
∴
An =
− sin nθ cos nθ
b11 b12
=
b
21 b22
b ij
A n 0 0
as lim
=
=0
0 0 n→∞ n
n
Now, lim
n→ ∞
10 A T B A = [ p] ⇒ ( A T BA )T = [ p] T = [ p]
⇒
A T B T A = A T (− B ) A = [ p]
⇒
[− p] = [ p] ⇒ p = 0.
matrix
∴ A2 = A; B 2 = B and ( A + B )2 = A + B
Now, consider ( A + B )2 = A + B
A2 + B 2 + AB + BA = A + B
⇒
⇒
A + B + AB + BA = A + B
AB = − BA
Q 2019 = (PAP T ) ( PAP T )
...(PAP T ) = PA2019P T
∴ P Q
Now,
P = P ⋅ PA
2019
P ⋅P = A
2019
T
2019
1 1 1 1 1 2
A =
=
0 1 0 1 0 1
2
A3 =
⇒
T
A2019 =
1 2 1 1 1 3
=
0 1 0 1 0 1
1 2019
0
1
B = A1 + 3 A33 + K + (2n − 1) A 22 nn −− 11
Now, B T = ( A1 + 3 A33
+ K + (2n − 1) A22nn −− 11 )T
= A1T + (3 A33 )T + K + ((2n − 1) A22nn −− 11 )T
= A1T + 3( A3T )3
+ K + (2n − 1) ( A 2T n − 1 )2 n − 1
=− A−
12 P is orthogonal matrix as P P = I
T
T
and
Σ aib i = Σb ic i = Σc i ai = 0
Now, from the given options, only
6 2 −3
1
2 3
6 satisfies these conditions.
7
3
−
6
2
6 2 −3
1
Hence, 2 3
6 is an orthogonal
7
3 −6 2
matrix.
14 We have,
11 Since, A, B and A + B are idempotent
⇒
a1 a2 a3
A = b1 b2 b3 will be orthogonal if
c 1 c 2 c 3
AA ′ = I , which implies
Σ a2i = Σb 2i = Σc 2i = 1
3 A33
− K − (2n −
1) A22nn −− 11
[Q A1 , A3 , K , A 2 n − 1 are skewsymmetric matrices
∴ ( A i )T = − A i ∀ i = 1, 3, 5, .... 2n − 1 ]
= − [ A + 3 A33 + K + (2n − 1) A22nn −− 11 ]
=−B
Hence, B is a skew-symmetric matrix.
a b c
a b c
c
c
15 A T A = b c a × b c a
a b
a b
a2 + b 2 + c 2 ab + bc + ca
= ab + bc + ca b 2 + c 2 + a2
ab + bc + ca ab + bc + ca
ac + ab + bc
ab + bc + ca
a2 + b 2 + c 2
A T A = I ⇒ a2 + b 2 + c 2 = 1
and
ab + bc + ca = 0
Since
a, b, c > 0,
∴ ab + bc + ca ≠ 0 and hence no real
value of a3 + b 3 + c 3 exists.
16 AA ′ = A ′ A, B = A −1 A ′.
BB ′ = ( A −1 A ′ ) ( A −1 ⋅ A ′ )′
= ( A −1 A ′ ) [( A ′ )′ ( A −1 )′ ]
= ( A −1 A ′ ) [ A( A ′ )−1 ]
[Q ( A −1 )′ = ( A ′ )−1 ]
= A
−1
−1
(A′ A) (A′ )
= A −1 ( AA ′ ) ( A ′ )−1
[Q A ′ A = AA ′ ]
= ( A −1 A ) [ A ′ ( A ′ )−1 ]
=I⋅I =I
17 Total number of matrices = 39. A is
symmetric, then aij = a ji .
Now, 6 places (3 diagonal, 3
non-diagonal), can be filled from any of
−1, 0, 1 in 36 ways. A is skew-symmetric,
then diagonal entries are ‘o’ and
a12 , a13 , a23 can be filled from any of
−1, 0, 1 in 33 ways. Zero matrix is
common.
∴Favourable matrices are
39 − 36 − 33 + 1.
Hence, required probability
39 − 36 − 33 + 1
=
39
DAY SIX
Determinants
Learning & Revision for the Day
u
Determinants
u
Properties of Determinants
u
Cyclic Determinants
u
Area of Triangle by using
Determinants
u
Minors and Cofactors
u
Adjoint of a Matrix
u
u
Inverse of a Matrix
Solution of System of Linear
Equations in Two and Three
Variables
Determinants
Every square matrix A can be associated with a number or an expression which is called its
determinant and it is denoted by det (A) or |A| or ∆ .
a11 a12 ... a1 n
a11 a12 ... a1 n
a
a22 ... a2 n
a
a22 ... a2 n
, then det ( A) = 21
If A = 21
M
M
M
M
M
M
M
M
a
a
...
a
a
a
...
a
n1
n2
nn
n1
n2
nn
●
a b
a b
If A =
, then | A| =
= ad − bc
c d
c d
●
a b c
a b c
If A = p q r , then | A| = p q r
u v w
u v w
=a
PRED
MIRROR
Your Personal Preparation Indicator
q r
p r
p q
[expanding along R1 ]
−b
+c
v w
u w
u v
There are six ways of expanding a determinant of order 3 corresponding to each of three rows
(R1 , R2 , R3 ) and three columns (C1 , C2 , C3 ).
• Rule to put + or − sign in the expansion of determinant
of order 3.
• A square matrix A is said to be singular, if | A| = 0
and non-singular, if | A| ≠ 0.
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
= a (qw − vr ) − b ( pw − ur ) + c ( pv − uq )
NOTE
u
+ − +
− + −
+ − +
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
56
SIX
40
Properties of Determinants
(i) If each element of a row (column) is zero, then ∆ = 0.
(ii) If two rows (columns) are proportional, then ∆ = 0.
(iii) | A T| = | A|, where A T is a transpose of a matrix.
(iv) If any two rows (columns) are interchanged, then ∆
becomes −∆.
(v) If each element of a row (column) of a determinant is
multiplied by a constant k, then the value of the new
determinant is k times the value of the original
determinant
(vi) det (kA) = k n det( A), if A is of order n × n.
(vii) If each element of a row (column) of a determinant is
written as the sum of two or more terms, then the
determinant can be written as the sum of two or more
determinants i.e.
a1 + a2 b c
a1 b c
a2 b c
p1 + p2 q r = p1 q r + p2 q r
u1 v w
u1 + u2 v w
u2 v w
(viii) If a scalar multiple of any row (column) is added to
another row (column), then ∆ is unchanged
a b c
a
b
c
i.e. p q r = p + ka q + kb r + kc , which is
u v w
u
v
w
obtained by the operation R2 → R2 + kR1
Product of Determinants
a1
If| A| = a2
a3
b1
b2
b3
c1
α1
c2 and| B| = α 2
c3
α3
β1
β2
β3
γ1
γ2 , then
γ3
a1α 1 + b1β1 + c1 γ1 a1α 2 +
| A| × | B| = a2α 1 + b2β1 + c2 γ1 a2α 2 +
a3α 1 + b3β1 + c3 γ1 a3α 2 +
a1α 3 + b1β3
a2α 3 + b2β3
a3α 3 + b3β3
b1β2 + c1 γ2
b2β2 + c2 γ2
b3β2 + c3 γ2
+ c1 γ3
+ c2 γ3 =| AB|
+ c3 γ3
[multiplying row by row]
We can multiply rows by columns or columns by rows or
columns by columns
NOTE
• | AB | = | A| |B | = |BA| = | AT B | =| ABT | = | AT BT |
• | An | = | A| n , n ∈ Z +
Cyclic Determinants
In a cyclic determinant, the elements of row (or column) are
arranged in a systematic order and the value of a determinant
is also in systematic order.
1
(i) 1
1
x
y
z
x2
y2 = ( x − y) ( y − z) (z − x)
z2
1
(ii) 1
1
x
y
z
1
(iii) 1
1
x2
y2
z2
(iv)
x3
y3 = ( x − y) ( y − z) (z − x)( x + y + z)
z3
x3
y3 = ( x − y) ( y − z) (z − x)( xy + yz + zx)
z3
a b c
b c a = − (a + b + c)(a 2 + b 2 + c2 – ab – bc – ca)
c a b
= − (a3 + b 3 + c3 − 3 abc)
abc
a a2
abc = b b 2
c c2
abc
a bc
(v) b ca
c ab
a3
b 3 = abc (a − b )(b − c)(c − a)
c3
Area of Triangle by
using Determinants
If A( x1 , y1 ), B( x2 , y2 ) and C( x3 , y3 ) are vertices of ∆ABC, then
Area of ∆ABC =
1
2
x1
x2
x3
y1
y2
y3
1
1
1
1
[ x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )]
2
x1 y1 1
If these three points are collinear, then x2 y2 1 = 0
x3 y3 1
and vice-versa.
=
Minors and Cofactors
The minor M ij of the element aij is the determinant
obtained by deleting the ith row and jth column of ∆.
a11 a12 a13
If
∆ = a21 a22 a23 ,
then M11
a31
a
= 22
a32
a32 a33
a23
a21
, M12 =
a33
a31
a23
etc.
a33
The cofactor Cij of the element aij is (−1)i + j M ij.
a11 a12 a13
a
a23
a
If ∆ = a21 a22 a23 , then C11 = 22
, C12 = − 21
a32 a33
a31
a31 a32 a33
a23
etc.
a33
The sum of product of the elements of any row (or column)
with their corresponding cofactors is equal to the value of
determinant.
i.e.
∆ = a 11 C 11 + a 12 C 12 + a 13 C 13
= a 21 C21 + a 22 C 22 + a 23 C 23
= a 31 C 31 + a 32 C 32 + a 33 C 33
But if elements of a row (or column) are multiplied with
cofactors of any other row (or column), then their sum is zero.
57
DAY
Adjoint of a Matrix
1. Matrix Method
If A = aij
In this method we first write the above system of equations in
matrix form as shown below.
a1 b1 c1 x d1
a b c y = d or AX = B
2
2
2
2
a
b
c
3
d3
3
3 z
[ ]
nxn
, then adjoint of A, denoted by adj ( A), is defined
as [Cij]Tnxn , where Cij is the cofactor of aij.
a11
If A = a21
a31
NOTE
a12
a22
a32
a13
C11
a23 , then adj ( A) = C21
a33
C31
d
a b
, then adj ( A) =
• If A =
c
− c
d
c12
C22
C32
C13
C23
C33
T
− b
.
a
Properties of Adjoint of a Matrix
Let A be a square matrix of order n, then
a1
where, A = a2
a3
●
●
(iii) adj (AB) = (adj B) (adj A)
●
(iv) adj ( A T ) = (adj A)T
c1
c2
c3
x
d1
, X = y and B = d
2
z
d3
Case I When system of equations is non-homogeneous
(i.e. when B ≠ 0).
(i) (adj A) A = A (adj A) = | A| ⋅ I n
(ii) |adj A | = | A|n − 1 , if| A| ≠ 0
b1
b2
b3
(v) adj (adj A) = | A|n – 2 A, if| A| ≠ 0
If| A| ≠ 0, then the system of equations is consistent and has
a unique solution given by X = A −1 B.
If| A| = 0 and (adj A), B ≠ 0, then the system of equations is
inconsistent and has no solution.
If| A| = 0 and (adj A) ⋅ B = O, then the system of equations
may be either consistent or inconsistent according as the
system have infinitely many solutions or no solution.
2
(vi) |adj (adj A)| = | A |( n – 1 ) , if| A| ≠ 0
Inverse of a Matrix
Let A be any non-singular (i.e.| A| ≠ 0) square matrix, then
inverse of A can be obtained by following two ways.
1. Using determinants
In this,
A −1 =
1
adj ( A)
| A|
Case II When system of equations is homogeneous
(i.e. when B = 0).
●
●
2. Cramer’s Rule Method
In this method we first determine
2. Using Elementary operations
In this, first write A = IA (for applying row operations) or
A = AI (for applying column operations) and then reduce A of
LHS to I, by applying elementary operations simultaneously
on A of LHS and I of RHS. If it reduces to I = PA or I = AP,
then P = A −1 .
Properties of Inverse of a Matrix
(i) A square matrix is invertible if and only if it is
non-singular.
(ii) If A = diag (λ 1 , λ 2 ,..., λ n ),
Solution of System of Linear
Equations in Two and Three
Variables
Let system of linear equations in three variables be
a1 x + b1 y + c1 z = d1 , a2 x + b2 y + c2 z = d2
and
a3 x + b3 y + c3 z = d3 .
Now, we have two methods to solve these equations.
a1
D = a2
a3
b1
b2
b3
c1
d1
c2 , D1 = d2
c3
d3
b1
b2
b3
a1
D2 = a2
a3
d1
d2
d3
c1
a1
c2 and D3 = a2
c3
a3
c1
c2 ,
c3
b1
b2
b3
d1
d2
d3
Case I When system of equations is non-homogeneous
●
●
then A −1 = diag (λ−11 , λ−12 ,..., λ−1n ) provided
λ i ≠ 0 ∀, i = 1, 2, … n.
If| A| ≠ 0, then system of equations has only trivial solution,
namely x = 0, y = 0 and z = 0.
If| A| = 0, then system of equations has non-trivial solution,
which will be infinite in numbers.
●
If D ≠ 0, then it is consistent with unique solution given by
D
D
D
x = 1 , y = 2 ,z = 3.
D
D
D
If D = 0 and atleast one of D1 , D2 and D3 is non-zero, then it
is inconsistent (no solution).
If D = D1 = D2 = D3 = 0, then it may be consistent or
inconsistent according as the system have infinitely many
solutions or no solution.
Case II When system of equations is homogeneous
If D ≠ 0, then x = y = z = 0 is the only solution, i.e. the
trivial solution.
If D = 0, then it has infinitely many solutions.
Above methods can be used, in a similar way, for the solution
of system of linear equations in two variables.
●
●
SIX
40
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 If x = cy + bz , y = az + cx and z = bx + ay , where x , y
and z are not all zero, then a 2 + b 2 + c 2 is equal to
(a) 1 + 2abc
(c) 1 + abc
(b) 1 − 2abc
(d) abc −1
2 Consider the set A of all determinants of order 3 with
entries 0 or 1 only. Let B be the subset of A consisting of
all determinants with value 1 and C be the subset of A
consisting of all determinants with value –1. Then,
(a) C is empty
(b) B and C have the same number of elements
(c) A = B ∪ C
(d) B has twice as many elements as C
1
3 If x , y and z are positive, then logy x
logz x
logx y logx z
logy z is
1
logz y
1
equal to
(a) 0
(c) −1
(b) 1
(d) None of these
4 If a , b and c are cube roots of unity, then
ea
eb
ec
(a) 0
e2a
e 2b
e2c
e3a −1
e 3b − 1 is equal to
e3c −1
(c) e 2
(b) e
(d) e 3
x −1 x + 3
−2x x − 4 , where p, q , r , s
x + 4 3x
and t are constants, then t is equal to
(a) 0
(b) 1
(c) 2
(d) –1
1
x
x +1
6 If f ( x ) =
2x
x ( x − 1)
( x + 1)x
,
3x ( x − 1) x ( x − 1)( x − 2) ( x + 1)x ( x −1)
then f ( 50) is equal to
(a) 0
(b) 50
(d) −50
(c) 1
7 If α , β and γ are the roots of the equation x + px + q = 0 ,
3
α β γ
then the value of the determinant β γ α is
γ α β
(a) 0
(b) −2
(c) 2
(d) 4
8 If ω is a cube root of unity, then a root of the following
x − ω − ω2
ω
ω2
(a) x = 0
(c) x = ω
a
bω
cω 2
bω 2
c
aω
aω
bω2 is
c
(a) a 3 + b 3 + c 2 − 3 abc
(c) 0
ω
ω2
x − ω −1
1
= 0 is
1
x − 1 − ω2
(b) x = − 1
(d) None of these
(b) a 2 b − b 2 c
(d) a 2 + b 2 + c 2
x − 4 2x
2x
10 If 2x x − 4 2x = ( A + Bx )( x − A )2 , then the
2x
2x
x −4
ordered pair ( A, B ) is equal to
(a) (−4,−5)
(c) (−4,5)
j
JEE Mains 2018
(b) (−4, 3)
(d) (4,5)
11 If x, y, z are non-zero real numbers and
1+ x
1
1
1 + y 1 + 2y
1
= 0,
1 + z 1 + z 1 + 3z
then x −1 + y −1 + z −1 is equal to
(b) −1
(d) −6
(a) 0
(c) −3
5 If px 4 + qx . 3 + rx 2 + sx + t
x 2 + 3x
= x +1
x −3
9 If ω is an imaginary cube root of unity, then the value of
b+c c+a a+b
a b c
12 If c + a a + b b + c = k b c a , then k is equal to
a+b b+c c+a
c a b
(a) 0
(b) 1
(c) 2
(d) 3
13 Let a, b and c be such that (b + c ) ≠ 0. If
a a + 1 a −1
a +1
b +1
c −1
−b b + 1 b − 1 +
a −1
b −1
c + 1 = 0,
( −1)n + 2 a (−1)n +1b ( −1)n c
c c −1 c + 1
then the value of n is
(a) zero
(c) an odd integer
(b) an even integer
(d) an integer
14 If one of the roots of the equation
7
6
x 2 − 13
2
x 2 − 13
2
= 0 is x = 2, then sum of all
2
x − 13
3
7
other five roots is
(a) – 2
(c) 2 5
(b) 0
(d) 15
15 If a, b and c are sides of a scalene triangle, then the value
a b c
of b c a is
c a b
(a) non-negative
(c) positive
JEE Mains 2013
(b) negative
(d) non-positive
j
59
DAY
x + ky + 3z = 0, 3x + ky − 2z = 0
2x + 4y − 3z = 0
xz
has a non-zero solution ( x , y , z ), then 2 is equal to
y
and
j
(a) −10
(c) −30
JEE Mains 2018
(b) 10
(d) 30
17 If A = [aij ]nxn and aij = (i 2 + j 2 − ij )( j − i ), n is odd, then
which of the following is not the value of tr ( A ).
(a) 0
(c) 2 | A |
(b) | A |
(d) None of these
18 If the coordinates of the vertices of an equilateral triangle
with sides of length a are ( x1, y1 ),( x 2 , y 2 ) and ( x 3 , y 3 ),
2
x1 y1 1
then x 2 y 2 1 is equal to
y3 1
x3
(a)
a4
4
j
(b)
3a2
4
(c)
5a4
4
NCERT Exemplar
3a4
(d)
4
19 The points A (a, b + c ), B (b, c + a ) and C (c, a + b ) are
(a)
(b)
(c)
(d)
vertices of an isoscele triangle
vertices of an equilateral triangle
collinear
None of the above
(a) x1 ∆
(c) (x1 + y1)∆
2
21 If A =
−4
z1
B
z 2 , then 2
B3
z3
C2
equals to
C3
(b) x1 x 3 ∆
(d) None of these
−3
, then adj ( 3A 2 + 12A ) is equal to
1
j
JEE Mains 2017
72 − 84
(a)
− 63 51
51 84
(c)
63 72
51 63
(b)
84 72
72 − 63
(d)
− 84 51
22 Which of the following is/are incorrect?
(i) Adjoint of a symmetric matrix is symmetric
(ii) Adjoint of a unit matrix is a unit matrix
(iii) A(adj A ) = (adj A ) =| A| I
(iv) Adjoint of a diagonal matrix is a diagonal matrix .
(a) (i)
(c) (iii) and (iv)
(b) (ii)
(d) None of these
1 α 3
23 If P = 1 3 3 is the adjoint of a 3 × 3 matrix A and
2 4 4
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JEE Mains 2013
| A| = 4, then α is equal to
(a) 4
(b) 11
(c) 5
to
j
(a) − 1
(b) 5
(c) 4
JEE Mains 2016
(d) 13
a b c
q −b y
25 If A = x y z , B = −p a −x and if A is invertible,
−c z
p q r
r
then which of the following is not true?
(a) | A | = −| B | and | adj A| ≠| adj B|
(b) | A | = − | B | and | adj A | = | adj B |
(c) A is invertible iff B is invertible
(b) None of the above
26 If A an 3 × 3 non-singular matrix such that AA′ = A′ A and
B = A −1A′, then BB′ is equal to
(a) I + B
(b) I
j
(c) B −1
JEE Mains 2014
(d) (B −1)′
−1
1
27 If
tan θ
− tan θ 1
tan θ a −b
, then
=
1 − tan θ
1 b a
(a) a = 1, b = 1
(b) a = sin 2θ, b = cos 2θ
(c) a = cos 2θ, b = sin2θ
(d) None of the above
1 0 0
elements x1, y1, z1,.... of the determinant
y1
y2
y3
24 If A =
3
28 If A = 2 1 0 and u1 , u 2 are column matrices such
20 If A1 , B1,C1 , ......... are respectively the co-factors of the
x1
∆ = x2
x3
− b
and A adj A = AAT , then 5a + b is equal
2
5a
16 If the system of linear equations
(d) 0
3 2 1
1
0
that Au1 = 0 and Au 2 = 1 , then u1 + u 2 is equal to
0
0
−1
(a) 1
0
−1
(b) 1
−1
−1
(c) −1
0
1
(d) −1
−1
29 The number of values of k, for which the system of
equations (k + 1)x + 8y = 4k
and k x + (k + 3)y = 3k − 1 has no solution, is
(a) infinite
(c) 2
(b) 1
(d) 3
j
JEE Mains 2013
30 The system of equations
αx + y + z = α − 1, x + αy + z = α − 1
and x + y + αz = α − 1
has no solution, if α is
(a) 1
(c) either –2 or 1
(b) not –2
(d) –2
31 If the system of linear equations
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JEE Mains 2013
x1 + 2x 2 + 3x 3 = 6, x1 + 3x 2 + 5x 3 = 9
2x1 + 5x 2 + ax 3 = b
is consistent and has infinite number of solutions, then
(a)
(b)
(c)
(d)
a = 8, b can be any real number
b = 15, a can be any real number
a ∈ R − {8 } and b ∈ R − {15 }
a = 8, b = 15
60
SIX
40
32 If the trivial solution is the only solution of the system of
equations
x − ky + z = 0, kx + 3y − kz = 0
and
3x + y − z = 0
Then, the set of all values of k is
(a) { 2 , − 3 }
(b) R − { 2 , − 3 } (c) R − { 2 } (d) R − { − 3 }
33 Let A, other than I or − I, be a 2 × 2 real matrix such that
A 2 = I , I being the unit matrix. Let tr ( A ) be the sum of
j JEE Mains 2013
diagonal elements of A.
Statement I tr ( A ) = 0
Statement II det ( A ) = − 1
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
34 The set of all values of λ for which the system of linear
equations 2x1 − 2x 2 + x 3 = λx1, 2x1 − 3x 2 + 2x 3 = λx 2 and
– x1 + 2x 2 = λx 3 has a non-trivial solution.
(a)
(b)
(c)
(d)
is an empty set
is a singleton set
contains two elements
contains more than two elements
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JEE Mains 2015
35 Statement I Determinant of a skew-symmetric matrix of
order 3 is zero.
Statement II For any matrix A , det( AT ) = det( A ) and
det( − A ) = − det( A ).
Where, det ( A ) denotes the determinant of matrix A.
j JEE Mains 2013
Then,
(a)
(b)
(c)
(d)
Statement I is true and Statement II is false
Both statements are true
Both statements are false
Statement I is false and Statement II is true
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
6 If the equations a ( y + z ) = x , b( z + x ) = y , c( x + y ) = z
1 If a 2 + b 2 + c 2 = −2 and
1 + a x (1 + b )x (1 + c )x
f ( x ) = (1 + a 2 )x 1 + b 2 x (1 + c 2 )x ,
(1 + a 2 )x (1 + b 2 )x 1 + c 2 x
2
2
2
have non-trivial solution, then
to
(a) 1
(b) 3
(c) 0
(d) 1
2 If A is a square matrix of order 3 such that | A| = 2, then
−1 −1
| (adj A ) | is
(a) 1
(b) 2
(c) 4
(d) 8
3 The equations (k − 1)x + ( 3k + 1)y + 2kz = 0,
(k − 1)x + ( 4k − 2)y + (k + 3)z = 0
and
2x + ( 3k + 1)y + 3(k − 1)z = 0
gives non-trivial solution for some values of k, then the
ratio x : y : z when k has the smallest of these values.
(a) 3 : 2 :1
(b) 3 : 3 : 2
(c) 1 : 3 : 1
(d) 1:1:1
4 If x = 1 + 2 + 4 +.... upto k terms, y = 1 + 3 + 9+.... upto k
terms and c = 1 + 5 + 25+ ... upto k terms. Then,
x
∆= 3
2k
(a) (20)k
2y
3
3k
4z
3 equals to
5k
(b) 5 k
(c) 0
1 + 2x
x
3
(b)
4
(d) −2
(c) – 1
system of equations in x , y and z.
x2 y2 z2
x2 y2 z2
x2 y2 z2
+ 2 − 2 = 1, 2 − 2 + 2 = 1 and − 2 + 2 + 2 = 1
2
a
b
c
a
b
c
a
b
c
has
(a) no solution
(b) unique solution
(c) infinitely many solutions (d) finitely many solutions
8 If S is the set of distinct values of b for which the following
system of linear equations
x + y + z = 1, x + ay + z = 1 and ax + by + z = 0
j
has no solution, then S is
JEE Mains 2017
(a) an infinite set
(b) a finite set containing two or more elements
(c) singleton set
(d) a empty set
9 If M is a 3 × 3 matrix, where M T M = I and
5 Product of roots of equation 2 − x
1
(a)
2
(b) 2
7 Let a, b and c be positive real numbers. The following
then f ( x ) is a polynomial of degree.
(a) 2
1
1
1
is equal
+
+
1+ a 1+ b 1+ c
4
(c)
3
(d) 2 k + 3 k + 5 k
1
1− x
2+ x 3+ x = 0
1+ x 1− x 2
1
(d)
4
det(M) = 1, then the value of det (M − I ) is
(a) –1
(b) 1
(c) 0
(d) None of these
10 If a1, a 2 , . . ., a n , . . . are in GP, then the determinant
log a n
∆ = log a n + 3
log a n + 6
(a) 2
(b) 4
log a n + 1
log a n + 4
log a n + 7
(c) 0
log a n
log a n
log a n
+ 2
+ 5
is equal to
+ 8
(d) 1
61
DAY
11 Let A be a square matrix of order 2 with | A| ≠ 0 such that
| A + | A| adj ( A )| = 0, then the value of | A −| A| adj ( A )| is
(a) 1
(c) 3
(b) 2
(d) 4
14 Area of triangle whose vertices are (a, a 2 ),(b, b 2 ),(c, c 2 ) is
1
, and the area of triangle whose vertices are
2
( p, p 2 ),(q , q 2 ) and (r , r 2 ) is 4, then the value of
12 Let P and Q be 3 × 3 matrices P ≠ Q . If P 3 = Q 3 and
(1 + ap )2
(1 + aq )2
(1 + ar )2
P 2Q = Q 2P , then the determinant of (P 2 + Q 2 ) is
(a) − 2
(c) 0
(b) 1
(d) − 1
3
1 + f (1) 1 + f ( 2)
13 If α , β ≠ 0, f (n ) = α + β and 1 + f (1) 1 + f ( 2) 1 + f ( 3)
1 + f ( 2) 1 + f ( 3) 1 + f ( 4)
n
(b) 4
j
1
(b)
αβ
(c) 1
JEE Mains 2014
(d) − 1
(1 + cp )2
(1 + cq )2 is
(1 + cr )2
(c) 8
(d) 16
l
m n
r and if (l − m )2 + ( p − q )2 = 9,
1 1 1
n
= K (1 − α )2 (1 − β )2 (α − β )2 , then K is equal to
(a) αβ
(a) 2
(1 + bp )2
(1 + bq )2
(1 + br )2
15 Let det A = p q
(m − n )2 + (q − r )2 = 16, (n − l )2 + (r − p )2 = 25, then the
value of (det A )2 equals to
(a) 36
(b) 100
(c) 144
(b) 169
ANSWERS
SESSION 1
SESSION 2
1 (b)
2 (b)
3 (a)
4 (a)
5 (a)
6 (a)
7 (a)
8 (a)
9 (c)
10 (c)
11 (c)
12 (c)
13 (c)
14 (a)
15 (b)
16 (b)
17 (d)
18 (d)
19 (c)
20 (a)
21 (b)
22 (d)
23 (b)
24 (b)
25 (a)
26 (b)
27 (c)
28 (d)
29 (d)
30 (d)
31 (d)
32 (b)
33 (b)
34 (c)
35 (a)
1 (a)
11 (d)
2 (c)
12 (c)
3 (d)
13 (c)
4 (c)
14 (d)
5 (a)
15 (c)
6 (b)
7 (d)
8 (c)
9 (c)
10 (c)
62
SIX
40
Hints and Explanations
SESSION 1
1 Given equation can be rewritteen as
x − cy − bz = 0
cx − y + az = 0
bx + ay − z = 0
Since, x, y and z are not all zero
∴The above system have non-trivial
solution.
1 −c −b
∴
c −1 a = 0
b a −1
⇒ 1(1 − a2 ) + c (−c − ab ) − b( ac + b ) = 0
⇒
1 − a2 − c 2 − abc − abc − b2 = 0
⇒
a2 + b 2 + c 2 = 1 − 2abc
1 e a e 2a
1 e a e 2a
b
2b
= 1 e
e
− 1 e b e 2b
c
2c
1 e e
1 ec e2c
− a b aω2
= ω ⋅ ω − b c bω2
− c a cω2
= 0 [Q a + b + c = 0 ⇒ e a + b + c = 1 ]
a b a
= − ω5 b c b = 0
c a c
5 Put x = 0 in the given equation, we get
t =
log x
log x
log x
=
log y
log x
log z
log y
log x
log y
log y
log y
log z
log z
log x
log z
log y
log z
log z
By taking common factors from the
rows, we get
1
∆ =
log x ⋅ log y ⋅ log z
log x log y
log x log y
log x log y
C2 , ( x + 1) from C3 and ( x − 1) from
R3 , we get
1
1
1
f ( x ) = x ( x2 − 1) 2 x x − 1 x
3x x − 2 x
e 3a
e a e 2a
3b
e
− e b e 2b
3c
e
ec e2c
1
0
0
= x ( x − 1) 2 x − ( x + 1) 1
3 x − 2( x + 1) 2
2
1
1
1
= e a ⋅ eb ⋅ ec
1 e a e 2a
e a 1 e 2a
b
2b
+ e b 1 e 2b
1 e
e
c
2c
1 e e
ec 1 e2c
C3 → C3 − C2 ,
C → C − C
2
1
2
=0
∴
f (50) = 0
7 Clearly, α + β + γ = 0
On applying C1 → C1 + C2 + C3 in the
given determinant, we get
α +β+ γ β γ
∴∆ = α +β+ γ γ α
α +β+ γ α β
⇒
∴
x
x
ω
x + ω2
ω2
1
x
1
x+ω
ω
x + ω2
1
2x
x−4
2x
2x
2x
x − 4
= ( A + Bx )( x − A ) 2
On applying C1 → C1 + C2 + C3 , we get
5x − 4
5x − 4
5x − 4
2x
x−4
2x
2x
2x
x − 4
= ( A + Bx )( x − A )2
On taking common (5x − 4) from C1 , we
get
2x
1 2 x
(5x − 4)1 x − 4 2 x
x − 4
1 2 x
= ( A + Bx )( x − A )2
On applying R2 → R2 − R1
and R3 → R3 − R1 , we get
2x
2x
1
∴ (5x − 4) 0 − x − 4
0
0
0
−
x
−
4
=0
[Q1 + ω + ω2 = 0]
ω2
=0
1
x+ω
On comparing, we get
A = − 4 and B = 5
1+ x
1
1
11 Let ∆ = 1 + y 1 + 2 y
1 =0
1 + z 1 + z 1 + 3z
On applying C1 → C1 − C3
and C2 → C2 − C3 , we get
∆ =
x=0
a(1 + ω ) bω2 aω
9 ∆ = b(ω + ω2 ) c bω2
c (ω2 + 1) aω
c
− aω2 bω2
= −b
c
− cω aω
x − 4
2x
2x
On expanding along C1 , we get
(5x − 4)( x + 4)2 = ( A + Bx )( x − A ) 2
8 On applying C1 → C1 + C2 + C3 , we get
1
x 1
1
10 Given,
= ( A + Bx )( x − A )2
0 β γ
= 0 γ α =0
0 α β
log z
log z
log z
Now, by taking common factor from the
1 1 1
columns, we get 1 1 1 = 0
1 1 1
e a e 2a
4 Let ∆ = e b e 2b
ec e2c
0 −1
3
1
0 − 4 = − 12 + 12 = 0
−3
4
0
6 On taking common factors x from
2 If we interchange any two rows of a
determinant in the set B, its value
becomes −1 and hence it is in C.
Likewise, for every determinant in C,
there is corresponding determinant in B.
So , B and C have the same number of
elements.
1
log x y log x z
3 Let ∆ = log y x
1
log y z
log z x log z y
1
2
[QC1 → C1 + C3 ]
aω
bω2
c
x
0
1
y
2y
1
−2z −2z 1 + 3z
On expanding along R1 , we get
∆ = x[2 y (1 + 3z ) + 2z]
+1[−2 yz + 4 yz] = 0
⇒ 2[ xy + 3 xyz + xz] + 2 yz = 0
⇒
xy + yz + zx + 3 xyz = 0
1 1 1
+ + = −3
⇒
x y z
63
DAY
b + c c + a a+ b
12 Let ∆ = c + a a + b b + c
⇒ ( x2 − 4)
a+ b b + c c + a
2 (a + b + c ) c + a a + b
= 2 (a + b + c ) a + b b + c
2 (a + b + c ) b + c c + a
[using C 1 → C 1 + C 2 + C 3 ]
a+ b + c c + a a+ b
=2 a+ b + c a+ b b + c
a+ b + c b + c c + a
[taking common 2 from C1 ]
a + b + c −b −c
= 2 a + b + c −c − a
a + b + c − a −b
[using C 2
a
=2 b
c
→
b
c
a
C2 − C1 , C3 → C3 − C1 ]
c
a
b
[using C1 → C1 + C2 + C3 ]
On comparing, k = 2
a a+ 1 a−1
13 − b b + 1 b − 1 + (−1)n
c c −1 c + 1
a a+ 1 a−1
= −b b + 1 b − 1
c c −1 c + 1
a+ 1 a−1 a
+ (−1)n b + 1 b − 1 −b
c −1 c + 1 c
[Q| A ′|= | A|]
a a+ 1 a−1
= −b b + 1 b − 1
c c −1 c + 1
a−1
b −1
c +1
a a+ 1 a−1
= [1 + (− 1)n +2 ] − b b + 1 b − 1
c c −1 c + 1
This is equal to zero only, if n + 2 is an
odd i.e. n is an odd integer.
7
14
2
x2 − 13
6
x2 − 13
3
1
1
x2 − 13 2 = 0
3
7
Now, on applying C2 → C2 − C1
and C3 → C3 − C1 , we get
1
0
0
( x2 − 4)
2
x2 − 15
0
=0
x2 − 13 16 − x2 20 − x2
On expanding along R1 , we get
( x2 − 4)( x2 − 15)( x2 − 20) = 0
Thus, other five roots are
−2, ± 15, ± 2 5
Hence, sum of other five roots is −2.
a b c
15 Let ∆ = b c a
c a b
=
=
=
=
a(bc − a2 ) − b (b2 − ac ) + c ( ab − c 2 )
abc − a3 − b 3 + abc + abc − c 3
− (a3 + b 3 + c 3 − 3 abc )
− (a + b + c ) (a2 + b 2
+ c 2 − ab − bc − ca)
1
= − (a + b + c )
2
{(a − b ) 2 + (b − c ) 2 + (c − a) 2 }
< 0, where a ≠ b ≠ c
18 If ( x1 , y 1 ),( x2 , y 2 ) and ( x3 , y 3 ) are the
vertices of
x1 y 1
1
= x2 y 2
2
x3 y 3
x2 − 13
2
=0
7
x + ky + 3z = 0; 3 x + ky − 2z = 0;
2 x + 4 y − 3z = 0
System of equation has non-zero
solution, if
1 k 3
3 k −2 = 0
2 4 −3
⇒ (−3k + 8) − k (−9 + 4) + 3(12 − 2k ) = 0
⇒ −3k + 8 + 9k − 4k + 36 − 6k = 0
⇒ −4k + 44 = 0 ⇒ k = 11
Let z = λ , then we get
…(i)
x + 11 y + 3λ = 0
…(ii)
3 x + 11 y − 2λ = 0
and
2 x + 4 y − 3λ = 0
…(iii)
On solving Eqs. (i) and (ii), we get
5λ
−λ
,y =
,z = λ
x=
2
2
xz
5λ2
⇒
=
= 10
2
2
y
λ
2 × −
2
17 We have,
∴
aij = (i 2 + j2 − ij )( j − i )
a ji = (i 2 + j2 − ij )(i − j )
= −(i 2 + j2 − ij )( j − i ) = −aij
a triangle, then
1
1
1
Area
…(i)
Also, we know that, if a is the length of
an equilateral triangle, then
Area =
3 2
a
4
…(ii)
From Eqs. (i) and (ii), we get
⇒
x1
3 2 1
a =
x2
4
2
x3
y1
y2
y3
1
1
1
x1
3 2
a = x2
2
x3
y1
y2
y3
1
1
1
On squaring both sides, we get
2
x1 y 1 1
3 4
a = x2 y 2 1
4
x3 y 3 1
19 Clearly, area of
a b +c
1
b c + a
2
c a+ b
a
1
= (a + b + c ) b
2
c
(∆ABC ) =
16 We have,
a+ 1 b + 1 c −1
a−1 b −1 c + 1
a
−b
c
a+ 1
a
+ (−1)n + 1 b + 1 − b
c −1 c
1
2
x2 − 13
1
1
1
1 1
1 1 =0
1 1
[QC2 → C2 + C1 ]
x1 z1
= x1 z3 − x3 z1
x3 z3
x
y1
C2 = − 1
= −( x1 y 3 − x3 y 1 )
x3 y 3
x z1
B3 = − 1
= −( x1 z2 − x2 z1 )
x2 z2
x
y1
and C3 = 1
= x1 y 2 − y 1 x2
x2 y 2
B C2
∴ 2
B3 C 3
x1 z3 − x3 z1
−( x1 y 3 − y 1 x3 )
=
−( x1 z2 − x2 z1 )
x1 y 2 − x2 y 1
20 Clearly, B2 =
=
x1 z3
− x1 z2 + x2 z1
− x1 y 3
x1 y 2 − x2 y 1
− x3 z1
y 1 x3
− x1 z2 + x2 z1 x1 y 2 − x2 y 1
− x1 y 3
x z
x z − x1 y 3
= 1 3
+ 1 3
− x1 z2 x1 y 2
x2 z1 − x2 y 1
− x3 z1 y 1 x3
− x3 z1
y 1 x3
+
+
− x1 z2 x1 y 2
x2 z1 − x2 y 1
+
On applying R1 → R1 + R2 + R3 , we get
⇒ A is a skew-symmetric matrix of
odd order.
= x12 (z3 y 2 − z2 y 3 ) − x1 x2 (z3 y 1 − z1 y 3 )
x2 − 4
2
x2 − 13
∴
= x1 [ x1 (z3 y 2 − z2 y 3 ) − x2 (z3 y 1 − z1 y 3 )
x2 − 4
x2 − 13
3
x2 − 4
2 =0
7
tr ( A ) = 0 and| A|= 0
− x1 x3 (z1 y 2 − z2 y 1 ) + x2 x3 (z1 y 1 − z1 y 1 )
= x1 ∆
+ x3 (z2 y 1 − z1 y 2 )]
64
SIX
40
21 We have, A =
− 4
2
− 3
1
a −b
=
b a
2b
10 + 3b = 13
15
∴ A2 = A ⋅ A
2 − 3 2 − 3
=
− 4 1 − 4 1
4 + 12 − 6 − 3 16 − 9
=
=
− 8 − 4 12 + 1 − 12 13
16 − 9
Now, 3 A2 + 12 A = 3
− 12 13
2 − 3
+ 12
− 4 1
48 − 27 24 − 36
=
+
− 36 39 − 48 12
72 − 63
=
− 84 51
51 63
∴adj (3 A2 + 12 A ) =
84 72
20b + 45b
= 13
15
65b
= 13 ⇒ b = 3
⇒
15
Now, substituting the value of b in Eq.
(iii), we get 5a = 2
⇒
Hence, 5a + b = 2 + 3 = 5
q −b y
25 Clearly,|B|= − p a − x
r
−c z
c
1 α 3
2 4 4
Find the determinant of P and apply the
formula
|adj A|=| A|n −1
5a − b
24 Given, A =
2
3
and A adj A = AA T
Clearly, A(adj A ) = A I2
[Q if A is square matrix of order n, then
A(adj A ) = (adj A ) ⋅ A = A I n ]
5a − b
=
I2
3
2
= (10a + 3b ) I2
a x
=−b y
c z
cos 2θ − sin 2θ a −b
⇒
=
sin 2θ cos 2θ b a
1 0 1 +
Au1 + Au2 = 0 + 1 = 0 +
0 0 0 +
z
(taken (–1) common from C2 )
p a x
= − q b y (QR1 ↔ R2 )
r c z
23 Given, adj A = P = 1 3 3
p
q
r
⇒
Since, A is non-singular matrix, i.e.
| A | ≠ 0, on multiplying both sides by
A −1 , we get
1
A −1 A(u1 + u2 ) = A −1 1
0
−1
(QC1 ↔ C2 and then C2 ↔ C3 )
= −| A T | = −| A|
Thus,| A|= −|B|
Hence,| A| ≠ 0 iff|B|≠ 0
∴ A is invertible iff B is invertible
Also,|adj A|= | A| = |B| = |adj B |
2
2
1 0 0
1
⇒ u1 + u2 = 2 1 0 × 1
3 2 1
0
1 0 0
Now,
|A | = 2 1 0
3 2 1
25a2 + b 2 15a − 2b
=
13
15a − 2b
...(ii)
Also we have, AA T = A T A and
B = A −1 A T
T
∴ A −1
Now, BB ′ = ( A −1 A T ) ( A −1 A T )T
= A −1 A T A ( A −1 )T
−1
−1 T
[Q( A ′ )′ = A]
= A AA ( A ) [Q AA = A T A ]
Q A(adj A ) = AA T
0
10a + 3b
∴
0
10a + 3b
25a2 + b 2 15a − 2b
=
13
15a − 2b
[using Eqs. (i) and (ii)]
⇒
15a − 2b = 0
2b
...(iii)
⇒
a=
15
and
10a + 3b = 13
...(iv)
On substituting the value of ‘a’ from Eq.
(iii) in Eq. (iv), we get
T
T
= IA T ( A −1 )T = A T ( A −1 )T
= ( A −1 A )T [Q( AB )T = B T A T ]
= IT = I
27 We have,
− tan θ 1
tan θ
1
tan θ
− tan θ
1
1
−1
a −b
=
b a
− tan θ
1
1
⋅
⇒
1 1 + tan2 θ
tan θ
− tan θ a −b
1
=
tan θ
1 b a
⇒
1 − tan2 θ −2 tan θ
1
2
1 + tan θ 2 tan θ
1 − tan2 θ
…(i)
= 1 (1 − 0) − 0 + 0 = 1
∴ adj of A
⇒| A|≠ 0.
...(i)
0
1
0
1
A(u1 + u2 ) = 1
0
26 Given, A is non-singular matrix
1 0
= (10a + 3b )
0 1
10
a
3
b
0
+
=
0
10a + 3b
5a − b 5a 3
and AA T =
2 − b 2
3
a −b
b a
⇒ a = cos 2θ and b = sin 2θ
(taken (–1) common from R2 )
q b y
= (+ ) p a x
r
−2 tan θ
1 + tan2 θ
=
1 − tan2 θ
2
1 + tan θ
28 On adding Au1 and Au2 , we get
q −b y
= − p −a x
r −c z
22 All the given statements are true.
1 − tan2 θ
1 + tan2 θ
⇒
2 tan θ
1 + tan2 θ
0 0
1 −2 1 1
= 0 1 −2 = −2 1 0
0 0 1 1 −2 1
0 0
1
= −2 1 0
[Q| A | = 1]
1 −2 1
From Eq. (i),
0
1
u1 + u2 = −2 1
1 −2
0 1
0 × 1
1 0
1 + 0 + 0 1
⇒ u1 + u2 = −2 + 1 + 0 = −1
1 − 2 + 0 −1
29 Given equations can be written in
matrix form as AX = B
8
k + 1
x
where, A =
, X = y
k
k
+
3
4 k
and B =
3 k − 1
For no solution,| A |= 0
and (adj A) B ≠ 0
65
DAY
Now, | A |=
k+1
8
=0
k
k+3
⇒
(k + 1) (k + 3) − 8 k = 0
⇒
k2 + 4 k + 3 − 8 k = 0
⇒
k2 − 4 k + 3 = 0
⇒
(k − 1) (k − 3) = 0
∴
⇒
∴
But α = 1 makes given three equations
same. So, the system of equations has
infinite solution. Hence, answer is
α = − 2 for which the system of
equations has no solution.
31 For consistency| A | = 0 and
k = 1, k = 3
k + 3 −8
Also, adj A =
− k k + 1
k + 3 −8 4 k
∴(adj A) B =
3 k − 1
−
k
k
+
1
(k + 3) (4 k ) − 8 (3 k − 1)
=
2
−4 k + (k + 1) (3 k − 1)
4 k 2 − 12k + 8
=
2
− k + 2k − 1
Put k = 3, then (adj A)
36 − 36 + 8 8
B =
= ≠ 0, true.
−9 + 6 − 1 −4
Hence, the required value of k is 3.
Alternate Method Condition for the
system of equations has no solution is
4k
a1
b
c
k+1
8
= 1 ≠ 1 ⇒
=
≠
a2
b2
c2
k
k + 3 3k − 1
kx + 3 y − kz = 0 and
3 x + y − z = 0 has trivial solution.
1 −k
1
k
3 −k ≠0
∴
3
1 −1
2− λ
2
–1
∴
35 Determinant of skew-symmetric matrix
of odd order is zero and of even order is
perfect square.
Also, det( A T ) = det( A )
det(− A ) = (−1)n det( A )
and
Hence, Statement II is false.
SESSION 2
1 Given that,
⇒ 1 (− 3 + k ) + k (− k + 3k )
+ 1 (k − 9) ≠ 0
⇒
k − 3 + 2 k2 + k − 9 ≠ 0
1 + a2 x (1 + b 2 )x (1 + c 2 )x
f ( x ) = (1 + a2 )x 1 + b 2 x (1 + c 2 )x
(1 + a2 )x (1 + b 2 )x 1 + c 2 x
⇒
⇒
⇒
On applying C1 → C1 + C2 + C3 , we get
1 + a2 x + x + b 2 x + x + c 2 x
f ( x ) = x + a2 x + 1 + b 2 x + x + c 2 x
x + a2 x + x + b 2 x + 1 + c 2 x
2 k 2 + 2 k − 12 ≠ 0
k2 + k − 6 ≠ 0
(k + 3) (k − 2) ≠ 0
k ≠ 2, − 3
k ∈ R − {2, − 3}
2
[false]
[true]
30 The system of given equations has no
solution, if
1 1
α 1 =0
1 α
On applying C1 → C1 + C2 + C3 , we get
α+2 1 1
α +2 α 1 =0
α +2 1 α
On applying R2 → R2 − R1 , R3 → R3 − R1 ,
−2
1
−(3 + λ ) 2 = 0
2
−λ
⇒ (2 − λ )(3λ + λ2 − 4) + 2(−2λ + 2)
+1(4 − 3 − λ ) = 0
⇒ (2 − λ )(λ2 + 3λ − 4) + 4(1 − λ )
+ (1 − λ ) = 0
⇒ (2 − λ )( λ + 4)( λ − 1) + 5 (1 − λ ) = 0
⇒
(λ − 1)[(2 − λ )(λ + 4) − 5] = 0
⇒
(λ − 1)(λ2 + 2λ − 3) = 0
⇒
(λ − 1)[(λ − 1)(λ + 3)] = 0
⇒
(λ − 1)2 (λ + 3) = 0
⇒
λ = 1,1,−3
(1 + b 2 )x
1 + b2 x
(1 + b 2 )x
a b
33 Let A =
such that A ≠ I , − I
c
d
∴
k =3
Hence, only one value of k exists.
1
1
1
(α + 2) 0 α − 1
0 =0
0
0
α −1
⇒ 1(3 a − 25) − 2(a − 10) + 3(5 − 6) = 0
⇒
a= 8
On solving, (adj A ) B = O , we get
b = 15
∴
Take
α
1
1
(adj A ) B = O
1 2 3
1 3 5 =0
2 5 a
32 Since, x − ky + z = 0,
Put k = 1, then
4 − 12 + 8 0
(adj A) B =
= , not true.
−1 + 2 − 1 0
k+1
8
=
k
k+3
⇒
k 2 + 4 k + 3 = 8k
⇒
k2 − 4 k + 3 = 0
⇒
(k − 1) (k − 3) = 0
∴
k = 1, 3
8
4⋅1
If k = 1, then
≠
,
1+ 3
2
8
4⋅3
,
and k = 3, then
≠
6 9−1
(α + 2)(α − 1)2 = 0
α = 1, − 2
a b
1 0
and
= 0 1
c
d
a2 + bc
⇒
ac + cd
ab + bd 1 0
=
bc + d 2 0 1
⇒ b(a + d ) = 0,c (a + d ) = 0
and a2 + bc = 1,bc + d 2 = 1
⇒
a = 1,d = −1,b = c = 0 or
a = −1,d = 1,b = c = 0
1 0
A=
, then
0 − 1
det( A ) = −1 and tr ( A ) = 1 − 1 = 0
34 Given system of linear equations are
2 x1 − 2 x2 + x3 = λx1
…(i)
⇒ (2 − λ) x1 − 2 x2 + x3 = 0
2 x1 − 3 x2 + 2 x3 = λx2
…(ii)
⇒ 2 x1 − (3 + λ )x2 + 2 x3 = 0
and
− x1 + 2 x2 = λx3
…(iii)
⇒
− x1 + 2 x2 − λx3 = 0
Since, the system has non-trivial
solution.
(1 + c 2 )x
(1 + c 2 )x
1 + c2 x
1 + (a2 + b 2 + c 2 + 2)x (1 + b 2 )x
= 1 + (a2 + b 2 + c 2 + 2)x (1 + b 2 )x
1 + (a2 + b 2 + c 2 + 2)x (1 + b 2 )x
1
= 1
1
(1 + b 2 )x (1 + c 2 )x
1 + b 2 x (1 + c 2 )x
(1 + b 2 )x 1 + c 2 x
(1 + c 2 )x
(1 + c 2 )x
1 + c2 x
[Qa2 + b 2 + c 2 = − 2, given]
On applying
R1 → R1 − R3 , R2 → R2 − R3 , we get
0
0
x −1
= 0
1− x
x −1
1 (1 + b 2 )x 1 + c 2 x
=
0
1− x
x −1
= ( x − 1) 2
x −1
Hence, f ( x ) is of degree 2.
66
SIX
40
−1
2 Clearly, adj A −1 = | A −1 |2 = 1 2
| A|
and |(adj A −1 )−1 |=
6 Here, b
c
1
|(adj A −1 )|
On applying, C2 → C2 − C1 and
C3 → C3 − C1 , we get
−1
a+ 1
a+ 1
b −(b + 1)
0
=0
c
0
− (1 + c )
= | A|2 = 22 = 4
3 For non - trivial solution,
k − 1 3k + 1
2k
k − 1 4k − 2 k + 3 = 0
2
3k + 1 3(k − 1)
⇒
On applying R1 →
−x + y = 0
...(i)
− x − 2 y + 3z = 0
...(ii)
2 x + y − 3z = 0
(iii)
From (i), we get x = y = λ (say), Then,
from Eq. (ii), we get
− λ − 2λ + 3z = 0
⇒
3z = 3λ
⇒
z=λ
∴
x: y : z = 1:1:1
k
4 Clearly, x = 2k − 1, y = 3 − 1
2
5k − 1
and z =
4
Q sum to n terms of a GP is
given by
a(r n − 1)
r − 1
2k − 1 3k − 1 5k − 1
Now, ∆ =
3
3
3
2k
3k
5k
2k 3k
= 3 3
2k 3k
R1
R2
, R2 →
a+ 1
b+1
R3
, we get
c+1
1
−
1 1
a+ 1
b
−1 0 = 0
b+1
c
0 −1
c+1
1
b
c
⇒
−
+
+
=0
a+ 1 b + 1 c + 1
1
1
c
+ 1−
+ 1−
=0
∴−
a+ 1
b+1
c+1
1
1
1
⇒
+
+
=2
a+ 1 b + 1 c + 1
For a = 1
∆ = ∆1 = ∆2 = ∆3 = 0
For b = 1 only
x + y + z = 1, x + y + z = 1
and
x+ y + z=0
i.e. no solution
(QRHS is not equal)
Hence, for no solution, b = 1 only.
9 Clearly,
det(M − I ) = det(M − I ) ⋅ det(M )
[Qdet(M ) = 1]
= det(M − I )det(M T )
and R3 →
k = 0 or 3
Now, when k = 0, then the equation
becomes
and
a a
−1 b = 0
c −1
5k
1 1
3 −3 1 1
5k
2k 3k
1
1
5k
= 0−3× 0= 0
1 + 2x
1
1− x
5 Let p( x ) = 2 − x 2 + x 3 + x = 0
x
1 + x 1 − x2
Clearly, product of roots
Constant term
=
Coefficient of x 4
Here, constant term is given by
1 1 1
P(0) = 2 2 3 = −1
0 1 1
and coefficient of x 4 is −2
1
∴ Product of roots is .
2
2
2
y2
7 Let x2 = X , 2 = Y , z2 = Z
a
b
c
Then, X + Y − Z = 1, X − Y + Z = 1,
−X + Y + Z = 1
Now, coefficient matrix is
1 1 −1
A = 1 −1 1
−1 1 1
Here,| A|≠ 0
∴ It has unique solution, viz., X = 1,
Y = 1 and Z = 1
Hence , x = ± a; y = ± b and z = ± c .
1 1 1
8 Here, ∆ = 1 a 1
a b 1
= 1 (a − b ) − 1 (1 − a) + 1 (b − a2 )
= − (a − 1)2
1 1 1
∆1 = 1 a 1
0 b 1
= 1 (a − b ) − 1 (1) + 1 (b ) = (a − 1)
1 1 1
∆2 = 1 1 1
a 0 1
= 1 (1) − 1 (1 − a) + 1 (0 − a) = 0
1 1 1
and ∆ 3 = 1 a 1
a b 0
= 1 ( − b ) − 1 (− a) + 1 (b − a2 )
= − a(a − 1)
[Q det( A T ) = det( A )]
= det(MM T − M T )
= det(I − M T )
[Q MM T = I ]
= − det(M − I )
T
= − det(M T − I )T
= − det(M − I )
⇒ 2det(M − I ) = 0
⇒ det(M − I ) = 0
10 Since, a1 , a2 , . . . , an, . . . are in GP, then,
log an , log an + 1 , log an + 2 , ...,
log an + 8 , ...are in AP.
Given that,
log an
log an + 1
∆ = log an + 3 log an + 4
log an + 6 log an + 7
a
∴ ∆ = a + 3d
a + 6d
a+ d
a + 4d
a + 7d
log an +2
log an + 5
log an + 8
a + 2d
a + 5d
a + 8d
where a and d are the first term and
common difference of AP.
On applying R2 → 2R2 , we get
a
a+ d
a + 2d
1
∆ =
2a + 6d 2a + 8d 2a + 10d
2
a + 6d
a + 7d
a + 8d
On applying R2
a
1
∆ =
a
2
a + 6d
→ R2 −
a+ d
a+ d
a + 7d
R3 , we get
a + 2d
a + 2d = 0
a + 8d
11 Let A =
. Then,
c d
a b
| A|= ad − bc = k (say)
d −b
and
adj ( A ) =
−c a
Now,| A +| A| adj ( A )|= 0
a + kd (1 − k )b
=0
⇒
(1 − k )c d + ka
⇒
⇒
(a + kd )(d + ka) − (1 − k )2 bc
ad + a2 k + k d 2 + k 2 ad
−(1 + k 2 − 2k )bc
⇒ (ad − bc ) + ( ad − bc )k 2 + k( a2 + d 2
+ 2bc ) = 0
67
DAY
⇒ (ad − bc ) + ( ad − bc ) k 2 + k ( a2 + d 2 )
+2(ad − k )) = 0
[Qbc = ad − k ]
⇒
(ad − bc ) + ( ad − bc )k 2
+ k (a + d )2 − 2k 2 = 0
3
⇒
(k + k − 2k 2 ) + k (a + d )2 = 0
⇒
k [(k − 1)2 + (a + d )2 ] = 0
⇒
k = 1 and a + d = 0
Now,| A −| A|adj A|
a − kd (1 + k )b
a−d
2b
=
=
(1 + k )c d − ak
2c
d−a
= −(a − d ) − 4bc
= −((a + d )2 − 4ad ) − 4bc
= 4ad − 4bc = 4k = 4
2
12 On subtracting the given equation, we
get
P 3 − P 2Q = Q 3 − Q 2 P
⇒
P 2 (P − Q ) = Q 2 (Q − P )
⇒ (P − Q ) (P 2 + Q 2 ) = 0
Now, if |P 2 + Q 2 | ≠ 0
(P 2 + Q 2 ) is invertible .
On post multiply both sides by
(P 2 + Q 2 )−1 , we get
P − Q = 0, which is a contradiction.
Hence,|P 2 + Q 2|= 0
3
1 + f (1) 1 + f (2)
13 Let ∆ = 1 + f (1) 1 + f (2) 1 + f (3)
1 + f (2) 1 + f (3) 1 + f (4)
...(i)
3
1+ α+ β
⇒ ∆ = 1 + α + β 1 + α 2 + β2
1 + α 2 + β2 1 + α 3 + β3
1 + α 2 + β2
1 + α 3 + β3
1 + α4 + β4
1⋅1 + 1⋅1 + 1⋅1 1⋅1 + 1⋅α + 1⋅β
= 1⋅1 + α ⋅1 + β ⋅1 1⋅1 + α ⋅α + α ⋅β
1 ⋅ 1 + 1 ⋅ α 2 + 1 ⋅ β2 1 ⋅ 1 + α 2 ⋅ α + β2 ⋅ β
1 ⋅ 1 + 1 ⋅ α 2 + 1 ⋅ β2
1 ⋅ 1 + α ⋅ α 2 + β ⋅ β2
1 ⋅ 1 + α 2 ⋅ α 2 + β2 ⋅ β2
1 1 1 1 1 1
1 1 1
= 1 α β 1 α β = 1 α β
1 α 2 β2 1 α 2 β2
1 α 2 β2
2
On expanding, we get
∆ = (1 − α )2 (1 − β )2 (α − β2 )
Hence, K (1 − α )2 (1 − β )2 (α − β )2
= (1 − α )2 (1 − β )2 (α − β )2
∴
K =1
(1 + ap )2 (1 + bp )2 (1 + cp )2
14 Let ∆ = (1 + aq )2 (1 + bq )2 (1 + cq )2
(1 + ar )2 (1 + br )2 (1 + cr )2
1 + 2ap + a p 1 + 2bp + b p
= 1 + 2aq + a2q 2 1 + 2bq + b 2q 2
1 + 2ar + a2 r 2 1 + 2br + b 2 r 2
1 + 2cp + c 2 p2
1 + 2cq + c 2q 2
1 + 2cr + c 2 r 2
2
2
2
2
1 p p2
1 2a a2
2
= 1 q q × 1 2b b 2
1 r r2
1 2c c 2
1 p p2 1 a a2
= 2 1 q q 2 × 1 b b 2
1 r r2 1 c c2
= 2(2 A1 × 2 A2 ) = 2(8 × 1) = 16
15 According to given conditions we get a
right angled triangle whose vertices are
(n, r ),(m,q ) and (l , p ) .
A(n, r)
4
B(m, q)
5
C(l, p)
3
l m n
Also, we have,| A|= p q r
1 1 1
2
l p 1
2
⇒| A| = m q 1 = [2ar (∆ABC )]
n r 1
2
2
1
= 2 × × 3 × 4 = 144
2
DAY SEVEN
Binomial Theorem
and Mathematical
Induction
Learning & Revision for the Day
u
u
Binomial Theorem
Binomial Theorem for Positive
Index
u
u
Properties of Binomial
Coefficient
Applications of Binomial
Theorem
u
u
Binomial Theorem for
Negative/Rational Index
Principle of Mathematical
Induction
Binomial Theorem
Binomial theorem describes the algebraic expansion of powers of a binomial. According to this
theorem, it is possible to expand ( x + y)n into a sum involving terms of the form axb yc , where
the exponents b and c are non-negative integers with b + c = n. The coefficient a of each term is
n
a specific positive integer depending on n and b, is known as the binomial coefficient .
b
Binomial Theorem for Positive Index
An algebraic expression consisting of two terms with (+ ) ve or (−)ve sign between them, is called
binomial expression.
If n is any positive integer,
then ( x + a)n = nC0 x n + nC1 x n − 1 a + L + n Cnan
=
Your Personal Preparation Indicator
n
∑ nCr ⋅ x n − r ar , where x and a are real (complex) numbers.
r=0
(i)
(ii)
(iii)
(iv)
PRED
MIRROR
The coefficient of terms equidistant from the beginning and the end, are equal.
n
( x − a)n = nC0 x n − nC1 x n − 1 a + L + (− 1)n Cnan
(1 + x)n = nC0 + nC1 x + nC2 x2 + L + nCn x n
Total number of terms in the expansion ( x + a)n is (n + 1).
(n + 1) (n + 2)
(v) If n is a positive integer, then the number of terms in ( x + y + z)n is
.
2
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
69
DAY
(vi) The number of terms in the expansion of
Properties of Binomial
Coefficients
n + 2
, if n is even
n
n
( x + a) + ( x − a) = 2
n+1
, if n is odd
2
(vii) The number of terms in the expansion of
In the binomial expansion of (1 + x)n,
(1 + x)n = nC0 + nC1 ⋅ x + nC2 ⋅ x2 + ... + nCr ⋅ x r + ... nCn ⋅ x n,
where, nC0, nC1,...,n Cn are the coefficients of various powers of
x are called binomial coefficients and it is also written as
n
,
if n is even
n
n
( x + a) − ( x − a) = 2
n+1
, if n is odd
2
n n
n
C0, C1, ..., Cn or , , … ,
0 1
n
General Term and Middle Term
(i) Let (r + 1)th term be the general term in the expansion of
( x + a)n.
Tr + 1 = nCr x n − r ar
n
●
n
(iii) The middle term in the expansion of (a + x)n.
n
(a) Case I If n is even, then + 1 th term is middle
2
term.
(n + 1)
(b) Case II If n is odd, then
th term and
2
(n + 3)
th terms are middle terms.
2
(iv) ( p + 1)th term from end = (n − p + 1)th term from
beginning.
(v) For making a term independent of x we put r = n in
general term of ( x + a)n, so we get nCnan, that is
independent of x.
NOTE
If the coefficients of rth, ( r + 1)th, ( r + 2)th term of ( 1 + x ) are
in AP, then n2 − ( 4 r + 1) n + 4 r 2 = 2
Tr + 1
Tr
=
Cr ⋅ x
n−r +1
=
⋅x
r −1
r
Cr − 1 ⋅ x
n
r
n
Let numerically, Tr + 1 be the greatest term in the above
Tr + 1
expansion. Then, Tr + 1 ≥ Tr or
≥ 1.
Tr
n−r +1
(n + 1)
…(i)
∴
| x | ≥ 1 or r ≤
| x|
r
(1 + | x |)
(i) Now, substituting values of n and x in Eq. (i), we get
r ≤ m + f or r ≤ m, where m is a positive integer and f is
a fraction such that 0 < f < 1.
(ii) When r ≤ m + f , Tm + 1 is the greatest term, when r ≤ m,
Tm and Tm +1 are the greatest terms and both are equal.
(iii) The coefficients of the middle terms in the expansion of
(a + x)n are called greatest coefficients.
n+ 1
•
Cr
n−1
•
Cr
n−r +1
=
r
Cr −1
n
n
n+ 1
Cr
Cr + 1
=
r +1
n+1
n
●
C0 + C1 + C2 + ...+ Cn = 2 n
●
C0 + C2 + C4 + ... = C1 + C3 + C5 +... = 2 n − 1
●
C0 − C1 + C2 − C3 + ... + (− 1)n ⋅ Cn = 0
●
C20 + C12 + C22 + ... + C2n = 2 nCn =
●
(−1)n/2 ⋅n Cn/2 , if n is even
C20 − C12 + C22 − C32 + .... =
0,
if n is odd
●
C0 ⋅ Cr + C1 ⋅ Cr + 1 + ... + Cn − r ⋅ Cn
Cr −1
= 2 nCn− r =
(2 n)!
(n !)2
(2 n)!
(n − r )!(n + r )!
●
C1 − 2C2 + 3C3 − .... = 0
●
C0 + 2C1 + 3C2 + ... + (n + 1) ⋅ Cn = (n + 2) 2 n − 1
●
C0 − C2 + C4 − C6 +K = 2 n ⋅ cos
●
If Tr and Tr + 1 be the rth and (r + 1)th terms in the expansion of
(1 + x)n, then
Cr + n Cr −1 =
Cr1 = n Cr2 ⇒ r1 = r2 or r1 + r2 = n
n
r ⋅ nCr = n ⋅
n
Greatest Term
●
●
(ii) If expansion is ( x − a)n, then the general term is
( − 1) r ⋅ n C r x n − r a r .
Cr = nCn− r
●
nπ
4
nπ
n
C1 − C3 + C5 − C7 + ... = 2 ⋅ sin
4
Applications of Binomial
Theorem
1. R-f Factor Relation
Here, we are going to discuss problems involving
( A + B)n = I + f , where I and n are positive integers
0 ≤ f ≤ 1,| A − B2| = k and| A − B| < 1.
2. Divisibility Problem
In the expansion, (1 + α )n. We can conclude that,
(1 + α )n − 1 is divisible by α, i.e. it is a multiple of α.
3. Differentiability Problem
Sometimes to generalise the result we use the
differentiation.
(1 + x)n = nC0 + nC1 x + nC2 x 2 + … + nCn x n
70
40
On differentiating w.r.t. x, we get
n(1 + x)n−1 = 0 + n C1 + 2 ⋅ x ⋅ nC2 + … + n ⋅ nCn ⋅ x n−1
Principle of Mathematical
Induction
Put x = 1, we get, n 2 n−1 = nC1 + 2 nC2 + … + n nCn
In algebra, there are certain results that are formulated in
terms of n, where n is a positive integer. Such results can be
proved by a specific technique, which is known as the
principle of mathematical induction.
Binomial Theorem for
Negative/Rational Index
Let n be a rational number and x be a real number such that
n(n − 1) 2 n(n − 1)(n − 2) 3
x < 1, then (1 + x)n = 1 + nx +
x +
x + ...
2!
3!
●
●
If n is a positive integer, then (1 + x)n contains (n + 1) terms
i.e. a finite number of terms. When n is any negative integer
or rational number, then expansion of (1 + x)n contains
infinitely many terms.
When n is a positive integer, then expansion of (1 + x)n is
valid for all values of x. If n is any negative integer or
rational number, then expansion of (1 + x)n is valid for the
values of x satisfying the condition| x| < 1.
(i) (1 + x)−1 = 1 − x + x2 − x3 + ...
(ii) (1 − x)−1 = 1 + x + x2 + x3 + K
(iii) (1 + x)−2 = 1 − 2 x + 3 x2 − 4 x3 + ...
(iv) (1 − x)−2 = 1 + 2 x + 3 x2 + 4 x3 + ...
First Principle of Mathematical Induction
It consists of the following three steps
Step I Actual verification of the proposition for the starting
value of i.
Step II Assuming the proposition to be true for k, k ≥ i and
proving that it is true for the value (k + 1) which is
next higher integer.
Step III To combine the above two steps. Let p(n) be a
statement involving the natural number n such that
(i) p(1) is true i.e. p(n) is true for n = 1.
(ii) p(m + 1) is true, whenever p(m) is true
i.e. p(m) is true ⇒ p(m + 1) is true. Then, p(n) is true
for all natural numbers n.
Product of r consecutive integers is divisible by r !
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 If (1 + ax )n = 1 + 8x + 24x 3 + ..., then the values of a and n
are
(a) 2, 4
(b) 2, 3
(c) 3, 6
(d) 1, 2
2 The coefficient of x n in the expansion of (1 + x )2n and
(1 + x )2n −1 are in the ratio
(a) 1 : 2
(c) 3 : 1
j
NCERT Exemplar
3 The value of (1. 002)12 upto fourth place of decimal is
(b) 1.0245
(d) 1.0254
n
(b)
(d)
C4
C4 + n C2 + n C2
n
n
C4 + n C2
C4 + n C2 + n C1 ⋅ n C2
1
+ x sin x
x
5 If the middle term of
10
is equal to 7
the value of x is
(a) 2nπ +
π
6
(c) nπ + (− 1)n
j
(b) nπ +
π
6
(a) e 2
(b) e
(c) e / 2
(d) 2e
7 If the number of terms in the expansion of
4
2
1 − + 2 , x ≠ 0 is 28, then the sum of the coefficients
x x
j
of all the terms in this expansion, is
JEE Mains 2016
(b) 2187
(c) 243
(d) 729
8 In the binomial expansion of (a − b ) , n ≥ 5, the sum of 5th
n
is
n
9
3 ln x , x > 0 is equal to 729, then x can be
j
JEE Mains 2013
(a) 64
4 The coefficient of x 4 in the expansion of (1 + x + x 2 + x 3 )n
(a)
(c)
3
+
3
84
n
(b) 1 : 3
(d) 2 : 1
(a) 1.0242
(c) 1.0004
6 If the 7th term in the binomial expansion of
π
6
(d) nπ + (− 1)
7
, then
8
NCERT Exemplar
π
3
and 6th terms is zero, then
5
n−4
n −5
(c)
6
(a)
a
is equal to
b
6
n −5
n−4
(d)
5
(b)
9 In the expansion of the following expression
1 + (1 + x ) + (1 + x )2 + ... + (1 + x )n , the coefficient of
x 4( 0 ≤ k ≤ n ) is
(a)
(c)
n +1
n
Ck + 1
Cn −k −1
(b) n Ck
(d) None of these
71
DAY
10 The coefficient of t 24 in the expansion of
21 If the sum of the coefficients in the expansion of
(1 + t 2 )12(1 + t 12 )(1 + t 24 ) is
(a)
12
C6 + 2
(b)
12
11 The coefficient of x
C5
53
(c)
12
12
(d)
C6
( x − 2y + 3z )n is 128, then the greatest coefficient in the
expansion of (1 + x )n is
C7
(a) 35
in the following expansion
∑ 100Cm ( x − 3)100−m ⋅ 2m is
(a)
C47
(b)
100
C53
(c) −
100
(d) −
100
C53
C100
(a) n = 2r
(c) n = 2r + 1
12 If p is a real number and if the middle term in the
8
p
expansion of + 2 is 1120, then the value of p is
2
(b) 517
23 If an = ∑
r =0
2
, is
x
(c) 569
(d) 581
(c) 12
(a)
AIEEE 2012
(b) an odd positive integer
(c) an even positive integer
(d) a rational number other than positive integers
16 If the (r + 1) th term in the expansion of 3
17 If x
2k
(b) 10
(c) 8
1
occurs in the expansion of x + 2
x
(a) n − 2k is a multiple of 2
(c) k = 0
21
15
20 1
(b)
6 81
20 1
(a)
7 27
1 20
(c)
9 9
JEE Mains 2013
(d) 1 : 32
1
3
(b) 2 21 − 210
1
(b) 20C10
2
(a) (n + 2)2n −1
(c) (n + 1)2n −1
is
(d) None of these
(c) 2 20 − 2 9
JEE Mains 2017
(d) 2 20 − 210
(c) 0
j
(d)
AIEEE 2007
20
(c) 7th
(d) 6th
C10
29 If n > ( 8 + 3 7 )10, n ∈ N, then the least value of n is
(a)
(b)
(c)
(d)
(8 + 3
(8 + 3
(8 + 3
(8 + 3
7 )10 − (8 − 3
7 )10 + (8 − 3
7 )10 − (8 − 3
7 )10 − (8 − 3
7 )10
7 )10
7 )10 + 1
7 )10 − 1
(a) 3
(b) 19
31 If A = 1000
1000
(c) 64
and B = (1001)
999
(d) 29
, then
(b) A = B
(d) None of these
32 If n − 1Cr = (k 2 − 3) ⋅ nCr + 1, then k belongs to
(a) (− ∞, − 2 ] (b) [2 , ∞)
(c) [ − 3 , 3 ] (d) ( 3 , 2 ]
33 The remainder left out when 82n − ( 62)2n + 1 is divided by
9, is
(b) 3th
C55
10
(b) (n + 1)2n
(d) (n + 2)2n
(a) A > B
(c) A < B
1
is
5
(a) 5th
21
30 49n + 16n − 1 is divisible by
20
20 The largest term in the expansion of ( 3 + 2x )50, where
x =
10
C0 + 2C1 + 3C2 + ...+ (n + 1)Cn will be
(b) n − 2k is a multiple of 3
(d) None of these
65
(d)
C10
28 If (1 + x )n = C0 + C1x + C2x 2 +...+ Cn X n , then the value of
, then
19 The greatest term in the expansion of 3 1 +
30
C0 − 20 C1 + 20 C2 − 20 C3 + ...+ 20C10 is
(a) − C10
n −3
(c) 1 : 4
(c)
C10
27 The sum of the series
20
2
independent of x in the expansion of x 2 + , is
x
(b) 7 : 64
60
26 The value of ( C1 − C1) + ( C2 − C2 )
21
20
18 The ratio of the coefficient of x 15 to the term
(a) 7 : 16
(b)
C11
j
(d) 6
j
30
(a) 2 21 − 211
has the same power of a and b, then the value of r is
(a) 9
(d) 0
(c) n
+ ( 21C3 −10 C3 ) + ( 21C4 −10 C4 ) + ... + ( 21C10 −10 C10 ) is
b
3
a
a
+
b
(b) −1
(a) 1
30 30 30 30
30 30
− + ... + is equal to
0 10 1 11
20 30
15 If n is a positive integer, then ( 3 + 1)2n − ( 3 − 1)2n is
(a) an irrational number
1 + rx
∑ ( −1)r (n Cr ) 1 + nx is equal to
25
(d) 24
j
(b) nan
(d) None of these
r =0
and x 2 are 3 and −6 respectively, them m is
(b) 9
n
1
r
,
then
∑ n C is equal to
n
Cr
r
r =0
n
24
14 If in the expansion of (1 + x )m (1 − x )n , the coefficient of x
(a) 6
(d) None of these
(b) n = 3r
(d) None of these
(a) (n − 1)an
1
(c) nan
2
6
13 The constant term in the expansion of 1 + x +
(a) 479
n
j NCERT Exemplar
(b) ±1
(d) None of these
(a) ±3
(c) ± 2
(c) 10
( 3r )th and (r + 2)th powers of x in the expansion of
(1 + x )2n are equal, then
m =0
100
(b) 20
22. If for positive integers r > 1, n > 2, the coefficient of the
100
(a) 0
(b) 2
(c) 7
(d) 8
72
40
34 If x is positive, the first negative term in the expansion of
(1 + x )
27 / 5
is
j
(a) 7th term
(b) 5th term
(c) 8th term
37 For each n ∈ N, 23n − 1 is divisible by
AIEEE 2003
(a) 8
(c) 32
(d) 6th term
35 Let P (n ) : n 2 + n + 1 ( n ∈ N ) is an even integer. Therefore,
(b) 16
(d) None of these
38 Let S (k ) = 1 + 3 + 5+...+( 2k − 1) = 3 + k 2.
Then, which of the following is true?
P (n ) is true
(a) for n > 1
(b) for all n
(c) for n > 2 (d) None of these
j
(a) (n + 1) ! − 2
(c) (n + 1) ! − 1
AIEEE 2004
(a) S(1) is correct
(b) S (k) ⇒ S (k + 1)
(c) S (k) ⇒
/ S (k + 1)
(d) Principle of mathematical induction can be used to
prove the formula
36 For all n ∈ N, 1 × 1! + 2 × 2 ! + 3 × 3! + ...+ n × n ! is equal
to
j
NCERT Exemplar
(b) (n + 1)!
(d) (n + 1)! − 3
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 The coefficient of x 2m + 1 in the expansion of
7 If a and d are two complex numbers, then the sum to
(n + 1) terms of the following series
aC0 − (a + d )C1 + (a + 2d )C2 − ... + ... is
1
E =
, x < 1 is
2
(1 + x )(1 + x )(1 + x 4 )(1 + x 8 )...(1 + x 2m )
(a) 3
(b) 2
(c) 1
a
2n
(c) 0
(d) 0
(a)
C
C
C
2 C1 − 2 + 3 – ... + ( − 1)n − 1 n is equal to
2
3
n
1
+
2
1
(c) 1 + +
2
(a) 1 −
(−1)n − 1
1
−... +
3
n
1
1
+ ... +
3
n −1
n
1 1
1
+ + ... +
2 3
n
(b) 1 +
(d) None of these
10
3 If the coefficient of x 5 in ax 2 +
1
is a times and equal
bx
10
1
to the coefficient of x −5 in ax − 2 2 , then the value
b x
of ab is
(a) (b)−3
(b) − (b)6
(c) (b)−1
(d) None of these
4 The sum of coefficients of integral powers of x in the
binomial expansion of (1 − 2 x )50, is
1 50
(a)
(3 + 1 )
2
1 50
(c)
(3 − 1 )
2
j
JEE Mains 2015
(a) 4
(b) 120
(c) 210
(b)
(a) 3n
(c) 3n + 2n
(b) 2n
(d) 3n − 2n
9 The sum of the series
1
n
3r
7r
15r
∑ ( −1)r nCr 2r + 22r + 23r + 24r + ... + m terms is
r =0
2mn − 1
2 (2n − 1)
2mn + 1
(c) n
2 +1
(a)
(b)
mn
2mn − 1
2n − 1
(d) None of these
10 The value of x, for which the 6th term in the expansion of
9 x −1 + 7
+
7
is 84, is equal to
(1/ 5 )log2 (3 x −1 + 1)
2
1
(b) 3
(c) 2
(d) 5
JEE Mains 2013
(d) 310
C02 + C12 + C22 + C32 + ... + Cn2 is equal to
n!
n !n !
(2n)!
(c)
n!
p =1 m = p
n m
is equal to
m p
11 If the last term in the binomial expansion of 21/ 3 −
6 If (1 + x )n = C0 + C1x + C2x 2 + ... + Cn x n , then
(a)
∑ ∑
(a) 4
5 The term independent of x in expansion of
j
(d) None of these
n
log2
2
1 50
(b)
(3 )
2
1 50
(d)
(2 + 1 )
2
x +1
x −1
−
2/ 3
is
x
− x 1/ 3 + 1 x − x 1/ 2
8
(b) na
(2n)!
n !n !
(d) None of these
1
is 5/ 3
3
1
2
n
log3 8
, then the 5th term from the beginning is
(a) 210
(c) 105
(b) 420
(d) None of these
12 The sum of the coefficients of all odd degree terms in the
expansion of
( x + x 3 − 1)5 + ( x − x 3 − 1)5,( x > 1) is
(a) −1
(b) 0
(c) 1
j
JEE Mains 2018
(d) 2
73
DAY
13 The greatest value of the term independent of x, as α
sin α
varies over R, in the expansion of x cos α +
x
(a)
20
(b)
C10
20
(c)
C15
20
C19
15 If the ratio of the fifth term from the beginning to the fifth
n
20
1
term from the end in the expansion of 4 2 + 4 is
3
6 : 1, then
is
(d) None of these
Statement I The value of n is 10.
14 Statement I For each natural number
n− 4
n,(n + 1)7 − n 7 − 1 is divisible by 7.
Statement II For each natural number n, n 7 − n is divisible
j AIEEE 2011
by 7.
2
Statement II
4
2⋅3
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true, Statement II is
correct explanation of Statement I.
(c) Statement I is true, Statement II is true; Statement II is
not a correct explanation of Statement I
(d) Statement I is true, Statement II is false
⋅ 3 −1
= 6
4 +n
j
NCERT Exemplar
4
(a) Statement I is true; Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true; Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
ANSWERS
SESSION 1
1 (a)
SESSION 2
2 (d)
3 (a)
4 (d)
5 (c)
6 (b)
7 (d)
8 (d)
9 (a)
10 (a)
11 (c)
12 (c)
13 (d)
14 (c)
15 (a)
16 (a)
17 (b)
18 (d)
19 (a)
20 (c)
21 (a)
22 (c)
23 (c)
24 (d)
25 (c)
26 (d)
27 (b)
28 (a)
29 (b)
30 (c)
31 (a)
32 (d)
33 (b)
34 (c)
35 (d)
36 (c)
37 (d)
38 (b)
1 (c)
11 (a)
2 (b)
12 (d)
3 (b)
13 (d)
4 (a)
14 (b)
5 (c)
15 (c)
6 (b)
7 (c)
8 (d)
9 (a)
10 (c)
Hints and Explanations
1 Given that,(1 + ax )n = 1 + 8 x + 24 x2 + ...
⇒ 1+
n
n(n − 1) 2 2
ax +
a x + ...
1
1⋅ 2
rewritten as (1 + 0. 002)
⇒ (1. 002)12 = 1 + 12 C1 ( 0. 002)
12
= 1 + 8 x + 24 x2 + ...
On comparing the coefficients of x, x2 ,
we get
na = 8,
n(n − 1) 2
a = 24
1⋅ 2
⇒ na(n − 1)a = 48
⇒
8(8 − a) = 48 ⇒ 8 − a = 6
⇒
a=2 ⇒ n= 4
2 Coefficient of x in (1 + x ) =
2n
n
and coefficient of x n
in (1 + x )2 n −1 =2 n −1 C n
∴ Required ratio
=
2n
Cn
2 n −1
Cn
2n
3 We have, (1. 002)12 or it can be
Cn
(2n )!
n ! n ! = 2 :1
=
(2n − 1)!
n !(n − 1)!
+ 12 C2 (0. 002)2 + 12 C3 (0. 002)3 + ...
We want the answer upto 4 decimal
places and as such we have left further
expansion.
∴(1. 002)12 = 1 + 12( 0. 002)
12⋅ 11
12⋅ 11⋅ 10
+
(0. 002)2 +
(0. 002)3 + ...
1⋅ 2
1⋅ 2⋅ 3
= 1 + 0. 024 + 2. 64 × 10−4 +1.76 × 10−6 + ...
= 1. 0242
4 (1 + x + x2 + x3 )n = {(1 + x )n (1 + x2 )n }
= (1 + nC1 x + nC2 x2 + nC3 x3
+ nC 4 x 4 +… nC n x n )
(1 + nC1 x2 + nC2 x 4 + K + nC n x2 n )
Therefore, the coefficient of x 4
= nC2 + nC2 nC1 + nC 4
= nC 4 + nC2 + nC1 nC2
5 1 + x sin x
10
x
Here, n = 10 [even]
10
⇒ Middle term =
+ 1 th = 6th
2
10− 5
1
C 5
( x sin x )5
x
7 63
⇒ 252(sin x )5 = 7 =
8
8
1
1
5
(sin x ) =
⇒ sin x =
⇒
32
2
⇒
sin x = sin π/6
π
∴
x = nπ + (− 1)n
6
T6 =
10
3
6 T7 = 9C 6 3 3 ( 3 ln x )6 = 729
84
⇒
⇒
84 × 33
× 33 × (ln x )6 = 729
84
= (ln x )6 = 1
x=e
74
40
⇒
7 Clearly number of terms in the
expansion of
n
1 − 2 + 4 is (n + 2)(n + 1) or n +2 C .
2
2
x x
2
1
1
[assuming and 2 distinct]
x
x
(n + 2)(n + 1)
∴
= 28
2
⇒ (n + 2)(n + 1) = 56 = (6 + 1)(6 + 2)
⇒
n=6
Hence, sum of coefficients
= (1 − 2 + 4)6 = 36 = 729
⇒
=
n +1
(1 + x )
−1
= x −1 [(1 + x )n +1 − 1]
(1 + x ) − 1
∴ The coefficient of x k in S.
= The coefficient of x k +1 in
[(1 + x )n +1 − 1]
=
n +1
C k +1
10 We have, (1 + t 2 )12 (1 + t 12 )(1 + t 24 )
= (1 + 12 C1t 2 + 12 C2t + ...+ 12 C 6t 12
+ ...+ 12 C12t 24 + ... )(1 + t 12 + t 24 + t 36 )
∴Coefficient of t 24 in
(1 + t 2 )12 (1 + t 12 )(1 + t 24 )
=
12
C6 +
12
C12 + 1 =
12
C6 + 2
∑
100− m
C m ( x − 3)
100
∴ Constant term
6 2
6
= 1 + 21 +
2
1
4
14 (1 + x )m (1 − x )n
m(m − 1)x
= 1 + mx +
+ ...
2
!
1 − nx + n(n − 1) x2 −...
2!
2
= 1 + (m − n )x
n2 − n
(m2 − m ) 2
+
− mn +
x + ...
2
2
Given, m − n = 3 ⇒ n = m − 3
n2 − n
m2 − m
and
− mn +
= −6
2
2
(m − 3)(m − 4)
− m(m − 3)
⇒
2
m2 − m
+
= −6
2
⇒ m2 − 7m + 12 − 2m2 + 6m
+ m2 − m + 12 = 0
⇒ −2m + 24 = 0 ⇒ m = 12
15 ( 3 + 1) =
2n
+
2n
⋅ 2 can be written
as [( x − 3) + 2]100 = ( x − 1)100 = (1 − x )100
∴Coefficient of x 53 in
100
C 53 = −100C 53
p
+ 2
2
2n
C 0( 3 )
2 n −2
C2 ( 3 )
+
2n
+ ...+
2 n −1
C1 ( 3 )
2n
2 n −2 n
C2 n ( 3 )
( 3 − 1)2 n =2 n C 0( 3 )2 n (−1)0
+ 2 n C1 ( 3 )2 n −1 (−1)1 + 2 n C2 ( 3 )2 n −2 (−1)2 + ...
+ 2 n C2 n ( 3 )2 n −2 n (−1)2 n
Adding both the binomial expansions
above, we get
( 3 + 1)2 n − ( 3 − 1)2 n = 2[2 n C1 ( 3 )2 n −1
+ ... + 2 n C2 n −1 ( 3 )2 n −(2 n −1 )]
8
n = 8 [even]
8
⇒ Middle term = + 1 th term
2
= 5 th term
T 5 = 8C 4 ( p / 2)8− 4 (24 )
which is most certainly an irrational
number because of odd powers of 3 in
each of the terms.
16 ∴General term is
Tr
Here,
⇒
2n
+ 2 n C3 ( 3 )2 n −3 + 2 n C 5( 3 )2 n − 5
m= 0
12 Given expression is
4 2
2 +
2
6 6 3
2
6 3
= 1 + 60 + 360 + 160 = 581
m
(1 − x )100 = (−1)53
x+ 1
x2
2
6
6
6
2
2
+ x + + ... + x +
x
x
2
6
11 The given sigma expansion
100
17 The general term in the expansion of
p=±2
6
6
13 1 + x + 2 = 1 + x + 2
x
x
1
8 Since, in a binomial expansion of
(a − b )n , n ≥ 5, the sum of 5th and 6th
terms is equal to zero.
∴ n C 4 an − 4 (−b )4 + n C 5an − 5(−b )5 = 0
n!
⇒
an − 4 ⋅ b 4
(n − 4)! 4!
n!
−
an − 5b 5 = 0
(n − 5)! 5!
n!
a
b
⇒
an − 5 ⋅ b 4
− = 0
n − 4 5
(n − 5)! 4!
a n−4
⇒
=
b
5
9 The given expression is
1 + (1 + x ) + (1 + x )2 + ...+ (1 + x )n being in
GP.
Let, S = 1 + (1 + x ) + (1 + x )2 + ...+ (1 + x )n
p 4 = 16
8 × 7 × 6 × 5 p4
×
× 24 = 1120
4 × 3 × 2 × 1 24
+1
a
= 21C r 3
b
=
21
Cr
r
7−
a 2
⋅
21 − r
b
3
a
r
2r 7
−
b3 2
Q Power of a = Power of b [given]
⇒
∴
7−
n −3
is given by
1
T r +1 = n −3 C r ( x )n −3 − r 2
x
= n −3 C r x n −3 −3 r
r
As x2 k occurs in the expansion of
n −3
x + 1 , we must have
x2
n − 3 − 3r = 2k for some non-negative
integer r.
⇒
3(1 + r ) = n − 2k
⇒ n − 2k is a multiple of 3.
18 T r + 1 = 15C r ( x2 )15 − r ⋅ 2
r
x
=
15
=
15
30 − 2 r
Cr x
⋅ 2 ⋅ x−r
30 − 3 r
Cr ⋅ x
r
⋅ 2r
…(i)
For coefficient of x , put 30 − 3 r = 15
15
⇒ 3 r = 15 ⇒ r = 5
∴ Coefficient of x15 =
C 5 ⋅ 25
15
For coefficient of independent of x
i.e. x 0 put 30 − 3r = 0
⇒
r = 10
∴ Coefficient of x 0 =
By condition ⇒
C10 ⋅ 210
15
Coefficient of x15
Coefficient of x 0
15
C 5 ⋅ 25
C ⋅ 25
= 15
= 15 10 10 = 1 : 32
10
C10 ⋅ 2
C10 ⋅ 2
15
19 Greatest term in the expansion of
(1 + x )n is T r
+1
(n + 1)x
where, r =
1+ x
1
Here, n = 20, x =
3
21
∴
r =
3 + 1
= [10.5 ( 3 − 1)] = (7.69) ≈ 7
Hence, greatest term is
7
20 1
20 1
3
.
=
7 3
7 27
20 Q
2x
(3 + 2 x )50 = 350 1 +
3
Here,
and
But
T r +1
Tr
50
2x
T r +1 = 350 50C r
3
r
2x
T r = 350 50C r −1
3
1
x = (given)
5
50
C 2 1
≥ 1 ⇒ 50 r ⋅ ≥ 1
C r −1 3 5
r 2
7
= r −
2 3
2
∴
r =9
⇒ 102 − 2r ≥ 15r ⇒ r ≤ 6
r −1
75
DAY
21 Sum of the coefficients in the expansion
of
( x − 2 y + 3z )n is (1 − 2 + 3)n = 2n
(put x = y = z = 1)
∴ 2n = 128 ⇒ n = 7
Therefore, the greatest coefficient in the
expansion of (1 + x )7 is 7 C3 or 7 C 4
because both are equal to 35.
22 In the expansion of (1 + x )2 n , the general
term =2 n C k x k , 0 ≤ k ≤ 2n
As given for r > 1, n > 2,
2n
C3 r =2 n C r +2
⇒ Either 3r = r + 2
or 3r = 2n − (r + 2)
(Q n C x = nC y ⇒ x + y = n or x = y )
⇒ r = 1 or n = 2r + 1
We take the relation only
n = 2r + 1
23 Let b =
n
∑
r
n
r=0 C r
n
= n∑
=
1
r=0
n
Cr
= nan −
n
n
∑
r=0
n−r
(1 + x )20 =20 C 0 + 20 C1 x + ...
+ 20C10 x10 + ... + 20 C20 x20
On putting x = −1 in the above
expansion, we get
0 = 20C 0 − 20C1 + ... − 20C a + 20C10
(Q C r = C n − r )
n
n− r
n
an
2
20
⇒
20
1
=
(0 + 0) = 0
1 + nx
25 Let
C10 = 2(20C 0 − 20C1 + ... + 20C10 )
1
⇒ 20C 0 − 20C1 + ... + 20C10 = 20C10
2
20
28 Since, x(1 + x )n = xC 0 + C1 x2
On differentiating w.r.t. x, we get
(1 + x )n + nx(1 + x )n −1
= C 0 + 2C1 x + 3C2 x2
+ ... + (n + 1)C n x n
Put x = 1, we get
C 0 + 2C1 + 3C2 + ... + (n + 1)C n
= 2n + n2n −1 = 2n −1 (n + 2)
∴ (8 + 3 7 )10 + (8 − 3 7 )10 is an integer,
hence this is the value of n.
49 + 16n − 1 = (1 + 48) + 16n − 1
n
n
= 1 + C1 (48) + C2 (48) + ...
n
30 30
30 30
+ − ... +
2
12
20 30
2
n
+ n C n (48)n + 16n − 1
= ( 48n + 16n ) + C2 ( 48)2 + n C3 ( 48)3 +
n
C 0 . C10 − C1 . C11
30
30
... + n C n (48)n
+ 30C2 . 30C12 − ... + 30C20 . 30C30
= Coefficient of x
in
(1 + x )30(1 − x )30
= Coefficient of x
in (1 − x )
= Coefficient of x
in
20
20
20
20
30 We have,
30 30 30 30
A = −
0 10 1 11
30
20
29 Let f = (8 − 3 7 )10, here 0 < f < 1
[Q n C 0− nC1 + nC2 − nC3 + ...(−1)n nC n = 0]
2 30
= 64n + 8 [ C2 ⋅ 6 + C3 ⋅ 6 ⋅ 8
2 n
r=0
= (−1)10 30C10
(for coefficient of x20, let r = 10)
=30 C10
2
3
n
+ n C 4 ⋅ 64 82 + ... + n C n ⋅ 6n ⋅ 8n −2 ]
Hence, 49n + 16n − 1 is divisible by 64.
n
31 Since, 1 + 1 < 3 for ∀n ∈ N
30
r 30
2 r
∑ (−1) C r ( x )
r+1
n
⇒ 0 < k2 − 3 ≤ 1
Qn ≥ r ⇒ r + 1 ≤ 1and n, r > 0
n
⇒ 3 < k2 ≤ 4
Hence, k ∈ [−2, − 3] ∪ ( 3,2)
⇒
k2 − 3 =
33 82 n − (62)2 n +1 = (1 + 63)n − (63 − 1)2 n +1
= (1 + 63)n + (1 − 63)2 n +1
= [1 + C1 ⋅ 63 + n C2 ⋅ ( 63)2 + ... + ( 63)n ]
n
+ [1 −2 n −1 C1 ⋅ 63 + (2 n +1 ) C2 ⋅ (63)2 − ...
n
+ (63)n −1 −(2 n +1 ) C1
+
20
20
Now,
=
n
(1001)999
(1000)1000
=
1 1001
1001 1000
1
1
1 +
1001
1000
1000
<
n n −1
Cr
r+1
+ (−1)(63)(2 n +1 )] ]
= 2 + 63[ C1 + C2 ( 63) + ...
+ C2 x3 + ... + C n x n +1
n
+ x ∑ r (−1)r nC r
r=0
C r = (k 2 − 3)
− 20C11 + ... + 20C20
20
r n
1 n
r n
=
(−1) ⋅ C r
1 + nx r∑
=0
B< A
n −1
n
⇒ 0 = 2( C 0 − C1 + ... − C 9 ) + C10
n
1 n
r n
=
(−1) C r (1 + rx )
1 + nx r∑
=0
or A =
27 We know that,
− C 9 + ... + C 0
24 Let E = ∑ (−1) C r 1 + rx
1 + nx
r=0
30
1
= (21 C1 + 21 C2 + ... + 21 C20 ) − (210 − 1)
2
1
= (21 C1 + 21C2 + ... + 21C21 − 1) − (210 − 1)
2
1
= (221 − 2) − (210 − 1) = 220 − 1 − 210 + 1
2
= 220 − 210
20
= nan − b ⇒ 2b = nan ⇒ b =
n
−(10C1 + 10 C2 + ... + 10 C10 )
20
Cr
∴
32 Since,
= (21 C1 + 21 C2 + ...+ 21 C10 )
⇒ 0 = C 0 − C1 + ... − C 9 + C10
Cr
n−r
n
+ (21 C3 −10 C3 )+ ...+ (21 C10 −10 C10 )
20
n
∑
∑ nC
r=0
n − (n − r )
r=0
n
−
(Q r > 1)
(1001)999 < (1000)1000
26 (21 C1 −10 C1 ) + (21 C2 −10 C2 )
1000
1
⋅3 < 1
1001
(2 n +1 )
C2 (63) − ...+ (−1)(63)(2 n )]
Hence, remainder is 2.
34 Since, (r + 1)th term in the expansion of
(1 + x )27 / 5
27 27
27
− 1 ...
− r + 1
5
r
5 5
=
x
r!
Now, this term will be negative, if the
last factor in numerator is the only one
negative factor.
27
32
− r + 1< 0 ⇒
<r
⇒
5
5
⇒ 6. 4 < r ⇒ least value of r is 7.
Thus, first negative term will be 8th.
35 Given, P (n ) : n2 + n + 1
At n = 1, P (1) : 3, which is not an even
integer.
Thus, P(1) is not true.
Also, n(n + 1) + 1 is always an odd
integer.
36 Let the statement P (n ) be defined as
P ( n ) : 1 × 1! + 2 × 2! + 3 × 3! K
+ n × n ! = (n + 1) ! − 1
for all natural numbers n.
Note that P(1) is true, since
P (1) : 1 × 1! = 1 = 2 − 1 = 2! − 1
Assume that P (n ) is true for some
natural number k, i.e.
P (k ) : 1 × 1! + 2 × 2! + 3 × 3! + ....
+ k × k ! = (k + 1)! − 1
…(i)
To prove P (k + 1) is true, we have
P (k + 1) : 1 × 1! + 2 × 2!
+ 3 × 3! + ... + k × k !
+ (k + 1) × (k + 1)!
= (k + 1) ! − 1 + (k + 1)! × (k + 1)
[by Eq. (i)]
= (k + 1 + 1)(k + 1)! – 1
= (k + 2)(k + 1)! – 1 = (k + 2)! – 1
76
40
Thus, P (k + 1) is true, whenever P (k ) is
true. Therefore, by the principle of
mathematical induction,P (n ) is true for
all natural numbers n.
37 Now, 23 n − 1 = (23 )n − 1 = (1 + 7)n − 1
=
x + x2 + ...+ x n − 1 ) dx
Tr
+ nC n ⋅ 7n − 1
= 7[ C1 + C2 7 + K + nC n ⋅ 7n − 1 ]
n
Hence, 7 divides 23n − 1 for all n ∈ N .
38 S (k ) = 1 + 3 + 5 + ... + (2k − 1) = 3 + k 2
Put k = 1 in both sides, we get
LHS = 1 and RHS = 3 + 1 = 4
LHS ≠ RHS
⇒
Put (k + 1) in both sides in the place of
k, we get
LHS = 1 + 3 + 5 + ... + (2k − 1) + (2k + 1)
RHS = 3 + (k + 1)2 = 3 + k 2 + 2k + 1
Let LHS = RHS
1 + 3 + 5 + ... + (2k − 1) + (2k + 1)
= 3 + k 2 + 2k + 1
⇒ 1 + 3 + 5+ ... + (2k − 1) = 3 + k 2
If S (k ) is true, then S (k + 1) is also true.
Hence, S (k ) ⇒ S (k + 1)
SESSION 2
+1
1
10
Since, x 5 occurs in T r
denominator by 1 − x, we have
1− x
E =
(1 − x )(1 + x )(1 + x2 )(1 + x 4 )
Again, let x −5 occurs in T r
r
1
1
10
10− r
− 2 2
a ⋅ x − 2 2 is C r (ax )
b x
b ⋅ x
r
=
10 − 3 r = − 5 ⇒ 15 = 3 r ⇒ r = 5
a5
So, the coefficient of x −5 is − 10C 5 10 .
b
According to the given condition,
a5
a5
10
C 5 5 = − a 10C 5 10
b
b
⇒
− b 5 = a ⇒ − b 6 = ab
∴
m
(1 − x2 )(1 + x2 )(1 + x 4 )...(1 + x2 )
1− x
m
(1 − x 4 )(1 + x 4 )...(1 + x2 )
m +1
1− x
=
= (1 − x )(1 − x2
)−1
2 m +1
(1 − x
)
m +1
∴Coefficient of x
2
+ x2
m +1
m +2
Tr
2
+ C3 ⋅ x3 – ...
1 − (1 − x )
= C1 − C2 ⋅ x
x
+ C3 ⋅ x2 − K
n
1
− C2 ⋅ x + C3 ⋅ x2 − K ) dx
=
⇒
− (1 − x )n
dx
1 − (1 − x )
11
∫0
10− r r
−
=10 C r (−1)r x 3 2
For independent for x, put
10 − r r
− = 0 ⇒ 20 − 2r − 3r = 0
3
2
⇒
20 = 5r ⇒ r = 4
10 × 9 × 8 × 7
∴ T 5 =10 C 4 =
= 210
4×3×2×1
6 We have,
(1 + x )n = C 0 + C1 x + C2 x2 + ... + C n x n
... (i)
n
2
1
1
1
and 1 + = C 0 + C1 + C2
x
x
x
=
50
C r (1)
(− 2 x
Cr ⋅ 2 ⋅ x
r /2
r
25
∑
)
( − 1)
r
50
r =0
(1 + 2 x )
50
11 − x n
C1 C2
C
−
+ 3 −K = ∫
dx
0 1 − x
1
2
3
1
1
Q
f ( x ) dx = ∫ f (1 − x ) dx
0
∫ 0
C2 r (2)2 r
…(i)
= C 0 + C1 ⋅ 2 x
+ C2 (2 x )2 + K + C 50(2 x )50
…(ii)
On adding Eqs. (i) and (ii), we get
(1 − 2 x )50 + (1 + 2 x )50
= 2[C 0 + C2 (2 x )2 + ... + C 50(2 x )50]
(1 − 2 x )50 + (1 + 2 x )50
⇒
2
= C 0 + C2 (2 x )2 + K + C 50 (2 x )50
On putting x = 1, we get
(1 − 2 1 )50 + (1 + 2 1 )50
2
= C 0 + C2 (2)2 + K + C 50 (2)50
⇒
1
+ ... + C n ... (ii)
x
1 /2 r
+ C2 (2 x )2 + K + C 50(2 x )50
2
⇒ 1 − (1 − x ) = C1 ⋅ x − C2 ⋅ x
∫ 0(C1
=
∴Sum of coefficients =
− C3 ⋅ x3 + ...
⇒
10
( x + 1)
= ( x1 /3 + 1) −
= ( x1 /3 − x −1 /2 )10
x
∴The genreal term is
T r +1 =10 C r ( x1 /3 )10− r (− x −1 /2 )r
n
50− r
For the integral power of x and r should
be even integer.
is 1.
2 Since, (1 − x ) = C 0 − C1 ⋅ x + C2 ⋅ x
⇒
+1
50
1
1
= [(1 + 2)50 + (1 − 2)50] = [350 + 1]
2
2
Alternate Method
We have,
(1 − 2 x )50 = C 0 − C1 ⋅ 2 x
+ ... )
n
n
1
C r (a)10 − r − 2 ( x )10 − 3 r
b
10
( x1 /3 )3 + 13
{( x )2 − 1}
= 2 /3
−
1 /3
x ( x − 1)
x − x +1
( x1 /3 + 1)( x2 /3 + 1 − x1 /3 )
=
10
x2 /3 − x1 /3 + 1
{( x )2 − 1}
−
x ( x − 1)
expansion of (1 − 2 x )50.
...(1 + x2 )
1− x
= (1 − x )(1 + x2
of
+1
10
m
=
+ 1.
∴
20 − 3 r = 5
⇒ 3 r = 15 ⇒ r = 5
So, the coefficient of x 5 is 10C 5(a)5(b )−5.
4 Let T r +1 be the general term in the
1 Multiplying the numerator and
=
10
r
1
C r ⋅ (a ⋅ x2 )10 − r ⋅
bx
r
1
= 10C r ⋅ (a)10 − r ( x )20 − 3 r
b
=
1 + 350
= C 0 + C2 (2)2 + K + C 50(2)50
2
10
( x − 1)
5 2 /3 x +11/3
−
x − x + 1 x − x1 /2
⇒
x2
xn
1 1
1
= x +
+…+
= 1 + + + ...+
2
n 0
2 3
n
3 General term is
= 1 + nC1 ⋅ 7 + nC2 ⋅ 72 + K
n
1
∫ 0(1 +
(− 1)50 + (3)50
2
= C 0 + C2 (2)2 + K + C 50(2)50
On multiplying Eqs. (i) and (ii) and taking
coefficient of constant terms in right hand
side = C 20 + C12 + C22 + ... + C 2n
n
1
In right hand side (1 + x )n 1 + or in
x
1
2n
(1 + x ) or term containing x n in
xn
(1 + x )2 n . Clearly, the coefficient of x n in
(2n )!
.
(1 + x )2 n is equal to 2 n C n =
n !n !
7 We can write,
aC 0 − (a + d )C1 + ( a + 2d )C2 − ...
upto (n + 1) terms
= a(C 0 − C1 + C2 − ... )
+ d (−C1 + 2C2 − 3C3 + ... )
... (i)
We know,
(1 − x )n = C 0 − C1 x + C2 x2
−...+ (−1)n C n x n
... (ii)
On differentiating Eq. (ii) w.r.t. x,
we get − n(1 − x )n −1 = −C1 + 2C2 x
− ... + (−1)n C n nx n −1
... (iii)
On putting x = 1 in Eqs. (ii) and (iii), we
get
C 0 − C1 + C2 − ... + (−1)n C n = 0
... (iv)
and −C1 + 2C2 − ... + (−1)n C n = 0
From Eq. (i),
... (v)
77
DAY
aC 0 − (a + d )C1 + (a + 2d )C2 − ... + upto
(n + 1) terms
= a⋅ 0 + d ⋅ 0 = 0
[from Eqs. (iv) and (v)]
n
m
8 Since, =
m p
n!
(n − m )! p ! (m − p )!
n n − p
=
p m − p
n
p=1 m = p
n
n
=
p=1
=
n
p
n
∑
p=1
=
p=1
∑
p=1
=3 −2
n
9
n
t=0
[ put m – p = t ]
p
1
3r
7r
C r r + 2 r + 3 r + ... upto m terms
2
2
2
n
n
1
3r
r n
r n
= ∑ (−1) C r ⋅ r + ∑ (−1) ⋅ C r 2 r
2
2
r=0
r=0
n
r
n
7
r=0
23 r
+ ∑ (−1)r n C r
n
n
(1 / 5)log2 (3
2
= C 5(9
x −1
⇒ 84 = 7C 5
⇒ 9
x −1
(−1)
(−1)
= 2−5 ⇒ n /2 =
2n /2
2
25
n
= 5 ⇒ n = 10
⇒
2
4
1
Now, T 5 = T 4 +1 = 10C 4 (21 /3 )10− 4 −
2
10! 1 /3 6
4 −1 /2 4
(2 ) (−1) (2
)
=
4! 6!
= 210(2)2 (1)(2−2 ) = 210
+ 7)
(3x −1
1
(3x −1 + 1)
(9x −1 + 7)
(3x −1 + 1)
+ 7 = 4(3
x −1
+ 1)
12 Key idea = (a + b )n + (a − b )n
= 2( n C 0an + n C2 an −2b 2 + n C 4 an − 4b 4 ... )
x3 − 1 )5 + ( x −
3
= 2( x + 10 x − 10 x +5x − 10 x + 5x)
7
4
Sum of coefficients of all odd degree
terms is
2(1 − 10 + 5 + 5) = 2
15 We know that, in the expansion of
(a + b )n , pth term from the end is
(n − p + 2)th term from the beginning.
So, 5th term from the end is
= (n − 5 + 2) th term from the beginning
= (n − 3) th term from the beginning
= (n − 4 + 1) th term from the beginning
…(i)
∴ We have,
n
n
1
4
1/4 + 1
2 + 4 = 2
3
31 / 4
Now, 5th term from the beginning is
T 4 + 1 = nC 4 (21 / 4 )n − 4 (3−1 / 4 )4
= nC 4 2
n− 4
4
3 −1
…(ii)
And 5th term from the end is
T( n − 4 ) + 1 = nC n − 4 (21 / 4 )n − n + 4 (31 / 4 )n − 4
= nC 4 2 ⋅ 3
−
n +4
4
[Q C r = n C n − r ] …(iii)
n
13 The general term in the expansion of
x cos α + sin α
x
20
C r ( x cos α )20− r
+ 1)1 / 5
x3 − 1 )5, x > 1
3
5
⇒ Divisible by 7.
So, both statements are true and
Statement II is correct explanation of
Statement I.
3
6
20
1
10
n
6
= 7λ + 7{k 6 + 3k 5 + ... + k} [Using Eq. (i)]
3
= 3−( 5/3 )log3 2 = 2−5
(−1)
5
is
sin α
x
r
= 20C r x20−2 r (cos α )20− r (sin α )r
1
= 9x −1 + 7 + x −1
(3
+ 1)1 / 5
+ 7)
7
= 2( x + 10 x ( x − 1) + 5x( x3 − 1)2 )
+1 )
7−5
log3 8
By mathematical induction
For n = 1,
P(1) = 0, which is divisible by 7.
For n = k
P (k ) = k 7 − k
+ 7 C7 ) − (k + 1)
= (k − k ) + 7{k + 3k + ... + k}
2n /2
+ 5C 4 x( x3 − 1 )4 )
7
x −1
(−1)n
n
5
7
x −1
2
=
n
= 2( 5C 0 x 5 + 5C2 x3 ( x3 − 1 )2
10 We have,
7
1
5/3
3
(x +
n
1
1
n /2
14 Let P (n ) = (n )7 − n
Let P (k ) be divisible by 7.
... (i)
∴ k 7 − k = 7λ, for some λ ∈ N
For n = k + 1,
P (k + 1) = (k + 1)7 − (k + 1)
= (7 C 0k 7 + 7 C1 k 6 + 7 C2 k 5 + ...+ 7 C 6 ⋅ k
2
We have
+ ...
1
3
7
= 1 − + 1 − + 1 − + ...
2
4
8
upto m terms
1
1
1
= n + 2 n + 3 n ... upto m terms
2
2
2
m
1
1
1− n
n
mn
2
2
= 2 −1
=
mn
1 − 1
2 (2n − 1)
n
2
∴T 6 = C 5( 9
x = 1,2
Thus,
r=0
7
( y − 3)( y − 9) = 0
y = 3, 9
3x = 3, 9
(put y = 3x )
Also, we have
r
+
⇒
⇒
⇒
= nC n (−1)n
n
9x −1 +7
y 2 − 12 y + 27 = 0
1
T n +1 = nC n (21 /3 )n − n −
2
∑ (−1)
log2
2
⇒
n
n 1
1
n
⋅ p = 2 1 + − 1
2
p 2
n
32x − 12(3x ) + 27 = 0
n
n − p
t
∑
n n −
2
p
n
∑
= 2n
n− p
20
⇒
n − p
m − p
n
∑
p m = p
∑
Thus, the greatest possible value of β is
10
1
C10 .
2
3x
+ 1
4
3
11 Last term of 21 /3 − 1 is
n n − p
p m − p
n
32x
+7=
9
⇒
∴Given series can be rewritten as
∑ ∑
⇒
5
For this term to be independent of x, we
get
20 − 2r = 0
⇒
r = 10
Let β = Term indeoendent of x
So, from the given condition, we have
Fifth term from the beginning
6
=
Fifth term from the end
1
n−4
n
⇒
C4 ⋅2
n
−1
4
2
n−8
⋅ 3 −1
−n + 4
C 4. ⋅ 2 ⋅ 3
n−4
⇒
4
⋅3
=
4
4 −n
−1 −
4
=
6
1
6
n−8
⇒
2
= 20C10(cos α )10(sin α)10
⇒
= 20C10(cos α sin α)10
⇒
(2 × 3) 4 = (2 ⋅ 3)1 /2
n−8
= 1/2
4
n = 2 + 8 ∴ n = 10
sin 2α
= C10
2
20
10
4
3
4
= 61 /2
n−8
⇒
DAY EIGHT
Permutations and
Combinations
Learning & Revision for the Day
u
u
u
Fundamental Principle of
Counting
Factorial Notation
Permutations
u
u
u
Circular Permutations
Combinations
Applications of Permutations
and Combinations
u
u
Prime Factors
Division of Objects into
Groups
Fundamental Principle of Counting
The fundamental principle of counting is a way to figure out the total number of ways in
which different events can occur. If a certain work A can be done in m ways and another
work B in n ways, then
(i) the number of ways of doing the work C, which is done only when either A or B is
done, is m + n. (addition principle)
(ii) the number of ways of doing the work C, which is done only when both A and B are
done, is mn. (multiplication principle)
Factorial Notation
The product of first n natural numbers is denoted by n! and read as ‘factorial n’.
Thus, n ! = n (n − 1) (n − 2) … 3 ⋅ 2 ⋅1
PRED
MIRROR
Properties of Factorial Notation
Your Personal Preparation Indicator
1. 0 ! = 1 ! = 1
2. Factorials of negative integers and fractions are not defined.
3. n ! = n (n − 1)! = n (n − 1) (n − 2)!
n!
4.
= n (n − 1) (n − 2)… (r + 1)
r!
Permutations
●
●
Permutation means arrangement of things. The number of permutations of n
different things taken r at a time is n Pr .
n!
n
,0 ≤ r ≤ n
Pr = n (n − 1) (n − 2)...(n − r + 1) =
(n − r )!
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
79
DAY
Properties of n Pr
(i) n P0 = 1, n P1 = n , n Pn = n ! (ii)
(iii) Pr = n
n
(v)
n −1
n −1
n
Pr + r ⋅n Pr − 1 =
n +1
Pr
(iv) Pr = (n − r + 1) Pr − 1
n
Pr − 1
Pr = (n − r )
n −1
n
Pr − 1
Important Results on Permutations
(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 .
(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) The number of permutations of n different things taken r
at a time, allowing repetitions is nr .
(iv) The permutations of n things of which p are identical of
one sort, q are identical of second sort, r are identical of
n!
third sort, is
, where p + q + r = n.
p!q ! r !
(v) Arrangements
(a) The number of ways in which m different things and n
different things (m + 1 ≥ n) can be arranged in a row, so
that no two things of second kind come together is
m ! ( m +1 )Pn .
(b) The number of ways in which m different things and n
different things (m ≥ n) can be arranged in a row so that
all the second type of things come together is
(m + 1)! n !.
(vi) Dearrangement The number of dearrangements
(No object goes to its scheduled place) of n objects, is
1
1
1
1
n!
−
+
− ... (− 1)n
.
2 ! 3 ! 4 !
n !
(vii) Sum of Digits
(a) Sum of numbers formed by taking all the given n digits
(excluding 0) is (sum of all the n digits)
× (n − 1)! × (111 ... n times).
(b) Sum of the numbers formed by taking all the given n
digits (including 0) is (sum of all the n digits)
× [(n − 1)! × (111 ... n times) − (n − 2) ×
{111 ...(n − 1) times}].
Circular Permutations
●
●
●
●
Number of circular permutations of n different things, taken
r at a time, when clockwise and anti-clockwise orders are
n
P
not different, is r .
2r
●
If different objects are arranged along a closed curve, then
permutation is known as circular permutation.
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.
If clockwise and anti-clockwise circular permutations are
(n − 1)!
.
considered to be same, then it is
2
Number of circular permutations of n different things,
taken r at a time, when clockwise and anti-clockwise
n
P
orders are taken as different, is r .
r
Important Results on
Circular Permutations
(i) The number of ways in which m different things and n
different things (where, m ≥ n) can be arranged in a circle,
so that no two things of second kind come together is
(m − 1)! m Pn .
(ii) The number of ways in which m different things and n
different things can be arranged in a circle, so that all the
second type of things come together is m ! n !.
(iii) The number of ways in which m different things and n
different things (where, m ≥ n) can be arranged in the form
of garland, so that no two things of second kind come
together is (m − 1)! m Pn / 2.
(iv) The number of ways in which m different things and n
different things can be arranged in the form of garland, so
that all the second type of things come together is m ! n !/ 2 .
Combinations
Combination means selection of things. The number of
combinations of n different things taken r at a time is
n
n
Cr or .
r
n
n
n n (n − 1) (n − 2)...(n − r + 1)
n!
P
Cr = =
=
= r
r
r!
r !(n − r )!
r!
Properties of n Cr
(i)
n
Cr = nCn − r
(iii)
n
Cr + Cr − 1 =
n
(ii) n Cx = nC y ⇒ x = y or x + y = n
n +1
Cr
Important Results on Combinations
(i) The number of combinations of n different things, taken r
at a time, when p particular things always occur is
n− p
Cr − p.
(ii) The number of combinations of n different things, taken r
at a time, when p particular things never occur is n − pCr .
(iii) The number of selections of zero or more things out of n
different things is n C0 + nC1 + … + nCn = 2 n .
(iv) The number of selections of one or more things out of
n different things is
n
C1 + nC2 + … + nCn = 2 n − 1.
(v) The number of selections of zero or more things out of n
identical things = n + 1.
(vi) The number of selections of one or more things out of n
identical things = n.
80
40
(vii) The number of selections of one or more things from
p + q + r things, where p are alike of one kind, q are alike
of second kind and rest are alike of third kind, is
[( p + 1) (q + 1) (r + 1)] − 1.
(viii) The number of selections of one or more things from p
identical things of one kind, q identical things of second
kind, r identical things of third kind and n different
things, is ( p + 1) (q + 1) (r + 1) 2 n − 1
(ix) If there are m items of one kind, n items of another kind
and so on. Then, the number of ways of choosing r items
out of these items = coefficient of x r in
(1 + x + x2 + K + x m ) (1 + x + x2 + K + x n )K
(x) If there are m items of one kind, n items of another kind
and so on. Then, the number of ways of choosing r items
out of these items such that atleast one item of each kind
is included in every selection = coefficient of x r in
( x + x2 + K + x m )( x + x2 + K + x n )K
(ii) Sum of divisors of nis
(iii) If p is a prime such that pr divides n! but pr +1 does not
divide n ! .
n n
Then, r = + 2 + K
p p
Division of Objects into Groups
Objects are Different
(i) The number of ways of dividing n different objects into 3
groups of size p, q and r ( p + q + r = n) is
n!
(a)
; p, q and r are unequal.
p!q ! r !
(b)
Applications of Permutations
and Combinations
Functional Applications
If the set A has m elements and B has n elements, then
(i) the number of functions from A to B is nm .
(ii) the number of one-one functions from A to B is n Pm , m ≤ n.
(iii) the number of onto functions from A to B is
n
n
nm − (n − 1)m + (n − 2)m − ..., m ≤ n.
1
2
(iv) the number of bijections from A to B is n!, if m = n.
Geometrical Applications
(i) Number of triangles formed from n points, when no three
points are collinear, is n C3 .
(ii) Out of n non-concurrent and non-parallel straight lines,
the points of intersection are n C2 .
(iii) The number of diagonals in a polygon of nsides is n C2 − n.
(iv) The number of total triangles formed by the n points on a
plane of which m are collinear, is n C3 − mC3 .
(v) The number of total different straight lines, formed by the
n points on a plane, of which m are collinear, is
n
C2 − mC2 + 1 .
Prime Factors
Let n = p1α 1 ⋅ p2α 2 ⋅ p3α 3 K prα r , where pi , i = 1, 2, K , r are distinct
primes and α i , i = 1, 2 , K , r are positive integers.
(i) Number of divisor of n is
(α 1 + 1) (α 2 + 1) (α 3 + 1)K (α r + 1).
( p1α 1 + 1 − 1) ( p2α 2 + 1 − 1) ( prα r + 1 − 1)
...
( p1 – 1)
( p2 − 1)
( pr − 1)
n!
;q = r
p ! 2 !(q !)2
(c)
n!
; p =q = r
3 !( p !)3
(ii) The number of ways in which n different things can be
distributed into r different groups, if empty groups are
allowed, is r n .
(iii) The number of ways in which n different things can be
distributed into r different groups, if empty groups are not
allowed, is
r
r
r n − (r − 1)n + (r − 2)n − … + (−1)r − 1 rCr − 1 ⋅ 1
1
2
Objects are Identical
(i) The number of ways of dividing n identical objects among
r persons such that each one may get atmost n objects, is
n + r − 1
. In other words, the total number of ways of
r −1
dividing n identical objects into r groups, if blank groups
are allowed, is n + r − 1 Cr − 1 .
(ii) The total number of ways of dividing n identical objects
among r persons, each one of whom, receives atleast one
item, is n − 1 Cr − 1 . In other words, the number of ways in
which n identical things can be divided into r groups
such that blank groups are not allowed, is n − 1 Cr − 1 .
(iii) Number of non-negative integral solutions of the equation
x1 + x2 + … + x r = n is equivalent to number of ways of
distributing n identical objects into r groups if empty
groups are allowed, which is n + r −1 Cr −1 .
(iv) Number of positive integral solutions of the equation
x1 + x2 + …+ x r = n is equivalent to the number of ways of
distributing n identical objects into r groups such that no
group empty, which is n −1 Cr −1 .
(v) Number of integral solutions of the equation
x1 + x2 + … + x r = n, where a ≤ x i ≤ b , ∀ i = ,1, 2,… , r , is
given by coefficient of x n in ( x a + x a + 1 + x a + 2 + …+ xb )r .
81
DAY
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 The value of 2n [1 ⋅ 3 ⋅ 5 K ( 2 n − 3) ( 2 n − 1)] is
(2n)!
(a)
n!
(2 n)!
(b)
2n
n!
(c)
(2 n)!
11 The total number of permutations of n ( > 1) different things
(d) None of these
2 The value of1⋅1! + 2 ⋅ 2! + 3 ⋅ 3! + … + n ⋅ n !, is
(a) (n + 1) !
(c) (n + 1) ! − 1
(b) (n + 1)! + 1
(d) None of these
from the digits of the number 223355888 by
rearrangement of the digits so that the odd digits occupy
even places?
(b) 36
(c) 60
(d) 180
4 A library has a copies of one book, b copies of each of
two books, c copies of each of three books and single
copies of d books. The total number of ways in which
these books can be arranged, is
(a + b + c + d)!
a! b ! c!
(a + 2b + 3c + d)
(c)
a! b ! c!
(a)
(b)
(a + 2b + 3c + d)!
a !(b !)2 (c !)3
(d) None of these
(c) 7!
(d)
4
C4 × 3 C3
6 In how many ways the letters of the word ‘ARRANGE’ can
be arranged without altering the relative positions of
vowels and consonants?
(a) 36
(c) 62
(b) 26
(d) None of these
7 If the letters of the word ‘SACHIN’ are arranged in all
possible ways and these words are written in dictionary
order, then the word ‘SACHIN’ appears at serial number
(a) 600
(b) 601
(c) 602
(d) 603
8 The number of integers greater than 6000 that can be
formed, using the digits 3, 5, 6, 7 and 8 without
j
repetition, is
JEE Mains 2015
(a) 216
(b) 192
(c) 120
(d) 72
9 The number of 5-digits telephone numbers having atleast
one of their digits repeated, is
(a) 90000
(b) 100000
(c) 30240
j
NCERT Exemplar
(d) 69760
10 If all the words (with or without meaning) having five
letters, formed using the letters of the word SMALL and
arranged as in dictionary, then the position of the word
j
SMALL is
JEE Mains 2016
(a) 46th
(c) 52nd
Pr
−1
a
=
(b)
nr − 1
n −1
(d) None of these
n
Pr + 1
Pr
, then
=
b
c
n
(a) b 2 = a (b + c)
(c) ab = a 2 + bc
(b) c 2 = a (b + c)
(d) bc = a 3 + b 2
13 The range of the function f ( x ) =
(a) {1, 2, 3 }
(c) {1, 2, 3, 4}
7 − x
Px − 3 is
j
AIEEE 2004
(b) {1, 2, 3, 4, 5, 6}
(d) {1, 2, 3, 4, 5}
14 Find the number of different words that can be formed
from the letters of the word TRIANGLE, so that no vowels
j NCERT Exemplar
are together.
(b) 14500
(c) 14400
(d) 14402
15 In a class of 10 students, there are 3 girls. The number of
letters of the word ARTICLE, so that vowels occupy the
j
even place is
NCERT Exemplar
(b) 144
n
12 If
(a) 14000
5 The number of words which can be formed out of the
(a) 1440
n (n n − 1)
n −1
n (n r − 1)
(c)
n −1
(a)
3 How many different nine-digit numbers can be formed
(a) 16
taken not more than r at a time, when each thing may be
repeated any number of times is
(b) 59th
(d) 58th
ways they can be arranged in a row, so that no 2 girls are
consecutive is k ⋅ 8!, where k is equal to
(a) 12
(b) 24
(c) 36
(d) 42
16 The sum of all the 4-digit numbers that can be formed
using the digits 1, 2, 3, 4 and 5 without repetition of the
digits is
(a) 399960
(b) 288860
(c) 301250
(d) 420210
17 If eleven 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) 10! × 2
(c) 9! × 2
(b) 10!
(d) None of these
18 Let f : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} that are onto and
f ( x ) ≠ x , is equal to
(a) 9
(c) 16
(b) 44
(d) None of these
19 There are 4 balls of different colours and 4 boxes of same
colours as those of the balls. The number of ways in
which the balls, one in each box, could be placed such
that a ball does not go to box of its own colour is
(a) 8
(b) 9
(c) 7
(d) 1
20 How many different words can be formed by jumbling the
letters in the word ‘MISSISSIPPI’ in which no two S are
j
adjacent?
AIEEE 2008
(a) 7 ⋅ 6 C4 ⋅ 8 C4
(c) 6 ⋅ 7 ⋅ 8 C4
(b) 8 ⋅ 6 C4 ⋅ 7 C4
(d) 6 ⋅ 8 ⋅ 7 C4
82
40
21 The number of ways in which seven persons can be
arranged at a round table, if two particular persons may
not sit together is
(a) 480
(c) 80
at a round table, if no two women are to sit together, is
j AIEEE 2003
given by
23 The value of
(c) 5 ! × 4 !
47
C4 +
5
∑
52 − r
(d) 7 ! × 5 !
(a)
(c)
24 If n C3 + nC4 >
(a) 312 − 1
C3 , then
(b) n > 7
(d) None of these
(b) 126
(d) None of these
balls. The number of ways of drawing 3 balls from the
box, if atleast one black ball is included, is
(c) 56
(d) 64
27 A student is allowed to select atmost n books from a
collection of ( 2n + 1) books. If the number of ways in
which he can select atleast one book is 63, then n is
equal to
(a) 3
(b) 4
(c) 6
(d) 5
28 Let A and B two sets containing 2 elements and 4
elements, respectively. The number of subsets of A × B
j
having 3 or more elements is
JEE Mains 2013
(a) 256
(b) 220
(c) 219
(d) 211
29 In an examination of 9 papers a candidate has to pass in
more papers than the number of papers in which he fails,
in order to be successful. The number of ways in which
he can be unsuccessful is
(a) 255
(b) 256
(c) 193
(d) 319
30 There are two urns. Urn A has 3 distinct red balls and
urn B has 9 distinct blue balls. From each urn, two balls
are taken out at random and then transferred to the
other. The number of ways in which this can be done, is
j
(a) 3
(b) 36
(c) 66
(b) 312
(c) (12)3 − 1
(d) (12)3
(a) N > 190
(c) 100 < N ≤ 140
(b) N ≤ 100
(d) 140 < N ≤ 190
35 Let Tn be the number of all possible triangles formed by
joining vertices of an n-sided regular polygon. If
j JEE Mains 2013
Tn +1 − Tn = 10, then the value of n is
26 A box contains 2 white balls, 3 black balls and 4 red
(b) 42
(d) 36
collinear. If N is the number of triangles formed by joining
j AIEEE 2011
these points, then
committee from four men and six women so that the
committee includes atleast two men and exactly twice as
j NCERT Exemplar
many women as men is
(a) 36
(c) 32
34 There are 10 points in a plane, out of these 6 are
25 The number of ways in which we can choose a
(a) 94
(c) 128
(b) 24
horses, cows and calves (not less than 12 each) ready to
be shipped. In how many ways, can the ship load be
made?
n +1
(a) n > 6
(c) n < 6
(d) 0
33 In a steamer, there are stalls for 12 animals and there are
(b) 52 C5
(d) None of these
C6
52
C4
(c) (m + 1) λ
all possible ways. If the number of times two particular
soldiers A and B are together on guard is thrice the
number of times three particular soldiers C, D, E are
together on guard, then n is equal to
(a) 18
C3 is equal to
r =1
47
(b) mλ
32 A guard of 12 men is formed from a group of n soldiers in
22 The number of ways in which 6 men and 5 women can sit
(b) 30
A = {a1, a 2 , a 3 , ..., a n } is λ times the number of m elements
subsets containing a 4 , then n is
(a) (m − 1) λ
(b) 120
(d) None of these
(a) 6! × 5 !
31 If the total number of m elements subsets of the set
AIEEE 2010
(d) 108
(a) 7
(c) 10
(b) 5
(d) 8
36 On the sides AB, BC, CA of a ∆ABC, 3, 4, 5 distinct points
(excluding vertices A , B , C) are respectively chosen. The
number of triangles that can be constructed using these
j
chosen points as vertices are
JEE Mains 2013
(a) 210
(c) 215
(b) 205
(d) 220
37 The number of divisors of the number of 38808
(excluding 1 and the number itself) is
(a) 70
(c) 71
(b) 72
(d) None of these
38 If a, b, c, d , e are prime integers, then the number of
divisors of ab 2c 2de excluding 1 as a factor is
(a) 94
(b) 72
(c) 36
(d) 71
39 The number of ways of distributing 8 identical balls in
3 distinct boxes, so that no box is empty, is
(a) 5
(c) 3 8
8
(b)
3
(d) 21
40 If 4 dice are rolled, then the number of ways of getting
the sum 10 is
(a) 56
(c) 72
(b) 64
(d) 80
83
DAY
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 From 6 different novels and 3 different dictionaries, 4 novels
Y together can throw a party inviting 3 ladies and 3 men,
so that 3 friends of each of X and Y are in this party, is
and 1 dictionary are to be selected and arranged in a row
on a shelf, so that the dictionary is always in the middle.
j JEE Mains 2018
The number of such arrangements is
j
(a) 485
(a) atleast 1000
(b) less than 500
(c) atleast 500 but less than 750
(d) atleast 750 but less than 1000
(a) 27378
(b) 163 × 15 2 × 14
(d) 162 × 15 × 14
(c)
52 !
17 !
(a) 36
(a) 36
a11 a12
: aij ∈ {0, 1, 2 }, a11 = a22 . Then, the number
a
a
21 22
(a) 27
(b) 24
(c) 10
(a)
(b) 629
30
(b)
C7
(d) 879
21
C8
(c)
21
(d)
C7
30
C8
number 1010 1111 1313 is
(a) 750
(d) 7200
(c) 630
13 The number of divisors of the form 4n + 1, n ≥ 0 of the
(d) 20
(c) 7600
(d) None of these
30 marks to 8 questions, giving not less than 2 marks to
j
any question, is
JEE Mains 2013
(b) 840
(c) 924
(d) 1024
14 Out of 8 sailors on a boat, 3 can work only one particular
side and 2 only the other side. Then, number of ways in
which the sailors can be arranged on the boat is
to contain atleast 3 men and 2 women. The number of
ways this can be done, if two particular women refuse to
serve on the same group, is
(b) 7800
(c) 120
12 The number of ways in which an examiner can assign
JEE Mains 2013
6 A group of 6 is chosen from 10 men and 7 women so as
(a) 8000
(d) None of these
colours, the number of ways in which one or more balls
can be selected from 10 white, 9 green and 7 black balls
j AIEEE 2012
is
(d) 309
j
(b) 90
(a) 880
5 Let S =
of non-singular matrices in the set S, is
(c) 0
11 Assuming the balls to be identical except for difference in
(d) None of these
(c) 308
(b) 18
xyz = 24 is
possible orders and these words are written out as in a
dictionary, then the rank of the word MOTHER is
(b) 261
(d) None of these
by four different even digits is
4 If the letters of the word MOTHER are written in all
(a) 240
(c) 27399
10 The total number of integral solutions ( x , y , z ) such that
players so that three players have 17 cards each and the
fourth players just one card, is
(b) 52!
(b) 27405
9 The number of numbers divisible by 3 that can be formed
3 The number of ways of dividing 52 cards amongst four
52 !
(17 !)3
(d) 484
from 1 to 30 so as to exclude every selection of four
consecutive numbers is
swimming and riding. How many prize lists could be
made, if there were altogether 6 prizes for different
values, one for running, 2 for swimming and 3 for riding?
(a)
(c) 469
8 The number of ways in which we can select four numbers
2 Sixteen men compete with one another in running,
(a) 16 × 15 × 14
(c) 163 × 15 × 14 2
(b) 468
JEE Mains 2017
(a) 2718
(b) 1728
(c) 7218
(d) None of these
15 In a cricket match between two teams X and Y , the team
X requires 10 runs to win in the last 3 balls. If the possible
runs that can be made from a ball be 0, 1, 2, 3, 4, 5 and 6.
The number of sequence of runs made by the batsman is
7 A man X has 7 friends, 4 of them are ladies and 3 are
men. His wife Y also has 7 friends, 3 of them are ladies
and 4 are men. Assume X and Y have no common
friends. Then, the total number of ways in which X and Y
(a) 12
(b) 18
(c) 21
(d) 36
ANSWERS
(a)
(c)
(a)
(b)
SESSION 1
1.
11.
21.
31.
SESSION 2
1. (a)
11. (d)
2.
12.
22.
32.
(c)
(a)
(a)
(c)
2. (b)
12. (c)
3.
13.
23.
33.
(c)
(a)
(c)
(b)
3. (a)
13. (c)
4.
14.
24.
34.
(b)
(c)
(a)
(b)
4. (d)
14. (b)
5.
15.
25.
35.
(b)
(d)
(a)
(b)
5. (d)
15. (d)
6.
16.
26.
36.
(a)
(a)
(d)
(b)
6. (b)
7.
17.
27.
37.
(b)
(c)
(a)
(a)
7. (a)
8.
18.
28.
38.
(b)
(b)
(c)
(d)
8. (a)
9.
19.
29.
39.
(d)
(b)
(b)
(d)
9. (a)
10.
20.
30.
40.
(d)
(a)
(d)
(d)
10. (c)
84
40
Hints and Explanations
SESSION 1
Four digit number can start from 6,7 or 8.
1 Clearly, [1 ⋅ 3 ⋅ 5 ...( 2n − 3)(2 n − 1 )]2
=
1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5⋅ 6...(2 n − 1 ) (2 n ) 2
2 ⋅ 4 ⋅ 6... 2 n
=
(2 n ) ! 2n
(2 n )!
=
2 (1 ⋅ 2 ⋅ 3... n )
n!
n
n
2 Clearly, given expression =
n
∑
r ⋅r!
r =1
=
n
∑
r =1
((r + 1) − 1) r ! =
n
∑
6, 7 or 8
n
((r + 1)! − r !)
r =1
= (2! − 1!) + (3!− 2!) + …+ ((n + 1)! − n !)
= (n + 1)! − 1
3 In a nine digits number, there are four
even places for the four odd digits 3, 3,
5, 5.
4!
5!
⋅
∴ Required number of ways =
2! 2! 2! 3!
= 60
4 Total number of books = a + 2b + 3c + d
∴ The total number of arrangements
(a + 2b + 3c + d )!
=
a!(b !) 2 (c !) 3
5 In a word ARTICLE, vowels are A, E, I
and consonants are C, L, R, T.
In a seven letter word, there are three
even places in which three vowels are
placed in 3! ways. In rest of the four
places, four consonants are placed in 4!
ways.
∴ Required number of ways
= 3! × 4! = 6 × 24 = 144
6 Clearly, the consonants in their
4!
positions can be arranged in
= 12
2!
ways and the vowels in their positions
3!
can be arranged in = 3 ways
2!
∴ Total number of arrangements
= 12 × 3 = 36
7 The letters of given word are A, C, H, I,
N, S.
Now, the number of words starting with
A = 5!
the number of words starting with C = 5!
the number of words starting withH = 5!
the number of words starting with I = 5!
and the number of words starting with
N = 5!.
Then, next word is SACHIN.
So, the required serial number is
= (5⋅ 5!) + 1 = 601.
8 The integer greater than 6000 may be of
4 digits or 5 digits. So,here two cases arise.
Case I When number is of 4 digit.
3
4
3
2
Thus, total number of 4-digit number,
which are greater than
6000 = 3 × 4 × 3 × 2 = 72
Case II When number is of 5 digit.
Total number of 5-digit number, which
are greater than 6000 = 5! = 120
∴ Total number of integers
= 72 + 120 = 192
9 Using the digits 0, 1, 2, ..., 9 the number
of five digits telephone numbers which
can be formed is 10 5 (since, repetition is
allowed). The number of five digits
telephone numbers, which have none of
the digits repeated = 10 P5 = 30240
∴ The required number of telephone
number = 105 − 30240 = 69760
10 Clearly, number of words start with
4!
= 12
2!
Number of words start with L = 4! = 24
4!
Number of words start with M =
= 12
2!
3!
Number of words start with SA=
=3
2!
Number of words start with SL = 3! = 6
Note that, next word will be ‘‘SMALL’’.
Hence, position of word ‘‘SMALL’’ is 58th.
A=
11 When we arrange things one at a time,
the number of possible permutations is
n. When we arrange them two at a time
the number of possible permutations are
n × n = n2 and so on. Thus, the total
number of permutations are
n (n r − 1)
n + n2 + K + n r =
[Q n > 1]
n−1
n
12 Q
Pr − 1
a
=
n
Pr
=
b
n
Pr + 1
c
b
=n−r +1
a
c
From last two terms
=n−r
b
b c
Hence, = + 1 ⇒ b 2 = a (b + c )
a b
From first two terms
13 Given that, f ( x ) =
7−x
14 In a word TRIANGLE, vowels are (A, E,
I) and consonants are (G, L, N, R, T).
First, we fix the 5 consonants in
alternate position in 5! ways.
_G_L_N_R_T_
In rest of the six blank position, three
vowels can be arranged in 6 P3 ways.
∴ Total number of different words
= 5! × 6 P3 = 120 × 6 × 5 × 4 = 14400
15 The 7 boys can be arranged in row in 7!
ways. There will be 6 gaps between
them and one place before them and one
place after them. The 3 girls can be
arranged in a row in 8 P3 = 8 ⋅ 7 ⋅ 6 ways
∴ Required number of ways
= 7! × 8 ⋅ 7 ⋅ 6 = 42 ⋅ 8!
16 Required sum
= (Sum of all the n-digits)
× n −1 Pr −1 × (111 ... r times)
= (1 + 2 + 3 + 4 + 5) 4 P3 × (1111)
= 15 × 24 × 1111 = 399960
17 Since, out of eleven members two
members sit together, the number of
arrangements = 9! × 2
(Q two members can be sit in two ways.)
18 Total number of functions
= Number of dearrangement of 5 objects
1
1
1
1
= 5! −
+
− = 44
2! 3! 4! 5!
19 Use the number of dearrangements
1 1 1
1
i.e. n ! 1 − + − + ... + (− 1)n
n !
1! 2! 3!
Here, n = 4
So, the required number of ways
1 1
1
= 4! − + = 12 − 4 + 1 = 9
2! 3 ! 4!
20 Given word is MISSISSIPPI.
Here, I = 4 times, S = 4 times,
P = 2 times, M = 1 time
_M_I_I_I_I_P_P_
∴ Required number of words
7 × 6!
7!
= 8C 4 ×
= 8C 4 ×
= 7 ⋅ 8C 4 ⋅ 6C 4
4!2!
4!2!
21 Clearly, remaining 5 persons can be
seated in 4! ways. Now, on five cross
marked places person can sit in 5 P2 ways.
P4
Px − 3 . The above
function is defined, if 7 − x ≥ 0 and
x − 3 ≥ 0 and 7 − x ≥ x − 3.
⇒
x ≤ 7, x ≥ 3 and x ≤ 5
∴
D f = {3, 4, 5}
Now,
f (3) = 4 P0 = 1
f (4) = 3 P1 = 3 and f (5) = 2 P2 = 2
∴
R f = {1, 2, 3}
P3
P5
P2
P1
So, number of arrangements
5!
= 4! × = 24 × 20 = 480 ways
3!
85
DAY
22 First, we fix the position of men, number
of ways in which men can sit = 5!.
Now, the number of ways in which
women can sit = 6 P5
29 Clearly, the candidate is unsuccessful, if
he fails in 9 or 8 or 7 or 6 or 5 papers.
∴ Numbers of ways to be unsuccessful
= 9C 9 + 9C 8 + 9C7 + 9C 6 + 9C 5
= 9C 0 + 9C1 + 9C2 + 9C3 + 9C 4
1
= ( 9 C 0 + 9C1 + K + 9C 9 )
2
1
= (29 ) = 28 = 256
2
M
M
M
M
M
5
C4 +
∑
52 − r
r =1
=
=
24
n
51
52
C3 +
C4
50
+
C3 +
C3 + nC 4 >
C3 +
50
49
n +1
C4 +
47
C3 +
51
C3 +
49
48
C3
C3 +
48
C3 + ( C3 +
47
47
47
C3
C4)
C3
C4 >
C3 ( nC r + n C r + 1 = n +1 C r +1 )
n +1
n−2
C
⇒ n +1 4 > 1 ⇒
> 1 ⇒n> 6
C3
4
⇒
n +1
C3 =
choose a committee
= Choose two men and four women
+ Choose three men and six women
= 4C2 × 6C 4 + 4C3 × 6C 6
= 6 × 15 + 4 × 1 = 90 + 4 = 94
26 The number of ways of drawing
1 black and 2 non-black balls is
3
C1 ⋅ 6C2 = 3 ⋅ 15 = 45
The number of ways of drawing
2 black and 1 non-black ball is
3
C2 ⋅ 6C1 = 3 ⋅ 6 = 18
The number of ways of drawing
3 black balls is 3 C3 = 1
∴ Number of ways = 45 + 18 + 1 = 64
27 He can select 1, 2, ... or n books.
The number of ways to select atleast one
book is
2n + 1
C1 + 2 n + 1 C2 + K + 2 n + 1 C n
1
= (2 n + 1 C1 + 2 n + 1 C2 + K + 2 n + 1 C n
2
+ 2 n + 1 C n + 1 + K + 2 n + 1 C2 n)
1 2n + 1 2n + 1
C 0 − 2 n + 1 C2 n + 1 )
= (2
−
2
[given]
= 22n − 1 = 63
⇒
22 n = 64 = 26 ⇒ n = 3
28 Given, n( A ) = 2, n(B ) = 4.
∴ n (A × B) = 8
The number of subsets of A × B having 3
or more elements = 8C 3 + 8C 4 +…+ 8C 8
= ( 8 C 0 + 8C1 + 8C2 + 8C3 +…+ 8C 8 )
− ( 8 C 0 + 8C1 + 8C2 )
8
8
8
= 2 − C 0 − C1 − 8C2
= 256 − 1 − 8 − 28 = 219
[Q 2n = nC 0 + nC1 +…+ nC n ]
3 distinct
red balls
9 distinct
blue balls
constructed using these chosen points as
vertices = 12C3 − 3C3 − 4C3 − 5C3
Here, we subtract those cases in which
points are collinear
= 220 − 1 − 4 − 10 = 220 − 15 = 205
Urn A
Urn B
38 The number of divisors of ab2c 2de
= (1 + 1 ) (2 + 1 ) (2 + 1 ) (1 + 1 ) (1 + 1 ) − 1
= 2 ⋅ 3 ⋅ 3 ⋅ 2 ⋅ 2 − 1 = 71
39 Required number of ways is equal to the
The number of ways in which 2 balls
from urn A and 2 balls from urn B can
be selected = 3C2 × 9C2 = 3 × 36 = 108
31 Total number of m elements subsets of
A = nC m
n +1
25 The number of ways in which we can
36 Required number of triangles that can be
∴ Number of divisors
= 4 × 3 × 3 × 2 − 2 = 72 − 2 = 70
∴ Total number of ways
= 5! × 6 P5 = 5! × 6!
47
⇒
n (n − 1)
× 3 = 10
3!
2
n − n − 20 = 0 ⇒ n = 5
37 Since, 38808 = 23 × 32 × 72 × 111
30
M
23
⇒
…(i)
and number of m elements subsets of A
each containing element a4 = n − 1 C m − 1
According to the question,
= λ ⋅ n −1 C m −1
n n −1
⇒
⋅
C m −1 = λ ⋅ n −1C m −1
m
n
⇒
λ=
⇒ n = mλ
m
32 Number of times A and B are together
n − 2
on guard is
.
10
number of positive integer solutions of
the equation x + y + z = 8 which equal
8 − 1 7
to
= = 21
3 − 1 2
40 Coefficient of x10 in ( x + x2 + ...+ x 6 )4
= Coefficient of x 6 in (1 + x + ...+ x 5 )4
= (1 − x 6 )4 (1 − x ) − 4 in (1 − 4 x 6 + ... )
4
1 + 1 x + ...
9
6
Hence, coefficient of x is − 4 = 80.
6
SESSION 2
1 Given, 6 different novels and 3 different
stall can be filled in 3 ways and so on.
∴ Number of ways of loading steamer
= 3C1 × 3C1 × K × 3C1 (12 times)
= 3 × 3 × K × 3 (12 times) = 312
dictionaries.
Number of ways of selecting 4 novels
6!
from 6 novels is 6 C 4 =
= 15
2! 4!
Number of ways of selecting
1 dictionary from 3 dictionaries is
3!
3
C1 =
=3
1!2!
Number of arrangement of
4 novels and 1 dictionary where
dictionary is always in the middle, is 4!
Required number of arrangement
15 × 3 × 4! = 45 × 24 = 1080
34 If out of n points, m are collinear, then
2 Number of ways of giving one prize for
Number of times C , D and E are together
n − 3
on guard is
.
9
According to the question,
n − 2
n − 3
= 3
10
9
⇒
n − 2 = 30 ⇒ n = 32
33 First stall can be filled in 3 ways, second
Number of triangles = C3 − C3
∴ Required number of triangles
= 10C3 − 6C3 = 120 − 20
⇒
N = 100
n
35 T n = nC3 , hence T n+1 =
Now,
⇒
⇒
⇒
n +1
m
C3
T n +1 − T n = 10
n +1
C3 − nC3 = 10
[given]
(n + 1) n (n − 1)
3!
n (n − 1) (n − 2)
−
= 10
3!
n (n − 1)
(n + 1 − n + 2) = 10
3!
running = 16
Number of ways of giving two prizes for
swimming = 16 × 15
Number of ways of giving three prizes
for riding = 16 × 15 × 14
∴ Required ways of giving prizes
= 16 × 16 × 15 × 16 × 15 × 14
= 163 × 152 × 14
3 For the first player, cards can be
distributed in the 52 C17 ways. Now, out
of 35 cards left 17 cards can be
distributed for second player in 52 C17
ways.
86
40
Similarly, for third player in 18 C17 ways.
One card for the last player can be
distributed in 1 C1 way.
Therefore, the required number of ways
for the proper distribution.
= 52C17 × 35C17 × 18C17 × 1 C1
52!
35!
18!
52!
=
×
×
× 1! =
35! 17! 18!17! 17!1!
(17!) 3
4 The number of words starting from
E = 5! = 120
The number of words starting from H
= 5! = 120
The number of words starting from ME
= 4! = 24
The number of words starting from MH
= 4! = 24
The number of words starting from MOE
= 3! = 6
The number of words starting from
MOH = 3! = 6
The number of words starting from
MOR = 3! = 6
The number of words starting from
MOTE = 2! = 2
The number of words starting from
MOTHER = 1! = 1
Hence, rank of the word MOTHER
= 2 (120) + 2 (24) + 3 (6) + 2 + 1
= 309
5 A matrix whose determinant is non-zero
is called a non-singular matrix.
Here, we have
a11 a 12
S =
; aij ∈ {0, 1, 2}, a11 = a22
a
a
22
21
Clearly, n(s ) = 27
[Qfor a11 = a22 , we have 3 choices,
for a12 , we have 3 choices and for a21 , we
have 3 choices]
a11 a12
Now,
=0
a21 a22
⇒ a11 ⋅ a22 − a12 ⋅ a21 = 0
⇒
(a11 ) 2 = a12 a21 = 0
[Q a11 = a22 ]
⇒
a12 a21 = 0, 1, 4
[Q a11 ∈ {0, 1, 2}]
Consider a12 a21 = 0, this is possible in 5
cases
a12 a21 = 1, this is possible in only 1 case
a12 a21 = 4, this is possible in only 1 case
Thus, number of singular matrices in
S are 7.
Hence, number of non-singular matrices
in S are 27 − 7 = 20.
6 Let the men be M1 , M2 , ..., M10 and
women be W1 , W2 , ..., W7 .
Let W1 and W2 do not want to be on the
same group. The six members group can
contain 4 men and 2 women or 3 men
and 3 women.
The number of ways of forming 4M , 2W
group is
10
C 4 ( 5C2 + 2C1 ⋅ 5C1 ) = 4200
where, 5C2 is the number of ways without
W1 and W2 and 5C1 is the number of ways
with W1 and without W2 or with W2 and
without W1 .
The number of ways of forming 3M , 3W
5
group is 10 C3 ( 5C3 + 2C1 C2 ) = 3600
where, 5C3 is the number of ways without
W1 and W2 and 5C2 is the number of ways
with W1 or W2 but not both.
∴ Number of ways = 4200 + 3600 = 7800
7 Given, X has 7 friends, 4 of them are ladies
and 3 are men while Y has 7 friends, 3 of
them are ladies and 4 are men.
∴ Total number of required ways
= 3C3 × 4C 0 × 4C 0 × 3C3
+ 3C2 × 4C1 × 4C1 × 3C2
+ 3C1 × 4C2 × 4C2 × 3C1
+ 3C 0 × 4C3 × 4C3 × 3C 0
= 1 + 144 + 324 + 16 = 485
8 The number of ways of selecting four
numbers from 1 to 30 without any
restriction is 30 C 4 . The number of ways
of selecting four consecutive numbers
[i.e. (1, 2, 3, 4), (2, 3, 4, 5), ...,
(27, 28, 29, 30)] is 27.
Hence, the number of ways of selecting
four integers which excludes selection
of consecutive four numbers is
30 × 29 × 28 × 27
30
C 4 − 27 =
− 27
24
= 27378
9 Possible even digits are 2,4,6,8,0.
Case I Number has digits 4,6,8,0.
(Here, sum of digits is 18, divisible by 3)
Number of arrangements = 3 × 3!
[Ist place can be filled using 4, 6, 8]
= 3 × 6 = 18
Case II Number has digits 2, 4, 6, 0
(Here, sum of digits is 12, divisible by 3)
[Qall white balls are mutually identical]
Number of ways to choose zero or more
from green balls = (9 + 1)
[Qall green balls are mutually identical]
Number of ways to choose zero or more
from black balls = (7 + 1)
[Qall black balls are mutually identical]
Hence, number of ways to choose zero
or more balls of any colour
= (10 + 1) (9 + 1) (7 + 1)
Also, number of ways to choose zero
balls from the total = 1
Hence, the number of ways to choose
atleast one ball
[irrespective of any colour]
= (10 + 1) (9 + 1) (7 + 1) − 1
= 880 − 1 = 879
12 Let x1 , x2 , ..., x 8 denote the marks assign
to 8 questions.
∴ x1 + x2 + ...+ x 8 = 30
Also, x1 , x2 , ...., x 8 ≥ 2
Let, u1 = x1 − 2, u2 = x2 − 2 ... u 8
= x8 − 2
Then, (u1 + 2 + u2 + 2 + ...+ u 8 + 2) = 30
⇒
u1 + u2 + ...+ u 8 = 14
∴ Total number of solutions
=
14 + 8 − 1
C 8 −1 =
21
c7
13 210 510 1111 1313 has a divisor of the form
2α ⋅ 5β ⋅ 11γ ⋅ 13δ , where
α = 0, 1, 2,...,10; β = 0, 1, 2,...,10;
γ = 0, 1, 2,...,11; δ = 0, 1, 2,...,13
It is of the form 4 n + 1, if
α = 0 ; β = 0, 1, 2,...,10;
γ = 0, 2, 4,...,10;
δ = 0, 1, 2,...,13.
∴ Number of divisors
= 11 × 6 × 14 = 924
14 Let the particular side on which 3
1st place cannot be filled by 0.
Number of arrangements = 3 × 3! = 18
∴ Number of numbers = 18 + 18 = 36
10 24 = 2 ⋅ 3 ⋅ 4, 2 ⋅ 2 ⋅ 6, 1 ⋅ 6 ⋅ 4, 1 ⋅ 3 ⋅ 8,
1 ⋅ 2 ⋅ 12, 1 ⋅ 1 ⋅ 24
[as product of three positive integers]
∴ The total number of positive integral
solutions of xyz = 24 is
3!
3!
equal to3! +
+ 3! + 3! + 3! + i.e. 30.
2!
2!
Any two of the numbers in each
factorisation may be negative. So, the
number of ways to associate negative sign
in each case is 3 C2 i.e. 3.
∴ Total number of integral solutions
= 30 + 3 × 30 = 120
11 The number of ways to choose zero or
more from white balls = (10 + 1)
particular sailors can work be named A
and on the other side by B on which 2
particular sailors can work. Thus, we are
left with 3 sailors only. Selection of one
sailor for side A = 3C1 = 3 and, then we
are left with 2 sailors for the other side.
Now, on each side, 4 sailors can be
arranged in 4! ways.
∴ Total number of arrangements
= 3 × 24 × 24 = 1728
15 Required number is the coefficient of x10
in (1 + x + x2 + ... + x 6 ) 3
= (1 − x7 )3 (1 − x )−3 = (1 − 3 x7 + ... )
3
4 2
1 + 1 x + 2 x + K
Hence, coefficient of x10 is
12
5
− 3 = 36.
10
3
87
DAY
DAY NINE
Unit Test 1
(Algebra)
cos x
1 If f ( x ) = 2 sin x
tan x
x
x2
x
1
f ′ (x )
is equal to
2x , then lim
x→ 0
x
1
(b) −1
(a) 1
(d) − 2
(c) 2
2 If log0. 5 ( x − 1) < log0. 25 ( x − 1), then x lies in the interval
(a) (2 , ∞)
(b) (3 , ∞)
(c) (− ∞, 0)
(d) (0, 3)
3 Sum of n terms of series 12 + 16 + 24 + 40 + … will be
(a)
(b)
(c)
(d)
2 (2 n
2 (2 n
3 (2 n
4 (2 n
− 1) +
− 1) +
− 1) +
− 1) +
a 2 x 2 + bx + c = 0, β is a root of a 2 x 2 − bx − c = 0 and
0 < α < β, then the equation of a 2 x 2 + 2bx + 2c = 0 has a
root γ, that always satisfies
(b) γ = β
(d) α < γ < β
1
an even number
6
of arithmetic means are inserted. If the sum of these
means exceeds their number by unity, then the number
of means are
5 Between two numbers whose sum is 2
(b) 10
(d) None of these
1 0
6 If A =
, then which of the following is not true?
