Download Study Modules XII Maths 2017 - Kendriya Vidyalaya Bilaspur

1
KENDRIYAVIDYALAYASANGATHAN
RAIPUR REGION
MODULES
for
CLASS-XII
MATHEMATICS
Session -2016-17
2
XII Class Chapter wise weightage
S.No
Chapter
1
Relations and Functions
(i)
Equivalence Relation
(ii)
Invertible functions
(iii)
Binary Operations
Weightage
6M
2
Inverse Trigonometric functions
2M
3
Matrices
2M
4
(i)
Practical Problems
(ii)
Equality of Matrices
Determinants
6M
(Matrix equation & Practical Problems)
5
Continuity and Differentiation
8M
6
Applications of Derivatives
4M
(i)
Increasing & Decreasing functions
(ii)
Tangents and Normal
7
Integration
4M
8
Application of Integrals
6M
9
Differential Equations
4M
(i)
Variable separable differential equation
(ii)
Homogeneous Differential equation
(iii)
Linear Differential equation
10
Vectors
4M
11
Three Dimensional Geometry
6M
(i)
Shortest distance
(ii)
Plane passing through intersection of two planes
12
Linear Programming problems
6M
13
Probability
6M
(i)
Bayes Theorem
(ii)
Probability distribution
(iii)
Multiplication rule
3
SAMPLE QUESTION PAPERS CAN BE DOWNLODED FROM
www.cbseacademic.nic.in
4
KENDRIYA VIDYALAYA SANGATHAN, RAIPUR REGION
BLUE PRINT AS PER SAMPLE PAPER PROVIDED BY CBSE
SUB:- MATHEMATICS
SL.
NO
NAME OF THE
CHAPTER
1
RELATIONS &
FUNCTIONS
INVERSE
TRIGONOMETRIC
FUNCTIONS
MATRICES
DETERMINANTS
CONTINUTY AND
DIFFERENTIABILITY
APPLCATION OF
DERIVATVES
INTEGRALS
APPICATION OF
INTEGRALS
DIFFERENTIAL
EQUATIONS
VECTORS
THREE DIMENSONAL
GEOMETRY
LINEAR PROGRAMMNG
PROBABILITY
TOTAL
2
3
4
5
6
7
8
9
10
11
12
13
1 Remembering – 20%
4. HOTS – 10%
Very
Short
Answer
(1 marks)
2(2)
CLASS-XII
Short
Answer
(2
marks)
6(1)
2(1)
2(1)
11(3)
10(3)
2(1)
4(1)
8(2)
2(1)
8(2)
2(1)
4(1)
2(1)
4(1)
2(1)
4(1)
4(1)
2(1)
16(8)
2. Understanding – 35%
5. Evaluation – 10%
8(3)
2(1)
6(1)
10(3)
6(1)
6(1)
12(3)
6(1)
6(2)
6(1)
6(1)
4(4)
Total
Marks
2(1)
1(1)
1(1)
L A – I ( L A - II
4marks)
(6
marks)
8(2)
44(11)
36(6)
7(3)
10(2)
6(1)
10(3)
100(29)
3. Application – 25%
5
MODULES FOR CLASS XII (M.L.L.)
Relations and Functions
Concept: - Types of relations
A relation R in a set A is called
(i) Reflexive, if a, a   R for every a  A
(ii) Symmetric, if a1 , a2   R  a2 , a1   R, for all a1 , a2  A
(iii) Transitive, if a1 , a2   R and a2 , a3   R implies that a2 , a3   R for all a1 , a 2 , a3  A
A relation R in a set A is said to be an equivalence relation if R is reflexive,
Symmetric and Transitive.
PRACTICE PROBLEMS
LEVEL –I
 Let T be the set of all triangles in plane with R a relation in T given by
R = {(T1,T2):T1 is congruent to T2}. Show that R is an equivalence relation.
 Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1,L2)
: L1 is a perpendicular to L2}. Show that R is symmetric but neither reflexive nor
transitive.

Show that the relation R in the set R of real numbers, defined as R  {( a, b) : a  b 2 } is
neither reflexive nor symmetric nor transitive.
LEVEL-II
 Show that the relation R in the set Z of integers given by R={(a, b) : 2 divides a - b}
is an equivalence relation
 Show that the relation R in the set A  {1, 2, 3, 4, 5} given by R = {(a, b) : a  b is

even }, is an equivalence relation. Show that all the element of {1,3,5} are related to
each other and all the element of {2,4} are related to each other. But no element of
{1,3,5} is related to any element of {2,4}.
Show that each of the relation R in the set A  {x  Z : 0  x  12} , given by
(i) R  {( a, b) : a  b is a multiple of 4}
(ii) R  {( a, b : a  b} , is an equivalence relation. Find the set of all elements related
to 1 in each case


LEVEL-III
Show that the relation R in the set A of points in a plane given by R = {(P, Q):
distance of the point P from the origin is same as the distance of the point Q from the
origin}, is an equivalence relation. Further, show that the set of all points P ≠ (0, 0) is
the circle passing through P with origin as centre.
Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1 is
similar to T2}, is equivalence relation. Consider three right angle triangles T1 with
sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangle among
T1, T2 and T3 are related?
6


Show that the relation R defined in the set A of all polygon as R = {(P1, P2) : P1 and
P2 have same number of sides}, is an equivalence relation. What is the set of all
elements in A related to the right angle triangle T with sides 3,4,and 5?
Let A = {1, 2, 3, 4…..,9} and R is the relation on A×A defined by (a, b) R (c, d) if
a + d = b + c for (a, b), (c, d) in A×A. Prove that R is an equivalence relation and also
obtain the equivalence class 2,5 .
Concept: One-one (injective), Onto (surjective) and bijective
Injective: - A function f : X  Y is define to be injective, if the image of distinct
element of X under f are distinct. For every x1 , x2  X , f x1   f x2   x1  x2
Surjective:- A function f : X  Y is said to onto (surjective) if every element of Y is the
image of some element of X under f, i.e. for every y  Y there exists an element x in X such
that f ( x )  y
Bijective: A function f : X  Y is said to be bijective if 𝑓 is one –one and onto
PRACTICE PROBLEMS
LEVEL – I
 Prove that the function f : R  R , given by f x   2 x is one - one and onto

Show that the function f : N  N given by f 1  f 2  1and f x   x  1 for every
x > 2 is onto but not one-one
Find the number of all one –one function from set A = {1, 2, 3} to itself.

LEVEL – II
Let A  R  {3} and B  R  {1} .consider the function f : A  B defined

 x2
by f ( x)  
 . Is f one-one and onto? Justify your answer.
 x3
Let A = {-1, 0, 1, 2}, B = {  4,  2, 0, 2} and f , g : A  B be a function defined by

f ( x)  x 2  x, x  A and g ( x)  2 x 
1
 1, x  A . Are f and g equal? Justify your
2
answer
LEVEL-III


Show that the function f : R  R given by f x   x 3 is not bijectve.
 x  1 , x  odd
Show that f : N  N given by f x   
is both one - one and onto.
 x  1 , x  even
Concept :- Composition of function and Inverse of Function
Composition of functions:- Let f : A  B and g : B  C be two function, then the
composition of f and g , denoted by gof , is defined as the function gof : A  C given by
gof  x  =g  f x , x  A
7
Inverse function:- A function f : X  Y is defined to be invertible. If there exists a function
g : Y  X Such that gof  I X and fog  I Y .The function g is called the inverse of f . If f is
invertible, then f must be one one and onto.
PRACTICE PROBLEMS
LEVEL – I
1
3 3

If f : R  R be given by f ( x)  (3  x ) , then fof (x) is………….

