Kerala Plus Two Maths Notes

PLUS

2 NOTES
Chapter Wise

Contents
Page

1 RELATIONS AND FUNCTIONS 4

2 INVERSE TRIGONOMETRIC FUNCTIONS 7

3 MATRICES 9

4 DETERMINANTS 13

5 CONTINUITY AND DIFFERENTIABILITY 20

6 APPLICATIONS OF DERIVATIVES 28

7 INTEGRALS 33

8 APPLICATION OF INTEGRALS 40

9 DIFFERENTIAL EQUATIONS 42

10 VECTOR ALGEBRA 46

11 THREE DIMENSIONAL GEOMETRY 51

12 LINEAR PROGRAMMING 54

13 PROBABILITY 57
Tap on the chapter name for direct access

3

Chapter 1
RELATIONS AND FUNCTIONS
Relations

• Any subset of A× A is a relation from A to A

A relation R in a set A is called

(i) reflexive, if (a,a) ∈ R, for every a ∈ A

(ii) symmetric, if (a,b) ∈ R implies that (b,a) ∈ R, for all a,b ∈ A.

(iii) transitive, if (a,b) ∈ R and (b,c) ∈ R implies that (a,c) ∈ R, for all a,b,c ∈ A

Equivalence Relation

• A relation R in a set A is said to be an Equivalence relation if R is reflexive,
symmetric and transitive

Questions and Answers

1. Determine whether the following relations are reflexive ,symmetric and transitive

(a) A relation R ={(1,1),(2,2),(3,3),(1,3),(3,2)} on the set A={1,2,3}
(1,1),(2,2),(3,3) ∈ R ⇒ R is reflexive
(1,3) ∈ R ,but (3,1) ∈/ R ⇒ R is not symmetric
(1,3) ∈ R and (3,2) ∈ R but (1,2) ∈ / R ⇒ R is not transitive
(b) A relation in the set A = {1,2,3,. . . . . .,13,14 } defined as R = {(x, y) : 3x − y = 0}
Here R={(1,3),(2,6),(3,9),(4,12)}
(1,1) ∈
/ R ,hence R is not reflexive
(1,3)∈ R but (3,1)∈
/ R ⇒ R is not symmetric
(1,3) ∈ R and (3,9) ∈ R but (1,9) ∈ / R ⇒ R is not transitive
(c) A relation in the set Z of integers defined as R ={(x, y) : x − y is an integer}
Since x - x = 0 is an integer, (x,x)∈ R for every x ∈ Z .So R is Reflexive
(x,y)∈ R ⇒ x-y is an integer
⇒ y-x is also an integer
⇒(y,x)∈ R. So R is Symmetric

Suppose (x,y)∈ R and (y,z)∈ R ⇒ x-y and y-z are integers
⇒ (x-y)+(y-z) is also an integer
⇒ x-z is an integer
⇒ (x,z)∈ R. So R is Transitive
Note:- R is reflexive ,symmetric and transitive and hence it is an equivalence rela-
tion.
4

Practice Problems

1. Show that the relation R in the set {1,2,3} given by R ={(1,1),(2,2),(3,3),(1,2),(2,3)} is
reflexive but neither symmetric nor transitive.

2. Show that the relation R in the set Z of integers given by R = {(a,b): 2 divides a−b}
is an equivalence relation

Types of Functions
One-One(Injective) function:-
A function f : A → B is said to be one-one if different elements have different
images.
That is, x1 6= x2 ⇒ f (x1 ) 6= f (x2 ), ie

f (x1 )=f (x2 ) ⇒ x1 =x2

Questions and Answers

(1) Check the injectivity (one-one) of the following functions

(a) f : R → R given by (b) f : R → R given by f(x)=3-4x.
f(x)=2x. Answer :-
Answer :-
f (x1 ) = f (x2 ) ⇒ 3 − 4x1 = 3 − 4x2
f (x1 ) = f (x2 ) ⇒ 2x1 = 2x2 ⇒ −4x1 = −4x2
⇒ x 1 = x2 ⇒ x1 = x2
∴ f is one-one
∴ f is one-one

(c) f : R → R given by f(x)=1+x2 . (d) Let A={1,2,3}, B={4,5,6,7}
and let f={(1,4),(2,5),(3,6)}
f (x1 ) = f (x2 )
1 + (x1 )2 = 1 + (x2 )2 be a function f : A→ B.
(x1 )2 = (x2 )2 Answer :-
(x1 )2 − (x2 )2 = 0 Every element in A is mapped to
(x1 − x2 )(x1 + x2 ) = 0 different elements in B.
x1 = x2 or x1 = −x2
Therefore f is one-one
∴ f is not one one

Practice Problems

Check whether the following functions are one-one or not

1. f : N → N given by f(x)=2x+5.

2. f : R → R given by f(x)=x2 .

5

x
3. f : [-1,1] → R given by f(x)=
x+2

Onto (Surjective) Functions :-

A function f : A → B is said to be Onto or Surjective, if every element of B
is image of some elements of A under f.
That is if for every element y ∈ B there exist an element x in A such that f(x)=y.

Write x in terms of y. That is x = g(y)

Questions and Answers

1. Check whether thefollowing functions f : R → R are onto or not

(b) f(x) = 4x+7
(a) f(x) = 3x Let y ∈ R (Codomain)
Let y ∈ R (Codomain)
f (x) = y
f (x) = y 4x + 7 = y
3x = y 4x = y − 7
y y−7
x = ∈ R(Domain) x= ∈ R(Domain)
3 4
Therefore corresponding to every Therefore corresponding to every
y ∈R there exist a real number y ∈R there exist a real number
y y−7
∈ R ⇒f is Onto ∈ R ⇒ f is Onto
3 4

(c) f(x)= x2
Let y ∈ R (Codomain)
f (x) = y ⇒ x2 = y
√
x=± y
√
If y = -1, there cannot exist a real number satisfying x2 = -1 or x = −1. Therefore
f is not onto

Practice Problems

1. Check whether the following functions are onto or not.

1
a. f : R+ → R+ defined by f(x) = where R+ is the set of all positive real numbers.
x
b. f : R →R defined by f(x) = 5x-3.

c. f : N → N defined by f(x) = x2

6

Chapter 2

INVERSE TRIGONOMETRIC
FUNCTIONS

Principal Value :-

The value of an inverse trigonometric function which lies in the range of principal
branch is called the principal value of that inverse trigonometric function

If y=sin−1 x then x=sin y

function Domain Principal Value  Branch
sin−1 : [−1, 1] → π π
−2, 2
cos−1 : [−1, 1] →  π[0,ππ]
cosec−1

: R-(−1, 1) → − 2 , 2 -{0}
sec−1 : R-(−1, 1) → [0, π]-{ π2
}
tan−1 : R → − π2 , π2
cot−1 : R → (0, π)

Questions and Answers

(1) Find the principal values of the following.
√
(a) . sin−1 ( √12 ) (b) cos−1 ( 23 )

Answers :-

  √ √
(a)sin −1 √1 =x ⇒ sin x = √12 (b) cos−1 ( 23 ) = x ⇒ cos x = 23
2
We have We have
√
π 1 π 3
sin( ) = √ cos( ) =
4 2 √6 2
1 π h π πi 3 π
∴ sin−1 ( √ ) = ∈ − , ∴ cos−1 ( ) = ∈ [0, π]

2 4 2 2 2 6

Principal value = π4 Principal value = π6

7

(2) Find the value of tan−1 (1) + cos−1 −1
+ sin−1 −1



2 2

π π
= 1 ⇒ tan−1 (1) =
tan (1)
4 4  
π 1 π 1 −1 −1 π
cos = ⇒ cos(π − ) = − ⇒ cos =π− (2)
3 2 3 2 2 3
π 1 π 1
sin = ⇒ sin(− ) = −
 6 2 6 2
−1 π
sin−1 =− (3)
2 6
(1) + (2) + (3) gives
   
−1 −1 −1 −1 −1 π π π 18π 3π
tan (1) + cos + sin = +π− − = =
2 2 4 3 6 24 4

(3) Prove that
√ 1 1
(a) sin−1 2x 1 − x2 = 2 sin−1 x , - √ ⩽ x ⩽ √


2 2
√ 1
(b) sin−1 2x 1 − x2 = 2 cos−1 x ,


√ ⩽x⩽1
2
(a) (b)

Put x = sin y, then sin−1 x = y Put x = cos y, then cos−1 x = y
 √   √ 
−1 2 −1 2
sin 2x 1 − x sin 2x 1 − x
 q   p 
= sin−1 2
2 sin y 1 − sin y = sin−1 2 cos y 1 − cos2 y
 q 
−1 2
 
= sin 2 cos y sin y
p
= sin−1 2 sin y cos2 y
= sin−1 (2 sin y cos y) = sin−1 (2 cos y sin y)
= sin−1 (sin 2y) = sin−1 (sin 2y)
= 2y = 2y
= 2 sin−1 x = 2 cos−1 x

Practice Problems

(1) Find the principal values of the following

(a) sin−1 ( 12 ) (b) tan−1 ( √13 ) (c) cosec−1 (2) (d) cos−1 (− 21 )

(2) Find the values of the following
√
−1 1 −1 1 (b) tan−1 ( 3) + sec−1 (−2)
(a) cos + 2 sin
2 2

8

Chapter 3

MATRICES

Definition
A matrix is an ordered rectangualar array of numbers or functions.
A matrix having m rows and n columns is called a matrix of order m×n
(read as m by n)

 
a11 a12 · · · a1n

 a21 a22 · · · a2n 

.. .. .. ..
 
