A Handbook for the B.Sc. (Honors) students in Computer Science & Engineering
B
asic Mathematics
B
y
Mohammad Abdul Halim
MS, B.Sc. in Mathematics (JU)
Senior Lecturer in Mathematics
Department of GED
Daffodil International University
Dhaka
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Copyright: All right reserved to the Author.
To reproduce any parts of this text is not permitted without written permission of the
Author.
Published in:
First edition: February, 2018
Second edition: May, 2018
Third edition: May, 2019
Cover Deigned by: M.A.Halim
Computer Compose by: M.A.Halim
Published By:
Contact:
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A few words
Our purpose is to give a sound introduction to the study of calculus for the engineering students,
especially for the B.Sc. (Honors) students in CSE, SWE, EEE, ETE, TE, Civil Engineering and
Architecture. We have utilized our teaching experience along with all open sources and relevant
books on the subject. I am grateful to my colleague and my family members for their useful
suggestions in preparing this Handbook for our beloved students.
Any suggestion and correction towards the improvement of this Handbook will be highly
appreciated.
M.A.Halim
Lecturer in Mathematics
Department of General Educational Development (GED)
Daffodil International University (DIU), Dhaka.
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Contents
 Algebra
Chapters
Topics
Pages
Chapter 01
Number System
06 – 11
Chapter 02
Surd(Radicals),Exponents(Indices)
12 – 18
Chapter 03
Logarithm
19 – 25
Chapter 04
Inequality
26 – 28
Chapter 05
Partial Fractions
29 – 35
Chapter 06
Polynomial Equations
35 – 41
Chapter 07
Binomial Theorem &Summations
42 – 44
 Calculus
Chapter 08
Function and its Domain &Range
45 – 49
Chapter 09
Basic Concept of Derivative
49 – 54
Chapter 10
Basic Concept of Integration
55 –
Appendix:
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Formulae, OBTL course outlines and some symbols used in the book
01 NUMBER SYSTEM
“Mathematics is the gate and key to the sciences.” – Roger Bacon
Number theory is a foundation of mathematics as basic as geometry and more basic than algebra. Pierre
de Fermat is usually given credit for being the father of number theory.
Digit: A mathematical symbol (0 – 9) stands for making a numeral is called a digit and numeral is a symbol
or name that stands for the number.
For example, the numeral 153 is made up of the digits "1", "5" and "3".
Number: A number is a mathematical object used to count, measure, and also label.
What is real number?
A number is said to be real if the square of it is non-negative otherwise complex. For example: The number
2
√2 is a real number due to its square (√2 ) = 2 (non-negative). On the other-hand √−2 is a complex
2
number due to its square (√−2 ) = −2 (non-negative). A real number geometrically represents a point on
x-axis.
Complex number: The complex number 𝑧 is the linear combination of two real numbers a & 𝑏 with a special
sign 𝑖 ,that is 𝑧 = 𝑎 + 𝑖𝑏 where 𝑎, 𝑏𝜖𝑅 𝑎𝑛𝑑 𝑖 = √−1. A complex number geometrically represents a point on
complex plane. For example, z  2  3i is a complex number.
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Note:
Modulus of complex number is
r  z  a 2  b2
Argument of the complex number is,   tan 1
Conjugate of z  a  ib is
b
a
z  a  ib
i
Euler Identity: e  cos   i sin   cis
Various form of z:
(a) cartesian form z  x  iy (b)Polar form,
z  r (cos   i sin  )
Rational number: A number of the form
(c) Exponential/Eulerian form,
p
, q  0 where
q
z  rei
p, q are co-primes.so every integer, fraction,
recurrence decimal point number and terminating decimal point number are rational.
Irrational number: A number which is not expressible as
Integer: A number of the form
p
q
p
, q  0 is
q
known as Irrational number.
where q  1 is called Integer.
Fraction: Fraction is the ratio of numerator and denominator that means Fraction 
Numerator
Denominator
.
Classification of Number System:
Number
Real Number
Rational
Complex Number
Irrational
Integer
Purely Imaginary
Fraction
Negative
Zero
Prime
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Purely Real
Positive/Natural Number
Neither prime nor Composite
Proper
Composite
Improper
Mixed
Factor: A factor of a number is a number which divides that number with remainder zero.
For example: The number 2 & 3 are the factor of the number 6 because 2 & 3 divides the number 6 with
remainder zero.
Multiplicand, multiplier, Product:
Multiplicand: The multiplicand is the number (factor) that gets multiplied.
Multiplier: The multiplier is the number (factor) that you are multiplying by.
Product: The product is the result of the multiplication.
Multiple: A multiple of a number is obtained by multiplying a number by a non-zero whole number (positive
integer). On other hand a multiple is a number that may be divided by another a certain number of times
without a remainder. For example, a x b = c; c is a multiple of a and b.
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Dividend, divisor, quotient, Remainder.
Dividend: The dividend is the number being divided.
Divisor (Factor): The divisor is the number which dividing the dividend exactly.
Quotient: The number obtained by dividing one quantity by another.
Remainder: The number left over when one integer is divided by another.
Prime number: A number which is divisible by only itself and one is called prime number. For example, the
number 2,3,5&7 etc. are prime numbers due to those numbers are divisible by factors one and itself.
Composite number: Composite number is the number which has the divisor except one and itself.
For example: The numbers 4,6 & 8 etc. are composite numbers.
Even number: A number which is divisible by 2 is called even number that means of the form 2 ∗ 𝑛.
Odd number: A number which is divisible by 2 is called even number that means of the form 2 ∗ 𝑛 + 1 𝑜𝑟 2 ∗
𝑛−1
Least Common Multiple (LCM): The smallest number which is exactly divisible by each of two or more
numbers is called the lowest common multiple (LCM).
Highest Common Factor (HCF/GCD): The highest common factor (HCF) is the largest number which
divides into two or more numbers exactly.
Note:
HCF of Fractions: HCF of fractions
LCM of Fractions: LCM of fractions
𝑎
,
𝑐
𝑏 𝑑
𝑎 𝑐
𝑏
,
𝑑
𝑎𝑛𝑑
𝑒
𝑓
𝑒
𝑎𝑛𝑑
𝑓
𝑖𝑠 =
𝑖𝑠 =
𝐻𝐶𝐹 𝑜𝑓 𝑎 ,𝑐 𝑎𝑛𝑑 𝑒
𝐿𝐶𝑀 𝑜𝑓 𝑏,𝑑 𝑎𝑛𝑑 𝑓
𝐿𝐶𝑀 𝑜𝑓 𝑎 ,𝑐 𝑎𝑛𝑑 𝑒
𝐻𝐶𝐹 𝑜𝑓 𝑏,𝑑 𝑎𝑛𝑑 𝑓
Co-primes (Relative Primes): Two prime numbers are said to be Co-primes if the GCD of two numbers is one.
Mathematical problems:
1. Find out the all factors of 540.
Solution:
2 540
3 270
3 90
3 30
2 10
5
Prime factorization of 540 is 540  2  2  3  3  3  5  22.33.51
The total number of factors is   2  1 . 3  1 .1  1  24
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Here 540  1 540
540  2  270
540  3 180
540  4 135
540  5 108
540  6  90
540  9  60
540  10  54
540  12  45
540  15  36
540  18  30
540  20  27
Therefor the factors of 540 are: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 27, 30, 36, 45, 54, 60,
90, 108, 135, 180, 270, and 540.
2. Find out the prime factorization of 540 using tree.
Solution: The tree diagram for the prime factors is as follows:
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In the above tree diagram prime factors are indicated by circles, so the prime factorization of
the number 540 is as of the form, 540  2  2  3  3  3  5  22.33.51
3.
What is the GCD & LCM of 240&540.
2 240
2 540
2 120
2 60
3 270
3 90
2 30
315
5
3 30
2 10
5
Therefore, the prime factorization of, 240  2  2  2  2  3  5  24.31.51
Also
540  2  2  3  3  3  5  22.33.51
Finally, the GCD of 240&540 is  22.31.51  4.3.5  60 and LCM  24.33.51  16.27.5  2160 (Ans)
4.
Find the H.C.F of 42,63 &140.
Solution:
Here, 42  2  21  2  3  7  21.31.71
63  3  21  3  3  7  32.71
140  2  70  2  2  35  2  2  5  7  22.51.71
Therefore, the HCF of 42,63 &140 is  7 (Ans)
5.
Find the H.C.F & L.C.M of
2
8 16
10
,
,
and
3
9 81
27
.
Solution:
Factorization of Numerators:
2  2  21
8  2  4  2  2  2  23
16  2  8  2  2  4  2  2  2  2  24
10  2  5  21  51
HCF of Numerators is  21  2
LCM of Numerators is  24  51  16  5  80
Factorization of Denominators:
3  3  31
9  3  3  32
81  3  27  3  3  9  3  3  3  3  34
10 | P a g e
27  3  9  3  3  3  33
HCF of Numerators is  31  3
LCM of Numerators is  34  81
6.
Finally, the HCF of
2
8 16
10
,
,
and
3
9 81
27
Evaluate
&
 16   4
Solution: We have
Now,
 16
4
is

HCF (2,8,16,10) 2

LCM (3,9,81, 27) 81
& LCM 
LCM (2,8,16,10) 80

HCF (3,9,81, 27) 3
(Ans)
.
i 2  1
 16 
 4  16 i 2 
 16
16i 2
4i 2 
22 i 2  4i  2i  8i 2  8   1   8
42 i 2 
Again,
4
7.

4i
2
42 i 2


2 2
2 i
4i
2
2i
Find the modulus and Argument of z  1  3 i and also its polar, exponential form.
1 3 i
Solution:



 
2
1 3 i
1 3 i
1 3 i
1  2 3 i  3i 2 1  2 3 i  3 2  2 3 i
1
3
z





 
i
2
4
4
2 2
2
1  3 i 1  3 i 1  3 i 12  3 2
1  3

 a  ib
Now

2
r  a 2  b2 
where

a
1 3
  1 1
4 4
So, the polar form is,

1
3
&b 
2
2
and
 
.
b
 
 3

2   tan 1  3   tan 1

  12 


  tan 1    tan 1 
a
2
2

z  r  cos   i sin    1.  cos
 i sin
3
3



2
2 
 
 i sin
   cos
&
3
3 
 
 3    3  23
Exponential form, z  e
i.
2
3
(Ans)
02
Radicals & Exponents
Radical: An expression containing the radical symbol
of a radical is
11 | P a g e
n
  is called a radical. The general form
a Where n is the index and a is the radicand.
Note: 1. The index n is omitted if n  2 .
2. Two or more radicals are called similar if the index and radicand are same.
Formulae for Radicals: The formulae for radicals are
1.
 a
n
n
a
2.
n
ab  n a . n b
3.
n
a na

b nb
4.
n
am 
5.
m n
,b0
 a
n
m
a  mn a
Simplification of Radicals: The radicals can be simplified as,
1. By removing the perfect nth powers of the radicand.
2. By reducing the index of the radical
3. By rationalizing of the denominator of the radicand.
Problem-1: Find the simplest form the followings:
a.
 6
3
3
3
b.
54
c.
5
5
32
d.
3
 27 
4
e.
3
5
Solution:
a. We have
 6
3
c. We have
5
e. We have
3
12 | P a g e
3
  6
3.
1
3
6
5
5
5
5
5 15
5


5
32
32 5 25 2
565
b. We have
3
54  3 33  2  3 33  3 2  3 3 2
d. We have
3
 27 
4