Consider f : N  N , g : N  N and h : N  R defined as f ( x)  2 x, g ( x)  3 y  4
and h(z )  sin z, x, y and z in N. Show that ho( gof )  (hog )of .

 4
4x
Let f : R     R be a function defined as f ( x) 
, Then show that
3x  4
 3
 4
4x
inverse of f is the map g : Range f  R    is g ( x) 
.
4  3x
 3
LEVEL – II

Consider f : R  4,  given by f ( x)  x 2  4 . Show that f is invertible with the
inverse f 1 of given by f 1 ( y) 

y  4 , where R is the set of all non-negative real
number.
Let f : R  R be defined as f ( x)  10 x  7 .Find the function g : R  R such that
gof  fog  1R .

Show that the function f : R  R defined by f  x  
x
, x  R is neither one-one
x 1
2
nor onto.
LEVEL – III

Consider f : R   5,  given by f ( x)  9 x 2  6 x  5 . Show that if f is

1
invertible with f ( x)  




y  6 1
.

3

Let f : N  R be a function defined as f ( x)  4 x 2  12 x  15 . Show that
f : N  S , where, S is the range of f , is invertible. Find the inverse of f .
Concept :- Binary Operations

A binary operation * on the set X is called commutative, if a * b  b * a , for every
a, b  X .

A binary operation *: A  A  A is said to be associative if
a * b * c  a * b * c, a, b, c, A .

Given a Binary operation *: A  A  A , an element e  A , if it exist, is called identity
for the operation *, if a * e  a  e * a, a  A .
8
Given a binary operation *: A  A  A with the identity element e in A, an element a  A is
said to be invertible with respect to the operation *,If there exist an element b in A such that
a * b  e  b * a and b is called the inverse of a and is donated by a-1.
PRACTICE PROBLEMS







LEVEL-I
Consider the binary operation  on the set{1,2,3,4,5} defined by a  b  min{ a, b} .
Write the operation table of the operation  .
Let * be the binary operation on N given by a*b = L.C.M of a and b.
Find
(i)
5*7
(ii)
20*16
Let * be a binary operation on the set Q of rational numbers as follows:(i) a*b = a - b
(ii) a*b = a2 + b2
(iii) a * b = a + ab
ab
(iv) a * b = (a - b)2
(v) a * b =
(vi) a * b = ab2
4
Find which of binary operation are commutative and which are associative?
LEVEL – II
Determine which of the following binary operation on the set R are associative and
which are commutative.
( a  b)
(a) a * b  1a, b  R
(b) a * b 
a, b  R
2
Let A  N  N and * be the binary operation on A defined by
a, b c, d   a  c, b  d  .Show that * is commutative and associative. Find the
identity element for *on A ,if any
Consider a binary operation *on N defined as a * b  a 3  b 3 .Choose the correct
answer.
(A)Is * both associative and commutative?
(B)Is * commutative but not associative?
(C)Is * associative but not commutative?
(D)Is * neither commutative nor associative
LEVEL – III
Consider the binary operation * : R  R  R and o: R  R  R defined as
a * b  a  b and a o b  a, a, b  R .Show that * is commutative but not associative,
o is associative but not commutative. Further, show that a, b, c  R,
a * (boc)  (a * b). [If it is so, we say that the operation * distributives over the

operation o].Does o distributive over *? Justify your answer.
Given a non-empty set X, let *: P ( X )  P( X )  P( X ) be defined as
A * B  ( A  B)  ( B  A), A, B  P( X ) .Show that the empty set  is the identify for
the operation * and all the element A of P(X) are invertible with A 1  A .
9

Define a binary operation * on the set {0,1,2,3,4,5,6} as
if a  b  7
a  b
a *b  
a  b  7 if a  b  7
Write the operation table of the operation * and prove that zero is the identity for this
operation and each element a  0 of the set is invertible with 7 - a being the inverse of
a.
INVERSE TRIGONOMETRIC FUNCTIONS
Introduction
Principal Value Branch Table
Principal Value
Functions
Domain
Branches
y = sin-1 x
[-1, 1]
  
 2 , 2 
y = cos-1 x
[-1, 1]
0, 
y = cosec-1 x
R – (-1, 1)
  
 2 , 2   0
y = sec-1 x
R – (-1, 1)
0,     
y = tan-1 x
R
  
 , 
 2 2
y = cot-1 x
R
0, 
2
Properties of Inverse Trigonometric Functions:
For suitable Values of domain, we have:
1.
(a) y = sin-1 x  x = sin y
(b) x = sin y  y = sin-1 x
2.
(a) sin (sin-1 x) = x
(b) sin-1 (sin x) = x
1
1
1
3.
(a) sin-1   = cosec-1 x
(b) cos-1   = sec-1 x (c) tan-1   = cot-1 x
x
x
x
4.
5.
6.
(a) cos-1 (-x) =  - cos-1 x
(a) sin-1 (-x) = - sin-1 x

(a) sin-1x + cos-1x =
2
(b) cot-1 (-x) =  - cot-1 x
(c) sec-1 (-x) =  - sec-1 x
(b) tan-1 (-x) = - tan-1 x
(c) cosec-1 (-x) = - cosec-1 x


(b) tan-1x + cot-1x =
(c) cosec-1x + sec-1x =
2
2
x y
1  xy
7.
1
1
1
(a) tan x  tan y  tan
1
1
1
(b) tan x  tan y  tan
8.
2 tan 1 x  sin 1
9.
(a) sin-1x + sin-1y = sin-1 x 1  y 2  y 1  x 2
2
2x
2x
1 1  x

cos
 tan 1
2
2
1 x
1 x
1 x2


x y
1  xy
10

(b) sin-1x - sin-1y = sin-1 x 1  y 2  y 1  x 2
10.

xy 
(a) cos-1x + cos -1y = cos -1 xy 
(b) cos -1x - cos -1y = cos -1

1  x 1  y  
1  x 1  y  
2
2
2
2
Important Solved Problems
1.
Write the principal value of tan 1 3  sec 1  2 .
Solution:




tan 1 3  sec 1  2  tan 1  tan   sec 1   sec 
3
3


 

   


  sec 1 sec           
3
3  3 
3
3
 
2.
Using principal value, evaluate the following:
 3 
sin 1  sin

5 

Solution:
 3  3 3    
sin 1  sin
as
  ,

5  5
5  2 2 

2
 3 

1 
sin 1  sin
  sin sin   
5 
5

 

3.
If tan 1 3  cot 1 x  , find x
2
Solution:

tan 1 3  cot 1 x 
2

cot 1 x   tan 1 3
2
2  2

1 
  sin  sin

5  5


tan-1x + cot-1x =

2
= cot 1 3  x  3
1  12 
1  3 
1  56 
Prove that: cos    sin    sin  
 13 
5
 65 
Solution:
4.
12
 12 
cos 1    x  cos x 
13
 13 
2
5
3
 12 
3
sin x  1  cos 2 x  1      and sin 1    y  sin y 
13
5
 13 
5
2
4
3
cos y  1  sin y  1    
5
5
sin (x + y) = sin xcosy + cosx sin y
2
11
 5  4   12  3  56
       
 13  5   13  5  65
=
 56 
 12 
3
 56 
 x  y  sin 1    cos 1    sin 1    sin 1  
 65 
 13 
5
 65 
 1  sin x  1  sin
1
Prove that cot 
 1  sin x  1  sin
Solution:
5.
x x
 