 
 . . . . 
am1 am2 · · · amn

[aij ]m×n represents a m × n general
 matrix
2 3 4
For example let A =
5 2 1
Then matrix A has 2 rows and 3 columns .Hence A is a matrix of order 2 × 3

Practice Problems

1. Write
 theorder of the following matrices. 
5 2   6
(a) 1 4
  (b) 3 −2 1 (c) −4

3 2 2

Types Matrices

• Row matrix:A matrix having only one row
• Column matrix: A matrix having only one column
• Square matrix : A matrix in which number of rows and number columns are
equal
• Identity matrix:In a square matrix ,if all diagonal elements are 1 and rest are
all 0.  
  1 0 0
  1 0
For example I1 = 1 I2 = I3 = 0 1 0

0 1
0 0 1

Questions and Answers

1. If a matrix has 18 elements ,what are the possible orders it can have?
Ans: 1 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, 18 × 1
9

2. Constructa 2 × 2 matrix
 whose elements are given by aij = i + 2j
a11 a12
Ans: A=
a21 a22
a11 = 1 + 2 × 1 = 3 ,a12 = 1 + 2 × 2 = 5
a21 = 2 + 2 × 1 = 4, a22 = 2 + 2 × 2 = 6
 
3 5
A=
4 6
3. Find
 thevalue of x, y,
 z, w satisfying
 the matrix equation
x z 1 −1 3 5
2 +3 =3
y w 0 0 4 6
     
2x 2z 3 −3 9 15
Ans: + =
2y 2w 0 0 12 18
2x + 3 = 9 2z − 3 = 15 2y = 12 2w = 18
Solving we get x = 3, z = 9, y = 6 and w = 9
   
5 2 3 6
4. Find X and Y if X+Y = and X−Y=
0 9 0 −1
     
5 2 3 6 8 8
X +Y +X −Y = + ⇒ 2X =
0 9 0 −1 0 8
   
1 8 8 4 4
X= =
2 0 8 0 4
     
5 2 3 6 2 −4
(X + Y ) − (X − Y ) = − ⇒ 2Y =
0 9 0 −1 0 10
   
1 2 −4 1 −2
Y = =
2 0 10 0 5

Practice Problems

1. If a matrix has 12 elements what are the possibe orders it can have
(i + 2j)2
2. Construct a 2 × 2 matrix [aij ] whose elements are given by aij =
2
   
a−b d −1 4
3. If = ,find a,b,c and d
2a − b c 0 5

Multiplication of matrices
The product of two matrices A and B is defined if number of columns of A is equal
to the number of rows of B
In general AB6= BA
If A is m × n matrix and B is n × p matrix ,then matrix AB is of order m × p

Questions and Answers
   
2 4 1 3
1. If A = and B = ,verify that AB 6= BA
 3 2    −2 5 
2 4 1 3 2 × 1 + 4 × (−2) 2 × 3 + 4 × 5
AB= =
3 2 −2 5 3 × 1 + 2 × (−2) 3 × 3 + 2 × 5
10

 
2 − 8 6 + 20
=
3 − 4 9 + 10
−6 26
AB =
    −1 19 
1 3 2 4 1×2+3×3 1×4+3×2
BA= =
−2 5 3 2 −2 × 2 + 5 × 3 −2  ×4+5×2
2+9 4+6
=
 −4 + 15
 −8 + 10
11 10
BA=
11 2
AB 6= BA

Transpose of a matrix
The transpose of a matrix A is obtained by interchanging the rows and columns
of A
 
2 3 4
2. Find the transpose of the matrix A =
 5 1 2
2 5
0 
A = 3 1
4 2

• Symmetric matrix:
A square matrix A is said to be symmetric if A0 =A

• Skew symmetric matrix:
A square matrix A is said to be skew symmetric if A0 = −A
 
4 5 2
3. Show that the matrix A = 5 3 1 is a symmetric matrix
2 1 4
 
4 5 2
A0 = 5 3 1
2 1 4
0
A =A,hence A is symmetric
 
0 2 3
4. Show that the matrix A = −2 0 5 is a skew symmetric matrix
−3 −5 0   
0 −2 −3 0 2 3
A0 = 2 0 −5= − −2 0 5
3 5 0 −3 −5 0
A0 = − A,hence A is skew symmetric

Any square matrix can be expressed as the sum of a symmetric and a skew

symmetric matrix
1 1
• A= (A+A0 )+ (A−A0 )
2 2
1 1
where (A+A )is symmetric and (A−A0 ) is skew symmetric
0
2 2

11

 
3 2 4
5. Express the matrix A =  5 1 6 as the sum of a symmetric and a skew sym-
−2 3 5
metric matrix
   
3 2 4 3 5 −2
A =  5 1 6 A0 = 2 1 3 
 −2 3 5 4 6 5 
6 7 2 0 −3 6
A+ A0 = 7 2 9  A− A0 =  3 0 3
2 9 10 −6 −3 0
 
6 7 2
1 1
(A+ A0 ) = 7 2 9  is symmetric
2 2
2 9 10 
0 −3 6
1 1
(A− A0 ) =  3 0 3is skew symmetric
2 2
−6 −3 0
1 1
A = (A+A0 )+ (A − A0 )
2 2
6. The diagonal elements of a skew symmetric matrix are . . . . . .
Answer: Zero
7. If A and B are symmetric matrices of the same order ,then
AB−BA is a . . . . . .
Answer: Skew symmetric matrix.

Practice Problems
 
1 2  
−1 4 −5
1. If A = 3 −1 and B= ,show that AB6= BA
2 1 0
4 2

2. Findthe transpose of the following matrices

  1 2
1 3 2
(a) (b) 3 0
4 0 8
6 1
3. Express the folowing
 matrix as the sum of a symmetric and a skew symmetric matrix
−2 5 3
 4 6 7
2 1 5
   
2 −6 5 −3
4. Consider the matrices A = and A+3B=
1 2 −2 −1
(a) Find the matrix B
(b) Find the matrix AB
(c) Find the transpose of B

12

Chapter 4

DETERMINANTS
Determinant
To every square matrix A= [aij ] of order n ,we can associate a number called determinant
of the square matrix A.
 
a b a b
If A = ,then determinant of A is written as |A| = = ad − bc
c d c d

Questions and Answers

2 4
1. Evaluate the detrminant
−5 1
2 4
= 2 × 1 − 4 × (−5) = 22
−5 1
cos θ − sin θ
2. Evaluate
sin θ cos θ
cos θ − sin θ
= cos θ × cos θ − (− sin θ) × sin θ
sin θ cos θ
= cos2 θ + sin2 θ = 1
3 x 3 2
3. Find the values of x for which =
x 1 4 1
3 x 3 2
=
x 1 4 1
2
3−x =3−8
x2 = 8√
x=± 8
1 1 −2
4. Evaluate 2 1 −3
5 4 −9
1 1 −2
1 −3 2 −3 2 1
2 1 −3 = 1 × −1× + (−2) ×
4 −9 5 −9 5 4
5 4 −9
= 1 (1 × (−9) − (−3) × 4) − 1 (2 × (−9) − (−3) × 5) − 2 (2 × 4 − 1 × 5)

= 1(−9 + 12) − 1(−18 + 15) − 2(8 − 5) = 3 + 3 − 6 = 0
 
1 2
5. If A = , then show that |2A| = 4|A|
4 2

13

   
1 2 2 4
2A = 2 =
4 2 8 4
2 4
|2A| = = 2 × 4 − 4 × 8 = 8 − 32 = −24 ((1))
8 4
1 2
|A| = = 1 × 2 − 2 × 4 = 2 − 8 = −6
4 2
4|A| = 4 × −6 = −24 ((2)
∴ from (1) and (2) we get |2A| = 4|A|

Practice Problems
x 2 6 2
1. Find the values of x for which =
18 x 18 6
3 −1 −2
2. Evalauate 0 0 2
3 −5 0
 
1 0 1
3. If A = 0 1 2, show that |3A| = 27|A|
0 0 4

Area of a triangle
Area of a triangle with vertices (x1 , y1 ), (x2 , y2 ) and (x3 , y3 ) are given by

x y 1
1 1 1
x2 y 2 1
2
x3 y 3 1
Since area is a positive quantity, we always take the absolute value of the determi-
nant

(1) Find the area of the triangle whose vertices are (3, 8), ( 4, 2) and (5, 1)

3 8 1
1
Area = −4 2 1
2
5 1 1
 
1 2 1 −4 1 −4 2
= 3 −8 +1
2 1 1 5 1 5 1
1
= [3(2 − 1) − 8(−4 − 5) + (−4 − 10)]
2
1

= × (3 + 72 − 14)
2
1
= × 61
2

(2) If area of triangle is 35 sq units with vertices (2,- 6), (5, 4) and (k, 4).
Find the value of k
14

2 −6 1
1
Area = 5 4 1
2
k 4 1
2 −6 1
±35 = 5 4 1
k 4 1
 
1 4 1 5 1 5 4
= 2 +6 +1
2 4 1 k 1 k 4
1
= [2(4 − 4) + 6(5 − k) + 1(20 − 4k)]
2
1
= × (30 − 6k + 20 − 4k)
2
1
±35 = (50 − 10k)
2
±70 = 50 − 10k

70 − 50 = −10k or −70 − 50 = −10k
20 = −10k or −120 = −10k ⇒ k = −2 or 12

Minors and Cofactors of a determinant
Minor of an element aij of a determinant is the determinant obtained by deleting
its ith row and j th column in which element aij lies
Minor of an element aij is denoted by Mij
Cofactor of an element aij , denoted by Aij is defined by
Aij = (−1)i+j Mij , where Mij is minor of aij

Adjoint of a Matrix

The adjoint of a square matrix A is defined as the transpose of the
cofactor matrix of A.
Remark
 
a11 a12
Let A =
a21 a22

 
2 3
1. Find adj A for A =

1 4 
4 −3
adj A =
−1 2

15

Singular and Nonsingular matrices

• A square matrix A is said to be singular if |A| = 0

• A square matrix A is said to be non singular if |A| =
6 0

 
1 2
2. Show that A= is a singular matrix
4 8
1 2
|A|= =1×8−2×4=0
4 8
|A| = 0, hence A is a singular matrix

Inverse of a matrix
If A is a non singular matrix,then inverse of the matrix A is given by
1
A−1 = adjA
|A|
 