3
  3 
4
27
3
3
4
 34  81
Problem-2: Find the simplest form the followings:
a.
b.
18
4
c.
6480
4
16
81
d.
5
 72
4
4 3
e.
2
Solution:
18  32  2  32  2  3 2
a. We have
b. We have
4
6480 
4
34  24  5 
34  4 24  4 5
4
 3 2 4 5  6 4 5
16 4 16 4 24 2
c. We have



81 4 81 4 34 3
d. We have
4
5
 72
4
 5  23  32   5 212  38
4
12
e. We have
4 3
8
 5 212  5 38  2 5  35  2
2  12 2
2
2
5
1
3
3
5


2
3

 

  22  2 5    3  3 5   4 5 2 2  3  5 3 3

 

 12 5 2 2  33  12 5 108
Problem-3: Find the simplest form the followings:
a.
6
81a

2
3
b. 7 4ab

 x  1
3
6
 y  2
3
2
c.
3
7
64x y
6
d.
Solution:
a. We have
4
6
6
81a2  6 34 a2  6 34 .6 a2
2
6
2
3

b. We have 7 3 4ab
1
3
 3 . a  3 . a  3 32 . 3 a  3 9a
c. We have
3
6
7

4x 2
y2
3
 x  1
3
6
 y  2
3
d. We have
13 | P a g e

6
3
 22  x
2
1
3
 x  1
x 1


2
6
3
 y  2  y  2
3
3
1
 y  2  4 x2  x 3  y 2
x
2
 49

3
2 2 ab

2
 49 3 23  2a2b2  98 3 2a2b2
64 x7 y 6  3 26  x7 y 6  3 26  3 x7  3 y 6
 23  x 3  y


Exercise: Find the simplest form the followings:
a.
b.
40
Solutions: a. 2 10
3
c.
648
b. 6 3 3
c. 7
6
343
d.
d. x  5
x  25
x 5
e.
3
f.
246
4 3
6ab 2
f .12 6ab2
e.2 3 2
Problem-4: Calculate the followings:
a.
18  50  72
b. 2 27  4 12
c.
d.
248  52  144
112
576
256


12
8
196
Solution:
b. We have 2 27  4 12
a. We have 2 27  4 12
 2 32  3  4 22  3
 2  32  2.52  23  32
 23 3  4 2 3
 3 2  5 2  3 22  2
 3 2 5 2 6 2
 6 3 8 3
2 2
 2 3
c. We have
248  52  144
112
576
256


12
8
196
b. We have
 248  52  24  32

112
22  7 2

26  32
28

12
8
2
 248  52  2  3

112 23  3 24


2  7 12
8

112 8  3 16

  32
2  7 12 8
 248  52  12
 248  64  248  26
 248  23  248  8
 28  24  16
Problem-5: Show that
Solution: L.H.S
5 2 6  3  2
.
 5 2 6
 5  2 3 2  5  2 3  2  3  2 3  2  2

14 | P a g e
 3
2
2 3 2 
 2
2


3 2

2
 3 2
= R.H.S
Problem-5: Find the cube root of 2744.
Solution: The cube root is
 3 2744
3
 23  73
 27
 14
Problem-6: If
(a  2)(b  3)
c 1
a *b *c 
Solution: We have
Using
a *b *c 
then find the value of 6*15*3 .
(a  2)(b  3)
c 1
a  6, b  15 & c  3 we
get
(6  2)(15  3)
3 1
6 *15*3 
8  45
4

 90
 2  32  5
 3 10
Problem-7: Show that
Solution: L.H.S =
1
1
1


3 3 3 3 3
3
 3
1
1
1


3.
3 3 3 3 3
3
3  3   3  3 
   

 3  3 3  3 
3
3
 3
 3

15 | P a g e

3
3 3 3 3

 


 

3 3
3 3 3


3
93
93
3 3
3 3 3


3
6
6
18  2 3  3  3  3  3
6

18
3
6
= R.H.S
Problem-8: If
x  1 2 & y  1 2


, then find the value of x2  y2 .
x  1 2 & y  1 2
Solution: We have

Now x2  y2

 1 2

  1  2 
2
 1 2 2 
 2
2
2
1 2 2 
 2
2
 1 2 2  2 1 2 2  2
6
Problem-9: If
1
x
13

144 12
Solution: We have
1
then find the value of x.
x
13

144 12
x  13 
 
144  12 
or , 1 
2
or ,
x 169

1
144 144
or ,
x 169  144

144
144
or ,
x
25

144 144
or , x  25
1
Problem-10: What will come in the place of question mark
(?) 4 
48
3
 ? 4
Solution: Let the required value is x
According to question we can write,
1
48
( x) 4 
3
 x4
3
1
or ,  x  4 ( x) 4  48
3 1

or ,  x  4 4  48
4
or ,  x  4  48
16 | P a g e
or , x  48
n
243  32n 1
Problem-11: Find the value of  n 5 n1 .
9 3
Solution: We have,
n
 243 5  32n1
n
n 1
9 3
n
35  5  32n1

32n  3n 1




3n  32n 1
32n  n 1
3n  2n 1
32n  n 1
33n 1
33n 1
33n  3
3n
3
 31
3
3   9
1
Problem-12: What will come in the place of question mark
Solution: Let the required value is x
According to question we can write,
86.49  5  (?)  12.3
86.49  5  ( x)  12.3
or, 5  ( x)  12.3  86.49
 123
8649 
or ,5  x  


100 
 10
2
2
 123
8649 
or , x  

 5
100 
 10
2
 123 93 
or , x  
  5
 10 10 
2
 30 
or , x     5
 10 
or, x   3  5
2
or , x  9  5
or , x  4
Problem-13: If
17 | P a g e
841  29 ,
then find the value of
841  8.41  0.0841  0.000841 .
Solution: Since
Now
841  29
841  8.41  0.0841  0.000841
 841 
 841 
 29 


841
841
841


100
10000
1000000
29 29
29


10 100 1000
29000  2900  290  29
1000
32219
1000
 32.219
18 | P a g e
841
841
841


100
10000
1000000
03
Logarithm
Exponents & Logarithms
Exponents: If a
is any number then the product of n numbers each of which is a , is defined as,
an
Where n is called an exponent or index and a is called a base.
5
2
Example: 2 , 3
, x 6 , p 7 etc.
Logarithms: If an expression is of the form
bx  N
1
where N  0 , b  0 & b  1
then the logarithm of N to the base b is defined as
x  logb  N 
 2
where N  0 , b  0 & b  1
The equations (1) & (2) are equivalent. The eq. (1) is in exponential form and the eq. (2) is in logarithmic
form.
Example: Since 23  8 , then 3 is the logarithm of 8 to the base 2 i.e., log2 8  3 .
Laws of Logarithms: The laws of logarithms are
6.
logb  MN   logb  M   logb  N 
7.
M
log b 
N
19 | P a g e

  log b  M   log b  N 

8.
logb  M P   P logb  M 
9.
logb b  1
Problem-1: Using logarithmic laws write the followings:
b.
 17 

 24 
log2  3  5
b. log 3 
 
7
c. log3 5
Solution:
a. We have log2  3  5  log2 3  log2 5
 17 
  log3 17  log 3 24
 24 
b. We have log3 
 
c. We have log3 57  7 log3 5
Common Logarithms: The system of logarithms whose base is 10 is called the common logarithm
system. When the base is omitted, it is understood that base 10 is to be used.
Thus, log 25  log10 25
Natural Logarithms: The system of logarithms whose base is the Eulerian constant e is called the
natural logarithm system. When we want to indicate the base of a logarithm is e we write ln .
Thus, ln 25  log e 25
NOTE: Since 101.5377  34.49 so log34.49  1.5377 . Here the digit 1 before decimal point is called the
characteristic and the digits . 5377 after decimal point is called the mantissa of the log.
Problem-2: Express each of the following exponential form in logarithmic form:
a.
1
b. 3 
9
2
4  16
2

c. 8
2
3

1
4
Solution:
b. We have 42  16
Using log of base 4 we get
log 4 42  log 4 16
or , 2 log 4 4  log 4 16
or , 2  log 4 16

c. We have 8
2
3
1

4
Using log of base 8 we get
1
 log8  
4
20 | P a g e
2
1
or ,  log8 8  log8  
3
4
log8 8
or , 
2

2
3
 log
1
a. We have 32 
1
9
Using log of base 3 we get
1
log3 32  log 3  
9
1
or ,  2 log3 3  log3  
9
1
or ,  2  log3  
9
Problem-3: Express each of the following logarithmic form in exponential form:
a.
log 5 25  2
b. log 2 64  6
c. log1 4
1
2
16
Solution:
a. We have log 5 25  2
By the definition of log we
get
b. We have log 2 64  6
By the definition of log we
get
25  52
64  26
c. We have log1 4
1
2
16
By the definition of log we
get
1 1
 
16  4 
2
Problem-4: Find the logarithm of 1728 to the base
Solution: We have 1728
After factorization by prime number we get,
1728  26  33
or , 26 
 3


or , 2 3
21 | P a g e
6
6
 1728
 1728
2 3
.
According to definition of logarithm we have,
6  log2 3 1728
 log2 3 1728  6




Problem-5: Find x if
1
log10 11  4 7  log10  2  x 
2
Solution: Given that,
1
log10 11  4 7  log10  2  x 
2
.
or , log10 11  4 7  log10  2  x 
11  4 7  2  x
or,
or ,
 11  4 7 
2
  2  x
2
or, 11  4 7  x2  4x  4
or, x2  4x  7  4 7


or , x 2  4 x  7  4 7  0
 x 

 4  42  4  1  7  4 7

2 1



 4  16  4 7  4 7

2
 4  16  28  16 7
2


 4  12  16 7
2
 4  2 3  4 7
2
  2  3  4 7
Problem-6: Prove that
2log x  2log x2  2log x3 
 2log xn  n(n  1)log x .
Solution: L.H.S  2log x  2log x2  2log x3   2log xn
 2log x  2log x2  2log x3 
 2log x  4log x  6log x 
22 | P a g e
 2log xn
 2n log x
 1  2  3 

 n  2 log x
n  n  1
 2 log x
2
 n  n  1 log x
 Pr oved 
 R.H .S
Problem-7 : Express the logarithm of
Solution: We have
a3
in terms of
5 2
c b
log a, log b & log c .
a3
5 2
c b
The logarithm of this part is,
 a3 
  log a3  log c5b 2
log 
 c5b2 


 
3
 
 
 log a 2  log c5  log b2
3
 log a  5log c  2log b
2
Problem-8: Find
x from the equation a x .c 2 x  b3 x 1
Solution: We have
a x .c 2 x  b3 x 1


or,ln a x .c2x  ln b3x1
or,ln a x  ln c2x  ln b3x1
or , x ln a  2 x ln c   3x  1 ln b
or , x ln a  2 x ln c   3x  1 ln b
or , x ln a  2 x ln c  3x ln b  ln b
or , x ln a  2 x ln c  3x ln b  ln b
or , x  ln a  2 ln c  3ln b   ln b


or, x ln a  ln c2  ln b3  ln b
or, x 

ln b
ln a  ln c2  ln b3
ln b
 a 
ln 

 c 2b 3 
Problem-09: Solve log10  3x  2   log10  x  1  1 .
23 | P a g e
(Expressed)
Solution: We have
log10  3 x  2   log10  x  1  1
or , log10  3 x  2  x  1  1


or, log10 3x2  3x  2x  2  1


or, log10 3x2  x  2  1
or, 3x2  x  2  101
or, 3x2  x  2  10
or, 3x2  x  12  0
  1 
 x
 12  4  3   12 
2 1