1 1
(a) lim
n→ ∞
1 − n 0 0
A =
n2
−1 0
1 0
(c) A − n =
, ∀ n ∈ N
−n 1
(a) 11
(b) 12
(c) 13
(d) 14
given by
4 Let a, b and c ∈ R and a ≠ 0. If α is a root of
(a) 12
(c) 8
Physics and Chemistry, 37 passed Mathematics,
24 Physics and 43 Chemistry. Atmost 19 passed
Mathematics and Physics, atmost 29 passed
Mathematics and Chemistry and atmost 20 passed
Physics and Chemistry. The largest possible number that
could have passed all three examinations is
8 The inequality | z − 4 | < | z − 2 | represents the region
8n
6n
8n
8n
(a) γ = α
(c) γ = (α + β) / 2
7 From 50 students taking examinations in Mathematics,
(b) lim
n→ ∞
1 − n 0 0
A =
n
−1 0
(d) None of these
(a) Re(z) > 0
(b) Re(z) < 0
(c) Re(z) > 3
(d) None of these
9 If 1, ω and ω 2 be the three cube roots of unity, then
(1 + ω )(1 + ω 2 )(1 + ω 4 ) K 2n factors is equal to
(b) −1
(d) None of these
(a) 1
(c) 0
10 If a < 0, then the positive root of the equation
x 2 − 2a | x − a | − 3 a 2 = 0 is
(a) a (− 1 − 6)
(c) a (1 − 6)
(b) a (1 −
(d) a (1 +
2)
2)
11 The common roots of the equations z 3 + 2z 2 + 2z +
1 = 0 and z 1985 + z 100 + 1 = 0 are
(a) −1, ω
(b) −1, ω2
(c) ω, ω
(d) None of these
2
12 Let z1, z 2 and z 3 be three points on | z | = 1. If θ1, θ 2 and θ 3
are the arguments of z1, z 2 and z 3 respectively, then
cos (θ1 − θ 2 ) + cos (θ 2 − θ 3 ) + cos (θ 3 − θ1 )
3
2
−3
(c) ≤
2
(a) ≥
(b) ≥ −
3
2
(d) None of these
88
40
13 If the roots of the equation
(a + b )x + 2 (bc + ad )x + (c + d ) = 0
are real, then a 2 , bd and c 2 are in
2
2
2
(a) AP
(c) HP
2
2
(b) GP
(d) None of these
14 150 workers were engaged to finish a piece of work in a
certain number of days. Four workers dropped the
second day, four more workers dropped the third day
and so on. It takes 8 more days to finish the work now.
Then, the number of days in which the work was
completed is
(a) 29 days
(b) 24 days
(c) 25 days
(d) 26 days
15 Let R be a relation defined by R = {( x , x ) : x is a prime
3
number < 10 }, then which of the following is true?
(a) R = {(1, 1), (2 , 8), (3 , 27), (4 , 64), (5 , 125),(6, 216), (7, 343),
(8, 512), (9, 729)}
(b) R = {(2 , 8), (3 , 27), (5 , 125), (7, 343)}
(d) None of the above
16 If b > a , then the equation ( x − a )( x − b ) − 1 = 0 , has
(c)
(b) 2
1
2
18 If f ( x , n ) =
(d) None of these
n
∑
r =1
r
logx , then the value of x satisfying the
x
equation f ( x , 11) = f ( x , 12) is
(a) 10
(c) 12
(b) 11
(d) None of these
19 The three numbers a , b and c between 2 and 18 are
such that their sum is 25, the numbers 2 ,a and b are
consecutive terms of an AP and the numbers b, c and18
are consecutive terms of a GP. The three numbers are
(a) 3, 8, 14
(c) 5, 8, 12
(b) 2, 9, 14
(d) None of these
20 If X is the set of all complex numbers z such that | z | = 1,
then the relation R defined on X by
2π
, is
| arg z1 − arg z 2 | =
3
(a) reflexive
(c) transitive
(b) symmetric
(d) anti-symmetric
21 If α and β are the roots of the equation ax − 2 bx + c = 0,
2
then α 3β 3 + α 2β 3 + α 3β 2 is equal to
c2
(a) 3 (c − 2 b)
a
bc 2
(c) 3
a
(b) 2 ≤ a ≤ 3
(c) 3 ≤ a ≤ 4
(d) a > 4
23 The integer k for which the inequality
x 2 − 2 ( 4 k − 1)x + 15 k 2 − 2k − 7 > 0 is valid for any x, is
(a) 2
(c) 4
(b) 3
(d) None of these
24 The maximum sum of the series
20 + 19
1
2
+ 18 + 18 + K is
3
3
(a) 310
(c) 320
(b) 290
(d) None of these
25 The number of common terms to the two sequences
17, 21, 25, ..., 417 and 16, 21, 26, ..., 466 is
(a) 21
(b) 19
(c) 20
(d) 91
26 If the sum of the first three terms of a GP is 21 and the
(a) 3 , 4
(c) 3 , 2
(b) 2 , 4
(d) None of these
1
2
3
+
+
+ ..., is
1 + 12 + 14
1 + 22 + 24
1 + 32 + 34
17 The value of x satisfying log2 ( 3x − 2) = log1/ 2 x is
1
3
(a) a < 2
27 The sum to n terms of the series
both the roots in [a , b]
both the roots in (−∞, a ]
both the roots in (b, ∞)
one root in (−∞, a) and other in (b, ∞)
(a) −
real and less than 3, then
sum of the next three terms is 168, then the first term and
the common ratio is
(c) R = {(2 , 8), (4 , 64), (6, 216), (8, 512)}
(a)
(b)
(c)
(d)
22 If the roots of the equation x 2 − 2ax + a 2 + a − 3 = 0 are
c2
(b) 3 (c + 2 b)
a
(d) None of these
n2 + n
2 (n 2 + n + 1)
n2 + n
(c)
2 (n 2 − n + 1)
(a)
(b)
n2 − n
2 (n 2 + n + 1)
(d) None of these
28 If C is a skew-symmetric matrix of order n and X is n × 1
column matrix, then X ′ C X is a
(a) scalar matrix
(c) null matrix
(b) unit matrix
(d) None of these
29 Which of the following is correct?
(a) Skew-symmetric matrix of an even order is always
singular
(b) Skew-symmetric matrix of an odd order is non-singular
(c) Skew-symmetric matrix of an odd order is singular
(d) None of the above
a1
b1
c1
30 If a 2 b2 c 2 = 5 , then the value of
a3
b3
c3
b2c3 − b3c2 a3c2 − a2c3 a2b3 − a3b2
∆ = b3c1 − b1c3 a1c3 − a3c1 a3b1 − a1b3 is
b1c2 − b2c1 a2c1 − a1c2 a1b2 − a2b1
(a) 5
(b) 25
(c) 125
(d) 0
31 The number of seven letter words that can be formed by
using the letters of the word ‘SUCCESS’ so that the two C
are together but no two S are together, is
(a) 24
(c) 54
(b) 36
(d) None of these
89
DAY
32 The greatest integer less than or equal to ( 2 + 1)6 is
(a) 196
(c) 198
(b) 197
(d) 199
33 If A and B are square matrices such that B = − A −1BA , then
(a)
(b)
(c)
(d)
AB + BA = O
(A + B)2 = A 2 − B 2
(A + B)2 = A 2 + 2 AB + B 2
(A + B)2 = A + B
xp + y
34 The determinant yp + z
0
(a) x, y, z are in AP
(c) x, y, z are in HP
x
y
xp + y
y
= 0 , if
z
yp + z
(b) x, y, z are in GP
(d) xy, yz, zx are in AP
35 The value of k, for which the system of equations
x + ky + 3z = 0, 3x + ky − 2z = 0 and 2x + 3y – 4z = 0
possess a non-trivial solution over the set of rationals, is
33
2
(d) None of these
33
2
(c) 11
(a) −
(b)
36 The number of groups that can be made from 5 different
green balls, 4 different blue balls and 3 different red balls,
if atleast 1 green and 1 blue ball is to be included is
(a) 3700
(c) 4340
(b) 3720
(d) None of these
37 If n is an integer greater than 1, then
2
3x 2
ascending power of x, is
42 The 8th term of 3x +
228096
x3
328179
(c)
x9
(a)
(b) 0
(c) a 2
(d) 2 n
38 If α , β and γ are the roots of the equation
x ( x + e ) = e ( x + 1). Then, the value of the determinant
1+ α
1
1
1
1+ β
1 is
1
1
1+ γ
2
(b) 1
(c) 0
(d) 1 +
1 1 1
+ +
α β γ
39 For all natural number n > 1, 24 n − 15 n − 1 is divisible by
(a) 225
(c) 325
(b) 125
(d) None of these
40 If x is so small that its two and higher power can be
neglected and if (1 − 2x )−1/ 2 (1 − 4x )−5 / 2 = 1 + kx, then k is
equal to
(a) − 2
(c) 10
(b) 1
(d) 11
2x + 1 4
41 If x = − 5 is a root of
2
2x
7
6
8
2 = 0 , then the
2x
other two roots are
(a) 3, 3.5
(b) 1, 3.5
(c) 3 , 6
(d) 2 , 6
(b)
228096
x9
(d) None of these
(a) 55 × 3 9
(c) 55 × 3 6
(b) 46 × 3 9
(d) None of these
44 In an examination a candidate has to pass in each of the
papers to be successful. If the total number of ways to
fail is 63, how many papers are there in the examination?
(a) 6
(b) 8
(c) 10
(d) 12
45 A is a set containing n elements. A subset P of A is
chosen. The set A is reconstructed by replacing the
elements of P. A subset Q of A is again chosen. The
number of ways of choosing P and Q, so that P ∩ Q
contains exactly two elements is
(a) 9 ⋅ n C2
(c) 2 ⋅ n Cn
(b) 3 n − n C2
(d) None of these
46 If the sets A and B are defined as
1
,0 ≠ x ∈ R }
x
B = {( x , y ): y = − x , x ∈ R }, then
A = {( x , y ) : y =
and
(a) A ∩ B = A
(c) A ∩ B = φ
(b) A ∩ B = B
(d) None of these
47 There are 16 points in a plane no three of which are in a
straight line except 8 which are all in a straight line. The
number of triangles can be formed by joining them equals
(a) 1120
(a) − 1
when expanded in
43 The greatest term in the expansion of ( 3 − 5x )11 when
1
x = , is
5
a − nC1(a − 1) + nC2 (a − 2) + ...+ ( − 1)n (a − n ) is equal to
(a) a
12
(b) 560
(c) 552
(d) 504
48 The value of the natural numbers n such that inequality
2n > 2n + 1 is valid, is
(a) for n ≥ 3
(b) for n < 3
(c) for mn
(d) for any n
2π
2π
49 Let w = cos + i sin and α = w + w 2 + w 4 and
7
7
β = w 3 + w 5 + w 6 , then α + β is equal to
(a) 0
(b) −1
(c) −2
(d) 1
f (x )
50 Let H( x ) =
, where f ( x ) = 1 − 2 sin2 x and
g( x )
g ( x ) = cos 2x , ∀ f : R → [ −1,1] and g : R → [ −1, 1].
Domain and range of H( x ) are respectively
(a) R and {1}
(b) R and {0, 1}
π
(c) R ~ {(2n + 1) } and {1},n ∈ I
4
π
(d) R ~ (2n + 1) and {0, 1},n ∈ I
2
90
40
55 The general term in the expansion of (a + x )n is
51 Let Tn denote the number of triangles which can be
n
formed using the vertices of a regular polygon of n sides.
If Tn +1 − Tn = 21, then n equals
(a) 4
(c) 7
Statement I The third term in the expansion of
m
2 x + 1 does not contain x . The value of x for
x 2
(b) 6
(d) None of these
52 If r is a real number such that | r | < 1 and if a = 5 (1 − r ),
which that term equal to the second term in the
expansion of (1 + x 3 ) 30 is 2.
then
(a) 0 < a < 5
(c) 0 < a < 10
Cr a n − r x r .
(b) − 5 < a < 5
(d) 0 ≤ a < 10
Statement II (a + x ) n =
Directions (Q. Nos. 53-57) Each of these questions
contains two statements : Statement I (Assertion) and
Statement II (Reason). Each of these questions also has
four alternative choices, only one of which is the
correct answer. You have to select one of the codes
( a), (b), (c) and (d ) given below.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
Statement II If p is a prime, then number of divisors of
p n is p n + 1 − 1.
n
Cr a n − r x r .
r = 0
56 Sets A and B have four and eight elements, respectively.
Statement I The minimum number of elements in
A ∪ B is 8.
Statement II A ∩ B = 5
a
b c
57 Let a ≠ 0, p ≠ 0 and ∆ = 0 p q
p q
0
Statement I If the equations ax 2 + bx + c = 0 and
px + q = 0 have a common root, then ∆ = 0.
Statement II If ∆ = 0, then the equations
ax 2 + bx + c = 0 and px + q = 0 have a common root.
53 Statement I The number of natural numbers which
divide 102009 but not 102008 is 4019.
n
∑
58 Assume X , Y , Z , W and P are matrices of order
2 × n, 3 × k , 2 × p, n × 3 and p × k , respectively.
Now, consider the following statements
1 0
2 0
54 Suppose A =
and B =
. Let X be a 2 × 2
0 −1
0 2
matrices such that X ′ AX = B.
I. PY + WY will be defined for k = 3 and p = n.
II. The order of the matrix 7 X − 5 Z is n × 2 (if p = n).
Choose the correct option.
Statement I X is non-singular and det ( X ) = ± 2 .
Statement II X is a singular matrix.
(a) Only I is true
(c) Both I and II are true
(b) Only II is true
(d) Neither I nor II is true
ANSWERS
1.
11.
21.
31.
41.
51.
(d)
(c)
(b)
(a)
(b)
(c)
2.
12.
22.
32.
42.
52.
(a)
(b)
(a)
(b)
(a)
(c)
3.
13.
23.
33.
43.
53.
(d)
(b)
(b)
(a)
(a)
(c)
4.
14.
24.
34.
44.
54.
(d)
(c)
(a)
(b)
(a)
(c)
5.
15.
25.
35.
45.
55.
(a)
(b)
(c)
(b)
(d)
(b)
6.
16.
26.
36.
46.
56.
(a)
(d)
(c)
(b)
(c)
(c)
7.
17.
27.
37.
47.
57.
(d)
(d)
(a)
(b)
(d)
(c)
8.
18.
28.
38.
48.
58.
(c)
(c)
(c)
(c)
(a)
(a)
9.
19.
29.
39.
49.
(a)
(c)
(c)
(a)
(b)
10.
20.
30.
40.
50.
(b)
(b)
(b)
(d)
(c)
91
DAY
Hints and Explanations
1 Applying R 1 → R 1 − R 3
cos x − tan x
f ( x) =
2 sin x
tan x
0
x2
2x
0
x
1
= (cos x − tan x)( x − 2 x )
= − x 2 (cos x − tan x)
∴ f ′ ( x) = − 2 x(cos x − tan x)
− x 2 (− sin x − sec2 x)
f ′ ( x)
= lim [−2 (cos x − tan x)
∴ lim
x→ 0
x→ 0
x
+ lim x(sin x + sec2 x)]
2
2
x→ 0
= −2 ×1= −2
2 Given, log 0.5( x − 1) < log 0.25( x − 1)
⇒
⇒
∴
log 0.5( x − 1) < log( 0.5 )2 ( x − 1)
1
log 0. 5 ( x − 1) < log 0.5( x − 1)
2
log 0.5 ( x − 1) < 0 ⇒ x − 1 > 1
x >2
3 Let,
S n = 12 + 16 + 24 + … + Tn
⇒
Again, S n = 12 + 16 + K + Tn
0 = (12 + 4 + 8 + 16 + K
upto n terms) −Tn
4 (2 n − 1 − 1)
∴ Tn = 12 +
= 2n + 1 + 8
2 −1
On putting n = 1, 2, 3,K , we get
T1 = 22 + 8, T2 = 23 + 8, T3 = 2 4 + 8 …
∴ S n = T1 + T2 + K + Tn
= (22 + 23 + … upto n terms)
+ (8 + 8 + … upto n terms)
22 (2 n − 1)
=
+ 8n
2 −1
= 4 (2 n − 1) + 8 n
4 Let f ( x) = a2 x2 + 2bx + 2c
Q We have, a2α 2 + bα + c = 0
and a2β2 − bβ − c = 0
∴ f (α ) = a2α 2 + 2bα + 2c = bα + c
= − a2α 2
2 2
f (β) = a β + 2bβ + 2c = 3 (bβ + c )
= 3 a2β2
But 0 < α < β ⇒ α , β are real number.
∴ f (α ) < 0, f (β) > 0
Hence, α < γ < β.
5 Let 2n arithmetic means be
A 1 , A 2 ,K , A 2 n between a and b .
a+b
Then, A 1 + A 2 +K + A 2 n =
× 2n
2
13 / 6
13 n
=
× 2n =
2
6
Given, A 1 + A 2 + K + A 2 n = 2 n +
13 n
2n + 1 =
⇒
6
⇒ 12 n + 6 = 13 n
∴
n=6
Hence, the number of means
= 2 × 6 = 12
1 0
6 A −1 =
−
1 1
1 0 1 0 1
⇒ ( A −1 )2 =
=
−1 1 −1 1 −2
1
10 If x ≥ a, then x2 − 2 a( x − a) − 3 a2 = 0
⇒ x 2 − 2 ax − a2 = 0
∴ x=
0
1
0 0
=
−1 0
n→ ∞
1 / n2
0
1
−n
A
=
lim
1
2
2
n
→
∞
−
n
n 1 / n
0 0
=
0 0
7 Given, n (M ∪ P ∪ C) = 50,
n (M ) = 37, n (P) = 24 , n (C) = 43
n (M ∩ P) ≤ 19, n (M ∩ C) ≤ 29,
n (P ∩ C) ≤ 20
∴ n (M ∪ P ∪ C) = n (M ) + n (P) + n (C)
− n ( M ∩ P ) − n ( M ∩ C)
− n (P ∩ C) + n (M ∩ P ∩ C)
⇒ 50 = 37 + 24 + 43 − n(M ∩ P)
− n (M ∩ C) − n (P ∩ C)
+ n (M ∩ P ∩ C)
⇒ n (M ∩ P ∩ C) = n (M ∩ P )
+ n (M ∩ C) + n (P ∩ C) − 54
∴ n (M ∩ P ∩ C)
≤ 19 + 29 + 20 − 54 = 14
8 |( x − 4) + iy|2 < |( x − 2) + iy|2
⇒
⇒
∴
2a ±
4 a2 + 4 a2
2
= a(1 ± 2 )
Since, x ≥ a
∴ x = a(1 + 2 ), it is impossible
1 0
Similarly, ( A −1 )n =
−n 1
0
1 / n
1
Now, lim A − n = lim
n→ ∞ n
n→ ∞
− 1 1 / n
and lim
= (1 + ω + ω2 + ω3 )n
= (0 + ω3 )n = ω3 n = 1
[let z = x + iy]
( x − 4)2 + y2 < ( x − 2)2 + y2
− 4 x < − 12 ⇒ x > 3
Re(z) > 3
9 (1 + ω)(1 + ω2 )(1 + ω 4 )(1 + ω 8) … to
2n factors
= (1 + ω)(1 + ω2 )(1 + ω)(1 + ω2 ) … to
2n factors
= [(1 + ω)(1 + ω) … to n factors]
[(1 + ω2 )(1 + ω2 ) … to n factors]
= (1 + ω)n (1 + ω2 )n
because a < 0
∴ x = a (1 − 2 )
If x < a, then x2 + 2 a ( x − a) − 3 a2 = 0
⇒
x 2 + 2 ax − 5 a2 = 0
∴ x = (− 1 ± 6 ) a
[impossible x < a and a < 0]
11 z3 + 2 z2 + 2 z + 1 = 0
⇒ (z + 1)(z2 + z + 1) = 0
⇒ z = – 1, ω, ω2
But z = − 1 does not satisfy the
second equation.
Hence, common roots are ω and ω2 .
12 We have,| z1| = | z2| = | z3| = 1
Now,| z1 + z2 + z3| ≥ 0
⇒ | z1|2 + | z2|2 + | z3|2 + 2 Re
(z1 z2 + z2 z3 + z3 z1 ) ≥ 0
⇒ 3 + 2 [cos(θ1 − θ2 ) + cos(θ2 − θ3 )
+ cos(θ3 − θ1 )] ≥ 0
⇒ cos(θ1 − θ2 ) + cos(θ2 − θ3 )
3
+ cos(θ3 − θ1 ) ≥ −
2
13 Here, D ≥ 0
∴4 (bc + ad )2 − 4 (a2 + b 2 )(c 2 + d 2 ) ≥ 0
⇒ b 2c 2 + a2d 2 + 2 abcd − a2c 2
− a2d 2 − b 2c 2 − b 2d 2 ≥ 0
⇒
(ac − bd )2 ≤ 0
⇒
ac − bd = 0
[since, square of any expression
cannot be negative]
∴ b 2d 2 = a2c 2
Hence, a2 , bd and c 2 are in GP.
14 Here, a = 150 and d = − 4
n
[2 × 150 + (n − 1) (− 4)]
2
= n (152 − 2 n)
Had the workers not dropped, then
the work would have finished is
(n − 8) days with 150 workers
working on each day.
Sn =
92
40
∴
⇒
⇒
∴
n(152 − 2 n) = 150 (n − 8)
n2 − n − 600 = 0
(n − 25) (n + 24) = 0
n = 25
[since, n cannot be negative]
15 Given, x is a prime < 10
∴ x = {2 , 3 , 5 , 7}
Now, from R = {( x, x 3 ) : x = 2, 3, 5 , 7}
= {(2 , 8), (3, 27), (5 , 125), (7, 343)}
16 Let f ( x) = ( x − a)( x − b ) − 1
We observe that the coefficient of x
in f ( x) is positive and
f (a) = f (b ) = − 1. Thus, the graph of
f ( x) is as shown in figure given
below
2
Y
y = f(x)
(a, 0)
X'
(b, 0)
O
(a, –1)
X
(b, –1)
Y'
It is evident from the graph that one
of the roots of f ( x) = 0 lies in (− ∞, a)
and the other root lies in (b , ∞).
log2 (3 x − 2) = − log 2 x
⇒ log2 (3 x − 2) = log2 x −1
⇒
3 x − 2 = x −1
⇒ 3 x2 − 2 x − 1 = 0
⇒ (3 x + 1) ( x − 1) = 0
1
⇒
x = 1 or x = −
3
∴ x =1
[since, negative of x cannot satisfy
the given equation]
18 f ( x, n) =
n
∑
r =1
=
n
∑
r =1
(log x r − log x x)
(log x r − 1) = log x (1⋅ 2 ... n) − n
= log x n ! − n
Given,
f ( x, 11) = f ( x, 12)
⇒ log x (11 !) − 11 = log x (12 !) − 12
12 !
⇒
log x
= 1 ⇒ log x (12) = 1
11 !
∴
x = 12
19 Given, a + b + c = 25
25 First series has common difference
20 Q | z| = 1 ⇒ z = cos θ + i sin θ
∴ arg (z) = θ
Let, arg (z1 ) = θ1 and arg (z2 ) = θ2
Then, z1 Rz2 ⇒|arg z1 − arg z2 | =
⇒
z1 Rz2 ⇒|θ1 − θ2 | =
2π
3
2π
3
2π
⇒ |θ2 − θ1 | =
3
⇒
z2 Rz1
Hence, it is symmetric.
2b
c
21 Here, α + β =
and αβ =
a
a
Now, (αβ)3 + α 2β2 ( β + α )
3
c2 2 b
c
= + 2
a
a a
=
c 2 (c + 2 b )
a3
22 According to the question,
17 Given, log2 (3 x − 2) = log1 /2 x
⇒
∴ Maximum sum,
31
2
S31 =
2 × 20 + (31 − 1) −
3
2
31
=
(40 − 20) = 310
2
From Eqs. (i) and (ii), we get
3b = 48 − 2 c
From Eq. (iii), we get
c 2 = 6 (48 − 2 c ) = 288 − 12 c
⇒
c 2 + 12 c − 288 = 0
⇒
(c + 24)(c − 12) = 0
⇒
c = 12 as c ≠ – 24
∴
b = 8 and a = 5
…(i)
Since, 2, a, b are in AP, therefore
…(ii)
2a = 2 + b
Since, b , c , 18 are in GP, therefore
…(iii)
c 2 = 18 b
D ≥ 0 and f (3) > 0
⇒
4 a2 − 4 (a2 + a − 3) ≥ 0
and
32 − 2 a (3) + a2 + a − 3 > 0
⇒ − a + 3 ≥ 0 and a2 − 5 a + 6 > 0
⇒
a ≤ 3 and a < 2 or a > 3
∴ a<2
23 Let f ( x) = x2 − 2 (4 k − 1) x + 15 k 2
− 2 k − 7, then f ( x) > 0
∴ D<0
⇒ 4 (4 k – 1)2 – 4 (15 k 2 – 2 k – 7) < 0
⇒ k2 − 6 k + 8 < 0 ⇒ 2 < k < 4
Hence, required integer value of k
is 3 .
2
24 Here, a = 20,d = −
3
As the common difference is
negative, the terms will become
negative after some stage. So, the
sum is maximum, if only positive
terms are added.
2
Now, Tn = 20 + (n − 1) − ≥ 0
3
⇒
60 − 2(n − 1) ≥ 0
⇒
62 ≥ 2 n ⇒ 31 ≥ n
Thus, the first 31 terms are
non-negative.
4 and second series has common
difference 5.
Hence, the series with common
terms has common difference is
equal to the LCM of 4 and 5 i.e. 20.
Since, the first common term is 21.
So, the series will be 21, 41, 61, …,
401 which has 20 terms.
26 Given, a1 + a2 + a3 = 21
⇒
a(1 + r + r 2 ) = 21
and
a4 + a5 + a6 = 168
⇒
ar 3 (1 + r + r 2 ) = 168
∴
r3 = 8 ⇒ r = 2
and
a (1 + 2 + 4) = 21
∴
a=3
r
27 Tr =
, r = 1, 2, 3, ..., n
1 + r2 + r 4
r
= 2
(r + r + 1) (r 2 − r + 1)
1
1
1
−
2 r 2 − r + 1 r 2 + r + 1
n
1 n
∴ ∑ Tr = ∑
2 r =1
r =1
=
1
1
r2 − r + 1 − r2 + r + 1
1 1 1
1 − + −
3
3
7
1
=
2
1
1
+
+
−
...
2
n − n + 1 n2 + n + 1
=
1
1
1− 2
2
n + n+
n2 + n
=
2
1 2(n + n + 1)
28 Here, X is n × 1, C is n × n and X ′ is
1 × n order matrix. Therefore, X ′ C X
is 1 × 1 order matrix. Let X ′ C X = K
Then, ( X ′ C X )′ = X ′ C′ X ′ ′
= X ′ (− C) X = − K
⇒
2K =O
∴
K =O
29 Since, the determinant of a
skew-symmetric matrix of an odd
order is zero. Therefore, the matrix
is singular.
30 We know that, if A is a square
matrix of order n and B is the matrix
of cofactors of elements of A. Then,
93
DAY
| B| = | A |n −1
∴ ∆ = | A |3 − 1 = 53 − 1 = 25
31 Considering CC as single letter,
U,CC, E can be arranged in 3 ! ways
Here, × U × CC × E ×
Hence, the required number of ways
= 3 !⋅ 4C3 = 24
32 Let ( 2 + 1)6 = I + F , where I is an
integer and 0 ≤ F < 1
Let
f = ( 2 − 1)6
1
Now, 2 − 1 =
2+1
⇒
0 < 2 −1 <1
Also, I + F + f = ( 2 + 1)6 + ( 2 − 1)6
= 2 [6C0 ⋅ 23 + 6C2 ⋅ 22 + 6C4 ⋅ 2 + 6C6]
= 2 (8 + 60 + 30 + 1) = 198
Hence, F + f = 198 − I is an integer.
But
0 < F + f <2
∴
F + f = 1 and I = 197
33 Given, B = − A −1 BA
⇒
− ( xp + 2 yp + z) 0
2
y
z =0
+ ...+ (− 1) Cn ]
+ [C1 − 2C2 + 3 C3 −... + (−1)n − 1 ⋅ nCn ]
x 3 + ex2 − ex − e = 0
Applying R 2 → R 2 − R 1
have
1 k
3 k
2 3
3
−2 = 0
−4
Applying
R2 → R2 − 3 R1 , R3 → R3 − 2R1 ,
1
0
k
− 2k
0 3 − 2k
43 (3 − 5 x)11 = 311 1 −
and R 3 → R 3 − R 2 ,
(1 + α )
1 1
∆ = −α
β 0
−β γ
0
3
− 11 = 0
− 10
= 1 + nC1 ⋅ 15 + nC2 ⋅ 152
+ nC3 ⋅ 153 + …
= 225 k
Hence, it is divisible by 225.
−
=
⇒ 20 k + 11 (3 − 2 k ) = 0
33
⇒
k =
2
36 Atleast one green ball can be
selected out of 5 green balls in
2 5 − 1, i.e. 31 ways. Similarly,
atleast one blue ball can be
selected from 4 blue balls in
2 4 − 1 = 15 ways and atleast one
red or not red can be selected in
23 = 8 ways.
= (1 + x) (1 + 10 x)
[neglecting higher power]
= 1 + 11 x
[neglecting higher power]
=1 + kx
k = 11
41 Applying R 1 → R 1 + R 2 + R 3 , we get
2 x + 10 2 x + 10 2 x + 10
2
2x
2
7
6
2x
=0
Taking 2 x + 10 common from R 1
and applying
C2 → C2 − C1 , C3 → C3 − C1 , we get
1
0
1
Now, T3 = 311 ⋅ 11C2 −
3
7
−1
11⋅ 10 1
= 311
× = 55 × 3 9
1⋅ 2
9
∴ Total number of ways to fail or pass
= nC0 + nC1 + K + nCn = 2 n
∴Total number of ways to fail
= 2n − 1
[since, there is only one way to pass]
According to the question,
2 n − 1 = 63 ⇒ 2 n = 2 6 ⇒ n = 6
45 Let A = {a1 , a2 , ..., an }, 1 ≤ i ≤ n
(i) ai ∈ P, ai ∈ Q
(ii) ai ∉ P, ai ∉ Q
(iii) ai ∉ P, ai ∈ Q (iv) ai ∈ P, ai ∉ Q
So, P ∩ Q contains exactly two
elements, taking 2 elements in (i) and
(n − 2) elements in (ii), (iii) and (iv).
∴ Required number of ways
= nC2 × 3 n − 2
=0
2x − 7
curves do not intersect anywhere.
∴
A∩B=φ
⇒
2 ( x + 5) (2 x − 2) (2 x − 7) = 0
∴ x = − 5 , 1, 3.5
Y
42 For ascending power of x, we take
the expression
12
2
2 + 3 x
3x
2
∴ T8 in 2 + 3 x
3x
12
2
46 It is clear from the graph that two
0
0
1
(11 + 1)
3
=3
1
−
+1
3
44 Let the number of papers be n.
40 (1 − 2 x)−1 /2 (1 − 4 x)−5/2
2 ( x + 5) 2 2 x − 2
11
1
1
= 3 11 1 −
Qx =
3
5
| x|(n + 1)
∴ Greatest term =
(| x| + 1)
39 Now, 2 4 n = (1 + 15)n
∴
5 x
3
11
= (1 + α ) (βγ − 0) + α (γ) + αβ
= αβ + βγ + γα + αβγ
= − e+e =0
⇒ − ( xp + 2 yp + z)( xz − y2 ) = 0
Hence, x, y and z are in GP.
35 For non-trivial solution, we must
12 ! 2
7
(3 x)
7 !5 ! 3 x 2
12 × 11 × 10 × 9 × 8 2 5 × 32
×
5 × 4×3 ×2
x3
228096
=
x3
= a⋅0 + 0 = 0
0
(3 x) 7
=
38 Given, equation is
2
12 − 7
5
=
n
[n C2 + nC3 ⋅ 15 + K ]
34 Applying R 3 → R 3 − pR 1 − R 2
x
y
37 LHS = a [C0 − C1 + C2 − C3
∴ 2 4 n − 1 − 15 n = 152
⇒
AB = − A ( A −1 BA )
⇒
AB = − I (BA )
∴ AB + BA = O
xp + y
yp + z
2
= 12C 7 2
3x
Hence, the required number of
ways = 31 × 15 × 8 = 3720
y=
1
x
X'
X
Y'
y = –x
94
40
47 ∴ Required number of ways
= C3 − C3 = 560 − 56 = 504
16
8
48 Check through options, the condition
2 n > 2 n + 1 is valid for n ≥ 3.
49 Here, α + β =
6
∑
k =1
wk =
w (1 − w 6)
1 −w
= −1
[Qw 7 = 1 ]
f ( x) 1 − 2 sin x
=
g ( x)
cos 2 x
cos 2 x
=
=1
cos 2 x
But cos 2 x ≠ 0
π
⇒ 2 x ≠ nπ + , n ∈ I
2
π
∴ x ∈ R ~ (2 n + 1) , n ∈ I
4
and range = {1}
2
50 Q H ( x) =
⇒
⇒
⇒
n +1
C3 − nC3 = 21
n
n
C2 + C3 − nC3 = 21
n
C2 = 21
⇒
⇒
⇒
∴ n=7
n(n − 1)
= 21
2
n2 − n − 42 = 0
(n − 7) (n + 6) = 0
[Q n ≠ − 6]
52 Since,| r | < 1 ⇒ − 1 < r < 1
Also, a = 5 (1 − r ) ⇒ 0 < a < 10
[Qat r = − 1, a = 10
and at r = 1 , a = 0]
53 The number of divisor of 10 m = 2 m5 m
is (m + 1)2 .
∴Number of divisors which divide
q
q
or a − + b − + c = 0
p
p
54 If X = O, then X ′ AX = O ⇒ B = O, a
contradiction.
Let det ( X ) = a, then det ( X ′ ) = a
∴ det ( X ′ AX ) = det (B)
⇒
a (−1) a = − 4
[Qdet ( X ′AX) = det ( X ′ ) det (A) det (X)]
∴ a=±2
As det ( X ) ≠ 0, X cannot be a
singular matrix.
1
x2
2
55 T3 = mC2 (2 x)m − 2
= mC2 (2)m − 2 ⋅ x m − 6
For independent of x, put
m −6 =0 ⇒ m =6
∴ T3 = 6C2 (2)6 − 2 = 15 × 16 = 240
According to the question,
30
Tn = nC3 and Tn +1 − Tn = 21
51
2
102009 but not 10 2008 is
(2010)2 − (2009)2 = 4019
⇒
⇒
C1 x 3 = 240
x3 = 8
x =2
56 n ( A ∪ B) = n ( A ) + n (B) − n ( A ∩ B)
= 4 + 8 − n( A ∩ B)
= 12 − n ( A ∩ B)
Since, maximum number of element
in n ( A ∩ B) = 4.
∴ Minimum number of element is
n ( A ∪ B) = 12 − 4 = 8
57 If λ is a common root of
ax + bx + c = 0
and px + q = 0, then aλ2 + bλ + c = 0,
pλ + q = 0 and pλ2 + qλ = 0
Eliminating λ, we obtained ∆ = 0.
For Statement II, expanding ∆ along
C1 , we obtain
aq 2 + p (bq − cp) = 0
2
Thus, ax2 + bx + c = 0 and
px + q = 0 have a common root.
58
I. Matrices P and Y are of the
orders p × k and 3 × k,
respectively.
Therefore, matrix PY will be
defined, if k = 3.
Consequently PY will be of the
order p × k . Matrices W and Y
are of the orders n × 3 and 3 × k ,
respectively.
Since, the number of columns in
W is equal to the number of rows
in Y , matrix WY is well-defined
and is of the order n × k .
Matrices PY and WY can be
added only when their orders are
same.
However, PY is of the order p × k
and WY is of the order n × k ,
therefore we must have p = n.
Thus, k = 3 and p = n are the
restrictions on n, k and p, so that
PY + WY will be defined.
II. Matrix X is of the order 2 × n.
Therefore, matrix 7 X is also of
the same order.
Matrix Z is of the order 2 × p,
i.e. 2 × n.
[Q n = p]
Therefore, matrix 5Z is also of
the same order.
Now, both the matrices 7 X and
5Z are of the order 2 × n.
Thus, matrix 7 X − 5 Z is welldefined and is of the order 2 × n.
DAY TEN
Real Function
Learning & Revision for the Day
u
Real Valued Function and
Real Function
u
u
Domain and Range of real
Function
Algebra of Real Functions
u
u
u
Inverse Function
Basic Functions
Nature of a Functions
Real Valued Function and Real Function
A Function f : A → B is said to be a real valued function if B ⊆ R (the set of real numbers), if
both A and B are subset of R (the set of real numbers) then f is called a real function.
NOTE Every real function is a real valued function but converse need not be true.
Domain and Range of Real Function
The domain of y = f ( x) is the set of all real x for which f ( x) is defined (real).
Range of y = f ( x) is collection of all distinct images corresponding to each real number in
the domain.
NOTE If f : A → B, then A will be domain of f and B will be codomain of f.
To find range
(i) First of all find the domain of y = f ( x) .
(ii) If domain has finite number of points, then range is the set of f − images of these points.
(iii) If domain is R or R − {some finite points}, express x in terms of y and find the values
of y for which the values of x lie in the domain.
(iv) If domain is a finite interval, find the least and the greatest values for range using
monotonicity.
Algebra of Real Functions
Let f : X → R and g : X → R be two real functions. Then,
The sum f + g : X → R defined as
( f + g) ( x) = f ( x) + g( x).
●
●
The difference f − g: X − R, defined as
( f − g)( x) = f ( x) − g( x)
PRED
MIRROR
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
TEN
96
●
●
●
●
●
The product fg : X → R,
defined as ( fg) ( x) = f ( x) g( x)
Some algebraic functions are given below
f + g and fg are defined only, if f and g have the same
domain. In case the domain of f and g are different,
domain of f + g or fg = Domain of f ∩ Domain of g.
The product cf : X→ R, defined as
(cf ) ( x) = cf ( x), where c is a real number.
f
f
f ( x)
The quotient is a function defined as ( x) =
,
g
g
g ( x)
provided g( x) ≠ 0, x ∈ X
If domain of y = f ( x) and y = g( x) are D1 and D2
respectively, then the domain of f ( x) ± g( x) or
f ( x)
is
f ( x) ⋅ g( x) is D1 ∩ D2 , while domain of
g ( x)
D1 ∩ D2 − { x : g( x) = 0}.
Equal or Identical Functions
Two functions f and g are said to be equal, if
(i) the domain of f = the domain of g
(ii) the range of f = the range of g
(iii) f ( x) = g( x), ∀ x ∈ domain
Inverse Functions
(i) Polynomial Function
(a) The function
f ( x) = a0 x n + a1 x n − 1 + a2 x n − 2 + K + an − 1 x + an
where, a0, a1 , a2 ,K , an are real numbers and n ∈ N is known
as polynomial function. If a0 ≠ 0, then n is the degree of
polynomial function.
(b) Domain of polynomial function is R.
(c) A polynomial of odd degree has its range (−∞, ∞) but a
polynomial of even degree has a range which is always
subset of R.
(ii) Constant Function The function f ( x) = k , where k is constant,
is known as constant function. Its domain is R and range is {k },
(iii) Identity Function The function f ( x) = x, is known as identity
function. Its domain is R and range is R.
(iv) Rational Function The function f ( x) =
p( x)
, where p( x) and
q ( x)
q ( x) are polynomial functions and q ( x) ≠ 0 , is called rational
function.
Its domain is R − { x | q ( x) = 0}.
(v) Irrational Function The algebraic functions containing one or
more terms having non-integral rational power of x are called
irrational functions.
e.g.,
y = f ( x) = 2 x − 3 x + 6
If f : A → B is a bijective function, then the mapping
f −1 : B → A which associate each element b ∈ B to a
unique element a ∈ A such that f (a) = b , is called the
inverse function of f .
f −1 (b ) = a ⇔ f (a) = b
The curves y = f ( x) and y = f −1 ( x) are mirror images of
each other in the line mirror y = x.
2. Piecewise Functions
●
f is invertible iff f is one-one and onto.
Piecewise functions are special type of algebraic functions.
●
Inverse of bijective function is unique and bijective.
●
●
●
●
●
●
−1
The solution of f ( x) = f ( x) are same as the solution
of f ( x) = x.
1
is called the
x
reciprocal function of x. Its domain is R − {0} and range is
R − {0}.
(vi) Reciprocal Function The function f ( x) =
(i) Absolute Valued Function ( Modulus Function) The function
x, x ≥ 0
is called modulus function.
f ( x) = | x | =
− x , x < 0
Y
If fo g = I = gof , then f and g are inverse of each other.
fof
−1
−1
−1 −1
= I B , f of = I A and ( f )
= f.
If f and g are two bijections such that (gof) exists, then
gof is also bijective function and (gof )−1 = f −1og −1 .
y = |x|
X¢
O
Basic Functions
Basic functions can be categorised into the following
categories.
1. Algebraic Functions
A function, say f ( x), is called an algebraic function, if it
consists finite number of terms involving powers and
roots of the independent variable x and the four
algebraic operations + ,−, × and ÷ .
Y¢
Its domain is R and range is [0, ∞).
Properties of Modulus Function
(a) | x | ≤ a ⇒ − a ≤ x ≤ a (a > 0)
(b) | x | ≥ a ⇒ x ≤ − a or x ≥ a (a > 0)
(c) | x ± y | ≤ | x | + | y |
(d) | x ± y | ≥ || x | − | y ||
X
97
DAY
(ii) Signum Function The function f ( x)
x
1, if
| x|
, if x ≠ 0
or
| x|
= − 1, if
= sgn ( x) = x
0,
if x = 0 0, if
Y
4
x > 0
x < 0
x = 0
3
2
1
called signum function.
X¢
Y
y=1
X¢
–1 0 1
–1
–2
–4 –3 –2
f (x) = (x)
Y¢
Its domain ∈ R and range ∈ I.
Y¢
Its domain is R and range is {− 1, 0, 1}.
(iii) Greatest Integer Function The symbol [ x] indicates the
integral part of x which is nearest and smaller than to x.
It is also known as floor of x.
x ∀ x ∈ I
The function f ( x) = [ x] =
is called
n, n ≤ x < n + 1, n ∈ I
greatest integer function.
Y
(vi) Transcendental function The function which is not
algebraic is called transcendental function.
(vii) Exponential Function The function f ( x) = ax , a > 0, a ≠ 1,
is called an exponential function.
Y
Y
0<a<1
1
1
X¢
X¢
X
O
If a > 1
2
1
–2
–1
X
4
–4
y=–1
X¢
3
–3
X
O
2
Y¢
0
1
–1
Y¢
Its domain is R and range is (0, ∞).
It is a one-one into function.
X
2
X
O
f (x) = [x]
(viii) Logarithmic Function The function f ( x) = log a x, ( x, a > 0)
and a ≠ 1 is called logarithmic function.
–2
Y
Y
Y¢
a>1
0<a<1
Its domain is R and range is I.
(iv) Fractional Part Function The symbol { x} indicates the
fractional part of x. i.e. { x} = x − [ x], x ∈ R
y = { x} = x − [ x]
0, ∀ x ∈ I
The function f ( x) = { x} =
x − n, n ≤ x < n + 1, n ∈ I
is called the fractional part function.
1
X¢
X
X¢
X
1
∴
Y
Y¢
Y¢
Its domain is (0, ∞) and range is R.
It is a one-one into function.
(ix) Trigonometric Functions Some standard trigonometric
functions with their domain and range, are given below
1
(a) Sine Function f ( x) = sin x,
X¢
–3
–2
–1
0
1
2
3
X
Y
(–3p/2, 1)
Y¢
(p/2, 1)
Its domain is R and range is (0, 1).
(v) Least Integer Function The symbol ( x) indicates the
integer part of x which is nearest and greater than x.
x, ∀ x ∈ I
The function f ( x) = ( x) =
n + 1, n < x ≤ n + 1 n ∈ I
is called least integer function.
X¢
–p
O
X
p
(3p/2, –1)
(–p/2, –1)
Y¢
Its domain is R and the range is [− 1, 1] .
TEN
98
(c) y = f ( x) = tan −1 x
(b) Cosine Function f ( x) = cos x,
Y
Y
(0, 1)
- 2p
X¢
y=x
π/2
y = tan –1 x
p
-p
-3p
2
y = tan x
-p/2
(–p, –1)
p/2
2p
3p
2
(p, –1)
X
X¢
–π/2
π/2
O
X
Y¢
Its domain is R and the range is [− 1, 1].
(c) Tangent Function f ( x) = tan x,
–π/2
Y
X¢
–3π/2
–π
–π/2
Y¢
π
O π/2
π π
Its domain is R and range is − , .
2 2
X
3π/2
Nature of a Function
A function f ( x) is said to be an odd function, if
Y¢
(2 n + 1) π
Its domain is R −
, n ∈ I and range is R.
2
(x) Inverse Trigonometric Function Some standard inverse
trigonometric functions with their domain and range, are
given below.
(a) y = f ( x) = sin −1 x
Y y=sin–1 x
π/2
f (− x) = − f ( x), ∀ x.
A function f ( x) is said to be an even function,
if f (− x) = f ( x), ∀ x.
Different Conditions for Even and Odd Functions
f( x )
y=sin x
−π/2 −1
1
O
π/2
X
Odd
Odd
Even
Even
Odd
Even
Neither odd
nor even
Neither odd
nor even
Even
Odd
Neither odd
nor even
Neither odd
nor even
f ( x) / g ( x)
(gof ) ( x )
( fog )( x )
Even
Odd
Odd
−π/2
Y′
Even
Even
Even
Even
Odd
Odd
Even
Even
Odd
Odd
Even
Even
NOTE
• Every function can be expressed as the sum of an even and
π
an odd function.
• Zero function f ( x ) = 0 is the only function which is both
π/2
even and odd.
y=x
• Graph of odd function is symmetrical about origin.
• Graph of even function is always symmetrical aboutY-axis.
1
π/2
–1
Odd
Even
Even
Y
X¢
Odd
f ( x )g ( x )
(b) y = f ( x) = cos −1 x
y= cos x
f ( x) − g ( x)
−1
π π
Its domain is [− 1, 1] and range is − , .
2 2
–1
f ( x) + g ( x)
Even
y=x
1
X¢
g ( x)
O
1
y=cos x
π
X
3. Periodic Function
●
Y′
●
Its, domain is [− 1, 1] and range is [0, π ].
A function f ( x) is said to be periodic function, if there exists
a positive real number T, such that f ( x + T ) = f ( x), ∀ x ∈ R.
The smallest value of T is called the Fundamental period of
f ( x).
99
DAY
Properties of Periodic Function
Periods of Some Important Functions
(i) If f ( x) is periodic with period T, then cf ( x), f ( x + c) and
f ( x) ± c is periodic with period T.
(ii) If f ( x) is periodic with period T, then kf (cx + d) has
T
period
.
|c|
(iv) If f (x) is periodic with period T1 and g ( x) is periodic
with period T2 , then f ( x) + g ( x) is periodic with period
equal to LCM of T1 and T2 , provided there is no positive
k, such that f (k + x) = g( x) and g (k + x) = f ( x).
(iv) If f ( x) is a periodic function with period T and g ( x) is
any function, such that range of f ⊆ domain of g, then
gof is also periodic with period T.
Function
Periods
sin x, cos x, sec x, cosec x, (sin x )2 n + 1 ,
(cos x )2 n + 1 , (sec x )2 n + 1 , (cosec x )2 n + 1
2π
tan x, cot x, tann x, cotn x, (sin x )2 n , (cos x )2 n ,
(sec x )2 n , (cosec x )2 n ,|sin x | ,|cos x | ,| tan x | ,
|cot x |,|sec x | ,|cosec x |
π
|sin x + cos x |, sin4 x + cos 4 x,
|sec x | + |cosec x | ,| tan x | + |cot x |
π /2
x − [ x]
1
Algebraic functions like
Period does
notexist
x, x , x + 5,c ,... etc.
2
2
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
6 The period of the function f ( x ) = sin3 x + cos 3 x is
1 Two sets A and B are defined as follows
A = {( x , y ) : y = e 2 x, x ∈ R } and
B = {( x , y ) : y = x 2, x ∈ R }, then
(a) A ⊂ B
(c) A ∪ B
(a) 2 π
2π
(c)
3
(b) B ⊂ A
(d) A ∩ B = φ
2 Domain of definition of the function f ( x ) = sin−1( 2x ) +
π
6
for real valued x , is
1 1
(a) − ,
4 2
1 1
(c) − ,
2 9
1
(b) − ,
2
1
(d) − ,
4
3 The domain of the function f ( x ) =
(a) (0, ∞)
(c) (− ∞, ∞) − (0)
(a)
(b)
(c)
(d)
(b) {3,−3 }, φ j NCERT Exemplar
(d) {4,−4 },{2,−2 }
x − 1, x ≥ 0
1
(b)
, x > −1
(x + 1)2
(c)
x + 1, x ≥ − 1
(d)
x − 1, x ≥ 0
1 1
j
(1, 2)
(−1, 0) ∪ (1, 2)
(1, 2) ∪ (2 , ∞)
(−1, 0) ∪ (1, 2) ∪ (2 , ∞)
5. If f : R → R is a function satisfying the property
(a) periodic with period 3
(b) periodic with period 4
(c) non- periodic
(d) periodic with period 5
(a) φ, {4,−4 }
(c) {4, − 4 }, φ
(a) −
(b) (− ∞, 0)
(d) (− ∞, ∞)
f ( x + 1) + f ( x + 3) = 2 for all x ∈ R , then f is
7 Let f : R → R be defined by f ( x ) = x 2 + 1.
Then, pre-images of 17 and −3, respectively are
whose graph is reflection of the graph of f ( x ) w.r.t. the
line y = x , then g ( x ) is equal to
1
is
|x|−x
3
+ log10 ( x 3 − x ), is
4 − x2
(d) None of these
8 Suppose f ( x ) = ( x + 1)2 for x ≥ −1. If g ( x ) is the function,
1
2
1
4
4 Domain of definition of the function
f (x ) =
(b) π
AIEEE 2003
x
9 The function f : R → − , defined as f ( x ) =
is
2 2
1+ x2
(a) invertible
(b) injective but not surjective
(c) surjective but not injective
(d) neither injective nor surjective
10 The inverse of the function f ( x ) =
x − 2
(a) loge
x − 1
JEE Main 2017
j
ex − e−x
+ 2 is given by
ex + e−x
1/ 2
x − 1
(b) loge
3 − x
1/ 2
x − 1
(d) loge
x + 1
x
(c) loge
2 − x
1/ 2
−2
TEN
100
11 If f ( x ) is an invertible function, and g ( x ) = 2f ( x ) + 5, then
the value of g −1 is
then domain of f ( x ) + g ( x ) is D1 ∩ D2 , then
1
2f −1 (x) + 5
x −5
(d) f −1
2
(a) 2f −1 (x) − 5
(b)
1
(c) f −1 (x) + 5
2
Statement I The domain of the function
f ( x ) = sin−1 x + cos −1 x + tan−1 x is [ −1, 1].
12 Let f : ( 2 , 3) → ( 0, 1) be defined by f ( x ) = x − [ x ], then
f −1 ( x ) is equal to
(a) x − 2
(b) x + 1
(c) x −1
(d) x + 2
13 For a real number x , [ x ] denotes the integral part of x.
The value of
99
1
is
+
2 100
(a) 49
1 1
2
1 1
+
+
+
+
+K +
2 2 100 2 100
(b) 50
(c) 48
19. Statement I The period of
(d) 51
1 − x
3
= x , [ where, x ≠ −1,1 and f ( x ) ≠ 0], then
1 + x
find | [f ( − 2)]| (where [.] is the greatest integer function)
(b) 1 − x
(c) 1
(d) 2
− 1 , x < 0
15 If g ( x ) = 1 + x − [ x ] and f ( x ) = 0 , x = 0 , ∀ x, then
1, x > 0
f {g ( x )} is equal to
(a) x
Statement II sin−1 x and cos−1 x is defined in | x | ≤ 1and
tan−1 x is defined for all x.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d)Statement I is false; Statement II is true
f ( x ) = 2 cos
14 If f 2 ( x )⋅ f
(a) 1/ x
18. If domain of f ( x ) and g ( x ) are D1 and D2 respectively,
(b) 1
(c) f (x)
(d) g (x)
1
1
( x − π ) + 4 sin ( x − π ) is 3π.
3
3
Statement II IfT is the period of f ( x ), then the period of
T
.
f (ax + b) is
|a |
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
16. The function f ( x ) = log ( x + x 2 + 1), is
(a) an even function
(b) an odd function
(c) a periodic function
(d) neither an even nor an odd function
17. Statement I f ( x ) = | x − 2 | + | x − 3 | + | x − 5 | is an odd
function for all values of x lie between 3 and 5.
Statement II For odd function f ( − x ) = − f ( x )
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
20. If the range of f ( x ) is collection of all outputs f ( x )
corresponding to each real number in the domain, then
1
Statement I The range of log
is ( − ∞, ∞).
2
1 + x
Statement II When 0 < x ≤ 1, log x ∈ ( − ∞, 0].
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
x −1
, where {}⋅ denotes the
x − 2 {x }
fractional part of x, is
1 Domain of f ( x ) =
(a) (− ∞, 0) ∪ (0, 2]
(c) (− ∞, ∞) ~ [0, 2)
(b) [1, 2)
(d) (− ∞, 0) ∪ (0, 1] ∪ [2, ∞)
2 Range of f ( x ) = [| sin x | + | cos x | ], where [⋅] denotes the
greatest integer function, is
(a) {0}
(b) {0, 1}
(c) {1}
(d) None of these
3 If [ x 2 ] + x − a = 0 has a solution, where a ∈ N and a ≤ 20,
then total number of different values of a can be
(a) 2
(b) 3
(c) 4
(d) 6
4 Total number of solutions of [ x ] = x + 2 {x }, where [⋅]
2
and {}⋅ denotes the greatest integer function and
fractional part respectively, is equal to
(a) 2
(c) 6
(b) 4
(d) None of these
101
DAY
5 If f ( x ) = sin x + cos x , g ( x ) = x 2 − 1, then g { f ( x )} is
invertible in the domain
π
(a) 0,
2
π π
(c) − ,
2 2
8 If f ( x ) and g ( x ) are two functions such that
f ( x ) = [ x ] + [ − x ] and g ( x ) = {x } ∀x ∈ R and h( x ) = f (g ( x ));
then which of the following is incorrect ?
([⋅] denotes greatest integer function and {}⋅ denotes
fractional part function).
π π
(b) − ,
4 4
(d) [0, π]
(a) f (x) and h (x) are inertial functions
(b) f (x) = g (x) has no solution
(c) f (x) + h (x) > 0 has no solution
(b) f (x) − h (x) is a periodic function
6 Let f ( x ) = x 10 + a ⋅ x 8 + b ⋅ x 6 + cx 4 + dx 2 be a polynomial
function with real corfficient. If f (1) = 1 and f ( 2) = −5, then
the minimum number of distinct real zeroes of f ( x ) is
(a) 5
(c) 7
9 The period of the function f ( x ) = [ 6x + 7] + cos πx − 6x ,
(b) 6
(d) 8
where [⋅] denotes the greatest integer function, is
−1
−1
(c) 2
(d) None of these
10 The number of real solutions of the equation
f og ( 23) equals
(a) 2
(c) 4
(b) 2 π
(a) 3
7 If f : R → R , f ( x ) = x 3 + 3, and g : R → R , g ( x ) = 2x + 1, then
log0. 5 | x | = 2| x | is.
(b) 3
(d) 5
(a) 1
(b) 2
(c) 0
(d) None of these
ANSWERS
SESSION 1
SESSION 2
1 (d)
2 (a)
3 (b)
4 (d)
5 (b)
6 (a)
7 (c)
8 (d)
9 (c)
10 (b)
11 (d)
12 (d)
13 (b)
14 (d)
15 (b)
16 (b)
17 (b)
18 (a)
19 (d)
20 (d)
1 (d)
2 (c)
3 (c)
4 (b)
5 (b)
6 (a)
7 (a)
8 (b)
9 (c)
10 (b)
Hints and Explanations
SESSION 1
1 Set A represents the set of points
lying on the graph of an
exponential function and set B
represents the set of points lying on
the graph of the polynomial.
Take e2 x = x 2 , then the two curves
does not intersect. Hence, there is
no point common between them.
2 For f ( x) to be defined,
π
sin (2 x) +
≥0
6
π
π
⇒
− ≤ sin −1 (2 x) ≤
2
6
π
π
sin − ≤ 2 x ≤ sin
⇒
6
2
1
1
⇒
− ≤ x≤
4
2
1 1
⇒
x∈ − ,
4 2
1
3 y=
| x| − x
−1
For domain,| x | − x > 0
⇒
| x| > x
i.e. only possible, if x < 0.
∴
x ∈ (− ∞, 0)
3
+ log10 ( x3 − x)
4 − x2
For domain of f ( x),
x3 − x > 0
⇒ x( x − 1)( x + 1) > 0
4 Given, f ( x) =
–
+
–1
–
0
+
1
⇒ x ∈ (−1, 0) ∪ (1, ∞)
and 4 − x2 ≠ 0
⇒ x≠±2
⇒
x ∈ (−∞, − 2) ∪ (−2 , 2) ∪ (2 , ∞)
So, common region is
(−1, 0) ∪ (1, 2) ∪ (2 , ∞).
6 We have,
f ( x + 1) + f ( x + 3) = 2 ...(i)
On replacing x by x + 2, we get
...(ii)
f ( x + 3) + f ( x + 5) = 2
On subtracting Eq. (ii) from Eq. (i),
we get
f ( x + 1) − f ( x + 5) = 0
⇒
f ( x + 1) = f ( x + 5)
Now, on replacing x by x − 1, we
get
f ( x) = f ( x + 4)
Hence, f is periodic with period 4.
3 sin x − sin 3 x
4
6 f ( x) =
3 cos x + cos 3 x
4
∴ Period of f ( x) = LCM of period of
+
{sin x, cos x, sin 3 x, cos 3 x}
LCM of {2 π , 2 π }
=
= 2π
HCF of {1, 3}
7 Let y = x2 + 1
⇒
∴
∴
x = ± y −1
f
−1
( x) = ±
x −1
−1
f (17) = ± 17 − 1 = ± 4
and f −1 (−3) = ± −3 − 1
= ± −4 ∉ R
∴
−1
f (−3) = φ
8 Let y = ( x + 1)2 for x ≥ − 1
⇒
± y = x +1⇒
⇒
⇒
y ≥ 0, x + 1 ≥ 0
x = y −1
⇒
f −1 ( y) =
⇒
−1
f ( x) =
y= x+1
y −1
x – 1, x ≥ 0
TEN
102
x
1 + x2
1
1
x
∴
f =
x
1
1+ 2
x
x
=
= f ( x)
1 + x2
1
∴
f = f (2)
2
1
or
f = f (3)
3
and so on.
So, f ( x) is many-one function.
Again, let y = f ( x)
x
⇒
y=
1 + x2
⇒
y + x2 y = x
⇒
yx2 − x + y = 0
As,
x ∈R
∴
(− 1)2 − 4 ( y)( y) ≥ 0
⇒
1 − 4 y2 ≥ 0
− 1 1
y∈
,
⇒
2 2
− 1 1
∴Range = Codomain =
,
2 2
9 We have, f ( x) =
So, f ( x) is surjective.
Hence, f ( x) is surjective but not
injective.
ex − e−x
10 Given, y = x
+2
e + e−x
⇒
⇒
⇒
e2 x − 1
+2
y = 2x
e +1
1− y y –1
e2 x =
=
y −3 3 − y
y − 1
1
x = loge
3 − y
2
y − 1
⇒ f −1 ( y) = loge
3 − y
⇒
x − 1
f −1 ( x) = loge
3 − x
13 Q[ x] denotes the integral part of x.
50
1
each
+
2 100
term will be one. Hence, the sum of
given series will be 50.
Hence, after term
1 − x
3
…(i)
= x
1 + x
1− x
On replacing x by
, we get
1+ x
14 f 2 ( x) ⋅ f
1 − x
1 − x
f2
f ( x) =
1 + x
1 + x
3
From Eqs. (i) and (ii),
3
1 + x
f 3 ( x) = x 6
1 − x
1 + x
f ( x) = x 2
1 − x
−4
f (− 2) =
⇒ [ f (− 2)] = − 2
3
| [ f (− 2)]| = 2
⇒
⇒
⇒
15 Q g ( x) = 1 + x − [ x] [put x = n ∈ Z ]
∴
g ( x) = 1 + x − x = 1
and g( x) = 1 + n + k − n = 1 + k
[put x = n + k ]
[where, n ∈ Z , 0 < k < 1 ]
−1, g( x) < 0
Now, f { g( x)} = 0, g( x) = 0
1, g ( x) > 0
Clearly,
g ( x) > 0 , ∀ x
So,
f { g( x)} = 1 , ∀ x
16 Given that, f ( x) = log ( x +
Now,
1 /2
11 We have, g( x) = 2 f ( x) + 5
Now, on replacing x by g −1 ( x), we get
g(g −1 ( x)) = 2 f (g −1 ( x)) + 5
⇒
x = 2 f (g −1 ( x)) + 5
x −5
⇒ f (g −1 ( x)) =
2
−1
−1 x − 5
⇒
g ( x) = f
2
12 f : (2 , 3) → (0, 1) and f ( x) = x − [ x]
∴ f ( x) = y = x − 2 ⇒ x = y + 2
⇒
f −1 ( x) = x + 2
x2 + 1 )
f (− x) = log (− x +
∴ f ( x) + f (− x) = log ( x +
1 /2
…(ii)
x2 + 1 )
x + 1)
2
+ log (− x + x 2 + 1 )
= log (1) = 0
Hence, f ( x) is an odd function.
−3 x + 10 , ∀ x ≤ 2
− x + 6, ∀ 2 < x ≤ 3
17 Here, f ( x) =
x, ∀ 3 < x ≤ 5
3 x − 10,
∀ x >5
∴
⇒
f ( x) = x, ∀ 3 < x < 5
f (− x ) = − x = − f ( x )
18 Since, sin −1 x is defined in [−1, 1] ,
cos −1 x is defined in [−1, 1] and
tan −1 x is defined in R.
Hence, f ( x) is defined in [−1, 1].
1
19 Period of 2 cos ( x − π ) and
3
1
2π 2π
or 6 π , 6 π
4 sin ( x − π ) are
,
3
1/3 1/3
∴ Period of their sum = 6π
1
is (0, 1) and
1 + x2
domain R
20 Range of
∴
1
log
∈ (− ∞, 0]
1 + x2
SESSION 2
1 We have,
x −1
≥ 0, here two
x − 2 { x}
cases arise
Case I x ≥ 1 and x > 2 { x}
⇒
x ≥2
∴
x ∈ [2, ∞).
Case II x ≤ 1 and x < 2 { x}
⇒
x < 1 and x ≠ 0.
∴ x ∈ ( − ∞, 0) ∪ (0, 1).
Finally, x = 1 is also a point of the
domain.
2 y = | sin x | + | cos x |
⇒ y2 = 1 + | sin 2 x |
⇒ 1 ≤ y2 ≤ 2 ⇒ y ∈[1, 2 ]
∴
f ( x) = [ y] = 1, ∀ x ∈ R
3 Since, [ x2 ] + x − a = 0
∴ x has to be an integer.
⇒
a = x 2 + x = x ( x + 1)
Thus, a can be 2, 6, 12, 20.
4 [ x]2 = x + 2{ x}
⇒ [ x] 2 = [ x] + 3 { x}
[ x] 2 − [ x]
⇒ { x} =
3
⇒ 0≤
[ x] 2 − [ x]
<1
3
1 − 13 1 + 13
, 0 ∪ 1,
⇒ [ x] ∈
2
2
⇒
⇒
∴
[ x] = − 1, 0, 1, 2
2
2
{ x} = , 0, 0,
3
3
1
8
x = − , 0, 1,
3
3
5 g{ f ( x)} = (sin x + cos x)2 − 1 is
invertible.
⇒ g{ f ( x)} = sin 2 x
We know that, sin x is bijective only
π π
when x ∈ − , .
2 2
Thus, g{ f ( x)} is bijective, if
π
π
− ≤ 2x ≤ .
2
2
π
π
∴
− ≤ x≤
4
4
103
DAY
6 Since, f ( x) is an even function.
∴Its graph is symmetrical about
Y -axis
Also, we have,
f (1) = 1 and f (2) = −5
⇒ f (−1) = 1 and f (−2) = −5
According to these information, we
have the following graph
Y
X1
• –2 •
–1
1
• 2 •
1
X
Y1
Thus, minimum number of zeroes
is 5.
7 Clearly, f −1og −1 (23) = (gof )−1 (23)
Here, gof ( x) = 2( x 3 + 3) + 1
= 2 x3 + 7
Now, let y = (gof )−1 (23)
⇒ (gof )( y) = 23
⇒ 2 y3 + 7 = 23
⇒
2 y3 = 16
⇒
y3 = 8
⇒
y =2
Hence, f −1og −1 (23) = 2
8 We have,
0, if x ∈ I
f ( x ) = [ x ] + [− x ] =
−1, if x ∉ I
0, if x ∈ I
g ( x) = { x} =
( x), if x ∉ I
and h( x) = f (g( x))
= f ({ x})
f (0), x ∈ I
=
f ({ x}), x ∉ I
0 , x ∈I
=
−1 , x ∉ I
Clearly, option (b) is incorrect.
9 We have, f ( x) = [6 x + 7] + cos πx − 6 x
= [6 x] + 7 + cos πx − 6 x
= 7 + cos πx − {6 x}
[Q{ x} = x − [ x]]
1
and
6
cos πx has the period 2, therefore the
1
period of f ( x) = LCM 2, which is
6
2.
Hence, the period is 2.
Now, as {6 x} has period
10 For the solution of given equation,
let us draw the graph of y = log 0. 5| x|
and y = 2| x|
y=2|x|
Y
•
•
X1
y=log0.5 (–x)
X
Y
1
y=log0.5 (x)
From the graph it is clear that there
are two solution.
DAY ELEVEN
Limits, Continuity
and Differentiability
Learning & Revision for the Day
u
Limits
u
Methods to Evaluate Limits
u
Important Results on Limit
u
Continuity
u
Differentiability
Limits
Let y = f ( x) be a function of x. If the value of f ( x) tend to a definite number as x tends to a, then
the number so obtained is called the limit of f ( x) at x = a and we write it as lim f ( x).
x→ a
●
●
If f ( x) approaches to l1 as x approaches to ‘a’ from left, then l1 is called the left hand limit of
f ( x) at x = a and symbolically we write it as f (a − 0) or lim− f ( x) or lim f (a − h)
x→ a
Similarly, right hand limit can be expressed as
h→ 0
f (a + 0) or lim+ f ( x) or lim f (a + h)
h→ 0
x→ a
●
lim f ( x) exists iff lim− f ( x) and lim+ f ( x) exist and equal.
x→ a
x→ a
x→ a
Fundamental Theorems on Limits
PRED
MIRROR
If lim f ( x) = l and lim g( x) = m (where, l and m are real numbers), then
x→ a
x→ a
(i) lim { f ( x) + g( x)} = l + m
[sum rule]
(ii) lim { f ( x) − g( x)} = l − m
[difference rule]
x→ a
x→ a
(iii) lim { f ( x) ⋅ g( x)} = l ⋅ m
[product rule]
x→ a
(iv) lim k ⋅ f ( x) = k ⋅ l
[constant multiple rule]
x→ a
f ( x) l
(v) lim
= ,m ≠0
x → a g ( x)
m
x→ a
x→ a
1
=0
f ( x)
(vii) lim f ( x) = lim f ( x)
x→ a
(viii) lim log{ f ( x)} = log {lim f ( x)}, provided lim f ( x) > 0
x→ a
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
[quotient rule]
(vi) If lim f ( x) = + ∞ or − ∞, then lim
x→ a
Your Personal Preparation Indicator
x→ a
x→ a
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
105
DAY
(ix) If f ( x) ≤ g( x), ∀x, then lim f ( x) ≤ lim g( x)
x→ a
g (x )
(x) lim[ f ( x)]
x→ a
= {lim f ( x)}
x→ a
lim g ( x )
x →a
x→ 0
x→ a
continuous at lim g( x) = m.
x→ a
(xii) Sandwich Theorem If f ( x) ≤ g( x) ≤ h( x)∀x ∈ (α , β) − {a}
and lim f ( x) = lim h( x) = l , then lim g( x) = l where a ∈(α , β)
x→ a
log a (1 + x)
= log a e; a > 0, ≠ 1
x
loge (1 + x)
loge (1 − x)
(ii) In particular, lim
= 1 and lim
= −1
x→ 0
x→ 0
x
x
(i) lim
x→ a
(xi) lim f {g( x)} = f {lim g( x)} = f (m) provided f is
x→ a
3. Logarithmic Limits
x→ a
x→ a
Important Results on Limit
Some important results on limits are given below
1. Algebraic Limits
x n − an
= nan −1 , n ∈ Q, a > 0
x→ a
x −a
1
(ii) lim n = 0, n ∈ N
x→ ∞ x
(iii) If m, n are positive integers and a0, b 0 are non-zero real
numbers, then
a x m + a x m −1 + ... + am −1 x + am
lim 0 n 1 n −1
x→ ∞ b x + b x
+ ... + b n −1 x + b n
0
1
a0
if m = n
b
00
if m<n
=
∞ if m > n, a0 b 0 > 0
−∞ if m > n, a0 b 0 < 0
(i) lim
(1 + x)n − 1
=n
x→ 0
x
(1 + x)m − 1 m
=
x → 0 (1 + x)n − 1
n
4. Expotential Limits
ax − 1
= loge a, a > 0
x
ex −1
e λx − 1
(ii) In particular, lim
= 1 and lim
=λ
x→ 0
x→ 0
x
x
0,
0 ≤ a <1
1,
a =1
(iii) lim ax =
x→ ∞
a
∞
,
>1
Does not exist, a < 0
(i) lim
x→ 0
5. 1∞ Form Limits
(i) If lim f ( x) = lim g( x) = 0, then
x→ a
x→ a
lim{1 + f ( x)}1 / g ( x ) = e
lim
x → a
f (x )
g (x )
x→ a
(ii) If lim f ( x) = 1 and lim g( x) = ∞, then
x→ a
x→ a
lim { f ( x ) − 1 } g ( x )
lim{ f ( x)}g ( x ) = e x → a
x→ a
In General Cases
x
(i) lim(1 + x)1 / x = e
x→ 0
(v) lim
(iv) lim
(iii) lim(1 + λx)1 / x = e λ
x→ 0
2. Trigonometric Limits
sin x
x
= 1 = lim
x → 0 sin x
x
tan x
x
lim
= 1 = lim
x→ 0
x → 0 tan x
x
sin −1 x
x
lim
= 1 = lim
x→ 0
x → 0 sin −1 x
x
tan −1 x
x
lim
= 1 = lim
x→ 0
x
→
0
x
tan −1 x
sin x °
π
sin x
cos x
(vi) lim
lim
=
= lim
=0
x→ 0
x→ ∞
x→ ∞
x
180
x
x
lim sin x or lim cos x oscillates between −1 to 1.
a
(v) lim (1 + ax) b / x = lim 1 +
x→ 0
x→ ∞
x
1
(ii) lim 1 + = e
x→ ∞
x
x
λ
(iv) lim 1 + = e λ
x→ ∞
x
bx
= e ab
(i) lim
x→ 0
(ii)
(iii)
(iv)
(v)
(vii)
x→ ∞
x→ ∞
sin P mx m
=
x → 0 sin P nx
n
P
(viii) lim
(ix) lim
x→ 0
tan P mx m
=
tan P nx n
P
1 − cos m x m2
cos ax − cos bx a2 – b 2
= 2 ; lim
=
x → 0 1 − cos n x
n x → 0 cos cx − cos dx c2 – d2
(x) lim
cos mx − cos nx n2 − m2
=
x→ 0
x2
2
(xi) lim
Methods To Evaluate Limits
To find lim f ( x), we substitute x = a in the function.
x→ a
If f (a) is finite, then lim f ( x) = f (a).
x→ a
0 ∞
If f (a) leads to one of the following form ; ; ∞ − ∞; 0 × ∞; 1 ∞ , 0
0 ∞
and ∞ 0 (called indeterminate forms), then lim f ( x) can be
x→ a
evaluated by using following methods
(i) Factorization Method This method is particularly used
when on substituting the value of x, the expression take
the form 0/0.
(ii) Rationalization Method This method is particularly used
when either the numerator or the denominator or both
involved square roots and on substituting the value of x,
0 ∞
the expression take the form , .
0 ∞
NOTE To evaluate x → ∞ type limits write the given expression in the
form N /D and then divide both N and D by highest power of x
occurring in both N and D to get a meaningful form.
TEN
106
L’Hospital’s Rule
Continuity of a Function in an Interval
If f ( x) and g( x) be two functions of x such that
A function f ( x) is said to be continuous in (a, b ) if it is
continuous at every point of the interval (a, b ). A function f ( x)
is said to be continuous in [a, b ], if f ( x) is continuous in (a, b ).
Also, in addition f ( x) is continuous at x = a from right and
continuous at x = b from left.
(i) lim f ( x) = lim g( x) = 0.
x→ a
x→ a
(ii) both are continuous at x = a.
(iii) both are differentiable at x = a.
(iv) f ′ ( x) and g′ ( x) are continuous at the point x = a, then
f ( x)
f ′ ( x)
provided that g(a) ≠ 0.
lim
= lim
x → a g ( x)
x → a g ′ ( x)
Results on Continuous Functions
(i) Sum, difference product and quotient of two
continuous functions are always a continuous
f ( x)
function. However, r ( x) =
is continuous at x = a
g ( x)
only if g(a) ≠ 0.
(ii) Every polynomial is continuous at each point of real
line.
(iii) Every rational function is continuous at each point
where its denominator is different from zero.
(iv) Logarithmic functions, exponential functions,
trigonometric functions, inverse circular functions and
modulus function are continuous in their domain.
(v) [ x] is discontinuous when x is an integer.
(vi) If g( x) is continuous at x = a and f is continuous at g(a),
then fog is continuous at x = a.
(vii) f ( x) is a continuous function defined on [a, b ] such that
f (a) and f (b ) are of opposite signs, then there is atleast
one value of x for which f ( x) vanishes, i.e. f (a) > 0,
f (b ) < 0 ⇒ ∃ c ∈ (a, b ) such that f (c) = 0.
Above rule is also applicable, if lim f ( x) = ∞ and lim g( x) = ∞.
x→ a
x→ a
If f ′ ( x), g′ ( x) satisfy all the conditions embeded in L’Hospital’s
rule, then we can repeat the application of this rule on
f ′ ( x)
f ′ ′ ( x)
f ′ ( x)
to get lim
.
= lim
x → a g ′ ( x)
x → a g ′ ′ ( x)
g ′ ( x)
Sometimes, following expansions are useful in evaluating
limits.
●
●
●
●
●
●
x2 x3 x 4 x 5
+
−
+
+ ... + (−1 < x ≤ 1)
2
3
4
5
x2 x3 x 4 x 5
log(1 − x) = − x −
−
−
−
− ...(−1 < x < 1)
2
3
4
5
2
3
4
x x
x
x
ex = 1 + +
+
+
+ ...
1 ! 2 ! 3 ! 4!
2
x
ax = 1 + x (loge a) +
(loge a)2 + ...
2!
x3 x 5 x7
sin x = x −
+
−
+ ...
3! 5! 7!
x2 x 4 x 6
cos x = 1 −
+
−
− ...
2 ! 4! 6 !
log(1 + x) = x −
Continuity
If the graph of a function has no break (or gap), then it is
continuous. A function which is not continuous is called a
1
discontinuous function. e.g. x2 and e x are continuous while
x
and [ x], where [ ⋅ ] denotes the greatest integer function, are
discontinuous.
Differentiability
The function f ( x) is differentiable at a point P iff there exists a
unique tangent at point P. In other words, f ( x) is differentiable
at a point P iff the curve does not have P as a corner point, i.e.
the function is not differentiable at those points where graph
of the function has holes or sharp edges. Let us consider the
function f ( x) = | x − 1|. It is not differentiable at x = 1. Since,
f ( x) has sharp edge at x = 1.
f (x) = – x + 1
Lf ' (x) = – 1
Continuity of a Function at a Point
X'
A function f ( x) is said to be continuous at a point x = a of its
domain if and only if it satisfies the following conditions
(i) f (a) exists, where (‘a’ lies in the domain of f
(ii) lim f ( x) exist, i.e. lim− f ( x) = lim+ f ( x)
x→ a
or
x→ a
x→ a
LHL = RHL
f (x) = x – 1
Rf ' (x) = 1
Y
O
1
2
3
X
Y'
(graph of f ( x) describe differentiability)
Differentiability of a Function at a Point
A function f is said to be differentiable at x = c, if left hand
(iii) lim f ( x) = f (a), f ( x) is said to be
x→ a
left continuous at x = a, if lim− f ( x) = f (a)
x→ a
right continuous at x = a, if lim+ f ( x) = f (a)
x→ a
and right hand derivatives at c exist and are equal.
●
Right hand derivative of f ( x) at x = a denoted by
f ′ (a + 0) or f ′ (a + ) is lim
h→ 0
f (a + h) − f (a)
⋅
h
107
DAY
●
Left hand derivative of f ( x) at x = a denoted by f ′ (a − 0)
or f ′ (a− ) is lim
h→ 0
●
●
f (a − h) − f (a)
.
−h
Thus, f is said to be differentiable at x = a, if
f ′ (a + 0) = f ′ (a − 0) = finite.
The common limit is called the derivative of f ( x) at x = a
f ( x) − f (a)
denoted by f ′ (a). i.e. f ′ (a) = lim
.
x→ a
x−a
Results on Differentiability
(i) Every polynomial, constant and exponential function is
differentiable at each x ∈ R.
(ii) The logarithmic, trigonometric and inverse trigonometric
function are differentiable in their domain.
(iii) The sum, difference, product and quotient of two
differentiable functions is differentiable.
(iv) Every differentiable function is continuous but
converse may or may not be true.
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 lim | x | [cos x ] , where [. ] is the greatest integer function, is
x→ 0
(a) 1
(c) Does not exist
9 lim
x→ 0
(b) 0
(d) None of these
(a) 4
2 Let f : R → [ 0, ∞ ) be such that lim f ( x ) exists and
x→ 5
(b) 0
(c) 1
(a) 0
(d) 2
2 x m
= (where [⋅] denotes greatest integer
x 5 n
function), then m + n (where m, n are relatively prime) is
(c) 5
(c) (1, − 1)
(b)
3
5
4
7
(d) −
(a) 0
lim
x→ 3
(a) − (2)3 / 2
(c) log2
x
2
(a) 1
x
4
(d) e
x
8
x
is
2n
x
(c)
sin x
2
3
(c)
(c) 0
(d) 1
3
2
(d) None of these
lim
x → π /2
(d) 3
j
JEE Mains 2013
j
JEE Mains 2017
1
(b) −
2
(d) 0
(c) 2
15
(c) (2)− 3 / 2
1 −1 x + 1 π
is
tan
−
2x + 1 4
x
(a) 1
(d) None of these
f ( x + T ) + 2f ( x + 2T )+ ...+nf ( x + nT )
lim n
is equal to
n→ ∞ f ( x + T ) + 4f ( x + 4T )+ ...+ n 2f ( x + n 2T )
(b)
(d) π
tan x
f (x )
, then lim
is equal to
x
x→ 0 x 2
1
(b) (2)1 / 2
x→ 0
8 If f is periodic with period T and f ( x ) > 0∀x ∈ R , then
(a) 2
JEE Mains 2014
1 − cos( x 2 − 10x + 21)
( x − 3)
14 The value of lim
4
7 The value of lim cos cos cos … cos
n→ ∞
sinx
(b)
x
(c) − π
(b) −1
20
7
1
(b)
2
cos x
x2
1
j
1
n
13 The limit of the following is
(c) −
1
2
(d) n +
(c) n
(b) 1
(a) 3
6 lim x + x + x − x is equal to
x→ ∞
n+1
n
sin( π cos 2 x )
is equal to
x2
π
(a)
2
(d) (0, 1)
3 ⋅ 2 n +1 − 4 ⋅ 5 n +1
5 lim
is equal to
n→ ∞
5 ⋅ 2n + 7 ⋅ 5n
(a) 0
x→ 0
(b)
sin x
12 If f ( x ) = x 3
2x
x2 +1
lim
− α x − β = 0 are respectively
x→ ∞ x + 1
(b) (−1, 1)
11 lim
(d) 6
4 The value of the constant α and β such that
(a) (1, 1)
(d)
x→ 0
x→ ∞
(b) 7
(c) 2
10 If lim
3 If lim
(a) 2
(b) 3
[(a − n ) nx − tan x ] sin nx
= 0, where n is non-zero
x2
real number, then a is equal to
[f ( x )]2 − 9
lim
= 0. Then, lim f ( x ) is equal to
x→ 5
x→ 5
| x − 5|
(a) 3
(1 − cos 2x )( 3 + cos x )
is equal to
x tan 4x
j JEE Mains 2015, 13
cot x − cos x
equals
( π − 2 x )3
1
(a)
24
1
(c)
8
1
(b)
16
1
(d)
4
TEN
108
16 If m and n are positive integers, then
(cos x )1/ m − (cos x )1/ n
equals to
lim
x→ 0
x2
(a) m − n
17 lim
x→ ∞
(b)
1 1
−
n m
(c)
m −n
2mn
(d) None of these
cot −1( x + 1 − x )
is equal to
x
−1 2 x + 1
sec
x −1
(a) 1
27 Let f ( x ) = 1 + | x − 2 | and g ( x ) = 1 − | x | , then the set of all
points, where fog is discontinuous, is
(a) {0, 2 }
π
(c)
2
(b) 0
(d) Does not exists
ln(cos 2x )
sin2 2x
and
, q = lim
2
x→ 0
x → 0 x (1 − e x )
3x
x −x
. Then p, q , r satisfy
r = lim
x →1
ln x
18 Let p = lim
(a) p < q < r
(b) q < r < p
(c) p < r < q
(d) q < p < r
19 Let p = lim (1 + tan2 x )1/ 2 x , then log p is equal to
x → 0+
(a) 2
j
1
(c)
2
(b) 1
3x − 4
20 The value of lim
x → ∞ 3x + 2
(a) e −1 / 3
21 If lim 1 +
x→ ∞
x +1
(b) e −2 / 3
a
b
+
x x2
JEE Mains 2016
1
(d)
4
3
is
(c) e −1
(d) e −2
= e 2 , then the values of a and b are
(b) a = 1, b ∈ R
(d) a = 1, b = 2
22 If f ′ ( 2) = 6 and f ′ (1) = 4 , then lim
h→ 0
(b) –3/2
(c) 3/2
(d) Does not exist
23 Let f (a ) = g (a ) = k and their nth derivatives f n (a ), g n (a )
exist and are not equal for some n. Further, if
f (a )g ( x ) − f (a ) − g (a )f ( x ) + g (a )
lim
= 4,
x→ a
g( x ) − f ( x )
then the value of k is
(a) 4
(b) 2
(c) 1
(d) 0
24 If f : R → R be such that f (1) = 3 and f ′ (1) = 6. Then,
f (1 + x )
lim
x→ 0
f (1)
(a) 1
1/ x
(b) e1 / 2
(c) e 2
(d) e 3
sin x , x is rational
25 Let f ( x ) =
, then set of points,
2
− sin x , x is irrational
2
where f ( x ) is continuous, is
(b) a null set
(d) set of all rational numbers
26 Let f :[a, b ] → R be any function which is such that f ( x ) is
rational for irrational x and f ( x ) is irrational for rational x.
Then, in [a, b ]
(d) an empty set
3π 7π
2π 5π
5π 7π
(b) , (c) , (d) ,
4 4
3 3
4 3
29 If f ( x ) is differentiable at x = 1 and lim
h→ 0
1
f (1 + h ) = 5, then
h
f ′ (1) is equal to
(a) 6
(b) 5
30 If f ( x ) = 3x
(c) 4
(d) 3
− 7x + 5x − 21x + 3x − 7, then the
f (1 − h ) − f (1)
value of lim
is
h→ 0
h3 + 3h
(a)
10
53
3
8
(b)
6
22
3
3
2
(c) 13
(d)
22
13
31 Let f ( 2) = 4 and f ′ ( 2) = 4. Then,
x f ( 2) − 2f ( x )
is given by
x −2
(a) 2
(b) –2
(c) – 4
(d) 3
32 If f is a real-valued differentiable function satisfying
| f ( x ) − f ( y )| ≤ ( x − y )2 ; x , y ∈ R and
f ( 0) = 0, then f (1) is equal to
(a) 1
(b) 2
(c) 0
(d) –1
log(a + x ) − log a
log x − 1
33 If lim
+ k lim
= 1, then
x→ 0
x
→
0
x
x −e
1
(a) k = e 1 −
a
(c) k = e (2 − a)
(b) k = e (1 + a)
(d) Equality is not possible
34 The left hand derivative of f ( x ) = [ x ] sin ( πx ) at x = k , k is
an integer, is
(a) (− 1)k (k − 1) π
(c) (− 1)k kπ
is equal to
π
(a) (2n + 1) ,n ∈ I
2
(c) {nπ,n ∈ I }
π 3π
(a) ,
2 4
lim
equal to
(c) {0}
JEE Mains 2013
points of discontinuity of f {g ( x )} in ( 0 , 2π ) is
x→ 2
f ( 2h + 2 + h 2 ) − f ( 2)
is
f (h − h 2 + 1) − f (1)
(b) {0, 1, 2 }
j
− 1 , x < 0
28 If f ( x ) = 0 , x = 0 and g ( x ) = sin x + cos x , then the
1 , x > 0
2x
(a) a ∈ R, b ∈ R
(c) a ∈ R, b = 2
(a) 3
(a) f is discontinuous everywhere
(b) f is continuous only at x = 0
(c) f is continuous for all irrational x and discontinuous for all
rational x
(d) f is continuous for all rational x and discontinuous for all
irrational x
(b) (− 1)k − 1 (k − 1) π
(d) (− 1)k − 1 kπ
ex,
x ≤0
, then
|1 − x | , x > 0
35 If f ( x ) =
(a)
(b)
(c)
(d)
f (x) is differentiable at x = 0
f (x) is continuous at x = 0, 1
f (x) is differentiable at x = 1
None of the above
36 If f ( x ) = xe
1
1
−
+
x
| x |
0
, x ≠ 0 , then f ( x ) is
,x = 0
109
DAY
(a)
(b)
(c)
(d)
continuous as well as differentiable for all x
continuous for all x but not differentiable at x = 0
neither differentiable nor continuous at x = 0
discontinuous everywhere
37 The set of points, where f ( x ) =
(a) (− ∞, − 1) ∪ (−1, ∞)
(c) (0, ∞)
x
is differentiable, is
1+ |x|
(b) (− ∞, ∞)
(d) (− ∞, 0) ∪ (0, ∞)
k x + 1 , 0 ≤ x ≤ 3
is differentiable,
mx + 2 , 3 < x ≤ 5
39 If the function g ( x ) =
then the value of k + m is
(c)
10
5
(b) 3
(c) 5
(d) 4
(a) f is everywhere differentiable
(b) f is everywhere continuous but not differentiable at
x = nπ,n ∈Z
(c) f is everywhere continuous but not differentiable at
π
x = (2n + 1) ,n ∈ Z
2
(d) None of the above
(a) g (x) is discontinuous at x = π
(b) g (x) is continuous for x ∈ [0, ∞]
(c) g (x) is differentiable at x = π
(d) g (x) is differentiable for x ∈ [0, ∞]
16
5
the greatest integer function. Then, the total number of
points, where f ( x ) is non-differentiable, is
42 If f ( x ) = | sin x | , then
min{f (t ): 0 ≤ t ≤ x } , x ∈ [ 0, π ]
then
g( x ) =
(sin x ) − 1 ,
x >π
(b)
41 If f ( x ) = [sin x ] + [cos x ], x ∈ [ 0, 2π ], where [. ] denotes
(a) 2
38 Let f ( x ) = cos x and
(a) 2
(a) f (x) is continuous and differentiable
(b) f (x) is continuous but not differentiable
(c) f is not continuous but differentiable
(d) f is neither continuous nor differentiable
(d) 4
sin(cos −1 x ) + cos(sin−1 x ), x ≤ 0
−1
−1
sin(cos x ) − cos(sin x ), x > 0
40 If f ( x ) =
then at x = 0
43 Statement I f ( x ) = | log x | is differentiable at x = 1 .
Statement II Both log x and – log x are differentiable at
x = 1.
(a) Statement I is false, Statement II is true
(b) Statemnt I is true, Statement II is true; Statement II is a
correct explanation of Statement I
(c) Statement I is true, Statement II is true; Statement II is
not a correct explanation of Statement I
(d) Statement I is true, Statement II is false
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 For each t ∈ R , let [t ] be the greatest integer less than or
1 2
15
equal to t. Then, lim+ x
+
+…+
x
x→ 0
x x
JEE Mains 2018
(b) is equal to 15
(d) does not exist (in R)
j
(a) is equal to 0
(c) is equal to 120
x
x 2 + 5x + 3
is equal to
x2 + x + 2
2 lim
x→ ∞
(a) e 4
(b) e 2
(c) e 3
(d) e
1 − cos (ax 2 + bx + c )
is equal to
x→ α
( x − α )2
lim
1
a2
(α − β)2 (b) −
(α − β)2 (c) 0
2
2
(d)
4 lim sin[ π n + 1] is equal to
n→ ∞
(b) 0
(d) None of these
5 If x > 0 and g is a bounded function, then
lim
n→ ∞
f ( x ) ⋅ e nx + g ( x )
is equal to
e nx + 1
a x + bx + c x
3
2/x
6 The value of lim
x→ 0
(a) (abc)3
(c) (abc)1 / 3
( where a , b , c > 0) is
(b) abc
(d) None of these
3x − x 3
1 − x 2
and g ( x ) = cos −1
, then
2
1 − 3x
1 + x 2
f ( x ) − f (a )
1
lim
, where 0 < a < , is equal to
x → a g ( x ) − g (a )
2
(a)
2
(a) ∞
(c) Does not exist
(b) f (x)
(d) None of these
7 If f ( x ) = cot −1
3 If α and β are the distinct roots of ax 2 + bx + c = 0, then
(a)
(a) 0
(c) g (x)
a2
(α − β)2
2
3
3
3
(b)
(c)
2 (1 + a 2 )
2 (1 + x 2 )
2
(d) −
3
2
8 If f : R → R is a function defined by
2x − 1
f ( x ) = [ x ] cos
π, where [ x ] denotes the greatest
2
integer function, then f is
(a)
(b)
(c)
(d)
continuous for every real x
discontinuous only at x = 0
discontinuous only at non-zero integral values of x
continuous only at x = 0
TEN
110
18 The function f ( x ) is discontinuous only at x = 0 such that
9 Let f be a composite function of x defined by
f 2 ( x ) = 1 ∀x ∈ R. The total number of such function is
1
1
f (u ) = 2
, u( x ) =
⋅
u +u−2
x −1
Then, the number of points x, where f is discontinuous, is
JEE Mains 2013
j
(a) 4
(b) 3
(c) 2
(d) 1
10 If f : R → R be a positive increasing function with
lim
x→ ∞
f ( 3x )
f ( 2x )
is equal to
= 1 . Then, lim
x → ∞ f (x )
f (x )
(a) 1
(b) 2/3
(c) 3/2
(d) 3
π 3π
11 Let f ( x ) = max {tan x , sin x , cos x }, where x ∈ − ,
.
2 2
Then, the number of points of non-differentiability is
(a) 1
(b) 3
(c) 0
(d) 2
12 Let S = {t ∈ R : f ( x ) = | x − π|⋅(e| x | − 1)} sin| x | is not
differentiable at t}. Then, the set S is equal to
(a) φ (an empty set)
(c) { π }
π
x +
3
n
(b) {0}
(d) {0, π }
j
JEE Mains 2018
n→ ∞
π
x n −1 +
3
n −1
, where n is an even integer,
Then which of the following is incorrect?
π
π
(a) If f: , ∞ → , ∞ , then f is both one-one and onto
3
3
(b) f (x) = f (− x) has infinitely many solutions
(c) f (x) is one-one for all x ∈ R (d) None of these
n −1
x
. Then,
14 Let f ( x ) = lim ∑
n→ ∞ k = 0 (kx + 1){(k + 1)x + 1}
(a) f is continuous but not differentiable at x = 0
(b) f is differentiable at x = 0
(c) f is neither continuous nor differentiable at x = 0
(d) None of the above
15 If x1, x 2 , x 3 ,..., x 4 are the roots of x n + ax + b = 0, then the
value of ( x1 − x 2 )( x1 − x 3 )...( x1 − x n ) is equal to
(b) nx1n − 1 + a
(d) nx1n
(a) nx1 + b
(c) nx1n − 1
16 If *lim x ln
x→ ∞
1
x
0
1
−b
1
x
0
c
−1 = −4, where a, b, c are real
a / x
numbers, then
(a) a = 1,b ∈ R , c = −1
(c) a = 1,b = 1,c ∈ R
17 If lim 1 + x +
x→ 0
(a) e
1/ x
f (x )
x
(b) e 2
(b) a ∈ R,b = 2,c = 4
(d) a ∈ R,b = 1,c = 4
1/ x
f (x )
= e 3 , then lim 1 +
x→ 0
x
(c) e 3
(b) 3
(d) None of these
19 Let f ( x ) = x | x | and g ( x ) = sin x
Statement I gof
is differentiable at x = 0 and its
derivative is continuous at that point.
Statement II gof is twice differentiable at x = 0.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true; Statement II is a
correct explanation of Statement I
(c) Statement I is true, Statement II is true; Statement II is
not a correct explanation of Statement I
(d) Statement I is true, Statement II is false
20 Statement I The function
f ( x ) = ( 3x
1
x = and
2
− 1) | 4x 2 − 12x + 5 | cos πx is differentiable at
5
.
2
π
= 0, ∀ n ∈ I .
2
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I.
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I.
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
Statement II cos(2n + 1)
n
13 If f ( x ) = lim
(a) 2
(c) 6
is equal to
(d) None of these
21 Define f ( x ) as the product of two real functions f1( x ) = x ,
x ∈ IR
sin 1 , if x ≠ 0
and f2 ( x ) =
as follows
x
0
, if x = 0
f ( x ) ⋅ f2 ( x ), if x ≠ 0
f (x ) = 1
0
, if x = 0
Statement I f ( x ) is continuous on IR.
Statement II f1( x ) and f2 ( x ) are continuous on IR.
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true; Statement II is
correct explanation of Statement I.
(c) Statement I is true, Statement II is true; Statement II is
not a correct explanation of Statement I
(d) Statement I is true, Statement II is false
22 Consider the function f ( x ) = | x − 2 | + | x − 5 | , x ∈ R .
Statement I f ′ ( 4) = 0
Statement II f is continuous in [ 2, 5 ] and differentiable in
( 2, 5) and f ( 2) = f ( 5 ).
(a) Statement I is true, Statement II is true; Statement II
is a correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II
is not correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
111
DAY
ANSWERS
SESSION 1
SESSION 2
1 (a)
2 (a)
3 (b)
4 (c)
5 (d)
6 (b)
7 (b)
8 (c)
9 (c)
10 (d)
11 (d)
12 (d)
13 (a)
14 (b)
15 (b)
16 (c)
17 (a)
18 (d)
19 (c)
20 (b)
21 (b)
22 (a)
23 (a)
24 (c)
25 (c)
26 (a)
27 (d)
28 (b)
29 (b)
30 (a)
31 (c)
32 (c)
33 (a)
34 (a)
35 (b)
36 (b)
37 (b)
38 (b)
39 (a)
40 (d)
41 (c)
42 (b)
43 (a)
1 (c)
11 (b)
21 (d)
2 (a)
12 (d)
22 (b)
3 (d)
13 (c)
4 (b)
14 (c)
5 (b)
15 (b)
6 (d)
16 (d)
7 (d)
17 (b)
8 (a)
18 (c)
9 (b)
19 (b)
10 (a)
20 (a)
Hints and Explanations
SESSION 1
n
1 RHL = lim ( x )0 = 1
= lim
x → 0+
n→ ∞
2
6 ⋅ − 20
5
n
0 − 20
20
=−
0+ 7
7
=
LHL = lim− (− x )0 = lim− 1 = 1
2
5⋅ + 7
5
RHL = LHL
lim | x |[cos x ] = 1
6 xlim
x+ x+ x − x
→∞
x→ 0
Q
∴
x→ 0
x→ 0
2 Given, lim
f ( x ) exists and
x→ 5
lim
[ f ( x )]2 − 9
| x − 5|
x→ 5
= lim
x→ ∞
x+
x − x
x+
x+
x +
x+
x
x→ ∞
x+
x +
x+
2
⇒
∴
8 Clearly, f ( x + T ) = f ( x + 2T ) = ...
x+
= lim
x→ 5
lim [ f ( x )] = 9
x→ 5
n→ ∞
=0
⇒ lim[ f ( x )]2 − 9 = 0
sin x
x / 2n
⋅
x sin( x / 2n )
sin x
=
x
= lim
= lim
x→ ∞
lim f ( x ) = 3, − 3
x
x
1 + x −1 / 2
1+
x
−1
+ x
−3 / 2
+1
=
1
2
= f ( x + nT ) = f ( x )
f ( x + T ) + 2 f ( x + 2T ) + ...
+ nf ( x + nT )
∴lim n
f ( x + T ) + 4 f ( x + 4T )
n→ ∞
2
2
+ ...+ n f ( x + x T )
nf ( x )(1 + 2 + 3+ ...+ n )
= lim
n→ ∞ f ( x )(1 + 22 + 32 + ...+ n2 )
n(n + 1)
n
3
2
= lim
=
n→ ∞ n( n + 1)(2n + 1)
2
6
x→ 5
But f : R → [0, ∞ )
∴ Range of f ( x ) ≥ 0
⇒
lim f ( x ) = 3
7 We know that,
cos A ⋅ cos 2 A ⋅ cos 4 A K cos 2n −1
A=
x→ 5
2 x
2x
x
3 Clearly, lim
= lim −
x→ ∞ x 5
x→ ∞ x 5 5
2
2
= −0=
5
5
∴ m+ n =7
x2 + 1
4 xlim
− α x − β = 0
→∞
x
+
1
⇒ lim
x→ ∞
x (1 − α ) − x (α + β ) + 1 − β
=0
x+1
∴
⇒
2
1 − α = 0, α + β = 0
α = 1, β = − 1
x
Take A = n ,
2
x
x
then cos n ⋅ cos n − 1 ...
2
2
n→ ∞
3 ⋅ 2n +1 − 4 ⋅ 5n +1
6 ⋅ 2n − 20 ⋅ 5n
= lim
n
n
n
→
∞
5⋅ 2 + 7 ⋅ 5
5⋅ 2n + 7 ⋅ 5n
9 We have, lim
x→ 0
2sin 2 x(3 + cos x )
tan 4 x
x×
× 4x
4x
2
(3 + cos x )
2sin x
= lim
× lim
x→ 0
x→ 0
x2
4
x→ 0
sin x
x
2 sin n
2
x
x
∴ limcos cos ...
n→ ∞
2
4
=
×
n
x
cos n − 1
2
= lim
n→ ∞
sin x
x
2n sin n
2
(1 − cos 2 x )(3 + cos x )
x tan 4 x
= lim
x
x
cos cos
4
2
5 Clearly,
lim
sin 2n A
2n sin A
1
tan 4 x
lim
x→ 0
4x
4
×1 = 2
4
Q lim sin θ = 1 and lim tan θ = 1
x→ 0
x → 2 θ
θ
= 2×
⋅ cos x
n
2
10 Since,
tan x sin nx
lim (a − n ) n −
⋅
=0
x→ 0
x
x
TEN
112
⇒ [(a − n ) n − 1] n = 0
⇒
(a − n ) n = 1
a= n+
sin θ
= π Qlim
= 1
θ → 0 θ
12 Q f ( x ) = x( x − 1)sin x − ( x3 − 2 x2 )
cos x − x3 tan x
= x sin x − x cos x − x tan x +
2 x2 cos x − x sin x
f ( x)
∴ lim 2 = lim
x→ 0 x
x→ 0
sin x − x cos x − x tan x
sin x
+ 2cos x −
x
2
3
1 − cos( x − 10 x + 21)
2
=
=
=
=
( x − 3)
( x − 3) ( x − 7)
2 sin
2
lim
x→3
( x − 3)
( x − 3) ( x − 7)
2 sin
( x − 7)
2
lim
⋅
x→3
( x − 3) ⋅ ( x − 7)
2
2
lim ( x − 7)
x→3
( x − 3) ( x − 7)
sin
1
2
⋅ lim
×
x→3
( x − 3) ( x − 7)
2
2
− (2)3 /2
x+1
π
−
x
4
2x + 1
x
+
1
−1
= lim tan −1
− tan (1)
x→ 0
2x + 1
x + 1 − 1
1
2x + 1
= lim ⋅ tan −1
x→ 0 x
x+1
1 +
2x + 1
x + 1 − 2x − 1
1
= lim ⋅ tan −1
x→ 0 x
2x + 1 + x + 1
1
−1
14 lim
tan
x→ 0
−x
1
= lim ⋅ tan −1
x→ 0 x
4x +
2
form 0
0
[using L’ Hospital rule]
1
= lim
x→ 0
1+
=
=
=
x2
(4 x + 2)2
=
=
x→ 0
cot x − cos x
( π − 2 x )3
1 cos x(1 − sin x )
lim ⋅
3
x → π /2 8
π
sin x − x
2
π
π
cos − h 1 − sin − h
2
2
1
lim ⋅
3
h→ 0 8
π
π
π
sin − h −
+ h
2
2
2
sin h (1 − cos h )
1
lim
8 h→ 0
cos h ⋅ h3
h
sin h 2sin2
1
2
lim
8 h → 0 cos h ⋅ h3
h
sin h ⋅ sin2
2
1
lim
4 h→ 0
h3 cos h
2
sin h
sin h
1
2 ⋅ 1 ⋅ 1
lim
4 h → 0 h h cos h 4
2
1 1
1
= ×
=
4 4 16
=
1/m
(cos x )
16 lim
x→ 0
− (cos x )1 / n
x2
1
(cos x )m
= lim
x→ 0
x2
−
−1
⋅ lim
x→ 0
1
1 − 2sin2 x m
2
= lim
2
x→ 0
x
1 1
= lim − 2 −
x→ 0
m n
−
1
(cos x )1 / n
1
n
x→1
= lim
x→1
= lim
x→1
x (1 − x )
1 + x − 1
ln
⋅ ( x − 1)
x −1
x (1 − x )
1 + ( x − 1)
ln
⋅ ( x − 1)(1 +
x −1
x
2 = m−n
x2
2mn
sin2
= lim( x + 1 −
x→ ∞
x→ ∞
1
x+1+
1
2
Hence, q < p < r .
1
19 Given, p = lim (1 + tan2
x→ 0
lim
= ex→0
tan
+
x
2x
1
lim
= e2 x→0
tan x
x
+
1
= e2
1
1
2
∴ log p = log e 2 =
3x − 4
3x + 2
20 xlim
→∞
x +1
3
6
= lim 1 −
x→ ∞
3x + 2
x +1
3
x +1
3
3x + 2 3x + 2
6 −6
= lim 1 −
x→ ∞
3x + 2
3x + 2
x→ ∞
=0
x
2+ 1
2x + 1
x =∞
lim
= lim
x→ ∞ x − 1
x→ ∞ 1 − 1 / x
cot ( x + 1 −
2x + 1
sec
x −1
x)
x
=
−1
=
x +1
3
= e −2 / 3
1 + a + b
21 Now, xlim
2
→∞
=e
−1
cot (0)
sec −1 (∞ )
π /2
=1
π /2
3x
ln(1 + cos 2 x − 1) cos 2 x − 1
= lim
⋅
x→ 0
(cos 2 x − 1)
3 x2
2x
x
a
b
= lim 1 +
+ 2
x→ ∞
x
x
x
x→ ∞
⋅
−2 ( x + 1 )
= lim e
x
∴ lim
2
−2( x + 1) − 2
=
Q xlim
→∞
3x + 2
3
and
−1
x ) 2x
+
(1∞ form)
2
x)
x
x)
=−
17 Clearly,
= lim
x−x
ln(1 + x − 1)
−6
−1
ln(1 + cos 2 x − 1)
18 Clearly, p = lim
2
x→ 0
4x + 2 − 4x
× −
2
(4 x + 2)
and r = lim
sin2 2 x
4 x2
⋅
= −4
2
4x
x (1 − e x )
3x + 2 − 6
= lim
x→ ∞
3x + 2
1
n
2
3
q = lim
15 xlim
→ π /2
3
=2−1=1
13 lim
x→3
=−
x→ 0
1
n
sin( π cos 2 x )
sin( π − π sin2 x )
11 lim
=
lim
x→ 0
x→ 0
x2
x2
2
sin( π sin x )
= lim
x→ 0
x2
[Q sin( π − θ) = sin θ ]
sin( π sin 2 x )
sin2 x
= lim
× ( π )
2
x→ 0
π sin x
x2
∴
(+2)
x2 + (4 x + 2)2
(+2)
−2 −1
=−
=
=
0 + (0 + 2)2
4
2
= lim −
2x
b
x2
b
+
x x2
a
x
a
+
a b
lim 2 x +
x x2
x →∞
Q lim (1 + x )1 / x = e = e 2 a
x→ ∞
a
b
But lim 1 +
+ 2
x→ ∞
x
x
⇒
⇒
and
e 2a = e 2
a=1
b ∈R
2x
= e2
113
DAY
22 lim
h→ 0
23 lim
x→ a
f (2h + 2 + h2 ) − f (2)
f (h − h2 + 1) − f (1)
f ′ (2h + 2 + h2 )(2 + 2h )
= lim
h → 0 f ′ ( h − h2 + 1)(1 − 2h )
f ′ (2) × 2
=
f ′ (1) × 1
6×2
=
=3
4×1
f (a) g ( x ) − f ( a) − g ( a) f ( x ) + g ( a)
=4
g( x) − f ( x)
Applying L’Hospital rule, we get
f (a) g ′ ( x ) − g (a)f ′ ( x )
lim
=4
x→ a
g ′( x ) − f ′( x )
k g ′( x ) − k f ′( x )
⇒ lim
=4
x→ a
g ′( x ) − f ′( x )
∴
k =4
f (1 + x )
f (1)
1/x
24 Let y =
⇒ log y =
1
[log f (1 + x ) − log f (1)]
x
0 < x < 3 π /4
x = 3 π / 4 , 7 π /4
− 1 , 3 π / 4 < x < 7 π /4
1,
28 f { g ( x )} = 0,
or 7 π /4 < x < 2
Clearly, [ f {g ( x )}] is not continuous at
3 π 7π
x=
,
.
4 4
f (1 + h ) − f (1)
29 f ′(1) = lim
h→ 0
h
f (1 + h )
f (1)
= lim
− lim
h→ 0
h→ 0
h
h
f (1 + h )
f (1)
Since, lim
must
= 5, so lim
h→ 0
h→ 0
h
h
f (1)
be finite as f ′(1) exists and lim
can
h→ 0
h
be finite only, if f (1) = 0 and
f (1)
lim
= 0.
h→ 0
h
f (1 + h )
∴
f ′ (1) = lim
=5
h→ 0
h
f (1 − h ) − f (1)
30 lim
h→ 0
h3 + 3h
= lim
1
⇒ lim log y = lim
f ′ (1 + x )
x→ 0
x→ 0
f (1 + x )
⇒
∴
lim log y =
x→ 0
h→ 0
−1
53
= f ′ (1) ⋅ =
3
3
f ′ (1) 6
=
f (1) 3
x f (2) − 2 f ( x )
x−2
x f (2) − 2 f (2) + 2 f (2) − 2 f ( x )
= lim
x→2
x−2
f (2)( x − 2) − 2 { f ( x ) − f (2)}
= lim
x→2
x−2
f ( x ) − f (2)
= f (2) − 2 lim
x→2
x−2
31 lim
x→2
lim y = e 2
x→ 0
25 Clearly, f ( x ) is continuous only when
sin2 x = − sin2 x ⇒ 2sin2 x = 0
⇒
x = nπ
26 We have,
rational, if x ∉ Q in[a,b]
f ( x ) =
irrational
, if x ∈ Q in[a,b]
Let C ∈ [a,b] and c ∈ Q . Then,
f (c ) = irrational
and lim f ( x ) = lim f (c + h ) = rational or
x→c
h→ 0
irrational
Thus, f is discontinuous everywhere.
1 + x, x < 0
27 g ( x ) =
1 − x,
1 + | x − 1|,
∴ f {g ( x )} =
1
+ |− x − 1|,
x< 0
x≥ 0
x, x < 0
1, x ≥ 0
x< 0
x≥ 0
It is a polynomial function, so it is
continuous in everywhere except at
x = 0.
Now, LHL = lim 2 − x = 2,
x→ 0
RHL = lim 2 + x = 2
x→ 0
f ( x ) − f (a)
Q f ′ ( x ) = lim
x→ a
x − a
= f (2) − 2 f ′ (2)
= 4−2× 4= −4
32 Q
∴
x≥ 0
1+ 1−
=
1 + x +
2 − x,
=
2 + x,
f (1 − h ) − f (1)
−1
⋅ 2
−h
h +3
Also, f (0) = 2 + 0 = 2
Hence, it is continuous everywhere.
⇒
| f ( x ) − f ( y )| ≤ ( x − y )
| f ( x ) − f ( y )|
lim
≤ lim | x − y |
x→ y
x→ y
|x − y|
2
| f ′ ( y )| ≤ 0 ⇒
f ′( y ) = 0
34 If x is just less than k, then [ x] = k − 1
∴
f ( x ) = (k − 1) sin π x
LHD of f ( x ) = lim
x→ k
(k − 1) sin πx − k sin πk
x−k
(k − 1) sin πx
,
= lim
x→ k
x−k
where x = k − h
(k − 1)sin π (k − h )
= lim
h→ 0
−h
= (k − 1)(−1)k π
x
x≤ 0
e ,
35 f ( x ) = 1 − x, 0 < x ≤ 1
x − 1, x > 1
f (0 + h ) − f (0)
h
1− h−1
= lim
= −1
h→ 0
h
f (0 − h ) − f (0)
Lf ′ (0) = lim
h→ 0
−h
e−h − 1
= lim
=1
h→ 0
−h
So, it is not differentiable at x = 0.
Rf ′ (0) = lim
h→ 0
Similarly, it is not differentiable at x = 1
but it is continuous at x = 0 and 1.
h
36 RHL = lim
(0 + h ) e −2 / h = lim 2 / h = 0
h→ 0
h→ 0 e
LHL = lim (0 − h ) e
1 1
− −
h h
=0
h→ 0
Hence, f ( x ) is continuous at x = 0.
Now, Rf ′ ( x ) = lim
(0 + h ) e
h→ 0
= lim e − 2 / h
h→ 0
1 1
− +
h h
1 1
− −
h
(0 − h ) e h
and Lf ′ ( x ) = lim
h→ 0
−h
h→ 0
∴
Lf ′ ( x ) ≠ Rf ′ ( x )
Hence, f ( x ) is not differentiable at x = 0.
g( x)
37 Since, f ( x ) = x =
[say]
1 + | x| h( x )
⇒
f ( y ) = Constant
⇒
f( y) = 0
(− ∞, 0) ∪ (0, ∞ ).
⇒
f (1) = 0
33 Let f ( x ) = log x
1
x
Therefore, given function
= f ′ (a) + k f ′ (e ) = 1
1
k
⇒
+
=1
a e
a − 1
⇒
k = e
a
⇒
f ′( x ) =
−0
= lim e − 0 = 1
It is clear that g ( x ) and h( x ) are
differentiable on (− ∞, ∞ ) and
[Q f (0) = 0, given]
−0
h
=∞
Now,
x
−0
f ( x ) − f (0)
1 + | x|
lim
= lim
=1
x→ 0
x→ 0
x−0
x
Hence, f ( x ) is differentiable on (− ∞, ∞ ).
x ∈ [0, π]
x> π
sin x − 1,
Also, g ( π − ) = g ( π ) = g ( π + ) = −1
and g ′( π − ) ≠ g ′( π + )
∴ g is continuous at x = π
but not differentiable at x = π
cos x ,
38 Clearly, g ( x ) =
TEN
114
39 Since, g ( x ) is differentiable ⇒ g ( x ) must
be continuous.
k x + 1, 0 ≤ x ≤ 3
g ( x) =
mx + 2, 3 < x ≤ 5
At, x = 3, RHL = 3 m + 2
and at x = 3, LHL = 2k
∴
2k = 3 m + 2
Also,
∴
k
, 0≤ x < 3
g ′( x ) = 2 x + 1
m , 3< x ≤ 5
L(g ′ (3)) =
k
4
and R {g ′ (3)} = m ⇒
k
=m
4
k = 4m
i.e.
On solving Eqs (i) and (ii), we get
8
2
k = ,m =
5
5
⇒
− log x, x < 1
43 f ( x ) =
log x, x ≥ 1
SESSION 2
1 We have,
1
15
1
lim x + + …+
x
x x
x→ 0 +
We know, [ x] = x − { x}
1 = 1 − 1
∴
x x x
n
n
n
Similarly, = −
x x x
+
π
, π, 2 π and [cos x] is
2
non-differentiable at
π 3π
x = 0, ,
and 2 π.
2 2
Thus, f ( x ) is definitely
3π
non-differentiable at x = π,
, 0.
2
Also,
π
π
f = 1, f − 0 = 0,
2
2
f (2 π ) = 1,
f (2 π − 0) = − 1
Thus, f ( x ) is also non-differentiable at
π
and 2π.
x=
2
x=
42 Let u( x ) = sin x
Q u( x ) = sin x is a continuous function
and v ( x ) = | x| is a continuous function.
∴ f ( x ) = vou ( x ) is also continuous
everywhere but v ( x ) is not differentiable
at x = 0
⇒ f ( x ) is not differentiable
where sin x = 0
⇒ x = nπ , n ∈ Z
Hence, f ( x ) is continuous everywhere
but not differentiable at
x = nπ, n ∈ Z .
a2
(α − β )2
2
1
4 nlim
sin nπ 1 + 2
→∞
function.
n→ ∞
1
4x + 1
= lim 1 + 2
x→ ∞
x + x + 2
lim
x →∞
=e
2
+
x x2
ax + b x + c x
y = lim
x→ 0
3
2/ x
ax + b x + c x
2
log
x→ 0 x
3
⇒ log y = lim
log(ax + b x + c x ) − log 3
x
Apply L’Hospital’s rule,
ax log a + b x log b + c x log c
ax + b x + c x
= 2 lim
x→ 0
1
log y = log (abc ) 2 /3
⇒
y = (abc ) 2 /3
x→ 0
x
( 4x + 1 )
x2 + x + 2
( 4 x +1 )x
x2 + x +2
1
= e4
x
1 + 1 = e
Q lim
x→ 0
x
1 − cos (ax + bx + c )
( x − α )2
2
3 Now, xlim
→α
6 Let
= 2 lim
x
1
4 +
x
1 +
f ( x ) ⋅ e nx + g ( x )
e nx + 1
f ( x)
g( x)
= lim
+ nx
n→ ∞
e + 1
1 + 1
nx
e
f ( x)
Finite
=
+
= f ( x)
1+ 0
∞
15
x
= 120 − 0 = 120
n
Q 0 ≤ < 1, therefore
x
n
n
0 ≤ x < x ⇒ lim x = 0
+
x→ 0
x
x
2
5 Given, x > 0 and g is a bounded
2 2
15
− + …
−
x x
x
4x + 1
= lim 1 + 2
x→ ∞
x
+ x+
n
1
1
= lim (−1)n sin π
−
+K ∞
3
n→ ∞
2n 8 n
Then, lim
x2 + 5x + 3
2 Now, xlim
2
→∞
x + x+2
1 /2
1
1
= limsin nπ 1 +
−
+ K∞
2
4
n→ ∞
2
n
8
n
π
π
= lim sin nπ +
−
+ K∞
n→ ∞
2n 8 n3
1 + 2 + 3+ ...+15) − x
= lim(
x→ 0 +
1 2
15
+ + ... +
x x
x
v( x) = |x |
∴ f ( x ) = vou ( x ) = v (u( x )) = v (sin x ) =|sin x|
=
Q lim sin x = 1
x → 0 x
a2
( x − β )2
2
=0
∴Given limit
1
1
= lim+ x −
x→ 0
x x
2
40 Clearly, f ( x ) = 2 1 − x , x ≤ 0
41 [sin x] is non-differentiable at
x→ α
− 1 / x, x < 1
f ′( x ) = 1
,
x>1
x
∴ f ′ (1− ) = − 1 and f ′ (1+ ) = 1
Hence, f ( x ) is not differentiable.
k+ m=2
0,
x> 0
∴ f ( x ) is discontinuous and hence
non-differentiable at x = 0.
= lim
ax2 + bx + c
2sin2
2
= lim
2
x→ α
(x − α)
a
2sin2 ( x − α )( x − β )
2
= lim
2
x→ α
a ( x − α )2 ( x − β )2
2
2
a ( x − β )2
2
3 x − x3
2
1 − 3x
7 f ( x ) = cot −1
1 − x2
and g ( x ) = cos −1
2
1 + x
On putting x = tan θ in both equations,
we get
3 tan θ − tan3 θ
f (θ) = cot −1
2
1 − 3 tan θ
⇒ f (θ) = cot −1 (tan 3 θ)
π
⇒ f (θ) = cot −1 cot − 3 θ
2
π
=
− 3θ
2
∴ f ′ (θ) = − 3
1 − tan2 θ
and g(θ) = cos −1
2
1 + tan θ
= cos −1 (cos 2θ) = 2θ
… (i)
115
DAY
∴ g ′ (θ) = 2
Now,
f ( x ) − f (a)
f ( x ) − f (a)
lim
= lim
x → a g ( x ) − g ( a)
x→ a
x−a
1
×
g ( x ) − g (a)
lim
x→ a
x−a
1
1
3
= f ′ (a) ⋅
= –3× = –
g ′ (a)
2
2
Hence, the composite function y = f ( x )
is discontinuous at three points,
1
x = , x = 1 and x = 2 .
2
10 Since, f ( x ) is a positive increasing
function.
∴ 0 < f ( x ) < f (2 x ) < f (3 x )
⇒ 0< 1<
8 Now, cos x is continuous, ∀x ∈ R
1
⇒ cos π x − is also continuous,
2
∀ x ∈ R.
Hence, the continuity of f depends upon
the continuity of [ x] , which is
discontinuous, ∀ x ∈ I.
So, we should check the continuity of f
at x = n, ∀ n ∈ I
LHL at x = n is given by
f(n− ) = lim− f ( x )
x→ n
1
= lim− [ x] cos π x −
x→ n
2
(2n − 1) π
= (n − 1) cos
=0
2
RHL at x = n is given by
f (n + ) = lim+ f ( x )
x→ n
1
= lim+ [ x] cos π x −
x→ n
2
⇒
x→ ∞
x→ ∞
is discontinuous at u = − 2 and u = 1.
When u = −2
1
⇒
= −2
x−1
1
x =
⇒
2
When u = 1
1
⇒
=1
x−1
⇒
x=2
= lim+ f ′( x )
x→ π
∴ f is differentiable at x = 0 and x = π
Hence, f is equal to the set {0, π}.
= lim
n→ ∞
f (3 x )
≤ lim
x → ∞ f ( x)
By Sandwich theorem,
lim
f (2 x )
=1
f ( x)
11 We have, f ( x ) = max{tan x,sin x,cos x},
π 3π
where x ∈ − ,
2 2
Let us draw the graph of y = tan x,
π 3π
y = sin x and y = cos x in − ,
2 2
y = tanx
y = tanx
y = f(x)
y = f(x)
(2n − 1) π
= (n ) cos
=0
2
Also, value of the function at x = n is
1
f (n ) = [n] cos π n −
2
(2n − 1)π
= (n ) cos
=0
2
∴ f (n + ) = f (n − ) = f (n )
Hence, f is continuous at
x = n, ∀ n ∈ I .
9 The function u ( x ) = 1 is
x−1
discontinuous at the point x = 1.
The function y = f (u )
1
= 2
u + u−2
1
=
(u + 2) (u − 1 )
f (2 x )
f ( x)
x→ ∞
O a
-π
2
π
2
y = f(x)
y = cos(x)
X
3π
2
π
–1
y = sinx
From the graph, it clear that f ( x ) is
π
non-differentiable at x = a, and π.
2
12 We have,
f ( x ) =| x − π|(
⋅ e − 1)sin| x|
−x
( x − π )(e − 1)sin x, x < 0
f ( x ) = −( x − π )(e x − 1)sin x, 0 ≤ x < π
( x − π )(e x − 1)sin x, x ≥ π
|x|
We check the differentiability at x = 0
and π.
We have,
f ′( x ) =
−x
x
( x − π )(e − 1)cos x + (e − 1)sin x
−x
+ ( x − π )sin xe (−1), x < 0
x
x
−[( x − π )(e − 1)cos x + (e − 1)sin x
x
+ ( x − π )sin xe ], 0 < x < π
( x − π )(e x − 1)cos x + (e x − 1)sin x
+ ( x − π )sin xe x , x > π
Clearly, lim− f ′( x ) = 0 = lim+ f ′( x )
x→ 0
x→ π −
13 We have, f ( x ) = lim
n→ ∞
f (2 x ) f (3 x )
<
f ( x)
f ( x)
lim 1 ≤ lim
lim f ′( x ) = 0
and
x→ 0
n
π
x 1 +
3x
π
1 +
3x
n −1
π
x n +
3
π
x n −1 +
3
= x, if x >
n
n −1
π
3
3x n
+ 1
π
π
π
Also, f ( x ) = lim
= ,
n→ ∞ 3
3
3 x n −1
+ 1
π
if x < π / 3
π
π
Note that f =
3
3
x, if x ≥ π
3
f ( x) = π
, if
π
x
<
3
3
From the given options, it is clear that
option (c ) is incorrect.
x
14 Let ak +1 =
(kx + 1){(k + 1)x + 1}
∴
1
1
=
−
kx
+
1
(
k
+
1
)x + 1
n −1
1
1
∴ f ( x ) = lim ∑
−
n→ ∞
(k + 1)x + 1
k = 0 kx + 1
1
= lim 1 −
n→ ∞
nx + 1
1 , if x ≠ 0
=
0 , if x = 0
Clearly, f ( x ) is neither continuous nor
differentiable at x = 0.
15 Clearly, x n + xa + b = ( x − x1 )
⇒
( x − x2 )...( x − x n )
x n + xa + b
= ( x − x2 )
x − x1
x n + ax + b
⇒ lim
x→ x 1
x − x1
( x − x3 )...( x − x n )
= ( x1 − x2 )( x1 − x3 )...( x1 − x n )
nx n −1 + a
x→ x 1
1
= ( x1 − x2 )( x1 − x3 )...( x1 − x n )
[using L’Hospital rule]
⇒ lim
⇒ nx1n −1 + a = ( x1 − x2 )
( x1 − x3 )...( x1 − x n ).
TEN
116
16 Let L = lim
x ln
x→ ∞
1
x
0
c
−1
a
x
−b
1
x
0
1
1,
(v) f ( x) = 1,
−1
Clearly, for limit to be exist, b = 1.
a c
Thus, L = lim x ln 1 + 3 −
x→ ∞
x
x
a c
= lim x 3 −
x→ ∞
x
x
x2
x3 x 4
Q ln(1 + x ) = x −
+
−
+ ...
2
3
4
= −c
Q
L = −4
∴
c =4
Hence, a ∈ R,b = 1 and c = 4.
f( x ) 1
lim 1 + x +
−1
x
x
⇒ e x →0
lim
⇒
e
= e3
= e3
1
f( x )
lim 1 +
−1
x
x
1/x
f ( x )
x
= e x →0
lim
f( x )
= e x →0 x = e 2
2
18 We have, f 2 ( x ) = 1 ∀x ∈ R
∴ f can take values +1 or −1
Since f is discontinuous only at x = 0
∴ f may be one of the followings
1, x ≤ 0
(i) f ( x) =
−1, x > 0
1, x < 0
(ii) f ( x) =
−1, x ≥ 0
−1, x ≤ 0
(iii) f ( x) =
1, x > 0
−1, x < 0
(iv) f ( x) =
1, x ≥ 0
1
Here, lim f2 ( x ) = lim sin
x→ 0
x→ 0
x
which does not exist.
So, f2 ( x ) is not continuous at x = 0.
Hence, Statement II is false.
19 f ( x ) = x | x | and g ( x ) = sin x
2
− sin x , x < 0
∴
gof ( x ) = sin ( x | x |) =
2
sin x , x ≥ 0
2
− 2 x cos x , x < 0
(gof )′ ( x ) =
2
2 x cos x , x ≥ 0
Clearly, L ( gof )′ (0) = 0 = R ( gof )′ (0)
So, gof is differentiable at x = 0 and also
its derivative is continuous at x = 0.
Now,
22 ∴ f ( x ) = | x − 2 | + | x − 5|
(2 − x ) + (5 − x ),
= ( x − 2) + (5 − x ),
( x − 2) + ( x − 5),
7 − 2 x,
= 3,
2 x − 7,
2
2
2
− 2 cos x + 4 x sin x ,
x< 0
( gof )′ ′ ( x ) =
2
2
2
cos
x
−
4
x
sin x2 ,
x≥ 0
∴
Now,
lim 1 +
x→ 0
x =0
L ( gof )′′ (0) ≠ R ( gof )′′ (0)
x<2
2≤ x≤ 5
x> 5
Y
y = 2x – 7
y = 7 – 2x
Hence, gof ( x ) is not twice differentiable
at x = 0.
Therefore, Statement I is true, Statement
II is false.
20 Statement I is correct as though
|4 x2 − 12 x + 5| is non-differentiable at
1
5
x = and but cos πx = 0 those points.
2
2
1
5
So, f ′ and f ′ exists.
2
2
1
21 Here, f ( x ) = x ⋅ sin x ,
,
0
x≠0
x=0
To check continuity at x = 0,
1
LHL = lim (− h ) sin − = 0
h→ 0
h
1
RHL = lim h sin = 0
h→ 0
h
f (0) = 0
So, f ( x ) is continuous at x = 0.
x<2
2≤ x ≤ 5
x> 5
Now, we can draw the graph of f very
easily.
∴ L ( gof )′ ′ (0) = − 2 and R ( gof )′′ (0) = 2
f( x )
1+
x2
= e3
f ( x)
lim 2 = 2
x→ 0 x
x →0
⇒
1/x
f ( x )
x
Hence, Statement I is correct.
sin 1 , x ≠ 0
f2 ( x ) =
x
0
, x=0
−1, x > 0
(vi) f ( x ) = −1, x < 0
1, x = 0
a
c
= lim x ln 3 + b −
x→ ∞
x
x
1 + x +
17 We have, lim
x→ 0
x >0
x <0
y=3
X′
X
0
2
5
Y′
Statement I f ′ (4) = 0
It is obviously clear that, f is constant
around x = 4, hence f ′ (4) = 0. Hence,
Statement I is correct.
Statement II It can be clearly seen that
(i) f is continuous, ∀ x ∈ [2, 5 ]
(ii) f is differentiable, ∀ x ∈ (2, 5)
(iii) f (2) = f (5) = 3
Hence, Statement II is also correct but
obviously not a correct explanation of
Statement I.
DAY TWELVE
Differentiation
Learning & Revision for the Day
u
Derivative (Differential
Coefficient)
u
u
Geometrical Meaning of
Derivative at a point
Methods of Differentiation
u
u
Second Order Derivative
Differentiation of a Determinant
Derivative (Differential Coefficient)
The rate of change of a quantity y with respect to another quantity x is called the derivative or
differential coefficient of y with respect to x. The process of finding derivative of a function
called differentiation.
Geometrical Meaning of Derivative at a Point
Geometrically derivative of a function at a point x = c is the slope of the tangent to the curve
y = f ( x) at the point P {c, f (c)}.
f ( x) − f (c)
df ( x)
Slope of tangent at P = lim
=
or f ′ (c).
x →c
x −c
dx x = c
Derivative of Some Standard Functions
●
●
●
●
●
●
●
d
(constant) = 0
dx
d
1
( x) =
dx
2 x
d
(sin x) = cos x
dx
d
(tan x) = sec2 x
dx
d
(cot x) = − cosec2 x
dx
d
1
(log x) = , for x > 0
dx
x
d x
(a ) = ax log a, for a > 0
dx
●
●
●
●
●
●
●
d
dx
d
dx
x n = nx n −1
n
1
n = − n +1
x
x
d
(cos x) = − sin x
dx
d
(sec x) = sec x tan x
dx
d
(cosec x) = − cosec x cot x
dx
d x
(e ) = e x
dx
d
1
(log a x) =
, for x > 0, a > 0, a ≠ 1
dx
x log a
PRED
MIRROR
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
TEN
118
●
●
●
●
●
●
(x) If u = f ( x) and v = g( x), then the differentiation of u
du
du dx
with respect to v is
=
.
dv dv
dx
Differentiation of a function
w. r. t another function
d
1
(sin −1 x) =
, for −1 < x < 1
dx
1 − x2
d
1
(cos −1 x) = −
, for −1 < x < 1
dx
1 − x2
d
1
(sec −1 x) =
, for| x| > 1
dx
| x| x 2 − 1
d
1
, for| x| > 1
(cosec −1 x) = −
dx
| x| x 2 − 1
d
1
(tan −1 x) =
, for x ∈ R
dx
1 + x2
d
1
(cot −1 x) = −
, for x ∈ R
dx
1 + x2
Differentiation Using Substitution
In order to find differential coefficients of complicated
expressions, some substitution are very helpful, which are
listed below
S. No.
Methods of Differentiation
(i) If y = f ( x) ± g( x), then
dy d
=
{ f ( x) ± g( x)}
dx dx
= f ′ ( x) ± g ′ ( x)
(ii) If y = c ⋅ f ( x), where c is any constant, then
dy d
=
(c ⋅ f ( x)) = c ⋅ f ′( x). [Scalar multiple rule]
dx dx
dy d
(iii) If y = f ( x) ⋅ g( x), then
=
{ f ( x) ⋅ g( x)}
dx dx
= f ( x) ⋅ g ′ ( x) + g ( x) ⋅ f ′ ( x)
[Product rule]
f ( x)
dy d f ( x) g( x) ⋅ f ′ ( x) − f ( x) ⋅ g′ ( x)
(iv) If y =
, then
=
=
g ( x)
dx dx g( x)
{g( x)}2
g ( x) ≠ 0
(v) If y = f (u) and u = g( x), then
dy df du
=
⋅
= f ′(u) ⋅ g′( x)
dx du dx
2
(vii) If given function cannot be expressed in the form
y = f ( x) but can be expressed in the form f ( x, y) = 0, then
to find derivatives of each term of f ( x, y) = 0 w.r.t x.
[differentiation of implicit function]
(viii) If y is the product or the quotient of a number of
complicated functions or if it is of the form ( f ( x))g ( x ),
then the derivative of y can be found by first taking log
on both sides and then differentiating it.
[logarithmic differentiation rule]
dy
When y = ( f ( x))g ( x ), then
= ( f ( x))g ( x )
dx
g ( x)
f ( x) ⋅ f ′( x) + log f ( x) ⋅ g′( x)
(ix) If x = φ (t ) and y = Ψ (t ), where t is parameter, then
dy dy/dt
[Parametric differentiation rule]
=
dx dx/dt
x = a sin θ or a cos θ
2
(i)
a –x
(ii)
x 2 – a2
x = a sec θ or a cosec θ
(iii)
x 2 + a2
x = a tan θ or a cot θ
(iv)
a+ x
or
a− x
(v)
a2 + x2
or
a2 − x2
a– x
a+ x
x = a cos 2θ
a2 – x 2
a2 + x 2
x2 = a2 cos 2θ
(vi)
x
a+ x
x = a tan2 θ
(vii)
( x – a)( x – b )
x = a sec2 θ – b tan2 θ
(viii)
ax – x 2
x = a sin2 θ
(ix)
x
a– x
x = a sin2 θ
(x)
( x – a) (b – x)
x = a cos2 θ + b sin2 θ
[Chain rule]
This rule can be extended as follows. If y = f (u),u = g(v)
dy df du dv
and v = h ( x), then
=
⋅
⋅ ,
dx du dv dx
d
(vi)
( f {g( x)}) = f ′ (g( x)) ⋅ g′( x)
dx
Substitution
Function
Usually this is done in case of inverse trigonometric functions.
Second Order Derivative
d dy
is called the second order derivative
dx dx
d2 y
of y w.r.t x. It is denoted by 2 or f ′′( x) or y′′ or y2 .
dx
If y = f ( x), then
Differentiation of a Determinant
dp
p q r
dx
dy
If y = u v w , then
= u
dx
l m n
l
p
du
+
dx
l
q
dv
dx
m
dq
dx
v
m
dr
dx
w
n
r
p
dw
+ u
dx
dl
n
dx
q
v
dm
dx
r
w
dn
dx
119
DAY
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
3π
is equal to
4
11 If y = (1 − x ) (1 + x 2 ) (1 + x 4 )...(1 + x 2 n ), then
1 If f ( x ) = | cos x |, then f ′
1
(a)
2
(b) 2
j
NCERT Exemplar
1
(c)
2
(d) 2 2
2 If f ( x ) = | x − 1 | and g ( x ) = f [f {f ( x )}], then for x > 2, g ′ ( x )
is equal to
(a) − 1, if 2 ≤ x < 3
(c) 1, if x > 2
(d) (−1)n − 1 (n − 1)!
(b) (−1)(n − 1)! (c) n! − 1
4 If f ( x ) = x , then the value of
n
f (1) −
(a) 2 n
f ′ (1) f ′′ (1) f ′ ′′ (1)
( −1)n f n (1)
is
+
−
+ ...+
1!
2!
3!
n!
(b) 0
(c) 2 n − 1
(d) None of these
x −x
d −1
5 If f ( x ) = 2
, where x ≠ 0, − 2, then
[f ( x )]
x + 2x
dx
2
(whenever it is defined) is equal to
(a)
−1
(1 − x)2
(b)
3
(1 − x)2
(c)
j
1
(1 − x)2
JEE Mains 2013
−3
(d)
(1 − x)2
6 If f ( x ) = 2+| x |−| x − 1|−| x + 1| , then
1
1
3
5
f ′ − + f ′ + f ′ + f ′ is equal to
2
2
2
2
(b) −1
(a) 1
(c) 0
(d) None of these
(c) 1
(d)
dy
10 If sin y = x sin(a + y ), then
is equal to
dx
(a)
sina
sin2 (a + y)
(c) sina sin2 (a + y)
(a)
(b)
(c)
(d)
sin (a + y)
sina
sin2 (a − y)
(d)
sina
2
(b)
x
(1 − x)2
(d) −2
AP
GP
Arithmetic-Geometric progression
None of the above
14 If y = f ( x ) is an odd differentiable function defined on
( − ∞, ∞ ) such that f ′ ( 3) = − 2, then f ′ ( − 3) is equal to
(a) 4
(c) −2
(b) 2
(d) 0
15 If f and g are differentiable function satisfying
g ′ (a ) = 2, g (a ) = b and fog = I (identity function). Then,
f ′ (b ) is equal to
1
2
(b) 2
(c)
2
3
(d) None of these
16 If y is an implicit function of x defined by
(b) 1
x+ y
xy
(b) xy
(c)
(b) −n 2 y
(a) n 2 y
19 If x =
3
2
(d) − log 2
(c) log 2
dy
+n
is equal to
, then
dx
x
y
(d)
y
x
d 2y
dy
is equal to
+x
2
dx
dx
(c) −y
(d) 2 x 2 y
2t
1−t
dy
and y =
is equal to
, then
2
2
1+ t
1+ t
dx
2
2t
t2 + 1
2t
(c)
1− t2
(a)
1
2
(d)
(c) 0
18 If y = ( x + 1 + x 2 )n , then (1 + x 2 )
π
f ′ is equal to
4
(b)
x
(1 + x 2 )
f (1) = f ( −1) and a, b, c are in AP, then f ′ (a ), f ′ (b ) and f ′ (c )
are in.
(a)
9 If f ( x ) = cos x ⋅ cos 2x ⋅ cos 4x ⋅ cos 8x ⋅ cos 16x , then
2
(b) −4
(a) 4
17 If x m y n = ( x + y ) m
π
8 If f ( x ) = | cos x − sin x |, then f ′ is equal to
2
(a)
(c)
13 Let f ( x ) be a polynomial function of second degree. If
(a) − 1
1
1
, where x ≠ 0
(b) for | x| > 1
| x|
x
1
(c) − for | x| > 1
x
1
1
(d) for x > 0 and − for x < 0
x
x
(a)
(b) −1
1
(1+ x)2
x 2 x − 2x x cot y − 1 = 0. Then, y ′ (1) is equal to
7 If f ( x ) = loge | x |, then f ′( x ) equals
(a) 1
(b)
12 If f : ( −1, 1) → R be a differentiable function with f ( 0) = 1
(a)
(d) −2
(c) 2
(a) −1
and f ′ ( 0) = 1. Let g ( x ) = [f ( 2f ( x ) + 2 )]2 . The, g′ ( 0) is
equal to
(b) 1, if 2 ≤ x < 3
(d) None of these
3 The derivative of y = (1 − x )( 2 − x )K (n − x ) at x = 1 is
(a) 0
is equal to
dy
at x = 0
dx
(b)
2t
t2 − 1
(d) None of these
π
2
20 For a > 0, t ∈ 0, , let x = a sin−1 t and y = a cos −1 t .
2
dy
Then, 1 + equals
dx
2
x
(a) 2
y
2
y
(b) 2
x
j
x + y
(c)
y2
2
2
JEE Mains 2013
x2 + y2
(d)
x2
TEN
120
x + 1
x − 1
dy
−1
+ sin
, then dx is equal to
x − 1
x + 1
21 If y = sec−1
1
x +1
(d) None of these
(a) 0
(a) 2
(c) 0
(b)
(c) 1
(a)
9
1 + 9x 3
(b)
3x x
1 − 9x 3
23 If y = sec(tan−1 x ), then
j
(c)
3x
1 − 9x 3
JEE Mains 2017
3
(d)
1 + 9x 3
dy
at x = 1 is equal to
dx
j
JEE Mains 2013
1
(a)
2
1
(b)
2
24 If y = tan−1
x
(b) −
1− x4
1− x2
(d) 2
, then dy is equal to
dx
1+ x 2 − 1− x 2
1+ x 2 + 1− x 2
x
(a)
(c) 1
(c)
2x
1− x4
(d) None of these
dy
25 If 1 − x + 1 − y = a ( x − y ), then
is equal to
dx
2
2
1− x2
1− y2
(a)
(b)
1− y2
1− x2
(c)
x2 − 1
1− y2
26 If y = sin−1( x 1 − x + x 1 − x 2 ), then
−2 x
(a)
(c)
a cos
1+ a
(a) −
(c)
2 x − x2
1
+
1− x2
27 If y =
1
+
1− x2
1
−1
2 x − x2
x
cos −1 x
and z = a cos
1
1 + a cos
1
(1 + a cos
−1
−1
(b)
−1
−
1
2 x − x2
(d) None of these
−1
x
, then
(b)
x
y2 − 1
1− x2
(d)
dy
is equal to
dx
1− x2
dy
is equal to
dz
1
1 + a cos
−1
)
28 Let g ( x ) be the inverse of f ( x ) such that f ′( x ) =
2
then
d (g ( x ))
is equal to
dx 2
1
1 + (g (x))5
(c) 5 (g (x))4 (1 + (g (x))5
(a)
29
d2y
(a) − 2
dx
−1
dy
dx
g ′(x)
1 + (g (x))5
(d) 1 + (g (x))5
d 2 y dy −3
(c) − 2
dx dx
d 2 y dy −2
(b) 2
dx dx
d2y
(d) 2
dx
−1
(a) f (0)g (0)
(b) 0
(c) can’t be determined
(d) None of the above
32 If
f ′( x ) f ( x )
= 0, where f ( x ) is continuously
f ′′( x ) f ′( x )
differentiable function with f ′( x ) ≠ 0 and satisfies f ( 0) = 1
f ( x ) −1
and f ′( 0) = 2, then lim
is
x→ 0
x
(a) 1
1
(c)
2
(b) 2
(d) 0
3x
33 If f ( x ) = 6
P
Then,
2
cos x
−1
P2
sin x
0 , where P is a constant.
P3
d2
{ f ( x )} at x = 0 is equal to
dx 2
(a) P
(c) P + P 3
(b) P + P 2
(d) independent of P
34 Which of the following statements is/are true?
Statement I If y = (log x )log x , then
dy
1 log (log x )
= (log x )log x
+
.
x
dx
x
Statement II If y = cos(a cos x + b sin x ) for some
constants a and b, then
y ′ = (a sin x − b cos x ) sin(a cos x + b sin x )
Only I is true
Only II is true
Both I and II are true
Neither I nor II is true
35 Statement I If u = f (tan x ), v = g (sec x ) and f ′ (1) = 2,
1
du
.
g′ ( 2 ) = 4 , then
=
dv x = π / 4
2
Statement II If u = f ( x ), v = g ( x ), then the derivative of f
du du /dx
with respect to g is
=
.
dv dv /dx
j
−3
1
,
1+ x 5
(b)
d 2x
is equal to
dy 2
that of y = g ( x ) is symmetrical about the origin and if
d 2h( x )
at x = 0 is
h( x ) = f ( x ) ⋅ g ( x ), then
dx 2
(a)
(b)
(c)
(d)
x
(d) None of these
x 2
(b) 1
(d) −1
31 If graph of y = f ( x ) is symmetrical about the Y-axis and
6x x
1
22 For x ∈ 0, , if the derivative of tan−1
is
4
1 − 9x 3
x ⋅ g ( x ), then g ( x ) equals
log(e / x 2 )
d 2y
−1 3 + 2 log x
is
,
then
+
tan
1 − 6 log x
dx 2
log(ex 2 )
30 If y = tan−1
AIEEE 2011
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d)Statement I is false; Statement II is true
121
DAY
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 For x ∈R ,f ( x ) = | log 2 − sin x | and g ( x ) = f (f ( x )), then
(a)
(b)
(c)
(d)
g is not differentiable at x = 0
g′ (0) = cos(log 2)
g′ (0) = − cos (log 2)
g is differentiable at x = 0 and g′ (0) = − sin(log 2)
(b) y ⋅ ∑ k cot kx
k =1
n
(c) y ⋅ ∑ k ⋅ tankx
(d)
k =1
∑ cotkx
2
7
(b)
1
2
(c) 2
(d)
7
2
1 +
(where, a > 0), then
4 If f ( x ) = (cos x + i sin x ) ⋅ (cos 2x + i sin 2x )
(cos 3x + i sin 3x ) ... (cos nx + i sin nx ) and f (1) = 1, then
f ′ ′ (1) is equal to
n (n + 1)
2
2
n (n + 1)
(c) −
2
n (n + 1)
(b)
2
(a)
5 If x 2 + y 2 = ae
2
is equal to
2 −π / 2
e
a
2
(c) − e − π / 2
a
2 π /2
e
a
(d) None of these
(b) −
(a)
6 If y = | sin x || x| , then the value of
(a)
(c)
2
−π
6
6
2
π
−
6
6
dy
π
at x = − is
dx
6
π
6
[6log 2 −
3 π]
2
(b)
[6log 2 +
6
[6log 2 +
3 π]
(d) None of these
3 π]
1
( 5) 2 x + 1
2
and g ( x ) = 5x + 4x loge 5 is
(b) (0, 1)
(c) (∞, 0)
(d) (0, ∞)
8 Let f ′ ′ ( x ) = − f ( x ), where f ( x ) is a continuous double
differentiable function and g ( x ) = f ′ ( x ).
2
x
x
If F ( x ) = f + g
2
2
equal to
(a) 0
(c) 10
(b) a 2
(c)
2
3a
(d)
g (x )
1
f ′ n +
3
then
1
f n +
3
2
and F ( 5) = 5, then F (10) is
2a
3
and g ( x + 1) = x + g ( x ) ∀ x ∈ R. If n ∈ I + ,
1
f′
3
is equal to
−
1
f
3
1 1
1
(b) 3 1 + + + ... +
3
5
2
n
− 1
(d) 1
1 1
1
(a) 3 1 + + + K +
2 3
n
(c) n
1− x
13 If the function f ( x ) = − 4e
2
+1+ x +
x2 x3
and
+
2
3
−7
g ( x ) = f −1( x ), then the value of g′ is equal to
6
1
5
6
(c)
7
1
5
6
(d) −
7
(b) −
(a)
7 The solution set of f ′ ( x ) > g ′ ( x ), where f ( x ) =
(a) (1, ∞)
a
3
12 Let f ( x ) = elne
, a>0 assuming y > 0, then y′ ′ ( 0)
3
2
dy 2
dx
at π is given by
d 2y
6
2
dx
(a)
(d) None of these
y
tan−1
x
JEE Mains 2013
(b) sin2 x cos (cos x)
(d) cos2 x sin(sin x)
11 If x = a cos t cos 2 t and y = a sin t cos 2 t
k =1
3 If 3f ( x ) − 2f (1 /x ) = x , then f ′ ( 2) is equal to
(a)
j
(a) cos2 x cos (sin x)
(c) sin2 x sin(cos x)
k =1
n
−1
81
−1
(d)
3
(b)
g ( x ) is equal to
n
(a) ∑ k ⋅ tankx
−1
9
−1
(c)
27
(a)
10 If f ( x ) = sin (sin x ) and f ′ ′ ( x ) + tan x f ′ ( x ) + g ( x ) = 0, then
2 If y = sin x ⋅ sin 2x ⋅ sin 3x ... sin nx , then y ′ is
n
9 If f ( 2) = 4, f ′ ( 2) = 3, f ′′ ( 2) = 1, then (f −1 )′′ (4) is equal to
14. If f ( x ) = ( x − 1)100 ( x − 2)2 ( 99 ) ( x − 3)3 ( 98 ) ...( x − 100)100 ,
then the value of
(a) 5050
(c) 3030
f ′ (101)
is
f (101)
(b) 2575
(d) 1250
15 The derivative of the function represented parametrically
as x = 2t − | t | , y = t 3 + t 2 | t | at t = 0 is
(b) 5
(d) 25
(a) −1
(c) 1
(b) 0
(d) does not exist.
TEN
122
ANSWERS
SESSION 1
SESSION 2
1 (a)
2 (a)
3 (b)
4 (b)
5 (b)
6 (d)
7 (b)
8 (a)
9 (a)
10 (b)
11 (a)
12 (b)
13 (a)
14 (c)
15 (a)
16 (a)
17 (d)
18 (a)
19 (b)
20 (d)
21 (a)
22 (a)
23 (a)
24 (a)
25 (b)
26 (c)
27 (c)
28 (c)
29 (c)
30 (c)
31 (b)
32 (b)
33 (d)
34 (c)
35 (a)
1 (b)
11 (d)
2 (b)
12 (c)
3 (b)
13 (a)
4 (c)
14 (a)
5 (c)
15 (b)
6 (a)
7 (d)
8 (b)
9 (c)
10 (d)
Hints and Explanations
SESSION 1
1 When π < x < π, cos x < 0, so that
2
|cos x |= − cos x,
f ′ (1)
f ′ ′ (1) f ′ ′ ′ (1)
+
−
1!
2!
3!
(− 1)n f n (1)
+ ... +
n!
n
n
n
= C 0 − C 1 + C 2 − nC 3 +
... + (−1)n nC n
= (1 − 1)n = 0
∴
f (1) −
i.e. f ( x ) = − cos x, f ′( x ) = sin x
3π
3π = 1
Hence, f ′
= sin
4
4
2
2 We have, f ( x ) = | x − 1|
[Q x > 2]
f [ f ( x )] = f ( x − 1) = |( x − 1) − 1 |
=| x − 2|
g ( x ) = f [ f { f ( x )}] = f ( x − 2)
=|( x − 2) − 1 |= | x − 3 |
x − 3, if x ≥ 3
=
− x + 3, if 2 ≤ x < 3
5 Let
⇒
⇒
∴
1, if x ≥ 3
∴ g ′ ( x ) =
−1, if 2 ≤ x < 3
3
dy
= − [(2 − x )(3 − x )K (n − x ) +
dx
(1 − x )(3 − x )K(n − x )
+ K + (1 − x )(2 − x )K (n − 1 − x )]
dy
⇒
= − [(n − 1)! + 0 + K + 0]
dx x = 1
x2 − x
x2 + 2 x
2y + 1
x=
;x≠ 0
−y + 1
y =
f −1 ( x ) =
(− x + 1 ) ⋅ 2 − (2 x + 1 ) (−1 )
d −1
[ f ( x )] =
dx
(− x + 1 )2
3
=
(− x + 1 )2
6 We have, f ( x ) = 2+| x|−| x − 1|−| x + 1|
∴
2 −
2 −
f ( x) =
2 +
2 +
= (−1)(n − 1)!
⇒
f (1) = 1 = nC 0
f ′ (1) n n
=
= C1
1!
1!
f ′ ′ (1) n (n − 1) n
=
= C2
2!
2!
f ′ ′ ′ (1) n (n − 1) (n − 2) n
=
= C3
3!
3!
M
M
M
M
f n (1) n ! n
=
= Cn
n!
n!
x + ( x − 1) + ( x + 1),
x + ( x − 1) − ( x + 1),
x + ( x − 1) − ( x + 1),
x − ( x − 1) − ( x + 1),
if x < −1
if − 1 ≤ x < 0
if 0 ≤ x < 1
if x ≥ 1
4 We have, f ( x ) = x n
⇒
2x + 1
−x + 1
if x < −1
x + 2,
− x, if − 1 ≤ x < 0
=
if 0 ≤ x < 1
x,
2 − x,
if x ≥ 1
1,
−1,
⇒ f ′( x ) =
1,
−1,
if x < −1
if −1 ≤ x < 0
if 0 ≤ x < 1
if
x≥1
1
1
3
5
Hence, f ′ − + f ′ + f ′ + f ′
2
2
2
2
= (−1) + 1 + (−1) + (−1) = −2
7 We have, f ( x ) = log e | x|
∴
log(− x )
− log(− x )
f ( x) =
− log x
log x
, x < −1
, −1 < x < 0
, 0< x < 1
, x >1
−
f ′( x ) =
−
1
, x< −1
x
1
, − 1< x < 0
x
⇒
1
, 0< x < 1
x
1
, x>1
x
1
Clearly, f ′( x ) = for| x|> 1
x
π
8 When 0 < x < , cos x > sin x
4
∴ cos x − sin x > 0
π
Also, when < x < π,cos x < sin x
4
∴ cos x − sin x < 0
∴ |cos x − sin x|= − (cos x − sin x ), when
π
< x< π
4
⇒ f ′ ( x ) = sin x + cos x
π
π
π
⇒ f ′ = sin + cos = 1 + 0 = 1
2
2
2
2sin x ⋅ cos x ⋅ cos 2 x ⋅ cos 4 x
⋅ cos 8 x ⋅ cos 16 x
9 f ( x) =
2sin x
sin 2 x cos 2 x cos 4 x ⋅ cos 8 x ⋅ cos 16 x
=
2 sin x
sin 32 x
= 5
2 sin x
123
DAY
1
⋅
32
32cos 32 x ⋅ sin x − cos x ⋅ sin 32 x
sin2 x
1
1
32 ×
−
×0
π
2
2
⇒ f ′ =
= 2
2
4
1
32
2
∴ f ′( x ) =
10 ∴
⇒
sin y = x sin( a + y )
x=
sin y
sin(a + y )
On differentiating w.r.t. y, we get
dx sin(a + y )cos y − sin y cos( a + y )
=
dy
sin2 (a + y )
dx
sin a
=
⇒
dy sin2 (a + y )
⇒
dy sin2 (a + y )
=
dx
sin a
11 Given, y = (1 − x ) (1 + x2 )
(1 + x 4 )...(1 + x2 n )
(1 − x ) (1 + x2 )...(1 + x2 n )
or y =
(1 + x )
1 − ( x )4 n
=
(1 + x )
(1 + x ) ⋅ (0 − 4n ⋅ x 4 n − 1 )
⇒
⇒
− (1 − x 4 n ) ⋅ 1
(1 + x )2
∴
∴
dy
= −1
dx x = 0
12 We have, f :(−1,1) → R
f (0) = −1, f ′(0) = 1
g ( x ) = [ f (2 f ( x ) + 2)]2
⇒ g ′( x ) = 2[ f (2 f ( x ) + 2)]
× f ′(2 f ( x ) + 2) × 2 f ′( x )
⇒ g ′(0) = 2[ f {2 f (0) + 2}]
× f ′{2 f (0) + 2} × 2 f ′(0)
= 2[ f (0)] × f ′( 0) × 2 f ′( 0)
= 2 × (−1) × 1 × 2 × 1 = −4
⇒
⇒
⇒
14 Since, f ( x ) is odd.
∴
f (− x ) = − f ( x )
f ′ ( g ( x )) g ′ ( x ) = 1 for all x
1
1
f ′(g (a)) =
=
g ′ (a) 2
1
[Q g (a) = b]
f ′( b ) =
2
16 x2 x − 2 x x cot y − 1 = 0
…(i)
Now, x = 1,
1 − 2 cot y − 1 = 0
⇒
cot y = 0
π
⇒
y =
2
On differentiating Eq. (i) w.r.t. x, we get
dy
2 x2 x (1 + log x ) − 2 [ x x (− cosec2 y )
dx
+ cot y x x (1 + log x )] = 0
π
At 1, , 2 (1 + log 1)
2
dy
− 2 1 (− 1)
+ 0 = 0
1 , π
dx
2
dy
⇒ 2 + 2
=0
dx 1, π
∴
=
⇒ (1 + x2 )
⇒
dy m + n n m m + n
− =
−
dx x + y
y
x
x+ y
⇒
dy my + ny − nx − ny
dx
y( x + y )
2t
1 − t2
,y =
2
1+ t
1 + t2
Put t = tan θ
2 tan θ
= sin 2θ and
1 + tan2 θ
1 − tan2 θ
= cos 2θ
1 + tan2 θ
dy dy / dθ − 2sin 2θ
∴
=
=
= − tan 2θ
dx dx / dθ
2cos 2θ
− 2 tan θ
=
1 − tan2 θ
− 2t
2t
=
= 2
1 − t2
t –1
y =
∴
1
sin
2 a
−1
t
−1
asin t ×
dy
1
=
cos −1
dt
2 a
asin
dy
=−
dx
acos
=−
t
−1
−1
a
cos
a
sin
−1
−1
−1
acos t ×
1 − t
1
2
1 − t 2
−1
acos t
× 1
sin −1 t
a
−1
t
t
t
t
−1
2
x + 1
mx + my − mx − nx
x( x + y )
dy
y
=
dx
x
dx
1 + x
x
dy
= n ( x + 1 + x2 )n
dx
d2 y
dy
x
⇒
( 1 + x2 ) +
2
2
dx 1 + x
dx
x
= n2 ( x + 1 + x2 )n − 1 1 +
1
+
x2
x − 1
−1
21 Q y = sec −1
+ sin x + 1
x − 1
18 d ( y ) = n ( x + 1 + x2 )n − 1
1 +
d y
dy
+ x
= n2 y
dx
dx2
dy
y2
acos t
∴ 1 + = 1 +
=1+ 2
−1
sin
t
dx
x
a
x2 + y 2
=
x2
=
∴
1 + x2 )n
2
∴ x=
17 Given that, x m y n = ( x + y )m + n
⇒
1 + x2
d y
dy
+ x
dx
dx2
19 We have, x =
and
Taking log on both sides, we get
m log x + n log y = (m + n )log( x + y )
On differentiating w.r.t. x, we get
dy
m n dy (m + n )
+
=
1 +
x
y dx ( x + y )
dx
1 + x2 )n
2
(1 + x2 )
dt
2
n2 ( x +
= n2 ( x +
20 Q dx =
2
dy
= −1
dx 1, π
13 Let f ( x ) = Ax2 + Bx + C
∴
f (1) = A + B + C
and f (−1) = A − B + C
Q
f (1) = f (−1) [given]
⇒ A+ B+C = A−B+C
⇒
2B = 0 ⇒ B = 0
∴
f ( x ) = Ax2 + C
⇒ f ′( x ) = 2 Ax
∴ f ′(a) = 2 Aa
f ′(b ) = 2 Ab and f ′(c ) = 2 Ac
Also, a, b, c are in AP.