  , x   0, 
x 2
 4
 1  sin x  1  sin x 
cot 1 

 1  sin x  1  sin x 
 1  sin x  1  sin x
1  sin x  1  sin
 cot 1 

1  sin x  1  sin
 1  sin x  1  sin x


x
1 1  sin x  1  sin x  2 1  sin x 1  sin x
  cot 

1  sin x   1  sin x 
x


x 

2 cos 2
2



2

2
1

sin
x
1

cos
x

1 
1
2   cot 1  cot x   x
 cot 1 

cot
  cot 


2 sin x
2 2
 sin x 



 2 sin x cos x 
2
2


1  1 
1  1 
1  1 
1  1 
Prove the following: tan    tan    tan    tan    .
 3
5
7
8 4
Solution:
6.
1
1 
1
1

LHS  tan 1  tan 1    tan 1  tan 1 
3
5 
7
8

 1 1 
 1 1 
 

  
 tan 1  3 5   tan 1  7 8 
1 1  1 
1 1  1 




 7 8
 3 5
8
 15 
4
3
 tan 1    tan 1    tan 1    tan 1  
 14 
 55 
7
 11 
 4 3 




 65 
 tan 1  7 11   tan 1   tan 1 1   RHS
4
1 4  3 
 65 


 7 11 
7.

1  x  1 
1  x  1 
If tan 
  tan 
  , then find the value of x.
 x2
 x2 4
Solution:
12

 x 1 
1  x  1 
tan 1 
  tan 

 x 2
 x  2 4
 x 1 
1  x  1 
1
tan 1 
  tan 
  tan 1
x

2
x

2




x 1 

1

1  x  1 
1
1  x  1 
1 
x

2
  tan 1  1 
tan 
  tan 1  tan 
  tan
x 1 

 x 2
 x  2
 2x  3 
1 

x2

x 1
1
1
1


 2x 2  1  0  x 2   x  
x  2 2x  3
2
2
1.
PRACTICE PROBLEMES:
Level-1
Write the Principal value of the following:
(i) tan-1  3 
(ii) sin-1   1 
(iii) cos-1  

2.
3.
4.
5.
1.
2.
3.
4.
2
Evaluate: cottan a  cot a.
Prove: 3sin-1x = sin-1(3x - 4x3)
Find x if sec-1(2) + cosec-1x = /2
Solve tan-12x + tan-13x = /4
1
1
Level-2
Write the principal value of the following:
2 
2 

1 
cos 1  cos
  sin  sin

3 
3 


 1  x 2  1
,
Write in the simplest form: tan 
x


Prove that sin-1(8/17) + sin-1(3/5) = tan-1(77/36).
Prove that 2 tan-1(1/2)+ tan-1(1/7) = tan-1(31/17)
1
2
2x
2
1 1  x
tan

cot

2x
3
1 x2
1
5.
Solve for x
6.
Find the value of x if sin[cot-1(x+1)]=cos(tan-1x).
3
17 
2 sin 1  tan 1

5
31 4
Level-3
7.

2.
3.
 1 x  1 x   1
1
1
Prove that : tan 
   cos x
 1 x  1 x  4 2
Prove that: sin-1(12/13) + cos-1(4/5) + tan-1(63/16) = .
Prove that: tan-11 + tan-12 + tan-13 = .
4.
Prove that: tan 1
1.
1 

2
 x y 
x
 
 tan 1 
y
 x y 4
x 0
13

 1  x 
 .

1

x


5.
1
Write in the simplest form: cos 2 tan 
6.
If t an 1 x  tan 1 y  tan 1 z 
7.
 cos   cos  

 
 2. tan 1 tan . tan  .
Prove that cos 1 

2
2

1  cos  . cos  
8.
 1  cos x  1  cos x   x
3
Prove that tan 1 
.
   ; If   x 
2
 1  cos x  1  cos x  4 2
9.
If cos 1 x  cos 1 y  cos 1 z   . Then show that x2 + y2 + z2 + 2xyz = 1.