1  1  24
2

1  25
2
1 5
2

  2, 3
Problem-10: Solve the equation
Solution: We have
or ,
or ,
ex 1
e x  1
ex 1
e x  1
 3
 3
ex 1
 3
1
1
ex
ex 1
1  ex
 3
ex
or,
e2 x  e x
1  ex
 3
or, e2x  ex  3  3ex
 
or , e x
 ex 
24 | P a g e
2
 4e x  3  0
   4 
  4 2  4  1  3
2 1

4  16  12
2

4 4
2

42
2
1, 3
Problem-11: Calculate the value of p from
Solution: We have
log10 4  2 log10 p  2
log10 4  2 log10 p  2
or, log10 4  log10 p2  2
or, log10 4 p2  2
or, 4 p2  102
or, 4 p2  100
or, p2  25
or , p   5
04
Inequality
Number Line: A straight line whose each point indicates a single number is called a number line.
Graphically it is denoted by
-2
-1
0
1
2
Interval: The set of all real numbers lie between two real numbers a and b, where 𝑎 ˂ 𝑏 is called
an interval.
Intervals are four kinds:
a) The set {𝑥𝜖ℛ: 𝑎
b) The set {𝑥𝜖ℛ: 𝑎
c) The set {𝑥𝜖ℛ: 𝑎
d) The set {𝑥𝜖ℛ: 𝑎
25 | P a g e
≤𝑥
<𝑥
<𝑥
≤𝑥
≤ 𝑏} is called a closed interval, denoted by [𝑎, 𝑏].
< 𝑏} is called an open interval, denoted by (𝑎, 𝑏).
≤ 𝑏} is called a left half open interval, denoted by (𝑎, 𝑏].
< 𝑏} is called a right half open interval, denoted by [𝑎, 𝑏).
Modulus/absolute Value: The modulus or absolute value of x is denoted by the symbol x
and is defined as follows,
 x
x 
 x
; x0
;x  0
Geometrically the modulus or absolute value of a number represents the distance of that number
from the origin. The absolute value of x is always positive or zero.
Inequality: An inequality is a statement that one real quantity or expression is greater or less than
another real quantity or expression.
The followings indicate the meaning of inequality signs:
1. a  b Means a is greater than b .
2.
3.
4.
a  b Means a is less than b .
a  b Means a is greater than or equal to b .
a  b Means a is less than or equal to b .
Problem-1: Solve
3x  4
2
x 1
Solution: The given inequality is
3x  4
2
x 1
or , 3x  4  2 x  2
or , 3x  2 x  4  2
or , x  2
Problem-2: Find the solution set of
Solution: The given inequality is
x3  3 x 2  4 x  12  0 .Write
3
the answer in interval notation.
2
x  3 x  4 x  12  0
or, x2  x  3  4  x  3  0


or,  x  3 x2  4  0
-3
or ,  x  3 x  2  x  2   0
From the number line it is clear that the inequality is satisfied if
x  2 or  3  x  2
The solution set is
  2,     3, 2
  3,     2, 2 
Problem-03: Solve the inequalities a)
Solution: (a) The given inequality is
When
2  3x
2  3x  7
2  3x  7
is nonnegative then
 2  3x   7
or,  3x  7  2
or ,  3x  5
or , x  
When
2  3x
5
3
is negative then
  2  3x   7
or ,  2  3x  7
or , 3x  7  2
26 | P a g e
b)
4x  3  5
-2
0
2
or , 3x  9
or, x  3
The solution is x  
5
3
and
x3
 5 
   ,     ,3
 3 
 5 
   , 3
 3 
(b) The given inequality is
4x  3  5
When 4x  3 is nonnegative then
4x  3  7
or , 4 x  7  3
or , x 
10
4
or , x 
5
2
When 4x  3 is negative then
  4 x  3  7
or,  4 x  3  7
or,  4 x  3  7
or ,  4 x  4
or , x  1
The solution is
x 1
and
x
5
3
5

  1,     , 
3

5

  1, 
3

Problem-4: Solve
x 2  3x  4  0
Solution: The given inequality is
x 2  3x  4  0
x2  4 x  x  4  0
x  x  4   1 x  4   0
-4
-2
0
 x  4  x  1  0
From the number line it is clear that the inequality is satisfied if
x  4 and x  1
The solution is
  4,     , 1
  4,1
Exercise:
1.
Solve the rational inequality
2.
Solve
3.
x2  x  2
x2  4 x  3
0 .
3x  6
 2  12
3
Solve each compound inequality. Write each solution in set-builder notation.
x 1  9
(b) x  3  4 and 2x  1  15
2 x  10 or
27 | P a g e
1
05
Partial Fractions
Rational Fraction: If P  x  & Q  x  are two polynomials in x and Q  x   0 then the
quotient
P  x
is called a rational fraction.
Q  x
Example:
x2  1
is a rational fraction.
x3  2 x  3
Proper Fraction: A fraction in which the degree of the numerator is less than the degree of
denominator is called a proper fraction.
Example:
x2  1
is a proper fraction.
x3  2 x  3
Improper Fraction: A fraction in which the degree of the numerator is greater or equal to the
degree of denominator is called an improper fraction.
Example:
x2  1
x3  1
are improper fractions.
&
x2  2 x  3
x2  2 x  3
Partial Fraction: A given fraction may be written as a sum of other fractions (called partial
fractions) whose denominator is less than the denominator of the given fraction.
Fundamental theorem: Any fraction may be written as the sum of partial fractions according the
following rules:
Case-1: When the fraction is Proper fraction:
a. When all factors are linear and different
i.e.,
f  x
 ?   ?

 x  a  x  b  x  a x  a
(1)
where the coefficients of the blank spaces cannot be zero.
28 | P a g e
NOTE: Using the Cover up method we can find the values of the blank spaces of (1).
Cover up method: This method is applicable only for linear factors.
If
f  x
A
B
then


 x  a  x  b  x  a x  b
For A: Cover  x  a  term in the denominator of the left-hand side and substitute x  a in
the remaining expression.
For B: Cover  x  b  term in the denominator of the left-hand side and substitute x  b in
the remaining expression.
b. When all factors are linear and some are repeated
i.e.,
f  x
?   ?  A 

n
n
n 1
 x  a  x  b   x  a   x  b   x  b 

B
 x  b
(2)
NOTE: Find the coefficients of the blank spaces by using Cover up method and then to find A
substitute any value for x except x   a & x   b .
c. When all factors are quadratic and different
i.e.,
f  x
Ax  B Cx  D
 2

2
2
 x  a  x  b  x  a x2  b
(3)
NOTE: To find the values of A , B , C & D multiplying both sides of (3) by  x 2  a  x 2  b 
and then substitute the appropriate values for x .
d. When all factors are quadratic and some are repeated
i.e.,
f  x
Ax  B
Cx  D C1 x  D1
 2


2
2
2
2
 x  a  x  b   x  a   x 2  b   x 2  b 
(4)
NOTE: To find the values of A , B , C , D , C1 & D1 multiplying both sides of (4) by
x
2
 a  x 2  b  and then substitute the appropriate value for x .
2
Case-2: When the fraction is improper fraction: To split an improper fraction into a partial
fraction, we will have to divide the numerator by denominator.
Example: if
29 | P a g e
3x 2  2 x  2
then
x 2  3x  2
x 2  3x  2
3x 2  3x  2
3
3x 2  9 x  6
6x  8
Since, Dividend   Divisor  Quotient   Re mainder
Rewriting the given improper fraction we get
3x 2  2 x  2
6x  8
 3 2
2
x  3x  2
x  3x  2
Now using the Cover up method anyone can solve the fraction.
5 x  11
Problem-1: Separate
Solution: We have



2 x2  x  6
5 x  11
2 x2  x  6
5 x  11
2 x 2  4 x  3x  6
5 x  11
2x  x  2  3 x  2
5 x  11
 x  2  2 x  3

3
1

x  2 2x  3
Problem-2: Separate
Solution: We have

Putting
x0
into partial fractions.
3x 2  x  2
 x  2 2 1  2 x 
into partial fractions.
3x 2  x  2
 x  2 2 1  2 x 
4
 x  2 2

1
3  A
1  2 x   x  2

in (1) we get,
1
3  A


2
2 1  2  0   0  2 
 0  2 1  2  0  0  2
3(0)2  0  2
4
1

2  4
A
or ,

 3
4
4
1
2
1
1 A
or ,    1  
2
3 2
30 | P a g e

(1)
or ,
A
1 1
 1 
2
3 2
or ,
A 6  2  3

2
6
or , A  
5
3
From (1) we get,
3x 2  x  2
4

 x  22 1  2 x   x  22
1
5

3 
3
1  2 x   x  2


4
 
 x  2
Problem-3: Separate
1  x  1  x 2 

1  x  1  x 2
3

x0
7 x
7 x
Solution: We have
Putting
1
1
5
1

 
3 1  2 x  3  x  2 

2

1  x 

Ax  B
1  x2 
in (1) we get,
70
3

1  0 1  0  1  0 

A(0)  B
1  0 
or , 7  3  B
or , B  4
Again putting
x 1
7 1
in (1) we get,

3
1  11  1 1  1
or ,
8
3 A 4
 
2 2 2
2
or , 2 
3 A 4

2
2
or ,
A 4
3
2
2
2
or ,
A 4 1

2
2
or , A  4  1
or , A   3
31 | P a g e

A(1)  B
1  1
into partial fractions.
(1)
From (1) we get,
7 x
1  x  1  x
2

3

1  x 
3



3 x  4
1  x2 
1  x 
Problem-4: Separate
Solution: We have


3x  4

1  x2 
x 1

x2  5 x2  3
x 1

x2  5 x2  3
Ax  B

into partial fractions.