2 5
3. Find the inverse of the matrix A =
3 2
2 5
|A| = = 2 × 2 − 5 × 3 = −11
3 2 
2 −5
adj A =
−3 2
1
A−1 = adjA
|A|    
1 2 −5 1 2 −5
= =−
−11 −3 2 11 −3 2
 
3 −4
4. Find the inverse of the matrix A =
2 −1
3 −4
|A| = = 3 × (−1) − (−4) × 2 = 5
2 −1 
−1 4
adj A =
−2 3
1
A−1 = adjA
|A|
 
1 −1 4
=
5 −2 3
 
1 3 3
5. Find A−1 if A = 1 4 3
1 3 4

1 3 3
4 3 1 3 1 4
|A| = 1 4 3 = 1 −3× +3×
3 4 1 4 1 3
1 3 4
= 1(16 − 9) − 3(4 − 3) + 3(3 − 4) = 7 − 3 − 3 = 1
1
A−1 = adjA
|A|

16

Minors
4 3 1 3 1 4
M11 = =7 M12 = =1 M13 = = −1
3 4 1 4 1 3
3 3 1 3 1 3
M21 = =3 M22 = =1 M23 = =0
3 4 1 4 1 3
3 3 1 3 1 3
M31 = = −3 M32 = =0 M33 = =1
4 3 1 3 1 4
Cofactors
A11 = M11 =7 A12 = -M12 = −1 A13 = M13 = −1
A21 = -M21 = −3 A22 =M22 =1 A23 = -M23 =0
A31 =M31 = −3 A32 = -M 32 =0  A33 = M33 =1
7 −1 −1
Cofactor matrix of A = −3 1 0
−3 0 1
 
7 −3 −3
adj A= −1 1 0
 −1 0  1  
7 −3 −3 7 −3 −3
1 1
A−1 = adjA = −1 1 0  = −1 1 0
|A| 1
−1 0 1 −1 0 1
 
3 1
6. If A = , show that A2 -5A+7I=0.Hence find A−1
−1 2  
2 3 1 3 1
A =
 −1 2 −1 2   
3 + 1 × (−1) 3+1×2 8 5
= =
−1 + 2 ×(−1)
 −1 +2 × 2 −5 3
3 1 15 5
5A =5 =
 −1 2  −5
 10
1 0 7 0
7I=7 =
0 1  0  7     
2 8 5 15 5 7 0 0 0
A -5A+7I = − + = =0
−5 3 −5 10 0 7 0 0

A2 − 5A + 7I = 0
7I = 5A − A2
7IA−1 = 5AA−1 − A2 A−1
∴ 7A−1 = 5I − AI
1
orA−1 = (5I − A)
7

     
−1 1 5 0 3 1 1 2 −1
A = − =
7 0 5 −1 2 7 1 3

17

Applications of determinants and matrices

Consider the system of equations
a1 x + b 1 y + c 1 z = d 1
a2 x + b 2 y + c 2 z = d 2
a3 x + b 3 y + c 3 z = d 3
Matrix form of the equations is AX=B
     
a1 b 1 c 1 x d1
where A = a2 b2 c2  , X =y , B=d2 
a3 b 3 c 3 z d3
6 0 solution of the system of equations is X=A−1 B
If |A| =
 
2 5
7. Let A =
3 2
(a) Find adj A
(b) Find A−1
(c) Using A−1 solve the system of equations
2x + 5y = 1
3x + 2y = 7
Answer:  
2 −5
(a) adj A =
−3 2
1
(b) A−1 = adjA
|A|
2 5
|A| = = 2 × 2 − 5 × 3 = −11
3 2    
−1 1 2 −5 1 2 −5
A = =−
−11 −3 2 11 −3 2
−1
(c)Since |A| =6 0solutionis X=A B  
1 2 −5 1 1 2 × 1 + (−5) × 7
X=− =−
11 −3
   2 7 11 −3 × 1 + 2 × 7
1 −33 3
=− =
11 11 −1
Solution is x = 3, y = −1
8. Solve the system of linear equations using matrix method
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
Answer:The
 above
 equations
  canbe written
 in matrix form as AX = B
2 3 3 x 5
A =1 −2 1  X = y  B = −4

3 −1 −2 z 3
2 3 3
|A| = 1 −2 1 = 2(4 + 1) − 3(−2 − 3) + 3(−1 + 6) = 40
3 −1 −2
−2 1 1 1 1 −2
M11 = =5 M12 = = −5 M13 = =5
−1 −2 3 −2 3 −1

18

3 3 2 3 2 3
M21 = = −3 M22 = = −13 M23 = = −11
−1 −2 3 −2 3 −1
3 3 2 3 2 3
M31 = =9 M32 = = −1 M33 = = −7
−2 1 1 1 1 −2
A11 =M11 =5 A12 =-M12 =5 A13 =M13 =5
A21 =-M21 =3 A22 =M22 =-13 A23 =-M23 =11
A31 =M31 =9 A32 =-M32 =1 A33 =M331 =-7
 
5 5 5
Cofactor matrix of A = 3 −13 11
9 1 7
 
5 3 9
adj A= 5 −13 1 
5 11 −7
 
5 3 9
1 1 
A−1 = adjA = 5 −13 1 
|A| 40
5 11 −7
X=A−1 B
  
5 3 9 5
1 
X= 5 −13 1   −4
40
5 11 −7 3     
5 × 5 + 3 × (−4) + 9 × 3 40 1
1  1 
X= 5 × 5 + (−13) × (−4) + 1 × 3 = 80  =  2 
40 40
5 × 5 + 11 × (−4) + (−7) × 3 −40 −1
Solution is x = 1, y = 2, z = −1

Practice Problems

1. Let 5x + 2y = 3
3x + 2y = 5
(a) Write the above system of equations in the form AX=B
(b) Find |A|
(c) Find A−1
(d) Solve the system of equations
2. Consider the system of equations
5x − 6y + 4z = 15
7x + 4y − 3z = 19
2x + y + 6z = 46
(a) Express the system in the form AX=B
(b) Find A−1

(c) Hence solve the system

19

Chapter 5
CONTINUITY AND
DIFFERENTIABILITY

Continuity

• Continuity at a point :-
f is a real function on a subset of the real numbers and let c be a point in
the domain of f. Then f is continuous at c if lim f (x) = f (c).
x→c

lim f (x) = lim+ f (x) = f (c).
x→c− x→c
Continuous Function :- A real function f is said to be continuous if it is contin-
uous at every point in the domain of f.

Questions and Answers

1. (a) Examine the continuity of the function f (x) = 2x + 3 at x=1.
(b) Examine the continuity of the function f (x) = 2x2 − 1 at x=3.

Answers :-

(a) (b)

lim f (x) = lim 2x + 3 lim f (x) = lim 2x2 − 1
x→1 x→1 x→3 x→3
=2×1+3 = 2 × 32 − 1
=5 = 17
f (1) = 2 × 1 + 3 f (3) = 2 × 32 − 1
=5 = 17

That is lim f (x) = f (1) ⇒ f is con- That is lim f (x) = f (3) ⇒ f is con-
x→1 x→3
tinuous at x=1 tinuous at x=3


kx2 if x < 2
2 (a) Find the values of k so that the function f(x)= is continuous at

3 if x ≥ 2
x=2. 
2x + 3 if x ≤ 2
(b) Find all points of discontinuity of f where f is defined by f (x) =
2x − 3 if x > 2

Answers :-

20

(b) For x<2 and x>2, f(x) is a poly-
nomial function and hence it is con-
(a) Since f is continuous at x=2,
tinuous.
lim f (x) = f (2)
x→2− At x=2, lim f (x) = lim (2x − 3)
x→2+ x→2
2
lim kx = 3
x→2 =2×2−3=1
2
k×2 =3 lim− f (x) = lim (2x + 3)
x→2 x→2
4k = 3
=2×2+3=7
3
k= lim+ f (x) 6= lim− f (x)
4 x→2 x→2

∴ f(x) is discontinuous at x=2.

Practice Problems

kx + 1 if x ≤ 5
1 Find the value of k so that the function f(x)= is continuous at x=5.
3x − 5 if x > 5

2 Is the function defined by

x + 5 if x ≤ 1
f(x)=
x − 5 if x > 1

a continuous function?

3 Find all points of discontinuity of f, where

sin x
(
if x < 0
f(x)= x
x + 1 if x ≥ 0

4 Find the values of a and b such that the function defined by

 5 if x ≤ 2
f(x)= ax + b if 2 < x < 10
21 if x ≥ 10


is a continuous function

Continuity of composite functions Let f and g be real functions such that
gof is defined at x=c. If f is continuous at c and g is continuous at f(c), then gof is
continuous at c.