So, 2 Aa, 2 Ab and 2 Ac are in AP.
Hence, f ′(a), f ′(b )and f ′(c )are also in AP.
d2 y
dy
(1 + x2 ) +
⋅x
2
dx
dx
⇒
1 + x2
15 Since, fog = I ⇒ fog ( x ) = x for all x
2
dy
=
dx
f ′ (− x )(−1) = − f ′ ( x )
f ′(− x ) = f ′( x ∴
)
f ′ (−3) = f ′ (3) = −2
2
⇒ ( 1 + x2 )
x − 1
x − 1 π
−1
= cos −1
+ sin x + 1 = 2
x
+
1
dy
π
= 0 Qsin −1 x + cos −1 x =
⇒
dx
2
22 Let y = tan −1 6 x x1
1 − 9x
2 ⋅ (3 x3 /2 )
= tan −1
3 /2 2
1 − (3 x )
−1
3 /2
= 2 tan (3 x )
Q2 tan −1 x = tan −1 2 x
1 − x2
dy
1
3 1 /2
∴
= 2⋅
⋅ 3 × ( x)
dx
2
1 + (3 x3 /2 )2
TEN
124
9
⋅ x
1 + 9 x3
9
∴g ( x ) =
1 + 9 x3
=
27 y =
⇒
−1
23 Given, y = sec (tan x )
Let tan −1 x = θ
⇒
x = tanθ
∴
y = secθ = 1 + x2
On differentiating both sides w.r.t. x, we
get
dy
1
=
⋅ 2x
dx 2 1 + x2
⇒
1 + x2 − 1 − x2
24 Given, y = tan
.
2
2
1 + x + 1 − x
−1
Put x = cos 2θ
2
cos θ − sin θ
y = tan −1
cos θ + sin θ
1 − tan θ
= tan
1 + tan θ
−1
π
= tan −1 tan − θ
4
π
π 1
=
−θ=
− cos −1 x2
4
4 2
dy
x
x
∴
= 0+
=
dx
1 − x4
1 − x4
25 On putting x = sinθ and y = sin φ , we get
Given equation becomes
cos θ + cos φ = a(sin θ − sin φ)
θ + φ
θ − φ
⇒ 2cos
cos
2
2
⇒
⇒
⇒
⇒
θ + φ
θ − φ
= a 2cos
sin
2
2
θ−φ
= cot −1 a
2
θ − φ = 2cot −1 a
sin −1 x − sin −1 y = 2cot −1 a
dy
1
1
−
=0
1 − x2
1 − y 2 dx
dy
=
dx
∴
1 − y2
1 − x2
26 On putting x = sin A and x = sin B
y = sin −1 (sin A 1 − sin2 B
+ sin B 1 − sin2 A )
−1
= sin (sin A cos B + sin B cos A )
= sin −1 [sin( A + B )]
= A + B = sin −1 x + sin −1 x
dy
1
1
⇒
=
+
2
dx
1− x
2 x − x2
−1
x
cos −1 x
1+ a
, z = acos
−1
x
32 Since,
z
y =
1+ z
dy (1 + z )1 − z (1)
=
dz
(1 + z ) 2
1
=
(1 + z ) 2
1
=
cos −1 x 2
(1 + a
)
28 Since g ( x ) is the inverse of f ( x )
∴
⇒
At x = 1
dy
1
=
dx
2
∴
⇒
acos
⇒
g ′′( x ) = 5 (g ( x ))4 ⋅ g ′ ( x ).
= 5 (g ( x ))4 (1 + (g ( x ))5 )
dy
⇒
−1
dx
dy
d2 x
= −
dx
dy 2
−2
d y dx
⋅
dx2 dy
⇒
( f ′( x ))2 − f ′′( x ) ⋅ f ( x )
=0
( f ′( x ))2
⇒
d f ( x)
=0
dx f ′( x )
⇒
f ( x)
= c , (constant)
f ′( x )
⇒
f ′( x )
=2
f ( x)
⇒
d
(log f ( x )) = 2
dx
⇒
log( f ( x )) = 2 x + k
On putting x = 0, we get 0 = k
−3
⇒ log( f ( x )) = 2 x
⇒
log(e / x2 )
30 Given, y = tan
2
log(ex )
−1
x→ 0
log e − log x2
y = tan −1
2
log e + log x
3 + 2 log x
+ tan −1
1 − 6 log x
1 − 2 log x
= tan −1
1 + 2 log x
3 + 2 log
+ tan −1
1 − 6 log
f ( x) = e2 x
Now, lim
3 + 2 log x
+ tan −1
1 − 6 log x
−1
( f ′( x ))2 − f ′′( x ) ⋅ f ( x ) = 0
2
d 2 y dy
= − 2
dx dx
∴
∴
On putting x = 0, we get
1
=c
2
f ( x) 1
⇒
=
f ′( x ) 2
f (g ( x )) = x
f ′(g ( x )) ⋅ g ′( x ) = 1
1
g ′( x ) =
= 1 + (g ( x ))5
f ′(g ( x ))
dy
29 Since, dx =
f ′( x ) f ( x )
=0
f ′′( x ) f ′( x )
x
x
−1
= tan (1) − tan (2 log x )
+ tan −1 (3) + tan −1 (2 log x )
−1
= tan (1) + tan −1 (3)
dy
d2 y
Now,
= 0 and 2 = 0
dx
dx
31 Since, y = f ( x ) is symmetrical about the
Y-axis
∴ f ( x ) is an even function.
Also, as y = g ( x ) is symmetrical about
the origin
∴ g ( x ) is an odd function.
Thus, h( x ) = f ( x ) ⋅ g ( x ) is an odd
function.
or
h( x ) = − h(− x )
Now, h ′( x ) = h ′ (− x )
and h "( x ) = − h ′′(− x )
⇒ h ′′(0) = − h ′′(0)
⇒ h ′′(0) = 0
f ( x) − 1
e2 x − 1
= lim
⋅ 2 = 2.
x
→
0
x
2x
33 f ′′( x )
d2
d2
d2
(3 x2 )
(cos x )
(sin x )
2
2
dx
dx
dx2
=
6
−1
0
P
P2
P3
6
= 6
P
− cos x − sin x
−1
0
P2
P3
6
∴ f ′′(0) = 6
P
−1 0
−1 0 = 0, which is
P2 P3
independent of P.
34 I. Let y = (log x ) log x
On taking log both sides, we get
log y = log ( log x ) log x
⇒ log y = log x log [log x]
[Q log m n = n log m]
On differentiating both sides w.r.t.
x, we get
1 dy
d
= (log x )
{log (log x )}
y dx
dx
d
+ log (log x )
log x
dx
1 1
1
= (log x )
+ log (log x )
log x x
x
1
= {1 + log (log x )}
x
125
DAY
⇒
dy
y
= {1 + log (log x )}
dx
x
(log x ) log x
=
{1 + log (log x )}
x
log(log x )
1
= (log x ) log x +
x
x
II. Let y = cos (a cos x + b sin x ).
On differentiating w.r.t. x, we get
d
{cos(acos x + b sin x}
dx
= − sin(a cos x + b sin x )
d
(acos x + b sin x )
dx
= − sin(acos x + b sin x )
[− asin x + b cos x]
= (asin x − b cos x )
sin (acos x + b sin x )
35 Given, u = f (tan x )
du
= f ′(tan x )sec2 x
dx
and v = g (sec x )
dv
⇒
= g ′(sec x )sec x tan x
dx
du (du / dx ) f ′(tan x ) 1
∴
=
=
⋅
dv (dv / dx ) g ′(sec x ) sin x
f ′(1)
du
=
⋅ 2
∴
dv x = π 4
g ′( 2 )
2
1
= ⋅ 2=
4
2
⇒
SESSION 2
1 We have, f ( x ) = |log 2 − sin x| and
g ( x ) = f ( f ( x )), x ∈ R
Note that, for x→ 0, log 2 > sin x
∴
⇒
f ( x ) = log 2 − sin x
g ( x ) = log 2 − sin( f ( x ))
= log 2 − sin(log 2 − sin x )
Clearly, g ( x ) is differentiable at x = 0 as
sin x is differentiable.
Now,
g ′ ( x ) = − cos (log 2 − sin x ) ( − cos x )
= cos x.cos(log 2 − sin x )
⇒ g ′ (0) = 1 ⋅ cos (log 2)
2 We have,
∴
y = sin x ⋅ sin 2 x ⋅ sin 3 x⋅....sin nx
y ′ = cos x ⋅ sin 2 x ⋅ sin 3 x...⋅ sin nx
+ sin x ⋅ (2cos 2 x ) sin 3x...sin nx
+ sin x ⋅ sin 2 x(3cos 3 x )...sin nx
+ ...+ sin x sin 2 x sin 3 x...(cos nx )
(by product rule)
⇒ y ′ = cot x ⋅ y + 2 ⋅ cot 2 x ⋅ y
+3 ⋅ cot 3 x ⋅ y + ...+ n ⋅ cot nx ⋅ y
⇒ y ′ = y [cot x + 2cot 2 x
+3cot 3 x + ...+ n cot nx]
⇒ y′ = y ⋅
n
∑ k cot kx
k =1
3 3 f ( x ) − 2 f (1 / x ) = x
…(i)
Let 1 / x = y , then
3 f (1 / y ) − 2 f ( y ) = 1 /y
⇒ −2 f ( y ) + 3 f (1 / y ) = 1 /y
⇒ −2 f ( x ) + 3 f (1 / x ) = 1 /x
…(ii)
On multiplying Eq. (i) by 3 and Eq. (ii)
by 2 and adding, we get
2
5 f ( x) = 3 x +
x
1
2
⇒
f ( x ) = 3 x +
5
x
⇒
⇒
1
3 −
5
1
f ′(2) = 3 −
5
f ′( x ) =
2
x2
2 1
=
4 2
When x = 0, we get from Eq. (i),
y ′ = −1
2
−2 − π /2
⇒
y ′′ (0) =
=
e
a
− ae π /2
6 Given, y = |sin x || x |
In the neighbourhood of
π
− , | x |and|sin x | both are negative
6
i.e. y = (− sin x )( − x )
On taking log both sides, we get
log y = (− x ) ⋅ log (− sin x )
On differentiating both sides, we get
1 dy
1
⋅
= (− x )
⋅ (− cos x )
− sin x
y dx
+ log(− sin x ) ⋅ (−1)
= − x cot x − log (− sin x )
= − [ x cot x + log ( − sin x )]
4 f ( x ) = (cos x + i sin x )
(cos 2 x + i sin 2 x )(cos 3 x + i sin 3 x )
… (cos nx + i sin nx )
= cos( x + 2 x + 3 x + ...+ nx ) + i sin
( x + 2 x + 3 x + ...+ nx )
n (n + 1)
n (n + 1)
= cos
x + i sin
x
2
2
n (n + 1)
⇒ f ′( x ) =
2
− sin n (n + 1) x + i cos n (n + 1) x
2
2
2
n (n + 1)
f ′′ ( x ) = −
2
n
(
n
+
1
)
n (n + 1)
cos
x + i sin
x
2
2
= −
2
n (n + 1)
⋅ f ( x)
2
∴ f ′ ′ (1) = −
= −
2
n (n + 1)
f (1)
2
On differentiating both sides w.r.t. x, we
get
1
xy ′− y
1 2 x + 2 yy ′
× 2
=
×
2
2
x + y2
x2
y
1 +
x
…(i)
Again, on differentiating both sides
w.r.t. x, we get
1 + ( y ′ )2 + yy ′′ = xy ′ ′+ y ′− y ′
⇒
1 + ( y ′ )2 = ( x − y )y ′′
1 + ( y ′ )2
y ′′=
x− y
−
x=−
π
6
π
(2) 6
[6 log 2 − 3 π]
=
6
7 Since, f ′( x ) > g ′( x )
1
⇒ 52 x
2
+1
log e 5 × 2 >
5x log e 5 + 4 log e 5
52 x ⋅ 5 > 5 x + 4
5⋅ 52 x – 5 x – 4 > 0
x
(5 – 1) (5⋅ 5 x + 4) > 0
5x > 1
x> 0
d
8 Given,
{ f ′( x )} = − f ( x )
dx
⇒
g ′( x ) = − f ( x )
⇒
⇒
⇒
∴
⇒
= f
On taking log both sides of the given
equation, we get
y
1
log ( x2 + y 2 ) = log a + tan −1
x
2
⇒
dy
∴
dx at
[Q g ( x ) = f ′ ( x ), given]
n (n + 1)
2
x + yy ′ = xy ′− y
dy
= − y [ x cot x + log (− sin x )]
dx
Also, given F ( x )
2
5 When x = 0, y > 0 ⇒ y = ae π /2
⇒
⇒
2
2
x + g x
2
2
⇒ F ′( x ) = 2 f
x f ′ x ⋅ 1
2
2 2
x
x 1
+ 2 g ⋅ g ′ ⋅ = 0
2 2
2
Hence, f ( x ) is constant. Therefore,
F(10) = 5 .
9 Let y = f ( x ), then x = f −1 ( y ).
Now,
Q
∴
d2 x
= ( f −1 )′′ ( y )
d y2
dx dy
=
dy dx
−1
d2 x
d d y
=
dy d x
d y2
=
d
dx
dy
dx
−1
−1
⋅
dx
dy
TEN
126
dy
= −
dx
−2
⋅
d 2 y dx
⋅
dx2 dy
−d 2 y
2
= dx 3
dy
dx
3 /2
2
dy
∴ 1 +
dx
2
dy
⇒ 1 +
dx
⇒ f ′ ( x ) = cos x ⋅ cos (sin x )
⇒ f ′ ′ ( x ) = − sin x ⋅ cos (sin x )
− cos 2 x ⋅ sin(sin x )
Now, g ( x ) = − [ f ′ ′ ( x ) + f ′ ( x ) ⋅ tan x]
= sin x ⋅ cos (sin x ) + cos 2 x ⋅sin(sin x )
− tan x ⋅ cos x ⋅ cos (sin x )
= sin x ⋅ cos (sin x ) + cos2 x ⋅ sin(sin x )
− sin x ⋅ cos (sin x )
= cos 2 x ⋅ sin (sin x )
π
3 cos
3
12 Clearly, f ( x ) = e g ( x )
Now, as g ( x + 1) = x + g ( x )
∴
e g (x + 1 ) = e x + g (x ) = e x ⋅ e g (x )
⇒
f ( x + 1) = e x f ( x )
On taking log both sides, we get
ln f ( x + 1) = ln (e x ⋅ f ( x ))
1
. f ′ ( x + 1)
⇒
f ( x + 1)
=1+
−asin 3 t
cos 2 t
dy
sin t ⋅ sin 2t
and
= a cos t cos 2t −
dt
cos 2t
acos 3 t
cos 2 t
dy dy / dt
=
= − cot3t
dx dx / dt
2
d y
dt
= 3cosec 2 3t ⋅
⇒
dx
dx2
−3cosec2 3t ⋅ cos 2t
=
asin 3t
3
= − cosec3 3t ⋅ cos 2t
a
⇒
1
. f ′( x )
f ( x)
f ′ ( x + 1) f ′ ( x )
−
=1
f ( x + 1)
f ( x)
⇒
11 We have,
cos t ⋅ sin 2t
dx
= a − sin t cos 2t −
dt
cos 2t
=
1
1
f ' 1 + f '
3
3
=1
−
1
1
f 1 +
f
3
3
f ′ 2 +
f 2 +
f ′ n +
f n +
1
f ′ 1 +
3
−
1
f 1 +
3
1
f ′ (n − 1) +
3
−
1
f (n − 1) +
3
1
3
=1
1
3
1
3
=1
1
3
on adding columnwise, we get
1
1
f ′ n +
f ′
3
3
−
=n
1
1
f n +
f
3
3
⇒
2a
3
f (g ( x )) = x ⇒ f ′ (g ( x )) ⋅ g ′ ( x ) = 1
1
g ′( x ) =
f ′ (g ( x ))
−7
g ′ =
6
1
7
f ′g−
6
1
=
−1 − 7
f ′ f
6
Q f (1) = − 4 + 1 + 1 + 1 + 1 = − 7
2 3
6
7
∴ f −1 = 1
6
⇒
d2 y
π
at t = is
6
dx2
a
10 f ( x ) = sin (sin x )
∴
3 /2
−1
=
27
x =2
=
∴
−3
= (1 + cot2 3 t )3 /2 cosec3 3t cos 2t
a
Since, y = 4 when x = 2
d2 y
dx2 x = 2
∴ ( f −1 )′′ (4) = −
3
dy
dx
=
13 Since, g ( x ) = f −1 ( x )
d2 y
dx2
=
1
5
1−x
2 − 1 + 1 + x + x2
Q f ′ ( x ) = − 4e
2
14 We have, f ( x ) = 100 ( x − i )i (101 − i )
11
i =1
100
⇒ log f ( x ) = Σ i (101 − i ) log ( x − i )
i =1
100
1
1
⋅ f '( x ) = ∑ i (101 − i )⋅
f ( x)
x−i
i =1
f ′ (101) 100 (101 − i )
= Σ i
⇒
f (101) i = 1 (101 − i )
100
100(101)
= Σ i=
= 5050
i =1
2
15 Given, x = 2t − |t|and y = t 3 + t 2 |t|
Clearly,x = t , y = 2t 3 when t ≥ 0
and x = 3t , y = 0 when t < 0
On eliminating the parameter t, we get
2 x3 , when x ≥ 0
y =
0, when x < 0
dy 6 x2 , when x > 0
=
dx 0, when x < 0
Q (LHD) at x = 0 = (RHD) at x = 0 = 0
∴ Its derivative at x = 0
(i.e. at t = 0 is 0)
Now,
DAY THIRTEEN
Applications of
Derivatives
Learning & Revision for the Day
u
u
Derivatives as the Rate of
Change
Increasing and Decreasing
Function
u
u
Tangent and Normal to a Curve
Angle of Intersection of Two
Curves
u
u
Rolle’s Theorem
Lagrange’s Mean Value
Theorem
Derivatives as the Rate of Change
dy
is nothing but the rate of change of y, relative to x. If a variable quantity y is some
dx
function of time t i.e. y = f (t ), then small change in time ∆t have a corresponding change ∆y
∆y
in y. Thus, the average rate of change =
∆t
When limit ∆t → 0 is applied, the rate of change becomes instantaneous and we get the rate
of change with respect to at the instant t , i.e.
∆y dy
lim
=
∆t → 0 ∆ t
dt
dy
is positive if y increases as t increase and it is negative if y decrease as t increase.
dt
PRED
MIRROR
Your Personal Preparation Indicator
Increasing and Decreasing Function
●
●
●
●
●
A function f is said to be an increasing function in ]a, b [, if x1 < x2 ⇒ f ( x1 ) ≤ f ( x2 ),
∀ x1 , x2 ∈] a, b [.
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
A function f is said to be a decreasing function in ] a, b [, if x1 < x2 ⇒ f ( x1 ) ≥ f ( x2 ),
∀ x1 , x2 ∈] a, b [.
f ( x) is known as increasing, if f ′ ( x) ≥ 0 and decreasing, if f ′ ( x) ≤ 0.
f ( x) is known as strictly increasing, if f ′ ( x) > 0 and strictly decreasing, if f ′ ( x) < 0.
Let f ( x) be a function that is continuous in [a, b ] and differentiable in (a, b ). Then,
(i) f ( x) is an increasing function in [a, b ], if f ′( x) > 0 in (a, b ).
(ii) f ( x) is strictly increasing function in (a, b ), if f ′( x) > 0 in (a, b ).
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
128
(iii) f ( x) is a decreasing function in [a, b ], if f ′( x) < 0 in (a, b ).
(iv) f ( x) is a strictly decreasing function in [a, b ], if f ′( x) < 0
in (a, b ).
Monotonic Function
A function f is said to be monotonic in an interval, if it is
either increasing or decreasing in that interval.
Results on Monotonic Function
(i) If f ( x) is a strictly increasing function on an interval
[a, b ], then f −1 exists and it is also a strictly increasing
function.
(ii) If f ( x) is strictly increasing function on an interval [a, b ]
such that it is continuous, then f −1 is continuous on
[ f (a), f (b )].
(iii) If f ( x) is continuous on [a, b ] such that f ′ (c) ≥ 0 [ f ′ (c) > 0]
for each c ∈(a, b ), then f ( x) is monotonically increasing
on [a, b ].
(iv) If f ( x) is continuous on [a, b ] such that f ′ (c) ≤ 0 ( f ′ (c) < 0)
for each c ∈[a, b ], then f ( x) is monotonically decreasing
function on [a, b ].
(v) Monotonic function have atmost one root.
(a) The slope of the
tangent is
dy
= tan ψ
dx ( x , y )
1
Y
Tangent
Normal
P (x1, y1)
A
C
B
X
graph desenber tangents
and normal
1
dy
(b) Equation of tangent is y − y1 =
dx ( x
( x − x1 )
1,
PA = y1 cosec ψ = y1
y1 )
1,
2
y1 )
Angle of Intersection of Two Curves
The angle of intersection of two curves is defined to be the angle
between their tangents, to the two curves at their point of
intersection.
The angle between the tangents of the two curves y = f1 ( x) and
y = f2 ( x) is given by
dy
dy
−
dx I ( x 1 , y1 ) dx II ( x 1 , y1 )
m1 − m2
or
tan φ =
dy
dy
1 + m1 m2
1+
dx I ( x , y ) dx II ( x , y )
1
1
1
1
If the angle of intersection of two curves is a right angle, the two
curves are said to intersect orthogonally and the curves are
called orthogonal curves.
π
dy dy
If φ = , m1 m2 = − 1 ⇒ = − 1
dx I dx II
2
y1 )
• Two curves touch each other, if m1 = m2 .
Rolle’s Theorem
Let f be a real-valued function defined in the closed interval
[a, b ], such that
(i) f ( x) is continuous in the closed interval [a, b ].
(ii) f ( x) is differentiable in the open interval (a, b ).
(iii) f (a) = f (b ), then there is some point c in the open interval
(a, b ), such that f ′ (c) = 0.
Geometrically,
Y
Y
2
dy
dx ( x 1 , y1 )
(d) Length of subtangent AC = y1 cot ψ =
1,
dy
(d) Length of subnormal is BC = y1 tan ψ = y1
dx ( x
(c) Length of tangent is
dy
1 +
dx
( x 1 , y1
( x − x1 )
dy
(c) Length of normal is PB = y1 sec ψ = y1 1 +
dx ( x 1 , y1 )
NOTE
ψ
ψ
O
1
dy
dx ( x
Orthogonal Curves
Tangent and Normal to
a Curve
(i) If a tangent is drawn to
the curve y = f ( x) at a
point P( x1 , y1 ) and this
tangent makes an angle
with
positive
ψ
X -direction, then
(b) Equation of normal is y − y1 = −
f(c)
O
y1
(dy / dx)( x 1 , y1 )
(ii) The normal to a curve at a point P( x1 , y1 ) is a line
perpendicular to tangent at P and passing through P,
then
1
(a) The slope of the normal is −
(dy / dx)( x 1 , y1 )
a
c
b
X
O a
b
X
graph of a differentiable function, satisfying the hypothesis of
Rolle’s theorem.
There is atleast one point c between a and b, such that the
tangent to the graph at (c, f (c)) is parallel to the X -axis.
Algebraic Interpretation of Rolle’s Theorem
Between any two roots of a polynomial f ( x), there is always a
root of its derivative f ′ ( x).
129
Lagrange’s Mean Value Theorem
Let f be a real function, continuous on the closed interval [a, b ]
and differentiable in the open interval (a, b ). Then, there is
atleast one point c in the open interval (a, b ), such that
Y
(b, f(b))
(a, f(a))
O a
(c, f(c))
c
f (b ) − f (a)
.
b −a
Geometrically For any chord of the curve y = f ( x), there is a
point on the graph, where the tangent is parallel to this chord.
Remarks In the particular case, where f (a) = f (b ).
f (b ) − f (a)
becomes zero.
The expression
b −a
Thus, when f (a) = f (b ), f ′ (c) = 0 for some c in (a, b ).
Thus, Rolle’s theorem becomes a particular case of the mean
value theorem.
X
b
f ′ (c) =
graph of a continuous functions explain Lagrange’s mean
value theorem.
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 If the volume of a sphere is increasing at a constant rate,
then the rate at which its radius is increasing, is
(a)
(b)
(c)
(d)
a constant
proportional to the radius
inversely proportional to the radius
inversely proportional to the surface area
(a) 10
(c) 8 cm/s
(d) 12 cm/s
3 An object is moving in the clockwise direction around the
unit circle x 2 + y 2 = 1. As it passes through the point
1 3
,
, its y-coordinate is decreasing at the rate of 3
2 2
units per second. The rate at which the x-coordinate
changes at this point is (in unit per second)
(a) 2
(b) 3 3
(c)
3
(d) 2 3
4 The position of a point in time ‘ t ’ is given by
x = a + bt − ct 2 , y = at + bt 2 . Its acceleration at time ‘ t ’ is
(a) b − c
(b) b + c
(c) 2 b − 2 c
(d) 2 b 2 + c 2
5 Water is dripping out from a conical funnel of
π
semi-vertical angle at the uniform rate of 2 cm 2 /s in the
4
surface area, through a tiny hole at the vertex of the
bottom. When the slant height of cone is 4 cm, the rate of
decrease of the slant height of water, is j NCERT Exemplar
2
cm/s
4π
1
(c)
cm/s
π 2
(a)
(b)
(b)
10
(c) 100
JEE Mains 2013
(d) 10 10
7 Oil is leaking at the rate of 16 cm 3 /s from a vertically kept
x = 10 + 6 t , x = 3 + t 2 . The speed with which they are
reaching from each other at the time of encounter is (x is
in centimetre and t is in seconds)
(b) 20 cm/s
35 cc/min. The rate of increase in the surface area
(in cm 2 /min) of the balloon when its diameter is 14 cm, is
j
2 Moving along the X -axis there are two points with
(a) 16 cm/s
6 A spherical balloon is being inflated at the rate of
1
cm/s
4π
(d) None of these
cylindrical drum containing oil. If the radius of the drum is
7 cm and its height is 60 cm. Then, the rate at which the
level of the oil is changing when oil level is 18 cm, is
(a)
−16
49π
(b)
−16
48 π
(c)
16
49π
(d)
−16
47 π
8 Two men A and B start with velocities v at the same time
from the junction of two roads inclined at 45° to each
other. If they travel by different roads, the rate at which
j NCERT Exemplar
they are being separated.
(a)
2−
(c)
2 ⋅v
(b)
2+
2 ⋅v
2 − 1⋅ v
(d)
2+
2 ⋅v
9 The interval in which the function f ( x ) = x 1 / x is
increasing, is
(a) (−∞, e)
(c) (−∞, ∞)
(b) (e, ∞)
(d) None of these
10 The function f ( x ) =
(a)
(b)
(c)
(d)
x
is
1+ | x |
strictly increasing
strictly decreasing
neither increasing nor decreasing
not differential at x = 0
11 The length of the longest interval, in which the function
3 sin x − 4 sin3 x is increasing, is
(a)
π
3
(b)
π
2
(c)
3π
2
(d) π
130
π
2
12 An angle θ, 0 < θ < , which increases twice as fast as its
sine, is
j
π
(a)
2
3π
(b)
2
π
(c)
4
NCERT Exemplar
π
(d)
3
13 The value of x for which the polynomial
2 x 3 − 9x 2 + 12 x + 4 is a decreasing function of x, is
(a) −1 < x < 1
(c) x > 3
14 If f ( x ) =
(a)
(b)
(c)
(d)
(b) 0 < x < 2
(d) 1 < x < 2
(a) a + b = 0 (b) a > 0
point having abscissa 2 is
(a) 5 2
(a) m ≤1
(b) 3 3
(b) m > − 1
(b) a = φ
(d) a = −2 ± 2 3
is parallel to the X-axis, is
4
, that
x2
AIEEE 2010
(d) y = 3
a −b
1 + ab
a+b
(c)
1 − ab
3
(d) 3 2
x = 3 t 2 + 1, y = t 3 − 1, at x = 1 is
22 Coordinates of a point on the curve y = x log x at which
the normal is parallel to the line 2 x − 2 y = 3, are
(d) m >1
x
loga − logb
1 + loga logb
loga + logb
(d)
1 − loga logb
(b)
at right angles, then the value of b is
(a) 6
(c) 4
(d) − 2
(c) m >1
32 If the curves y 2 = 6x , 9x 2 + by 2 = 16 intersect each other
21 The slope of the tangent to the curve
(b) (e,e)
(d) (e − 2 , − 2 e − 2 )
(c)
31 If the curves y = a and y = b intersects at angle α , then
x
(a)
j
x
at the
1− x2
30 If m is the slope of a tangent to the curve e = 1 + x 2 , then
20 The equation of the tangent to the curve y = x +
(a) (0, 0)
(c) (e 2 , 2 e 2 )
(c) a < 0, b < 0 (d) a = − 2b
29 The length of subnormal to the curve y =
tan α is equal to
(c) ∞
(d) (−6, − 7)
y
ax
the curve y =
, then
1+ x
(a) 0
(c) (6, − 7)
(d) λ > 4
19 Line joining the points (0, 3) and (5, −2) is a tangent to
1
(b)
2
(b) (−6, 7)
then
y = a , is
(c) y = 2
(b) square of the ordinate
(d) None of these
28 If the line ax + by + c = 0 is normal to curve xy + 5 = 0,
(b) 2a
(d) None of these
(b) y = 1
(d) 4 , 2
2
1
3
, b = − and c = 3
2
4
1
3
(b) a = , b = − and c = − 3
2
4
1
1
(c) a = , b = − and c = 3
2
4
(d) None of the above
(a) (6, 7)
18 The sum of intercepts on coordinate axes made by
(a) y = 0
(c) −4 ,14
(b) 4 ,14
circle x 2 + y 2 + 16x + 12y + c = 0 at
λ sin x + 6 cos x
is monotonic increasing, if
2 sin x + 3 cos x
(a) a = 1 ± 3
(c) a = −1 ± 3
(a) 4 ,− 14
27 The tangent at (1, 7) to the curve x 2 = y − 6 touches the
f (x) is decreasing in (−∞, 1)
f (x) is decreasing in (−∞, 2)
f (x) is increasing in (−∞, 1)
f (x) is increasing in (−∞, 2)
(a) a
(c) 2 a
of the form ax − 5y + b = 0 , then a and b are
(a) square of the abscissa
(c) constant
16 If f ( x ) = − 2x 3 + 21x 2 − 60x + 41, then
tangent to the curve x +
24 If the normal to the curve y 2 = 5x − 1 at the point (1,−2) is
at any point of a curve is
(d) None of these
(c) λ < 4
(d) (0, 0)
26 The product of the lengths of subtangent and subnormal
π π
(b) ,
4 2
(b) λ < 1
(c) (2, 0)
(a) a = −
decreasing in 0 ≤ x ≤ 2 π, is
(a) λ > 1
1
(b) − , 0
2
P( −2, 0) and cuts the Y -axis at a point Q, where its
gradient is 3. Then,
an increasing function
a decreasing function
both increasing and decreasing function
None of the above
17 Function f ( x ) =
1
(a) , 0
2
25 The curve y = ax + bx + cx + 5 touches the X -axis at
15 If f ( x ) = sin x − cos x , the interval in which function is
(a)
(b)
(c)
(d)
y = e 2 x , meets X -axis at the point
3
1
− log(1 + x ), x > 0, then f is
x +1
5π 3π
(a) ,
6 4
3π 5π
(c)
,
2 2
23 The tangent drawn at the point (0, 1) on the curve
7
2
9
(d)
2
(b)
33 Angle between the tangents to the curve y = x 2 − 5x + 6
at the points (2, 0) and (3, 0) is
π
2
π
(c)
4
(a)
π
6
π
(d)
3
(b)
131
x2
y2
+
= 1 and y 3 = 16x intersect at right
α
4
j JEE Mains 2013
angles, then the value of α is
34 If the curves
(a) 2
(b)
4
3
(c)
1
2
(d)
3
4
2
and f ( x ) is continuous in [1, 2], then ∫ f ′ ( x ) dx is equal to
1
(b) 0
(c) 1
(d) 2
36 If the function f ( x ) = x 3 − 6x 2 + ax + b satisfies Rolle’s
2 3 + 1
theorem in the interval [1, 3] and f ′
= 0, then
3
(a) a = −11
(b) a = − 6
(c) a = 6
theorem holds for the function f ( x ) = loge x on the interval
j AIEEE 2007
[1, 3] is
(d) a = 11
5
then ∫ f ( x ) dx is equal to
39 The abscissa of the points of the curve y = x 3 in the
interval [ −2, 2], where the slope of the tangents can be
obtained by mean value theorem for the interval [ −2, 2],
are
(a) ±
2
3
(b) +
(c) ±
3
2
(d) 0
3
40 In the mean value theorem, f (b ) − f (a ) = (b − a )f ′ (c ), if
3
(b) −1
(b)
(c) log3 e
37 If f ( x ) satisfies the conditions for Rolle’s theorem in [3, 5],
(a) 2
1
loge 3
2
(d) loge 3
(a) 2 log3 e
35 f ( x ) satisfies the conditions of Rolle’s theorem in [1, 2]
(a) 3
38 A value of C for which the conclusion of mean value
(d) −
(c) 0
4
3
a = 4 , b = 9 and f ( x ) = x , then the value of c is
(a) 8.00
(b) 5.25
(c) 4.00
(d) 6.25
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 A kite is moving horizontally at a height of 151.5 m. If the
speed of kite is 10 m/s, how fast is the string being let out,
when the kite is 250 m away from the boy who is flying the
j NCERT Exemplar
kite? The height of boy is 1.5 m.
(a) 8 m/s
(b) 12 m/s
(c) 16 m/s
(d) 19 m/s
2 The normal to the curve y ( x − 2) ( x − 3) = x + 6 at the
point, where the curve intersects the Y -axis passes
through the point
1
1
(a) − , −
2
2
1
1
(c) , −
2
3
1
(b) ,
2
1
(d) ,
2
1
2
1
3
3 The values of a for which the function
(a + 2)x 3 − 3 ax 2 + 9 ax − 1 = 0 decreases monotonically
throughout for all real x, are
(a) a < − 2
(c) −3 < a < 0
(b) a > − 2
(d) − ∞ < a ≤ − 3
x
x
and g ( x ) =
, where 0 < x ≤ 1, then in
sin x
tan x
this interval
4 If f ( x ) =
(a)
(b)
(c)
(d)
both f (x) and g (x) are increasing functions
both f (x) and g (x) are decreasing functions
f (x) is an increasing function
g (x) is an increasing function
x
5 f ( x ) = ∫ | log2 [log3 {log4 (cos t + a )}]| dt . If f ( x )
0
is increasing for all real values of x, then
(a) a ∈ (−11
,)
(c) a ∈ (1, ∞)
(b) a ∈ (15
, )
(d) a ∈ (5 , ∞)
6 If f ( x ) satisfy all the conditions of mean value theorem in
[ 0 , 2 ]. If f ( 0) = 0 and | f ′ ( x )| ≤
(a) f (x) < 2
(c) f (x) = 2 x
1
for all x in [ 0, 2 ], then
2
(b) | f (x)| ≤ 1
(d) f (x) = 3 for atleast one x in [0, 2]
π
2
and g ( x ) = f (sin x ) + f (cos x ), then g ( x ) is decreasing in
7 If f ′ (sin x ) < 0 and f ′ ′ (sin x ) > 0 , ∀x ∈ 0 ,
π π
(a) ,
4 2
π
(b) 0,
4
π
(c) 0,
2
π π
(d) ,
6 2
8 If f ( x ) = ( x − p )( x −q )( x −r ), where p < q < r , are real
numbers, then application of Rolle’s theorem on f leads to
(a) (p + q + r )(pq + qr + rp) = 3
(b) (p + q + r )2 = 3 (pq + qr + rp)
(c) (p + q + r )2 > 3 (pq + qr + rp)
(d) (p + q + r )2 < 3 (pq + qr + rp)
9 If f ( x ) is a monotonic polynomial of 2m −1 degree, where
m ∈ N, then the equation
[f ( x ) + f ( 3x ) + f ( 5x ) + ...+ f ( 2m − 1)x ] = 2m − 1 has
(a) atleast one real root
(c) exactly one real root
(b) 2m roots
(d) (2m + 1) roots
10 A spherical balloon is filled with 4500 π cu m of helium
gas. If a leak in the balloon causes the gas to escape at
the rate of 72π cu m/min, then the rate (in m/min) at
which the radius of the balloon decreases 49 min after
j AIEEE 2012
the leakage began is
(a)
9
7
(b)
7
9
(c)
2
9
(d)
9
2
132
11 The normal to the curve x 2 + 2xy − 3y 2 = 0 at (1, 1)
(a)
(b)
(c)
(d)
j JEE Mains 2015
(a) does not meet the curve again
(b) meets the curve again in the second quadrant
(c) meets the curve again in the third quadrant
(d) meets the curve again in the fourth quadrant
15 The angle of intersection of curves,
y = [| sin x | + | cos x | ] and x 2 + y 2 = 5, where [⋅]
denotes greatest integral function is
12 If f and g are differentiable functions in ( 0, 1) satisfying
f ( 0) = 2 = g (1), g ( 0) = 0 and f (1) = 6, then for some
j JEE Mains 2014
c ∈] 0, 1[
(a) 2f ′ (c) = g ′ (c)
(c) f ′ (c) = g ′ (c)
(a)
(b) 2f ′ (c) = 3g ′ (c)
(d) f ′ (c) = 2g ′ (c)
1
1 − x,
x<
2
2
(a) f (x) =
(b) f (x) =
2
1
1
− x , x ≥
2
2
increasing function of x. Then,
0
∫
b
0
(d) None of these
applicable to
(b) f ′ (1) = 1
(d) 2f (0) = 1 + f ′ (0)
14 Let a + b = 4, a < 2 and g ( x ) be a monotonically
a
1
(b) tan−1
2
16 In [0, 1], Lagrange’s mean value theorem is not
by the line y = x at the point where x = 1, then
f ( x ) = ∫ g ( x ) dx +
π
4
(c) tan−1 (2)
13 If y = f ( x ) is the equation of a parabola which is touched
(a) 2f ′ (0) = 3f ′ (1)
(c) f (0) + f ′ (1) + f ′ ′ (1) = 2
increases with increase in (b − a)
decreases with increase in (b − a)
increases with decreases in (b − a)
None of the above
(c) f (x) = x | x |
g ( x ) dx
sin x , x ≠ 0
x
1,
x=0
(d) f (x) = | x |
ANSWERS
SESSION 1
SESSION 2
1 (d)
2 (c)
3 (b)
4 (d)
5 (a)
6 (a)
7 (c)
8 (a)
9 (a)
10 (a)
11 (a)
12 (d)
13 (d)
14 (b)
15 (d)
16 (b)
17 (d)
18 (a)
19 (b)
20 (d)
21 (a)
22 (d)
23 (b)
24 (a)
25 (a)
26 (b)
27 (d)
28 (c)
29 (d)
30 (a)
31 (b)
32 (d)
33 (a)
34 (b)
35 (b)
36 (d)
37 (d)
38 (a)
39 (a)
40 (d)
1 (a)
11 (d)
2 (b)
12 (d)
3 (d)
13 (b)
4 (c)
14 (a)
5 (d)
15 (c)
6 (b)
16 (a)
7 (b)
8 (c)
9 (a)
10 (c)
Hints and Explanations
SESSION 1
1 Given that, dV = k (say)
dt
4 3
dV
dR
= 4 πR 2
Q V = πR ⇒
3
dt
dt
dR
k
=
⇒
dt
4 πR 2
Rate of increasing radius is inversely
proportional to its surface area.
2 They will encounter, if
10 + 6t = 3 + t 2
2
⇒
t − 6t − 7 = 0 ⇒ t = 7
At t = 7 s, moving in a first point
d
v 1 = (10 + 6t ) = 6 cm/s
dt
At t = 7 s, moving in a second point
d
v 2 = (3 + t 2 ) = 2t = 2 × 7 = 14 cm/s
dt
∴ Resultant velocity
= v 2 − v 1 = 14 − 6 = 8 cm/s
3 The equation of given circle is
x + y =1
On differentiating w.r.t. t, we get
dy
dx
2x
+ 2y
=0
dt
dt
dy
dx
⇒
x
+ y
=0
dt
dt
2
2
1
3
But we have, x = , y =
2
2
dy
1 dx
3
= − 3, then
+
(− 3) = 0
dt
2 dt
2
dx
=3 3
⇒
dt
and
4 Given point is x = a + bt − ct 2
Acceleration in x direction and point
d2 x
is y = at + bt 2 = 2 = − 2c
dt
and point is y = at + bt 2 acceleration
in y direction
d2 y
= 2 = 2b
dt
∴ Resultant acceleration
2
=
2
2
d y
d x
2 + 2
dt
dt
2
= (−2c )2 + (2b )2 = 2 b 2 + c 2
5 If S represents the surface area, then
dS
= 2 cm2 / s
dt
r
h
l
π/4
133
S = πrl = πl ⋅ sin
π
π 2
l =
l
4
2
Therefore,
dS 2 π
dl
dl
=
l⋅
= 2 πl ⋅
dt
dt
dt
2
dl
2
=
dt
2π ⋅ 4
1
2
=
=
cm / s
2 2π 4π
when l = 4 cm,
From ∆LOM,
OL2 + OM 2 − LM 2
cos 45° =
2 ⋅ OL ⋅ OM
x2 + x2 − y 2 2 x2 − y 2
1
=
=
⇒
2 x2
2
2⋅ x ⋅ x
2
2
2
⇒
2x = 2x − y
⇒
(2 − 2 ) x2 = y 2
∴
6 V = 4 πr 3 ⇒ dV = 4 π3r 2 dr
3
dt 3
dt
dr
dr
5
⇒
=
dt
dt 28 π
Surface area of balloon, S = 4 πr 2
dS
dr
∴
= 8πr
dt
dt
5
= 8π ×7×
= 10 cm2 / min
28 π
⇒ 35 = 4 π (7)2
7 Let h be height of oil level at any instant
t and V be the volume of oil in
cylindrical drum.
Given, h = 60 cm, r = 7 cm
dV
and
= − 16 cm3 /s
dt
H
h
So, height of oil is decreasing at the rate
16
of
cm/s.
49π
8 Let L and M be the positions of two men
A and B at any time t.
Let OL = x and LM = y
Then, OM = x
dy
dx
Given,
= v and we have to find
dt
dt
M
y
x
O
A
x
L
For decreasing, f ′ ( x ) < 0
π
π
3π
< x − <
2
4
2
(within 0 ≤ x ≤ 2 π)
π
π
π
π
3π
π
< x −
+ <
+
⇒ +
2
4
4
4
2
4
3π
7π
⇒
< x<
4
4
9 Given, f ( x ) = x1 / x
1
(1 − log x ) x1 / x
x2
f ′ ( x ) > 0, if 1 − log x > 0
⇒ log x < 1 ⇒ x < e
∴ f ( x ) is increasing in the interval
(− ∞, e ).
x
10 Given,
f ( x) =
1 + | x|
| x|
(1 + | x |) ⋅ 1 − x ⋅
x
∴ f ′( x ) =
(1 + | x |)2
1
=
> 0, ∀ x ∈ R
(1 + | x |)2
f ′( x ) =
Since, sin x is increasing in the interval
− π , π ⋅
2 2
π
π
π
π
≤ 3x ≤
⇒ − ≤ x≤
2
2
6
6
Thus, length of interval
π π
π
=
− − =
6 6
3
∴
−
12 Given, 2 d (sinθ) = dθ
dt
dt
dθ dθ
=
⇒ 2 × cos θ
dt
dt
⇒ 2cos θ = 1 ⇒ cos θ =
16 Given,
…(i)
f ( x ) = − 2 x 3 + 21 x2 − 60 x + 41
On differentiating Eq. (i) w.r.t. x, we get
f ′ ( x ) = − 6 x2 + 42 x − 60
= − 6 ( x2 − 7 x + 10 )
= − 6 ( x − 2) ( x − 5)
If x < 2, f ′ ( x ) < 0 i.e. f ( x ) is decreasing.
17 Q f ( x ) = λ sin x + 6cos x
…(i)
2sin x + 3cos x
On differentiating w.r.t. x, we get
(2sin x + 3cos x )
(λ cos x − 6sin x )
− (λ sin x + 6cos x )
(2cos x − 3sin x )
f ′( x ) =
(2sin x + 3cos x )2
The function is monotonic
increasing, if f ′ ( x ) > 0
⇒
⇒ f ′ ( x ) = 6 x2 − 18 x + 12
f ′ ( x ) < 0 for function to be decreasing
⇒ 6( x2 − 3 x + 2) < 0
⇒ ( x2 − 2 x − x + 2) < 0
( x − 2)( x − 1) < 0 ⇒ 1 < x < 2
1
− log(1 + x )
x+1
On differentiating w.r.t. x, we get
3λ (sin2 x + cos 2 x ) − 12
(sin2 x + cos 2 x )> 0
⇒ 3λ − 12 > 0
1
π
⇒θ =
2
3
13 Let f ( x ) = 2 x3 − 9 x2 + 12 x + 4
⇒
45°
On differentiating w.r.t. x, we get
f ′ ( x ) = cos x + sin x
1
1
= 2
cos x +
sin x
2
2
π
π
= 2 cos cos x + sin sin x
4
4
π
= 2 cos x −
4
Hence, they are being separated from
each other at the rate 2 − 2 v.
14 Given curve is f ( x ) =
B
15 Q f ( x ) = sin x − cos x
11 Let f ( x ) = 3 sin x − 4 sin3 x = sin 3 x
dV
dh
Q
V = πr 2 h ⇒
= πr 2
dt
dt
(since, r is constant all the time)
dh
dh
16
⇒
=−
⇒ −16 = π(7)2
dt
dt
49π
dh
16
⇒
=−
dt at h =18
49π
⇒ f ′ ( x ) = − ve, when x > 0
∴ f ( x ) is a decreasing function.
On differentiating w.r.t. t, we get
dy
dx
= 2− 2
dt
dt
Q dx = v
= 2 − 2v
dt
⇒
1
1
−
( x + 1)2 1 + x
1
1
⇒ f ′( x ) = −
+
x + 1 ( x + 1)2
y = 2 − 2x
⇒ f ( x ) is strictly increasing.
r
f ′( x ) = −
⇒
18
x +
(Q sin2 x + cos 2 x = 1)
λ> 4
y =
a ; (i )
1
2
x
1
+
2
y
dy
=−
∴
dx
x
Hence, tangent at ( x, y ) is
y
Y − y =−
( X − x)
x
⇒ X y + Y x = xy ( x +
⇒
X y +Y x =
dy
=0
y dx
y)
xy a
(using Eq. (i))
134
⇒
X
+
a x
Y
=1
a y
dy
dy
= 2e 2 x ⇒
= 2e 0 = 2
dx
dx ( 0, 1 )
Clearly, its intercepts on the axes are
a ⋅ x and a ⋅ y .
Sum of intercepts =
=
a( x +
a⋅ a = a
and (5, −2) is y = 3 − x. If this line is
ax
tangent to y =
, then
( x + 1)
(3 − x ) ( x + 1) = ax should have equal
roots.
Thus, (a − 2)2 + 12 = 0 ⇒ no value of a
⇒ a ∈ φ.
20 We have, y = x + 42
x
On differentiating w.r.t. x, we get
dy
8
= 1− 3
dx
x
Since, the tangent is parallel to X -axis,
therefore
dy
= 0 ⇒ x3 = 8
dx
∴
x = 2 and y = 3
21 Given curve is x = 3t 2 + 1, y = t 3 − 1
For x = 1, 3t 2 + 1 = 1 ⇒ t = 0
dy
dx
∴
= 6t ,
= 3t 2
dt
dt
dy
dy dt 3t 2 t
Now,
=
=
=
dx dx 6t
2
dt
0
dy
= =0
dx ( t = 0 ) 2
22 Given curve is y = x log x
On differentiating w.r.t. x, we get
dy
= 1 + log x
dx
The slope of the normal
−1
1
=−
=
(dy /dx ) 1 + log x
The slope of the given line 2 x − 2 y = 3 is
1.
Since, these lines are parallel.
−1
∴
=1
1 + log x
⇒
⇒
and
y − 1 = 2( x − 0) ⇒ y = 2 x + 1
y)
19 Equation of line joining the points (0, 3)
∴
Equation of tangent at (0, 1) with slope
2 is
log x = − 2
x = e −2
y = − 2e − 2
∴ Coordinates of the point are
(e − 2 , − 2e − 2 ).
23 Given curve is y = e 2 x
On differentiating w.r.t. x, we get
This tangent meets X -axis.
∴
y = 0 ⇒ 0 = 2x + 1 ⇒ x = −
1
2
∴ Coordinates of the point on X -axis is
− 1 , 0 .
2
24 We have, y 2 = 5x − 1
At (1, − 2),
…(i)
dy 5
−5
=
=
dx 2 y (1 , −2 ) 4
∴ Equation of normal at the point (1, − 2)
is
−5
[ y − (− 2)] + x − 1 = 0
4
∴
4 x − 5y − 14 = 0
As the normal is of the form
ax − 5y + b = 0
…(ii)
On comparing this with Eq. (ii), we get
a = 4 and b = − 14
25 Since, we have the curve
y = ax3 + bx2 + cx + 5 touches X -axis at
P(−2, 0), then X -axis is the tangent at
(−2, 0). The curve meets Y-axis in (0, 5).
dy
= 3ax2 + 2bx + c
dx
dy
(given)
⇒ = 0 + 0 + c = 3
dx 0, 5
⇒
c =3
…(i)
dy
and
=0
dx ( −2, 0 )
[from Eq. (i)]
⇒ 12a − 4b + c = 0
…(ii)
⇒ 12a − 4b + 3 = 0
and ( −2, 0) lies on the curve, then
0 = − 8a + 4b − 2c + 5
(Q c = 3)
⇒ 0 = − 8a + 4b − 1
…(iii)
⇒ 8a − 4b + 1 = 0
From Eqs. (ii) and (iii), we get
1
3
a = − ,b = −
2
4
26 Length of subtangent = y dx
dy
dy
and length of subnormal = y
dx
∴
Product = y 2
⇒ Required product is the square of
the ordinate.
27 The tangent to the parabola
x2 = y − 6 at (1, 7) is
1
x(1) = ( y + 7) − 6 ⇒ y = 2 x + 5
2
which is also a tangent to the given
circle.
i.e. x2 + (2 x + 5)2 + 16 x
+ 12 (2 x + 5) + c = 0
⇒ (5x2 + 60 x + 85 + c = 0) must have
equal roots.
Let the roots be α = β.
60
α +β=−
∴
5
⇒
α =−6
∴
x = − 6 and
y = 2x + 5 = − 7
28 Given curve, is xy = − 5 < 0
dy
dy
−y
+ y = 0⇒
=
>0
dx
dx
x
[as xy = − 5 < 0]
−1
x
Slope of normal =
= <0
dy
y
dx
[as xy = −5 < 0]
Hence, slope of normal will be negative.
The line ax + by + c = 0
⇒
by = − ax − c
−a
c
⇒
y =
x−
b
b
−a
is negative.
Slope of normal
b
−a
a
⇒
< 0⇒ > 0
b
b
⇒ a > 0, b > 0 or a < 0, b < 0
29 Given, y = x 2
1− x
⇒
At
x
x = 2, y = − 2, therefore
point is ( 2, − 2 ).
dy
1 + x2
=
dx (1 − x2 )2
∴
dy
⇒
dx (
=
2, − 2 )
1+ 2
=3
(1 − 2)2
∴ length of subnormal at ( 2, − 2 )
= |(− 2 )(3)| = 3 2
30 We have, e y = 1 + x2
⇒
⇒
⇒
ey
dy
=2x
dx
dy
2x
=
dx 1 + x2
2x
,
1 + x2
2| x|
2| x|
|m|=
=
≤1
|1 + x2 | 1 + | x|2
m=
Q
| x|2 + 1 − 2| x|≥ 0 ≥
⇒
(| x|− 1 )2 ≥ 0
⇒
⇒
| x|2 + 1 ≥ 2| x|
2 | x|
1≥
1 + | x|2
135
Now, apply the condition of
perpendicularity of two curves,
i.e. m1 m2 = − 1
4
and get α = with the help of equation
3
of curves.
∴
⇒ 3 2 +
Also,
⇒
⇒
3
y1
9 x + by = 16
dy
18 x + 2by
=0
dx
dy
−9 x
=
dx
by
2
−9 x1
Slope of tangent at ( x1 , y 1 ) is m2 =
by 1
Since, these are intersection at right
angle.
27 x1
∴ m1 m2 = − 1 ⇒
=1
by 12
27 x1
[Q y 12 = 6 x1 ]
⇒
=1
6bx1
9
⇒
b =
2
33 Q
∴
Now,
and
Now,
⇒
1 Let AB be the position of boy who is
flying the kite and C be the position of
the kite at any time t.
C
y
2
a = 11
37 Since, f ( x ) satisfies all the conditions of
Rolle’s theorem in [3, 5].
Let f ( x ) = ( x − 3)( x − 5) = x2 − 8 x + 15
∫
Now,
5
3
f ( x )dx =
∫
5
3
( x2 − 8 x + 15)dx
5
=
50
4
− 18 = −
3
3
38 Using mean value theorem,
f (3) − f (1)
3−1
⇒
1 log e 3 − log e 1
=
c
2
dy
m2 =
= 6 − 5= 1
dx (3, 0 )
∴
c =
m1 m2 = − 1 × 1 = − 1
2
2
34 Slope of the curve, x + y = 1 is
α
4
f (b ) − f (a)
Q f ′ (c ) =
b − a
2
= 2 log 3 e
log e 3
39 Given that, equation of curve
y = x = f ( x)
So, f (2) = 8 and f (− 2) = − 8
f (2) − f (– 2)
Now, f ′( x ) = 3 x2 ⇒ f ′( x ) =
2 − (− 2)
3
⇒
Now, slope of the curve, y = 16 x is
16
.
m2 =
3 y2
1.5 m
3
∴
8 − (− 8)
= 3 x2
4
x= ±
B
D
x
Let BD = x and AC = y , then AE = x
Given, AB = 1.5 m, CD = 151.5 m
∴
CE = 150 m
dx
Given,
= 10 m/s
dt
dy
when
dt
y = 250 m
125
=
− 100 + 75 − (9 − 36 + 45)
3
f ′ (c ) =
E
A
Here, we have to find
3
8 x2
x
−
+ 15x
2
3
3
=
1
2
x
f (b ) − f (a) 3 − 2 1
Also, f ′ (c ) =
=
=
b −a
9− 4 5
1
1
25
= ⇒C =
= 6.25
∴
5
4
2 c
1
1
− 12 2 +
+ a= 0
3
3
y = x − 5x + 6
dy
= 2x − 5
dx
dy
m1 =
= 4 − 5= −1
dx (2 , 0 )
−4 x
.
αy
9 = 3; f ′ ( x ) =
f (b ) =
⇒ 12 + 1 + 4 3 − 24 − 4 3 + a = 0
2
π
Hence, angle between the tangents is .
2
m1 =
4=2
1
4
1
⇒ 3 4 + +
− 12 2 +
+ a= 0
3
3
3
Slope of tangent at ( x1 , y 1 ) is m1 =
2
2
On differentiating w.r.t. x, we get
f ′ ( x ) = 3 x2 − 12 x + a
By the definition of Rolle’s theorem
1
f ′ (c ) = 0 ⇒ f ′ 2 +
=0
3
dy
=6
dx
dy
3
=
dx
y
∴ f (a) =
SESSION 2
f (2) = f (1)
36 Q f ( x )= x − 6 x + ax + b
32 We have, y 2 = 6 x
⇒
f ′ ( x )dx = [ f ( x )]21 = f (2) − f (1) = 0
3
m1 − m2
1 + m1 m2
log a − log b
=
1 + log a log b
2y
2
1
[Q f ( x ) satisfies the
conditions of Rolle’s theorem]
∴ tan α =
⇒
∫
x
151.5 m
Slope of tangent of second curve,
dy
m2 =
= b x log b
dx
dy
⇒
m2 =
= log b
dx ( 0 , 1 )
35
f ( x) =
40
150 m
curves is (0, 1).
Now, slope of tangent of first curve,
dy
m1 =
= ax log a
dx
dy
⇒
= m1 = log a
dx ( 0 , 1 )
1.5 m
31 Clearly, the point of intersection of
2
3
Now, from ∆CAE, y 2 = x2 + 1502
On differentiating, we get
dy
dx
2y
= 2x
dt
dt
⇒
dy
x dx
x
= ⋅
= ⋅ 10
dt
y dt
y
In ∆ACE, x = 2502 − 1502
…(i)
(Q y = 250)
= 200 m
∴ From Eq. (i), we get
dy 200
=
× 10 = 8 m/s
dt
250
2 Given curve is
y ( x − 2)( x − 3) = x + 6
…(i)
Put x = 0 in Eq. (i), we get
y(− 2) (− 3) = 6 ⇒ y = 1
So, point of intersection is (0, 1).
x+ 6
Now, y =
( x − 2)( x − 3)
1 ( x − 2)( x − 3) − ( x + 6)
dy
( x − 3 + x − 2)
⇒
=
dx
( x − 2)2 ( x − 3)2
136
dy
6 + 30 36
⇒
=
=
=1
dx ( 0, 1 )
4 × 9 36
∴ Equation of normal at (0, 1) is given by
−1
y −1=
( x − 0) ⇒ x + y − 1 = 0
1
1 1
which passes through the point , .
2 2
3 Let f ( x ) = (a + 2) x3 − 3 ax2 + 9 ax − 1
decreases
monotonically for all x ∈ R, then
f ′ ( x ) ≤ 0 for all x ∈ R
⇒ 3(a + 2) x2 − 6 ax + 9 a ≤ 0
for all x ∈ R
⇒ (a + 2) x2 − 2 ax + 3 a ≤ 0
for all x ∈ R
⇒ a + 2 < 0 and discriminant ≤ 0
⇒ a < − 2, − 8 a2 − 24a ≤ 0
⇒ a < − 2 and a (a + 3) ≥ 0
⇒ a < − 2, a ≤ − 3 or a ≥ 0 ⇒ a ≤ − 3
∴ − ∞ < a≤ − 3
sin x − x cos x
4 Now, f ′( x ) =
sin2 x
cos x (tan x − x )
=
sin2 x
∴
f ′ ( x ) > 0 for 0 < x ≤ 1
So, f ( x ) is an increasing function.
tan x − x sec2 x
Now, g ′ ( x ) =
tan2 x
sin x cos x − x sin 2 x − 2 x
=
=
sin2 x
2sin2 x
d
Again, (sin 2 x − 2 x ) = 2 cos 2 x − 2
dx
= 2 (cos 2 x − 1) < 0
So, sin 2 x − 2 x is decreasing.
sin 2 x − 2 x < 0
⇒
∴
g ′ ( x) < 0
So, g ( x ) is decreasing.
5 f ′( x ) = | log 2 [log 3 {log 4 (cos x + a)}] |
Clearly, f ( x ) is increasing for all values
of x, if
log 2 [log 3 {log 4 (cos x + a)}] is defined
for all values of x.
⇒ log 3 [log 4 (cos x + a)] > 0, ∀ x ∈ R
log 4 (cos x + a) > 1, ∀ x ∈ R
⇒
cos x + a > 4, ∀ x ∈ R
⇒
∴
a> 5
f (2) − f (0)
6 Since,
= f ′( x )
2− 0
f (2) − 0
df ( x ) f (2)
⇒
= f ′( x ) ⇒
=
2
dx
2
f (2)
⇒
f ( x) =
x+C
2
Q f (0) = 0 ⇒ C = 0
f (2)
…(i)
∴
f ( x) =
x
2
f (2) 1
1
Also, | f ′ ( x )|≤ ⇒
…(ii)
≤
2
2
2
From Eq. (i), | f ( x )|
f (2)
f (2)
1
=
x =
| x| ≤ | x|
2
2
2
[from Eq. (ii)]
In interval [0, 2], for maximum x,
1
| f ( x )| ≤ ⋅ 2 ⇒ | f ( x )| ≤ 1
[Q x = 2]
2
7 g ′( x ) = f ′(sin x ) ⋅ cos x − f ′(cos x ) ⋅ sin x
⇒ g ′ ′ ( x ) = − f ′ (sin x ) ⋅ sin x
+ cos2 x f ′ ′ (sin x )
+ f ′ ′ (cos x ) ⋅ sin2 x − f ′ (cos x ) ⋅ cos x
π
> 0, ∀ x ∈ 0,
2
π
⇒ g ′( x ) is increasing in 0, .
2
π
Also, g ′ = 0
4
π π
⇒ g ′ ( x ) > 0, ∀ x ∈ ,
4 2
π
and g ′ ( x ) < 0, ∀ x ∈ 0,
4
π
Thus, g ( x ) is decreasing in 0, .
4
8 We have, f ( x ) = ( x − p )( x − q )( x − r )
⇒ f ( p ) = 0 = f (q ) = f (r )
⇒ p,q and r are three distinct real
roots of f ( x ) = 0
So, by Roolle’s theorem, f '( x ) has one
real root in the interval ( p,q ) and other
in the interval (q, r ). Thus, f '( x ) has two
distinct real roots.
Now, f ( x ) = ( x − p )( x − q )( x − r )
⇒
f ( x ) = x3 − x2 ( p + q + r )
+ x( pq + qr + rp ) − pqr
⇒ f '( x ) = 3 x2 − 2( p + q + r )x
+ ( pq + qr + rp )
As f '( x ) has distinct real roots
∴ 4( p + q + r )2 − 12( pq + qr + rp ) > 0
⇒ ( p + q + r )2 > 3( pq + qr + rp )
9 Given that, f ( x ) is monotonic.
⇒ f ′( x ) = 0 or f ′( x ) > 0, ∀ x ∈ R
⇒ f ′( px ) < 0 or f ′( px ) > 0, ∀ x ∈ R
So, f ′( px ) is also monotonic.
Hence, f ( x ) + f (3 x ) + ... + f [(2m − 1)x]
is a monotonic.
Polynomial of odd degree (2m − 1), so it
will attain all real values only once.
10 Since, the balloon is spherical in shape,
hence the volume of the balloon is
4
V = πr 3.
3
On differentiating both the sides w.r.t. t,
we get
dV
4
dr
= π 3 r 2 ×
dt
3
dt
dr dV / dt
…(i)
=
⇒
dt
4 πr 2
dr
at the rate t = 49 min,
dt
dV
we require
the radius (r ) at that
dt
dV
stage.
= − 72 π m3 / min
dt
Also, amount of volume lost in 49 min
= 72 π × 49 m3
∴ Final volume at the end of 49 min
= (4500 π − 3528 π ) m3
= 972 π m3
If r is the radius at the end of 49 min,
4
then πr 3 = 972 π ⇒ r 3 = 729
3
⇒
r =9
Radius of the balloon at the end of 49
min = 9 m
From Eq. (i),
dV
dt t = 49
dr dV / dt
dr
=
⇒
=
2
dt t = 49 4π (r 2 ) t = 49
dt
4 πr
Now, to find
72 π
2
dr
=
= m/min
dt t = 49 4 π(9 2 ) 9
11 Given equation of curve is
x2 + 2 xy − 3 y 2 = 0
On differentiating w.r.t. x, we get
2 x + 2 xy ′ + 2 y − 6 yy ′ = 0
x+ y
⇒ y′ =
3y − x
dy
At x = 1, y = 1, y ′ = 1 i.e.
=1
dx (1 , 1 )
Equation of normal at (1, 1) is
1
y − 1 = − ( x − 1) ⇒y − 1 = − ( x − 1)
1
⇒ x+ y =2
On solving Eqs. (i) and (ii)
simultaneously, we get
x2 + 2 x (2 − x ) − 3 (2 − x )2 = 0
⇒ x2 + 4 x − 2 x2 − 3(4 + x2 − 4 x ) = 0
⇒
− x2 + 4 x − 12 − 3 x2 + 12 x = 0
⇒
− 4 x2 + 16 x − 12 = 0
4 x2 − 16 x + 12 = 0
⇒
x2 − 4 x + 3 = 0
⇒
( x − 1) ( x − 3) = 0
⇒
⇒
x = 1, 3
Now, when x = 1, then y = 1
and when x = 3, then y = − 1
∴
P = (1, 1) and Q = (3, − 1)
Hence, normal meets the curve again at
(3, − 1) in fourth quadrant.
Alternate Method
Given, x2 + 2 xy − 3 y 2 = 0
⇒ ( x − y ) ( x + 3y ) = 0
⇒ x − y = 0 or x + 3 y = 0
Equation of normal at (1, 1)
y − 1 = − 1 ( x − 1) ⇒ x + y − 2 = 0
It intersects x + 3 y = 0 at (3, − 1) and
hence normal meets the curve in fourth
quadrant.
137
x+y=2
x + 3y = 0
y=x
Y
14 a + b = 4 ⇒ b = 4 − a
and b − a = 4 − 2a = t
Now, ∫
a
0
g ( x ) dx +
0
dx +
(1, 1)
X′
∫
b
X
O
(3, –1)
Y′
0
∫
0
g( x)
g ( x ) dx = I (a)
dI (a)
= g (a) − g (4 − a)
da
As a < 2 and g ( x ) is increasing.
⇒ 4 − a > a ⇒ g (a) − g (4 − a) < 0
Now,
and
dI (a) dI (a) dt
dI (a)
=
= − 2⋅
d (a)
dt da
dt
dI (a)
>0
dt
Thus, I (a) is an increasing function of t.
Hence, the given expression increasing
with (b − a).
⇒
15 We know that,
1 ≤ |sin x | + |cos x | ≤ 2
√5
P
O
[equation of parabola]
As it touches y = x at x = 1.
∴
y = a+ b + c
and
y =1 ⇒ a+ b + c =1
dy
Now,
= 2ax + b
dx
dy
= 2a + b ⇒ 2a + b = 1
⇒
dx at x = 1
–√5
y=1
x
O
x 2+y 2=5
–√5
⇒
±2
Q(−2, 1)
+ y 2 = 5 w.r.t. x,
⇒
dy
x
=−
dx
y
dy
= −2
dx (2, 1 )
dI (a)
<0
da
⇒
13 Let y = ax2 + bx + c
[from y = x, slope = 1]
Now, f ( x ) = ax2 + bx + c
⇒ f ′ ( x ) = 2ax + b ⇒ f ′ ′ ( x ) = 2a
∴
f (0 ) = c , f ′ (0 ) = b , f ′ ′ (0 ) = 2a,
f ′ (1 ) = 2a + b = 1
∫
4−a
a
⇒
12 Given, f (0) = 2 = g (1), g (0) = 0
and
f (1) = 6
, ).
f and g are differentiable in (01
Let
h( x ) = f ( x ) − 2g ( x )
…(i)
⇒ h(0) = f (0) − 2g (0)
⇒
h(0) = 2 − 0 ⇒ h(0) = 2
and h(1) = f (1) − 2g (1) = 6 − 2(2)
⇒
h(1) = 2, h( 0) = h(1) = 2
Hence, using Rolle’s theorem, we get
h ′ (c ) = 0, such that c ∈ (01
, )
On differentiating Eq.(i) at c, we get
f ′ (c ) − 2g ′ (c ) = 0 ⇒ f ′ (c ) = 2g ′ (c )
(say)
g ( x ) dx =
x2 + 1 = 5 ⇒ x =
Now,
P(2, 1) and
On differentiating x2
we get
dy
2x + 2y
=0
dx
y = [|sin x | + |cos x |] = 1
Let P and Q be the points of intersection
of given curves.
Clearly, the given curves meet at points
where y = 1, so we get
dy
=2
dx ( −2, 1 )
Clearly, the slope of line y = 1 is zero
and the slope of the tangents at P and Q
are (−2) and (2), respectively.
Thus, the angle of intersection is
tan −1 (2).
16 There is only one function in option (a),
1
∈ (0,1) but in
2
other parts critical point 0 ∉( 0,1). Then,
we can say that functions in options (b),
(c) and (d) are continuous on [0, 1] and
differentiable in (0, 1).
1
1 − x,
x<
2
2
Now, for f ( x ) =
2
1
1
− x , x ≥
2
2
whose critical point
1
Here, Lf ′ = − 1
2
1
1 1
and Rf ′ = 2 − (−1) = 0
2
2 2
∴
1
1
Lf ′ ≠ Rf ′
2
2
⇒ f is non-differentiable at x =
1
∈ (0,1).
2
∴ Lagrange mean value theorem is not
applicable to f ( x ) in [0, 1].
DAY FOURTEEN
Maxima and
Minima
Learning & Revision for the Day
u
Maxima and Minima of a Function
u
Concept of Global Maximum/Minimum
Maxima and Minima of a Function
Greatest value/
A function f ( x) is said to attain a maximum at x = a,
Y
Local
Local
absolute
if there exists a neighbourhood (a − δ, a + δ), x ≠ a
maximum
maximum
maximum
C
i.e. f ( x) < f (a), ∀ x ∈ (a − δ, a + δ),
A
x ≠ a ⋅ h > 0 (very small quantity)
Local minimum
D
X
X′
In such a case f (a) is said to be the maximum value
of f ( x) at x = a.
A function f ( x) is said to attain a minimum at
B
x = a, if there exists a neighbourhood (a − δ, a + δ) Least value/
minimum
absolute
Y′
such that f ( x) > f (a), ∀x ∈(a − δ, a + δ), x ≠ a.
Graph of a continuous function explained local maxima (minima) and absolute maxima
(minima). In such a case f (a) is said to be the minimum value of f ( x) at x = a.
The points at which a function attains either the maximum or the minimum values are known
as the extreme points or turning points and both minimum and maximum values of f ( x) are
called extreme values. The turning points A and C are called local maximum and points B and
D are called local minimum.
PRED
MIRROR
Critical Point
●
●
A point c in the domain of a function f at which either f ′ (c) = 0 or f is not differentiable is
called a critical point of f. Note that, if f is continuous at point c and f ′ (c) = 0, then there
exists h > 0 such that f is differentiable in the interval (c − h, c + h).
The converse of above theorem need not be true, that is a point at which the derivative
vanishes need not be a point of local maxima or local minima.
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
Method to Find Local Maxima or Local Minima
First Derivative Test
Let f be a function defined on an open interval I and f be continuous at a critical point c in I.
Then,
(i) If f ′ ( x) changes sign from positive to negative as x increases through c, i.e. if f ′ ( x) > 0 at
every point sufficiently close to and to the left of c and f ′ ( x) < 0 at every point
sufficiently close to and to the right of c, then c is a point of local maxima.
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
139
DAY
(ii) If f ′ ( x) changes sign from negative to positive as x
increases through point c, i.e. if f ′ ( x) < 0 at every point
sufficiently close to and to the left of c and f ′ ( x) > 0 at
every point sufficiently close to and to the right of c,
then c is a point of local minima.
(iii) If f ′ ( x) does not change sign as x increases through c,
then c is neither a point of local maxima nor a point of
local minima. Infact, such a point is called point of
inflection.
Y
Point of local maxima
f′ (
0
Point of non-differentiability
and point at local maxima
f′(c1)=0
0
x)
>0
f′ (
c1
O
●
●
●
●
Point of non-differentiability
but it is a point of local
minima
f ′(c2)=0
Point
of local
minima
X′
●
x) <
x)>
f ′(
Important Results
c2
c3
●
X
c4
If n is even and f n (a) < 0 ⇒ x = a is a point of local
maximum.
If n is even and f n (a) > 0 ⇒ x = a is a point of local
minimum.
If n is odd ⇒ x = a is a point of neither local maximum nor
a point of local minimum.
ax + b
The function f ( x) =
has no local maximum or
cx + d
minimum regardless of values of a, b , c and d.
The function f (θ) = sin m θ ⋅ cos n θ attains maximum values at
m
θ = tan −1
.
n
If AB is diameter of circle and C is any point on the
circumference, then area of the ∆ ABC will be maximum, if
triangle is isosceles.
Y′
Graph of f around c explained following points.
(iv) If c is a point of local maxima of f, then f (c) is a local
maximum value of f . Similarly, if c is a point of local
minima of f, then f (c) is a local minimum value of f .
Second or Higher Order Derivative Test
(ii) Find f ′ ′ ( x) and at x = a1 ,
(a) if f ′ ′ (a1 ) is positive, then f ( x) is minimum at x = a1 .
(b) if f ′ ′ (a1 ) is negative, then f ( x) is maximum at
x = a1 .
(iii) (a) If at x = a1 , f ′ ′ (a1 ) = 0, then find f ′ ′ ′ ( x). If
f ′ ′ ′ (a1 ) ≠ 0 , then f ( x) is neither maximum nor
minimum at x = a.
(b) If f ′ ′ ′ (a1 ) = 0, then find f iv ( x).
iv
(c) If f ( x) is positive (minimum value) and f ( x) is
negative (maximum value).
(iv) If at x = a1 , f (a1 ) = 0 , then find f ( x) and proceed
similarly.
iv
v
Point of Inflection
At point of inflection
(i) It is not necessary that 1st derivative is zero.
(ii) 2nd derivative must be zero or 2nd derivative changes
sign in the neighbourhood of point of inflection.
nth Derivative Test
Let f be a differentiable function on an interval I and a be an
interior point of I such that
(i) f ′ (a) = f ′′(a) = f ′′′(a) = ... = f n −1 (a) = 0 and
(ii) f n (a) exists and is non-zero.
●
●
(i) Find f ′ ( x) and equate it to zero. Solve f ′ ( x) = 0 let its
roots be x = a1 , a2 ,...
iv
Concept of Global
Maximum/Minimum
Let y = f ( x) be a given function with domain D and
[a, b ] ⊆ D, then global maximum/minimum of f ( x) in [a, b ] is
basically the greatest / least value of f ( x) in [a, b ].
Global maxima/minima in [a, b ] would always occur at critical
points of f ( x) within [a, b ] or at end points of the interval.
Global Maximum/Minimum in [a, b]
In order to find the global maximum and minimum of f ( x) in
[a, b ].
Step I
Find out all critical points of f ( x) in [a, b ]
[i.e. all points at which f ′ ( x) = 0]
points are c1 , c2 ,..., c n .
Step II
and let these
Find the value of f (c1 ), f (c2 ) ,..., f (c n ) and also at the
end points of domain i.e. f (a) and f (b ).
Step III
Find M1 → Global maxima or greatest value
and M2 → Global minima or least value.
where, M1 = max { f (a), f (c1 ), f (c2 ),..., f (c n ), f (b )}
and M2 = min { f (a), f (c1 ), f (c2 ),..., f (c n ), f (b )}
Some Important Results on Maxima
and Minima
(i) Maxima and minima occur alternatively i.e. between
two maxima there is one minimum and vice-versa.
(ii) If f ( x) → ∞ as x → a or b and f ′ ( x) = 0 only for one
value of x (say c) between a and b, then f (c) is
necessarily the minimum and the least value.
(iii) If f ( x) → − ∞ as x → a or b, then f (c) is necessarily the
maximum and greatest value.
(iv) The stationary points are the points of the domain,
where f ′ ( x) = 0.
140
40
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 If f is defined as f ( x ) = x +
1
, then which of following is
x
true?
j
NCERT Exemplar
(a) Local maximum value of f (x) is − 2
(b) Local minimum value of f (x) is 2
(c) Local maximum value of f (x) is less than local minimum
value of f (x)
(d) All the above are true
2 If the sum of two numbers is 3, then the maximum value
of the product of the first and the square of second is
j
(a) 4
(b) 1
(c) 3
NCERT Exemplar
(d) 0
3 If y = a log x + bx + x has its extremum value at x = 1
2
and x = 2, then (a, b ) is equal to
1
(a) 1,
2
1
(b) , 2
2
−1
(c) 2 ,
2
−2 −1
(d) ,
3 6
4 The function f ( x ) = a cos x + b tan x + x has extreme
π
values at x = 0 and x = , then
6
2
,b = −1
3
2
(c) a = − , b = 1
3
2
,b = −1
3
2
(d) a = , b = 1
3
(a) a = −
(b) a =
5 The minimum radius vector of the curve
(b) 5
(c) 7
(d) None of these
6 The function f ( x ) = 4x 3 − 18x 2 + 27x − 7 has
(a) one local maxima
j
NCERT Exemplar
(b) one local minima
(c) one local maxima and two local minima
(d) neither maxima nor minima
7 The function f ( x ) =
(a)
(b)
(c)
(d)
k − 2x , if x ≤ − 1
.
2x + 3, if x > − 1
If f has a local minimum at x = −1, then a possible value
j
of k is
AIEEE 2010
(d) −1
(b) 28
(c) 38
(d)
1
2
11 If f ( x ) = x 2 + 2bx + 2c 2 and g ( x ) = − x 2 − 2cx + b 2 such
that minimum f ( x ) > maximum g ( x ), then the relation
between b and c is
(a) 0 < c < b 2
(c) | c | > | b | 2
(b) | c | < | b | 2
(d) No real values of b and c
12 Let f ( x ) be a polynomial of degree four having extreme
f (x )
values at x = 1 and x = 2. If lim 1 + 2 = 3, then f ( 2) is
x→ 0
x
j JEE Mains 2015
equal to
(a) −8
(c) 0
(b) −4
(d) 4
13 If a differential function f ( x ) has a relative minimum at
x = 0, then the function φ( x ) = f ( x ) + ax + b has a relative
minimum at x = 0 for
(b) all b, if a = 0
(d) all a > 0
(a) all a and all b
(c) all b > 0
14 The denominator of a fraction is greater than 16 of the
square of numerator, then least value of fraction is
(b) − 1/ 8
(d) 1/16
b
x
15 The function f ( x ) = ax + , b, x > 0 takes the least value
at x equal to
(a) b
(b)
a
(c)
(d)
b
b
a
(d) 18
j
AIEEE 2011
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
17 The absolute maximum and minimum values of the
function f given by f ( x ) = cos 2 x + sin x , x ∈[ 0, π ]
j
9 The minimum value of 9x + 4y , where xy = 16 is
(a) 48
(c) 2
Statement I x = 0 is point of minima of f .
Statement II f ′ (0) = 0.
8 Let f : R → R be defined by f ( x ) =
(b) 0
(b) 1
tan x
, x ≠0
x
1 , x = 0
no point of local minima
no point of local maxima
exactly one point of local minima
exactly one point of local maxima
(a) 1
(a) 3
16 Let f be a function defined by f ( x ) =
x2 − 2
has
x2 − 4
1
(c) −
2
attains its maximum and minimum at p and q
respectively such that p 2 = q , then a is equal to
(a) −1/4
(c) 1/12
4
9
+
= 1 is of length
x2 y2
(a) 1
10 If the function f ( x ) = 2x 3 − 9 ax 2 + 12 a 2 x + 1, where a > 0
(a) 2.25 and 2
(c) 1.75 and 1.5
NCERT Exemplar
(b) 1.25 and 1
(d) None of these
141
DAY
18 The maximum value of f ( x ) =
(a) −
1
4
(b) −
1
3
x
on [ −1,1] is
4+ x + x2
(c)
1
6
(d)
1
5
(a) e
1
1
(b) log
e
e
2
(c) e log e (d) None of these
20 If x is real, the maximum value of
3x 2 + 9x + 17
is
3x 2 + 9x + 7
AIEEE 2007
1
(d)
4
j
(a) 41
17
(c)
7
(b) 1
f ( x ) = sec x + log cos 2 x , 0 < x < 2π are respectively
(a)
(b)
(c)
(d)
NCERT Exemplar
(1, − 1) and {2 (1 − log 2), 2 (1 + log 2)}
(1, − 1) and {2 (1 − log 2), 2 (1 − log 2)}
(1, − 1) and (2, − 3)
None of the above
(a) π
(b) 2 π
(c) 3 π
(b) (0, 0)
j
NCERT Exemplar
π
(d)
2
(c) (1, 0)
3
(d) (0, 1)
(a)
h
2 2
(b)
JEE Main 2013
h2
(d)
4
j
2
h
2
(c)
h
2
2
25 A straight line is drawn through the point P ( 3, 4) meeting
the positive direction of coordinate axes at the points A
and B. If O is the origin, then minimum area of ∆OAB is
equal to
(a) 12 sq units
(c) 24 sq units
(b) 6 sq units
(d) 48 sq units
roots, where p > 0 and q > 0. Then, which one of the
following holds?
2
− (a − 2)x − a + 1 = 0 assume
(b) 1
(c) 3
(d) 0
30 The minimum intercepts made by the axes on the
tangent to the ellipse
x2 y2
+
= 1 is
16
9
(b) 7
(c) 1
(d) None of these
sphere is maximum, if
(a) h =
4R
3
(b) h =
R
3
(c) h =
2R
(d) None of these
3
32 The volume of the largest cone that can be inscribed in a
sphere of radius R is
(b)
(c)
(d)
j
NCERT
3
of the volume of the sphere
8
8
of the volume of the sphere
27
2
of the volume of the sphere
7
None of the above
33 Area of the greatest rectangle that can be inscribed in
x2 y2
the ellipse 2 + 2 = 1 is
a
b
(a)
ab
(b)
a
b
(c) 2ab
minimum value of the sum at x equal to
(a) 2
(b) 1
(c) –1
(d) ab
p
3
p
3
27 If A ( x1, y1 ), B ( x 2 , y 2 ) and C ( x 3 , y 3 ) are the vertices of a
∆ ABC. A parallelogram AFDE is drawn with D, E and F
on the line segment BC, CA and AB, respectively. Then,
maximum area of such parallelogram is
j
AIEEE 2003
(d) –2
35 The greatest value of
f ( x ) = ( x + 1)1/ 3 − ( x − 1)1/ 3 on [0, 1] is
p
p
and −
3
3
p
and maxima at −
3
p
(c) The cubic has minima at −
and maxima at
3
p
p
(d) The cubic has minima at both
and −
3
3
(b) The cubic has minima at
(d) 46, −6
34 The real number x when added to its inverse gives the
26 Suppose the cubic x 3 − px + q has three distinct real
(a) The cubic has maxima at both
roots of the equation x
the least value is
(a)
24 The maximum area of a right angled triangle with
hypotenuse h is
(c) 40, −6
(b) 46, 6
29 The value of a , so that the sum of the squares of the
(a) 25
23 The point of inflection for the curve y = x 5 / 2 is
(a) (1, 1)
(a) 36, 3
31 The curved surface of the cone inscribed in a given
22 The difference between greatest and least values of the
π π
function f ( x ) = sin 2x − x , on − ,
is
2 2
that x = 3 t 2 − 18 t + 7 and y = 2 t 3 − 15 t 2 + 24 t + 10,
∀ x ∈ [ 0, 6].
Then, the maximum and minimum values of y = f ( x ) are
(a) 2
21 The maximum and minimum values of
j
1
( area of ∆ABC)
4
1
( area of ∆ABC)
(d)
8
(b)
28 If y = f ( x ) is a parametrically defined expression such
19 In interval [1, e ], the greatest value of x 2 log x is
2
1
( area of ∆ABC)
2
1
(c) ( area of ∆ABC)
6
(a)
(a) 1
(b) 2
(c) 3
j
AIEEE 2002
(d) 1/3
36 The coordinate of a point on the parabola y 2 = 8x whose
distance from the circle x 2 + ( y + 6) 2 = 1 is minimum, is
(a) (2,− 4)
(b) (2,4)
(c) (18,−12)
(d) (8, 8)
37 The volume of the largest cylinder that can be inscribed
in a sphere of radius r cm is
(a)
4 πr 3
3
(b)
4 πr 3
3 3
(c)
4 πr 3
2 3
(d)
4 πr 3
5 2
38 Maximum slope of the curve y = − x 3 + 3x 2 + 9x − 27 is
(a) 0
(b) 12
(c) 16
(d) 32
TEN
142
39 If ab = 2 a + 3b, a > 0, b > 0, then the minimum value of ab
is
40 The perimeter of a sector is p. The area of the sector is
maximum, when its radius is
(a) 12
1
(c)
4
(b) 24
(a)
p
(b)
(d) None of these
1
p
(c)
p
2
(d)
p
4
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 The minimum radius vector of the curve
a2
b2
+
= 1 is of
x2
y2
` 75 while at 40 km/h, it is ` 65. Then, the most
economical speed (in km/h) of the bus is j JEE Mains 2013
(a) 45
length
(a) a − b
(c) 2a + b
(b) a + b
(d) None of these
2 f ( x ) = x 2 − 4 | x | and
min {f (t ) : − 6 ≤ t ≤ x }, x ∈ [ − 6, 0]
, then g ( x ) has
g( x ) =
max {f (t ) : 0 < t ≤ x }, x ∈ ( 0, 6]
(a) exactly one point of local minima
(b) exactly one point of local maxima
(c) no point to local maxima but exactly one point of local
minima
(d) neither a point of local maxima nor minima
4x − x 3 + log (a 2 − 3 a + 3),
x − 18 ,
3 f (x ) =
0≤x <3
x ≥3
Complete the set of values of a such that f ( x ) has a local
maxima at x = 3, is
(a) [−1, 2]
(c) [1, 2]
(b) (−∞, 1) ∪ (2, ∞)
(d) (−∞, − 1) ∪ (2, ∞)
4 The point in the interval [ 0, 2π] , where f ( x ) = e x sin x has
maximum slope is
π
4
(c) π
(a)
π
2
(d) None of these
(b)
5 The total number of local maxima and local minima of the
( 2 + x )3 , − 3 < x ≤ − 1
function f ( x ) = 2 / 3
is
x , −1 < x < 2
(a) 0
(b) 1
(c) 2
(d) 3
the form of a circular sector, then the maximum area (in
j
sq m) of the flower-bed is
JEE Mains 2017
(b) 10
(c) 25
(a) 1, 4
(c) 2, 4
(d) 30
b
v
where v km/h is the average speed of the bus. When the
bus travels at 30 km/h, the cost comes out to be ` 75
7 The cost of running a bus from A to B , is ` av + ,
(c) 60
(d) 40
(b) 0, 4
(d) None of these
1
3
| x + x 2 + 3x + sin x | 3 + sin , x ≠ 0
, then
x
0,
x =0
number of points [where, f ( x ) attains its minimum value]
is
9 Let f ( x ) =
(a) 1
(c) 3
(b) 2
(d) infinite many
10 A wire of length 2 units is cut into two parts which are
bent respectively to form a square of side = x units and a
circle of radius = r units. If the sum of the areas of the
square and the circle so formed is minimum, then
(a) 2 x = (π + 4)r
(c) x = 2r
11 Let f ( x ) = x 2 +
h( x ) =
(b) (4 − π)x = πr
(d) 2x = r
1
1
and g ( x ) = x − , x ∈ R − {−1, 0, 1}. If
2
x
x
f (x )
, then the local minimum value of h( x ) is
j
g( x )
JEE Mains 2018
(a) 3
6 If 20 m of wire is available for fencing off a flower-bed in
(a) 12.5
(b) 50
| x 2 − 2 | , − 1 ≤ x < 3
x
8 If f ( x ) =
,
3 ≤ x < 2 3, then the points,
3
2 3≤x ≤4
3 − x ,
where f ( x ) takes maximum and minimum values, are
(b) −3
(c) −2 2
(d) 2 2
n2
12 The largest term in the sequence a n = 3
is given by
n + 200
529
49
49
(c)
543
(a)
(b)
8
89
(d) None of these
13 All possible values of the parameter a so that the function
f ( x ) = x 3 − 3(7 − a )x 2 − 3( 9 − a 2 )x + 2 has a negative
point of local minimum are
(a) all real values
(c) (0, ∞)
(b) no real values
(d) (−∞,0)
143
DAY
R
14 The circle x 2 + y 2 = 1 cuts the
X -axis at P and Q. Another circle
with centre at Q and variable
radius intersects the first circle at
R above the X -axis and the line
segment PQ at S. Then, the
maximum area of the ∆QSR is
(a) 4 3 sq units
4 3
(c)
sq units
9
15 Given, P ( x ) = x 4 + ax 3 + bx 2 + cx + d such that x = 0 is
x2+y2=1
the only real root of P ′ ( x ) = 0. If P ( − 1) < P (1), then in the
j AIEEE 2009
interval [ − 1, 1] .
Q
T
S
O
(a) P (− 1) is the minimum and P (1) is the maximum of P
(b) P (− 1) is not minimum but P (1) is the maximum of P
(c) P (− 1) is the minimum and P (1) is not the maximum of P
(d) Neither P (− 1) is the minimum nor P (1) is the maximum
of P
P
(b) 14 3 sq units
(d) 15 3 sq units
ANSWERS
SESSION 1
SESSION 2
1 (d)
2 (a)
3 (d)
4 (a)
5 (b)
6 (d)
7 (d)
8 (d)
9 (a)
10 (c)
11 (c)
12 (c)
13 (b)
14 (b)
15 (d)
16 (b)
17 (b)
18 (c)
19 (a)
20 (a)
21 (b)
22 (a)
23 (b)
24 (d)
25 (c)
26 (b)
27 (a)
28 (d)
29 (b)
30 (b)
31 (a)
32 (b)
33 (c)
34 (b)
35 (b)
36 (a)
37 (b)
38 (b)
39 (b)
40 (d)
1 (b)
11 (d)
2 (d)
12 (c)
3 (c)
13 (b)
4 (b)
14 (c)
5 (c)
15 (b)
6 (c)
7 (c)
8 (b)
9 (a)
10 (c)
Hints and Explanations
SESSION 1
dy
1
1 Let y = x + 1 ⇒
=1− 2
x
dx
x
dy
Now,
= 0 ⇒ x2 = 1
dx
⇒
x=±1
d2 y
2
⇒
= 3 , therefore
dx2
x
d2 y
(at x = 1) > 0
dx2
d2 y
and
(at x = − 1) < 0
dx2
Hence, local maximum value of y is at
x = − 1 and the local maximum value
= − 2.
Local minimum value of y is at x = 1
and local minimum value = 2.
Therefore, local maximum value − 2 is
less than local minimum value 2.
2 Let two numbers be x and (3 − x).
Then, product P = x(3 − x )2
dP
= −2 x(3 − x ) + (3 − x )2
dx
dP
d 2P
= (3 − x )(3 − 3 x ) and 2 = 6 x − 12
dx
dx
dP
For maxima or minima, put
=0
dx
⇒ (3 − x )(3 − 3 x ) = 0 ⇒x = 3, 1
At x = 3,
d 2P
= 18 − 12 = 6 > 0 [minima]
dx2
At x = 1,
d 2P
= −6 < 0
dx2
So, P is maximum at x = 1.
∴ Maximum value of P = 1 (3 − 1)2 = 4
dy
a
= + 2bx + 1
dx
x
3 Q
⇒
dy
= a + 2b + 1 = 0
dx x =1
⇒
a = − 2b − 1
dy
a
and
= + 4b + 1 = 0
dx x =2 2
− 2b − 1
+ 4b + 1 = 0
2
1
−1
⇒ – b + 4b + = 0 ⇒ 3b =
2
2
−2
1
−1
and a = − 1 =
⇒ b =
6
3
3
⇒
4 f ′( x ) = − a sin x + b sec2 x + 1
Now,
π
f ′ (0) = 0 and f ′ = 0
6
⇒ b + 1 = 0 and −
a 4b
+
+ 1= 0
2
3
⇒
b = − 1, a = −
2
3
5 The given curve is 42 + 92 = 1
x
y
Put x = r cos θ, y = r sin θ, we get
r 2 = (2 sec θ)2 + (3 cosec θ)2
So, r 2 will have minimum value
( 2 + 3)2 .
or r have minimum value equal to 5.
6 f ( x )= 4 x 3 − 18 x2 + 27 x − 7
f ′ ( x ) = 12 x2 − 36 x + 27
= 3(4 x2 − 12 x + 9) = 3(2 x − 3)2
3
f ′ ( x ) = 0 ⇒ x = (critical point)
2
3
Since, f ′ ( x ) > 0 for all x < and for all
2
3
x>
2
3
Hence, x = is a point of inflection i.e.,
2
neither a point of maxima nor a point
of minima.
3
x = is the only critical point and f
2
has neither maxima nor minima.
7 For y =
x2 − 2
− 4x
dy
⇒
=
x2 − 4
dx ( x2 − 4)2
TEN
144
dy
> 0, for x < 0
dx
dy
and
< 0, for x > 0
dx
Thus, x = 0 is the point of local maxima
1
for y. Now, ( y ) x = 0 = (positive). Thus,
2
x = 0 is also the point of local
x2 − 2
maximum for y = 2
.
x −4
⇒
8 If f ( x ) has a local minimum at x = −1,
then
lim f ( x ) = lim− f ( x )
x→ −1 +
⇒
⇒
x→ −1
lim+ 2 x + 3 = lim− 1 < −2 x
x→ −1
x→ −1
−2 + 3 = k + 2 ⇒ k = −1
Y
f(x)=2x+3
f(x)=1<–2x
(–1,1)
X′
O
(–1,0)
X
Y′
9 Let S = 9 x + 4 y
Since, xy = 16 is given.
16
64
or S = 9 x +
∴
y =
x
x
On differentiating both sides, we get
dS
64
...(i)
= 9− 2
dx
x
dS
64
8
Q
= 0⇒ 2 = 9⇒ x = ±
dx
x
3
Again, on differentiating Eq. (i)
d 2S 128
w.r.t. x, we get 2 = 3
dx
x
8
Hence, it is minimum at x = and
3
minimum value of S is
8
S min = 9 + 4(6) = 48
3
10 We have,
f ( x ) = 2 x3 − 9 ax2 + 12a2 x + 1
f ′( x ) = 6 x2 − 18 ax + 12a2
f ′ ′( x ) = 12 x − 18 a
For maximum and minimum,
6 x2 − 18 ax + 12a2 = 0
⇒
x2 − 3 ax + 2a2 = 0
⇒
x = a or x = 2a
At x = a maximum and at x = 2a
minimum.
Q
p2 = q
∴
a2 = 2a ⇒ a = 2 or a = 0
But a > 0, therefore a = 2
11 Minimum of f ( x ) = − D
4a
− ( 4b 2 − 8 c 2 )
=
4
= 2c 2 − b 2
and maximum of g ( x ) = −
( 4 c 2 + 4b 2 )
4 (− 1)
= b2 + c 2
Since, min f ( x ) > max g ( x )
⇒
2c 2 − b 2 > b 2 + c 2
⇒
c 2 > 2b 2
⇒
|c |> 2 |b |
12 Central Idea Any function have
extreme values (maximum or
minimum) at its critical points, where
f ′( x ) = 0.
Since, the function have extreme values
at x = 1 and x = 2.
∴
f ′( x ) = 0 at x = 1 and x = 2
⇒
f ′(1) = 0 and f ′(2) = 0
Also it is given that
f ( x)
f ( x)
lim 1 + 2 = 3 ⇒ 1 + lim 2 = 3
x→ 0
x→ 0 x
x
f ( x)
⇒ lim 2 = 2
x→ 0 x
⇒ f ( x ) will be of the form
ax 4 + bx3 + 2 x2
[Q f ( x ) is of four degree polynomial]
Let f ( x ) = ax 4 + bx3 + 2 x2 ⇒ f ′( x )
= 4ax3 + 3bx2 + 4 x
⇒ f ′(1) = 4a + 3b + 4 = 0
...(i)
and f ′(2) = 32a + 12b + 8 = 0
...(ii)
⇒
8a + 3b + 2 = 0
On solving Eqs. (i) and (ii), we get
1
a = ,b = − 2
2
x4
∴ f ( x) =
− 2 x3 + 2 x2
2
⇒ f (2) = 8 − 16 + 8 = 0
13 φ′( x ) = f ′( x ) + a
Q
φ′ (0) = 0 ⇒ f ′ (0) + a = 0
⇒
a= 0
[Q f ′ (0) = 0]
Also, φ′ (0) > 0 [Q f ′ ′ (0) > 0]
⇒ φ( x ) has relative minimum at
x = 0 for all b, if a = 0
14 Let the number be x, then
f (x ) =
x
x2 + 16
On differentiating w.r.t. x, we get
( x2 + 16) ⋅ 1 − x (2 x )
f ′( x ) =
( x2 + 16)2
2
2
x + 16 − 2 x
16 − x2
…(i)
=
= 2
2
2
( x + 16)
( x + 16)2
Put f ′ ( x ) = 0 for maxima or minima
f ′ ( x ) = 0 ⇒ 16 − x2 = 0
⇒
x = 4, − 4
Again, on differentiating w.r.t. x, we
get
( x2 + 16)2 (−2 x ) − (16 − x2 )
f ′ ′( x ) =
x = 4, f ′ ′ ( x ) < 0
2( x2 + 16) 2 x
At
( x2 + 16)4
∴ f ( x ) is maximum at x = 4.
and at x = − 4, f ′ ′ ( x ) > 0, f ( x ) is
minimum.
∴ Least value of
−4
1
f ( x) =
=−
16 + 16
8
15 Given, f ( x ) = ax + b
x
On differentiating w.r.t. x, we get
b
f ′( x ) = a − 2
x
For maxima or minima, put f ′ ( x ) = 0
b
⇒
x=
a
Again, differentiating w.r.t. x, we get
2b
f ′ ′( x ) = 3
x
b
At x =
, f ′ ′ ( x ) = positive
a
b
.
⇒ f ( x ) is minimum at x =
a
b
.
∴ f ( x ) has the least value at x =
a
tan x
16 f ( x ) = x , x ≠ 0
x=0
1,
tan x
As
> 1, ∀ x ≠ 0
x
∴ f (0 + h ) > f (0) and f (0 − h ) > f (0)
At x = 0, f ( x ) attains minima.
f (h ) − f (0)
Now, f ′ (0) = lim
h→ 0
h
tan h
−1
tan h − h
= lim h
= lim
h→ 0
h→ 0
h
h2
[using L’ Hospital’s rule]
sec 2 h − 1
= lim
[Q tan2 θ = sec 2 θ − 1]
h→ 0
2h
tan2 h
1
= lim
⋅h= ⋅0 = 0
h→ 0
2h2
2
Therefore, Statement II is true.
Hence, both statements are true but
Statement II is not the correct
explanation of Statement I.
17 Given, f ( x ) = cos2 x + sin x, x ∈ [0, π]
Now,
f ′ ( x ) = 2 cos x (− sin x ) + cos x
= − 2sin x cos x + cos x
For maximum or minimum put
f ′( x ) = 0
⇒ −2sin x cos x + cos x = 0
⇒ cos x (− 2sin x + 1 ) = 0
⇒
⇒
cos x = 0 or sin x =
x=
π π
,
6 2
1
2
For absolute maximum and absolute
minimum, we have to evaluate
145
DAY
π
π
f (0), f , f , f ( π )
6
2
At x = 0,
f (0) = cos 2 0 + sin 0 = 12 + 0 = 1
π
π
π
π
At x = , f = cos 2 + sin
6
6 6
6
2
At x =
π
,
2
3
1 5
=
.
+ = = 125
2 4
2
π
π
π
f = cos 2 + sin
= 02 + 1 = 1
2
2
2
At x = π,
f ( π ) = cos 2 π + sin π = (−1)2 + 0 = 1
Hence, the absolute maximum value of
π
f is 1.25 occurring at x = and the
6
absolute minimum value of f is 1
π
occurring at x = 0, and π.
2
Note If close interval is given, to determine
global maximum (minimum), check the
value at all critical points as well as end
points of a given interval.
x
18 Q f ( x ) =
4 + x + x2
On differentiating w.r.t. x, we get
4 + x + x2 − x ( 1 + 2 x )
f ′( x ) =
(4 + x + x2 )2
For maximum, put f ′ ( x ) = 0
4 − x2
⇒
= 0 ⇒ x = 2, − 2
(4 + x + x2 )2
Both the values of x are not in the
interval [−1, 1] .
−1
−1
=
∴ f ( −1 ) =
4−1+ 1
4
f (1 ) =
1
1
= (maximum)
4+ 1+1
6
19 Given, f ( x ) = x2 log x
On differentiating w.r.t. x, we get
f ′ ( x ) = (2 log x + 1 ) x
For a maximum, put f ′ ( x ) = 0
⇒ (2 log x + 1 ) x = 0
⇒
x = e −1 /2 , 0
Q
0 < e −1 / 2 < 1
None of these critical points lies in the
interval [1, e ] .
So, we only compute the value of f ( x ) at
the end points 1 and e.
We have, f (1 ) = 0, f (e ) = e 2
Hence, greatest value of f ( x ) = e 2
10
20 Let
f (x ) = 1 +
7
2
3 x + 3 x +
3
10
2
3
1
3 x + +
2
12
So, the maximum value of f ( x ) at
3
x = − is
2
=1+
f −
3
=1+
2
10
= 1 + 40 = 41
1
3
12
21 Given, f ( x ) = sec x + log cos2 x
⇒
f ( x ) = sec x + 2 log(cos x )
Therefore,
f ′ ( x ) = sec x tan x − 2 tan x
= tan x (sec x − 2)
f ′( x ) = 0
⇒
tan x = 0 or sec x = 2 ⇒ cos x =
1
2
Therefore, possible values of x are
π
5π
.
x = 0, x = π and x = or x =
3
3
2
Again,f ′ ′ ( x ) = sec x (sec x − 2)
+ tan x (sec x tan x )
= sec3 x + sec x tan2 x − 2 sec2 x
= sec x (sec2 x + tan2 x − 2 sec x )
⇒ f ′ ′ (0 ) = 1 (1 + 0 − 2) = − 1 < 0
Therefore, x = 0 is a point of maxima.
f ′ ′ ( π ) = − 1 (1 + 0 + 2) = − 3 < 0
Therefore, x = π is a point of maxima.
π
f ′ ′ = 2 (4 + 3 − 4) = 6 > 0
3
π
Therefore, x = is a point of minima.
3
5π
f ′ ′
= 2 (4 + 3 − 4) = 6 > 0
3
5π
Therefore, x =
is a point of minima.
3
Maximum value of y at x = 0 is
1 + 0 = 1.
Maximum value of y at x = π is
− 1 + 0 = − 1.
π
Minimum value of y at x = is
3
1
2 + 2 log = 2 (1 − log 2).
2
5π
Minimum value of y at x =
is
3
1
2 + 2 log = 2 (1 − log 2).
2
22 Given, f ( x ) = sin 2 x − x
⇒ f ′ ( x ) = 2 cos 2 x − 1
1
2
π
π
π
π
or
⇒ 2x = − or ⇒ x = −
3
3
6
6
π
π
π
Now, f − = sin(− π ) +
=
2
2
2
Put f ′ ( x ) = 0 ⇒ cos 2 x =
π
2π
π
3
π
f − = sin −
=−
+
+
6
6
6
2
6
π
2π
π
3
π
f = sin
=
−
−
6
6
6
2
6
π
π
π
and f = sin( π) − = −
2
2
2
π
π
Clearly, is the greatest value and −
2
2
is the least.
π
π
Therefore, difference =
+
= π
2
2
23 Given, y = x 5/2
d 2 y 15 1 /2
dy
5
x
∴
= x 3 /2 , 2 =
dx
4
dx 2
2
dy
d y
At x = 0,
=0
= 0,
dx
dx2
3
d y
and 3 is not defined,
dx
when x = 0, y = 0
∴(0, 0) is a point of inflection.
24 Area of triangle, ∆ = 1 x h2 − x2
2
x ( −2 x )
d∆ 1 2
2
= h − x +
=0
dx 2
2 h2 − x2
⇒
x=
h
2
C
h
A
⇒
∴
x
√h2 – x2
B
d ∆
h
< 0 at x =
dx2
2
2
∆ =
1
h
h2 h2
×
h2 −
=
2
2
2
4
25 Let the equation of drawn line be
y
x
+
= 1, where a > 3,
a
b
b > 4, as the line passes through (3, 4)
and meets the positive direction of
coordinate axes.
4a
3
4
We have, + = 1 ⇒ b =
a b
(a − 3)
Now, area of ∆AOB,
1
2a2
∆ = ab =
2
(a − 3)
d∆ 2a (a − 6)
=
da
(a − 3)2
Clearly, a= 6 is the point of minima for ∆.
2 × 36
Thus, ∆ min =
= 24 sq units
3
26 Let f ( x ) = x3 − px + q
Maxima
−√p/3
Minima
√p/3
TEN
146
f ′ ( x ) = 3 x2 − p
f ′( x ) = 0
p
p
x=
,−
3
3
f ′′ ( x ) = 6 x
p
p
, f ′′ ( x ) = 6
>0
3
3
[minima]
Then,
Put
⇒
Now,
At x =
and at x = −
p
f ′′ ( x ) < 0
3
[maxima]
27 We have, AF || DE and AE || FD
A
y
F
29 Let α and β be the roots of the equation
E
x2 − (a − 2) x − a + 1 = 0
x
B
D
C
Now, in ∆ABC and ∆EDC,
∠DEC = ∠BAC , ∠ACB is common.
⇒
∆ABC ≅ ∆EDC
b − y
x
c
Now,
= ⇒ x = (b − y )
b
b
c
Now, S = Area of parallelogram
AFDE = 2 (area of ∆AEF )
1
⇒
S = 2 xy sin A
2
c
= (b − y )y sin A
b
dS c
= sin A (b − 2 y )
b
dy
Sign scheme of
+
dS
,
dy
b/2
d2 y
<0
dx2
⇒ t = 1 is a point of local maxima.
d2 y
At (t = 4), 2 > 0
dx
⇒ t = 4 is a point of local minima.
d2 y
dy
At (t = 3),
and 2 are not defined
dx
dx
and change its sign.
d2 y
is unknown in the vicinity of t = 3,
dx2
thus t = 3 is a point of neither maxima
nor minima.
Finally, maximum and minimum values
of expression y = f ( x ) are 46 and −6,
respectively.
At (t = 1),
So, z has minima at a = 1.
So, α2 + β2 has least value for a = 1.
This is because we have only one
stationary value at which we have
minima.
Hence, a = 1.
30 Any tangent to the ellipse is
–
Hence, S is maximum when y =
Then, α + β = a − 2, αβ = − a + 1
Let
z = α 2 + β2
= (α + β )2 − 2αβ
= (a − 2)2 + 2 (a − 1)
= a2 − 2a + 2
dz
⇒
= 2a − 2
da
dz
Put
= 0 , then
da
⇒
a=1
d 2z
∴
= 2> 0
da2
b
⋅
2
c b b
× sin A
b 2 2
1 1
1
= bc sin A = (area of ∆ABC )
2
22
∴ S max =
28 We have,
dy
= 6 t 2 − 30 t + 24 = 6 (t − 1) (t − 4)
dt
dx
and
= 6 t − 18 = 6 (t − 3)
dt
dy (t − 1) (t − 4)
Thus,
=
dx
(t − 3)
which indicates that t = 1, 3 and 4 are
the critical points of y = f ( x ).
d2 y
d dy dt
Now,
=
⋅
dx2
dt dx dx
t 2 − 6 t + 11
1
=
×
(t − 3)2
6 (t − 3)
y
x
cos t + sin t = 1, where the point of
4
3
contact is (4cos t , 3sin t )
y
x
or
+
= 1,
4 sec t
3 cosec t
It means the axes Q (4sec t , 0) and
R (0, 3 cosec t ).
∴The distance of the line segment QR is
QR2 = D = 16 sec2 t + 9 cosec2 t
So, the minimum value of D is (4 + 3)2
or QR = 7.
31 Let S be the curved surface area of a
cone.
C
h
O
R
B
A
r
OA = AC − OC = h − R
In ∆OAB, R2 = r 2 + (h − R )2
⇒ r = 2Rh − h2
∴
S = πrl = π ( 2Rh − h2 )( h2 + r 2 )
= ( π 2Rh − h2 )( 2Rh )
Let S 2 = P
∴ P = π2 2R ( 2 Rh2 − h3 )
Since, S is maximum, if P is maximum,
then
dP
= 2 π2 R (4Rh − 3h2 ) = 0
dh
4R
∴
h = 0,
3
dP
Again, on differentiating
, we get
dh
2
d P
= 2 π2 R ( 4 R − 6 h )
dh2
4R
d 2P
< 0 at h =
dh2
3
32 Let OC = x, CQ = r
Now, OA = R [given]
Height of the cone = h = x + R
∴ Volume of the cone
1
= V = π r 2h
3
A
R
O
x
P
…(i)
h
r
Q
C
Also, in right angled ∆OCQ,
OC 2 + CQ 2 = OQ 2
⇒
x2 + r 2 = R2
…(ii)
⇒
r 2 = R2 − x2
From Eqs. (i) and (ii),
1
V = π (R2 − x2 )( x + R )
…(iii)
3
[Q h = x + R ]
On differentiating Eq. (iii) w.r.t. x, we
get
dV
1
= π [(R2 − x2 ) − 2 x( x + R )]
dx 3
dV
π
⇒
= (R2 − x2 − 2 x2 − 2 xR )
dx
3
dV
π
= (R2 − 2 xR − 3 x2 )
⇒
dx
3
dV
π
⇒
= (R − 3 x )(R + x )
…(iv)
dx
3
dV
For maxima, put
=0
dx
π
(R − 3 x )(R + x ) = 0
⇒
3
R
R
or x = − R ⇒ x =
x=
⇒
3
3
[since, x cannot be negative]
147
DAY
On differentiating Eq. (iv) w.r.t. x, we get
d V
π
= [(−3)(R + x ) + (R − 3 x )]
dx2
3
π
π
= (−2R − 6 x ) = − (2 R + 6 x )
3
3
6R
R d 2V
−π
At x = ,
=
2R +
3 dx2
3
3
4π
=−
R<0
3
So, V has a local maxima at x = R / 3.
Now, on substituting the value of x in
Eq. (iii), we get
π
R2
R
V = R2 −
R +
3
9
3
2
π 8R 4R
8 4
3
= ⋅
⋅
=
πR
3 9
3
27 3
8
× Volume of sphere
⇒ V =
27
2
33
Any point on parabola is (at 2 ,2at ),
i.e., (2t 2 , 4t ).
For its minimum distance from the
circle means its distance from the centre
(0, − 6) of the circle.
Let
z = (2t 2 )2 + (4t + 6)2
= 4(t 4 + 4t 2 + 12t + 9)
dz
= 4(4t 3 + 8t + 12)
∴
dt
⇒
16(t 3 + 2t + 3) = 0
⇒
(t + 1 )(t 2 − t + 3) = 0
⇒
t = −1
d 2z
2
⇒
= 16(3t + 2)> 0, hence minimum.
dt 2
So, point is (2, − 4).
37 We know that, volume of cylinder,
V = πR 2 h
B′
Y
(–a cos θ, b sin θ) (a cos θ, b sin θ)
A
B
X′
O
D
C
(–a cos θ, –b sin θ) (a cos θ, –b sin θ)
Y′
35 Given that, f ( x ) = ( x + 1)1 /3 − ( x − 1)1 /3
On differentiating w.r.t. x, we get
1
1
1
f ′( x ) =
−
3 ( x + 1)2 /3 ( x − 1)2 /3
=
( x − 1)2 /3 − ( x + 1)2 /3
3( x2 − 1)2 /3
Clearly, f ′( x ) does not exist at x = ± 1.
Now, put f ′ ( x ) = 0, then
( x − 1)2 /3 = ( x + 1)2 /3 ⇒ x = 0
At x = 0
f ( x ) = (0 + 1)1 /3 − (0 − 1)1 /3 = 2
Hence, the greatest value of f ( x ) is 2.
36 Q y 2 = 8 x. But y 2 = 4ax
⇒ 4a= 8
⇒ a= 2
r
h/2
B
C
A
R
2
h
In ∆OCA, r 2 = + R2
2
⇒
R2 = r 2 −
On differentiating w.r.t. x, we get
g ′( x ) = − 6 x + 6
For maxima or minima put g ′ ( x ) = 0
⇒
x=1
Now, g ′ ′ ( x ) = − 6 < 0 and hence, at
x = 1, g ( x ) (slope) will have maximum
value.
∴[g ( 1 )] max = − 3 × 1 + 6( 1 ) + 9 = 12
39 Given,
ab = 2 a + 3b
⇒
⇒ (a − 3) b = 2 a
2a
b =
a−3
Now, let z = ab =
2 a2
a−3
40 Q Perimeter of a
h2
4
sector = p
h2
V = π r2 −
∴
h
4
π
V = πr 2 h − h3 …(i)
⇒
4
On differentiating Eq. (i) both sides
w.r.t. h, we get
dV
3 πh2
= πr 2 −
dh
4
d2 V
−3 πh
=
⇒
dh2
2
For maximum or minimum value of V ,
dV
3 πh2
= 0 ⇒ πr 2 −
=0
dh
4
2
4r
2
⇒ h=
r
⇒ h2 =
3
3
d2 V
Now, 2
= − 3 πr < 0
dh h = 2 r
3
Thus, V is maximum when h =
2r
, then
3
2
h2
1 2r
2 2
= r2 −
= r
4
4 3
3
4 πr 3
2
Max V = πR h =
3 3
R2 = r 2 −
38 Let f ( x ) = − x 3 + 3 x2 + 9 x − 27
The slope of this curve
f ′ ( x ) = − 3 x2 + 6 x + 9
g ( x ) = f ′ ( x ) = − 3 x2 + 6 x + 9
On differentiating w.r.t. x, we get
dz 2 [(a − 3) 2 a − a2 ] 2 [a2 − 6a]
=
=
(a − 3)2
da
(a − 3)2
dz
For a minimum, put
=0
da
⇒ a2 − 6a = 0
⇒
a = 0, 6
d 2z
At a = 6, 2 = positive
da
When a = 6, b = 4
∴ (ab )min = 6 × 4 = 24
A′
h/2
X
Area of rectangle ABCD
= (2a cos θ) (2b sin θ) = 2ab sin 2θ
Hence, area of greatest rectangle is
equal to 2ab when sin 2θ = 1.
1
34 Let
f ( x) = x +
x
1
f ′( x ) = 1 − 2
x
For maxima and minima, put f ′ ( x ) = 0
1
⇒ 1 − 2 = 0⇒ x = ± 1
x
2
Now, f ′ ′ ( x ) = 3
x
At
x = 1, f ′ ′ ( x ) = + ve
[minima]
and at x = − 1, f ′ ′ ( x ) = −ve [maxima]
Thus, f ( x ) attains minimum value at
x = 1.
C′
Let
Let AOB be
the sector with
radius r.
If angle of the
sector be θ radians,
then area of sector,
1
A = r 2θ
2
A
r
O θ
s
r
and length of arc, s = r θ ⇒ θ =
B
…(i)
s
r
∴Perimeter of the sector
p = r + s + r = 2 r + s …(ii)
s
On substituting θ = in Eq. (i), we get
r
2A
1
s
1
A = r 2 = rs ⇒ s =
2 r 2
r
Now, on substituting the value of s in
Eq. (ii), we get
2A
2
p = 2 r +
⇒ 2 A = pr − 2 r
r
On differentiating w.r.t. r, we get
dA
2
= p − 4r
dr
For the maximum area, put
dA
=0
dr
⇒
p − 4r = 0
p
⇒
r =
4
TEN
148
SESSION 2
1 Let radius vector is r.
∴ r 2 = x2 + y 2
a2 y 2
⇒ r2 = 2
+ y2
y − b2
a2
b2
Q 2 + 2 = 1
y
x
For minimum value of r,
−2 yb 2 a2
d (r 2 )
=0 ⇒
+ 2y = 0
dy
( y 2 − b 2 )2
y 2 = b( a + b )
x2 = a(a + b )
r 2 = (a + b )2 ⇒ r = a + b
⇒
∴
⇒
Clearly, f ′( x ) changes its sign at x = − 1
from positive to negative and so f ( x ) has
local maxima at x = − 1.
Also, f ′(0) does not exist but f ′ (0− ) < 0
and f ′ (0+ ) > 0. It can only be inferred
that f ( x ) has a possibility of a minimum
at x = 0. Hence, it has one local maxima
at x = − 1 and one local minima at x = 0
So, total number of local maxima and
local minima is 2.
6 Total length = 2r + r θ = 20
y = g ( x ) , clearly g ( x ) has neither a
point of local maxima nor a point of
local minima.
r
y = x2 – 4|x|
–2
0
2
–6 –4
4
6
3 Clearly, f ( x ) in increasing just before
x = 3 and decreasing after x = 3. For
x = 3 to be the point of local maxima.
f (3) ≥ f (3 − 0)
⇒ − 15 ≥ 12 − 27 + log (a2 − 3 a + 3)
⇒
0 < a2 − 3 a + 3 ≤ 1 ⇒ 1 ≤ a ≤ 2
4 (Slope) f ′( x ) = e x cos x + sin xe x
f ′ ′ ( x ) = 2e x {sin ( x + π/4)
+ cos ( x + π/4)}
= 2e x ⋅ sin ( x + π/ 2)
For maximum slope, put f ′ ′ ( x ) = 0
⇒ sin ( x + π/2 ) = 0
⇒
cos x = 0
∴
x = π / 2, 3 π / 2
f ′ ′ ′ ( x ) = 2e x cos ( x + π/ 2)
f ′ ′ ′ ( π/2 ) = 2e x ⋅ cos π = − ve
Maximum slope is at x = π /2.
2
3 (2 + x ) , − 3 < x ≤ − 1
5 f ′( x ) = 2 −1 /3
x , − 1< x < 2
3
Y
A
X′
(–2, 0) (–1, 0) O
X
∴
Y′
⇒
Y
2 y = |x2 – 2|
y= x
3
1
A = 10r − r 2
dA
= 10 − 2r
dr
For maxima or minima, put
⇒
10 − 2r = 0 ⇒ r = 5
dA
= 0.
dr
1 2 20 − 2 (5)
(5)
2
5
1
= × 25 × 2 = 25sq m
2
7 Let c = av + b
…(i)
v
When v = 30 km/ h, then c = ` 75
b
…(ii)
75 = 30 a +
∴
30
When v = 40 km/h, then c = ` 65
b
…(iii)
∴
65 = 40 a +
40
On solving Eqs. (ii) and (iii), we get
1
and b = 1800
a=
2
On differentiating w.r.t. v in Eq. (i),
dc
b
= a− 2
dv
v
For maximum or minimum c,3
dc
b
=0 ⇒ v =±
dv
a
⇒
(–3, 0)
1
× 1800 = 2 × 30
2
c = 60
rθ
⇒
A max =
d 2c 2b
= 3
dv 2
v
at v =
b
a
+b
= 2 ab
a
b
c =2
r
20 − 2r
⇒
θ=
r
Now, area of flower-bed,
1
A = r 2θ
2
20 − 2r
1
⇒
A = r 2
2
r
∴
= e x 2 sin ( x + π/4)
c =a
| x2 − 2 |, − 1 ≤ x < 3
8 f ( x ) = x , 3 ≤ x < 2 3
3
3 − x, 2 3 ≤ x ≤ 4
θ
2 Bold line represents the graph of
b
the speed is most
a
economical.
∴Most economical speed is
So, at v =
b dx
,
>0
a dv 2
X′
–1
√2 √3
O
4
2√3
X
y=3 – x
Y′
From the above graph,
Maximum occurs at x = 0 and
minimum at x = 4.
3
2
| x + x + 3 x + sin x |
x≠0
9 f ( x ) = 3 + sin 1 ,
x
0,
x=0
Let g ( x ) = x3 + x2 + 3 x + sin x
g ′ ( x ) = 3 x2 + 2 x + 3 + cos x
2x
= 3 x2 +
+ 1 + cos x
3
1
8
= 3 x + + + cos x > 0
3
9
1
and 2 < 3 + sin < 4
x
2
Hence, minimum value of f ( x ) is 0 at
x = 0.
Hence, number of points = 1
10 According to given information, we
have
Perimeter of square + Perimeter of circle
= 2 units
⇒
4 x + 2 πr = 2
1 − 2x
...(i)
⇒
r =
π
Now, let A be the sum of the areas of
the square and the circle. Then,
A = x2 + πr 2
(1 − 2 x ) 2
= x2 + π
π2
(
1
−
2
x )2
⇒ A( x ) = x2 +
π
149
DAY
Now, for minimum value of A( x ),
dA
=0
dx
2 (1 − 2 x )
⇒ 2x +
⋅ (− 2) = 0
π
2 − 4x
⇒
x=
π
⇒
πx + 4 x = 2
2
...(ii)
⇒
x=
π+ 4
Now, from Eq. (i), we get
2
1 − 2⋅
π+ 4
r =
π
π + 4− 4
1
...(iii)
=
=
π(π + 4)
π+ 4
From Eqs. (ii) and (iii), we get
x = 2r
11 We have,
f ( x ) = x2 +
1
1
and g ( x ) = x −
x2
x
f ( x)
g( x)
⇒ h( x ) =
⇒
a7 =
=
49
is the greatest term.
543
13 f ( x ) = x 3 − 3(7 − a)x2 − 3(9 − a2 )x + 2
f ′ ( x ) = 3 x2 − 6(7 − a)x − 3(9 − a2 )
For real root D ≥ 0,
⇒ 49 + a2 − 14a + 9 − a2 ≥ 0
58
⇒a≤
14
For local minimum
f ′ ′ ( x ) = 6 x − 6(7 − a) > 0
⇒
7− x
x − 1 + 2
1
2
x
x
∴ h( x ) =
=
1
1
x−
x−
x
x
1
2
⇒ h( x ) = x − +
1
x
x−
x
1
1
2
x − > 0, x − +
∈ [2 2, ∞ )
1
x
x
x−
x
1
x − < 0,
x
2
x − 1 +
∈ (−∞, 2 2]
1
x
x−
x
∴ Local minimum value is 2 2.
x2 +
When
Thus constradictory, i.e., for real roots
58
and for negative point of local
a≤
14
minimum a > 7.
No possible values of a.
12 Consider the function
x2
( x + 200)
(400 − x 3 )
f ′( x ) = x 3
=0
( x + 200)2
3
when x = (400)1 /3 , (Q x ≠ 0)
x = (400)1 /3 − h ⇒ f ′ ( x ) > 0
x = (400)1 /3 + h ⇒ f ′ ( x ) < 0
(− 1, 0 ).
The equation of circle centre at Q with
variable radius r is
R
Since, 7 < (400) < 8, either a7 or a8 is
the greatest term of the sequence.
49
543
and
a8 =
8
89
P
[on solving Eqs. (i) and (ii)
A = Area of ∆QSR
1
= × QS × RT
2
1
r
= r . 4 − r 2
2 2
1 /3
a7 =
S
…(i)
( x + 1 )2 + y 2 = r 2
This circle meets the line segment QP at
S, where QS = r
It meets the circle x2 + y 2 = 1 at …(ii)
r2 − 2 r
R
,
4 − r2
2
2
∴ f ( x ) has maxima at x = (400)1 /3
Q
T O
∴
4 4 − r2
− (8r − 3 r 3 )
(−4r )
4 − r2
16 (4 − r )
2
r =
8
d2 A
, then 2 < 0
3
dr
Hence, A is maximum when r =
has x must be negative
⇒
7 − a< 0
⇒
a> 7
Q
8r − 3 r 3
dA
= 0, when r (8 − 3 r 2 ) = 0 giving
dr
8
r =
3
4 4 − r 2 (8 − 9r 2 )
d2 A
⇒ 2 =
dr
14 From the given figure coordinate of Q is
2
f (x ) =
49
8
>
543 89
and
[since, RT is the y-coordinate of R]
1
= [r 2 4 − r 2 ]
4
r 2 (− r )
dA 1
= 2r 4 − r 2 +
dr
4
4 − r 2
{2r (4 − r 2 ) − r 3 }
=
4 4 − r2
8
.
3
Then, maximum area
8
8 4 3
sq unit
=
4− =
4×3
3
9
15 Given,
P ( x ) = x 4 + a x3 + bx2 + cx + d
⇒ P ′ ( x ) = 4 x3 + 3 ax2 + 2 bx + c
Since, x = 0 is a solution for
P ′ ( x ) = 0, then
c =0
…(i)
∴ P ( x ) = x 4 + ax3 + bx2 + d
Also, we have P (− 1) < P (1)
⇒ 1− a+ b + d <1+ a+ b + d
⇒
a> 0
Since, P ′ ( x ) = 0, only when x = 0 and
P ( x ) is differentiable in (− 1, 1), we
should have the maximum and
minimum at the points x = − 1, 0 and 1.
Also, we have P (− 1) < P (1)
∴ Maximum of P ( x ) = Max {P (0), P (1)}
and minimum of P ( x ) = Min
{P (− 1), P (0)}
In the interval [0, 1],
P ′ ( x ) = 4 x3 + 3ax2 + 2 bx
= x (4 x2 + 3ax + 2 b )
Since, P ′( x ) has only one root x = 0,
then 4 x2 + 3 ax + 2 b = 0 has no real
roots.
∴
⇒
(3a)2 − 32 b < 0
9 a2
<b
32
∴
b>0
Thus, we have a > 0 and b > 0.
∴ P ′ ( x ) = 4 x3 + 3 ax2 + 2 bx > 0,
∀ x ∈ (0, 1)
Hence, P ( x ) is increasing in [0, 1].
∴ Maximum of P ( x ) = P (1)
Similarly, P ( x ) is decreasing in [−1 , 0].
Therefore, minimum P ( x ) does not
occur at x = − 1.
DAY FIFTEEN
Indefinite
Integrals
Learning & Revision for the Day
u
Integral as an
Anti-derivative
u
Fundamental Integration
Formulae
u
Methods of Integration
Integral as an Anti-derivative
A function φ ( x) is called a primitive or anti-derivative of a function f ( x), if φ′ ( x) = f ( x). If f1 ( x)
and f2 ( x) are two anti-derivatives of f ( x), then f1 ( x) and f2 ( x) differ by a constant. The collection
of all its anti-derivatives is called indefinite integral of f ( x) and is denoted by ∫ f ( x) dx.
d
{φ ( x) + C} = f ( x) ⇒ ∫ f ( x) dx = φ ( x) + C
dx
where, φ ( x) is an anti-derivative of f ( x), f ( x) is the integrand and C is an arbitrary constant
known as the constant of integration. Anti-derivative of odd function is always even and of
even function is always odd.
Thus,
Properties of Indefinite Integrals
●
●
●
∫ { f ( x) ± g( x)} dx = ∫ f ( x) dx ± ∫ g( x) dx
∫ k ⋅ f ( x) dx = k ⋅ ∫ f ( x) dx, where k is any non-zero real number.
∫ [k f ( x) + k f ( x) +...+ k f ( x)] dx = k ∫ f ( x) dx + k ∫ f ( x) dx + ... + k ∫ f ( x) dx,
1 1
2 2
1
n n
1
2
2
n
PRED
MIRROR
Your Personal Preparation Indicator
n
where k1 , k2 ,... k n are non-zero real numbers.
Fundamental Integration Formulae
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
There are some important fundamental formulae, which are given below
1. Algebraic Formulae
(i)
∫
xn + 1
x ndx =
+ C, n ≠ − 1
n+1
(ii)∫ (ax + b )n dx =
1 (ax + b )n
⋅
a
n+1
+1
+ C, n ≠ − 1
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
151
DAY
∫
1
dx = log| x | + C
x
1
1
dx = (log| ax + b |) + C
ax + b
a
(v)
(vi)
(iii)
(iv)
∫
(ix)
∫
1
1
a+ x
dx =
log
+C
2
2
a −x
2a
a− x
(iii)
x
∫
1
1
x−a
dx =
log
+C
x2 − a2
2a
x+a
∫a
(iv)
∫a
(bx + c )
∫a
2
1
∫
x − a2
2
x2 + a2
(xi)
∫
a2 − x2
(xiv) ∫
∫
∫
(xvii) ∫
dx =
1 a (bx + c )
⋅
+ C, a > 0 and a ≠ 1
b loge a
Methods of Integration
Following methods are used for integration
The method of reducing a given integral into one of the
standard integrals, by a proper substitution, is called method
of substitution.
x
dx = sin −1 + C
a
−1
−1
x −a
2
dx =
1
x
cosec −1 + C
a
a
1
1
x
a2 − x2 dx = x a2 − x2 + a2 sin −1 + C
a
2
2
1
1
x2 − a2 dx = x x2 − a2 − a2 log| x + x2 − a2 | + C
2
2
1
1
x2 + a2 dx = x x2 + a2 + a2 log| x + x2 + a2 | + C
2
2
∫ sin x dx = − cos x + C
(ii) ∫ cos x dx = sin x + C
(iii) ∫ tan x dx = − log|cos x | + C = log|sec x | + C
(iv) ∫ cot x dx = log|sin x | + C = − log| cosec x | + C
(i)
x
π
(v) ∫ sec xdx = log| sec x + tan x |+ C = log tan + + C
4 2
x
(vi) ∫ cosec x dx = log| cosec x − cot x | + C = log tan + C
2
∫ sec x dx = tan x + C
(viii) ∫ cosec x dx = − cot x + C
(ix) ∫ sec x ⋅ tan x dx = sec x + C
(x) ∫ cosec x ⋅ cot x dx = − cosec x + C
2
2
+C
ax
+ C, a > 0 and a ≠ 1
loge a
To evaluate an integral of the form
∫ f {g( x)} ⋅ g′ ( x) dx,
we
substitute g( x) = t , so that g′ ( x)dx = dt and given integral
reduces to ∫ f (t ) dt .
[ f ( x )]n +1
+C
n+ 1
• ∫ [ f ( x )]n ⋅ f ′( x ) =
NOTE
• If ∫ f ( x ) dx = φ( x ), then ∫ f ( ax + b) dx =
1
φ( ax + b) + C
a
(i) To evaluate integrals of the form
∫ ax
∫
2
dx
dx
or ∫
or
2
+ bx + c
ax + bx + c
ax 2 + bx + c dx
b
c
We write, ax2 + bx + c = a x2 + x +
a
a
2. Trigonometric Formulae
(vii)
+ b)
x2 + a2 | + C
1
x
dx = sec −1 + C
2
2
a
a
x −a
x
dx =
1 ( ax
⋅e
a
dx = log| x +
1
2
dx =
1. Integration by Substitutions
x
dx = cos + C
2
2
a
a −x
x
+C
x2 − a2 | + C
−1
∫
(xiii) ∫
(xvi)
1
( ax + b )
x
dx = log| x +
1
∫
(xv)
x
∫e
(x)
(xii)
∫ e dx = e
(i)
(ii)
1
1
x
dx = tan −1 + C
a
+ x2
a
−1
1
x
(viii) ∫ 2
dx = cot −1 + C
a
a + x2
a
(vii)
3. Exponential Formulae
2
b
c b2
= a x + + −
2 a
a 4c
This process reduces the integral to one of following forms
dX
dX
dX
,
=∫ 2
,
or ∫ 2
X − A2 ∫ X 2 + A2
A − X2
dX
dX
dX
∫ A 2 − X 2 , ∫ x2 − A 2 , ∫ X 2 + A 2
or
∫
A2 − X 2 dX , ∫
X 2 − A2 dX , ∫
A2 + X 2 dX
(ii) To evaluate integrals of the form
(px + q)
dx or
2
+ bx + c
∫ ax
or
∫ (px + q)
∫
(px + q)
ax 2 + bx + c
dx
ax 2 + bx + c dx
We put px + q = A {differentiation of (ax2 + bx + c)} + B,
where A and B can be found by comparing the coefficients of
like powers of x on the two sides.
152
2. Integration using Trigonometric
Identities
●
In this method, we have to evaluate integrals of the form
●
∫ sin mx ⋅ cos nx dx or ∫ sin mx ⋅ sinnx dx or
∫ cos mx ⋅ cos nx dx or ∫ cos mx ⋅ sinnx dx
To evaluate integrals of the form
1
1
∫ a sin x + b cos x dx or ∫ a + b sin x dx
1
1
or ∫
dx or ∫
dx
a + b cos x
a sin x + b cos x + c
x
2
(i) Put sin x =
and cos x =
x
x
1 + tan2
1 + tan2
2
2
x
x
x
(ii) Replace 1 + tan2 by sec2 and put tan = t .
2
2
2
2 tan
In this method, we use the following trigonometrical identities
(i) 2 sin A ⋅ cos B = sin ( A + B) + sin ( A − B)
(ii) 2 cos A ⋅ sin B = sin ( A + B) − sin ( A − B)
(iii) 2 cos A ⋅ cos B = cos ( A + B) + cos ( A − B)
(iv) 2 sin A ⋅ sin B = cos ( A − B) − cos ( A + B)
●
(v) 2 sin A ⋅ cos A = sin 2 A
1 + cos 2 A
(vi) cos2 A =
2
1 − cos 2 A
(vii) sin2 A =
2
(ix) sin2 A + cos2 A = 1
To evaluate integrals of the form ∫ sin p x cos q x dx
Where p, q ∈ Q, we use the following rules :
(i) If p is odd, then put cos x = t
(ii) If q is odd, then put sin x = t
(iii) If both p, q are odd, then put either sin x = t or cos x = t
(iv) If both p, q are even, then use trigonometric identities
only.
p + q − 2
(v) If p, q are rational numbers and
is a negative
2
integer, then put cot x = t or tan x = t as required.
To evaluate integrals of the form ∫
dx
dx
or
a + b cos 2 x
dx
,
x
a sin x + b cos 2 x
dx
dx
∫ ( asin x + bcos x) 2 or ∫ a + bsin2 x + c cos 2 x
∫ a + bsin
2
or ∫
a sin x + b cos x
dx,
c sin x + d cos x
d
we write a sin x + b cos x = A ( c sin x + d cos x)
dx
+B( c sin x + d cos x)
To evaluate integral of form∫
a sin x + b cos x + c
dx.
p sin x + q cos x + r
d
We write a sin x + b cos x + c = A (p sin x + q cos x + r)
dx
+ B (p sin x + q cos x + r) + C
To evaluate integral of the form∫
3. Integration of Different Types
of Functions
●
1 − tan2
Where A and B can be found by equating the coefficient of
sin x and cos x on both sides.
(viii) cos2 A − sin2 A = cos 2 A
●
x
2
2
(i) Divide both the numerator and denominator by
cos2 x.
(ii) Replace sec2 x by 1 + tan2 x in the denominator, if any.
(iii) Put tan x = t , so that sec2 x dx = dt
Where A, B and C can be found by equating the coefficient of
sin x, cos x and the constant term.
●
To evaluate integrals of the form ∫
or ∫
x2 − 1
dx
x 4 + kx 2 + 1
x2 + 1
dx
x 4 + kx 2 + 1
We divide the numerator and denominator by x2 and make
2
1
perfect square in denominator as x ± and then put
x
1
1
x + = t or x – = t as required.
x
x
●
Substitution for Some Irrational Integrand
(i)
a− x
a+ x
,
, x = a cos 2θ
a+ x
a− x
(ii)
x
a+ x
,
, x (a + x),
a+ x
x
1
, x = a tan2 θ
x (a + x)
or x = a cot2 θ
(iii)
x
a− x
,
, x (a − x),
a− x
x
or x = a cos2 θ
1
, x = a sin2 θ
x (a − x)
153
DAY
x
,
x−a
(iv)
x−a
, x ( x − a) ,
x
dx
∫ ( x − α)(β − x) , ∫
(v)
∫
(vi)
(vii)
∫
(ix)
x − α
dx
β − x
( x − α )(β − x) dx, put x = α cos2 θ + β sin2 θ
2
dx
+ bx + c)
px + q
dx
( px + q ) ax2 + bx + c
∫ ( px
dx
2
+ q ) (ax + b )
2
, put px + q = t 2
, put px + q =
first put x =
1
t
1
t
and then a + bt 2 = z2
4. Integration by Parts
(i) If u and v are two functions of x, then
du
∫ uI vII dx = u ∫ v dx − ∫ dx ⋅ ∫ v dx dx
We use the following preference in order to select the
first function
I
Inverse function
→
L
Logarithmic function
→
A
Algebraic function
→
T
Trigonometric function
→
E
Exponential function
→
(ii) If one of the function is not directly integrable, then we
take it as the first function.
(iii) If both the functions are directly integrable, then the
first function is chosen in such a way that its derivative
vanishes easily or the function obtained in integral sign
is easily integrable.
(iv) If only one which is not directly integrable, function is
there e.g. ∫ log x dx, then 1 (unity) is taken as second
function.
Some more Special Integrals Based on
Integration by Parts
(i)
∫e
x
(ii)
∫e
ax
{ f ( x) + f ′ ( x)}dx = f ( x)e x + C
sin (bx + c) dx =
e ax
a + b2
{a sin (bx + c) − b cos (bx + c)} + k
2
(iii)
e ax
a + b2
{a cos (bx + c) + b sin (bx + c)} + k
Here, c and k are integration constant.
∫e
ax
cos(bx + c)dx =
2
5. Integration by Partial Fractions
To evaluate the integral of the form
dx
, put ax + b = t 2
( px + q ) ax + b
∫ (ax
(viii) ∫
1
, x = a sec2 θ
x ( x − a)
P( x )
∫ Q( x) dx, where P( x), Q( x)
are polynomial in x with degree of P( x) < degree of Q( x) and
Q( x) ≠ 0, we use the method of partial fraction.
The partial fractions depend on the nature of the factors of
Q( x).
(i) According to nature of factors of Q( x), corresponding
form of partial fraction is given below:
If Q ( x) = ( x − a1 ) ( x − a2 ) ( x − a3 ) .... ( x − an ), then
we assume that
A2
A3
P ( x)
A1
An
=
+
+
+ ... +
,
Q( x) ( x − a1 ) ( x − a 2 ) ( x − a3 )
( x − an )
where the constants A1 , A2 , K , An can be determined
by equating the coefficients of like power of x or by
substituting x = a1 , a2 , K , an .
(ii) If Q ( x) = ( x − a)k ( x − a1 ) ( x − a2 ) K ( x − ar ), then we
assume that
P( x )
A1
A2
Ak
=
+
+ ... +
Q( x) ( x − a) ( x − a)2
( x − a)k
B1
B2
Br
+
+
+ ... +
( x − a1 ) ( x − a2 )
( x − ar )
where the constants A1 , A2 ,..., Ak , B1 , B2 ..., Br can be
obtained by equating the coefficients of like power of
x.
(iii) If some of the factors in Q( x) are quadratic and
non-repeating, corresponding to each quadratic factor
ax2 + bx + c (non-factorisable), we assume the partial
Ax + B
fraction of the type 2
, where A and B are
ax + bx + c
constants to be determined by comparing coefficients
of like powers of x.
(iv) If some of the factors in Q( x) are quadratic and
repeating, for every quadratic repeating factor of the
type (ax2 + bx + c)k where ax2 + bx + c cannot be further
factorise, we assume
A2 k − 1 x + A2 k
A1 x + A2
A3 x + A4
+
+ ... +
2
2
2
ax + bx + c (ax + bx + c)
(ax2 + bx + c)k
If degree of P ( x) > degree of Q( x), then we first divide P( x) by
P( x )
P ( x)
is expressed in the form of T ( x) + 1 , where
Q( x)
Q( x)
P ( x)
is a proper rational function
T ( x) is a polynomial in x and 1
Q( x)
(i.e. degree of P1 ( x) < degree of Q( x))
Q( x) so that
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 If ∫
2
dx
x6
to
=
p
(
x
),
then
∫ x + x 7 dx is equal
x + x7
j JEE Mains 2013
(a) log | x | − p (x) + C
(b) log | x | + p (x) + C
(c) x − p (x) + C
(d) x + p (x) + C
∫
x 3 −1
dx is equal to
( x 4 + 1)( x + 1)
8 If ∫ x + x 2 + 5 dx = P {x + x 2 + 5} 3 / 2
(a)
3
10
∫ ( x + 1)( x + 2) ( x + 3) dx is equal to
7
4 The integral ∫
dx
3
11
12
equals
x + 1 4
(a)
+C
x4
j
(c) − (x + 1) + C
(b) (x 4 + 1) 4 + C
4
(a) (sinα,cosα)
(c) (− sinα,cosα)
sin8 x − cos 8 x
∫ 1 − 2 sin2 x cos 2 x dx is equal to
14
(b) (cosα,sinα )
(d) (− cosα,sinα)
(b) cos x
(d) cot x
2
(log x − 1)
dx is eqal to
2
∫ 1 + (log x )
x
+C
(log x)2 + 1
x
(c) 2
+C
x +1
∫
1
sin 2 x + C
2
(d) − sin2x + C
(b) −
1
sin 2 x + C
2
( 3 x + 2 − x 2 )( 6 1 − x 2 − x 2 )
3
1− x 2
cos 6x + 6 cos 4x + 15 cos 2x + 10
dx = f ( x ) + C,
2
x + 5 cos x cos 3x + cos x cos 5x
then f (10) is equal to
(b) 10
∫x
(a)
(c)
15
5
x 2 −1
3
2x 4 − 2x 2 + 1
dx
2
xe x
+C
1+ x 2
log x
(d)
+C
(log x)2 + 1
(d) 2 cos10
x10
+C
2 (x + x 3 + 1)2
− x10
(d)
+C
5
2 (x + x 3 + 1)2
(b)
5
dx is equal to
2x 4 − 2x 2 + 1
+C
x2
4
2
2x − 2x + 1
+C
2x
∫ (1 + x
(c) 2 sin10
2x 12 + 5x 9
dx is equal to
( x 5 + x 3 + 1)3
j
JEE Mains 2016
) p 2 + q 2 (tan−1 x )2
(b)
(d)
2x 4 − 2x 2 + 1
+C
x
4
2
2x − 2x + 1
+C
2x 2
is equal to
1
log[q tan−1 x + p 2 + q 2 (tan−1 x)2 ] + C
q
(b) log[q tan−1 x + p 2 + q 2 (tan−1 x)2 ] + C
2
(c)
(p 2 + q 2 tan−1 x)3 / 2 + C
3q
(a)
(b)
dx ;x ∈( 0,1) equals
∫ 10 cos
(a)
f (x )
dx = log log sin x + C, then f ( x ) is equal to
log sin x
(a) sinx
(c) logsinx
(a)
1
x 3π
1
x 3π
log tan −
log tan +
+ C (d)
+C
2
2
2
8
2
8
− x5
+C
(x + x 3 + 1)2
x5
(c)
+C
5
2 (x + x 3 + 1)2
x 4 + 1 4
(d) −
+C
x4
sin x
dx = Ax + B log sin ( x − α ) + C,
sin( x − α )
then the value of ( A , B ) is
7
(c)
13 The integral ∫
5 If ∫
6 If ∫
1
x π
1
x
log tan − + C (b)
log cot + C
2 8
2
2
2
(a) 20
JEE Mains 2015
1
1
1
(d) −5
(a) 21/ 6 x + C (b) 21/12 x + C (c) 21/ 3 x + C (d) None of these
x ( x 4 + 1) 4
4
(c) −3
(a)
(c)
2
1
(b) −4
dx
is equal to
cos x − sin x
8
(a)
4
∫
+ C, then the value of 3PQ is
(a) sin2x + C
(x + 2)
(x + 2)
−
+C
10
8
(x + 1)2 (x + 2)8 (x + 3)2
(b)
−
−
+C
2
8
2
10
(x + 2)
(c)
+C
10
2
(x + 1)
(x + 2)8 (x + 3)2
(d)
+
+
+C
2
8
2
10
x x2 + 5
(a) −1
9
1
1
log(1 + x 4) + log(1 + x 3) + C
4
3
1
1
(b) log(1 + x 4) − log(1 + x 3) + C
4
3
1
(c) log(1 + x 4) − log(1 + x) + C
4
1
(d) log(1 + x 4) + log(1 + x) + C
4
Q
+
(d) None of the above
155
DAY
16 In the integral ∫
cos 8x + 1
dx = A cos 8x + k , where
cot 2x − tan 2x
24
k is an arbitrary constant, then A is equal to
j
(a) −
17
∫
∫
(b)
1
16
(c)
1
8
(sin θ + cos θ )
dθ is equal to
sin 2 θ
(a)
(b)
(c)
(d)
18
1
16
j
log | cosθ − sinθ + sin 2 θ |
log | sinθ − cosθ + sin 2 θ|
sin−1 (sinθ − cosθ) + C
sin−1 (sinθ + cosθ) + C
equal to
1 g (x)
(b)
2 f (x)
2
(c)
1 g (x)
log
+ C
2 f (x)
25
20 If ∫
21
∫ cos
−
x sin
−
11
7
1
(c) , − 1
5
(b)
4
22
∫
1
(d) − , 0
5
4
tan7 x + C
7
3
28 If ∫
+C
+C
+C
∫ g ( x ) e dx is equal to
x
(b) f (x) + g (x) + C
(d) f (x) − g (x) + C
29
j
∫ tan
−1
(b) cos x
(c) 2 tan−1
tan (x / 2)
+C
2
(d) None of these
(d) cot x
j
NCERT Exemplar
(b) x tan −1 x − x + C
(d) x − (x + 1) tan −1 x + C
30 If I n = ∫ (log x )n dx , then I n + n I n − 1 is equal to
(a) x (log x)n
(b) (x log x)n
(a) g −1(x)
(c) xf −1(x) − g −1(x)
32
(c) (log x)n −1
(d) n (log x)n
∫
23 If the integral
∫ tan x − 2 dx = x + a ln | sin x − 2 cos x | + k ,
(b) xf −1(x) − g (f −1(x))
(d) f −1(x)
( x + 3) e x
dx is equal to
( x + 4) 2
1
+C
(x + 4) 2
ex
(c)
+C
x+ 4
(a)
(b) − 2
(d) 2
(c) tanx
xdx is equal to
(a) (x + 1) tan−1 x − x + C
(c) x − x tan−1 x + C
NCERT Exemplar
tan (x / 2) + 1
−1 tan (x / 2) + 1
+ C (b) tan
+C
2
2
then a is equal to
JEE Mains 2013
31 If ∫ f ( x ) dx = g ( x ), then ∫ f −1( x ) dx is equal to
(a) 2 tan−1
5 tan x
j
1 − 6 cos 2 x
f (x )
dx =
+ C, then f ( x ) is equal to
6
2
sin x cos x
(sin x )6
(a) sinx
(d) log| cos 7 x | + C
dx
is equal to
2 + sin x + cos x
(a) − 1
(c) 1
3 −1 x
sin
a
2
2/ 3
(d)
3/ 2
1 3
[x Ψ (x 3)] − ∫ x 2 Ψ (x 3)dx + C
3
1
(b) [x 3 Ψ (x 3)] − 3 ∫ x 3 Ψ (x 3)dx + C
3
1
(c) [x 3 Ψ (x 3) − ∫ x 2 Ψ (x 3)dx] + C
3
1
(d) [x 3 Ψ (x 3)] − ∫ x 3 Ψ (x 3)dx + C
3
4
−
−7
tan 7 x + C
4
2 −1 x
sin
a
3
3/ 2
(b)
(a)
4
(c)
3 −1 x
sin
a
2
+C
27. If ∫ f ( x ) dx = Ψ( x ), then ∫ x 5f ( x 3 ) dx is equal to
x dx is equal to
(a) log | sin7 x | + C
x 2 − x + 1
1
log 2
+C
2
x + x + 1
3/ 2
(a) f (x) ⋅ g (x) + C
(c) e x cos x + C
2
(b) 2
(d) None of these
3
7
(d)
3
then ∫ f ( x ) cos x dx +
tan x
2
dx = x −
tan−1
1 + tan x + tan2 x
A
2 tan x + 1
+ C, then the value of A is
A
(a) 1
(c) 3
x 2 − x − 1
1
log 2
+C
2
x + x + 1
26 If an anti-derivative of f ( x ) is e x and that of g ( x ) is cos x,
2
g (x)
(d) log
+C
f (x)
(b)
x
dx is equal to
a −x3
∫
(c)
I 4 + I 6 = a tan5 x + bx 5 + C, where C is a constant of
integration, then the ordered pair (a, b ) is equal to
1
(b) , 0
5
x 2 + x + 1
1
log 2
+C
2
x − x + 1
x 2 − x + 1
(c) log 2
+C
x + x + 1
19 Let I n = ∫ tann x dx (n > 1). If
1
(a) − , 1
5
x 2 −1
dx is equal to
+ x2 +1
x
(a) sin−1
a
f ( x ) ⋅ g ' ( x ) − f ' ( x ) ⋅ g ( x )
((log g ( x ) − log f ( x )) dx is
f ( x ) ⋅ g( x )
g (x)
(a) log
+C
f (x)
4
(a)
JEE Mains 2013
1
(d) −
8
JEE Mains 2017
∫x
33 If ∫
ex
+C
(x + 4) 2
ex
(d)
+C
x+ 3
(b)
−1
x 2 − x + 1 cot −1 x
e
dx = A ( x ) e cot x + C, then A ( x ) is
2
x +1
equal to
(a) − x
j
(b) x
(c)
1− x
JEE Mains 2013
(d)
1+ x
156
34 If g ( x ) is a differentiable function satisfying
35
d
{ g ( x )} = g ( x ) and g( 0) = 1, then
dx
2 − sin 2x
∫ g ( x ) 1 − cos 2x dx is equal to
(a) g (x) cot x + C
g (x)
(c)
+C
1 − cos 2 x
(b) − g (x) cot x + C
∫x
3
dx 3
is equal to
( x n + 1)
(a)
3 xn
ln
+C
n x n + 1
(b)
1 xn
ln
+C
n x n + 1
(c)
3 x n + 1
ln
+C
n xn
x n +1
(d) 3n ln n + C
x
(d) None of these
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 If ∫ f ( x ) dx = f ( x ), then ∫ {f ( x )} 2 dx is equal to
(a)
1
{f (x)}2
2
(b) {f (x)}3
0
2 If f ( x ) = sin x − x 2
2 − cos x
(c)
x 2 − sin x
0
2x − 1
{f (x)}3
3
(d) {f (x)}2
cos x − 2
1 − 2x ,
0
6
x
− x 2 sin x + sin 2 x + C
3
x3
(b)
− x 2 sin x − cos 2 x + C
3
x3
(c)
− x 2 cos x − cos 2 x + C
3
(d) None of the above
(a)
∫e
(a)
(b)
(c)
(d)
cosθ + sinθ
(b) (cosθ + sinθ) 2 log
+C
cosθ − sinθ
1 − cos 2a x
dx is equal to
1 + sin 2a x
7
1
π
− e 2ax cos + ax + C
4
a
1 2ax
π
−
e cot + ax + C
4
2a
1 2ax
π
−
e cos + ax + C
4
2a
1 2ax
π
− e cosec + ax + C
4
a
∫x
(a) log
(c)
1
log | sec (x 2 + 1) | + C
2
cosθ − sinθ
(cosθ − sin θ) 2
log
+C
2
cosθ + sinθ
(d)
1
π
1
sin 2 θ logtan + θ − log sec 2 θ + C
4
2
2
∫ (x
x2
dx is equal to
sin x + cos x ) 2
sin x + cos x
+C
x sin x + cos x
x sin x − cos x
(b)
+C
x sin x + cos x
(c)
sin x − x cos x
+C
x sin x + cos x
(d) None of the above
2 sin ( x + 1) − sin 2 ( x + 1)
dx is equal to
2 sin ( x 2 + 1) + sin 2 ( x 2 + 1)
1
sec (x 2 + 1) + C
2
(c)
(a)
4 If x 2 ≠ nπ − 1, n ∈ N. Then, the value of
2
cos θ + sin θ
∫ cos 2 θ log cos θ − sin θ dθ is equal to
cosθ + sinθ
(a) (cosθ − sinθ) 2 log
+C
cosθ − sinθ
3
3
2
x 2 + 1
(b) log sec
+C
2
(d) None of these
8 If f ( x ) = ∫
x 2 dx
(1 + x )(1 + 1 + x 2 )
2
5
sin2 x cos 2 x
dx
x + cos x sin2 x + sin3 x cos 2 x + cos 5 x )2
is equal to
3
j
JEE Mains 2018
and f ( 0) = 0, then the
value of f (1) is
(a) log (1 +
2)
(c) log (1 +
2) +
5 The integral
∫ (sin
(b)
(where C is a constant of integration)
then∫ f ( x ) dx is equal to
2 ax
−1
+C
3 (1 + tan3 x)
−1
(d)
+C
1 + cot 3 x
1
+C
3 (1 + tan3 x)
1
(c)
+C
1 + cot 3 x
(a)
9 If I = ∫
(b) log (1 +
π
2
(a) 2/3
(c) 1/3
( x + 1) ( x − 1)
2
π
4
(d) None of these
dx
3
2) −
4
=k
3
1+ x
+ C, then k is equal to
1− x
(b) 3/2
(d) 1/2
157
DAY
10
dx
∫ (sin x + 2)(sin x − 1) is equal to
1
x
2 tan +
2
2
−1
2
2
(a)
tan
−
+C
x
3
3 tan − 1 3 3
2
2
(b)
+C
tan x + 1
2
2 tan x − 1
2
1
2
−1
2
(c) −
+
tan
+C
3 tan x − 1
3 3
3
2
2 tan x − 1
2
2
2
2
(d) −
+
tan−1
+C
3 tan x − 1
3 3
3
2
11
∫
(a) (x − 1) e
13
1 x2
6 2 + 3 x 2
3/ 2
(b)
1 x2
6 2 + 3 x 2
7/ 2
(c)
x+
x
x+
+C
(b) xe
x+
1
x
+C
(d) − xe
99
JEE Mains 2014
1
x
+C
x+
1
x
+C
x ) dx is equal to
cos(100 x) (sin x)100
+C
100
100
cos(100 x)(cos x)
(d)
+C
100
sin(100x) (sin x)
+C
100
100
cos(100 x)(cos x)
(c)
+C
100
(a)
∫
dx is equal to
j
100
14
1
x
1
∫ (sin(101 x ) ⋅ sin
(b)
−1
x
( 2018)2 x
( 2018)sin ( 2018 ) dx is equal to
1 − ( 2018)2 x
(a) (log2018 e)2 (2018)sin
x
dx is equal to
( 2 + 3 x 2 )5 / 2
3/ 2
x+
(c) (x + 1) e
2
1 x2
(a)
5 2 + 3 x 2
1
x
12 The integral ∫ 1 + x − e
−1
( 2018 ) x
+C
2
x + sin −1 ( 2018 ) x
+C
2
x − sin −1 ( 2018 ) x
+C
(b) (log2018 e) (2018)
(c) (log2018 e) (2018)
−1
x
(2018)sin ( 2018)
(d)
+C
(log2018 e)2
+C
15
+C
∫ (∫ e
x
2 1
log x + − 2 ) dx is equal to
x x
(a) e x log x + C1x + C2
+C
(c)
(d) None of the above
logx
+ C1x + C2
x
(b) logx +
1
+ C1x + C2
x
(d) None of these
ANSWERS
SESSION 1
SESSION 2
1 (a)
2 (c)
3 (a)
4 (d)
5 (b)
6 (d)
7 (a)
8 (d)
9 (d)
10 (b)
11 (a)
12 (a)
13 (b)
14 (d)
15 (a)
16 (a)
17 (c)
18 (c)
19 (b)
20 (c)
21 (c)
22 (a)
23 (d)
24 (d)
25 (b)
26 (c)
27 (a)
28 (c)
29 (a)
30 (a)
31 (b)
32 (c)
33 (b)
34 (b)
35 (a)
1 (a)
11 (b)
2 (d)
12 (b)
3 (b)
13 (a)
4 (b)
14 (a)
5 (b)
15 (a)
6 (d)
7 (c)
8 (b)
9 (b)
10 (a)
158
Hints and Explanations
1 Let I =
x6
∫x+
7
x
dx =
x6
∫ x (1 +
6
x )
(1 + x 6 ) − 1
=∫
dx
x (1 + x 6 )
dx
dx
⇒ I =∫
−∫
x
x + x7
∫ (x
= cot x
x −1
dx
+ 1)( x + 1)
3
4
7
∫
=
∫
x ( x + 1) − ( x + 1)
dx
( x 4 + 1)( x + 1)
=
∫ x
=
1
log( x 4 + 1) − log( x + 1) + C
4
3
=
x3
1
−
dx
+ 1 x + 1
4
=
=
9
7
t
t
( x + 2)
( x + 2)
=
− +C=
−
+C
10 8
10
8
dx
dx
4 ∫
=∫
3
3
2
4
4
1 4
x ( x + 1)
x 5 1 + 4
x
8
(log x ) + 1 − 2 log x
dx
[(log x ) 2 + 1] 2
d
10
8
x
∫ dx (log x )
∫
⇒
1/4
x4 + 1
+ C = −
x4
Put x − α = t ⇒ dx = dt
sin (t + α )
∴I =∫
dt
sin t
cos t
∫ sin α ⋅ sin t
dt
= cos α ⋅ t + sin α log sin t + C
= x cos α + sin α
log {sin( x − α )} + C
∴
A = cos α, B = sin α
∫
x2 + 5 dx
x2 + 5 = t − x
⇒
⇒
⇒
x2 + 5 = t 2 + x2 − 2 xt
5 = t 2 − 2 xt
2 xt = t 2 − 5
1
5
x = t −
2
t
1
5
dx = 1 + 2
2
t
⇒
2
I = ∫t
1 /2
=
∫
=
∫
=
∫
1/4
+C
3
6
3
[ x 2 + ( 2 − x2 )2 − 2 x 2 − x 2 ]
6
2
3
x + 2− x2
1/6 3
2
3
21 / 6 3 1 − x 2
=
∫
(sin 4 x + cos 4 x ) (sin 4 x − cos 4 x )
dx
2
x + cos 2 x )2 − 2sin2 x cos 2 x
dx
= 2x + C
Clearly, f (10) = 20
2 x12 + 5x 9
dx
5
+ x3 + 1) 3
13 Let I =
∫ (x
∫ (1 +
∫ (sin
dx
⋅ cos x + 10 ⋅ (2cos 2 x )
dx = 2∫ dx
10cos 2 x + 5cos x ⋅ cos 3 x
+ cos x cos 5x
=
I =
1 − x2
2cos 5x ⋅ cos x + 10 ⋅ cos 3 x
1
x 3π
log tan +
+C
2
8
2
sin x − cos x
dx
2
x cos 2 x
2− x2 − x
dx
(cos 6 x + cos 4 x ) + 5 (cos 4 x
+ cos 2 x ) + 10(cos 2 x + 1)
=∫
dx
10cos 2 x + 5cos x cos 3 x
+ cos x cos 5x
=
∫ 1 − 2 sin
dx
cos 6 x + 6cos 4 x
x
+
cos
+
15
2
10
dx
12 Let I = ∫
2
10cos x + 5cos x cos 3 x
+ cos x cos 5x
∫x
I =
3
(2 − x 2 ) − x 2
=
10 Let
1− x2
3
1
π
x
π
log tan +
+ + C
4
2
2
8
8
1 − x2
dx
x + 2− x2
=
8
1 − x2
= 21 / 6 ∫ dx = 21 / 6 x + C
1
5
⋅ 1 + 2 dt
2
t
1 1 /2
(t + 5t −3 / 2 )dt
2∫
12
10
1 3 /2
5
= t 3 /2 −
+C
+C= t −
23
3
t
t
Clearly, 3PQ = −5
dx
9 Let I = 1 ∫
2 1 cos x − 1 sin x
2
2
1
x + π dx
=
sec
∫ 4
2
( x + 2 − x2 )( 1 − x 2 − x2 )
3
3
=
5 Let I = ∫ sin x dx
sin( x − α )
= ∫ cos α dt +
x+
⇒
Now,
Hence, the integral becomes
− t 3dt
∫ t 3 = − ∫ dt = − t + C
1
= − 1 + 4
x
dx
+ 1
x + 5=t
and
1
= t4
x4
4
− 5 dx = 4t 3dt
x
dx
= − t 3dt
x5
2
Put x +
Put 1 +
⇒
=
x
+C
(log x ) 2 + 1
8 Let I =
= ∫ (t − 1)⋅ t dx = ∫ (t − t )dx
7
∫
∫
sin2 x
+C
2
1
x + 2 − x 2 6 (2 − 2 x 2 − x 2 )
2
3
2
∫ − cos 2x dx
=−
1
(log x )2 + 1 − 2 x log x ⋅
x
=∫
dx
[(log x ) 2 + 1]2
Put x + 2 = t
⇒ x = t − 2 and dx = dt
∴ I = ∫ (t − 1)t 7 (t + 1)dx
10
(log x − 1)
∫ 1 + (log x )2 dx
4
3 Let I = ∫ ( x + 1)( x + 2) 7 ( x + 3)dx
2
=
11 Let I =
2
x3 + x 4 − x 4 − 1
dx
( x 4 + 1)( x + 1)
=
= ∫ (sin2 x − cos 2 x )dx
f '( x )
Q ∫ f ( x ) dx = log( f ( x )) + C
= log| x| − p( x ) + C
2 Let I =
= ∫ (sin 4 x − cos 4 x )dx
f ( x)
= log(log sin x ) + C
log sin x
d
∴ f ( x) =
(log sin x )
dx
6 We have, ∫
dx
15
2 x12 + 5x 9
dx
(1 + x − 2 + x − 5 ) 3
2 x − 3 + 5x − 6
dx
x − 2 + x − 5)3
Now, put 1 + x − 2 + x − 5 = t
⇒ (− 2 x − 3 − 5x − 6 ) dx = dt
⇒
(2 x − 3 + 5x − 6 ) dx = − dt
dt
I = − ∫ 3 = − ∫ t − 3 dt
∴
t
159
DAY
=−
=
14 Let I =
t −3 +1
1
+C= 2 +C
−3+ 1
2t
x10
+C
2( x + x3 + 1) 2
5
x2 − 1
∫x
dx
2x4 − 2x 2 + 1
x2 − 1
=∫
dx
2
1
x5 2 − 2 + 4
x
x
1
1
−
x3 x 5
=∫
dx
2
1
2− 2 + 4
x
x
2
1
Now, putting 2 − 2 + 4 = t, we get
x
x
1
1
4 3 − 5 dx = dt
x
x
1 dt
∴I = ∫
4
t
3
1
= ⋅2 t + C =
4
g( x)
=t
f ( x)
f ( x )⋅ g '( x ) − g ( x )⋅ f '( x )
⇒
dx = dt
( f ( x )) 2
f ( x )⋅ g '( x ) − g ( x ) ⋅ f '( x )
⇒
f ( x )⋅ g ( x )
g( x)
⋅
⋅ dx = dt
f ( x)
Put
f ( x )⋅ g '( x ) − g ( x )⋅ f '( x )
dt
⇒
dx =
t
f ( x )⋅ g ( x )
Now, I =
1
∫ t ⋅ log t dt =
2
=
g ( x )
1
log
+ C
2
f ( x )
19 We have, I n = ∫ tan n x dx
∫ tan
= ∫ tan
= ∫ tan
∴ I n + I n +2 =
2x4 − 2x 2 + 1
+C
2x 2
=
15 Put q tan −1 x = t
q
1
dt
dx =
dx = dt ⇒
1 + x2
q
1 + x2
dt
1
= log [ t + p2 + t 2 ]
∴∫
q p2 + t 2 q
⇒
=
1
log [q tan −1 x
q
+
16 LHS =
∫
p + q (tan
2
2
−1
x) ] + C
2
2
2cos 4 x
dx
cos 2 2 x − sin2 2 x
cos 2 x sin 2 x
2cos 2 4 x × cos 2 x sin 2 x
dx
cos 4 x
1
= ∫ cos 4 x × sin 4 xdx = ∫ sin 8 xdx
2
−1 cos 8 x
=
+ k
2
8
−1
Hence, we get A =
16
sin θ + cos θ
17 Let I = ∫
dθ
1 − (1 − 2sin θ cos θ )
=
∫
=
Put
sin θ + cos θ
∫
1 − (sin θ − cos θ) 2
dθ
sin θ − cos θ = t
⇒ (cos θ + sin θ) dθ = dt
dt
∴ I =∫
= sin −1 ( t ) + C
1 − t2
−1
= sin (sin θ – cos θ) + C
f ( x ) ⋅ g '( x ) − f '( x )⋅ g ( x )
f ( x )⋅ g ( x )
18 Let I = ∫
g ( x )
log
dx
f ( x )
(log t )2
+C
2
∫ tan
n +2
n
x dx +
n
x(1 + tan x ) dx
n
x sec2 x dx
x dx
2
n +1
tan
x
+C
n+1
Put n = 4, we get I 4 + I 6 =
tan 5 x
+C
5
1
and b = 0
5
tan x
20 Let I = ∫
1 + tan x + tan2 x
∴
a=
sin x
cos x
dx = ∫
dx
sin2 x
sin x
1+
+
cos 2 x cos x
sin 2 x
=∫
dx
2 + sin 2 x
dx
= ∫ dx − 2 ∫
2 + sin 2 x
= x−2∫
sec2 x
dx
2
2 sec x + 2 tan x
Let tan x = t
⇒ sec2 x dx = dt
2
dt
= x− ∫ 2
2 t +t +1
dt
= x−∫
2
2
t + 1 + 3
2
2
⇒I = x−
2
tan −1
3
2 tan x + 1
+C
3
Hence, we get A = 3.