Cx  D
 x 2  5   x 2  3
Multiplying both sides of (1) by


(1)
 x2  5 x2  3 we get,


x  1   Ax  B x2  3  Cx  D x2  5
or, x  1  Ax3  3Ax  Bx2  3B  Cx3  5Cx  Dx2  5D
or, x  1   A  C  x3   B  D  x2  5C  3A x  3B  5D
Equating the coefficients of like term we get,
A  C  0 ; B  D  0 ; 5C  3 A  1 ;  3B  5D  1
A  C ; B   D ; 5C  3 A  1 ;  3B  5D  1
Since
A C
so
5C  3 A  1  5C  3  C   1
or , 5C  3C  1
or , 8C  1
or , C 
Again
1
8
and
A 
1
8
B  D so 3B  5 D  1   3   D   5 D  1
or, 3D  5D  1
or , 8D  1
or , D 
From (1) we get,
32 | P a g e
1
8
and
B 
1
8

1
1 1
1
 x
x
8 8
8
 8
x2  5 x2  3
x2  5
x2  3
x 1

 
 

1 x 1
1 x 1
 
 
8 x2  3 8 x2  5


Problem-5: Separate 22x
2

 x 1
x  2x  3
Solution: We have

into partial fractions.
2 x2  x  1
x2  2 x  3
2
2
7  3x
x2  2 x  3
7  3x
2
x  3x  x  3
7  3x
x  x  3  1 x  3
2
2
7  3x
 x  3 x  1
2
1
4

x 1 x  3
2
1
4

x 1 x  3
Exercise:
x2
1. Resolve
 x  1 x  3
2. Resolve
 x  2  x  1
3. Resolve
4. Resolve
1
into partial fractions.
into partial fractions.
x
 x  2 x  12
42  19 x
 x  1  x  4
2
into partial fractions.
into partial fractions.
5. Find the decomposition of
6. Resolve
7. Resolve
8. Resolve
33 | P a g e
x2  5x  7
x2  x  2

1

x2  5 x2  3

.
into partial fractions.
6 x3  5x2  7
3x 2  2 x  1
x 4  5 x3  7
x2  5x  6
into partial fractions.
into partial fractions.
06
Polynomial Equations
Expression:
An expression is a finite combination of mathematical symbols that is well-formed according to the rules
that depend on the context.
For example: An algebraic expression can be represented as:
Fun Facts:
 An expression does not contain equal to sign or any inequalities signs.
 When we add inequality or equality sign to an expression, it becomes an equation.
 Both sides of an equation are an expression.
 In expression power of the variable is any number.
Polynomial:
A polynomial is an expression consisting of variables (also called indeterminates) and coefficients, that
involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents
of variables. For example: A polynomial of a single indeterminate, x, is x2 − 4x + 7.
Zeros:
Zeros are the values of the variables that vanishes the expression or polynomial.
For example: 1 & 3 are the zeroes of the polynomial x2 − 4x + 3.
Equation & Identity:
Equation is a mathematical statement that the values of two expressions are equal and indicated by the sign
=. Identity is also an equation but it number of roots are more than its degree. For example, the equality of
two expression x 2  4 x  3 is called an equation. On the other hand, the equality of two expression
x2  x  x  x  1 is called Identity due to it has more roots from its degree.
Roots/solutions of an equation:
The roots /solutions of an equation are the values of the variables that satisfies the equation or Identities.
For example, the equation x 2  4 x  3  0 has two roots as 1 and 3. But the identity x2  x  x  x  1 has infinitely
many roots.
Remainder theorem:
It states that the remainder of the division of a polynomial f ( x) by a linear polynomial x  r is equal to f (r ) .
For example: For the polynomial f ( x)  x2  5x  6 ,the division of the polynomial
f (3)  18 (Remainder).
34 | P a g e
f ( x)
by ( x  3) yields 18 ,so
Factor theorem:
The factor theorem states that a polynomial
has a factor ( x  k ) if and only if
f ( x)
of the polynomial. For example, the polynomial
x  4 x  3 has
2
a factor
( x  1) for
f (k )  0 where
account of
k is the root
f (1)  0
if we say
f ( x)  x  4x  3 .
Quadratic Equations:
An equation of the form ax2  bx  c  0 , a  0 is called quadratic equation because quadratic comes from
Latin quadratus which mean "square" . The constants a , b & c are called the coefficients of the equation
and may be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient
and the constant or free term.
Solution of the quadratic Equation:
General quadratic equation is ax2  bx  c  0 , a  0
Multiplying the above equation by 4a we get,
2
4a.ax 2  4a.bx  4a.c  4a.0
4a 2 x 2  4abx  4ac  0
 2ax 2  2. 2ax  b  b2  b2  4ac  0
 2ax 2  2. 2ax  b  b2  b2  4ac
 2ax  b 2  b2  4ac
 2ax  b   
b 2  4ac
2ax  b  b2  4ac
x
b  b2  4ac
2a
(Here a must be positive)
Discriminant: Discriminant is a function of the coefficients of a polynomial equation whose value gives
information about the roots of the polynomial.
Here discriminant, D  b 2  4ac
Nature of the roots are:
Cubic Equation:
Rene de cartes sign rules
Mathematical problems
1.
Solve the equation x 2  5 x  6  0 by using factoring Method
Solution:
Factorization Method:
We have, x 2  5 x  6  0
x2  3x  2 x  6  0
x( x  3)  2  x  3  0
( x  2)  x  3  0
or
Therefore x  2  0
or
x  3
So x  2
2nd Method:
We have, x 2  5 x  6  0
35 | P a g e
x3 0
x
5  25  4.1.6 5  25  24 5  1 5  1



2.1
2
2
2
Taking(+ve) we get x  2 and taking (-ve)
2.
x  3 .
(Ans)
Solve the equation x  3 x  3 x  1  0 using Remainder theorem.
Solution: Given equation is x3  3 x 2  3 x  1  0 .
Let f ( x)  x3  3x2  3x 1 .For x  1 , f (1)  13  3.12  3.1 1  0 ,so one factor of f(x) is (x-1).
Now,
3
2
x3  3 x 2  3 x  1  0
x2 ( x 1)  2x( x  1)  1( x  1)  0


( x  1) x2  2x  1  0
2
( x 1)( x  1)  0
( x  1)( x  1)( x  1)  0
Therefore,
x 1  0
x 1
3.
x 1  0
or
x 1
or
x 1  0
or
(Ans)
x 1
or
Solve the equation 4 x  24 x  23x  18  0 having that the roots are in arithmetical progression.
Solution: We have, 4 x3  24 x 2  23x  18  0
In accordance with the question, assume that the roots are    ,  &    .
Now,
3
2
24
 3  6    2.
4
18
9
9
9
9
And              2   2    2 4   2    4   2     2  4 
4
2
2
4
4
9
5
2
2 16  9 25
   4   

 
4
4
4
2
1
9
Therefore, the roots are  , 2 &
(Ans)
2
2
        


4.




Solve the equation 3x3  26 x 2  52 x  24  0 having that the roots are in geometrical progression.
Solution: We have, 3x3  26 x 2  52 x  24  0
In accordance with the question, assume that the roots are

Now,
And
r

r
 r  
. . r  

r
, &  r
.
26 
26
1
 26
  r 
   1  r  
3
r
3
r
 3
24
  . .  8   3  8    2.
3
1
26 1
13 1  r  r 2 13
The value   2 implies 2   1  r     1  r  
  3  3r  3r 2  13r
r

or
3r  10r  3  0
or
3r 2  9r  r  3  0
or
or
3
r
3
r
3
2
3r  r  3  1 r  3  0
 r  3 3r  1  0
r  3  0 or 3r  1  0
r  3 or r 
1
3
2
3
Therefore, the roots are , 2 & 6 . (Ans)
5.
Solve the equation 2 x3  x 2  22 x  24  0 having that the roots are in the ratio of 3:4.
Solution: Given that, 2 x3  x 2  22 x  24  0
In accordance with the question, assume that the roots are 3 , 4 &  .
36 | P a g e
Now, 3  4    
And 3 .4 .  
1
2
 7   
1
2
......(i)
24
1
12 2  12   2  1   
2
2
From (i), we get
7 
1

2

1
7 3  1 1

  14 3  2   2 14 3   2  2  0
2
2
2
14    2  0
3
2
3
14 1
7 1
 1
 1  1
7 1
f    14 3   2  2 ,then f     14        2  2     2     2       2  2  2  0 .
8 4
4 4
 2
 2  2
 4 4
So,  2  1 is a one factor of f   .
Let


Therefore, 7 2  2  1  4  2  1  2  2  1  0   2  1 7 2  4  2  0


  2  1 7  4  2  0
2
 2  1  0 or 7  4  2  0
2
  
  4    4   4.7.2
1
or  
2
2.7
  
1
4  16  84
4  68
or  

 not real 
2
14
14
2
7
2
1
2
1
2
7
2
From (i)          4
3
2
Therefore, the roots of the equation are  , 2 & 4 .
6.
Solve the equation 24 x3  14 x 2  63 x  45  0 having that one root being double another.
Solution: Given equation is 24 x3  14 x 2  63 x  45  0 .
Let us consider the roots according to the question are 2 , &  .
Now, 2      
14
24
7
......(i)
12
21
 2 2  3   ......(ii)
8
 3   
2 2    2 
63
24
And 2 . .  
45
45
15
 2       
24
48
16 2
From (i) & (ii), we get
(iii)
21
21
7

7

2 2  3   3     2 2     9 2   
8
8
 12

4

7
21
 2 2    9 2  
4
8
7
21
2
   7  
4
8
 14  56 2  21
 2  8 2  3
 8 2  2  3  0

  2  
 2 2  4.8.  3
2.8

2  4  4.8.3 2  4  96 2  100 2  10



16
16
16
16
2  10 12 3
  and
16
16 4
7
From equation (i), we get    3
12
3
7
3 7 9 7  27
20
5
 
For   ,    3.   
4
12
4 12 4
12
12
3
Taking (+ve), we get   
37 | P a g e
for (-ve)  
.
2  10
8
1
 
16
16
2
.
For   
1
2
, 
7
 1  7 3 7  18 25
 3.      

12
12
12
 2  12 2
3
1
But in the equation (iii) for   ,    15   15   5 and for    ,    15   15 .
4
It is found that for   
7.
1
2
16.
9
16
9
2
3
16.
1
4
the third equation is not satisfied, so the roots are
4
3
4
,
3
2
&
5
3
. (Ans.)
From an equation whose roots are 1, 2, 3 &4.
Solution: The roots of the equations are 1, 2, 3 &4.
Therefore,  x  1 x  2  x  3 x  4   0
 x2  3x  2 x2  7x 12  0
x 4  7 x3  12 x 2  3 x3  21x 2  36 x  2 x 2  14 x  24  0
x 4  10 x3  35 x 2  50 x  24  0
8.
(Ans)
x 4  16 x3  86 x 2  176 x  105  0 whose
Solve the equation
two roots being 1 & 7.
Solution: Given equation is x 4  16 x3  86 x 2  176 x  105  0 .
Here two roots x  1& x  7 .
So  x 1 x  7  0  x2  8x  7  0
Write the given equation with the help of
x
2
x
2
 
2
 
x 2  8 x  7  0 ,we

2
get
 8x  7  8x x  8x  7  15 x  8x  7  0
 x2  8x  7 x2  8x 15  0
There other two roots are in the quadratic equation
x 2  8 x  15  0
2
x  5 x  3 x  15  0
x  x  5  3  x  5  0
 x  5 x  3  0
or
x 3 0 x  5
or
x  3.
Therefore
Finally, the four roots of the given equation are 1,3,5 &7 (Ans.)
Solve the equation 6 x 4  13x3  35 x 2  x  3  0 whose one root being 2 
Solution: Given equation is 6 x 4  13x3  35 x 2  x  3  0 .
According to the question x  2  3 .
Now, x  2   3
x 5  0
9.
Squaring the above equation, we get  x  22   
Write the given equation with the help of
6x
2
x
2


2

2
 x2  4 x  4  3  x2  4x  1  0.
x 2  4 x  1  0 ,we
 
2
.

get
 4x  1  11x x  4x  1  3 x  4x  1  0
 x2  4x 16x2 11x  3  0
Therefore x2  4x  1  0
x
3
3
  4  
or
 42  4.1.1
2.1
6x2  11x  3  0