21

Standard Results
d d
(c) = 0, c is a constant (sin x) = cos x
dx dx
d d
(x)=1 (cos x) = − sin x
dx dx
d n d
(tan x) = sec2 x
(x ) = nxn−1 dx
dx
 
d 1 −1 d
= 2 (sec x) = sec x tan x
dx x x dx
d √ 1 d
( x) = √ (cosecx) = −cosecx cot x
dx 2 x dx
d x d
(e ) = ex (cot x) = −cosec2 x
dx dx
d d x
(logx) = x1 (a ) = ax log a
dx dx

• Product Rule :-
d d d
(f (x)g(x)) = f (x) · (g(x)) + g(x) · (f (x))
dx dx dx
• Quotient Rule :-
  g(x) · d (f (x)) − f (x) · d (g(x))
d f (x) dx dx
= 2
dx g(x) (g(x))
Derivative of Composite Functions

Function of a function or Chain rule :-

If y is a function of u and u is a function of x then
dy dy du
= ·
dx du dx

Differentiate the following functions with respect to x

(1) sin (x2 ) (2)(2x + 1)3

f (x) = sin x2 f (x) = (2x + 1)3



d d d d
sin(x2 ) (2x + 1)3


(f (x)) = (f (x)) =
dx dx dx dx
d d
= cos(x2 ) x2 = 3(2x + 1)3−1


(2x + 1)
dx dx
= cos(x2 ) × 2x = 3(2x + 1)2 × 2

= 6(2x + 1)2

22

√
(3) cos( x) (4) sin(cos x)
√
f (x) = cos x f (x) = sin (cos x)
d d √ d d
(f (x)) = cos x (f (x)) = (sin(cos x))
dx dx dx dx
√ d √ d
= − sin( x) × ( x) = cos(cos x) × (cos x)
dx dx
√ 1 = cos(cos x) × − sin x
= − sin( x) × √
2 x = − sin x × cos(cos x)

(5) Find the derivative of sin (ax + b) w.r.t. x
Answer :-

y = sin (ax + b)
dy d
= (sin (ax + b))
dx dx
d
= cos (ax + b) × (ax + b)
dx
= cos (ax + b) × a
= a cos (ax + b)

Practice Problems

1. Show that the following functions are continuous
a)f (x) = cos x2 b)f (x) = | cos x| c)f(x)=sin |x|

2 Differentiate
√ the following √
a) tan x b)cos(sin x) c)cos (sin x) d)tan ( x) e)sin2 x

Derivative of implicit function

When a relationship between x and y which cannot be easily expressed in the
form of y = f(x),then y is called an implicit function of x

Questions and Answers

dy dy
(1)Find , if x2 + y 2 = a2 (2) Find if y + sin y = cos x
d dx d 2 d
dx
d
(x2 + y 2 ) = (a ) (y + sin y) = (cos x)
dx dx dx dx

dy dy dy
2x + 2y =0 + cos y = − sin x
dx dx dx
dy dy
2y = −2x (1 + cos y) = − sin x
dx dx
dy 2x dy sin x
=− =−
dx 2y dx (1 + cos y)
23

dy dy
(3)Find , if 2x+3y = sin y (4) Find if ax + by 2 = cos y
dx dx
Answer :- Answer :-
d d d d
ax + by 2 =


(2x + 3y) = (sin y) (cos y)
dx dx dx dx
d d dy d dy
y 2 = − sin y ·


(2x) + (3y) = cos y · a+b·
dx dx dx dx dx
dy dy dy dy
2+3· = cos y · a + b · 2y · = − sin y ·
dx dx dx dx
dy dy dy dy
3· − cos y = −2 2by · + sin y · = −a
dx dx dx dx
dy dy
(3 − cos y) = −2 (2by + sin y) = −a
dx dx
dy −2 dy −a
= =
dx 3 − cos y dx 2by + sin y

Practice Problems

dy
1. Find in the following
dx
(a) x2 + xy + y 2 = 100 (b) xy + y 2 = tan x + y (c) sin2 x + cos2 y = 1

Derivatives of Inverse Trigonometric Functions

Standard Results
d 1 d −1
sin−1 x = √ (cos−1 x) = √


dx 1 − x2 dx 1 − x2
d 1 d −1
(tan−1 x) = (cot−1 x) =
dx 1 + x2 dx 1 + x2
d 1 d −1
(sec−1 x) = √ (cosec−1 x) = √
dx x x2 − 1 dx x x2 − 1

Questions and Answers

dy dy
(1) Find if y= x sin−1 x. (2) Find if y = tan−1 x2
dx dx
y=
dy d tan−1 x2
x sin−1 x


=
dx dx dy 1
d = × 2x
sin−1 x + sin−1 x · 1


=x· dx 1 + (x2 )2

dx
1 2x
=x· √ + sin−1 x =
2 1 + x4
1−x

24

d x d 1 d x
∗ e = ex ∗ log x = ∗ a = ax log a
dx dx x dx

(1)Find the derivative of sin(log x) Logarithmic Differentiation
with respect to x Find the derivative of xx with re-
spect to x
y = sin(log x)
dy d y = xx
= (sin(log x))
dx dx log y = log(xx ) = x × log x
d 1 dy 1
= cos(log x) × log x = x × + log x × 1 = 1 + log x
dx y dx x
dy 1
= cos(log x) × dy
dx x = y(1 + log x) = xx (1 + log x)
dx

Practice Problems

1. Differentiate the following with respect to x.
1 − x2
 
−1 −1
a) etan x b) sin (x3 ) c) sin (d) xsin x
1 + x2

Derivative of functions in Parametric Forms
A relation expressed between two variables x and y in the form x=f(t), y=g(t) is
said to be parametric form


dy
dy dt
= 
dx dx
dt

Questions and Answers

dy dy
(1) Find if x = at2 , y = 2at. (2) Find , if x=acos θ, y=bsin θ
dx dx
dy dx dy dx
= 2a and = 2at =bcos θ, = −a sin θ
dt dt dθ dθ
   
dy dy
dy dt 2a dy dθ b cos θ
∴ = = = =
dx dx 2at dx dx −a sin θ

dt dθ
1 −b
= = cot θ
t a

25

Practice Problems

dy
1. Find in the following
dx
a) x= sint, y=cos2t
b) x=asec θ, y=btan θ
c) x=acos3 θ, y=asin3 θ

Second Order Derivative
d2 y
If y = f(x),then second order derivative of y is denoted as 2 and is defined as
dx
d2 y
 
d dy
2
=
dx dx dx

Questions and Answers

(1) Find the second order derivatives of the functions

(b) x cos x
2
(a) x + 3x + 4 Answer:
Answer:
y = x cos x
2
y = x + 3x + 4 dy
= x × (− sin x) + cos x
dy dx
= 2x + 3 = −x sin x + cos x
dx
d2 y 2
dy
=2 = (−x cos x + sin x × (−1)) − sin x
dx2 dx2
= −x cos x − 2 sin x

d2 y
(2) If y = A sin x + B cos x ,Show that +y =0
dx2
y = A sin x + B cos x
dy
= A cos x + B × (− sin x)
dx
d2 y
= A × (−sinx) + B × (− cos x)
dx2
= −(A sin x + B cos x) = −y
d2 y
∴ +y =o
dx2

26

Practice Problems

(1) Find the second order derivatives of the following functions

(a) x20 (b) log x (c) x2 log x (d) tan−1 x

d2 y dy
(2) If y = sin−1 x ,Show that (1 − x2 ) − x =0
dx2 dx
(3) If y = (tan−1 x)2 Show that (x2 + 1)y2 + 2x(x2 + 1)y1 = 2

27

Chapter 6

APPLICATIONS OF DERIVATIVES

dy
gives the rate of change of y with respect to x
dx

Questions and Answers

1. The radius of a circle is increasing uniformly at the rate of 3cm/s. Find the rate at which
the area of the circle is increasing when the radius is 10cm?
Answer :-
dr
=3cm/s,
dt
A = πr2
dA dr
= π × 2r
dt dt
= π × 2 × 10 × 3
= 60πcm2 /s

2. A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s.
At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area
increasing?
dr
=5cm/s,
dt
A = πr2
dA dr
= π × 2r
dt dt
=π×2×8×5
= 80πcm2 /s

3. The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is
increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rates of
change of (a) the perimeter, and (b) the area of the rectangle
Answer:-
Since the length x is decreasing and the width y is increasing with respect to time, we
dx dy

have = −5 cm/min and = 4cm/min
dt dt
(a) The perimeter P of a rectangle is given by

P = 2(x + y)
dP dx dy
∴ = 2( + ) = 2(−5 + 4) = −2cm/min
dt dt dt
28

(b) The area A of the rectangle is given by

A = xy
dA dy dx
∴ =x +y
dt dt dt
= 8 × 4 + 6 × (−5)
= 32 − 30 = 2cm2 /min

Practice Problems

1. The radius of a circle is increasing at the rate of 0.7cm/s. What is the rate of increase of
its circumference ?
2. An edge of a variable cube is increasing at the rate of 3cm/s. How fast is the volume of
the cube increasing when the edge is 10cm long ?
3. The volume of a cube is increasing at the rate of 8 cm3 /s. How fast is the surface area
increasing when the length of an edge is 12 cm?

Increasing and Decreasing Function

Let f be a continuous function on [a, b] and differentiable on the open interval (a,b).
Then
(a) f is increasing in [a,b] if f 0 (x) > 0 for each x ∈ (a, b)
(b) f is decreasing in [a,b] if f 0 (x) <0 for each x ∈ (a, b)
(c) f is a constant function in [a,b] if f 0 (x) = 0 for each x ∈ (a, b)

Questions and Answers

1. Show that the function f(x)=x3 − 3x2 + 6x − 5 is strictly increasing on R.
Answer :-

f 0 (x) = 3x2 − 6x + 6
= 3(x2 − 2x + 2)
= 3 (x − 1)2 + 1
 

= 3(x − 1)2 + 3 > 0

Hence f is strictly increasing on R
2. Find the intervals in which f(x)=x2 − 4x + 6 is
(i)increasing (ii)decreasing.
Answer :-

f 0 (x) = 2x − 4

∴ f 0 (x) = 0 ⇒ 2x − 4 = 0
⇒ 2x = 4
⇒x=2

29

Interval Sign of f 0 (x) Nature of the function f
(−∞, 2) f 0 (x) < 0 f is decreasing
0
(2, ∞) f (x) > 0 f is increasing

(3) Find the intervals in which the function f given by f (x) = 4x3 − 6x2 − 72x + 30 is
(a) increasing (b) decreasing

f (x) = 4x3 − 6x2 − 72x + 30
f 0 (x) = 12x2 − 12x − 72
= 12(x2 − x − 6)
= 12(x − 3)(x + 2)

f 0 (x) = 0 gives x = - 2, 3.

−∞ -2 3 ∞

The points x = -2 and x = 3 divides the real line into three disjoint intervals, namely,
(−∞, −2), (−2, 3) and (3, ∞)

Interval Sign of f 0 (x) Nature of the function f
(−∞, −2) (−)(−) > 0 f is increasing
(−2, 3) (−)(+) < 0 f is decreasing
(3, ∞) (+)(+) > 0 f is increasing

Practice Problems

1. Show that the function given by f (x) = 3x + 17 is increasing on R.

2. Show that the function f(x)=x3 − 6x2 + 15x + 4 is strictly increasing on R.

3. Find the intervals in which the following functions are strictly increasing or decreasing
(i) x2 + 2x − 5 (ii)x3 − 6x2 + 9x + 15

4 Show that the function given by f (x) = sin x is

(a) increasing in (0, π2 )
(b) decreasing in ( π2 , π)
(c) neither increasing nor decreasing in (0, π)

30

Maxima and Minima
(First Derivative Test)
Let f be a function defined on an open interval I.
Let f be continuous at a critical point c in I. Then

(i) If f 0 (x) changes sign from positive to negative as x increases through c
then c is a point of local maxima.