2
, x, y, z > 0 then find the value of xy + yz + zx.
14
MATRICES & DETERMINANTS
SCHEMATIC DIAGRAM
INTRODUCTION:
MATRIX: If mn elements can be arranged in the form of m row and n column in a
rectangular array then this arrangement is called a matrix.
Order of a matrix: A matrix having m row and n column is called a matrix of 𝑚 × 𝑛 order.
Addition and subtraction of matrices: Two matrices A and B can be added or subtracted if
they are of the same order i.e. if A and B are two matrices of order 𝑚 × 𝑛 then A ± B is also
a matrix of order m×n.
Multiplication of matrices: The product of two matrices A and B can be defined if the
number of rows of B is equal to the number of columns of A i.e. if A be an 𝑚 × 𝑛 matrix and
B be an 𝑛 × 𝑝 matrix then the product of matrices A and B is another matrix of order 𝑚 × 𝑝.
Transpose of a Matrix: If A = [aij] be an m × n matrix, then the matrix obtained by
interchanging. the rows and columns of A is called the transpose of A. Transpose of the
matrix A is denoted by 𝐴′ 𝑜𝑟 𝐴𝑇 .
Properties of transpose of the Matrices: For any matrices A and B of suitable orders, we
have
(i)
A 
T T
 A (𝑖𝑖) (𝐾𝐴)𝑇 = 𝐾𝐴𝑇 (𝑖𝑖𝑖) (𝐴 + 𝐵)𝑇 = 𝐴𝑇 + 𝐵𝑇 (𝑖𝑣) (𝐴𝐵)𝑇 = 𝐵𝑇 𝐴𝑇
Symmetric Matrix: A square matrix M is said to be symmetric if 𝐴𝑇 = 𝐴
𝑥 𝑦 𝑧
𝑎 𝑏
e.g.[
] , [𝑦 𝑢 𝑣 ]
𝑏 𝑐
𝑧 𝑣 𝑤
Note: there will be symmetry about the principal diagonal in Symmetric Matrix.
Skew symmetric Matrix: A square matrix M is said to be skew symmetric if 𝐴𝑇 = −𝐴
0
𝑒 𝑓
e.g.[ −𝑒 0 𝑔]
−𝑓 −𝑔 0
Note: All the principal diagonal element of a skew symmetric Matrix are zero.
Determinant: For every Square Matrix we can associate a number which is called the
Determinant of the square Matrix.
Determinant of a matrix of order one
Let A = [a ] be the matrix of order 1, then determinant of A is defined to be equal to a.
Determinant of a matrix of order two
𝑎 𝑏
Let A= [
] be a Square Matrix of order2× 2 then the determinant of A is denoted by |𝐴|
𝑥 𝑦
𝑎 𝑏
and defined by |𝐴| = |
|= ay - bx
𝑥 𝑦
Determinant of a matrix of order 3× 𝟑: Let us consider the determinant of a square matrix
of order 3× 3,
15
𝑎 𝑏 𝑐
|𝐴|= |𝑝 𝑞 𝑟|
𝑥 𝑦 𝑧
Expansion along first row |𝐴| = 𝑎(𝑞𝑧 − 𝑦𝑟) − 𝑏(𝑝𝑧 − 𝑥𝑟) + 𝑐(𝑝𝑦 − 𝑞𝑥)
We can expand the determinant with respect to any row or any column.
Minors and cofactors:
Minor of an element 𝑎𝑖𝑗 of a determinant is the determinant obtained by
deleting its ith row and jth column in which element 𝑎𝑖𝑗 lies. Minor of an element 𝑎𝑖𝑗 is
denoted by 𝑀𝑖𝑗 .
Cofactors: cofactors of an element 𝑎𝑖𝑗 𝑑𝑒𝑛𝑜𝑡𝑒𝑑 by 𝐴𝑖𝑗 and is defined by 𝐴𝑖𝑗 = (−1)𝑖+𝑗 𝑀𝑖𝑗
where 𝑀𝑖𝑗 is the minor of 𝑎𝑖𝑗 .
𝛼 𝛽
Adjoint of a Matrix: Let A = [
] be a Matrix of order 2 × 2
𝛾 𝛿
𝜹 −𝜷
Then adj(A) = [
]
−𝜸 𝜶
𝑥 𝑦 𝑧
Again letA = [𝑝 𝑞 𝑟] be a Matrix of order 3 × 3
𝑎 𝑏 𝑐
𝑞 𝑟
𝑝 𝑞 𝑇
𝑝 𝑟
|
| −|
| |
|
𝑎 𝑐
𝑏 𝑐
𝑎 𝑏
𝑦 𝑧
𝑥 𝑦
𝑥 𝑧
Then adj (A) = − |𝑏 𝑐 | |𝑎 𝑐 | − |𝑎 𝑏 |
=
𝑦 𝑧
𝑥 𝑧
𝑥 𝑦
|
|
−
|
|
|
[ 𝑞 𝑟
𝑝 𝑟
𝑝 𝑞| ]
𝑞𝑐 − 𝑏𝑟
[−(𝑦𝑐 − 𝑏𝑧)
𝑦𝑟 − 𝑞𝑧
𝑇
−(𝑝𝑐 − 𝑎𝑟)
𝑝𝑏 − 𝑎𝑞
𝑞𝑐 − 𝑏𝑟
𝑥𝑐 − 𝑎𝑧
−(𝑏𝑥 − 𝑎𝑦)] = [ 𝑎𝑟 − 𝑝𝑐
𝑝𝑏 − 𝑎𝑞
−(𝑥𝑟 − 𝑝𝑧)
𝑥𝑞 − 𝑝𝑦
𝑏𝑧 − 𝑎𝑦
𝑥𝑐 − 𝑎𝑧
𝑎𝑦 − 𝑏𝑥
𝑦𝑟 − 𝑞𝑧
𝑝𝑧 − 𝑥𝑟 ]
𝑥𝑞 − 𝑝𝑦
Inverse of a Matrix: Inverse of a Square Matrix A is defined as 𝑨−𝟏 =
𝒂𝒅𝒋(𝑨)
|𝑨|
Note: If A be a given Square Matrix of order n then
(i)
A(adj(A) = adj(A)A=|𝑨|𝑰where I is the Identity Matrix of order n.
(ii)
A square Matrix A is said to be singular and non-singular according as |𝑨| =
𝟎 𝒂𝒏𝒅 |𝑨| ≠ 𝟎
(iii) |𝒂𝒅𝒋(𝑨)|= |𝑨|𝒏−𝟏 (𝑭𝒐𝒓 𝒂 𝒔𝒒𝒖𝒂𝒓𝒆 𝑴𝒂𝒕𝒓𝒊𝒙 𝒐𝒇 𝒐𝒓𝒅𝒆𝒓 𝟑 × 𝟑 |𝒂𝒅𝒋(𝑨)| = |𝑨|𝟐 )
IMPORTANT SOLVED PROBLEMS
−2
Q1. If A=[ 4 ] , 𝐵 = [1 3 −6] , Verify that (AB)′ = B′A′
5
Solution: - We have
−2
If A=[ 4 ] , 𝐵 = [1 3 −6]
5
−2
−2 −6 12
Then
AB =[ 4 ] [1 3 −6] = [ 4 12 −24]
5
5 15 −30
16
Now
1
5] , B′= [ 3 ]
−6
A′ = [−2 4
1
−2
4
5
B′A′ = [ 3 ] [−2 4 5] = [−6 12
15 ] = (AB)′
−6
12 −24 −30
Clearly (AB)′ = B′A′
2
10
−1
Q2. If 𝑥 [ ] + 𝑦 [ ] = [ ] then find the value of x and y.
3
5
1
2
10
−1
Sol. Given 𝑥 [ ] + 𝑦 [ ] = [ ]
3
5
1
−𝑦
2𝑥
−
𝑦
2𝑥
10
10
[ ] + [ 𝑦 ] = [ ] or [
]=[ ]
3𝑥 + 𝑦
3𝑥
5
5
So 2x – y = 10 and 3x + y = 5 ; On solving we get x = 3 and y = -4
𝑐𝑜𝑠𝑥 −𝑠𝑖𝑛𝑥 0
Q3. If F(x) = [ 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 0] prove that F(x) F(y) = F(x+y)
0
0
1
𝑐𝑜𝑠𝑦 −𝑠𝑖𝑛𝑦 0
𝑐𝑜𝑠𝑥 −𝑠𝑖𝑛𝑥 0
Sol. Given F(x) = [ 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 0] so F(y) = [ 𝑠𝑖𝑛𝑦 𝑐𝑜𝑠𝑦 0]
0
0
1
0
0
1
𝑐𝑜𝑠𝑥 −𝑠𝑖𝑛𝑥 0 𝑐𝑜𝑠𝑦 −𝑠𝑖𝑛𝑦 0
Hence F(x) .F(y) = [ 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥 0] [ 𝑠𝑖𝑛𝑦 𝑐𝑜𝑠𝑦 0] =
0
0
1
0
0
1
𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 − 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦 −𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦 − 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦 0
[𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦 + 𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦 −𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦 + 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 0]
0
0
1
cos(𝑥 + 𝑦) −sin(𝑥 + 𝑦)
= [ sin(𝑥 + 𝑦) cos(𝑥 + 𝑦)
0
0
Hence F(x) F(y) = F(x+y)
0
0]
1
Q4. Express the given Matrix as the sum of a symmetric and skew symmetric matrix
6 −2 2
A= [−2 3 −1]
2 −1 3
6 −2 2
Sol. Here 𝐴𝑇 = [−2 3 −1]
2 −1 3
12 −4 4
6 −2 2
1
1
P = 2 (𝐴 + 𝐴𝑇 ) = 2 [−4 6 −2] = [−2 3 −1]
4 −2 6
2 −1 3
1
𝑇
𝑇
Now 𝑃 = 𝑃 so
P = 2 (𝐴 + 𝐴 ) is a symmetric Matrix.
0 0 0
0 0
Also let Q = = 2 (𝐴 − 𝐴 = 2 [0 0 0] = [0 0
0 0 0
0 0
𝑇
𝑄 = −𝑄 Hence Q is an Skew Symmetric Matrix.
6 −2 2
0 0 0
6
Now P + Q = [−2 3 −1] + [0 0 0] = [−2
2 −1 3
0 0 0
2
1
𝑇)
1
0
0]
0
−2 2
3 −1] = 𝐴
−1 3
17
Thus A is represented as the sum of a symmetric and skew symmetric matrix.
Q5. Using the property of determinant prove that:
𝑎−𝑏−𝑐
2𝑎
2𝑎
| 2𝑏
𝑏−𝑐−𝑎
2𝑏 | = (𝑎 + 𝑏 + 𝑐)3
2𝑐
2𝑐
𝑐−𝑎−𝑏
Sol.Applying𝑅1 → 𝑅1 + 𝑅2 + 𝑅3 𝑤𝑒 𝑔𝑒𝑡
𝑎+𝑏+𝑐 𝑎+𝑏+𝑐 𝑎+𝑏+𝑐
L.H.S = | 2𝑏
𝑏−𝑐−𝑎
2𝑏 |
2𝑐
2𝑐
𝑐−𝑎−𝑏
Taking common a + b + c from first Row we get
1
1
1
L.H.S =(a + b+ c) |2𝑏 𝑏 − 𝑐 − 𝑎
2𝑏 |
2𝑐
2𝑐
𝑐−𝑎−𝑏
Now applying 𝐶2 → 𝐶2 − 𝐶1 , 𝐶3 → 𝐶3 − 𝐶1 𝑤𝑒 𝑔𝑒𝑡
1
0
0
0
L.H.S =(a + b+ c) |2𝑏 −(𝑎 + 𝑏 + 𝑐)
|
2𝑐
0
−(𝑎 + 𝑏 + 𝑐)
Expanding along first Row L.H.S = (a+b+c)[(𝑎 + 𝑏 + 𝑐)2 − 0] = (𝑎 + 𝑏 + 𝑐)3 = R.H.S
Hence proved
Q6. Solve the system of equations x + 2y – 3z = - 4, 2x + 3y + 2z = 2, 3x – 3y – 4z = 11
Sol. The given system of equation can be written as A X = B where
𝑥
1 2 −3
−4
𝑦
A = [2 3
]
𝑋
=
[
]
𝐵
=
[
2
2]
𝑧
3 −3 −4
11
1 2 −3
Now |𝐴| = |2 3
2 | = −6 + 28 + 45 = 67
3 −3 −4
−6 17 13
−6 14 −15 𝑇
𝒂𝒅𝒋(𝑨)
adj(A) = [ 17 5
𝑨−𝟏 = |𝑨|
=
5 −8] 𝑵𝒐𝒘
9 ] = [ 14
−15 9 −1
13 −8 −1
−6 17 13
𝟏
[ 14
5 −8]
𝟔𝟕
−15 9 −1
24 + 34 + 143
−6 17 13 −4
𝟏
𝟏
Hence 𝑋 = 𝑨−𝟏 𝑩 = 𝟔𝟕 [ 14
]
[
]
=
[
−56 + 10 − 88]
5 −8 2
𝟔𝟕
60 + 18 − 11
−15 9 −1 11
𝑥
201
3
𝟏
So[𝑦] = 𝟔𝟕 [−134] = [−2]
𝑧
67
1
Hence x = 3 , y = -2 , z = 1
Flow chart:
Step 1.Write the given system of equation in the form of A X = B
Step2. Find |𝑨|
Step3. Find adj(A)
Step4. Find 𝑨−𝟏 =
𝒂𝒅𝒋(𝑨)
|𝑨|
−𝟏
Step5. Find 𝑿 = 𝑨 𝑩
18
Step6 Find the value of x , y and z
ASSIGNMENTS
(i). Order, Addition, Multiplication and transpose of matrices:
LEVEL I
1. If a matrix has 6 elements, what are the possible orders it can have?
2. Construct a 3 × 2 matrix whose elements are given by aij = |i – 3j |
3.
If A =
4. If A =
, B=
,
then find A – 2 B.
and B =
, write the order of AB and BA.
3 6
5 2
5. Find the matrices X and Y if 𝑋 + 𝑌 = [
] 𝑎𝑛𝑑 𝑋 − 𝑌 = [
]
0 −1
0 9
0 𝑥
[𝑥 1] [ 1
6. Solve:
] [ ] = 0.
−2 −3 3
LEVEL II
1. For the following matrices A and B, verify (AB)T = BTAT, where A =
, B=
.
2. Give example of matrices A & B such that AB = O, but BA ≠ O, where O is a zero matrix and
A, B are both non zero matrices.
3. If B is skew symmetric matrix, write whether the matrix (ABAT) is symmetric or skew symmetric.
4. If A =
and I =
, find a and b so that A2 + aI = bA
LEVEL III
1. If
A =
, then find the value of A2 –3A + 2I
2. Express the matrix A as the sum of a symmetric and a skew symmetric matrix, where:
A=
3. If A = [
𝑐𝑜𝑠𝜃
−𝑠𝑖𝑛𝜃
𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝑛𝜃
] 𝑡ℎ𝑒𝑛 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 = 𝐴𝑛 = [
𝑐𝑜𝑠𝜃
−𝑠𝑖𝑛𝑛𝜃
(ii) Cofactors &Adjoint of a matrix
LEVEL I
1. Find the co-factor of 𝑎12
in A =
2. Find the adjoint of the matrix A =
LEVEL II
Verify A(adjA) = (adjA) A =
1. A =
I if
2.
(iii) Inverse of a Matrix & Applications
A=
𝑠𝑖𝑛𝑛𝜃
] , 𝑛𝜖𝑁
𝑐𝑜𝑠𝑛𝜃
19
LEVEL I
1.
2.
If A =
, write A-1 in terms of A
If A is square matrix satisfying A2 = I, then what is the inverse of A ?
3.
For what value of k , the matrix A =
LEVEL II
is not invertible ?
1. If A =
, show that A2 –5A – 14I = 0. Hence find A-1
2. If A, B, C are three non-zero square matrices of same order, find the condition
on A such that AB = AC  B = C.
3. Find the number of all possible matrices A of order 3 × 3 with each entry 0 or 1.
LEVEL III
1.
2.
3.
4.
If A =
, find A-1 and hence solve the following system of
equations: 2x – 3y + 5z = 11,
3x + 2y – 4z = - 5, x + y – 2z = - 3.
Using matrices solve the following system of equations:
(i) x + 2y - 3z = - 4 ,
2x + 3y + 2z = 2 ,
3x - 3y – 4z = 11
(ii) 4x + 3y + 2z = 60 ,
x + 2y + 3z = 45 ,
6x + 2y + 3z = 70
Find the product AB, where A =
,B=
the equations x – y = 3, 2x + 3y + 4z = 17, y + 2z = 7
1