4  16  4 4  12 4  4.3 4  2 3



 2 3
2
2
2
2
11  112  4.6.3 11  121  72 11  49 11  7



2.6
12
12
12
11  7
4
1
11  7
18
3
    and for (-ve) x 
  .
Taking(+ve) x 
12
12
3
12
12
2
1 3
Finally, the four roots of the given equation are 2  3 , 2  3 ,  & .
3 2
or
x
10. How many real, positive, negative & imaginary roots of the equation 6 x 4  13x3  35 x 2  x  3  0 have.
Solution: Let f ( x)  6x4 13x3  35x2  x  3 .
38 | P a g e
In the above function f(x) the sign of the terms are + - - - +. There are two change in the
signs, so it has two positive roots of the given equation.
Replacing x by -x in f(x) , we get f ( x)  6   x 4  13   x 3  35   x 2    x   3  6  x 4  13  x 3  35  x 2   x   3 .
In the above function f(-x) the sign of the terms are + + - + +. There are two change in the
signs, so it has two negative roots of the given equation. It has no complex root since its degree is
4. (Ans.)
11. Solve the equation
x2  6 x  9  4 x2  6 x  6
Solution: Given that
Let u  x
2
 6x  9
2
.
.
2
x  6x  9  4 x  6x  6
then the given equation reduces to u  4 u  3 .
u 2  16  u  3  16u  48
Squaring both sides, we get
u 2  16u  48  0
u 2  12u  4u  48  0
u  u  12   4  u  12   0
 u  12  u  4   0
Therefore, u  12  0 or
u40
2
x2  6 x  9  4  0
x  6x  9 12  0 or
2
x  6x  3  0 or
x  6x  5  0
x2  6x  3  0 or
x2  6 x  5  0
Therefore, x 
  6  
 6 2  4.1.  3
2.1
Taking (+ve) & (-ve) we get x 
12. Solve the equation
[Putting value of u]
2

6  36  12 6  4 3

 3 2 3
2
2
64
64
5& x 
1.
2
2
1  x2  1  x2
1  x2  1  x2
x
6  36  20 6  16 6  4


2
2
2
(Accuracy Test is highly needed.)
3
1  x2  1  x2
Solution: Given equation is
and
1  x2  1  x2
3.
1  x 2  1  x 2  3  1  x 2  1  x 2 


1  x2  1  x2  3 1  x2  3 1  x2
1  x2  3 1  x2  3 1  x2  1  x2
4 1  x2  2 1  x2
4 1  x2  2 1  x2
2 1  x2  1  x2
Squaring both-sides, we get


4 1  x2  1  x2  4  4 x2  1  x2  3  5x2  x2 
13. If  &  are the roots of the equation
2 x2  4 x  1  0
(Accuracy Test is highly needed.)
then form the equation whose roots are  2  
and  2   .
Solution: The given equation is
So,
2 x2  4 x  1  0
whose roots are  &  .
4
1
      2 &   .
2
2
Therefore, the equation whose roots are  2   and  2   is

 


x2   2     2   x   2    2    0
39 | P a g e
3
3
x .
5
5
 


x2   2   2     x   2 2   3   3    0

 

x2       2     x   2 2       3        0
2
3
1
1
x 2  4  1  2 x    8  3    0
2
4
1
1
x2  5x    5    0
2
4
4 x2  20x  1  20  2  0
(Ans.)
14. The quadratic equation x  4x 1  2k ( x  5) where k is a constant, has two equal roots. Calculate the
possible value of k.
Solution: Given equation is x2  4x 1  2k ( x  5)
4 x 2  20 x  23  0
2
x 2  4 x  1  2kx  10k
x 2  4 x  2kx  10k  1  0
x2   4  2k  x  10k  1  0
The two roots of the above equation, is equal if b2  4a c  0 .
  4  2k 2  4.1.10k  1  0
 4  2k 2  40k  4  0
16  16k  4k 2  40k  4  0
4k 2  24k  20  0
k 2  6k  5  0
k 2  5k  k  5  0
k  k  5   1 k  5   0
 k  5 k  1   0
Therefore,
k 5  0
or
k 1  0  k  5
k 1
or
(Ans.)
15. Find the values of k for which the equation 2 x  5 x  3  k  0 has two real distinct roots.
Solution: Given equation is 2 x 2  5 x  3  k  0
The roots of the equation will be distinct if b2  4a c  0
2
52  4.2. 3  k   0
25  8  3  k   0
25  24  8k  0
Therefore,
40 | P a g e
1  8k  0
8k  1
1
k
8
(Ans.)
07
Binomial Theorem
Binomial Theorem:
The Binomial Theorem is a quick way of expanding a binomial expression that has been raised to some
power. The formal expression of the Binomial Theorem is as follows:
 x  y n  nC0 xn y0  nC1xn1y  nC2 xn2 y 2  nC3xn3 y3 
 xn  nC1xn1 y  nC2 xn2 y2  nC3 xn3 y3 
 x n  nx n 1 y 
n

 r  x
r 0
n
nr
 nCr x n r y r 
 nCr xn r yr 
n(n  1) n  2 2 n(n  1)(n  2) n 3 3
x
y 
x
y 
2!
3!
yr

 nCn x n n y n
 yn
n(n  1)(n  2)(n  3)
r!
(n  r  1) n  r r
x
y 
 n
, where nCr    is a binomial coefficient.
r
 
Note:
1.
There are (n+1) terms in the expansion of  x  y n
2.
The (r+1)-th term of  x  y n is Tr 1 
3.
Pascal Triangle rules for calculating nCr is as follows:
4.
Well used expansion:
n(n  1)(n  2)(n  3)
r!
(n  r  1) n  r r
x
y
(a) 1  x 1  1  x  x2  x3   xr   
(b) 1  x 1  1  x  x2  x3   (1)r xr   
(c) 1  x 2  1  2 x  3x2  4 x3   (r  1) xr   
(d) 1  x 2  1  2 x  3x2  4 x3    1r (r  1) xr   
(r  1)(r  2) r
x  
2
r (r  1)(r  2) r
  1
x  
2
(e) 1  x 3  1  3x  6 x2  10 x3  
(f)
41 | P a g e
1  x 3  1  3x  6 x2  10 x3 
 yn
Problem 01: Expand 1  x 4 .
Solution: We have 1  x 4  1  ( x)4
 1  4( x) 
4.3
4.3.2
4.3.2.1
(  x )2 
( x)3 
(  x )4
1.2
1.2.3
1.2.3.4
(As desired)
 1  4 x  6 x 2  4 x3  x 4
3
Problem 02: Expand 1  x  2 to four terms.
Solution: We have
3
1  x  2  1    x 


3
2
(As desired)
3
3
1
 1  x  x 2  x3
2
8
16
Problem 03: Expand  2  3x  4 to four terms.

Solution: We have  2  3x  4  2 1 
 
3x  

2 
4
Here, the 10
th
16
3
4
1

 3x   4 5  3x   4 5 (6)  3x 
1    4   
  
 
16 
2!
3!
 2
 2
 2

3

1 
45 2 135 3 
x 
x 
1  6 x 
16 
2
2

2


1  3x 
1  
16 
2

Problem 04: Find the 10th term of 1  3a 2
Solution: We have 1  3a 2

(As desired)
16
3 .
.
 16  16  16

   1  2 
3  3
3



term, T10  T91 
9!
 16  13  10 
   
9  3  3  3 
3 
9!
 16

  9  1 16
 3
 1 9 3a 2 9
 3
 
 8
 
 3  a18
16.13.10.7.4.1.( 2).( 5).( 8) 18
 3 9
a
9
3  9!
16.13.10.7.4.1.(2).( 5).( 8) 18

a
9!
1040 18

a
81
(As desired)
Problem 05: Find the general term in the expansion of
Solution: We have

1
3 1  3x
1
 1  3x  2
1
3 1  3x
.
.
Here, the (r+1)th term,
 1  1  1
 1 
     1   2    3 
3  3  3

 3 
Tr 1 
r!
 1  4  7  10 
        
3  3  3  3 
  3  
r!
r
  3 
r
42 | P a g e
 1r 
1  4  7  10 
   
 3  3  3  3 
r!
 1

   r  1
 3
 1 n  r 3x r
  
 3r  2 


3 
r

  x
 3r  2 


 3  x r
 




  3 
r

3r  r !
1.4.7.10...  3r  2 
r!
x
Problem 06: Expand
Solution: We have
1.4.7.10...  3r  2 
  x
r
(As desired)
xr
.
x2  5x  6
x
x
x
x
3
2
3
2







x 2  5 x  6 x 2  3 x  2 x  6 x  x  3  2  x  3  x  3 x  2  x  3 x  2 x  3 x  2
 3  x  3
 x
 1  
 2

1
1
 2  x  2
 x
 1  
 3
x 5 x 2 19 x 2



6 36
216
1
1
  x 
 3 3 1   
  3 


1
 x
  1  
 3
1
 x
 1  
 2
  x  x  2  x 3
  1        
  3  3  3
 
1




(As desired)



  x 
 2 2 1   
  2 
 x  x 2  x 3
 1        
 2  2  2

Problem 07: Find the term independent of x in  2 x 
Solution: We have  2 x 
1
.
9
1 

3x2 
.
9
1 

3x2 
r
 1 
1
1
 9Cr  29  r x 9  r   r x 2r  9Cr  29  r    r x 9 3 r
2
3


 3
 3x 
The (r+1) th term, Tr 1  9Cr  2 x 9  r 
(1)
According to the question the expected term is independent of x that means the term with x 0 , so from (1)
we get x 9 3r  x 0  9  3 r  0  r  3.
From the desired term is
1 1792
1
T31  T4  9C3  29  3    3 9C3  26 