(ii) If f 0 (x) changes sign from negative to positive as x increases through c
then c is a point of local minima.

(iii) If f 0 (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.
Such a point is called point of inflection

Second Derivative Test
Let f be a function defined on an interval I
and c ∈ I. Let f be twice differentiable at c. Then

(i) x = c is a point of local maxima if f 0 (c) = 0 and f 00 (c) < 0
The value f (c) is local maximum value of f .

(ii) x = c is a point of local minima if f 0 (c) = 0 and f 00 (c) > 0
In this case, f (c) is local minimum value of f .

(iii) The test fails if f 0 (c) = 0and f 00 (c) = 0.

1. Find local maximum and local minimum values of the function f given by
f (x) = 3x4 + 4x3 − 12x2 + 12

f (x) = 3x4 + 4x3 − 12x2 + 12
f 0 (x) = 12x3 + 12x2 − 24x
= 12x(x2 + x − 2)
= 12x(x + 2)(x − 1)
0
f (x) = 0 atx = 0, −2, 1
f 00 (x) = 36x2 + 24x − 24
f 00 (0) = −24 < 0
f 00 (−2) = 72 > 0
f 00 (1) = 36 > 0

Therefore, by second derivative test, x = 0 is a point of local maxima and

local maximum value of f at x = 0 is f (0) = 12
x = 1 and x = -2 are the points of local minima and
local minimum values of f at x = 1 and - 2 are
f (1) = 7 and f (-2) = -20, respectively.

(2) Find the absolute maximum value and the absolute minimum value of the function
f (x) = x3 , x ∈ [−2, 2]
31

f (x) = x3
f 0 (x) = 3x2
f 0 (x) = 0 ⇒ 3x2 = 0
⇒x=0
f (0) = 0, f (−2) = (−2)3 = −8, f (2) = 23 = 8

Absolute maximum value is 8 at x=2 and Absolute minimum value is -8 at x=-2

Practice Problems

(1) Find the local maxima and local minima, if any, of the following functions. Find also the
local maximum and the local minimum values, as the case may be:
(i) f (x) = x2 (ii) g(x) = x3 − 3x

(2) Find the absolute maximum and minimum values of a function f given by

(a) f (x) = 2x3 − 15x2 + 36x + 1 on the interval [1, 5]
(b) f (x) = sin x + cos x, x ∈ [0, π]

32

Chapter 7

INTEGRALS
Integration is the inverse process of differentiation.
The Integral calculus is divided into two types.

• Indefinite Integrals and
• Definite Integrals

Indefinite integrals Z
The indefinite integral of a function f (x) with respect to x is denoted as f (x)dx.
Z  
d
f (x) dx = F (x) + C where F (x) = f (x) , C is constant of integration.
dx
Z Z Z
• (f (x) ± g(x)) dx = f (x) dx ± g(x) dx
Z Z
• k f (x) dx = k f (x) dx where k is any constant

Z Z
• dx = x + C • sec x tan xdx = sec x + C

x2
Z Z
• xdx = +C • cosec2 xdx = − cot x + C
2
Z
xn+1
Z
n • cosec x cot xdx = −cosecx + C
• x dx = + C, n 6= −1
n+1
Z
1
Z
• sin x dx = − cos x + C • dx = log |x| + C
x
Z Z
• cos xdx = sin x + C • ex dx = ex + C

ax
Z Z
• 2
sec xdx = tan x + C • ax dx = + C,a> 0
log a

Questions and Answers
Z Z Z
1. (sin x + cos x) dx = sin x dx + cos x dx= − cos x + sin x + C

33

x3
Z Z Z
2 x 2
2. (2x + e )dx = 2 x dx + ex dx = 2 + ex + C
3
x3 + 5x2 − 4
Z
3. dx
x2

x3 + 5x2 − 4 x3 5x2
Z Z Z Z
4
dx = dx +dx − dx
x2 x2 x 2 x2
Z Z Z
= xdx + 5 dx − 4 x−2 dx
x2 x−1
= + 5x − 4 +C
2 −1
x2 1
= + 5x + 4 + C
2 x

Integration by substitution
Questions and Answers
Z
1. e2x+3 dx
Z Z
2x+3 dt
e dx = et
2
Z
1 Let 2x + 3 = t
= et dt
2 Then 2dx = dt
1 dt
= et + C or dx =
2 2
e2x+3
= +C
2
−1
etan x
Z
2. dx
1 + x2 −1
etan x
Z Z
dx = et dt
1 + x2 Let tan−1 x = t
1
= et + C Then dx = dt
−1
1 + x2
= etan x + C
Z Z
1 1
3. dx = dx
x + x log x Z x(1 + log x) Let 1 + log x = t,
1 1
= dt then dx = dt
t x
= log |t| + C = log |1 + log x| + C

Note:-
Z 0
f (x)
• dx = log |f (x)| + C
f (x)

Z
2x
4. dx = log |(1 + x2 )| + C
1 + x2
Z
cos x
5. dx = log |1 + sin x| + C
1 + sin x
34

Practice Problems
sin−1 x tan−1 x
Z Z
1. √ dx 3. dx
1 − x2 1 + x2
√
sin(tan−1 x)
Z
cos x
Z
2. √ dx 4. dx
x 1 + x2

Integration using Partial Fractions

Form of Rational Function Form of Partial Fraction
px + q A B
(i) +
(x − a)(x − b) (x − a) (x − b)
px + q A B
(ii) +
(x − a)2 (x − a) (x − a)2

Questions and Answers
Z
1
1. dx
(x + 1)(x + 2)
1 A B
By using partial fractions, = +
(x + 1)(x + 2) (x + 1) (x + 2)
Then 1 = A(x + 2) + B(x + 1)
Equating the coefficient of x and constant on both sides,we get
A+B =0 and 2A+B=1
Solving we get A =1 and B=-1
1 1 −1
i.e. = +
(x + 1)(x + 2) (x + 1) (x + 2)
−1
Z Z
1 1
∴ dx = + dx
(x + 1)(x + 2) (x + 1) (x + 2)
= log |x + 1| − log |x + 2| + C
x+1
= log +C
x+2
Z
2x + 3
2. 2
dx
x − 5x + 6
Z Z
2x + 3 2x + 3
dx = dx
x2 − 5x + 6 (x − 3)(x − 2)
2x + 3 A B
Using partial fractions, = +
(x − 3)(x − 2) x−3 x−2
Then 2x + 3 = A(x − 2) + B(x − 3)

In the above equation,put x=2 ,we get B = -7 and put x=3 we get A =9
−7
Z Z
2x + 3 9
∴ dx = + dx
(x − 3)(x − 2) x−3 x−2
= 9 log |x − 3| − 7 log |x − 2| + C

35

Practice Problems
Z
x
1. dx
(x + 1)(x + 2)
Z
x
2. dx
(x + 1)(x + 2)
Z
2x
3. 2
dx
x + 3x + 2

Integration by parts

Z Z Z  Z 
d
• f (x)g(x)dx = f (x) g(x)dx − f (x) g(x)dx dx
dx

Questions and Answers
Z
1. xex dx
Z Z Z  Z 
x x d x
xe dx = x e dx − x e dx dx
dx
Z
= xe − ex dx
x

= xex − ex + C
Z
2. x cos xdx
Z Z Z  Z 
d
x cos xdx = x cos xdx − x cos xdx dx
dx
Z
= x sin x − sin xdx

= x sin x + cos x + C
Z
3. Find log xdx
Z Z
log x dx = log x × 1dx
Z Z  Z 
d
= log x 1dx − log x 1dx dx
dx
Z
1
= log x × x − × xdx
x
Z
= x log x − 1dx

= x log x − x + C

36

Z
4. ex sin xdx
Z Z Z  Z 
x x d x
e sin xdx = e sin xdx − e
sin xdx dx
dx
Z
= −ex cos x − ex (− cos x)dx
Z
= −e cos x + ex cos xdx
x

Z
= −e cos x + e sin x − ex sin xdx
x x

Z
2 ex sin xdx = ex (sin x − cos x) + C
Z
1
∴ ex sin xdx = ex (sin x − cos x) + C
2
Z
5. Find x2 ex dx
Z Z
2 x 2 x
x e dx = x e − 2xex dx
Z
= x e − 2 xex dx
2 x

 Z 
2 x x x
= x e − 2 xe − e dx

= x2 ex − 2xex + 2ex + C

Z
• ex [ f (x) + f 0 (x) ] dx = ex f (x) + C

Z
6. ex (sin x + cos x)dx
Z
ex [ f (x) + f 0 (x) ] dx = ex f (x) + C
(x) = sin x, f 0 (x) = cos x)
(f Z
∴ ex (sin x + cos x)dx = ex sin x + C
Z
7. ex sec x(1 + tan x)dx
Z
ex [ f (x) + f 0 (x) ] dx = ex f (x) + C
Z Z
e sec x(1 + tan x)dx = ex [sec x + sec x tan x] dx
x

(f (x) = sec x, f 0 (x) = sec x tan x)
= ex sec x + C

37

One Zmark Questions and Answers Z
1
1. tan x dx = log | sec x| + C 6. √ dx = sin−1 x + C
1−x 2
Z Z
1
2. cot x dx = log | sin x| + C 7. dx = tan−1 x + C
Z Z 1 + x2
1 1 −1 x
3. sec xdx = log | sec x + tan x| + C 8. dx = tan +C
Z Z a2 + x 2 a a
1 1 x
4. cosecxdx = log |cosecx − cot x| + C 9. 2
dx = tan−1 + C
Z Z 9+x 3 3
2x
5. log x dx = x log x − x + C 10. 2
dx = log |1 + x2 | + C
1+x