21 Here, m + n =
− 3 −11
+
= −2
7
7
I = ∫ cos −3 /7 x(sin( − 2 + 3 /7 ) x )dx
= ∫ cos −3 /7 x sin −2 x sin3 /7 xdx
=
cosec2 x
dx =
3 /7
x
sin3 / 7 x
∫ cos
cosec2 x
dx
3 /7
x
∫ cot
Put cot x = t ⇒ − cosec2 x dx = dt
dt
−7 4 / 7
t +C
∴
I = − ∫ 3 /7 =
t
4
7
= − tan − 4 / 7 x + C
4
dx
22 Let I = ∫
2 + sin x + cos x
dx
⇒ I =∫
x
x
2 tan
1 − tan2
2 +
2
2+
x
x
1 + tan2
1 + tan2
2
2
x
sec2 dx
2
=∫
x
x
x
2 + 2 tan2 + 2 tan + 1 − tan2
2
2
2
x
sec2 dx
2
I =∫
x
x
tan2 + 2 tan + 3
2
2
x
1
x
Put tan = t ⇒ sec2 dx = dt
2
2
2
2dt
∴I = ∫ 2
t + 2t + 3
dt
2dt
= 2∫ 2
= 2∫
t + 2t + 1 + 2
(t + 1)2 + ( 2 ) 2
= 2⋅
1
t + 1
tan −1
+C
2
2
tan x + 1
2
⇒ I = 2 tan
+C
2
5 tan x
23 Given, ∫
dx
tan x − 2
−1
= x + a ln|sin x − 2 cos x | + k
Now, let us assume that
5 tan x
I =∫
dx
tan x − 2
…(i)
On multiplying by cos x in numerator
and denominator, we get
5 sin x
I =∫
dx
sin x − 2 cos x
Let 5 sin x = A (sin x − 2 cos x )
+ B (cos x + 2 sin x )
⇒ 0 ⋅ cos x + 5 sin x = ( A + 2B ) sin x
+ (B − 2 A ) cos x
On comparing the coefficients of sin x
and cos x, we get
A + 2B = 5 and B − 2 A = 0
⇒
A = 1 and B = 2
⇒
5 sin x = (sin x − 2 cos x )
+ 2 (cos x + 2 sin x )
160
⇒I =
5 sin x
∫ sin x − 2 cos x dx
(sin x − 2cos x ) + 2 (cos x + 2 sin x )
dx
(sin x − 2 cos x )
d (sin x − 2 cos x )
⇒ I = ∫ 1 dx + 2 ∫
(sin x − 2 cos x )
=
∫
I = x + 2 log|(sin x − 2 cos x )| + k
…(ii)
where, k is the constant of integration.
On comparing the value of I in Eqs. (i)
and (ii), we get a = 2
24
∫
1 − 1
x2
dx
2
1
x
+
−
1
x
1
1
Put x + = t ⇒ 1 − 2 dx = dt
x
x
∴
t − 1
dt
1
∫ t 2 − 1 = 2 log t + 1 + C
x2 − x + 1
1
= log 2
+C
2
x + x + 1
25 Let I = ∫ 3 x 3 dx
a −x
Put x = a(sin θ) 2 / 3
2
⇒ dx = a(sin θ)−1 / 3 cos θ dθ
3
2
a1 / 2 (sin θ) 1 / 3 ⋅
3
a(sin θ)−1 /3 ⋅ cos θ
∴ I =∫
dθ
a3 − a3 sin2 θ
2 a3 / 2 ⋅ cos θ
2
= ∫ 3 /2
dθ = ∫ dθ
3 a cos θ
3
3 /2
2
2 −1 x
= θ + C = sin + C
a
3
3
26
∫
=
f ( x ) cos x dx +
x
∫ g ( x ) e dx
−
29 Let I = ∫ tan −1 x (1) dx
= tan −1 x ∫ 1dx
d
− ∫ (tan −1
dx
x⋅x−∫
= tan −1
= x tan
−1
e
(sin x − cos x ) + C
2
27 Given, ∫ f ( x ) dx = Ψ( x )
I =
∫
x )∫ (1) dx dx
1
1
×
. xdx
1+ x 2 x
1
x
x− ∫
dx
2 (1 + x ) x
x−
x 5 f ( x3 ) dx
x3 = t
dt
2
…(i)
⇒ x dx =
3
1
1
∴ I = ∫ t f ( t ) dt = [t Ψ (t ) − ∫ Ψ (t ) dt ]
3
3
1
= [ x3 Ψ ( x3 ) − 3 ∫ x2 Ψ( x3 ) dx] + C
3
[from Eq. (i)]
1 3
3
2
3
= x Ψ ( x ) − ∫ x Ψ ( x ) dx + C
3
= ∫ 1 ⋅ e cot
= xe cot
x2
∫
⇒
= ( x + 1)tan −1 x −
x+C
30 I n = ∫ ( log x ) dx = x (log x )
n
n
− n ∫ ( log x ) n –1 ⋅
1
⋅ x dx
x
31 Consider, ∫ f −1 ( x )dx = ∫ f −1 ( x )⋅1dx
d −1
( f ( x ))⋅ xdx
dx
−1
Now, let f ( x ) = t , then
d −1
dt
( f ( x )) =
dx
dx
d −1
( f ( x ))⋅ dx = dt
⇒
dx
∴ ∫ f −1 ( x )dx = x ⋅ f −1 ( x ) − ∫ f (t )dt
32 Let I = ∫ e x + 3 2 dx
( x + 4)
x
x + 4 − 1
= ∫ex
dx
( x + 4) 2
1
1
= ∫ex
−
dx
x + 4 ( x + 4) 2
x
x
dx −
∫x
2
− ∫ x ⋅ e cot
e cot
−1
x
−1
−1
x
e cot x dx
+1
x
1
−
dx
2
1+ x
dx + C = xe cot
−1
x
+C
g ′( x )
dx = ∫ 1 dx
g( x)
log e {g ( x )} = x + log C1
g ( x ) = C1 e x
Now,
g(0) = 1 ⇒ C1 = 1
∴
g( x) = e x
2 − sin 2 x
∫ g ( x ) 1 − cos 2 x dx
= ∫ e (cosec x − cot x ) dx
2
x
= − e cot x + C
= − g ( x ) cot x + C
x
3 x2 dx
dx3
dx = ∫ 3 n
dx
n
+ 1)
x ( x + 1)
x n −1 dx
dx
= 3∫
= 3∫ n n
n
x( x + 1)
x ( x + 1)
∫ x (x
3
On putting x n = t , we get
3
dt
3 1
1
I = ∫
=
−
dt
n t (t + 1) n ∫ t t + 1
3
= [log t − log(t + 1)] + C
n
3
t
= log
+C
t + 1
n
= f −1 ( x )⋅ x − ∫
[Qf −1 ( x ) = t ⇒ x = f (t )]
= x ⋅ f −1 ( x ) − g (t )
= x ⋅ f −1 ( x ) − g ( f −1 ( x ))
−1
−1
⇒
35 Let I =
= x tan −1 x − t + tan −1 t + C
= x tan −1 x − x + tan −1 x + C
∴ A( x ) = x
34 We have, d {g ( x )} = g ( x )
dx
⇒
g ′( x ) = g ( x )
2
t
dt
1 + t2
−1
x
∫1+
−
1
t2
2t dt
2 ∫ (1 + t 2 ) ⋅ t
x−∫
= x tan −1
x2 + 1
x
cot
x
33 LHS = ∫ 2
− 2
dx
e
x + 1 x + 1
∴
Put x = t 2 ⇒ dx = 2t ⋅ dt
∴ I = x tan −1
1
+C
x+ 4
x
[Q∫ e ( f ( x ) + f '( x )dx = e x ⋅ f ( x ) + C ]
= ex ⋅
⇒
∴ I n + n I n − 1 = x (log x ) n
x
ex
=
(2cos x ) + C
2
x
= e cos x + C
Put
∫ sin
x
e
(cos x + sin x )
2
Let
1 − 6cos 2 x
dx
6
x cos 2 x
1
dx
=∫
dx − 6∫
sin 6 x cos2 x
sin 6 x
(say)
= I1 − I2
sec2 x
Here, I1 = ∫
dx
sin 6 x
1
1
=∫
⋅ sec2 xdx =
⋅ tan x
sin 6 x
sin 6 x
(−6)
−∫
⋅ cos x tan xdx
sin7 x
tan x
tan x
=
+ I2 ⇒ I1 − I2 =
+C
sin 6 x
sin 6 x
tan x
Thus,
I =
+C
sin 6 x
Hence, f ( x ) = tan x
28 Let I =
=
xn
3
log n
+C
n
x + 1
SESSION 2
1 We have,
f ( x)
d
{ f ( x )} = f ( x )
dx
1
d [ f ( x ) ] = dx
f ( x)
⇒
⇒
⇒
⇒
⇒
∴
∫ f ( x ) dx =
∫
log { f ( x )} = x + log C
f ( x ) = Ce x
{ f ( x )}2 = C 2e 2 x
{ f ( x )}2 dx = ∫ C 2e 2 xdx
=
C 2e 2 x 1
= { f ( x )}2
2
2
161
DAY
2 We have,
0
f ( x ) = sin x − x2
2 − cos x
x − sin x cos x − 2
0
1 − 2x
2x − 1
0
2
0
sin x − x2 2 − cos x
⇒ f ( x ) = x − sin x
0
2x − 1
2
cos x − 2
1 − 2x
0
[interchanging rows and columns]
⇒ f ( x ) = (− 1)3
0
x2 − sin x cos x − 2
2
sin x − x
0
1 − 2x
2 − cos x
2x − 1
0
∫x
=
∫ x tan
∴
∫ tan
⇒I =
x2 + 1
dx
2
x2 + 1 x2 + 1
d
2 2
x2 + 1
= log sec
+C
2
5 We have,
I =
⇒ I =−
1 2t
π
1
e cot + t + ∫ e 2 t
4
2a
a
π
1
cot + t dt − ∫ e 2 t
4
a
π
cot + t dt + C
4
1 2t
π
e cot + t + C
4
2a
1 2 ax
π
∴ I =−
e cot + ax + C
4
2a
=
⋅ cos ( x 2 + 1)
⇒
tan2 x sec2 x
dx
tan3 x )2
∫ (1 +
=
1
π
sin 2θ log tan + θ
4
2
1
− log sec 2θ + C
2
x2
7 Let I =
∫ ( x sin x + cos x )
=
∫ ( x sin x + cos x )
∴
2
x cos x
2
dx
⋅
x
dx
cos x
Q d ( x sin x + cos x ) = x cos x
dx
−1
x
I =
.
( x sin x + cos x ) cos x
1
+ ∫
( x sin x + cos x )
tan2 θ sec 2 θ
∫ sec θ(1 + sec θ)dθ
2
1 − cos 2 θ
∫ cos θ (1 + cos θ) dθ
f (0) = log (0 +
⇒
∴
− tan −1 (0) + C
f (1) = log (1 +
9 Let I =
∴
∴
2) –
π
+0
4
dx
∫
2
x + 1
1 − x
1+ x
2
=t ⇒
dx = dt
1− x
(1 − x ) 2
1 dt
3
I = ∫ 2 /3 = [ t 1 /3 ] + C
2 t
2
3 1+ x
=
+ C
2 1− x
(1 − x )2
10
1 + 0)
C=0
Put
π + θ
2
1
π
⇒ ∫ sec 2θ dθ = log tan + θ
4
2
d
π
2sec 2θ =
log tan + θ
4
dθ
1
π
∴I = sin 2θ log tan + θ − ∫ tan2θ dθ
4
2
1 + x2 )
= log (sec θ + tan θ) – θ + C
= log ( x + 1 + x2 ) − tan – 1 x + C
∫ sec θ dθ = log tan 4
and
x )(1 +
2
= ∫ sec θ dθ – ∫ dθ
∫ (sin
cos θ + sin θ
π
log
= log tan + θ
4
cos θ − sin θ
2 sin ( x 2 + 1) − 2 sin ( x 2 + 1)
⋅ cos ( x2 + 1)
dx
2 sin ( x2 + 1) + 2 sin ( x 2 + 1)
=
6 Since,
2 sin ( x 2 + 1) − sin 2 ( x 2 + 1)
x
dx
2 sin ( x 2 + 1) + sin 2( x 2 + 1)
∫x
dx
3
4 We have,
=
∴ f ( x) =
Put tan x = t
⇒ 3 tan2 x sec2 xdx = dt
1
dt
−1
+C
∴I = ∫
⇒I =
3 (1 + t )2
3(1 + t )
−1
⇒
I =
+C
3(1 + tan3 x )
x 2 dx
∫ (1 +
Put x = tanθ ⇒ dx = sec2 θ dθ
= (1 + x2 ) dθ
dx
sin2 x cos 2 x
dx
3
x + cos 3 x )2
2
2
sin x cos x
=∫
dx
cos 6 x(1 + tan3 x )2
⇒ I =−
∫
5
2
sin2 x cos 2 x
3
{sin x(sin2 x + cos2 x ) +
=
⇒ I =
1
π
− ∫ e 2 t cot + t dt
4
a
sin x ⋅ cos x
x + cos 3 x ⋅ sin2 x
2
cos 3 x(sin2 x + cos2 x )}2
1 2 t 1 − cos 2t
e
dt , [where, ax = t ]
a∫
1 + sin 2t
1 cosec2 π + t − cot π + t dt
4
4
2
1
π
e 2 t cosec2 + t dt
⇒ I =
II
4
2a ∫ I
∫
=
1 − cos 2ax
dx
1 + sin 2ax
1 2t
e
a∫
π
π
1 − 2 sin + t ⋅ cos + t
4
4
dt
2 π
2 sin + t
4
1
⇒ I = ∫ e2 t
a
∫ (sin
8 f ( x) =
+ sin3 x ⋅ cos 2 x + cos 5 x )2
[taking (−1) common from each column]
⇒
f ( x) = − f ( x) ⇒ f ( x) = 0
∴ ∫ f ( x ) dx = ∫ 0 dx = C
3 Let I = ∫ e 2 ax
cos x − x (− sin x )
dx
cos 2 x
−x
=
+ tan x + C
( x sin x + cos x )cos x
− x + x sin2 x + sin x ⋅ cos x
=
+C
( x sin x + cos x ) ⋅ cos x
sin x − x cos x
=
+C
x sin x + cos x
1 − cos ( x2 + 1)
dx
1 + cos ( x2 + 1)
=
k =
3
3
2
dx
∫ (sin x + 2) (sin x − 1)
1
dx
1
dx
−
3 ∫ (sin x − 1) 3 ∫ (sin x + 2)
1
dx
= ∫
x
3
2 tan
2 − 1
1 + tan2 x
2
1
dx
− ∫
x
3
2 tan
2 + 2
1 + tan2 x
2
x
Put tan = t
2
=
⇒
1 2 x
sec dx = dt
2
2
∴
1
2dt
1
2dt
−
3 ∫ 2t − 1 − t 2 3 ∫ 2t + 2t 2 + 2
162
dt
dt
2
1
−
2
2
3 ∫ (t − 1)2 3 ∫
3
1
t + +
2
2
t + 1
2 1
1 2
2
=
−
tan −1
+C
3 (t − 1) 3 3
3
2
2
1
2
=
−
tan −1
x
3
3 3
tan − 1
2
tan x + 1
2 2
+C
3
2
2
=
x
3 tan − 1
2
=−
−
11
2
tan −1
3 3
x 1
2 tan +
2 2
+C
3
x2
∫ (2 + 3 x2 )5/2 dx
dx = −
2t
dt
x ( t − 3)2
= ∫e
1 dt
1
1 x
=
+C=
2 ∫ t 4 6t 3
6 2 + 3 x2
2
x+
1
x−
1 x+x
dx
e
x
1
dx +
x
= ∫e
x+
1
= ∫e
x+
1
dx + xe
x
x+
1
x
x+
1
− ∫e
x
x+
=
x
dx
1
x
+C
13 Let I = ∫ (sin(101 x )⋅sin 99 x )dx
= ∫ sin(100 x + x )sin
99
=
xdx
⋅ sin x )sin
= ∫ sin 100 x ⋅ (cos x ⋅ sin 99 x )dx
99
xdx
II
+ ∫ cos(100 x ) ⋅ sin
100
xdx
sin(100 x ) ⋅ sin100 x
100
− ∫ sin100 x ⋅ cos(100 x )dx
+ ∫ cos(100 x ) ⋅ sin100 xdx
3 /2
+C
sin(100 x ) ⋅ sin100 x
=
+C
100
1 − (2018) 2 x
⋅ (2018) sin
−1
(2018 )x
(2018) t
+C
ln2 (2018)
= (log 2018 e ) 2 ⋅ (2018) sin
= ∫ (sin(100 x ) ⋅ cos x + cos(100 x )
I
(2018)x
∫
⋅(2018)x l n(2018)dx = dt
1
(2018) t dt
∴I =
ln (2018) ∫
1
(2018)t
=
⋅
+C
ln (2018) ln 2018
= ∫ e x +1 / xdx + xe x +1 / x − ∫ e x + 1 / xdx
x+
−1
x
(2018) 2 x
⋅ (2018) sin (2018 ) dx
1 − (2018) 2 x
Put sin −1 (2018)x = t
1
⇒
1 − (2018x ) 2
1
1
x+
1 x+x
dx = e x
Q ∫ 1 − x2 e
= xe
∫
1
1 x+x
dx
e
x2
1
x+
d
− ∫ ( x ) e x dx
dx
∫ x 1 −
dx + x e
x
14 Let I =
1
=
2
x2 − 2t
dt
∴∫
⋅
dt = − 2∫ 4
( tx )5 x( t 2 − 3)2
4t
=−
∫ 1 +
sin100 x
= sin(100 x )⋅
− ∫ cos(100 x )
100
sin100 x
⋅100 ⋅
dx + ∫ cos(100 x ) ⋅ sin100 xdx
100
On substituting 2 + 3 x2 = t 2 x2
2
x2 = 2
⇒
( t − 3)
∴
12
−1
(2018 ) x
+C
15 Let I = ∫ (∫ e x (log x + 2 − 12 )dx )dx
x x
1 1
1
= ∫ (∫ e x log x + + − 2 dx )dx
x x x
log x + 1 dx + e x 1 − 1 dx
∫ x x2
x
1
= ∫ e x logx + e x + C1 dx
x
=
∫ ∫ e
x
[Q∫ e x ( f ( x ) + f '( x ))dx = e x ⋅ f ( x ) + C ]
1
= ∫ e x logx + dx + ∫ C1 dx
x
= e x log x + C1 x + C2
DAY SIXTEEN
Definite
Integrals
Learning & Revision for the Day
u
Concept of Definite Integrals
u
Walli’s Formula
u
Leibnitz Theorem
u
Inequalities in Definite Integrals
u
Definite Integration as the Limit
of a sum
Concept of Definite Integrals
Let φ ( x) be an anti-derivative of a function f ( x) defined on [a, b ] i.e.
definite integral of f ( x) over [a, b ] is denoted by
∫
b
a
∫
b
a
d
[φ ( x)] = f ( x). Then,
dx
f ( x) dx and is defined as [φ (b ) − φ (a)] i.e.
f ( x) dx = φ (b ) − φ (a). The numbers a and b are called the limits of integration, where a is
called lower limit and b is upper limit.
NOTE
• Every definite integral has a unique value.
PRED
MIRROR
• The above definition is nothing but the statement of second fundamental theorem of integral
calculus.
Your Personal Preparation Indicator
Geometrical Interpretation of Definite Integral
b
In general,
∫ f ( x)dx represents an algebraic sum of areas of the region bounded by the curve
a
y = f ( x), the X -axis, and the ordinates x = a and x = b as show in the following figure.
Y′
X′
Y
y=f(x)
a
b
X
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
164
(xi) Some important integrals, which can be obtained with
the help of above properties.
Evaluation of Definite Integrals by
Substitution
When the variable in a definite integral is changed due to
substitution, then the limits of the integral will accordingly be
changed.
b
b
Step II Find the limits of integration in new system of
variable. Here, the lower limit is g(a), the upper limit
is g(b ) and the integral is now ∫
g (b )
g ( a)
a
Properties of Definite Integrals
∫ f ( x)dx = 0
β
∫
(iii)
∫
β
∫
β
(iv)
α
β
f ( x) dx = ∫ f (t ) dt
α
β
f ( x) dx = ∫
c1
α
(v) (a) ∫ f ( x) dx =
α
(b)
f ( x) dx + ∫
c2
f ( x) dx + ... +
c1
∫
β
cn
f ( x) dx
α
−α
∫
α
0
∫
β
α
f ( x) dx =
f (α + β − x) dx
∫
α
0
f (α − x) dx
f ( x) dx
2 α f ( x) dx, if f (− x) = f ( x)
∫0
i.e. f ( x) is an even function
=
if f (− x) = – f ( x)
0,
i.e. f ( x) is an odd function
(vii)
2α
α
α
0
0
0
∫ f ( x)dx = ∫ f ( x)dx + ∫ f (2α − x)dx
(viii)
∫
(ix)
∫
a
x2
∫
xn
f ( x)dx + ... +
x1
∫
b
f ( x)dx +
xn−1
∫ f ( x)dx.
xn
Leibnitz Theorem
Walli’s Formula
α
f ( x) dx = − ∫ f ( x) dx
α
β
∫
x1
f ( x)dx = ∫ f ( x)dx +
α
where, α < c1 < c2 < ... < c n < β
(vi)
π
log 2.
8
d ψ (x )
d
d
f (t ) dt = ψ ( x) f (ψ ( x))− {φ ( x)} { f (φ( x)}
dt
dx ∫ φ( x )
dx
α
(ii)
log(1 + tan x)dx =
If function φ( x) and ψ ( x) are defined on [α , β] and differentiable
on [ α , β ] and f (t ) is continuous on [ψ (α ), φ (β)], then
α
(i)
∫
f (t ) dt .
StepIII Evaluate the integral, so obtained by usual method.
0
1
(xii) If a function f ( x) is discontinuous at points x1 , x2 ,..., x n
in (a, b ), then we can define sub-intervals
(a, x1 ),( x1 , x2 ),...,( x n −1 , x n ), ( x n , b ) such that f ( x) is
continuous in each of these sub-intervals and
Substitute g( x) = t so that g ′ ( x) dx = dt
Step I
0
π /4
π
0
f [ g ( x )] ⋅ g ′ ( x ) dx, we use the following steps
a
π /2
∫
(b)
For example, to evaluate definite integral of the form
∫
π /2
∫ log sin xdx = ∫ log cos xdx = 2 log 2 .3
(a)
2α
0
β
α
2 α f ( x) dx, if f (2α − x) = f ( x)
f ( x) dx = ∫ 0
if f (2α − x) = − f ( x)
0,
1
f ( x) dx = (β − α )∫ f [(β − α) x + α]dx
0
(x) If f ( x) is a periodic function with period T, then
(a)
(b)
(c)
∫
∫
∫
α + nT
α
βT
αT
T
f ( x) dx = n ∫ f ( x) dx, n ∈ I
0
α + nT
∫
π /2
0
sin n x dx =
∫
π /2
0
cos n x dx
(n − 1) (n − 3)(n − 5) K 5 ⋅ 3 ⋅ 1 π
× , n = 2 m (even)
n(n − 2)(n − 4)K 6 ⋅ 4 ⋅ 2
2
=
(n − 1) (n − 3) (n − 5) K 6 ⋅ 4 ⋅ 2 , n = 2 m + 1 (odd)
n(n − 2) (n − 4) K 5 ⋅ 3 ⋅ 1
where, n is positive integer.
π /2
∫ sin
0
m
x ⋅ cos n x dx
(m − 1)(m − 3)...(2 or 1).(n − 1)(n − 3)...(2 or 1) π
,
(m + n)(m + n − 2)...(2 or 1)
2
when both m and n are even positive integers
= (m − 1)(m − 3)...(2 or 1) ⋅ (n − 1)(n − 3)...(2 or 1) ,
(m + n)(m + n − 2)...(2 or 1)
when either m or n or both are odd
positive integers
Inequalities in Definite Integrals
β
β
(i) If f ( x) ≥ g( x) on [α , β], then ∫ f ( x) dx ≥ ∫ g( x) dx
α
α
T
f ( x) dx = ( β – α) ∫ f ( x) dx, α, β ∈ I
β + nT
This is a special type of integral formula whose limits from 0
to π /2 and integral is either integral power of cos x or sin x or
cos x sin x.
0
f ( x) dx =
∫
β
α
f ( x) dx, n ∈ I
β
(ii) If f ( x) ≥ 0 in the interval [α , β], then ∫ f ( x) dx ≥ 0
α
(iii) If f ( x), g( x) and h( x) are continuous on [a, b ] such that
b
b
b
a
a
a
g( x) ≤ f ( x) ≤ h( x), then ∫ g( x) dx ≤ ∫ f ( x)dx ≤ ∫ h( x)dx
165
DAY
(iv) If f is continuous on [α , β] and l ≤ f ( x) ≤ M , ∀
β
x ∈[α , β], then l (β − α ) ≤ ∫ f ( x) dx ≤ M (β − α )
α
(v) If f is continuous on [α , β],
∫
then
β
α
The converse is also true, i.e. if we have an infinite series of
the above form, it can be expressed as definite integral.
α
Some Particular Cases
n
f ( x) dx ≤ k (β − α )
(i) lim
Definite Integration as the
Limit of a Sum
(ii) lim
∫
then
α
as n → ∞
β
f ( x) dx ≤ ∫ | f ( x)| dx
(vi) If f is continuous on [α , β] and| f ( x)| ≤ k , ∀ x ∈ [α , β],
β
b–a
→0
n
where, h =
n→ ∞
∑
r =1
pn
n→ ∞
∑
r =1
r
=0
h
r
β = lim = p
n→ ∞ h
interval [a, b ], then ∫ f ( x) dx = lim h ∑ f (a + rh)
n→ ∞
(Qr = 1)
n→ ∞
n –1
a
β
1 r
f =
f ( x) dx
n n ∫ α
where, α = lim
Let f ( x) be a continuous function defined on the closed
b
n −1
1
1 r
1 r
f or lim ∑
f = ∫ f ( x) dx
n→ ∞
0
n n
n
n
r =1
and
r=0
(Qr = pn)
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1
∫
3π /4
π /4
dx
is equal to
1 + cos x
(a) − 2
(b) 2
7
JEE Mains 2017
j
2 If f ( x ) is continuous function, then
(a)
(b)
(c)
(d)
3
∫
π /4
0
(a)
4
∫
1
0
∫
∫
∫
∫
2
−2
5
−3
5
2
0
2 f (x) dx =
f (x) dx =
−3
5
∫
f (x) dx =
f (x) dx =
−3
[ tan x +
2π
∫
∫
4
−4
6
−2
∫
6
f (x − 1) dx
9
f (x − 1) dx
π
2
(c)
5 If I1( n ) = ∫
(b) π /4
π
2
0
π
2
(d) 2 π
(c) π / 2
(d) π
π
2
sin( 2n − 1)
sin nx
dx and I 2 ( n ) = ∫
dx ,
0
sin x
sin2 x
2
n ∈ N, then
6
(a) I 2 ( n + 1) − I 2 ( n ) = I1 ( n )
(b) I 2 ( n + 1) − I 2 ( n ) = I1 ( n + 1)
(c) I 2 ( n + 1) + I1( n ) = I 2 ( n )
(d) I 2 ( n + 1) + I1 ( n + 1) = I 2 ( n )
∫
1
−1
{x 2 + x − 3} dx, where {x } denotes the fractional part
of x, is equal to
1
(1 + 3 5 )
3
1
(c)
(3 5 − 1)
3
(a)
1
(1 + 3 5 )
6
1
(d) (3 5 − 1)
6
(b)
| 2 t | dt = f ( x ), then for any x ≥ 0 , f ( x ) is equal to
∫
π /2
−π / 2
(b) 4 − x 2
1
(d) (4 − x 2 )
4
cos x
dx is equal to
1 + ex
(a) 1
(c) −1
1+ x
sin 2 tan−1
dx is equal to
1− x
(a) π / 6
x
−2
(b) 2 + 2
(d) − 2 − 3 + 5
(a) 4 + x 2
1
(c) (4 + x 2 )
2
f (x − 1) dx
cot x ] dx is equal to
(b)
[ x 2 ] dx is equal to
8 If ∫
[f (x) − f (− x)] dx
10
2
0
(a) 2 − 2
(c) 2 − 1
(d) − 1
(c) 4
∫
(b) 0
(d) None of these
10 Let a, b and c be non-zero real numbers such that
∫
3
0
3
( 3 ax 2 + 2bx + c ) dx = ∫ ( 3 ax 2 + 2 bx + c ) dx , then
1
(a) a + b + c = 3
(c) a + b + c = 0
11 The value of ∫
a
1
(b) a + b + c = 1
(d) a + b + c = 2
[ x ] f ′ ( x ) dx , a > 1, where [ x ] denotes the
greatest integer not exceeding x, is
(a)
(b)
(c)
(d)
[a]f (a ) − {f (1) + f (2) + . . . + f ([a])}
[a]f ([a]) − {f (1) + f (2) + . . . + f (a )}
af ([a]) − {f (1) + f (2) + . . . + f (a )}
af (a ) − {f (1) + f (2) + . . . + f ([a])}
12 The correct evaluation of ∫
(a) 2 + 2
(c) − 2 + 2
π /2
0
π
sin x − dx is
4
(b) 2 −
(d) 0
2
166
13
∫
2x
sin
dx, where [ ⋅ ] denotes the greatest
π
3π /2
0
integer function is equal to
π
(sin1 + cos1)
2
π
(c)
(sin1 − cos1)
2
n→ ∞
(b)
of ∫
f ( x ) = f (a − x ) and g ( x ) + g (a − x ) = a, then
∫
(a)
x [ x ] dx is
5
4
(b) 0
15 The value of ∫
(a)
16
∫
π /2
(d)
π
log 2
2
(b)
17 The value of ∫
e
e
(a) 3/2
2
−1
n
k =1
1
0
NCERT Exemplar
f (x) dx (b)
1
0
∫
(c) 3
satisfies ∫
n
2
0
f (x) dx (c)
∫
n
0
(d) n
f (x) dx
f ( x ) dx =
2
∫
1
0
f (x) dx
28
n
, ∀ n ∈ I, then ∫ f (| x | ) dx is
−3
2
1
7π / 3
7π / 4
(b) 4
2
1
tan2 x dx is
2
2 x dx and
π
2
(b)
π
−π
JEE Mains 2015
(d) 6
π
π4
(c) − 1(d)
4
32
[f ( x ) + f ( − x )] [g ( x ) − g ( − x )] dx is
(b) π
(a) 0
j
JEE Mains 2013
(b)
32
∫
π
−π
(a)
π
2
(d) None of these
3π
0
π
f (cos 2 x ) dx and Q = ∫ f ( cos 2 x ) dx , then
0
(a) P − Q = 0
(c) P − 3 Q = 0
NCERT Exemplar
1
log( 2 + 1)
2
(d) None of these
(d) None of these
n
1
is
+ ... +
(n + 3) 3 ( 2n + 3)
2n 3
π
3
π
(c)
4
31 If P = ∫
j
(c) 1
n
n
30 The value of lim
+
n → ∞ (n + 1)
2n + 1 (n + 2) 2 ( 2n + 2)
(a)
3
(b) log 2
2
(d) log 2
(b)
(c) 1
[( x + π )3 + cos 2 ( x + 3π )] dx is equal to
+
sin 2x
dx is equal to
sin x +cos x
1
log( 2 + 1)
2
(c) − log( 2 + 1)
(a)
∫
(b) I 3 = I 4
(d) I 2 > I1
(c) 2 log 2
0
2
3
(a) log 2 2
∫
log x 2
dx is equal to
2 log x + log( 36 − 12 x + x 2 )
4
2 x dx , then
21 The value of ∫
π /2
−π /2
−3π /2
π
2
function, then the value of the integral
3
0
(a) I 3 > I 4
(c) I1 > I 2
22
∫
(d) None of these
2
(d) −
29 If f : R → R and g : R → R are one to one, real valued
35
(b)
2
0
2
(c) − 1
(b) 1
π4
π
(a)
+
2
32
5
20 If I1 = ∫ 2 x dx ,I 2 = ∫ 2 x dx ,I 3 = ∫
1
π
2
(a) 2
19
(a)
2
17
(c)
2
I4 = ∫
(c) 4 π
JEE Mains 2018, 13
π
(d)
4
j
equal to
1
j
f (k − 1 + x ) dx is
19 f ( x ) is a continuous function for all real values of x and
n +1
sin x
dx is
1 + 2x
π
(b)
2
27 The integral ∫
(d) 5
−π / 2
[cot x ] dx , where [] denotes the greatest integer
(a)
loge x
dx is
x
∫
0
function, is equal to
(d) 2 ( 2 − 1)
(c) 2
π
0
NCERT Exemplar
2
π /2
(a) π
∫
j
a a
(b) ∫ f (x)d x
2 0
a
(d) a ∫ f (x)d x
0
26
j
(b) 5/2
18 The value of ∑
∫
3
4
(d) π log2
(c) log2
(b) 2 ( 2 + 1)
(a) 2 2
f ( x ) ⋅ g ( x ) d x is equal to
25 The value of ∫
1 − sin 2x is equal to
0
(a)
3
2
8 log (1 + x )
dx is
1+ x2
1
0
π
log 2
8
(c)
a
0
a
(a)
2
a
(c) ∫ f (x)d x
2
0
(b) 4 − 2 ln 2
(d) 2 − 2 ln 2
24 If f and g are continuous functions in [0, 1] satisfying
14 If [. ] denotes the greatest integer function, then the value
1.5
1 1
2
3n
+
+ ...+
is
n n + 1 n + 2
4n
(a) 5 − 2 ln 2
(c) 3 − 2 ln 2
π
(sin1 − sin 2)
2
π
(d) ( sin1 + sin 2)
2
(a)
23 The value of lim
(b) P − 2 Q = 0
(d) P − 5 Q = 0
2x (1 + sin x )
dx is equal to
1 + cos 2 x
π2
4
(b) π 2
(c) 0
(d)
π
2
167
DAY
33 If f ( x ) is differentiable and ∫
t2
0
x f ( x ) dx =
4
f is equal to
25
(a)
2
5
(b) −
34 If f ( x ) =
1
x2
∫
x
4
5
2
(c) 1
(b)
32
9
∫
x2
0
dx
2
1+ x2
1
and I 2 = ∫
2
1
(b) I 2 > I1
(b)
y
(d) I1 > 2 I 2
1+ y
j
JEE Mains 2013
2
38 Let f : R → R be a differentiable function having f ( 2) = 6,
1
f ′ ( 2) = . Then, lim
x→ 2
48
(a) 18
∫
4t3
dt is equal to
x −2
f (x )
6
(b) 12
(c) 36
4
4
3
3
3
39 lim
n→ ∞
(a)
1
30
40 If I = ∫
(b) 0
π /2
0
dx
1 + sin3 x
(c)
1
4
(d)
π
2 2
(b) I >
(c) I <
(d) I < 2 π
2 π
1
5
dx , then
(a) 0 < I < 1
1 cos x
sin x
dx and J = ∫
dx . Then, which one
0
x
x
of the following is true?
41 If I = ∫
1
0
b
b
a
a
π /3
π /6
dx
π
is .
6
1 + tan x
JEE Mains 2013
(a) Statement I is true; Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true; Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d)Statement I is false; Statement II is true
1
200
0
0
44 Statement I If ∫ e sin x dx = λ , then ∫
Statement II If
∫
na
0
∫
f ( x ) dx = n
a
0
e sin x dx = 200λ
f ( x ) dx ,
n ∈I
and
f (a + x ) = f ( x )
(d) 24
[1 + 2 + 3 + K + n ] [1 + 2 + 3 + K + n ]
is
−
n5
n5
equal to
4
x 2
where, f ( x ) =
x , for 1 ≤ x ≤ 2
j
(d) y 2
1+ y 2
4 ( 2 − 1)
,
3
, for 0 ≤ x < 1
f ( x ) dx =
Statement II ∫ f ( x ) dx = ∫ f (a + b − x ) dx
dx
, then
x
(c) I1 = I 2
(a) y
0
43 Statement I The value of the integral ∫
d y
is equal to
, then
dx 2
1 + t2
0
2
∫
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for statement I
(c) Statement I is true; Statement II is false
(d)Statement I is false; Statement II is true
2
dt
y
2
and J > 2
3
2
(d) I < and J > 2
3
(b) I >
Statement II f ( x ) is continuous in [0, 2].
(b) 2 / 9
(d) 2 / 3
(a) I1 > I 2
(c)
32
3
is
(a) 0
(c) 1 / 3
2
and J < 2
3
2
(c) I < and J < 2
3
(a) I >
42 Statement I
sin t dt
x3
x→ 0
37 If x = ∫
5
2
(d) None of these
35 The value of lim
36 If I1 = ∫
(d)
[ 4 t 2 − 2f ′ ( t )] dt , then f ′ ( 4) is equal to
(a) 32
(c)
2 5
t , then
5
(a) Statement I is true; Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true; Statement II is true, Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d)Statement I is false; Statement II is true
1
1
0
0
45 Statement I ∫ e − x cos 2 x dx < ∫ e − x cos 2 x dx
b
b
a
a
2
Statement II ∫ f ( x ) dx < ∫ g ( x ) dx , ∀ f ( x ) < g ( x )
(a) Statement I is true; Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true; Statement II is true, Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d)Statement I is false; Statement II is true
168
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1
x
1 If f ( x ) is a function satisfying f + x 2f ( x ) = 0 for all
cosec θ
non-zero x, then ∫
(a) sinθ + cosec θ
(c) cosec2 θ
(b) sin2 θ
(d) None of these
4 3
3
e sin x
d
; x > 0. If ∫
e sin x dx = f (k ) − f (1), then
f (x ) =
1
x
dx
x
the possible value of k, is
2 If
(a) 15
(b) 16
(c) 63
(d) 64
2
3 If g (1) = g ( 2), then ∫ [ f {g ( x )}]−1 f ′ {g ( x )} g ′ ( x ) dx is
1
equal to
(a) 1
(b) 2
(c) 0
(d) None of these
π
4 For 0 ≤ x ≤ , the value of
2
sin2 x
cos 2 x
−1
sin
(
t
) dt + ∫
cos −1( t ) dt is
∫
0
(a)
0
π
4
(b) 0
(c) 1
1
x
x
5 If F ( x ) = f ( x ) + f , where f ( x ) = ∫
1
6
(b) 0
∫
The expression
∫
n
(c) 1
JEE Mains 2013
π
(d) −
4
log t
dt . Then, F (e )
1+ t
(d) 2
, where [ x ] and {x } are integral
{x } dx
0
and fractional part of x and n ∈ N, is equal to
(a)
1
n −1
(b)
1
n
(c) n
(d) n − 1
f (a )
ex
, I1 = ∫
x g [ x (1 − x )] dx and
x
f
(− a )
1+ e
f (a )
I
I2 = ∫
g [ x (1 − x )] dx , then the value of 2 is
f (− a )
I1
7 If f ( x ) =
(a) 2
(b) – 3
x
8 The value of lim
x→ ∞
(a) 1 / 3
9 The integral ∫
0
(c) 4 3 − 4
{∫ e dt }
0
x
∫
0
2
2
e 2 t dt
(b) 2 / 3
π
(a) π − 4
(c) –1
t
x f (sin x ) dx is equal to
π
0
∫
(c) π
π π /2
f (sin x ) dx
2 ∫0
π
(d) π ∫ f (cos x ) dx
f (sin x ) dx
π /2
0
11 If f ( x ) = ∫
(b)
f (cos x ) dx
x2 + 1
x2
0
e − t dt , then f ( x ) increases in
2
(a) (2, 2)
(c) (0, ∞)
(b) no value of x
(d) (− ∞, 0)
(n + 1)(n + 2) K 3n
n2n
1/ n
12 lim
n→ ∞
18
(a) 4
e
9
(c) 2
e
is equal to
j
(d) 1
2
is
(c) 1
(d) None of these
x
x
1 + 4 sin2 − 4 sin dx is equal to
2
2
j
JEE Mains 2014
2π
−4−4 3
3
π
(d) 4 3 − 4 −
3
(b)
JEE Mains 2016
27
(b) 2
e
(d) 3 log 3 − 2
f (x ) = ∫
x
5π /4
( 3 sin u + 4 cos u ) du on the interval
5π 4π
,
4 3
is
3
2
3
1
(c)
+
2
2
3 +
(a)
(b) − 2 3 +
3
1
+
2
2
(d) None of these
x
[ x ] dx
0
n
π
0
13 The least value of the function
j
is equal to
(a) 1/2
∫
(a) π ∫
f ( x ) dx is equal to
sin θ
10
14 If g ( x ) = ∫ f (t ) dt , where f is such that,
0
1
≤ f ( t ) ≤ 1, for
2
1
t ∈[ 0, 1] and 0 ≤ f ( t ) ≤ , for t ∈[1, 2]. Then, g( 2) satisfies
2
the inequality
3
1
≤ g (z) <
2
2
3
5
(c) < g(2) ≤
2
2
(a) −
(b)
1
3
≤ g(2) ≤
2
2
(d) 2 < g(2) < 4
15 Let n ≥ 1,n ∈ z . The real number a ∈( 0, 1) that minimizes
1
the integral ∫ | x n − a n | dx is
0
1
(a)
2
(b) 2
(c) 1
(d)
1
3
16 The value of
1/ n
π
2π
nπ
3π
lim tan tan tan K tan
n→ ∞
2n
2n
2n
2n
(a) 1
(c) 3
(b) 2
(d) Not defined
is
169
DAY
17 If f ( x ) =
x −1 2
, f ( x ) = f ( x ), ..., f k
x +1
+1
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d)Statement I is false; Statement II is true
( x ) = f {f k ( x )},
k = 1, 2, 3,K and g ( x ) = f 1998 ( x ), then ∫
1
1 /e
g ( x ) dx is equal
to
(a) 0
(c) −1
(b) 1
(d) e
20 Consider sin6 x and cos 6 x is a periodic function with π.
18 If f ( x ) is a function satisfying f ′ ( x ) = f ( x ) with
f ( 0) = 1 and g ( x ) is a function that satisfies
Statement I
1
f ( x ) + g ( x ) = x . Then, the value of ∫ f ( x ) g ( x ) dx , is
2
π
,
8
0
2
e
5
−
2
2
e2
3
(c) e −
−
2
2
(a) e −
2
e
3
−
2
2
e2
5
(d) e +
+
2
2
(b) e +
1
dx
dx
=
1 + x n ∫ 0 (1 − x n )1/ n
b
b
a
a
0
( sin6 x + cos 6 x ) dx lie in the interval
π
.
2
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d)Statement I is false; Statement II is true
∞
0
π /2
Statement II sin6 x + cos 6 x is periodic with period π / 2.
19 If n > 1, then
Statement I ∫
∫
Statement II ∫ f ( x ) dx = ∫ f (a + b − x ) dx
ANSWERS
SESSION 1
SESSION 2
1 (b)
2 (d)
3 (c)
4 (b)
5 (b)
6 (b)
7 (d)
8 (a)
9 (a)
10 (c)
11 (a)
12 (b)
13 (d)
14 (d)
15 (d)
16 (d)
17 (b)
18 (c)
19 (b)
20 (c)
21 (b)
22 (a)
23 (c)
24 (b)
25 (d)
26 (d)
27 (c)
28 (b)
29 (a)
30 (a)
31 (c)
32 (b)
33 (a)
34 (c)
35 (d)
36 (b)
37 (a)
38 (a)
39 (d)
40 (b)
41 (c)
42 (d)
43 (d)
44 (d)
45 (a)
1 (d)
11 (d)
2 (d)
12 (b)
3 (c)
13 (b)
4 (a)
14 (b)
5 (a)
15 (a)
6 (d)
16 (a)
7 (a)
17 (c)
8 (d)
18 (c)
9 (d)
19 (b)
10 (c)
20 (b)
170
Hints and Explanations
SESSION 1
=
∫
3π / 4
(cosec2 x − cosec x cot x )dx
π/4
π /2
=
∫
0
=
∫
0
=
∫
π /2
π /2
0
[sin nx − sin (n − 1) x]
dx
sin2 x
2
2
sin( 2n − 1) x ⋅ sin x
dx
sin2 x
∴ I =
∫
0
sin (2n − 1) x
dx = I1( n)
sin x
=
∫
0
=
∫
0
= [− cot x + cosec x]3ππ/ /44
⇒ I2 ( n + 1 ) − I2 ( n ) = I1 ( n + 1 )
= [(1 +
∫
2 ) − (− 1 +
2 )] = 2
6
1
−1
5
Therefore, ∫
π /4
∫
0
=
∫
0
−3
[ tan x +
π /4
= 2
∫
=∫
f ( x ) dx =
6
6
sin x cos x
∫
0
dx
dx
7
∫
2
[ x2 ] dx =
0
2
∫
+
1
∫
1
1
∫
=
[ x + x] dx
3
∫
2
= [ x]1 2 + [2 x]
3
2
∫
⇒
2 dx +
0
(3 ax2 + 2bx + c ) dx
1
(3 ax2 + 2bx + c ) dx
∫
2
3
8
∫
1
3
1
∴
a
∫
+
2
3
3 dx
3
2
π
4
∫
2
1
f ′ ( x ) dx
2f ′ ( x ) dx + . . . +
∫
a
[a] f ′ ( x ) dx
[ a]
= [ f ( x )]21 + 2 [ f ( x )]32 + . . . + [a][ f ( x )][aa ]
= f (2) − f (1) + 2 f (3) − 2 f (2) + . . .
+ [a] f (a) − [a] f ([a])
= [a] f (a) − { f (1) + f (2) + . . . + f ([ a])}
[ x2 ] dx
sin x − π
dx
4
π /4
π
=−∫
sin x − dx
0
4
∫
12 Let I =
π /2
0
+
∫
π /2
π/4
π
sin x − dx
4
π/4
∫
0
|2t|dt +
−2
∫
x
0
π /2
9 Let I =
∫
=
∫
−π / 2
0
− π /2
2t dt
x
13
π /2
2x
2x
sin dx = ∫ sin dx
0
π
π
π
3π / 2
2x
2x
+ ∫ sin dx + ∫
sin dx
π /2
π
π
π
∫
3 π /2
0
cos x
dx
1 + ex
= 0 + sin 1 ∫
cos x
dx
1 + ex
=
+
∫
0
cos x
dx …(i)
1 + ex
π
π /2
dx + sin 2 ∫
3π / 2
π
dx
π
(sin 1 + sin 2)
2
1. 5
π /2
π /2
π
π
= cos x −
− cos x −
4 0
4 π / 4
1
1
= 1−
−
− 1 = 2 − 2
2 2
|2t | dt = f ( x )
t 2
t 2
+2
2
2
−2
0
x2
= − 2 [0 − 2] + 2
− 0 = 4 + x2
2
1 − x dx
(3 ax2 + 2bx + c ) dx
(3 ax2 + 2bx + c ) dx
11 Since, ∫ 1 [ x] f ′( x ) dx =
= −2
2
=
−2
3
1
(3 ax2 + 2bx + c ) dx = 0
0
+ [3 x]2 3
0
1 + x
dx
∴ ∫ sin 2 tan −1
0
1 − x
1
0
x
=
1
1
∫
∫
1
[ x2 ] dx
∫
=1
3 ax3
2bx2
3 + 2 + cx = 0
0
a+ b + c = 0
⇒
1 dx
1
3
∫
1
∫
=
[ x2 + x] dx
π /2
0
cos x dx = [sin x]
+
5 − 1
2
2
π /2
(1 + e x ) cos x
dx
(1 + e x )
(3 ax2 + 2bx + c ) dx
= 2−1+ 2 3−2 2+ 6−3 3
π θ
= sin 2 tan –1 tan −
2 2
π θ
= sin 2 −
2 2
= sin ( π − θ) = sin θ
= 1 − cos 2 θ = 1 − x2
1
1
= x 1 − x2 + [sin −1 x]
2
0 2
0
3
π /2
π / 2 cos x
e x cos x
dx + ∫
dx
0
1 + ex
1 + ex
2
2
( 5 − 1 )/ 2
∫
2
∫
0 dx +
0
3
0
π /2
= 5− 3 − 2
θ
= sin 2 tan −1 cot
2
0
∫
[ x2 ] dx +
+
1 + x
4 ∫ 0 sin 2 tan −1
dx
1 − x
Put x = cos θ, then
1 + cos θ
sin 2 tan −1
1 − cos θ
∫
∫
1
5 −1
[ x2 ] dx
0
∫
⇒
[ x2 + x] dx
+
π
2
10
=
5−1 1
= (1 + 3 5)
2
6
2
+
3
=
1
=
−1
−
−
Put sin x − cos x = t
⇒ (cos x + sin x ) dx = dt
0
dt
∴ I= 2∫
⇒ I = 2 [sin −1 t ] −01
−1
1 − t2
1
0
∫
2
= + 1 − 0 − 1 1 −
3
1 − (sin x − cos x ) 2
= 2 [0 − (− π / 2)] =
2
f ( x − 1) dx
sin x + cos x
π /4
−1
1
x
x
=
+
−
2 −1
3
3
cot x ] dx
sin x + cos x
{ x + x} dx
2
( x2 + x − [ x2 + x]) dx
−1
f (t − 1) dt
−2
−2
∫
=
Let
x = t −1
∴
dx = dt
When x tends to − 3 and 5, then
t tends to − 2 , 6.
1
1
∫
{ x + x − 3} dx =
2
2 Since, f is continuous function.
3 Let I =
On putting x = − x in 1st integral, we get
x
0
π / 2 e cos x
cos x
∫ − π /2 1 + e x dx = ∫ 0 1 + e x dx
5 I2 ( n ) − I2 ( n − 1 )
dx
1 Let I = ∫π / 4
1 + cos x
3 π / 4 1 − cos x
=∫
dx
π/4
1 − cos 2 x
3 π / 4 1 − cos x
dx
=∫
π/4
sin2 x
3π / 4
14 Here, ∫ 0
I =
∫
1
0
x [ x2 ] dx
x ⋅ 0 dx +
∫
2
1
x ⋅ 1 dx +
∫
1. 5
2
x ⋅ 2dx
171
DAY
0
2
1
{2 − 1} + {(1.5) 2 − 2}
2
1
1 1 3
= + 2. 25 − 2 = + =
2
2 4 4
1 8 log (1 + x )
15 Let I = ∫ 0
dx
(1 + x2 )
=
When x = 0 ⇒ tan θ = 0
∴
θ=0
When x = 1 = tanθ ⇒ θ =
∫
∴I =
0
= 8∫
∫
⇒
∫
π/4
π/4
I = 8∫
0
⇒
π
4
19
log (1 + tan θ) d θ
…(i)
π /2
0
+∫
∫
π /4
0
(cos x − sin x ) dx
17
∫
e
log e x
dx =
x
−1
1
=
∫
e
=
∫
−1
−1
0
π /2
π /4
(sin x − cos x ) dx
log e x
dx
∫ e −1
x
e 2 log x
e
+ ∫
dx
1
x
2
e log x
− log e x
e
dx + ∫
dx
1
x
x
− z dz +
3
= 2∫
1
∫
2
2
0
z dz
[put log e x = z ⇒ (1 / x ) dx = dz)
f ( x ) dx + ...+ ∫
1
∫
f ( x ) dx +
0
3
f (|x|) dx +
−3
∫
∫
5
n
n −1
f ( x ) dx
∫
5
3
f (|x|)dx
f ( x ) dx + ∫ f ( x ) dx
2
1
∫ 4 f ( x ) dx
∫
I3 =
x3
2
2 x dx
3
2 x dx,
2
1
3
<2
1
0
2
x3
for 0 < x < 1
2
∫
⇒
dx <
2
1
2
3
x
∫
1
0
2
x2
∫
∫
2π
7 π /3
7π / 4
7π / 4
2π
=−
∫
∫
dx >
1
2
x2
dx
tan2 x dx
= − [log sec x]
2 +1
2−1
2π
7π / 3
2π
2π
7π / 4
|tan x |dx
tan x dx
+ [log sec x] 72 ππ /3
7π
= − log sec 2 π − log sec
4
7π
+ log sec
− log sec 2 π
3
π
= − log 1 − log sec
4
π
+ log sec
− log 1
3
1
= log 2 + log 2 = log 2 + log 2
2
3
= log 2
2
( 2 + 1)2
1
log
1
2
=
2
=
log ( 2 + 1 )]
2
1
∴ l=
log ( 2 + 1)
2
= lim
n→ ∞
∫
0
=
∫
0
3n
∑
r =1
+
3n
2
+ ...+
n+ 2
n + 3n
1 r
n n + r
x
dx
1+ x
3
=
1
1 −
dx
+
x
1
3
= [ x − ln (1 + x )] 30 = 3 − ln 4
= 3 − 2 ln 2
∫
=∫
=∫
24 Ql =
7π / 3
∫
tan x dx + ∫
|tan x |dx +
7π / 4
1
log
2
dx
2
I2 < I1 and I 4 > I3
21 I =
=
n n + 1
2 x dx
x2
1
[log ( 2 + 1) − log ( 2 − 1 )]
2
1 1
23 nlim
→∞
2
2 x dx
3
and
=
1
0
2 x > 2 x for x > 1
∫
∴
2
1
∫
and I 4 =
Since, 2
1
0
∫
=
5
1 2
9 16 35
= 2 0 + + + +
=
2
2 2
2
2
20 Given that, I1 =
π
π
− log sec − + tan −
4
4
3
2
∫
π /2
1
π
π
log sec x − + tan x −
2
4
4 0
1
π
π
=
log sec + tan
2
4
4
f ( x ) dx
3
2
4
+ ∫ f ( x ) dx +
3
I2 =
π
sin2 − x
2
=∫
dx
0
π
π
sin − x + cos − x
2
2
π /2
cos 2 x
l =∫
dx
0
sin x + cos x
1 π /2
dx
Thus, 2l =
∫
2 0 cos x − π
4
1 π /2
π
=
∫ sec x − 4 dx
2 0
=
f (| x |) dx =
−3
and
= [sin x + cos x] π0 / 4
+ [− cos x − sin x] ππ // 24
π
π
= sin + cos − (sin 0 + cos 0)
4
4
π
π
π
π
+ − cos − sin − − cos − sin
2
2
4
4
1
1
1
1
=
+
− 1 + −1 − −
−
2
2
2
2
2
=
− 1 + [−1 + 2]
2
= ( 2 − 1) + (−1 + 2 ) = 2( 2 − 1)
e
∫
f ( x ) dx
0
(sin x − cos x ) dx
+∫
2
n
2
π
[Qcos x − sin x > 0 when x ∈ 0, and
4
cos x − sin x < 0 when x ∈ ( π / 4, π / 2)]
=
f ( x ) dx
k −1
= 2 ∫ f ( x )dx +
0
(cos x − sin x ) 2 dx
π /4
5
k
1
1 − sin2 xdx
π /2
∫
f ( x ) dx
k −1
sin2 x
dx
sin x + cos x
π /2
22 We have, l = ∫0
π /2
f ( t ) dt , where
f ( x ) dx +
0
∫
=
log (1 + tan θ) d θ
1
∫
=
= π log 2
0
π /4
∑ ∫
k =1
Q π / 4 log (1 + tan θ)dθ = π log 2
∫
8
0
∫
=∫
k
k −1
k
∫
I =
n
∴
π
= 8 ⋅ ⋅ log 2
2
16 Let l =
I =
f (k − 1 + x ) dx
0
t = k −1+ x
⇒ dt = dx
8 log (1 + tan θ)
⋅ sec2 θd θ
1 + tan2 θ
0
1
18 Let I =
Put x = tan θ ⇒ dx = sec2 θd θ
π/4
2
z2
z2
1
5
+
= + 2=
∴I = −
2
2
2
−1
0 2
x2
+ [ x2 ] 1 .25
2
1
= 0+
a
0
a
f ( x )⋅ g ( x ) dx
f (a − x ) g ( a − x )dx
0
a
f ( x ){ a − g ( x )} dx
0
a
a
0
a
0
= a ∫ f ( x )d x− ∫ f ( x ) ⋅ g ( x )dx
= a ∫ f ( x)d x − I
0
a a
∴ l = ∫ f ( x ) dx
2 0
25 Let I =
⇒ I =
∫
∫
π /2
− π /2
π /2
− π /2
sin2 x
dx
1 + 2x
−π π
sin2
+ − x dx
2
2
−π
1+ 22
Q b f ( x ) dx =
∫a
∫
b
a
+
π
2
−x
dx
f (a + b − x ) dx
172
⇒ I =
∫
⇒
π /2
− π /2
sin2 x
dx
1 + 2− x
⇒ I =
2x ⋅ sin2 x
∫ − π /2 1 + 2x dx
⇒ 2I =
π /2
I =
π /2
⇒ 2I =
∫
− π /2
=
∫
0
π /2
∫
sin2 x dx =
⇒
π
I =
∫
0
I =
∫
0
π /2
0
=
∫
2sin2 x dx
…(i)
…(ii)
∫
∫
=
π
[cot x] dx +
0
π
∫
π
0
[− cot x] dx
(− 1) dx
0
=
=
∫
4
2
∫
4
∫
4
2
2
log xdx
[log x + log(6 − x )]
4
log(6 − x )
⇒ l =∫
dx
2
log (6 − x ) + log x
4
2
4
∫
⇒
2l =
∫
⇒
2l = 2 ⇒ l = 1
∫
− π /2
−3 π / 2
2
4
2
r
n 1 +
n
r =1
∫
1
dx
0
(1 + x ) 2 x + x2
=
∫
1
dx
0
(1 + x ) (1 + x )2 − 1
∫
=∫
3π
0
1
0
π
33 Using Newton-Leibnitz’s formula, we get
d
t 2 { f (t 2 )} (t )2 − 0
dt
d
⋅ f (0) (0) = 2t 4
dt
⇒ t 2 { f (t 2 )} 2t = 2t 4
⇒ f (t 2 ) = t
4
2
put t = ± 2
f = ±
⇒
25
5
5
4 2
∴
f =
25 5
[neglecting negative sign]
34 We have,
1 x
[4t 2 − 2 f ′(t )] dt
x2 ∫ 4
1
∴ f ′ ( x ) = 2 [4 x2 − 2 f ′ ( x )]
x
2 x
− 3 ∫ [4 t 2 − 2 f ′(t )] dt
x 4
1
⇒ f ′ (4) =
[64 − 2 f ′ (4)] − 0
16
32
∴ f ′ (4) =
9
f (cos 2 x ) dx and
f (cos 2 x ) dx
...(ii)
Also, P = 3∫ f (cos 2 x )dx = 3Q
π
0
∴
P − 3Q = 0
π 2 x(1 + sin x )
32 Let I = ∫ − π
dx
1 + cos 2 x
∫
35
2x
dx
1 + cos 2 x
π
−π
+
dx = [ x]24
[( x + π )3 + cos 2 x] dx
⇒ I = 0+ 4∫
π
0
∫
π
−π
⇒ I = 4∫
x sin x
dx
1 + cos 2 x
0
x2
0
sin t dt
form 0
0
sin x ⋅ 2 x
= lim
x→ 0
3 x2
sin x
2
= lim
3 x→ 0 x
2
2
= ⋅1 =
3
3
x3
x→ 0
…(i)
2 x sin x
is an even function
1 + cos 2 x
π
∫
lim
2 x sin x
dx
1 + cos 2 x
2x
Q 1 + cos 2 x is an odd function
and
1
dt
1 + t2
= π2
π
3
Q
=
−1
1
f ( x) =
...(i)
0
∫
π
π
= 2 π[tan −1 t ]1−1 = 2 π +
4
4
r
r
2 +
n
n
S =
31 P =
3
π 3 π
∫ −3π /2 − 2 − 2 − x + π
π 3π
+ cos 2 −
−
− x dx
2
2
Q b f ( x ) dx = b f ( a + b − x ) dx
∫a
∫a
−π / 2
Put cos x = t
1
= sec −1 2 − sec −1 1 =
log x + log (6 − x )
dx
log x + log(6 − x )
2l =
⇒ I =
n
= lim ∑
n→ ∞
0
n
(n + r ) r (2n + r )
S = [sec (1 + x )]
Q b f ( x ) dx = b f ( a + b − x ) dx
∫a
∫a
On adding Eqs.(i) and (ii),we get
28 Let I =
n
30 S = nlim
→∞∑
−1
2 log x dx
2[log x + log (6 − x )]
∫
φ ( x ) dx = 0
sin x
dx
1 + cos 2 x
⇒ − sin x dx = dt
(Qφ( x ) is an odd function)
log x 2
dx
log x2 + log(36 − 12 x + x2 )
2 log x
dx
2 log x + log(6 − x ) 2
⇒ l =
π
−π
0
π
∫
∴ I = − 2π
r =1
− 1 if x ∉ Z
Q[ x] + [− x] = 0, if x ∈ Z
= [− x] π0 = − π
π
∴I =−
2
27 l =
⇒I = 2π
∴ φ (− x ) = [ f (− x ) + f ( x )] [g (− x )
− g ( x )] = − φ ( x )
∫
x sin x
dx
1 + cos 2 x
[from Eq. (i)]
29 Let φ ( x ) = [ f ( x ) + f (− x )] [g ( x ) − g (− x )]
∴
π
0
sin x
dx − I
1 + cos 2 x
π
∫
⇒ I = 4π
π
2
I =
On adding Eqs. (i) and (ii), we get
2I =
(1 + cos 2 x ) dx
π sin (− π )
= −
+
2
2
∴
[− cot x] dx
0
−4 ∫
3 π sin (−3 π )
− −
+
= π
2
2
[cot ( π − x )] dx
π
−3 π / 2
π
2
=
[cot x] dx
π
− π /2
π sin x
dx
1 + cos 2 x
π
0
2cos 2 x dx
− π /2
π /2
26 Let
− π /2
⇒ I = 4∫
sin 2 x
= x +
2 −3 π /2
(1 − cos 2 x ) dx
sin 2 x
⇒ 2I = x −
2 0
π
∴
I =
4
[ − ( x + π ) 3 + cos 2 x] dx
−3 π / 2
∫
=
2x ⋅ sin2 x
∫ − π /2 1 + 2x dx
−3 π / 2
∫
π /2
I =
− π /2
∫
( π − x ) sin ( π − x )
dx
1 + cos 2 ( π − x )
36 We have, (1 + x2 ) > x2 , ∀ x
1 + x2 > x, ∀ x ∈ (1, 2)
⇒
1
⇒
∴
⇒
1 + x2
∫
<
2
dx
1
1+ x
2
1
, ∀ x ∈ (1, 2)
x
<
∫
2
1
dx
⇒ I1 < I2
x
I2 > I1
173
DAY
37 We have, x =
∫
y
dt
0
1+ t
∴
By Leibnitz rule, we get
dy
1
1=
⋅
1 + y 2 dx
d2 y
=
dx2
f( x )
38 lim
x→2 ∫ 6
y
4t 3
dt = lim
x→2
x−2
∫
=
∫
1
=
4t 3 dt
4
1 n r
1
1
= lim ∑ − lim × lim
n→ ∞ n
n→ ∞ n
n→ ∞ n
r =1 n
∫
0
n
∑
r =1
r
n
x3 dx
⇒
2
1
≤
2
∫
π /2
0
1
dx ≤
2
1
1 + sin3 x
∫
π /2
0
≤1
∫
∴ l =
π /6
∫
sin x
f ( x ) dx
π /3
dx
π
1 + tan − x
2
∫
π /3
tan x dx
π /6
1 + tan x
...(ii)
On adding Eqs. (i) and (ii), we get
2l =
π /6
⇒
dx
1
x
0
z(2sin z cos z ) dz
+
∫
∫
x
a
=
∫
b
a
f (a + b − x ) dx is a true statement by
∫
200
0
e
2
− x < − x2 ⇒ e − x < e − x
1
∫
0
e − x cos 2 xdx
2
∫
b
a
g ( x ) dx
SESSION 2
1 We have, f 1 + x2 f ( x ) = 0
x
1
⇒ f ( x) = − 2 f
x
0
∫
I =
∫
=
=
⇒
∫
cosec θ
sinθ
cosec θ
sinθ
sin θ
cosec θ
I =−
∫
II
π /2
u sin 2u du
f ( x ) dx
− 1 f 1 dx
2
x x
f (t ) dt = − I
− u cos 2u sin 2u
−
+
2
4 π / 2
cos 2 x sin 2 x
= −x⋅
+
− { 0 + 0}
2
4
cos 2 x sin 2 x π
− −x⋅
+
− + 0
4
2
4
π
=
4
x log t
5 Since, f ( x ) = ∫ 1
dt
1+ t
1
and f ( x ) = f ( x ) + f
x
1
∴ F (e ) = f (e ) + f
e
e log t
⇒ F (e ) = ∫
dt +
1 1 + t
1
x
f (t ) dt, where t =
cosec θ
sin θ
z sin 2z dz −
0 I
x
45 For 0 < x < 1,x > x2
⇒
∫
− u (2cos u sin udu )
x
dx ≠ 200λ
sin x
=
x
x
π /2
cos 2z sin 2z
= − z⋅
+
2
4 0
is 2 π.
sin x
b
π /2
∫
b
a
0
t = cos 2 u in 2nd integral, we get
dt = 2sin z cos z and
dt = − 2cos u sin udu
Hence, Statement I is false but ∫ f ( x )dx
3
∫
[ f {g ( x )}] −1 f ′ {g ( x )} {g ′ ( x )} dx
4 Put t = sin2 z in 1st integral and
∴I =
If f ( x ) ≥ g ( x ), then ∫ f ( x ) dx ≥
1 + sin x
2
1
Put
f {g ( x )} = z
⇒ f ′ {g ( x )} g ′ ( x ) dx = dz
When x = 1, then z = f {g (1)}
When
x = 2, then z = f {g (2)}
f { g(2 )} 1
∴ I =∫
dz = [log z] ff {{ gg((21 )})}
f { g(1 )}
z
⇒ I = log f {g (2)} − log f {g (1)} = 0
[Q g (2) = g (1)]
dx
2l = [ x] ππ //36
π
1 π π
l = − =
2 3 6 12
⇒
3
∫
π /3
∫
3 Let I =
...(i)
π /6
1
x
dx < ∫
dx,
0
x
x
because in x ∈ (0, 1), x > sin x
1
2
⇒
I<∫
x dx = [ x3 /2 ]10
0
3
2
I<
⇒
3
1 cos x
1
1
and J = ∫
dx < ∫
0
0
x
x
41 Since, I =
2
1
dx
1 + tan x
0
π
π
≤I≤
2 2
2
1
∫
x dx
⇒ ∫ e − x cos 2 x dx<
dx
≤
∴
1
π /3
∫
43 Let l =
∴
40 Since, x ∈ 0, π ⇒ 1 ≤ 1 + sin3 x ≤ 2
⇒
2
f ( x ) dx +
44 Since, period of e
x5
1
− 0=
5
5
0
∫
1
0
property of definite integrals.
1
=
∫
x3
x3 /2
+
3
0 3 / 2 1
⇒l =
= 18
1 + 24 + 34
1 + 23 + 33
4
3
+ . . . + n − lim + . . . + n
39 nlim
5
5
→∞
n→ ∞
n
n
0
x
2
4 3x
3 sin x3
sin x3
∴
∫ 1 x e dx = ∫ 1 x3 e dx
Put x3 = t
⇒
3 x2 dx = dt
sin t
64 e
∴
f (t ) = ∫
dt
1
t
64
= [ f (t )]1
= f (64) − f (1)
On comparing, we get
k = 64
4
1 2 3 /2
+ (2 − 1)
3 3
1 4 2 2 4 2 − 1
= +
− =
3
3
3
3
4 { f ( x )}
f ′ ( x ) = 4 { f (2)}3 f ′ (2)
1
1
= 4 × (6)3 ×
48
Q f (2) = 6 and f ′ (2) = 1 , given
48
∫
sin x
2 d f ( x) = e
2
x→2
=
dx = 2 [ x1 /2 ]10 = 2
=
( x − 2)
1
2
dx
x2 dx +
0
3
1
x 4 dx − lim ×
n→ ∞ n
⇒ 2I = 0 ⇒ I = 0
1
1
[by Leibnitz’s rule]
1
−
J<2
2
form 0
0
= lim
0
0
f( x )
6
x
∴ ∫ f ( x ) dx =
⋅ 1 + y2 = y
1 + y2
1
42 Since, f ( x ) is continuous in [0, 2].
dy
⇒
= 1 + y2
dx
d2 y
y
dy
=
⋅
⇒
dx2
1 + y 2 dx
∴
∫
=
2
1
x
∫
1 /e
1
log t
dt
1+ t
1
in second integration
t
e log t
e − log t
1
∴ F (e ) = ∫
dt + ∫
d
11 + t
1
1 t
1+
t
Put t =
174
=
=
=
=
∫
e
1
∫
∫
∫
e
1
e
1
e
1
log t
dt −
1+ t
log t
dt +
1+ t
log t
dt +
1+ t
∫
t log t − dt
×
(1 + t ) t 2
e
1
∫
∫
e
1
e
1
log t
dt
t
8 xlim
→∞
log t
dt
t (1 + t )
= lim
1
1
1
Qt (1 + t ) = t − t + 1
= lim
e
1
∫
0 dx +
2
1
1 dx +
∫
3
2
∫
n
n −1
(n − 1) dx
= 1 (2 − 1) + 2 (3 − 2) + 3 (4 − 3) +
K + (n − 1) {n − (n − 1)}
= 1 + 2 + 3 + K + (n − 1)
n (n − 1)
=
2
n
n
n
and ∫ { x} dx = ∫ ( x − [ x] ) dx =
0
0
2
∴
∫
∫
n
[x] dx
0
n
=n −1
2∫ e
x
e
1 + ex
form
e
x2
2e
1
= lim
= lim =
2
x→ ∞
x→ ∞ x
2 xe x
x2
…(i)
=
∫
π
3
0
x
= x + 4cos
20
∫
I =∫
I =∫
I =
10 Let
⇒
⇒
f (a) = 1 − t
1 −t
∫
t
Q I =
∫
a
∫
⇒ I1 =
b
1 −t
t
xg [ x(1 − x )] dx
f ( x ) dx =
∫
b
a
…(iii)
f ( a + b − x ) dx
(1 − x )g [ x(1 − x )] dx …(iv)
On adding Eqs. (iii) and (iv), we get
1 −t
t
I
∴ 2 =2
I1
g [ x(1 − x )] dx = I2
π
2
π
π
0
π
∫
0
1
(n + 1) ⋅ (n + 2) K (3n ) n
n2 n
On taking log on both sides, we get
1 + 1 1 + 2
n
n
1
log l = lim log
n→ ∞ n
... 1 + 2n
n
⇒
log l = lim
n→ ∞
…(i)
⇒
( π − x )f [sin ( π − x )] dx
( π − x ) f (sin x ) dx
π
0
…(ii)
π f (sin x ) dx
f (sin x ) dx
1
n
1
2
log 1 + + log 1 + + ...
n
n
3
x f (sin x ) dx
∫
)> 0
n + 1 n + 2
n + 2n n
= lim
K
n→ ∞
n
n
n
x
− x + 4cos
2 π
⇒
2n
+ log 1 +
n
1 2n
r
log l = lim ∑ log 1 +
n→ ∞ n
n
r =1
log l =
∫
2
0
log (1 + x ) dx
⇒ log l = log (1 + x )
2
⋅ x−
…(iii)
−
2 f ( x ) dx, if f (2a − x ) = f ( x )
∫0
,if f (2a − x ) = − f ( x )
0
[given]
∫
π /2
0
f (sin x ) dx
π
π
Put
− x=t ⇒ x=
−t
2
2
Put dx = − dt in Eq. (iii), we get
π π /2
I =
f (cos t ) dt
2 ∫ −π / 2
π
/
2
π
=
f (cos x ) dx
2 ∫ −π / 2
= π
∫
π /2
f (cos x ) dx
0
[Q f (cos x ) is an even function]
11 On differentiate the given interval by
using Newton-Leibnitz formula,
we get
1
∫ 1 + x ⋅ x dx
n
⇒ log l = [log (1 + x ) ⋅ x]20
2a
Q
f ( x ) dx =
∫0
⇒ I = π
f (− a) = t
Now, I1 =
0
)
2x < 0
x< 0
a
f (a) = 1 − f (− a)
⇒
π
(1 − e
⋅ 2x
2 x2 + 1
(n + 1) ⋅ (n + 2) ... ( n + 2n ) n
= lim
n→ ∞
n2 n
π
3
0
+1
)
1
π /3
π
= 2 xe
− ( x + 2x + 1 )
2 2
2
For f ′ ( x ) > 0,
2
then 2 x (1 − e 2 x
⇒
⇒
⋅ 2 x − e −(x
4
1
0
1 − 2sin x dx − π 1 − 2sin x dx
∫ π
2
2
3
= 4 3 − 4−
+ 1 )2
12 Let l = nlim
→∞
2
…(ii)
f (a) + f (− a) = 1
Let
0
0
1 − 2sin x dx = π |1 − 2sin x | dx
∫0
2
2
π
0
⇒I =
On adding Eqs. (i) and (ii), we get
⇒
dt
0
x→ ∞
2I =
e −a
1
=
1 + e −a 1 + e a
and f (− a) =
∫
e 2x
t2
On adding Eqs. (i) and (ii), we get
ea
f (a) =
1 + ea
∴
form 0
0
2
{x } dx
0
7 Given that, f ( x ) =
2 I1 =
x
2
2
2 ∫ e t dt (e x )
0
x
∫
2 dx + K
+
2
e2 t d t
x→ ∞
= e −(x
2
x − a,
x≥ a
to break given
| x − a|=
−( x − a) x < a
integral in two parts and then integrate
separately.
n
0
x
0
2
9 Use the formula,
6 We have, ∫ 0 [ x] dx
∫
∫
log t
log t
dt − ∫
dt
1 (1
t
+ t)
e
(log t )2
=
2 1
1
= [(log e)2 − (log 1)2 ]
2
1
=
2
=
x
2
∫ e t dt
0
f ′( x ) = e − ( x
2
+ 1 )2
d
⋅ ( x 2 + 1)
dx
− e −(x
2 2
)
d
⋅ ( x ) 2
dx
⇒ log l = 2 ⋅ log 3 −
∫
2
0
∫
2
0
x+ 1−1
dx
1+ x
1
1 −
dx
1 + x
⇒
log l = 2 ⋅ log 3 − [ x − log 1 + x ] 20
⇒
log l = 2 ⋅ log 3 − [2 − log 3]
⇒
log l = 3 ⋅ log 3 − 2
⇒
log l = log 27 − 2
⇒
l = e log 27 − 2 = 27 ⋅ e − 2 =
27
e2
13 We have, f ′( x ) = 3 sin x + 4 cos x
5π 4 π
Since, in ,
, f ′ ( x ) < 0, so assume
4 3
4π
the least value at the point x =
.
3
Thus,
4π
f
=
3
∫
4 π /3
5π / 4
(3 sin u + 4cos u ) du
175
DAY
= [– 3cos u + 4 sin u] 45 ππ //34
3
1
−2 3+
2
2
=
14 We have, g ( x ) =
⇒
x
∫
∫
g (2) =
⇒ g (2) =
∫
1
0
f ( t )dt +
n→ ∞
2
0
f ( t ) dt
∫
2
1
f ( t ) dt
We know that, m ≤ f ( x ) ≤ M for x ∈ [a, b]
⇒ m (b − a) ≤
b
∫
a
f ( x ) dx ≤ M (b − a)
1
≤ f ( t ) ≤ 1, for t ∈[0, 1]
2
1
and 0 ≤ f ( t ) ≤ , for t ∈ [1, 2]
2
1
1
⇒
(1 − 0) ≤ ∫ f ( t ) dt ≤ 1 (1 − 0)
0
2
2
1
and 0 (2 − 1) ≤ ∫ f ( t ) dt ≤ (2 − 1)
1
2
1
2
1
1
⇒ ≤ ∫ f ( t ) dt ≤ 1 and 0 ≤ ∫ f ( t ) dt ≤
0
1
2
2
1
2
1
3
≤
f ( t ) dt + ∫ f ( t ) dt ≤
⇒
1
2 ∫0
2
1
3
≤ g(2) ≤
∴
2
2
∴
15 Let f (a) =
=
∫
a
0
∫
1
0
(a − x ) dx +
n
n
⇒
a
∫
a
( x n − an ) dx
x
x
= an x −
+
− an ⋅
n + 1 0 n + 1
n +1
x
2an +1
1
− an +
n+1
n+1
n
1
= 2an +1
− an +
n + 1
n+1
∴
⇒
⇒
2
= 2an +1 −
n −1
⇒ f ′(a) = n (2a − 1) a
Thus, only critical point in (0, 1) is
a = 1/2
1
Also, f ′(a) < 0 for a ∈ 0,
2
1
and f ′(a) > 0 for a ∈ , 1 .
2
∴
1
f (a) is minimum for a = .
2
2
∴
g( x) = −
∫
1
⇒
∫
1
1 /e
∫
1
− 1 dx
x
= − [log e x] 11 /e
1
g ( x ) dx = − log e 1 − log e
e
= − [0 + 1] = − 1
f (0) = 1
f ( x) = e x
0
0
x
…(ii)
f ( x )g ( x ) dx
e x ( x2 − e x ) dx
[from Eqs. (i) and (ii)]
=
∫
1
x2e
0
x
dx −
∫
1
0
e 2 x dx
1 2x 1
[e ] 0
2
1
= [ x2e x − 2 xe x + 2e x ]10 − (e 2 − 1)
2
1
1
= [( x2 − 2 x + 2)e x ]10 − e 2 +
2
2
= [(1 − 2 + 2)e 1 − (0 − 0 + 2)e 0 ]
1
1
− e2 +
2
2
e2
1
e2 3
=e −2−
+ =e −
−
2
2
2
2
= [x e
2
x
−
∫ 2 xe
dx]10 −
x
nx n − 1 dx = 2 tan θ sec2 θ dθ
∞
0
dx
2
=
1 + xn
n
∫
2
n
∫
=
18 Given, f ′( x ) = f ( x )
and
Let
∫
1
1
x
2
1 /e
=
∴∫
1
x
g ( x ) dx =
1 /e
∫
Now,
⇒
− 1
+ 1
499 times
⇒
g ( x ) = x2 − e
1
19 In LHS, put x n = tan2 θ
1
f ( x ) = f [ f ( x )] = f −
x
−1
=
= x
1
−
x
g ( x ) = f 1998 ( x ) = f 2of 1996 ( x )
g ( x ) = f 2 [ f 1996 ( x )]
g ( x ) = f 2 ( x ) [Q f 1996 ( x )
= {( f 4of 4of 4o K f 4 )} ( x ) = x]
14243
4
a
1
an +1
− an −
+ an +1
n + 1
n+1
x−1
x+1
x
f 2 ( x ) = f [ f ( x )] = f
x
x−1
−1
x+1
=
=−
x−1
+1
x+1
⇒
⇒
an +1
= an +1 −
+
n + 1
r =1
17 We have, f ( x ) =
1
n +1
∑
P =1
n
1
n
2 π /2
( ln tan x + lncot x )dx
π ∫0
2 π /2
=
ln 1 dx = 0
π ∫0
ln P = 0
| x − a |dx
n
⇒
2 ln P =
∴
f ( x ) + g ( x ) = x2
Also,
1
rπ
ln tan
2n
n
1
πx
= ∫ ln tan
dx
0
2
2 π /2
…(i)
ln tan x dx
⇒ ln P =
π ∫0
2 π /2
and ln P =
…(ii)
ln cot x dx
π ∫0
On adding Eqs. (i) and (ii), we get
∴ ln P = lim
f ( t ) dt
0
1/n
n
rπ
16 Let P = nlim
tan
rΠ
→∞
=1
2n
…(i)
π /2
0
π /2
0
tan1 − 2 + 2 / n θ d θ
tan (2 / n) −1 θ d θ
In RHS, put x n = sin2 θ
nx n − 1 dx = 2 sin θ cos θ dθ
dx
2 π /2 1
∴∫
= ∫
0
(1 – x n ) 1 / n
n 0 cos 2/n θ
⇒
1
2
sin n
−1
θ cos θ dθ =
2
n
∫
π /2
0
tan (2 / n) − 1 θ dθ
20 sin 6 x + cos 6 x = (sin2 x ) 3 + (cos 2 x ) 3
= (sin2 x + cos2 x ) 3 − 3 sin2
x cos 2 x (sin2 x + cos 2 x )
2
2
= 1 − 3 sin x cos x
3
Q period = π
= 1 − sin2 2 x
4
2
So, the least and greatest value of
1
sin 6 x + cos 6 x are and 1.
4
π
1
Hence, − 0 ×
2
4
π /2
π
< ∫ (sin 6 x + cos 6 x )dx < − 0 × 1
0
2
π /2
π
π
6
6
∴ < ∫ (sin x + cos x ) dx <
0
8
2
DAY SEVENTEEN
Area Bounded
by the Curves
Learning & Revision for the Day
u
Curve Area
u
Area between a Curve and Lines
u
Area between Two Curves
Curve Area
The space occupied by a continuous curve, which is bounded under the certain
conditions, is called curve area or the area of bounded by the curve.
Area between a Curve and Lines
1. The area of region shown in the following figure, bounded by the curve y = f ( x)
β
β
α
α
defined on [α , β], X-axis and the lines x = α and x = β is given by ∫ y dx or ∫ f ( x) dx.
Y
y = f (x) R
Q
y
X′
O
S
P
x=α dx x=β
X
Y′
2. If the curve y = f ( x) lies below X-axis, then area of region bounded by the curve
y = f ( x), X-axis and the lines x = α and x = β will be negative as shown in the
following figure. So, we consider the area as
β
∫α
y dx
or
Y
β
∫α
PRED
MIRROR
Your Personal Preparation Indicator
f ( x) dx
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
X′
x=β
x=α dx
O
y
y = f (x)
Y′
X
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in
JEE, your Accuracy Level should be
above 85 & Prep Level should be
above 75.
177
DAY
3. Area of region shown in the following figure, bounded by
the curve x = f ( y), Y-axis and the lines y = c and y = d is
given by
d
d
∫ c f ( y) dy or ∫ c
Area between Two Curves
1. (i) Area of region shown in the following figure,
bounded between the curves y = f ( x), y = g( x), where
f ( x) ≤ g( x), and the lines x = a, x = b (a < b ) is given by
x dy
Y
Y
y=d
S
R
x
dy
y=c
X′
y=
x = f (y)
y = f (x)
Q
P
x=a
O
X
O
)
g (x
X
x=b
b
Area = ∫ [ g( x) − f ( x)] dx
a
Y′
(ii) If f ( x) ≥ g( x) in [a, c] and f ( x) ≤ g( x) in [c, b ] where
a < c < b , then Area
If the position of the curve under consideration is on the
left side of Y-axis, then area is given by
c
= ∫ [ f ( x) − g( x)] dx +
d
∫c
f ( y) dy .
4. If the curve crosses the X-axis number of times, the area of
region shown in the following figure, enclosed between
the curve y = f ( x), X-axis and the lines x = α and x = β is
given by
a
b
∫ c [g( x) − f ( x)] dx
2. Area of region shown in the following figure, bounded
by the curves y = f ( x), y = g( x), X -axis and lines x = a,
x = b is given by
c
Area = ∫ f ( x) dx +
a
Y
b
∫ c g( x) dx
Y
y=
y = g(x)
)
f(x
α1
O
x =α
α1
∫α
f ( x) dx +
α2
∫ α1
α2 x = β
f ( x) dx +
β
∫ α2
X
X′
f ( x) dx
x=a
O
x=c
X
x=b
Y′
Area and the shape of some important curves
S.No.
Curves
(i)
x2
y2
x
y
f ( x, y) : 2 + 2 ≤ 1, + ≥ 1
a
b
a
b
Point of intersection
Area of shaded region
Y
B (0, b)
A (a , 0), B(0 , b )
⇒ or
Area = ab
x2
y2
x
y
+ 2 ≤1≤ +
2
a b
a
b
(π – 2)
sq units
4
(–a, 0) A′
X′
A (a, 0)
X
(0, 0)
Y′
Y
A
(ii)
Parabola y2 = 4ax and
its latusrectum x = a
A (a , 2a), B(a , − 2a)
Area =
8 2
a sq units
3
X′
O
Y′
(a
,2
a)
y2=4ax
(a, 0)
B (a, –2a)
X
178
Y
A
(iii)
4a 4a
f ( x, y) : y2 = 4ax and y = mx O(0, 0), A 2 ,
m m
Area =
2
8a
sq units
3 m3
X′
X
O (0, 0)
Y′
x2=4ay
(iv)
f ( x, y) : x2 = 4ay
y = 4 bx
2
O(0, 0),
2 /3
A (4a
b
1 /3
1 /3 2 /3
, 4a b
)
Area =
16
(ab ) sq units
3
X′
(4a2/3 b1/3,4a1/3 b2/3)
A
Y
X
(0, 0) O
y2=4bx
Y′
Y
(v)
A (b – a, 2√ab)
Area bounded by
y2 = 4a ( x + a)
y = 4 b (b − x)
2
and
A(b − a , 2 ab )
Area =
B(b − a , − 2 ab )
8
ab (a + b ) sq units
3
X′
A′ (b, 0)
(0, 0)
X
B′(–a, 0) O
B (b – a, – 2√ab)
Y′
(vi)
Common area bounded
by the ellipses
2
2
x
y
1
and
+
=
b 2 a 2 a 2b 2
x2
y2
1
+ 2 = 2 2, 0 < a < b
2
a
b
a b
Y
x=y=
x=y=−
1
a + b2
2
1
a +b
2
2
Area = Area of region PQRS
= 4 × Area of OLQM
4
a
=
tan −1 sq units
b
ab
P
m Q
L
R
S
X′
O
(0, 0)
X
Y′
x=b
Y
(vii)
If α , β > 0, α > β the area
between the hyperbola
xy = p2 , the X-axis and
the ordinates x = α, x = β
—
α
Area = p2 log
β
X′
x=a
X
(0, 0) O
Y′
179
DAY
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 Let f : [ − 1, 2] → [ 0, ∞ ) be a continuous function such that
f ( x ) = f (1 − x ) for all x ∈ [ − 1, 2]. Let R1 = ∫
2
−1
x f ( x ) dx, and
R 2 be the area of region bounded by y = f ( x ),x = − 1,
x = 2 and the X -axis. Then
(a) R1 = 2R 2
(b) R1 = 3R 2
(c) 2R1 = R 2
(d) 3R1 = R 2
2 Let A1 be the area of the parabola y 2 = 4ax lying between
the vertex and the latus rectum and A2 be the area
between the latus rectum and the double ordinate x = 2a.
Then A1 / A2 is
(a) (2 2 − 1) /7
(c) (2 2 + 1)
(b) (2 2 + 1) /7
(d) (2 2 − 1)
x 2 + y 2 = 9 by x = 1 is
−1
(b) 9 sec
(3) −
8
(b) 1 : 1
(c) 1 : 2
(d) 2 : 1
lines x = 1, x = b is ( b 2 + 1 − 2 ) for all b > 1, then f ( x )
(b) x + 1
(c) x / x + 1 (d) None of these
8
curve y = 1 + 2 , X -axis and the ordinates x = 2, 4 into
x
two equal parts, then a is equal to
(b) 2 2
(c) 2
(d)
2
7 Let the straight line x = b divide the area enclosed by
y = (1 − x )2, y = 0 and x = 0 into two parts R1 ( 0 ≤ x ≤ b )
and R 2 (b ≤ x ≤ 1) such that R1 − R 2 = 1/ 4. Then b equals
(a) 3/4
(b) 1/2
(c) 1/3
(d) 1/4
8 The curve y = a x + bx passes through the point (1, 2)
and the area enclosed by the curve, the X -axis and the
line x = 4 is 8 square units. Then a − b is equal to
(a) 2
(b) − 1
(c) − 2
(d) 4
9 If y = f ( x ) makes positive intercepts of 2 and 1 unit on
x and y-coordinates axes and encloses an area of
2
3
sq unit with the axes, then ∫ xf ′ ( x ) dx , is
0
4
(a)
3
2
(c)
7
3
(d)
14
9
11 The area bounded by the curve y = ln ( x ) and the lines
y = 0, y = ln ( 3) and x = 0 is equal to
j
JEE Mains 2013
(b) 3 ln (3) −2 (c) 3 ln (3)+2 (d) 2
(a) 3
12 The area of the region bounded by the curve ay 2 = x 3,
the Y -axis and the lines y = a and y = 2a, is
3 2
a (2 ⋅ 2 2 / 3 − 1) sq unit
5
3
(c) a 2 (2 2 / 3 + 1) sq unit
5
(a)
(b)
NCERT Exemplar
2
a (2 2 / 3 − 1) sq unit
5
(d) None of these
(a) 4.5 sq units
(c) 3.5 sq units
(b) 6.3 sq units
(d) None of these
14 The area between the curve y = 4 − | x | and X -axis is
(a) 16 sq units
(c) 12 sq units
(b) 20 sq units
(d) 18 sq units
2
6 If the ordinate x = a divides the area bounded by the
(a) 8
14
3
15 The area under the curve y = | cos x − sin x | , 0 ≤ x ≤
equals to
(a) x + 1
(b)
y = | x − 3 | and y = 2, is
5 The area bounded by the curve y = f ( x ), X -axis and the
2
12
9
13 The area of the region bounded by
and y = cos 2 x between x = 0, π / 3 and X -axis, is
2 :1
(a)
(d) None of these
4 The ratio of the area bounded by the curves y = cos x
(a)
bounded by the parabola y = 9x 2 and the lines x = 0,
j JEE Mains 2013
y = 1 and y = 4, is
j
3 The area of smaller segment cut off from the circle
1
(a)
(9 sec−1 3 − 8 )
2
(c) 8 − 9 sec−1 (3)
10 The area of the region (in sq units), in the first quadrant,
(b) 1
(c)
5
4
(d) −
3
4
and above X-axis is
(b) 2 2 − 2
(a) 2 2
j
(c) 2 2 + 2
π
2
JEE Mains 2013
(d) 0
16 The area of the region enclosed by the curves y = x ,
x = e, y =
1
and the positive X-axis is
x
(a) 1 sq unit (b)
3
5
1
sq units (c) sq units (d) sq unit
2
2
2
17 The area bounded by y = sin x , X -axis and the lines
x = π is
(a) 2 sq units
(c) 4 sq units
(b) 3 sq units
(d) None of these
18 For 0 ≤ x ≤ π, the area bounded by y = x and
y = x + sin x , is
(a) 2
(c) 2 π
(b) 4
(d) 4 π
19 The area (in sq unit) of the region described by
{( x , y ): y 2 ≤ 2x and y ≥ 4x − 1} is
7
(a)
32
15
(c)
64
5
(b)
34
9
(d)
32
j
JEE Mains 2015
180
20 The area bounded by the curves y 2 = 4x and x 2 = 4y , is
(a) 0
32
(b)
3
16
(c)
3
8
(d)
3
21 If the area bounded by y = ax 2 and x = ay 2, a > 0 is 1,
then a is equal to
1
3
(d) None of these
y
22 The area bounded between the parabolas x = and
4
x 2 = 9 y and the straight line y = 2, is
(b)
10 2
3
(c)
20 2
3
(d) 10 2
23 The area enclosed the curves y = x 3 and y = x is
5
(a) sq units
3
5
(c)
sq unit
12
5
(b) sq units
4
12
(d)
sq units
5
x 2 + y 2 ≤ 4x , x ≥ 0, y ≥ 0 } is
4
3
4 2
(c) π −
3
j
JEE Mains 2016
(b) π −
(b)
π 4
−
2 3
(c)
π 2
−
2 3
j
JEE Mains 2014
π 2
(d) +
2 3
3π + 2
9π − 2
2π − 9
(d)
9π + 2
(b)
27 The area bounded by the curves y = 2 − | 2 − x | and
| x | y = 3 is
(a)(5 − 4 ln 2)/ 3
(c)(4 − 3 ln 3)/ 2
y = x , 2y − x + 3 = 0, X-axis and lying in the first
j JEE Mains 2013
quadrant is
(a) 9
(b) 36
27
(d)
4
(c) 18
31 The area bounded by the curves y = x 2 and
(b) (3 π − 2)/ 3
(d) None of these
32 The area bounded by y = tan x , y = cot x , X -axis in
0≤x ≤
π
is
2
(b) log2
(d) None of these
y = loge x and y = 2x is equal to
two parts. Then, the ratio of the areas of these parts is
3π − 2
10 π + 2
6π − 3
(c)
11π − 5
30 The area (in sq units) bounded by the curves
33 Area of the region bounded by curves x = 1/ 2, x = 2,
26 The parabola y 2 = 2x divides the circle x 2 + y 2 = 8 in
(a)
(b) (π − 2)/ 2
(d) None of these
(a) 3 log 2
(c) 2 log 2
25 The area of the region described by
A = {( x , y ): x 2 + y 2 ≤ 1and y 2 ≤ 1 − x } is
(a) (π − 2)/ 4
(c) (π + 2)/ 2
(a) (3 π + 2)/ 3
(c) (3 π − 2)/ 6
8
3
π 2 2
(d) −
2
3
(a) π −
π 4
+
2 3
(b) 2 sq units
(d) 4 sq units
y = 2 /(1 + x 2 ) is
24 The area (in sq units) of the region {( x , y ) : y 2 ≥ 2x and
(a)
(a) 1 sq unit
(c) 2 2 sq units
x 2 + y 2 = 1 and the lines | y | = x + 1 is
2
(a) 20 2
y = − | x | + 1, is
29 The area of the smaller region bounded by the circle
1
(b)
3
(a) 1
(c)
28 The area bounded by the curves y = | x | − 1 and
(b) (2 − ln 3)/ 2
(d) None of these
( 4 − 2 )/log 2 + b − c log 2. Then, b + c equals
(b) − 1
(d) None of these
(a) 2
(c) 4
34 The area bounded by the curve y = ( x + 1)2, y = ( x − 1)2
and the line y =
1
sq unit
6
1
(c) sq unit
4
(a)
1
is
4
2
sq unit
3
1
(d) sq unit
3
(b)
35 The area of bounded region by the curve y = loge x and
y = (loge x )2, is
(a) 3 − e
1
(c)
(3 − e)
2
(b) e − 3
1
(d)
(e − 3)
2
181
DAY
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 Let f ( x ) = x 2, g ( x ) = cos x and h( x ) = f (g ( x )). Then area
bounded by the curve y = h( x ), and X -axis between
x = x1 and x = x 2, where x1, x 2 are roots of the equation
18x 2 − 9πx + π 2 = 0, is
(a) π/12
(b) π/ 6
(c) π/ 3
(d) None of these
9 The area enclosed between the curves y = loge ( x + e ),
x = log(1/ y ) and the X -axis is equal to
(a) 2e
(b)
1
4
(c)
1
3
(d)
f (1 + x )
= 3. The area
x
bounded by the curve y = f ( x ), y-axis and the line y = 3
is equal to
x→0
1
2
(a) e
3 A point P ( x , y ) moves in such a way that [| x | ] + [| y | ] = 1,
[⋅] = G.I.F. Area of the region representing all possible
positions of the point P is equal to
(a) 8
(b) 4
(c) 16
(d) None of these
(c) 12 π
6 Area of the region bounded by the parabola ( y − 2)2
= x − 1, the tangent to it at the point with the ordinate 3,
and the X -axis is given by
(b) 9 sq units
(d) None of these
(a)
(a) 2
5
(b)
2
59
(c)
12
(d) 2 π 2
(b) 3
(c) 6
(d) 9
maximum, then b is equal to
(b) − 1
(a) 1
(c) b = ± 1 (d) None of these
15 Let An be the area bounded by the curve y = (tan x )n and
the lines x = 0 and x = π / 4. Then, for n > 2
1
1
< An <
2n
2n − 2
1
1
(c)
< An <
2n + 2
2n − 2
(a)
x } is
JEE Mains 2017
(d)
(c) π 2
14 If area bounded by the curves y = x − bx and by = x 2 is
8 The area (in sq units) of the region
7
(a)
3
π2
2
2
(b) π / (3 3 + π)
(d) None of these
j
(b)
f ( x ) − f ( x /7) = x /7 ∀x ∈ R . The area bounded by the
curve y = f ( x ) and the coordinate axes is
tangent and normal at (1, 3 ) and X -axis in A1 and A2.
A
Then 1 equals to
A2
{( x , y ) : x ≥ 0, x + y ≤ 3, x 2 ≤ 4y and y ≤ 1 +
π2
4
13 Let f ( x ) be continuous function such that f ( 0) = 1,
7 Let the circle x 2 + y 2 = 4 divide the area bounded by
(a) π / (3 3 − π)
(c) π / (3 − π 3 )
(b) 2 x , x ≥ 1
(d) None of these
y = cos −1 (sin x ) − sin−1 (cos x ) and the lines y = 0,
3π
x =
, x = 2π is
2
3π
5π
π
π
sq units (b)
sq units (c)
sq units (d)
sq unit
8
8
2
8
(a)9/2 sq units
(c)18 sq units
(d) None of these
12 The area bounded by the curve
x 2 + y 2 ≤ 1 } is
(a)
(a) 2 2 x , x ≥ 1
(c) 2 x , x ≥ 1
(d) 6π
5 The area of the region R = {( x , y ) : | x | ≤ | y | and
(c) 3e
curve y = f ( x ), which cuts the last two lines above the
2
first line for all a ≥ 1, is equal to [( 2a )3/ 2 − 3a + 3 − 2 2 ].
3
Then f ( x ) =
4 The area bounded by the curve xy = 4 ( 2 − x )and Y-axis is
(b) 4 π
(b) 2e
11 The area bounded by the lines y = 2, x = 1, x = a and the
2
(a) 2 π
(d) None of these
f ( x / y ) = f ( x ) − f ( y ) and lim
[ −1, 1] and X-axis, is (in sq unit)
1
5
(c) 2/e
10 Let f ( x ) be a real valued function satisfying the relation
2 The area bounded by f ( x ) = min(| x | ,| x − 1| ,| x + 1| ) in
(a)
(b) 2
3
2
1
1
< An <
2n + 1
2n − 1
1
1
(d)
< An <
2n + 2
2n
(b)
ANSWERS
SESSION 1
SESSION 2
1
11
21
31
(c)
(d)
(b)
(b)
1 (a)
11 (a)
2
12
22
32
(b)
(a)
(c)
(b)
2 (d)
12 (a)
3
13
23
33
(b)
(d)
(c)
(c)
3 (a)
13 (b)
4
14
24
34
(d)
(a)
(b)
(d)
4 (b)
14 (c)
(c)
(b)
(a)
(a)
6 (b)
16 (b)
26 (b)
7 (b)
17 (c)
27 (c)
8 (d)
18 (a)
28 (b)
9 (d)
19 (d)
29 (b)
10 (d)
20 (c)
30 (a)
5 (c)
15 (c)
6 (b)
7 (a)
8 (b)
9 (b)
10 (c)
5
15
25
35
182
Hints and Explanations
SESSION 1
5
1 Given, R1 =
2
∫− 1
2
∫− 1 f ( x ) dx and f (1 −
and R2 =
Consider, R1 =
=
=
x f ( x ) dx,
x) = f ( x)
2
∫− 1 x f ( x ) dx
2
∫− 1 (1 −
b
∫1
Now, on differentiating both sides w.r.t.
b, we get
2b
f (b ) =
∀b > 1
⇒
2 b2 + 1
⇒
x ) f (1 − x ) dx
f ( x ) = x / x2 + 1.
∫− 1 (1 −
x ) f ( x ) dx
=∫
= R2 − R1
∴ R2 = 2R1
4
2
2 A 1 = 2 ∫ 2 a x dx
0
A
2
= 4 a × [ x3 /2 ] a0 = 8a2 / 3
3
a
Now, ∫ x f ′ ( x ) dx = [ xf ( x )]20 −
3
4
3
x f ′ ( x ) dx = 2 × 0 − 0 × 1 −
4
3
=−
4
⇒
∫0
2
(2, 3)
C
10 Required area =
B (4, 3/2)
y=9x2
y
4
∫1
3
4
dy =
3
3 Required area, A = 2 ∫
1
O
M
D
x=a
=
∫2
8
8
1 + 2 dx ⇒ x – = 2
x 2
x
⇒ a = ± 2 2⇒ a =2 2
(3, 0)
(1, 0)
(0, 0)
X′
a
a
7 We have, R1 − R2 =
X′
y=1
Area of region, ACDM
9 − x2 dx
Y
y=4
X
N
b
1 y 3 /2
3 3 / 2 1
Y
2
2 3 /2 2 a 8a
[ x ]a =
(2 2 − 1)
3
3
(2 2 + 1)
1
=
∴ A 1 / A2 =
7
2 2−1
∫ 0 f ( x ) dx
∫ 0 x f ′( x ) dx = [2 f (2) − 0 f (0)] −
4
1 + 8 dx = x − 8 = 4
x 2
x2
2
2
⇒
2 a x dx
=4 a×
∫0 (1 − x )
1
− ∫ (1 −
b
2
2
2
14
= (43 /2 − 13 /2 ) = × 7 =
9
9
9
dx
x )2 dx =
X
Y′
[Q a > 0]
1
4
X
11 Required area
=
log 3
∫0
xdy =
log 3
∫0
e ydy
3
= [e y] log
= [e log 3 − e 0] = 3 − 1 = 2
0
12 We have,
x=1
R1
Y
Y′
C y=2a
2a
N
3
x 9 − x2 + 9 sin −1 x
3 1
1
= 2⋅
2
π
= 9 −
2
1
= 9 cos −1 − 8
3
−1
= [ 9 sec (3) − 8]
4 Here, A 1 =
π /3
∫0
= [sin
A2 =
R2
1
8 − 9 sin −1
3
π
1
= 9 − sin −1 −
2
3
and
[given]
2
Y
a
A2 = 2 ∫
(2, 0) and (0, 1).
∴ 0 = f (2) and 1 = f (0)
2
3
Also, ∫ f ( x ) dx =
0
4
0
6 Area of region AMNB
2
2a
9 Clearly, y = f ( x ) passes through
f ( x ) dx = b 2 + 1 − 2
π /3
∫0
=
3
4
…(i)
(a x + bx ) dx = 8
16a
+ 8b = 8
3
On solving Eqs. (i) and (ii),
we get a = 3 and b = −1
∴ a−b = 4
⇒
X
Y′
1 2 (1 − b )3
1
1
−
=
⇒ (1 − b )3 =
8
3
3
4
1
∴ b =
2
∫0
M
By=a
X′ O
(1,0)
⇒
4
cos 2 x dx
sin 2 x
=
2 0
∴ A1 : A2 = 2:1
y=(1 – x)2
At x = 1, y = 2, we get 2 = a + b
3
=
2
π /3
O (b,0)
8 y = a x + bx ( x ≥ 0).
cos x dx
x] π0 / 3
8
a
Required area =
Area BMNC =
2 a 1 /3
=
∫a
=
3 a3
5
=
1 5
5
3 3 3 3
a a (2) − 1
5
a
1
…(ii)
y 2 /3 dy =
2a
∫a
1
3 a3
5
xdy
[ y 5/3 ]2aa
5
5
(2a)3 − a3
2
3
= a2 2⋅ 23 − 1 sq unit
5
183
DAY
e1
1
1
(1 × 1) + ∫
dx = + [log | x |] e1
1
2
x
2
1
1
3
= + [log| x|] e1 = + 1 = sq units
2
2
2
13 The curve y = | x − 3| meets the line
=
y = 2, when x = 1 and x = 5
Y
y=|x – 3|
Hence, required area
y + 1 y2
−
dy
4
2
2
1
=
∫−1
=
1 y2
1
+ y
− ( y 3 )1−1 /2
4 2
−1 /2 6
=
1 1
1
1 1 1
+ 1 − − − 1 +
8 2 6
4 2
8
1
17 We have,
y=2
X¢
sin x, if x ≥ 0
y = sin x =
−
sin x, if x < 0
and| x|= π ⇒ x = ± π
X
01 3 5
1 3 3 1 9
+ −
4 2 8 6 8
1 15 3
9
= ×
−
=
4 8 16 32
Y
Y¢
The area of the shaded region
1
= × 2 × 4 = 4 sq units
2
1
X¢
–p
x=–p
Y
A (0, 4)
20 For the point of intersection of
Y¢
Y
x=p
y2=4x
x2=4y
π
Now, required area = 2∫ sin xdx
2
y=4–x
C
(0, 0) (4, 0)
(4, 4)
= 2[− cos x] π0
x
4+
X
p
y 2 = 4 x and x2 = 4 y
4 + x, x < 0
y =
4 − x, x > 0
y=
y=|sin x|
O
14 y = 4 − | x|represents two curves as,
B
X¢
(– 4, 0)
=
= −2[cos π − cos 0]
= −2[−1 − 1] = 4 sq units
X
X¢
(0, 0)D
X
(4, 0)
18 Given, curves y = x and y = x + sin x,
which intersect at (0, 0) and ( π, π ).
∴ Area, A =
Y¢
The area of the shaded portion
1
= × 8 × 4 = 16 sq units
2
=
=
|cos x − sin x|dx
π /4
∫0
π
xdx
∫ π / 4 (sin x − cos x )dx
+
= [sin x + cos x] π0 / 4
+ [− cos x − sin x] ππ //24
= ( 2 − 1) − (1 − 2 ) = 2 2 − 2
2
parabola y 2 = 2 x
…(i)
∴ Area bounded between curves
4
x2
= ∫ 4x −
dx
0
4
4
and y ≥ 4 x − 1 represents a region to the
left of the line y = 4 x − 1
…(ii)
The point of intersection of the curves
(i) and (ii) is given by
x3
x3 /2
= 2 ⋅
−
3
12
0
2
(4 )3
4
3 /2
= ⋅ (4 ) −
3
12
32 16 16
=
−
=
3
3
3
(4 x − 1)2 = 2 x ⇒ 16 x2 + 1 − 8 x = 2 x
16 Given, y = x, x = e and y = 1 , x ≥ 0
x
Since, y = x and x ≥ 0 ⇒ y ≥ 0
∴ Area to be calculated in 1st quadrant
shown in figure
Y
⇒ 16 x2 − 10 x + 1 = 0
1 1
⇒
x= ,
2 8
So, the points where these curves
1
1 1
intersect are ,1 and ,−
2
8 2
Y
x
in y 2 = 4 x
4
x2
⇒ = 4 x ⇒ x 4 = 43 x ⇒ x = 0, 4
4
y 2 ≤ 2 x represents a region inside the
π /2
2
Substitute y =
{( x, y ): y 2 ≤ 2 x and y ≥ 4 x − 1}
(cos x − sin x )dx
Y¢
sin x dx = [− cos x] π0
∫0
19 Given region is
π /2
∫0
π
= − cos π + cos 0
= − (−1) + 1 = 2
15 Required area
=
π
∫ 0 ( x + sin x ) dx − ∫ 0
21 The intersection point of two curves is
1 , 1 .
a a
y=4x–1
Y
y=x
y2=2x
1
)
A
1
(1,
1/2
X¢
X¢
B
O
D (1, 0)
Y¢
C (e, 0)
x=e
y=1/x
X
∴Required area = Area of ∆ODA
+ Area of DABCD
(1/2,1)
–1 –1/2 1/4 1/2 1
1, – 1
–1/2
8 2
–1
Y¢
y=ax2
y2=x/a
X¢
X
O
X
Y¢
∴ Area, A =
1/a
∫0
x
− ax2 dx
a
184
1 2 3 /2 1 / a a 3 1 / a
⋅ [x ] 0 − [x ] 0
3
a 3
1
1
2
a = ⇒a=
3
3
⇒
Required area =
1=
⇒
22
Y
X¢
2
= 2∫
=
0
2
y
3 y −
2
3 /2
5 y dy = 5 y
3
2
2
∫0 Ycircle − ∫0 Yparabola dx
=
πr
−
4
2
2
∫0
2 2
x
8
x
+ 2
8 − x2 + sin −1
2
2
2
2 2
16
π
=
+ 2 2 π − 2 + 4 ×
3
4
4
= + 2π sq units
3
= 8π sq units
4
Now, A2 = 8 π − A1 = 6 π −
3
A1
and the required ratio,
is
A2
(–1, 0)
(1,0)
X
4
+ 2π
2 + 3π
3
=
=
4 9π − 2
6π −
3
Y¢
y=0
1
1 2
πr + 2∫ (1 − y 2 )dy
0
2
1
y3
1
π 4
= π(1)2 + 2 y −
= +
2
3 0 2 3
x are given by x = 0
26
Y
2=
y
2
x +
B
Y
y= x
X¢
X
8
C
Y
y=–3/x
y 2 =2x
A (2Ö2, 0)
O
: x<2
x
x≥2
4
−
:
−
3
/x: x< 0
3
and y =
⇒ y =
x
3
/
: x> 0
| x|
x
27 y = 2 − |2 − x| =
4
y=x
3
X
2
1
X¢
∴A =
1
1
∫0
( x3 −
x 4 2 x3 / 2
x ) dx =
−
3 0
4
1 2
5
sq unit
= − =
4 3 12
24 We have, x2 + y 2 ≤ 4 x and y 2 ≥ 2 x
To find point of intersection, substitute
y 2 = 2 x in x2 + y 2 = 4 x
x + y = 4x ⇒ x + 2x = 4x
2
⇒
⇒
⇒
2
2
x2 − 2 x = 0
(2,2)
(2,0)
Y¢
(4,0)
The point B is a point of intersection
(lying in the first quadrant) of the given
parabola and the circle, whose
coordinates can be obtained by solving
the two equations y 2 = 2 x and
⇒ x2 + 2 x = 8 ⇒ ( x − 2) ( x + 4) = 0
x = 0 or x = 2
y = 0 or y = 2
O
Let the area of the smaller part be A 1
and that of the bigger
A1
part be A 2 . We have to find
.
A2
x2 + y 2 = 8.
⇒ x( x − 2) = 0
Y
X¢
O
Y¢
Y¢
X
2
Y
X¢
and x = 1 .
8 − x2 dx
Clearly, area of the circle = π (2 2 )2
y=2
23 Clearly, the intersection of two curves
2 2
∫2
2
⇒ A 1 = 2 2 ⋅ ⋅ x3 /2
3
0
2( x )1 /2 dx
Required area =
X¢
2
2 xdx +
∫ 0
A = {( x, y ): x2 + y 2 ≤ 1and y 2 ≤ 1 − x}
dy
20 2
10 3 /2
(2 − 0) =
3
3
y = x3 and y =
2
=
25 Given,
X
0
⇒ A1 = 2
2
y=1 x2
9
y=2
Required area = 2 ∫
− y 2 ) dx
3
π×4
2
=
− 2 x2
4
3
0
2 2 3 /2
8
= π−
(2 − 0) = π −
3
3
y=4x2
Y¢
2
2
∫0 ( y 1
⇒ x = 2, − 4
x = − 4 is not possible as both the points
of intersection have the same positive
x-coordinate. Thus, C ≡ (2, 0).
Now, A 1 = 2 [Area (OBCO )
+ Area (CBAC )]
2 2
2
= 2 ∫ y 1 dx + ∫
y 2 dx ,
2
0
where, y 1 and y 2 are respectively the
values of y from the equations of the
parabola and that of the circle.
1Ö3 2
3
4
y=4–x
Y¢
⇒
y=3/x
X
x = 3/ x ⇒ x = 3
and 4 − x = 3 / x ⇒ x = 3 ( x > 2).
∴ Required area
2
3
3
3
= ∫ x − dx + ∫ 4 − x − dx
3
2
x
x
= (4 − 3 log 3)/2
28 The region is clearly square with
vertices at the points (1, 0), (0, 1), (−1, 0)
and (0, − 1).
Y
(0,1)
y=|x| –1
X¢
X
(–1,0)
y=–|x|+1
(1,0)
(0,–1)
Y¢
So, its area = 2 × 2 = 2 sq units
185
DAY
Y
35 Required area, A
=
(0,–1)
0
e
∫ 1 [ log
x − (log x ) 2 ] dx
Y
[ 1 − x − ( x + 1)] dx
y=(logex)2
2
−1
1
or Area = 2 [area of sector
AOBA − ∆AOB ]
π 1 ( π − 2)
=2
−
=
4 2
2
X¢
O
p/4
X
p/2
(e, 1)
=
...(i)
x
and 2 y − x + 3 = 0
y=logex
Y¢
X¢
Now, required area
30 Given curves are
y =
y=cot x
= 2∫
Y
X
O
–y=x+1
( x − 1)3 x
1 1 1
= 2
− = 2 −
− +
24 8 3
3
4
0
1
= sq unit
3
π ,1 .
4
y=cot x
A
(–1,0)
1 /2
32 Clearly, the two curves will intersect at
B (0,1)
y=x+1
∴ Required area
1 /2
1
= 2∫ ( x − 1)2 − dx
0
4
1
2 x3
3π − 2
= 4 tan −1 x −
= π − 2/3 =
3
3 0
29 Due to symmetry, required area
π /4
∫0
π /2
∫π / 4 cot x dx
tan xdx +
Y¢
= [log sec x] π0 / 4 + [log sin x] ππ // 24
...(ii)
X
(1, 0)
O
A=
= [log 2 − 0] + [0 + log 2]
e
∫ 1 log
x dx −
e
∫ 1 ( log
x ) 2 dx
= [ x log x − x] e1 − [ x (log x )2
= 2 log 2 = log 2 sq units
−2( x log x − x )] e1
33 Required area
= [e − e − (− 1)] − [e (1) 2
y=2x
− 2e + 2e − (2)]
= 1 − (e − 2) = 3 − e
SESSION 2
1
On solving Eqs. (i) and (ii) , we get
½ 1
2 x − ( x )2 + 3 = 0
⇒
( x )2 − 2 x − 3 = 0
⇒
=
( x − 3)( x + 1 ) = 0
∴ y =3
3
3
∫0 ( x2
= ∫ {( 2 y + 3 ) − y } dy
0
3
= 9+ 9− 9= 9
0
31 Curves x = y and y = 2 / (1 + x ) are
2
2
⇒
4− 2 5
3
− log 2 +
log 2
2
2
=
4− 2
+ b − c log 2
log 2
2
x dx = π / 12
2 Graph of
f ( x ) = min (| x |, | x − 1 |, | x + 1 |) is
Y
1/2
b = 3 / 2, c = 5/ 2 ⇒ b + c = 4
X¢
–1
–1/2
O
1/2
1
Y¢
34
Required area
1
1
1
= 2 × × 1 × = sq unit
2
2 2
y=(x–1)2
y=(x+1)2
symmetrical about Y-axis.
π /3
∫π / 6 cos
∴ Required area =
− log e x ) dx
=
− x1 ) dy
2
y3
= y2 + 3 y −
3
x
18 x2 − 9 πx + π2 = 0
⇒ (6 x − π ) (3 x − π ) = 0
⇒ x1 = π / 6, x2 = π / 3
2
[Q x = − 1 is not possible]
∴ Required area =
2
∫1 /2 (2
1 h( x ) = f (cos x ) = cos2 x
2
2x
=
− x log x + x
log
2
1 /2
x = 3,
⇒
y=logex
Y
Y
3
Y
2
1
y= 4
X¢
O
1
1
2
Area = 2 ∫
− x2 dx
0 1 + x2
X
O
–1 –1/2 1/2 1
Y¢
X
–2
O
–2
[| x |] + [| y |] = 1
2
X
X
186
⇒
[| x |] = 1, [| y |] = 0
or
[| x |] = 0, [| y |] = 1
⇒
1 ≤ | x | < 2, 0 ≤ | y | < 1
6 ( y − 2) 2 = x − 1
…(i)
8 On solving x2 = 4 y and x + y = 3
Curve (i) is a parabola with vertex at the
point A(1, 2), axis y − 2 = 0 i.e. y = 2
and concavity towards positive X -axis.
For the point, say P, at which ordinate
y = 3 and x = 2
Equation of tangent at P (2, 3) is
dy
y −3=
( x − 2)
dx (2, 3 )
or 0 ≤ | x | < 1 or 1 ≤ | y | < 2
⇒ x ∈ (− 2, − 1] ∪ [1, 2), y ∈ (− 1, 1)
or x ∈ (− 1, 1), y ∈ (− 2, − 1] ∪ [1, 2)
∴ Required area = 8 × 1 × 1 = 8
4 In the equation of curve xy 2 = 4(2 − x ) ,
the degree of y is even. Therefore, the
curve is symmetrical about X-axis and
lies in 0 < x ≤ 2.
The area bounded by the curve and the
1
or
y − 3 = ( x − 2) i.e.
2
x − 2y + 4 = 0
⇒
⇒ ( x + 6) ( x − 2) = 0 ⇒ x = 2, y = 1
Solving y = 1 + x and y = 3 − x,
we get 1 +
Y-axis is 2∫ y dx
x = 3 − x ⇒ x = 1, y = 2
4y=x2
Y
(0,3) (1,2)
(2,1)
X¢
)
2,3
0
P(
2 2 − x
2− x
dx = 4∫
dx
0
x
x
Put x = 2sin2 θ ⇒ dx = 4sin θ ⋅ cos θ dθ
O1
2 (3,0)
2
= 2∫ 2
0
∴Required area = 4∫
= 8∫
= 8∫
π /2
0
π /2
0
π /2
Y¢
2 − 2sin2 θ
O
∴ Area =
2cos 2 θ dθ
3
∫0
1
(1 + cos 2θ)dθ
7
=
= 4∫
0
O
p
3
( y 1 − y 2 ) dx
( 1 − x − x ) dx
1/ 2
1
1
x2
= 4 x 1 − x2 + sin −1 x −
2
2 0
2
Y
P(x1, y1)
(–1/Ö2, 1/Ö2)
X¢
Q (x 2, y 2)
(–1/Ö2, –1/Ö2)
2
Normal
B(1, Ö3)
A2
A1
X
A C(4, 0)
9
5
2
y = log e ( x + e )
y =e
…(i)
−x
…(ii)
Required area
=
∞
0
∫1 − e log( x + e ) dx + ∫0
e − x dx
Let area of portion OAB = A 1
and area of portion ABC = A 2
The equation of tangent at (1, 3 ) is
x+
3y = 4
X
(1/Ö2, –1/Ö2)
Y¢
–e (1– e)
[Q xx1 + yy 1 = a2 is the tangent for
O
X
= [ x log ( x + e )]10−e
0
x
−∫
dx + [− e − x ] ∞0
1 −e x + e
Now, the area of the
1
∆OBC = × OB × BC
2
1
= × 2× 2 3 = 2 3
2
The area of portion OAB i.e. A 1 =
= 0 + 1 − [ x − e log ( x + e )]10 − e
r 2θ
2
4 ⋅ π /3 2 π
.
=
2
3
Now,A2 = ∆OBC − OAB = 2 3 − 2 π / 3
A1
2 π /3
=
A2
6 3 − 2π
3
2π
π
.
=
=
6 3 − 2π 3 3 − π
=
1
1 π 1
1
=4
×
+ ×
−
2 2
2 2 4 4
π
= sq units
2
1
the circle x2 + y 2 = a2 at ( x1 , y 1 )]
y=x
(1/Ö2, 1/Ö2)
O
x2
dx
4
Y
2
y=–x
2
0
2
Y
2
2
1
x ) dx + ∫ (3 − x )dx − ∫
1
2
3
x + 2 x3 /2 + 3 x − x − x
3
2 1 12 0
0
2
1
8
= 1 + + (6 − 2) − 3 − −
3
2 12
Tangent
region
= 4 (Area of the shaded region in
first quadrant)
1/ 2
∫0 (1 +
[ y 2 − 4 y + 5 − (2 y − 4)] dy
5 Required area = Area of the shaded
1/ 2
Required Area
3
π
0
X
1
y3
=
− 3 y2 + 9 y
3
0
= 9 − 27 + 27 − 0 = 9.
2 θ 2
= 8 θ + sin
2 0
π
=8
+ 0 − 0 = 4π
2
= 4∫
X
x+
y=
3
(1,2)
2sin2 θ
⋅ 4 sin θ ⋅ cos θ dθ
0
y=1+Öx
…(ii)
Y
2
x2
+ x=3
4
x2 + 4 x − 12 = 0
we get,
=1+ 1=2
10 f ( x / y ) = f ( x ) − f ( y )
…(i)
x = y = 1 ⇒ f (1) = 0
f ( x + h) − f ( x)
f ′ ( x ) = lim
h→ 0
h
x + h
f
x
[using Eq. (i)]
= lim
h→ 0
h
f (1 + h / x ) 1 3
= lim
⋅ =
h→ 0
h/ x
x
x
187
DAY
π
cos −1 (sin x ) = cos −1 cos 2 π +
− x
2
π
= 2π +
− x
2
and sin −1 (− cos x )
⇒ f ( x ) = 3 ln x + c
f (1) = 0 ⇒ c = 0.
∴ y = f ( x ) = 3 log e x
3
O
∴ y = 2 π + π /2 − x + 3 π /2 − x
= 4π − 2x
3
∫− ∞ e
y/3
dy
13 f ( x ) − f ( x /7) = x /7
Y
∴ An =
=
Adding, we get
y=2
x
1
1
1 + + K+ n − 1
7
7
7
x
1
= 1 − n
6
7
f ( x ) − f ( x / 7n ) =
a
X
Y′
a
2
3 /2
− 3a + 3 − 2 2]
1
On differentiating both sides w.r.t. a, we
get
2 3
f (a) − 2 = (2a)1 /2 ⋅ 2 − 3
3 2
⇒ f (a) − 2 = 2 2a − 2
⇒
f (a) = 2 2a
⇒
f ( x ) = 2 2 x, x ≥ 1
12 3 π ≤ x ≤ 2 π ⇒ − 2 π ≤ − x ≤ − 3 π
2
2
∆′(b) =
Taking limit as n → ∞, we get
x
x
f ( x ) − f (0) = ⇒ f ( x ) = 1 +
6
6
[Q f (0) = 1]
0
x
Required area = ∫ 1 + = 3
−6
6
⇒
⇒
y = x − bx and by = x
∆(b ) =
b /b + 1
∫0
–
x2
− x + bx2 dx
b
b /(b2 + 1 )
b 2 + 1 x3
x2
=
−
2
b 3
0
π /4
∫0
π /4
∫0
tan n x dx
tan n − 2 x(sec2 x − 1) dx
π /4
For n > 2, 0 < x < π / 4
⇒ 0 < tan x < 1 ⇒ tan n − 2 x > tan n x
2
Solving these, we get x = 0, b /(1 + b 2 )
∴
+
–1 0 1
π /4
tan n − 1 x
=
−∫
tan n − 2 x dx
0
n
−
1
0
1
An =
− An − 2
⇒
n−1
1
⇒
An + An − 2 =
n−1
⇒
14 Given curves are
2
–
+
15 In (0, π / 4), tan x > 0
f ( x / 7) − f ( x / 72 ) = x / 72
M M M M M M
f ( x / 7n − 1 ) − f ( x / 7n ) = x / 7n .
∫ [ f ( x ) − 2]dx = 3 [(2a)
(b 2 + 1)3
⇒ ∆(b ) is max. for b = 1, − 1.
f ( x / 72 ) − f ( x / 73 ) = x / 73
1
=
− 2b 2 (b 2 + 1) × 2b
1
⋅
6
(b 2 + 1)4
2b (1 − b ) (1 + b )
π2
Required area = ∫
(4 π − 2 x )dx =
3 π /2
4
11 According to given condition, we have
X′
∆ ′ (b ) =
2π
= 3 [e y/3 ]3− ∞ = 3e
y=f(x)
1
b2
2
6 (b + 1)2
2b (b 2 + 1)2
3π
3π
= sin −1 sin
− x =
− x
2
2
1
Required area =
=
A n − 2 > An
1
2 An <
n−1
1
An <
2n − 2
+ An =
Also, A n
+2
⇒
2A n >
∴
1
n+1
1
n+1
1
1
< An <
2n + 2
2n − 2
DAY EIGHTEEN
Differential
Equations
Learning & Revision for the Day
u
Solution of Differential Equation
u
Solutions of Differential
Equations of First Order and
First Degree
u
Linear Differential Equation
u
Inspection Method
An equation involving independent variables, a dependent variable and the derivatives of
dependent variable w.r.t. independent variables, is called differential equation.
A differential equation which contains only one independent variable, is called an ordinary
differential equation.
Order and Degree of a Differential Equation
●
●
The order of a differential equation is the order of the highest derivative occurring in the
differential equation.
The degree of differential equation is the degree of the highest order derivative, when
differential coefficients are made free from radicals and fractions.
Solution of Differential Equation
A solution of the differential equation is a relation between the dependent and independent
variables of the equation not containing the derivatives, but satisfying the given differential
equation.
General Solution
A general solution of a differential equation is a relation between the variables (not
involving the derivatives) which contains the same number of the arbitrary constants as the
order of the differential equation.
Particular Solution
Particular solution of the differential equation is obtained from the general solution by
assigning particular values to the arbitrary constant in the general solution.
PRED
MIRROR
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
189
DAY
Solutions of Differential
Equations of the First Order and
First Degree
NOTE
• Sometimes given equation becomes a linear differential
dx
+ Rx = S, where R and S are
dy
function of y (or constants).
R dy
• The integrating factor in this case is IF = e ∫
equation of the form
and the solution is given by x ⋅ (IF ) = ∫ ( S × IF ) dy + C.
A differential equation of first order and first degree may be of
the following types:
Differential Equation with
Variable Separable
‘Variable separable method’ is used to solve such an equation
in which variable can be separated completely, e.g. the term
f ( x) with dx and the term g( y) with dy
i.e.
g( y)dy = f ( x)dx
Reducible to Variable Separable Form
dy
= f (ax + by + c) can be
dx
reduced to variable separable form by the substitution
ax + by + c = z.
dz
dy dz
dz
1
− a = f (z) ⇒
= a + bf (z)
⇒ a+b
=
⇒
dx
b
dx
dx dx
Differential equations of the form
Homogeneous Differential Equation
A function f ( x, y) is said to be a homogeneous function of
degree n if f (λx, λy) = λn f ( x, y) for some non-zero constant λ.
●
Let f ( x, y) and g( x, y) be two homogeneous functions of
same degree, then a differential equation expressible in the
form
dy f ( x, y)
=
dx g( x, y)
is called a homogeneous differential equation.
●
To solve a homogeneous differential equation of the type
dy f ( x, y)
y
=
= h , we put y = vx and to solve a
x
dx g( x, y)
homogeneous differential equation of type
x
dx f ( x, y)
=
= h , we put x = vy.
y
dy g( x, y)
Linear Differential Equation
A differential equation of the form
dy
+ Py = Q,
dx
where, P and Q are the functions of x (or constants), is called a
linear differential equation.
To solve such an equation, first find integrating Factor,
P dx
IF = e ∫ . Then, the solution of the differential equation is
given by y (IF) = ∫ Q(IF) dx + C.
Bernoulli’s Differential Equation
(Reducible to Linear Form)
Let the differential equation be of the form
dy
dy
+ P y = Q yn ⇒ y−n
+ P y−n
dx
dx
dy dz
Here, put y − n + 1 = z ⇒(− n + 1) y − n
=
dx dx
dz
⇒
+ (1 − n) Pz = (1 − n) Q
dx
+1
=Q
which is a linear differential equation in z.
The solution of this equation is given by
ze ∫
(1 − n )P dx
( 1 − n ) Pdx
= ∫ ( 1 − n ) ⋅ Q ⋅ e ∫
dx + C
Inspection Method
If we can write the differential equation in the form
f { f1 ( x, y)} d{ f1 ( x, y)} + φ { f2 ( x, y)} d{ f2 ( x, y)} + ... = 0, then
each term can be easily integrated separately.
For this use the following results.
(i) d ( x ± y) = dx ± dy
x ydx − x dy
(iii) d =
y
y2
(ii) d ( xy) = x dy + ydx
x y dx − x dy
(iv) d tan −1 =
y
x 2 + y2
x dy + y dx
(v) d [log( xy)] =
xy
x y dx − x dy
(vi) d log =
y
xy
1
x dx + y dy
log ( x2 + y2 ) =
2
x 2 + y2
1 x dy + y dx
(viii) d − =
xy
x 2 y2
(vii) d
e x ye x dx − e x dy
(ix) d =
y
y2
(x) d( x2 + y2 ) =
x dx + y dy
x 2 + y2
1
x + y x dy − y dx
(xi) d log
=
2
x − y
x 2 − y2
190
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 The order of the differential equation whose general
solution is given by
y = c1 e 2 x + c 2 + c 3 e x + c 4 sin( x + c 5 ), is
(a) 5
(b) 4
(c) 3
Then, y( −3) is equal to
(d) 2
(a) 3
(c) 1
d y
d y
dy
+3
= x 2 log 2 , is
2
dx
dx
dx
2
2
(a) 1
(c) 3
10 If
(b) 2
(d) None of these
(b) 2
(d) 0
dy
= y + 3 > 0 and y( 0) = 2, then y(log 2) is equal to
dx
(a) 5
representing the family of curves y = 2c( x + c ), where
c is a positive parameter, are respectively equal to
2
(b) 2, 3
(d) 1, 2
4 The differential equation of all circles in the first quadrant
which touch the coordinate axes is of order
(a) 1
(c) 3
A
+ B,
r
where A and B are arbitrary constants, is
d 2 v 1 dv
+
=0
dr 2 r dr
2
d v 2 dv
(c)
+
=0
dr 2 r dr
(a)
(b)
d 2 v 2 dv
−
=0
dr 2 r dr
(d) None of these
x dy − y dx
= 0, is
x2 + y2
(a)
(b)
(c)
(d)
variable radii and a fixed centre at (0, 1)
variable radii and a fixed centre at (0, − 1)
fixed radius 1 and variable centre along the X-axis
fixed radius 1 and variables centre along the Y-axis
dy ax + 3
=
dx 2y + f
represents a circle, then the value of a is
(b) − 2
(a) 2
(d) − 4
(c) 3
dy
dy
13 Solution of the differential equation y − x
is
= y2 +
dx
dx
(a) Cy = (1 − x)(1 − y)
(c) Cy = (1 + x)(1 − y)
(b) Cx = (1 + x)(1 − y)
(d) Cx = (1 − x)(1 + y)
dy
= ax + by , is
dx
ae −by + be ax + C = 0
aeby + be ax + C = 0
aeby + be − ax + C = 0
None of the above
15 The solution of the differential equation
dy
= e x − y + x 2e − y is
dx
x3
+C
3
y
x
3
(c) e = e + x + C
(a) e y = e x +
(d) None of the above
7 The differential equation of the family of parabolas with
focus at the origin and the X -axis as axis, is
2
determines family
of circles with,
(a)
(b)
(c)
(d)
x2 + y2 + C
(a) y = x tan
2
2
2
x + y + C
(b) x = y tan
2
C − x2 − y2
(c) y = x tan
2
dy
dy
(a) y + 4 x
= 4y
dx
dx
2
dy
dy
(c) y + y = 2 xy
dx
dx
1− y
dy
=
dx
y
14 Solution of the equation ln
6 The solution of the differential equation
x dx + y dy +
(d) 7
2
12 If the solution of the differential equation
(b) 2
(d) None of these
5 The differential equation of the family of curves v =
(c) − 2
(b) 13
11 The differential equation
3 Order and degree of the differential equation,
(a) 1, 3
(c) 2, 4
(b) 3 y12 y 2 − (1 + y12 )y 3 = 0
(d) 3 y1 y 22 − (1 + y12 )y 3 = 0
9 If x dy = y (dx + ydy ), y (1) = 1 and y ( x ) > 0.
2 The degree of the differential equation
2
(a) 3 y1 y 2 − (1 + y12 )y 3 = 0
(c) 3 y1 y 22 + (1 + y12 )y 3 = 0
2
dy
dy
(b) y = 2 x
−y
dx
dx
2
dy
dy
(d) y + 2 xy
+ y=0
dx
dx
8 The differential equation whose solution is
x 2 + y 2 + 2ax + 2by + c = 0, where a, b, c are arbitrary
constants, is
16 If ( 2 + sin x )
(b) e y = e x + 2 x + C
(d) y = e x + C
dy
π
+ ( y + 1) cos x = 0 and y( 0) = 1, then y
2
dx
is equal to
1
(a) −
3
j
4
(b)
3
1
(c)
3
JEE Mains 2017
(d) −2 / 3
7
2
17 If a curve passes through the point 2, and has slope
1
1 − 2 at any point ( x , y ) on it, then the ordinate of the
x
191
DAY
point on the curve, whose abscissa is − 2, is
j
3
(a) −
2
3
(b)
2
5
(c)
2
JEE Mains 2013
5
(d) −
2
dy
18 Solution of the equation ( x + y )2
= 4, y ( 0) = 0 is
dx
x + y
(a) y = 2 tan−1
2
x + y
(c) y = 4 tan−1
2
19 The solution of
x + y
(b) y = 4 tan−1
4
(d) None of these
dy
+ 1 = e x + y is
dx
(a) e − ( x + y ) + x + C = 0
(c) e x + y + x + C = 0
(b) e − ( x + y ) − x + C = 0
(d) e x + y − x + C = 0
20 Solution of the equation xdy = ( y + xf ( y / x ) / f ′( y / x )) dx
is
(a) f (y / x) = C x , C ∈R
(b) f (y / x) = x + C, C > 0
(c) f (y / x) = C x , C > 0
(d) None of these
dy
21 If x
= y (log y − log x + 1), then the solution of the
dx
equation is
x
(a) log = Cy
y
y
(c) x log = Cy
x
y
(b) log = Cx
x
22 The general solution of y 2 dx + ( x 2 − xy + y 2 ) dy = 0, is
x
(a) tan−1 + log y + C = 0
y
x
(b) 2 tan−1 + log x + C = 0
y
(b) y − 2 x = c 2 x 2 / y
(d) None of these
4
5
(b)
2e 2 − 1 / 2
(d)
e2 + 1/ 2
(c)
2
5
IIT 2012
π
π
(b) y ′ =
4 18
π
4π
π2
(d) y ′ =
+
3
3
3 3
29 The solution of the equation
dy
+ y tan x = x m cos x , is
dx
(m + 1) y = x m + 1 cos x + C (m + 1) cos x
my = (x m + C ) cos x
y = (x m + 1 + C ) cos x
None of the above
30 Let the population of rabbits surviving at a time t be
t
2
dP (t ) 1
= P (t ) − 200.
dt
2
j
(b) 300 − 200 e
(d) 400 − 300 e
JEE Mains 2014
−
t
2
−
t
2
(d)
j
( xy 5 + 2y ) dx − xdy = 0, is
(a) 9x + 4 x 9 y 4 = 9y 4 C
(c) x 8 (9 + 4 y 4 ) = 10y 4 C
8
4
5
JEE Mains 2013
(b) 9x 8 − 4 x 9 y 4 − 9y 4 C = 0
(d) None of these
32 Let Y = y ( x ) be the solution of the differential equation
dy
π
+ y cos x = 4x , x ∈( 0, π ). If y = 0, then
2
dx
j
JEE Mains 2018
y( π / 6) is equal to
sin x
(a) −
26 The equation of the curve passing through the origin and
satisfying the differential equation
dy
(1 + x 2 )
+ 2xy = 4x 2 is
dx
j
2
31 The solution of differential equation
satisfies the differential equation y (1 + xy )dx = xdy , then
1
f − is equal to
2
j
JEE Mains 2016
(b) −
π
π
(a) y =
4 8 2
π
π2
(c) y =
3
9
2
t
25 If a curve y = f ( x ) passes through the point (1,−1) and
2
5
y ′ − y tan x = 2x sec x and y( 0) = 0, then
(c) 600 − 500 e 2
24 If ( x 2 + y 2 )dy = xydx and y ( x 0 ) = e, y (1) = 1, then x 0 is
(a) −
28 If y ( x ) satisfies the differential equation
(a) 400 − 300 e
is
e2 − 1 / 2
1
sin 2 x + C
2
1
(b) y (1 + x 3 ) = Cx + sin 2 x
2
1
(c) y (1 + x 3 ) = Cx − sin 2 x
2
x 1
(d) y (1 + x 3 ) = − sin 2 x + C
2 4
(a) y (1 + x 3 ) = x +
If P( 0) = 100, then P (t ) is equal to
23 Solution of the differential equation x 2dy + y ( x + y )dx = 0
(c)
dy
3x 2
sin2 x
, is
+
y =
3
dx 1 + x
1+ x3
governed by the differential equation
(c) log(y + x 2 + y 2 ) + log y + C = 0
x
(d) sin−1 + log y + C = 0
y
(a) e 3
(b) 3 (1 + x 2 )y = 2 x 3
(d) 3 (1 + x 2 ) y = 4 x 3
27 The solution of the differential equation
(a)
(b)
(c)
(d)
x
(d) y log = Cx
y
(a) y + 2 x = c 2 x 2 y
(c) y + 2 x = c 2 x 2 / y
(a) (1 + x 2 )y = x 3
(c) (1 + x 2 )y = 3 x 3
8
8
π 2 (b) − π 2
9 3
9
(c) −
4 2
π
9
(d)
4
π2
9 3
33 Statement I The slope of the tangent at any point P on a
parabola, whose axis is the axis of x and vertex is at the
origin, is inversely proportional to the ordinate of the point
P.
Statement II The system of parabolas y 2 = 4ax satisfies
a differential equation of degree 1 and order 1.
j
JEE Mains 2013
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
192
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
34 Statement I The elimination of four arbitrary constants in
y = (c1 + c 2 + c 3 ec 4 )x results into a differential equation
dy
of the first order x
= y.
dx
Statement II Elimination of n independent arbitrary
constants results in a differential equation of the nth order.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is not
a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 The differential equation which represents the family of
curves y = c1ec 2 x where c1 and c 2 are arbitrary constant, is
(a) y ′′ = y ′y
(c) yy ′′ = (y ′)2
(b) yy ′′ = y ′
(d) y ′ = y 2
2 The population p ( t ) at time t of a certain mouse species
dp(t )
= 0.5 (t ) − 450. If
dt
p( 0) = 850, then the time at which the population
becomes zero is
satisfies the differential equation
(a) 2 log18
1
(c) log18
2
(b) log9
(d) log 18
π
6
3 A curve passes through the point 1, . Let the slope of
y
y
+ sec , x > 0.
x
x
Then the equation of the curve is j JEE Advanced 2013
the curve at each point (x, y) be
1
y
(a) sin = log x +
x
2
y
(b) cosec = log x + 2
x
2y
(c) sec
= log x + 2
x
2y
1
(d) cos
= log x +
x
2
1
1 ey
(d) 1 + −
y
e
6 The equation of the curve y = f ( x ) passing through the
origin which satisfies the differential equation
dy
= sin(10x + 6y ) is
dx
1 −1 5 tan 4 x
− 5 x
tan
3
4 − 3 tan 4 x
1 −1 5 tan 4 x
(b) y = tan
+ 5 x
3
4 − 3 tan 4 x
1 −1 5 tan 4 x
(c) y = tan
− 5x
3
4 + 3 tan 4 x
(d) None of the above
(a) y =
7 Let I be the purchase value of an equipment andV (t ) be
(a) I −
df ( x )
and g ( x ) is a given non-constant
f ′( x ) denotes
dx
differentiable function on R with g ( 0) = g ( 2) = 0. Then
the value of y( 2) is
(b) 4
(d) 0
5 Consider the differential equation
1
y dx + x − dy = 0. If y(1) = 1, then x is given by
y
2
1
2 ey
(b) 4 − −
y
e
the value after it has been used for t yr. The value V (t )
depreciates at a rate given by differential equation
dV (t )
= − k (T − t ), where k > 0 is a constant and T is the
dt
total life in years of the equipment. Then, the scrap value
V (T ) of the equipment is
4 Let y ′( x ) + y ( x )g ′( x ) = g ( x )g ′( x ), y( 0) = 0, x ∈ R , where
(a) 2
(c) 5
1
1 ey
(a) 1 − +
y
e
1
y
1
(c) 3 − + e
y
e
(c) e − kT
kT 2
2
k (T − t )2
2
1
(d) T 2 −
k
(b) I −
8 The degree of the differential equation satisfying the
relation 1 + x 2 + 1 + y 2 = λ ( x 1 + y 2 − y 1 + x 2 ), is
(a) 1
(c) 3
(b) 2
(d) None of these
9 If length of tangent at any point on the curve y = f ( x )
intercepted between the point of contact and X -axis is of
length 1, the equation of the curve is
193
DAY
(a)
1 − y 2 + ln (1 − 1 − y 2 ) / y = ± x + C
14 Let f be a real-valued differentiable function on R (the set
of all real numbers) such that f (1) = 1 . If the y-intercept of
the tangent at any point P ( x , y ) on the curve y = f ( x ) is
equal to the cube of the abscissa of P, then the value of
f ( −3) is equal to
(b) 1 − y 2 − ln (1 − 1 − y 2 ) / y = ± x + C
(c)
1 − y 2 + ln (1 + 1 − y 2 ) / y = ± x + C
(d) None of the above
(a) 3
π
cos x dy = y (sin x − y ) dx , where, 0 < x < , is
2
f (1) = 1 and lim
t→x
(b) y sec x = tan x + C
(d) tan x = (sec x + C ) y
1
2x 2
+
3x
3
1
2
(c) −
+ 2
2x x
(b) ellipse and circles
(d) None of these
(b) −
(d)
1
4x 2
+
3x
3
1
x
16 Let a solution y = y ( x ) of the differential equation
x x 2 − 1 dy − y y 2 − 1 dx = 0 satisfy y( 2) =
12 A curve passes through (2, 0) and the slope of tangent at
a point P ( x , y ) is equal to (( x + 1) + y − 3) / ( x + 1). Then
j IIT 2004
equation of the curve is
2
2
.
3
π
Statement I y ( x ) = sec sec−1 x − .
6
(b) y = x 2 − 2 x
(d) None of these
Statement II y ( x ) is given by
13 The solution of the differential equation
x3 x5
+
+ ...
dx − dy
3! 5!
is
=
2
4
x
x
dx + dy
1+
+
+ ...
2! 4!
x+
(a) 2 ye 2 x = Ce 2 x + 1
(c) ye 2 x = Ce 2 x + 2
t 2f ( x ) − x 2f (t )
= 1 for each x > 0. Then f ( x )
t −x
(a)
(1 − x 2 )y ′ + xy = ax are
(a) y = x 2 + 2 x
(c) y = 2 x 2 − x
(d) 1
is
11 The curves satisfying the differential equation
(a) ellipse and parabola
(c) ellipse and hyperbola
(c) 9
15 Let f ( x ) be differentiable in the interval ( 0, π ) such that
10 The solution of differential equation
(a) sec x = (tan x + C ) y
(c) y tan x = sec x + C
(b) 6
1 2 3
1
=
− 1− 2 .
y
x
x
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
(b) 2 ye 2 x = Ce 2 x − 1
(d) None of these
ANSWERS
SESSION 1
1.
11.
21.
31.
SESSION 2
1. (c)
11. (c)
(b)
(c)
(b)
(a)
2.
12.
22.
32.
(d)
(b)
(a)
(b)
2. (a)
12. (b)
3.
13.
23.
33.
(a)
(c)
(a)
(b)
3. (a)
13. (b)
(a)
(a)
(a)
(a)
5. (c)
15. (a)
25. (d)
6. (c)
16. (c)
26. (d)
7. (b)
17. (a)
27. (d)
8. (d)
18. (a)
28. (a)
9. (a)
19. (a)
29. (a)
10. (d)
20. (c)
30. (a)
4. (d)
14. (c)
5. (d)
15. (a)
6. (a)
16. (c)
7. (a)
8. (a)
9. (a)
10. (a)
4.
14.
24.
34.
194
Hints and Explanations
SESSION 1
1 Given,
y = c 1 ⋅ e 2 x + c 2 + c 3 e x + c 4 sin( x + c 5 )
= c 1 ⋅ e c 2 e 2x + c 3 e x
+ c 4 (sin x cos c 5 + cos x sin c 5 )
= Ae 2 x + c 3 e x + B sin x + D cos x
Here, A = c 1 e c 2 , B = c 4 cos c 5 and
D = c 4 sin c 5
Since, equation consists four arbitrary
constants.
So, the order of differential equation
is 4.
On integrating, we get
y
1 2
C
( x + y 2 ) + tan −1 =
x
2
2
C − x2 − y 2
−1 y
⇒
= tan
x
2
∴
3 Given, y 2 = 2c ( x + c )
7 Equation of family of parabolas with
focus at (0, 0) and axis as X -axis is
y 2 = 4 a( x − a)
…(i)
On differentiating Eq. (i) w.r.t. x, we get
2 yy 1 = 4 a
On differentiating both side w.r.t. x, we
get
dy
dy
2y
= 2c ⇒ c = y
dx
dx
On putting in Eq. (i),
3 /2
dy
dy
y 2 = 2 xy
+ 2 y 3 /2
dx
dx
3
dy
dy
⇒ 8 y = y 2 − 2 xy
dx
dx
2
3
Which is the differential equation of
order one and degree 3.
4 Clearly, the equation of family of circle
which touch both the axes is
( x − a)2 + ( y − a)2 = a2 , where a is a
parameter.
Since, there is only one parameter,
therefore order of differential equation
representing this family is 1.
5 We have, v = A + B , where A and B are
r
parameters
On differentiating twice w.r.t. r, we get
dv
−A
... (i)
= 2
⇒
dr
r
d 2v 2 A
and
... (ii)
= 3
dr 2
r
Now, on substituting the value of A
from Eq. (i) in Eq. (ii), we get
d 2v
2
dv −2 dv
= 3 − r 2
=
dr 2
r
dr
r dr
d 2 v 2 dv
+
= 0, which is the
⇒
dr 2
r dr
required differential equation.
xdy − y dx
6 We have, x dx + y dy +
=0
x2 + y 2
⇒
yy 1
y 2 = 2 yy 1 x −
2
∴
...(i)
y
1
d ( x2 + y 2 ) + d tan −1 = 0
2
x
⇒
C − x2 − y 2
y = x tan
2
2 Since, the equation is not a polynomial
in all differential coefficients, so its
degree is not defined.