x
2
Using matrices solve the following system of equations:

x
1

x
and use it to solve
1 1
 4
y z
1 3
 0
y z
1 1
 2
y z
5.
A trust caring for indicate children gets rupees 30,000/- every month from its donors.
The trust spends half of the funds received for medical and educational care of the
children and for that it charges 2% of the spent amount from them and deposits the
balance note in a private bank to get the money multiplied so that in future the trust
goes on functioning regularly. What % of interest should the trust get from the bank to
get a total of rupees 1800/- every month? Use matrix method to find the rate of
interest? Do u think people should donate to such trusts?
6.
Using elementary transformations, find the inverse of the matrix
(iv)
LEVEL I
20
1. Evaluate
Cos15 Sin15
Sin 75 Cos75
2. What is the value of
, where I is identity matrix of order 3?
3. If A is non singular matrix of order 3 and
= 3, then find
4. For what valve of a,
is a singular matrix?
LEVEL II
If A is a square matrix of order 3 such that
= 64, find
If A is a nonsingular matrix of order 3 and
= 7, then find
LEVEL III
1.
2.
3
If A =
and
= 125, then find a.
A square matrix A, of order 3, has
= 5, find
1.
2.
(v)
Properties of Determinants
LEVEL I
1.
Find positive value of x if
2.
1.
2.
=
Evaluate
LEVEL II
Using properties of determinants prove the following:
bc
a
a
b
ca
b  4abc
c
c
ab
1  a 2  b2
2ab
 2b
2
2
2ab
1 a  b
2a
 1  a 2  b2
2b
 2a
1 a 2  b2

3.