Practice Problems

Integrate the following with respect to x
1
 
1. 2x3 1
2. √ 3. e tan−1 x +
x
16 − x2 1 + x2

Definite Integral
Let f be continuous function defined on the closed interval [a, b] and
Z b
F be antiderivative of f ,then f (x)dx = [F (x)]ba = F (b) − F (a)
a

Questions and Answers
Z 3
1 3
1. dx = [ log |x| ]32 = log 3 − log 2 = log
2 x 2
Z 5
2. ex dx = [ex ]54 = e5 − e4
4
Z 1
1 1 π π
dx = sin−1 x 0 = sin−1 (1) − sin−1 (0) = − 0 =

3. √
0 1 − x2 2 2
Z a Z a
f (x)dx = f (a − x)dx
0 0

Z π
2 sin4 x
4. Evaluate dx
0 sin4 x + cos4 x
Z π
sin4 x 2
Let I =
4 4
dx (1)
0 sin x + cos x
Z π
2 sin4 ( π2 − x)

I= 4 π 4 π
dx
0 sin ( 2 − x) + cos ( 2 − x)
Z π
2 cos4 x
I= 4 4 dx (2)
0 cos x + sin x
Adding (1) and (2) we get

38

Z π
2 sin4 x + cos4 x
2I = dx
0 sin4 x + cos4 x
Z π
2
= 1dx
0
π
= [x]02
π
2I =
2
π
∴I=
4

Practice Problems
Z π Z π √ Z 1
2
2
2 sin x tan−1 x
1. cos xdx 2. √ √ dx 3. dx
0 0 sin x + cos x 0 1 + x2

........................

39

Chapter 8

APPLICATION OF INTEGRALS

Area under simple curves

• Let f(x) be a continuous function defined on[a,b].
Then the area bounded by the curve y=f(x),the x-axis and the ordinates x=a
and x=b is given by

Z b Z b
Area = f (x) dx = y dx
a a

• The Area of the region bounded by the curve x=g(y),the yaxis and
the lines y = c and y = dZ is given by Z
d d
Area = g(y) dy = x dy
c c

Questions and Answers

1. Find the area enclosed by the circle x2 + y 2 = a2

40

The whole area of the circle = 4 × area AOBA
Z a
4 × area AOBA = 4 y dx
Z0 a √
=4 a2 − x2 dx
0
a
x√ 2 2

a −1 x
=4 a − x2 + sin ( )
2 2 a 0
2
  
a a −1
=4 × 0 + sin 1 − 0
2 2
 2 
a π
=4 = πa2
2 2

Area enclosed by the circle x2 + y 2 = a2 is πa2
x2 y 2
2. Find the area enclosed by the ellipse 2 + 2 = 1
a b
Z a = 4 × area AOBA
The whole area of the ellipse
4 × area AOBA = 4 y dx
0
b √ 2
Z a
=4 a − x2 dx
0 a
a
b x√ 2 a2

2 −1 x
=4 a − x + sin ( )
a 2 2 a 0
2
  
b a a −1
=4 × 0 + sin 1 − 0
a 2 2
 2 
b a π
=4 = πa b
a 2 2
x2 y 2
Area enclosed by the ellipse 2 + 2 = 1 is πab
a b
3. Find the area of the region bounded by the curve y 2 = x and the lines x = 1, x =
4 and the x axis Z
4
Required area is 2 ydx
Z 4 Z 4 1
√
2 ydx = 2 xdx
1 1

h 3 i4 h 3 3
i
= 2 × 32 x 2 = 43 × 4 2 − 1 2
1
4 28
= ×7=
3 3

Practice Problems

1. Find the area of the region bounded by y2 = x and the lines x=1,x=2 and the xaxis
2. Find the area of the region bounded by the curve x2 = 4y, y = 2, y = 4 and the y
axis in the first quadrant

41

Chapter 9

DIFFERENTIAL EQUATIONS
Order and Degree

The order of a differential equation is the order of the highest order derivative.

Questions and Answers

(1) Write the order of the following differential equations

dy dy
(i) = ey , Here highest order derivative is . ∴ order=1
dx dx
 3
d2 y dy d2 y
( ii) + + y = 0, Order=2 (Since is present)
dx2 dx dx2

Degree of a differential equation when it is a polynomial equation in derivatives, is
the highest power of the highest order derivative present in the equation

Questions and Answers

(1) Write the order and degree of the following differential equations
dy dy
i. = ex , in this equation highest order derivative is . Therefore Order=1. Its
dx dx
power is also 1. Therefore Degree=1.
 3
d2 y dy
ii. 2
+2 +y =0
dx dx
d2 y
Order=2, Degree=1 (Since degree of 2 is 1)
dx
3
 
dy dy
iii. 3
+ sin =0
dx dx
Order=3, Degree is not defined since it is not a polynomial equation

in its derivatives.

42

Practice Problems

1. Determine 
the order and degree
d2 y

dy
a) 2 − 3 + 2y = 0
dx dx
 2
dy dy
b) + − sin2 y = 0
dx dx
d2 y
c) 2 + y = 0
dx

Method of Solving First Order First Degree Differential Equation

• Variable Separable
Differential equations which can be expressed in the form f (x)dx = g(y)dy is called
variable separable
Example :-
dy x+1
1. Solve =
dx 2−y

(2 − y)dy = (x + 1)dx
Z Z
(2 − y)dy = (x + 1)dx
y2 x2
2y − = +x+c
2 2
x2 + y 2 + 2x − 4y + c = 0

dy 1 + y2
2. Find the general solution of = .
dx 1 + x2
dy 1 + y2
=
dx 1 + x2
dy dx
2
=
1+y 1 + x2
Z Z
dy dx
2
=
1+y 1 + x2
tan−1 y = tan−1 x + c

Practice Problems

1. Find the general solution of

dy
(i) = ex−y
dx

dy 1 − cos x
(ii) =
dx 1 + cos x

43

Linear Differential Equations
dy
Differential equations of the form +Py =Q,where P and Q are constants or
dx
functions of x
To solve this equation first find integrating factor
R
I.F = e P dx

Then general solution is

Z
y× I.F= Q × I.Fdx+ C

Questions and Answers

dy
1 Find the general solution of + 3y = e−2x
dx
Answer :-
dy
+ 3y = e−2x
dx
This is a linear differential equation of the form
dy
+ Py = Q where P=3 and Q=e−2x
dx
R R
I.F = e P dx = e 3dx
= e3x
Z
Solution is y × I.F = Q × I.Fdx + C
Z
ye = e−2x × e3x dx + C
3x

Z
= ex dx + C

= ex + C

Solution is ye3x = ex + C
dy y
2 Find the general solution of + = x2
dx x
Answer :-
dy y
+ = x2
dx x
This is a linear differential equation of the form
dy 1
+ Py = Q where P= and Q=x2
dx x

44

Z
1
R dx
I.F = e P dx = e x
= elogx = x
Z
Solution is y × I.F = Q × I.Fdx + C
Z
y × x = x2 × xdx + C
Z
= x3 dx + C
x4
= +C
4
x4
Solution is yx = +C
4
dy
3 Find the general solution of x + 2y = x2
dx
Answer :-
dy
x + 2y = x2
dx
dy y
+2 =x
dx x
This is a linear differential equation of the form
dy 2
+ Py = Q where P= and Q=x
dx x
Z
2
R dx
I.F = e P dx = e x
2
= e2logx = elogx = x2
Z
Solution is y × I.F = Q × I.Fdx + C
Z
y × x = x × x2 dx + C
2

Z
= x3 dx + C
x4
= +C
4
x4
Solution is yx2 = +C
4

Practice Problems

dy
1 Find the general solution of + 2y = sin x
dx

dy
2 Find the general solution of x + 2y = x2 log x
dx

45

Chapter 10

VECTOR ALGEBRA
• Vector :-Quantity having both magnitude and direction

Addition of Two vectors
→
−
Let →
−
a =a1 î + a2 ĵ + a3 k̂ and b =b1 î + b2 ĵ + b3 k̂, then
→
− →
−
a + b =(a1 + b1 ) î + (a2 + b2 ) ĵ + (a3 + b3 ) k̂

Questions and Answers

→
− →
− →
−
1. If →
−
a =2î + 3ĵ + 3k̂ and b =3î − 2ĵ + k̂. Find →
−
a + b and →
−
a − b
Answer :-
→
− →
−
a + b = (2 + 3) î + (3 − 2) ĵ + (3 + 1) k̂
= 5î + 1ĵ + 4k̂
→
− →
−
a − b = (2 − 3) î + (3 − −2) ĵ + (3 − 1) k̂
= −î + 5ĵ + 2k̂

→
−
a
→
−
Unit vector in the direction of a is â= → , where |→
−a |=
p
a21 + a22 + a23
|−
a|

(2) Consider →
−
a =5î + 3ĵ + 2k̂

(a) Find the magnitude of →
−
a (b) Find the unit vector in the direction of →
−
a

(a) →
−
a =5î + 3ĵ + 2k̂ (b) Unit vector in the direction of →
−
a
q is
→
−
| a | = a21 + a22 + a23 →
−a
√ a= → −
|a|
b
= 52 + 32 + 22

√
= 38 5î + 3ĵ + 2k̂
= √
38

46

(3) Find the vector in the direction of a having magnitude 7 units
A vector of 7 units in the direction of →
−
a = 7b
a

35î + 21ĵ + 14k̂
7b
a= √
38

(4) Find the unit vector in the direction of the sum of the vectors
→
− →
−
a =2î + 2ĵ − 5k̂ and b =2î + ĵ + 3k̂
The sum of the given vector is
→
− →
−
a + b = (2 + 2) î + (2 + 1) ĵ + (−5 + 3) k̂
= 4î + 3ĵ − 2k̂
→
− √
|→
− p
a + b | = 42 + 32 + (−2)2 = 29