11 Clearly,
⇒
⇒
y = 2 x y 1 − y y 12
y y 12 = 2 x y 1 − y
8 x2 + y 2 + 2ax + 2by + c = 0
...(i)
On differentiating Eq. (i) three times, we
get
2 x + 2 yy 1 + 2a + 2by 1 = 0
⇒
x + yy 1 + a + by 1 = 0
...(ii)
...(iii)
1 + y 12 + yy 2 + by 2 = 0
...(iv)
3 y 1 y 2 + yy 3 + by 3 = 0
On eliminating b from Eqs. (iii) and (iv),
we get
3 y 1 y 22 − y 3 − y 12 y 3 = 0
9 We have, x dy = y (dx + y dy ), y > 0
∴
x dy − y dx
= dy
y2
⇒
x
=− y +C
y
Now,
∴
y(1) = 1 ⇒ C = 2
x
+ y =2
y
For x = − 3, − 3 + y = 2 y
⇒ y2 − 2 y − 3 = 0
⇒ ( y + 1) ( y − 3) = 0
∴
y =3
[Q y > 0]
dy
10 Here,
= y + 3 > 0 and y(0) = 2
dx
dy
⇒
∫ y + 3 = ∫dx
2
⇒
log e | y + 3 | = x + C
But
y(0) = 2
∴
log e |2 + 3 | = 0 + C
⇒
C = log e 5
⇒
log e | y + 3 | = x + log e 5
When x = log e 2, then
⇒ log e | y + 3 | = log e 2 + log e 5
= log e 10
∴
y + 3 = 10 ⇒ y = 7
y
1 − y2
dy = ∫ dx
− 1 − y2 = x + C
⇒ ( x + C )2 + y 2 = 1
Hence, the centre is (− C , 0) and radius
is 1.
dy
ax + 3
12 We have,
=
dx 2 y + f
⇒
(ax + 3) dx = (2 y + f ) dy
On integrating, we obtain
x2
a⋅
+ 3 x = y 2 + fy + C
2
a
⇒ − x2 + y 2 − 3 x + fy + C = 0
2
a
This will represent a circle, if − = 1
2
[Qcoefficient of x2 = coefficient of y 2 ]
⇒
a = −2
dy
13 ( x + 1) = y − y 2
dx
dy
dx
⇒
=
y (1 − y ) x + 1
1
1
dx
⇒ +
dy =
y
1
−
y
x+1
⇒ log y − log(1 − y )
= log( x + 1) + log C
y
⇒
= C ( x + 1)
1− y
⇒
[integrating]
∫
( x + 1)(1 − y ) = Cy .
dy
= e ax .e by
dx
⇒
e −by dy = e axdx
1
1
− e −by = e ax + C
⇒
b
a
⇒ be ax + ae −by + C ′ = 0
14 We have,
[C ′ = abC ]
15 We have,
dy
= e x − y + x2e − y = e − y ( x2 + e x )
dx
On separating the variables, we get
e y dy = ( x2 + e x )dx
On integrating both sides, we get
x3
ey =
+ ex + C
3
16 We have,
dy
+ ( y + 1)cos x = 0
dx
dy
cos x
=−
dx
y+1
2 + sin x
dy
cos x
= −
dx
y + 1 ∫ 2 + sin x
(2 + sin x )
⇒
⇒
∫
⇒
ln( y + 1) = − ln(2 + sin x ) + ln C
⇒ ( y + 1)(2 + sin x ) = C
Now,
y(0) = 1
195
DAY
⇒
(2)(2 + 0) = C
⇒
C =4
Thus, ( y + 1) (2 + sin x ) = 4
π
π
Now, at x = ,
( y + 1) 2 + sin = 4
2
2
1
⇒ dy = 1 − 2 dx
x
1
[integrating]
⇒ y = x+
+C
x
Since, the curve passing through the
7
point 2, .
2
7
1
i.e.
=2+ + C ⇒ C =1
2
2
1
y = x+
+1
…(i)
∴
x
Now, at x = − 2
1
Ordinate, y = − 2 − + 1 = − 3 / 2
2
2 dy
18 We have, ( x + y )
=4
dx
dy dv
On putting x + y = v and 1 +
,
=
dx dx
we get
dv
v 2
− 1 = 4
⇒
dx
2 dv
v
= v 2 + 4.
⇒
dx
v2 + 4 − 4
∴
dv = dx
v2 + 4
⇒ v − 2 tan −1 (v / 2) = x + C
x+ y = x+ C
⇒ x + y − 2 tan −1
2
∴
y(0) = 0 ⇒ C = 0.
x+ y
y = 2 tan −1
2
dy
19 We have, + 1 = e x + y
dx
On putting x + y = z and 1 +
dy dz
= ,
dx dx
we get
dz
= ez
dx
⇒ e − zdz = dx
On integrating both sides, we get
e −z
= x+C
−1
⇒
x + e −( x + y ) + C = 0
dy
y
f ( y / x)
20
= +
dx
x f ′( y / x )
dy
dv
On putting y / x = v and
=v+ x ,
dx
dx
we get
∴
⇒ log f (v ) = log x + log C ,C > 0
⇒ f ( y / x ) = C x ,C > 0.
dy y
y
= log + 1
x
dx x
Put y = tx,
dy
dt
and
=t + x
dx
dx
dt
Then, we get t + x
= t log t + t
dx
dt
dx
⇒
=
t log t
x
⇒
∴
log log t = log x + log C
y
log = Cx
x
22 Given, dx +
dy
x2 − xy + y 2
=0
y2
2
⇒
Put
and
x
x
dx
+ − + 1= 0
dy
y
y
x
v = or x = v y
y
dx
dv
=v + y
dy
dy
Then, we get v + y
⇒
1
+ log v = − log x + C
2v 2
2
y
x
⇒ 2 + C = log + log x = log y .
2y
x
⇒−
21 Given,
( y + 1)(2 + 1) = 4
4
1
y = −1 =
3
3
dy
1
17 Given,
=1− 2
dx
x
Now,
f ′(v )
dx
dv =
f (v )
x
dv
+ v2 − v + 1 = 0
dy
dy
dv
+
=0
v2 + 1
y
Now,
⇒
and
⇒
⇒
25 We have,
y (1 + xy ) dx = xdy
dy
y (1 + xy ) y
⇒
=
= + y2
dx
x
x
dy
dv
On putting y = vx and
=v+ x ,
dx
dx
we get
dv
v+ x
= v + v2 x 2
dx
dv
⇒
= xdx
v2
dv
⇒
∫ v 2 = ∫ xdx
1
x2
⇒
− =
+C
v
2
x
x2
⇒
− =
+C
2
y
Put (1, −1 ), then
∴
dy
y( x + y )
+
=0
dx
x2
On putting
y = vx, we get
dv
v+ x
+ v (1 + v ) = 0
dx
dx
dv
or
+
=0
x
v (v + 2)
dx 1 1
1
or
+ −
dv = 0
x 2 v v + 2
∴
On integrating, we get
1
log x + [log v − log(v + 2)] + log C = 0
2
or log(v + 2) = log x2 vC 2
or
y + 2 x = C 2 x2 y .
dy
xy
24 Given,
= 2
dx
x + y2
dy
dv
Put y = vx ⇒
= v + x , we get
dx
dx
xdv
v
⇒ v+
=
dx
1 + v2
dv
v
v3
⇒
x
=
−v = −
2
dx 1 + v
1 + v2
2
1+ v
dx
dv = −
⇒
v3
x
1
2
x2 1
=
+
2
2
1
4
− ,y =
2
5
4
=
5
C =
⇒ tan −1 (v ) + log y + C = 0
[integrating]
−1 x
∴ tan + log y + C = 0
y
23 We have
x = 1, y = 1
C = −1 / 2
x = x0, y = e
x 02 1
− =1
2e 2 2
x 02 = 3e 2 .
−
x
y
Now, put x =
1
f −
2
dy
+ 2 xy = 4 x 2
dx
dy
2x
4x 2
+
y =
2
dx 1 + x
1 + x2
26 Given, (1 + x 2 )
⇒
∫
2x
dx
∴ IF = e 1 + x
= e log (1 + x ) = 1 + x 2
and the solution is
y ⋅ (1 + x 2 ) = ∫ 4 x 2 dx + C
2
y (1 + x 2 ) =
2
4 x3
+C
3
Now, x = 0, y = 0
⇒
C =0
4 x3
y (1 + x 2 ) =
∴
3
⇒ 3 y (1 + x 2 ) = 4 x3
27 We have,
dy
3 x2
sin2 x
+
y =
3
dx 1 + x
1 + x3
Since, it is a linear equation with
3 x2
P =
1 + x3
∴IF = e ∫
P dx
= e log (1 + x
3
)
= 1 + x3
196
and the solution is
sin2 x
y (1 + x3 ) = ∫
(1 + x3 ) dx
1 + x3
1 − cos 2 x
=∫
dx
2
1
sin 2 x
+C
∴ y (1 + x3 ) = x −
2
4
dy
28 Given, − y tan x = 2 x sec x, y (0) = 0
dx
− tan xdx
IF = e ∫
= e − log sec x
∴
IF = cos x
cos x ⋅ y = ∫ 2 x sec x ⋅ cos xdx
⇒ cos x ⋅ y = x2 + C
⇒ y (0) = 0 ⇒ C = 0
∴
y = x2 sec x
and y ' = 2 x ⋅ sec x + x2 sec x ⋅ tan x
π
π2
π2
y =
⋅ 2=
4 16
8 2
π
π
π2
y ′ = ⋅ 2 +
⋅ 2
4
2
16
π
π2
2 π2
y =
⋅2 =
3
9
9
π
π
π2
y ′ = 2 ⋅ 2 +
⋅2⋅ 3
3
3
9
=
4 π 2 π2 3
+
3
9
29 This is the linear equation of the form
dy
+ Py = Q .
dx
where, P = tan x and Q = x m cos x
Now, integrating factor (IF)
P dx
tan x dx
= e∫
= e∫
= e log s ec x = sec x
and the solution is given by,
P dx
P dx
= Q ⋅e ∫
dx + C
y ⋅e ∫
⇒ y ⋅ sec x =
⇒
y sec x =
∫
∫x
m
⋅ cos x ⋅ sec xdx + C
xm +1
+C
m+1
∴ (m + 1)y = x m + 1 cos x
+ C (m + 1) cos x
30 Given differential equation is
dP 1
− P (t ) = − 200, which is a linear
dt
2
differential equation.
−1
Here, P ( t ) =
and Q ( t ) = − 200
2
1
t
−
∫ − dt
Now, IF = e 2 = e 2
and the solution is
P ( t ). IF = ∫ Q (t ) IF dt + K
P ( t )⋅ e
P ( t )⋅ e
−
−
t
2
t
2
−
= − ∫ 200 e
= 400 e
−
t
2
dt + K
t
2
+ K
t
⇒ P (t ) = 400 + Ke 2
If P(0) = 100, then K = − 300
t
⇒ P (t ) = 400 − 300 e 2
31 We have, ( xy 5 + 2 y )dx = xdy
dy
− 2 y = xy 5
dx
dy 2 y
⇒
−
= y5
dx
x
dy 2 y −4
⇒ y −5
−
=1
dx
x
Put,
y −4 = t
dt
−5 dy
⇒
−4 y
=
dx dx
dy
−1 dt
⇒
y −5
=
dx
4 dx
From Eqs. (i) and (ii), we get
1 dt 2t
−
−
=1
4 dx
x
8t
dt
⇒
+
= −4
dx
x
⇒
x
8
Now, IF = e ∫ x
dx
…(i)
34 Let c 1 + c 2 + c 3 e c = A
= e 8 log x = x 8
x8
4 ⋅ x9
=−
+C
4
y
9
⇒ 9 x 8 + 4 x 9 ⋅ y 4 = 9 y 4C
dy
32 We have, sin x + y cos x = 4 x
dx
⇒ sin xdy + y cos xdx = 4 xdx
⇒ d ( y sin x ) = 4 xdx
On integrating both sides, we get
y sin x = 2 x2 + C
π
Since, it is passes through ,0
2
π2
+ C ⇒ C = − π2 / 2
2
⇒ y sin x = 2 x2 − π2 / 2
π2
⇒ y = 2 x2cosec x −
cosec x
2
π2
π π2
⇒ y( π / 6) = 2 cosec −
cosec π/6
6
2
36
∴
0=
π2
π2
= 2 2 −
⋅2
2
36
=−
[constant]
4
…(ii)
and the solution is
t ⋅ x 8 = ∫ (−4)x 8dx + C
⇒
Statement II y 2 = 4 ax
…(i)
dy
⇒
2y
= 4a
dx
y dy
[from Eq. (i)]
⇒
a= ⋅
2 dx
y dy
⇒
y2 = 4 x ⋅ ⋅
2 dx
dy
⇒
y 2 = 2 xy
dx
dy
⇒
y = 2x ⋅ ,
dx
which has order = 1 and degree = 1
8 π2
9
33 Statement I Let the equation of parabola
whose axis is the axis of x and vertex at
the origin is
y 2 = 4 ax
dy
⇒
2y
= 4a
dx
dy 2a
⇒
=
dx
y
dy
1
⇒
∝
dx
y
[where, a → parameter]
Then, y = Ax
dy
= A
dx
dy
⇒
y = x
dx
dy
∴
x
= y
dx
⇒
SESSION 2
1 Given, y = c 1 e c
⇒
2
x
y ′ = c 1 c 2e c 2 x ⇒ c 2 =
y′
y
y ′
and y ′′ = c 1 c 22e c 2 x ⇒ y ′′ = y .
y
2
⇒ yy ′′ = ( y ′) 2
2 Given,
dp ( t )
= 0.5 p (t ) − 450
dt
2dp ( t )
⇒
= dt
p ( t ) − 900
2dp ( t )
⇒ ∫
= dt
p ( t ) − 900 ∫
p ′ (t ) =
⇒ 2 log | p (t ) − 900 | = t + C
To find the value of C, let’s substitute
t = 0 and p(0) = 850
⇒ 2 log | p (0) − 900 | = 0 + C
⇒ C = 2 log |850 − 900 |
⇒ C = 2 log 50
Now, 2 log | p (t ) − 900 | = t + 2 log 50
Now, put p (t ) = 0, then
2 log |0 − 900 | = t + 2 log 50
900
⇒ t = 2 log
= 2 log 18
50
3 Given slope at ( x, y ) is
dy
y
= + sec( y / x )
dx
x
y
Let
= t ⇒ y = xt
x
dy
dt
and
=t+ x
dx
dx
dt
Now, t + x
= t + sec(t )
dx
197
DAY
1
∫ cos t dt = ∫ x dx
sin t = ln x + C
sin( y / x ) = ln x + C
QThis curve passes through (1, π / 6)
1
sin( π / 6) = ln(1) + C ⇒ C =
2
y
1
Thus, sin = ln x +
x
2
4 We have, y ′( x ) + y ( x )g ′( x ) = g ( x )g ′( x )
Linear differential equation with
integrating factor e g( x )
⇒ y ( x ) ⋅ e g( x ) = ∫ g ( x ) ⋅ g ′( x ) ⋅ e g( x )dx
⇒ y ( x ) ⋅ e g( x ) = e g( x )(g ( x ) − 1) + C
Since, y(0) = 0 and g(0) = 0, therefore
C =1
⇒ y ( x ) = (g ( x ) − 1) + e − g( x )
⇒ y (2) = (g (2) − 1) + e − g(2 ) = 0,
as g(2) = 0.
5 Here, dx + 12 ⋅ x = 13
dy
y
y
[linear differential equation in x]
∫y
1
IF = e
Clearly,
2
dy
=e
−
1
y
Now, complete solution is,
x ⋅e
−
1
−
1
x ⋅e
⇒
=
y
1
∫
−
1
⋅ e y dy
y3
∫
−
1 1
⋅
⋅ e y dy
y y2
=
y
1
Put
1
−
=t ⇒
y
⇒
xe
⇒
xe
⇒
xe
⇒
xe
−
1
−
1
y
y
−1
−
y
=
y
−1
1
dy = dt
y2
∫ − t ⋅ e dt
t
= − {t ⋅ e t − ∫ 1 ⋅ e tdt } + C
= − te t + e t + C
1
1
=
1
−
1 −y
⋅e
+e y +C
y
−1
+ e + C [Q y (1) = 1]
1
C =−
e
1
1
1
−
−
1 −y
1
y
xe
= ⋅e
+e y −
y
e
⇒
e
⇒
⇒
=e
−1
1
x=
∴
1
1
+ 1 − ⋅e y
y
e
6 On putting 10 x + 6 y = t and
dy dt
= , we get
dx dx
2 tan t / 2
dt
= 6sin t + 10 = 6
+ 10
2
dx
1 + tan t / 2
10 + 6
⇒
2
dz
[z = tan(t / 2)]
= dx
5z2 + 6z + 5
1
dz
⇒
= ∫ dx
2
5∫
3
2
z + + (4 / 5)
5
1
5z + 3
tan −1
= x+C
⇒
4
4
= ∫ (cosecθ − sin θ)dθ
⇒
sec t /2
dt =dx
10 tan2 t /2 + 12 tant / 2 + 10
⇒ 5tan t / 2 + 3 = 4 tan 4( x + C )
⇒ 5tan(5x + 3 y ) + 3 = 4 tan 4( x + C )
1
Now, x = 0, y = 0 ⇒ C = tan −1 (3 / 4)
4
Hence, equation of the curve is
25tan 4 x
5tan(5x + 3 y ) =
4 − 3 tan 4 x
1
5tan 4 x
⇒ y = tan −1
− 5x .
3
4 − 3 tan 4 x
⇒
dt
d {V (t )} = − k (T − t ) dt
∫
T
0
d {V (t )} =
∫
T
0
…(i)
− k (T − t )dt
T
(t − T )2
⇒ V (T ) − V (0) = k
2
0
k
2
⇒ V (T ) − I = [(T − T ) − (0 − T )2 ]
2
∴
[Qwhen t = 0, then V (t ) = I]
k
V (T ) = I − T 2
2
8 On putting x = tan A, and y = tan B in
the given relation, we get
cos A + cos B = λ (sin A − sin B )
A − B 1
⇒
tan
=
2 λ
1
⇒ tan −1 x − tan −1 y = 2 tan −1
λ
On differentiating w.r.t. x, we get
dy
1
1
−
⋅
=0
1 + x2 1 + y 2 dx
dy 1 + y 2
=
dx 1 + x2
⇒
Clearly, it is a differential equation of
degree 1.
9 y 1 + (dx / dy ) 2 = 1
⇒ y2 1 +
dy
∴
=±
dx
⇒∫
1− y
y
2
⇒ C ± x = log
dx
=1
dy
y
2
1 − y2
dy = ± x + C.
On putting y = sinθ and dy = cos θ dθ, we
get
cos θ ⋅ cos θ dθ
C± x=∫
sin θ
1 − 1 − y2
y
+ 1 − y2 .
cos x dy = y sin x dx − y 2 dx
1 dy 1
⇒
− tan x = − sec x
y 2 dx y
1
1 dy dz
Put
− =z ⇒ 2
=
y
y dx dx
dz
∴
+ (tan x ) z = − sec x
dx
This is a linear differential equation.
tan x dx
Now, IF = e ∫
= e log sec x = sec x
Hence, the solution is
z ⋅ (sec x ) = ∫ − sec x ⋅ sec x dx + C1
10 Given,
⇒
7 Given, d {V (t )} = − k (T − t )
∴
⇒ C ± x = log cosecθ − cot θ + cos θ
−
1
sec x = − tan x + C1
y
∴
sec x = y (tan x + C )
where C = − C1 .
dy
11 We have, (1 − x2 ) + xy = ax
dx
dy
x
ax
[L. D. E.]
⇒
+
y
=
dx 1 − x2
1 − x2
1
IF = e ∫
=e 2
1
, if − 1 < x < 1
2
= 1− x
1
, if x < −1or x > 1.
2
x −1
If −1 < x < 1, then solution is
1
x
y⋅
= a∫
dx + C
(1 − x2 )3 /2
1 − x2
a
1
y⋅
=
+C
1 − x2
1 − x2
( x / 1 − x2 )dx
⇒
−
log 1 − x2
y = a + C 1 − x2
⇒ ( y − a) 2 = C 2 (1 − x2 )
⇒ C 2 x2 + ( y − a)2 = C 2
which is an ellipse.
If x < −1 or x > 1, solution is
1
x
y⋅
= C + a∫
dx
2
2
x −1
(1 − x ) x2 − 1
x
a
dx = C +
= C − a∫ 2
( x − 1)3 /2
x2 − 1
⇒
y = a + C x2 − 1
⇒
( y − a)2 = C 2 x2 − c 2
⇒ C 2 x2 − ( y − a)2 = C 2
which represents a hyperbola.
12 Given,
dy ( x + 1) 2 + y − 3
y −3
=
= ( x + 1) +
dx
x+1
x+1
dy dY
Putting x + 1 = X , y − 3 = Y ,
=
,
dx dX
the equation becomes
dY
Y
= X+
dX
X
198
dY
1
− ⋅Y = X
dX X
or
IF = e ∫
( −1 / X )dX
[L. D. E.]
= e − log X = X −1 =
1
X
∴ The solution is
1
1
Y ⋅ = C + ∫ X ⋅ dx = C + X
X
X
( y − 3)
or
=C + x+1
( x + 1)
x = 2, y = 0
0−3
⇒
= C + 2+ 1
2+ 1
⇒
C = −4
∴ The equation of the curve is
y −3
= x − 3 or y = x2 − 2 x.
x+1
Now,
3
5
x
x
x+
+
+K
dx − dy
3
!
5!
13 We have,
=
x2
x4
dx + dy
1+
+
+K
2!
4!
On applying componendo and
dividendo, we get
x3
x5
x2
x4
+
+ K + 1 +
+
+ K
x+
3!
5!
2!
4!
x3
x4
x2
x4
+
+ K − 1 +
+
+ K
x+
3!
4!
2!
4!
=
⇒
(dx − dy ) + (dx + dy )
(dx − dy ) − (dx − dy )
x2
x3
+
+ K
1 + x +
2! 3!
x2 x3
− 1 − x +
−
+ K
2! 3!
=
dx
2dx
=−
dy
−2dy
⇒
ex
dx
=−
−e − x
dy
⇒
dy = e −2 xdx
On integrating both sides, we get
e −2 x
y =
+ C1
−2
⇒ 2 ye 2 x = −1 + 2C1 e 2 x
⇒ 2 ye 2 x = Ce 2 X − 1, where C = 2C1
dy
14 Equation of tangent is Y − y = ( X − x )
dx
dy
Y-intercept of tangent is y − x
dx
dy
From given condition, y − x
= x3 ,
dx
dy 1
...(i)
⇒
− ⋅ y = − x2
dx x
1
−
dx
1
Now,
IF = e ∫ x = e − loge x =
x
and solution is
1
x2
⋅ y = − ∫ xdx = −
+C
x
2
− x3
⇒ f ( x) =
+ Cx
2
1 3
Now, f (1) = 1 ⇒ C = 1 + =
2 2
− x3 3
+ x
∴ f ( x) =
2
2
27 9
− =9
⇒ f (−3) =
2 2
t 2 f ( x ) − x2 f ( x ) + x2 f ( x ) − x2 f (t )
=1
t−x
f ( x ) − f (t )
lim (t + x )f ( x ) + x2
=1
t→ x
t − x
dy
− 2 xy
1
dx
= − 4,
x4
x
where y = f ( x )
d y
1
=− 4
dx x2
x
1
⇒
x −2 y = 3 + C
3x
1
2
⇒
x −2 y = 3 +
3x
3
1
2 x2
.
⇒
y =
+
3x
3
⇒
x2
16 Given,
⇒
y y2 − 1
dy
=
dx
x x2 − 1
∫y
dy
y2 − 1
=
∫x
dx
x2 − 1
sec −1 y = sec −1 x + C
2
Now, x = 2, y =
3
π
π
π
⇒
=
+ C ⇒C = −
6
3
6
π
∴ y = sec sec −1 x −
6
⇒
⇒
⇒
1
1
3
= cos cos −1 − cos −1
y
x
2
3
1
= cos cos −1
+
y
2x
15 lim
t→ x
2 xf ( x ) + x2 (− f ′( x )) = 1
[Q f (1) = 1]
1−
1
x2
⋅ 1−
∴
1
3 1
1
=
+
1− 2
y 2x
x
2
3
4
199
DAY
DAY NINTEEN
Unit Test 2
(Calculus)
1 Let f : ( 2, 3) → ( 0, 1) be defined by f ( x ) = x − [ x ], then
(b) increasing for every value of x
(c) decreasing for every value of x
(d) None of the above
−1
f ( x ) is equal to
(a) x − 2
2
∫
(b) x + 1
dx
sin x − cos x +
2
(c) x − 1
1
x
π
(b)
tan + + C
2 8
2
1
x
π
(d) −
cot + + C
2 8
2
(c) e −1
(b) −
(a) − 1
3
4
(c)
4
3
(d) 1
1
(b) 1 −
sq units
2
π
1
(d)
+
− 1 sq units
4 2
2
(c) 2
(a)
(a)
(b)
(c)
(d)
x
x2
x2
x2
+ y
+ y2
− y2
− y2
2
(a) 1 −
x3
is
3
1
1
(a) increasing for x > and decreasing for x <
4
4
(c)
1
60
(d)
3
20
(b) 0
(d) not defined
2
3
a = 0, b =
(b)
8
7
1
1
cos 2x − cos 3x is
2
3
(c)
9
4
(d)
f (b ) − f (a )
= f ′ (c ), if
b −a
3
8
1
and f ( x ) = x ( x − 1)( x − 2), then value of c is
2
15
6
(b) 1 +
15
(c) 1 −
21
6
(d) 1 +
21
14 If the sides and angles of a plane triangle vary in such a
way that its circumradius remains constant. Then,
da
db
dc
is equal to (where, da, db and dc
+
+
cos A cos B cos C
are small increments in the sides a, b and c,
respectively).
+ xy − x + y = C
− xy + x + y = C
+ 2 xy − x + y = C
− 2 xy + x − y = C
8 The function f ( x ) = x 4 −
1
40
13 In the mean value theorem
( 2x − y + 1) dx + ( 2y − x + 1) dy = 0 is
2
(b)
the function f ( x ) = cos x +
(d) None of these
7 The general solution of the differential equation
xn
n!
12 The difference between the greatest and least values of
2 − sin θ
6 The value of ∫
log
d θ is
−π /2
2 + sin θ
(b) 1
1
20
(a) −2
(c) 2
π /2
(a) 0
(d) y − 1 −
| x − 1| + | x − 3| at the point x = 2 is
5 The area bounded by y = sin−1 x , x = 1/ 2 and X-axis is
1
(a)
+ 1 sq units
2
π
(c)
sq units
4 2
xn
n!
11 The differential coefficient of the function
4 If the normal to the curve y = f ( x ) at the point ( 3, 4)
makes an angle 3 π / 4 with the positive X -axis, then f ′ ( 3)
is equal to
(c) y −
x 1/ 4 − x 1/ 5
is
x →1
x 3 −1
(a)
(d) 1
xn
n!
10 The value of lim
is equal to
(b) e 2
x2
x3
xn
dy
is equal to
+
+ ... +
, then
2!
3!
n!
dx
(b) y +
(a) y
n ( n − 1)
3 lim
n→ ∞
(a) e
9 If y = 1 + x +
is equal to
1
x
π
(a) −
tan + + C
2 8
2
1
x
π
(c)
cot + + C
2 8
2
n 2 − n + 1
n 2 − n − 1
(d) x + 2
(a) 1
15
∫
1.5
0
(b) 2
(c) 0
(d) 3
2
[ x ] dx , where [.] denotes the greatest integer
function, is equal to
(a) 2 +
2
(b) 2 −
2
(c) −2 +
2
(d) −2 −
2
200
16 The integrating factor of the differential equation
dy
= y tan x − y 2 sec x , is
dx
(a) tanx
(b) sec x
(a 2 + b 2 ) (u 2 + v 2 ) is equal to
(c) − sec x
(d) cot x
17 The area bounded by the straight lines x = 0 , x = 2 and
the curve y = 2x , y = 2x − x 2 is
4
1
−
3 log 2
4
(c)
−1
log 2
3
4
+
log 2 3
3
4
(d)
−
log 2 3
(a)
(b)
19 If I1 = ∫
I2 = ∫
k
1− k
k
1− k
(c)
20 If I1 = ∫
and I 2 = ∫
sec 2 z
3
2
(c) 1
(b)
1
sq unit
6
I1
is equal to
I2
1
2
(d) None of these
(b)
21 If f and g are continuous functions on [ 0, π ] satisfying
f ( x ) + f ( π − x ) = g ( x ) + g ( π − x ) = 1, then
π
∫ [f ( x ) + g ( x )] dx is equal to
(b) 2 π
(c)
(a) a = 1, b = 2
(c) a = 2, any b
(d)
3π
2
2x − 3 ⋅ [ x ], x ≥ 1
πx
sin , x < 1
2
x→
(b)
f ′ (x)
(b)
2y + 1
f ′ (x)
(c)
1− 2 y
π
2
is
(c)
π
4
(d)
π
8
π
dy
is equal to
dx
x, then
h is increasing whenever f is increasing
h is increasing whenever f is decreasing
h is decreasing whenever f is increasing
Nothing can be said in general
(b) 4
f ′ (x)
(d)
4 + 2y
(c) 5
(d) 6
sin x
π
, then f ′ 3 is equal to
1
+
cos
x
32 If f ( x ) = tan−1
(a)
24 If h( x ) = f ( x ) − (f ( x ))2 + (f ( x ))3 for every real numbers
(a)
(b)
(c)
(d)
t dt
sin( 2x − π )
1
2 (1 + cos x)
1
(c)
4
continuous at x = 2
differentiable at x = 1
continuous but not differentiable at x = 1
None of the above
f ′ (x)
(a)
2y − 1
x
π
2
31 Let f ( x ) = 2 cos x + , 0 < x ≤ 2π (where, [.]
4
(a) 3
(where, [x] denotes the greatest integer ≤ x ) is
23 If y = f ( x ) + f ( x ) + f ( x ) + ... ∞ , then
2
∫
denotes the greatest integer ≤ x ). The number of points
of discontinuity of f ( x ) are
22 The function f ( x ) =
(a)
(b)
(c)
(d)
(b) a = 2, b = 4
(d) any a, b = 4
30 The value of limπ
(a) ∞
π
2
(b) h is differentiable for all x
(d) None of these
e 2 x − 1
, x <0
is continuous
29 The function f ( x ) =
bx 2
ax + 2 − 1, x ≥ 0
and differentiable for
0
(a) π
5
sq unit
6
(d) None of these
(a) h is continuous for all x
(c) h ′ (x) = 2 for all x > 1
f {x ( 3 − x )} dx , where f is a continuous
(a)
[g (b)]2 − [g (a)]2
2
28 If h ( x ) = min{x , x 2 }, for every real number of x. Then,
x f {x ( 3 − x )} dx
function and z is any real number, then
(d)
27 If y be the function which passes through (1 , 2) having
(c)
(b) 2 I1 = I 2
(d) None of these
2 − tan2 z
[f (b)] − [f (a)]
2
(a) 6 sq units
sin {x (1 − x )} dx , then
sec 2 z
(b) g (b) − g (a)
2
slope ( 2x + 1). Then, the area bounded between the
curve and X-axis is
x sin {x (1 − x )} dx and
2 − tan2 z
b
d [f ( x )]
= g ( x ) for a ≤ x ≤ b, then ∫ f ( x ) ⋅ g ( x ) dx is
a
dx
equal to
26 If
2
(b) 1 sq unit
(d) None of these
(a) I1 = 2 I 2
(c) I1 = I 2
(b) (a 2 + b 2 ) e 2 ax
(d) (a 2 − b 2 ) e 2 ax
(a) 2e ax
(c) e 2 ax
(a) f (b) − f (a)
18 The area of the region bounded by y = | x − 1| and y = 1 is
(a) 2 sq units
(c) 1 / 2 sq unit
25 If u = ∫ e ax cos bx dx and v = ∫ e ax sin bx dx , then
(b)
1
2
(d) None of these
33 Let f ( x ) be a polynomial function of the second degree. If
f (1) = f ( −1) and a1, a 2 , a 3 are in AP, then
f ′ (a1 ), f ′ (a 2 ) and f ′ (a 3 ) are in
(a) AP
(c) HP
(b) GP
(d) None of these
34 If y = cos −1(cos x ), then y ′ ( x ) is equal to
(a)
(b)
(c)
(d)
1 for all x
−1 for all x
1in 2nd and 3rd quadrants
−1in 3rd and 4th quadrants
201
DAY
35 In ( −4 , 4) the function f ( x ) = ∫
(a) no extreme
(c) two extreme
x
−10
(t 4 − 4) e −4t dt has
44 I1 = ∫ sin−1x dx and I 2 = ∫ sin−1 1 − x 2 dx , then
(a) I1 = I 2
(b) one extreme
(d) four extreme
(c) I1 + I 2 =
36 If f ( x ) = kx 3 − 9x 2 + 9x + 3 is monotonically
increasing in each interval, then
(a) k < 3
(c) k > 3
(b) k ≤ 3
(d) None of these
2
x
+ 1
−1
−1 x
+
tan
tan
dx is equal to
2
∫ −1
x +1
x
3
(a) π
(c) 3 π
(b) 2 π
(d) None of these
199 + 299 + 399 + ... + n 99
is equal to
n→ ∞
n100
38 lim
1
(b)
100
39 If ∫ ( sin 2x + cos 2x ) dx =
1
(c)
99
1
(d)
101
1
sin( 2x − c ) + a, then the
2
value of a and c is
π
and a = k
4
π
π
(b) c = − and a =
4
2
π
(c) c = and a is an arbitrary constant
2
(d) None of the above
(a) c =
40
cos 2x
∫ (cos x + sin x )
2
dx is equal to
(a) log cos x + sin x + C
(c) log(cos x + sin x) + C
1
2
1
(c) k = −
4
(a) k = −
42
(b) log(cos x − sin x) + C
1
(d) −
+C
cos x + sin x
cos 4x + 1
dx = k cos 4x + C, then
cot x − tan x
41 If ∫
∫
π /4
0
πx
2
1
8
(d) None of these
(b) k = −
dx
∫ sin( x − a ) sin( x − b ) is equal to
(a)
sin(x − a)
1
log
+C
sin(a − b)
sin(x − b)
(b)
sin(x − a)
−1
log
+C
sin(a − b)
sin(x − b)
(c) log sin(x − a) ⋅ sin(x − b) + C
sin(x − a)
(d) log
+C
sin(x − b)
( x 2 + 1)
43 ∫ e
dx is equal to
( x + 1) 2
x
x − 1 x
(a)
e + C
x + 1
x + 1
(b) e x
+C
x − 1
(c) e x (x + 1) (x − 1) + C
(d) None of these
π
2I1
(d) I1 + I 2 =
π
2
dx
is equal to
cos 4 x − cos 2 x sin2 x + sin4 x
π
2
π
(c)
3
(a)
37 The integral
9
(a)
100
45
(b) I 2 =
(b)
π
4
(d) None of these
46 If f ( x ) = | x − 1| + | x − 3 | + | 5 − x | , ∀ x ∈ R . If f ( x ) is
increases, then x belongs to
(a) (1, ∞)
47 The value of ∫
(b) (3, ∞)
(c) (5, ∞)
(d) (1, 3)
dx
is
1/ 2
(1 + x ) − (1 + x )1/ 3
(a) 2 λ1 / 2 + 3 λ1 / 3 + 6λ1 / 6 + 6 ln | λ1 / 6 − 1| + C
(b) 2 λ1 / 2 − 3 λ1 / 3 + 6λ1 / 6 + 6 ln | λ1 / 6 − 1| + C
(c) 2 λ1 / 2 + 3 λ1 / 3 − 6λ1 / 6 + 6 ln | λ1 / 6 − 1| + C
(d) 2 λ1 / 2 + 3 λ1 / 3 + 6λ1 / 6 − 6 ln | λ1 / 6 − 1| + C
(where, λ = 1 + x)
48 The value of ∫
(a) −
(b)
x
dx is
x −1
x1/ 4 − 1
4 3/4
x
+ 4 x1 / 4 + 2 ln 1 / 4
+C
3
x +1
x1/ 4 − 1
4 3/4
x
+ 4 x1 / 4 + 2 ln 1 / 4
+C
3
x +1
(c) −
(d)
4
x1/ 4 − 1
4 3/4
x
− 4 x1 / 4 + 2 ln 1 / 4
+C
3
x +1
x1/ 4 − 1
4 3/4
x
− 4 x1 / 4 + 2 ln 1 / 4
+C
3
x +1
49 The value of n for which the function
(( 5)x − 1)3
, x ≠0
x
x2
f ( x ) = sin ⋅ log 1 +
n
3
3
15 (log 5) ,
x =0
may be continuous at x = 0, is
(a) 5
(b) 3
(c) 4
(d) 2
50 The longest distance of the point (4, 0) from the curve
2x (1 − x ) = y 2 is equal to
(a) 3 units
(c) 5 units
(b) 4.5 units
(d) None of these
51 The normal to the curve 5x 5 − 10x 3 + x + 2y + 4 = 0 at
P ( 0, − 2) meets the curve again at two points at which
equation of tangents to the curve is equal to
(a) y = 3 x + 2
(c) 3 y + 2 x + 7 = 0
(b) y = 2 (x − 1)
(d) None of these
202
52
∫
2
−1
{|x − 1| + [ x ]} dx , where [ x ] is greatest integer is equal
58 Statement I The tangents to curve y = 7x 3 + 11 at the
points, where x = 2 and x = − 2 are parallel.
Statement II The slope of the tangents at the points,
where x = 2 and x = − 2, are equal.
to
(a) 8
(b) 9
(c)
5
2
(d) 4
59 Statement I The derivative of
53 The equation of the curve passing through the point
1 π and having slope of tangent at any point ( x , y ) as
,
2 8
( y / x ) − cos 2 ( y / x ) is equal to
e
(a) y = x tan−1 log
2x
2x
(c) y = x 2 tan−1
e
e
(b) x = y tan−1 2
x
(d) None of these
54 If x dy = y (dx + y dy ), y (1) = 1 and y ( x ) > 0, then y ( − 3)
is equal to
(a) 3
55
∫
π /4
0
(b) 2
(c) 1
(d) 0
sin x + cos x
dx is equal to
9 + 16 sin 2x
1
log 3
20
1
(c)
log 5
20
(a)
e
56 If f ( x ) =
(a) 0
Statement II The solution of a linear equation is obtained
by multiplying with its integrating factor.
x 2 −1
x2 +1
61 Statement I ∫ 2
dx = sec −1
+C
4
x 2
( x + 1) x + 1
(b) log 3
3
sin x , | x | ≤ 2
, then ∫ f ( x ) dx is equal
−
2
2,
otherwise
(c) 2
2
(b)
π
is
4
2 +1
(c)
2 −1
(d)
t t −a
2
=
1
t
sec −1
+C
a
a
| x − 1| + | x − 2 | + | x − 3 | is 2.
Statement II The function | x − 1| + | x − 2 | + | x − 3 | is
differentiable on R ~ {1, 2}.
63 Statement I If f ( x ) = max {x 2 − 2x + 2,| x − 1| }, then the
greatest value of f ( x ) on the interval [0, 3] is 5.
Statement II Greatest value, f ( 3) = max {5 , 2) = 5
(d) 3
57 The area bounded by curves y = cos x and y = sin x and
ordinates x = 0 and x =
dt
62 Statement I The absolute minimum value of
(d) None of these
(b) 1
3
cos t 2dt , ( x > 0) at x = 1 is cos 1.
2
d φ (x )
f (t ) dt = f ( φ ( x )) − f ( ψ ( x ))
dx ∫ ψ ( x )
60 Statement I The solution of the equation
dy
1
x
+ 6y = 3xy 4 / 3 is y ( x ) =
.
dx
( x + cx 2 ) 3
cos x
x
1/ x
Statement II
Statement II ∫
to
(a)
f (x ) = ∫
2 ( 2 − 1)
Direction (Q. Nos. 58-66) Each of these questions
contains two statements : Statement I (Assertion) and
Statement II (Reason). Each of these questions also has
four alternative choices, only one of which is the correct
answer. You have to select one of the codes ( a), (b), (c)
and (d) given below.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
64 Statement I The point of contact of the vertical tangents
to x = 2 − 3 sin θ, y = 3 + 2 cos θ are ( − 1, 3) and ( 5 , 3).
Statement II For vertical tangent, dx /d θ = 0
65 Let f : R → R be differentiable and strictly increasing
function throughout its domain.
Statement I If | f ( x ) | is also strictly increasing function,
then f ( x ) = 0 has no real roots.
Statement II At ∞ or − ∞ f ( x ) may approach to 0, but
cannot be equal to zero.
66 Statement I The area by region | x + y | + | x − y | ≤ 2 is
4 sq units.
Statement II Area enclosed by region
| x + y | + | x − y | ≤ 2 is symmetric about X-axis.
ANSWERS
1
11
21
31
41
51
61
(d)
(b)
(a)
(d)
(b)
(b)
(d)
2
12
22
32
42
52
62
(d)
(c)
(c)
(b)
(a)
(c)
(c)
3
13
23
33
43
53
63
(b)
(c)
(a)
(a)
(a)
(a)
(b)
4
14
24
34
44
54
64
(d)
(c)
(a)
(d)
(c)
(a)
(a)
5
15
25
35
45
55
65
(d)
(b)
(c)
(c)
(a)
(a)
(b)
6
16
26
36
46
56
66
(a)
(b)
(c)
(c)
(b)
(c)
(b)
7
17
27
37
47
57
(b)
(d)
(c)
(b)
(a)
(c)
8
18
28
38
48
58
(a)
(b)
(a)
(b)
(b)
(a)
9
19
29
39
49
59
(c)
(b)
(c)
(a)
(a)
(c)
10
20
30
40
50
60
(c)
(a)
(c)
(c)
(c)
(b)
203
DAY
Hints and Explanations
1 Given, f : (2, 3) → (0, 1)
=
and f ( x) = x − [ x]
∴ f ( x) = y = x − 2
⇒
x = y + 2 = f −1 ( y )
−1
⇒ f ( x) = x + 2
=
dx
π
π
2 sin x ⋅ sin − cos x ⋅ cos + 1
4
4
1
dx
=
∫
x π
2
1 − cos 2 +
2
8
1
π
2 x
=
∫ cosec 2 + 8 dx
2 2
x π
− cot +
2
1
8
=
+C
1/2
2 2
=−
1
2
n (n − 1 )
n( n − 1 )
−1
dy
dx
3π
−1
dy
=
⇒
=1
dx (3, 4 )
4
dy
dx (3, 4 )
∴
f ′ (3) = 1
= Area of rectangle OABC − Area
of curve OBCO
p
0,
2
O
B
A
1
x=
Ö2
−1
f (θ) is an odd function of θ .
I =∫
∴
π /2
−π /2
2 − sin θ
log
dθ = 0
2 + sin θ
x = X + h, y = Y + k
dY 2 X − Y + 2 h − k + 1
=
dX
X − 2 Y + h − 2k − 1
Again, put 2 h − k + 1 = 0
and
h − 2k − 1 = 0
On solving,
h = − 1, k = − 1
dY 2 X − Y
∴
=
dX
X − 2Y
On putting Y = vX , we get
dv 2 X − vX
2−v
⇒ v+ X
=
=
dX
X − 2 vX 1 − 2 v
⇒X
∴
∴
dv 2 − 2 v + 2 v
2 (v − v + 1)
=
=
dX
1 − 2v
1 − 2v
2
X
x3
3
⇒ f ′ ( x) = 4 x 3 − x 2
For increasing,
4 x 3 − x 2 > 0 ⇒ x 2 (4 x − 1) > 0
Therefore, the function is increasing
for x > 1 / 4.
1
Similarly, decreasing for x < .
4
8 f ( x) = x 4 −
9 Given,
x2
x3
xn
+
+K+
2! 3!
n!
dy
x2
xn −1
⇒
=0+1+ x+
+ ... +
dx
2!
(n − 1) !
y =1 + x +
⇒
⇒
2
dX
(1 − 2 v)
=
dv
X
2 (v2 − v + 1)
v2 − v + 1 = t
(2 v − 1) dv = dt
dt
dX
=−
X
2t
On integrating,
log X = log t −1 /2 + log C1 /2
∴
X = t −1 / 2 C 1 / 2
⇒
X = (v2 − v + 1)−1 /2 C1 /2
2
⇒ X (v2 − v + 1) = C
Y
p
4 C
⇒ x 2 + y2 − xy + x + y = C
2 − sin θ
= − log
= − f (θ)
2 + sin θ
Put
⇒
5 Required area
0,
⇒ y2 + x 2 − xy + x + y = C − 1
+ [cos y] π0 / 4
2 − sin θ
2 + sin θ
Put
1
1 +
n
(
n
−
1
)
e
= lim
= −1 = e 2
n( n − 1 )
n→ ∞
e
1
1 −
n
(
n
−
)
1
⇒ tan
⇒ ( y + 1) 2 − ( y + 1) ( x + 1) + ( x + 1) 2
=C
sin y dy
6 f (− θ) = log
⇒
4 Slope of the normal =
π /4
0
(2 x − y + 1) dx + (2 y − x + 1) dy = 0
dy 2 x − y + 1
,
⇒
=
dx
x − 2y − 1
n (n − 1 )
n (n − 1) + 1
= lim
n → ∞ n (n − 1) − 1
4 2
∫
7. Given,
x π
cot + + C
2
8
n2 − n + 1
3 lim 2
n→ ∞ n − n − 1
4 2
π
−
π
1
=
+
− 1 sq units
4 2
2
2 Let
I =∫
π
( y + 1) 2 ( y + 1)
⇒ ( x + 1) 2
−
+ 1
2
( x + 1)
( x + 1)
=C
dy x n
x2
+
=1+ x+
+K
dx n !
2!
xn −1
xn
+
+
(n − 1) ! n !
dy
xn
= y−
dx
n!
1 −3 / 4 1 −4 / 5
x
− x
x1 / 4 − x1 / 5
4
5
10 lim
=
lim
x →1
x →1
x3 − 1
3 x2
1 1
−
1
4 5
=
=
3
60
11 Given, f ( x) = | x − 1| + | x − 3|
x <1
−( x − 1) − ( x − 3),
f ( x) = ( x − 1) − ( x − 3), 1 ≤ x < 3
( x − 1) + ( x − 3),
x >3
4 − 2 x , x < 1
= 2, 1 < x < 3
2 x − 4 , x ≥ 3
In the neighbourhood of x = 2,
f ( x) = 2
Hence, f ′ ( x) = 0
12 The given function is periodic with
period 2π . So, the difference
between the greatest and least values
of the function is the difference
between these values on the interval
[0, 2π ].
Now,
f ′ ( x) = − (sin x + sin 2 x − sin 3 x)
= − 4 sin x sin(3 x/ 2) sin( x/2)
204
Hence, x = 0, 2 π /3, π and 2π are the
critical points.
1 1 7
Also, f (0) = 1 + − =
2 3 6
13
1
2π
f =−
, f (π ) = –
3
12
6
7
and f (2 π ) =
6
∴Required difference
7 13 27 9
= − − =
=
6 12 12 4
13 From mean value theorem,
f ′ (c ) =
f (b ) − f (a)
b −a
Given, a = 0 ⇒ f (a) = 0
1
3
and b = ⇒ f (b ) =
2
8
f ′ ( x) = ( x − 1)( x − 2) + x( x − 2)
+ x( x − 1)
f ′ (c ) = (c − 1)(c − 2) + c (c − 2)
+ c (c − 1)
= c 2 − 3 c + 2 + c 2 − 2c + c 2 − c
⇒ f ′ (c ) = 3 c 2 − 6 c + 2
According to mean value theorem,
f (b ) − f (a)
f ′ (c ) =
b −a
(3 /8) − 0 3
⇒ 3 c2 − 6 c + 2 =
=
(1 / 2) − 0 4
5
⇒ 3 c2 − 6 c + = 0
4
6 ± 36 − 15
6 ± 21
⇒ c=
=
2 ×3
6
=1 ±
21
=1 −
6
Q 1 +
21
6
21 1
∉ 0,
6 2
∴
da
db
dc
+
+
cos A cos B cos C
= 2 R (dA + dB + dC) = 0
1.5
15 ∴ ∫
=
=
∫
∫
[x2 ] dx
0
1
0
1
0
[x2 ] dx +
0 dx + ∫
a = 2 R sin A
On differentiating, we get
da = 2 R cos A dA
da
= 2 R dA
cos A
db
Similarly,
= 2 R dB
cos B
dc
and
= 2 RdC
cos C
2
1
∫
2
1
[x 2 ] dx +
1 dx +
∫
1. 5
2
∫
1. 5
2
[ x 2 ] dx
2 dx
∫
k
1−k
16 The differential equation is
dy
− y tan x = − y2 sec x
dx
This is Bernoulli’s equation, which
can be reducible to linear equation.
On dividing the equation by y2 , we
get
1 dy 1
…(i)
− tan x = − sec x
y2 dx y
1
1 dy dY
= Y ⇒− 2
=
y
y dx dx
Put
Eq. (i) reduces to
dY
−
− Y tan x = − sec x
dx
dY
⇒
+ Y tan x = sec x, which is a
dx
linear equation.
tan x dx
Hence, IF = e ∫
= sec x
2
17 A = ∫ [2 x − (2 x − x2 )] dx
0
2
2x
x3
3
4
=
− x2 +
=
−
3 0 log 2 3
log 2
18 y = x − 1, if x > 1 and y = − ( x − 1) ; if
x <1
Y
C
x=1
D
y=1
B
X¢
O
x=2
X
A
x+y=1
Y¢
y=x –1
20 I1 = ∫
2 − tan 2 z
sec 2 z
1
1
= × BC × AD = × 2 × 1
2
2
= 1 sq unit
19 I1 = ∫
k
1−k
(1 − x) sin [ x(1 − x)] dx
[by property]
k
x
1−k
(3 − x) f ( x(3 − x)) dx
[by property]
I1 = 3
∫
2 − tan 2 z
sec 2 z
∫
−
f ( x (3 − x)) dx
2 − tan 2 z
sec 2 z
I1 = 3 I2 − I1 ⇒
x f ( x (3 − x)) dx
I1 3
=
I2 2
π
21 Let I = ∫ [ f ( x) + g( x)] dx
0
∫
=∫
I =
π
0
π
0
=2
∫
[ f (π − x) + g(π − x)] dx
[1 − f ( x) + 1 − g( x)] dx
π
0
dx −
∫
π
0
[f ( x) + g ( x)] dx
⇒ 2I = 2 π
∴ I =π
22 At x = 1, f ( x) = 1,
lim f ( x) = lim+ 2 x − 3 [ x] = 1
x→1
πx
and lim− f ( x) = lim− sin
=1
x→1
x→1
2
Hence, f ( x) is continuous at x = 1.
Now,
2h − 1 − 1
f (1 + h) − 1
lim
= lim+
h → 0+
h→ 0
h
h
1 − 2h − 1
= lim+
= −2
h→ 0
h
f (1 + h) − 1
and lim−
h→ 0
h
π π h
sin +
−1
2
2
= lim−
h→ 0
h
πh
cos
−1
2
= lim−
=0
h→ 0
h
Hence, f ( x) is not differentiable at
x = 1.
x → 1+
23 Given that,
y = f ( x) +
Area of bounded region,
A = Area of ∆ ABC
∫
sin x (1 − x) dx
1. 5
2
= 2 −1 + 3 −2 2 =2 − 2
sin x (1 − x) dx −
= I2 − I1 or 2 I1 = I2
= 0 + ( x)1 + 2 ( x)
2
14 We know that, in a triangle
∠A + ∠ B + ∠ C = π
∴
dA + dB + dC = 0
If R is circumradius, then
a
b
c
=
=
=2 R
sin A sin B sin C
=
f (x ) +
f ( x) + ... ∞
y = f ( x) + y
On squaring both sides, we get
y2 − y = f ( x)
On differentiating w.r.t. x, we get
dy
(2 y − 1)
= f ′ ( x)
dx
dy
f ′ ( x)
∴
=
dx 2 y − 1
205
DAY
⇒ x( x − 1) ≥ 0 ⇒ x ≤ 0 or
24 We have,
h( x) = f ( x) − [ f ( x)] 2 + [ f ( x)]3
h′ ( x) = f ′ ( x) − 2 f ( x) f ′ ( x)
+ 3 [ f ( x)] 2 ⋅ f ′ ( x)
= f ′ ( x) [1 − 2 f ( x) + 3 [ f ( x)]2 ]
2
1
2
= 3 f ′ ( x ) f ( x ) − +
3
9
Hence, h′ ( x) and f ′ ( x) have same
sign.
25 We have,
u = ∫e
ax
cos bx dx = e
−
=
e
ax
ax
sin bx
⋅
b
a
e ax ⋅ sin bx dx
b ∫
⋅ sin bx a
− v
b
b
…(i)
⇒ bu + av = e ax ⋅sin bx
Similarly, bv − au = − e ax ⋅cos bx …(ii)
On squaring and adding Eqs. (i) and
(ii), we get
(a2 + b 2 ) (u2 + v2 ) = e 2 ax
b
26 Let I = ∫ f ( x) ⋅ g( x) dx
a
Put f ( x) = t ⇒ f ′ ( x) dx = dt
⇒ g( x) dx = dt
f (b )
f (b )
t2
[ f (b )] 2 − [ f (a)] 2
I = ∫ t dt =
=
f ( a)
2 f ( a)
2
dy
= 2 x + 1 ⇒ y = x2 + x + C
dx
y = x 2 + x,
[QC = 0 by putting x = 1, y = 2]
27 Given,
⇒
∴
x ,
h( x) = x 2 ,
x ,
∴
(0, 0)
= lim
⇒
⇒
0
x3
x2
2
∫ −1 ( x + x) dx = 3 + 2
−1
0
⇒
−1 1
1
=
+
= sq unit
3
2
6
28 If x ∈ R − (0, 1), then
x ≤ x 2 ⇒ x(1 − x) ≤ 0
x
π /2
Then,
f ′( x) = 2ax + b
Also,
f (1) = f (−1)
⇒ a + b + c = a −b + c
⇒
b =0
∴
f ′ ( x) = 2 ax ;
⇒
f ′ (a1 ) = 2 aa1 , f ′ (a2 ) = 2 aa2 ,
f ′ (a3 ) = 2 aa3
As a1 , a2 , a3 are in AP,
f ′ (a1 ), f ′ (a2 ), f ′ (a3 ) are in AP.
34 Given, y = cos −1 (cos x)
⇒ y′ ( x ) =
t dt
1
1 − cos2 x
⇒
t 2
2
π /2
y = lim
x → π /2 sin( 2 x − π )
y = lim
x → π /2
x2 π 2
−
2
8
sin(2 x − π )
1 (4 x 2 − π 2 )
y = lim
x → π /2 8 sin(2 x − π )
1 (2 x − π )(2 x + π )
y = lim
x → π /2 8
sin(2 x − π )
lim (2 x + π )
1 x→
π /2
y=
sin(2 x − π )
8
lim
x → π /2
2x − π
sin θ
Q lim
=1
θ → 0 θ
1
1
y = × 2π = π
8
4
sin x =
sin x
|sin x|
1 , 1st and 2nd quadrants
=
−1 , 3rd and 4th quadrants
sin(2 x − π )
x
∴Required area
=
33 Let f ( x) = ax2 + bx + c
[L′ Hospital’s rule]
⇒ 2 = a + 0 ⇒ a = 2, b any number
∫
π 1
f′ =
3 2
Hence,
bh
− 1 + 1
ah +
2
h→ 0
h
−2e −2 h
⇒ lim
= lim ( a + bh)
h→ 0
−1 h → 0
⇒
V
(–1/2, –1/4) Y¢
sin x
1 + cos x
x x
1
= tan −1 tan
=
⇒ f ′ ( x) =
2 2
2
32 f ( x) = tan −1
2
⇒
(–1, 0)
π
f ( x) = 2 cos x + is an
4
integer for
π π 3π 5π 3π 7π 9π
x+
= ,
,
,
,
,
4 2 4 4 2
4 4
π π
5π 3π
⇒
x = , , π,
,
, 2π
4 2
4 2
∴
h′ ( x) = 1, ∀ x > 1
x→ π / 2
X
discontinuous at all integer numbers.
x ≥1
∴ f (0 − ) = f (0 + ) = f (0) = − 1
Also, f is differentiable at x = 0,
therefore Lf ′ (0) = Rf ′ (0)
f (0 − h) − f (0)
lim
⇒
h→ 0
−h
f (0 + h) − f (0)
= lim
h→ 0
h
−2 h
e
− 1 + 1
⇒ lim
h→ 0
−h
Y
X¢
x ≤0
29 Since, f is continuous at x = 0.
30 y = lim
equation of parabola whose vertex is
−1 −1
V
, .
2 4
31 Using the fact that [ x] is
0 < x <1
h( x) is continuous for everywhere but
not differentiable at x = 0 and 1.
i.e.
1,
x < 0
2 x, 0 < x < 1
h′ ( x) = not exist ,
x = 0
1,
x > 1
x = 1
not exist,
2
1
1
⇒ x + = y + , which is an
2
4
x ≥ 1,
35
f ( x) =
∫
x
− 10
(t 4 − 4) e −4 tdt
⇒
f ′ ( x) = ( x 4 − 4) e −4 x
Now, f ′ ( x) = 0 ⇒ x = ± 2
Now,
f ′ ′ ( x) = − 4 ( x 4 − 4) e −4 x + 4 x 3e −4 x
At x = 2 and x = − 2 , the given
function has extreme value.
36 f ′ ( x) = 3 kx2 − 18 x + 9
= 3 [ kx 2 − 6 x + 3] > 0, ∀x ∈ R
∴ ∆ = b 2 − 4 ac < 0, k > 0
i.e. 36 − 12 k < 0 or k > 3 .
3
x
x2 + 1
37 Let I = ∫ tan −1
−1
x2 + 1
+ tan −1
dx
x
206
∫ {tan
3
=
−1
−1
= ∫ e x [ f ( x) + f ′ ( x)] dx
x
2
x + 1
+ cot
∫
=
3
−1
−1
π
dx = 2 π
2
π
Q tan −1 x + cot −1 x = , ∀ x ∈ R
2
n
38 lim
n→ ∞
∑
r =1
r 99
100 =
n
∫
1
0
x −1
2
∴ f ( x) = x + 1 , f ′ ( x) = ( x + 1) 2
−
x
1
= ex
+C
x + 1
x
2
dx
x + 1
99
x dx
100 1
x
1
=
= 100
100
0
− cos 2 x sin 2 x
39
I =
+
+ k
2
2
1
π
π
=
sin 2 x cos − cos 2 x sin + k
4
4
2
1
π
=
sin 2 x − + k
4
2
π
∴
c = ;a=k
4
(cos x − sin x)(cos x + sin x)
40 ∫
dx
(cos x + sin x) 2
cos x − sin x
=∫
dx
cos x + sin x
44 I1 = ∫ sin −1 x dx.
Let sin −1 x = θ ⇒ x = sin θ
⇒
dx = cos θ dθ
∴
I1 = ∫ θ cos θ dθ
= θ sin θ − ∫ sin θ dθ = θ sin θ + cos θ
= x sin −1 x + 1 − x 2
+
∴
dx
∫ sin( x − a) sin( x − b )
Put
1
sin{( x − b ) − ( x − a)}
=
dx
sin(a − b ) ∫ sin( x − a) sin( x − b )
1
=
[ cot( x − a) dx
sin(a − b ) ∫
and
43
=
1
sin( x − a)
log
+C
sin(a − b )
sin( x − b )
∫
e ( x − 1 + 2)
dx
( x + 1) 2
x
2
x −1
2
= ∫ ex
+
dx
2
+
+
x
1
(
x
1
)
I1 + I2 = x (cos −1 x + sin −1 x)
π x
=
2
I =
∫
π /4
0
sec2 x sec2 x dx
1 − tan2 x + tan 4 x
Put tan x = t ⇒ sec2 x dx = dt
1
1 + t2
I =∫ 4
dt
∴
0 t
− t2 + 1
1
1
1 + 2
1+ 2
1
1
t
t
=∫
dt = ∫
dt
2
0 2
0
1
1
t −1 + 2
t
−
+
1
t
t
1
1
sin 4 x dx = − cos 4 x + C
2∫
8
1
∴
k =−
8
− ∫ cot ( x − b ) dx
∫ − cos φ dφ
= φ cos φ − sin φ
= x cos −1 x − 1 − x 2
∴
=
42
∫ φ sin φ dφ = φ cos φ
by cos 4 x,
= ∫ cos 2 x ⋅ sin 2 x dx
⇒ dx = 6 t 5 dt
t3
6 t 5dt
Then, I = ∫ 3
=6 ∫
dt
2
( t − 1)
(t − t )
( t 3 − 1) + 1
= 6∫
dt
( t − 1)
=6
]
46 Given,
z=t −
∫ t
2
+t +1+
1
dt
t − 1
t3 t2
=6 +
+ t + ln|t − 1 | + C
3
2
45 Divide numerator and denominator
1
∫ t dt = log t + C
= log (sin x + cos x) + C
2 cos2 2 x
41 ∫
⋅ sin x cos x dx
cos2 x − sin2 x
∴
Put 1 + x = t 6
1 − x2 dx
Let cos φ = x
⇒ − sin φ dφ = dx
∴ I2 = −
x <1
1 < x <3
3 < x <5
x >5
It is clear that f ′ ( x) > 0, when
x ∈ (3 , ∞).
dx
47 Let I = ∫
(1 + x)1 /2 − (1 + x)1 /3
= ∫ cos −1 x dx
Put sin x + cos x = t
⇒ (cos x − sin x) dx = dt
I2 = ∫ sin −1
and
− 3 ,
− 1 ,
⇒ f ′ ( x) =
1,
3 ,
1
t
1
dz = 1 + 2 dt
t
0
dz
I =∫
dz = [tan −1 z] 0− ∞
− ∞ 1 + z2
π
= tan −1 (0) − tan −1 (−∞) =
2
f ( x) = | x − 1| + | x − 3| + |5 − x|,
∀ x ∈R
x <1
9 − 3 x ,
7 − x , 1 ≤ x < 3
∴ f ( x) =
x + 1, 3 ≤ x < 5
3 x − 9 ,
x ≥5
= 2 (1 + x)1 /2 + 3 (1 + x)1 /3
+ 6 (1 + x)1 / 6+ 6 ln|
(1 + x)1 / 6 − 1 | + C
1 /2
1 /3
= 2 λ + 3 λ + 6 λ1 / 6 + 6
ln| λ1 / 6 − 1| + C [where, λ = 1 + x ]
I =
48 Let
∫x
x1 / 4
dt
−1
1 /2
x = t 4 ⇒ dx = 4 t 3dt
t 4 − 1 + 1
t ⋅ 4 t 3dt
∴I = ∫ 2
= 4∫ 2
dt
t −1
(t − 1)
Put
1
= 4∫ t 2 + 1 + 2
dt
t − 1
t 3
1
t −1
= 4 + t + ln
+C
2
t +1
3
4
x1 / 4 − 1
= x3 / 4 + 4 x 1 / 4 + 2 ln 1 / 4
+C
3
x +1
49 Clearly,
= lim
x→ 0
= lim
x→ 0
(5 x − 1)3
x2
x
sin ⋅ log 1 +
n
3
(5 x − 1)3
1
⋅ lim
2
x
→
0
sin
x/n
x x
+ ...
x
/
n
n 3
3
(5 x − 1)
= lim
⋅3 n
x→ 0
x
= 3 n (log 5)3
…(i)
Since, the value of the function at
x = 0 is 15 (log 5)3 .
∴ 3 n(log 5)3 = 15 (log 5)3
⇒
n=5
207
DAY
50 Let the distance of point (4 , 0) from
the point ( x, y) lying on the curve be
D2 = ( x − 4)2 + y2
⇒ D2 = ( x − 4)2 + 2 x − 2 x 2
= x2 + 16 − 8 x + 2 x − 2 x 2
…(i)
= − x 2 − 6 x + 16
On differentiating Eq. (i), we get
dD
…(ii)
2D
= − 2x − 6
dx
= − 2( x + 3)
For maximum or minimum value,
dD
put
=0
dx
∴
x = −3
Again, on differentiating Eq. (ii), we
get
d2 D
= negative on putting x = − 3
dx2
∴The longest distance is
D2 = − 9 − 6 (−3) + 16
= − 9 + 18 + 16 = 25
∴
D = 5 units
51 The given curve is
…(i)
5 x 5 − 10 x 3 + x + 2 y + 4 = 0
On differentiating Eq. (i), we get
dy
25 x 4 − 30 x 2 + 1 + 2
=0
dx
dy −25 x 4 + 30 x 2 − 1
⇒
=
dx
2
dy
1
⇒
= − at P
dx
2
∴ Slope of normal is 2.
Therefore, its equation is
( y + 2) = 2( x − 0)
…(ii)
⇒
y = 2x − 2
On solving Eqs. (i) and (ii), we get
5 x 5 − 10 x 3 + x + 4 x − 4 + 4 = 0
⇒
5 x[ x 4 − 2 x 2 + 1] = 0
⇒
5 x( x2 − 1)2 = 0
⇒
x =0
or x2 = 1 or x = 0, 1, − 1
∴
y = − 2, 0, − 4
Since, the other two points are
(1, 0), (−1, − 4) .
The tangents at these points are
( y − 0) = 2 ( x − 1)
and ( y + 4) = 2 ( x + 1) or y = 2 ( x − 1)
2
52 Let I = ∫ {| x − 1| + [ x]} dx
−1
∫
=∫
=
2
−1
1
−1
∫
− ( x − 1) dx + ∫
(| x − 1|) dx +
2
−1
1
2
[ x] dx
( x − 1) dx
+
∫
2
−1
[ x] dx
I = I1 + I2
where,
I1 = −
55 Let I = ∫
1
1
[( x − 1)2 ]1−1 + [( x − 1)2 ]21
2
2
1
{[( x − 1)2 ]21 − [( x − 1)2 ]1−1 }
2
1
5
…(i)
= {1 + 4} =
2
2
and I2 =
∫
−1
− dx +
∫
1
0
0 ⋅ dx +
= −1 + 0 + 1=0
From Eqs. (i) and (ii), we get
5
5
I = I1 + I2 = + 0 =
2
2
∫
2
1
…(ii)
0
dt
25 − 16 t 2
=
∫
=
1
10
=
1 1
⋅ [log (5 + 4 t ) − log (5 − 4 t )]
10 4
−1
=
1
1
(log 9 − log 1) =
log 3
40
20
dx
−1
∫
0
−1
1
1
+
dt
5 − 4t 5 + 4t
0
53 According to the question,
dy y
y
= − cos2
x
dx x
xdy − ydx
y
= − cos2
dx
⇒
x
x
y ( xdx − ydx)
dx
⇒ sec2
=−
x
x2
x
dx
y
2 y
…(i)
sec
d =−
x x
x
sin x + cos x
dx
9 + 16 sin 2 x
Put sin x − cos x = t
Then, (sin x + cos x) dx = dt
0
dt
∴I = ∫
− 1 9 + 16 (1 − t 2 )
=
0
π /4
0
56.
∫
3
−2
f ( x) dx =
∫
2
−2
f ( x) dx +
54. Given,
xdy − y dx
= dy
y2
x
=− y+C
y
f ( x) dx
∫
∴
3
−2
f ( x) dx = 0 + 2 (3 − 2) = 2
57. Required area, A = ∫
x2
x1
y dx
y=sin x
p/4
O
y=cos x
–1
=
∫
π /4
0
p/2
cos x dx −
∫
π /4
0
sin x dx
π /4
0
= [sin x]
− [− cos x] π0 / 4
1
1
=
− 0 +
− 1 = 2 − 1
2
2
58 The equation of the given curve is
x
⇒ d = − dy
y
⇒
3
2
Since, ecos x sin x is an odd function.
On integrating both sides of Eq. (i),
we get
y
tan = − log x + C
x
When x = 1 / 2 and y = π / 8, then
1
1 = − log + C = − [− log 2] + C
2
1 − log 2 = C
y
∴ tan = − log x + 1 − log 2
x
e
= − log 2 x + log e = log
2 x
y
e
⇒
= tan −1 log
x
2 x
e
⇒
y = x tan −1 log
2 x
∫
[integrating]
As y (1) = 1 ⇒ C = 2
x
∴
+ y =2
y
Again, for x = − 3,
− 3 + y2 = 2 y
⇒ ( y + 1)( y − 3) = 0
Also, y > 0
[neglecting y = − 1 ]
⇒
y =3
…(i)
y = 7 x 3 + 11
dy
2
2
⇒
= 7 × 3 x = 21 x
dx
[differentiating w.r.t. x ]
∴Slope of tangent at x = 2 is
dy
= 21 (2)2 = 84
dx x = 2
Slope of tangent at x = − 2 is
dy
= 21 (−2)2 = 84
dx x = − 2
It is observed that the slopes of the
tangents at the points where, x = 2
and x = − 2 are equal. Hence, the two
tangents are parallel.
208
Hence, both the statements are true
and Statement II is correct
explanation of Statement I.
d
59 f ′ ( x) = cos ( x ) 2
( x)
dx
62 Let f ( x) = | x − 1 | + | x − 2| + | x − 3 |
x ≤1
6 − 3 x ,
4 − x , 1 < x ≤ 2
=
2 < x ≤3
x,
3 x − 6 ,
x >3
2
1 d 1
− cos
x dx x
1 cos x
1 1
=
+ cos 2 2
x x
2
x
1
⇒ f ′ (1) = cos 1 + cos 1
2
3
= cos 1
2
60 Given equation can be rewritten as
x
y 4 /3
dy
+ 6 y −1 / 3 = 3 x
dx
y −1 /3 = v
dy
dv
y −4 / 3
= −3
dx
dx
dv 2
− v = −1
dx x
Put
⇒
∴
Here, IF = e ∫
−
2
dx
x
= x −2
1
∴Solution is x −2 v = + C
x
⇒
v = x + Cx 2
1
⇒
y( x ) =
( x + Cx 2 ) 3
61
∫ (x
=
x2 − 1
2
∫
+ 1)
x4 + 1
dx
1
x2
1−
dx
1
1
2
x + x + 2
x
x
dt
1
=∫
put t = x +
2
x
t t − 2
=
1
2
sec −1
x +1
2
x 2
+C
y
y=3x–6
6
5
4
3
2
1
X¢
–6 –5 –4 –3 –2 –1 O
–1
–2
–3
–4
–5
–6
y=x
1 2 3 4 5 6
π
, x =2 −3
2
= − 1, y = 3 + 0 = 3
i.e. (− 1, 3) and
3π
At θ =
, x =2 + 3 =5
2
and y = 3 + 0 = 3
i.e. (5 , 3).
At θ =
65 Suppose f ( x) = 0 has a real root say
x
y=4–x
y=6–3x
Y¢
x = a, then f ( x) < 0 for all x < a.
Thus,| f ( x)| becomes strictly
decreasing on (− ∞, a).
So, Statement I is true.
66 As the area enclosed by | x | + | y | ≤ a
is the area of square i.e. 2 a2 .
∴ Area enclosed by
| x + y| +| x − y| ≤ 2
is area of square shown as
Clearly, the function has absolute
minimum at x = 2.
So, the absolute minimum is equal to
2.
Also, the curve is taking sharp turn
at x = 1, 2 and 3.
∴ f is not differentiable at x = 1, 2
and 3.
63 Given,
f ( x) = max {( x − 1) 2 + 1,| x − 1|}
= ( x − 1) 2 + 1
[say]
∴
f ′ ( x) = 2 ( x − 1) = 0
⇒
x = 1 ∈ [0, 3]
Now, f (0) = 2, f (1) = 1, f (3) = 5
∴Greatest value of
f ( x) = max { f (0), f (1), f (3)} = 5
dx
64 For vertical tangent,
=0
dθ
∴ − 3 cos θ = 0 ⇒ cos θ = 0
π 3π
⇒
θ= ,
2 2
Y
y=x
1
–1
X
1
1
y=–x
∴ Required area
1
= 4 × 2 × 1 = 4 sq units
2
Also, the area enclosed by
| x + y| +| x − y| ≤ 2
is symmetric about X-axis, Y-axis,
y = x and y = − x .
Hence, both the statements are true
but Statement II is not the correct
explanation of Statement I.
DAY TWENTY
Trigonometric
Functions and
Equations
Learning & Revision for the Day
u
u
u
u
Angle on Circular System
Trigonometric Functions
Trigonometric Identities
Trigonometric
Ratios/Functions of Acute
Angles
u
u
u
Trigonometric Ratios of
Compound and Multiple
Angles
Transformation Formulae
Conditional Identities
u
u
u
Maximum and Minimum
Values
Trigonometric Equations
Summation of Some
Trigonometric Series
Angle on Circular System
If the angle subtended by an arc of length l at the centre of a circle of radius r is θ,
l
then θ =
r
l
If the length of arc is equal to the radius of the circle, then the angle
subtended at the centre of the circle will be one radian. One radian is
denoted by 1c and 1c = 57°16′ 22′′ approximately.
q
B
O
(Figure shows the angle whose measure are one radian.)
2 π C = 360°, 1C =
180°
π
, 1° =
180
π
C
Also, let P( x, y) be any on the circle with ∠AOP = θ radian,
i.e. length of arc AP = θ
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
Y
B
Trigonometric Functions
Let X ′ OX and YOY′ be the coordinate axes. Taking O as the
centre and a unit radius, draw a circle, cutting the coordinate
axes at A, B, A′ and B′ as shown in the figure
PRED
MIRROR
A
P(x, y)
1
X¢
A¢
q y
x M
O
B¢
Y¢
q
A
X
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
210
Then, the six trigonometric functions, can be defined
as
OM
=x
OP
PM
(ii) sinθ =
=y
OP
OP 1
(iii) sec θ =
= , x ≠0
OM x
OP 1
(iv) cosec θ =
= , y ≠0
PM y
(i) cos θ =
Trigonometric Ratios/Functions of
Acute Angles
The ratios of the sides of a triangles with respect to its acute angles are
called trigonometric ratios or T-ratios.
In a right angled ∆ ABC, if ∠CAB = θ, then
BC Perpendicular
=
AC
Hypotenuse
C
po
te n
us
e
AB
Base
2. cos θ =
=
AC Hypotenuse
BC Perpendicular
=
AB
Base
1
AC
Hypotenuse
4. cosec θ =
=
=
sin θ BC Perpendicular
PM
y
= , x ≠0
OM x
OM x
(vi) cot θ =
= , y ≠0
PM
y
Hy
3. tan θ =
(v) tan θ =
1
AC Hypotenuse
=
=
cos θ AB
Base
1
AB
Base
6. cot θ =
=
=
tan θ BC Perpendicular
An equation involving trigonometric functions
which is true for all those angles for which the
functions which is true for all those angles for which
the functions are defined is called trigonometric
identity.
Some identities are given below
90°
q
A
Base
5. sec θ =
Trigonometric Identities
Perpendicular
1. sin θ =
B
Sign for Trigonometric Ratios in four Quadrants
sin θ cos θ
Quadrant
(iii) cosec2 θ − cot2 θ = 1
cot θ
sec θ
cosec θ
+
+
+
+
+
+
II. (90 ° , 180 ° )
+
−
−
−
−
+
III. (180 ° , 270 ° )
−
−
+
+
−
−
IV. (270 ° , 360 ° )
−
+
−
−
+
−
I. (0, 90° )
(i) sin2 θ + cos2 θ = 1
(ii) sec2 θ − tan2 θ = 1
tan θ
Angle
Trigonometric ratios of some useful angles between 0°and 90°
0°/0
sin θ
0
cos θ
1
tan θ
0
cot θ
∞
sec θ
cosec θ
π
15°/
12
π
18°/
10
3−1
5−1
4
2 2
3+1
2 2
10 + 2 5
4
3−1
5−1
3+1
10 + 2 5
3+1
10 + 2 5
3−1
5−1
1
2 2
3+1
10 + 2 5
∞
2 2
3−1
4
5+ 1
22.5°/
π
π
30°/
8
6
2− 2
36°/
π
5
1
2
10 − 2 5
3
2
5+ 1
4
2−1
1
3
10 − 2 5
2+1
3
4−2 2
4+ 2 2
2
2+
2
2
4
5+1
5+ 1
45°/
π
4
54°/
3π
10
60°/
π
3
67.5°/
3π
8
2
1
2
5+ 1
4
3
2
2+
1
2
10 − 2 5
1
2
2− 2
10 − 2 5
3
2+1
10 − 2 5
1
3
2−1
4+ 2 2
1
4
5+ 1
10 − 2 5
1
2
3
5−1
2
10 − 2 5
2
2
10 − 2 5
2
5−1
2
3
5+ 1
4
4
2
2
4−2 2
72°/
2π
5
10 + 2 5
4
5−1
4
75°/
5π
π
90°/
12
2
3+1
2 2
3−1
2 2
10 + 2 5
3+1
5−1
3−1
1
0
∞
5−1
3−1
10 + 2 5
3+1
5+ 1
2 2
3−1
∞
2 2
3+1
1
4
10 + 2 5
0
211
DAY
Trigonometric ratios of allied angles
cosec θ
cos θ
sec θ
tan θ
cot θ
−θ
θ
− sin θ
sin θ
− cosec θ
cos θ
secθ
− tan θ
− cot θ
90° − θ
cos θ
sec θ
sin θ
cosec θ
cot θ
tan θ
90° + θ
cos θ
sec θ
− sin θ
− cosec θ
− cot θ
− tan θ
180° − θ
sin θ
cosec θ
− cos θ
− sec θ
− tan θ
− cot θ
180° + θ
− sin θ
− cosec θ
− cos θ
− sec θ
tan θ
cot θ
270° − θ
− cos θ
− sec θ
− sin θ
− cosec θ
cot θ
tan θ
270° + θ
− cos θ
− sec θ
sin θ
cosec θ
− cot θ
− tan θ
360° − θ
− sin θ
− cosec θ
cos θ
sec θ
− tan θ
− cot θ
Trigonometric Ratios of
Compound and Multiple Angles
Compound Angles
Transformation Formulae
(i) 2 sin A cos B = sin( A + B) + sin ( A − B)
(ii) 2 cos A sin B = sin ( A + B) − sin ( A − B)
(iii) 2 cos A cos B = cos ( A + B) + cos ( A − B)
(i) sin ( A + B) = sin A cos B + cos A sin B
(iv) 2 sin A sin B = cos ( A − B) − cos ( A + B)
(ii) sin ( A − B) = sin A cos B − cos A sin B
C+D
C−D
cos
2
2
C+D
C−D
(vi) sin C − sin D = 2 cos
sin
2
2
C+D
C−D
(vii) cos C + cos D = 2 cos
cos
2
2
C+D
C−D
(viii)cos C − cos D = − 2 sin
sin
2
2
(iii) cos ( A + B) = cos A cos B − sin A sin B
(iv) cos ( A − B) = cos A cos B + sin A sin B
tan A + tan B
(v) tan ( A + B) =
1 − tan A tan B
(vi) tan ( A − B) =
tan A − tan B
1 + tan A tan B
(vii) cot ( A + B) =
cot A cot B − 1
cot A + cot B
cot A cot B + 1
(viii) cot ( A − B) =
cot B − cot A
1 + tan A
(ix) (a)
= tan
1 − tan A
π
+ A
4
1 − tan A
(b)
= tan
1 + tan A
π
− A
4
Multiple Angles
(i) sin 2 A = 2 sin A cos A =
2 tan A
1 + tan2 A
(ii) cos 2 A = cos2 A − sin2 A =
1 − tan2 A
1 + tan2 A
= 2 cos2 A − 1 = 1 − 2 sin2 A
2 tan A
(iii) tan 2 A =
1 − tan2 A
(iv) sin 3 A = 3 sin A − 4 sin3 A
(v) cos 3 A = 4 cos3 A − 3 cos A
3 tan A − tan3 A
(vi) tan 3 A =
1 − 3 tan2 A
(v) sin C + sin D = 2 sin
(ix) sin ( A + B)sin ( A − B) = sin2 A − sin2 B
(x) cos ( A + B)cos ( A − B) = cos2 A − sin2 B
(xi) cos A cos 2 A cos 4 A cos 8 A … cos 2 n − 1 A
=
1
sin (2 n A)
2 n sin A
Conditional Identities
If A + B + C = 180 °, then
(i) sin 2 A + sin 2 B + sin 2 C = 4sin A sin B sin C
(ii) cos 2 A + cos 2 B + cos 2 C = − 1 − 4cos A cos B cos C
A
B
C
(iii) sin A + sin B + sin C = 4 cos cos cos
2
2
2
A
B
C
(iv) cos A + cos B + cos C = 1 + 4 sin sin sin
2
2
2
(v) tan A + tan B + tan C = tan A tan B tan C
(vi) tan
A
B
B
C
C
A
tan + tan tan + tan tan = 1
2
2
2
2
2
2
Maximum and Minimum Values
1. −1 ≤ sin x ≤ 1,|sin x| ≤ 1
2. −1 ≤ cos x ≤ 1,|cos x| ≤ 1
3. |sec x| ≥ 1,|cosec x| ≥ 1
4. tan x,cot x take all real values
212
NOTE
• Maximum value of a cos θ ± b sin θ = a2 + b2
• Minimum value of a cos θ ± b sin θ = − a2 + b2
• Maximum value of a cos θ ± b sin θ + c = c +
a2 + b2
• Minimum value of a cos θ ± b sin θ + c = c − a2 + b2
Trigonometric Equations
An equation involving one or more trigonometrical ratios of
unknown angle is called a trigonometrical equation.
e.g.
sin θ + cos2 θ = 0
Principal Solution
The value of the unknown angle (say θ) which satisfies the
trigonometric equation is known as principal solution, if
0 ≤ θ < 2 π.
General Solution
Since, trigonometrical functions are periodic function,
solution of trigonometric equation can be generalised with
the help of the periodicity of the trigonometrical functions.
The solution consisting of all possible solutions of a
trigonometric equation is called its general solution.
Some general trigonometric equations and their solutions
Equations
π
π
sin x = sin α − ≤ α ≤
2
2
cos x = cos α (0 ≤ α ≤ π )
π
π
tan x = tan α − < α <
2
2
Solutions
Equations
x = nπ + (− 1)n α
n ∈I
sin2 x = sin2 α
x = 2 nπ ± α
n ∈I
cos2 x = cos2 α
x = nπ + α, n ∈ I
tan2 x = tan2 α
Solutions
x = nπ ± α, n ∈ I
sin x = 0
x = n π, n ∈ I
sin x = 1
x = (4n + 1) π / 2, n ∈ I
cos x = 0
x = (2 n + 1) π / 2, n ∈ I
cos x = 1
x = 2 nπ, n ∈ I
tan x = 0
x = nπ, n ∈ I
cos x = − 1
x = (2 n + 1) π, n ∈ I
sin x = sin α and cos x = cos α
x = 2 nπ + α, n ∈ I
Summation of Some Trigonometric Series
(i) sin α + sin (α + β) + sin (α + 2 β) + K to n terms
n β
sin
2
β
=
⋅ sin α + (n − 1)
2
β
sin
2
(ii) cos α + cos (α + β) + cos (α + 2 β) + … to n terms
n β
sin
2
β
=
⋅ cos α + (n − 1)
2
β
sin
2
213
DAY
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 If
2 sin θ
1 − cos θ + sin θ
is equal to
= x , then
1 + cos θ + sin θ
1 + sin θ
(a) x
2 If
(b)
1
x
(c) 1 − x
(d) 1 + x
3π
< α < π , then cosec2 α + 2 cot α is equal to
4
(a) 1 + cotα
(c) −1 − cotα
(b) 1 − cotα
(d) −1 + cotα
(b) 26
(c) 28
(d) 36
4 If tan A + sin A = m and tan A − sin A = n, then
(m 2 − n 2 )2
mn
is equal to
(a) 4
(b) 3
(c) 16
(d) 9
2
a sin3 α + 3a cos 2 α sin α = n, then
(m + n )2 / 3 + (m − n )2 / 3 is equal to
(a) 2 a
2
1/ 3
(b) 2 a
(c) 2 a
(b) cos
7 If sin θ = −
π
8
(c)
1
8
(d) 2 a
(d)
3
1+ 2
2 2
4
θ
and θ lies in the third quadrant, then cos is
5
2
equal to
1
(a)
5
1
(b) −
5
8 The expression
2
5
(c)
NCERT Exemplar
2
(d) −
5
tan A
cot A
can be written as
+
1 − cot A 1 − tan A
JEE Mains 2013
(a) sin A cos A + 1
(c) tan A + cot A
(b) secA cosec A + 1
(d) secA + cosec A
9 If tan θ, 2 tan θ + 2 and 3 tan θ + 3 are in GP, then the
value of
(a)
12
5
7 − 5 cot θ
9 − 4 sec2 θ − 1
(b) −
33
28
is
33
100
(d)
12
13
1
, then
2
tan (α + 2 β) tan ( 2 α + β) is equal to
10 If sin (α + β) = 1 and sin (α − β) =
(a) 1
(c) zero
(b) –1
(d) None of these
sec A + sec B + sec C is
(b) 4
(a) b = a + c
(c) c = a + b
(c) 5
(c) 45°
(d) 22
1
°
2
(b) b = c
(d) a = b + c
4
5
π
and sin (α − β ) =
, where 0 ≤ α , β ≤ .
5
13
4
Then, tan 2α is equal to
AIEEE 2010
14 If cos (α + β) =
25
16
(b)
56
33
(c)
19
12
(d)
20
7
rational function is
z2 − x2 − y2
xy
z2 + x2 + y2
(c)
xy
(d) 6
z2 − x2 − y2
2 xy
z2 + x2 + y2
(d)
2 xy
(a)
(b)
16 Let α and β be such that π < α − β < 3π. If
21
27
and cos α + cos β = −
, then the
65
65
α − β
value of cos
is
AIEEE 2004
2
sin α + sin β = −
(a) −
3
130
3
130
(b)
(c)
6
65
(d) −
6
65
17 If A + B + C = π and cos A = cos B cos C, then
tan B tan C is equal to
(a)
1
2
(b) 2
(c) 1
(d) −
1
2
18 If tan α = (1 + 2− x )−1, tan β = (1 + 2x +1 )−1, then α + β equals
π
6
(b)
π
4
(c)
19 If 0 < x < π and cos x + sin x =
(a)
11 In an acute angled triangle, the least value of
(a) 3
(b) 30°
π
P
Q
2
2
2
of ax 2 + bx + c = 0, where a ≠ 0, then
(a)
(c)
1
, then A is equal to
2
15 If sin α = x , sin β = y , sin(α + β) = z , then cos(α + β) as a
2/ 3
π
3π
5π
7π
6 1 + cos 1 + cos
1 + cos is
1 + cos
8
8
8
8
equal to
(a) 1
(a) 15°
(a)
5 If a cos α + 3a cos α sin α = m and
3
cos A + cos B =
3
and
2
13 In a ∆ PQR , ∠R = . If tan and tan are the roots
3 The least value of cosec2 x + 25 sec2 x is
(a) 0
12 If 0 < A < B < π, sin A + sin B =
− (4 + 7 )
1+ 7
(b)
3
4
(c)
π
3
(d)
π
2
1
, then tan x is equal to
2
1− 7
4
(d)
4− 7
3
20 cosec10° − 3 sec10° is equal to
(a) 1
(b) 1/2
(c) 2
(d) 4
21 The value of sin12° sin 48° sin 54° is equal to
(a)
1
16
(b)
1
32
(c)
1
8
(d)
1
4
214
22 The value of1 + cos 56° + cos 58° − cos 66° is equal to
(a) 2 cos 28 ° cos 29° cos 33 ° (b) 4 cos 28 ° cos 29° sin 33 °
(c) 4 cos 28 ° cos 29° cos 33 ° (d) 2 cos 28 ° cos 29° sin 33 °
π
π
n
+ cos
=
. Then,
2n
2n
2
(a) n = 6
(c) n = 5
(c) cos A
25 The maximum value of sin x +
π
interval 0, is attained at
2
π
(a) x =
12
π
(b) x =
6
(b)
(d) sinA
π
(d) x =
2
m+1
m
(d) None of these
2π
4π
= z cos θ +
, then the
3
3
1 1 1
value of + + is equal to
x y z
(b) 2
(c) 0
(d) 3 cosθ
π
π
28 The maximum value of cos − x − cos 2 + x is
3
3
2
(a) −
3
2
(b)
1
2
(c)
3
2
(d)
3
2
29 If sum of all the solution of the equation
2
3
(b)
13
9
(c)
8
9
(d)
4xy
30 sin θ =
is true, if and only if
( x + y )2
2
(a) x − y ≠ 0
(c) x + y ≠ 0
20
9
AIEEE 2002
(b) x = − y
(d) x ≠ 0, y ≠ 0
(b) 1
(c) 3
32 If A = sin x + cos x , then for all real x
13
≤ A≤1
16
3
13
(c) ≤ A ≤
4
16
(a)
4
(b) 1 ≤ A ≤ 2
(d)
3
≤ A≤1
4
(b) 2 π
(c)
5π
2
(d) None of these
36 The number of solution of cos x = |1 + sin x |, 0 ≤ x ≤ 3π is
(a) 2
(b) 3
(c) 4
(d) 5
37 If sin 2x + cos x = 0, then which among the following
is/are true?
I. cos x = 0
II. sin x = −
π
,n ∈ Z
2
1
2
7π
,n ∈ Z
6
IV. x = nπ + (– 1)n
(a) I is true
(c) I,II and III are true
(b) I and II are true
(d) All are true
38 The number of solutions of the equation sin 2 x − 2 cos x
+ 4 sin x = 4 in the interval [ 0 , 5π ] is
(a) 3
(b) 5
(c) 4
JEE Mains 2013
(d) 6
39 The possible values of θ ∈( 0, π ) such that
sin (θ ) + sin ( 4θ ) + sin (7θ ) = 0 are
2π
,
9
2π
(c)
,
9
(a)
(a) 6
equation cos( p sin x ) = sin( p cos x ) in x has a solution in
[ 0, 2π ] is
2
(a) π
π
,
4
π
,
4
4π π
, ,
9 2
π 2π
,
,
2 3
3π
,
4
3π
,
4
8π
9
35 π
36
π 5π π 2π 3π 8π
,
, ,
,
,
4 12 2 3 4 9
2π π π 2π 3π 8π
(d)
, , ,
,
,
9 4 2 3 4 9
(b)
the equation 2 sin2 x + 5 sin x − 3 = 0 is
31 The smallest positive integral value of p for which the
(a) 2
(d) [−1, 3]
(c) [0,1]
40 The number of values of x in the interval [ 0, 3π ] satisfying
π
π
1
8 cos x ⋅ cos + x ⋅ cos − x −
6
6
2
= 1 in [ 0, π ] is kπ, then k is equal to
(a)
3
and less than or equal to 3
2
5π 5π
35 If x ∈ −
,
, then the greatest positive solution of
2 2
1 + sin4 x = cos 2 3x is
III. x = ( 2n + 1)
27 If x cos θ = y cos θ +
(a) 1
(b) [−11
,]
(a) [0, 3]
A+B
B−A
26 If cos A = m cos B and cot
= λ tan
, then λ is
2
2
m
m−1
m+1
(c)
m−1
3
2
then the interval of S is
π
π
+ cos x + in the
6
6
π
(c) x =
3
(a)
(b) less than or equal to
34 If f : R → S , defined by f ( x ) = sin x − 3 cos x + 1, is onto,
(b) n = 1, 2 , 3, . . . , 8
(d) None of these
(b) tanA
3
2
(d) None of the above
24 The value of tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A is
(a) cot A
(a) greater than −
(c) greater than or equal to −
23 Let n be a positive integer such that
sin
33 cos 2 θ + 2 cos θ is always
(d) 5
(b) 1
(c) 2
(d) 4
41 If α is a root of 25 cos 2 θ + 5 cos θ − 12 = 0,
then sin 2 α is equal to
(a)
24
25
(b) −
42 Statement I cos
24
25
(c)
13
18
π
< α < π,
2
(d) −
13
18
π
2π
4π
⋅ cos
⋅ cos
= − 1/ 8
7
7
7
Statement II cos θ cos 2 θ cos 4 θ . . .
1
π
cos 2 n − 1θ = − n for θ = n
2
2 −1
(a) Statement I is true, Statement II is true; Statement II is
the correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not the correct explanation for Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
215
DAY
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1
(sink x + cos k x ), where x ∈ R , k ≥ 1, then
k
JEE Mains 2014
f4 ( x ) − f6 ( x ) is equal to
1 If fk ( x ) =
(a)
1
6
1
3
(b)
(c)
1
4
(d)
1
12
2 If tan α , tan β and tan γ are the roots of the equation
x 3 − px 2 − r = 0, then the value of
(1 + tan2 α ) (1 + tan2 β) (1 + tan 2 γ ) is equal to
(b) 1 + (p − r ) 2
(d) None of these
(a) (p − r ) 2
(c) 1 − (p − r ) 2
∞
∞
π
3 For 0 < φ < , if x = ∑ cos 2 n φ, y = ∑ sin2 n φ and
2
n = 0
n = 0
z=
∞
∑
cos
2n
φ sin
2n
φ , then
n= 0
(a) xyz = xz + y
(c) xyz = x + y + z
(b) xyz = xy − z
(d) xyz = yz + x
π
4 If sin x + cos y + 2 = 4 sin x cos y and 0 ≤ x , y ≤ ,
2
then sin x + cos y is equal to
4
4
(a) –2
(b) 0
(c) 2
(d) 3/2
5 If 0 ≤ x < 2π, then the number of real values of x, which
satisfy the equation
cos x + cos 2x + cos 3x + cos 4x = 0, is
(a) 3
(c) 7
(b) 5
(d) 9
(b) 1 − 2a 2 − 2b 2
(d) 2 − a 2 − b 2
7 If α = sin θ + cos θ, then which of the following is true?
(a) α > 1
14
(b) α ≤ 1
(c) α = 0
8 Let n be an odd integer. If sin nθ =
n
∑
(d) α < 0
br sin θ for all real
r
r=0
| sin θ ⋅ cos θ | +
(a) 0
2 + tan2 θ + cot 2 θ = 3,θ ∈[ 0, 4π] is/are
(b) 1
(c) 2
(d) 3
π π
11 The number of values of θ in the interval − ,
2 2
sec 2 θ
4
satisfying the equation ( 3 )
= tan θ + 2 tan2 θ is
(a) 1
these
(b) 2
(c) 3
(d) None of
12 The equation e sin x − e − sin x − 4 = 0 has
(a)
(b)
(c)
(d)
AIEEE 2012
infinite number of real roots
no real roots
exactly one real root
exactly four real roots
13 Find the general solution of the equation
π 5π
−
4 12
3π
(c) 2n π ± π −
12
(b) 2n π ±
π 5π
+
4 12
(d) None of these
14 At how many points the curve y = 81sin
2
+ 81cos
will intersect X-axis in the region − π ≤ x ≤ π ?
(a) 4
these
(b) 6
(c) 8
x
2
x
− 30
(d) None of
15 In a ∆PQR, if 3 sin P + 4 cos Q = 6 and
4 sin Q + 3 cos P = 1, then the angle R is equal to
θ, then
(a) b0 = 1, b1 = 3
(c) b0 = − 1, b1 = n
(b) cos a , cos b , cos c
(d) cos 2 a , cos 2 b,cos 2 c
10 The number of solutions of the equation
(a) 2n π ±
cos 2 (α − β ) − 4ab cos(α − β ) is equal to
8
(a) sina , sinb, sinc
(c) sin 2 a , sin 2 b, sin 2 c
( 3 − 1 ) cos θ + ( 3 + 1 ) sin θ = 2.
6 If sin (θ + α ) = a and sin (θ + β) = b, then
(a) 1 − a 2 − b 2
(c) 2 + a 2 − b 2
9 If x sin a + y sin 2 a + z sin 3 a = sin 4 a
x sin b + y sin 2 b + z sin 3 b = sin 4 b
x sin c + y sin 2 c + z sin 3 c = sin 4 c
Then, the roots of the equation
z
y + 2
z − x
t 3 − t 2 −
t +
= 0, a , b , c ≠ nπ, are
2
4
8
(b) b0 = 0, b1 = n
(d) b0 = 0, b1 = n 2 − 3n − 3
5π
(a)
6
π
(b)
6
π
(c)
4
AIEEE 2012
3π
(d)
4
216
ANSWERS
(a)
(d)
(c)
(a)
(b)
SESSION 1
1.
11.
21.
31.
41.
SESSION 2
1. (d)
11. (b)
2.
12.
22.
32.
42.
(c)
(a)
(b)
(d)
(a)
3.
13.
23.
33.
2. (b)
12. (b)
(d)
(c)
(a)
(c)
4.
14.
24.
34.
3. (c)
13. (b)
(c)
(b)
(a)
(d)
4. (c)
14. (c)
5.
15.
25.
35.
(c)
(b)
(a)
(b)
5. (c)
15. (b)
6.
16.
26.
36.
(c)
(a)
(c)
(b)
7.
17.
27.
37.
6. (b)
(b)
(b)
(c)
(d)
8.
18.
28.
38.
7. (b)
(b)
(b)
(c)
(a)
8. (b)
9.
19.
29.
39.
(c)
(a)
(b)
(a)
10.
20.
30.
40.
9. (b)
(a)
(d)
(c)
(d)
10. (a)
Hints and Explanations
SESSION 1
1 Given,
∴ x=
2 sin θ
= x
1 + cos θ + sin θ
2 sin θ (1 + sin θ − cos θ)
(1 + sin θ)2 − cos2 θ
2 sin θ (1 + sin θ − cos θ)
= 2
1 + sin2 θ + 2 sin θ − (1 − sin2 θ)
=
1 + sin θ − cos θ
1 + sin θ
2 We have,
cosec2 α + 2 cot α
= 1 + cot2 α + 2 cot α
= |1 + cot α |
3π
But
<α < π
4
⇒
cot α < − 1
⇒
1 + cot α < 0
Hence, |1 + cot α | = − (1 + cot α )
3 cosec2 x + 25 sec2 x
= 1 + cot2 x + 25 (1 + tan2 x)
= 26 + cot2 x + 25 tan2 x
= 26 + 10 + (cot x − 5 tan x)2 ≥ 36
4 Since, tan A + sin A = m
and
∴
and
Also,
Now,
tan A − sin A = n
m + n = 2 tan A
m − n = 2 sin A
mn = (tan A + sin A )
(tan A − sin A )
2
= tan A − sin2 A
(m2 − n2 )2 (m + n)2 (m − n)2
=
mn
mn
=
(2 tan A )2 (2 sin A )2
tan2 A − sin2 A
=
16 tan2 A sin2 A
= 16
sin2 A tan2 A
5 Given,
a cos3 α + 3 a cos α sin2 α = m
and a sin3 α + 3 a cos2 α sin α = n
∴(m + n) = a cos3 α + 3 a cos α sin2 α
+ 3 a cos2 α sin α + a sin3 α
= a (cos α + sin α )3
and similarly,
(m − n) = a(cos α − sin α )3
∴ (m + n)2 /3 + (m − n)2 /3
= a2 /3 {(cos α + sin α )2
+ (cos α − sin α )2 }
2 /3
= a {2 (cos2 α + sin2 α )}
= 2 a2 /3
6 Given expression
π
3π
= 1 + cos 1 + cos
8
8
3π
π
1 − cos
1 − cos
8
8
π
3π
= 1 − cos2 1 − cos2
8
8
π
3π
π
π
= sin
sin2
= sin2 cos2
8
8
8
8
1
1
2 π
= sin
=
4
4 8
2
4
7 Given that, sin θ = − and θ lies in
5
the 3rd quadrant.
∴
cos θ = − 1 −
16
3
=−
25
5
Now, cos
θ
=±
2
=±
1 + cos θ
2
1−
2
3
5 =±
1
5
θ
1
. Since, if θ
=−
2
5
θ
lies in 3rd quadrant, then will be
2
in 2nd quadrant.
θ
1
Hence, cos = −
2
5
But we take cos
8 Given expression is
tan A
cot A
sin A
+
=
1 − cot A 1 − tan A cos A
sin A
cos A
×
+
sin A − cos A sin A
cos A
×
cos A − sin A
1
sin3 A − cos3 A
sin A − cos A cos A sin A
sin2 A + sin A cos A + cos2 A
=
sin A cos A
=
1 + sin A cos A
sin A cos A
= 1 + sec A cosec A
=
9 We have,
(2 tan θ + 2)2 = tan θ (3 tan θ + 3)
⇒
4 tan2 θ + 8 tan θ + 4
= 3 tan2 θ + 3 tan θ
2
⇒
tan θ + 5 tan θ + 4 = 0
⇒
(tan θ + 4) (tan θ + 1 ) = 0
⇒
tan θ = − 4
(Q tan θ ≠ − 1 )
217
DAY
∴
7 − 5 cot θ
9 − 4 tan2 θ
=
5
4 = 33
9 − 4 (− 4) 100
7+
10 Since, sin (α + β) = 1
∴
α +β=
π
2
…(i)
1
2
π
…(ii)
⇒
α −β =
6
On solving Eqs. (i) and (ii), we get
π
π
and β =
α =
3
6
Now, tan (α + 2 β) tan (2 α + β)
2 π
5 π
= tan
tan
3
6
and sin (α − β) =
π
π
= tan π − tan π −
3
6
π
π
= − cot − cot
3
6
1
=
× 3 =1
3
11 sec A , sec B and sec C are positive in
an acute angled triangle.
Also, arithmetic mean ≥ harmonic
mean
sec A + sec B + sec C
⇒
3
3
≥
cos A + cos B + cos C
P Q R π
+
+
=
2
2
2 2
π
P+Q π
⇒
=
Q∠ R =
4
2
2
P
Q
b
tan + tan
−
2
2
a =1
⇒
=1 ⇒
P
Q
c
1 − tan tan
1−
2
2
a
⇒ c=a+b
4
14 cos (α + β) =
5
⇒
α + β ∈1st quadrant
5
sin (α − β) =
13
⇒
(α − β) ∈1st quadrant
Now, as 2α = (α + β) + (α − β)
tan (α + β) + tan (α − β)
∴ tan 2α =
1 − tan (α + β) tan (α − β)
3
5
+
56
4
12
=
=
3 5
33
1− ⋅
4 12
Also,
15 We have,
z = sin(α + β) = x 1 − y2 + y 1 − x 2
⇒
+ 2 xy 1 − x 2 1 − y2
Now,
cos(α + β) = 1 − x 2 1 − y2 − xy
We have, in ∆ABC,
cos A + cos B + cos C ≤
∴
⇒
3
2
1
2
≥
cos A + cos B + cos C 3
⇒
sec A + sec B + sec C
≥2
3
sec A + sec B + sec C ≥ 6
12 Clearly, (sin A + sin B)2
+ (cos A + cos B) 2 = 2
∴ 2 + 2 (sin A sin B
+ cos A cos B) = 2
⇒ cos(B − A ) = 0 ⇒ B = A + 90 °
Second equation gives,
1
cos A − sin A =
2
⇒ cos( A + 45 ° ) = cos 60 °
∴
A = 15 °
P
Q
b
+ tan = −
2
2
a
P
Q c
and tan tan
=
2
2 a
13 Since, tan
z2 = x 2 + y2 − 2 x 2 y2
=
z2 − x 2 − y2 + 2 x 2 y2
− xy
2 xy
=
z − x − y
2 xy
2
2
2
16 Given that, sin α + sin β = −
21
…(i)
65
27
…(ii)
65
On squaring and adding Eqs. (i) and
(ii), we get
sin2 α + sin2 β + 2 sin α sin β
+ cos2 α + cos2 β + 2 cos α cos β
2
2
21
27
= − + −
65
65
and cos α + cos β = −
⇒ 2 + 2 (cos α cos β + sin α sin β)
441
729
=
+
4225 4225
1170
⇒ 2 [1 + cos (α − β)] =
4225
1170
2 α − β
cos
⇒
=
2 4 × 4225
⇒
9
α − β
cos2
=
2 130
3
α − β
cos
=−
2
130
π α − β 3π
Q π < α − β < 3π ⇒ <
<
2
2
2
∴
17 Since, A + B + C = π
∴
A = π − (B + C)
We have, cos A = cos B cos C
⇒ cos [π − (B + C)] = cos B cos C
⇒
− cos (B + C) = cos B cos C
⇒
− [cos B cos C − sin B sin C]
= cos B cos C
⇒
sin B sin C = 2 cos B cos C
⇒
tan B tan C = 2
18 We know,
tan α + tan β
1 − tan α tan β
1
Q
tan α =
1 + 2− x
1
and
tan β =
1 + 2x + 1
1
1
+
1
1 + 2x + 1
1+ x
2
∴ tan (α + β) =
1
1
1−
⋅
1 1 + 2x + 1
1+ x
2
⇒ tan (α + β)
2 x + 2 ⋅ 22 x + 2 x + 1
=
1 + 2 x + 2 ⋅ 2 x + 2 ⋅ 22 x − 2 x
tan (α + β) =
⇒ tan (α + β) = 1
π
⇒
α +β=
4
π
4
19 We have, cos x − =
1
2 2
π
tan x − = 7
4
tan x − 1
= 7
tan x + 1
⇒
⇒
∴ tan x =
7+1
1− 7
=
− (4 + 7 )
3
20 cosec 10 ° − 3 sec 10 °
=
1
3
−
sin 10 ° cos 10 °
cos 10 °
3
4
=
−
sin 10 °
2
2
sin
20 °
=
4 sin(30 ° − 10 ° )
=4
sin 20 °
21 Now, sin 12 ° sin 48 ° sin 54°
=
1
(cos 36 ° − cos 60 ° ) cos 36 °
2
218
=
1 5 + 1 1 5 + 1
−
2 4
2 4
1 5 − 1 5 +
2 4 4
5 −1
4 1
=
=
=
32
32 8
=
1
22 1 + cos 56 ° + cos 58 ° − cos 66 °
= 2 cos2 28 ° + 2 sin 62 ° sin 4°
= 2 cos2 28 ° + 2 cos 28 ° cos 86 °
= 2 cos 28 ° (cos 28 ° + cos 86 ° )
= 2 cos 28 ° (2 cos 57 ° cos 29 ° )
= 4 cos 28 ° cos 29 ° sin 33 °
23 We have,
π
π
n
sin + cos =
2 n
2 n
2
On squaring both sides, we get
π
π
π n
sin2 + cos2 + sin =
n 4
2 n
2 n
⇒
⇒
⇒
π n
sin = − 1
n 4
π n − 4
sin =
n
4
n = 6 only
24 tan A + 2 tan 2 A + 4 tan 4 A
1 − tan2 4 A
+ 8
2 tan 4 A
= tan A + 2 tan 2 A
4 tan2 4 A + 4 − 4 tan2 4 A
+
tan 4 A
= tan A + 2 tan 2 A + 4 cot 4 A
1 − tan2 2 A
= tan A + 2 tan 2 A + 4
2 tan 2 A
2 tan2 2 A + 2 − 2 tan2 2 A
= tan A +
tan 2 A
= tan A + 2 cot 2 A
1 − tan2 A
= tan A + 2
2 tan A
tan2 A + 1 − tan2 A
= cot A
tan A
π
π
25 We have, sin x + + cos x +
6
6
=
π
1
= 2
sin x +
6
2
1
π
+
cos x +
6
2
π π
= 2 cos x +
−
6
4
π
= 2 cos x −
12
Hence, maximum value will be at
π
x= .
12
26 We have, cos A = m cos B
cos A m
=
⇒
cos B
1
cos A + cos B m + 1
⇒
=
cos A − cos B m − 1
A+ B
B− A
2 cos
cos
m+1
2
2
⇒
=
A+ B
B− A
m −1
2 sin
sin
2
2
A + B m + 1
B− A
=
⇒ cot
tan
m − 1
2
2
A+ B
B− A
= λ tan
2
2
m+1
∴
λ =
m −1
2π
27 We have, x cos θ = y cos θ +
3
4π
= z cos θ +
= k (say)
3
But cot
⇒
cos θ =
k
2π k
, cos θ +
=
x
3
y
4π k
cos θ +
=
3
z
k
k
k
2π
+
+
= cos θ + cos θ +
∴
x
y
z
3
4π
+ cos θ +
3
π
π
= cos θ − cos − θ − cos + θ
3
3
π
= cos θ − 2 cos cos θ
3
1
⇒ cos θ − 2 × cos θ = 0
2
1
1 1
⇒
+
+ =0
x
y z
π
π
28 cos2 − x − cos2 + x
3
3
π
π
= cos − x + cos + x
3
3
π
π
cos 3 − x − cos 3 + x
π
π
= 2 cos cos x 2 sin sin x
3
3
2π
3
= sin
sin 2 x =
sin 2 x
3
2
Hence, maximum value of given
3
.
expression is
2
and
29 Key idea Apply the identify
cos( x + y) cos( x − y) = cos2 x − sin2 y
and cos 3 x = 4 cos3 x − 3 cos x
We have,
π
cos + x
6
=1
8 cos x
π
1
cos − x −
6
2
π
1
⇒ 8 cos x cos2 − sin2 x − = 1
6
2
1
3
⇒
8 cos x − sin2 x − = 1
4
2
3 1
⇒
8 cos x − − 1 + cos2 x = 1
4 2
⇒
−3 + 4 cos2 x
8 cos x
=1
4
⇒
2(4 cos3 x − 3 cos x) = 1
1
⇒
2 cos 3 x = 1 ⇒ cos 3 x =
2
π 5π 7π
[0 ≤ 3 x ≤ 3 π ]
⇒
3x = ,
,
3 3 3
π 5π 7π
⇒
x= ,
,
9 9 9
π 5 π 7 π 13 π
Now, Sum = +
+
=
9
9
9
9
13 π
⇒
kπ =
9
13
Hence, k =
9
sin2 θ ≤ 1
4 xy
∴
≤1
( x + y)2
30 Q
⇒ x2 + y2 + 2 xy − 4 xy ≥ 0
⇒
( x − y)2 ≥ 0
which is true for all real values of
x and y provided x + y ≠ 0,
4 xy
otherwise
will be
( x + y)2
meaningless.
π
31 Clearly, p sin x = ± p cos x
2
π
⇒ sin x ± cos x =
2p
π
π
π
⇒ sin x ± =
⇒
≤1
4 2 2 p
2 2p
π
π
but 1 <
<2
2 2
2 2
Hence, the smallest positive integral
value of p for which equation has
solution is p = 2.
For positive p, p ≥
32 A = sin2 x + cos 4 x
⇒
A = 1 − cos2 x + cos 4 x
219
DAY
= cos 4 x − cos2 x +
π
sin x = sin π +
6
1 3
+
4 4
2
1
3
= cos2 x − +
2
4
[Qsin (π + θ) = − sin θ]
7π
sin x = sin
6
7π
x = nπ + (−1)n
[Q n ∈ Z ]
6
... (i)
2
1
1
where, 0 ≤ cos2 x − ≤
2
4
3
≤ A ≤1
∴
4
... (ii)
33 cos 2 θ + 2 cos θ = 2 cos2 θ
− 1 + 2 cos θ
2
1
3
3
= 2 cos θ + − ≥ −
2
2
2
2
1
Q2 cos θ + ≥ 0, ∀ θ
2
and the maximum value of
cos 2 θ + 2 cos θ is 3.
34 Given that, f ( x) = sin x − 3 cos x + 1
Q
−2 ≤ sin x − 3 cos x ≤ 2
[Q a2 + b 2 ≤ a sin x + b cos x
≤ a2 + b 2 ]
⇒ −1 ≤ sin x − 3 cos x + 1 ≤ 3
∴ Range of f ( x) = [−1, 3]
35 We have, sin 3 x + sin x = 0
2
4
⇒ sin2 x {(3 − 4 sin2 x)2 + sin2 x} = 0
∴ sin x = 0 ⇒ x = nπ
Hence, greatest positive solution
is 2π .
36 Clearly, 1 + sin x ≥ 0
∴ The given equation becomes
π
1
cos x − sin x = 1 ⇒ cos x + =
4
2
π π 7 π 9 π 15 π
,…
⇒
x+
= ,
,
,
4 4 4 4
4
3π
7π
⇒
x = 0,
, 2π,
, ...
2
2
Q
0 ≤ x ≤ 3π
3π
∴
x = 0,
, 2π
2
⇒
38 sin 2 x − 2 cos x + 4 sin x = 4
⇒ (sin x − 1 ) (2 cos x + 4) = 0
⇒ sin x = 1, cos x ≠ − 2
π
∴ x = nπ + (−1 )n ,
2
where n ∈ z
Hence, the required value of
π 5π 9π
in interval [0, 5π ] .
x= ,
,
2 2 2
sin θ + sin 4 θ + sin 7 θ = 0
⇒ sin 4 θ + (sin θ + sin 7 θ) = 0
⇒ sin 4 θ + 2 sin 4 θ ⋅ cos 3 θ = 0
⇒
sin 4 θ {1 + 2 cos 3 θ} = 0
1
2
As, 0 < θ < π
∴ 0 < 4θ < 4 π
∴ 4 θ = π, 2π, 3π
Also, 0 < 3 θ < 3 π
2 π 4π 8 π
⇒
3θ =
,
,
3 3 3
π π 3 π 2 π 4π 8 π
⇒
θ= , ,
,
,
,
4 2 4 9 9 9
40 Given equation is
2 sin2 x + 5 sin x − 3 = 0.
⇒ (2 sin x − 1)(sin x + 3) = 0
1
⇒
sin x =
[Q sin x ≠ − 3]
2
Y
y=sin x
y=1/2
p
O p/6
37 sin 2 x + cos x = 0
⇒ 2 sin x cos x + cos x = 0
[Qsin 2 x = 2 sin x cos x]
⇒ cos x (2 sin x + 1) = 0
1
⇒ cos x = 0 or sin x = −
2
π
When cos x = 0, then x = (2 n + 1)
2
1
When sin x = − ,
2
π
then
sin x = − sin
6
sin 4 θ = 0, cos 3 θ = −
2p
3p
X
It is clear from figure that the curve
intersect the line at four points in
the given interval.
Hence, number of solutions are 4.
41 Since, α is a root of
∴
⇒
⇒
π
2π
4π
⋅ cos
⋅ cos
7
7
7
π
π
2π
4π
⋅ cos
2 sin ⋅ cos ⋅ cos
7
7
7
7
=
π
2 sin
7
π
8π
sin π +
sin
1
1
1
7
7
=
=
=−
π
π
8
8
8
sin
sin
7
7
(Q2 sin x cos x = sin 2 x)
π
Again, θ = n
⇒ 2n θ = π + θ
2 −1
42 cos
39 We have,
⇒
π
< α < π i.e. in second quadrant.
2
4
∴
cos α = −
5
3
⇒
sin α =
5
Now, sin 2 α = 2 sin α cos α
3 4
24
= 2 × × − = −
5 5
25
But
25 cos2 θ + 5 cos θ − 12 = 0
25 cos2 α + 5 cos α − 12 = 0
(5 cos α − 3)(5 cos α + 4) = 0
4
3
cos α = − and
5
5
∴
sin (2 n θ) = − sin θ
We know,
cos θ ⋅ cos 2θ ⋅ K ⋅ cos 2 n − 1 θ
1 sin (2 n θ)
= n
2
sin θ
1 sin θ
1
=− n
=− n
2 sin θ
2
So, Statements I and II both are true
and Statement II is a correct
explanation for Statement I.
SESSION 2
1
(sin k x + cos k x),
k
where x ∈ R and k ≥ 1
Now, f4 ( x) − f6 ( x)
1
= (sin 4 x + cos 4 x)
4
1
− (sin 6 x + cos 6 x)
6
1
2
= (1 − 2 sin x ⋅ cos2 x)
4
1
− (1 − 3 sin2 x ⋅ cos2 x)
6
1 1
1
= − =
4 6 12
1 Given, fk ( x) =
2 From the given equations, we have
Σ tan α = p
Σ tan α tan β = 0
and tan α tan β tan γ = r
Now,
(1 + tan2 α ) (1 + tan2 β) (1 + tan2 γ)
220
= 1 + Σ tan2 α + Σ tan2 α tan2 β
+ tan2 α tan2 β tan2 γ
= 1 + (Σ tan α )2 − 2 Σ tan α tan β
+ (Σ tan α tan β)2
− 2 tan α tan β tan γ Σ tan α
+ tan2 α tan2 β tan2 γ
2
= 1 + p − 2 pr + r 2 = 1 + ( p − r )2
1
1
1
3 x=
=
,y=
2
2
1 − cos φ sin φ
cos2 φ
1
and
z=
1 − sin2 φ ⋅ cos2 φ
1
1
Clearly,
+
=1
x
y
1
1
⇒
xy = x + y and = 1 −
z
xy
⇒
and
∴
xy = x + y
xy = xyz − z
xyz = xy + z = x + y + z
4 The given equation is
sin x + cos y + 2 = 4 sin x cos y
⇒ (sin2 x − 1 )2 + (cos2 y − 1 )2
+ 2 sin2 x + 2 cos2 y − 4 sin x
cos y = 0
⇒ (sin2 x − 1 )2 + (cos2 y − 1 )2
+ 2 (sin x − cos y)2 = 0
which is possible only when
sin2 x − 1 = 0, cos2 y − 1 = 0,
sin x − cos y = 0
⇒ sin2 x = 1, cos2 y = 1, sin x = cos y
π
As 0 ≤ x, y ≤
2
∴We get sin x = cos y = 1 and so
sin x + cos y = 1 + 1 = 2
4
4
5 Given equation is
cos x + cos 2 x + cos 3 x + cos 4 x = 0
⇒ (cos x + cos 3 x)
+ (cos 2 x + cos 4 x) = 0
⇒ 2 cos 2 x cos x + 2 cos 3 x cos x = 0
⇒
2 cos x(cos 2 x + cos 3 x) = 0
5x
x
⇒
2 cos x 2 cos
cos = 0
2
2
⇒
⇒
Now,
⇒
5x
x
⋅ cos = 0
2
2
5x
cos x = 0 or cos
=0
2
x
or cos = 0
2
cos x = 0
π 3π
[Q0 ≤ x < 2 π ]
x= ,
2 2
5x
cos
=0
2
cos x ⋅ cos
5 x π 3 π 5 π 7 π 9 π 11 π
= ,
,
,
,
,
,...,
2
2 2 2 2 2
2
π 3π
7π 9π
⇒ x= ,
, π,
,
5 5
5 5
[Q0 ≤ x < 2 x ]
x
and
cos = 0
2
x π 3π 5π
⇒
= ,
,
,...
2 2 2 2
[Q0 ≤ x < 2 π ]
⇒
x=π
π 3π
π 3π 7π 9π
Hence, x = ,
, π, ,
,
,
2 2
5 5 5 5
⇒
6 We have, sin (θ + α ) = a
and
sin (θ + β) = b
⇒ θ + α = sin −1 a and θ + β = sin −1 b
∴ α − β = sin −1 a − sin −1 b
π
π
= − cos −1 a − + cos −1 b
2
2
= cos −1 b − cos −1 a
= cos −1 {ab + 1 − b 2 1 − a2 }
⇒ cos (α − β) = ab
+ (1 − a2 − b 2 + a2b 2 )
⇒ cos2 (α − β) = a2b 2
+ 1 − a2 − b 2 + a2b 2
Putting θ = 0 in Eq. (i), we get 0 = b 0
Again, Eq. (i) can be written as
sin n θ =
+ 4 ab 1 − a2 − b 2 + a2b 2 − 1
− 4 a2b 2 − 4 ab 1 − a2 − b 2 + a2b 2
= 1 − 2 a − 2b
2
2
7 Given, α = sin 8 θ + cos14 θ
Since, sin 8 θ ≥ 0 and cos14 θ ≥ 0.
So,
α ≥0
Also, sin 8 θ + cos14 θ = 0 is not
possible.
Since, sin θ = 0 ⇒ cos θ ≠ 0
and
cos θ = 0 ⇒ sin θ ≠ 0
So, α > 0
Again, sin2 θ ≤ 1 ⇒ (sin2 θ )4 ≤ sin2 θ
⇒
sin 8 θ ≤ sin2 θ
Also, cos2 θ ≤ 1
⇒ (cos2 θ )7 ≤ cos2 θ
⇒
cos14 θ ≤ cos2 θ
So,
α = sin 8 θ + cos14 θ
≤ sin2 θ + cos2 θ = 1
α ≤ 1 and α > 0
8 Given, sin n θ =
n
∑
r=0
b r sin r θ = b 0
+ b1 sin θ + b2 sin θ
+ K + b n sin n θ …(i)
2
b r sin r θ
r =1
sin n θ
=
sin θ
n
∑
b r sin r − 1 θ
r =1
On taking limit as θ → 0, we get
sin n θ
lim
= b1
θ → 0 sin θ
sin nθ θ
⇒ lim n
= b1
θ→ 0
nθ sin θ
⇒ n = b1
Hence, b 0 = 0; b1 = n
9 Equation first can be written as
x sin a + y × 2 sin a cos a + z
× sin a (3 − 4 sin2 a)
= 2 × 2 sin a cos a cos 2 a
⇒ x + 2 y cos a + z (3 + 4 cos2 a − 4)
= 4 cos a (2 cos2 a − 1 ) as sin a ≠ 0
⇒ 8 cos3 a − 4z cos2 a − (2 y + 4)
cos a + (z − x) = 0
z
3
⇒ cos a − cos2 a
2
y + 2
z − x
−
cos a +
=0
4
8
+ 2 ab (1 − a2 − b 2 + a2b 2 )
Now, cos 2 (α − β) − 4 ab cos (α − β)
= 2 cos2 (α − β) − 1 − 4 ab cos (α − β)
= 4 a2b 2 + 2 − 2 a2 − 2b 2
n
∑
which shows that cos a is a root of
the equation
z
y + 2
z − x
t3 − t2 −
t +
=0
2
4
8
Similarly, from second and third
equation we can verify that cos b and
cos c are the roots of the given
equation.
10 Given,|sin θ.cos θ|
+ 2 + tan2 θ + cot2 θ = 3
⇒ |sin θ ⋅ cos θ |
+ (tan θ + cot θ)2 = 3
⇒ | sin θ ⋅ cos θ |
+ | tan θ + cot θ| = 3
⇒
| sin θ ⋅ cos θ | +
sin 2 θ + cos2 θ
= 3
sin θ ⋅ cos θ
⇒ |sin θ ⋅ cos θ| +
1
= 3
| sin θ ⋅ cos θ |
We know that,
|sin θ ⋅ cos θ | +
1
≥2
| sin θ ⋅ cos θ |
Hence, there is no solution of this
equation.
221
DAY
2
11 ( 3 )sec
θ
= (tan2 θ + 1 )2 − 1
= (sec2 θ )2 − 1
Put sec θ = x
Then, ( 3 )x = x2 − 1
(x ≥ 1 )
2
Let
y = ( 3 )x = ( x 2 − 1 )
(x > 1 )
Now, graphs of y = ( 3 )x and
y = x 2 − 1 intersect at one point
y
y=( 3)x
y=x2–1
1
x¢
O
1
x
_1
y¢
So, esin x = 2 + 5 is not possible for
any x ∈ R
and esin x = 2 − 5 is also not possible
for any x ∈ R
Hence, we can say that the given
equation has no solution.
13 Given, ( 3 − 1) cos θ + ( 3 + 1)
sin θ = 2 ...(i)
( 3 − 1) = r cos α
and
( 3 + 1) = r sin α
Therefore, there are two values of θ
π π
in − , .
2 2
12 Given equation is
esin x − e − sin x = 4 ⇒ e sin x −
Now, let y = e sin x
Then, we get
1
y− =4 ⇒
y
∴
y=
⇒ e sin x
y2 − 4 y − 1 = 0
4 ± 16 + 4
2
=2 ± 5
1
=4
e sin x
⇒ y =2 ± 5
Since, sine is a bounded function i.e.
− 1 ≤ sin x ≤ 1. Therefore, we get
e −1 ≤ e sin x ≤ e
1
e sin x ∈ , e
⇒
e
Also, it is obvious that 2 + 5 > e
and 2 − 5 <
1
1
⇒ 2 ± 5 ∉ ,e
e
e
2
2
⇒
r2 = 3 + 1 − 2 3 + 3 + 1 + 2 3
⇒
r2 = 8 ⇒ r = 2 2
and
3 +1
r sin α
1+1/ 3
=
=−
r cos α
3 −1
1 −1 / 3
i.e. x = 2, then y = 3
Thus, sec2 θ = 2 ⇒ sec θ = ± 2
⇒
5π
tan α = tan
12
π π
= tan +
4 6
5π
⇒ α =
12
Also, Eq. (i) we have,
r (cos α cos θ + sin α sin θ) = 2
⇒ 2 2 cos(α − θ) = 2
1
⇒
cos(θ − α ) =
2
π
⇒
θ − α = 2n π ±
4
π 5π
∴
θ = 2n π ± +
4 12
14 Given, y = 81sin
2
x
+ 81cos
2
− 30
x
[For intersection X −axis, put
y = 0]
⇒ 81sin
2
x
2 sin 2 x
⇒ 81
+ 811 − sin
2
x
− 30 = 0
+ 81 − 30 ⋅ 81
sin 2 x
=0
[multiplying by 81 sin2 x ]
2 (sin 2 x )
⇒ 81
− 3 ⋅ 81sin
2
x
− 27 ⋅ 81sin
⇒ (81sin
2
x
− 3) (81sin
2
x
x
= (3)1 or [(3 4 )]sin
2
x
= (3)3
1
3
or sin x = ±
2
2
π
5π
π
2π
or x = ± , ±
⇒ x =± ,±
6
6
3
3
Clearly, the graph of
2
y = 81sin x + 81cos x − 30 intersects
the X-axis at eight points in
− π ≤ x ≤ π.
15 Given A ∆PQR such that
r (cos α + sin α )
= ( 3 − 1)2 + ( 3 + 1)2
Then,
2
⇒ sin x = ±
2
Let
2
⇒ [(3)4 ]sin
2
x
+ 81 = 0
− 27) = 0
... (i)
3 sin P + 4 cos Q = 6
... (ii)
4 sin Q + 3 cos P = 1
On squaring and adding the Eqs. (i)
and (ii), we get
(3 sin P + 4 cos Q)2
+ (4 sin Q + 3 cos P)2 = 36 + 1
2
⇒ 9(sin P + cos2 P)
+ 16(sin2 Q + cos2 Q)
+ 2 × 3 × 4(sin P cos Q
+ sin Q cos P) = 37
⇒ 24[sin(P + Q)] = 37 − 25
1
⇒
sin(P + Q) =
2
Since, P and Q are angles of ∆PQR,
therefore,
0 ° < P, Q < 180 °
⇒
P + Q = 30 ° or 150°
⇒
R = 150 ° or 30°
Hence, two cases aries here.
Case I When, R = 150 °
R = 150 ° ⇒ P + Q = 30 °
⇒
0 < P, Q < 30 °
1
⇒ sin P < , cos Q < 1
2
3
⇒ 3 sin P + 4 cos Q < + 4
2
11
⇒ 3 sin P + 4 cos Q <
<6
2
⇒ 3 sin P + 4 cos Q < 6 not possible
Case II R = 30 °
Hence, R = 30 ° is the only possibility.
DAY TWENTY ONE
Properties of
Triangle, Height
and Distances
Learning & Revision for the Day
u
u
Properties Related to Triangle
Some Important Theorems
u
Circles Connected with Triangle
u
Angle of Elevation and Depression
Properties Related to Triangle
In any ∆ABC,
(i) perimeter, 2s = a + b + c
(ii) sum of all angles of a triangle is 180°, i.e. ∠A + ∠B + ∠C = 180 °
PRED
MIRROR
(iii) a + b > c, b + c > a, c + a > b
(iv) a − b < c, b − c < a, c − a < b
(v) a > 0, b > 0, c > 0
Your Personal Preparation Indicator
Relations between the Sides and Angles of Triangle
For a triangle ∆ABC with sides a, b , c and opposite angles are respectively A, B and C, then
a
b
c
(i) Sine Rule In any ∆ ABC,
=
=
sin A sin B sin C
(ii) Cosine Rule
(a) cos A =
b +c −a
2 bc
2
2
2
a2 + b 2 − c2
(c) cos C =
2 ab
(b) cos B =
c + a −b
2ca
2
2
2
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
223
DAY
(iii) Projection Rule
(a) a = b cos C + c cos B
(b) b = c cos A + a cos C
The length of medians AD, BE and CF of a ∆ABC are (shown as
in given figure)
(c) c = a cos B + b cos A
A
(iv) Napier’s Analogy
B −C b −c
A
C−A c−a
B
(a) tan
=
cot (b) tan
=
cot
2
b +c
2
2
c+a
2
(c) tan
A − B a −b
C
=
cot
2
a+b
2
B
(v) Half Angle of Triangle
A
(s − b ) (s − c)
(a) sin =
2
bc
B
(s − c) (s − a)
(b) sin =
2
ca
(c) sin
C
(s − a) (s − b )
=
2
ab
(d) cos
(e) cos
B
=
2
s(s − b )
ca
(f) cos
(g) tan
A
(s − b ) (s − c)
=
2
s(s − a)
(i) tan
C
(s − a) (s − b )
=
2
s(s − c)
A
=
2
C
=
2
(h) tan
s(s − a)
bc
s(s − c)
ab
B
(s − a) (s − c)
=
2
s(s − b )
E
F
D
1
2 b 2 + 2 c2 − a2 ,
2
1
CF =
2 a2 + 2 b 2 − c2
2
AD =
and
C
BE =
1
2 c2 + 2 a2 − b 2
2
Circles Connected with Triangle
1. Circumcircle
The circle passing through the vertices of the ∆ABC is called
the circumcircle. (shown as in given figure)
A
1
1
1
bc sin A = ca sin B = a b sin C
2
2
2
= s (s − a) (s − b ) (s − c)
(vi) Area of a Triangle ∆ =
Some Important Theorems
1. m-n Theorem (Trigonometric Theorem)
If in a ∆ABC, D divides AB in the ratio m : n, then
(shown as in given figure)
O
R
B
C
Its radius R is called the circumradius,
a
b
c
abc
and
R=
=
=
=
2 sin A 2 sin B 2 sin C 4 ∆
NOTE
• The mid-point of the hypotenuse of a right angled triangle
is equidistant from the three vertices of the triangle.
C
• The mid-point of the hypotenuse of a right angled triangle
is the circumcentre of the triangle.
a b
A
A
m
• Distance of circumcentre from the side AC is R cos B.
• Radius of circumcircle of a n-sided regular polygon with
q
D
B
n
each side a is R =
B
a
π
cosec .
2
n
2. Incircle
(i) (m + n) cot θ = n cot A − m cot B
The circle touching the three sides of the triangle internally is
called the inscribed circle or the incircle of the triangle. Its
(ii) (m + n) cot θ = m cot α − n cot β
A
2. Appolonius Theorem
If in ∆ABC, AD is median, then
A
r
B
C
D
B
AB + AC = 2 ( AD + BD 2 )
2
2
2
C
radius r is called inradius of the circle.
∆
(i) r =
s
A
B
C
(ii) r = (s − a) tan = (s − b ) tan = (s − c) tan
2
2
2
ONE
224
A
B
C
sin sin
2
2
2
B
C
C
A
A
B
a sin sin
b sin sin
c sin sin
2
2
2
2
2
2
(iv) r =
=
=
A
B
C
cos
cos
cos
2
2
2
p/n
Similarly, r2 and r3 denote the radii of the escribed circles
opposite to angles B and C, respectively.
r1 , r2 and r3 are called the exradius of ∆ABC.
B
C
a cos cos
∆
A
A
B
C
2
2
(i) r1 =
= s tan = 4R sin cos cos =
A
s−a
2
2
2
2
cos
2
C
A
b cos cos
∆
B
B
C
A
2
2
(ii) r2 =
= s tan = 4R sin cos cos =
B
s −b
2
2
2
2
cos
2
A
B
c cos cos
∆
C
C
A
B
2
2
(iii) r3 =
= s tan = 4R sin cos cos =
C
s −c
2
2
2
2
cos
2
r
NOTE
(iii) r = 4R sin
NOTE Radius of incircle of a n-sided regular polygon with each side
a is r =
a
π
cot . (shown as in given figure)
2
n
O
p/n
• r1 + r2 + r3 = 4 R + r
• r1 r2 + r2 r3 + r3 r1 =
a
• ( r1 − r ) ( r2 − r ) ( r3 − r ) = 4 Rr 2 •
3. Orthocentre and Pedal Triangle
●
The point of intersection of perpendiculars drawn from the
vertices on the opposite sides of a triangle is called
orthocentre. (shown as in given figure)
1
1
1 1
+
+
=
r1
r2
r3 r
Angle of Elevation and Depression
Let O be the observer’s eye and OX be the horizontal line
through O. (shown as in following figures)
If a object P is at a higher level than eye, then ∠POX is called
the angle of elevation.
A
P
E
F
Horizontal line
O
O
●
●
●
D
e
Lin
C
The ∆DEF formed by joining the feet of the altitudes is
called the pedal triangle.
O
of
X
q
t
B
r1 r2 r3
r
h
sig
Lin
eo
fs
igh
t
q
X
Horizontal line
P
Orthocentre of the triangle is the incentre of the pedal
triangle.
If a object P is at a lower level than eye, then ∠POX is called
the angle of depression.
Distance of the orthocentre of the triangle from the angular
points are 2R cos A, 2R cos B, 2R cos C and its distances
from the sides are 2 R cos B cos C, 2 R cos C cos A,
2 R cos A cos B.
Important Results on Heights and
Distances
Results shown by the following figures.
(i) a = h (cot α − cot β)
4. Escribed Circle
D
The circle touching BC and the two sides AB and AC produced
of ∆ABC, is called the escribed circle opposite to A. Its radius
is denoted by r1 . (shown as in given figure)
B
r1
A
C
h
A
b
a
a
B
C
r
225
DAY
α + β
(ii) If AB = CD, then x = y tan
2
(iv) H =
A
h cot β
cot α
D
y
B
D
E
b
a
C
E
B
x
H
h
a
b
A
H sin (β − α )
h cot α
(iii) h =
and H =
cos α sin β
cot α − cot β
C
(v) a = h (cot α + cot β), h = a sin α sin β cosec (α + β)
and d = h cot β = asin α cos β cosec (α + β)
A
A
E
h
H
a
h
B
a
B
b
b
C
C
D
D
d
a
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
6 In a ∆ABC, (c + a + b ) (a + b − c ) = ab. The measure of ∠C
1 If a , b and c are sides of a triangle, then
(a) a + b > c
(b)
a − b > c (if c is smallest)
(c) a + b < c
(d) None of the above
is
2 If in a ∆ABC, A = 30 °, B = 45 ° and a = 1, then the
values of b and c are respectively
(a)
2,
(c)
3,
3 +1
2
3 −1
2
(b)
2,
(d)
2,
3 −1
2
3 + 2
2
3 If A = 75 °, B = 45 °, then b + c 2 is equal to
(a) 2a
(b) 2a + 1
(c) 3a
4 If in a ∆ ABC, 2 b 2 = a 2 + c 2 , then
(a)
c2 −a2
2 ca
c 2 − a 2
(c)
ca
(b)
2
(d) 2 a − 1
sin 3B
is equal to
sin B
c2 −a2
ca
c 2 − a 2
(d)
2 ca
2
(b) 90°
π
. Then, the greatest
2
(c) 120°
π
6
(d) None of these
7 In ∆ ABC, if tan
A
C 1
tan = , then a , b and c are in
2
2 2
(a) AP
(c) HP
(b) GP
(d) None of these
8 In a ∆ABC, a : b : c = 4 : 5 : 6 . The ratio of the radius of the
circumcircle to that of the incircle is
(a)
16
9
(b)
16
7
(c)
11
7
(d)
7
16
s
s
s
− 1 − 1 − 1 is equal to
a
b c
9 In any ∆ ABC, 4
r
R
3r
(c)
R
(b)
2r
R
(d) None of these
10 In a ∆ ABC,R = circumradius and r = inradius.
angle of the triangle is
(a) 60°
(b)
(a)
5 The sides of a triangle are sin α , cos α and
1 + sin α cos α for some 0 < α <
π
3
2π
(c)
3
(a)
(d) 150°
The value of
(a)
R
r
a cos A + b cos B + c cos C
is
a+b+c
(b)
R
2r
(c)
r
R
(d)
2r
R
ONE
226
11 Let the orthocentre and centroid of a triangle be A( −3, 5)
and B( 3, 3), respectively. If C is the circumcentre of this
triangle, then the radius of the circle having line segment
AC as diameter, is
(a)
10
(b) 2 10
(c) 3
5
2
(d)
3 5
2
r
r
12 In a triangle 1 − 1 1 − 1 = 2 , then the triangle is
r2
r3
(a) right angled
(c) isosceles
(b) equilateral
(d) None of these
13 There are two stations A and B due North, due South of a
tower of height 15 m. The angle of depression of A and B
12
3
as seen from top of the tower are cot −1 , sin−1 ,
5
5
then the distance between A and B is
(a) 48 m
(b) 56 m
(c) 25 m
(d) None of these
14 A house of height 100 m subtends a right angle at the
window of an opposite house. If the height of the window
be 64 m, then the distance between the two houses is
(a) 48 m
(b) 36 m
(c) 54 m
(d) 72 m
15 The angle of elevation of the top of a tower from the top
and bottom of a building of height ‘a’ are 30° and 45°,
respectively. If the tower and the building stand at the
same level, then the height of the tower is
a (3 +
(a)
2
3)
(b) a ( 3 + 1) (c) a 3
(d) a ( 3 − 1)
16 A person walking along a straight road observes that at
two points 1 km apart, the angles of elevation of a pole in
front of line are 30° and 75°. The height of the pole is
(a) 250 ( 3 + 1) m
(c) 225 ( 2 − 1) m
(b) 250 ( 3 − 1) m
(d) 225 ( 2 + 1) m
17 A man is walking towards a vertical pillar in a straight
path, at a uniform speed. At a certain point A on the path,
he observes that the angle of elevation of the top of the
pillar is 30°. After walking for 10 min from A in the same
direction, at a point B, he observes that the angle of
elevation of the top of the pillar is 60°. Then, the time
taken (in minutes) by him, from B to reach the pillar, is
j
(a) 6
(b) 10
(c) 20
JEE Mains 2016
(d) 5
18 A bird is sitting on the top of a vertical pole 20 m high
and its elevation from a point O on the ground is 45°. It
flies off horizontally straight away from the point O. After
1s, the elevation of the bird from O is reduced to
30°.Then, the speed (in m/s)of the bird is j JEE Mains 2014
(a) 40 ( 2 − 1) (b) 40 ( 3 − 2 ) (c) 20 2
(d) 20 ( 3 − 1)
19 The shadow of a pole of height ( 3 + 1) m standing on
the ground is found to be 2 m longer, when the elevation
is 30° than when elevation was α , then α is equal to
(a) 15°
(b) 30°
(c) 45°
(d) 75°
20 If the angles of elevation of the top of a tower from three
collinear points A , B and C on a line leading to the foot of
the tower are 30°,45° and 60° respectively, then the ratio
j JEE Mains 2015
AB : BC is
(a) 3 :1
(b) 3 : 2
(c) 1: 3
(d) 2 : 3
21 From the top of a tower 100 m height, the angles of
depression of two objects 200 m apart on the horizontal
plane and in a line passing through the foot of the tower
and on the same side of the tower are 45 ° − A and
45 ° + A. Then, the angle A is equal to
(a) 25°
(b) 30°
(c) 22
1°
2
(d) 45°
22 Two vertical poles 20 m and 80 m stands apart on a
horizontal plane. The height of the point of intersection of
the lines joining the top of each pole to the foot of the
other is
(a) 15 m
(b) 16 m
(c) 18 m
(d) 50 m
23 Let a vertical tower AB have its end A on the level
ground. Let C be the mid-point of AB and P be a point on
the ground such that AP = 2AB. If ∠BPC = β, then tan β is
j JEE Mains 2017
equal to
(a)
6
7
(b)
1
4
(c)
2
9
(d)
4
9
24 From the tower 60 m high angles of depression of the top
and bottom of a house are α and β, respectively. If the
60 sin(β − α )
, then x is equal to
height of the house is
x
(a) sin α sin β
(c) sin α cos β
(b) cos α cos β
(d) cos α sin β
25 A vertical tower stands on a declivity which is inclined at
15° to the horizon. From the foot of the tower a man
ascends the declivity for 80 ft and then, finds that the
tower subtends an angle of 30°. The height of tower is
(a) 20( 6 − 2 ) ft
(c) 40( 6 + 2 ) ft
(b) 40( 6 − 2 ) ft
(d) None of these
26 A ladder rest against a wall at an ∠α to the horizontal. Its
foot is pulled away through a distance a1, so that it slides
a distance b1 down the wall and rests inclined at ∠β with
the horizontal. It foot is further pulled aways through a 2 ,
so that it slides a further distance b 2 down the wall and is
now, inclined at an ∠γ. If a1 a 2 = b1b 2 , then
(a)
(b)
(c)
(d)
α + β + γ is greater than π
α + β + γ is equal to π
α + β + γ is less than π
nothing can be said about α + β + γ
27 ABCD is a trapezium such that AB and CD are parallel
and BC⊥CD. If ∠ADB = θ, BC = p and CD = q , then AB is
j
equal to
JEE Mains 2013
(a)
(p 2 + q 2 ) sinθ
p cosθ + q sinθ
(b)
p 2 + q 2 cosθ
p cosθ + q sinθ
(c)
p2 + q2
p cosθ + q 2 sinθ
(d)
(p 2 + q 2 ) sinθ
(p cosθ + q sinθ)2
2
227
DAY
28 The angle of elevation of the top of the tower observed
from each of the three points A, B, C on the ground
forming a triangle is the same ∠α. If R is the
circumradius of the ∆ABC, then the height of the tower is
(a) R sin α
(c) R cot α
(b) R cos α
(d) R tan α
and the angle of elevation of the top of the tower from A
or B is 30°. The height of the tower is
(a)
2a
3
(b) 2a 3
(c)
a
3
(d)
3
30 ABCD is a square plot. The angle of elevation of the top
29 A tower stands at the centre of a circular park. A and B
are two points on the boundary of the park such that
AB( = a ) subtends an angle of 60° at the foot of the tower
of a pole standing at D from A or C is 30° and that from B
is θ, then tan θ is equal to
(a)
6
(b) 1/ 6
(c)
3/ 2
(d)
2 /3
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 For a regular polygon, let r and R be the radii of the
inscribed and the circumscribed circles. A false
statement among the following is
r
R
r
(b) there is a regular polygon with
R
r
(c) there is a regular polygon with
R
r
(d) there is a regular polygon with
R
(a) there is a regular polygon with
1
2
1
=
2
2
=
3
3
=
2
=
45°. After ascending 2 km towards the mountain up an
incline of 30°, the elevation changes to 60°. The height of
mountain is
(b) GP
(d) None of these
3 If A0 , A1, A2 , A3 , A4 and A5 are the consecutive vertices of
a regular hexagon inscribed in a unit circle. Then, the
product of length of A0 A1 × A0 A2 is
3
(b) 2 3
(c) 2
(d) 3 3
4 Two ships leave a port at the same time.One goes
24 km/h in the direction North 45° East and other travels
32 km/h in the direction South 75° East. The distance
between the ships at the end of 3 h is approximately
(a) 81.4 km
(b) 82 km
(c) 85 km
(d) 86.4 km
5 A round balloon of radius r subtends an angle α at the
observer, while the angle of elevation of its centre is β.
The height of the centre of balloon is
(a) r cosec α sin
(c) r sin
β
2
α
cosec β
2
(b) r sin β cosec
(d) r cosec
(b) 50
(d) 50 2
7 At the foot of a mountain the elevation of its summit is
∆ABC are in
(a)
tower stands at the mid-point of QR. If the angles of
elevation of the top of the tower at P, Q and R are
respectively 45°, 30° and 30°, then the height of the
j JEE Mains 2018
tower (in m) is
(a) 100
(c) 100 3
2 If cos A + cos B + 2 cos C = 2, then the sides of the
(a) AP
(c) HP
6 PQR is a triangular park with PQ = PR = 200 m. A TV
α
2
α
sin β
2
(a) 2.732 km
(c) 2.03 km
(b) 1.732 km
(d) 1.045 km
8 In a cubical hall ABCDPQRS with each side 10 m, G is
the centre of the wall BCRQ and T is the mid-point of the
side AB. The angle of elevation of G at the point T is
(a) sin−1 (1/ 3 )
(c) cot −1 (1/ 3 )
(b) cos−1 (1/ 3 )
(d) None of these
9 A tree stands vertically on a hill side which makes an angle
of 15° with the horizontal.From a point on the ground 35 m
down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°.The height of the tree is
(a) 31 2 m
(c) 35 2 m
(b) 33 2 m
(d) 34 2 m
10 A flag staff stands in the centre of a rectangular field
whose diagonal is 1200 m, and subtends angles 15° and
45° at the mid-points of the sides of the field. The height
of the flag staff is
(b) 300 ( 2 +
(a) 200 m
(c) 300 ( 2 −
3) m
(d) 400 m
3) m
ONE
228
ANSWERS
SESSION 1
1. (a)
11. (c)
21. (c)
2. (a)
12. (a)
22. (b)
3. (a)
13. (b)
23. (c)
4. (d)
14. (a)
24. (d)
5. (c)
15. (a)
25. (b)
6. (c)
16. (a)
26. (c)
7. (d)
17. (d)
27. (a)
8. (b)
18. (d)
28. (d)
9. (a)
19. (c)
29. (c)
10. (c)
20. (a)
30. (b)
SESSION 2
1. (c)
2. (a)
3. (a)
4. (d)
5. (b)
6. (a)
7. (a)
8. (a)
9. (c)
10. (c)
Hints and Explanations
SESSION 1
= −1 +
1 Clearly, ( a + b + c )( a + b − c )
=( a+
a2 + c 2
2
= −1 +
(ac )2
b )2 − c
= a + b − c + 2 ab > 0
∴
a+
b> c
= −1 +
2 Given, A = 30 °, B = 45°, a = 1
⇒
=
3+1
2
3 Given, A = 75° and B = 45°
⇒
C = 60 °
a
b
c
, we
By sine rule,
=
=
sin A sin B sin C
a
b
c
get
=
=
sin 75° sin 45° sin 60 °
sin 45°
sin 60 °
Now, b + c 2 =
a+ 2
a
sin 75°
sin 75°
1
3
2 a+ 2
2
=
a
3+1
3+1
=
4
2 2
2 2
2
2 3a
a+
= 2a
3+1
3+1
sin 3B 3 sin B − 4 sin3 B
=
sin B
sin B
= 3 − 4 sin2 B
= 3 − 4 (1 − cos 2 B )
(a + c 2 ) 2
4 (ac ) 2
2
2
5 Let a = sinα, b = cos α
b = 2
and
c = 2 sin 105° = 2cos 15°
3 + 1
∴
c =2
=
2 2
2
(a2 + c 2 )2 − 4a2c 2
c 2 − a2
=
2
4 (ac )
2 ac
and c = 1 + sin α cos α
Here, we can see that the greatest side isc.
a2 + b 2 − c 2
∴ cos C =
2ab
sin2 α + cos2 α − 1 − sin α cos α
⇒ cos C =
2ab
sin α cos α
⇒ cos C = −
2sin α cos α
1
⇒ cos C = − = cos 120°
2
⇒
∠C = 120°
6 We have,
2s (2s − 2c ) = ab ⇒
s (s − c ) 1
=
ab
4
C 1
=
2
4
C 1
we get, cos =
2 2
since, C must be acute
2
⇒
cos 2
C
= 60° ⇒ C
2
7 We have, tan A tan
2
(s − b )(s − c )
⇒
s(s − a)
⇒
2π
3
C 1
=
2 2
(s − a)(s − b ) 1
=
s(s − c )
2
=
s−b 1
= ⇒ 2 s − 2b − s = 0
s
2
⇒
a + c − 3b = 0
8 We have, R = abc and r = ∆
4∆
s
R abc s
abc
=
⋅ =
∴
r
4∆ ∆ 4(s − a)(s − b )(s − c )
⇒
[Q2b 2 = a2 + c 2 (given)]
[Qa + b > c , as sides of triangle]
∴
C = 105°
Using sine rule,
a
b
c
, we get
=
=
sin A sin B sin C
1
b
c
=
=
1
1 sin 105°
2
2
4 (a2 + c 2 − b 2 )2
4 (ac )2
Since, a : b : c = 4 : 5 : 6
a b c
(say)
⇒
= = =k
4 5 6
R
(4k )(5k )(6k )
Thus,
=
15k
15k
r
4
− 4k
− 5k
2
2
15k − 6k
2
3
120k ⋅ 2
16
= 3
=
k ⋅ 7 ⋅ 5⋅ 3
7
9 We have, 4 s − 1 s − 1 s − 1
a
b
c
(s − a) (s − b ) (s − c )
abc
s( s − a) (s − b ) (s − c )
= 4⋅
s ⋅ abc
4∆ ∆
∆2
r
= 4⋅
=
⋅ =
s ⋅ abc abc s
R
a cos A + b cos B + c cos C
10 We have,
a+ b + c
2 R sin A cos A + 2 R sin B cos B
+ 2 R sin C cos C
=
2s
R
=
⋅ (sin 2 A + sin 2B + sin 2C )
2s
R
=
⋅ 4 sin A sin B sin C
2s
2R abc
abc
=
⋅
=
s 8 R3
4sR2
4 ∆R
r
abc
∆
=
=
QR =
,r =
∆
2
R
4∆
s
4⋅ ⋅ R
r
= 4⋅
229
DAY
11 Key Idea Orthocentre, centroid and
circumcentre are collinear and centroid
divide orthocentre and circumcentre in
2:1 ratio.
We have orthocentre A(−3, 5) and centroid
B(3,3). Let C be the circumcentre.
C
B(3, 3)
A(–3, 5)
…(ii)
⇒
d = 36 tanθ
From Eqs. (i) and (ii), we get
d 2 = 36 × 64 ⇒ d = 48 m
15 Let CD be a tower of height h and AB is
building of height a.
Clearly, AB = (3 + 3)2 + (3 − 5)2
D
= 36 + 4 = 2 10
We know that, AB : BC = 2:1 ⇒ BC = 10
AC = AB + BC = 2 10 + 10
= 3 10
Since, AC is a diameter of circle.
AC
3 10
5
∴ r =
=3
⇒ r =
2
2
2
12 Since, 1 − r1 1 − r1 = 2
r2
r3
1 − s − b 1 − s − c = 2
∴
s − a
s − a
(b − a)(c − a)
=2
⇒
(s − a)2
2
bc − ab − ac + a
=2
⇒
2
a + b + c − a
2
2bc − 2ab − 2ac + 2a2
= b 2 + c 2 + a2 + 2bc − 2ab − 2ac
2
2
⇒ a = b + c2
So, triangle is right angled.
13 Given, cot α = 12 and sin β = 3
5
5
C
30°
h L
Now,
2(bc − ab − ac + a2 )
⇒
=1
(b + c − a)2
(100 − 64)
d
In ∆CDE, tan(90°− θ) =
B
a
45°
C
A
(h − a)
...(i)
= 3 (h − a)
tan30°
h
In ∆ACD, tan 45° =
⇒ h = CA ⇒
CA
h = LB [Q LB = CA] ...(ii)
∴
LB =
From Eqs. (i) and (ii), we get
h ( 3 − 1) = 3 a
∴
h=
3a
3−1
=
3 ( 3 + 1) a
2
... (i)
... (ii)
From Eqs. (i) and (ii)
x+ y
= 3y
3
⇒
x + y = 3y
⇒
h−a
LB
In ∆BL D, tan30° =
Now, from ∆ACD and ∆BCD,
we have
h
tan30° =
x+ y
h
and
tan 60° =
y
x+ y
⇒
h=
3
and
h = 3y
x − 2y = 0 ⇒ y =
x
2
Q Speed is uniform
∴ Distance y will be cover in 5 min.
(Q Distance x covered in 10 min.)
x
⇒ Distance will be cover in 5 min.
2
A B
18 In ∆OA 1 B 1 , tan 45° = 1 1
OB1
20
= 1 ⇒ OB1 = 20
⇒
OB1
A1
A2
3 + 3
=
a
2
⇒
20 m
16 Let OP be the pole of height x m.
P
45°
30°
45°
O
20
B1
x
B2
In ∆OA 2 B2 ,
A
15 m
b
B
D
d
In ∆DAC and ∆DBC ,
AD = 15 cot α, BD = 15 cot β
⇒
d = 15 (cot α + cot β )
12 4
= 15
+ = 56 m
5
3
64
14 In ∆DAB, tanθ =
d
75°
30°
B
1 km
a
A
Using sine rule in APB,
⇒ PB = 1000 ×
1/2
tan30° =
O
sin 30°
PB
=
sin 45°
AB
= 500 2 m
1/ 2
3+1
2 2
⇒
B1 B2 = 20 3 − 20
⇒
B1 B2 = 20( 3 − 1)m
Distance 20( 3 − 1)
=
Time
1
= 20( 3 − 1) m/s
= 250 ( 3 + 1) m
17 According to given information, we have
the following figure
C
⇒ B1 B2 + OB1 = 20 3
Now, Speed =
In ∆ PBO, x = 500 2 ⋅ sin 75°
= 500 2 ×
20
⇒ OB2 = 20 3
OB2
19 Let OP be a tower with height ( 3 + 1 ) m
and AB = 2 m.
P
D
90°–q
q
d
D
100 m
(Ö3 + 1) m
E
q
⇒
a
30°
B
O
In ∆AOP,
64 m
A
A
Pillar h
d
d = 64cotθ
B
A
…(i)
30°
x
B
3+1
OA
OA = ( 3 + 1 ) 3
tan 30 ° =
60°
y
C
⇒
ONE
230
3+1
OB
OB = ( 3 + 1 ) cot α
and in ∆BOP,
⇒
tan α =
Now,
OA − OB = (3 +
3 ) − ( 3 + 1 ) cot α
2=3+
⇒
⇒
∴
3 − ( 3 + 1 ) cot α
cot α = 1
α = 45°
20 According to the given information the
figure should be as follows.
Let the height of tower = h
E
⇒ BD = 100 cot (45° − A )
⇒ BC + CD = 100 cot (45° − A ) …(ii)
∴ From Eqs. (i) and (ii), we get
CD = 100 [cot (45° − A ) − cot(45° + A )]
1 + tan A 1 − tan A
= 100
−
1 − tan A 1 + tan A
4 tan A
= 100
2
= 200 tan 2 A
1 − tan A
⇒ 200 = 200 tan 2 A ⇒ tan2 A = 1
⇒ tan 2 A = tan 45° ⇒ 2 A = 45°
1°
∴ A = 22
2
22 Let PQ and RS be the poles of height 20
m and 80 m subtending angles α and β
at R and P , respectively. Let h be the
height of the point T , the intersection
of QR and PS.
h
S
30°
A
45°
B
60°
C
D
ED
In ∆EDA, tan30° =
AD
1
ED
h
⇒
=
=
⇒ AD = h 3
AD
AD
3
h
In ∆EDB, tan 45° =
⇒ BD = h
BD
h
h
In ∆EDC, tan 60° =
⇒ CD =
CD
3
AB
AD − BD
AB
h 3−h
Now,
=
⇒
=
h
BC
BD − CD
BC
h−
3
h ( 3 − 1)
AB
AB
3
⇒
=
⇒
=
BC
BC
1
h ( 3 − 1)
3
AB : BC = 3 :1
∴
Q
80 m
T
a
R
h
b
V
20 m
D
a
b
…(i)
d = 60cotβ
DC
tanα =
EC
⇒
DC = d tanα
⇒ 60 − h = d tanα
(Q BC = EA = h )
⇒ 60 − h = 60cot β tan α [from Eq. (i)]
cos β sin α
⋅
⇒
h = 60 1 −
sin β cos α
60sin(β − α )
⇒
h=
cos α sin β
60sin (β − α ) 60sin (β − α )
(given)
⇒
=
x
cos α sin β
In ∆DEC,
x = cos α sin β
30°
45°+A
B
200 m
100
BC
⇒
BC = 100 cot (45° + A )
and in ∆ADB,
100
tan (45° − A ) =
BD
In ∆ACB, tan(45° + A ) =
…(i)
8
75°
C
0 ft
15°
B
In ∆ABC, on applying sine rule
BC
AB
=
sin75° sin30°
80sin 30° 40 × 2 2
⇒
AB =
=
sin 75°
3+1
= 40( 6 − 2 ) ft
b
a
26 Clearly, a1 = tan α + β
P
2h
and
Now, in ∆ABP,
AB
h
1
tan(α + β ) =
=
=
AP 2h 2
C
Tower
75°
C
A
100 m
B
⇒
h/2
h
h/2
D
d
A
B
45°–A
45°–A
b
A
h = 80 − 4 h
h = 16 m
A
60
tower.
h
2
Again, let ∠CPA = α
45°+A
C
d
h
Again, h cot α + h cot β = 20 cotα
⇒
(h − 20)cot α = − hcotβ
cot α
h
⇒
=
=4
cot β 20 − h
⇒
∴
a
E
25 Let BC be the declivity and BA be the
Then, PR = h cot α + h cot β
= 20cot α = 80cot β
cot α
⇒ cotα = 4cotβ ⇒
=4
cot β
AC = BC =
100 m.