3
= (1 + pxyz)(x - y)(y - z) (z - x)
LEVEL III
1. Using properties of determinants, solve the following for x :
a.
b.
= 0
= 0
c.
= 0
2. If a, b, c, are positive and unequal, show that the following determinant is negative:
21
=
3.
a2 1
ab
ac
ab
b2  1
bc  1  a 2  b 2  c 2
ca
cb
c2  1
4.
a
b
c
a  b b  c c  a  a 3  b 3  c3  3abc
bc ca ab
5.
b 2c 2
c 2a 2
a 2b2
 bc
6.
bc b  c
ca c  a  0
ab a  b
b 2  bc c 2  bc
a 2  ac
 ac
c 2  ac  (ab  bc  ca ) 3
a 2  ab b 2  ab
 ab
= 2abc( a + b + c)3
7.
8.
If a, b, c are real numbers, and
bc ca a b
ca a b bc  0
a b bc ca
Show that either a + b +c = 0 or a = b = c.
ANSWERS
1. Order, Addition, Multiplication and transpose of matrices:
LEVEL I
1. 1  6, 6  1 , 2  3 , 3  2
3.
skew symmetric
2.
3.
LEVEL II
4. a = 8, b = 8
LEVEL III.
4. 2  2, 3  3
22
1.
2.
+
(ii). Cofactors & Adjoint of a matrix
LEVEL I
1. 46
2.
(iii)Inverse of a Matrix & Applications
LEVEL I
-1 = -
1. A
-1 =
A
2. A
A
3. k = 17
LEVEL II
1.
3. 512
LEVEL III
1. x = 1, y = 2, z = 3.
2. (i) x = 3, y = -2, z = 1. (ii) x = 7, y = 4, z = 10 3. AB = 6I,
x = , y = - 1, z =
4. x = ½, y = -1, z = 1.
6.
(iv). To Find The Difference Between
LEVEL I
1. 0
2. 27
1. 8
2. 49
3.24
LEVEL II
LEVEL III
1. a = 3
2. 125
(v). Properties of Determinants
LEVEL I
1. x = 4
2.
+
+ +
LEVEL II


2. [Hint: Apply C1 C1–bC3 and C2 C2+aC3]
4.
23
CONTINUITY AND DIFFERENTIABILITY
Concept :- Continuity
Suppose 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  or
x c
lim f c  h   f c   lim f c  h 
h 0
h 0
PRACTICE PROBLEMS
LEVEL – I

Examine whether the function f given by f x   x 2 is continuous at x = 0

Discuss the continuity of the function f given by f x   x at x = 0

Show that every polynomial function is continuous
LEVEL- II
Show that the function f defined by f x   1  x  x , where x is any real number, is a



continuous function .
Find the relationship between a and b so that the function f defined by
ax  1, if x  3
f x   
is continuous at x  3 .
bx  3 if x  3
 ( x 2  2 x ), if
For what value of  is the function defined by f x   
if
4 x  1
x0
x0
continuous at x  0 ? What about continuity at x  1 ?
LEVEL-III
 1  sin x
if
 3 cos 2 x


Let f  x   
a
if

 b1  sin x  if
2

   2 x 
a and b.
3



x
x
x

2

2
If f x  be a continuous function at x 

the
value
2𝑥+2 −16
{
of
k,
so
that
𝑘
, 𝑖𝑓 𝑥 = 2
is continuous at 𝑥 = 2.

Evaluate:
𝑡𝑎𝑛𝑥−𝑠𝑖𝑛𝑥
𝑥→0 𝑠𝑖𝑛3 𝑥
lim
, find
x 1
x  1 is continuous at x = 1,find the value of a
x 1
, 𝑖𝑓 𝑥 ≠ 2
4𝑥 −16
2
2
 3ax  b if

If the function f x    11
if
5ax  2b if

and b.
Find

√1+2𝑥 − √1−2𝑥
𝑠𝑖𝑛𝑥
𝑥→0
lim
a
function
𝑓(𝑥) =
24
Concept :- Differentiation of Implicit function, logarithmic functions, functions in parametric
forms, second order derivatives, Rolle’s theorem and lagrange’s Mean value theorem
log m  n   log m  log n
m
log    log m  log n
n
log m n  n log m
log a b 
log c b
log c a
log a a  1
PRACTICE PROBLEMS
LEVEL – I
𝑑𝑦

Find 𝑑𝑥

Find 𝑑𝑥

Find 𝑑𝑥

𝑑𝑦
Find 𝑑𝑥
: 𝑥 3 + 𝑥 2 𝑦 + 𝑥𝑦 2 + 𝑦 3 = 81

Find 𝑑𝑥
𝑑𝑦
: 𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑦 = 1


Verify Rolle’s theorem for the function 𝑦 = 𝑥 2 + 2, 𝑎 = −2 𝑎𝑛𝑑 𝑏 = 2
If 𝑓: [−5, 5] → 𝑅 is differentiable function and if 𝑓′(𝑥) does not vanish anywhere,
then prove that 𝑓(−5) ≠ 𝑓(5)
LEVEL - II

Differentiate

Find

d2y
Find 2 , if y  x 3  tan x .
dx

If x  a sin t , y  a cos

x 1

 sin x 
1  2

Differentiate the following w.r.t x (i) tan 1 
 (ii) sin 
x 
 1  cos x 
1  4 

d2y
dy
 5  6y  0 .
If y  3e  2e , prove that
2
dx
dx

 1 
1

Differentiate the function :-  x    x  x  .
x


Differentiate the function :- x cos x x  x sin x  x .
𝑑𝑦
𝑑𝑦
:
2𝑥 + 3𝑦 = 𝑠𝑖𝑛𝑦
:
𝑦 = 𝑠𝑖𝑛2 (3𝑥 + 1)3
:
𝑦 = tan(𝑥 + 𝑦)
x  3x 2  4 w.r.t
3x 2  4 x  5
x.
dy
of the function:- x y  y x  1 .
dx
1
2x
1
t
, show that
dy
y
 .
dx
x
3x
x

1
1
25

Differentiate the function :- x sin x  sin x 

Differentiate the function :- x x cos x 

Find










cos x
.
x2 1
.
x2 1
dy
of the function:- x y  y x  1 .
dx
dy
y
x
Find
of the function:- cos x   cos y  .
dx
x  a  sin  , y  a1  cos  .
t

x  a cos t  log tan , y  a sin t .
2

Verify Mean Value Theorem if 𝑓(𝑥) = 𝑥 2 − 4𝑥 − 3 in the interval [a, b], where a= 1
and b = 4.
LEVEL –III
dy
Find
, if y x  x y  x x  a b .
dx


x
2
If y  tan 1 x ,show that
2
 1 y2  2 x x 2  1y1  2
2
  dy 
1   
  dx 
2
2
If x  a    y  b   c 2 , for some 𝑐 > 0 prove that 
d2y
dx 2
independent of a and b
2



3
2
dy cos 2 a  y 

dx
sin a
dy
1

If x 1  y  y 1  x  0 ,for, -1< x <1 , Prove that
dx
1  x 2
If cos y  x cosa  y  with cos a  1 ,prove that

If x  cos(log y ),  1  x  1 , Show that 1  x 2

d
2
y
d
2
dx
y
2
x
dy
y0
dx
dy
 0.
dx

If y  sin 1 x , show that 1  x 2

Find

If x 

Differentiate

t
dy
 1
For a positive constant a find
,where y  a t , x   t  
dx
 t
dx
2
x
dy
of the function:- y x  x y .
dx
sin 3 t
cos 2t
,y 
cos 3 t
cos 2t
. Find
x  3x 2  4
3x 2  4 x  5
dy
dx
w.r.t x.
1
a
is a constant
26