Then required unit vector is →
−
→
−
a + b 4î + 3ĵ − 2k̂
→
− = √
|→
−
a + b| 29

Product of Two Vectors

• Dot Product(Scalar Product) :-
→
− →
−
The scalar product of two nonzero vectors → −
a and b denoted by →
−
a · b
→
− →
−
is defined as → −
a · b = |→ −a || b | cos θ
→
−
where θ is the angle between → −
a and b
→
−
Let →
−
a =a1 î + a2 ĵ + a3 k̂ and b =b1 î + b2 ĵ + b3 k̂, then

→
− →
−
a · b =a1 b1 + a2 b2 + a3 b3

a·b
∗ The angle θ between the two non zero vectors a and b is given by cos θ =
|a| |b|

Questions and Answers

→
−
1. Consider →
−
a = î + ĵ − k̂ and b = î − ĵ + k̂
→
− →
−
(a) Find →
−
a · b (b) Find the angle between →
−
a and b

√
(b)|→
− p
a | = 12 + 12 + (−1)2 = 3
(a) →
− p √
| b | = 12 + (−1)2 + 12 = 3

→
− →
−
a · b =a a +b b +c c
1 2 1 2 1 2 →
− →
−
a · b = −1
= 1 × 1 + 1 × (−1) + (−1) × 1 →
− →
−
a · b −1 −1 −1
= −1 cos θ = − = 3 ⇒ θ = cos
→
→
−
| a || b | 3

47

a·b
∗ The projection of a vector a on another vector b is given by
|b|

→
−
2. Find the projection of the vector →
−
a =2î + 3ĵ + 2k̂ on the vector b =î + 2ĵ + k̂
→
−
The projection of vector →
−a on the vector b is

  
→
− →
− 2î + 3ĵ + 2k̂ î + 2ĵ + k̂
a · b
→
− = √
b 12 + 22 + 12
2×1+3×2+2×1
= √
6
10
=√
6

3. Show that the points A(−2î + 3ĵ + 5k̂),B(î + 2ĵ + 3k̂) and C(7î − k̂) are collinear

−→
AB = (1 + 2)î + (2 − 3)ĵ + (3 − 5)k̂ = 3î − ĵ − 2k̂
−−→
BC = (7 − 1)î + (0 − 2)ĵ + (−1 − 3)k̂ = 6î − 2ĵ − 4k̂
−→
AC = (7 + 2)î + (0 − 3)ĵ + (−1 − 5)k̂ = 9î − 3ĵ − 6k̂
−→ p √
|AB| = 32 + (−1)2 + (−2)2 = 14
−−→ p √ √
|BC| = 62 + (−2)2 + (−4)2 = 56 = 2 14
−→ p √ √
|AC| = 92 + (−3)2 + (−6)2 = 126 = 3 14
−→ −→ −−→
|AC| = |AB| + |BC|

Hence A,B,C are collinear

Note:-
→
− →
−
Two non zero vectors →
−
a and b are perpendicular ⇔ →
−
a · b =0

(4) If a = 5bi − b
j − 3b j − 5b
k and b = bi + 3b k ,the show that the vectors
a + b and a − b are perpendicular
Answer
a + b = 5bi − bj − 3b j − 5b
k + bi + 3b k = 6bi + 2bj − 8b
k
a − b = (5bi − bj − 3b
k) − (bi + 3b
j − 5b
k) = 4bi − 4b
j + 2b
k
(a + b) · (a − b = 6 × 4 + 2 × (−4) + (−8) × 2 = 24 − 8 − 16 = 0
Hence a + b and a − b are perpendicular

Practice Problems

1 Find the angle between the vectors î − 2ĵ + 3k̂ and 3î − 2ĵ + k̂

2 Find the projection of the vector î + 3ĵ + 7k̂ on the vector 7î − ĵ + 8k̂

3 Show that the point A(1,2,7),B(2,6,3) and C(3,10,-1) are collinear.
48

4 Show that the vector 2î − ĵ + k̂, î − 3ĵ − 5k̂ and 3î − 4ĵ − 4k̂ form the vertices of a right
angled triangle

Cross Product(Vector Product) :-
→
−
Let →
−
a =a1 î + a2 ĵ + a3 k̂ and b =b1 î + b2 ĵ + b3 k̂, then

î ĵ k̂
→
− →
−
a × b = a1 a2 a3

b1 b2 b3

Questions and Answers
→
−
1 Consider →
−a =2î + ĵ and b =3î + ĵ + 4k̂
→
−
(a) Find →
−a × b
→
−
(b) Find unit vector perpendicular to both →
−
a and b

(a) (b)Unit vector which is perpendicu-
→
−
lar to both →
−
a and b is
i j k →
− →
−
a × b
→
− = →
−
→
−
a × b = →
−a × b
2 1 0
4î − 8ĵ − k̂
3 1 4 =p
42 + (−8)2 + (−1)2
= î (1 × 4 − 0 × 1) 4î − 8ĵ − k̂
= √
− ĵ (2 × 4 − 3 × 0) 81
+ k̂ (2 × 1 − 3 × 1) 4î − 8ĵ − k̂
=
= 4î − 8ĵ − k̂ 9

Practice Problems
→
− →
− →
−
1. Consider the vectors →−
a =2î − ĵ + 2k̂ and b =−î + ĵ − k̂. Find →
−a · b and → −
a × b.
→
−
2. Find the unit vector perpendicular to both of the vectors → −a =î−2ĵ + k̂ and b =2î+4ĵ +5k̂

Area of The Triangle

→
−
If →
−
a and b represent the adjacent sides of triangle, then its area is given by
→
−
Area = 12 →
−
a × b

49

Questions and Answers

→
−
1. Find the area of the triangle having adjacent sides→
−
a = 2î + ĵ + 3k̂, b =3î + 5ĵ − 2k̂

1 →
− →
−
Area = a × b
2

i j k
→
− →
−
a × b = 2 1 3

3 5 −2
= i(−2 − 15) − j(−4 − 9) + k(10 − 3)
= −17î + 13ĵ + 7k̂
→
− →
− p
a × b = (−17)2 + 132 + 72
√
= 507
√
507
Area =
2

Area of the Parallelogram
→
−
If →
−
a and b represent the adjacent sides of parallelogram, then its area is
given by
→
−
Area = →
−
a × b

Questions and Answers

→
−
1. Find the area of the parallelogram having adjacent sides→
−
a = 3î + ĵ + 4k̂, b =î − ĵ + k̂.
→
−
Area = →
−
a × b

i j k
→
− →
−
a × b = 3 1 4

1 −1 1
= i(1 − −4) − j(3 − 4) + k(−3 − 1)

= 5î + ĵ − 4k̂
→
− →
− p
a × b = 52 + 12 + (−4)2
√
= 42
√
Area = 42

50

Chapter 11

THREE DIMENSIONAL GEOMETRY

Relation between Direction cosines and direction ratios of a line

∗ If a,b,c are direction ratios of a line then direction cosines are given by
a b c
l=√ ,m = √ ,n = √
a2 + b 2 + c 2 a2 + b 2 + c 2 a2 + b 2 + c 2

∗ Direction ratios of a line joining P(x1 , y1 , z1 ) and Q(x2 , y2 , z2 ) may be taken as
x2 − x1 , y2 − y1 , z2 − z1

Questions and Answers

(1) If a line has direction ratios 2,-1,-2 find its direction cosines
a 2 2 2
l=√ =p =√ =
a2 + b 2 + c 2 22 + (−1)2 + (−2)2 9 3
b −1
m= √ =
a2 + b 2 + c 2 3
c −2
n= √ =
a2 + b 2 + c 2 3
2 −1 −2
∴ direction ratios are , ,
3 3 3
(2) Find the direction cosines of the line passing through the two points (-2,4,-5) and (1,2,3)
Answer
Direction ratios are 1-(-2),2-4,3-(-5)
that is 3,-2,8
Direction cosines are
a 3 3
l=√ =p =√
2
a +b +c2 2 2 2
3 + (−2) + (8) 2 77
b −2 −2
m= √ =p =√
a2 + b 2 + c 2 32 + (−2)2 + (8)2 77
c 8 8
n= √ =p =√
a2 + b 2 + c 2 32 + (−2)2 + (8)2 77

Practice Problems

(1) If a line has direction ratios −18, 12, −4 find its direction cosines

(2) Find the direction cosines of the line passing through the two points (3,5,-4) and (-1,1,2)

51

Line

Equation of a line passing through a given point and parallel to a given
vector
Vector equation of a line passes through the given point with position vector
→
− →
−
a and parallel to the given vector b is

→
− →
−
r =→
−
a +λ b

In Cartesian form, if the given point is A(x1 , y1 , z1 ) and parallel to the line
with direction ratios a,b,c is
x − x1 y − y1 z − z1
= =
a b c

Questions and Answers
1. Find the vector equation and Cartesian equation of the line through the point (5, 2, 4)
and which is parallel to the vector 3î + 2ĵ − 8k̂
Answer :-
→
−
Given (x1 , y1 , z1 ) = (5, 2, 4) and b = 3î + 2ĵ − 8k̂
Vector equation is
→
− →
−
r =→ −
a +λ b
 
= 5î + 2ĵ + 4k̂ + λ 3î + 2ĵ − 8k̂
Cartesian equation is
x − x1 y − y1 z − z1
= =
a b c
x−5 y−2 z−4
= =
3 2 −8
x−5 y−2 z−4
= =
3 2 −8
Practice Problems
(1) Find the vector equation of a line passing through th point (1,2,3) and is parallel to the
j − 2b
vector 3bi + 2b k
(2) Find the equation of the line in vector and in cartesian form that passes through the
point with position vector2bi − b
j + 4b j−b
k and is in the direction bi + 2b k

Distance between two Skew Lines
→
− →
−
Consider →−r =→ −
a1 + λ b1 and →
−
r =→ −
a2 + µ b2 then shortest distance between

two lines is given by
→
− → − − →
b1 × b2 · (→
a2 − −
a1 )
SD= →
− → −
b1 × b2

52

Questions and Answers

1. Find the shortest
 distancebetween the lines l1 and l2 whose vector equations are
→
−
r = î + ĵ + λ 2î − ĵ + k̂ and →
−
r = 2î + ĵ − k̂ + µ 3î − 5ĵ + 2k̂