Distance between the objects,
CD = 200 m
d
⇒
P
23 Let AB = h, then AP = 2h and
21 Let AB be the tower with height
24 In ∆ABD, tanβ = 60
h
AC
1
Also, in ∆ACP, tanα =
= 2 =
AP
2h 4
Now, tan β = tan[(α + β ) − α]
1 1
1
−
tan(α + β ) − tan α
2
4
=
=
= 4
9
1 + tan(α + β )tan α 1 + 1 × 1
2 4
8
2
=
9
…(i)
2
b1
a2
β + γ
= tan
2
b2
…(ii)
G
b1
F
b2
E
A
a2
g
B
a
b
a1
C
D
231
DAY
Since, a1 a2 = b1 b2
a
b
α + β
∴ 1 = 2 ⇒ tan
=
2
b1
a2
1
β + γ
tan
2
⇒
α + β
β + γ = 1
tan
2
2
α+β β+ γ
π
+
=
2
2
2
α + β + γ = π −β < π
∴
⇒
27 Let AB = x
D
θ
p
C
√p 2
+q
p
π–(θ+α)
A
sin θ cos α + cos θ sin α
( p2 + q 2 )sin θ
=
q sin θ + p cos θ
2
SESSION 2
1
P
a
π
a
π
and
= sin
= tan
2R
n
2r
n
r
π
= cos
R
n
r
1
n = 3 gives, =
R 2
r
1
n = 4 gives, =
R
2
r
3
n = 6 gives, =
R
2
∴
R
C
OP
⇒ OP = OA tanα
OA
OP = R tanα
(QOA = R, given)
In ∆OAP, tanα =
⇒
29 Let h be the height of a tower
C
Alternate Solution
Applying sine rule in ∆ABD,
a = b = c
sin A sin B sin C
O
90º
30º
60º
P
C
a
p
2
a
p +q
AB
=
sin θ sin{ π − (θ + α)}
⇒
⇒
s−c 1
a+ b −c
1
= ⇒
=
s
3
a+ b + c 3
⇒
3 a + 3b − 3c = a + b + c
⇒
a + b = 2c
Hence, a, c and b are in AP.
A 0 A 1 = 2 × 1 cos 60 ° = 1 = A1 A2
[using cosine rule]]
A4
A3
30°
a
a
B
2
⇒
3 Clearly,
D
30°
A
⇒
⇒
Since, ∠AOB = 60°
Also,
OB = OA = radii
∴
∠OBA = ∠OAB = 60°
So, ∆OAB is an equilateral
∴
OA = OB = AB = a
h
In ∆OAC, tan30° =
a
1
h
a
⇒
=
⇒ h=
a
3
3
30 Let PD be a pole.
q
⇒
A
a
B
A+ B
A−B
C
cos
= 4 sin2
2
2
2
A−B
A+ B
cos
= 2 cos
2
2
Qcos A + B ≠ 0
2
A
B
A
B
cos cos + sin ⋅ sin
2
2
2
2
A
B
A
B
= 2 cos cos − sin ⋅ sin
2
2
2
2
A
B
A
B
3 sin ⋅ sin = cos cos
2
2
2
2
A
B 1
tan ⋅ tan =
2
2 3
(s − b ) (s − c )
(s − c ) (s − a) 1
⋅
=
s (s − a)
s (s − b )
3
2 2 cos
h
( p + q )sin θ
p cos θ + q sin θ
π–(θ+α)
a
B
2
Öp 2
+q
O
a
R
p
nr
R 2p
n
∴
− pq cos θ + p2 sin θ
p cos θ + q sin θ
2
2
R
p
x −q
q cos θ − p sin θ
= q − p
q sin θ + p cos θ
q 2 sin θ + pq cos θ
q
p + q
p
2
make equal angles at the vertices of the
triangle, therefore foot of tower is at the
circumcentre.
B
q
Qin ∆BDC ,cot α = p
q cot θ − p
= q − p
.
q + p cot θ
2
, sin α =
2
a
p
⇒ tan(θ + α ) =
q−x
⇒ q − x = p cot(θ + α )
⇒
x = q − p cot(θ + α )
cot θ cot α − 1
= q − p
cot α + cot θ
q cot θ − 1
p
= q − p
q + cot θ
p
D
p +q
x
In ∆DAM, tan( π − θ − α ) =
AB =
q
2
A
q
x=
Qcos α =
α
x–q
M
⇒
p2 + q 2 sin θ
AB =
DP =
⇒
28 Let OP be the tower. Since, the tower
q
α
a
3
DP
In ∆PDB, tan θ =
BD
a/ 3
1
tan θ =
=
⇒
2a
6
p2 + q 2
AB
=
sin θ sin(θ + α )
tan
⇒
⇒
⇒
C
A
q
a
a
O
A5
B
DP
In ∆DAP, tan 30° =
AD
A2
60°
0°
12
A0
A1
ONE
232
and
Let height of tower TM be h.
TM
In ∆PMT , tan 45° =
PM
h
⇒
1=
PM
⇒
PM = h
h
; QM = 3h
In ∆TQM, tan30° =
QM
( A 0 A1 )2 + ( A1 A2 )2 – ( A 0 A2 )2
cos 120 ° =
2 ⋅ A 0 A 1 ⋅ A1 A2
1 + 1 − ( A 0 A2 )2
2⋅1⋅1
⇒ A 0 A2 = 3
=
∴ A 0 A1 × A 0 A2 = 1 × 3 = 3
ships at the end of 3 h.
Then, OA = (24 × 3) = 72 km and
OB = (32 × 3) = 96 km
km
x km
O
72
75° 96 km
E
B
S
Hence, the distance between the ships at
the end of 3 h is
86.53 km ≈ 86.4 km (approx.).
5 In ∆APC,
α
r
α
sin =
⇒ AC = r cosec
2
2
AC
r C
a
H
P
A
b
B
In ∆ ABC , sin β =
BC
AC
⇒ H = AC sin β ⇒ H = r sin β cosec
6
α
2
9 Let PAQ be the hill,AB be the tree and
[say]
45º
200 m
T
30º 90º
Q
30º
M
R
PD be the horizontal. Let P be the point
of observation.
B
35 m A
C 60°
A
D
45°
2 km
30°
B
O
Let C be the point where elevation
is 60°.
Then, ∠BAC = 30° and AC = 2 km
BC
∴
= sin30°
AC
1
⇒
BC = 2 × = 1
2
AB
i.e.
BC = 1 km and
= cos 30°
AC
2 3
⇒
AB =
= 3
2
i.e.
AB = 3 km
PD = OP − OD
= OP − BC = h − 1
and
CD = BO = AO − AB = h − 3
h−1
PD
In ∆DCP,
= tan 60° ⇒
= 3
CD
h− 3
⇒
3 h − 3 = h − 1 ⇒ ( 3 − 1) h = 2
h=
2
×
3−1
3+1
3+1
= 3+1
= 1.732 + 1 = 2732
.
km
200 m
GH
5
1
=
=
TG
5 3
3
θ = sin −1 (1 / 3 )
H
∴
P
H
B
Then, sinθ =
⇒
Now,
A
T
Let θ be the required angle of elevation
of G at T .
P
Let AB = x km
We have, ∠NOA = 45° and ∠SOB = 75°
∴ ∠AOB = 180°−(45°+75° ) = 60°
Using cosine formula on ∆AOB, we get
(72)2 + (96)2 − x2
cos 60° =
2 × 72 × 96
⇒
x2 = 14400 − 6912 = 7488
⇒
x = 7488 = 86.53 km
G
C
q
4h2 = (200)2 ⇒ h = 100 m
the mountain.
Then,
∠OAP = 45°
∴
∠OPA = 45°
Hence, ∆OAP is an isosceles.
Let AO = OP = h km height of
mountain.
Q
D
7 Let P be the summit and A be the foot of
A
45°
W
⇒
R
P
In ∆PMQ, PM 2 + QM 2 = PQ 2
h2 + ( 3h )2 = (200)2
4 Let A and B be the positions of the two
N
S
8 Let H be the mid-point of BC. Since,
∠ TBH = 90° , therefore,
TH 2 = BT 2 + BH 2 = 52 + 52 = 50
Also since,
∠ THG = 90° , TG 2 = TH 2 + GH 2
= 50 + 25 = 75
Q
45 °
15°
P
C
D
Produce BA to meet PD at C.
Let AB = H m.
Then, ∠DPA = 15° , PA = 35m
∠CPB = 60° and ∠PCA = 90°.
∴
∠APB = (60°−15° ) = 45°
In ∆PAC, ∠PAC = 180°− (15° + 90° ) = 75°
∴ ∠PAB = (180°−75° ) = 105°
and ∠PBA = 180°−(45° + 105° ) = 30°
Applying sine rule on ∆PAB, we get
PA
AB
35
H
=
⇒
=
sin ∠PBA sin ∠APB
sin 30° sin 45°
∴
35 × 2 = H × 2 ⇒ H = 35 2 m
Hence, the height of the tree is 35 2 m.
10 Let OP be the flag staff of height h
standing at the centre O of the field.
P
45°
45°
D
C
h
F
90° 15°
90°
E
O
A
B
In ∆ OEP, OE = h cot15° = h (2 +
3)
and in ∆ OFP, OF = h cot 45° = h
∴ EF = h 1 + (2 + 3 ) 2 = 2h 2 +
Since, BD = 1200 m
⇒
2EF = 4 h (2 +
∴ h=
300
2+
3 ) = 1200
= (300 2 − 3 ) m
3
3
DAY TWENTY TWO
Inverse Trigonometric
Function
Learning & Revision for the Day
u
Inverse Trigonometric Function
u
Properties of Inverse Trigonometric Function
Inverse Trigonometric Function
Trigonometric functions are not one-one and onto on their natural domains and ranges, so
their inverse do not exists in the whole domain. If we restrict their domain and range, then
their inverse may exists.
y = f ( x) = sin x. Then, its inverse is x = sin −1 y.
NOTE
1
y
• sin −1 y ≠ (sin y ) −1
• sin −1 y ≠ sin
The value of an inverse trigonometric functions which lies in its principal value branch is
called the principal value of that inverse trigonometric function.
Domain and range of inverse trigonometric functions
Function
Domain
x
[− 1, 1]
cos −1 x
[−1, 1]
tan −1 x
R
sin
cot
−1
−1
x
sec −1 x
−1
cosec x
Range
(Principal Value Branch)
[0, π ]
π π
− ,
2 2
R
(0, π )
R − (−1, 1)
π
[0, π ] −
2
R − (−1, 1)
PRED
MIRROR
π π
− ,
2 2
π π
− 2 , 2 − {0}
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
234
Properties of Inverse
Trigonometric Functions
π
;
(−1 ≤ x ≤ 1)
2
π
(ii) tan −1 x + cot −1 x = ;
x ∈R
2
π
(iii) sec −1 x + cosec −1 x = ;
(x ≤ − 1 or x ≥ 1)
2
−1
−1
2. (i) sin (− x) = − sin x;
(− 1 ≤ x ≤ 1)
(ii) cos −1 (− x) = π − cos −1 x;
(− 1 ≤ x ≤ 1)
(iii) tan −1 (− x) = − tan −1 ( x);
(− ∞ < x < ∞)
(iv) cot −1 (− x) = π − cot −1 x; (− ∞ < x < ∞)
(v) sec −1 (− x) = π − sec −1 x;
x ≤ − 1 or x ≥ 1)
(vi) cosec −1 (− x) = − cosec −1 x; (x ≤ − 1 or x ≥ 1)
3. (i) sin −1 (sin x) is a periodic function with period 2π.
π π
x∈ − ,
x,
2 2
π − x, x ∈ π , 3 π
2 2
−1
sin (sin x) =
3π 5π
x − 2 π, x ∈ ,
2 2
5π 7π
3 π − x, x ∈ ,
2 2
(ii) cos –1 (cos x) is a periodic function with period 2π.
x ∈ [0, π ]
x,
2 π − x, x ∈ [π, 2 π ]
cos −1 (cos x) =
x − 2 π, x ∈ [2 π, 3 π]
4π − x, x ∈ [3 π, 4π ]
−1
−1
1. (i) sin x + cos x =
(iii) tan −1 (tan x) is a periodic function with period π.
π π
x∈ − ,
x,
2 2
π 3π
x − π,
x∈ ,
2 2
−1
tan (tan x) =
3π 5π
x − 2 π, x ∈
,
2 2
5π 7π
x − 3 π, x ∈ ,
2 2
(iv) cot −1 (cot x) is a periodic function with period π.
cot −1 (cot x) = x; 0 < x < π
(v) sec −1 (sec x) is a periodic function with period 2π .
π
π
sec −1 (sec x) = x; 0 ≤ x < or
<x≤π
2
2
(vi) cosec −1 (cosec x) is a periodic function with
period 2 π.
π
π
cosec −1 (cosec x) = x; − ≤ x < 0 or 0 < x ≤
2
2
1
4. (i) sin −1 = cosec −1 x, if x ∈ (−∞, − 1] ∪ [1, ∞)
x
−1 1
(ii) cos = sec −1 x, if x ∈ (−∞, − 1] ∪ [1, ∞)
x
−1
if x > 0
1 cot x,
(iii) tan −1 =
x − π + cot −1 x, if x < 0
5. (i) sin −1 x = cos −1 1 − x2
1 − x2
x
= tan −1
= cot −1
1 − x2
x
1
−1 1
= sec −1
= cosex , if x ∈(0, 1)
2
x
1− x
(ii) cos −1 x = sin −1 1 − x2
1 − x2
x
−1 1
= cot −1
= tan −1
= sec
x
1 − x2
x
1
= cosec −1
, if x ∈(0, 1)
1 − x2
x
(iii) tan −1 x = sin −1
1 + x2
1
−1 1
= cos −1
= cot
x
1 + x2
1 + x2
= sec −1 ( 1 + x2 ), if x ∈ (0, ∞)
= cosec −1
x
6. (i) sin −1 x + sin −1 y
sin −1 ( x 1 − y2 + y 1 − x 2 ); | x|,| y| ≤ 1 and
x2 + y2 ≤ 1 or ( xy < 0 and x 2 + y2 > 1)
−1
2
2
π − sin ( x 1 − y + y 1 − x ); 0 < x, y ≤ 1
=
and x 2 + y2 > 1
−1
2
2
− π − sin ( x 1 − y + y 1 − x );
2
2
− 1 ≤ x, y < 0 and x + y > 1
(ii) sin −1 x − sin −1 y
sin −1 ( x 1 − y2 − y 1 − x 2 ); x , y ≤ 1
2
2
2
2
and x + y ≤ 1 or ( xy > 0 and x + y > 1)
−1
2
2
π − sin ( x 1 − y − y 1 − x );
=
2
2
0 < x ≤ 1, − 1 ≤ y < 0 and x + y > 1
−1
2
2
− π − sin ( x 1 − y − y 1 − x );
− 1 ≤ x < 0, 0 < y ≤ 1 and x 2 + y2 > 1
(iii) cos −1 x + cos −1 y
cos −1 { xy − (1 − x2 ) (1 − y2 )}; | x |,| y | ≤ 1
and x + y ≥ 0
=
2
2
−1
2 π − cos { xy − (1 − x ) (1 − y )};
| x |,| y | ≤ 1 and x + y ≤ 0
−1
−1
(iv) cos x − cos y
cos −1 { xy + (1 − x2 ) (1 − y2 )}; | x |,| y | ≤ 1
and x ≤ y
=
−1
2
2
− cos { xy + (1 − x ) (1 − y )};
− 1 ≤ y ≤ 0, 0 < x ≤ 1 and x ≥ y
235
DAY
(v) tan −1 x + tan −1 y
−1 x + y
xy < 1
tan 1 − xy ;
x + y
x > 0, y > 0, xy > 1
= π + tan −1
;
1 − xy
−1 x + y
; x < 0, y < 0, xy > 1
− π + tan
1 − xy
(vi) tan −1 x − tan −1 y
−1 x − y
tan 1 + xy ;
xy > − 1
−1 x − y
xy < − 1, x > 0, y < 0
= π + tan
;
1 + xy
−1 x − y
; xy < − 1, x < 0, y > 0
− π + tan
1 + xy
7. (i) 2 sin −1 x
1
1
−
≤x≤
sin −1 {2 x (1 − x 2 )};
2
2
1
2
−1
= π − sin (2 x 1 − x );
≤ x ≤1
2
− π − sin −1 (2 x 1 − x 2 ); − 1 ≤ x < − 1
2
−1
2
cos (2 x − 1);
0 ≤ x ≤1
(ii) 2 cos −1 x =
−1
2
2
π
cos
(
2
x
1
);
1 ≤ x <0
−
−
−
(iii) 2 tan −1
2x
−1
−1 < x <1
tan 1 − x 2 ;
2x
x >1
x = π + tan −1
;
1 − x 2
2x
− π + tan −1
; x < −1
1 − x 2
(iv) 2 tan −1
(v) 2 tan −1
NOTE
−1 2 x
−1 ≤ x ≤1
sin 1 + x 2 ;
2x
x = π − sin −1
; x >1
1 + x 2
2x
− π − sin −1
; x < −1
1 + x 2
−1 1 − x 2
cos 1 + x2 ; 0 ≤ x < ∞
x=
2
− cos −1 1 − x ; − ∞ < x ≤ 0
1 + x2
• If sin −1 x + sin −1 y = θ, then cos −1 x + cos −1 y = π − θ
• If cos −1 x + cos−1 y = θ, then sin −1 x + sin −1 y = π − θ
−1
1
sin −1 (3 x − 4 x3 ),
if
≤x≤
2
2
1
−1
−1
3
(i)
3
sin
x
=
π
−
sin
(
3
x
−
4
x
),
if
<
x
≤
1
8.
2
− π − sin −1 (3 x − 4 x3 ), if − 1 ≤ x < −1
2
1
cos −1 (4 x3 − 3 x),
if ≤ x ≤ 1
2
−1
1
(ii) 3 cos −1 x = 2 π − cos −1 (4 x3 − 3 x), if
≤x≤
2
2
1
2 π + cos −1 (4 x3 − 3 x),
if − 1 ≤ x ≤
2
3
−1
1
−1 3 x − x
if
<x<
tan 1 − 3 x2 ,
3
3
3
3
x
−
x
1
(iii) 3 tan −1 x = π + tan −1
, if x >
1 − 3 x2
3
3
−1
−1 3 x − x
− π + tan 1 − 3 x2 , if x <
3
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
`
1 The principal value of sin
(a)
3π
5
20
2 If
∑
(b)
7π
5
sin−1 xi = 10π , then
i =1
(a) 20
−1
33π
cos
is
5 j NCERT Exemplar
(c)
π
10
(d) −
π
10
20
∑
xi is equal to
i =1
(b) 10
(c) 0
(d) None of these
3 The domain of the function defined by
f ( x ) = sin−1
(a) [1, 2]
(c) [0, 1]
x − 1 is
NCERT Exemplar
(b) [– 1, 1]
(d) None of these
j
4 The value of cos ( 2 cos
(a) 1
−1
(b) 3
x + sin
−1
x ) at x =
1
is
5
(d) −
(c) 0
2 6
5
5 The value of cos[tan−1 {sin(cot −1 x )}] is
(a)
1
x +2
2
(b)
x2 + 2
x2 + 1
(c)
x2 + 1
x2 + 2
(d)
1
x2 + 1
1
has
3
6 The equation tan−1 x − cot −1 x = tan−1
(a)
(b)
(c)
(d)
no solution
unique solution
infinite number of solutions
two solutions
j
NCERT Exemplar
236
1
2x
. Then,
, where | x | <
1 − x 2
3
7 Let tan−1 y = tan−1 x + tan−1
a value of y is
j
3x − x 3
(a)
1− 3x 2
3x − x 3
(c)
1+ 3x 2
JEE Mains 2015
3x + x 3
(b)
1− 3x 2
3x + x 3
(d)
1+ 3x 2
(a) 0
(b)
π
4
(c)
π
2
π2
2
(c) π 2
5 π2
4
(d) None of these
(a)
(b)
19 The trigonometric equation sin−1 x = 2 sin−1 a , has a
solution for
8 If θ = tan− 1 a , φ = tan− 1 b and ab = − 1, then (θ − φ) is
equal to
18 The maximum value of ( sec −1x )2 + (cosec −1x )2 is
(d) None of these
20 If x ∈ −
9 The range of
(a) [− π, π)
(b) (− π, π)
(c) [ − π, π]
(a)
10 The number of solutions of the equation
cos (cos −1 x ) = cosec (cosec −1x ) is
(a) 2
(b) 3
(c) 4
11 The value of cot cosec−1
5
(a)
17
(d) 1
5
2
+ tan−1 is
3
3
6
(b)
17
1
2
12 If tan (cos −1 x ) = sin cot −1 , then x is equal to
(a) ±
5
3
(b) ±
5
3
(c) ±
5
3
(d) None of these
π
and tan−1 x − tan−1 y = 0, then
2
x 2 + xy + y 2 is equal to
13 If cos −1 x + cos −1 y =
1
(b)
2
(a) 0
14 If sin
−1
3
(c)
2
1
(d)
8
π
x
−1 5
+ cosec = , then the value of x is
5
4 2
j
AIEEE 2007
(a) 1
(b) 3
(c) 4
(d) 5
3π
15 If sin−1 x + sin−1 y + sin−1 z =
, then the value of
2
101
101
202
202
∑ ( x303 + y 303) ( x 404 + y 404) is
(x
+ y ) (x
+y )
(a) 0
(b) 1
(c) 2
(d) 3
16 The root of the equation
x − 1
−1 2 x − 1
−1 23
tan−1
+ tan
= tan is
36
x + 1
2x + 1
3
(a) −
8
1
(b) −
2
3
(c)
4
17 The number of solutions of
tan−1 x ( x + 1) + sin−1 x 2 + x + 1 =
(b) 1
(c) 2
1
2
π π
, , then the value of
2 2
x
2
(b) 2 x
(c) 3x
(d) x
π
π
+ θ + tan − + θ = a tan 3 θ, then a
3
3
21 If tan θ + tan
(a) 1/3
(c) 3
(b) 1
(d) None of these
22 If cos −1 x = tan−1 x, then sin (cos −1 x ) is equal to
(a) − x
(b) x 2
23 The real solution of
tan−1 x ( x + 1) + sin−1 x
(a) 2, 3
x
5
(a) 1
2
+ x +1 =
5
4
(d) 3, 1
π
, then the value of x is
j
2
AIEEE 2007
(b) 3
−1
1
x2
π
is
2
(c) −1, 0
(b) 1, 0
24 If sin−1 + cosec −1 =
(d) −
(c) x 3
−1
(c) 4
(d) 5
−1
25 If cosec x + cos y + sec z
≥ α 2 − 2π α + 3π , where α is a real number,
then
(a) x = 1, y = −1
(c) x = 2, y = 1
(b) x = −1, z = −1
(d) x = 1, y = −2
26 The solution of sin−1 x ≤ cos −1 x is
1
(a) −1,
2
1
(c) 1,
2
1
(b) −1,
2
1
(d) 1,
2
27 If m and M are the least and the greatest value of
4
(d)
3
π
is
2
j
(a) 0
(d) | a | ≥
is equal to
AIEEE 2008
4
(d)
17
j
3
(c)
17
AIEEE 2003
(b) all real values of a
3 sin 2 x
tan x
−1
tan− 1
+ tan
is
4
5 + 3 cos 2 x
f ( x ) = | 3 tan−1 x − cos −1( 0) | − cos −1( − 1) is
π π
(d) − ,
2 2
j
1
1
(a) < | a | <
2
2
1
(c) | a | ≤
2
NCERT Exemplar
(d) infinite
(cos −1 x )2 + (sin−1 x )2 , then
(a) 10
(b) 5
M
is equal to
m
(c) 4
(d) 2
28 The number of real solutions of the equation
1 + cos 2x = 2 cos −1 (cos x ) in
π
, π is
2
j
(a) 0
(b) 1
(c) 2
NCERT Exemplar
(d) infinite
237
DAY
31 The value of x for which sin[cot −1(1 + x )] = cos (tan−1 x ),
29 The sum of the infinite series
2 − 1
3 − 2
1
−1
sin−1
+ sin−1
+ sin
2
6
12
n − (n − 1)
is
+ K + sin−1
n (n + 1)
π
(a)
8
π
(b)
4
π
(c)
2
(d) π
30 A root of the equation
10
17
(b) −1
(c) −
7
17
j
1
(a) −
2
(b) 1
(c) 0
JEE Mains 2013
1
2
(d)
32 If 0 < x < 1, then 1 + x 2 [{x cos (cot −1 x )
+ sin(cot −1 x )} 2 − 1]1/ 2 is equal to
(a)
x
1+ x2
(c) x 1 + x 2
(b) x
1+ x2
(d)
33 If x, y and z are in AP and tan−1 x, tan−1 y and tan−1 z are
1 π
17x 2 + 17x tan 2 tan−1 − − 10 = 0 is
5 4
(a)
is
also in AP, then
(d) 1
j
(a) x = y = z
(c) 6x = 3 y = 2 z
AIEEE 2012
(b) 2 x = y = 6z
(d) 6x = 4 y = 3 z
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 If sin−1 x −
x2
x3
x4
x6
+
−... + cos −1 x 2 −
+
−...
2
4
2
4
π
= , where 0 < | x | < 2, then x is equal to
2
(a)
1
2
(c) −
(b) 1
1
2
(d) −1
2 If the mapping f ( x ) = ax + b, a > 0 maps [ − 1, 1] onto
−1
−1
−1
[ 0, 2], then cot [cot 7 + cot 8 + cot 18 ] is equal to
(a) f (− 1)
(b) f (0)
(c) f (1)
(d) f (2)
1
1
−1
3 If S = tan−1 2
+ tan 2
+K
n + n + 1
n + 3 n + 3
1
+ tan−1
, then tan S is equal to
1
+
(
n
+
19
)
(
n
+
20
)
j
JEE Mains 2013
20
401 + 20n
20
(c) 2
n + 20n + 1
n
n 2 + 20n + 1
n
(d)
401 + 20n
(a)
4 If f ( x ) = e
π
cos −1 sin x +
3
7π
(a) f −
=e
4
π
11
3π
7π
(c) f −
= e 12
4
(b)
, then
13 π
8π
(b) f
= e 18
9
11 π
7π
(d) f −
= e 13
4
3π
and f (1) = 2 ,
2
f ( p + q ) = f ( p ) ⋅ f (q ), ∀ p, q ∈ R , then
(x + y + z )
is equal to
x f (1) + y f ( 2 ) + z f ( 3 ) − f (1)
x + y f (2 ) + z f (3 )
5 If sin−1 x + sin−1 y + sin−1 z =
(a) 0
(c) 2
(b) 1
(d) 3
6 If [cot −1 x ] + [cos −1 x ] = 0, where x is a non-negative real
number and [.] denotes the greatest integer function,
then complete set of values of x is
(a) (cos 1, 1]
(c) (cos 1, cot 1)
(b) (cot 1, 1)
(d) None of these
7 cot −1 ( cos α ) − tan−1 ( cos α ) = x , then sin x is equal to
AIEEE 2002
α
(d) cot
2
j
α
α
(a) tan (b) cot 2 (c) tan α
2
2
2
8 If (tan−1 x ) 2 + (cot −1 x ) 2 =
(a) −2
(b) −3
5π 2
, then the value of x is
8
(c) −1
(d) 2
−1
9 The solution set of tan (sin x ) > 1 is
2
1 1
(a) − 1, −
, 1
∪
2 2
(c) (− 1, 1 ) ~ {0}
1 1
(b) −
,
~ {0 }
2 2
(d) None of these
10 If θ and φ are the roots of the equation
8x 2 + 22x + 5 = 0, then
(a)
(b)
(c)
(d)
both sin−1 θ and sin−1 φ are equal
both sec−1 θ and sec−1 φ are real
both tan−1 θ and tan−1 φ are real
None of the above
11 2 tan−1( − 2) is equal to
−3
(a) cos−1
5
π
3
(c) − + tan−1 −
4
2
3
5
3
(d) − π + cot −1 −
4
(b) π + cos−1
12 Let x ∈( 0, 1 ). The set of all x such that sin−1 x > cos −1 x , is
the interval
1 1
(a) ,
2 2
(c) (0, 1)
j
1
(b)
, 1
2
3
(d) 0,
2
JEE Mains 2013
238
ANSWERS
SESSION 1
1 (d)
2 (a)
3 (a)
4 (d)
5 (c)
6 (b)
7 (a)
8 (c)
9 (a)
10 (a)
11 (b)
12 (b)
13 (c)
14 (b)
15 (d)
16 (d)
17 (b)
18 (b)
19 (c)
20 (d)
21 (c)
22 (b)
23 (c)
24 (b)
25 (a)
26 (b)
27 (a)
28 (a)
29 (c)
30 (d)
31 (a)
32 (c)
33 (a)
1 (b)
11 (c)
2 (d)
12 (b)
3 (c)
4 (b)
5 (c)
6 (b)
7 (a)
8 (c)
9 (a)
10 (c)
SESSION 2
Hints and Explanations
SESSION 1
1 cos 33 π = cos 6 π + 3 π = cos 3 π
5
5
5
π 3π
π
= sin −
= sin −
2
10
5
33 π
∴ sin −1 cos
5
π
π
= sin −1 sin −
= −
10
10
2 Since, − π ≤ sin −1 x ≤ π
2
2
π
, 1 ≤ i ≤ 20
2
x i = 1, 1 ≤ i ≤ 20
sin −1 x i =
∴
⇒
20
Thus,
∑
x i = 20
i =1
3 Given, f ( x ) = sin −1 x − 1
For domain of f ( x )
⇒
∴
− 1≤
x−1≤1
0 ≤ (x − 1)≤ 1 ⇒1 ≤ x ≤ 2
x ∈ [1, 2]
4 cos (2 cos −1 x + sin −1 x )
= cos [2(cos −1 x + sin −1 x ) − sin −1 x]
= cos ( π − sin −1 x ) = − cos (sin −1 x )
1
Q x = 1
= − cos sin −1
5
5
2
1
= − cos cos −1 1 −
5
2 6
−1 2 6
= − cos cos
=−
5
5
5 We have, cos[tan −1 {sin(cot −1 x )}]
Let cot −1 x = α
⇒
cotα = x
⇒ cosec α = 1 + x2
⇒
sinα =
⇒
1
7 Given,
1 + x2
α = sin
−1
1
1 + x2
Hence, cos[tan {sin(cot −1 x )}]
1
= cos tan −1 sin sin −1
2
1 + x
−1
1
= cos tan −1
2
1
+
x
x2 + 1
−1
= cos cos
x2 + 2
1
−1
=β
Q let tan
1 + x2
1
=
tan
β
2
+
x
1
2
1
x + 2
=
+
=
sec
β
1
1 + x2
x2 + 1
x2 + 1
cos β =
x2 + 2
=
x2 + 1
x2 + 2
6 Given, tan −1 x − cot −1 x = tan −1 1
3
π
…(i)
⇒ tan x − cot x =
6
π
But tan −1 x + cot −1 x =
…(ii)
2
On adding Eqs. (i) and (ii), we get
π π 2π
2 tan −1 x = + =
6 2
3
π
−1
⇒ tan x =
3
π
⇒
x = tan ⇒ x = 3
3
It has unique solution.
−1
−1
2x
tan −1 y = tan −1 x + tan −1
,
1 − x2
1
where| x | <
3
2x
x + 1 − x2
⇒ tan y = tan
1 − x 2 x
2
1 − x
−1
−1
−1 x + y
Q tan x + tan y = tan 1 − xy ,
−1
−1
xy < 1
3
−1 x − x + 2 x
= tan
1 − x2 − 2 x2
3 x − x3
tan −1 y = tan −1
1 − 3 x2
3 x − x3
1 − 3 x2
Alternate Method
1
1
1
| x |<
⇒−
< x<
3
3
3
π
π
Let x = tanθ ⇒ − < θ <
6
6
∴
tan −1 y = θ + tan −1 (tan 2θ)
= θ + 2θ = 3θ
⇒
y = tan3θ
3 tan θ − tan3 θ
⇒
y =
1 − 3 tan θ
3 x − x3
y =
⇒
1 − 3 x2
⇒
y =
8 Given that, θ = tan −1 a
and
φ = tan −1 b
and
ab = −1
∴tan θ tan φ = ab = − 1
DAY
⇒
⇒
⇒
tan θ = − cot φ
π
tan θ = tan + φ
2
π
θ−φ=
2
−1
−1
−1
9 f ( x ) =| 3 tan x − cos (0 ) | − cos (−1 )
⇒
⇒
−1
π
π
< tan −1 x <
2
2
−3 π
3π
< 3 tan −1 x <
2
2
π
−2 π < 3 tan −1 x − < π
2
π
0 ≤ 3 tan −1 x −
< 2π
2
π
− π ≤ 3 tan −1 x −
− π< π
2
−1
cosec (cosec x )
= x, ∀ x ∈ (−∞, − 1] ∪ [1, ∞ )
⇒ cos (cos −1 x ) = cosec (cosec −1 x ) for
x = ± 1 only.
Hence, there are two roots.
11 Since, cosec 5 = tan −1 3
3
4
2
−1 3
+ tan −1
∴ cot tan
4
3
3 + 2
= cot tan −1 4 3
1
1 −
2
−1
= cot tan −1
⇒ sin
−1
⇒ sin
−1
⇒ sin −1
⇒ sin
sin φ =
1 + cot φ
2
⇒
19 Given that, sin −1 x = 2sin −1 a
Q
⇒
x
−1 3
= sin
5
5
⇒
x=3
sin −1 x = sin −1 y = sin −1 z =
π
2
2
⇒
⇒
⇒
x= y = z=1
( x101 + y 101 ) ( x202 + y 202 )
∴ Σ 303
( x + y 303 ) ( x 404 + y 404 )
(1 + 1) (1 + 1)
=Σ
= Σ1 = 3
(1 + 1) (1 + 1)
π
π
≤ sin −1 x ≤
2
2
π
π
− ≤ 2sin −1 a ≤
2
2
π
π
− ≤ sin −1 a ≤
4
4
π
π
sin − ≤ a ≤ sin
4
4
1
1
−
≤ a≤
2
2
1
|a| ≤
2
−
tan x
3 sin 2 x
−1
20 tan −1
+ tan
= tan −1
x − 1 2x − 1
+
x + 1 2x + 1
x
−
1
2
x
−
1
1 −
x + 1 2 x + 1
=
23
= tan −1
36
2
5
tan x
6 tan x
−1
= tan −1
+ tan
2
4
8 + 2 tan x
2 x2 − 1 23
=
3x
36
⇒
3 tan x
tan x
−1
= tan −1
+ tan
2
4
4 + tan x
⇒ 24 x − 12 − 23 x = 0
4
3
x= ,−
⇒
3
8
But x cannot be negative.
4
∴
x=
3
2
1 − x2
1
− 1 ⇒ tan θ =
x2
x
1
2
1 − x2
x
2
= sin sin −1
5
17 For existence, x ( x + 1) ≥ 0
and
⇒
⇒
x2 + x + 1 ≤ 1
x2 + x ≤ 0
x( x + 1) ≤ 0
4
5 + 3 cos 2 x
tan x + tan −1
4
6 tan x
2
1
+
tan
x
2
3 ( 1 − tan x )
+
5
1 + tan2 x
16 tan −1
Now, tan (cos −1 x ) = sin cot −1
⇒ tan tan −1
∴ I max
⇒
1
Let cos x = θ ⇒ sec θ =
x
⇒
tan θ = sec2 θ − 1
tan θ =
5
4
2
x
π
−1 4
+ sin =
5
5
2
x
π
=
−1 4
− sin
5
5
2
x = cos −1 4
5
5
π2
+ 2 (sec −1 x )2
4
π
π2
π2
− (sec −1 x ) +
−
2
16 16
2
π2
π
=
+ 2 sec −1 x −
8
4
2
2
9π
π
5π2
=
+2
=
8
4
16
=
15 Given, sin x + sin −1 y + sin −1 z = 3 π
−1
⇒
−1
2
1
= (sec −1 x + cosec −1 x )2
− 2 sec −1 x cosec −1 x
π2
π
=
− 2 sec −1 x − sec −1 x
2
4
−1
12 Let cot −1 1 = φ ⇒ 1 = cot φ
⇒
x − tan
∴
17
12
1
2
17 = 6
6 17
2
18 Let I = (sec −1 x )2 + (cosec −1 x )2
−1
14 Q sin −1 x + cosec −1 5 = π
10 cos (cos −1 x ) = x, ∀ x ∈[− 1, 1] and
= cot tan −1
From Eqs. (i) and (ii), we get
x( x + 1) = 0 ⇒ x = 0, − 1
But x = −1 is not satisfied the given
equation.
y = 0⇒ x = y
π
−1
Also, cos x + cos −1 y =
2
π
π
⇒ 2 cos −1 x = ⇒ cos −1 x =
2
4
1
1
2
⇒
⇒ x =
x=
2
2
3
2
2
Hence, x + xy + y = 3 x2 =
2
We know that, −
⇒
2
=
⇒ (1 − x2 )5 = 2x
x
5
On squaring both sides, we get
(1 − x2 )5 = 4 x2
5
⇒
9 x2 = 5 ⇒ x = ±
3
13 Qtan
π
= 3 tan −1 x − − π
2
⇒
1 − x2
⇒
= tan −1
…(i)
…(ii)
tan x
3 tan x
+
4
4
+ tan2 x
2
3 tan x
1−
4(4 + tan2 x )
tan x 3 tan x
< 1
⋅
as
4 tan2 x
4
16 tan x + tan3 x
= tan −1
16 + tan2 x
= tan −1 (tan x ) = x
240
21 tan θ + tan π + θ + tan − π + θ
= a tan 3 θ
3 + tan θ
x
4
sin −1 = cos −1
5
5
x
3
⇒ sin −1 = sin −1
5
5
1 − 3 tan θ
∴
3
⇒ tan θ +
+
tan θ − 3
1+
3 tan θ
3
RHS = α2 − 2 π α + 3 π
⇒
⇒
x = tan
sin θ
cos θ =
cos θ
cos θ = tan θ ⇒
⇒ cos 2 θ = sin θ
⇒ sin2 θ + sin θ − 1 = 0
−1± 1+ 4
⇒ sin θ =
2
5−1
⇒ sin θ =
2
5−1
2
2
∴ x = cos θ = sin θ =
2
and sin (cos −1 x ) = sin θ
5−1
=
= x2
2
π
x + x+1=
2
2
1
1 + (x + x )
x2 + x + 1 =
1
⇒ cos −1
1 + ( x2 + x )
=
π
2
− sin −1
⇒
π
− 1 ≤ x ≤ sin
4
1
x ∈ − 1,
2
x2 + x + 1
x2 + x + 1
1
x2 + x + 1
=
⇒
x2 + x + 1
x2 + x + 1
x + x + 1 = 1 ⇒ x = − 1, 0
2
5
4
x
4
π
⇒ sin −1 + sin −1 =
5
5
2
x
π
4
sin −1 =
− sin −1
5
5
2
2
= tan
π2
+ 2 (sin −1 x )2 − π sin −1 x
4
2
π
π2
=
+ 2 sin −1 x −
8
4
π2
5π2
Here, m =
,M =
8
4
M
∴
= 10
m
=
28 Given, 1 + cos 2 x = 2 cos −1 (cos x )
∴
2cos 2 x = 2 x
⇒
2 |cos x | = 2 x
π
For x ∈ , π , |cos x| = − cos x
2
− 2 cos x = 2 x
⇒
− cos x = x
−1
∴ S ∞ = tan
r − r −1
1 + r r − 1
−1
r − tan −1 (r − 1)]
−1
n − tan −1
0
n −0
∞ =
π
2
1
2×
−1 1
5 = 5
30 Now, tan 2 tan =
1
5
12
1−
25
Given equation can be rewritten as
π
1
17 x2 − 17 x tan − 2 tan −1 − 10 = 0
5
4
5
1−
2
12
− 10 = 0
⇒ 17 x − 17 x ⋅
5
1+
12
⇒
17 x2 − 7 x − 10 = 0
⇒
( x − 1) (17 x + 10) = 0
Hence, x = 1 is a root of the given
equation.
1
31 sin sin −1
(1 + x )2 + 1
1
= cos cos −1
2
1+ x
1
1
=
2
(1 + x ) + 1
1 + x2
2
27 π − sin −1 x + (sin −1 x )2
2
24 Since, sin −1 x + cosec −1 5 = π
⇒
π
2
1
−1
= cos −1
⇒
Qsin −1 x + cos −1 x = π , ∀x ∈ [−1,1]
2
π
π
π
−1
−1
⇒ sin x ≤ ⇒ − ≤ sin x ≤
4
2
2
− π , π
−1
Q
range
of
sin
x
is
2 2
2
+ sin −1
⇒ cos
π
≥ 2 sin −1 x
2
⇒
n
∑ [tan
= tan −1
…(i)
π π
QLHS, cosec x ∈ − , − {0}
2 2
cos −1 y ∈ [0, π]
π
and sec −1 z ∈ [0, π] −
2
5π
LHS ≤
…(ii)
∴
2
From Eqs. (i) and (ii), we get only
possibility is sign of equality
x = 1, y = − 1, z = − 1
⇒
−1
r =1
26 Given, cos −1 x ≥ sin −1 x
23 Given, tan −1 x ( x + 1 ) + sin −1
⇒ cos −1
=
−1
x=θ
n
∑ tan
r =1
2
π
5π 5π
≥
+
2
2
2
= α −
x = cos θ = tan θ
∴ Sn =
π
π
π
α +
+ 3π −
2
2
2
= α2 − 2
⇒ 3 tan 3 θ = a tan 3 θ
⇒ a=3
22 Let cos
x=3
cosec −1 x + cos −1 y + sec −1 z ≥ α2
− 2π α + 3π
3(3 tan θ − tan3 θ )
⇒
= a tan 3 θ
1 − 3 tan2 θ
−1
r − (r − 1)
r (r + 1)
r − (r − 1)
= tan −1
1 + r ( r − 1 )
29 Q T r = sin −1
25 Given,
= a tan 3 θ
8 tan θ
= a tan 3 θ
⇒ tan θ +
1 − 3 tan2 θ
−1
∴
cos x = − x
Hence, no solution exist.
⇒
⇒
⇒
⇒
(1 + x )2 + 1 = 1 + x2
2x + 1 = 0
1
x=−
2
32 We have, 0 < x < 1
Now, 1 + x2 [{ x cos(cot −1 x )
+ sin(cot −1 x )}2 − 1]1 /2
= 1+ x
2
x
x
+
1 + x2
1 /2
2
− 1
1 + x2
1
1 /2
2
1 + x2
= 1 + x
− 1
1 + x2
2
= 1 + x2 [1 + x2 − 1)1 /2
= x 1 + x2
DAY
33 Since, x, y and z are in AP.
∴
2y = x + z
Also, tan −1 x, tan −1 y and tan −1 z are in
AP.
∴2 tan −1 y = tan −1 x + tan −1 z
⇒
⇒
2y
x+ z
tan −1
= tan −1
2
1 − y
1 − xz
x+ z
x+ z
=
1 − y 2 1 − xz
[Q2y = x + z]
⇒
y 2 = xz
Since x, y and z are in AP as well as in
GP.
∴
x= y =z
= cot tan −1
65
195
= 1 + 2 = f (2)
[Q f ( x ) = x + 1]
(n + 1 ) − (n + 0)
1
+ (n + 0) (n + 1 )
(n + 2) − (n + 1 )
+ tan
1 + (n + 1 ) (n + 2)
(n + 20) − (n + 19)
+ K + tan −1
1 + (n + 19) (n + 20)
−1
2
3
2
4
x
2x
=
x 2+ x
1+
2
4
x
x6
2
and x −
+
−K
2
4
2
x
2 x2
=
=
2
x
2 + x2
1+
2
π
Q sin −1 α + cos −1 α =
2
2x
2 x2
∴
=
,x ≠ 0
2 + x 2 + x2
=
2 + x2 = 2 x + x2
x=1
= tan −1 (n + 1 ) − tan −1 n
⇒
f (− 1) = 0 and f (1) = 2
⇒
− a+ b = 0
∴
f ( x) = x + 1
Now, cot[cot −1 7 + cot −1 8 + cot −1 18]
1
1
tan −1 + tan −1
7
8
= cot
1
+ tan −1
18
1+ 1
−1
−1 1
7
8
= cot tan
+ tan
18
1 − 1⋅ 1
7 8
15
1
= cot tan −1 + tan −1
55
18
3
1
= cot tan −1 + tan −1
11
18
and 0 < cot −1 x < π
Given, [cot −1 x] + [cos −1 x] = 0
⇒ [cot −1 x] = 0 and [cos −1 x] = 0
⇒ 0 < cot −1 x < 1 and 0 ≤ cos −1 x < 1
∴ x ∈ (cot 1, ∞ )
and
x ∈ (cos 1, 1 ) ⇒ x ∈(cot 1, 1 )
7 Given that,
cot −1 ( cos α ) − tan −1
( cos α ) = x …(i)
We know that,
cot −1 ( cos α ) + tan −1 ( cos α ) =
n + 20 − n
= tan −1
1 + n(n + 20)
Qcot −1 x + tan −1
20
= tan −1 2
n + 20n + 1
20
⇒ tan S = tan tan −1 2
n + 20n + 1
20
n2 + 20n + 1
4 Given, f ( x ) = e
⇒
π
cos −1 sin x +
3
cos
8π
f
=e
9
=e
a+ b =2
a=b =1
0 ≤ cos −1 x ≤ π
= tan −1 (n + 20) − tan −1 n
∴ f ′( x ) = a > 0
So, f ( x ) is an increasing function.
3
=3−1=2
1+ 1+ 1
+ tan −1 (n + 2) − tan −1 (n + 1 )
+ ... + tan −1 (n + 20) − tan −1 (n + 19)
⇒ tan S =
2 Q f ( x ) = ax + b
Then,
6Q
3 S = tan −1
1 Now, x − x + x − …
and
=1+ 1+ 1−
1
= cot tan −1 = cot(cot −1 3) = 3
3
SESSION 2
⇒
∴
Then, f (2) = f (1 )f (1 ) = 2 ⋅ 2 = 4
and put p = 1, q = 2
Then, f (3) = f (1 )f (2) = 2 ⋅ 22 = 8
x+ y + z
∴ x f(1 ) + y f(2 ) + z f(3 ) − f(1 )
x
+ y f(2 ) + z f(3 )
3 + 1
−1
= cot tan 11 18
3
1
1 −
⋅
11 18
−1
11 π
cos −1 sin
9
cos
8π
f
=e
9
⇒
8π π
sin
+
9
3
−1
13 π
cos
18
cos
7π
Also, f −
=e
4
=e
−1
=e
13 π
18
7π π
sin −
+
4
3
17 π
cos −1 sin −
12
=e
cos −1 cos
π
π
12
= e 12
5 Q− π ≤ sin −1 x ≤ π , − π ≤ sin −1 y ≤ π
2
2
2
π
π
≤ sin −1 z ≤
2
2
Given that,
2
and −
sin −1 x + sin −1 y + sin −1 z =
3π
2
which is possible only when
sin −1 x = sin −1 y = sin −1 z =
⇒ x= y = z=1
Put p = q = 1
π
2
π
3
2
…(ii)
π
x=
2
On adding Eqs. (i) and (ii), we get
π
2cot −1 ( cos α ) =
+ x
2
π
x
⇒
cos α = cot +
4
2
x
cot − 1
2
cos α =
⇒
x
1 + cot
2
x
x
cos − sin
2
2
⇒
cos α =
x
x
cos + sin
2
2
On squaring both sides, we get
x
x
cos 2 + sin2
2
2
⇒
cos α =
x
x
cos 2 + sin2
2
2
x
x
− 2sin cos
2
2
x
x
+ 2sin cos
2
2
1 − sin x
⇒
cos α =
1 + sin x
α
1 − tan2
2 = 1 − sin x
⇒
2 α
1 + sin x
1 + tan
2
On applying componendo and
dividendo rule, we get
α
sin x = tan2
2
242
8 Given,
(tan −1 x )2 + (cot −1
⇒ (tan −1 x + cot −1
5π2
x )2 =
8
x )2 − 2 tan −1
x ⋅ cot −1 x =
or −
⇒
5π2
8
⇒
π
π
< sin −1 x < −
2
4
1
1
x∈
, 1 or x ∈ − 1, −
2
2
1
1
x ∈ − 1, −
, 1
∪
2
2
⇒
2
π
π
⇒ − 2 tan −1 x − tan −1 x
2
2
5π2
π
−1
−1
=
Q tan x + cot x =
8
2
π2
π
⇒
− 2 ⋅ tan −1 x
4
2
5π2
+ 2 (tan −1 x )2 =
8
3 π2
−1
2
−1
⇒ 2 (tan x ) − π tan x −
=0
8
π 3π
⇒ tan −1 x = − ,
4 4
π
⇒ tan −1 x = −
4
⇒
x= −1
neglecting tan −1 x
3π
as principal value of
=
4
π π
tan −1 x ∈ − ,
2 2
9 tan2 (sin −1 x ) > 1
⇒
π
π
< sin −1 x <
4
2
10 8 x2 + 22 x + 5 = 0
1
5
x= − ,−
⇒
4
2
1
5
Q−1 < − < 1 and − < − 1
4
2
1
5
∴ sin −1 − exists but sin −1 −
4
2
does not exist.
5
1
sec −1 − exists but sec −1 − does
2
4
−3
− 2θ = cos −1
5
3
= π − cos −1
5
3
5
−1 4
π + tan
3
3
π + cot −1
4
π
3
π+
− tan −1
2
4
π
−1 3
− tan
2
4
π
3
−1
+ tan −
4
2
⇒
2θ = − π + cos −1
⇒
2θ = −
=−
=−
=−
=−
12
Y
not exist.
So,
and
tan −1
tan −1
− 1
4
− 5 both exist.
2
11 Let tan −1 (− 2) = θ
⇒
⇒
π
y = cos –1 x
X′
(–1, 0)
π
2
y = sin–1 x
π
4
O
tan θ = − 2
π
θ ∈ − , 0
2
⇒
2θ ∈ (− π, 0 )
Now, cos ( − 2θ) = cos 2θ
1 − tan2 θ
=
1 + tan2 θ
−3
=
5
X
1,0
(1, 0)
2
Y′
1
∴ sin −1 x > cos −1 x, ∀ x ∈
, 1
2
243
DAY
DAY TWENTY THREE
Unit Test 3
(Trigonometry)
1 The range of f ( x ) = sin−1 x + tan−1 x + sec −1x , is
π
(a) ,
4
π
(c) ,
4
3π
4
3π
4
π 3π
(b) ,
4 4
(a) 3
(c) 5
(d) None of these
2 From a point a m above a lake the angle of elevation of a
cloud is α and the angle of depression of its reflection is
β. The height of the cloud is
a sin (α + β)
m
sin (α − β)
a sin (β − α)
(c)
m
sin (α + β)
(b)
(a)
2n
a sin(α + β)
m
sin( β − α)
(d) None of these
i =1
(a) n
n (n + 1)
(c)
2
(d) None of these
4 The value of x for which cos (cos 4) > 3x − 4x, is
2
2 + 6π − 8
(a) 0,
3
2 − 6π − 8
(b)
, 0
3
(a) α = 0
x + cos
value of y is
9 If cos
(a)
1
2
(b)
−1
(c) positive
1 − x + cos
3
2
−1
(d) zero
1 − y = π, then the
(c) 1
(d)
1
4
9
40
π
(c) α =
4
(b)
9
41
(c)
3
10
(d)
21
31
11 The angular depressions of the top and the foot of a
chimney as seen from the top of a second chimney,
which is 150 m high and standing on the same level as
4
the first are θ and φ, respectively. If tan θ = and
3
5
tan φ = , then the distance between their tops is
2
(b) 110 m
(d) None of these
60° and at a place Q due South of it is 30°. If the distance
PQ be 200 m, then the height of the hill is
π
(d) β =
3
6 The angles of elevation of the top of a tower at the top
and the foot of a pole of height 10 m are 30° and 60°,
respectively. The height of the tower is
(a) 10 m
(c) 20 m
(b) negative
−1
12 The elevation of the hill from a place P due West of it is
6π − 8
3
5 If α ≤ sin−1 x + cos −1x + tan −1x ≤ β, then
π
(b) β =
2
(a) − 3
(a) 120 m
(c) 100 m
(c) (− 2, 2)
6π − 8 2 +
,
3
when α , β and γ are real numbers satisfying
α + β + γ = π , is
(a)
(b) 2n
−1
2 −
(d)
8 The minimum value of the expression sin α + sin β + sin γ,
5 cos θ + 4 sin θ = 3, then cos (α + β ) is equal to
3 If ∑ cos xi = 0, then ∑ xi is equal to
i =1
(b) 4
(d) None of these
10 If α and β are the roots of the equation
2n
−1
7 The maximum value of 12 sin θ − 9 sin2 θ is
(b) 15 m
(d) None of these
(a) 109.54 m
(c) 110.6 m
(b) 108.70 m
(d) None of these
7π
2π
1
− sin
cos
is equal to
5
5
2
13 cos −1
13 π
20
33 π
(c)
20
(a)
(b)
21 π
20
(d) None of these
244
14 sec 2 θ =
4 ab
, where a , b ∈ R is true if and only if
(a + b ) 2
(a) a + b ≠ 0
(c) a = b
(b) a = b, a ≠ 0
(d) a ≠ 0, b ≠ 0
(a) 5
15 If x = cos −1(cos 4) and y = sin−1(sin 3), then which of the
following conditions holds?
(a) x − y = 1
(c) x + 2 y = 2
(b) x + y + 1 = 0
(d) tan(x + y) = − tan 7
16 If log2 x ≥ 0 , then log1/ π sin−1
2x
+ 2 tan−1 x is equal
1+ x2
(b) 0
(d) None of these
5 sin 2A + 3 sin A + 4 cos A is equal to
24
5
48
(d)
5
(b) −
(a) 0
24
5
2
(b)
3 3
2
(c)
9 3
x
x
is equal to
= cosec x − sin x , then tan2
2
2
(a) 2 − 5
(c) (9 − 4 5 ) (2 +
5)
(b) 5 − 2
(d) (9 + 4 5 ) (2 − 5 )
θ
(a) 2 cos n
2
θ
(c) 2 cos n + 1
2
θ
(b) 2 cos n − 1
2
(d) None of these
sin2 A : sin2 B : sin2 C is equal to
9 5 8
:
:
10 10 10
9 8 7
(c)
:
:
10 10 10
(a)
1
3 3
(d) None of these
(b)
9 7 8
:
:
10 10 10
(d) None of these
29 If 0 ≤ x ≤ 3π, 0 ≤ y ≤ 3π and cos x ⋅ sin y = 1, then the
19 If sin x + a sin x + 1 = 0 has no real number solution,
2
possible number of values of the ordered pair ( x , y ) is
(a) 6
then
(a) | a | ≥ 2
(c) | a | < 2
(b) | a | ≥ 1
(d) None of these
n+1
(a)
tan α
n −1
n
(c)
tan α
1+ n
(a)
(a)
(b)
(c)
(d)
21 sin6 x + cos 6 x lies between
(c) 0 and 1
1
and 2
4
π
22 If n =
, then tan α ⋅ tan 2 α ⋅ tan 3 α K tan ( 2n − 1) α is
4α
equal to
23 The ratio of the greatest value of 2 − cos x + sin2 x to its
least value is
(b)
9
4
(c)
13
4
(d)
2
3
2
sin θ, then the value of (m + n ) 2 / 3 + (m − n ) 2 / 3 is
(b) a 2 / 3
(c) a 3 / 2
nπ
2
(c) 3nπ
(d) None of these
x
π
sin2 x = x 2 + x −2 , 0 < x ≤ has
2
2
no real solution
a unique real solution
finitely many real solutions
infinitely many real solutions
subtend the same angle α at a point O on the road. If the
height of the largest pole is h and the distance of the foot
of the smallest pole from O is a, the distance between
two consecutive poles is
h cos α − a sin α
(n − 1) sin α
h cos α − a sin α
(c)
(n + 1) sin α
(d) 4 a 2 / 3
h cos α + a sin α
(n − 1) sin α
a cos α − h sin α
(d)
(n − 1) sin α
(b)
33 The solution of the equation
17
4
24 If m = a cos θ + 3 a cos θ sin θ, n = a sin θ + 3 a cos θ
3
(b)
(a)
(b) −1
(d) None of these
1
4
(d) 15
32 The n poles standing at equal distance on a straight road
(d) None of these
(a) 1
(c) ∞
nπ
3
31 The equation 2 cos 2
(d) None of these
(b)
(c) 8
1 + sin4 2x = cos 2 6x is
1+ n
(b)
tan α
1− n
1
and 1
4
(b) 12
30 The general solution of the equation
20 If sin θ = n sin ( θ + 2 α ), then tan ( θ + α ) is equal to
(a) 2a 2 / 3
(d) None of these
28 In ∆ABC, tan A + tan B + tan C = 6, tan B tan C = 2 , then
18 The minimum value of 27 cos x + 81 sin x is
(a)
a 2 − 5a + 6
a2
a 2 + 5a + 6
a2
(where, number of 2’ s is n) is equal to
17 If A lies in the third quadrant and 3 tan A − 4 = 0, then
(a)
(b)
27 If 0° < θ < 180° , then 2 + 2 + 2 + ... + 2 (1 + cos θ )
(a) log1 / π (4 tan−1 x)
(c) −1
(a)
(c)
26 If tan
to
(c)
π
1
cosec 2 α = 0, 0 < α < , then
a
2
cos 2 α + 5 sin α cos α + 6 sin2 α is equal to
25 If a sin2 α −
8
tan−1( x + 1) + tan−1 ( x − 1) = tan−1 is
31
1
4
(c) − 8
(a)
1
or − 8
4
(d) None of these
(b)
245
DAY
π
2
I. The general value of θ satisfying the equations
sin2 θ = sin2 α, cos 2 θ = cos 2 α and tan2 θ = tan2 α is
given by θ = n π ± α
II. The general value of θ satisfying equations
sin θ = sin α and cos θ = cos α simultaneously is given
by θ = 2nπ ± α , n ∈ Z .
34 The solution of cos −1 x 3 + cos −1 x = , is
(a) −
(c)
1
2
1
2
(b) −
1
1
or
2
2
(d) None of these
35 The number of solutions of the equation
tan x + sec x = 2 cos x lying in the interval [ 0, 2 π ], is
(a) 0
(b) 1
(c) 2
(d) 3
36 If cot 2 x + cosec x − a = 0 has atleast one solution, then
complete set of value of a is
(a) [− 1, ∞ )
(b) (− 1, 1)
(c) (− ∞, 1)
(d) (− ∞, 1]
37 From the top of the cliff 400 m high the top of a tower
was observed at an angle of depression 45° and from
the foot of the tower the top of the cliff was observed at
an angle of elevation 60°. The height of the tower is
(a) 169.06 m
(c) 180.0 m
(b) 179. 2 m
(d) None of these
38 If cos 5 θ = P cos θ − 20 cos 3 θ + Q cos 5 θ + R, then
P + Q + R is equal to
(a) 21
(b) 18
(c) 15
(d) 31
(b) obtuse angled
(d) isosceles
40 If the area of a ∆ ABC be λ, then a 2 sin 2 B + b 2 sin 2A is
equal to
(a) 2 λ
(c) 4λ
(c) 2
(d) −1
42 If sin x (1 + sin x ) + cos x (1 + cos x ) = A and
sin x (1 − sin x ) + cos x (1 − cos x ) = B, then the value of
A 2 − 2A − sin 2x = B 2 + 2B − sin 2x is
(a) 1
(b) 2
(c) 3
(d) 0
43 If the sides a, b and c of a ∆ABC are roots of the equation
x 3 − 15x 2 + 47x − 82 = 0, then the value of
cos A cos B cos C
is
+
+
a
b
c
(a)
225
164
(b)
131
164
(c)
131
82
(d)
169
82
Statement I The number of solutions of the pair of
equations
2 sin2 θ − cos 2θ = 0
and
2 cos 2 θ − 3 sin θ = 0
in the interval [ 0, 2π ] is two.
Statement II If 2 cos 2 θ − 3 sin θ = 0, then θ cannot lie in
III or IV quadrant.
θ
2
49 Statement I If 2 sin = 1 + sin θ + 1 − sin θ,
θ
π
3π
lies between 2nπ + and 2nπ +
.
2
4
4
π
3π
θ
Statement II If ≤ θ ≤
, then sin > 0 .
4
4
2
50 Suppose a person is standing on a ground and see the
angle c is
(a) 30° or 150°
(c) 60° or 120°
tower, then they make an angle of elevation.
(b) 45° or 135°
(d) None of these
45 Let f (θ, α ) = 2 sin2 θ + 4 cos(θ + α )
π
sin θ sin α + cos 2 (θ + α ). Then, the value of f ,
3
(b) 1
(c) 2
π
is
4
(d) 3
46 Which among the following is/are correct statement/
statements?
1 1
Statement II sin−1( 2x 1 − x 2 ) = 2 sin−1 x , if x ∈ −
,
.
2 2
then
44 If in a ∆ABC, a 4 + b 4 + c 4 − 2b 2c 2 − 2c 2a 2 = 0. Then, the
(a) 0
1
≤ x ≤ 1, then
2
x
3 − 3x 2
π
cos −1 x − sin−1 +
is equal to − .
2
2
3
said to be solution of an equation.
is
(b) 1
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
48 Suppose any angle is satisfied both equations, then it is
(b) λ
(d) None of these
41 If cos 2x + 2 cos x = 1, then the value of sin2 x ( 2 − cos 2 x )
(a) 0
Directions (Q. Nos. 47-50) Each of these questions contains
two statements : Statement I and Statement II. Each of these
questions also has four alternative choices, only one of which
is the correct answer. You have to select one of the codes
(a ), (b), (c) and (d ) given below.
47 Statement I If
cos A cos B cos C
, then the triangle is
39 In ∆ABC, if
=
=
a
b
c
(a) right angled
(c) equilateral
(a) I is correct
(b) II is correct
(c) Both I and II are correct (d) Both I and II are incorrect
Statement I A tower subtends angles α , 2α and 3 α
respectively at points A , B and C all lying on a horizontal
line
through
the
foot
of
tower,
then
AB
AB : BC = sin 3 α : sin α.
a
b
c
Statement II In ∆ABC, if
, then ∆ABC
=
=
sin A sin B sin C
is an equilateral triangle.
246
ANSWERS
1 (c)
2 (b)
3 (b)
4 (d)
5 (a)
6 (b)
7 (b)
8 (c)
9 (c)
10 (b)
11 (c)
12 (a)
13 (d)
14 (b)
15 (d)
16 (c)
17 (a)
18 (c)
19 (c)
20 (b)
21 (a)
22 (a)
23 (c)
24 (a)
25 (d)
26 (c)
27 (a)
28 (a)
29 (a)
30 (b)
31 (a)
32 (a)
33 (a)
34 (c)
35 (c)
36 (a)
37 (a)
38 (a)
39 (c)
40 (c)
41 (b)
42 (d)
43 (b)
44 (b)
45 (a)
46 (a)
47 (b)
48 (a)
49 (b)
50 (c)
Hints and Explanations
⇒ 3 x2 − 4 x − (2 π − 4) < 0
1 f ( x ) = sin −1 x + tan −1 x + sec −1 x,
Hence, domain of f ( x ) is ± 1. So, the
π 3π
range is { f (1), f (− 1)} i.e. ,
.
4 4
2 In ∆ PMC, tanα =
h−a
PM
⇒
am
h
M
β
Q
3
< x<
2+
6π − 8
3
5 Since, sin x + cos x + tan x = π
−1
−1
Now, −
π
π
≤ tan−1 x ≤
2
2
⇒
0≤
∴
α = 0, β = π
h–a
α
6π − 8
−1
2
+ tan−1 x
C
P
2−
7 12 sin θ − 9 sin2 θ
O
π
+ tan−1 x ≤ π
2
6 Let AB and CD be the pole and tower,
respectively.
h
D
= − (3 sin θ − 2)2 + 4 ≤ 4
Hence, maximum value is 4.
8 Since, sin α + sin β + sin γ
α
β
γ
cos cos
2
2
2
α β
γ
Also, since each of , and is less
2 2
2
π
α
β
γ
than . So, cos , cos and cos are all
2
2
2
2
positive.
= 4cos
Hence, minimum value of
α
β
γ
4cos cos cos is positive.
2
2
2
9 ∴ cos −1 x + cos −1 1 − x
= cos −1 [ x
C'
⇒
…(i)
PM = (h − a) cotα
h+ a
and in ∆ PMC ′, tan β =
PM
⇒ h + a = PM tan β
∴ h = (h − a) cot α tan β − a
[from Eq. (i)]
⇒ h (1 − cot α tan β ) = − a(cot α tan β + 1)
a(sin α cos β + cos α sin β )
⇒
h=
sin β cos α − sin α cos β
a sin (α + β )
⇒ h=
m
sin (β − α )
3 Since, 0 ≤ cos −1 x i ≤ π
∴
∴
∴
cos −1 x i = 0 ∀ i
x i = 1, ∀ i
2n
∑x
i
= 2n
i =1
4 Now, cos −1 (cos 4) = cos −1 [cos(2 π − 4)]
= 2π − 4
∴
cos −1 (cos 4) > 3 x2 − 4 x
∴
2 π − 4 > 3 x2 − 4 x
E
30°
B
10 m
60°
C
A
CD
In ∆ ACD, tan 60° =
AC
CD
...(i)
⇒ AC =
3
DE
DE
In ∆ DBE, tan30° =
=
BE
CA
1
DE
⇒
=
3 AC
1
DE
[from Eq. (i)]
⇒
=
3 CD/ 3
DE + EC
CD
=3 ⇒
=3
⇒
DE
DE
EC 10
=
= 5m
⇒ DE =
2
2
∴ CD = DE + EC = 10 + 5 = 15 m
1 − x − 1 − x x]
π
= cos (0) =
2
π
∴ π = + cos −1 1 − y
2
π
= cos −1 1 − y ⇒ 1 − y = 0
⇒
2
∴
y =1
−1
10 Since, α and β are the roots of the
equation
5 cos θ + 4sin θ = 3.
∴
5 cos α + 4sin α = 3
…(i)
and
5 cos β + 4sin β = 3
…(ii)
On subtracting Eq. (ii) from Eq. (i), we
get
5(cos α − cos β ) + 4(sin α − sin β ) = 0
α +β
α −β
⇒ − 5 × 2 ⋅ sin
⋅ sin
2
2
α +β
α −β
+ 4 ⋅ 2 ⋅ cos
⋅ sin
=0
2
2
α +β
α + β
⇒ 4 cos
− 5 sin
=0
2
2
247
DAY
⇒ tan
∴
α +β 4
=
2
5
α + β
1 − tan2
2
cos (α + β) =
α + β
1 + tan2
2
16
9
1−
9
25
25
=
=
=
16 41 41
1+
25 25
A
150 – h
R
h
B
= cos
Now, PR = PQ − RQ = 150 − h
PR
In ∆ PAR, tanθ =
AR
4 150 − h
⇒
=
3
x
and in ∆ PBQ, tan φ =
∴
= cos
= cos
PQ
BQ
…(ii)
150 − h 4 2
8
= × =
3 5 15
150
[dividing Eq. (i) by Eq. (ii)]
h
8
⇒ 1−
=
⇒ h = 70 m
150 15
From Eq. (i),
⇒
= cos
…(i)
5 150
=
2
x
⇒
1
h
=
⇒ y =h 3
y
3
2
Q
x
AB
BQ
13 Now, 1 cos 7 π − sin 2 π
h
φ
∴ 5 sin 2 A + 3 sin A + 4 cos A
24 12 12
=
−
−
=0
5
5
5
In right angled ∆ PBQ,
PQ 2 = PB 2 + BQ 2 = x 2 + y 2
h2
10
+ 3 h2 = h2
⇒ (200) 2 =
3
3
3
⇒
h2 = 200 2 ×
10
3
∴
AB = 200
= 109.54 m
10
P
d
⇒
⇒
chimney, AB = h.
150 − 70 4
=
x
3
x = 60 m and
In ∆ PAR, AP 2 = AR2 + PR2
5
5
π
7π
π
2π
cos
− sin sin
4
5
4
5
π
7π
π
7π
cos
+ sin sin
4
5
4
5
7 π − π = cos 23 π
5
20
4
2 π − 17 π = cos 17π
20
20
1
7π
2π
17 π
∴ cos −1
− sin =
cos
5
5
20
2
14 Since, sec 2 θ ≥ 1 ⇒
4 ab
≥ 1 and
(a + b ) 2
a, b ≠ 0
(a − b ) 2
≥0
(a + b ) 2
⇒
a, b ≠ 0 and −
⇒
⇒
a, b ≠ 0 and − (a − b ) 2 ≥ 0
a = b and a ≠ 0
15 Given,
PR = 150 − 70 = 80 m
x = cos −1 (cos 4)
⇒
x = cos −1 cos (2 π − 4)
⇒
x = 2 π − 4 and
⇒
y = sin −1 sin ( π − 3) ⇒ y = π − 3
y = sin −1 (sin 3)
⇒
d 2 = 60 2 + 80 2
∴
x + y = 3π − 7
∴
d = 100 m
∴
tan ( x + y ) = − tan 7
12 Let the height of the hill be h and let A
be its top.
16 Since, log 2 x ≥ 0 ⇒ x ≥ 1
For x ≥ 1, we have
2x
sin −1
= π − 2 tan −1 x
2
1 + x
A
∴
h
B
60°
x
P
30°
y
200 m
Q
Since, BQ and BP represents South and
West, respectively.
2 tan A
24
=
1 + tan2 A 25
∴ sin 2 A =
Again, in ∆ AQB, tan30° =
11 Let AR = x and the height of the
θ
AB
PB
h
h
3=
⇒ x=
x
3
In ∆ APB, tan 60° =
2x
log 1 / π sin −1
+ 2 tan −1 x
1 + x2
= log 1 / π { π − 2 tan −1 x + 2 tan −1 x}
= log 1 / π π = − 1
17 Q 3 tan A − 4 = 0 ⇒ tan A = 4 / 3
⇒
3
4
and sin A = −
5
5
[since, A lies in III quadrant]
cos A = −
18 27 cos x + 81 sin x = 33 cos x + 34 sin x
≥ 2 ⋅ 33 cos x ⋅ 34 sin x
[Q AM ≥ GM]
1
= 2 ⋅ 3 (3 cos x + 4 sin x )/2 ≥ 2 ⋅ 32
( − 5)
[Q − 5 ≤ 3 cos x + 4 sin x ≤ 5]
−
5
= 2 ⋅ 3 2 = 2 ⋅ 3− 2 ⋅ 3
2
=
9 3
−
1
2
19 Let sin x = t
∴
⇒
t 2 + at + 1 = 0 ⇒ t +
|a |= t +
1
=−a
t
1
≥2
t
⇒
|a | ≥ 2
[Q AM ≥ GM]
Hence, for no real solution|a | < 2.
1 sin (θ + 2α )
20 Given,
=
n
sin θ
On applying componendo and
dividendo, we get
1 + n sin (θ + 2α ) + sin θ
⇒
=
1 − n sin (θ + 2α ) − sin θ
⇒
1 + n 2 sin (θ + α ) cos α
=
1− n
2 cos(θ + α ) sin α
= tan (θ + α ) cot α
1+ n
tan α = tan(θ + α )
⇒
1− n
21 (sin2 x ) 3 + (cos2 x ) 3
= (sin 2 x + cos 2 x ) 3 − 3 sin 2 x cos 2 x
(sin 2 x + cos 2 x )
3
= 1 − 3 sin2 x cos 2 x = 1 − (sin 2 x ) 2
4
3
∴ Maximum value = 1 − × 0 = 1
4
3
1
and minimum value = 1 − × 1 =
4
4
22 Now, tan α ⋅ tan(2n − 1) α
π
= tan α tan
− 1 α = tan α cot α = 1
2
α
Hence, the value of given expression is
1.
23 2 − cos x + sin2 x = 2 − cos x
+ 1 − cos 2 x
= 3 − (cos x + cos x )
2
2
1
1
= 3 − cos x + −
2
4
Hence, the maximum value occurs at
1
cos x = − and its value
2
248
1
1
13
= 2 − − + 1 − =
2
4
4
and minimum value occurs at cos x = 1
and its value = 2 − 1 + (1 − 1) = 1.
13
∴ Required ratio =
4
m
3
24 Given, = cos θ + 3 cos θ sin2 θ
a
n
and = sin3 θ + 3cos 2 θ sin θ
a
m
n
∴
+ = (sin θ + cos θ)3
a
a
⇒
m + n
a
m − n
and
a
∴
m + n
a
1 /3
= sin θ + cos θ
θ
= 2 1 + cos n − 1
2
P
θ
θ
= 2 ⋅ 2cos 2
= 2cos n
2 ⋅ 2n − 1
2
h
28 In ∆ ABC, A + B + C = π
∴ tan A + tan B + tan C
= tan A tan B tan C
⇒
6 = 2 tan A ⇒ tan A = 3
∴
tan B + tan C = 3
and
tan B tan C = 2
⇒ tan B = 1 or 2 and tan C = 2 or 1
Now, sin2 A =
1 /3
2 /3
= cos θ − sin θ
2 /3
(m + n ) 2 /3 + (m − n ) 2 /3 = 2a 2 /3
25 Given, a sin2 α − 1 cosec 2α = 0
a
1
⇒ sin α =
a
∴ cos 2 α + 5 sin α cos α + 6sin2 α
1
5
1
6
=1−
+
1− +
a
a
a
a
5 a−1 −1 6
= 1+
+
a
a
a+ 5 a−1+ 5
=
a
x
x
1 + tan2
2 tan
x
2
2
26 tan =
−
x
x
2
2 tan
1 + tan2
2
2
x
2 x
⇒ 2 tan2
1 + tan
2
2
2
x
x
= 1 + tan2 − 4 tan2
2
2
⇒ 2 y (1 + y ) = (1 + y )2 − 4 y
put tan2 x = y
2
⇒
= (9 − 4 5) (2 +
5
5
5
5)
27 Now, 2 + 2 + 2 + ...+ 2(1 + cos θ)
θ
= 2 + 2 + ...+ 2 + 2 cos
2
KKKKKKK
KKKKKKK
θ
= 2 + 2 cos n − 1
2
tan2 B
1
4
sin B =
=
,
1 + tan2 B 1 + 1 1 + 4
1 4
5 8
, =
,
2 5 10 10
tan2 C
8 5
and sin2 C =
,
=
1 + tan2 C 10 10
=
∴ sin A : sin B : sin C
9 5 8
9 8 5
=
:
:
or
:
:
10 10 10 10 10 10
2
2
y 2 + 4y − 1 = 0
− 4 ± 16 + 4
= −2±
∴ y =
2
Since, y ≥ 0, we get
( 5 − 2)2 2 +
y = 5−2=
⋅
5−2 2+
tan2 A
9
=
1 + tan2 A 10
2
m − n
+
a
= 2 (sin2 θ + cos 2 θ)
2
2
29 Maximum value of sin θ and cos θ is 1.
Q cos x ⋅ sin y = 1
⇒ cos x = 1, sin y = 1
or
cos x = − 1, sin y = − 1
π 5π
x = 0, 2 π, y = ,
⇒
2 2
3π
or
x = π, 3 π, y =
2
∴ Required number of ordered pair
=2×2+ 2×1= 6
30 Given, (1 − cos 6 x ) + sin 2 x = 0
2
4
⇒ sin2 6 x + sin 4 2 x = 0
⇒
sin2 6 x = 0 and sin 4 2 x = 0
⇒ 6x = nπ and 2x = nπ
nπ
⇒
x=
6
nπ
and
x=
2
31 Since, x2 + x −2 ≥ 2
a
a
O
[Q AM ≥ GM]
therefore the equation is valid only if
x
2cos 2 sin2 x = 2
2
x
⇔ cos = cosec x
2
x
i.e. iff cosec x = cos = 1
2
which cannot be true.
32 Consider A 1 , A 2,..., A n be the foot of the
n poles subtending angle α to O, such
that OA 1 = a, if d be the distance
between two consecutive poles
A1 A2 A3
An
OA 2 = a + d
OA3 = a + 2d
M
M
M
OA n = a + (n − 1) d
Now, in ∆ POA n ,
h
tan α =
OA n
OA n = h cotα
⇒ a + (n − 1) d = h cot α
h cotα − a
d =
n−1
h cos α − asin α
=
(n − 1)sin α
( x + 1) + ( x − 1)
−1 8
= tan
31
1 − ( x + 1) ( x − 1)
33 tan −1
Provided ( x + 1) ( x − 1) < 0
x2 < 1
i.e.
tan −1
⇒
…(i)
2x
8
= tan −1
31
1 − ( x2 − 1)
2x
8
=
2 − x2 31
⇒
⇒
4 x2 + 31 x − 8 = 0
⇒
(4 x − 1) ( x + 8) = 0
1
1
or − 8 ⇒ x =
x=
4
4
⇒
Since, x = − 8 is not satisfied the
Eq. (i).
34 Obviously x > 0 and x 3 < 1
i.e. x <
If x >
1
3
1
, then cos −1 ( x 3 ) will be
3
undefined. If x < 0, then x 3 < 0. Hence,
cos −1 x >
π
π
and cos −1 x 3 >
2
2
which is not satisfied the equation.
1
∴
x ∈ 0,
3
Given, cos −1 ( x 3 )
π
=
− cos −1 x = sin −1 x
2
⇒
cos −1 ( x 3 ) = cos −1 1 − x2
⇒ x 3 = 1 − x2
⇒
1
2
1
1
Q x = − ∉ 0,
2
3
3 x2 = 1 − x2 ⇒ x = ±
x=
1
2
249
DAY
35 Q
∴
1 + sin x
= 2 cos x, cos x ≠ 0
cos x
= 8 cos 5 θ − 10 cos 3 θ + 3 cos θ
1 + sin x = 2 (1 − sin2 x )
= 8 cos 5 θ − 10 cos 3 θ + 3 cos θ
⇒ 1 + sin x = 2 (1 + sin x ) (1 − sin x )
⇒ (1 + sin x ) [1 − 2 (1 − sin x )] = 0
1
⇒ sin x = , − 1
2
π
π 3π
∴
x= ,π− ,
6
6 2
π 5π
3π
x= ,
,x≠
[Q cos x ≠ 0]
6 6
2
36 Given,
cot2 x + cosec x − a = 0
⇒ cosec2 x + cosec x − 1 − a = 0
− 2cos θ (4 cos 2 θ − 1) (1 − cos 2 θ)
= 8 cos θ − 10 cos 3 θ + 3 cos θ
5
∴
cosec x ≥ 1 or ≤ − 1
1 3
1
≥ or ≤ −
2 2
2
cosec x +
⇒
2
1
1
5
1
⇒ cosec x + ≥ ⇒ + a ≥
2
4
4
4
∴
3
On comparing the coefficient of
cos 5 θ, cos θ and constant term,
we get P = 5, Q = 16 and R = 0
∴ P + Q + R = 21
cos A cos B cos C
=
=
a
b
c
cos A
cos B
cos C
=
=
k sin A k sin B
k sin C
39 Q
∴
⇒
cot A = cot B = cot C
⇒
A = B = C = 60°
40 a2 sin 2B + b2 sin 2 A
= 2a2 sin B ⋅ cos B + 2b 2 sin A ⋅ cos A
a2b
b2 a
=
cos B +
cos A
R
R
ab
abc
=
(acos B + b cos A ) =
R
R
1
= 2 bc sin A = 4 bc sin A
2
37 Let h be the height of the tower QR.
A
45°
60°
45°
x3 − 15x2 + 47 x − 82 = 0
∑ a = 15
∑ ab = 47
abc = 82
cos A cos B cos C
Now,
+
+
a
b
c
2
2
2
2
2
b +c −a
c + a − b 2 a2 + b 2 − c 2
+
+
2abc
2abc
2abc
(by cosine rule)
a2 + b 2 + c 2
=
2abc
(∑ a)2 − 2 ∑ ab
=
2abc
225 − 94 131
=
=
2⋅ 82
164
4
− 1 + cos 2 θ]
= 16 cos θ − 20 cos θ + 5 cos θ
5
a≥ − 1
400 m
P
43 Given equation is
− 2cos θ [4 cos θ − 4 cos θ
2
2
1
1 5
⇒ cosec x + = 1 + a + = + a
2
4 4
⇒ (B + 2)⋅ B = sin 2 x
[from Eq. (i)]
⇒
B 2 + 2B − sin 2 x = 0
⇒
( A − 2)2 + 2( A − 2) − sin 2 x = 0
⇒
A2 − 2 A − sin 2 x = 0
− 2sin2 θ (3 − 4sin2 θ) ⋅ cos θ
44 Given, a4 + b 4 + c 4 − 2b2c 2 − 2c 2 a2 = 0
⇒
(a2 + b 2 − c 2 )2 = 2a2b 2
(a2 + b 2 − c 2 )2 1
=
⇒
4a2b 2
2
2
2
2
a + b −c
1
⇒
=±
2ab
2
1
cos C = ±
⇒
2
π
3π
C = or
⇒
4
4
So, the angle is 45° or 135°.
= 4λ
Q
41 Given, cos 2 x + 2cos x = 1
h
60°
B
Then,
R
⇒
PA = 400 − h
AP
= 1 ⇒ AP = PQ
PQ
In ∆ APQ,
⇒
−1 + 5
2
−1 − 5
, As − 1 ≤ cosx ≤ 1
neglecting
2
cos x =
⇒ 400 3 − h 3 = 400
⇒ (400 3 − 400) = h 3
400( 3 − 1)
3
5 − 1
∴ cos 2 x =
2
∴
= cos 3θ ⋅ cos 2θ − sin 3θ ⋅ sin 2θ
= (4cos 3 θ − 3cos θ)(2cos 2 θ − 1)
− ( 3 sin θ − 4sin3 θ) × (2sin θ ⋅ cos θ)
= 2 sin2 θ + 2cos(θ + α ) cos (θ − α ) − 1
= 2 sin2 θ + 2cos 2 θ − 2 sin2 α − 1
2
6−2 5 3− 5
=
4
2
2
2
sin x(2 − cos x )
3 − 5
3 − 5
= 1 −
2 −
2
2
5 − 1 5 + 1
=
=1
2 2
=
38 cos 5 θ = cos(3θ + 2θ)
sin θ sin α + 2 cos 2 ( θ + α ) − 1
= 2sin θ + 2 cos ( θ + α )
[2 sin θ sin α + cos ( θ + α )] − 1
= 2 sin2 θ + 2 cos(θ + α )
[sin θ sin α + cos θ cos α] − 1
2
= 1 − 2 sin2 α = cos 2α
π π
π
∴
f , = cos 2 × = 0
3 4
4
=
=h
400(3 − 3 )
m
3
= 169.06 m
45 f (θ, α ) = 2sin2 θ + 4cos (θ + α )
−1 − 5
and
< −1
2
400 − h = PQ
400
Again in ∆ABR, tan 60° =
BR
[QBR = PQ ]
400
∴
3=
400 − h
⇒
2cos 2 x − 1 + 2cos x − 1 = 0
cos 2 x + cos x − 1 = 0
⇒
⇒
42 From the given parts of question, we get
⇒
and
cos x + sin x = A − 1 = B + 1
A = B+2
... (i)
A ⋅ B = (sin x + cos x + 1)
(sin x + cos x − 1)
= (sin x + cos x )2 − 1
1 + sin 2 x − 1 = sin 2 x
46
I. The general value of θ satisfying any
of the equations
sin2 θ = sin2 α,cos2 θ = cos2 α
and tan2 θ = tan2 α is given by
θ = n π ± α.
II. The general value of θ satisfying
equations sin θ = sin α and
cos θ = cos α simultaneously is given
by θ = 2nπ + α, n ∈ Z .
So, Statement I is correct and
Statement II is incorrect.
250
47 Statement I Put x = cos θ, then 0 ≤ θ ≤ π
3
LHS = cos −1 (cos θ) − sin −1
1
3
2 cos θ + 2 sin θ
π
−1
= θ − sin
sin θ +
3
π
π
=θ−θ−
=−
3
3
Statement II Put x = sin θ,
π
π
then − ≤ θ ≤
4
4
LHS = sin −1 (2 sin θ cos θ)
= sin −1 (sin 2θ)
= 2θ = 2 sin −1 x
48 Given, 2sin2 θ − cos 2θ = 0
1
⇒ 4sin2 θ = 1 ⇒ sin θ = ±
2
So,
sin θ =
1
2
π
5π
or
6
6
which also satisfy 2cos 2 θ − 3sin θ = 0.
Hence, the number of solutions are two.
∴
49 Given, 2 sin θ = cos θ + sin θ
2
2
2
2
θ
θ
cos – sin
2
2
2
+
⇒
⇒
1
does not satisfy
2
the second equation]
θ
θ
cos + sin > 0
2
2
θ π
sin + > 0
2 4
θ
θ
and cos − sin < 0
2
2
⇒
θ π
cos + < 0
2 4
2nπ +
π θ
π
< +
< 2nπ + π
2 2
4
∴
2nπ +
π θ
3π
< < 2nπ +
4 2
4
50 Statement I ∠APB = 2α − α = α
∠BPC = 3α − 2α = α
and
P
θ
θ
θ
θ
= cos + sin + cos – sin
2
2
2
2
and
[Q sin θ = −
θ=
a a
H
3a
2a
a
A
B
C
Q
Hence, PB is an angle bisector of ∠APC.
AB
AP
Then,
=
BC
CP
H cosec α
=
H cosec 3α
sin 3α
=
sin α
Statement II But Statement II is not
always true.
DAY TWENTY FOUR
Cartesian System
of Rectangular
Coordinates
Learning & Revision for the Day
u
u
u
Rectangular Coordinates
Distance Formula
Section Formulae
u
u
u
Area of a Triangle
Area of Some Geometric Figures
Coordinates of Different Points
of a Triangle
u
u
u
Translation of Axes
Slope of a Line
Locus and its Equation
Rectangular Coordinates
Y
Let XOX ′ and YOY′ be two perpendicular axes in the
plane intersecting at O (as shown in the figure). Let P be
any point in the plane. Draw PM perpendicular to OX .
The ordered pair ( x, y) is called the rectangular or
cartesian coordinates of point P.
P (x, y )
y
X¢
Distance Formula
x M
O
X
Y¢
Let P ( x1 , y1 ) and Q ( x2 , y2 ) be the two points.
Then, PQ = d = ( x2 − x1 )2 + ( y2 − y1 )2 or
( x1 − x2 )2 + ( y1 − y2 )2
Distance between the points (0, 0) and ( x, y) is
x 2 + y2 .
Section Formulae
The coordinates of a point which divide the line segment joining two points P ( x1 , y1 ) and
Q ( x2 , y2 ) in the ratio m1 : m2 are
m x + m2 x1 m1 y2 + m2 y1
(i) 1 2
,
m1 + m2
m1 + m2
m x − m2 x1 m1 y2 − m2 y1
(ii) 1 2
,
m1 − m2
m1 − m2
[internal division]
[external division]
When m1 and m2 are of opposite signs, then division is external.
PRED
MIRROR
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
252
●
●
●
(iii) A triangle having vertices (at 12 , 2 at 1 ), (at 22 , 2 at 2 ) and
(at 32 , 2 at 3 ), then area of triangle = a2 [(t 1 − t 2 )(t 2 − t 3 )(t 3 − t 1 )]
Mid-point of the line joining ( x1 , y1 ) and ( x2 , y2 ) is
x1 + x2 y1 + y2
,
.
2
2
Coordinates of any point on one line segment which divide
the line segment joining two points P ( x1 , y1 ) and Q ( x2 , y2 )
in the ratio λ : 1 are given by
x1 + λ x2 y1 + λ y2
,
, (λ ≠ − 1)
λ+1
λ+1
X -axis and Y-axis divide the line segment joining the points
y
x
( x1 , y1 ) and ( x2 , y2 ) in the ratio of − 1 and − 1 respectively.
y2
x2
If the ratio is positive, then the axis divides it internally and if
ratio is negative, then the axis divides externally.
Area of a Triangle
Area of a triangle whose three vertices has coordinates
( x1 , y1 ), ( x2 , y2 ) and ( x3 , y3 ) as shown in the figure below is
given by
A (x1, y1)
(iv) Area of triangle formed by coordinate axes and the lines
c2
.
ax + by + c is =
2 ab
Coordinates of Different
Points of a Triangle
1. Centroid
The centroid of a triangle is the point of intersection of its
medians. It divides the medians in the ratio 2 : 1. If A( x1 , y1 ),
B ( x2 , y2 ) and C ( x3 , y3 ) are the vertices of ∆ABC, then the
x + x2 + x3 y1 + y2 + y3
coordinates of its centroid G are 1
,
3
3
2. Orthocentre
The orthocentre of a triangle is the point of intersection of its
altitudes. If A ( x1 , y1 ), B ( x2 , y2 ) and C ( x3 , y3 ) are the vertices of a
∆ABC, then the coordinates of its orthocentre O are
x1 tan A + x2 tan B + x3 tan C y1 tan A + y2 tan B + y3 tan C
,
tan A + tan B + tan C
tan A + tan B + tan C
3. Circumcentre
B (x2, y2)
Area of a ∆ABC =
The circumcentre of a triangle is the point of intersection of
the perpendicular bisectors of its sides. It is the centre of the
circle passing through the vertices of a triangle and so it is
equidistant from the vertices of the triangle.
C (x3, y3)
1 ( x1 y2 + x2 y3 + x3 y1 )
2 − ( y1 x2 + y2 x3 + y3 x1 )
x1
1
1
= | x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )| = | x2
2
2
x3
y1
y2
y3
1
1|
1
It should be noted that area is a positive quantity and its unit
is square of unit of length.
In the inverse problems, i.e. when area of a triangle is given to
be a square units, then we have
x1 y1 1
x1 y1 1
1
1
∆ABC = | x2 y2 1 | = a ⇒ | x2 y2 1 | = ± a
2
2
x3 y3 1
x3 y3 1
NOTE If area of ∆ABC is zero. It mean points are collinear.
Area of Some Geometrical Figures
(i) Suppose a and b are the adjacent sides of a parallelogram
and θ be the angle between them as shown in the figure
below, then area of parallelogram ABCD = ab sin θ.
D
(ii) Area of convex quadrilateral with
vertices ( x1 , y1 ),( x2 , y2 ), ( x3 , y3 ),( x 4 , y4 )
b
in that order is
θ
A
1 x1 − x3 y1 − y3
.
2 x2 − x 4 y2 − y4
a
C
b
a
B
Here, OA = OB = OC, where O is the centre of circle and A, B
and C are the vertices of a triangle. The coordinates of the
circumcentre are also given by
x sin 2 A + x2 sin 2 B + x3 sin 2C
,
S 1
sin 2 A + sin 2 B + sin 2C
y1 sin 2 A + y2 sin 2 B + y3 sin 2C
sin 2 A + sin 2 B + sin 2C
Incentre
The point of intersection of the internal bisectors of the angles
of a triangle is called its incentre.
If A ( x1 , y1 ), B ( x2 , y2 ) and C ( x3 , y3 ) are the vertices of a ∆ABC
such that BC = a, CA = b and AB = c, then the coordinates of
ax + bx2 + cx3 ay1 + by2 + cy3
the incentre are I 1
,
.
a+b +c
a+b +c
Excentre
Coordinate of excentre opposite of ∠ A is given by
− ax1 + bx2 + cx3 − ay1 + by2 + cy3
I1 ≡
,
and similarly for
−a+b +c
−a+b +c
excentres (I2 and I3 ) opposite to ∠B and ∠C are given by
ax − bx2 + cx3 ay1 − by2 + cy3
I2 ≡ 1
,
a −b + c
a −b + c
253
DAY
and
ax + bx2 − cx3 ay1 + by2 − cy3
I3 ≡ 1
,
.
a + b −c
a + b −c
In an equilateral triangle, orthocentre, centroid,
circumcentre, incentre, coincide.
Important Results
●
●
●
●
●
●
Circumcentre of the right angled ∆ABC, right angled at A is
B+C
.
2
Orthocentre of the right angled ∆ABC, right angled at A is
A.
Orthocentre, centroid, circumcentre of a triangle are
collinear.
Centroid divides the line joining the orthocentre and
circumcentre in the ratio 2 : 1.
The circumcentre of right angled triangle is the mid-point
of the hypotenuse.
A triangle is isosceles, if any two of its medians are equal.
Translation of Axes
1. To Alter the Origin of Coordinates Without
Altering the Direction of the Axes
Let origin O (0, 0) be shifted to a point (a, b ) by moving the X
and Y-axes parallel to themselves. If the coordinates of point
P with reference to old axis are ( x1 , y1 ), then coordinates of
this point with respect to new axis will be ( x1 − a, y1 − b ).
2. To Change the Direction of the Axes of
Coordinates without Changing Origin
Let OX and OY be the old axes and OX ′ and OY′ be the
new axes obtained by rotating the old OX and OY
through an angle θ, then the coordinates of P( x, y) with
respect to new coordinate axes will be given by
x↓
y↓
x′ →
cos θ
sin θ
y′ →
− sin θ
cos θ
(i) x and y are old coordinates, x ′, y′ are new coordinates.
(ii) The axes rotation in anti-clockwise is positive and
clockwise rotation of axes is negative.
3. To Change the Direction of the Axes of
Coordinates by Changing the Origin
If P( x, y) and the axes are shifted parallel to the original axis,
so that new origin is (α, β) and then the axes are rotated about
the new origin (α, β) by angle φ in the anti-clockwise ( x′ , y′ ),
then the coordinates of P will be given by
x = α + x′ cos φ − y′ sin φ
y = β + x′ sin φ + y′ cos φ
Slope of a Line
The tangent of the angle that a line makes with the positive
direction of the X -axis is called the slope or gradient of the
line. The slope of a line is generally denoted by m.
Thus, m = tan θ.
Slope of a Line in Terms of
Coordinates of any Two Points on it
Let P( x1 , y1 ) and Q( x2 , y2 ) be two points on a line making an
angle θ with the positive direction of X -axis. Then, its slope m
is given by
m=
y2 − y1 Difference of ordinates
=
x2 − x1
Difference of abscissa
Parallel and Perpendicular
Lines on the Coordinate Axes
A line parallel to X -axis makes an angle of 0° with X -axis.
Therefore, its slope is tan 0 ° = 0. A line parallel to Y-axis i.e.
perpendicular to X -axis makes an angle of 90° with X -axis, so
π
its slope is tan = ∞. Also, the slope of a line equally inclined
2
with axes is 1 or −1 as it makes an angle of 45° or 135° with
X -axis.
Locus and its Equation
It is the path or curve traced by a moving point satisfying the
given condition.
Equation to the Locus of a Point
The equation to the locus of a point is the algebraic relation
which is satisfied by the coordinates of every point on the locus
of the point.
Steps to Find the Locus of a Point
The following steps are used to find the locus of a point
Step I Assume the coordinates of the point say
(h, k ) whose locus is to be find.
Step II Write the given condition involving (h, k ).
Step III Eliminate the variable(s), if any.
Step IV Replace h → x and k → y. The equation, so obtained is
the locus of the point which moves under some
definite conditions.
254
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 Length of the median from B to AC where A( −1, 3),
B (1, − 1), c ( 5, 1) is
(a) 18
(b) 10
(c) 2 3
(d) 4
2 Three points ( p + 1, 1), ( 2p + 1, 3) and ( 2p + 2, 2p ) are
collinear if p is equal to
(a) − 1
(b) 1
(c) 2
(d) 0
3 The point A divides the join of P ≡ ( −5,1) and Q ≡ ( 3, 5) in
the ratio k :1. The two values of k for which the area of
∆ABC, where B ≡ (1, 5), C ≡ (7, − 2) is equal to 2 sq units
are
30
(a) 7,
9
31
(b) 7,
9
31
(c) 4 ,
9
31
(d) 7,
3
4 If ∆1 is the area of the triangle with vertices (0, 0),
(a tan α , b cot α ), (a sin α , b cos α ), ∆ 2 is the area of the
triangle with vertices (a, b ), (a sec2 α , b cosec 2α ),
(a + a sin2 α , b + b cos 2 α ) and ∆ 3 is the area of the
triangle with vertices ( 0 , 0), (a tan α , − b cot α ),
(a sin α , b cos α ). Then,
(a) ∆1, ∆ 2 , ∆ 3 are in GP
(c) Cannot be discussed
(b) ∆1, ∆ 2 , ∆ 3 are not in GP
(d) None of these
5 If the area of the triangle formed by the points O( 0, 0),
1
sq units, then x =
A(a x , 0) and B( 0, a 6 x ) is
2a 5
2
(a) 1, 5
(b) −1, 5
(c) 1, − 5
(d) −1, − 5
6 The value of k for which the distinct points (k , 2 − 2k ),
(1 − k , 2k ) and ( −4 − k ,6 − 2k ) are collinear is (are)
(a) −1or 1 / 2
(c) Only −1
(b) Only 1 / 2
(d) can not be found
7 If the line 2x + y = k passes through the point which
divides the line segment joining the points (1, 1) and
(2, 4) in the ratio 3 : 2, then k is equal to
(a)
29
5
(b) 5
(c) 6
(d)
11
5
8 A line L intersects the three sides BC, CA and AB of a
∆ABC at P , Q and R, respectively. Then,
BP CQ AR
is
⋅
⋅
PC QA RB
equal to
(a) 1
(c) −1
(b) 0
(d) None of these
9 If the coordinates of the vertices of a triangle are
integers, then the triangle cannot be
(a) equilateral
(c) scalene
(b) isosceles
(d) None of these
10 Let O( 0, 0), P ( 3, 4) and Q ( 6, 0) be the vertices of the
∆OPQ. The point R inside the ∆OPQ is such that ∆OPR,
∆PQR and ∆OQR are of equal area. Then, R is equal to
4
(a) , 3
3
4
(c) 3,
3
2
(b) 3,
3
4
(d)
3
11 The number of points having both coordinates as
integers that lie in the interior of the triangle with vertices
j JEE Mains 2015
( 0, 0), ( 0, 41) and ( 41, 0) is
(a) 901
(b) 861
(c) 820
(d) 780
12 If (0, 0), (1,1) and (1,0) be the middle points of the sides
of a triangle, its incentre is
(a) (2 +
(c) (2 −
2 , 2 + 2)
2 , 2 − 2)
(b) [(2 +
(d) [(2 −
2 , − (2 + 2)]
2 , − (2 − 2)]
13 Vertices of a triangle are (1,2), (2,3) and (3,1) Its
circumcentre is
(a) (11/6, 13/6)
(c) (13/6, 11/6)
(b) (11/6, 2)
(d) None of these
14 If a vertex of a triangle be (1,1) and the middle points of
two sides through it be ( − 2, 3) and (5, 2), then the
centroid of the triangle is
(a) (3, 5/3)
(c) (5/3, 3)
(b) (3, 5)
(d) None of these
15 The centroid of the triangle is (3,3) and the orthocentre is
( −3, 5) then its circumcentre is
(a) (0, 4)
(b) (0, 8)
(c) (6, 2)
(d) (6, − 2)
16 Let the orthocentre and centroid of a triangle be A( −3, 5)
and B( 3, 3) respectively. If C is the circumcentre of this
triangle, then the radius of the circle having line segment
j
JEE Mains 2018
AC as diameter, is
(a) 10
(b) 2 10
(c) 3
5
2
(d)
3 5
2
17 If G is the centroid of ∆ABC with vertices A(a, 0),
B( −a, 0)and C(b, c ), then
(a) 1
(b) 2
( AB 2 + BC 2 + CA 2 )
is equal to
(GA 2 + GB 2 + GC 2 )
(c) 3
(d) 4
18 Let a, b, c and d be non-zero numbers. If the point of
intersection of the lines 4ax + 2ay + c = 0 and
5bx + 2by + d = 0 lies in the fourth quadrant and is
j
equidistant from the two axes, then
JEE Mains 2014
(a) 2bc − 3ad = 0
(c) 2ad − 3bc = 0
(b) 2bc + 3ad = 0
(d) 3bc − 2ad = 0
255
DAY
19 The origin is shifted to (1,2). The equation
y 2 − 8x − 4y + 12 = 0 changes to y 2 = 4ax , then a is
equal to
(a) 1
(b) 2
(c) −2
(d) −1
20 If the axes are rotated through an angle of 60°, the
coordinates of a point in the new system are ( 2, − 3),
then its original coordinates are
5
2
(a) , −
3
3
5 3
(c) ,
2 2
5 2
(b) − ,
3 3
5
3
(d) − , −
2
2
21 By rotating the axes through 180°, the equation
x − 2y + 3 = 0 changes to
(a) x + 2 y − 3 = 0
(c) x − 2 y − 3 = 0
(b) x − 2 y + 3 = 0
(d) None of these
π
be a fixed angle. If P = (cos θ, sin θ ) and
2
Q = {cos(α − θ ), sin(α − θ )}, then Q is obtained from P by
22 Let 0 < α <
(a) clockwise rotation around the origin through angles α
(b) anti-clockwise rotation around origin through angle α
(c) reflection in the line through the origin with slope tanα
(d) reflection in the line through the origin with slope tan α / 2
23 The point ( 4 , 1) undergoes the following transformations
(i) Reflection in the line x − y = 0
(ii) Translation through a distance of 2 units along
positive direction of X-axis.
(iii) Projection on X-axis.
The coordinate of the point in its final position is
(a) (3, 4)
(b) (3, 0)
(c) (1, 0)
(d) (4, 3)
24 If the points are A ( 0, 4) and B ( 0, − 4), then find the locus
of P ( x , y ) such that | AP − BP | = 6 .
(a) 9 x 2 − 7 y 2 + 63 = 0
(c) 9 x 2 + 7 y 2 + 63 = 0
(b) 9 x 2 + 7 y 2 − 63 = 0
(d) None of these
25 ABC is a variable triangle with the fixed vertex C(1, 2) and
A , B having the coordinates (cos t , sin t ), (sin t , − cos t )
respectively, where t is a parameter. The locus of the
centroid of the ∆ABC is
(a)
(b)
(c)
(d)
2
3 (x
3 (x 2
3 (x 2
3 (x 2
+ y ) − 2x − 4y − 1 = 0
+ y 2 ) − 2x − 4y + 1 = 0
+ y 2 ) + 2x + 4y − 1 = 0
+ y 2 ) + 2x + 4y + 1 = 0
2
26 If A ( 2,−3)and B( −2, 1) are two vertices of a triangle and
third vertex moves on the line 2x + 3y = 9, then the locus
of the centroid of the triangle is
(a) 2 x − 3 y = 1
(c) 2 x + 3 y = 1
(b) x − y = 1
(d) 2 x + 3 y = 3
27 Let A ( − 3, 2) and B( − 2, 1 ) be the vertices of a ∆ABC. If
the centroid of this triangle lies on the line
3x + 4y + 2 = 0, then the vertex C lies on the line
j
(a) 4 x + 3 y + 5 = 0
(c) 4 x + 3 y + 3 = 0
JEE Mains 2013
(b) 3 x + 4 y + 3 = 0
(d) 3 x + 4 y + 5 = 0
a
k
28 The coordinates of points A and B are (ak , 0) and , 0 ,
where (k ≠ ±1) if p moves in such a way that PA = kPB,
the locus of P is
(a) k 2 (x 2 + y 2 ) = a 2
(c) x 2 + y 2 + a 2 = 0
(b) x 2 + y 2 = k 2 a 2
(d) x 2 + y 2 = a 2
29 If A( −a, 0) and B(a, 0) are two fixed points, then the locus
of the point at which AB subtends a right angle is
(a) x 2 + y 2 = 2a 2
(c) x 2 + y 2 + a 2 = 0
(b) x 2 − y 2 = a 2
(d) x 2 + y 2 = a 2
30 A point moves in such a way that the sum of its distances
from two fixed points (ae, 0) and ( −ae, 0) is 2a. Then the
locus of the points is
x2
y2
+ 2
=1
2
a
a (1 − e 2 )
x2
y2
(b) 2 − 2
=1
a
a (1 − e 2 )
x2
y2
(c) 2
+ 2 =1
2
a (1 − e ) a
(d) None of the above
(a)
Directions (Q. Nos. 31-35) Each of these questions
contains two statements : Statement I and Statement II. Each
of these questions also has four alternative choices, only one
of which is the correct answer. You have to select one of the
codes (a ), (b), (c) and (d ) given below.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
31 Statement I If A ( 2a, 4 a ) and B ( 2a, 6 a ) are two vertices of
a equilateral ∆ABC and the vertex C is given by
( 2a + a 3, 5a ).
Statement II An equilateral triangle all the coordinates of
three vertices can be rational.
32 Statement I If the circumcentre of a triangle lies at the
origin and centroid is the middle point of the line joining
the points (2, 3) and (4, 7), then its orthocentre lies on the
line 5x − 3y = 0.
Statement II The circumcentre, centroid and the
orthocentre of a triangle lie on the same line.
33 Statement I If the origin is shifted to the centroid of the
triangle with vertices (0, 0), (3, 3) and (3, 6) without
rotation of axes, then the vertices of the triangle in the
new system of coordinates are ( −2, 0), (1, 3) and (1, − 3).
Statement II If the origin is shifted to the point ( 2, 3)
without rotation of the axes, then the coordinates of the
point P (α − 1, α + 1) in the new system of coordinates are
(α − 3, α − 2).
256
34 Let the equation of the line ax + by + c = 0.
Statement I If a, b and c are in AP, then ax + by + c = 0
pass through a fixed point (1, − 2).
Statement II Any family of lines always pass through a
fixed point.
The bisector of the acute angle between L 1 and L 2
intersects L 3 at R.
Statement I The ratio PR : RQ equals 2 2 : 5.
Statement II In any triangle, bisector of an angle divides
the triangle into two similar triangles.
35 The lines L 1 : y − x = 0 and L 2 : 2x + y = 0 intersect the
line L 3 : y + 2 = 0 at P and Q, respectively.
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 If the coordinates of two points A and B are ( 3, 4) and
( 5, − 2), respectively. Then, the coordinates of any point P,
if PA = PB and area of ∆PAB = 10 , are
(a) (7,5), (1, 0)
(c) (7, 2), (−1, 0),
(b) (7, 2), (1, 0)
(d) None of these
2 Lat A(a, b ) be a fixed point and O be the origin an
coordinates. If A1 is the mid-point at OA, A2 is the
mid-point at AA1, A3 is the mid-point at AA2 and so on.
Then, the coordinates of An are
(a) (a (1 − 2 − n ), b (1 − 2 − n ))
(b) (a (2 − n − 1), b (2 − n − 1))
− ( n − 1)
− ( n − 1)
(c) (a (1 − 2
), b (1 − 2
)) (d) None of these
3 The coordinates of points A, B, C are ( x1, y1 ), ( x 2 , y 2 ) and
( x 3 , y 3 ) and point D divides AB in the ratio l : k . If P
divides line DC in the ratio m : (k + l ), coordinates of P are
kx + lx 2 + mx 3 ky1 + ly 2 + my 3
(a) 1
,
k+l+m
k+l+m
lx1 + mx 2 + kx 3 ly1 + my 2 + ky 3
(b)
,
l+m+k
l+m+k
mx1 + kx 2 + lx 3 my1 + ky 2 + ly 3
(c)
,
m+k+l
m+K+l
(d) None of the above
(t1
(t1
(t1
(t1
+ t2
+ t2
− t2
+ t2
+ t 3 + t1 + t 2 t 3 )}
+ t 3 + t1 t 2 t 3 )}
− t 3 − t1 t 2 t 3 )}
− t 3 − t1 t 2 t 3 )}
25
sq unit if the
6
coordinates of base are B(1, 3) and C( −2, 7), the
coordinates of A are
8 ABC is an isosceles triangle of area
11 5
(a) (1, 6), − ,
6 6
5
11
(c) , 6 , − , 4
6 6
1
5
(b) − , 5 , 4,
2 6
5
11
(d) 5, , , 4
6 6
9 If A( 6, − 3), B( −3, 5), C( 4, − 2), P(α , β), then the ratio of the
(b) | α − β |
(d) | α + β − 2 |
points, then | OP1| ⋅ | OP2 | cos( ∠POP
1
2 ) is equal to
(a) x1 y 2 + x 2 y1
(c) (x1 − x 2 )2 + (y1 − y 2 )2
(b) (x12 + y12 ) (x 22 + y 22 )
(d) x1 x 2 + y1 y 2
11 If points ( 0, 0), ( 2, 2 3 ) and (a, b ) are vertices of an
equilateral triangle, then (a, b ) is equal to
− 72 x + 54 y + 225 = 0
− 72 x + 54 y + 225 = 0
+ 72 x − 54 y + 225 = 0
− 72 x − 54 y − 225 = 0
(a) (0, − 4)
(b) (0, 4)
(c) (4, 0)
(d) (− 4, 0)
12 If the equation of the locus of a point equidistant from the
point on one side of the line PQ such that ∠RPQ − ∠RQP
is 2α, then
locus of R is x 2 − y 2 + 2 xy cot 2 α − a 2 = 0
locus of R is x 2 + y 2 + 2 xy cotα − a 2 = 0
locus of R is a hyperbola, if α = π/4
locus of R is a circle, if α = π/4
−1
6 If the axis be turned through an angle tan 2, then what
does the equation 4xy − 3x 2 = a 2 become?
(a) X 2 − 4Y 2 = a 2
(c) X 2 + 4Y 2 = − a 2
{−a , a
{−a , a
{−a , a
{−a , a
10 If O be the origin and if P1( x1, y1 ) and P2 ( x 2 , y 2 ) be two
5 Two points P (a , 0) and Q ( −a , 0) are given, R is a variable
(a)
(b)
(c)
(d)
(a)
(b)
(c)
(d)
(a) | α + β |
(c) | α + β + 2 |
where A( 0, 0) and B( 4, − 3) are points, is
− 5y 2
+ 5y 2
+ 5y 2
+ 5y 2
{at1 t 2 , a (t1 + t 2 )}, {at 2 t 3 , a (t 2 + t 3 )}, {at 3 t1, a (t 3 + t1 )} is
areas of the triangles PBC and ABC is
4 The locus of a point P which moves such that 2PA = 3 PB,
(a) 5 x 2
(b) 5 x 2
(c) 5 x 2
(d) 5 x 2
7 The orthocentre of the triangle whose vertices are
(b) X 2 + 4Y 2 = a 2
(d) None of these
points (a1, b1 ) and (a 2 , b2 ) is (a1 − a 2 )x + (b1 − b2 )y + c = 0,
then the value of c is
(a) a12 − a22 + b12 − b22
1
(c) (a12 + a22 + b12 + b22 )
2
(b) a12 + b12 − a22 − b22
1
(d) (a22 + b22 − a12 − b12 )
2
13 If ( 2, 1), ( 5, 2) and ( 3, 4) are vertices of a triangle, its
circumcentre is
13 9
(a) ,
2 2
9 13
(c)
4, 4
13 9
(b) ,
4 4
9 13
(d) ,
2 2
257
DAY
15 The area of a triangle is 5 and its two vertices are A( 2, 1)
14 A point moves is such a way that the sum of squares of
and B( 3, − 2). The third vertex lies on y = x + 3. Then,
third vertex is
its distances from A( 2, 0) and B( − 2, 0) is always equal to
the square of the distance between A and B, then the
locus of point P is
(a) x 2 + y 2 − 2 = 0
(c) x 2 + y 2 + 4 = 0
7 13
(a) ,
2 2
3
3
(c) − , −
2
2
(b) x 2 + y 2 + 2 = 0
(d) x 2 + y 2 − 4 = 0
5 5
(b) ,
2 2
(d) (0, 0)
ANSWERS
SESSION 1
SESSION 2
1 (b)
2 (c)
3 (b)
4 (b)
5 (d)
6 (c)
7 (c)
8 (c)
9 (a)
10 (c)
11 (d)
12 (c)
13 (c)
14 (c)
15 (c)
16 (c)
17 (c)
18 (c)
19 (b)
20 (c)
21 (c)
22 (d)
23 (b)
24 (a)
25 (b)
26 (c)
27 (b)
28 (d)
29 (d)
30 (a)
31 (c)
32 (a)
33 (a)
34 (a)
35 (c)
1 (b)
11 (c)
2 (a)
12 (d)
3 (a)
13 (b)
4 (b)
14 (d)
5 (a)
15 (a)
6 (a)
7 (b)
8 (c)
9 (d)
10 (d)
Hints and Explanations
SESSION 1
3 Coordinates of A, dividing the join of
1 Let BD be the median from B to AC,
where D is the mid-point of AC.
B (1, –1)
h
A (–1, 3)
D (2, 2)
C (5, 1)
According to mid-point formula
coordinates of D are
−1 + 5, 3 + 1 = (2,2)
2
2
∴ Length of median BD
= (2 − 1)2 + (2 + 1)2 = 1 + 9 = 10
P ≡ (−5,1) and Q ≡ (3, 5) in the ratio k :1
3 k − 5 5k + 1
are given by A
,
.
k+1 k+1
Also, area of ∆ABC is given by
1
∆ = Σ x1 ( y 2 − y 3 )
2
1
= |[ x1 ( y 2 − y 3 ) + x2 ( y 3 − y 1 )
2
+ x3 ( y 1 − y 2 )]|
5k + 1
1 3 k − 5
(7) + 1 −2 −
⇒
2 k + 1
k+1
5k + 1
+ 7
− 5 = 2
k+1
5k + 1
1 3 k − 5
⇒
(7) + −2 −
2 k + 1
k+1
5k + 1
+ 7
− 5 = ± 2
k+1
2 Condition of collinearity
p+1 1 1
∆ = 2p + 1 3 1 = 0
2p + 2 2p 1
⇒ ( p + 1)(3 − 2 p ) − 1(2 p + 1 − 2 p − 2)
+ 1(4 p2 + 2 p − 6 p − 6) = 0
2
⇒ − 2 p + p + 3 + 1 + 4 p2 − 4 p − 6 = 0
⇒
2 p2 − 3 p − 2 = 0
2
⇒ 2p − 4p + p − 2 = 0
1
p = 2, −
∴
2
⇒
⇒
or
14 k − 66 = 4 k + 4
10 k = 70 ⇒ k = 7
14 k − 66 = −4 k − 4
⇒
⇒
18 k = 62
31
k =
9
Therefore, the values of k are
31
.
9
7 and
0
0
1
4 We have, ∆ 1 = 1 a tan α b cot α 1
2
asin α
b cos α 1
1
= ab|sin α − cos α|
2
a
b
1
1
and ∆ 2 =
asec2 α
b cosec 2α 1
2
a + asin2 α b + b cos 2 α 1
On applying C1 → C1 − aC3 and
C2 → C2 − bC3 , we get
∆2 =
0
0
1
1
ab tan2 α cot2 α 1
2
sin2 α cos 2 α 1
=
and ∆ 3 =
0
1
a tan α
2
asin α
1
ab |sin2 α − cos 2 α|
2
0
1
− b cot α 1
b cos α 1
=
1
ab |sin α + cos α|
2
1
ab∆ 2
2
Suppose, ∆ 1 , ∆ 2 and ∆ 3 are in GP.
1
Then, ∆ 1 ∆ 3 = ∆22 ⇒ ab∆ 2 = ∆22
2
1
⇒
∆ 2 = ab
2
So that, ∆ 1 ∆ 3 =
258
1
1
ab(sin2 α − cos 2 α ) = ab
2
2
⇒
sin2 α − cos 2 α = 1
π
i.e. α = (2m + 1) , m ∈ I .
2
A
⇒
R
Q
But for this value of α, the vertices of the
given triangles are not defined.
Hence, ∆ 1 , ∆ 2 and ∆ 3 cannot be in GP for
any value of α.
5 We have,
1
sq units
2a5
2
1
1
× ax × a6 x = a−5
2
2
2
ax + 6 x = a−5
x2 + 6 x + 5 = 0
x = −1, − 5
area of ∆OAB =
⇒
⇒
⇒
⇒
P
B
C
Also, as P lies on L, we have
λx + x2
λy + y 2 + n = 0
l 3
+ m 3
λ+1
λ+1
lx + my 2 + n BP
…(i)
− 2
=
=λ
⇒
lx3 + my 3 + n PC
Similarly, we obtain
lx + my 3 + n
CQ
=− 3
QA
lx1 + my 1 + n
…(ii)
lx + my 1 + n
AR
=− 1
RB
lx2 + my 2 + n
…(iii)
and
6 Points are collinear so
k
2 − 2k 1
⇒ 1 − 2k 4k − 2 0 = 0
−4 − 2k
4
0
12 If A( x1 , y 1 ), B( x2 , y 2 ), C ( x3 , y 3 ) be the
vertices of the triangle and if (0, 0), (1, 1)
and (1,0) are the middle points of
AB , BC and CA respectively, then
x1 + x2 = 0, x2 + x3 = 2, x3 + x1 = 2
y1 + y2 = 2 , y2 + y3 = 2 , y3 + y1 = 0
So, A(0, 0), B(0, 2) and C(2, 0) are the
vertices of the ∆ABC.
Now, a = BC = 2 2 , b = CA = 2,
c = AB = 2
The coordinates (α,β) of the in-centre are
given by
ax + bx2 + cx3
α = 1
= 2 − 2,
a+ b +c
ay 1 + by 2 + cy 3
β=
=2− 2
a+ b + c
9 Let A = ( x1 , y 1 ), B = ( x2 , y 2 ), C = ( x3 , y 3 ) be
[applying C2 → C2 − C1 and
C3 → C3 − C2 ]
⇒ 4 − 8k + (4k − 2)(4 + 2k ) = 0
⇒
2k 2 + k − 1 = 0
so k = − 1 and 1 / 2
1
But for k = , points are (1/2, 1),
2
9
(1/2, 1) and − , 5
5
Which is a contradiction as given points
are distinct.
7 Using section formula, the coordi- nates
of the point P, which divides AB
internally in the ratio 3 : 2 are
3 × 2 + 2 × 1 3 × 4 + 2 × 1
,
3+ 2
3+ 2
8 14
≡ P ,
5 5
Also, since the line L passes through
P, hence substituting the coordinates
8 14
of P , in the equation of
5 5
L : 2 x + y = k , we get
8
14
2 + = k ⇒ k = 6
5 5
8 Let A( x1 , y 1 ), B( x2 , y 2 ) and C ( x3 , y 3 ) be the
vertices of ∆ABC and lx + my + n = 0 be
the equation of the line. If P divides BC
in the ratio λ :1, then the coordinates of
λx + x2 λy 3 + y 2
P are 3
,
.
λ+1
λ+1
the vertices of a triangle and
x1 , x2 , x3 and y 1 , y 2 , y 3 be integers.
So, BC 2 = ( x2 − x3 )2 + ( y 2 − y 3 )2 is a
positive integers.
i.e. The in-centre is (2 − 2, 2 − 2 ).
If the triangle is equilateral, then
AB = BC = CA = a
[say]
13 ( x − 1)2 + ( y − 2)2
= ( x − 2)2 + ( y − 3)2
= ( x − 3)2 + ( y − 1)2
⇒ x + y = 4, 4 x − 2 y = 5
⇒
x = 13 / 6 , y = 11 / 6
13 11
∴ Circumcentre = ,
6 6
∠A = ∠B = ∠C = 60°.
1
∴ Area of the triangle = sin A ⋅ bc
2
and
1
= a2 sin 60°
2
a2 3
=
2 2
3 2
a
4
which is irrational.
[since, a2 is a positive integer]
Now, the area of the triangle in terms of
the coordinates
1
= [ x1 ( y 2 − y 3 ) + x2 ( y 3 − y 1 )
2
+ x3 ( y 1 − y 2 )]
which is a rational number.
This contradicts that the area is an
irrational number, if the triangle is
equilateral.
=
10 If the centroid is joined to the vertices,
we get three triangles of equal area.
4
∴
R = G = 3,
3
11 Required points (x, y) are such that, it
satisfy x + y < 41 and x > 0, y > 0.
(41,0)
∴Total number of integral coordinates
= 41 −1 C3 −1 = 40C2
40!
=
= 780
2! 38!
On multiplying Eqs. (i), (ii) and (iii), we
get
BP CQ AR
⋅
⋅
= −1
PC QA PB
k
2 − 2k 1
1− k
2k
1=0
−4 − k 6 − 2k 1
P
Number of positive
(0,41)
integral solution of
the equation
x + y + k = 41 will
be number of
integral coordinates (0,0)
in the bounded
region.
14
x1 + 1
= −2
2
⇒
x1 = − 5
A (1, 1)
(–2, 3)E
F (5, 2)
B (x1, y1)
C (x2, y2)
y1 + 1
= 3 ⇒ y1 = 5
2
B = (−5, 5)
x2 + 1
= 5 ⇒ x2 = 9
2
y2 + 1
= 2 ⇒ y2 = 3
2
∴
C = (9, 3)
1 + 9 − 5 1 + 5 + 3
G =
,
3
3
= (5 / 3,3)
259
DAY
triangle be A(−3, 5) and B(3, 3)
respectively and C circumcentre
C
B(3, 3)
A (–3, 5)
We know that,
AB : BC = 2 : 1
AB = (3 + 3)2 + (3 − 5)2
= 36 + 4 = 2 10
∴
BC = 10
…(iii)
19 Let P ( x, y ) be the original position of the
point w.r.t the original axes. Let us move
the origin at new position to (h, k).
Hence, the position of the same point P
in the new system is
x′ = x − h
y′ = y − k
Here, (h, k ) = (1, 2)
∴
x ′ = ( x − 1)
y ′ = ( y − 2)
As per the given situation
y 2 − 8 x − 4 y + 12 = ( y − 2)2 − 4a( x − 1)
⇒ y 2 − 8 x − 4 y + 12 = y 2 − 4 y
+ 4 − 4ax + 4a
Comparing respective coefficients, we
have
4a = 8
∴
a=2
20 Let P (a′, y ′ ) be the coordinates of the
AC = AB + BC
= 2 10 + 10 = 3 10
point obtained by rotating the axes
through an angle of 60°.
∴The transformation matrix can be
written as
x ′ cos θ sin θ x
=
y ′ − sin θ cos θ y
Since, AC is a diametre of circle
AC 3 10
5
∴
r =
=
=3
2
2
2
17 Coordinates of point G is G b , c
3 3
( AB ) + (BC ) + (CA )2
Let E =
(GA )2 + (GB )2 + (GC )2
2
⇒ E =
2ad − 3bc = 0
2
4a2 + (a + b )2 + c 2 + (a − b )2 + c 2
2
2
b − a + c + b + a
3
3
3
2
2
c
2b
2c
+ + +
3
3
3
2
2
4 a2 + 2 c 2 + 2 a2 + 2 b 2
2 b2
6c2
4 b2
+ 2 a2 +
+
9
9
9
6a2 + 2b 2 + 2c 2
⇒ E=
1
(6b 2 + 18a2 + 6c 2 )
9
2(3a2 + b 2 + c 2 )
⇒ E=
=3
1
6(3a2 + b 2 + c 2 )
9
18 Let coordinate of the intersection point
in fourth quadrant be (α, − α).
Since, (α, − α) lies on both lines
4 ax + 2ay + c = 0 and
5bx + 2by + d = 0.
∴ 4 aα − 2aα + c = 0
−c
...(i)
⇒
α =
2a
⇒ E=
and 5bα − 2bα + d = 0
−d
⇒
α =
3b
From Eqs. (i) and (ii),we get
…(ii)
1
3
x
2 2
2
=
⇒
− 3 3 1 y
−
2
2
x
3y
+
2 2
2
=
⇒
− 3 3 x
y
−
+
2
2
⇒ x + 3 y = 4 and 3 x − y = 2 3
Solving the above equations,
5 3
we have ( x, y ) = ,
2 2
21 Let P ( x ′, y ′ ) be the coordinates of the
point P ( x, y ) after rotation of axes at an
angle of 180°
x ′ cos θ sin θ x
⇒
=
y ′ − sin θ cos θ y
Since, here θ = 180°
x ′ −1 0 x
⇒
=
y ′ 0 −1 y
x′ − x
=
y ′ − y
∴
x = − x ′ and y = − y ′
Hence, the new equation of curve,
x − 2 y + 3 = 0 is (− x ′ ) − 2(− y ′ ) + 3 = 0
⇒ − x′ + 2 y ′ + 3 = 0
⇒ x′ − 2 y ′ − 3 = 0
or x − 2 y − 3 = 0 in general
⇒
is inclined at angle α − 2θ with X -axis.
The bisector of angle POQ is inclined at
angle
α − 2θ
α
+ θ = with X-axis.
2
2
Y
Q
P
θ
16 We have, orthocentre and centroid of a
⇒
22 OP is inclined at angle θ with X -axis OQ
–2
Centroid divides the join of orthocenter
and circumcenter in the ratio of 2 : 1
Let the circumcenter of ∆ is (α, β )
2α − 3 2β + 5
⇒ G (3, 3) = G
,
3
3
2α − 3
2β + 5
∴
= 3 and
=3
3
3
or
α = 6, β = 2
∴Circumcentre of ∆ is C(6, 2).
α
−c
−d
=
⇒ 3bc = 2ad
2a
3b
15 Since, we know that
O
θ
X
23 Image of (4, 1) in the line x = y is
(1, 4) on translating this point along
positive direction of X-axis by 2 units,
this point is transformed into (3, 4) and
projection of the point (3, 4) on X-axis is
(3, 0).
24 BP − AP = ± 6 or BP = AP ± 6
⇒ x2 + ( y + 4) 2 =
x 2 + ( y − 4) 2 ± 6
On squaring and simplifying, we get
4y − 9 = ± 3
x2 + ( y − 4) 2
Again on squaring, we get
9 x2 − 7 y 2 + 63 = 0
25 Let G (α,β ) be the centroid in any
position. Then,
1 + cos t + sint 2 + sint − cos t
(α, β ) =
,
3
3
1 + cos t + sin t
∴ α =
3
2 + sin t − cos t
and β =
3
⇒ 3α − 1 = cos t + sin t
…(i)
and 3 β − 2 = sin t − cos t
…(ii)
On squaring and adding Eqs. (i) and (ii),
we get
(3α − 1)2 + (3 β − 2)2 = (cos t + sin t )2
+ (sin t − cos t )2
= 2(cos 2 t + sin2 t ) = 2
∴ The equation of the locus of the
centroid is (3 x − 1)2 + (3 y − 2)2 = 2
⇒ 9 ( x2 + y 2 ) − 6 x − 12 y + 3 = 0
⇒ 3 ( x2 + y 2 ) − 2 x − 4 y + 1 = 0
26 The third vertex lies on 2 x + 3 y = 9
9 − 2x
i.e. x,
3
A (2, –3)
B (–2, 1)
(
C x, 9 – 2x
3
(
260
∴ Locus of centroid is
9 − 2x
+ 1
2 − 2 + x −3 +
3
,
= (h, k )
3
3
x
3 − 2x
and k =
∴
h=
3
9
⇒ 9 k = 3 − 2(3 h ) ⇒ 9 k = 3 − 6 h
(h − ae)2 + k 2 +
Hence, locus of a point is 2 x + 3 y = 1.
27 Let third vertex be C ( x1 , y 1 ).
−3 − 2 + x1 2 + 1 + y 1
,
lies
3
3
on line
3x + 4y + 3 = 0
(PA ) = k(PB )
⇒ (PA )2 = k 2 ( PB )2
⇒ (α − ak )2 + β2
2
a
= k 2 α − + β2
k
⇒ α2 + β2 − 2akα + a2 k 2 = k 2α2
2ak 2
+ k 2β2 −
α + a2
k
⇒ (1 − k 2 )α2 + (1 − k 2 )β 2 = (1 − k 2 ) a2
⇒ (1 − k 2 ) {α2 + β2 } = (1 − k 2 )a2
{Q k ≠ ± 1}
∴
α 2 + β2 = a 2
Replace α by x and β by y, we have
x2 + y 2 = a2
29 Let P (h, k ) represents all those points
subtending a right angle at A and B
Y
A(–a, 0)
O
B (a, 0)
X
∴
m AP ⋅ m PB = − 1
k − 0 k − 0
⇒
= −1
h + a h − a
⇒
k 2 = − (h2 − a2 )
⇒
k 2 + h2 = a2
Replace k → y and h → x, we get
x2 + y 2 = a2
30 Since, A(ae, 0) and B(− ae, 0) be the
given points and let P (h, k ) be any point
whose distance from A and B is constant
i.e. 2a.
i.e. |PA | + |PB | = 2a
⇒ (h − ae)2 + k 2
+ (h + ae )2 + k 2 = 2a …(i)
and L 3 : y + 2 = 0 as shown below,
Y
= − 2eh …(iii)
{Q a − b = ( a + b ) ( a − b )}
Adding Eqs. (i) and (iii), we have
2 (h − ae )2 + k 2 = (2a − 2eh )
Squaring both sides, we have
(h − ae )2 + k 2 = (a − eh )2
⇒ h2 + a2e 2 − 2aeh + k 2
= a2 + e 2 h2 − 2aeh
⇒ h2 − e 2 h2 + k 2 = a2 − a2e 2
⇒ h2 (1 − e 2 ) + k 2 = a2 (1 − e 2 )
Replacing h by x and k by y, we get the
locus of point P (h, k ) which is the locus
of an ellipse.
y2
x2
+ 2
=1
2
a
a (1 − e 2 )
31 Statement I : AB = BC = CA
∴ A, B ,C are the vertices of triangle ABC.
Statement II : Let A( x1 , y 1 ), B ( x2 , y 2 ) and
C ( x3 , y 3 ) are all rational coordinates.
x1 y 1 1
1
3
x2 y 2 1 =
∴ Area (∆ABC ) =
2
4
x3 y 3 1
cannot be all rational.
32 The orthocentre lies on the line joining
Y¢
35 Here, L 1 : y − x = 0 and L 2 : 2 x + y = 0
⇒ (h − ae)2 + k 2 − (h + ae )2 + k 2
[( x1 − x2 )2 + ( y 1 − y 2 )2 ]
LHS = rational, RHS = irrational
Hence, ( x1 , y 1 ), ( x2 , y 2 ) and ( x3 , y 3 )
P (h , k )
X¢
− 4aeh
2a
⇒ 2 (h − ae )2 + k 2 = 2(a − eh )
28 Let P(α, β ) be any point such that
As a,b and c are in AP.
⇒ 2b = a + c ⇒ a − 2b + c = 0
On comparing with ax + by + c = 0, it
passes through fixed points (1, − 2).
(h + ae )2 + k 2
=
⇒ 2h + 3k = 1
∴Centroid
⇒ Family of coincident lines,
if L 1 and L 2 are coincident.
Let us assume
{(h − ae )2 + k 2 } − {(h + ae )2 + k 2 }
= − 4aeh …(ii)
On dividing Eqs. (ii) by (i), we have
{(h − ae )2 + k 2 } − {(h + ae )2 + k 2 }
the points (0, 0) and (3, 5) i.e.
5x − 3 y = 0.
Also, Statement II is true.
33 Statement II is true as the coordinates of
the point P in new system are
(α − 1 − 2, α + 1 − 3).
In Statement I, the centroid is (2, 3), so
the coordinates of the vertices in the
new system of coordinates are (− 2, − 3),
(1, 0), (1, 3).
34 Statement II is false as L 1 + λL 2 = 0
⇒ Family of concurrent lines, if L 1 and
L 2 are intersect.
⇒ Family of parallel lines,
if L 1 and L 2 are parallel.
L1
y=x
X¢
X
O (0, 0)
P
L3
Q
(–2, –2)
R –2
Angle
|PO |=
Y¢
Bisector
4+ 4
L2
(1, –2)
y=–2
y = –2x
= 2 2;|OQ |= 1 + 4 =
5
Since, OR is angle bisector
OP
PR
=
OQ
RQ
⇒
PR
2 2
=
RQ
5
Hence, Statement I is true.
But, it does not divide the triangle in
two similar triangles.
Hence, Statement II is false.
SESSION 2
1 Let the coordinates of P be ( x, y ).
Then, PA = PB ⇒ PA2 = PB 2
⇒ ( x − 3)2 + ( y − 4)2 = ( x − 5)2 + ( y + 2)2
⇒
x − 3y − 1 = 0
…(i)
Now, area of ∆PAB = 10
x y 1
1
⇒
3 4 1 = ± 10
2
5 −2 1
⇒
6 x + 2 y − 26 = ± 20
⇒
6 x + 2 y − 46 = 0
or
6x + 2y − 6 = 0
⇒
3 x + y − 23 = 0
or
3x + y − 3 = 0
…(ii)
On solving, x − 3 y − 1 = 0 and
3 x + y − 23 = 0, we get
x = 7, y = 2
On solving x − 3 y − 1 = 0
and 3 x + y − 3 = 0, we get
x = 1, y = 0
Thus, the coordinates of P are (7,2)
or (1, 0).
261
DAY
2 The coordinates of A1 are a , b
2 2
a+ a b + b
2,
2
The coordinates of A2 are
2
2
X¢
a a
a b b
b
= + 2 + 3 , + 2 + 3
2 2
2 2 2
2
Continuing in this manner we observe
that the coordinates of A n are
a + a + a + ...+ a ,
2 22 23
2n
b b
b
b
+
+
+ ... + n
2 22 23
2
1
1
= a 1 − n , b 1 − n
2
2
C (x3, y3)
(k + I)
lx + kx1 ly 2 + ky 1
D 2
,
k + l
k + l
Coordinates of point P are
lx + kx1 + mx3 ly 2 + ky 1 + my 3
P 2
,
k + l + m
k + l + m
4 Let P (h, k ) be any point such that
2(PA ) = 3( PB )
4(PA )2 = 9(PB )2
⇒ 4(h2 + k 2 ) = 9 {(h − 4)2 + (k + 3)2 }
⇒ 4(h2 + k 2 ) = 9(h2 + k 2 − 8h + 6k + 25)
⇒ 5h + 5k − 72h + 54k + 225 = 0
2
∴ Required locus is
5x2 + 5y 2 − 72 x + 54 y + 225 = 0
5 Let ∠RPQ = θ and ∠RQP = φ
θ − φ = 2α
Let RM ⊥ PQ , so that RM = k ,
MP = a − h
MQ = a + h
RM
k
Then, tanθ =
=
MP
a− h
and
and
tan φ =
X
P
(a, 0)
=
2
⇒
⇒ a2 − h2 + k 2 = 2hk cot 2α
Hence, the locus is
Using Eq. (i), we have
2
8 k − 43 1
25
⇒
+ + (k − 5)2 =
6
2
9
x2 − y 2 + 2 xy cot 2α − a2 = 0
6 Here, tan θ = 2
So,
1
2
, sin θ =
5
5
cos θ =
RM
k
=
MQ
a+ h
X − 2Y
5
2X + Y
5
The equation 4 xy − 3 x = a reduces to
4( X − 2Y ) (2 X + Y )
⋅
5
5
2
2
2
X − 2Y
2
− 3
=a
5
⇒
4(2 X 2 − 2Y 2 − 3 XY )
− 3( X − 4 XY + 4Y ) = 5a
5X 2 − 20Y 2 = 5a2
2
⇒
∴
2
2
X − 4Y = a
2
2
1
1
25
(5) h + + (k − 5)2 =
2
2
6
On squaring, we get
2
1
25
⇒ h + + (k − 5)2 =
2
9
k (a + h ) − k (a − h )
a2 − h2 + k 2
and y = X sin θ + Y cos θ =
P
Coordinates of point D are
∴
M
Again, now 2α = θ − φ
∴
tan 2α = tan (θ − φ)
tan θ − tan φ
=
1 + tan θ tan φ
x = X cos θ − Y sin θ =
m
2
θ
O
For x and y, we have
= (a(1 − 2− n ), b(1 − 2− n ))
⇒
φ
Q
(–a, 0)
Y¢
The coordinates of A3 are
a+ a + a b + b + b
2 22 ,
b 2 22
=
2
2
k
I D
A (x1, y1) B (x2, y2)
(h − 1)2 + (k − 3)2
= (h + 2)2 + (k − 7)2
⇒ −2h + 1 − 6k + 9
= 4h + 4 − 14k + 49
…(i)
⇒ 6h − 8k + 43 = 0
Since, the area of triangle is 10 sq unit
{given}
1
ar (∆ABC ) = |BC || AC |
2
R(h, k)
a a b b
= + 2 , + 2
2 2 2 2
3
⇒
Y
2
7 Let the vertices be C , A and B ,
respectively. The altitude from A is
y − a ( t2 + t 3 )
= – t1
x − at 2 t 3
⇒ xt 1 + y = at 1 t 2 t 3 + a ( t2 + t 3 )
…(i)
The altitude from B is
xt 2 + y = at 1 t 2 t 3 + a (t 3 + t 1 ) …(ii)
On subtracting Eq. (ii) from Eq. (i), we
get x = − a
Hence, y = a( t 1 + t 2 + t 3 + t 1 t 2 t 3 )
So, the orthocentre is
{− a, a( t 1 + t 2 + t 3 + t 1 t 2 t 3 )}.
8 Given that, the triangle ABC is isosceles
∴ | AB | = | AC |
Let the coordinate of A are A(h, k )
⇒ (4k − 20)2 + 9(k − 5)2 = 25
⇒
25⋅ (k − 5)2 = 25
⇒
(k − 5)2 = 1
⇒
|k − 5| = 1
⇒
k − 5= ± 1
∴ k = 1 + 5 or k = − 1 + 5
⇒ k = 6 or k = 4
5
Using Eq. (i), we have h =
6
Using Eq. (i), we have
11
h=−
6
Therefore, the vertex A of the isosceles
5
11
∆ABC is A , 6 or A − , 4 .
6
6
α
9 ar (∆ PBC ) = 1 − 3
2
4
β 1
5 1
−2 1
1
⇒ ar (∆PBC ) = | 7α + 7β − 14|
2
7
= |α + β − 2|
2
6 −3 1
1
Also, ar (∆ABC ) =
−3 5 1
2
4 −2 1
1
7
⇒ ar (∆ABC ) = |42 − 21 − 14| =
2
2
ar (∆PBC )
= |α + β − 2|
ar (∆ ABC )
10 By PROJECTION FORMULA, we have
cos ∠ P1 OP2 =
|OP1 |2 + |OP2|2 − |P1 P2|2
2 |OP1 ||OP2|
Let E = |OP1 ||OP2|cos ∠ P1 OP2
Y
∴ (h − 1)2 + (k − 3)2
P2 (x2, y2)
C1(x1, y1)
= (h + 2)2 + (k − 7)2
O
X
262
( x12 + y 12 ) + ( x22 + y 22 )
− ( y 2 − y 1 )2 ]
⇒ E =
2
2 x1 x2 + 2 y 1 y 2
2
∴|OP1 ||OP2|cos ∠ P1 OP2 = x1 x2 + y 1 y 2
⇒ E =
11 The points A(0, 0), B(2, 2 3 ) and C (a, b )
are the vertices of an equilateral triangle
if
|AB| = |BC| = |CA|
⇒ |AB|2 = |BC|2 = |CA|2
⇒ 4 + 12 = (a − 2)2 + (b − 2 3 )2
= a2 + b 2
Now, (a − 2)2 + (b − 2 3 )2 = a2 + b 2
3b = 4
a = 4 − 3b
Also, a2 + b 2 = 16
(4 − 3b )2 + b 2 = 16
⇒ 4b 2 − 8 3b + 16 = 16
⇒
⇒
If
⇒
4b(b − 2 3 ) = 0
b = 0 or b = 2 3
b =0
a= 4
⇒
a= −2
[using Eq. (i)]
12 Let (h, k ) be the point on the locus. Then
by the given conditions
(h − a1 )2 + (k − b1 )2
= (h − a2 )2 + (k − b2 )2
⇒ 2h(a1 − a2 ) + 2k (b1 − b2 ) + a22 + b22
− a12 − b12 = 0
1
⇒ h(a1 − a2 ) + k (b1 − b2 ) + (a22 + b22
2
− a12 − b12 ) = 0 …(i)
Since, the locus of (h, k ) is the line
(a1 − a2 )h + (b1 − b2 )k + c = 0 …(ii)
∴ Comparing Eqs. (i) and (ii), we get
1
c = (a22 + b22 − a12 − b12 )
2
13 Circumcentre of a triangle is the point
a2 + b 2 − 4a − 4 3 b + 16 = a2 + b 2
a+
Case II |PC| = |RC|
or if b = 2 3
− {( x2 − x1 )2 + ( y 2 − y 1 )2 }
⇒ E =
2
[ x12 + x22 + y 12 + y 22 − ( x2 − x1 )2
…(i)
[using Eq. (i)]
which is equidistant from the vertices of
a triangle.
Let the circumcentre of triangle be
C ( x, y ) and the three vertices of the
triangle are represented by
P (2, 1), Q (5, 2), R(3, 4)
∴ According to given condition,
we have
|PC| = |QC| = |RC|
Case I |PC| = |QC|
( x − 2)2 + ( y − 1)2 = ( x − 5)2 + ( y − 2)2
…(i)
⇒
6 x + 2 y = 24
( x − 2)2 + ( y − 1)2 = ( x − 3)2 + ( y − 4)2
…(ii)
⇒
2 x + 6 y = 20
Solving Eqs. (i) and (ii) for x and y, we
have
∴Co-ordinates of circumcentre are
13 9
C ( x, y ) = C ,
4 4
14 Let P (h, k ) be the point such that
|PA|2 + |PB|2
⇒ (h − 2)2 + k 2 + (h + 2)2 + k 2
⇒
2h 2 + 8 + 2k 2
⇒
h2 + k 2
∴Locus of P is x2 + y 2 = 4
= |AB|2
= 42 + 0
= 16
=4
15. Let the third vertex be ( p, q ).
⇒
Now,
q = p+3
∆ = |5|
∆ =± 5
p q 1
1
2 1 1 =± 5
2
3 −2 1
…(i)
⇒
q + 3 p − 7 = ± 10
…(ii)
⇒
3 p + q = 17
and
3p + q = − 3
…(iii)
Solving Eqs. (i) and (ii) and solving
Eqs. (i) and (iii), we get points
7 , 13 and − 3 , 3
2 2
2 2
DAY TWENTY FIVE
Straight Line
Learning & Revision for the Day
u
Concept of Straight Line
u
Angle between Two Lines
u
Conditions for Concurrence
of Three Lines
u
Distance of a Point from a
Line
Concept of Straight Line
Any curve is said to be a straight line, if for any two points taken on the curve, each and
every point on the line segment joining these two points lies on the curve.
Y
A (x1, y1)
q
X
X¢
Y¢
The slope of a line AB is m = tan θ =
B (x2, y2)
y2 − y1
x2 − x1
Various Forms of Equations of a Line
The equation of a line in the general form can be written as ax + by + c = 0
1. Slope Intercept Form The equation of a line with slope m and making an intercept c on
Y-axis is y = mx + c
2. Point Slope Form The equation of a line which passes through the point ( x1 , y1 ) and has
the slope m is y − y1 = m ( x − x1 ).
3. Two Points Form The equation of a line passing through two points ( x1 , y1 ) and ( x2 , y2 ) is
y − y1
( y − y1 ) = 2
( x − x1 ).
x2 − x1
4. Intercept Form of a Line The equation of a line which cuts off intercepts a and b
x y
respectively from the X and Y-axes is + = 1.
a b
PRED
MIRROR
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
FIVE
264
5. Normal or Perpendicular Form The equation of the
straight line upon which the length of the perpendicular
from the origin is p and this perpendicular makes an
angle α with positive direction of X -axis in
anti-clockwise sense is
Conditions for Concurrence of
Three Lines
1. Three lines are said to be concurrent, if they pass through a
common point i.e. they meet at a point.
2. If three lines are concurrent, the point of intersection of two
lines lies on the third line.
3. The lines a1 x + b1 y + c1 = 0, a2 x + b2 y + c2 = 0
a1 b1 c1
and a3 x + b3 y + c3 = 0, are concurrent iff a2 b2 c2 = 0
a3 b3 c3
x cos α + y sin α = p, where 0 ≤ α ≤ 2 π.
6. General Equation of a Line to the Normal Form The
general equation of a line is
Ax + By + C = 0
Now, to reduce the general equation of a line to normal
form, we first shift the constant term on the RHS and
make it positive, if it is not so and then divide both sides
by (coefficient of x)2 + (coefficient of y)2
A
B
−C
x+
y=
2
2
2
2
A +B
A +B
A2 + B2
A
B
Now, take cos α =
, sin α =
and
2
2
2
A +B
A + B2
−c
, which gives the required normal form.
p=
2
A + B2
⇒
7. Intersection of lines Let the equation of lines be
a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0, then their point
b c − b2c1 c1 a2 − c2 a1
of intersection is 1 2
,
a1b2 − a2b1 a1b2 − a2b1
8. Distance Form or Parametric form The equation of the
straight line passing through ( x1 , y1 ) and making an
angle θ with the positive direction of X -axis is
x − x1
y − y1
=
= r,
cos θ
sin θ
where r is the distance of any point ( x, y) on the line
from the point ( x1 , y1 ).
This is the required condition of concurrence of three lines.
Distance of a Point from a Line
(i) The length of the perpendicular from a point ( x1 , y1 ) to a line
ax1 + by1 + c
ax + by + c = 0 is
.
a2 + b 2
(ii) Distance between two parallel lines
ax + by + c1 = 0 and ax + by + c2 = 0 is
Condition of Parallel Lines
Let m1 , m2 be slope of two lines, then lines are parallel, if
m1 = m2 .
Equation of any line parallel to ax + by + c = 0 can be taken
as ax + by + λ = 0
Condition of Perpendicular Lines
Let m1 , m2 be slope of two lines, then the lines are
perpendicular, if m1 m2 = − 1
If one line is parallel to X − axis, then its perpendicular line
is parallel to Y − axis
Equation of the line perpendicular to ax + by + c = 0 is taken
as bx − ay + λ = 0
Straight line ax + by + c = 0 and a′ x + b ′y + c′ = 0 are right
angle if aa′ + bb ′ = 0
a2 + b 2
.
Important Results
●
●
●
Angle between Two Lines
The acute angle θ between the lines having slopes m1 and m2
m2 − m1
is given by tan θ =
.
1 + m1 m2
|c2 − c1|
●
●
The foot of the perpendicular (h, k ) from ( x 1, y1) to the line
(ax 1 + by1 + c)
h − x1 k − y1
ax + by + c = 0 is given by
=−
=
a2 + b 2
a
b
Foot of perpendicular from (a, b ) on
a + b a + b
x − y = 0 is
,
.
2
2
Foot of perpendicular from (a, b ) on
a − b b − a
x + y = 0 is
,
.
2
2
Image (h, k ) from (x1 , y1 ) w.r.t. the line mirror ax + by + c = 0
is given by
h − x1 k − y1 −2 (ax1 + by1 + c)
=
=
a
b
a2 + b 2
Area of the parallelogram formed by the lines
a1 x + b 1 y + c 1 = 0; a2 x + b2 y + c2 = 0
(d − c )(d − c )
a1 x + b 1 y + d1 = 0; a2 x + b2 y + d2 = 0 is 1 1 2 2 .
a1b 2 − a2b 1
Equation of Internal and External Bisectors of
Angles between Two Lines
The bisectors of the angles between two straight lines are the
locus of a point which is equidistant from the two lines. The
equation of the bisector of the angles between the lines
…(i)
a1 x + b1 y + c1 = 0
and
…(ii)
a2 x + b2 y + c2 = 0
a x + b1 y + c1
a x + b2 y + c2
are given by, 1
=± 2
a12 + b12
a22 + b22
265
DAY
where,
(i) if a1 a2 + b1b2 > 0, the positive sign for obtuse and negative
sign for acute.
(ii) a1 a2 + b1b2 < 0, negative sign for obtuse and positive sign
for acute.
Equation of Family of Lines Through
the Intersection of Two given Lines
The equation of the family of lines passing through the
intersection of the lines
a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 is
( a1 x + b1 y + c1 ) + λ ( a2 x + b2 y + c2 ) = 0 .
where λ is a parameter.
Important Properties
(i) The position of a point ( x1 , y1 ) and ( x2 , y2 ) relative to the
line ax + by + c = 0
(ax1 + by1 + c)
(a) If
> 0, then points lie on the same side.
ax2 + by2 + c
(b) If
ax1 + by1 + c
< 0, then the points lie on opposite side.
ax2 + by2 + c
(ii) The equations of the straight lines which pass through a
given point ( x1 , y1 ) and make a given angle α with the
given straight line y = mx + c are
m ± tan α
( y − y1 ) =
( x − x1 ).
1 m tan α
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 The equation of the line, the reciprocals of whose
intercepts on the axes are a and b, is given by:
x y
+ =1
a b
(c) ax + by = ab
(a)
(b) ax + by = 1
(d) ax − by = 1
2 The equation of the straight line passing through the
point (4, 3) and making intercepts on the coordinate axes
whose sum is − 1, is
x
y
x
y
+ = – 1,
+ = −1
2 3
−2 1
x y
x
y
(b)
− = − 1,
+ = −1
2 3
−2 1
(a)
x
y
x
y
+ = 1, + = 1
2 3
2 1
x y
x
y
(d)
− = 1,
+ =1
2 3
−2 1
(c)
3
8
(c) −
3
2
(d) −
3
16
6 For which values of a and b, intercepts on axes by line
ax + by + 8 = 0 are equal and opposite in sign of
intercepts on axis by line 2x − 3y + 6 = 0
8
,b = − 4
3
8
(c) a = , b = 4
3
(a) a =
−8
,b = − 4
3
−8
(d) a =
,b = 4
3
(b) a =
7 Equation of the line passing through the points of
intersection of the parabola x 2 = 8y and the ellipse
x2
j JEE Mains 2013
+ y 2 = 1 is
3
8 A ray of light along x + 3y = 3 gets reflected upon
reaching X-axis, the equation of the reflected ray is
point divided the portion of line between axes in the ratio
5 : 3 internally, is
(b) 20x + 9y + 96 = 0
(d) 20x − 9y − 96 = 0
4 A straight line through a fixed point ( 2,3) intersects the
coordinate axes at distinct points P and Q. If O is the
origin and the rectangle OPRQ is completed, then the
locus of R is
(a) 3 x + 2 y = 6
(c) 3 x + 2 y = xy
(b) −
(a) y − 3 = 0 (b) y + 3 = 0 (c) 3 y + 1 = 0 (d) 3 y − 1 = 0
3 The equation of a line passing through ( − 4,3) and this
(a) 9x + 20y + 96 = 0
(c) 9x − 20y + 96 = 0
(a) − 3
(b) 2 x + 3 y = xy
(d) 3 x + 2 y = 6xy
5 If the x-intercept of some line L is double as that of the
line, 3x + 4y = 12 and the y-intercept of L is half as that
j
of the same line, then the slope of L is
JEE Mains 2013
(a) y = x + 3
(c) y = 3 x − 3
(b)
(d)
3y = x − 3
3y = x − 1
9 The range of values of α such that ( 0, α ) lie on or inside
the triangle formed by the lines
3x + y + 2 = 0, 2x − 3y + 5 = 0 and x + 4y − 14 = 0 is
(a) 1 / 2 ≤ α ≤ 1
(c) 5 ≤ α ≤ 7
(b) 5 / 3 ≤ α ≤ 7 / 2
(d) None of these
10 The lines x + y = | a | and ax − y = 1 intersect each other in
the first quadrant. Then, the set of all possible values of a
is the interval
(a) (−11
, ]
(c) [1, ∞)
(b) (0, ∞)
(d) (−1, ∞)
11 Area of the parallelogram formed by the lines
y = mx , y = mx + 1, y = nx , y = nx + 1 is equal to
FIVE
266
|m + n |
(m − n)2
1
(c)
|m + n |
2
|m + n |
1
(d)
|m − n |
(a)
21 If p is the length the perpendicular from the origin on the
(b)
line
12 If PS is the median of the triangle with vertices P( 2, 2),
Q( 6,−1) and R(7, 3), then equation of the line passing
j JEE Mains 2014
through (1, − 1) and parallel to PS is
(a) 4 x − 7 y − 11 = 0
(c) 4 x + 7 y + 3 = 0
(b) 2 x + 9y + 7 = 0
(d) 2 x − 9y − 11 = 0
13 If A( 2,−1) and B( 6, 5) are two points, then the ratio in
which the foot of the perpendicular from ( 4,1) to AB
divides it, is
(a) 8 : 15
(b) 5 : 8
(c) −5 : 8
(d) −8 : 5
x y
+ = 1 passes through the point
5 b
(13, 32). The line K is parallel to L and has the equation
x y
+ = 1. Then, the distance between L and K is
c 3
14 The line L given by
(a)
23
15
(b)
17
(c)
17
15
(d)
23
17
15 The nearest point on the line 3x − 4y = 25 from the
origin is
(a) (− 4,5)
(b) (3, − 4)
(c) (3,4)
(d) (3,5)
16 Two sides of rhombus are along the lines, x − y + 1 = 0
and 7x − y − 5 = 0 if its diagonals intersect at ( − 1, − 2),
then which one of the following is a vertex of this
j
rhombus
JEE Mains 2016
(a) (−3,−9)
1 − 8
(c) ,
3 3
(b) (−3,−8)
− 10 −7
(d)
,
3 3
18 For all real values of a and b lines
( 2a + b )x + (a + 3 b )y + (b − 3a ) = 0 and
mx + 2y + 6 = 0 are concurrent, then m is equal to
(c) − 4
(d) − 5
19 If p is the length of perpendicular from origin to the line
which intercepts a and b on axes, then
(a) a 2 + b 2 = p 2
(c)
1
1
2
+ 2 = 2
a2
b
p
1
p2
1
1
1
(d) 2 + 2 = 2
a
b
p
(b) a 2 + b 2 =
20 A straight line through the origin O meets the parallel
lines 4x + 2y = 9 and 2x + y = −6 at points P and Q,
respectively. Then, the point O divides the segment PQ
in the ratio
(a) 1 : 2
(b) 3 : 4
(c) 2 : 1
(b) 1
(d) None of these
22 If p and p′ be perpendiculars from the origin upon the
straight lines x sec θ + y cosec θ = a and
x cos θ − y sin θ = a cos 2 θ respectively, then the value of
the expression 4p 2 + p′ 2 is
(a) a 2
(b) 3a 2
(c) 2a 2
(d) 4a 2
23 If p1, p2 , p3 , are lengths of perpendiculars from points
(m 2 , 2m ), (mm′ , m + m′ ) and (m′ 2 ,2m′ ) to the line
sin2 α
x cos α + y sin α +
= 0, then p1, p2 , p3 are in
cos α
(a) AP
(c) HP
(b) GP
(d) None
24 The equation of bisector of acute angle between lines
3x − 4y + 7 = 0 and 12x + 5y − 2 = 0 is
(a) 21x + 77 y − 101 = 0
(c) 31x + 77 y + 101 = 0
(b) 11x − 3 y + 9 = 0
(d) 11x − 3 x − 9 = 0
25 A ray of light coming along the line 3x + 4y − 5 = 0 gets
reflected from the line ax + by − 1 = 0 and goes along the
line 5x − 12y − 10 = 0, then
64
112
,b =
115
15
64
8
(c) a =
,b = −
115
115
(a) a =
64
8
,b =
115
115
64
8
(d) a = −
,b = −
115
115
(b) a = −
x + 2y = 1, 3x + y + 5 = 0, x − y + 2 = 0 . The altitude
through B is
cx + 4 y + 1 = 0 are concurrent, then a, b, c are in
(a) AP
(b) GP
(c) HP
(d) None of these
(b) −3
(a) 0
(c) data is inconsistent
26 The sides BC, CA , AB of ∆ABC are respectively
17 If the lines ax + 2y + 1 = 0, bx + 3y + 1 = 0,
(a) − 2
x y
+ = 1 and a 2 , p 2 , b 2 are in AP then a 4 + b 4 =
a b
(d) 4 : 3
(a) x − 3 y + 1 = 0
(c) 3 x − y + 4 = 0
(b) x − 3 y + 4 = 0
(d) x − y + 2 = 0
27 A variable straight line drawn through the point of
x y
x y
+ = 1 and + = 1 meets
a b
b a
the coordinates axes at A and B, the locus of the
mid-point of AB is
intersection of the lines
(a) 2xy (a + b) = ab (x + y)
(b) 2xy (a − b) = ab (x − y)
(c) 2xy (a + b) = ab (x − y)
(d) None of the above
28 The base BC of ∆ABC is bisected at the point (p,q) and
equations of AB and AC are px + qy = 1 and qx + py = 1
respectively, then equation of the median passing
through A is
(a) (2 pq − 1) (px + qy − 1) = (p 2 + q 2 − 1) (qx + py − 1)
(b) (2 pq + 1) (px + qy − 1) = (p 2 + q 2 − 1) (qx + py − 1)
(c) (2 pq + 1) (px + qy − 1) = (p 2 + q 2 + 1) (qx + py − 1)
(d) None of the above
267
DAY
29 If P is a point ( x , y ) on the line y = − 3x such that P and
the point ( 3, 4) are on the opposite sides of the line
3x − 4y = 8, then
8
8
,y< −
15
5
8
8
(c) x =
,y = −
15
5
(a) x >
(b) x >
8
8
,y<
5
15
(d) None of these
α
α
,2+
be any point on a line, then the
2
2
range of values of α for which the point P lies between
the parallel lines x + 2y = 1 and 2x + 4y = 15 is
30 If P 1 +
4 2
5 2
<α<
3
6
−4 2
(c)
<α< 0
3
(a) −
(b) 0 < α <
5 2
6
(d) None of these
2
31 If (a, a ) falls inside the angle made by the lines
x − 2y = 0, x > 0 and y = 3x ( x > 0), then a belongs to :
(b) (3, ∞)
(d) (− 3, − 1 / 2)
(a) (01
, / 2)
(c) (1 / 2,3)
32 The lines passing through ( 3, − 2) and inclined at angle
60° with 3x + y = 1 is
(a) y + 2 = 0
(c)x + y = 2
(b) x + 2 = 0
(d) x − y = 3
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
33 Statement I Consider the points A( 0, 1) and B( 2, 0) and P
be a point on the line 4x + 3y + 9 = 0, then coordinates
− 12 17
of P such that | PA − PB | is maximum, is
, ⋅
5
5
Statement II | PA − PB | ≤ | AB |
34 Statement I If point of intersection of the lines
4x + 3y = λ and 3x − 4y = µ, ∀ λ , µ ∈ R is ( x1, y1 ), then
the locus of ( λ , µ ) is x + 7y = 0, ∀ x1 = y1.