If x  acos t  t sin t  and y  asin t  t cos t  , find

If y  500e 7 x  600e 7 x ,show that

d2y
dx 2
d2y
 49 y
dx 2
Examine the validity and conclusion of the Lagrange’s mean value theorem for the
function 𝑓(𝑥) = √𝑥 2 − 4 in the interval [2, 4]
Applications of Derivatives
Concept :Increasing or Decreasing function:- Let I be an open interval contained in the
domain of a real valued function f . Then f is said to be
Increasing on I if x1  x2 in I  f x1   f x2 , x1 , x2  I
(i)
Strictly increasing on I if x1  x2 in I  f x1   f x2 , x1 , x2  I
(ii)
(iii)
Decreasing on I if x1  x2 in I  f x1   f x2 , x1 , x2  I
Strictly decreasing on I if x1  x2 in I  f x1   f x2 , x1 , x2  I
(iv)
Theorem ;- Let f be continuous on a, b and differentiable on the open interval a, b
then
(i)
(ii)
f is increasing in a, b if f ' x  0 for each x  a, b
f is decreasing in a, b if f ' x  0 for each x  a, b
(iii)
f is a constant function in a, b if f ' x   0 for each x  a, b
(i)
f is strictly increasing in a, b if f ' x  0 for each x  a, b
Theorem ;- Let f be continuous on a, b and differentiable on the open interval a, b
then
(ii)
(iii)
f is decreasing in a, b if f ' x  0 for each x  a, b
f is a constant function in a, b if f ' x   0 for each x  a, b
PRACTICE PROBLEMS



LEVEL-I
Show that f(x) = 2x + 3 is strictly increasing.
Prove that f(x) = cosx is
(i)
Strictly decreasing in (0,  )
(ii)
Strictly increasing in (  ,2  )
(iii)
Neither increasing nor decreasing in (0,2  ).
Find the least value of a such that the function given by f(x) = x2 + ax + 1 is strictly
increasing on (1,2)
LEVEL-II
27







Find the intervals in which the function f is given by 𝑓(𝑥) = 4 𝑥 3 − 6𝑥 2 − 72𝑥 + 30
is (a) strictly increasing (b) strictly decreasing.
 
Find the intervals in which the function given by f  x   sin 3x, x  0,  is
 2
(a) increasing (b) decreasing
Find the intervasl in which the function f given by f x   sin x  cos x,0  x  2 is
strictly increasing or strictly decreasing.
Find the interval in which𝑦 = 𝑥 2 𝑒 −𝑥 is increasing.
The length of rectangle is decreasing at the rate of 5cm/min and the width y is
increasing at the rate of 4cm/min .When x = 8cm and y = 6cm Find the rate of change
of (a) the perimeter and (b) the area of rectangle.
3
A balloon, which always remains spherical, has a variable diameter 2 x  1 find the
2
rate of change of its volume with respect to x.
A particle moves along the curve 6 y  x 3  2 .Find the point on the curve at which the
y-coordinate is changing 8 times as fast as the x-coordinate.
LEVEL – III
4 sin x  2 x  x cos x
is
2  cos x

Find the intervals in which the function f given by f  x  

(i) increasing
(ii) decreasing
A water tank has the shape of an inverted right circular cone with its axis vertical and
vertex lower most. Its semi vertical angle is tan 1 0.5 .Water is poured into it at a
constant rate of 5 cubic meters per hour. Find the rate at which the level of the water
is rising at the instant when the depth of water in the tank is 4 m.
1
 Find the intervals in which the function f given by f x   x 3  3 , x  0 is
x
(i) increasing (ii) decreasing
Concept :Tangents and Normal , Approximation
Tangents and Normal:- The slope of the tangent to the curve y  f x  at the
point x0 , y 0  is given by
dy 
dx   x0 , y0 
The slope of the normal to the curve y  f x  at the point x0 , y 0  is given by
−1
dy 
dx   x0 , y0 
The equation of a tangent at x0 , y 0  to the curve y  f x  is given by
y  y0  f ' x0 x  x0 
28
The equation of a normal at x0 , y 0  to the curve y  f x  is given by y  y 0 
1
 x  x0 
f '  x0 
Tangent line parallel to x-axis then equation of the tangent y  y0
Tangent line parallel to y-axis then equation of the tangent x  x0
Approximations:
(i)
The differential of x, denoted by dx , is defined by dx  x .
(ii)
The differential of y, denoted by dy , is defined by dy  f ' x0  dx or
 dy 
dy   x .
 dx 
PRACTICE PROBLEMS
LEVEL –I
 Find the slope of the tangent to the curve y  3x  4 x at x = 4
2
 Find the slope of the tangent to the curve 𝑦 = 𝑥 2 − 3𝑥 + 2 at the point whose xcoordinate is 3

 Find the slope of the normal to the curve x  a cos 3  , y  a sin 3  , at ,  
4
LEVEL –II
2
 Find the point at which the tangent to the curve 𝑦 = √4𝑥 − 3 − 1 has its slope .
3
 Find the equation of the tangent line to the curve y  x 2  2 x  7 which is (i) parallel to
the line 2 x  y  9  0 (ii) Perpendicular to the line 5 y  15x  13
 Find the point on the curve x 2  y 2  2 x  3  0 at which the tangents are the parallel to the
x-axis.
LEVEL-III

Find the equation of the normal to the curve y  x  2 x  6 which are parallel to the line
3
x + 14y + 4 = 0.