Answer :-
→
− →
−
Let →
−
a1 = î + ĵ, →
−
a2 = 2î + ĵ − k̂, b1 = 2î − ĵ + k̂, b2 = 3î − 5ĵ + 2k̂
→
−
a2 − →
−
a1 = (2 − 1) î + (1 − 1) ĵ + (−1 − 0) k̂
= î − k̂

i j k
→
− → −
b1 × b2 = 2 −1 1

3 −5 2

= 3î − ĵ − 7k̂
→
− → − p √
b1 × b2 = 32 + (−1)2 + (−7)2 = 59

→
− → − → − →
−
   
b1 × b2 · ( a2 − a1 ) = 3î − ĵ − 7k̂ · î − k̂
= 3 + 0 + 7 = 10

→
− → − →
b1 × b2 · (−
a2 − →
−
a1 )
SD = →
− → −
b1 × b2
10
=√
59

Practice Problems

(1) Find the shortest
 distancebetween the lines l1 and l2 whose vector equations are
→
−
r = î + ĵ + λ 2î − ĵ + k̂ and →
−
r = 2î + ĵ − k̂ + µ 3î − 5ĵ + 2k̂

(2) Find the shortest distance between the lines
x+1 y+1 z+ x−3 y−5 z−7
= = 1 and = =
7 −6 1 5 −2 1

53

Chapter 12

LINEAR PROGRAMMING
Questions and Answers

1. Consider the following Linear progamming problem
Minimise z = 3x + 9y
subject to the constraints
x + 3y ≤ 60
x + y ≥ 10
x≤y
x ≥ 0, y ≥ 0
(a) Draw its feasible region
(b) Write the corner points of the feasible region
(c) Find the minimum value of z subject to the constraints
x + 3y = 60 x + y = 10 x=y

x 0 60 x 0 10 x 0 10
(a)
y 20 0 y 10 0 y 0 10

(b)Corner points of the feasible region are
A(0,20),B(0,10),C(5,5) and D(15,15)
(c)

Corner Points Value of z = 3x + 9y
A(0,20) z = 3 × 0 + 9 × 20 = 180
B(0,10) z = 3 × 0 + 9 × 10 = 90
C(5,5) z = 3 × 5 + 9 × 5 = 60
D(15,15) z = 3 × 15 + 9 × 15 = 180

54

Minimum value of z is 60 at (5,5)
2. Consider the following LPP
Maximise z = 200x + 500y
subject to the constraints
x + 2y ≥ 10
3x + 4y ≤ 24
x ≥ 0, y ≥ 0
(a) Draw its feasible region
(b) Write the corner points of the feasible region
(c) Find the maximum value of z subject to the constraints
x + 2y = 10 3x + 4y = 24
(a) x 0 10 x 0 8
y 5 0 y 6 0

(b)Corner points of the feasible region are
A(4,3),B(0,6),C(0,5)
(c)

Corner Points Value of z = 200x + 500y
A(4,3) z = 200×4+500×3 = 2300
B(0,6) z = 200×0+500×6 = 3000
C(0,5) z = 200×0+500×5 = 2500

Maximum value of z is 3000 at (0,6)
3. Solve the folowing LPP graphically
Maximise z = 75x + 25y subject to the constraints
4x + y ≤ 120
x + y ≤ 90
x ≥ 0, y ≥ 0

4x + y = 120 x + y = 90
x 0 30 x 0 90
y 120 0 y 90 0

55

Corner points of the feasible region are
A(10,80),B(0,90),C(30,0) and D(0,0)

Corner Points Value of z = 75x + 25y
A(10,80) z = 75×10+25×80 = 2750
B(0,90) z = 75 × 0 + 25 × 90 = 2250
C(30,0) z = 75 × 30 + 25 × 0 = 2250
D(0,0) z = 75 × 0 + 25 × 0 = 0

Maximum value of z is 2750 at (10,80)

Practice problems

1. Solve the following linear programming problem graphically:
Maximise z = 4x + y
subject to the constraints:
x + y ≤ 50
3x + y ≤ 90
x ≥ 0, y ≥ 0
2. Solve the inear programming problem graphically
Maximise z = 4x + y
subject to the constraints
x + y ≤ 50
3x + y ≤ 90
x ≥ 0, y ≥ 0
3. Minimise z = 3x + 2y
subject to the constraints:

x+y ≥8
3x + 5y ≤ 15
x ≥ 0, y ≥ 0

56

Chapter 13

PROBABILITY

Key Notes :-

n(A)
P(A) =
n(S)
P (A0 ) = 1 − P (A)
P(A or B) = P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
P(A and B) = P (A ∩ B) = P (A) + P (B) − P (A ∪ B)
P(not A and not B) = P (A0 ∩ B 0 ) = 1 − P (A ∪ B)
P(not A or not B) = P (A0 ∪ B 0 ) = 1 − P (A ∩ B)
Conditional Probability :- If A and B are two events, then the conditional
probability of A given B is given by

P (A ∩ B)
P(A/ B) =
P (B)

Questions and Answers

7 9 4
(1) If P(A)= , P(B)= and P (A ∩ B) = evaluate P (A/B)
13 13 13
P (A ∩ B)
P (A/B) =
P (B)
4/13
=
9/13
4
=
9

6 5 7
(2) If P(A)= , P(B)= and P (A ∪ B) = find
11 11 11
(i) P (A ∩ B) (ii) P (B/A)

P (A ∩ B)

(i) P (A ∩ B) = P (A) + P (B) (ii) P (B/A) =
− P (A ∪ B) P (A)
6 5 7 4/11
= + − =
11 11 11 6/11
4 4
= =
11 6

57

(3) If P(A)=0.8, P(B)=0.5, and P(B/A)=0.4.
Find (i) P(A∩B) (ii)P(A/B)

P (A ∩ B) P (A ∩ B)
(i) P (B/A) = (ii) P (A/B) =
P (A) P (B)
P (A ∩ B) = P (B/A) · P (A) 0.32
=
= 0.4 × 0.8 0.5
= 0.64
= 0.32

Independent Events :-

Two events are said to be independent if the occurrence of one event does
not affect the occurrence of other.

Two events A and B are independent if P(A∩B)=P(A)P(B)

Questions and Answers

1. An unbiased die is thrown twice. Let the event A be ’odd number on the first throw’ and
B the event ’odd number on the second throw’. Check the independence of the event A
and B.
S = {(1, 1), (1, 2), ..., (6, 6)}
A = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, ), (3, 4),
(3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
B = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3),
(6, 3), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
A ∩ B = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
P(A) = 18
36
P(B) = 18
36
9
P(A∩B) = 36
18 18 1 1 1
P (A) · P (B) = × = × =
36 36 2 2 4
9 1
P (A ∩ B) = =
36 4
∴ P(A∩B) = P(A)·P(B)
Thus A and B are independent events.
3 1
2. If P(A) = and P(B)= , find P(A∩B) if A and B are independent events.

5 5
P (A ∩ B) = P (A) · P (B)
3 1
= ×
5 5
3
=
25
58

Baye’s Theorem :-

Let {E1 ,E2 ,E3 ,...,En } be a partition of the sample space S and A be any
event associated with S. Then
P (Ei ) · P (A/Ei )
P(Ei /A)=
P (E1 ) · P (A/E1 ) + P (E2 ) · P (A/E2 ) + ... + P (En ) · P (A/En )

Questions and Answers

1. Bag I contains 3 red and 4 black balls while another bag II contains 5 red and 6 black
balls. One ball is drawn at random from one of the bags and it is found to be red. Find
the probability that it was drawn from bag II.
Answer :-
Let E1 be the event of choosing the bag I and E2 be the event of choosing the bag II and
A be the event of drawing a red ball. Then P(E1 )=P(E2 )= 21
3
Also P(A/E1 )=P(drawing a red ball from Bag I) =
7
5
P(A/E2 )=P(drawing a red ball from Bag II) =
11
Now the probability of drawing a ball from Bag II, being given that it is red, is P(E2 /A)

By using Baye’s Theorem

P (E2 ) · P (A/E2 )
P (E2 /A) =
P (E1 ) · P (A/E1 ) + P (E2 ) · P (A/E2 )
1
× 5
= 1 23 111 5
2
× 7 + 2 × 11
5
11 35
= 33+35 =
7×11
68

2. In a factory which manufactures bolts, machines A,B and C manufacture respectively
25%, 35%, and 40% of the bolts of their output 5,4 and 2 percent are respectively defective
bolts. A bolt is drawn from the product and is found to be defective. What is the
probability that it is manufactured by machine B ?
Answer :-
Let A : Event of getting defective bolts.
E1 : Bolt is manufactured by machine A
E2 : Bolt is manufactured by machine B
E3 : Bolts is manufactured by machine C
25 35 40
P(E1 )= 100 , P(E2 )= 100 and P(E3 )= 100

59

5 4 2
P(A/E1 )= 100 , P(A/E2 )= 100 and P(A/E3 )= 100

P (E2 ) · P (A/E2 )
P (E2 /A) =
P (E1 ) · P (A/E1 ) + P (E2 ) · P (A/E2 )
35 4
100
× 100
= 25 5 35 4 40 2
100
× 100 + 100 × 100 + 100 × 100
140
=
125 + 140 + 80
28
=
69

Practice Problems

1. Given that E and F are events such that P(E)=0.6, P(F)=0.3 and P(E∩F)=0.2. Find
P(E/F) and P(F/E).
3
2. Let E and F are events with P(E)= 53 , P(F)= 10 and P(E∩F)= 15 . Are E and F independent.

3. Let A and B are two independent events with P(A)=0.3 and P(B)=0.4. Find (i)P(A∩B)
(ii)P(A∪B) (iii)P(A/B)

4. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One
of the two bags is selected at random and a ball is drawn from the bag which is found to
be red. Find the probability that the ball is drawn from the first bag.

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗∗

60