27
9
3
(As desired)
Problem 08: Evaluate the followings:
(a)
3
4

i  2 j  
i 1 j 1
(b)
3  4

3
3
3
3


i

2
j

i

2

i

4

i

6

i

8

4
i

20

4
i

20
1  4  1  2  3  20  3  24  60  84
















 i 1
i 1  j 1
i

1
i

1
i

1


n
  2i  3 j 
i, j 1
(c)
(d)
n

i 1
n
5
j 2
 2i  3 j 
 
i 1
43 | P a g e
5
3
j 2
k 1
 2i  3 j  k 


 
08
Function
The mathematical term function was included to this subject about 17 century to express the dependency
of one variable on another one. The word function means doing something. Doing something means it has
a unique output. Otherwise it is not done properly. The mathematical term function’s meaning is coincident
to our practical perception because it has only unique output for single input. The term "function" was
introduced by Gottfried Leibniz, in a 1673 and Leonhard Euler introduced the familiar notation "f(x)" for the
value of a function.
Definition of function: If x and y are two variables related to one another in a such way that each values
of x determines exactly one value of y, then we say that y is a function of x and it is simply written as
y  f (x) ,where x is an independent variable (Argument) and y is a dependent variable. Values of y or f(x)
is called functional value.
For example: y  2 x  3 is an example of a function in single independent variable x and single dependent
variable y.
Domain of a function: Set of all values of x for which the given function is defined.
Mathematically, Df
 x : y  f(x)  R and x  R
Range of a function: Set of all values of y corresponding to x such that the given function is defined.
Mathematically, R f
 y : y  f(x)  R and x  D f 
Function is represented by the following ways:
44 | P a g e
1.
Graphically
2.
By table
3.
Formula
4.
Equation
5.
By description
Various Types of Functions
Even Function: A function 𝑓(𝑥) is said to be even function if 𝑓(− 𝑥) = 𝑓(𝑥).
For example: 𝑓(𝑥) = 𝑠𝑖𝑛4 𝑥 is an even function.
 Prove that 𝑓(𝑥) = 𝑠𝑖𝑛2 𝑥 × 𝑐𝑜𝑠 6 𝑥 is an even function.
Proof:
Given function is,
𝑓(𝑥) = 𝑠𝑖𝑛2 𝑥 × 𝑐𝑜𝑠 6 𝑥
𝑓(− 𝑥) = {sin(−𝑥)}2 × {cos(−𝑥)}6
𝑓(− 𝑥) = {𝑠𝑖𝑛3 𝑥} 𝑐𝑜𝑠 6 𝑥
𝑓(− 𝑥) = 𝑠𝑖𝑛3 𝑥 × 𝑐𝑜𝑠 6 𝑥
𝑓(− 𝑥) = 𝑓(𝑥)
Above relation shows that given function 𝑓(𝑥) is an even function.
Odd Function:A function 𝑓(𝑥) is said to be odd function if 𝑓(− 𝑥) = − 𝑓(𝑥).
For example: 𝑓(𝑥) = 𝑠𝑖𝑛3 𝑥 is an odd function.
 Prove that 𝑓(𝑥) = 𝑠𝑖𝑛3 𝑥 × 𝑐𝑜𝑠 6 𝑥 is an even function.
Proof:
Given function is,
𝑓(𝑥) = 𝑠𝑖𝑛3 𝑥 × 𝑐𝑜𝑠 6 𝑥
𝑓(− 𝑥) = {sin(−𝑥)}3 × {cos(−𝑥)}6
𝑓(− 𝑥) = {− 𝑠𝑖𝑛3 𝑥} 𝑐𝑜𝑠 6 𝑥
𝑓(− 𝑥) = − 𝑠𝑖𝑛3 𝑥 × 𝑐𝑜𝑠 6 𝑥
𝑓(− 𝑥) = − 𝑓(𝑥)
Above relation shows that given function 𝑓(𝑥) is an even function.
Explicit Function:A function in which the dependent variable can be written explicitly in terms of the
independent variable. For example, the following are explicit functions: y = x2 – 3, and y = log2 x.
Implicit Function:A function in which the dependent variable is not isolated on one side of the equation.
For example, the equation 𝑥 2 + 𝑥𝑦 + 𝑦 2 = 1 represents an implicit function.
Homogeneous Function:A function f(x, y) is said to be homogeneous of degree n if it is possible to
expressed as
𝑦
𝑥
𝑥
𝑦
𝑓(𝑥, 𝑦) = 𝑥 𝑛 ∅ ( ) 𝑜𝑟 𝑓(𝑥, 𝑦) = 𝑦 𝑛 ∅ ( ) .
A function is homogeneous of degree n if, when each of its arguments is multiplied by any number
t > 0, the value of the function is multiplied by 𝑡 𝑛 that is 𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡 𝑛 𝑓(𝑥, 𝑦).
 Prove that 𝑓(𝑥, 𝑦) =
45 | P a g e
𝑥 4 +𝑦 4
𝑥−𝑦
is a homogeneous function of degree 3.
Proof: Given function is,
𝑓(𝑥, 𝑦) =
𝑥 4 +𝑦 4
𝑥−𝑦
𝑦4
𝑓(𝑥, 𝑦) =
𝑓(𝑥, 𝑦) =
𝑥 4 (1+ 4 )
𝑥
𝑦
𝑥
𝑦4
𝑥 3 (1+ 4 )
𝑥
𝑦
(1− )
𝑥
𝑥(1− )
𝑦4
𝑦
𝑦
(1+ 4 )
𝑥
𝑥
𝑥
(1− )
𝑓(𝑥, 𝑦) = 𝑥 ∅ ( ) , where ∅ ( ) =
3
𝑦
𝑥
.
Therefore, the given function f(x) is a homogeneous function of degree 3.
2nd process:
Given function is,
𝑓(𝑥, 𝑦) =
𝑥 4 +𝑦 4
𝑥−𝑦
𝑥4
𝑓(𝑥, 𝑦) =
𝑓(𝑥, 𝑦) =
𝑦 4 (1+ 4 )
𝑦
𝑥
𝑦
𝑥4
𝑦 3 (1+ 4 )
𝑦
𝑥
(1− )
𝑦
𝑦(1− )
𝑥4
𝑥
𝑥
(1+ 4 )
𝑦
𝑦
𝑦
(1− )
𝑓(𝑥, 𝑦) = 𝑦 ∅ ( ), where ∅ ( ) =
3
𝑥
𝑦
.
Therefore, the given function f(x) is a homogeneous function of degree 3.
3rd Process:
Given function is,
𝑓(𝑥, 𝑦) =
𝑓(𝑡𝑥, 𝑡𝑦) =
𝑓(𝑡𝑥, 𝑡𝑦) =
𝑓(𝑡𝑥, 𝑡𝑦) =
𝑥 4 +𝑦 4
𝑥−𝑦
𝑡 4 𝑥 4 +𝑡 4 𝑦 4
𝑡𝑥−𝑡𝑦
𝑡 4 (𝑥 4 +𝑦 4 )
𝑡(𝑥−𝑦)
𝑡 3 (𝑥 4 +𝑦 4 )
(𝑥−𝑦)
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑡 3 𝑓(𝑥, 𝑦)
Therefore, the given function f(x) is a homogeneous function of degree 3.
Periodic function: A function 𝑦 = 𝑓(𝑥) is called a periodic function of period T if it satisfies the condition
𝑓(𝑥 + 𝑻) = 𝑓(𝑥) where T is least positive real number. It means that the function 𝑓(𝑥) possesses the same
value after an interval T.
Example: 𝑠𝑖𝑛𝑥 is a periodic function with period 2π.
Note:
1. The fundamental period of sinx, cosx, secx, cosecx is 2π and also tanx, cotx is π.
2. The period of a function of the form 𝑦 = (𝐴 𝒕𝒓𝒊𝒈𝒐𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝒓𝒂𝒕𝒊𝒐𝒔)𝒏 (𝐵𝑥 + 𝐶) is obtained by
dividing the fundamental period of that function by coefficient of x.
That is, 𝑝𝑒𝑟𝑖𝑜𝑑 =
46 | P a g e
𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑝𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝐵×𝑛
.
3. The period of the sum or the difference of two functions is the LCM (least common multiple) of the
individual periods.
4. LCM of fractions
𝑎
𝑏
,
𝑐
𝑑
𝑎𝑛𝑑
𝑒
𝑓
𝑖𝑠 =
𝐿𝐶𝑀 𝑜𝑓 𝑎 ,𝑐 𝑎𝑛𝑑 𝑒
𝐻𝐶𝐹 𝑜𝑓 𝑏,𝑑 𝑎𝑛𝑑 𝑓
Problem01: Find the period of the function𝑓(𝑥) = sin4 (2𝑥 + 5) + 𝑐𝑜𝑠 4 (3𝑥 + 4) .
Solution:
Given function is
𝑓(𝑥) = sin4 (2𝑥 + 5) + 𝑐𝑜𝑠 4 (4𝑥 + 3)
𝑝𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑠𝑖𝑛𝑥
Now the period of the function sin4 (2𝑥 + 5) is =
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥 ×𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑠𝑖𝑛𝑥
𝑝𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑐𝑜𝑠𝑥
Again, the period of the function 𝑐𝑜𝑠 4 (3𝑥 + 4) is =
Here, the LCM of
𝜋
2
𝑎𝑛𝑑
𝜋
6
=
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥 ×𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑐𝑜𝑠𝑥
2𝜋
𝜋
= .
2×4
2
2𝜋
𝜋
=
3×4
= .
6
𝜋
is .
6
Therefore 𝑓(𝑥) is a periodic function with period
𝜋
6
(as desired)
.
Problem02: Find the period of the function𝑓(𝑥) = sin(2𝑥 + 5) + tan(3𝑥 + 7) .
Solution:
Given function is
𝑓(𝑥) = sin(2𝑥 + 5) + tan(3𝑥 + 7)
Now, the period of the function sin(2𝑥 + 5) is =
𝑝𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑠𝑖𝑛𝑥
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥×𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑠𝑖𝑛𝑥
𝑝𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑡𝑎𝑛𝑥
Again, the period of the function tan(3𝑥 + 7) is =
=
𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥×𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡𝑎𝑛𝑥
𝜋
Therefore, the period of the function 𝑓(𝑥) is = The LCM of 𝜋 and
3
=𝜋.
2𝜋
2
=
=𝜋
𝜋
3
=
Problem 01: Find the domain and Range of the following functions:
f ( x)  2 x  5


(b) f ( x)  x  3x  2  R (c) f ( x ) 
2
Solution:
(a)
Domain: Given function is,
f ( x )  2 x  5 R , iff 2 x  5  0
2x  5
x 
5
2
5
2
Therefore the domain of the given function is D f { x:x } .
5
D f  [ , )
2
Range: Suppose y  f ( x) 
Here
y2  2x  5
2x  5  y2
47 | P a g e
2x  5 , y  0
[Squaring both sides]
3
(as desired)
Domain & Range
(a)
𝜋
2x 1
x5
2x  y2  5
y2  5
 R for all y  R .
2
Now, the range of the given function is R f  {y : y  0}  {y : y  R}
x
R f  [0,  )  ( , )
R f  [0,  )
(Ans)
Note: The range, most of the positive square root function is R f  [0,  ) .
(b)
Domain: Given Rational function is,
f ( x) 
2x 1
 R iff x  5  0
x5
 x5
That means given function f ( x) produces real output for all real values of x accept x  5 .
Therefore, the domain of the given function f ( x) is,
Df  R  5
For Range: Suppose
y  f ( x) 
2x 1
x 5
xy  5 y  2x  1
2x 1
x5
 y
xy  2x  5 y  1
x  y  2  5 y  1
5y 1
x
R
y2
It shows
if y  2  0  y  2
x produces real values for all values of y accept y  2 .
Therefore the domain of the given function f ( x) is,
R f  R  2
(c)
Domain:Given function is,
f ( x)   x 2  3x  2   R if and if x  R .
That means x takes all real values from the real number set R and make sense.
Therefore the domain of this function is the real number set.
So, the domain of the given function is, D f  R .
Range:Given function is,
48 | P a g e
y  f ( x)  x 2  3 x  2
[Say]
y  x  3x  2
2
x2  3x   2  y   0
In the above equation the value of x will be real if and only if
32  4.1.(2  y )  0
9  4(2  y)  0
9 8  4y  0
1 4 y  0
4 y  1
1
y
4
Therefore, the range of the given function is R f  [ 
09
1
,  ) . (Ans)
4
Differentiation
If y and x are two variables related to one another, then the rate of change of y in terms of x is denoted by
dy
dx
and is calculated by the first principle rule
First Principle Rules:
f  x  h   f ( x)
dy
 lim h  0
dx
h
.
f  x  h   f ( x)
dy
 lim h  0
dx
h
Problem 01: Find the derivative of 𝑦 = 𝑓(𝑥) the followings by using First principle rule:
(a)
f ( x)  xn
(e)
f ( x)  cos x
49 | P a g e
(b)
(f)
f ( x)  e x
c)
f ( x)  a x
d)
f ( x)  ln x
f ( x)  sin a x
(g)
f ( x)  tan x
(h)
f ( x)  x
(i)
(j)
f ( x)  c
f ( x)  xn
(a) Given function is,
Now,
f ( x)  cg ( x)
(k)
f ( x)  sin 1 x
(l)
f ( x)  tan 1 x
.
f  x  h   f ( x)
 x  h  x n
dy
 lim h  0
 lim h  0
dx
h
h
n
n
  h 
 h
 hn
n
x n 1   n  x n
 x 1     x
1    1
x
x
 x 
n



 lim h  0
 lim h  0
 x lim h  0
h
h
h
2
n  n  1 n  2  h 3
 hn
 h  n  n  1  h 
1 n  
1    1
  
  
x
2!  x 
3!
 x
 x
 x n lim h  0 
 x n lim h  0
h
h
n  n  1 n  2   h 
 h  n  n  1  h 
n  
  
  
x
2!  x 
3!
n

 x
 x lim h  0
h
2
3

  1  n  n  1  h  n  n  1 n  2   h2 
 
h n   
 
 x3 
2!  x 2 
3!
  x 
 
n
 x lim h  0
h

 