Statement II If 4λ + 3 µ > 0 and 3λ − 4 µ > 0, then ( x1, y1 )
is in first quadrant.
35 Let θ1 be the angle between two lines 2x + 3y + c1 = 0
and − x + 5y + c 2 = 0 and θ 2 be the angle between two
lines 2x + 3y + c1 = 0 and − x + 5y + c 3 = 0, where
c1, c 2 , c 3 are any real numbers.
Statement I If c 2 and c 3 are proportional, then θ1 = θ 2 .
j JEE Mains 2013
Statement II θ1 = θ 2 for all c 2 and c 3 .
36 Statement I Each point on the line y − x + 12 = 0 is
Directions (Q. Nos. 33-36) Each of these questions contains
two statements : Statement I (Assertion) and Statement II
(Reason). Each of these questions also has four alternative
choices, only one of which is the correct answer. You have to
select one of the codes (a ), (b), (c) and (d ) given below.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
equidistant from the lines 4y + 3x − 12 = 0,
3y + 4x − 24 = 0.
Statement II The locus of a point which is equidistant
from two given lines is the angular bisector of the two
lines.
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 A line 4x + y = 1 through the point A ( 2, − 7) meets the
line BC whose equation is 3x − 4y + 1 = 0. The equation
to the line AC, so that AB = AC is
(a) 52 x − 89y − 519 = 0
(c) 52 x − 89y + 519 = 0
(b) 52 x + 89y − 519 = 0
(d) 52 x + 89y + 519 = 0
2 If the lines y = mr x , r = 1,2,3 cut off equal intercepts on the
transversal x + y = 1, then 1 + m1, 1 + m2 , 1 + m3 are in:
(a) AP
(c) HP
(b) GP
(d) None of these
3 In triangle ABC, equation of the right bisectors of the
sides AB and AC are x + y = 0 and x − y = 0
respectively. If A ≡ ( 5,7) then equation of side BC is
(a) 7 y = 5 x
(b) 5x = y
(c) 5 y = 7 x
(d) 5y = x
4 Let k be an integer such that the triangle with vertices
(k , − 3k ), ( 5, k ) and ( − k , 2) has area 28 sq units. Then,
the orthocentre of this triangle is at the point
1
(a) 2, −
2
3
(b) 1,
4
3
(c) 1, −
4
1
(d) 2,
2
5 A variable line through the point ( p, q ) cuts the x and y
axes at A and B respectively. The lines through A and B
parallel to Y − axis and the X − axis respectively meet at
P. If the locus of P is 3x + 2y − xy = 0, then
(a) p = 2, q = 3
(c) p = − 2,q = − 3
(b) p = 3, q = 2
(d) p = − 3,q = − 2
6 If the three lines x − 3y = p, ax + 2y = q and ax + y = r
from a right angled triangle, then
(a)a 2 − 9a + 18 = 0
(b) a 2 − 6a − 12 = 0
(c)a 2 − 6a − 18 = 0
j
JEE Mains 2013
FIVE
268
(d) a 2 − 9a + 12 = 0
7 A light ray emerging from the point source placed at
P(1, 3) is reflected at a point Q in the axis of x. If the
reflected ray passes through the point R( 6, 7), then the
j JEE Mains 2013
abscissa of Q is
(a) 1
7
(c)
2
(a) 25 x − 62 y + 86 = 0
(c) 25 x − 62 y = 86
12 Equation of the straight line which belongs to the system
(a) 4 x + 11y − 15 = 0
(c) 7 x + y − 8 = 0
8 Locus of the image of the point ( 2, 3) in the line
j
JEE Mains 2015
(b) straight line parallel toY-axis
(c) circle of radius 2
(d) circle of radius 3
and its one vertex (1,2), then equations of a side passing
through this vertex are
(a) 7 x
(b) 7 x
(c) 7 x
(d) 7 x
9 A square of side a lies above the X -axis and has one
vertex at the origin. The side passing through the origin
π
makes an angle α where, 0 < α < with the positive
4
direction X -axis. The equation of its diagonal not passing
through the origin is
y (cos α − sinα) − x (sinα − cos α) = a
y (cosα + sinα) − x (sinα − cosα) = a
y (cosα + sinα) + x (sinα + cosα) = a
y (cosα + sinα) + x (sinα − cosα) = a
+
−
+
+
23 y − 53 = 0, 23 x − 7 y − 9 = 0
23 y − 53 = 0,23 x + 7 y − 9 = 0
23 y + 53 = 0,23 x − 7 y + 9 = 0
23 y + 53 = 0, 23 x + 7 y + 9 = 0
14 The equations of the straight lines through ( − 2, − 7) and
having intercept of length 3 between the lines
4x + 3y = 12 and 4x + 3y = 3 is
(a) 7 x − 24 y − 182 = 0
(c) 7 x + 24 y − 182 = 0
(b) 7 x + 24 y + 182 = 0
(d) None of these
15 If angle between lines ax + by + p = 0 and
π
and these lines with other line
4
x sin α − y cos α = 0 are concurrent, then a 2 + b 2 =
x cos α + y sin α = p is
10 A variable line ‘L’ is drawn through O( 0,0) to meet the line
L1 : y − x − 10 = 0 and L2 : y − x − 20 = 0 at the point A and
B respectively. A point P is taken on ‘L’ such that
2
1
1
. Locus of ‘P’ is
=
+
OP OA OB
(a) 3 x + 3 y = 40
3 x + 3 y + 40 = 0
(c) 3 x − 3 y = 40
(b) 3 x − 4 y + 1 = 0
(d) None of these
13 One diagonal of a square is along the line 8x − 15y = 0
( 2x − 3y + 4) + k ( x − 2y + 3) = 0, k ∈ R , is a
(a)
(b)
(c)
(d)
(b) 62 x − 25 y + 86 = 0
(d) 5 x − 2 y − 7 = 0
of straight lines a( 2x + y − 3) + b ( 3x + 2y − 5) = 0 and is
farthest from the point ( 4, − 3) is
(b) 3
5
(d)
2
(a) straight line parallel to X-axis
x − y + 1 + λ 2 ( 2x − y − 2) = 0. Equation of a straight line
that belongs to both families is
(b)
(d) 3 y − 3 x = 40
11 Consider the family of lines
(a) 1
(b) 2
(c) 3
(d) 4
16 Straight lines y = mx + c1 and y = mx + c 2 , where m ∈R +
meet the X − axis at A1 and A2 respectively and Y − axis
at B1 and B2 respectively. It is given that point A1, A2 , B1
and B2 are concyclic. Locus of intersection of lines A1B2
and A2B1 is
(a) y = x
(c) xy = 1
(b) y + x = 0
(d) xy + 1 = 0
ANSWERS
5x + 3y − 2 + λ 1 ( 3x − y − 4) = 0 and
SESSION 1
SESSION 2
1 (b)
2 (d)
3 (c)
4 (c)
5 (d)
6 (d)
7 (d)
8 (b)
9 (b)
10 (c)
11 (d)
12 (b)
13 (b)
14 (d)
15 (b)
16 (c)
17 (a)
18 (a)
19 (d)
20 (b)
21 (c)
22 (a)
23 (b)
24 (b)
25 (c)
26 (b)
27 (a)
28 (a)
29 (a)
30 (a)
31 (c)
32 (a)
33 (d)
34 (b)
35 (a)
36 (a)
1 (d)
11 (d)
2 (c)
12 (b)
3 (a)
13 (a)
4 (d)
14 (b)
5 (a)
15 (b)
6 (a)
16 (b)
7 (d)
8 (c)
9 (b)
10 (d)
269
DAY
Hints and Explanations
SESSION 1
2
3
+
= 1 ⇒ 2k + 3h = hk
h
k
So, locus is 3 x + 2 y = xy
5 We have, x + y = 1
4
3
For line L, x-intercept = 2 × 4 = 8
1
3
y-intercept = × 3 =
2
2
x
y
3
∴Line L is +
= 1, Slope, m = −
8 3/2
16
∴
1 If a1 , b1 are intercepts of the line on the
axes, then
1 / a1 = a,1 / b1 = b
⇒
a1 = 1 / a, b1 = 1 / b
∴ Equation of the line is
x / a1 + y / b1 = 1 or ax + by = 1
2 Let x −intercept = a
and y −intercept = b
Since, a + b = − 1 ⇒ b = − (a + 1)
y
x
∴ Equation of line is −
=1
a a+ 1
Clearly,
4a + 4 − 3a
4
3
−
=1 ⇒
=1
a a+ 1
a (a + 1)
⇒ a + 4 = a2 + a ⇒ a = ± 2
Hence, equation of line is
x
y
x
y
−
= 1 or
+
= 1.
2
3
−2
1
y
3 Let the equation of line be x + = 1
a
b
∴ According to given condition,
we have
3a 5b
C (− 4, 3) ≡ C ,
8 8
32
24
and b =
a= −
⇒
3
5
∴ Equation of line is
5y
3x
−
+
=1
32
24
Y
3
(0, b)
O
(a, 0)
X
⇒
ax + by = −8
y
x
+
= 1 (intercept form)
8
8
−
−
a
b
x
y
Also, 2 x − 3 y = − 6 ⇒ − +
=1
3
2
According to given condition,
we have
8
8
− = − (− 3) and − = − 2
a
b
8
⇒
a = − and b = 4
3
⇒
7 On solving both the equations, we get
8y
+ y2 = 1
3
⇒
3 y2 + 8 y − 3 = 0
⇒ (3 y − 1) ( y + 3) = 0
1
⇒ y = − 3, here y ≠ − 3
3
1
2
At y = , x = ± 2
3
3
So, the points of intersection are
2 1
2 1
2 , and −2 , .
3 3
3 3
B(0, 1)
3y = x – 1
4 Equation of PQ is
Q(0, k)
X¢
A
y+3x+2=0
5
7
≤α≤
3
2
10 As x + y =|a|and ax − y = 1.
Intersect in first quadrant.
So, x and y-coordinates are positive.
1 + |a|
a|a|− 1
≥ 0 and y =
≥0
∴ x=
1+ a
a+ 1
⇒
1 + a ≥ 0 and a|a|− 1 ≥ 0
⇒
a ≥ −1 and a|a|≥ 1 …(i)
If −1 ≤ a < 0 ⇒ − a2 ≥ 1 [not possible]
If a ≥ 0 ⇒ a2 ≥ 1 ⇒ a ≥ 1 ⇒ a ∈ [1, ∞ )
11 Let lines OB : y = mx, CA : y = mx + 1
BA : y = nx + 1 and OC : y = nx
So, the point of intersection B of OB and
1
.
AB has x-coordinate
m−n
Y
A
C
X¢
B
D
X
O
Y¢
Now, area of a parallelogram
OBAC = 2 × Area of ∆OBA
1
1
1
= 2 × × OA × DB = 2 × ×
2
2 m−n
1
1
=
=
m − n |m − n|
depending upon whether m > n or m < n.
y
x
+
=1
h
k
Y
0
–5=
x
2
7/2 3y–
B x+4y
–14=0
5/3
X
Y
C
–2
8 Take any point B (0,1 ) on given line.
⇒ − 9 x + 20 y − 96 = 0
⇒ 9 x − 20 y + 96 = 0
the triangle,
6 ax + by + 8 = 0
From option (d); 3 y − 1 = 0 is the
required equation which satisfied the
intersection points.
5
9 From figure, for (0,α ) to be inside or on
12 Coordinate of
7 + 6 3 − 1 13
S =
,
= , 1
2
2 2
A( 3, 0)
R(h, k)
B¢(0, –1)
(2,3)
O P(h,0)
X
Y¢
Since, it is passes through the points
(2,3)
Equation of AB ′ is
y−0 =
[since, S is mid-point of line QR]
P (2,2)
−1 − 0
0− 3
( x − 3)
⇒
− 3y = − x + 3
⇒
x − 3y = 3
⇒
3y = x − 3
Q
(6,–1)
S
R
(7,3)
FIVE
270
−2
.
9
Required equation of line passes through
(1, − 1) and parallel to PS is
−2
y + 1=
( x − 1)
9
⇒
2x + 9y + 7 = 0
21 p =
A (1,2)
Slope of the line PS is
0
1=
7x–y–5=0
y+
x–
B
D
E(–1,–2)
P (4, 1)
C
⇒ Equation of BD is
⇒
∴
A
(2, – 1)
λ
D
1
B
(6, 5)
Let line AB is divided by PD in the ratio
λ :1, then intersecting point
6λ + 2 5λ − 1
D
,
lies on
λ+1 λ+1
2 x + 3 y − 11 = 0.
6λ + 2
5λ − 1 − 11 = 0
⇒ 2
+ 3
λ+1
λ+1
⇒
16λ − 10 = 0 ⇒ λ :1 = 5: 8
14 Since, the line L is passing through the
point (13,32).
13 32
32
8
Therefore,
+
=1 ⇒
=−
5
b
b
5
⇒ b = − 20
The line K is parallel to the line L, its
equation must be
y
x y
x
−
= a or
−
=1
5 20
5a 20a
x y
On comparing with +
= 1, we get
c
3
20a = −3, c = 5a
−3
−3 −3
and c = 5 ×
a=
=
20
20
4
Hence, the distance between lines
−3
−1
|a − 1|
23
20
=
=
=
1
1
17
17
+
25 400
400
16 Coordinates of A ≡ (12
, )
∴
Slope of AE = 2
⇒ Slope of BD = −
1
2
x + 2y + 5 = 0
1 − 8
Coordinates of D = ,
3 3
ax + 2 y + 1 = 0,bx + 3 y + 1 = 0,
cx + 4 y + 1 = 0 are concurrent
a 2 1
∴
b 3 1 =0
c 4 1
⇒
− a + 2b − c = 0 ⇒ 2b = a + c
∴ a, b, c are in AP.
(2a + b )x + (a + 3b )y + (b − 3 a) = 0
and mx + 2 y + 6 = 0 are concurrent for
all real values of a and b, so they must
represent the same line for some values
of a and b. Therefore, we get
2a + b (a + 3b ) b − 3 a
=
=
m
2
6
On taking last two ratios,
a + 3b
−3 a + b
3
=
⇒b = − a
2
6
4
On taking first two ratios,
2(2a + b ) 2{2a − (3 / 4) a}
m=
=
a + 3b
a + 3(− 3 /4) a
10
=−
= −2
5
19 The length of perpendicular from (0,0) to
∴
y
x
+
= 1, is p =
a
b
ab
a2 + b 2
a2 , p2 , b 2 are in AP.
2a2b 2
= a2 + b 2
⇒
a2 + b 2
22 Since, p = length of the perpendicular
from (0,0) on x secθ + y cosec θ = a
a
asin 2θ
=
2
sec2 θ + cosec2 θ
∴ p=
…(i)
⇒ 2 p = asin 2θ
Also, p′ = length of perpendicular from
(0,0) on x cos θ − y sinθ = acos 2θ
acos 2θ
∴
p′ =
cos 2 θ + sin2 θ
…(ii)
= acos 2θ
On squaring and adding Eqs. (i), (ii), we
get
4 p2 + p ′2 = a2
2
23 p1 = m2 cos α + 2m sin α + sin α
cos α
18 Given equations,
15 The desired point is the foot of the
perpendicular from the origin on the
line 3 x − 4 y = 25.
The equation of a line passing through
the origin and perpendicular to
3 x − 4 y = 25is 4 x + 3 y = 0.
Solving these two equations we get
x = 3, y = − 4.
Hence, the nearest point on the line
from the origin is (3, − 4).
y + 2 −1
=
2
x+1
17 It is given that the lines
line
=
⇒ a4 + b 4 = 0 i.e. a = b = 0
This is impossible, therefore given
information is inconsistent.
13 Let P(41
, ) and PD⊥ AB .
Equation of AB is 3 x − 2 y − 8 = 0
∴ Equation of PD is 2 x + 3 y − 11 = 0
1
(1 / a2 ) + (1 / b 2 )
1
1 1
+
a2 b 2
1
1
1
+ 2 = 2
a2
b
p
20 Now, distance of origin from
4 x + 2 y − 9 = 0 is
|−9|
9
=
2
2
20
4 +2
and distance of origin from
2 x + y + 6 = 0 is
|6|
6
=
2
2
5
2 +1
9 / 20 3
∴ Required ratio =
=
4
6/ 5
p2 = |mm ′ cos α + (m + m ′ )sin α +
p3 = m ′2 cos α + 2m ′ sin α +
sin2 α
cos α
sin2 α
cos α
mm ′ cos 2 α + (m + m ′ )sin α cos α
+ sin2 α
p2 =
cos α
(m cos α + sin α )2
p1 =
cos α
(m ′ cos α + sin α )2
p3 =
cos α
∴ p2 = p1 p3 ⇒ p22 = p1 p3
Hence, p1 , p2 and p3 are in GP.
24 Make constant terms of both equation
positive.
3x − 4y + 7 = 0
and
− 12 x − 5y + 2 = 0
Since, a1 a2 + b1 b2 = − 36 + 20 < 0
∴ Bisector of acute angle is given by
with positive sign
− 12 x − 5y + 2
3x − 4y + 7
= +
9 + 16
144 + 25
⇒ 39 x − 52 y + 91 = − 60 x − 25y + 10
⇒ 99 x − 27 y + 81 = 0
∴
11 x − 3 y + 9 = 0
25 Equation of bisectors of the given lines
are
3x + 4y −
32 + 42
∴
5x − 12 y − 10
5
=±
52 + (− 12)2
(39 x + 52 y − 65) = ± (25x
− 60 y − 50)
271
DAY
⇒
or
⇒
or
∴
or
14 x + 112 y − 15 = 0
64 x − 8 y − 115 = 0
14
112
x+
y −1= 0
15
15
64
8
x−
y −1= 0
115
115
14
112
a=
,b =
15
15
64
8
a=
,b = −
115
115
−
⇒ (2 pq − 1)( px + qy − 1)
= ( p2 + q 2 − 1)(qx + py − 1)
At (3, 4), L 1 = 9 − 16 − 8 = − 15 < 0
For the point P ( x, y ), we should have
L 1 > 0.
3x − 4y − 8 > 0
⇒
3 x − 4 (− 3 x ) − 8 > 0
⇒
x > 8/15 and − y − 4 y − 8 > 0
⇒
y < − 8/ 5
4+
⇒
− 5+
∴
2 <0 ⇒
6α
2
−4 2
5 2
<α<
3
6
y=3x
P(a, a2)
y=x/2
X
O
⇒
Equation of line AB
px + qy = 1
Equation of line AC
qx + py = 1
The equation of line passing through the
intersection point of above lines is
px + qy − 1 + λ (qx + py − 1) = 0
which passes through ( p,q )
∴ p2 + q 2 − 1
+ λ ( pq + pq − 1) = 0 …(i)
p2 + q 2 − 1
⇒
λ= −
2 pq − 1
∴ Substituting the value of λ in Eq. (i),
of the line through A is ( px + qy − 1)
y=
1
32
l2(m2)
y + 2= 0
y + 2 = 3 ( x − 3)
33 Equation of line AB is
⇒
Here,
0−1
( x − 0)
2− 0
x + 2y − 2 = 0
|PA − PB | ≤ | AB |
Thus, for|PA − PB | to be maximum,
A, B and P must be collinear.
34 The point of intersection of lines
4 x + 3 y = λ and 3 x − 4 y = µ is
4λ + 3µ
x1 =
25
3λ − 4µ
and y 1 =
25
4λ + 3µ 3λ − 4 µ
=
Q
x1 = y 1 ⇒
25
25
⇒
λ + 7µ = 0
Hence, locus of a point (λ, µ ) is
x + 7 y = 0.
= 1 = tan 45°
⇒
θ1 = 45°
Also, the angle between the lines
2 x + 3 y + c 1 = 0 and
− x + 5y + c 3 = 0 is θ2 .
1/ 5 + 2/3
13 / 15
tan θ2 =
=
∴
1 − 2 / 15
13 / 15
= 1 = tan 45°
⇒
θ2 = 45°
Here, we observe that the value of c 1 , c 2
and c 3 is not depend on measuring the
angle between the lines.
So, c 2 and c 3 are proportional or for all c 2
and c 3 θ1 = θ2
x+
C
.
1
< a < 3.
2
Ö3
D( p,q)
( x − 3)
2 x + 3 y + c1 = 0
and − x + 5 + c 2 = 0 is θ1 .
1/ 5 + 2/3
13 / 15
∴
tan θ1 =
=
1 − 2 / 15
13 / 15
2
Y
⇒
or
− 3± 3
1 m (− 3 ) 3
35 Here, angle between the lines
31 Clearly, a2 − a > 0, a2 − 3a < 0
A
B
4 2
α +
3
<0
5 2
α −
6
3α
y + 2=
y −1=
2
2
the parallel lines x + 2 y = 1 and
2 x + 4 y = 15, therefore
α
α
1 +
+ 2 2 +
−1
2
2
<0
α
α
2 1 +
+ 4 2 +
− 15
2
2
∴ ( x + y ) ab = 2 xy (a + b )
28
[Q y = − 3 x]
30 Since, P 1 + α , 2 + α lies between
y
y
x
x
+
− 1 + λ +
− 1 = 0
b
a
b
a
meets the coordinate axes at
1+ λ
1+ λ
A
, 0 and B 0 ,
1
λ
1
λ
+
+
a b
b
a
The mid-point of AB is given by
1+ λ
1+ λ
2x =
, 2y =
1
λ
1
λ
+
+
a b
b
a
1
1
⇒ (1 + λ ) +
y
x
1
λ
1
λ
=2
+ + 2 +
a b
b
a
1 1
= 2 (1 + λ ) +
a b
⇒
[QP ( x, y ) lies on y = − 3 x]
27 The intersection of given lines is
Now, equation of lines l 1 and l 2 are
y + 2 = tan(θ ± 60° )( x − 3)
tan θ ± tan 60°
⇒ y + 2=
( x − 3)
1 m tan θ tan 60°
⇒
29 Let L1 = 3 x − 4 y − 8
26 The required line is given by
x + 2 y − 1 + λ ( x − y + 2) = 0 …(i)
It is perpendicular to 3 x + y + 5 = 0
5
∴ 3 (1 + λ ) + 2 − λ = 0 ⇒ λ = −
2
From Eq. (i), we get
x − 3y + 4 = 0
p2 + q 2 − 1
(qx + py − 1) = 0
2 pq − 1
60°
36 Equation of bisector of
60°
l1(m1)
A(3,–2)
(m=tan θ)
l
Let l 1 and l 2 are the equations of the
lines inclined at an angle of 60° with the
line l.
∴Slopes of lines are tan(θ ± 60° )
4 y + 3 x − 12 = 0
and 3 y + 4 x − 24 = 0 is
4 y + 3 x − 12
3 y + 4 x − 24
=±
16 + 9
9 + 16
⇒
y − x + 12 = 0
and 7 y + 7 x − 36 = 0
So, the line y − x + 12 = 0 is the
angular bisector.
FIVE
272
⇒
C = (7,5)
Equation of BC is 7 y = 5x.
SESSION 2
1
A (2, –7)
4 Given, vertices of triangle are
(k , − 3k ), (5, k ) and (− k , 2).
k
− 3k 1
1
∴
5
k
1 = ± 28
2
−k
2
1
4x + y = 1
B 3x – 4y + 1 = 0 C
⇒
⇒
− 3k 1
k
1 = ± 56
2
1
k
5
−k
⇒
Let m be the slope of AC, then
3
3
+ 4
m−
4
tan B = tanC ⇒ 4
=
1 − 3 1 + 3m
4
19 4 m − 3
52
⇒ −
=
⇒ m= −
8
4 + 3m
89
or
⇒ k (k − 2) + 3k (5 + k ) + 1(10 + k 2 )
= ± 56
⇒ k 2 − 2k + 15k + 3k 2 + 10 + k 2
= ± 56
⇒
5k 2 + 13k + 10 = ± 56
⇒
5k 2 + 13k − 66 = 0
or
5k 2 + 13k − 46 = 0
⇒
k =2
[Qk ∈ I ]
Thus, the coordinates of vertices of
triangle are A(2, − 6), B(5, 2) and C(− 2, 2).
∴ Equation of AC is
52
y + 7= −
( x − 2)
89
⇒ 52 x + 89 y + 519 = 0
2 AB = BC ⇒ B is mid-point of AC.
Y
y=m3 x
C (–2, 2)
D
B (5, 2)
(2, 1/2)
y=m2 x
E
C
y=m1 x
X¢
X
O
B
A (2, –6)
A
x+y=1
1
m1
Clearly, A =
,
m1 + 1 m1 + 1
1
m2
B=
,
m2 + 1 m2 + 1
1
m3
C=
,
m
+
1
m
3
3 + 1
2
1
1
Now,
=
+
m2 + 1 m1 + 1 m3 + 1
∴ m1 + 1, m2 + 1, m3 + 1 are in HP.
3 Clearly, B = reflection of A (57
, ) in the
line x + y = 0
A(5,7)
x+y=0
and y = q − mp or mp = q − y
…(ii)
On eliminating m from Eqs. (i) and (ii),
we get
p− x
p
=
q
q − y
x–y=0
Y¢
Now, equation of altitude from vertex A
−1
is y − (− 6) =
( x − 2)
2−2
− 2 − 5
⇒
x=2
…(i)
Equation of altitude from vertex C is
−1
y −2=
[ x − (− 2)]
2 − (− 6)
5− 2
…(ii)
⇒ 3 x + 8 y − 10 = 0
On solving Eqs. (i) and (ii), we get
1
x = 2 and y = .
2
1
∴ Orthocentre = 2,
2
5 Let the equation of the variable line be
y − q = m( x − p ), where m is a variable
mp − q
A ≡
, 0
m
Then,
B
C
⇒
B ≡ (− 7, − 5)
C = reflection of A (57
, ) in the line
x− y =0
and B ≡ (0,q − mp )
Let P ≡ ( x, y ), then
mp − q
or m( p − x ) = q
x=
m
…(i)
pq − qx − py + xy = pq
py + qx = xy
p
q
+
=1
x
y
This is the locus of P.
But locus of P is 3 x + 2 y = xy (given)
2
3
or
+
=1
x
y
∴
and
p =2
q =3
6 Case I Let line l 1 ≡ x − 3 y = p and
l 2 ≡ ax + 2 y = p are perpendicular, then
1
a
× − = −1
3
2
⇒
a= 6
Case II Let line l 2 ≡ ax + 2 y = p and
l 3 ≡ ax + y = r are perpendicular, then
−a
× − a = −1
2
⇒
a2 = − 2 [not possible]
Case III Let line l 3 ≡ ax + y = r and
l 1 ≡ x − 3 y = p are perpendicular, then
1
− a × = − 1 ⇒ a = 3. So, formation of
3
quadratic equation in a, whose roots are
3 and 6, is
⇒
a2 − (6 + 3) a + (6 ⋅ 3) = 0
a2 − 9 a + 18 = 0
7 Here, QS ⊥ OX
S
R
(6, 7)
P
(1, 3)
q
Q(k, 0)
X¢
X
It means QS bisect the ∠PQR.
Then, ∠PQS = ∠RQS
⇒
∠RQX = ∠PQO = θ
[let]
⇒
∠XQP = 180° − θ
y − y1
Slope of QR = tanθ = 2
x2 − x1
=
7− 0
6− k
…(i)
Slope of QP = tan(180° − θ) = − tan θ
y − y1
3− 0
…(ii)
= 2
=
x2 − x1
1− k
273
DAY
∴ From Eqs. (i) and (ii),
7
3
=−
6− k
1− k
⇒
=
π/4
a
5
Hence, the coordinate of Q is , 0 .
2
8 Key Idea First of all find the point of
inter- section of the lines 2 x − 3 y + 4 = 0
and x − 2 y + 3 = 0 (say A). Now, the
line (2 x − 3 y + 4) + k ( x − 2 y + 3) = 0 is
the perpendicular bisector of the line
joining points P(2, 3) and image P ′(h, k ).
Now, AP = AP ′ and simplify.
Given line is
(2 x − 3 y + 4) + k ( x − 2 y + 3) = 0,
k ∈ R …(i)
This line will pass through the point of
intersection of the lines
…(ii)
2x − 3y + 4 = 0
and
x − 2y + 3 = 0
…(iii)
On solving Eqs. (ii) and (iii), we get
X′¢
Thus, point of intersection of lines (ii)
and (iii) is (1, 2).
Let M be the mid-point of PP ′, then AM
is perpendicular bisector of PP ′ (where,
A is the point of intersection of given
lines).
2x
–3
y+
4
=0
P (2, 3)
M
A
x–
=0
+3
2y
Clearly,
⇒
P¢ (h, k)
AP = AP ′
(2 − 1)2 + (3 − 2)2
= (h − 1)2 + (k − 2)2
⇒
2=
h2 + k 2 − 2h − 4 k + 1 + 4
⇒
2=
h2 + k 2 − 2h − 4 k + 5
⇒ h + k 2 − 2h − 4 k + 5 = 2
⇒ h2 + k 2 − 2h − 4 k + 3 = 0
Thus, the required locus is
x2 + y 2 − 2 x − 4 y + 3 = 0
which is an equation of circle with
radius = 1 + 4 − 3 = 2
2
9 Slope of the diagonal = tan 3 π + α
4
− 1 + tan α
=
1 + tan α
3π
+α
4
a
O
X
13
D
Y′¢
C
The equation is
y − a sin α sin α − cos α
=
x − a cos α sin α + cos α
p
4
⇒ y (cos α + sin α )
− x (sin α − cos α ) = a
0
y= A
15
–
8x
m(q)
10 Let equation of the line OAB be
x
y
=
=r
cos θ sin θ
⇒
x = r cos θ, y = r sin θ.
For A, let OA = r1 then A (r1 cos θ, r1 sin θ)
lies on L1 .
x = 1, y = 2
(1, 2)
2 x + y − 3 = 0 and 3 x + 2 y − 5= 0.
i.e. Through (1,1).
The line of this family which is farthest
from (4, − 3) is the line through (1,1) and
perpendicular to the line joining (1,1)
and (4, − 3).
∴ Required line is y − 1 = 3 / 4( x − 1)
i.e. 3 x − 4 y + 1 = 0
Y
7 − 7k = − 18 + 3k
10k = 25
5
k =
2
⇒
sin α − cos α
sin α + cos α
B
L=0
P
A
O
L1: y=x+10
L2 : y=x+20
sin θ − cos θ
1
1
⇒
=
=
10
OA r1
1
1 sin θ − cos θ
Similarly,
=
=
OB
r2
20
Let P = (h, k ) = (r cos θ, r sin θ)
2
2
1
1
Then,
= =
+
OP
r
r1
r2
2 sin θ − cos θ sin θ − cos θ
⇒
=
+
r
10
20
⇒
40 = 3r sin θ − 3r cos θ
= 3k − 3h
⇒ Locus of P is 3 x − 3 y + 40 = 0
⇒
3 y − 3 x = 40
11 Lines 5x + 3 y − 2 + λ1 (3 x − y − 4) = 0 are
B(1, 2)
As can be seen from the figure, AB and
AD are the line segments inclined at an
angle of 45° with the diagonal line
AC (8 x − 15y = 0)
Now, slope of line AB = m AB
π
⇒ m AB = tan θ −
4
where, θ is the angle of inclination of
diagonal with the positive X–axis.
tan θ − 1 8 / 15− 1
=
∴ m AB =
1 + tan θ 1 + 8 / 15
7
⇒ m AB = −
23
∴ Equation of line AB is
7
y − 2 = − ( x − 1)
23
⇒ 23 y − 46 = − 7 x + 7
∴ 7 x + 23 y = 53
Also, slope of line
23
AD (⊥ AB ) = m BC =
7
23
∴
y −2=
( x − 1)
7
⇒
23 x − 7 y = 23 − 14
∴
23 x − 7 y = 9
14 Let m be the slope of the line and angle θ
it makes with the parallel line.
concurrent at the point of intersection of
the lines 5x + 3 y − 2 = 0 and
3 x − y − 4 = 0 i.e. at A (1, − 1).
(–2, –7)
θ
Similarly, lines
x − y + 1 + λ2 (2 x − y − 2) = 0 are
concurrent at B (3,4).
The line, that belongs to both families is
AB, whose equation is
−1− 4
y − 4=
( x − 3)
1−3
i.e.
5x − 2 y − 7 = 0.
12 a(2 x + y − 3) + b (3 x + 2 y − 5) = 0 passes
through the point of intersection of the
lines
p
4
4x+3y=12
3
θ
9/5
4x+3y=3
∴
sin θ =
3
3
or tan θ =
5
4
Hence, slope of the parallel lines is −
4
.
3
FIVE
274
4
3 = tan θ = 3
4m
4
1−
3
3m + 4
3
7
=±
⇒ m= −
⇒
4
24
3 − 4m
7
The line is y + 7 = −
( x + 2) or
24
7 x + 24 y + 182 = 0.
∴
m+
15 According to the question,
a cos α
− +
π
b
sin α
tan =
acos α
4
1+
b sin α
b cos α − asin α
⇒
1=
acos α + b sin α
⇒ acos α + b sin α = b cos α − asin α
⇒ (a − b )cos α = − (b + a)sin α
b−a
…(i)
⇒
tanα =
b+ a
Intersection point of ax + by + p = 0 and
y = x tanα given by is ax + bx tanα = p
p
⇒
x=
a + b tanα
p tan α
and
y=
a + b tan α
Intersection point of x cos α + y sin α = p
and y = x tanα is given by
x cos α + x tan α sin α = p
⇒
x = p cos α, y = p cos α tan α
y = p sinα
According to the question,
p
… (ii)
x=
= p cos α
a + b tan α
p tan α
and y =
…(iii)
= p sin α
a + b tan α
p
p
∴
=
b − a secα
a+ b
b + a
p
p
⇒
=
b − a
sec2 α
a+ b
b + a
p
=
1 + tan2 α
⇒
b+ a
=
ab + a2 + b 2 − ab
=
⇒
1
b − a
1+
b + a
[using Eq. (i)]
b+ a
(b + a)2 + (b − a)2
a2 + b 2 = 2(a2 + b 2 )
⇒
(a2 + b 2 )2 = 2(a2 + b 2 )
⇒
(a2 + b 2 )(a2 + b 2 − 2) = 0
⇒
a + b ≠0
∴
a2 + b 2 = 2
2
2
2
16 A1 B1 ≡ y = mx + c 1
A2 B2 ≡ y = mx + c 2
Y
B2
B1
O
∴
A2
A1
X
c
A1 = − 1 , 0 , B1 = (0,c 1 ),
m
c
A2 = − 2 , 0 , B2 = (0,c 2 )
m
Since A1 , A2 , B1 , B2 are concyclic,
c1c2
= c1c2
m2
OA1 OA2 = OB1 OB2 ⇒
∴
∴
m2 = 1 ⇒ m = 1(m > 0)
A1 = (− c 1 , 0), A2 = (− c 2 ,0),
B1 = (0,c 1 ), B2 = (0,c 2 )
y
x
Now, A1 B2 ≡ −
+
=1
c1
c2
y
x
and A2 B1 ≡ −
+
=1
c2
c1
For point of intersection, consider
y
y
x
x
+
=− +
c1 c2
c2 c1
−
(c 1 − c 2 ) x + (c 1 − c 2 ) y = 0 ⇒ x + y = 0
DAY TWENTY SIX
The Circle
Learning & Revision for the Day
u
u
u
u
u
Concept of Circle
Line and Circle
Equation of Tangents
Equation of Normal
Pair of Tangents
u
u
u
u
Common Tangents of Two
Circles
Director Circle
Chord of Contact
Pole and Polar
u
u
u
u
Angle of Intersection of Two
Circles
Family of Circles
Radical Axis
Coaxial System of Circles
Concept of Circle
Circle is the locus of a point which moves in a plane, such that its distance from a fixed point
in the plane is a constant. The fixed point is the centre and the constant distance is the radius.
Standard Form of Equation of a Circle
The equation of a circle whose centre is at (h, k ) and radius r is given, is ( x − h)2 + ( y − k )2 = r 2 . It
is also known as the central form of the equation of a circle.
The equation x2 + y2 + 2 gx + 2 fy + c = 0 always represents a circle, whose centre is (− g, − f ) and
radius is g2 + f 2 − c . This is known as the general equation of a circle.
PRED
MIRROR
Equation of Circle when the End Points of a
Diameter are Given
If A and B are end points of a diameter of a circle whose coordinates are ( x1 , y1 ) and ( x2 , y2 ),
respectively.
Then, the equation of circle is ( x − x1 ) ( x − x2 ) + ( y − y1 )( y − y2 ) = 0
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
NOTE Parametric equations of a circle is ( x − h) 2 + ( y − k ) 2 = r 2 , where x = h + r cosθ, y = k + r sin θ,
No. of Correct Questions (z)—
(Without referring Explanations)
0 ≤ θ ≤ 2 π.
Intercept on Axes
The length of intercepts made by the circle x2 + y2 + 2 gx + 2 fy + c = 0 with X and Y-axes are
2 g2 − c and 2 f 2 − c , respectively.
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
SIX
276
y − b = m ( x − a) ± r 1 + m2
Position of a Point w.r.t. a Circle
and the coordinates of the points of contact are
mr
r
,b +
a ±
.
1 + m2
1 + m2
A point ( x1 , y1 ) lies outside, on or inside a circle
S = x2 + y2 + 2 gx + 2 fy + c = 0
according as S1 >, =, or < 0,
S1 = x12 + y12 + 2 gx1 + 2 fy1 + c
where,
Greatest and least distance of a point A( x1 , y1 ) from a circle
with centre C and radius r as shown in the figure below, is
| AC + r | and| AC − r |.
(ii) Point of intersection of the tangent drawn to the circle
x2 + y2 = r 2 at the point P(α ) and Q(β) is
r cos α + β
2
α +β
2
and k =
h=
.
α −β
α −β
cos
cos
2
2
r sin
P (α)
r
C
A (x1, y1)
R
(h , k )
Line and Circle
Q (β)
Let y = mx + c be a line and x2 + y2 = r 2 be a circle. If r is the
radius of circle and p =
c
is the length of the
1 + m2
perpendicular from the centre on the line, then
(i) p > r ⇔ the line passes outside the circle.
(ii) p = r ⇔ the line touches the circle or the line is a tangent
to the circle.
(iii) p < r ⇔ the line intersect the circle at two points or the
line is secant of the circle.
(iv) p = 0 ⇔ the line is a diameter of the circle.
●
The length of the intercept cut-off from the line
y = mx + c by the circle x2 + y2 = r 2 is
Y
(x2, y2) Q
(i) Equation of tangent for x2 + y2 = r 2 at ( x1 , y1 ) is
xx1 + yy1 = r 2 .
(ii) The equation of the tangent at the point P( x1 , y1 ) to a circle
x2 + y2 + 2 gx + 2 fy + c = 0 is
xx1 + yy1 + g( x + x1 ) + f ( y + y1 ) + c = 0.
3. Parametric form
(i) Parametric coordinates of circle x2 + y2 = r 2 is
(r cos θ, r sin θ), then equation of tangent at (r cos θ, r sin θ)
is x cos θ + y sin θ = r .
(ii) Equation of the tangent to the circle ( x − h)2 + ( y − k )2 = r 2
at (h + r cos θ, k + r sin θ) = r 2 is
( x − h) cos θ + ( y − k )sin θ = r .
Length of the Tangents
M
P (x1, y1)
X¢
2. Point Form
X
O
The length of the tangent from the point P( x1 , y1 ) to the circle
x2 + y2 + 2 gx + 2 fy + c = 0 is equal to
y = mx + c
Y¢
PQ = 2
r 2 (1 + m2 ) − c2
1 + m2
Equation of Tangents
x12 + y12 + 2 gx1 + 2 fy1 + c = S1
Equation of Normal
The normal at any point on a curve is a straight line which is
perpendicular to the tangent to the curve at that point.
1. Point Form
The equation of normal to the circle
A line which touch only one point of a circle is called its
tangent as shown in the following figure. This tangent may be
in slope or point form as given below.
x2 + y2 + 2 gx + 2 fy + c = 0 or x2 + y2 = a2
x − x1
y − y1
at any point ( x1 , y1 ) is
=
x1 + g y1 + f
1. Slope Form
or
(i) The equation of tangents of slope m to the circle
( x − a)2 + ( y − b )2 = r 2 are given by
x
y
=
x1
y1
277
DAY
(v) When two circles are separately as shown in the figure
below, four common tangents are possible.
2. Parametric Form
The equation of normal to the circle x2 + y2 = a2
x
y
or y = x tan θ.
(a cos θ, a sin θ) is
=
cos θ sin θ
at point
C1
C2
Pair of Tangents
From a given point, two tangents can be drawn to a circle
which are real and distinct, coincident or imaginary according
as the given point lies outside, on or inside the circle.
The combined equation of the pair of tangents drawn from a
point P( x1 , y1 ) to the circle x2 + y2 = a2 is SS1 = T 2.
where, S = x2 + y2 − a2 , S1 = x12 + y12 − a2 and T = xx1 + yy1 − a2
Common Tangents of Two Circles
Let the centres and radii of two circles be C1 , C2 and r1 , r2 ,
respectively.
(i) When one circle contains other as shown in the figure
below, no common tangent is possible. Condition
C1C2 < r1 − r2 .
C1
C2
Condition, C1C2 > r1 + r2
Director Circle
The locus of the point of intersection of two perpendicular
tangents to a given circle is known as its director circle. The
equation of the director circle of the circle
x2 + y2 = a2 is x2 + y2 = 2 a2 .
Chord of Contact
●
●
●
(ii) When two circles touch internally as shown in the figure,
one common tangent is possible.
The chord joining the points of contact of the two tangents
from a point, which is outside is called the chord of contact
of tangents.
The equation of the chord of contact of tangents drawn
from a point ( x1 , y1 ) to the circle x2 + y2 = a2 is xx1 + yy1 = a2
or T = 0.
If AB is a chord of contact of tangents from C to the circle
x2 + y2 = r 2 and M is the mid-point of AB as shown in
figure. Then,
B
O
C1
C2
A
M
α
r
C
(x1, y1)
A
Condition, C1C2 = r1 − r2
(iii) When two circles intersect as shown in the figure below,
two common tangents are possible.
Angle between two tangents ∠ ACB is 2 tan −1
r
.
S1
Chord Bisected at a Given Point
The equation of the chord of the circle
C2
C1
A
x2 + y2 = a2
bisected at the point ( x1 , y1 ) is given by
Condition,| r1 − r2| < C1C2 < r1 + r2
(iv) When two circles touch externally as shown in the figure
below, three common tangents are possible.
C1
B
C2
A
Condition, C1C2 = r1 + r2
A divides C1C2 externally in the ratio r1 : r2 . B divides C1C2
internally in the ratio r1 : r2 .
xx1 + yy1 − a2 = x12 + y12 − a2
or
T = S1
Of all the chords which passes through a given point M (a, b )
inside the circle the shortest chord is one whose middle point
is (a, b ).
Pole and Polar
If through a point P( x1 , y1 ) (inside or outside a circle) there be
drawn any straight line to meet the given circle at Q and R as
shown in the following figure, the locus of the point of
intersection of the tangents at Q and R is called the polar of P
and P is called the pole of the polar.
SIX
278
The polar of a point P( x1 , y1 ) with respect to the circle
x2 + y2 = a2 as shown in the below figure is xx1 + yy1 = a2 or
T = 0.
R
P (x1, y1)
T
Q
1. Conjugate Points
Two points A and B are conjugate points with respect to a
given circle, if each lies on the polar of the other with respect
to the circle.
2. Conjugate Lines
If two lines be such that the pole of one line lies on the other,
then they are called conjugate lines with respect to the given
circle.
2. The equation of the family of circles passing through the
point A( x1 , y1 )and B( x2 , y2 ) is
x y 1
( x − x1 )( x − x2 ) + ( y − y1 )( y − y2 ) + λ x1 y1 1 = 0
x2 y2 1
⇒
( x − x1 )( x − x2 ) + ( y − y1 )( y − y2 ) + λL = 0
where, L = 0 represents the line passing through A( x1 , y1 )
and B( x2 , y2 ) and λ ∈ R.
3. The equation of the family of circles touching the circle
S = x2 + y2 + 2 gx + 2 fy + c = 0 at point P( x1 , y1 ) is
x2 + y2 + 2 gx + 2 fy + c
+ λ { xx1 + yy1 + g( x + x1 ) + f ( y + y1 ) + c} = 0
or
S + λL = 0
where, L = 0 is the equation of the tangent to S = 0 at ( x1 , y1 )
and λ ∈ R.
4. The equation of a family of circles passing through the
intersection of the circles
S1 = x2 + y2 + 2 g1 x + 2 f1 y + c1 = 0
and S2 = x2 + y2 + 2 g2 x + 2 f2 y + c2 = 0 is
S1 + λS2 = 0,where (λ ≠ −1) is an arbitrary real number.
Angle of Intersection of Two Circles
Radical Axis
The angle of intersection of two circles is defined as the angle
between the tangents to the two circles at their point of
r 2 + r22 − d2
intersection is given by cos θ = 1
2 r1 r2
The radical axis of two circles is the locus of a point which
moves in such a way that the lengths of the tangents drawn
from it to the two circles are equal.
where, d is distance between centres of the circles.
Orthogonal Circles
Two circles are said to be intersect orthogonally, if their angle
of intersection is a right angle.
(Radius of Ist circle)2 + (Radius of IInd circle)2
= (Distance between centres)2
⇒
2(g1 g2 + f1 f2 ) = c1 + c2
which is the condition of orthogonality of two circles. The
circles having radii r1 and r2 intersect orthogonally. Then,
2 r1 r2
length of their common chord is
.
2
r1 + r22
Family of Circles
1. The equation of a family of circles passing through the
intersection of a circle x2 + y2 + 2 gx + 2 fy + c = 0 and line
L ≡ lx + my + n = 0 is
x2 + y2 + 2 gx + 2 fy + c + λ (lx + my + n) = 0 or S + λL = 0
where, λ is any real number.
The radical axis of two circles S1 = 0
and S2 = 0 is given by, S1 − S2 = 0.
(i) The equations of radical axis and the common chord of
two circles are identical.
(ii) The radical axis of two circles is always perpendicular to
the line joining the centres of the circles.
(iii) Radical centre The point of intersection of radical axis of
three circles whose centres are non-collinear, taken in
pairs is called their radical centre.
Coaxial System of Circles
A system of circles is said to be coaxial system of circles, if
every pair of the circles in the system has the same radical
axis.
1. If the equation of a member of a system of coaxial circles is
S = 0 and the equation of the common radical axis is L = 0,
then the equation representing the coaxial system of circle
is S + λL = 0, where λ ∈ R.
2. If S1 = 0 and S2 = 0 are two circles, then S1 + λS2 = 0
S1 + λ(S1 − S2 ) = 0 or S2 + λ(S1 − S2 ) = 0, λ ∈ R
represent a family of coaxial circles having S1 − S2 = 0 as
the common radical axis.
279
DAY
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 Equation of a circle whose two diameters are along the
lines 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 and passes
through the origin is
(a) x 2 − y 2 +
2 Points (2, 0), (0, 1), (4, 5), and (0, a) are concyclic. Then
a is equal to
14
or 1
3
14
or −1
(c) −
3
(b) 14 or
(d) None of these
equation x 2 + 2ax − 4 = 0 and their ordinates are the roots
of the equation x 2 + 2bx − 9 = 0. Then equation of the
circle with AB as diameter is
(a) x + y − ax − bx + 13 = 0
(b) x 2 + y 2 + ax + by − 13 = 0
(c) x 2 + y 2 + 2ax + 2by − 13 = 0
(d) None of the above
2
coordinate axes in concyclic points, then
(b) a1b1 = a2 b2
(d) None of these
(2, 0) and whose centre is the limit of the point of
intersection of the lines 3x + 5y = 1 and
( 2 + c )x + 5 c 2 y = 1 as c → 1 is
6 The lines 3x − y + 3 = 0 and x − 3y − 6 = 0 cut the
coordinate axes at concyclic points. The equation of the
circle through these points is
(a) x + y − 5 x − y − 6 = 0 (b) x + y + 5 x + y + 6 = 0
(c) x 2 + y 2 + 2 xy = 0
(d) None of these
2
2
2
7 The circle passing through (1, − 2) and touching the
X -axis to at ( 3, 0) also passes through the point
j
(a) (− 5, 2)
(c) (5, − 2)
JEE Mains 2013
8 AB is chord of the circle x + y = 25 . The tangents of
2
A and B intersect at C. If ( 2 , 3) is the mid-point of AB,
then area of the quadrilateral OACB is
(a) 50
13
3
(b) 50
3
13
11 Circles are drawn through the points (a, b ) and (b, − a )
such that the chord joining the two points subtends an
angle of 45° at any point of the circumference. Then, the
distance between the centres is
(c) 50 3
3 times the radius of either circle
(b) 2 times the radius of either circle
1
(c)
times the radius of either circle
2
(d) 2 times the radius of either circle
the tangents at the points B(1, 7) and D( 4, − 2) on the circle
meet at C, then find the area of the quadrilateral ABCD.
(a) 78
(b) 75
(c) 79
(d) 85
13 Find the equation of a circle concentric with the circle
(a)
(b)
(c)
(d)
x 2 + y 2 − 6x + 12 y − 15 = 0
x 2 + y 2 − 6x − 12 y + 15 = 0
x 2 + y 2 − 6x + 12 y + 15 = 0
None of the above
14 The equation of the locus of the mid-points of the chords
of the circle 4x 2 + 4y 2 − 12x + 4y + 1 = 0 that subtend an
angle of 2π / 3 at its centre is
(a) x 2 + y 2 + 3 x − y + 31 / 16 = 0
(b) x 2 + y 2 − 3 x + y + 31 / 16 = 0
(c) x 2 + y 2 − 3 x + y − 31 / 16 = 0
(d) None of the above
15 Let PQ and RS be tangents at the extremities of the
(b) (2, − 5)
(d) (−2, 5)
2
(a) x 2 + y 2 − 9x + 2ky + 14 = 0
(b) 3 x 2 + 3 y 2 + 27 x − 2ky + 42 = 0
(c) x 2 + y 2 − 9x − 2ky + 14 = 0
(d) x 2 + y 2 − 2kx − 9y + 14 = 0
x 2 + y 2 − 6x + 12y + 15 = 0 and has double of its area.
25 (x 2 + y 2 ) − 20x + 2 y + 60 = 0
25 (x 2 + y 2 ) − 20x + 2 y − 60 = 0
25 (x 2 − y 2 ) − 20x − 2 y − 60 = 0
None of the above
2
(d) None of these
12 Let A be the centre of the circle x 2 + y 2 − 2x − 4y − 20 = 0. If
5 The equation of circle which passes through the points
(a)
(b)
(c)
(d)
2y
− 1= 0
3
10 Circles are drawn through the point (2, 0) to cut intercept
(a)
4 If the lines a1x + b1y + c1 = 0 and a 2 x + b2 y + c 2 = 0 cut the
(a) a1a2 = b1b2
(c) a1b1 = a2 b1
(b) x 2 + y 2 −
of length 5 units on the X -axis. If their centres lie in the
first quadrant, then their equation for k > 0 is
1
3
3 The abscissae of two points A and B are the roots of the
2
(1, 0) and the third vertex lies above the X-axis. Find the
equation of its circumcircle.
2y
+ 1= 0
3
y
(c) x 2 − y 2 −
=0
3
(a) x 2 + y 2 + 2 x − 44 y = 0
(b) 17 x 2 + 17 y 2 − 2 x + 44 y = 0
(c) 17 x 2 + 17 y 2 + 2 x − 44 y = 0
(d) None of the above
(a)
9 Two vertices of an equilateral triangle are ( − 1, 0) and
(d)
50
3
diameter PR of the circle of radius r. If PS and RQ
intersect at a point X on the circumference of the circle,
then 2r equals
(a) (PQ ⋅ RS)
(c) 2 PQ ⋅ RS / (PQ + RS)
(b) (PQ + RS) / 2
(PQ 2 + RS 2 )
(d)
2
SIX
280
16 If the line ax + by = 0 touches the circle
26 The length of the common chord of two circles
x 2 + y 2 + 2x + 4y = 0 and is normal to the circle
x 2 + y 2 − 4x + 2y − 3 = 0, (a, b ) is given by
(a) (2, 1)
(b) (1,−2)
(d) (−1, 2)
(c) (1, 2)
17 A circle touches the hypotenuse of a right angle triangle
at its middle point and passes through the mid-point of
the shorter side. If a and b (a < b ) be the length of the
sides, then the radius is
b 2
a + b2
a
b
(c)
a2 + b 2
4a
(a)
(b)
b
a2 − b 2
2a
(d) None of these
18 The number of common tangents to the circles
x 2 + y 2 − 4x − 6y − 12 = 0 and x 2 + y 2 + 6x + 18y + 26 = 0
is
JEE Mains 2015
(a) 1
(b) 2
(c) 3
(d) 4
19 If the two circles ( x − 1)2 + ( y − 3)2 = r 2 and
x 2 + y 2 − 8x + 2y + 8 = 0 intersect in two distinct points,
then
(a) 2 < r < 8
(c) r = 2
(b) r < 2
(d) r > 2
20 Let C be the circle with centre at (1, 1) and radius 1. If T is
the circle centred at (0, k) passing through origin and
touching the circle C externally, then the radius ofT is
equal to
JEE Mains 2014
(a)
3
2
3
2
1
(d)
4
(b)
1
(c)
2
also touches the circle x 2 + y 2 − 8x + 6y + 20 = 0, find
its point of contact.
(b) x = 3, y = − 1
(d) None of these
22 If the tangent at (1, 7) to the curve x 2 = y − 6 touches the
circle x 2 + y 2 + 16x + 12y + c = 0, then the value of c is
(a) 195
(b) 185
JEE Mains 2018
(d) 95
(c) 85
23 For the circle x + y = r , find the value of r for which
2
2
2
the area enclosed by the tangents from the point P ( 6 , 8)
to the circle and the chord of contact is maximum.
(a) r = 4
(b) r = 5
(c) r = 3
(d) r = 1
24 From any point on the circle x + y = a tangents are
2
2
2
drawn to the circle x 2 + y 2 = a 2 sin2 α. The angle
between them is
(a) α / 2
(b) α
(c) 2α
(d) None of these
25 If one of the diameters of the circle, given by the
equation x 2 + y 2 − 4x + 6y − 12 = 0, is a chord of a circle
S, whose centre is at ( −3, 2), then the radius of S is
(a) 5 2
(b) 5 3
(c) 5
(a)
4 c 2 + 2 (a − b)2
(b)
4 c 2 − (a − b)2
(c)
4 c − 2 (a − b)
(d)
2c 2 − 2 (a − b)2
2
2
27 If P , Q and R are the centres and r1, r2 and r3 are the
corresponding radii of the three circles form a system of
coaxial circle, then r12 ⋅ QR + r22 ⋅ RP + r32 ⋅ PQ is equal to
(a) PQ ⋅ QR ⋅ RP
(b) −PQ ⋅ QR ⋅ RP
PQ
(d)
× RP
QR
(c) PQ + QR + RP
28 The condition that the chord x cos α + y sin α − p = 0 of
x 2 + y 2 − a 2 = 0 may subtend a right angle at the centre of
circle, is
(a) a 2 = 2 p 2 (b) p 2 = 2a 2
(c) a = 2 p
(d) p = 2a
29 The equation of the circle of minimum radius which
contains the three circles x 2 + y 2 − 4y − 5 = 0,
x 2 + y 2 + 12x + 4y + 31 = 0
and x 2 + y 2 + 6x + 12y + 36 = 0 is
2
2
31
23
5
(a) x − + y −
949
= 3 −
18
12
36
2
2
2
2
2
2
2
23
31
5
(b) x +
949
+ y − = 3 +
12
18
36
31
23
5
(c) x +
949
= 3 +
+ y +
18
12
36
(d) None of the above
30 The centres of two circles C1 and C2 each of unit radius
21 The tangent to the circle x 2 + y 2 = 5 at the point (1, − 2),
(a) x = 2, y = 1
(c) x = 5, y = 7
( x − a )2 + ( y − b )2 = c 2 and ( x − b )2 + ( y − a )2 = c 2 is
JEE Mains 2016
(d) 10
are at a distance of 6 units from each other. Let P be the
mid-point of the line segment joining the centres of C1
and C2 and C be a circle touching circles C1 and C2
externally. If a common tangent to C1 and C passing
through P is also a common tangent to C2 and C, then
the radius of the circle C is
(a) 16
(b) 4
(c) 8
(d) 2
Direction (Q. Nos. 31-35) Each of these questions
contains two statements : Statement I (Assertion) and
Statement II (Reason). Each of these questions also has four
alternative choices, only one of which is the correct answer.
You have to select one of the codes (a ), (b), (c) and (d) given
below.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
31 Tangents are drawn from the point (17, 7) to the circle
x 2 + y 2 = 169.
Statement I The tangents are mutually perpendicular.
Statement II The locus of the points from which mutually
perpendicular tangents can be drawn to the given circle
is x 2 + y 2 = 338.
281
DAY
32 Consider the radius should be zero in limiting points.
Statement I Equation of a circle through the origin and
belonging to the coaxial system, of which the limiting
points are (1, 1) and ( 3, 3) is
2x 2 + 2y 2 − 3x − 3y = 0.
Statement II Equation of a circle passing through the
point, (1, 1) and (3, 3) is x 2 + y 2 − 2x − 6y + 6 = 0.
33 Statement I The circle of smallest radius passing through
two given points A and B must be of radius 1 AB.
2
Statement II A straight line is a shortest distance
between two points.
34 Consider L1 ≡ 2x + 3y + p − 3 = 0,
L2 ≡ 2x + 3y + p + 3 = 0 , where p is a real number
and C ≡ x 2 + y 2 + 6x − 10y + 30 = 0.
Statement I If line L1 is a chord of circle C, then L2 is not
always a diameter of circle C.
Statement II If line L1 is a diameter of circle C, then L2 is
not a chord of circle C.
35 Statement I The only circle having radius 10 and a
diameter along line 2x + y = 5 is
x 2 + y 2 − 6x + 2y = 0.
Statement II 2x + y = 5 is a normal to the circle
JEE Mains 2013
x 2 + y 2 − 6x + 2y = 0
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 If the line y = x + 3 meets the circle x 2 + y 2 = a 2 in A and
B, then equation of the circle on AB as diameter is
(a) x 2 + y 2 + 3 x − 3 y − a 2 + 9 = 0
(b) x 2 + y 2 − 3 x + 3 y − a 2 + 9 = 0
(c) x 2 + y 2 + 3 x + 3 y − a 2 + 9 = 0
(d) None of the above
(a) 0
touches both the axes. Equation of the circle having
centre at (6, 5) and touching the circle C1 externally is
(a) x 2 + y 2 − 12 x − 10y + 52 = 0
(b) x 2 + y 2 − 12 x − 10y + 12 = 0
(c) x 2 + y 2 − 12 x − 10y − 52 = 0
(d) None of the above
(b) 1
(d) None of these
4 Tangents PA and PB are drawn to x 2 + y 2 = a 2 from the
point P ( x1, y1 ). Equation of the circumcircle of triangle
PAB is
(b) x 2 + y 2 + xx1 − yy1 = 0
(d) x 2 + y 2 + xx1 + yy1 = 0
5 If a > 2b > 0 then the positive value of m for which
y = mx − b (1 + m 2 ) is a common tangent tox 2 + y 2 = b 2
and ( x − a )2 + y 2 = b 2 is
2b
a 2 − 4b 2
(b)
a 2 − 4b 2
2b
(c)
2b
1 − 2b
(d)
b
a − 2b
6 The straight line 2x − 3y = 1 divides the circular region
x 2 + y 2 ≤ 6 into two parts.
(d) 4
parallel to the side CD and AB = 2CD. Let AD be
perpendicular to AB and CD. If a circle is drawn inside
the quadrilateral ABCD touching all the sides, then its
radius is
(a) 3
x 2 + y 2 + 2gx + 2fy + c = 0 for all values of θ, then
(g 2 + f 2 + c ) / (g + f + c ) is equal to
(a)
(c) 2
(b) 2
(c) 3 / 2
(d) 1
8 If one of the diameters of the circle
3 The line ( x − 2) cos θ + ( y − 2) sin θ = 1 touches the circle
(a) x 2 + y 2 − xx1 − yy = 0
(c) x 2 + y 2 − xx1 + yy1 = 0
(b) 1
7 Let ABCD be a quadrilateral with area 18, with side AB
2 A circle C1 of radius 2 units lies in the first quadrant and
(a) 5
(c) –15
1 1 1
3 5 3 1
If S = 2, , , , , − , , , then the number
4
2
4
4
4 8 4
of point(s) in S lying inside the smaller part is
S = x 2 + y 2 − 2x − 6y + 6 = 0 is the common chord to the
circle C with centre ( 2, 1), then the radius of the circle is
(a) 1
(b) 3
(c) 3
(d) 2
9 A rational point is a point both of whose coordinates are
rational numbers. Let C be any circle with centre ( 0, 2 ).
Then, the maximum number of rational points on the
circle is
(a) 0
(c) Infinitely many
(b) 2
(d) None of these
10 Let L1 be a line passing through the origin and L2 be the
line x + y = 1. If the intercepts made by the circle
x 2 + y 2 − x + 3y = 0 on L1 and L2 are equal, then L1 is
(a) x + y = 0
(c) x + 7 y = 0
(b) x + y = 2
(d) x − 7 y = 0
11 If a n , n = 1, 2, 3, 4 represent four distinct positive real
numbers other than unit such that each pair of the
logarithm of a n and the reciprocal of logarithm denotes a
point on a circle, whose centre lies on Y-axis. Then, the
product of these four members is
(a) 0
(b) 1
(c) 2
(d) 3
SIX
282
12 The set of values of a for which the point ( 2a , a + 1) is an
16 A ray of light incident at the point (3, 1) gets reflected
from the tangent at ( 0, 1) to the circle x 2 + y 2 = 1. The
reflected ray touches the circle. The equation of the line
along which the incident ray moves is
interior point of the larger segment of the circle
x 2 + y 2 − 2x − 2y − 8 = 0 made by the chord x − y + 1 = 0,
is
5 9
(a) ,
9 5
9
(c) 0,
5
5
(b) 0,
9
9
(d) 1,
5
(a) 3 x + 4 y − 13 = 0
(c) 4 x + 3 y − 13 = 0
17 The number of integral values of λ for which
x 2 + y 2 + λx + (1 − λ )y + 5 = 0 is the equation of a circle
whose radius cannot exceed 5, is
13 Three concentric circles of which biggest circle is
x 2 + y 2 = 1,have their radii in AP. If the line y = x + 1 cuts
all the circles in real and distinct points, then the interval
in which the common difference of AP will lie, is
1
(a) 0, 1 −
2
(a) 14
circles x 2 + y 2 − 2λ i x = c 2 , (i = 1, 2, 3) are in G.P., then
the lengths of the tangents drawn to them from any point
on the circle x 2 + y 2 = c 2 are in
(b) GP
(d) None of these
(b) P ∈ [−1, 0) ∪ [0, 1] ∪ [1, 2)
(d) None of these
x 2 + y 2 − 4x − 8y − 5 = 0 at (a, b ). Then k, (a, b ) is
(a) 15, (5, 0)
(c) Both (a) and (b)
(b) −35, (−1, 8)
(d) None of these
20 If the equation of the circle obtained by reflecting the
circle x 2 + y 2 − a 2 = 0 in the line y = mx + c is
x 2 + y 2 + 2gx + 2 fy + c = 0, then
by x 2 + y 2 + 2gx + c + λ( x 2 + y 2 + 2fy + k ) = 0 Subtend a
tight angle at the origin, if
c
k
−
=2
g2 f 2
c
k
(c) 2 − 2 = 2
g
f
(d) 10
19 The line 3x − 4y − k = 0 touches the circle
15 The limiting points of the coaxial system of circles given
2cm
4c 2
2
,
a
+
c
=
1+ m2
1+ m2
2cm 2
4c 2
(b) g = −
,a + c =
2
1+ m
1+ m2
4c
4
c2
(c) f =
, a2 + c =
2
1+ m
1+ m2
(d) None of the above
(a) g =
c
k
+
= −2
g2 f 2
c
k
(d) 2 + 2 = 2
g
f
(a) −
(c) 16
less than or equal to x), lying inside the region bounded
by the circles x 2 + y 2 − 2x − 15 = 0 and
x 2 + y 2 − 2x − 7 = 0, then
(a) P ∈ [−1, 2) − {0, 1}
(c) P ∈ (−1, 2)
14 If the distance from the origin of the centres of the three
(a) AP
(c) HP
(b) 18
18 The point ([P + 1], [P ]) (where [ x ] is the greatest integer
1
(b) 0,
2
1
1
(d) 0, 1 −
2
2
(c) (11
, )
(b) 4 x − 3 y − 10 = 0
(d) 3 x − 4 y − 5 = 0
(b)
ANSWERS
SESSION 1
SESSION 2
1 (c)
2 (a)
3 (c)
4 (a)
5 (b)
6 (a)
7 (c)
8 (b)
9 (b)
10 (c)
11 (d)
12 (b)
13 (a)
14 (b)
15 (a)
16 (c)
17 (c)
18 (c)
19 (a)
20 (d)
21 (b)
22 (d)
23 (b)
24 (c)
25 (b)
26 (c)
27 (b)
28 (a)
29 (c)
30 (c)
31 (a)
32 (b)
33 (b)
34 (c)
35 (c)
1 (a)
11 (b)
2 (a)
12 (c)
3 (a)
13 (d)
4 (a)
14 (b)
5 (a)
15 (d)
6 (c)
16 (a)
7 (b)
17 (d)
8 (b)
18 (d)
9 (b)
19 (c)
10 (c)
20 (a)
283
DAY
Hints and Explanations
2
1
∴ Centre = , −
5 25
SESSION I
1 Equation of two diameters are
2 x − 3 y + 4 = 0 and 3 x + 4 y − 5 = 0
∴ Centre is (−1 / 17, 22 / 17)
Circle passes through origin
∴ Equation of the circle is
2
2
2
2
x + 1 + y − 22 = 1 + 22
17
17
17
17
2
44
2
2
x−
y =0
⇒ x + y +
17
17
2
2
⇒ 17 x + 17 y + 2 x − 44 y = 0
2 Let circle be
x + y + 2gx + 2fy + c = 0
This passes through (2, 0), (0, 1), (4, 5)
So, 4 + 4g + c = 0, 1 + 2 f + c = 0,
41 + 8g + 10 f + c = 0
Solving these equations, we get
g = −13 / 6, f = −17 / 6, c = 14 / 3.
So, circle is
3( x2 + y 2 ) − 13 x − 17 y + 14 = 0
Since (0, a) also lies on it, we get
∴
3a2 − 17a + 14 = 0
So, a = 1 or 14 / 3
2
9 Length of side of triangle is 2.
∴ Equation of circle is
2
2
x − 2 + y + 1 = 2 −
5
25
2
2
1
+ 2
5
25
⇒ 25 ( x2 + y 2 ) − 20 x + 2 y − 60 = 0
Hence, the points are concyclic.
∴ L1 L2 + λx y = 0
⇒ (3 x − y + 3) ( x − 3 y − 6) + λ x y = 0
3( x2 + y 2 ) + (λ − 10)xy − 15x − 3 y − 18 = 0
Now for circle coefficient of x y = 0
∴ 3( x + y ) − 15x − 3 y − 18 = 0
⇒
2
Y-axis in A(− c 1 / a1 ,0) and B (0,−c 1 / b1 )
and the line
a2 x + b2 y + c 2 = 0 cut axes in
C (− c 2 / a2 ,0) and D(0, − c 2 / b2 ) ,
So, AC and BD are chords along X and
Y-axes intersecting at origin O.
Since A, B , C , D are concyclic so
OA ⋅ OC = OB ⋅ OD
c c c c
or − 1 − 2 = − 1 2
a1 a2 b1 b2
⇒
7 Let the equation of circle be
( x − 3) 2 + ( y − 0) 2 + λy = 0
B¢ O
(–3, 0)
A(3, 0)
X¢
P (1, –2)
(2 + c )x + 5 c 2 y = 1
…(ii)
and
From Eqs. (i) and (ii), we get
(1 − c )x + 5 (1 − c 2 ) y = 0
c = 1, x + 10 y = 0
A
(2, 0)
X
B
(7, 0)
(5/2)
X
Y¢
As it passes through (1, − 2).
∴ (1 − 3) 2 + (−2) 2 + λ (−2) = 0
⇒
4 + 4 − 2λ = 0
⇒
λ=4
∴Equation of circle is
( x − 3)2 + y 2 + 4 y = 0
Now, by hit and trial method, we see
that point (5, − 2) satisfies equation of
circle.
Since centre lies in the first quadrant,
h = 9/2
∴ Equation of circle is
( x − 9 / 2) 2 + ( y − k ) 2 = (h − 2) 2
+ k 2 = 25 / 4 + k 2
2
2
or x + y − 9 x − 2ky + 14 = 0
11 Let P ( x, y ) be any point on the
circumference of the circle.
b−y
Then, m1 = Slope of PA =
a− x
−a − y
and
Slope
of
m2 =
PB =
b−x
8 Area of quadrilateral OACB ,
P (x, y)
45°
O
(a, b) A
5
135
°
(2, 3)
A
B
q
…(i)
C(h, k)
Y
a1 a2 = b1 b2 .
5 Given lines are 3 x + 5y = 1
10 Here, clearly, x-coordinate of centre is
Y
x1 and x2 are roots of x2 + 2ax − 4 = 0
4 The line a1 x + b1 y + c 1 = 0 cut X and
So, the equation of circle is
2y
x2 + y 2 −
− 1 = 0.
3
x2 + y 2 − 5x − y − 6 = 0
3 Let A and B be ( x1 , y 1 ), ( x2 , y 2 ).
Then, x1 + x2 = −2a, x1 x2 = −4.
y 1 , y 2 are roots of x2 + 2bx − 9 = 0.
Then, y 1 + y 2 = −2b , y 1 y 2 = −9.
Equations of circle on AB as diameter is
( x − x1 )( x − x2 ) + ( y − y 1 )( y − y 2 ) = 0
⇒ x2 + y 2 − ( x1 + x2 )x − ( y 1 + y 2 )y
+ x1 x2 + y 1 y 2 = 0
⇒
x2 + y 2 + 2ax + 2by − 13 = 0
2
3
x- coordinate of the mid-point of AB or
AB ′ i.e. 9 / 2 or −1 / 2.
λ − 10 = 0 ⇒ λ = 10
2
λ= −
∴
6 Here, 3 × 1 = (−1) (−3) = 3
2
The equation of circle passing through
(−1, 0) and (1, 0) is
( x + 1) ( x − 1) + y 2 + yλ = 0.
The coordinates of third vertex will be
(0, 3 ), which is passing by the circle.
C
A = OB ⋅ BC = 52 cotθ
3
= 50
Qsin θ =
13
13
5
P
B
(x, y)
(b, –a)
We have, ∠APB = 45° or 135°
m1 − m2
⇒
= tan 45° or tan135°
1 + m1 m2
b − y −a − y
−
a− x
b−x
⇒
= ±1
b − y −a − y
1+
×
a− x
b−x
SIX
284
⇒
(b − y )(b − x ) + (a + y )(a − x )
= ±1
(a − x )(b − x ) − (b − y )(a + y )
⇒
x2 + y 2 = a2 + b 2
⇒ { x − (a + b )}2 + { y − (b − a)}2
Q
2
= (a + b )2 + (a − b )2 = 2 a2 + b 2
a2 − b 2
=0
4
Let r be the radius of this circle. Then,
+
r
∴ Distance between the centre
(– r, 0)P
by
b2
x2 + y 2 − a −
x−
2a
2
⇒
X
2
1/2 a
2
r
Oa
r2 =
R (r, 0)
= 2 (Radius of either circle)
=
12 The tangent at B (1, 7) is y = 7
and D (4, − 2) is 3 x − 4 y − 20 = 0.
Then, meet at C(16, 7).
Area of quadrilateral ABCD
= AB ⋅ BC = 75
13 Centre of given circle
x + y − 6 x + 12 y + 15 = 0 is (3, − 6).
2
2
∴ Radius = (3) 2 + (−6) 2 − 15 = 30
Area of circle = πr 2 = π ( 30 )2 = 30 π
Area of required circle
= 2 (Area of given circle)
∴
πR2 = 2 × 30 π = 60 π
⇒
R2 = 60 ⇒ R = 2 15
∴ Equation of required circle is
( x − 3) 2 + ( y + 6) 2 = (2 15) 2
⇒ x2 + 9 − 6 x + y 2 + 36 + 12 y = 60
⇒
x2 + y 2 − 6 x + 12 y − 15 = 0
14 Let Q (h, k ) be the mid-point of
chord AB.
∴ ∠ACQ = ∠BCQ = π / 3
Coordinate of centre are
(3 / 2,−1 / 2) and radius = 3 / 2
∴
RS
∴
∠XRP
PQ
∴
PQ . RS
C
Centre of circle (ii) is C2 (−3, − 9) and
radius
[say]
= 9 + 81 − 26 = 8(r2 )
Solving (i) and (ii), we get 4a2 = b 2 .
From the choices, only solution is (1, 2)
17 The equation of the circle touching
2
b
x y
+ y − + λ + − 1 = 0
a b
2
…(i)
Y
B (0, b)
r
P (a/2, b/2)
B
A (a, 0)
X¢
C1 C2 =
⇒
C1 C2 = 25 + 144 = 13
52 + 12 2
Also, r1 + r2 = 5 + 8 = 13
∴
C1 C2 = r1 + r2
Thus, both circles touch each other
externally. Hence, there are three
common tangents.
19 Centres and radii of given circles are
C1 (1, 3), r1 = r , C2 (4, − 1)
and r2 = 42 + 12 − 8 = 3
⇒ r − 3 < (4 − 1) 2 + (−1 − 3) 2 < r + 3
⇒
⇒
⇒
r − 3 < 5< r + 3
r − 3 < 5 and 5 < r + 3
r < 8 and 2 < r ⇒ 2 < r < 8
X
touch each other externally, then
distance between the centre is equal to
sum of their radii, to get required
radius.
Let the coordinate of the centre of T be
(0, k ).
Q
/
(a
Y
2,
0)
x y
+ =1
a b
C
C1 (1,1)
15 Taking diameter PR as X -axis and
centre O as origin, tangents at P and R
are given by
…(i)
x = −r
and
…(ii)
x=r
Let coordinates of X on circle be
(r cos α, r sin α )
⇒
20 Use the property, when two circles
Q (h, k)
Now, CQ = r cos π / 3 = r / 2 = 3 / 4.
⇒ (h − 3 / 2) 2 + (k + 1 / 2) 2 = (3 / 4) 2
∴ Locus of Q (h, k ) is
x2 + y 2 − 3 x + y + 31 / 16 = 0.
Now, C1 C2 = (2 + 3) 2 + (3 + 9) 2
Since, r1 − r2 < C1 C2 < r1 + r2
2
A
r
p/3
…(i)
x2 + y 2 − 4 x − 6 y − 12 = 0
…(ii)
x + y 2 + 6 x + 18 y + 26 = 0
Centre of circle (i) is C1 (2, 3) and radius
[say]
= 4 + 9 + 12 = 5(r1 )
x2 + y 2 − 4 x + 2 y − 3 = 0, so its centre
(2, − 1) lie on line.
∴ 2a − b = 0 i .e ., b = 2a
…(i)
Also line touches the circle
x2 + y 2 + 2 x + 4 y = 0
− a − 2b
…(ii)
= 5
∴
(a2 + b 2 )
a
2
b 2 (a2 + b 2 )
b
⇒r =
a2 + b 2
16 a2
4a
2
16 Line ax + by = 0 is normal to the circle
x y
a b
+
= 1 at P , is x −
2 2
a b
1
b2
b 2 a2 − b 2
−
a− +
4
2a
16 4
18 Given equations of circles are
1
1
∠XOR = α
2
2
1
= 2r tan α;
2
= 90º − α / 2
= 2r tan(90º −α / 2)
= 2r cot α / 2
= 4r 2 ⇒ 2r = (PQ . RS )
∠XPR =
Now, AB = 5, BC = 15
2
2
x − a + y − b + b x + y − 1 = 0
2
2
2 a b
X=r
X=–r
= a +b
The centres of these circles are O (0, 0)
and C (a + b,b − a).
2
On putting the value of λ in Eq. (i), we get
S
T
(0,k)
Y¢
a
Since, it passes through Q ,0 .
2
Therefore, λ =
b2
2
X¢
(1– k)
1
O
C2
[from Eq. (i)]
Y¢
(1,0)
X
285
DAY
24 Let angles between the tangents = 2θ,
Distance between their centre
k + 1 = 1 + (k − 1) 2
then
[QC1 C2 = k + 1]
(a sin α)
⇒ k + 1 = 1 + k 2 + 1 − 2k
⇒ k + 1=
k + 2 − 2k
B
a θ
θ
2
O
(0, 0)
⇒ k + 1 + 2k = k + 2 − 2k
1
k =
⇒
4
2
2
C
So, the radius of circle T is k i.e.
1
.
4
21 Equation of tangent at (1, − 2) is
x − 2 y − 5 = 0. For IInd circle centre
(4, − 3) and r = 5.
x−4 y +3
Point of contact is
=
1
−2
(4 + 6 − 5)
=−
= −1
12 + 22
y+3
x−4
⇒
= −1,
= −1
1
−2
∴
x = 3, y = − 1
22 Equation of tangent at (1, 7) to the curve
x = y − 6 is
y + 7
x =
−6
2
sin θ = OB / OA = ( asin α / a) = sin α
So, θ = α.
Required angle = 2θ = 2α
Centre (2, − 3)
Radius = (2) + (−3) + 12
= 4 + 9 + 12 = 5
Distance between two centres c 1 (2, − 3)
and c 2 (−3, 2)
d = (2 + 3)2 + (−3 − 2)2
= 25 + 25 =
x + y + 16 x + 12 y = C = 0
⇒
⇒
2(−8) − (−6) + 5
(2) 2 + (1) 2
100 − c = 5
c = 95
2
100 – r
θ
P (6, 8)
θ
–r
2
r
r
The equation of the common chord of
these circles is
S1 − S2 = 0
⇒
( x − a) 2 − ( x − b ) 2 + ( y − b ) 2
− ( y − a)2 = 0
(2x − a − b )(b − a)
+ (2 y − b − a)(a − b ) = 0
⇒ 2x − a − b − 2y + b + a = 0
⇒
x− y = 0
The centre coordinates of circles
S 1 and S 2 are C1 (a,b ) and C2 (b, a),
respectively.
10
0
O
…(i)
…(ii)
⇒
23 Now, OP = 62 + 82 = 10
A
S 1 ≡ ( x − a) 2 + ( y − b ) 2 = c 2
and S 2 ≡ ( x − b ) 2 + ( y − a) 2 = c 2
1
PA ⋅ PB ⋅ sin 2θ
2
r
(100 − r 2 ) 3 /2
100
Put f ′ (r ) = 0
3
(100 − r 2 ) 1 /2 ( − 2r 2 ) + (100 − r 2 ) 3 /2
⇒
2
⇒
⇒
100 − r 2 (−3 r 2 + 100 − r 2 ) = 0
r = ± 10 or r = 5
Hence, r = 5
⇒
x2 ( p2 − a2 cos 2 α ) + y 2 ( p2 − a2 sin2 α )
−2 xya2 sin α cos α = 0
The lines given by this equation are at
right angle if
( p2 − a2 cos 2 α ) + ( p2 − a2 sin2 α ) = 0
⇒ 2 p2 = a2 (sin2 α + cos 2 α )
⇒ a2 = 2 p2
29 The coordinates of the centres and radii
of three given circles are as given
P
C1 (0, 2)
M
C2
(b, a)
Q
Now, C1 M =
|a − b |
a−b
=
1+ 1
2
In right ∆C1 PM ,
(a − b ) 2
PM = C1 P 2 − C1 M 2 = c 2 −
2
∴ PQ = 2PM = 2 c 2 −
=
C3 (–3, –6)
C2 (– 6, –2)
C1
(a, b)
= (100 − r 2 )sin θ ⋅ cos θ
=
joining the origin to the points of
intersection of x cos α + y sin α = p and
x2 + y 2 − a2 = 0 is a homogeneous
equation of second degree in given by
2
x cos α + y sin α
x2 + y 2 − a2
=0
p
C (h, k)
c
Let f (r ) = ∆PAB =
+ (g 3 − g 1 ) + (g 1 − g 2 ) }
= − (g 1 − g 2 ) ⋅ (g 2 − g 3 ) ⋅ (g 3 − g 1 )
= − PQ ⋅ QR ⋅ RP
P
B
PA = S 1 = 100 − r 2
+ g 32 (g 1 − g 2 ) − c {(g 2 − g 3 )
28 The combined equation of the lines
26 The equations of two circles are
2
(8)2 + (6)2 − c =
50
52 + ( 50 )2
Radius of circle S =
= (g 12 − c ) (g 2 − g 3 )
+ (g 22 − c ) (g 3 − g 1 ) + (g 32 − c ) (g 1 − g 2 )
= g 12 (g 2 − g 3 ) + g 22 ( g 3 − g 1 )
2
= 25 + 50 = 5 3
⇒
2x − y + 5 = 0
Since, 2 x − y + 5 = 0 touches the circle
∴
Now, r12 ⋅ QR + r22 ⋅ RP + r32 ⋅ PQ
2
2
x2 + y 2 + 2gx + c = 0.
where, g is a variable and c is a constant, be a coaxial system of circle having
common radical axis as X-axis.
Let x2 + y 2 + 2g i x + c = 0; i = 1,2,3 be
three members of the given coaxial
system of circles.
Then, the coordinates of their centres
and radii are
P (− g 1 , 0), Q (− g 2 , 0), R ( − g 3 , 0)
and r12 = g 12 − c , r22 = g 22 − c , r32 = g 32 − c
25 We have, x + y − 4 x + 6 y − 12 = 0
2
2
2
A
27 Let equation of circle be
(a − b ) 2
2
4c 2 − 2(a − b ) 2
Circle
Centre
Radius
Ist Circle
C1 (0,2)
r1 = 3
IInd Circle
C2 (−6, − 2)
r2 = 3
IIIrd Circle
C3 (−3,−6)
r3 = 3
Let C (h, k ) be the centre of the circle
passing through the centres of the
circles Ist, IInd and IIIrd.
Then,
⇒
CC1 = CC2 = CC3
CC12 = CC22 = CC32
⇒ (h − 0) + (k − 2) 2 = (h + 6) 2 + (k + 2)2
= (h + 3) 2 + (k + 6) 2
2
SIX
286
⇒
⇒
− 4k + 4 = 12h + 4k + 40
= 6 h + 12k + 45
12h + 8 k + 36 = 0
and
6h − 8k − 5 = 0
⇒
3h + 2k + 9 = 0
and
6h − 8k − 5 = 0
⇒
−31
−23
h=
,k =
18
12
2
2
A, B and C whose diameter is AB and C2
be another circle which passes through
A and B, then centres of C1 and C2 must
lie on perpendicular bisector of AB.
Indeed centre of C1 is mid-point M of
AB and centre of any other circle lies
somewhere else on bisector.
2 AB = (6 − 2) 2 + (5 − 2) 2 = 5
30 (r + 1) 2 = α2 + 9; r 2 + 8 = α2
A
C2
α
r
M
A
B
C2
C1
Then, M1 A > AM
[hypotenuse of right angled ∆AMM1 ]
1
⇒ Radius of C2 > AB
2
So, C1 is the circle whose radius is least.
Thus, Statement I is true but does not
actually follow from Statement II which
is certainly true.
31 Since, the tangents are perpendicular.
So, locus of perpendicular tangents to
the circle x2 + y 2 = 169 is a director
circle having equation x2 + y 2 = 338.
32 Equation of circle, when the limiting
points are (1, 1) and (3, 3) is
( x − 1)2 + ( y − 1)2 = 0
( x − 3) + ( y − 3) = 0
2
and
⇒
2
i.e. ( x + 3) + ( y − 5) = 4
∴ Centre is (3, − 5).
If L1 is a diameter of a circle, then
2(3) + 3(−5) + p − 3 = 0
⇒
p = 12
∴
L1 is 2 x + 3 y + 9 = 0
and
L2 is 2 x + 3 y + 15 = 0.
2
2
2
Distance of centre of circle C from L2
=
=
2(3) + 3(−5) + 15
x + y − 6 x − 6 y + 18
+ λ( x2 + y 2 − 2 x − 2 y + 2) = 0
It passes through origin, therefore
λ=−9
4 ∠PAO = ∠PBO = π / 2.
A
35 Statement I Centre of circle = (3, − 1 )
2
3 The line ( x − 2)cos θ + ( y − 2)sin θ = 1
6
<2
13
and x + y − 6 x − 6 y + 18 = 0
Equation of the coaxial system of circle
is
2
Equation of required circle is
( x − 6) 2 + ( y − 5) 2 = 9.
⇒ x2 + 36 − 12 x + y 2 + 25 − 10 y = 9
⇒ x2 + y 2 − 12 x − 10 y + 52 = 0
∴ P , A, O , B are concyclic.
Hence, L 2 is a chord of circle C.
2
x
O
22 + 32
x2 + y 2 − 2 x − 2 y + 2 = 0
2
A
(2, 2)
touches the circle
( x − 2)2 + ( y − 2)2 = 1
2
i.e. x + y 2 − 4 x − 4 y + 7 = 0
for all values of θ.
Comparing it with the circle
x2 + y 2 + 2gx + 2 fy + c = 0
we get, g = −2, f = −2, c = 7
g2 + f 2 + c
4 + 4 + 7 15
=
=
=5
∴
−2 − 2 + 7
3
g+ f +c
( x + 3) + ( y − 5) = 9 + 25 − 30
2
r 2 + 2r + 1 = r 2 + 8 + 9
2r = 16 ⇒ r = 8
B
(6, 5)
y
34 Equation of circle C is
C
AC = 2
BC = 3
⇒
C
5
= CP =
949 + 3 .
36
Hence, its equation is
2
2
2
x + 31 + y + 23 = 3 + 5 949
18
12
36
intersections of circle x2 + y 2 − a2 = 0
and line x − y + 3 = 0, ( AB ) is
( x2 + y 2 − a2 ) + λ( x − y + 3) = 0
Since, AB is diameter, centre
− λ , λ lies on it.
2 2
λ λ
∴ − − + 3 = 0 i.e. λ = 3
2 2
Hence, equation of required circle is
x2 + y 2 + 3 x − 3 y + 9 − a 2 = 0
M1
Thus, required circle has its centre at
−31 , −23 and radius
18 12
⇒
1 Equation of circle through points of
33 Let C1 be a circle which passes through
CC1 =
C1
SESSION 2
Statement II is also true but it is not a
correct explanation for Statement I.
0 + 31 + 2 + 23
18
12
5
=
949
36
Now, CP = CC1 + C1 P
5
CP =
949 + 3
⇒
36
∴
Hence, required circle is
2 x2 + 2 y 2 − 3 x − 3 y = 0.
Now, 2(3) + (−1 ) = 5 = 5,
O
P
[true]
Statement II Centre = (3, − 1 ), which
B
lies on given line.
Simplify it and get the result.
∴ Equation of circumcircle of ∆ABP is
x( x − x1 ) + y ( y − y 1 ) = 0
i.e. x2 + y 2 − xx1 − yy 1 = 0
287
DAY
5 The line y = mx − b 1 + m2
…(i)
is tangent to the circle x + y = b
It is also tangent to the circle
…(ii)
( x − a) 2 + y 2 = b 2
It is length of ⊥ on (i) from centre (a, 0)
= radius of circle (ii)
ma − b (1 + m2 )
=b
⇒
1 + m2
2
⇒
2
Then, x12 + y 12 − 2 2 y 1 = x22 + y 22
−2 2 y 2 = x32 + y 32 − 2 2 y 3
2
ma − b (1 + m2 )
= ± b (1 + m2 ) …(iii)
⇒
x12 + y 12 = x22 + y 22 = x32 + y 32
and y 1 = y 2 = y 3 ⇒ x12 = x22 = x33
⇒ There exists two rational points of
the form (a, b ) and (− a, b ), a, b ∈ Q.
10 The chords are equal length, then the
distances of the centre from the lines
are equal. Let L1 be y − mx = 0 centre is
1 , − 3.
2
2
– ve sign givess m = 0, which is none of
the given options.
∴ Taking + ve sign on RHS of (iii), we
get
ma = 2b (1 + m2 )
∴
⇒
⇒
m = 2b / (a − 4b ).
2
2
6 L : 2 x − 3 y − 1; S : x2 + y 2 − 6
If L1 > 0 and S 1 < 0
Then, point lies in the smaller part are as
L : 2x – 3y – 1
S : x2 + y2 – 6
3
1
1
∴ 2, and , − lie inside.
4
4
4
7 1 ( x + 2 x ) 2r = 18
2
(3 x − 2r )2 = 4r 2 + x2
On solving for r, we get r = 2
x–r
D r
C
x–r
r
r
2x – r
r
A
x
r
P
B
2x – r
8 S = x + y − 2 x − 6 y + 6 = 0,
2
2
Centre = (1, 3)
Let radius of circle C = r .
Then, C = ( x − 2)2 + ( y − 1)2 = r 2
= x2 + y 2 − 4 x − 2 y + 5 − r 2 = 0
Common chord of circles S and C is
2x − 4y + 1 − r2 = 0
It is a diameter of circle S.
∴ 2 − 12 + 1 − r 2 = 0 ⇒ r = 3
9 S = x2 + y 2 − 2 2 y + c = 0
Let us assume that there are more than
two rational points on the circle. Let
( x1 , y 1 ), ( x2 , y 2 ), ( x3 , y 3 )
x i , y i ∈ Q , i = 1, 2, 3 be three rational
points.
−
⇒
3 m
−
2
2
1 3
− −1
2 2
=
2
m2 + 1
7m2 − 6 m − 1 = 0
1
m = 1, −
7
Hence, L1 be y +
1
x = 0 ⇒ x + 7y = 0
7
11 Let (0,b ) be the centre and r be the
radius of the given circle, then its
equation is
( x − 0) 2 + ( y − b ) 2 = r 2
⇒ x2 + y 2 − 2 yb + b 2 − r 2 = 0
…(i)
It is given that the point
1
Pn log an,
; n = 1,2,3, 4 lie on the
log
an
circle given by Eq. (i).
Therefore, (log an )2 +
1
2b
,
−
(log an ) 2 (log an )
+ b2 − r 2 = 0
n = 1,2,3, 4
Since, log a1 , log a2 , log a3 and log a4 are
roots of the equation.
Then, λ4 + (b 2 − r 2 )λ2 − 2bλ + 1 = 0
∴ Sum of the roots = 0
⇒ log a1 + log a2 + log a3 + log a4 = 0
⇒ log (a1 a2 a3 a4 ) = 0 ⇒ a1 a2 a3 a4 = 1
12 The point (2a, a + 1) will be an interior
point of the larger segment of the circle
x2 + y 2 − 2 x − 2 y − 8 = 0
(i) The point (2a, a + 1) is an interior
point.
(ii) The point (2a, a + 1) and the
centre (1,1) are on the same side
of the chord x − y + 1 = 0.
∴
(2a) 2 + (a + 1) 2 − 2(2a)
− 2(a + 1) − 8 < 0
and (2a − a − 1 + 1)(1 − 1 + 1) > 0
⇒
5a2 − 4 a − 9 < 0 and a > 0
⇒
(5a − 9)(a + 1) < 0 and a > 0
⇒
−1 < a <
9
and a > 0
5
9
a ∈ 0,
5
⇒
13 The equation of the biggest circle is
x2 + y 2 = 12
Clearly, it is centred at O(0, 0) and has
radius 1. Let the radii of the other two
circles be 1 − r , 1 − 2r , where r > 0.
Thus, the equations of the concentric
circles are
…(i)
x2 + y 2 = 1
x2 + y 2 = (1 − r ) 2
…(ii)
x2 + y 2 = (1 − 2r ) 2
…(iii)
Clearly, y = x + 1 cuts the circle (i) at
(1, 0) and (0, 1). This line will cut
circles (ii) and (iii) in real and distinct
points, if
1
1
< 1 − r and
< 1 − 2r
2
2
1
1
< 1 − r and
< 1 − 2r
⇒
2
2
1
1
1
⇒
r < 1−
and r < 1 −
2
2
2
1
1
⇒
r < 1 −
2
2
1
1
⇒
r ∈ 0, 1 −
[Qr > 0]
2
2
14 Given circles are
x2 + y 2 − 2λ i x − c 2 = 0, i = 1, 2, 3
∴ Centres are (λ1 , 0), (λ2 , 0), (λ3 , 0)
Distances of the centres from origin are
in G.P.,
∴ λ22 = λ1 λ3
Let ( x1 y 1 ) be any point on the circle
x2 + y 2 = c 2 .
∴ x12 + y 12 − c 2 = 0
Lengths of tangents from ( x1 , y 1 ) to the
three given circles are
li =
x12 + y 12 − 2λ j x1 − c 2 =
−2λ j x1
∴ l l = (−2λ1 x1 ) (−2λ3 x1 )
= 4λ1 λ3 x12 = 4λ22 x12 = λ42
∴ l 1 , l 2 and l 3 are in G.P.
2
1
2
3
15 The equation representing the coaxial
system of circle is
x2 + y 2 + 2gx + c
⇒
+ λ( x2 + y 2 + 2 fy + k ) = 0
2g
2f λ
x2 + y 2 +
x+
1+ λ
1+ λ
y+
c + kλ
=0
1+ λ
…(i)
The coordinates of the centre of this
circle are
− g ,− f λ
…(ii)
1+ λ 1+ λ
SIX
288
and radius
g + f λ − (c + kλ )(1 + λ )
(1 + λ )2
2
=
2 2
For the limiting points,
we must have
Radius = 0
⇒ g 2 + f 2 λ2 − (c + kλ )(1 + λ ) = 0
⇒ λ2 ( f 2 − k ) − λ(c + k )
+ (g 2 − c ) = 0 …(iii)
Let λ1 and λ2 be the roots of this
equation.
c+k
Then,
λ1 + λ2 = 2
f −k
and
λ1 λ2 =
g2 − c
f2 − k
17 S = x2 + λx + (1 − λ )y + 5 = 0
...(iv)
Thus, the coordinates of limiting points
L1 and L2 are,
− g − f λ1
L1
,
and
1 + λ1 1 + λ1
− g − f λ2
,
L2
1 + λ2 1 + λ2
[from Eq. (iv)]
Now, L1 L2 will subtend a right angle at
the origin.
If slope of OL1 × slope of OL2 = −1
⇒
i.e.
mx + y − 3m − 1 = 0
It touches the circle so ( p = r )
3m + 1
=1
1 + m2
⇒
9m2 + 6m + 1 = 1 + m2
⇒
m = 0, − 3 / 4.
m = 0 gives the slope for the tangent
y = 1, so equation of reflected ray is
3
y − 1 = − ( x − 3)
4
i.e., 3 x + 4 y − 13 = 0
f λ1
fλ
× 2 = −1
g
g
is a circle of radius not exceeding 5.
λ2 (1 − λ )2
∴
+
− 5≤ 5
4
4
2
2
λ
(1 − λ )
and
+
− 5≥ 0
4
4
∴
λ2 + (1 − λ )2 − 20 ≤ 100
and
λ2 + (1 − λ )2 − 20 > 0
1 − 239
1 + 239
≤ λ≤
⇒
2
2
i.e. − 7.2 ≤ λ ≤ 8.2
1 − 39
1 + 39
and λ <
or λ >
2
2
i.e.
λ − 2.62 or λ > 3.62
∴ Possible integral values of λ are
−7, − 6,−5, − 4, − 3, 4, 5, 6, 7, 8
∴ In all, there are 10 possible integral
values of λ.
⇒
f 2 λ1 λ2 = − g 2
18 [P + 1] = [P ] + 1. Let [P ] = n, then n is
⇒
g2 − c
2
f 2 2
= −g
f − k
integer.
([P + 1],[P ]) = (n + 1, n ) lies inside the
region of circles.
S 1 = x2 + y 2 − 2 x − 15 = 0,
C1 = (1, 0), r1 = 4.
and S 2 = x2 + y 2 − 2 x − 7 = 0,
C2 = (1, 0), r2 = 2 2
⇒
⇒
f 2 (g 2 − c ) + g 2 ( f 2 − k ) = 0
c
k
+ 2 =2
g2
f
16 Tangent at (0, 1) to the circle x2 + y 2 = 1
is y = 1. Incident ray, incident at (3, 1)
is y − 1 = m ( x − 3)
Incident and reflected rays are equally
and reflected rays are equally inclined
to the line is y = 1, so slope of reflected
ray is −m.
∴ Equation of reflected ray is
y − 1 = − m( x − 3)
Both circles are concentric.
∴
(n + 1)2 + n2 − 2(n + 1) − 7 > 0
and (n + 1)2 + n2 − 2(n + 1) − 15 < 0
⇒ 4 < n2 < 8
which is not possible for any integer.
19 L = 3 x − 4 y − k = 0 touches the circle
S = x2 + y 2 − 4 x − 8 y − 5 = 0 at (a,b )
Tangent at (a, b ) is
ax + by − 2( x + a) − 4( y + b ) − 5 = 0
T = x(a − 2) + y (b − 4) − 2a − 4b − 5 = 0
Comparing L = 0, T = 0, we get
a − 2 b − 4 2a + 4b + 5
=
=
−4
k
3
Also, using p = r , we get
6 − 10 − k
= ±5 ⇒ k = 15or −35
5
Using k = 15, we get
4a + 3b = 20 and 9a − 12b = 45
Solving a = 5, b = 0 and using
k = −35, we get a = −1,b = 8
We can verify that (5, 0) and (−1, 8) both
satisfy equation of circle. Hence
k ,(a, b ) = 15 (5, 0) or −35, (−1, 8)
20 Centre of the circle x2 + y 2 = a2 is
(0, 0).
Let its reflection about the line
y = mx + c be (h, k ).
Then, (h / 2, k / 2) lies on this line.
h
k
m + c = k and − × m = −1
2
h
⇒
mk = − h
On solving these, we get
2cm
2c
h=
,k =
1 + m2
1 + m2
Also, radius of reflected circle is a.
∴ Equation of relfected circle is
2
2
x + 2cm + y − 2c = a2
1 + m2
1 + m2
4
cm
4c
⇒ x2 + y 2 +
x−
1 + m2
1 + m2
4c 2 (1 + m2 )
+
− a2 = 0
(1 + m2 )2
But given equation of reflected
circle is
x2 + y 2 + 2gx + 2 fy + c = 0
On comparing, we get
g
f
c
=1
=
=
2 cm
−2c
4c 2
2
a
−
2
2
1+ m
1+ m
1 + m2
2cm
2c
4c 2
,f =−
,
∴ g =
2
2
1+ m
1 + m 1 + m2
− a2 = c
DAY TWENTY SEVEN
Parabola
Learning & Revision for the Day
u
u
u
u
Conic Section
Concept of Parabola
Line and a Parabola
Equation of Tangent
u
u
u
u
Equation of Normal
Equation of a Pair of Tangents
Equations of Chord of Contact
u
u
Conormal Points
Diameter
Director Circle
Conic Section
A conic is the locus of a point whose distance from a fixed point bears a constant ratio to its
distance from a fixed line. The fixed point is the focus S and the fixed line is the directrix, l.
The constant ratio is the eccentricity denoted by e.
●
If 0 < e < 1, then conic is an ellipse.
●
If e = 1, then conic is a parabola.
●
If e > 1, then conic is a hyperbola.
General Equation of Conic Section
A second degree equation ax2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0 represents
Case I When the focus lies on the directrix
a h g
(i) Pair of straight lines, if ∆ = h b f = 0
g f c
(ii) If e > 1, then the lines will be real and distinct intersecting at fixed point.
(iii) If e = 1, then the lines will coincident passing through a fixed point.
(iv) If e < 1, then the lines will be imaginary.
Case II When the focus does not lie on the directrix
(i) Circle : a = b , h = 0, e = 0 and ∆ ≠ 0
(ii) Parabola : h2 = ab , ∆ ≠ 0, e = 1
(iii) Ellipse : h2 < ab , ∆ ≠ 0, 0 < e < 1
(iv) Hyperbola : h2 > ab , 0 ∆ ≠ 0, e > 1
(v) Rectangular hyperbola : a + b = 0, ∆ ≠ 0, e > 1, h2 > ab
PRED
MIRROR
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
290
Concept of Parabola
If S lies on L, parabola reduces to a straight line through S and
perpendicular to L.
Parabola is the locus of a point which moves in a plane such
that its distance from a fixed point (focus, S ) is equal to its
distance from a fixed straight line (directrix, L).
Definitions Related to Parabola
1. Vertex The intersection point of parabola and axis.
2. Centre The point which bisects every chord of the conic
passing through it.
3. Focal chord Any chord passing through the focus.
4. Double ordinate A chord perpendicular to the axis of a
conic.
5. Latusrectum A double ordinate passing through the
focus of the parabola.
6. Focal distance The distance of a point P( x, y) from the
focus S is called the focal distance of the point P.
Let S ≡ ( x1 , y1 ) and L ≡ ax + by + c = 0.
Then, equation of parabola is
X¢
ax+by + c = 0
(a2 + b 2 ) [( x − x1 )2 + ( y − y1 )2 ] = (ax + by + c)2 .
Y
P (x, y)
M
S
(x1, y1)
O
X
Y¢
Some related terms of parabolas (in standard form)
S. No. Related Terms
y 2 = 4 ax
y 2 = − 4 ax
x 2 = 4 ay
S
Z
A
S
S
A
Z
A
Z
x 2 = − 4 ay
Z
A
S
1.
Vertex
A(0, 0)
A(0, 0)
A(0, 0)
A(0, 0)
2.
Focus
S (a, 0)
S (− a, 0)
S (0, a)
S (0, − a)
3.
Equation of axis
y=0
y=0
x=0
x=0
4.
Equation of directrix
x + a= 0
x − a= 0
y + a= 0
y − a= 0
5.
Eccentricity
e =1
e =1
e =1
e =1
6.
Extremities of latusrectum
(a, ± 2a)
(− a, ± 2a)
(± 2a, a)
(± 2a, − a)
7.
Length of latusrectum
4a
4a
4a
4a
8.
Equation of tangent at vertex
y=0
y=0
x=0
x=0
x = at
y = 2at
x = − at
y = 2at
x = 2at
2
y = at
x = 2at
2
y = − at
Focal distance of any point P (h, k ) on
the parabola
h+ a
h−a
k+ a
k−a
Equation of latusrectum
x=a
x + a= 0
y=a
y + a= 0
2
9.
Parametric equation
10.
11.
Results on Parabola y 2 = 4 ax
(i) Length of latusrectum = 2 (Harmonic mean of focal
segment)
(ii) If y1 , y2 and y3 are the ordinates of the vertices of triangle
inscribed in the parabola y2 = 4ax,
1
then its area =
( y1 − y2 ) ( y2 − y3 ) ( y3 − y1 )
8a
2
(iii) For the ends of latusrectum of the parabola y2 = 4ax, the
values of the parameter are ± 1.
Position of a Point
A point (h, k ) with respect to the parabola S lies inside, on or
outside the parabola, if S1 < 0, S1 = 0 or S1 > 0.
291
DAY
Line and a Parabola
(i) The line y = mx + c meets the parabola y2 = 4ax in two
a
points real, coincident or imaginary according to c > ,
m
a
a
or c <
respectively.
c=
m
m
(ii) Length of the chord intercepted by the parabola on the
line y = mx + c is =
4 a(1 + m2 ) (a − mc)
m2
(iii) Length of the focal chord making an angle α with the
X -axis is 4a cosec2α.
(iv) If t 1 and t 2 are the end points of a focal chord of the
parabola y2 = 4ax, then t 1t 2 = − 1
(vii) The orthocentre of any triangle formed by three tangents
to a parabola lies on the directrix.
(viii) The length of the subtangent at any point on a parabola is
equal to twice the abscissae of the point.
(ix) Two tangents can be drawn from a point to a parabola.
Two tangents are real and distinct or coincident or
imaginary according as given point lies outside, on or
inside the parabola.
Equation of Normal
A line which is perpendicular to the tangent of the parabola is
called the normal to the parabola.
Equation of normal to parabola in different cases are given
below;
y
In point ( x1 , y1 ) form, ( y − y1 ) = − 1 ( x − x1 ).
2a
In slope m form, y = mx − 2 am − am3 .
In parametric t form, y + tx = 2 at + at 3 .
●
Equation of Tangent
●
●
A line which intersects the parabola at only one point is called
the tangent to the parabola.
Equation of tangent to parabola in different cases are given
below;
●
●
In point ( x1 , y1 ) form, yy1 = 2 a ( x + x1 )
a
In slope (m) form, y = mx +
m
●
In parametric (t ) form, ty = x + at 2
●
The line y = mx + c touches a parabola iff c =
a
and the
m
a 2 a
coordinates of the point of contact are 2 , .
m m
Results on Tangent
(i) Points of intersection of tangents at two points P(at 12 , 2at 1 ),
Q (at 22 , 2 at 2 ) on the parabola y2 = 4ax is R{at 1t 2 , a(t 1 + t 2 )}
(where, R is GM of x-coordinates of P, Q and AM of
y-coordinates of P, Q).
NOTE Point
of intersection of normals of t1
[ a (t12 + t 22 + t1t 2 + 2), − at1t 2 (t1 + t 2 )].
(iv) If the tangents at the points P and Q on a parabola meet T,
then ST is the GM between SP and SQ
i.e. ST 2 = SP ⋅ SQ
(v) If the tangent and normal at any point P of the parabola
intersect the axis at T and G, then ST = SG = SP, where S is
the focus.
(vi) Any tangent to a parabola and the perpendicular on it
from the focus meet on the tangent at the vertex.
are
Results on Normal
(i) If the normals at two points P and Q of a parabola
y2 = 4ax intersects at a third point R on the curve, then
the product of the ordinates of P and Q is 8 a2 .
(ii) Normal at the ends of latusrectum of the parabola y2 = 4ax
meet at right angles on the axis of the parabola.
(iii) Tangents and normals at the extremities of the
latusrectum of a parabola y2 = 4ax constitute a square,
their points of intersection being (−a, 0)and ( 3 a, 0).
(iv) The normal at any point of a parabola is equally inclined
to the focal distance of the point and the axis of the
parabola.
(v) The normal drawn at a point P(at 12 , 2 at 1 ) to the parabola
y2 = 4ax meets again the parabola at Q (at 22 , 2 at 2 ), then
2
t2 = − t1 − .
t1
(ii) Angle
between
tangents
at
two
points
θ
P(at 12 , 2at 1 ), Q(at 22 , 2at 2 ) on the parabola y2 = 4ax is given
t − t1
by tan θ = 2
.
1 + t1 t2
(iii) Locus of the point of intersection of perpendicular
tangents to the parabola is its directrix.
and t 2
Y
2,
t1
P (a
X¢
)
2at 1
R
A
B
Y¢
X
Q (a
t2 2, 2
at2 )
(vi) The normal chord of a parabola at a point whose ordinate
is equal to the abscissae, subtends a right angle at the
focus.
(vii) Three normals can be drawn from a point to a parabola.
292
Equation of a Pair of Tangents
Conormal Points
The equation of pair of tangents drawn from an external
point P( x1 , y1 ) to the parabola is SS1 = T 2 .
The points on the parabola at which the normals pass through a
common point are called conormal points. The conormal points are
called the feet of the normals.
where, S = y2 − 4ax, S1 = y12 − 4ax1 and
T = yy1 − 2 a ( x + x1 )
Y
A
Equations of Chord of Contact
P
1. The equation of chord of contact is
yy1 − 2 a ( x + x1 ) = 0
2. The equation of chord of parabola, whose mid-point
( x1 , y1 ) is T = S1 , i.e. yy1 − 2 a( x + x1 ) = y12 − 4ax1
3. Length of the chord of contact is
( y12 − 4 ax1 )( y12 + 4a2 )
.
a
4. Area of the ∆PAB formed by the pair of tangents and
their chord of contact is
( y2 − 4ax1 )3 /2
.
A= 1
2a
l=
NOTE
• Equation of the chord joining points
P ( at12 , 2at1 ), Q ( at 22 , 2at 2 ) is (t1 + t 2 ) y = 2x + 2 at1 t 2 .
• For PQ to be focal chord, t1 t 2 = − 1.
• Length of the focal chord having t1 , t 2 as end points is
a(t 2 − t1 ) 2 .
Director Circle
The locus of the point of intersection of perpendicular
tangents to a conic is known as director circle. The
director circle of a parabola is its directrix.
y2 = 4ax
X¢
O
X
B
C
Y¢
Points A, B and C are called conormal points with respect to point P.
1. The algebraic sum of the slopes of the normals at conormals
point is 0.
2. The sum of the ordinates of the conormal points is 0.
3. The centroid of the triangle formed by the conormal points on a
parabola lies on its axis.
Diameter
Diameter is the locus of mid-points of a system of parallel chords of
parabola.
1. The tangent at the extremities of a focal chord intersect at right
angles on the directrix and hence a circle on any focal chord as
diameter touches the directrix.
2. A circle on any focal radii of a point P(at 2 , 2 at ) as diameter
touches the tangent at the vertex and intercepts a chord of
length a 1 + t 2 on a normal at the point P.
3. The diameter bisecting chords of slope m to the parabola y2 = 4 ax
2a
is y = .
m
DAY
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 Equation of the parabola whose vertex is ( −1, − 2) , axis is
vertical and which passes through the point (3, 6) is
(a) x 2 + 4 x + 28 y − 136 = 0
(b) x 2 + 2 x − 2 y − 3 = 0
(c) y 2 + 4 y − 16x − 12 = 0
(d) None of the above
(c) 40/3
(d) 40/9
3 Latus rectum of the parabola whose axis is parallel to the
Y -axis and which passes through the points (0, 4), (1, 9),
and ( −2, 6) is equal to
(a) 1 / 2
(b) 1
(c) 2
(d) None of these
4 If the line x − 1 = 0 is the directrix of the parabola
y 2 − kx + 8 = 0, then one of the value of k is
(a) 1 / 8
(b) 8
(c) 4
(d) 1 / 4
5 The locus of trisection point of any double ordinate of the
parabola y 2 = 4 a x is
(a) y 2 = 9ax
(c) 9y 2 = 4 ax
(b) y 2 = ax
(d) None of these
x 2 = 8y . If the point P divides the line segment OQ
internally in the ratio1: 3, then the locus of P is
JEE Mains 2015
(b) y 2 = x
(d) x 2 = 2 y
7 At any points P on the parabola y − 2y − 4x + 5 = 0, a
2
tangent is drawn which meets the directrix at Q the locus of
1
the points R which divides QP externally in the ratio : 1, is
2
(a) (x + 1) (1 − y)2 + 4 = 0
(c) (1 − y)2 − 4 = 0
(b) x + 1 = 0
(d) None of these
8 The line x − b + λy = 0 cuts the parabola y 2 = 4ax at
P (at1 , 2at1 ) and Q (at 22 , 2at 2 ). If b ∈[ 2a, 4a ] and λ ∈R, then
t1 t 2 belongs to
2
(a) [−4, −2]
(c) [−3, − 2]
(b) [4, −3]
(d) None of these
9 The centre of the circle passing through the point ( 0, 1) and
touching the curve y = x 2 at ( 2, 4) is
−16 27
(a)
,
5 10
−16 53
(c)
,
5 10
−16 53
(b)
,
7 10
(d) None of these
10 Equation of common tangents to parabolas y = x 2 and
y = − x 2 + 4x − 4 is/are
(a) y = 4 (x − 1) ; y = 0
(c) y = 0, y = − 10 (x + 5)
the tangents of the parabola y 2 = 4ax , then
(a) m1 + m2 = 0
(c) m1m2 = − 1
(b) y = 0 , y = − 4 (x − 1)
(d) None of these
(b) m1m2 = 1
(d) m1 + m2 = 1
13 Set of values of h for which the number of distinct
common normals of ( x − 2)2 = 4 ( y − 3) and
x 2 + y 2 − 2x − hy − c = 0 where, (c > 0) is 3, is
(a) (2, ∞)
(b) (4, ∞ )
(c) (2, 4)
(d) (10, ∞)
14 Tangent and normal are drawn at P(16,16) on the
parabola y 2 = 16x , which intersect the axis of the
parabola at A and B respectively. If C is the centre of
the circle through the points P, A and B and ∠CPB = θ,
then a value of tan θ is
JEE Mains 2018
(a)
6 Let O be the vertex and Q be any point on the parabola
(a) x 2 = y
(c) y 2 = 2 x
(b) π / 4
(d) π /2
12 If the lines y − b = m1 ( x + a ) and y − b = m2 ( x + a ) are
at an angle tan−1 3. Then its length is equal to
(b) 80/9
to the parabola y 2 = 4x is
(a) π / 6
(c) π / 3
2 A focal chord of the parabola y 2 = 8x in inclined to X -axis
(a) 80/3
11 Angle between the tangents drawn from the point (1, 4)
1
2
(b) 2
(c) 3
(d)
4
3
15 P is a point on the parabola y 2 = 4x and Q is a point
on the line 2x + y + 4 = 0. If the line x − y + 1 = 0 is
the perpendicular bisector of PQ, then the coordinates
of P is
(a) (8, 9), (10, 11)
(c) (7, 8), (9, 8)
(b) (1, − 2), (9, − 6)
(d) None of these
16 The locus of the vertices of the family of parabolas
y =
a3x 2 a2x
+
− 2a is
3
2
(a) xy = 105 / 64
(c) xy = 35 / 16
(b) xy = 3 / 4
(d) xy = 64 / 105
17 The parabola y 2 = λx
and 25 [( x − 3)2 + ( y + 2)2 ] = ( 3x − 4y − 2)2
are equal, if λ is equal to
(a) 1
(b) 2
(c) 3
(d) 6
18 A line is drawn from A ( −2, 0) to intersect the curve
y 2 = 4x in P and Q in the first quadrant such the
1
1
1
+
< , then slope of the line is always
AP AQ 4
(a) <
(c) ≥
3
3
(b) > 3
(d) None of these
19 Vertex A of a parabola y 2 = 4ax is joined to any point P
on it and line PQ is drawn at right angle to AP to meet
the axis at Q. Then, the projection of PQ on the axis is
always equal to
(a) 3 a
(c) 3a
(b) 2a
(d) 4a
294
20. The slope of the line touching both the parabolas y 2 = 4x
Directions (Q. Nos. 31-35) Each of these questions
JEE Mains 2014
2
(d)
3
contains two statements : Statement I (Assertion) and
Statement II (Reason). Each of these questions also has
four alternative choices, only one of which is the correct
answer. You have to select one of the codes (a ), (b), (c) and (d )
given below.
and x 2 = −32y is
1
(a)
2
3
(b)
2
1
(c)
8
21 If the normals at the end points of variable chord PQ of
the parabola y 2 − 4y − 2x = 0 are perpendicular, then
the tangents at P and Q will intersect on the line
(a) x + y = 3 (b) 3 x − 7 = 0 (c) y + 3 = 0 (d) 2 x + 5 = 0
22 Find the length of the normal drawn from the point on the
axis of the parabola y 2 = 8x whose distance from the
focus is 8.
(a) 10
(b) 8
(c) 9
(d) None of these
23 If x + y = k is a normal to the parabola y 2 = 12x , p is the
length of the perpendicular from the focus of the
parabola on this normal, then 3k 3 + 2p 2 is equal to
(a) 2223
(c) 2222
(b) 2224
(d) None of these
24 If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through
the points of intersection of the parabolas y 2 = 4ax and
x 2 = 4ay , then
(a) d 2 + (2b + 3c)2 = 0
(c) d 2 + (2b − 3c)2 = 0
(b) d 2 + (3b + 2c)2 = 0
(d) d 2 + (3b − 2c)2 = 0
25 Slopes of the normals to the parabola y 2 = 4ax
intersecting at a point on the axis of the parabola at a
distance 4a from its vertex are in
(a) HP
(c) AP
(b) GP
(d) None of these
26 The area of the triangle formed by the tangent and the
normal to the parabola y 2 = 4ax , both drawn at the same
end of the latusrectum and the axis of the parabola is
(a) 2 2 a 2
(c) 4 a 2
(b) 2 a 2
(d) None of these
27 If the tangent at the point P( 2, 4) to the parabola y 2 = 8x
meets the parabola y 2 = 8x + 5 at Q and R, then
mid-point of QR is
(a) (2, 4)
(c) (7, 9)
(b) (4, 2)
(d) None of these
28 The equation of the common tangent touching the circle
( x − 3)2 + y 2 = 9 and the parabola y 2 = 4x above the
X -axis is
(a) 3 y = 3 x + 1
(c) 3 y = x + 3
(b) 3 y = − (x + 3)
(d) 3 y = − (3 x + 1)
29 If tangents drawn from point P to the parabola y 2 = 4x
are inclined to X -axis at angles θ1 and θ 2 such that
cot θ1 + cot θ 2 = 2, then locus of the point P is
(a) y = 2
(b) y = 8
(c) y = 1
(d) None of these
30 Tangents to the parabola y = 4x are drawn from the
2
point (1, 3). The length of chord of contact is
(a) 5
(b) 13
(c) 65
(d) None of these
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
31 Statement I The perpendicular bisector of the line
segment joining the points ( −a, 2at ) and (a, 0) is tangent
to the parabola y 2 = −4ax , where t ∈ R .
Statement II Number of parabolas with a given point
as vertex and length of latusrectum equal to 4 is 2.
32 Consider the equation of the parabola is y 2 = 4ax .
Statement I Length of focal chord of a parabola
having focus (2, 0) making an angle of 60° with X-axis
is 32.
Statement II Length of focal chord of a parabola
y 2 = 4ax making an angle α with X-axis is 4a cosec 2 α.
33 Consider the equation of the parabola is y 2 = 4ax .
Statement I Area of triangle formed by pair of tangents
drawn from a point (12, 8) to the parabola having focus
(1,0) and their corresponding chord of contact is
32 sq units.
Statement II If from a point P ( x 1, y 1) tangents are
drawn to a parabola, then area of triangle formed by
these tangents and their corresponding chord of
( y 2 − 4ax 1 )3/ 2
contact is 1
sq units.
4 a
34 Statement I The latusrectum of a parabola is 4 units,
axis is the line 3x + 4y − 4 = 0 and the tangent at the
vertex is the line 4x − 3y + 7 = 0 , then the equation of
directrix of the parabola is 4 x − 3 y + 8 = 0 .
Statement II If P is any point on the parabola and PM
and PN are perpendiculars from P on the axis and
tangent at the vertex respectively, then
(PM ) 2 = (latusrectum) (PN ).
35 A circle, 2x 2 + 2y 2 = 5 and a parabola, y 2 = 4 5x .
Statement I An equation of a common tangent to
these curves is y = x + 5.
5
where, m ≠ 0 is
m
the common tangent, then m satisfies m 4 − 3m 2 + 2 = 0.
Statement II If the line, y = mx +
JEE Mains 2013
DAY
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 A ray of light moving parallel to the X -axis gets reflected
from a parabolic mirror whose equation is
( y − 2)2 = 4 ( x + 1). After reflection, the ray must pass
through the point
(a) (0, 2)
(c) (0, −2)
the parabola y 2 = 8x . The minimum length of AB is
(b) 4
(d) None of these
3 If a line x + y = 1 cuts the parabola y 2 = 4x at points A
and B and normals at A and B meet on C. The normals to
the parabola from C, other than above two, meet the
parabola in D, the coordinates of D are
(a) (2, 1)
(c) (4, 4)
(b) (−4, 4)
(d) None of these
4 A chord PP′ of a parabola cuts the axis of the parabola at
A. The feet of the perpendiculars from P and P′ on the
axis are M and M′ respectively. If V is the vertex, then
VM , VA, VM ′ are in
(a) AP
(c) HP
(b) GP
(d) None of these
5 The set of points on the axis of the parabola y = 4x + 8
from which the 3 normals to the parabola are all real and
different, is
(b) {(k, 0)| k > − 2 }
(d) None of these
6 Normals drawn to y 2 = 4ax at the points where it is
intersected by the line y = mx + c, intersect at the point
P. Foot of another normal drawn to the parabola from the
point P may be
(a) (a / m 2 , − 2a / m)
(b) (9a / m 2 , − 6a / m)
(c) (4a / m 2 , − 4a / m)
(d) None of the above
10 The equation of the curve obtained by reflecting the
parabola y 2 = 4x about the line x − y + 13 = 0 is
(a) (2 y − x − 13)2 = 4 (y + 13)
(b) (2 y + x − 13)2 = 4 (y − 13)
(c) (2 y − x − 13)2 = 4 (y − 13)
(d) None of the above
11 Let P be the point on the parabola, y 2 = 8x which is at a
minimum distance from the centre C of the circle,
x 2 + ( y + 6)2 = 1, Then the equation of the circle, passing
through C and having its centre at P is
JEE Mains 2016
(a) x 2 + y 2 − 4 x + 8 y + 12 = 0
(b) x 2 + y 2 − x + 4 y − 12 = 0
x
(c) x 2 + y 2 − + 2 y − 24 = 0
4
(d) x 2 + y 2 − 4 x + 9y + 18 = 0
touches the curve y = 4 − x 2 and the line y = | x | is
JEE Mains 2017
(a) 4 ( 2 − 1)
(c) 2 ( 2 + 1)
(b) 4 ( 2 + 1)
(d) 2 ( 2 − 1)
13 If y1, y 2 are the ordinates of two points P and Q on the
parabola and y 3 is the ordinate of the point of
intersection of tangents at P and Q, then
(a) y1, y 2 , y 3
(b) y1, y 3 , y 2
(c) y1, y 2 , y 3
(d) y1, y 3 , y 2
are in AP
are in AP
are in GP
are in GP
14 The number of points with integral coordinates that lie in
the interior of the region common to the circle
x 2 + y 2 = 16 and the parabola y 2 = 4x is
7 Sides of an equilateral triangle ABC touch the parabola
y 2 = 4ax , then points A, B, C lie on
(a) y 2 = 3 (x + a)2 + 4ax
(b) y 2 = (x + a)2 + ax
(c) y 2 = 3 (x + a)2 + ax
(d) None of the above
(a) 8
(c) 16
(b) 10
(d) None of these
15 The tangent and normal at the point P = (16, 16) to the
8 Minimum distance between the curves y 2 = 4x and
x 2 + y 2 − 12x + 31 = 0 is
(a) 21
(c) 2 7 − 5
(b) 2
(d) None of these
12 The radius of a circle, having minimum area, which
2
(a) {(k, 0)| k ≤ − 2 }
(c) {(k, 0) | k > 0}
y = x 2 at the point whose abscissa is x 0 , 1 ≤ x 0 ≤ 2, the
Y -axis and the straight line y = x 02 has the greatest area if
x0 =
(a) 1
(c) 3/2
(b) (2, 0)
(d) (−1, 2)
2 Mutually perpendicular tangentsTA and TB are drawn to
(a) 16
(c) 8
9 The triangle formed by the tangent to the parabola
(b) 5
(d) None of these
parabola y 2 = 16x intersect the X -axis at the points Q
and R respectively. The equation to the circum circle of
∆PQR is
(a) x 2 + y 2 − 8 x − 384 = 0
(b) x 2 + y 2 − 2 x − 8 y − 352 = 0
(c) x 2 + y 2 + 2 y − 544 = 0
(d) None of the above
296
ANSWERS
1 (b)
2 (b)
3 (a)
4 (c)
5 (c)
6 (d)
7 (a)
8 (a)
9 (c)
10 (a)
11 (c)
12 (c)
13 (d)
14 (b)
15 (b)
16 (a)
17 (d)
18 (b)
19 (d)
20 (a)
26 (c)
27 (a)
28 (c)
29 (a)
30 (c)
6 (c)
7 (a)
8 (b)
9 (b)
10 (d)
SESSION 1
SESSION 2
21 (d)
22 (b)
23 (a)
24 (a)
25 (c)
31 (c)
32 (d)
33 (c)
34 (d)
35 (b)
1 (a)
11 (a)
2 (c)
12 (a)
3 (c)
13 (b)
4 (b)
14 (d)
5 (c)
15 (a)
Hints and Explanations
SESSION 1
2
∴
1 Axis is vertical i.e. parallel to Y-axis so
Here a = 2, α = tan −1 3 i.e. tanα = 3.
∴Length of focal chord
= 4 × 2 × (1 + 1 / 9) = 80 / 9.
3 Let vertex be (b, c ). Then equation of
parabola is ( x − b )2 = 4a ( y − c ). It
passes through the points (0, 4),
(1, 9) and (−2, 6).
∴
b 2 = 4a (4 − c )
(1 − b 2 ) = 4a (9 − c )
and (−2 − b )2 = 4a(6 − c ).
Solving these equations, latus rectum
4a = 1 / 2.
6 Any point on the parabola x2 = 8 y is (
4t ,2t 2 ). Point P divides the line segment
joining of O (0, 0) and Q (4t ,2t 2 ) in the
ratio 1: 3. Apply the section formula for
internal division.
Equation of parabola is x2 = 8 y
Let any point Q on the parabola (i) is
(4t , 2t 2 ).
Let P (h, k ) be the point which divides
the line segment joining (0,0) and
(4t ,2t 2 ) in the ratio 1:3.
Y
3
X¢
Q (4t,2
∴ α = at 2 ,β =
2(2at ) + 1(–2at )
3
Y (at 2, 2at)
A
X¢
⇒
B
Y¢ (at 2, –2at)
3β
2
β = at ⇒ t =
3
2a
∴
h=
1 × 4t + 3 × 0
⇒h = t
4
1 × 2t 2 + 3 × 0
t2
⇒k =
4
2
1 2
[Qt = h]
k = h
2
and k =
⇒ 2k = h2 ⇒ 2 y = x2 , which is
required locus.
7 Given, ( y − 1) 2 = 4( x − 1). P has
X
Y
2
P (at1 , 2at1)
X¢
O
R (b, 0)
Q(
at
coordinates x = 1 + t 2 , y = 1 + 2 t .
Tangent at P is
( x − 1) − ( y − 1) t + t 2 = 0.
So, the directrix is x = 0.
∴
1
Q = 0, t + 1 −
t
X
2
, 2a
t2 )
Slope of PR = Slope of RQ
b
a
∴ Minimum value of t 1 t 2 = −4
and maximum value of t 1 t 2 = −2
⇒
Y¢
⇒
P (α, β)
through (b, 0).
Y¢
X
(0, 0) O
∴
5 Let P (α, β ) be the trisection point.
8 Line x − b + λy = 0 always passes
2
2
Directrix is x − 8 / k = − k / 4
or x = 8 / k − k / 4 = 1
⇒ k 2 + 4k − 32 = 0 ⇒ k = − 8 or 4.
∴ One value of k is 4.
⇒ ( x + 1) (1 − y ) 2 + 4 = 0
t 2)
4 y − kx + 8 = 0 ⇒ y = k ( x − 8 / k ).
2
If R( x, y ) divides QP externally in the
ratio 1 : 2.
2
2
∴ x = − (1 + t 2 ) and y = 1 − ⇒ t =
t
1− y
4
∴
x+ 1+
=0
(1 − y ) 2
)
2 Length of focal chord = 4a cosec α.
2
Hence, the locus of P is 9 y 2 = 4 ax .
1: P (
h,
k
its equation should be
( x + 1)2 = 4a ( y + 2)
It passes through (3, 6) so 4a = 2.
Hence the equation of the required
parabola is x2 + 2 x − 2 y − 3 = 0,
3β
α = a ⇒ 9β2 = 4 aα
2a
t1 t2 = −
9 The slope of the tangent to y = x2 at
(2, 4) is 4 and the equation of the
tangent is 4 x − y − 4 = 0
Equation of the circle is
( x − 2) 2 + ( y − 4) 2 + λ (4 x − y − 4) = 0 …(i)
Since, it passes through (0, 1).
13
Hence, λ =
5
On putting the value of λ in Eq. (i), we get
5( x − 2)2 + 5( y − 4)2 + 13(4 x − y − 4) = 0
⇒ 5( x2 + 4 − 4 x ) + 5( y 2 + 16 − 8 y )
+52 x − 13 y − 52 = 0
⇒ 5x2 + 5y 2 + 32 x − 53 y + 48 = 0
32
53
48
x−
y+
⇒ x2 + y 2 +
5
5
5
−16 53
So, the centre of the circle is
,
.
5 10
297
DAY
10 Tangent to parabola is y = mx − am2 .
∴Tangents to two given parabolas are
y = mx − (m2 / 4) and
y = m( x − 2) + (m2 / 4)
These are identical ⇒ m = 0 or 4.
∴Common tangents are y = 0 and
y = 4 x − 4.
11 y = mx + 1 / m passes through (1, 4).
⇒ m2 − 4m + 1 = 0.
m1 − m2
∴ tanθ =
1 + m1 m2
=
=
(m1 + m2 )2 − 4m1 m2
1 + m1 m2
16 − 4
1+ 1
2 3
= 3
2
=
⇒ θ = π / 3.
12 Both lines pass through (−a, b ) which is
a point on the directrix x = − a.
Therefore, tangents drawn from (−a, b )
are perpendicular, so m1 m2 = − 1.
13 The equation of any normal of
( x − 2)2 = 4( y − 3) is
( x − 2) = m ( y − 3) − 2m − m3 .
h
If it passes through 1, , then
2
h
1 − 2 = m − 3 − 2m − m3
2
⇒ 2m + m (10 − h ) − 2 = 0 = f (m )
3
[say]
This equation will given three distinct
values of m.
If f ′ (m ) = 0 has two distinct roots,
where
f (m ) = 2m3 + m (10 − h ) − 2
Now, f ′ (m ) = 6 m2 + (10 − h )
h − 10
Put
f ′(m ) = 0 ⇒ m = ±
6
So, the values of m are real and distinct,
if h > 10 i.e. h ∈ (10, ∞ ).
15 Any point on the parabola is P = ( t 2 , 2 t ).
Q is its image of the line x − y + 1 = 0.
∴
x−t
y − 2t
=
= − ( t 2 − 2 t + 1)
1
−1
⇒
Q = ( 2t − 1, t 2 + 1)
2
Since, it lies on the line
2x + y + 4 = 0
∴
4t − 2 + t 2 + 1 + 4 = 0
⇒ t 2 + 4t + 3 = 0 ⇒ t = − 1, − 3
So, the possible positions of P are (1, − 2)
and (9, − 6).
3
2
2
3
3
35
⇒ x +
a
= 3 y +
4a
a
16
3
35
Vertex P (h, k ) = −
,−
a
4a 16
3
16k
⇒ a= −
,a= −
4h
35
⇒ Locus of vertex P is xy = 105 / 64.
6)
ge
nt
Ta
n
X¢
q
x – 2y + 16=0
C (4,0)
17 Let us recall that two parabolas are
equal, if the length of their latusrectum
are equal.
Length of the latusrectum of
y 2 = λx is λ.
The equation of the second parabola is
25 {( x − 3)2 + ( y + 2)2 } = (3 x − 4 y − 2)2
|3 x − 4 y − 2|
⇒ ( x − 3)2 + ( y + 2)2 =
32 + 42
Clearly, it represents a parabola having
focus at (3, − 2) and equation of the
directrix as 3 x − 4 y − 2 = 0.
∴ Length of the latusrectum
= 2 (Distance between focus
and directrix)
A(–16, 0)
3 × 3 − 4 × (−2) − 2
32 + (−4)2
=6
parabola.
B (24,0)
P
X
2x + y – 48=0
X¢
(–2, 0)
A
A
N
Q
So, the coordinates of Q and N are
(4 a + at 2 , 0) and (at 2 , 0), respectively.
So, length of projection
= 4 a + at 2 − at 2 = 4 a
20 Let the tangent to parabola be
y = mx + a / m, if it touches the other
curve, then D = 0, to get the value of m.
For parabola, y 2 = 4 x
Let y = mx +
1
be tangent line and it
m
touches the parabola x2 = −32 y .
1
x2 = −32 mx +
m
∴
⇒
x2 + 32mx +
32
=0
m
D =0
Q
∴
32
1
(32m ) − 4 = 0 ⇒ m3 =
m
8
∴
m=
2
1
2
q
O
Q
X
rectangle.
Hence, tangents meet on the directrix.
Now, ( y − 2) 2 = 2( x + 2)
5
Vertex = (−2, 2) and directrix, x = −
2
⇒ 2x + 5 = 0
22 Here, a = 2 normal at t is
Y¢
X
21 The tangents and normals form a
Y
Normal
Equation of the line PQ is
t
y − 2at = − ( x − at 2 ).
2
Y
y2 = 4ax
P
Y¢
On putting y = 0, we get x = 4a + at 2
Thus, the two parabolas are equal, if λ = 6.
,1
19 Let P ≡ (at 2 ,2at )
X¢
18 Let P (−2 + r cos θ, r sin θ) and P lies on
16
P(
because cos θ is decreasing and tan θ
π
is increasing in 0,
2
⇒
m> 3
2
16 y = 1 a3 x2 + a x − 2a
=2
Y
4cos θ
8
⇒ r1 r2 =
sin2 θ
sin2 θ
r1 + r2
1
1
∴
=
+
r1 r2
AP
AQ
1
⇒ cos θ <
⇒ tanθ > 3
2
⇒ r1 + r2 =
4
+2
3
Hence, tanθ =
=2
4
1− ×2
3
14 Equation of tangent at P(16, 16) is
x − 2 y + 16 = 0
r 2 sin2 θ − 4(−2 + r cos θ) = 0
⇒
4
3
Slope of PB = − 2
Slope of PC =
Y¢
xt + y = 2 t 3 + 4 t. Focus = (2, 0).
298
So, the point on the axis is (10, 0).
Normal passes through (10, 0).
∴
10 = 2 t 2 + 4 ⇒ t 2 = 3
So, the normal is at the point (6, 4 3 ).
So, the required length is
⇒ y 1 = 4 and 8 = −4 x1 + y 12
⇒ y 1 = 4 and x1 = 2
Hence, required mid-point is (2, 4).
28 As common tangent is above X -axis, its
points of intersection are (0, 0) and
(4a, 4a).
Both points lie on the line
2bx + 3cy + 4d = 0
⇒
d =0
and
2b + 3c = 0
(Q a ≠ 0)
∴ d 2 + (2b + 3c )2 = 0
25 .The normal y = mx − 2am − am
3
passes through (4a, 0).
∴ m3 − 2m = 0 ⇒ m = 0, ±
∴
2
Slopes of normals are − 2, 0, 2
which are in AP.
26 The coordinate of end of the
latusrectum is (a, 2a). The equation of
the tangent at (a, 2a) is y ⋅ 2a = 2a( x + a),
i.e. y = x + a. The normal at (a, 2a) is
y + x = 2a + a, i.e. y + x = 3 a.
On solving y = 0 and y = x + a,
we get
x = – a, y = 0
On solving y = 0 and y + x = 3 a,
we get
x = 3 a, y = 0
The area of the triangle with vertices
(a, 2a), ( − a, 0), (3 a, 0)
1
= × 4 a × 2a = 4 a2
2
27 Equation of tangent to y 2 = 8 x at
(2, 4) is
4 y = 4( x + 2) i.e. x − y + 2 = 0
…(i)
Let mid-point of QR be ( x1 , y 1 ). Then,
equation of QR(T = S 1 ) is
yy 1 − 4( x + x1 ) − 5 = y 12 − 8 x1 − 5
…(ii)
⇒ 4 x − yy 1 − 4 x1 + y 12 = 0
On comparing Eqs. (i) and (ii), we get
y 2 − 4 x1
4 y1
=
= 1
1
1
2
65
31 Image of (a, 0) with respect to tangent
yt = x + at 2 is (−a, 2at ).
So, perpendicular bisector of (a, 0) and
(−a, 2at ) is the tangent line yt = x + at 2
to the parabola.
23 The equation of normal to the parabola
24 Solving y 2 = 4ax and x2 = 4ay (a ≠ 0),
= 45 + 20 =
slope is positive.
(10 − 6) 2 + (4 3 ) 2 = 16 + 48 = 8
y 2 = 12 x with slope − 1 is
y = − x − 2(3) (−1) − 3 (−1)3
⇒
y=− x+ 9 ⇒ x+ y=9
∴
k=9
Since, the focus of the parabola is (3, 0).
3− 9
∴
p=
2
⇒
2 p2 = 36
∴
3 k 3 + 2 p2 = 3(9)3 + 36
= 2223
Hence, length of chord of contact
O
(3, 0)
Hence, Statement I is true.
Statement II Infinitely many parabolas
are possible.
y = mx + 1 / m is a tangent to the
parabola.
It touches the circle ( x − 3)2 + y 2 = 9 if
3m − 0 + 1 / m
1 + m2
Hence, Statement II is false.
32 Let AB be a focal chord slope of
AB =
=3
2t
= tanα
t2 − 1
Y
⇒ (3m + 1 / m )2 = 9(1 + m2 )
⇒ 6 + 1 / m2 = 9 i.e. m2 = 1 / 3.
As m > 0, m = 1 / 3.
∴ Equation of common tangent above
X -axis is
1
y =
x+ 3
3
⇒
3 y = x + 3.
2
A (at1 , 2at1)
O
S
⇒
x1 tan2 θ − y 1 tan θ + 1 = 0
tan θ1 + tan θ2 = y 1 / x1 ,
tan θ1 tan θ2 = 1 / x1
Given that, cotθ1 + cotθ2 = 2
⇒ tan θ1 + tan θ2 = 2 tan θ1 tan θ2
⇒
y 1 / x1 = 2 / x1
⇒ Locus of P ( x1 , y 1 ) is y = 2.
30 Equation of chord of contact of (1,3) to
the parabola y 2 = 4 x is
3 y = 2( x + 1)
…(i)
Solving Eq. (i) and parabola, we get
4
( x + 1)2 = 4 x
9
⇒
x2 − 7 x + 1 = 0
∴
x1 + x2 = 7, x1 x2 = 1
⇒
( x1 − x2 )2 = 49 − 4 = 45
Also,
y1 − y2 =
⇒
( y 1 − y 2 )2 = 20
2
( x1 − x2 )
3
X
B
a ,– 2a
t2 t2
29 Let P ( x1 , y 1 ). Equation of any tangent
making angle θ with X -axis
(slope = tanθ) is
1
(Q y = mx + a / m )
y = x tan θ +
tan θ
It passes through P ( x1 , y 1 )
1
∴ x1 tan θ +
= y1
tan θ
a
(a, 0)
X¢
Y¢
α 1
tan =
2 t
⇒
⇒
t = cot
α
2
1
Length of AB = a t +
t
2
= 4 acosec2α
When a = 2, α = 60°
∴ Length of AB = 4 ( 2) cosec2 (60° )
32
=
3
33 Statement II Area of triangle formed by
these tangents and their corresponding
( y 2 − 4 a x1 )3 /2
chord of contact is 1
.
2 |a|
Hence, Statement II is false.
Statement I x1 = 12, y 1 = 8
∴ Area =
( y 12 − 4 ax1 )3 /2
2
(64 − 48)3 /2
= 32
2
Hence, Statement I is true.
=
34 Let P ( x , y ) be any point on the parabola
and let PM and PN are perpendiculars
from P on the axis and tangent at the
vertex respectively, then
299
DAY
3 Here A, B and D are co-normal points,
Y
N
P (x, y)
A
M
3x +
O
X
4y –
4=
0
4x
–3
y+
7
=
0
X¢
Y¢
(PM ) = (Latusrectum) (PN )
2
4 x + 3y + 7
3 x + 4y − 4
⇒
= 4
2
2
2
2
3 +4
4 + (−3)
2
⇒
P(t1)
x2 + y 2 =
5
m
So, the perpendicular from centre to the
tangent is equal to radius.
5
5
m
∴
=
2
2
1+ m
m 1 + m2 = 2
5
Both statements are correct as
m=±1
satisfies the given equation of Statement
II.
But, Statement II is not a correct
explanation of Statement I.
SESSION 2
1 Equation of axis is y = 2 which is
parallel to X -axis.
Therefore, reflected ray will pass
through the focus, which is (0, 2)
2 Tangents TA,TB are perpendicular
⇒
⇒
∴
AB is focal chord
AB is latusrectum.
AB = 8
AM
P¢(t2)
5
2
m2 (1 + m2 ) = 2
m 4 + m2 − 2 = 0
(m2 + 2) (m2 − 1) = 0
m=±1
[Qm2 + 2 ≠ 0, as m ∈ R]
y =± x±
M¢
V
Let common tangent be y = mx +
⇒
A(at 1 t 2 , a (t 1 + t 2 )), B (at 1 t 3 , a(t 1 + t 3 ))
and C (at 1 t 3 , a(t 2 + t 3 )).
Triangle ABC is equilateral.
a(t 3 − t 2 )
1
m AB =
= , and
at 1 (t 3 − t 2 ) t 1
m AC = 1 / t 2 .
1 / t1 − 1 / t2
∴
3=
1 + 1 / t1 t2
at 12 2at 1
at 22 2at 2
k
0
1
1
1
[Q P , A, P ′ are collinear]
⇒ k + at 1 t 2 = 0
⇒ VM . VM ′ = (at 1 t 2 )2 = k 2 = VA2
⇒ VM , VA and VM ′ are in GP.
=
|t 2 − t 1 |
|1 + t 1 t 2|
⇒ (t 2 − t 1 )2 = 3(1 + t 1 t 2 )2
⇒ (t 1 + t 2 )2 − 4t 1 t 2 = 3 + 6t 1 t 2 + 3(t 1 t 2 )2
Let A be ( x, y ). Then
y2
x 3 x2
= 3 + 10 + 2
a2
a
a
⇒ y 2 = 3a2 + 10ax + 3 x2
= 3(x + a)2 + 4ax
8 Circle is x2 + y 2 − 12 x + 31 = 0, C (6, 0),
r =
35 Equation of circle can be rewritten as
∴
4 VM = at 12 , VM ′ = at 22 and VA = k , then
Y 2 = 4 AX
3x + 4y − 4
∴
A = 1, Y =
,
5
4 x − 3y + 7
X =
5
So, the directrix is X + A = 0.
4 x − 3y + 7
⇒
+ 1= 0
5
⇒
4 x − 3 y + 12 = 0
⇒
⇒
⇒
⇒
Let A, B and D be ( x1 , y 1 ), ( x2 , y 2 ) and
( x3 , y 3 ) respectively. AB is the chord
x + y = 1.
Solving x + y = 1 and y 2 = 4 x,
we get
⇒
y 2 = 4 (1 − y )
2
⇒ y + 4y − 4 = 0
∴
y1 + y2 = − 4
y1 + y2 + y3 = 0
⇒
y3 = 4
∴ 16 = 4 x3 ⇒ x3 = 4.
Hence, D is (4, 4).
5.
Shortest distance will take place along
the common normal. Normal to y 2 = 4 x
at A(t 2 , 2t ) is
y = − tx + 2t + t 3 . It must passes
through (6, 0).
∴ t 3 − 4t = 0 ⇒ t = 0 or ± 2.
∴Distances between the curves along
common normal are 6 − 5, and 5.
Hence, minimum distance between the
curves = 5.
9 Area A = 2 x20 × x 0 × 1 = x30
2
5 Let P(k, 0) be a point on the axis on the
parabola y 2 = 4( x + 2)
Equation of normal at (−2 + t 2 , 2t ) is
t ( x + 2) + y = 2t + t 3
⇒
y + tx = t 3 .
This passes through (k , 0)
∴ t 3 − kt = 0 or t = 0, t 2 = k
For three real and distinct normals
k > 0.
∴ Set of all such point
= {(k , 0)| k > 0}.
6 Let y = mx + c intersect y 2 = 4ax at
A(t 1 ) and B (t 2 ). Then
2
m=
t1 + t2
⇒ t1 + t2 = 2 / m
Normals at A and B meet at P. Let
another normal from P meet the
parabola at C (t 3 ).
Then A, B and C are co-normal points.
∴
t1 + t2 + t3 = 0
⇒
t 3 = −2 / m
4a
4a
∴ C may be 2 , −
.
m
m
7 Let the sides of the triangle touch the
parabola y 2 = 4ax at t 1 , t 2 and t 3 .
Tangent at t 1 , t 2 , t 3 meets in
(0, x02 ) R
P (0, x02 )
Q (0, –x02 )
Since 1 ≤ x 0 ≤ 2, then area is max. at
x 0 = 2.
10 Let A( x1 , y 1 ) be any point on the
parabola
A (x1, y1)
x – y + 13 = 0
( h, k)
B
y 2 = 4 x and B (h, k ) be the reflection of
A with respect to the line
x − y + 13 = 0. Then,
h + x1
k + y1
−
+ 13 = 0
2
2
k − y1
and
. (1) = − 1
h − x1
Then, x1 = k − 13, y 1 = h + 13
300
∴ (h + 13) 2 = 4(k − 13)
∴ Locus of the point B is
( x + 13) 2 = 4( y − 13).
x=
− 1 + 17
2
14 Let ( p, q ), p, q ∈ Z be an interior point
0, 4
11 Normal at P (at 2 , 2at ) is
y= – x
y + tx = 2at + at
Given it passes (0, −6)
⇒
−6 = 2at + at 3
⇒
− 6 = 4t + 2t 3
⇒ t 3 + 2t + 3 = 0
⇒
t = −1
3
r
r
y= x
(0, 4 – r)
[Qa = 2]
4− r − 0
=r
2
4 − r = ± 2r
y2=8x
r =
4
2+ 1
4
< 0
Q
1− 2
r = 4 ( 2 − 1)
P(at2, 2at)
13 Let P ( x1 , y 1 ) and Q ( x2 , y 2 ).
C(0, – 6)
So, P (a, − 2a) = (2, − 4)
[Q a = 1]
Radius of circle
= CP = 22 + (−4 + 6)2 = 2 2
Equation of circle is
( x − 2)2 + ( y + 4)2 = (2 2 )2
x2 + y 2 − 4 x + 8 y + 12 = 0
12 x2 + x − 4 = 0
x=
− 1±
Tangents at P and Q to the parabola
y 2 = 4ax are
yy 1 = 2a ( x + x1 )
and yy 2 = 2a ( x + x2 ).
∴
2
2a( y − y )
=
4a
y1 + y2
y3 =
2
2
1
⇒
⇒
1 + 16
y ( y 1 − y 2 ) = 2a( x1 − x2 )
y 1 , y 3 , y 2 are in AP.
of both the curves.
Then,
p2 + q 2 − 16 < 0
2
and q − 4 p < 0, p ≥ 0.
⇒ p > (q / 2)2 and p2 < 16 − q 2 .
q = 0 ⇒ p = 1, 2, 3
q = 1 ⇒ p = 1, 2, 3
q = 2 ⇒ p = 2, 3
q = 3 ⇒ p = has no value.
∴There only 8 points (1, 0), (2, 0),
(3, 0), (1, 1), (2, 1), (2, 2) (3, 1), (3, 2) in
upper half.
Due to symmetry about X -axis.
(1, − 1), (2, − 1), (2, − 2), (3, − 1), (3, − 2) are
also interior points. Hence in all, three
are 13 interior integral points.
15 Clearly, QR is the diameter of the
required circle.
P
(16, 16)
Q(–16, 0) S(4, 0)
R(24, 0)
2
2
16 y = 8( x + 16) ⇒ Q = (−16, 0)
y − 16 = − 2( x − 16) ⇒ R = (24, 0)
∴ Equation of required circle is
( x + 16) ( x − 24) + y 2 = 0
⇒
x2 + y 2 − 8 x − 384 = 0
DAY TWENTY EIGHT
Ellipse
Learning & Revision for the Day
u
u
Concept of Ellipse
Tangent to an Ellipse
u
u
u
Normal to an Ellipse
Auxiliary Circle
Eccentric Angle of a Point
u
Diameter and Conjugate
Diameters
Concept of Ellipse
Ellipse is the locus of a point in a plane which moves in such a way that the ratio of its
distances from a fixed point (focus) in the same plane to its distance from a fixed straight line
(directrix) is always constant, which is always less than unity.
Equations of Ellipse in Standard Form
Different forms of an ellipse and their equations are given below
1. Ellipse of the Form
x2 y 2
+
= 1, 0 < b < a
a2 b2
If the coefficient of x2 has the larger denominator, then its major axis lies along the X-axis and
it is said to be horizontal ellipse as shown below.
Various elements of horizontal ellipse are as follows
(i) Centre, O (0, 0)
(ii) Coordinates of the vertices : A (a, 0) and A1 (−a, 0)
(iii) Equation of the major axis, y = 0
(iv) Equation of the minor axis, x = 0
X¢
(v) Focal distance of a point ( x, y) is a ± ex.
(vi) Major axis, A A1 = 2 a, Minor axis, B B1 = 2b
(vii) Foci are S (ae, 0) and S1 (− ae, 0).
a
a
(viii) Equations of directrices are l : x = , l ′ : x = −
e
e
2b 2
(ix) Length of latusrectum, LL1 = L′ L1 ′ =
a
b2
(x) Eccentricity, e = 1 − 2
a
(xi) Sum of focal distances of a point ( x, y) is 2a.
PRED
MIRROR
Your Personal Preparation Indicator
Y
L'
A1
l'
S1
B P
L
O
S
B1 L1
L'1
Y¢
X
A
l
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
302
2. Ellipse of the Form
x2 y 2
+
= 1, 0 < a < b
a2 b2
If the coefficient of x2 has the smaller denominator, then
its major axis lies along the Y-axis and it is said to be
vertical ellipse as shown below.
(i) Centre O(0, 0)
(ii) Coordinates of the vertices B (0, b ) and B1 (0, − b ).
(iii) Equation of the major axis, x = 0
(iv) Equation of the minor axis, y = 0
(v) Focal distance of a point ( x, y) is b ± ey.
(vi) Major axis, BB1 = 2b , Minor axis, AA1 = 2 a
(vii) Foci are S (0, be) and S1 (0, − be).
(viii) Equation of directrices are
b
b
l : y = , l′ : y = − .
e
e
(ix) Length of latusrectum,
2 a2
LL1 = L′ L1 ′ =
b
Y
X¢
A1
L
S
O
L'1
X
A
S1
B1
The equation of tangent to an ellipse for different forms are given
below.
xx
yy
(i) In point ( x1 , y1 ) form, 21 + 21 = 1.
a
b
(ii) In slope ‘m’ form, y = mx ± a2 m2 + b 2 . and the point of
a2 m
b2
contact is ±
,m
a2 m2 + b 2
a2 m2 + b 2
x
y
(iii) In parametric (a cos θ, b sin θ) form, cos θ + sin θ = 1.
a
b
Results on Tangent to an Ellipse
(i) The line lx + my + n = 0 is a tangent to the ellipse
x2
y2
+ 2 = 1, then n2 = a2 l 2 + b 2 m2
2
a
b
(ii) The line y = mx + c touches an ellipse, iff c2 = a2 m2 + b 2 and
a2 m
b2
the point of contact is ±
,+
.
c
c
l
B
L1
Tangent to an Ellipse
L'
l'
X¢
a2
b2
(xi) Sum of focal distances of a point ( x, y) is 2b.
(x) Eccentricity, e = 1 −
x 2 y2
Results on Ellipse 2 + 2 = 1
a b
(i) The equations x = a cos θ, y = b sin θ taken together
are called the parametric equations of the ellipse
x 2 y2
+
= 1, where θ is the parameter.
a2 b 2
(ii) A point ( x1 , y1 ) with respect to ellipse ‘S’ lie inside,
on or outside the ellipse, if S1 < 0, S1 = 0 or S1 > 0
x2
y2
where,
S1 ≡ 21 + 12 − 1
a
b
(iii) Locus of mid-points of focal chords of an ellipse
x2
y2
x2
y 2 ex
,
(
)
is
+
=
1
a
>
b
+
=
⋅
a2
b2
a2
b2
a
(iv) Let P(a cos θ, b sin θ) and Q(a cos φ, b sin φ) be any two
x 2 y2
points of the ellipse 2 + 2 = 1 (a > b ). Then,
a
b
equation of the chord joining these two points is
x
θ + φ y
θ + φ
θ − φ
cos
+ sin
= cos
2
a
2
b
2
(iii) The equation of pair of tangents drawn from an external
point P( x1 , y1 ) to the ellipse is SS1 = T 2 ,
x2
y2
x2
y2
where S ≡ 2 + 2 − 1, S1 ≡ 12 + 12 − 1
a
b
a
b
xx1
yy1
and T ≡ 2 + 2 − 1
a
b
xx
yy
(iv) The equation of chord of contact of tangents is 21 + 21 = 1
a
b
or T = 0.
(v) The equation of chord of an ellipse, whose mid-point is
xx
yy
x2
y2
( x1 , y1 ), is T = S1 , i.e. 21 + 21 = 12 + 12
a
b
a
b
(vi) The locus of the point of intersection of perpendicular
tangents to the ellipse is a director circle, i.e.
x2 + y2 = a2 + b 2 .
(vii) The point of intersection of the tangents at α and β is
α+β
α + β
b sin
a cos
2
2 .
,
α
−
β
α
−
β
cos
cos
2
2
Normal to an Ellipse
The equations of normal in different form to an ellipse are given
below
a2 x b 2 y
1. In point ( x1 , y1 ) form,
−
= a2 − b 2
x1
y1
m (a2 − b 2 )
2. In slope ‘m’ form, y = mx ±
at the points
a2 + b 2 m2
±
a2
a +b m
2
2
2
,±
.
a + b 2 m2
b2 m
2
3. In parametric (a cos θ, b sin θ) form,
ax sec θ − by cosec θ = a2 − b 2 .
303
DAY
4. The point of intersection of normals to the ellipse at two
points (a cos θ1 , b sin θ1 ) and (a cos θ2 , b sin θ2 ) are (λ , µ),
θ + θ2
cos 1
2
(a 2 − b 2 )
where λ =
⋅ cos θ1 ⋅ cos θ2 ⋅
(a)
θ − θ2
cos 1
2
θ + θ2
sin 1
2
(a2 − b 2 )
and
µ=−
⋅ sin θ1 ⋅ sin θ2 ⋅
(b )
θ − θ2
cos 1
2
Results on Normal to an Ellipse
(i) The line lx + my + n = 0 is a normal to the ellipse
x2
y2
a2
b2
(a2 − b 2 )2
+ 2 = 1, then 2 + 2 =
2
a
b
l
m
n2
(ii) Four normals can be drawn from a point to an ellipse.
(iii) If the line y = mx + c is a normal to the ellipse
m2 (a2 − b 2 )2
x 2 y2
is the condition of
+ 2 = 1, then c2 = 2
2
a + b 2 m2
a
b
normality of the line to the ellipse.
(iv) The points on the ellipse, the normals at which to the
ellipse pass through a given point are called conormal
points.
(v) Tangent at an end of a latusrectum (Ist quadrant) is
1 − e2
ex
+
y = 1, or ex + y = a and normal is
a
b
ax
by
−
= a2 − b 2 or x − ye = ae3
e
1 − e2
Auxiliary Circle
The circle described on
the major axis of an
ellipse as diameter is
called an auxiliary circle.
x 2 y2
If 2 + 2 = 1 is an ellipse,
a
b
then its auxiliary circle is
x2 + y2 = a2
Y
X'
A'
C
P
N
x2 + y 2 = a2
Y'
(iii) The sum of the eccentric angles of the conormal points
x 2 y2
on the ellipse 2 + 2 = 1 is an odd multiple of π.
a
b
Diameter and Conjugate Diameters
The locus of the mid-points of a system of parallel chords is
called a diameter. If y = mx + c represents a system of
x2
y2
parallel chords of an ellipse 2 + 2 = 1, then the line
a
b
b2
y = − 2 x is the equation of the diameter.
am
The two diameters are said to be conjugate diameters, when
each bisects all chords parallel to the other.
If y = mx and y = m1 x be two conjugate diameters of an
b2
ellipse, then m m1 = − 2
a
Results on Conjugate Diameters
(i) The area of a parallelogram formed by the tangents at
the ends of conjugate diameters of an ellipse is constant
and is equal to the product of the axes.
(ii) The sum of the squares of any two conjugate
semi-diameters of an ellipse is constant and equal to the
sum of the squares of the semi-axes of the ellipse, i.e.
CP2 + CD2 = a2 + b 2 .
(iii) The product of the focal distance of a point on an ellipse
is equal to the square of the semi-diameter which is
conjugate to the diameter through the point.
Q
φ
(ii) A circle cut an ellipse in four points real or imaginary.
The sum of the eccentric angles of these four concyclic
points on the ellipse is an even multiple of π.
Important Points
A
X
x2
y2
+
=1
a2 b 2
Eccentric Angle of a Point
x 2 y2
+
= 1. Draw PN
a2 b 2
perpendicular from P on the major axis of the ellipse and
produce NP to meet the auxiliary circle in Q. Then, ∠XCQ = φ
is called the eccentric angle of the point P on the ellipse.
Let P be any point on the ellipse
So, the coordinates of Q and P are (a cos φ, a sin φ) and
(a cos φ, b sin φ), where φ is an eccentric angle.
Results on Eccentric Angles
(i) Eccentric angles of the extremities of latusrectum of the
x2
y2
b
ellipse 2 + 2 = 1 are tan −1 ±
.
ae
a
b
(i) The eccentric angles of the ends of a pair of conjugate
diameter of an ellipse differ by a right angle.
(ii) The tangents at the ends of a pair of conjugate diameters
of an ellipse form a parallelogram.
Conjugate Points Two points P and Q are conjugate points
with respect to an ellipse, if the polar of P passes through Q
and the polar of Q passes through P.
Conjugate Lines Two lines are said to be conjugate lines with
respect to an ellipse, if each passes through the pole of the
polar.
Pole and Polar Let P be a point inside or outside an ellipse.
Then, the locus of the point of intersection of tangents to the
ellipse at the point, where secants drawn through ‘P’ intersect
the ellipse is called the polar of point P with respect to the
ellipse and the point P is called the pole of the polar. The
x 2 y2
polar of a point ( x1 , y1 ) with respect to the ellipse 2 + 2 = 1
a
b
xx
yy
is 21 + 21 = 1
a
b
304
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 The equation
x2
y2
+
= 1 will represent an ellipse if
8−a a −2
(a) a ∈ (1, 4)
(c) a ∈ (2, 8)
(b) a ∈(− ∞, 2) ∪ (8, ∞)
(d) None of these
2 Equation of ellipse whose minor axis is equal to the
distance between the foci and whose latusrectum is 10,
is given by (take origin as centre and major axis along
X-axis)
(a) 2 x 2 + y 2 = 100
(c) 2 x 2 + y 2 = 50
(b) x 2 + 2 y 2 = 100
(d) None of these
9 Tangents are drawn to the ellipse x 2 + 2y 2 = 4 from any
arbitrary point on the line x + y = 6. The corresponding
chord of contact will always pass through the fixed point
1 2
(a) ,
3 3
1 1
(c) ,
3 3
2 1
(b) ,
3 3
(d) no such fixed point exist.
10 The equation of the circle passing through the foci of the
ellipse
x2
y2
+
= 1 and having centre at (0, 3) is
16
9
JEE Mains 2013
(b) x + y − 6y + 7 = 0
(d) x 2 + y 2 − 6y + 5 = 0
j
3 Let AB be a rod of length 4 units with A on x and B on
Y-axis. Rod AB slides on axes. If point P divides AB in
the ratio 1: 2, locus of P is an ellipse. The eccentricity of
this ellipse is
(a)
3
4
(b)
3
2
(c)
2
3
(d) None
4 Equation of the ellipse whose axes are the axes of
coordinates and which passes through the point ( −3, 1)
2
and has eccentricity
is
5
(a) 5 x 2 + 3 y 2 − 48 = 0
(c) 5 x 2 + 3 y 2 − 32 = 0
(b) 3 x 2 + 5 y 2 − 15 = 0
(d) 3 x 2 + 5 y 2 − 32 = 0
5 If the angle between the straight lines joining foci and the
end of minor axis of the ellipse
x2 y2
+
= 1 is 90°, then its
a 2 b2
eccentricity is
(a)
1
2
(b)
1
4
(c)
1
2
(d) None
6 The ellipse 4x 2 + 9y 2 = 36 and the straight line
y = mx + c intersect in real points only if
(a) 9m 2 ≤ c 2 − 4
(c) 9m 2 ≥ c 2 − 4
(b) 9m 2 > c 2 − 4
(d) None of these
7 If the line 3x + 4y = 7 touches the ellipse 3x 2 + 4y 2 = 1
then the point of contact is
1 1
(a)
,
7 7
1 − 1
(c)
,
7 7
1 − 1
(b)
,
3 3
(d) None of these
8 The equation of common tangent of the curves
x 2 + 4y 2 = 8 and y 2 = 4x are
(a) x − 2 y + 4 = 0, x + 2 y + 4 = 0
(b) 2 x −y + 4 = 0, 2 x + y + 4 = 0
(c) 2 x − y + 2 = 0, 2 x + y + 2 = 0
(d) None of the above
(a) x + y − 6y − 7 = 0
(c) x 2 + y 2 − 6y − 5 = 0
2
2
2
2
11 If tangent at any point P on the ellipse 7x 2 + 16y 2 = 12
cuts the tangent at the end points of the major axis at the
points A and B, then the circle with AB as diameter
passes through a fixed point whose coordinates are
(a) (±
(c) (0, ±
a 2 − b 2 , 0)
(b) (±
a2 − b 2 )
a 2 + b 2 , 0)
(d) (0, ±
a2 + b 2 )
12 The length of the common tangent to the ellipse
x2
y2
+
= 1 and the circle x 2 + y 2 = 16 intercepted by
25
4
the coordinate axes is
(a) 5
(b) 2 7
(c)
7
3
(d)
14
3
13 The distance of the centre of ellipse x 2 + 2y 2 − 2 = 0 to
those tangents of the ellipse which are equally inclined
from both the axes, is
(a)
3
2
(b)
3/2
(c)
2 /3
(d)
3
2
14 PQ is a chord of the ellipse through the centre. If the
square of its length is the HM of the squares of major
and minor axes, find its inclination with X -axis.
π
4
2π
(c)
3
(a)
(b)
π
2
(d) None of these
15 If straight line ax + by = c is a normal to the ellipse
4x 2 + 9y 2 = 36, then 4a 2 + 9b 2 is equal to
169a 2 b 2
c2
13a 2 b 2
(c)
c2
(a)
(b)
25a 2b 2
c2
(d) None of these
305
DAY
16 If the normal at the point P(θ) to the ellipse
5x 2 + 14y 2 = 70 intersects it again at the point Q(2θ), then
cos θ is equal to
(a)
2
3
(b) −
2
3
(c)
1
3
(d) −
1
3
17 An ellipse is drawn by taking a diameter of the circle
( x − 1)2 + y 2 = 1 as its semi-minor axis and a diameter of
the circle x 2 + ( y − 2)2 = 4 as its semi-major axis. If the
centre of the ellipse is at the origin and its axes are the
coordinate axes, then the equation of the ellipse is
(a) 4 x 2 + y 2 = 4
(c) 4 x 2 + y 2 = 8
(b) x 2 + 4 y 2 = 8
(d) x 2 + 4 y 2 = 16
18 The area (in sq units) of the quadrilateral formed by the
tangents at the end points of the latusrectum to the
x2
y2
JEE Mains 2015
ellipse
+
= 1 is
9
5
(a)
27
4
(b) 18
(c)
27
2
(d) 27
19 The locus of the foot of perpendicular drawn from the
centre of the ellipse x 2 + 3y 2 = 6 on any tangent to it is
(a) (x − y ) = 6x + 2 y
(c) (x 2 + y 2 )2 = 6x 2 + 2 y 2
2
2 2
2
2
JEE Mains 2014
(b) (x − y ) = 6x 2 − 2 y 2
(d) (x 2 + y 2 )2 = 6x 2 − 2 y 2
2
2 2
x2
+ y 2 = 1 at
27
( 3 3 cos θ , sin θ ) (where, θ ∈( 0, π /2)). Then, the value of θ
such that the sum of intercepts on axes made by this
tangent is minimum, is
20 Tangent is drawn to the ellipse
(a)
π
3
(b)
π
6
(c)
π
8
(d)
π
4
21 If the tangent at a point (a cos θ, b sin θ ) on the ellipse
x2
y2
+ 2 = 1 meets the auxiliary circle in two points, the
2
a
b
chord joining them subtends a right angle at the centre,
then the eccentricity of the ellipse is given by
(a) (1 + cos2 θ)−1 / 2
(c) (1 + sin2 θ)−1 / 2
(b) (1 + sin2 θ)
(d) (1 + cos2 θ)
22 Chord of contact of tangents drawn from the point P (h, k )
to the ellipse x 2 + 4y 2 = 4 subtends a right angle at the
centre. Locus of the point P is
(a) x 2 + 16y 2 = 20
(c) 16x 2 + y 2 = 20
(b) x 2 + 8 y 2 = 10
(d) None of these
23 Tangent at a point P on the ellipse x 2 + 4y 2 = 4 meets the
X -axis at B and AP is ordinate of P. If Q is a point on AP
produced such that AQ = AB, then locus of Q is
(a) x 2 + xy − 4 = 0
(c) x 2 + xy − 1 = 0
(b) x 2 − xy + 4 = 0
(d) x 2 − xy + 1 = 0
24 A parabola is drawn whose focus is one of the foci of the
x2 y2
+
= 1 (where, a > b ) and whose directrix
a 2 b2
passes through the other focus and perpendicular to the
major axis of the ellipse. Then, the eccentricity of the
ellipse
ellipse for which the latusrectum of the ellipse and the
parabola are same, is
2 −1
(a)
(b) 2 2 + 1
2+1
(c)
2
(d) 2 2 − 1
2
x
y
+
= 1 tangent PQ is
a 2 b2
drawn. If the point Q be at a distance 1/p from the point
P, where p is distance of the tangent from the origin, then
the locus of the point Q is
25 At a point P on the ellipse
x2 y2
1
+
= 1+ 2 2
a2 b 2
a b
x2 y2
1
(c) 2 + 2 = 2 2
a
b
a b
x2 y2
1
−
= 1− 2 2
a2 b 2
a b
x2 y2
1
(d) 2 − 2 = 2 2
a
b
a b
(a)
(b)
26 If CP and CD are semi-conjugate diameters of an ellipse
x2 y2
+
= 1, then CP 2 + CD 2 is equal to
14
8
(a) 20
(b) 22
(c) 24
(d) 26
27 If θ and φ are eccentric angles of the ends of a pair of
conjugate diameters of the ellipse
x2
y2
+ 2 = 1, then θ − φ
2
a
b
is equal to
π
2
(a) ±
(b) ± π
(c) 0
(d) None of these
28 The coordinates of all the points P on the ellipse
x2 y2
+
= 1, for which the area of the ∆PON is maximum,
a 2 b2
where O denotes the origin and N, the foot of the
perpendicular from O to the tangent at P, is
a2
(a) ±
2
a + b2
a2
(b) ±
a2 − b 2
2a 2
(c) ±
a2 + b 2
2a 2
(d)
,
a2 − b 2
a +b
b2
,±
2
2
a −b
2b 2
,±
a2 + b 2
2b 2
a2 − b 2
b2
,±
2
2
29 If α, β, γ are the eccentric angles of three points on the
ellipse 4x 2 + 9y 2 = 36, the normals at which are
concurrent, then sin(α + β) + sin(β + γ ) + sin( γ + α) is
equal to
(a)
2
3
(b)
4
9
(c) 0
(d) None
30 A triangle is drawn such that it is right angled at the
x2 y2
+
= 1(where, a > b ) and its
a 2 b2
other two vertices lie on the ellipse with eccentric angles
α + β
1 − e 2 cos 2
2
is equal to
α and β. Then,
α − β
cos 2
2
centre of the ellipse
(a)
a2
a + b2
2
(b)
a2 + b 2
a2
(c)
a2
a − b2
2
(d)
a2 − b 2
a2
306
Directions (Q. Nos. 31-35) Each of these questions contains
two statements : Statement I (Assertion) and Statement II
(Reason). Each of these questions also has four alternative
choices, only one of which is the correct answer. You have to
select one of the codes (a ), (b), ( c) and (d) given below.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is not
a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
31 Let S1 ≡ ( x − 1)2 + ( y − 2)2 = 0 and
S 2 ≡ ( x + 2)2 + ( y − 1)2 = 0 be the equations of two circles.
Statement I Locus of centre of a variable circle touching
two circles S1 and S 2 is an ellipse.
Statement II If a circle S1 = 0 lies completely inside the
circle S 2 = 0, then locus of centre of variable circle S = 0
which touches both the circles is an ellipse.
3 3
,1 is a point on the ellipse
2
32 Statement I If P
4x 2 + 9y 2 = 36. Circle drawn AP as diameter touches
another circle x 2 + y 2 = 9, where A ≡ ( − 5, 0).
Statement II Circle drawn with focal radius as diameter
touches the auxiliary circle.
33 Statement I The product of the focal distances of a
point on an ellipse is equal to the square of the
semi-diameter which is conjugate to the diameter
through the point.
Statement II If y = mx and y = m1x are two conjugate
b2
diameters of an ellipse, then mm1 = − 2 .
a
34 Statement I The condition on a and b for which two
x2
y2
+
= 1 passing
2
2a
2b 2
through (a, − b ) are bisected by the line x + y = b is
a 2 + 6ab − 7b 2 ≥ 0 .
distinct chords of the ellipse
Statement II Equation of chord of the ellipse
x2 y2
+
= 1 whose mid-point is ( x 1, y 1), is T = S1.
a2 b 2
35 Statement I An equation of a common tangent to the
parabola y 2 = 16 3x and the ellipse 2x 2 + y 2 = 4 is
y = 2x + 2 3.
4 3
, (m ≠ 0) is a
m
common tangent to the parabola y 2 = 16 3 x and
the ellipse 2 x 2 + y 2 = 4, then m satisfies
m 4 + 2m 2 = 24.
Statement II If the line y = mx +
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 The eccentricity of an ellipse whose centre is at origin is
1
. If one of its directrices is x = − 4, then the equation of
2
3
the normal to it at 1, is
2
JEE Mains 2017
(a) 4 x + 2 y = 7
(c) 2 y − x = 2
(b) x + 2 y = 4
(d) 4 x − 2 y = 1
2 The locus of the mid-points of a focal chord of the ellipse
x2 y2
+
= 1 is
a 2 b2
(a)
x 2 y 2 ex
+
=
a2 b 2
a
(c) x 2 + y 2 = a 2 + b 2
(b)
x 2 y 2 ex
−
=
a2 b 2
a
(d) None of these
3 Foot of normal to the ellipse 4x 2 + 9y 2 = 36 having slope
2 may be
9 − 8
(a) ,
5 5
− 9 8
(c)
,
5 5
9 8
(b) ,
5 5
(d) None of these
4 On the ellipse 4x 2 + 9y 2 = 1, the points at which the
tangents are parallel to the line 8x = 9y are
− 2 1
2 1
(a) , ,
,
5 5 5 5
− 3 − 1 3 1
(c)
,
, ,
5 5 5 5
− 2 1 2 − 1
(b)
, , ,
5 5 5 5
(d) None of these
5 The normal at a point P on the ellipse x 2 + 4y 2 = 16 meets
the X -axis at Q. If M is the mid-point of the line segment
PQ, then the locus of M intersects the latusrectums of the
given ellipse at the points
3 5
2
(a) ±
,±
2
7
1
(c) ± 2 3 , ±
7
3 5
19
(b) ±
,±
2
4
4 3
(d) ± 2 3 , ±
7
6 The line passing through the extremity A of the major axis
and extremity B of the minor axis of ellipse x 2 + 9y 2 = 9
meets its auxiliary circle at the point M. Then the area of
the triangle with vertices at AM and the origin O is
(a)
31
10
(b)
29
10
(c)
21
10
(d)
27
10
307
DAY
1 1
1
+
4 a 2 b 2
1
1
(c) 2 + 2
a
b
7 The ellipse x 2 + 4y 2 = 4 is inscribed in a rectangle
(a)
alingent with the coordinate axes, which in turn is
inscribed in another ellipse that passes through the
point ( 4, 0). Then the equation of the ellipse is
(a) x 2 + 12 y 2 = 16
(c) 4 x 2 + 64 y 2 = 48
(b) 4 x 2 + 48 y 2 = 48
(d) x 2 + 16y 2 = 16
(a)
(b) x 2 + 2 3 y = 3 − 3
(d) None
x
y
+
= 1 is inscribed in a rectangle R
9
4
whose sides are parallel to the co-ordinate axes.
Another ellipse E 2 passing through the point ( 0, 4)
circumscribes the rectangle R. The eccentricity of the
ellipse E 2 is
3
(b)
2
1
(c)
2
3
(d)
4
π
in points whose eccentric angles differ by , then r 2 is
3
equal to
3
(4 p 2 + q 2 )
4
2
(c) (4 p 2 + q 2 )
3
(b)
4
(4 p 2 + q 2 )
3
(d) None of these
11 The sum of the square of the reciprocals of two
perpendicular diameters of an ellipse
(b)
b
a
(c)
a2
b2
(d)
b2
a2
α
β
chord of the ellipse 16x 2 + 25y 2 = 400, then tan tan
2
2
is equal to
(a) − 4
(b)
−1
4
(c)
−3
8
(d) None
14 The tangent and normal drawn to the ellipse x 2 + 4y 2 = 4 at
the point P(θ) meets the X -axis at the points A and B. If
AB = 2, then cos 2 θ is equal to
10 If the line px + qy = r intersects the ellipse x 2 + 4y 2 = 4
(a)
a
b
13 If α, β are the eccentric angles of the extremities of a focal
2
9 The ellipse E 1 :
2
(a)
2
(d) None of these
constant ratio of the area of triangle formed by joining the
corresponding points on the auxiliary circle of the vertices of
the first triangle. This ratio is
points of the latusrectum of the ellipse x 2 + 4y 2 = 4.
The equations of parabolas with latusrectum PQ are
2
1 1
1
+
2 a 2 b 2
12 The area of a triangle inscribed in an ellipse bears a
8 Let P ( x1, y1 ) and Q ( x 2 , y 2 ), y1 < 0, y 2 < 0, be the end
(a) x 2 − 2 3 y = 3 + 3
(c) both (a) and (b)
(b)
x2 y2
+
= 1, is
a 2 b2
(a)
4
9
(b)
8
9
(c)
2
9
(d) None
x2 y2
+
= 1 (a > b ) with foci at S and S′ and
a 2 b2
vertices at A and A′. A tangent is drawn at any point P on
the ellipse and let R , R ′ , B, B′ respectively be the feet of the
perpendiculars drawn from S , S ′ , A, A′ on the tangent at P.
Then, the ratio of the areas of the quadrilaterals S ′ R ′ RS and
A′ B′ BA is
15 Given an ellipse
(a) e : 2
(c) e : 1
(b) e : 3
(d) e : 4
ANSWERS
SESSION 1
SESSION 2
1 (d)
2 (b)
3 (b)
4 (d)
5 (c)
6 (c)
7 (a)
8 (a)
9 (b)
10 (a)
11 (a)
12 (d)
13 (b)
14 (a)
15 (b)
16 (b)
17 (d)
18 (d)
19 (c)
20 (b)
21 (c)
22 (a)
23 (a)
24 (a)
25 (a)
26 (b)
27 (a)
28 (a)
29 (c)
30 (b)
31 (d)
32 (a)
33 (b)
34 (a)
35 (b)
1 (d)
11 (a)
2 (a)
12 (b)
3 (b)
13 (b)
4 (b)
14 (a)
5 (c)
15 (c)
6 (d)
7 (a)
8 (c)
9 (c)
10 (a)
308
Hints and Explanations
SESSION 1
2
1
It passes through p(−31
, ) and e =
2
y
x
+
= 1 will represent an
8− a a−2
ellipse if
8 − a > 0, a − 2 > 0, 8 − a ≠ a − 2
⇒ 2 < a < 8, a≠ 5
⇒ a ∈ (2, 8) − {5}
2 2b = 2ae
⇒
a2e 2 = b 2 = a2 (1 − e 2 )
1
⇒
e2 =
2
2b 2
= 10 ⇒ b 2 = 5a
a
⇒ a2 (1 − 1 / 2) = 5a
⇒
a= 10
∴
b 2 = 50, a2 = 100
Hence, equation of ellipse is
x2
y2
+
=1
100 50
2
2
⇒
x + 2 y = 100
2
.
5
9
1
…(i)
+
=1
a2 b 2
b2
and
e2 = 1 − 2
a
2
b2
b2 3
⇒
= 1− 2 ⇒ 2 =
5
a
a
5
9
5
From Eq. (i), 2 + 2 = 1
a
3a
27 + 5
=
1
⇒
3a2
32
32
and then b 2 =
⇒
a2 =
3
5
∴Equation of ellipse is
3 x2 5y 2
+
= 1 ⇒ 3 x2 + 5y 2 = 32
32
32
∴
5 Let equation of ellipse be
2
2
y
x
+
=1
a2 b 2
B
3 Let A = (a, 0), B(0, b ).
B(0,b) Y
F2 O
2:1
O
P(h,k)
A(a,0)
X
∴
x2
y2
+
=1
64 16
9 9
4 Let the equation of ellipse is
(-3,1) Y
P
X
Y¢
4 x2 + 9(mx + c )2 − 36 = 0
⇒
(9m2 + 4)x2 +
18cmx + 9c 2 − 36 = 0
Roots are real, so
18 × 18c 2 m2 − 4 ( 9m2 + 4)
(9c 2 − 36)≥ 0
⇒
9m2 − c 2 + 4 ≥ 0
⇒
9m2 ≥ c 2 − 4
7 Equation of ellipse is 3 x2 + 4 y 2 = 1
y2
x2
+ 2 = 1, (a > b )
2
a
b
O
a2e 2 = b 2 = a2 (1 − e 2 )
1
1
e2 =
⇒e =
2
2
y = mx +
1
m
It touches the ellipse
x2
y2
+
=1
8
2
1
= 8m2 + 2 [Qc 2 = a2 m2 + b 2 ]
m2
⇒ 8m 4 + 2m2 − 1 = 0
1
1
⇒ m2 = i.e. m = ±
4
2
Hence, required tangent are
x − 2 y + 4 = 0, x + 2 y + 4 = 0
⇒
9 Any point on the line can be taken as
P (h, 6 − h )
Equation of chord of contact of P w.r.t.
x2 + 2 y 2 = 4 is hx + 2(6 − h )y = 4 [T = 0]
⇒ h( x − 2 y ) + 4(3 y − 1) = 0
which passes through the fixed point
2 1
1
2
y = , x = i.e., through , .
3 3
3
3
10 Given equation of ellipse is
x2
y2
+
= 1.
16
9
Here, a = 4, b = 3, e = 1 −
∴ Foci is (± ae , 0)
7
= ± 4 ×
, 0 = (±
4
9
7
=
16
4
7, 0)
Y
r
we get
16 64
3
3
=
(1 − e 2 ) ⇒ e 2 = i.e. e =
9
9
4
2
X¢
⇒
8 Any tangent to parabola y 2 = 4 x is
6 Solving 4 x2 + 9 y 2 = 36 and y = mx + c ,
[Q AB = 4]
⇒ Locus of P is
Then, F1 = (ae , 0), F2 = (− ae , 0), B = (0,b ).
π
∠F1 BF2 =
2
− b − b = − 1
⇒
ae − ae
⇒
P divides AB in the ratio 1:2.
2a
b
h = ,k=
3
3
9h2
2
2
∴ a +b =
+ 9k 2 = 16
4
∴
F1
1 1
Clearly, P
,
lies on the ellipse,
7 7
therefore it is a point of contact.
Tangent at P ( x1 , y 1 ) is
...(i)
3 xx1 + 4 yy 1 = 1
On comparing Eq. (i) with the given
4y
3x
1
tangent 3 x + 4 y = 7, 1 = 1 =
⇒
3
4
7
1 1
P ( x1 , y 1 ) = P
,
7 7
X¢
X
Y¢
∴
Radius of the circle,
r = (ae )2 + b 2 = 7 + 9
= 16 = 4
Now, the equation of circle is
( x − 0)2 + ( y − 3)2 = 16
⇒
x2 + y 2 − 6 y − 7 = 0
11 Equation of tangent at any point
y
x
cos θ + sin θ = 1.
a
b
The equation of tangents at the end
P (acos θ,b sin θ) is
DAY
309
1
cos 2 θ sin2 θ
=
+
r2
a2
b2
points of the major axis are
x = a, x = − a.
∴The intersection point of these
tangents are
θ
θ
A = a,b tan , B = − a,b cot .
2
2
∴
Equation of circle with AB as diameter
θ
( x − a)( x + a) + y − b tan
2
From Eqs. (i) and (ii), we get
y − b cot θ = 0
2
θ
θ
⇒ x2 − a2 + y 2 − by tan + cot
2
2
+ b2 = 0
⇒ x2 + y 2 − a2 + b 2 − 2by cosec θ = 0
which is the equation of family of
circles passing through the point of
intersection of the circle
x2 + y 2 − a2 + b 2 = 0 and y = 0
So, the fixed point is (± a2 − b 2 , 0).
2
2
12 The tangent to x + y = 1 is
25
4
y
x
cos θ + sin θ = 1. If it is also tangent
5
2
to the circle, then
1
100
16 =
=
cos 2 θ sin2 θ 4 + 21sin2 θ
+
25
4
3
25
2
, cos 2 θ =
⇒ sin θ =
28
28
If the tangents meets the axes at A and
B, then
5
2
A =
, 0 and B = 0,
cos θ
sinθ
25
4
+
cos 2 θ sin2 θ
4
196
14
= 28 + ⋅ 28 =
⇒ AB =
3
3
3
∴ AB 2 =
13 Given, equation of ellipse is
x2
y2
+
= 1. General equation of tangent
2
1
to the ellipse of slope m is
y = mx ± 2m2 + 1
Since, this is equally inclined to axes, so
m = ± 1.
Then, tangents are
y = ± x± 2+ 1 = ± x± 3
Distance of any tangent from origin
=
|0 + 0 ±
12 + 12
3|
= 3/2
14 The straight line x = r cos θ , y = r sin θ
meets the ellipse
given.
y2
x2
+ 2 = 1, where r is
2
a
b
But PQ 2 = 4 r 2 = HM of 4 a2 , 4b 2
1
1
1
∴
=
+ 2
2b
r 2 2a2
…(i)
…(ii)
cos 2 θ sin2 θ
1
1
+
= 2 + 2
a2
b2
2a
2b
π
It is possible, when θ =
4
2
2
15 Given ellipse is x + y = 0, (a = 3, b = 2)
9
4
Equation of normal is
3 x sec θ − 2 y cosec θ = 9 − 4 = 5
Comparing it with given normal
ax + by = c ,
3sec θ − 2 cosec θ 5
=
=
a
b
c
3c
2c
⇒ cos θ = , sinθ = −
5a
5b
4c 2
9c 2
+
=1
⇒
25a2 25b 2
25a2b 2
⇒
9b 2 + 4a2 =
c2
2
2
16 Given ellipse is x + y = 1
∴
14
5
P(θ) is ( 14 cos θ, 5 sin θ)
Equation of normal at P(θ) is
14 x
5y
−
= 14 − 5 = 9
cos θ sin θ
Q(2θ) lies on this normal, therefore
14 cos 2θ 5sin 2θ
−
=9
cos θ
sin θ
⇒ 14 cos 2θ sin θ − 10sin θ cos 2 θ
= 9 cos θ sin θ
⇒ 18 cos 2 θ − 9 cos θ − 14 = 0
⇒ (3 cos θ + 2) (6 cos θ − 7) = 0
−2 7
or
⇒ cos θ =
6
3
−2
7
as cos θ ≠
⇒ cos θ =
3
6
17 Given,
(i) An ellipse whose semi-minor axis
coincides with one of the diameters
of the circle ( x − 1)2 + y 2 = 1.
(ii) The semi-major axis of the ellipse
coincides with one of the diameters
of circle x2 + ( y − 2)2 = 4.
(iii) The centre of the ellipse is at origin.
(iv) The axes of the ellipse are
coordinate axes.
Now, diameter of circle ( x − 1)2 + y 2 = 1
is 2 units and that of circle
x2 + ( y − 2)2 = 4 is 4 units.
Semi-minor axis of ellipse, b = 2 units
and semi-major axis of ellipse,
a = 4 units.
Hence, the equation of the ellipse is
y2
x2
x2
y2
+ 2 =1 ⇒
+
=1
2
a
b
16
4
⇒
x2 + 4 y 2 = 16
18 Area of quadrilateral formed by tangents
at the ends of latusrectum =
2a2
.
e
2
3
Here, a2 = 9, b 2 = 5, e =
3
∴ Area = 2 × 9 × = 27
2
19 x2 + 3 y 2 = 6
x2 y 2
...(i)
+
= 1 (a2 = 6, b 2 = 2)
6
2
Equation of any tangent to Eqs. (i) is
...(ii)
y = mx ± 6m2 + 2
⇒
Equation of perpendicular line drawn
from centre (0, 0) to Eqs. (ii) is
1
...(iii)
y=−
x
m
Eliminating m from Eqs. (ii) and (iii),
required locus of foot of perpendicular is
x
x2
y = − x ± 6 2 + 2
y
y
⇒
( y 2 + x2 )2 = 6 x2 + 2 y 2
20 The equation of tangent from the point
(3 3 cos θ,sin θ) to the curve is
x cos θ y sin θ
+
= 1.
3 3
1
Thus, sum of intercepts
= 3 3secθ + cosecθ = f (θ)
[say]
3 3 sin θ − cos θ
sin2 θ cos 2 θ
Put
f ′ (θ) = 0
1
⇒
sin3 θ = 3 /2 cos 3 θ
3
1
π
⇒
tanθ =
⇒θ=
6
3
Now, f ′ ′ (θ) > 0, for θ = π / 6
[i.e. minimum]
π
Hence, value of θ is .
6
⇒
f ′ (θ) =
3
3
21 Equation of tangent at (acos θ,b sin θ) to
y
x
cos θ + sin θ = 1 …(i)
a
b
The joint equation of the lines joining
the points of intersection of Eq. (i) and
the auxiliary circle x2 + y 2 = a2 to the
origin, which is the centre of the circle,
2
y
x
is x2 + y 2 = a2 cos θ + sin θ
b
a
the ellipse is
Since, these lines are at right angles.
cos 2 θ
sin2 θ
∴ 1 − a2 2 + 1 − a2 2 = 0
a
b
[Qcoefficient of x2 + coefficient of y 2 = 0]
310
⇒
a2
sin2 θ 1 − 2 + 1 = 0
b
⇒ sin2 θ (b 2 − a2 ) + b 2 = 0
⇒
(1 + sin2 θ)(a2e 2 ) = a2
⇒ e = (1 + sin2 θ)−1 /2
x 2 + 4 y 2 = 4 is
hx + 4ky − 4 = 0
...(i)
Making x2 + 4 y 2 − 4 = 0 homogeneous
with Eq. (i), we get
2
hx + 4ky
x2 + 4 y 2 − 4
=0
4
Chord of contact subtends right angles
at the centre, so
coefficient of x2 + coefficient of y 2 = 0
1
1 + 4 − h2 − 4k 2 = 0
∴
4
⇒ locus of P (h, k ) is x2 + 16 y 2 = 20
23 Ellipse is x2 + 4 y 2 = 4. P = (2 cos θ,sin θ)
Equation of tangent at P is
2cos θx + 4 y sin θ = 4
It meets X -axis at B.
∴ B = (2sec θ, 0), A = (2 cos θ, 0)
2sin2 θ
AQ = AB = 2 |sec θ − cos θ| =
cos θ
y − b sin θ
x − acos θ
=
asin θ
−b cos θ
2
y2
24 Equation of the ellipse is x2 + 2 = 1.
a
b
Equation of the parabola with focus
S (ae , 0) and directrix
x + ae = 0 is y 2 = 4 aex.
Now, length of latusrectum of the ellipse
2b 2
and that of the parabola is 4 ae.
is
a
For the two latusrectum to be equal,
Y
(ae, 0)
Y¢
2b 2
2a2 (1 − e 2 )
= 4 ae ⇒
= 4 ae
a
a
2
2
⇒ 1 − e = 2e ⇒ e + 2e − 1 = 0
X
sin θ sin φ = − cos θ cos φ
cos(θ − φ) = 0
π
θ − φ= ±
2
⇒
(a cos q, b sin q)
P
X¢
28 Equation of tangent at P is
X
O
x cos θ y sin θ
+
= 1.
a
b
Y
Y¢
The distance of the tangent from the
origin is
ab
p=
b 2 cos 2 θ + a2 sin2 θ
1
=
p
2
2
P (a cos q, b sin q)
X¢
x = acos θ −
and
y = b sin θ +
2
2
X
O
2
L
b cos θ + a sin θ
ab
⇒
⇒
N
b cos θ + a sin θ
ab
2
1
=
=
p
∴ Locus of Q is xy = 4 − x2 .
⇒
x2 + xy − 4 = 0
S
1/p
Q
Now, the coordinates of the point Q are
given as follows,
x − acos θ
− asin θ
b 2 cos 2 θ + a2 sin2 θ
y − b sin θ
b cos θ
=
b 2 cos 2 θ + a2 sin2 θ
1 − h2
⇒ hk = 4
4
⇒
⇒
Y
⇒
Let Q = (h, k ). Then,
2sin2 θ 2(1 − cos 2 θ)
=
h = 2cos θ, k =
cos θ
cos θ
S' (–ae, 0)
O
b2
a2
b sin θ − 0
b sin φ − 0
b2
=− 2
⇒
×
a
a cos θ − 0 a cos φ − 0
Then, m1 m2 = −
25 Equation of the tangent at P is
22 Chord of contact of P (h, k ) to the ellipse
X¢
Q (acos φ, b sin φ) be ends of these two
diameters.
−2 ± 8
= −1 ± 2
2
Hence, e = 2 − 1 as 0 < e < 1 for ellipse.
Therefore, e =
2
2
asin θ
ab
b cos θ
ab
2
2
x + y = 1 + 1
a
b
a2b 2
is the required locus.
26 CP and CD are semi-conjugate diameters
x2
y2
+
= 1 and let
14
8
eccentric angle of P is φ, then eccentric
π
angle of D is + φ, therefore the
2
coordinates of P and D are (acos φ,b sin φ)
and
π
π
14 cos + φ , 8 sin + φ
2
2
of an ellipse
i.e.CP 2 + CD 2 = (a2 cos 2 φ + b 2 sin2 φ)
+ ( a2 sin2 φ + b 2 cos 2 φ)
= a2 + b 2 = 14 + 8 = 22
27 Let y = m1 x and y = m2 x be a pair of
conjugate diameters of an ellipse
y2
x2
+ 2 = 1 and let P (acos θ,b sin θ) and
2
a
b
Normal
Y¢
ON =
=
1
cos θ sin2 θ
+
a2
b2
ab
2
b 2 cos 2 θ + a2 sin2 θ
Equation of the normal at P is
axsecθ – by cosec θ = a2 − b 2 .
a2 − b 2
OL =
2
2
a sec θ + b 2 cos ec2 θ
=
(a2 − b 2 )sin θ cos θ
a2 sin2 θ + b 2 cos 2 θ
where, L is the foot of perpendicular
from O on the normal.
1
Area of ∆PON = × ON × OL
2
[QNP = OL ]
(a2 − b 2 ) ab tan θ
(a2 − b 2 ) ab
=
= 2
a2 tan2 θ + b 2
a tan θ + b 2 cot θ
which is minimum when
a2 tan θ + b 2 cot θ is maximum.
Thus, for area to be minimum, tanθ =
∴ cos θ = ±
a
a2 + b 2
, sinθ = ±
b
a2 + b 2
So, the required coordinates is
a2
b2
,±
±
⋅
2
2
2
2
a +b
a +b
29 If α,β, γ, δ are eccentric angles of four
co-normal points.
Then, α + β + γ + δ = (2n + 1)π
b
a
DAY
311
and Σsin(α + β ) = 0
Now, sin (α + β )
= sin [2nπ + π − ( y + δ )] = sin(γ + δ)
Similarly, sin(β + γ ) = sin(α + δ )
and sin(γ + α) = sin(β + δ)
∴2[sin(α + β ) + sin(β + γ ) + sin(γ + α)] = 0
30. Equation of the chord with eccentric
angles of the extremities as α and β is
x
α + β y
α + β
cos
+ sin
2 b
2
a
α −β
= cos
2
α+β
α −β
Let
= ∆ 1 and
= ∆2 ,
2
2
y
x
so that cos ∆ 1 + sin ∆ 1 = cos ∆ 2
a
b
As the triangle is right angled,
homogenising the equation of the curve,
we get
2
y 2 x cos ∆ 1
y sin ∆ 1
x2
+ 2 −
+
=0
2
a
b
acos ∆ 2 b cos ∆ 2
1
cos 2 ∆ 1
⇒ 2 − 2
a cos 2 ∆ 2
a
1
sin2 ∆ 1
+ 2 − 2
=0
b cos 2 ∆ 2
b
[as coefficient of x2 + coefficient of
y 2 = 0]
2
2
2
⇒ b (cos ∆ 2 − cos ∆ 1 )
+ a2 (cos 2 ∆ 2 − sin2 ∆ 1 ) = 0
2
2
⇒ cos ∆ 2 (a + b 2 ) − b 2 cos 2 ∆ 1
− a2 + a2 cos 2 ∆ 1 = 0
⇒ cos 2 ∆ 2 (a2 + b 2 ) = a2 (1 − e 2 cos 2 ∆ 1 )
α + β
1 − e 2 cos 2
2
a2 + b 2
⇒
=
α −β
a2
cos 2
2
31 Let C1 and C2 be the centres and
R1 and R2 be the radii of the two circles.
Let S 1 = 0 lie completely inside in the
circle S 2 = 0.
Let ‘C’ and ‘r’ be the centre and radius of
the variable centre.
Then,
CC2 = R 2 − r
and
C1 C = R1 + r
[constant]
∴ C1 C + C 2C = R1 + R2
So, the locus of C is an ellipse.
Therefore, Statement II is true.
Hence, Statement I is false
(two circles are intersecting).
2
2
32 The ellipse is x + y = 1.
9
4
So, auxiliary circle is x2 + y 2 = 9 and
(− 5, 0) and ( 5, 0) are foci.
Hence, Statement I is true, then
Statement II is also true.
33 Let PCP ′ and DCD ′ be the conjugate
diameters of an ellipse and let the
eccentric angle of P is φ, then coordinate
of P is (acos φ, b sin φ).
So, coordinate of D is (− asin φ, b cos φ).
Let S and S ′ be two foci of the ellipse.
Then, SP ⋅ S ′ P
= (a − ae cos φ) ⋅ (a + ae cos φ)
P (a cos φ, b sin φ)
D
S¢
C
S
D¢
⇒ b 2 = a2 (1 − e 2 )
= a2 − a2e 2 cos 2 φ
2
2
2 2
⇒ a − b = a e
= a2 − (a2 − b 2 )cos 2 φ
= a2 sin2 φ + b 2 cos 2 φ = CD 2
P¢
34 Let (t ,b − t ) be a point on the line
x + y = b, then equation of chord whose
mid-point is (t ,b − t ), is
(b − t )y
tx
+
−1
2a2
2b 2
(b − t )2
t2
= 2 +
− 1 …(i)
2a
2b 2
Since, point (a, − b ) lies on Eq. (i), then
b(b − t )
(b − t )2
ta
t2
−
= 2 +
2
2
2a
2b
2a
2b 2
2
2
2
⇒ t (a + b ) − ab (3 a + b )t + 2a2b 2 = 0
Since, t is real.
∴
B 2 − 4 AC ≥ 0
2 2
⇒ a b (3 a + b ) 2 − 4(a2 + b 2 ) 2a2b 2 ≥ 0
⇒
9 a2 + 6 ab + b 2 − 8 a2 − 8b 2 ≥ 0
∴
a2 + 6 ab − 7b 2 ≥ 0
35. Statement I Given, a parabola
y 2 = 16 3 x and an ellipse 2 x2 + y 2 = 4.
To find the equation of common tangent
to the given parabola and the ellipse.
This can be very easily done by
comparing the standard equation of
tangents. Standard equation of tangent
with slope ‘m’ to the parabola
y 2 = 16 3 x is
4 3
…(i)
m
Standard equation of tangent with slope
x2
y2
‘m’ to the ellipse
+
= 1 is
2
4
…(ii)
y = mx ± 2m2 + 4
y = mx +
If a line L is a common tangent to both
parabola and ellipse, then L should be
tangent to parabola i.e. its equation
should be like Eq. (i) and L should be
tangent to ellipse i.e. its equation should
be like Eq. (ii) i.e. L must be like both of
the Eqs. (i) and (ii).
Hence, comparing Eqs. (i) and (ii), we get
4 3
= ± 2m2 + 4
m
On squaring both sides, we get
m2 (2m2 + 4) = 48
4
⇒
m + 2m2 − 24 = 0
⇒
(m2 + 6)(m2 − 4) = 0
⇒
m2 = 4 [Qm2 ≠ −6]
⇒
m=±2
On substituting m = ± 2 in Eq. (i), we get
the required equation of the common
tangent as
y = 2 x + 2 3 and y = −2 x − 2 3
Hence, Statement I is correct.
Statement II We have already seen
4 3
that, if the line y = mx +
is a
m
common tangent to the parabola
x2
y2
+
= 1,
y 2 = 16 3 x and the ellipse
2
4
then it satisfies the equation
m 4 + 2m2 − 24 = 0.
Hence, Statement II is also correct but is
not able to explain the Statement I. It is
an intermediate step in the final answer.
SESSION 2
1 Let equation of ellipse be
x2 y 2
+
=1
a2 b 2
Equation of directrix is x = ±
a
e
a
1
= 4 ⇒ a = 4e ⇒ a= 4 × = 2 Qe =
e
2
b2
2
2
e = 1 − 2 ⇒b = 3
a
x2 y 2
∴ Equation of ellipse = +
=1
4
3
Equation of normal
a2 x b 2 y
4x 3y
−
= a2 − b 2 ,
−
=1
x1
y1
1
3
2
∴
1
2
3
Q( x1 , y 1 ) = 1,
2
⇒
4x −2y = 1
2 Let P (h, k ) be the mid-point of the focal
chord. Then its equation is
hx ky
h2
k2
...(i)
+ 2 = 2 + 2
2
a
b
a
b
Eq. (i) is focal chord so it passes
through (ae , 0)
h ea h2 k 2
∴
= 2 + 2
a2
a
b
y 2 ex
x2
∴ Locus of P is 2 + 2 =
a
b
a
3 Let foot of normal be P ( x1 , y 1 ). Then
a2 y
slope of normal = 2 1 = 2 ⇒ 9 y 1 = 8 x1
b x1
P lies on the ellipse 4 x2 + 9 y 2 = 36
312
64 x12
81
2
⇒ 4 x12 + 9
= 36 ⇒ x1 =
25
81
9
8
On taking x1 = , we get y 1 =
5
5
9 8
∴One possible foot of normal is ,
5 5
4 Slope of tangent m = 8
9
Let P ( x1 , y 1 ) be the point of contact.
Then, equation of tangent is
4 xx1 + 9 yy 1 = 1
4 x1 8
∴
−
= ⇒ x1 = − 2 y 1
9 y1 9
4 x + 9 y = 1 ⇒ 16 y + 9 y = 1
1
y1 = ±
5
∴Required points of contact are
− 2 , 1 and 2 , − 1
5 5
5 5
2
1
2
1
2
1
2
2
7 x2 + 4 y 2 = 4 ⇒ x + y = 1
Y
4
1
⇒ a = 2, b = 1 ⇒ P = (2, 1)
y2
x2
Required Ellipse is 2 + 2 = 1
a
b
x2
y2
+
=
1
(
2
,
1
) lies on it
⇒
42
b2
(–3,2)
P(2,1)
(–3,–2)
1
V′
A′
2
4
1
+
=1
16 b 2
1
1 3
4
= 1 − = ⇒ b2 =
⇒
b2
4 4
3
x2
y2
∴
+
=1
16 4
3
⇒
5 Normal is 4 x sec φ − 2 y cosec φ = 12
Q ≡ (3cos φ, 0), M = (α,β)
3 cos φ + 4 cos φ 7
α=
= cos φ
2
2
⇒
P (4 cos φ, 2 sin φ)
x2 + 12 y 2 = 16
2
2
8 x + y = 1 and b2 = a2 (1 − e 2 )
4
1
M
R
Q(x2, y2)
2
⇒ cos φ = α and β = sin φ
7
cos 2 φ + sin2 φ = 1
4 2
4 2
α + β2 = 1 ⇒
x + y2 = 1
⇒
49
49
⇒ Latusrectum x = ± 2 3
6 Equation of auxiliary circle is
x2 + y 2 = 9
x y
Equation of AM is + = 1
3 1
...(i)
...(ii)
X
O
(3,0)
(3,–2)
9
4
9 1
+ 2 = 1, 2 + = 1 ⇒ a2 = 12
a2
b
a
4
12
2
2
2
= 1 − e2
a = b (1 − e ) ⇒
16
12 14 1
1
⇒ e2 = 1 −
=
= ⇒e =
16 16 4
2
2
2
10 Ellipse is x + y = 1
4
1
Let A(θ) = (2 cos θ, sin θ)
π
Then, B = 2 cos θ + , sin
3
θ + π
3
Equation of chord AB is
x
π
y
π
cos θ + +
sin θ +
2
6
1
6
π
3
=
6
2
or given chord is px + qy = r
Comparing the coefficient, we get
π
π
cos θ + sin θ +
3
6
6
=
=
2p
q
2r
e =
less than 0)
Co-ordinates of mid-point of PQ are
1
R ≡ 0, −
2
PQ = 2 3 = length of latusrectum.
⇒ Two parabolas are possible whose
3 1
vertices are 0, −
− and
2 2
⇒
⇒
3 p2 3q 2
+ 2 =1
r2
4r
3(4 p2 + q 2 ) = 4r 2
3
r 2 = (4 p2 + q 2 )
4
11 Let POP ′, QOQ ′ be two perpendicular
diameters.
Let ∠AOP = θ, then ∠AOQ =
Q
B(0,1)
A(3,0)
x2 + 2 3 y = 3 − 3
9 Let required ellipse is
On solving Eqs. (i) and (ii), we get
− 12 9
M
,
5 5
Now, area of ∆AOM
1
27
sq unit.
= ⋅ OA ⋅ MN =
2
10
θ
y2
x2
E2 : 2 + 2 = 1
a
b
It passes through (0, 4)
16
0 + 2 = 1 ⇒ b 2 = 16
b
It also passes through (± 3, ± 2)
A
O
Hence, the equations of the parabolas
are x2 − 2 3 y = 3 + 3 and
P′
π
+θ
2
P
3 1
−
0,
2 2
M
O
P(x1, y1)
3
1
⇒ P 3, −
2
2
1
and Q = − 3, − (given y 1 and y 2
2
⇒
48
1
1
+ y 2 = 1 ⇒ y = ± ± 2 3, ±
49
7
7
N
(3, 2)
= cos
Q (3 cos φ, 0)
12 9
– ,
5 5
2
V
A 2 (4, 0)
2
1
⇒
4
Q′
Then, P = (OP cos θ, OP sin θ)
OP 2 cos 2 θ OP 2 sin2 θ
+
=1
∴
a2
b2
2
2
cos θ sin θ
1
...(i)
⇒
=
+
OP 2
a2
b2
Also
Q = (OQ cos(90° + θ),
OQ sin(90° + θ))
DAY
313
sin2 θ cos 2 θ
1
= 2 +
OQ 2
a
b2
1
1
1
1
+
= 2 + 2
⇒
OP 2 OQ 2
a
b
1
1
1 1
1
∴
+
= 2 + 2
(POP ′ )2 (QOQ ′ )2
4 a
b
[QPOP ′ = 2OP , QOQ ′ = 2OQ ]
∴
12 Let P(θ1 ), Q(θ2 ), R(θ3 ) be the vertices of
∆PQR, inscribed in the ellipse
y2
x2
+ 2 =1
2
a
b
Then, P ′, Q ′, R ′, corresponding points on
auxiliary circle x2 + y 2 = a2 are
P ′(a cos θ1 , a sin θ1 ),
Q ′(acos θ2 , a sin θ2 ) and
R ′(acos θ3 , asin θ3 )
∆ 1 = Area of ∆PQR
a cos θ1 b sin θ1 1
1
=
a cos θ2 b sin θ2 1
2
a cos θ3 b sin θ3 1
cos θ1
1
= ab cos θ2
2
cos θ3
sin θ1
sin θ2
sin θ3
∆ 2 = area of P ′ Q ′ R ′
cos θ1 sin θ1
1
= a2 cos θ2 sin θ2
2
cos θ3 sin θ3
∴
1
1
1
Equation of the line S ′ R ′ is
1
y = − ( x + ae )
m
⇒ x + my + ae = 0
⇒
⇒
⇒
⇒
∆1
b
= = Constant
∆2
a
2 − 3 cos 2 θ = 4 cos 2 θ
2
9 cos 4 θ − 40 cos 2 θ + 16 = 0
(9 cos 2 θ − 4) (cos 2 θ − 4) = 0
4
cos 2 θ = as cos 2 θ ≠ 4
9
y = mx +
13 Equation of chord P(α), Q(β) is
...(i)
Here, a = 5, b = 4. PQ is a focal chord.
a2 − b 2
9
3
e2 =
= ⇒e =
a2
25
5
One of the foci = (ae , 0) = (3, 0)
Eq. (i) passes through (3, 0)
am +b
2
2
2
− aem +
Y¢
a2 m2 + b 2
1 + m2
1
(S ′ R ′+ SR ) × SD
2
Then,
∆1 =
⇒
a2 m2 + b 2
∆ 1 = 2ae
2
1+ m
Area of A ′ B ′ BA is
∆2 =
1
( A ′ B ′+ AB ) × AE
2
Equation of A ′ B ′ is
y =−
1
( x + a)
m
x + my + a = 0
Therefore,
and
AB =
…(i)
∴ S ′ R ′RS is a trapezium and its area
1
∆ 1 = (SR + S ′ R ′ ) × SD
2
Y
B¢ R¢
P R
B
E
D
X¢
X
O
A¢
S
S¢
A (a, 0)
(ae, 0)
a2 m2 + b 2
1 + m2
and S ′ R ′ =
⇒
15 Equation of tangent at P is
1 + m2
aem +
SR =
…(ii)
ae + ae
SD =
Therefore,
2
1
1
1
α +β y
α + β
x
cos
sin
+
2 b
2
a
α − β
= cos
2
α +β
3
α −β
cos
= cos
2
2
5
α
β
α
β
⇒ 5 cos cos + sin sin
2
2
2
2
α
β
α
β
= 3 cos cos − sin sin
2
2
2
2
α
β
α
β
⇒ 8sin sin = − 2 cos cos
2
2
2
2
α
β
1
⇒ tan tan = − .
2
2
4
2
2
14 Any point P(θ) on the ellipse x + y = 1
4
1
is P(2 cos θ, sin θ)
Equation of tangent at P(θ) is
2 x cos θ + 4 y sin θ = 4
∴
A = (2 sec θ, 0)
Equation of normal of P(θ) is
2x
y
−
= 4−1=3
cos θ sin θ
3
∴
B = cos θ, 0
2
3
AB = 2 ⇒ 2 sec θ − cos θ = 2
2
∴
and A ′ B ′ =
…(iii)
a+ a
AE =
1 + m2
ma +
a2 m2 + b 2
1 + m2
− ma +
a2 m2 + b 2
1 + m2
1
( A ′ B ′+ AB ) × AE
2
a2 m2 + b 2
∆ 2 = 2a
2
1+ m
Then, ∆ 2 =
⇒
Hence,
∆1 : ∆2 = e : 1
DAY TWENTY NINE
Hyperbola
Learning & Revision for the Day
u
u
Concept of Hyperbola
Equations of Hyperbola
in Standard Form
u
Tangent to a Hyperbola
u
u
Normal to a Hyperbola
u
u
Pole and Polar
u
Director Circle
Asymptotes
Rectangular or Equilateral Hyperbola
Concept of Hyperbola
Hyperbola is the locus of a point in a plane which moves
in such a way that the ratio of its distance from a fixed point
(focus) in the same plane to its distance from a fixed line
(directrix) is always constant which is always greater
than unity.
M
N
T
P
S
SP
Mathematically,
= e, where e > 1.
PN
M¢
Terms Related to Hyperbola
Some important terms related to hyperbola are given below.
1. Vertices The points A and A′, where the curve meets the line joining the foci
S and S ′, are called the vertices of the hyperbola.
2. Transverse and conjugate axes Transverse axis is the one which lie along the line
passing through the foci and perpendicular to the directrices and conjugate axis and
conjugate axis is the one which is perpendicular to the transverse axis and passes
through the mid-point of the foci i.e. centre.
3. Centre The mid-point C of AA′ bisects every chord of the hyperbola passing through
it and is called the centre of the hyperbola.
4. Focal chord A chord of a hyperbola which is passing through the focus is called a
focal chord of the hyperbola.
5. Directrix A line which is perpendicular to the axis and it lies between centre and
a
vertex. The equation of directrix is x = ± .
e
6. Double ordinates If Q be a point on the hyperbola draw QN perpendicular to the axis
of the hyperbola and produced to meet the curve again at Q′. Then, QQ′ is called a
double ordinate of Q.
PRED
MIRROR
Your Personal Preparation Indicator
u
No. of Questions in Exercises (x)—
u
No. of Questions Attempted (y)—
u
No. of Correct Questions (z)—
(Without referring Explanations)
u
Accuracy Level (z / y × 100)—
u
Prep Level (z / x × 100)—
In order to expect good rank in JEE,
your Accuracy Level should be
above 85 & Prep Level should be
above 75.
315
DAY
7. Latusrectum The double ordinate passing through focus
is called latusrectum.
Y
Directrix
M'
S' (–ae, 0) A'
(– a, 0)
Latusrectum
Centre
X¢
Directrix
Latusrectum
M
P
Q
A S(ae, 0)
(a, 0)
C
(viii) Length of conjugate axis, 2b
(ix) Equation of transverse axis, y = 0
(x) Equation of conjugate axis, x = 0
X
N
Double
ordinate
(xi) Focal distances of a point on the hyperbola
is ex ± a.
(xii) Difference of the focal distances of a point on the
hyperbola is 2a.
Q¢
x = –a/e
x=a/e
Y¢
NOTE
(vii) Length of transverse axis, 2a
• The vertex divides the join of focus and the point of
intersection of directrix with axis internally and externally in
the ratio e : 1.
x2 y 2
• Domain and range of a hyperbola 2 − 2 = 1 are x ≤ − a or
a
b
x ≥ a and y ∈ R, respectively.
• The line through the foci of the hyperbola is called its
transverse axis.
• The line through the centre and perpendicular to the
transverse axis of the hyperbola is called its conjugate axis.
x 2 y2
2. Conjugate Hyperbola − 2 + 2 = 1
a
b
The hyperbola whose transverse and conjugate axes are
respectively the conjugate and transverse axes of a given
hyperbola is called the conjugate hyperbola of the given
hyperbola. The conjugate hyperbola of the hyperbola
x 2 y2
x 2 y2
− 2 = 1 is − 2 + 2 = 1.
2
a
b
a
b
Y
l
X¢
x 2 y2
1. Hyperbola of the Form 2 − 2 = 1
a b
When the hyperbola is in the given form, then it is also called
the equation of auxiliary circle.
Y
S1
L
A1
A
S
O
X
X
O
If the centre of the hyperbola is at the origin and foci are on
the X-axis or Y-axis, then that types of equation are called
standard equation of an ellipse.
X¢
L
A
Equations of Hyperbola in
Standard Form
L¢
S
L1
l¢
A1
L¢1
S1
L¢
Y¢
(i) Centre, O(0, 0)
(ii) Foci, S (0, be), S1 (0, − be)
(iii) Vertices, A (0, b ), A1 (0, − b )
b
b
(iv) Directrices l : y = , l ′ : y = −
e
e
(v) Length of latusrectum LL1 = L′ L1 ′ =
2 a2
b
2
L¢1
l¢
l
L1
Y¢
(vii) Length of transverse axis, 2b
(i) Centre, O(0, 0)
(viii) Length of conjugate axis, 2a
(ii) Foci : S (ae, 0), S1 (− ae, 0)
(ix) Equation of transverse axis, x = 0
(iii) Vertices : A(a, 0), A1 (− a, 0)
a
a
(iv) Directrices l : x = , l ′ : x = −
e
e
(v) Length of latusrectum, LL 1 = L′ L′1 =
b
(vi) Eccentricity, e = 1 +
a
a
(vi) Eccentricity, e = 1 + as a2 = b 2 (e2 − 1)
b
(x) Equation of conjugate axis, y = 0
(xi) Focal distances of a point on the hyperbola is ey ± b .
2 b2
a
2
or b 2 = a2 (e2 − 1)
(xii) Difference of the focal distances of a point on the
hyperbola is 2b.
NOTE If the centre of the hyperbola is ( h, k ) and axes are parallel to
the coordinate axes, then its equation is
( x − h) 2 ( y − k ) 2
−
= 1.
a2
b2
316
Results on Hyperbola
x2 y2
−
=1
a2 b2
or T = 0.
1. The foci of a hyperbola and its conjugate are concyclic
and form the vertices of a square.
2. The points of intersection of the directrix with
the transverse axis are known as foot of the directrix.
3. Latusrectum (l ) = 2e (distance between the focus and the
foot of the corresponding directrix).
4. The parametric equation of a hyperbola is x = a sec θ and
y = b tan θ, where θ ∈(0, 2 π ).
5. The position of a point (h, k ) with respect to the hyperbola S
lie inside, on or outside the hyperbola, if
x2
y2
S1 > 0, S1 = 0 or S1 < 0 where, S1 ≡ 12 − 12 − 1
a
b
A line which intersects the hyperbola at only one point is
called the tangent to the hyperbola.
(ii) In slope ‘ m’ form, y = mx ±
(iii) In parametric form,
a m −b
2
2
x
y
sec θ − tan θ = 1 at (a sec θ, b tan θ).
a
b
(iv) The line y = mx + c touches the hyperbola, iff
c2 = a 2 m 2 − b 2 and the point of contact is
1. Two tangents can be drawn from a point to a hyperbola.
2. The point of intersection of tangents at t 1 and t 2 to the curve
2 c t1 t2
2c
xy = c2 is
,
.
t1 + t2 t1 + t2
3. The tangent at the point P(a sec θ1 , b tan θ1 )
Q (a sec θ2 , b tan θ2 ) intersect at the point
and
θ + θ2
θ1 − θ2
b sin 1
a cos
2
2
.
R
,
θ1 + θ2
θ + θ2
cos 1
cos
2
2
4. The equation of pair of tangents drawn from an external
point P( x1 , y1 ) to the hyperbola is SS1 = T 2
and
2
1
2
2
1
2
x
y
x
y
− 2 − 1, S1 =
−
2
a
b
a
b
xx1
yy1
T = 2 − 2 −1
a
b
where S =
A line which is perpendicular to the tangent of the hyperbola
is called the normal to the hyperbola.
a2 x b 2 y
+
= a2 + b 2 .
x1
y1
m(a2 + b 2 )
(ii) In slope ‘m’ form y = mx m
a2 − m2b 2
a2
a −b m
2
2
2
,m
and the point of
a −b m
b2 m
2
2
2
(iii) In parametric form, ax cos θ + by cot θ = a2 + b 2 at
(a sec θ, b tan θ)
Results on Normals
2. Four normals can be drawn from any point to a
hyperbola.
Results on Tangent
2
Normal to a Hyperbola
1. If the straight line lx + my + n = 0 is a normal to the
x2 y2
a2 b2
(a 2 + b 2 )2
hyperbola 2 − 2 = 1, then 2 − 2 =
a
b
l
m
n2
a2 m
b2
, ± , where c = a2 m2 − b 2 .
±
c
c
2
7. Equation of chord joining the points (a sec α , b tan α ) and
(a sec β, b tan β) is
x
α − β y
α + β
α + β
cos
− sin
= cos
.
2 b
2
2
a
intersection is ±
xx1
yy
− 21 = 1
a2
b
2
6. The equation of chord of the hyperbola, whose
xx
yy
x2
y2
mid-point is ( x1 , y1 ) is T = S1 i.e. 21 − 21 = 12 − 12 .
a
b
a
b
(i) In point ( x1 , y1 ) form
Tangent to a Hyperbola
(i) In point ( x1 , y1 ) form,
xx1
yy
− 21 = 1
2
a
b
5. The equation of chord of contact is
−1
3. The line y = mx + c will be a normal to the hyperbola
x 2 y2
m2 (a2 + b 2 )2
.
− 2 = 1, if c2 = 2
2
a
b
a − b 2 m2
4. If the normals at four points P ( x1 , y1 ), Q ( x2 , y2 ), R ( x3 , y3 )
x2
y2
and S ( x 4 , y4 ) on the hyperbola
− 2 = 1 are
2
a
b
concurrent, then
1
1
1
1
( x1 + x2 + x3 + x 4 )
+
+
+
= 4.
x1
x2
x3
x4
Pole and Polar
Let P be a point inside or outside a hyperbola. Then, the
locus of the point of intersection of two tangents to the
hyperbola at the point where secants drawn through P
intersect the hyperbola, is called the polar of point P with
respect to the hyperbola and the point P is called the pole of
the polar.
317
DAY
(i) Polars of two points P( x1 y1 ) and Q( x2 , y2 ) of hyperbola
x 2 y2
xx
yy
xx
yy
−
= 1 are 21 − 21 = 1 and 22 − 22 = 1
a2 b 2
a
b
a
b
(ii) If the polar of P( x1 , y1 ) passes through Q( x2 , y2 ), then
x1 x2 y1 y2
− 2 = 1 and the polar of Q( x2 , y2 ) passes through
a2
b
x x
y y
P( x1 , y1 ), then 1 2 2 − 1 2 2 = 1
a
b
Conjugate Points and Conjugate Lines
(i) Two points are said to be conjugate points with respect to
a hyperbola, if each lies on the polar of the other.
(ii) Two lines are said to be conjugate lines with respect to a
x 2 y2
hyperbola 2 − 2 = 1, if each passes through the pole of
a
b
the other.
(iii) Two lines l1 x + m1 y + n1 = 0 and l2 x + m2 y + n2 = 0 are
x 2 y2
conjugate lines with respect to the hyperbola 2 − 2 = 1,
a
b
if a2 l1 l2 − b 2 m1 m2 = n1 n2 .
Director Circle
Y
The locus of the point of
intersection of the tangents
P (h, k)
to the hyperbola
90°
2
2
x
y
− 2 = 1, which are
X¢
X
2
C
a
b
perpendicular to each
other, is called a director
circle.
Y¢
The equation of director circle is x2 + y2 = a2 − b 2 .
The circle x2 + y2 = a2 is known the auxiliary circle of both
hyperbola.
Asymptotes
An asymptote to a curve is a straight line, which touches on it
two points at infinity but which itself does not lie entirely at
infinity.
Results on Asymptotes
1. A hyperbola and its conjugate hyperbola have the same
asymptotes.
2. The angle between the asymptotes of hyperbola
x 2 y2
b
−
= 1 is 2 tan −1 .
a
a2 b 2
3. Asymptotes always passes through the centre of the
hyperbola.
4. The bisectors of the angle between the asymptotes are the
coordinate axes.
5. The asymptotes of a hyperbola are the diagonals of the
rectangle formed by the lines drawn through the
extrimities of each axis parallel to the other axis.
Rectangular or Equilateral
Hyperbola
A hyperbola whose asymptotes include a right angle is said to
be rectangular hyperbola or if the lengths of transverse and
conjugate axes of any hyperbola is equal, then it is said to be a
rectangular hyperbola.
Rectangular Hyperbola of the
Form x 2 − y 2 = a2
1. Asymptotes are perpendicular lines i.e. x ± y = 0
2. Eccentricity, e = 2,
Y
L¢
3. Centre, O (0, 0)
4. Foci, S and S1 (± 2 a, 0)
a
5. Directrices, x = ±
2
X¢
L
A1
A
S1
6. Latusrectum = 2a
L¢1
7. Point form, x = x1 , y = y1
S
O
l¢
l
X
L1
Y¢
Equation of tangent, xx1 − yy1 = a2
x
y
Equation of normal, 1 + 1 = 2
x
y
8. Parametric form, x = a sec θ, y = a tan θ
Equation of tangent, x sec θ − y tan θ = a
x
y
Equation of normal,
+
= 2a
sec θ tan θ
Rectangular Hyperbola of
the Form xy = c 2
1. Asymptotes are perpendicular lines i.e. x = 0 and y = 0
2. Eccentricity, e = 2
Y
3. Centre, (0, 0)
4. Foci
S ( 2c, 2c), S1 (− 2c, − 2c)
5. Vertices, V1 (c, c), V2 (− c, − c)
V1
X¢
X
6. Directrices, x + y = ± 2 c
7. Latusrectum = 2 2 c
S
V2
S1
8. Point form, x = x1 , y = y1
Equation of tangent, xy1 + yx1 = 2c2
x
y
⇒
+
=2
x1 y1
Equation of Normal, xx1 − yy1 = x12 − y12
c
9. Parametric form : x = ct , y =
t
Equation of tangent, x + yt 2 = 2 ct
1
Equation of normal, t 2 x − y = c t 3 −
t
Y¢
318
DAY PRACTICE SESSION 1
FOUNDATION QUESTIONS EXERCISE
1 If the vertices of hyperbola are ( 0 , ± 6) and its
eccentricity is
5
, then
3
j
10 Length of the latusrectum of the hyperbola
xy − 3x − 4y + 8 = 0 is
NCERT Exemplar
y2 x2
−
= 1.
36 64
II. The foci of hyperbola are ( 0, ± 10).
(a) Both I and II are true
(c) Only II is true
(b) Only I is true
(d) Both I and II are false
3
latusrectum is 8 and eccentricity is
, is
5
(b) 4 x 2 − 5 y 2 = 100
(d) − 5 x 2 + 4 y 2 = 100
x2 y2
−
= 1 which
a 2 b2
passes through the points ( 3, 0) and ( 3 2, 2), is
3 The eccentricity of the hyperbola
j
(a)
13
3
13
3
(b)
4 Eccentricity of hyperbola
(a)
1+ k
(b) 1− k
(c)
13
9
NCERT Exemplar
13
(d)
3
1+
1
k
(d) 1 −
1
k
5 The equation of the hyperbola whose foci are ( − 2 , 0) and
( 2 , 0) and eccentricity is 2, is given by
(a) − 3 x 2 + y 2 = 3
(c) 3 x 2 − y 2 = 3
j
AIEEE 2011
(b) x 2 − 3 y 2 = 3
(d) − x 2 + 3 y 2 = 3
(b)
16 2
3
(c)
3
32
(d)
64
3
7 The difference between the length 2a of the transverse
axis of a hyperbola of eccentricity e and the length of its
latusrectum is
(a) 2a (3 − e 2 )
(c) 2a (e 2 − 1)
(b) 2a | 2 − e 2 |
(d) a (2e 2 − 1)
x2 y2
8 If eccentricity of hyperbola 2 − 2 = 1 is e and e′ is the
a
b
eccentricity of its conjugate hyperbola, then
(a) e = e ′
1
1
(c) 2 +
e
(e ′)2
(a)
4
3
4
3
(b)
2
3
(c)
12 A general point on the hyperbola
(d) None of these
x2 y2
−
= 1 is
a 2 b2
(a) (a sin θ,b cos θ) (where, θ is parameter)
(b) (a tan θ,b sec θ) (where, θ is parameter)
et + e −t et − e −t
(c) a
,b
(where, t is parameter)
2
2
(d) None of the above
x2
y2
−
= 1. Which of the
cos 2 α sin2 α
following remains constant when α varies?
13 For the hyperbola
(b) Abscissae of vertices
(d) Eccentricities
14 The product of the perpendicular from two foci on any
tangent to the hyperbola
(a) a 2
x2 y2
−
= 1, is
a 2 b2
(c) −a
(b) b 2
2
(d) −b
2
15 If a line 21x + 5y = 116 is tangent to the hyperbola
(a) (− 6, 3)
3x 2 − 4y 2 = 32 is
8 2
3
(d) None of these
7x 2 − 5y 2 = 232, then point of contact is
6 The length of transverse axis of the hyperbola
(a)
(c) 8
(a) Directrix
(c) Abscissae of foci
x2 y2
+ 2 = 1 (k < 0) is
k
k
(c)
(b) 4 2
and conjugate axis is equal to half of the distance
j NCERT Exemplar
between the foci is
2 The equation of the hyperbola, the lenght of whose
(a) 5 x 2 − 4 y 2 = 100
(c) −4 x 2 + 5 y 2 = 100
(a) 4
11 The eccentricity of the hyperbola whose latusrectum is 8
I. The equation of hyperbola is
(b) ee′ = 1
(d) None of these
9 A hyperbola, having the transverse axis of length 2 sin θ,
is confocal with the ellipse 3x 2 + 4y 2 = 12. Then, its
equation is
(a) x 2 cosec2 θ − y 2 sec2 θ = 1 (b) x 2 sec2 θ − y 2 cosec2 θ = 1
(c) x 2 sin 2 θ − y 2 cos2 θ = 1 (d) x 2 cos 2 θ − y 2 sin2 θ = 1
(b) (6, − 2)
(c) (8, 2)
(d) None of these
x2 y2
16 If e1 is the eccentricity of the ellipse
+
= 1 and e 2 is
16 25
the eccentricity of the hyperbola passing through the foci
of the ellipse and e1 e 2 = 1, then equation of the hyperbola is
x2 y2
−
=1
9 16
2
2
x
y
(c)
−
=1
9 25
(a)
(b)
x2 y2
−
= −1
16 9
(d) None of these
17 If the line 2x + 6y = 2 touches the hyperbola
x 2 − 2y 2 = 4, then the point of contact is
(a) (−2, 6)
1 1
(b) (−5, 2 6) (c) ,
2 6
(d) (4, −
6)
18 If a hyperbola passes through the point P( 2, 3 ) and
has foci at ( ± 2, 0), then the tangent to this hyperbola at
j
JEE Mains 2017
P also passes through the point
(a) (3 2 , 2 3 ) (b) (2 2 , 3 3 ) (c) ( 3 , 2 )
(d) (−
2, − 3)
19 The common tangent to 9x − 4y = 36 and x + y = 3 is
2
2
2
2
(a) y − 2 3 x −
39 = 0
(b) y + 2 3 x +
(c) y − 2 3 x +
39 = 0
(d) None of the above
39 = 0
319
DAY
20 The locus of a point P(α , β) moving under the condition
that the line y = α x + β is a tangent to the hyperbola
x2 y2
−
= 1, is
j AIEEE 2005
a 2 b2
(a) a hyperbola (b) a parabola (c) a circle (d) an ellipse
21 The straight line x + y = 2p will touch the hyperbola
4x 2 − 9y 2 = 36, if
(a) p 2 = 2
(b) p 2 = 5
(c) 5 p 2 = 2
(d) 2 p 2 = 5
22 The locus of the point of intersection of perpendicular
tangents to the hyperbola
(a) x 2 + y 2 = 2
(c) x 2 − y 2 = 3
x2 y2
−
= 1 is
3
1
(b) x 2 + y 2 = 3
(d) x 2 + y 2 = 4
x2 y2
−
= 1 and N is foot
9
4
of the perpendicular from P on the transverse axis. The
tangent to the hyperbola at P meets the transverse axis
at T . If O is the centre of the hyperbola, then OT ⋅ ON is
equal to
23 If P is a point on the hyperbola
(a) 9
(b) 4
(c) e 2
(d) None of these
points of intersection of perpendicular
24 The locus of the
2
2
x
y
tangents to
−
= 1 is
16
9
(a) x 2 − y 2 = 7
(c) x 2 + y 2 = 25
(b) x 2 − y 2 = 25
(d) x 2 + y 2 = 7
25 Tangents are drawn from points on the hyperbola
x2 y2
−
= 1 to the circle x 2 + y 2 = 9. The locus of the
9
4
mid-point of the chord of contact is
x2
y2
x2
y2
(a) x + y =
(b) (x 2 + y 2 ) 2 =
−
−
9
4
9
4
x2 y2
x2
y2
2
2 2
2
2 2
(c) (x + y ) = 81 − (d) (x + y ) = 9
−
4
9 4
9
2
2
x2 y2
−
= 1 meets X -axis at
4
2
P and Y -axis at Q. Lines PR and QR are drawn such that
OPRQ is a rectangle (where, O is the origin). Then, R
j
lies on
JEE Mains 2013
26 A tangent to the hyperbola
4
2
+ 2 =1
x2
y
2
4
(c) 2 + 2 = 1
x
y
(a)
2
4
− 2 =1
x2
y
4
2
(d) 2 − 2 = 1
x
y
(b)
29 Let P (a sec θ, b tan θ ) and Q (a sec φ , b tan φ ), where
π
x2 y2
, be two points on 2 − 2 = 13. If (h, k ) is the
2
a
b
point of intersection of normals at P and Q, then k is
θ+φ=
a2 + b 2
a
a2 + b 2
(c)
b
30 If the chords of contact of tangents from two points
( x1, y1 ) and ( x 2 , y 2 ) to the hyperbola 4x 2 − 9y 2 − 36 = 0
x x
are at right angles, then 1 2 is equal to
y 1y 2
(a)
9
4
(b) −
3
− 1 sq unit
2
3
+ 1 sq units
2
28 Given the base BC of ∆ABC and if ∠ B − ∠C = K, a
constant, then locus of the vertex A is a hyperbola.
(a) No
(b) Yes
(c) Both
81
16
(d) None
(d) −
81
16
(a) 9x 2 − 8 y 2 + 18 x − 9 = 0 (b) 9x 2 − 8 y 2 − 18 x + 9 = 0
(c) 9x 2 − 8 y 2 − 18 x − 9 = 0 (d) 9x 2 − 8 y 2 + 18 x + 9 = 0
32 If chords of the hyperbola x 2 − y 2 = a 2 touch the
parabola y 2 = 4ax . Then, the locus of the middle points
of these chords is
(a) y 2 = (x − a)x 3
(c) x 2 (x − a) = x 3
(b) y 2 (x − a) = x 3
(d) None of these
33 The locus of middle points of chords of hyperbola
3x 2 − 2y 2 + 4x − 6y = 0 parallel to y = 2x is
(a) 3 x − 4 y = 4
(c) 4 x − 3 y = 3
(b) 3 y − 4 x + 4 = 0
(d) 3 x − 4 y = 2
34 The equation of the chord joining two points
( x1, y1 ) and ( x 2 , y 2 ) on the rectangular hyperbola xy = c 2
j
AIEEE 2002
is
x
+
x1 + x 2
y1
x
(c)
+
y1 + y 2
x1
(a)
y
=1
+ y2
y
=1
+ x2
x
y
+
=1
x1 − x 2
y1 − y 2
x
y
(d)
+
=1
y1 − y 2
x1 − x 2
(b)
35 Match the vertices (v ) and foci (f ) of hyperbola given in
Column I with their corresponding equation given in
Column II and choose the correct option from the codes
given below.
Column I
x − 2y − 2 2x − 4 2y − 6 = 0 with vertex at the point A.
Let B be one of the end points of its latusrectum. If C is
the focus of the hyperbola nearest to the point A , then
the area of the ∆ABC is
(b)
(d)
(c)
x 2 − y 2 = 9, then the equation of the corresponding pair
of tangents is
2
2
(a) 1 −
sq unit
3
2
(c) 1 +
sq units
3
9
4
31 If x = 9 is the chord of contact of the hyperbola
27 Consider a branch of the hyperbola
2
(a 2 + b 2 )
a
(a 2 + b 2 )
(d) −
b
(b) −
(a)
Column II
2
A.
v (± 2, 0), f (± 3, 0)
1.
y
x2
−
=1
25 39
B.
v (0, ± 5), f (0, ± 8)
2.
y2
x2
−
=1
9
16
C. v (0, ± 3), f (0 ± 5)
3.
x2
y2
−
=1
4
5
Codes
A
(a) 3
(c) 3
B
1
2
C
2
1
A
(b) 1
(d) 2
B
3
1
C
2
3
320
DAY PRACTICE SESSION 2
PROGRESSIVE QUESTIONS EXERCISE
1 If chords of the hyperbola x 2 − y 2 = a 2 touch the
parabola y 2 = 4ax . Then, the locus of their middle point is
(a) y (x − a) = 2 x
(c) y 2 (x − a) = x 4
2
(b) y (x − a) = x
(d) y 2 (x + a) = x 3
2
2
3
2 The equation of a tangent to the hyperbola 3x 2 − y 2 = 3
parallel to the line y = 2x + 4 is
(a) y = 3 x + 4
(c) y = 2 x − 2
(b) y = 2 x + 1
(d) y = 3 x + 5
π
x2 y2
be two points on the hyperbola 2 − 2 = 1. If
2
a
b
(h, k ) is the point of intersection of normals at P and Q,
then k is equal to
θ+φ =
a 2 + b 2
a 2 + b 2
a2 +b2
(b) −
(c)
(d) −
a
b
b
4 The equation of the asymptotes of the hyperbola
3x 2 + 4y 2 + 8xy − 8x − 4y − 6 = 0 is
(a) 3 x 2 + 4 y 2 + 8 xy − 8 x − 4 y − 3 = 0
(b) 3 x 2 + 4 y 2 + 8 xy − 8 x − 4 y + 3 = 0
(c) 3 x 2 + 4 y 2 + 8 xy − 8 x − 4 y + 6 = 0
(d) 4 x 2 + 3 y 2 + 2 xy − x + y + 3 = 0
(b)
π
3
(c)
π
4
(d)
π
6
x2 y2
−
=1
a 2 b2
such that OPQ is an equilateral triangle, O being the
centre of the hyperbola. Then, the eccentricity e of the
hyperbola satisfies
6 If PQ is a double ordinate of the hyperbola
(a) 1 < e <
2
3
2
3
2
(d) e >
3
(b) e =
3
(c) e =
2
intersects its transverse and conjugate axes at L and M
respectively. If locus of the mid-point of LM is hyperbola,
then eccentricity of the hyperbola is
(b)
e
(e 2 − 1)
(c) e
(d) None of these
8 Tangents are drawn to the hyperbola 4x 2 − y 2 = 36 at the
points P and Q. If these tangents intersect at the point
T (0, 3), then the area (in sq units) of ∆PTQ is
j
(a) 45 5
(c) 60 3
(b) 54 3
(d) 36 5
y 2 = 8x and xy = −1 is
(a) 3 y = 9x + 2
(c) 2 y = y + 8
(b) y = 2 x + 1
(d) y = x + 2
11 A series of hyperbolas is drawn having a common
transverse axis of length 2a. Then, the locus of a point P
on each hyperbola, such that its distance from the
transverse axis is equal to its distance from an
asymptote, is
(a) (y 2 − x 2 )2 = 4 x 2 (x 2 − a 2 ) (b) (x 2 − y 2 )2 = x 2 (x 2 − a 2 )
(c) (x 2 − y 2 ) = 4 x 2 (x 2 − a 2 ) (d) None of these
5+1
(a)
, 2
2
5 −1
(c)
, 1
2
(b) (1, 2]
5 −1
5 + 1
(d)
, 1 ∪ 1,
2
2
13 A rectangular hyperbola whose centre is C is cut by any
circle of radius r in four points P , Q , R and S. Then,
CP 2 + CQ 2 + CR 2 + CS 2 is equal to
(a) r 2
(c) 3 r 2
(b) 2 r 2
(d) 4 r 2
14 A point P moves such that sum of the slopes of the
7 The normal at P to a hyperbola of eccentricity e ,
e + 1
(a)
e − 1
(b) x 2 + y 2 = a 2
(d) None of these
a tangent to the ellipse x 2 + α 2 y 2 = α 2 and the portion of
the tangent intercepted by the hyperbola α 2 x 2 − y 2 = 1
subtends a right angle at the center of the curves is
( y − mx )(my + x ) = a 2 and
(m 2 − 1)( y 2 − x 2 ) + 4mxy = b 2 is
π
2
(a) x 2 + y 2 = b 2
(c) x 2 + y 2 = (a 2 − b 2 )
12 The exhaustive set of values of α 2 such that there exists
5 The angle between the rectangular hyperbolas
(a)
x2 y2
−
= 1 (a > b ) so that the fourth vertex
a 2 b2
S of parallelogram PQSR lies on circumcircle of ∆PQR,
then locus of P is
the hyperbola
10 The equation of the common tangent to the curves
3 Let P (a sec θ, b tan θ ) and Q (a sec φ, b tan φ ), where
a2 +b2
(a)
a
9 If tangents PQ and PR are drawn from variable point P to
JEE Mains 2018
normals drawn from it to the hyperbola xy = 4 is equal to
the sum of ordinates of feet of normals. Then, the locus of
P is
(a) a parabola
(c) an ellipse
(b) a hyperbola
(d) a circle
15 A triangle is inscribed in the rectangular hyperbola
xy = c 2 , such that two of its sides are parallel to the lines
y = m1x and y = m2 x . Then, the third side of the triangle
touches the hyperbola
c 2 (m1 + m2 )2
(a) xy =
4m1 m2
c 2 (m1 − m2 )2
(b) xy =
4m1 m2
c 2 (m1 − m2 )2
(c) xy =
m1 m2
c 2 (m1 + m2 )2
(d) xy =
m1 m2
321
DAY
ANSWERS
SESSION 1
SESSION 2
1. (a)
11. (c)
21. (d)
2. (b)
12. (c)
22. (a)
3. (d)
13. (c)
23. (a)
4. (d)
14. (b)
24. (d)
5. (c)
15. (b)
25. (c)
31. (b)
32. (b)
33. (a)
34. (a)
35. (a)
1. (b)
11. (a)
2. (b)
12. (a)
3. (d)
13. (d)
4. (a)
14. (a)
5. (a)
15. (a)
6. (a)
16. (b)
26. (d)
7. (b)
17. (d)
27. (b)
8. (c)
18. (b)
28. (b)
9. (a)
19. (a)
29. (d)
10. (b)
20. (a)
6. (d)
7. (b)
8. (a)
9. (c)
10. (d)
30. (d)
Hints and Explanations
SESSION 1
5
Y
E = 2a −
1 Since, the vertices are on the Y-axis
(with origin as the mid-point), the
y2
x2
equation is of the form 2 − 2 = 1
a
b
As vertices are (0, ± 6)
25
∴ a = 6, b 2 = a2 (e 2 − 1) = 36
− 1
9
= 64
So, the required equation of the
y2
x2
hyperbola is
−
= 1 and the foci
36 64
are (0, ± ae ) = (0, ± 10).
2 Let the equation of the hyperbola be
x2 y 2
…(i)
−
=1
a2 b 2
we have, length of the latusrectum = 8
2b 2
⇒
= 8 ⇒ b 2 = 4a
a
⇒ a2 (e 2 − 1) = 4a ⇒ a(e 2 − 1) = 4
9
⇒ a − 1 = 4 ⇒ a = 5
5
Putting a = 5 in b 2 = 4a, we getb 2 = 20
Hence, the equation of the required
x2 y 2
hyperbola is
−
= 1.
25 20
2
y2
3 Given that the hyperbola x2 − 2 = 1
a
b
is passing through the points (3, 0) and
(3 2, 2), so we get a2 = 9 and b 2 = 4.
Again, we know that b 2 = a2 (e 2 − 1).
This gives 4 = 9 (e 2 − 1)
13
13
⇒
e2 =
⇒e =
9
3
4 Given equation can be rewritten as
y2
x2
−
= 1(− k > 0)
2
k
(− k )
(− k )
k
e2 = 1 + 2 = 1 − 2
k
k
1
⇒
e = 1−
k
2b 2
2
= 2a2 − a2e 2 |
a
a
∴ Difference = 2a | 2 − e 2 |
8 Given equation of hyperbola is
X′
O
(–2, 0)
X
(2, 0)
Y′
Let equation of hyperbola be
y2
x2
− 2 =1
2
a
b
where, 2ae = 4 and e = 2
⇒
a=1
Q
a2e 2 = a2 + b 2 ⇒ 4 = 1 + b 2
∴
b2 = 3
Thus, equation of hyperbola is
x2
y2
−
= 1 or 3 x2 − y 2 = 3
1
3
6 The given equation may be written as
2
2
2
2
x
y
x
y
−
= 1⇒
−
=1
2
32
8
(2 2 )2
4 2
3
3
On comparing the given equation with
the standard equation
x2 y 2
−
= 1, we get
a2 b 2
4 2
a2 =
3
2
and b 2 = (2 2 )2
∴ Length of transverse axis of a
4 2 8 2
hyperbola = 2 a = 2 ×
=
3
3
7 Let equation of hyperbola be
y2
x2
− 2 =1
2
a
b
Length of transverse axis is 2a and
2b 2
Length of latusrectum is
.
a
Now, difference
x2 y 2
−
= 1 and equation of conjugate
a2 b 2
x2 y 2
hyperbola is 2 − 2 = 1
b
a
Since, e and e ′ are the eccentricities of
the respective hyperbola, then
b2
a2
e 2 = 1 + 2 ,(e ′ )2 = 1 + 2
a
b
1
1
a2
b2
+
=1
∴ 2 + 2 = 2
e
e′
a + b 2 a2 + b 2
9 Here, a = sinθ
Since, foci of the ellipse are (± 1, 0).
∴ ±1 =
a2 + b 2 ⇒ b 2 = cos 2 θ
Then, equation is
x2
y2
−
=1
2
sin θ cos 2 θ
⇒
x2 cos ec2θ − y 2 sec2 θ = 1
10 Given equation can be rewritten as
( x − 4) ( y − 3) = 4 which is a rectangular hyperbola of the type xy = c 2 .
∴
c=2
Then,
a= b = c 2 = 2 2
∴ Length of latusrectum
2b 2
=
= 2a = 4 2
a
11 Let equation of hyperbola be
x2 y 2
−
= 1.
a2 b 2
2
2b
b2
Given,
= 8⇒
=4
a
a
1
and 2b = (2ae ) ⇒ 2b = ae
2
b2
⇒ 4b 2 = a2e 2 ⇒ 4 2 = e 2
a
⇒ 4 (e 2 − 1) = e 2 [Qb 2 = a2 (e 2 − 1)]
4
2
⇒ 3e 2 = 4 ⇒ e 2 =
⇒ e =
3
3
322
12 Now, taking option (c).
13 The given equation of hyperbola is
x2
y2
−
=1
cos 2 α sin2 α
2
2
Here, a = cos α and b 2 = sin2 α
So, the coordinates of foci are (± ae , 0).
∴
e = 1+
y = −x + 2 p
17 As we know, equation of tangent at
2x
e t + e −t
Let x = a
⇒
= e t + e − t …(i)
2
a
2y
and
…(ii)
= e t − e −t
b
On squaring and subtracting Eq. (ii)
from Eq. (i), we get
4 x2 4 y 2
x2 y 2
− 2 = 4 ⇒ 2 − 2 =1
a2
b
a
b
b2
sin2 α
⇒ e = 1+
2
a
cos 2 α
⇒ e = 1 + tan2 α = sec α
Hence, abscissae of foci remain constant
when α varies.
14 Let equation of tangent to hyperbola
y2
x2
− 2 = 1 is y = mx + a2 m2 − b 2
2
a
b
i.e. mx − y + a2 m2 − b 2 = 0
∴ Required product
mae + a2 m2 − b 2
=
m2 + 1
( x1 , y 1 ) is xx1 − 2 yy 1 = 4, which is
same as 2 x + 6 y = 2
2y
x1
4
=− 1 =
⇒ x1 = 4
∴
2
2
6
and y 1 = − 6
18 Let the equation of hyperbola be
2
2
y
x
− 2 = 1.
a2
b
∴
ae = 2 ⇒ a2e 2 = 4
⇒
a2 + b 2 = 4 ⇒ b 2 = 4 − a2
y2
x2
∴
−
=1
a2
4 − a2
Since, ( 2, 3 ) lie on hyperbola.
2
3
∴
−
=1
a2
4 − a2
⇒
8 − 2a2 − 3a2 = a2 (4 − a2 )
⇒
8 − 5a2 = 4a2 − a4
⇒
a4 − 9a2 + 8 = 0
⇒
(a2 − 8)(a2 − 1) = 0
⇒
a2 = 8, a2 = 1
∴
a=1
Now, equation of hyperbola is
x2
y2
−
= 1.
1
3
∴ Equation of tangent at ( 2, 3 ) is
given by
− mae +
a2 m2 − b 2
m +1
2
a2 m2 − b 2 − m2 a2e 2
=
m2 + 1
m2 a2 (1 − e 2 ) − b 2 − m2b 2 − b 2
=
=
2
m2 + 1
m +1
[Qb 2 = a2 (e 2 − 1)]
2
=b
15. Here, a2 = 232 , b2 = 232
7
5
21
116
21
and y = −
with slope − .
x+
5
5
5
2
116
2
2
2
Now, a m − b =
5
[since, line is tangent]
If ( x1 , y 1 ) is the point of contact, then
tangent is S 1 = 0
∴ 7 x1 x − 5y 1 y = 232
On comparing it with
21 x + 5y = 116, we get x1 = 6, y 1 = − 2
So, the point of contact is (6, − 2).
16 Here, e 1 = 1 − 16 = 3
25 5
5
Q e1 e2 = 1 ⇒ e2 =
3
Since, foci of ellipse are (0, ± 3).
Hence, equation of hyperbola is
x2
y2
−
= − 1.
16
9
y
3y
= 1 ⇒ 2x −
=1
3
3
which passes through point (2 2, 3 3 ).
2x −
19 Suppose the common tangent is
y = mx + c to
9 x2 − 4 y 2 = 36 and x2 + y 2 = 3
…(i)
∴ c 2 = a2 m2 − b 2 = 4 m2 − 9
and c 2 = 3 + 3 m2
…(ii)
From Eqs. (i) and (ii), we get
4 m2 − 9 = 3 m2 + 3 ⇒ m2 = 12
⇒ m=2 3
∴
c = 3 + 3 × 12 = 39
Hence, the common tangent is
y = 2 3 x + 39
20 Since, the line y = αx + β is tangent to
2
2
y
x
− 2 = 1.
a2
b
∴ β2 = a2α2 − b 2. .
So, locus of (α, β ) is y 2 = a2 x2 − b 2
⇒ a2 x2 − y 2 − b 2 = 0
Since, this equation represents a
hyperbola, so locus of a point P(α, β ) is a
hyperbola.
the hyperbola
21 Given equation of hyperbola is
x2 y 2
−
= 1.
9
4
Here,
a2 = 9, b 2 = 4
and equation of line is
...(i)
If the line y = mx + c touches the
hyperbola
x2 y 2
−
= 1, then c 2 = a2 m2 − b 2 …(ii)
a2 b 2
From Eq. (i), we get m = −1,c = 2 p
On putting these values in Eq. (ii), we
get
( 2 p )2 = 9(1) − 4 ⇒ 2 p2 = 5
22 We know that the locus of the point of
intersection of perpendicular tangents
to the hyperbola
x2 y 2
−
= 1 is a circle x2 + y 2 = a2 − b 2
a2 b 2
Thus, locus of the point of intersection
of perpendicular tangents to the
hyperbola
x2 y 2
−
= 1 is a circle
3
1
x2 + y 2 = 3 − 1 ⇒ x2 + y 2 = 2
23 The point on the hyperbola is P ( x1 , y 1 ),
then N is ( x, 0).
xx1
yy 1
−
=1
9
4
9
This meets X-axis at T , 0
x1
9
∴ OT ⋅ ON =
⋅ x1 = 9
x1
∴ Tangent at ( x1 , y 1 ) is
24 The equation of tangent in slope form to
x2
y2
−
= 1 is
16
9
16 m2 − 9
the hyperbola
y = mx +
Since, it passes through (h, k ).
∴
k = mh +
16 m2 − 92
⇒ (k − mh )2 = (16 m2 − 92 )
⇒ k 2 + m2 h2 − 2m kh − 16 m2 + 9 2 = 0
It is quadratic in m and let the slope of
two tangents be m1 and m2 , then
k2 + 9
m1 m2 = 2
h − 16
⇒
−1=
k2 + 9
⇒ h2 + k 2 = 7
h2 − 16
The required locus is x2 + y 2 = 7
25 Any point on the hyperbola is
(3sec θ , 2 tan θ). The chord of contact to
the circle is 3 x sec θ + 2 y tan θ = 9 …(i)
If ( x1 , y 1 ) is the mid-point of the chord,
then its equation is
…(ii)
xx1 + yy 1 = x12 + y 12
From Eqs. (i) and (ii), we get
3sec θ 2 tan θ
9
=
= 2
x1
y1
x1 + y 12
1 2
x2 y 2
( x1 + y 12 ) 2 = 1 − 1
81
9
4
Hence, the locus is
x2
y2
( x2 + y 2 ) 2 = 81
−
4
9
Eliminating θ,
323
DAY
tan B − tan C
1 + tan B ⋅ tan C
y
y
−
λ+ x λ− x
=
y2
1+ 2
λ − x2
2
2
26 Given hyperbola is x − y = 1.
∴ tan K =
4
2
Here, a2 = 4 , b 2 = 2 ⇒ a = 2, b = 2
The equation of tangent is
y
x
sec θ − tan θ = 1
a
b
x
y
⇒
sec θ −
tan θ = 1
2
2
So, the coordinates of P and Q are
P (2cos θ, 0) and Q (0, − 2 cot θ),
⇒
λ2 − x2 + y 2 = − 2 xy cot K
⇒ x2 − 2 cot K ⋅ xy − y 2 = λ2
which is a hyperbola.
ax
by
θ is
+
= a2 + b 2 and normal
sec θ
tan θ
π
ax
by
at φ =
− θ is
+
= a2 + b2
2
cos ecθ cot θ
Eliminating x, we get
1
1
2
2
by
−
= (a + b )
sin θ cos θ
1 − 1
cos θ sin θ
(a2 + b 2 )
⇒ by = −(a2 + b 2 ) or k = −
b
30 The equation of hyperbola is
( x − 2 )2
( y + 2 )2
−
=1
4
2
For A( x, y ),
Y
2
9 x1 4 x2
x x
9
81
⋅
× ⋅
= − 1 ⇒ 1 2 = − = −
4
4 y1 9 y2
y1 y2
16
B
X¢
X
C
A
Y¢
2
3
=
4
2
x− 2=2⇒ x=2+
e = 1+
∴
For C ( x, y ), x − 2 = ae =
∴
x=
6+
2
6
2
Now, AC =
6 + 2−2− 2 = 6 −2
b2 2
and BC =
= =1
a 2
1
∴ Area of ∆ABC = × AC × BC
2
3
1
= × ( 6 − 2) × 1 =
− 1 sq unit
2
2
28 Let B (−λ , 0), C (λ , 0) and A ( x, y ).
Given, K = ∠B − ∠ C
Y
x2 y 2
...(i)
−
=1
9
4
The equation of the chords of contact of
tangents from ( x1 , y 1 ) and ( x2 , y 2 ) to the
given hyperbola are
x x1
y y1
...(ii)
−
=1
9
4
x x2 y y 2
...(iii)
and
−
=1
9
4
Lines (ii) and (iii) are at right angles.
4 x2 − 9 y 2 = 36 ⇒
27 The given equation can be rewritten as
31 Let (h, k ) be the point whose chord of
contact w.r.t. hyperbola x2 − y 2 = 9 is
x = 9. We know that chord of (h, k )
w.r.t. hyperbola x2 − y 2 = 9 is T = 0
⇒ hx − ky − 9 = 0
But it is the equation of line x = 9. This
is possible only whenh = 1, k = 0. Again,
equation of pair of tangents is T 2 = SS 1
⇒ ( x − 9)2 = ( x2 − y 2 − 9)(1 − 9)
⇒ x2 − 18 x + 81 = ( x2 − y 2 − 9)(−8)
⇒ 9 x2 − 8 y 2 − 18 x + 9 = 0
32 Equation of chord of hyperbola
x2 − y 2 = a2 with mid-point as (h, k ) is
given by
h
(h2 − k 2 )
xh − yk = h2 − k 2 ⇒ y = x −
k
k
This will touch the parabola y 2 = 4ax, if
h2 − k 2
a
⇒ ak 2 = − h3 + k 2 h
−
=
k h/k
A (x, y)
∴ Locus of the mid-point is
x 3 = y 2 ( x − a)
B
-l
O
l
C
X
33 Let (h, k ) is mid-point of chord.
Then, its equation is
⇒
3h − 4k = 4
Thus, locus of mid-point is 3 x − 4 y = 4
34 The mid-point of the chord is
29 Equation of normal at
respectively.
Let coordinates of R are (h, k).
∴
h = 2cos θ, k = − 2cotθ
k
− 2
− 2h
⇒
=
⇒ sin θ =
h sinθ
2k
On squaring both sides, we get
2h2
2h2
sin2 θ =
⇒ 1 − cos 2 θ =
4 k2
4 k2
2h2
h2
h2
2h2
⇒ 1−
+
=1
=
⇒
2
2
4k
4
4
4k
h2 2
2
4
⇒
+ 1 = 1 ⇒ 2 + 1 = 2
4 k2
k
h
4
2
−
=1
⇒
h2 k 2
4
2
Hence, R lies on 2 − 2 = 1.
x
y
3hx − 2 ky + 2( x + h ) − 3( y + k )
= 3h2 − 2 k 2 + 4h − 6k
⇒ x(3h + 2) + y (−2 k − 3)
= 3h2 − 2 k 2 + 2 h − 3k
Since, this line is parallel to y = 2 x.
3h + 2
= 2 ⇒ 3h + 2 = 4k + 6
∴
2k + 3
x1 + x2 , y 1 + y 2 .
2
2
The equation of the chord T = S 1 .
y + y2
x1 + x2
∴ x 1
+ y
2
2
x1 + x2 y 1 + y 2
=2
2
2
⇒
⇒
x ( y 1 + y 2 ) + y ( x1 + x2 )
= ( x1 + x2 )( y 1 + y 2 )
y
x
+
=1
x1 + x2
y1 + y2
35 A. Since, vertices (± 2, 0) and foci
(± 3, 0) lie on X -axis, as coefficient
of Y-axis is zero.
Hence, equation of hyperbola will
be of the form
y2
x2
…(i)
− 2 =1
2
a
b
Here, it is given that a = 2 and c = 3
Q
c 2 = a2 + b 2
⇒
9 = 4 + b2 ⇒ b2 = 5
Put the values of a2 = 4 and b 2 = 5 in
Eq. (i), we get
x2
y2
…(ii)
−
=1
4
5
B. Since, vertices (0, ± 5) and foci
(0, ± 8) lie on Y-axis as coefficient of
X-axis is zero. Hence,equation of
hyperbola will be of the form
y 2 x2
…(i)
−
=1
a2 b 2
Here, it is given that
(0, ± a) = (0, ± 5) and foci
(0, ± c ) = (0, ± 8)
⇒
a = 5and c = 8
Q
c 2 = a2 + b 2 ⇒ b 2 = 39
Put a2 = 25 and b 2 = 39 in Eq. (i), we
get
y2
x2
−
=1
25 39
C. Since, vertices (0, ± 3) and foci
(0, ± 5) lie on Y- axis as coordinate
of x is zero.
Hence, equation of hyperbola will
be of the form
y2
x2
…(i)
− 2 =1
2
a
b
324
Here, it is given that vertices
(0, ± 3) = (0, ± a) and foci
(0, ± 5) = (0, ± c )
⇒
a = 3 and c = 5
Q
c 2 = a2 + b 2
⇒
25 = 9 + b 2 ⇒ b 2 = 16
2
Put a = 9 and b 2 = 16 in Eq. (i), we
get
y2
x2
−
=1
9
16
SESSION 2
− 4 (− 4)2 − λ ( 4)2 = 0
⇒
12λ + 64 − 12 − 64 − 16λ = 0
⇒
− 4λ − 12 = 0 ⇒ λ = −3
∴ Required equation is
3 x2 + 4 y 2 + 8 xy − 8 x − 4 y − 3 = 0
5 Given equations are
…(i)
( y − mx )(my + x ) = a2
and (m2 − 1)( y 2 − x2 ) + 4 mxy = b 2 …(ii)
On differentiating Eq.(i), we get
dy
( y − mx ) m
+ 1
dx
1 If (h, k ) is the mid-point of the chord,
then its equation by T = S 1 is
hx − ky = h2 − k 2
k 2 − h2
h
⇒
y = x+
k
k
If it touches the parabola y 2 = 4ax, then
we get
k 2 − h2
a⋅ k
=
⇒ ak 2 = hk 2 − h3
k
h
⇒
ay 2 = xy 2 − x3
So, required locus is y 2 ( x − a) = x3 .
dy
+ (my + x )
− m = 0
dx
dy
(my + x + my − m2 x )
dx
⇒
+ y − mx − m2 y − mx = 0
dy
− y + m2 y + 2mx
⇒
=
= m1 [say]
dx
2my + x − m2 x
On differentiating Eq.(ii), we get
dy
(m2 − 1) 2 y
− 2 x
dx
2 Equation of the hyperbola is
x2
y2
−
=1
1
3
Equation of tangent in terms of slope
y = mx ± (m2 − 3 )
dy
+ 4 m x
+ y = 0
dx
3 x2 − y 2 = 3 ⇒
m=2
y = 2x ± 1
Given,
∴
⇒
dy
[2 y (m2 − 1) + 4mx]
dx
= − 4 my + 2 x (m2 − 1)
dy
−2my + m2 x − x
=
dx
m2 y − y + 2mx
⇒
3 Equation of the tangents at
y
x
sec θ − tan θ = 1
a
b
∴ Equation of the normal at P is
ax + b cosecθ y = (a2 + b 2 )sec θ ...(i)
Similarly, the equation of normal at
Q (asec φ,b tan φ) is
= m2
m1 m2 = − 1
P (asec θ,b tan θ) is
ax + b cosecφ y = (a + b )sec φ
2
2
...(ii)
On subtracting Eq. (ii) from Eq. (i), we
get
a2 + b 2
sec θ − sec φ
y =
⋅
b
cosecθ − cosec φ
a2 + b 2
So, k = y =
b
π
secθ − sec − θ
2
⋅
π
cosecθ − cosec − θ
2
=
Q
So, angle between the hyperbola =
2
2
MP = MQ = l .
Given that ∆OPQ is an equilateral, then
OP = OQ = PQ
⇒
(OP )2 = (OQ )2 = (PQ )2
2
a
a2
⇒ 2 (b 2 + l 2 ) + l 2 = 2 (b 2 + l 2 ) + l 2 = 4l 2
b
b
a2 2
2
(
b
+
l
)
=
3l 2
⇒
b2
a 2
2
Y
b Öb + l , l
P
2
X¢
O
4 The equation of the asymptotes of the
hyperbola
3 x2 + 4 y 2 + 8 xy − 8 x − 4 y − 6 = 0
is 3 x2 + 4 y 2 + 8 xy − 8 x − 4 y + λ = 0
It should represent a pair of straight
lines.
∴ abc + 2 fgh − af 2 − bg 2 − ch2 = 0
3 ⋅ 4 ⋅ λ + 2 ⋅ (−2) (−4) 4 − 3 (−2)2
π
⋅
2
6 Q PQ is the double ordinate. Let
a + b sec θ − cosecθ
a + b
⋅
=−
b
b
cosecθ − sec θ
2
[say]
Y¢
⇒
⇒
M
X
Q
a 2
2
b Öb + l , –l
a2
a2 = l 2 3 − 2
b
a2b 2
2
l =
>0
(3b 2 − a2 )
∴
⇒
⇒
3b 2 − a2 > 0
3b 2 > a2
3a2 (e 2 − 1) > a2
2
3
7 Equation of normal at P (asec φ,b tan φ) is
a xcos φ + by cot φ = a2 + b 2 .
Then, coordinates of L and M are
a2 + b 2
a2 + b 2
⋅ sec φ,0 and 0,
tan φ
b
a
⇒
e 2 > 4/3 ∴ e >
respectively.
Let mid-point of ML is Q (h, k ),
(a2 + b 2 )
then
h=
sec φ
2a
2 ah
...(i)
∴ sec φ = 2
(a + b2 )
2
2
(a + b )
and
k =
tan φ
2b
2bk
...(ii)
∴ tan φ = 2
(a + b2 )
From Eqs. (i) and (ii), we get
4a2 h2
4b 2 k 2
sec2 φ − tan2 φ = 2
−
(a + b 2 )2 (a2 + b 2 )2
Hence, required locus is
y2
x2
−
=1
2
2
2
2
2
a +b
a + b2
2a
2b
Let eccentricity of this curve is e 1 .
2
2
a2 + b 2
a2 + b 2 2
⇒
=
(e 1 − 1)
2b
2a
⇒ a2 = b 2 (e 12 − 1) ⇒a2 = a2 (e 2 − 1)(e 12 − 1)
[Qb 2 = a2 (e 2 − 1 )]
2 2
2
2
⇒
e e1 − e − e1 + 1 = 1
e
⇒
e 12 (e 2 − 1) = e 2 ⇒ e 1 =
(e 2 − 1)
8 Tangents are drawn to the hyperbola
4 x2 − y 2 = 36 at the point P and Q.
Tangent intersects at point T (0, 3)
Y
T (0, 3)
O
(–3Ö5, –12)Q
S(0, –12)
X
P(3Ö5, –12)
Clearly, PQ is chord of contact.
∴Equation of PQ is − 3 y = 36
⇒ y = − 12
Solving the curve 4 x2 − y 2 = 36 and
y = − 12, we get x = ± 3 5
1
Area of ∆PQT = × PQ × ST
2
1
= (6 5 × 15) = 45 5
2
325
DAY
9 Fourth vertex of parallelogram lies on
circumcircle
⇒ Parallelogram is cyclic.
⇒ Parallelogram is a rectangle.
⇒ Tangents are perpendicular
⇒ Locus of P is the director circle
i.e.,
x2 + y 2 = a2 − b 2
10 Any point on parabola y 2 = 8 x is
(2t 2 , 4 t ). The equation of tangent at that
point is
...(i)
yt = x + 2t 2
Given that, xy = −1
...(ii)
On solving Eqs. (i) and (ii), we get
y ( yt − 2t 2 ) = −1 ⇒ ty 2 − 2t 2 y + 1 = 0
Q It is common tangent. It means they
are intersect only at one point and the
value of discriminant is equal to zero.
i.e.
4 t 4 − 4 t = 0 ⇒ t = 0, 1
∴ The common tangent is y = x + 2,
(when t = 0, it is x = 0 which can touch
xy = −1 at infinity only)
11 The equation of a hyperbola of the
y2
x2
− 2 = 1 where, λ is a
2
a
λ
parameter. The asymptotes of this
y
x
hyperbola = ± ⋅ Suppose ( x ′ , y ′ ) is
a
λ
a point P on the hyperbola which is
equidistant from the transverse axis and
asymptote.
y ′2
x ′2
Then,
...(i)
− 2 =1
2
a
λ
x ′/ a − y ′/λ
and
...(ii)
y′ =
1
1
+
a2
λ2
2
2
y′
x′
i.e.
= 2 − 1 [from Eq. (i)] ...(iii)
λ2
a
y ′2 2 x ′ y ′
1
1
x ′2
and y ′2 2 + 2 = 2 + 2 −
a
λ a
λ
aλ
series is
[from Eq. (ii)] ...(iv)
On simplification the second relation gives
4 x ′2 y ′2 a2
( y ′2 − x ′2 )2 =
λ2
= 4 x ′2 ( x ′2 − a2 ) [using Eq. (iii)]
So, the locus of P is
( y 2 − x2 )2 = 4 x2 ( x2 − a2 ).
12 The equation of tangent at point
P (α cos θ,sin θ)
x
y
cos θ + sin θ = 1
...(i)
α
1
Let it cut the hyperbola at points P and Q.
Homogenising hyperbola α2 x2 − y 2 = 1
with the help of Eq. (i), we get
2
x
α2 x2 − y 2 = cos θ + y sin θ
α
This is a pair of straight lines OP OQ.
Given ∠POQ = π/2.
Coefficient of x2 + Coefficient
of y 2 = 0
cos 2 θ
− 1 − sin2 θ = 0
α2
cos 2 θ
or α2 −
− 1 − 1 + cos 2 θ = 0
α2
α2 (2 − α2 )
or cos 2 θ =
α2 − 1
2
Now, 0 ≤ cos θ ≤ 1
α2 (2 − α2 )
or
0≤
≤1
α2 − 1
5+ 1
After solving, we get α2 ∈
, 2
2
13 Let equation of the rectangular
hyperbola be xy = c 2
…(i)
and equation of circle be
…(ii)
x2 + y 2 = r 2 .
From Eq. (i) and Eq.(ii) eliminating y,
we get
...(iii)
x 4 − r 2 x2 + c 4 = 0
Let x1 , x2 , x3 and x 4 are the roots of Eq. (iii).
4
∑x
∴ Sum of roots =
i
=0
i =1
Sum of products of the roots taken two at
a time = Σ x1 x2 = − r 2
From Eq. (i) and Eq. (ii) eliminating x,
we get
...(iv)
y 4 − r 2 y2 + c 4 = 0
Let y 1 , y 2 , y 3 and y 4 are the roots of
Eq. (iv).
∴
4
∑y
i
= 0 and
Σ
y1 y2 = − r 2
i =1
Now, CP 2 + CQ 2 + CR2 + CS 2
= x1 + y 1 + x2 + y 2
2
2
2
2
+ x + y
2
3
2
3
+ x4 + y 4
2
2
= ( x1 + x2 + x3 + x 4 )
2
2
2
+ ( y1 + y
2
+ y3 + y 4 )
2
2
4
[Q ( ∑ x i ) = 0, Σ x1 x2 = − r 2 ,
∑y
i
From Eqs. (ii) and (iii), we get
h2
t 12 + t 22 + t 32 + t 24 =
4
h2
Given that,
=k
4
Hence, locus of (h, k ) is x2 = 4 ay , which
is a parabola.
15 Let P ct 1 , c , Q ct 2 , c , R ct 3 , c be
= 0, Σ y 1 y 2 = − r 2 ]
i =1
= 4r2
14 Any point on the hyperbola xy = 4 is
2t , 2 . Now, normal at 2t , 2 is
t
t
2
y − = t 2 ( x − 2t ).
[its slope is t 2 ]
t
If the normal passes through P (h, k ),
then
2
k − = t 2 ( h − 2t )
t
…(i)
⇒ 2t 4 − ht 3 + tk − 2 = 0
Roots of Eq. (i) give parameters of feet of
normals passing through (h, k ). Let roots
be t 1 , t 2 , t 3 and t 4 , then
t1
t2
t3
the vertices of a ∆PQR inscribed in the
rectangular hyperbola xy = c 2 such that
the sides PQ and QR are parallel to
y = m1 x and y = m2 x, respectively.
∴ m1 = Slope of PQ
and m2 = Slope of QR
1
m1 = −
⇒
t1 t2
1
t2 t3
and
m2 = −
∴
m
m1
t
= 3 ⇒ t3 = 1 t1
m2 t 1
m2
2
2
2
4
= ( ∑ x i )2 − 2 Σ x1 x2
i
=
1
4
+ ( ∑ y i )2 − 2 Σ y 1 y 2
i =1
= (0 + 2r 2 ) + (0 + 2r 2 )
i =1
4
h
…(ii)
2
t1 t2 + t2 t3 + t1 t3 + t1 t 4
+ t 2 t 4 + t 3 t 4 = 0 …(iii)
t1 t2 t3 + t1 t2 t 4 + t1 t3 t 4
k
…(iv)
+ t2 t3 t 4 = −
2
and
t 1 t 2 t 3 t 4 = − 1 …(v)
On dividing Eq. (iv) by Eq. (v),
we get
1
1
1
1
k
+
+
+
=
t1
t2
t3
t4
2
2
⇒
y 1 + y 2 + y 3 + y 4 = k Q y =
t
t1 + t2 + t3 + t 4 =
or α2 −
The equation of PR is
x + yt 1 t 3 = c (t 1 + t 3 )
m
m
⇒ x + y 1 t 12 = c t 1 + 1 t 1
m2
m2
m
m
⇒ x + y 1 t 12 = c 1 + 1 t 1
m2
m2
⇒
m1
x+ y
t1
m2
2
c (m1 + m2 ) m1
= 2
⋅ t1
m
2
m
m
2
1
2
⇒ x + yt 2 = 2λt , where,
c (m1 + m2 )
m1
t =
⋅ t 1 and λ =
m2
2 m1 m2
Clearly, it touches the hyperbola,
2
or
c (m1 + m2 )
xy =
2 m1 m2
c 2 (m1 + m2 )2
xy =
4 m1 m2
326
DAY THIRTY
Unit Test 4
(Coordinate Geometry)
1 The parabola y 2 = 4x and the circle ( x − 6)2 + y 2 = r 2 will
have no common tangent, then
(a) r >
(b) r <
20
(c) r > 18
20
(d) r ∈( 20, 28 )
nx + ly + m = 0 are concurrent, if
substends 60° angle at the other focus, then its
eccentricity e is
2
(b)
3
(c)
5
(d)
6
4 Set of values of m for which a chord of slope m of the
circle x 2 + y 2 = 4 touches the parabola y 2 = 4x , is
(a) −∞, −
2 − 1
∪
2
2 − 1
,∞
2
(d) R
5 If ω is one of the angles between the normals to the
x2
y2
+ 2 = 1 at the points whose eccentric angles
2
a
b
π
2 cot ω
are θ and + θ, then
is equal to
2
sin 2θ
ellipse
1− e
2
(b)
e2
1+ e
2
(c)
e2
1 − e2
(d)
e2
1 + e2
6 If in a ∆ABC (whose circumcentre is origin), a ≤ sin A ,
then for any point ( x , y ) inside the circumcircle of ∆ABC
(a) xy <
(c)
1
8
1
1
< xy <
8
2
(b) | xy | >
(a)
(b)
(c)
(d)
(− ∞,5 2 )
(4 2 − 14 , 5 2 )
(4 2 − 14 , 4 2 +
None of the above
14 )
Tangents drawn to parabola at A and B meet Y -axis at A 1
and B 1 , respectively. If the area of trapezium AA1B1B is
equal to 24 a 2 , then angle subtended by A1B1 at the focus
of the parabola is equal to
(c) (−1, 1)
e2
r
r
,− 3 +
− 5 +
is an interior point of the major
2
2
segment of the circle x 2 + y 2 = 16, cut off by the line
x + y = 2, is
9 AB is a double ordinate of the parabola y 2 = 4 ax .
(b) (−∞,1) ∪ (1, ∞)
(a)
(b) 1
(d) 3
8 The range of values of r for which the point
(b) l + m − n = 0
(d) None of these
3 If the latusrectum of a hyperbola through one focus
(a)
the quadrilateral formed by the lines x = 0, y = 0,
3x + y − 4 = 0 and 4x + y − 21 = 0, then the possible
number of positions of the point A is
(a) 0
(c) 2
2 The lines lx + my + n = 0, mx + ny + l = 0 and
(a) l + m + n = 0
(c) l − m + n = 0
7 If A (n, n 2 ) (where, n ∈ N) is any point in the interior of
1
8
(d) None of these
(a) 2 tan−1 (3)
(c) 2 tan−1 (2)
(b) tan−1 (3)
(d) tan−1 (2)
10 If two tangents can be drawn to the different branches of
hyperbola
x2 y2
−
= 1 from the point (α , α 2 ), then
1
4
(a) α ∈ (− 2, 0)
(c) α ∈ (− ∞, − 2)
(b) α ∈ (− 3, 0)
(d) α ∈ (− ∞, − 3)
11 A hyperbola has the asymptotes
x + 2y = 3 and x − y = 0 and passes through ( 2, 1). Its
centre is
(a) (1, 2)
(c) (1, 1)
(b) (2, 2)
(d) (2, 1)
327
DAY
12 The equation of the ellipse having vertices at ( ± 5 , 0) and
foci at ( ± 4, 0) is
( 3, 3 3 ), then the third vertex is
x2
y2
(a)
+
=1
25
16
2
(c) 9x + 25 y 2 = 225
(b) 4 x 2 + 5 y 2 = 20
(d) None of these
13 The foci of an ellipse are ( 0 , ± 1) and minor axis is of unit
length. The equation of the ellipse is
(a) 2 x 2 + y 2 = 2
(c) 4 x 2 + 20y 2 = 5
(b) x 2 + 2 y 2 = 2
(d) 20x 2 + 4 y 2 = 5
14 The radius of the circle passing through the foci of the
ellipse
x2
y2
+
= 1 and having its centre at (0, 3) is
16
9
(a) 3
(c)
(b) 4
7
(d)
2
12
ax + by − 1 = 0 form a triangle with origin as orthocentre,
then (a, b ) is given by
(a) (− 3, 3)
(c) (− 8, 8)
(b) (6, 4)
(d) (0, 7)
16 If P (1, 0), Q ( − 1, 0) and R ( 2, 0) are three given points,
then the locus of S satisfying the relation
SQ 2 + SR 2 = 2 SP 2 is
1
(b) a ∈ (− ∞, − 3) ∪ , 1
3
1
(d) a ∈ , ∞
3
x = 2y is
32 x
9
32 y
(d) x =
9
(b) y =
(b) x 2 + 4 y 2 = 9
(d) 4 x 2 + y 2 = 81
(b) 6x − y + 32 = 0
(d) x + 6y − 32 = 0
21 The length of the latusrectum of the parabola
(c)
12
13
between the pair of tangents drawn from (a, a ) to the
π
circle x 2 + y 2 − 2x − 2y − 6 = 0 lies in the range , π
3
is
(a) (0, ∞)
(c) (− 2, − 1) ∪ (2, 3)
(b) − 2
(b) (− 3, − 1) ∪ (3, 5)
(d) − (3, 0 ) ∪ (1, 2)
(c) 2
(d) 1
anti-clockwise direction about origin, then coordinates of
P in new position are
1 7
(a)
,
2 2
1 7
(c) −
,
2 2
7
1
(b) −
,−
2
2
1
7
(d)
,−
2
2
28 The number of integral values of λ for which the equation
x 2 + y 2 − 2λx + 2λy + 14 = 0 represents a circle whose
radius cannot exceed 6, is
(b) 10
(c) 11
(d) 12
(d)
which passes through the focus of the parabola y 2 = 16x
are
(a) ± 2
(c) –1/2, 2
(b) 1/2, –2
(d) ± 1
both the parabolas y 2 = | x | is
(b) (− 1, 1)
(d) None of these
(a) (0, 1)
(c) (− 1, 0)
31 The parameters t and t′ of two points on the parabola
y 2 = 4ax , are connected by the relation t = k 2t ′. The
tangents at these points intersect on the curve
(a) y 2 = ax
169 [( x − 1)2 + ( y − 3)2 ] = ( 5x − 12 y + 17)2 is
28
13
25 The exhaustive range of values of a such that the angle
30 The range of values of n for which (n, − 1) is exterior to
by it with the points (1, 5) and (3, – 7) is 21 sq units.
Then, locus of the point is
(b)
(b) (x − q) 2 = 4 py
(d) (x − q) 2 = 4 px
29 The slopes of tangents to the circle ( x − 6) + y 2 = 2
20 A point moves such that the area of the triangle formed
14
13
the X-axis. The locus of the outer end of the diameter
through A is
2
of tangent to the hyperbola x 2 − y 2 = 9 is
(a)
(b) (1, 2)
(d) None of these
24 A variable circle through the fixed point A ( p, q ) touches
(a) 9
19 The locus of poles with respect to the parabola y 2 = 12x
(a) 6x + y − 32 = 0
(c) 6x − y − 32 = 0
(a) (3, 6)
(c) (4, 8)
27 If a point P ( 4, 3) is rotated through an angle 45° in
18 The diameter of 16x 2 – 9y 2 = 144 which is conjugate to
(a) 4 x 2 + y 2 = 36
(c) x 2 + 4 y 2 = 36
A ( − 1, 4), B ( 6, − 2) and C ( − 2, 4). If D, E and F are the
points which divide each AB, BC and CA respectively, in
the ratio 3 : 1 internally. Then, the centroid of the ∆DEF is
(a) 0
3x − y + 1 = 0 and x + 2y − 5 = 0 containing the origin, if
16x
9
16y
(c) x =
9
23 Let ABC is a triangle with vertices
are concyclic, if k is equal to
17 The point (a 2 , a + 1) lies in the angle between the lines
(a) y =
(b) (− 3, 3)
(d) None of these
26 The four distinct points ( 0, 0), ( 2, 0), ( 0, − 2) and (k , − 2)
a straight line parallel to X-axis
a circle through the origin
a circle with centre at the origin
a straight line parallel toY-axis
1
(a) a ∈ (− 3, 0) ∪ , 1
3
1
(c) a ∈ − 3,
3
(a) (3, − 3)
(c) (− 3, 3 3 )
(a) (x − p) 2 = 4qy
(c) (x − p) 2 = 4qx
15 If the straight lines 2x + 3y − 1 = 0 , x + 2y − 1 = 0 and
(a)
(b)
(c)
(d)
22 If two vertices of an equilateral triangle are ( 0, 0) and
16
13
1
(c) y 2 = ax k +
k
(b) y 2 = k 2 x
2
(d) None of these
328
32 Triangle ABC is right angled at A. The circle with centre A
and radius AB cuts BC and AC internally at D and E
respettvely if BD = 20 and DC = 16 then the length of AC
equals.
(a) 6 21
(b) 6 26
(c) 30
(d) 32
33 If the line y − 3x + 3 = 0 cuts the parabola y 2 = x + 2 at
A and B, and if P ≡ ( 3,0), then PA ⋅ PB is equal to
2+ 3
(a) 2
1
(c)
(b)
4 (2 + 3 )
3
(d)
4 (2 − 3 )
3
2 (2 − 3 )
3
x2 y2
+
= 1 and P is
25 16
any point on it, then difference of maximum and minimum
of SP ⋅ S ′ P is equal to
(b) 9
(c) 15
(d) 25
35 The locus of a point which moves, such that the chord of
contact of the tangent from the point to two fixed given
circles are perpendicular to each other is
(a) circle
(c) ellipse
(b) parabola
(d) None of these
x
+ y 2 = 1 at
27
π
( 3 3 cos θ, sin θ) where,θ ∈ 0, . Then, the value of θ
2
such that sum of intercept on axes made by this tangent
is minimum, is
36 Tangent is drawn to the ellipse
π
3
(b)
π
6
(c)
π
8
(d)
π
4
37 The condition for the line px + qy + r = 0 to be tangent to
the rectangular hyperbola x = ct , y =
(a) p < 0, q > 0
(c) p > 0, q < 0
c
is
t
(b) p > 0, q > 0
(d) None of these
38 If the line x + 3y + 2 = 0 and its perpendicular line are
conjugate w.r.t. 3x 2 − 5y 2 = 15 , then equation of
conjugate line is
(a) 3 x − y = 15
(c) 3 x − y + 10 = 0
(b) 3 x − y + 12 = 0
(d) 3 x − y = 4
39 The product of the lengths of perpendicular drawn from
any point on the hyperbola x 2 − 2y 2 − 2 = 0 to its
asymptotes, is
(a) 1/2
(b) 2/3
(c) 3/2
(d) 2
x2 y2
40 Tangents at any point on the hyperbola 2 − 2 = 1 cut
a
b
the axes at A and B respectively, if the rectangle OAPB,
where O is the origin is completed, then locus of the point
P is given by
a 2 b2
(a) 2 − 2 = 1
x
y
(c)
a2
b2
− 2 =1
2
y
x
(a) x + y + 1 = 0
(b) 5 x + y + 17 = 0
(c) 4 x + 9y + 30 = 0
(d) x − 5 y − 7 = 0
a2
b2
(b) 2 + 2 = 1
k
y
(d) None of these
a
b
c
−2 =
+
where a, b, c > 0 the family of lines
c
b
bc
a x + by + c = 0 passes through the point.
(a) (1, 1)
(b) (1, − 2)
(c) (−1, 2)
(d) (− 1, 1)
43 A series of ellipses E 1, E 2 , E 3 ,....., E n are drawn such that
E n touches E n −1 at the extremeties of the major axis of
E n −1 and the foci of E n coincide with the extremeties of
minor axis of E n −1. If eccentricity of the ellipse B
independent of x then the value of eccentricity is.
5 −1
2
5+1
4
(a)
(c)
2
(a)
3x − 2y + 5 = 0 are equation of on altitude and on angle
bisector respectively drawn from B, the equation of BC is.
42 If
34 If S and S′ are the foci of the ellipse
(a) 16
41 In a triangle ABC, if A( 2,−1) and 7x − 10y + 1 = 0 and
(b)
(d)
5+1
3
5 −1
4
44 If one of the diagonal of a square is along the line x = 2y
and one of its vertices is (3, 0), then its sides through this
vertex are given by the equations
(a)
(b)
(c)
(d)
y
y
y
y
− 3 x + 9 = 0, 3 y + x − 3 = 0
+ 3 x + 9 = 0, 3 y + x − 3 = 0
− 3 x + 9 = 0, 3 y − x + 3 = 0
− 3 x + 3 = 0, 3 y + x + 9 = 0
45 Given A( 0,0) and B( x , y ) with x ∈( 01
, ) and y > 0. Let the
slope of line AB equals m1. Point C lies on the line x = 1
such that the slope of BC equals m2 where 0 < m2 < m1. If
the area of the triangle ABC can be expressed as
(m1 − m2 )f ( x ), then the largest possible value of f ( x ) is.
(a) 1 / 8
(b) 1 / 2
(c) 1 / 4
(d) 1
46 A circle is inscribed into a rhombus ABCD with one angle
60°. The distance from the centre of the circle to the
nearest vertex is equal to 1. If P is any point of the circle,
then
| PA|2 + | PB|2 +| PC| 2 +| PD| 2 is equal to.
(a) 8
(b) 9
(c) 10
(d) 11
47 The equation of common tangent touching the circle
( x − 3) 2 + y 2 = 9 and the parabola y 2 = 4x above the
X-axis is
(a)
(c)
3y = 3x + 1
3y = x + 3
(b)
(d)
3 y = − (x + 3)
3 y = − (3 x + 1)
x 2 11y 2
+
=1
16
256
− 2x − 15 = 0, then φ is equal to
48 If the tangent at the point φ on the ellipse
touches the circle x
π
2
π
(c) ±
3
(a) ±
2
+y
2
π
4
π
(d) ±
6
(b) ±
329
DAY
Directions (Q.Nos. 49-55) Each of these questions
contains two statements : Statement I (Assertion) and
Statement II (Reason). Each of these questions also has four
alternative choices, only one of which is the correct answer.
You have to select one of the codes (a), (b), (c) and (d) given
below.
(a) Statement I is true, Statement II is true; Statement II is a
correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
Statement II A cyclic quadrilateral is a square, if its
diagonals are the diameters of the circle.
52 If a circle S = 0 intersects a hyperbola xy = c 2 at four
points.
Statement I If c = 2 and three of the intersection points
2
are (2, 2), (4, 1) and 6, , then coordinates of the fourth
3
1
point are , 16 .
4
Statement II If a circle intersects a hyperbola at
t1, t 2 , t 3 , t 4 , then t1 ⋅ t 2 ⋅ t 3 ⋅ t 4 = 1.
53 The auxiliary circle of an ellipse is described on the major
49 If p, x1, x 2 , x 3 and q , y1, y 2 , y 3 form two arithmetic
progression with common differences a and b.
Statement I The centroid of triangle formed by points
( x1, y1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ), lies on a straight line.
Statement II The point (h, k ) given by
x + x2 + K + xn
y + y2 + K + yn
and k = 1
always lies
h= 1
n
n
on the line b ( x − p ) = a ( y − q ) for all values of n.
50 Statement I A is a point on the parabola y 2 = 4ax . The
normal at A cuts the parabola again at point B.
If AB subtends a right angle at the vertex of the parabola,
1
then slope of AB is
.
2
Statement II If normal at (at12 , 2at1 ) cuts again the
2
parabola at (at 22 , 2at 2 ), then t 2 = − t1 − .
t1
51 Suppose ABCD is a cyclic quadrilateral inscribed in a
circle.
Statement I If radius is one unit and AB ⋅ BC ⋅ CD ⋅ DA ≥ 4,
then ABCD is a square.
axis of an ellipse.
Statement I The circle x 2 + y 2 = 4 is auxiliary circle of an
x2
y2
ellipse
+ 2 = 1 (where, b < 2).
4
b
Statement II A given circle is auxiliary circle of exactly
one ellipse.
x2
y2
+ 2 = 1, which
2
a
b
is not an extremity of major axis, meets a directrix atT .
54 The tangent at a point P on the ellipse
Statement I The circle on PT as diameter passes through
the focus of the ellipse corresponding to the directrix on
which T lies.
Statement II PT subtends a right angle at the focus of the
ellipse corresponding to the directrix on whichT lies.
55 Statement I If the perpendicular bisector of the line
segment joining P (1, 4) and Q (k , 3) has y-intercept − 4,
then k 2 − 16 = 0.
Statement II Locus of a point equidistant from two given
points is the perpendicular bisector of the line joining the
given points.
ANSWERS
1 (b)
2 (a)
3 (b)
4 (a)
11 (c)
12 (c)
13 (d)
14 (b)
21 (b)
22 (c)
23 (b)
24 (a)
5 (a)
6 (a)
7 (b)
8 (b)
9 (d)
10 (c)
15 (c)
16 (d)
17 (a)
18 (b)
19 (a)
20 (a)
25 (b)
26 (c)
27 (a)
28 (c)
29 (d)
30 (b)
31 (c)
32 (b)
33 (c)
34 (b)
35 (a)
36 (b)
37 (b)
38 (b)
39 (b)
40 (a)
41 (b)
42 (d)
43 (a)
44 (a)
45 (a)
46 (d)
47 (c)
48 (c)
49 (a)
50 (d)
51 (c)
52 (d)
53 (c)
54 (a)
55 (a)
330
Hints and Explanations
e2 − 1
2e
3 e 2 − 2e − 3 = 0
1 Any normal of parabola is
y = − t x + 2t + t .
⇒
Y
X¢
⇒
y 2 = 4x
A
X
⇒
t = 0, t 2 = 4
Thus,
A ≡ (4, 4)
0 + 0 − 4< 0
Y
⇒ m2 −
X
S
(ae, 0)
L¢ (ae, –b2/a)
2 − 1
2
⇒ m −
⇒m ∈ −∞,−
m +
2 − 1
∪
2
5 The equations of the normals to the
2
2
y
x
+ 2 = 1 at the points whose
a2
b
π
eccentric angles are θ and + θ are
2
ax secθ − by cosec θ = a2 − b 2
and − ax cosec θ − by sec θ = a2 − b 2 ,
respectively.
Since, ω is the angle between these two
normals.
Therefore,
a
a
tan θ +
cot θ
b
b
tan ω =
a2
1− 2
b
ellipse
=
Y¢
ab (tan θ + cot θ)
b 2 − a2
2 ab
⇒ tan ω =
sin 2θ (b 2 − a2 )
=
∴
2 cot ω
=
sin 2θ
2a2 1 − e 2
2ab
= 2 2
2
2
(a − b ) sin 2θ
a e sin 2θ
e2
1 − e2
S
2 − 1
>0
2
2 − 1
,∞
2
A
(n, n2)
3x
+
y–
X¢
O
0
L (ae, b2/a)
x=0
2 − 1
>0
2
=
Y
R
21
focus S (ae , 0) of the
hyperbola
y2
x2
− 2 = 1.
2
a
b
It substends angle 60° at the otherfocus
S ′ (− ae , 0).
2 − 1 2 1 + 2
m +
>0
2
2
y–
3 Let LSL ′ be a latusrectum through the
We have, ∠LS ′ L ′ = 60°
∴
∠LS ′ S = 30°
LS
In ∆LS ′ S , tan30° =
S ′S
1
b2 / a
⇒
=
3
2ae
1
b2
= 2
⇒
3 2a e
7 Origin is on the left of PS, as
<2
+
+ (l + m + n ) = 0
l + m+ n=0
30°
60° O
m
⇒ m2 −
∴ 1 (lx + my + n ) + 1 (mx + ny + l )
+ 1 (nx + ly + m ) = 0
⇒ x (l + m + n ) + y (l + m + n )
S¢
(–ae, 0)
m +1
2
4x
4 + 16 > r
r < 20
2 Since, lines are concurrent.
X¢
1
i.e.
sin A
1
⇒ 2R ≤ 1 ⇒ R ≤
2
So, for any point ( x, y ) inside the
circumcircle,
1
x2 + y 2 <
4
x2 + y 2
Using AM ≥ GM,
≥ | xy |
2
1
⇒
| xy | <
8
⇒ 4m 4 + 4m2 − 1 > 0
Thus, for no common tangent,
∴
1
.
m
This will be a chord of the circle
x2 + y 2 = 4, if length of the
perpendicular from the centre (0, 0) is
less than the radius.
Y¢
AC =
e = 3
the parabola y 2 = 4 x isy = mx +
If it passes through (6,0), then
− 6t + 2t + t3 = 0
⇒
(e − 3 ) ( 3 e + 1) = 0
∴
4 The equation of tangent of slope m to
C
(6, 0)
O
6 Given, a ≤ sin A ⇒ a ≤ 1
=
3
4=
0
Q
P
y=0
X
Y¢
∴At point A (n, n2 ),
3 n + n2 − 4 > 0
⇒
n2 + 3 n − 4 > 0
⇒
(n + 4) (n − 1) > 0
⇒
n − 1> 0
…(i)
Now, as A and O lies on the same sides
of QR.
and
4 x + y − 21 = 0 + 0 − 21 < 0
∴At point A (n, n2 ),
4 n + n2 − 21 < 0
⇒
n2 + 4 n − 21 < 0
⇒
(n + 7) (n − 3) < 0
[Qn ∈ N ]…(ii)
⇒
0< n< 3
From Eqs. (i) and (ii),
1< n< 3 ⇒ n = 2
Hence, A (2, 4) is only one point.
8 The given point is an interior point, if
2
2
r
r
− 5+
+ − 3 +
− 16 < 0
2
2
⇒ r 2 − 8 2 r + 18 < 0
⇒ 4 2 − 14 < r < 4 2 +
14
Since, the point is on the major segment,
the centre and the point are on the same
side of the line x + y = 2.
331
DAY
⇒
r
−3+
2
− 5+
r
− 2< 0
2
⇒ r<5 2
Hence,
4 2 − 14 < r < 5 2
9 Let A ≡ (at , 2at1 ), B ≡ (at , − 2at1 ).
2
1
2
1
Equation of tangents at A and B are
yt 1 = x + at 12 and yt 1 = x − at 12 ,
respectively.
Now, A1 ≡ (0, at 1 ), B1 = ( 0, − at 1 )
Area of trapezium
1
A A1 B1 B = ( AB + A1 B1 ) ⋅ OC
2
Y
A
A1
O
S
C
X
Here, a = 5 and ae = 4
4
⇒
e =
5
Now, b 2 = a2 (1 − e 2 )
2
4
= 25 1 − = 9
5
Hence, the equation of the ellipse is
x2
y2
+ 2 =1
2
5
3
⇒ 9 x2 + 25y 2 = 225
13 Given, 2b = 1 ⇒ b = 1 and a ⋅ e = 1
2
1
Since, a 2 (1 − e 2 ) =
4
1
5
⇒ a 2 − 12 =
⇒ a2 =
4
4
Hence, the equation of the ellipse
y2
x2
x2
y2
+ 2 = 1 is
+
= 1 or
2
b
a
1 / 4 5/4
20 x 2 + 4 y 2 = 5.
B1
14 Since, a 2 (1 − e 2 ) = 9
B
⇒ 16 − a 2e 2 = 9 ⇒ ae = 7
1
⇒ 24 a = ⋅ (4 at 1 + 2at 1 ) ( at 12 )
2
⇒ t 13 = 8 ⇒ t 1 = 2 ⇒ A 1 = (0, 2a)
If ∠ OSA 1 = θ, then
2a
tanθ =
= 2 ⇒ θ = tan −1 (2)
a
So, foci are at ( 7, 0) and ( − 7, 0) .
2
15 Equation of AO is
O
0
1=
ax+by – 1=0 C
Similarly,
(2 x + 3 y − 1) + µ (ax − ay − 1) = 0
will be equation of BO for µ = − 1.
Thus, BO is perpendicular to AC.
⇒
x+ 2 y =3
…(i)
and
x− y =0
…(ii)
On solving Eqs. (i) and (ii), we get
x = 1, y = 1
So, the centre of hyperbola is (1, 1).
12 The line joining foci and vertices is
X-axis and the centre is (0, 0). So, axes
of the ellipse coincide with coordinate
axes.
3 a 2 − (a + 1) + 1 > 0
and
a 2 + 2 (a + 1) − 5 < 0
i.e. a (3 a − 1) > 0 and a 2 + 2a − 3 < 0
i.e. a(3a − 1) > 0 and (a − 1)(a + 3) < 0
1
a ∈ (− ∞, 0) ∪ , ∞
⇒
3
and
a ∈ (− 3, 1)
∴
1
a ∈ (− 3, 0) ∪ , 1
3
18 Diameters y = m1 x and y = m2 x are
conjugate diameters of the hyperbola
x2 y 2
b2
–
= 1, if m1 m2 = 2 .
a2 b 2
a
1
2
2
Here, a = 9, b = 16 and m1 =
2
b2
Q
m1 m2 = 2
a
1
16
(m2 ) =
⇒
2
9
32
⇒
m2 =
9
32 x
Thus, the required diameter is y =
.
9
19 Let the pole be (h, k ), so that polar is
ky = 6( x + h )
6x
6h
y =
+
k
k
Since, it is tangent to the hyperbola,
⇒
∴
∴
x+ y =0
Since, AO is perpendicular to BC.
a
∴ (− 1) − = − 1 ⇒ a = − b
b
11 Given equations of asymptotes are
∴
x2 − y 2 = 9
y–
B
So, the asymptotes are y = ± 2 x.
∴ 2α < α2 ⇒ α < 0 or α > 2
and − 2 α < α2
α < − 2 or α > 0
∴
α ∈ (−∞ , − 2) or (2, ∞ )
on the same side of both the lines.
A
2
x+
1
4
Since, (α, α2 ) lies on the parabola y = x2 ,
then (α, α2 ) must lie between the
x2
y2
asymptotes of hyperbola
−
=1
1
4
in 1st and 2nd quadrant.
0
10 Given that, x − y = 1
2 x + 3 y − 1 + λ ( x + 2 y − 1) = 0.
Since, it passes through (0, 0), then
λ = − 1.
y-1
=
2
2x
+3
2
∴ Required radius
= ( 7 − 0) 2 + (3 − 0) 2 = 4
17 Since, origin and the point (a 2 , a + 1) lie
⇒
⇒
2 – a − 1
−
⋅
= −1
3+ a 2
2 − a = − 6 − 2a
a = − 8 and b = 8
16 Let the coordinates of a point S be ( x, y ).
Since, SQ 2 + SR2 = 2SP 2
⇒ ( x + 1) 2 + y 2 + ( x − 2) 2 + y 2
= 2 [( x − 1) 2 + y 2 ]
⇒ 2x + 3 = 0
Hence, it is a straight line parallel to
Y-axis.
c 2 = 9 m2 − 9
36 h2 324
= 2 −9
k2
k
Qc = 6 h , m = 6
k
k
2
2
⇒
4 h + k = 36
Hence, the locus is 4 x2 + y 2 = 36.
⇒
20 Let ( x, y ) be the required point.
Then,
x y 1
1
1 5 1 = 21
2
3 −7 1
⇒ (5 + 7)x − (1 − 3)y + (−7 − 15) = 42
⇒ 12 x + 2 y − 22 = 42
⇒ 6 x + y − 32 = 0
21 Given parabola is
5x − 12 y + 17
( x − 1) 2 + ( y − 3) 2 =
13
Focus = (1, 3), directrix is
5x − 12 y + 17 = 0
∴ Length of latusrectum
5 − 36 + 17
28
=2
=
13
13
2
332
22 Since, ∠AOM is 30°.
Y
3
B
M
2a2 − 4 a − 6 > 0
a < − 1 or a > 3
θ
2 2
Now, in ∆PAC, tan =
2
2
2a − 4 a − 6
So,
⇒
A (3, 3√3)
3
3√3
Y
2√2
X
O (0, 0)
P
θ/2
C
Hence, the required point B is (− 3, 3 3 ).
O
23 Let ( x1 , y 1 ), ( x2 , y 2 ) and ( x3 , y 3 ) are
coordinates of the points D, E and F
which divide each AB , BC and CA
respectively in the ratio 3 : 1 (internally).
A (–1, 4)
(x1, y1) D
E (x2, y2)
B (6, –2)
C (–2, 4)
3 × 6 − 1 × 1 17
=
4
4
−2×3+ 4×1
2
1
y1 =
=− =−
4
4
2
5
Similarly, x2 = 0, y 2 =
2
5
and
x3 = − , y 3 = 4
4
Let ( x, y ) be the coordinates of centroid
of ∆DEF .
1 17
5
+ 0 − = 1
∴ x =
3 4
4
1
1 5
and y = − + + 4 = 2
3 2 2
∴ x1 =
So, the coordinates of centroid are (1, 2).
24 The circle touching the X -axis is
x 2 + y 2 + 2gx + 2 fy + g 2 = 0.
Since, it passes through ( p, q ) .
∴
p2 + q 2 + 2gp + 2 fq + g 2 = 0 ...(i)
If ( x, y ) is the other end of the diameter,
then
p + x = − 2g , q + y = − 2 f
Now, Eq. (i) gives
p2 + q 2 − p ( p + x ) − q (q + y )
( p + x )2
+
=0
4
2
⇒ ( x + p ) = 4 px + 4qy
⇒ ( x − p ) 2 = 4qy
25 Given that,
x + y − 2x − 2y − 6 = 0
Centre = C (1, 1), radius = 2 2
2
2
Since, point (a, a) must lie
outside the circle.
X
π
As given that, < θ < π
3
π θ π
< <
⇒
6 2 2
2 2
1
∴
>
3
2a2 − 4 a − 6
⇒
F (x3, y3)
A (a , a )
∴
⇒
⇒
∴
a − 2a − 3 < 2 3
2
a2 − 2a − 3 < 12
2
a − 2a − 15 < 0
− 3< a< 5
a ∈ (− 3, − 1) ∪ (3, 5)
26 Let the general equation of the circle be
x 2 + y 2 + 2g x + 2 f y + c = 0.
QThe equation of circle passing through
(0, 0), (2, 0) and (0, − 2).
∴
c=0
…(i)
4 + 4g + c = 0
…(ii)
and
4 − 4f + c = 0
…(iii)
On solving Eqs. (i), (ii) and (iii), we get
c = 0, g = − 1, f = 1
∴ The equation of circle becomes
x2 + y 2 − 2 x + 2 y = 0
Since, it is passes through (k , − 2),
k 2 + 4 − 2k − 4 = 0
⇒ k 2 − 2k = 0 ⇒ k = 0, 2
We have already take a point (0, − 2), so
we take only k = 2.
27 Slope of line OP = 3 , let new position is
4
Q ( x, y ).
y
Slope of OQ = ,
x
also x2 + y 2 = OQ 2 = 25 = (OP 2 )
y 3
−
x 4
∴ tan 45° =
3y
1+
4x
4y − 3x
⇒
± 1=
4x + 3y
⇒
or
4x + 3y = 4y − 3x
− 4x − 3y = 4y − 3x
1
...(i)
⇒
x= y
7
or
...(ii)
− x = 7y
1
Correct relation is x = y as new
7
point must lies in Ist quadrant.
x 2 + 49 x 2 = 25
1
7
x=+
,y =
2
2
∴
⇒
28 Since, ( radius ) 2 ≤ 36
⇒ λ2 + λ2 − 14 ≤ 36
⇒ λ2 ≤ 25 ⇒ − 5≤ λ ≤ 5 ⇒ λ
= 0, ± 1, ± 2, ± 3, ± 4, ± 5
Hence, number of integer values of λ is
11.
29 Tangent to the circle with slope m is
y = m ( x − 6) ±
2 (1 + m2 ).
Since, it passes through (4, 0).
∴
4 m2 = 2 + 2m2
⇒
m=±1
30 Since, 1 − |n| > 0
⇒ |n| < 1 or n ∈ (− 1, 1)
31 Tangents at t and t ′ meet on the point
( x, y ) given by
...(i)
x = att ′ = ak 2 t ′ 2
and y = a (t + t ′ ) = a (k 2t ′ + t ′ )
= at ′ (k 2 + 1)
...(ii)
From Eqs. (i) and (ii),
ak 2 ⋅ y 2
k2 y 2
=
x= 2 2
2
a (k + 1)
a (k 2 + 1) 2
⇒ y2 =
ax (k 2 + 1) 2
1
= ax k +
k2
k
2
32. In ∆ ABC
AC 2 + AB 2 = BC 2
AC 2 + r 2 = 362
and
CF × CE = BC × CD
⇒ ( AC + r )( AC − r ) = 36 × 16
⇒
AC 2 − r 2 = 36 × 16
From Eqs. (i) and (ii), we get
2 AC 2 = 36(36 + 16)
⇒
AC 2 = 18 × 52
⇒
AC = 6 26
...(i)
...(ii)
C
16
E
D
20
r
r
A
B
F
33 Let PA = r1 , PB = − r2
Put ( 3 + r cos θ, r sin θ) in y 2 = x + 2
⇒ r 2 sin2 θ = ( 3 + r cos θ) + 2
⇒ r 2 sin2 θ − r cos θ − ( 3 + 2) = 0
PA ⋅ PB = − r1 . r2 =
3+2
sin2 θ
= ( 3 + 2)(1 + cot2 θ)
333
DAY
1
= ( 3 + 2) 1 +
3
4(2 + 3 )
PA ⋅ PB =
3
Y
[Qtanθ = 3]
A
X¢
X
P( 3,0)
O
(-2,0)
A′∈ (−43
, )
Coordinate of B is intersection point of
7 x − 10 y + 1 = 0 and 3 x − 2 y + 5 = 0
i.e. (−3,−2)
∴Equation of BC is
3+ 2
y −3 =
( x + 4)
−4 + 3
35 Let the equations of two given circles
are
…(i)
x2 + y 2 + 2g 1 x + 2 f1 y + c 1 = 0
and x2 + y 2 + 2g 2 x + 2 f2 y + c 2 = 0
…(ii)
Now, the equations of the chords of
contacts from P (h, k ) to Eqs. (i) and (ii)
are
x ( h + g 1 ) + y ( k + f1 )
+ g 1 h + f1 k + c 1 = 0
and x (h + g 2 ) + y (k + f2 ) + g 2 h
+ f2 k + c 2 = 0
According to the given condition,
(h + g 1 ) (h + g 2 )
×
= −1
( k + f1 ) ( k + f2 )
⇒ h2 + (g 1 + g 2 ) h + g 1 g 2
+ k 2 + k ( f1 + f2 ) + f1 f2 = 0
Hence, the locus of point is
x2 + y 2 + (g 1 + g 2 )x + ( f1 + f2 )y
+ g 1 g 2 + f1 f2 = 0
which is the equation of a circle.
36 Equation of tangent at (3 3 cos θ, sin θ)
2
x
+ y 2 = 1 is
27
+ y sin θ = 1.
3 3
This intersect on the coordinate axes at
(3 3 sec θ, 0) and (0,cosec θ)
[say]
On differentiating w.r.t. θ, we get
f ′ (θ) = 3 3 sec θ tan θ − cosec θ cot θ
3 3 sin3 θ − cos 3 θ
sin2 θ cos 2 θ
⇒
⇒
5x + y + 17 = 0
B(–3,–2)
0
3x–2y+5=0
y+1=
= a2 (1 − e 2 cos 2 θ)
= a2 − a2e 2 cos 2 θ
= 25 − 9cos 2 θ
Maximum = 25 − 9(0) = 25
[θ = 90° ]
Minimum = 25 − 9(1) = 16
[θ = 0° ]
Maximum–Minimum = 25 − 16 = 9
∴ Sum of intercepts on axes is
3 3 secθ + cosec θ = f (θ)
dy
−c
dx
= 2 and
=c
dt
t
dt
dy
−1
∴
= 2
dx
t
But equation of tangent is px + qy + r
= 0.
p
p 1
1
− =− 2 ⇒
= >0
∴
q
t
q t2
p
⇒
>0
q
A¢
7x–10
34 (SP )(S ′ P ) = a (1 − e cos θ) a (1 + e cos θ)
=
3 x − 2 y + 5 = 0. A′ is given by
x − 2 y + 1 −2(6 + 2 + 5)
=
=
= −2
3
−2
13
Then,
Y¢
x cos θ
41 Image of A(2,−1) with respect to line
37 Given, x = ct , y = c /t
B
to the ellipse
For maxima and minima, put f ′ (θ) = 0
3 3 sin3 θ − cos 3 θ = 0
1
⇒
tanθ =
3
π
⇒
θ=
6
π
At θ = , f ′ ′ (θ) > 0. So, f (θ) is
3
π
minimum at θ = .
6
p > 0, q > 0 or p < 0, q < 0
38 Since, the lines x + 3 y + 2 = 0 and
3 x − y + k = 0 are conjugate w.r.t.
x2
y2
−
= 1.
5
3
∴ 5 (1) (3) − 3 (3) ( − 1) = 2k
⇒
k = 12
Hence, equation of conjugate line is
3 x − y + 12 = 0.
42 We have, a − 2 = b + c
39 Given, equation can be rewritten as
x2
y2
−
=1
2
1
Here, a 2 = 2, b 2 = 1
The product of length of perpendicular
drawn from any point on the hyperbola
to the asymptotes is
a 2 b2
2 (1) 2
=
= .
a2 + b 2 2 + 1 3
⇒
⇒
⇒
c
bc
a − 2 bc = b + c
a = b + c + 2 bc
( a )2 = ( b + c )2
⇒
( b + c )2 − ( a )2 = 0
⇒
or
⇒
b 2 h2
= sin2 θ
a 2k2
b 2 h2
h2
+ 2 =1
⇒
2 2
a k
a
[Q sin2 θ + cos 2 θ = 1]
a 2 b2
⇒
− 2 =1
h2
k
a2
b2
Hence, the locus of P is 2 − 2 = 1.
x
y
b
b+ c − a=0
b+ c + a=0
ax + by + c = 0
(rejected)
passes through fixed point (−11
, )
43 Here,
b ′ = a, ae
′ =b
(b ′ )2 = (a′ )2 − (a′ e )2
⇒
40 The equation of tangent is
y
x
sec θ −
tan θ = 1.
a
b
So, the coordinates of A and B are
(a cos θ, 0) and ( 0, − b cot θ), respectively.
Let coordinates of P are (h, k ) .
∴h = a cos θ, k = − b cot θ
k
b
⇒
=−
h
a sinθ
C
A(2,–1)
a2 =
b2
− b2
e2
Y
(0, a¢)
a¢e
X¢
{
(0,b)
X
a
(a, 0) or
(b¢, 0)
⇒
Y¢
b2
Q1 − e 2 = 2
a
⇒
b2
e 2 = 2 (1 − e 2 )
a
⇒
e 2 = (1 − e 2