  1  n  n  1  h  n  n  1 n  2   h2 
 
h n   
 
 x3 
2!  x 2 
3!
  x
 
 x n lim h  0 
h

 


  1  n  n  1  h  n  n  1 n  2   h2 
 
 x n lim h  0n   
 
 x3 
2!  x2 
3!
  x
 



 


1
 x n n    n x n 1
 x
(As desired)
(b) Given function is, f ( x)  ex .
f  x  h   f ( x)
dy
 lim
 lim
h 0
dx
e
x
h
 e x lim h  0h 
h h2


2! 3!
h

(c) Given function is, f ( x)  a x .
f  x  h   f ( x)
dy
 lim
 lim
h 0
dx
a
x
a
x
h 0
e xh e x
e x.e h  e x
 lim h  0
h
h
h2 h3
1 h 


e h 1
2! 3!
lim h  0
 e x lim h  0
h
h
1
h
h 0
  1
 e x lim h  0
 h h2
 e x lim h  0 1  

 2! 3!

lim h  0
h2
h3
 ln a 2   ln a 3 
2!
3!
h
2

h
2 h
 a x lim h  0 ln a   ln a  
 ln a 3 

2!
3!

 a xln a
(d) Given function is,
50 | P a g e
h

   e x


h2 h3


2! 3!
h

(As desired)
a xh a x
a x.a h  a x
 lim h  0
h
h
3
h2
2 h
3
1  h ln a   ln a    ln a  
a h 1
x
2!
3!
lim h  0
 a lim h  0
h
h
h ln a 
  1
 1





(As desired)
f ( x)  ln x
.
f  x  h   f ( x)
dy
ln( x  h)  ln x
 lim h  0
 lim h  0
 lim h  0
dx
h
h
2
3
h 1h
1 h
 h
ln 1  
      
x
x 2 x
3 x 

 lim h  0
 lim h  0
h
h
(e) Given function is,
f ( x)  cos x

 xh
 h
ln 
ln 1  

 x   lim
 x
h 0
h
h
 1 1 h 1 h2
 lim h  0 


 x 2 x 2 3 x3

(As desired)
.
f  x  h   f ( x)
dy
cos( x  h)  cos x
 lim h  0
 lim h  0
 lim h  0
dx
h
h
 2x  h 
h
h
2sin 
sin  
 sin  
 2x  h 
 2 
 2    lim
2
  lim h  0 h
h  0 sin 
h
 
 2 
 
2
 2x  0 
  sin 
  1   sin x
 2 
(f) Given function is,
 1
 
 x

(As desired)
f ( x)  sin ax
.
f  x  h   f ( x)
dy
sin a( x  h)  sin ax
sin(ax  ah)  sin ax
 lim h  0
 lim h  0
 lim h  0
 lim h  0
dx
h
h
h
 2ax  ah 
 ah 
2cos 
 sin  
2


 2 
h
 2ax  ah 
 ah 
 2ax  ah 
 ah 
 2ax  ah 
 ah 
2 cos 
2 cos 
a cos 
 sin  
 sin  
 sin  
2
2 
2
2 
2








 2 
 lim h  0
 lim h  0
 lim h  0
h
h
 ah 
 
 2 
 2ax  ah 
 a lim h  0 cos 
  lim h  0
2


(g) Given function is,
f ( x)  tan x
 ah 
sin  
 2ax  ah 
 2   a lim
  a cos ax
h  0 cos 
2
 ah 


 
 2 
(As desired)
.
sin ( x  h) sin x
sin ( x  h) cos x  sin x cos( x  h)

f  x  h   f ( x)
dy
tan( x  h)  tan x
cos ( x  h) cos x
cos ( x  h) cos x
 lim h  0
 lim h  0
 lim h  0
 lim h  0
dx
h
h
h
h
 lim h  0
 lim h  0
sin ( x  h) cos x  sin x cos( x  h )
sin ( x  h  x )
sin (h )
 lim h  0
 lim h  0
h cos ( x  h) cos x
h cos ( x  h ) cos x
h cos ( x  h ) cos x
sin (h)
1
1
1
1
 lim h  0
 lim h  0


 sec2 x
h
cos ( x  h) cos x
cos ( x  h) cos x cos ( x ) cos x cos 2 x
(h) Given function is,
f ( x)  x
.
f  x  h   f ( x)
dy
xh x
h
 lim h  0
 lim h  0
 lim h  0  lim h  0 1  1
dx
h
h
h
(i) Given function is,
f ( x)  c
f ( x)  c g ( x )
51 | P a g e
(As desired)
.
cg  x  h   cg ( x)
g  x  h   g ( x)
dy
dg
 lim h  0
 c lim h  0
c
dx
h
h
dx
(k) Given function is,
(As desired)
.
f  x  h   f ( x)
dy
c c
0
 lim h  0
 lim h  0
 lim h  0  lim h  0 0  0
dx
h
h
h
(j) Given function is,
(As desired)
f ( x)  sin 1 x .
(As desired)
sin
dy
 lim h  0
dx
Say
1
 x  h  sin1( x)
(i)
h
sin 1  x   y  x  sin y
and
sin 1  x  h   y  k  x  h  sin  y  k  .
From equation(i), we get
dy
yk  y
k
 lim h  0
 lim k  0
dx
h
sin  y  k   sin  y 
k
 lim k  0
 lim k  0
sin  y  k   sin  y 
 x  h  sin  y  k   h  0  k  0 
k
 
k
2
 lim k  0
 2y  k   k 
 2y  k   k 
2 cos 
cos 
 sin  
 sin  
 2  2
 2   2
k
 
1
1
1
1
1
2
 lim k  0
 lim k  0   



2
2
 2y  k 
 k  cos  y 
cos
y
1

sin
y
1

x2
cos 
sin  

 2 
2
(i)
Given function is,
tan
dy
 lim h  0
dx
1
f ( x)  tan 1 x .
 x  h   tan 1( x)
(i)
h
tan 1  x   y  x  tan y
Say
(As desired)
and
tan 1  x  h   y  k  x  h  tan  y  k  .
From equation(i), we get
dy
yk  y
k
 lim h  0
 lim k  0
dx
h
tan  y  k   tan  y 
 x  h  sin  y  k   h  0  k  0 
k cos  y  k  cos  y 
k
k
 lim k  0
 lim k  0
 lim k  0
sin  y  k  sin  y 
sin  y  k  cos  y   sin  y  cos  y  k 
sin  y  k  cos  y   sin  y  cos  y  k 

cos  y  k  cos  y 
cos  y  k  cos  y 
 lim k  0
k cos  y  k  cos  y 
k cos  y  k  cos  y 
k
 lim k  0
 lim k  0
 lim k  0 cos  y  k  cos  y 
sin  y  k  y 
sin  k 
sin k
 cos  y  cos  y   cos2 y 
Problem 02: Find
dy
dx
1
sec2 y

1
1  tan 2 y

1
(As desired)
1  x2
for f ( x)  sin(x2 ) using first principle rule.
Solution: Given function is f ( x)  sin(x2 ) .


2
2
sin x 2  2 xh  h2  sin  x 
f  x  h   f ( x)
sin  x  h   sin  x 
dy
 lim h  0
 lim h  0
 lim h  0
dx
h
h
h
 lim h  0


 
sin x 2  2 xh  h2  sin x 2
h
2
 x 2  2 xh  h2  x 2   x 2  2 xh  h 2  x 2 
 sin 

2cos 

 

2
2




 lim h  0
h
 2 x 2  2 xh  h2   2 xh  h2 
 2 x2  2 xh  h2   2 xh  h2 
 sin 

 sin 

2cos 
cos 

 


 

2
2
2
2

 
  lim

 
  2x  h
 lim h  0


h 0
h  2x  h
h
2
 2 xh  h2 

sin 


2
 2 x 2  2 xh  h2 

  lim
2
2
  lim h  0
 lim h  0 cos 
h  0 2 x  h   cos x 1 2x  2x cos x


h  2x  h
2


2
d
2
2 d
2
2
sin x  cos x 
x  cos x  2 x  2 x cos x 2
Note:
dx
dx
 

52 | P a g e

 
 
(As desired)
Problem 03: Find
dy
dx
for y  x using first principle rule.
Solution: Given function is f ( x)  sin(x2 ) .
f  x  h   f ( x)
dy
 lim h  0
 lim h  0
dx
h
 lim h  0
h

h
xh  x
Problem 04: Find
dy
dx

xh  x
 lim h  0
h
1

xh  x
 lim h  0

xh  x
h


xh  x
xh  x

  lim
h 0
h

1
1

x x 2 x
xhx
xh  x

(As desired)
for the following functions:
(a)
1
y  e x  2sin x  log x
2
(b)
y  x2 tan1 x
(c)
(e)
1
y  e tan x
(f)
x y  yx 1
(g)
y  a x x3 sin1 x
(d)
y
ex
1 x
x  a cos  b sin  , y  b sin 
1
2
(a) Given function is y  e x  2sin x  log x
Taking derivative with respect to x on both sides we get
dy d  x
1

  e  2sin x  log x 
dx dx 
2


 
d x
d
1 d
1
e  2  sin x  
 log x   e x  2cos x 
dx
dx
2 dx
2x
(As desired)
(b) Given function is y  x2 tan1 x
Taking derivative with respect to x on both sides we get




 
dy d 2 1
d
d 2
1
x2

x tan x  x2
tan 1 x  tan 1 x
x  x2 
 tan 1 x  2x   2x tan 1 x 
2
dx dx
dx
dx
1 x
1  x2
(As desired)
(c) Given function is y  a x x3 sin1 x .
Taking derivative with respect to x on both sides we get




 
 
dy d x 3 1
d
d 3
d x

a x sin x  a x x3
sin 1 x  a x sin 1 x
x  x3 sin 1 x
a
dx dx
dx
dx
dx

1
 a x x3   

2
 1 x

a x x3
  3x 2a x sin 1 x  a x x3 sin 1 x ln a  3x 2a x sin 1 x  a x x 3 sin 1 x ln a 

1  x2

(As desired)
ex
(d) Given function is y 
.
1 x
Taking derivative with respect to x on both sides we get
 
d x
d
1 x 
e  e x  1  x  1  x  e x  e x  0  1
 
   1  x  e x  e x  xe x
dy d  e x  
dx
dx

 

2
2


dx dx  1  x 
1  x 
1  x 
1  x 2
1  x 2
(e) Given function is y  etan
1
x
(As desired)
.
Taking derivative with respect to x on both sides we get


1
1
dy d  tan 1 x  tan 1 x d
1
etan x
 e

tan 1 x  etan x 

e
dx dx 
dx

1  x2 1  x2
53 | P a g e
(As desired)
(f) Given equation is x y  y x  1 .
Taking derivative with respect to x on both sides we get




d y
d
x  y x  1
dx
dx
d y
x  yx  0
dx
 
 
d y
d x
x 
y 0
dx
dx
d  ln x y  d  ln y x 
e
  e
0
dx 
 dx 

eln x
xy
y




x d
d
ln x y  eln y
ln y x  0
dx
dx
d
d
 y ln x   y x  x ln y   0
dx
dx
 1 dy

dy 
 1
x y  y.  ln x.   y x  x. .  ln y.1  0
dx 
 x
 y dx

 x dy

dy 
y
x y   ln x.   y x  .  ln y   0
dx 
x
 y dx

x y 1 y  x y ln x.
x y ln x.
dy
dy
 y x 1x  y x ln y  0
dx
dx

dy
dy
 y x 1x   x y 1 y  y x ln y
dx
dx

 x y ln x  y x1x  dydx    x y 1y  y x ln y 


x y 1 y  y x ln y
dy

dx
x y ln x  y x 1x


(g) Given parametric equations are,
x  a cos  b sin 
Taking derivative with respect to  on both sides we get
dx d
d
d
d

 a cos    b sin    a  cos   b  sin    a   sin    b  cos   b cos  a sin 
d d
d
d
d
And
y  b sin 
Taking derivative with respect to  on both sides we get
dy
d
d

 b sin    b  sin    b cos
d d
d
54 | P a g e
(As desired)
By chain rule:
dy dy d
1
b cos 

.
 b cos  .

dx d dx
b cos   a sin  b cos   a sin 
10
(As desired)
Integration
Integration is the process by which we can calculate an integral function or find the area of a zig-zag region. If
function then it’s derivative
dy
 f ( x)
dx
can express as the form
dy  f ( x) dx
y  f ( x) is
a
which defines the differential of the dependent
variable or the deferential of the function f ( x) .
Differential of a function:
The differential of a function is obtained by d f ( x)  f ( x) dx that differentiate the function with respect to its variable x and
multiply it also by dx.
Definition:
If
is the integral function of f ( x) then the expression  ( x)  c is called the indefinite integral of
 f (x) dx  ( x)  c .The symbol  f (x) dx stands to mean that f (x) is to be integrated with respect to x.
 ( x)
In addition, integration is the reverse process of differentiation.
55 | P a g e
f ( x) and is denoted by
Standard integrals:
1.
2.
56 | P a g e
 tan x dx  ln sec x
 cot x dx  ln sin x
3.
4.
dx
1
x
 tan 1
2
a a
a
dx
1
xa
 x2  a 2  2a ln x  a
x
2
7.
a
2
8.

9.

10.

11.
x
12.
13.
5.
6.
14.
15.
16.
17.
18.
57 | P a g e
x 
 sec x dx  ln sec x  tan x  ln tan  2  4 
 x
 cos ecx dx  ln cos ecx  cot x  ln tan  2 
dx
1
ax
 ln
2
 x 2a a  x
dx
 ln x  x 2  a 2
2
2
x a
dx
 ln x  x 2  a 2
2
2
x a
dx
x
 sin 1
a
a2  x2




dx
1
x
 sec1
a
x2  a2 a

a
 ln  x 
2

x a 

x 2  a 2 dx 
x x2  a2 a2
 ln x  x 2  a 2
2
2

x 2  a 2 dx 
x x2  a2
2
2
x a2  x2 a2
x
 sin 1

2
2
a
ax
e (a sin bx  b cos bx)
ax
 e sin bx dx 
a 2  b2
eax (a cos bx  b sin bx)
ax
 e cos bx dx 
a 2  b2
f '( x)
 f ( x) dx  ln f ( x)
f '( x)
 f ( x) dx  f ( x)
a 2  x 2 dx 
1
dx  2 x
x
19.

20.
 e  f ( x)  f '(x) dx  e
x
x
f ( x)
2
2
Mathematical Problems
1.
f '( x)
dx  ln f ( x) .
f ( x)
f '( x)
Proof: we have to prove that 
dx  ln f ( x)
f ( x)
Prove that
Now letting
So,


f ( x)  z  d f ( x)  d  z   f '( x) dx  1.dz  f '( x) dx  dz .
f '( x)
dz
dx   ln z  c
f ( x)
z
 ln f ( x)  c (As desired)
2.
 tan x dx  ln sec x .
Proof: we have to prove that  tan x dx  ln sec x
Prove that
Now,
 sin x
dx
cos x
1
  ln cos x   ln
sec x
sin x
 tan x dx   cos x dx   
   ln 1  ln sec x     0  ln sec x 
 ln sec x  c (As desired)
3.
 cot x dx  ln sin x .
Proof: we have to prove that  cot x dx  ln sin x
Prove that
Now,
4.
(As desired)
 sec x dx  ln sec x  tan x .
Proof: we have to prove that  sec x dx  ln sec x  tan x
Prove that
Now,
5.
cos x
 cot x dx   sin x dx  ln sin x  c
 sec x dx  
sec x(secx  tanx)
dx  ln sec x  tan x  c (As desired)
sec x  tan x
 cos ecx dx  ln cos ecx  cot x .
Proof: we have to prove that  cos ecx dx  ln cos ecx  cot x
Prove that
Now,
 cos ecx dx 
6. Prove that
x
dx
2
 a2

cos ecx(cos ecx  cot x)
dx  ln cos ecx  cot x  c (As desired)
cos ecx  cot x
1
xa
ln
2a x  a
Proof: we have to prove that
58 | P a g e
x
1
2
 a2
dx 
1
xa
ln
2a x  a
Now,
x
1
2
 a2
a
7. Prove that
dx 

  x  a  x  a  dx  2a  x  a dx   x  a dx
 1
1 
Now,
a
2
 x2

1
1  xa 
ln x  a  ln x  a   c 
ln
c
2a 
2a  x  a 

 x2


1 
1
2a
1

1
(As desired)
1
ax
ln
2a a  x
Proof: we have to prove that
dx

1 
 1
  2a
2a  dx



x a x a



dx
2
1
dx 
( x  a )( x  a )
a
dx
2
1
dx 
(a  x)(a  x)
 x2


1
ax
ln
2a a  x
1 
 1
 2a

2
 a  dx

a

x
a

x




1
  a  x  a  x  dx  2a  a  x dx   a  x dx
1 
1
1
1  ax 
 1

dx 
 a  x dx  2a ln a  x  ln a  x   c  2a ln a  x   c
2a   a  x
x
8. Prove that
dx
2
a
 1
1
2a


2
1 
1
 x
tan 1  
a
a

1
(As desired)
.
x
Proof: We have to show that
Let
1 
dx
2
 a2

1
 x
tan 1   .
a
a
x  a tan   d  x   d  a tan    1.dx  a sec2 x dx  dx  a sec2 x dx .
Now,
x
dx
2
 a2

a
a sec2  d
2
tan 2   a 2

a sec2  d
1 sec2  d 1
1
1
x

d    c  tan 1    c
2
a
a
a
a
a
sec 
 a  tan   1 
2
2


9. Evaluates the followings:
(a)
 xx
2

 x  1 dx
 e
(b)

x
 5a x  2 dx
(c)

x2 x 7
x
dx
Solutions:
(a) Here
 xx
2

 x  1 dx 

(b) Here
(c) Here
  x  x  x dx   x dx   x dx   x dx
3
2
x 4 x3 x 2


c
4
3
2
3
(As desired)
a
 ex  5a x  2 dx   e x dx  5 a x dx  2 dx  ex  5. ln a  2x  c (As desired)
x

x2 x 7
x
 x 2 x
7 
dx  


 dx 
 x
x
x 


 1

7 
  x2  2 
dx 

x 



2

3
59 | P a g e
2
 x
3

 2 x  14 x  c

 1 1

 x 2  2  7  dx 

x 


1
x 2 dx  2 dx  7


 1

 x 2  2  7  dx

x 


3

1
x2
dx 
 2 x  7.2 x  c
3
x
2
(As desired)
(As desired)
10. Evaluate the followings:
(a)  sin(2 x  3) dx
(e)  sin 2 ( x) dx
(i)  cos 2 ( x) dx
(m)  sin 2 (2 x  5) dx
(b)  cos2 (5 x  3) dx
(f) ln x dx

(g)  cos ( x) dx
(j) e5 x 3 dx

(k)  sin
(n) a 5 x 3 dx

(d)   cos ( x)  4sin  9 x   ln x  dx
3
3
(c ) sin ( x) dx
(h)
2

sin x  cos x
dx
(1  sin 2 x )
(l)
x
2
4
( x) dx
dx
 5x  6

(o)  cos ( x) dx
4
(p)
x
2
dx
 5x  7
Solutions:
(a) Here  sin(2 x  3) dx
1
Let 2 x  3  z  d  2 x  3  d  z    2.1  0  dx  1. dz  2 dx  dz  dx  dz
2
Now,
1
1
1
cos(2 x  3)
 sin(2 x  3) dx   sin(z) . 2 dz  2  sin(z) dz  2   cos z   c   2  c (As desired)

H.W: (j) e5 x 3 dx (n)
a
5 x 3
dx
1
1
2sin 2 ( x) dx   1  cos 2 x  dx

2
2
(b) Here  sin 2 ( x) dx 
 cos 2 A  1  2sin 2 A  2cos2 A  1
1
1
x 1 sin 2 x
x sin 2 x
dx   cos 2 x dx   .
c  
 c (As desired)

2
2
2 2
2
2
4
H.W: (b)  cos2 (5 x  3) dx (i)  cos 2 ( x) dx (m)  sin 2 (2 x  5) dx

(f ) Here  ln( x) dx
We know that  uv dx  u  v dx    du  v dx  dx
 dx

Now  ln x dx   ln x.1 dx  ln x  1 dx    d  ln x   1 dx  dx
 dx

1

1 
 ln x  dx     dx   x ln x    . x  dx
x

x 
 x ln x   dx  x ln x  x  c (As desired)


(c ) Here sin 3 ( x) dx  sin 2 x .sin x dx 
 1  cos x .sin x dx   cos x 1.   sin x  dx
2
2
Let cos x  z  d  cosx   d  z     sin x  dx  1. dz    sin x  dx  dz






Now, sin 3 ( x) dx  ( z 2  1)dz  z 2 dz  1. dz  z 2 dz  dz
60 | P a g e

z3
cos3 x
 zc 
 cos x  c (As desired)
3
3

H.W: (g) cos3 ( x) dx
(k) Here
 sin
4
( x) dx 

1
2sin 2 x

4

2
dx 


1
1
2
1  cos 2 x  dx   1  2 cos 2 x  cos 2 2 x dx

4
4

1 
1  cos 4 x 
1 
1 cos 4 x 
1  2cos 2 x 
 dx   1  2cos 2 x  
 dx

4 
2
4 
2
2 


1 3
cos 4 x 
1 3
1
1 cos 4 x
dx
  2cos 2 x 
 dx   dx   2cos 2 x dx  

4 2
2 
4 2
4
4
2

3
1
1
3x 1  sin 2 x  1  sin 4 x 
dx   cos 2 x dx   cos 4 x dx 
 
 
c

8
2
8
8 2 2  8 4 

3 x sin 2 x sin 4 x


 c (As desired)
8
4
32

H.W: (o) cos 4 ( x) dx
(h) Here

sin x  cos x
sin x  cos x
sin x  cos x
dx  
dx  
dx
(1  sin 2 x )
(sin 2 x  cos2 x  2sin x cos x)
(sin x  cos x)2

(l) Here
x
2
sin x  cos x
dx   dx  x  c (As desired)
sin x  cos x
dx
dx
dx
 2
dx  
 5x  6
x  3x  2 x  6
x  x  3 x  2  x  3

dx
1 
1 
 1
 1
 

dx   

dx

 x  3 x  2   x  2 x  3 
 x  2 x  3 
x2
 1 
 1 
 
dx   
dx  ln x  2  ln x  3  c  ln
 c (As desired)


x3
 x  2
 x  3
H.W: (d)
61 | P a g e
  cos ( x)  4sin 9 x   ln x  dx , (p)  x
2
2
dx
 5x  7
62 | P a g e