3
Prove that the curves x  y and xy  k cut at right angle if 8k  1

Find the equation of the tangent to the curve y 
2
3 x  2 which is parallel to the line
4x  2 y  5  0
INTEGRATION
INTRODUCTION
IF f(x) is derivative of function g(x), then g(x) is known as antiderivative or integral of
f(x)
29
𝒅
i.e.,
⇔
(g(x)) = f(x)
𝒅𝒙
∫ 𝒇(𝒙)𝒅𝒙 = 𝒈(𝒙) + C
STANDARD SET OF FORMULAS
* Where c is an arbitrary constant.
1.
∫ 𝒙𝒏 𝒅𝒙
2.
∫ 𝒅𝒙
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
=
𝒙𝒏+𝟏
𝒏+𝟏
=
𝟏
∫ 𝒙dx
(n  -1)
+𝒄
x+ c
= log |x| + c
= 𝒔𝒊𝒏 𝒙 + 𝒄
∫ 𝒄𝒐𝒔 𝒙 𝒅𝒙
= − 𝒄𝒐𝒔 𝒙 + 𝒄
∫ 𝒔𝒊𝒏 𝒙 𝒅𝒙
= 𝒕𝒂𝒏 𝒙 + 𝒄
∫ 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙
𝟐
∫ 𝒄𝒐𝒔𝒆𝒄 𝒙 𝒅𝒙 = −𝒄𝒐𝒕 𝒙 + 𝒄
∫ 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙 𝒅𝒙 = 𝒔𝒆𝒄 𝒙 + 𝒄
∫ 𝒄𝒐𝒔𝒆𝒄 𝒙 𝒄𝒐𝒕 𝒙 𝒅𝒙 = −𝒄𝒐𝒔𝒆𝒄 𝒙 + 𝒄
= 𝒆𝒙 + 𝒄
∫ 𝒆𝒙 𝒅𝒙
= 𝒍𝒐𝒈 |𝒔𝒆𝒄 𝒙 | + 𝒄 = −𝒍𝒐𝒈 |𝒄𝒐𝒔𝒙| + 𝒄
∫ 𝒕𝒂𝒏 𝒙 𝒅𝒙
= 𝒍𝒐𝒈 |𝒔𝒊𝒏 𝒙 | + 𝒄 = −𝒍𝒐𝒈 |𝒄𝒐𝒔𝒆𝒄 𝒙 | + 𝒄
∫ 𝒄𝒐𝒕 𝒙 𝒅𝒙
= 𝒍𝒐𝒈 |𝒔𝒆𝒄 𝒙 + 𝒕𝒂𝒏 𝒙 | + 𝒄
∫ 𝒔𝒆𝒄 𝒙 𝒅𝒙
= 𝒍𝒐𝒈 |𝒄𝒐𝒔𝒆𝒄 𝒙 − 𝒄𝒐𝒕 𝒙 | + 𝒄
∫ 𝒄𝒐𝒔𝒆𝒄 𝒙 𝒅𝒙
𝟏
15.
∫√
16.
∫ 𝟏+𝒙𝟐 𝒅𝒙
17.
∫
𝟏− 𝒙𝟐
𝟏
= sin -1 x + c
dx
𝟏
𝒙√ 𝒙𝟐 − 𝟏
= 𝒕𝒂𝒏−𝟏 𝒙 + 𝒄
= 𝐬𝐞𝐜 −𝟏 𝒙 + 𝒄
𝒅𝒙
18.
∫ 𝒂𝒙 𝒅𝒙
19.
∫ √𝒙 𝒅𝒙
𝒂𝒙
=
𝟏
𝒍𝒐𝒈 𝒂
+ 𝒄
= 𝟐√ 𝒙 + 𝒄
INTEGRALS OF LINEAR FUNCTIONS
1.
∫(𝒂𝒙 + 𝒃)𝒏 𝒅𝒙
2.
∫ 𝒂𝒙+𝒃 𝒅𝒙
3.
∫ 𝒔𝒊𝒏 (𝒂𝒙 + 𝒃 )𝒅𝒙
𝟏
=
(𝒂𝒙+𝒃)𝒏+𝟏
=
+ 𝒄
(𝒏+𝟏)𝒂
𝒍𝒐𝒈 (𝒂𝒙+𝒃)
+ 𝒄
𝒂
−𝒄𝒐𝒔 (𝒂𝒙+𝒃)
=
𝒂
+ 𝒄
In the same way if ax +b comes in the place of x, in the standard set of formulas, then
divide the integral by a
SPECIAL INTEGRALS
1.
𝟏
∫ 𝒙𝟐 −𝒂𝟐 𝒅𝒙
𝟏
2.
∫ 𝒂𝟐 −𝒙𝟐 𝒅𝒙
3.
x
4.
∫√
𝒙𝟐 −𝒂𝟐
5.
∫√
𝒙𝟐 +𝒂𝟐
2
𝟏
=
=
𝟐𝒂
𝟏
𝟐𝒂
𝒙−𝒂
𝒍𝒐𝒈 |𝒙+𝒂| + c
𝒂+𝒙
𝒍𝒐𝒈 |𝒂−𝒙| + 𝒄
1
1
x
dx  tan 1  c
2
a
a
a
𝟏
𝟏
𝒅𝒙
= 𝒍𝒐𝒈| 𝒙 + √𝒙𝟐 − 𝒂𝟐 | + 𝒄
𝒅𝒙
= 𝒍𝒐𝒈 |𝒙 + √𝒙𝟐 + 𝒂𝟐 |+ c
30
𝟏
𝒙
= 𝐬𝐢𝐧−𝟏 (𝒂) + 𝒄
𝒅𝒙
6.
∫√
7.
∫ √𝒙𝟐 + 𝒂𝟐 𝒅𝒙
𝒂𝟐 −𝒙𝟐
𝒙
= 𝟐 √𝒙𝟐 + 𝒂𝟐 +
𝒂𝟐
𝒙
𝒂𝟐
𝒙
𝟐
𝒂𝟐
8.
∫ √𝒙𝟐 − 𝒂𝟐 dx
= 𝟐 √𝒙𝟐 − 𝒂𝟐 −
9.
∫ √𝒂𝟐 − 𝒙𝟐 dx
= 𝟐 √𝒂𝟐 − 𝒙𝟐 +
𝒍𝒐𝒈 | 𝒙 + √𝒙𝟐 + 𝒂𝟐 | + 𝒄
𝟐
𝟐
𝒍𝒐𝒈 | 𝒙 + √𝒙𝟐 − 𝒂𝟐 | + 𝒄
𝒙
𝐬𝐢𝐧−𝟏 (𝒂) + 𝒄
INTEGRATION BY PARTS
𝒅𝒖
∫ 𝒖. 𝒗 𝒅𝒙
1.
= 𝒖. ∫ 𝒗 𝒅𝒙 − ∫( 𝒅𝒙 ∫ 𝒗 𝒅𝒙) 𝒅𝒙
OR
The integral of product of two functions = (first function) x integral of the second
function – integral of [(differential coefficient of the first function ) × (integral of the
second function)]
We can choose first and second function according to I L A T E where I → inverse
trigonometric function 𝑳 → 𝒍𝒐𝒈𝒂𝒓𝒊𝒕𝒉𝒎𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏, 𝐀 → 𝒂𝒍𝒈𝒆𝒃𝒓𝒂𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝐓 →
𝒕𝒓𝒊𝒈𝒐𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒂𝒏𝒅 𝐄 → 𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒕𝒊𝒂𝒍 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏
 e  f ( x)  f
x
2.
1

( x) dx  e x f ( x)  c .
Working Rule for different types of integrals
1. Integration of trigonometric function
Working Rule
(a) Express the given integrand as the algebraic sum of the functions of the
following forms
(i) Sin k𝜶, (ii) cos k𝜶, (iii) tan k𝜶, (iv) cot k𝜶, (v) sec k𝜶, (vi) cosec k𝜶
(vii) sec2 k𝜶, (viii) cosec2 k𝜶, (ix) sec k𝜶 tan k𝜶 (x) cosec k𝜶 cot k𝜶
For this use the following formulae whichever applicable
(i) 𝒔𝒊𝒏𝟐 𝒙 =
𝟑
(iii) 𝒔𝒊𝒏 𝒙 =
𝟏−𝒄𝒐𝒔 𝟐𝒙
(ii) 𝒄𝒐𝒔𝟐 𝒙 =
𝟐
𝟑 𝒔𝒊𝒏 𝒙−𝒔𝒊𝒏 𝟑𝒙
𝟑
(iv) 𝒄𝒐𝒔 𝒙 =
𝟒
(v)
x
=
x–1
(vi) cot2 x
(vii) 2sin x sin y = cos (x – y ) – cos ( x + y)
(viii) 2 cos x cosy = cos (x + y) + cos (x – y )
(ix) 2 sin x cos y = sin (x + y ) + sin (x – y )
(x) 2cos x sin y = sin (x + y ) – sin ( x – y )
tan2
sec2
𝟏+𝒄𝒐𝒔 𝟐𝒙
𝟐
𝟑 𝒄𝒐𝒔 𝒙+𝒄𝒐𝒔 𝟑𝒙
𝟒
= cosec2x – 1
2. Integration by substitution
(a) Consider I =  f ( x)dx
Put x = g(t) so that
dx
 g 1 (t ).
dt
We write dx = g1(t). Thus I =  f ( x)dx   f ( g (t )) g 1 (t )dt
(b) When the integrand is the product of two functions and one of them is a function
g (x) and the other is kg’(x), where k is a constant then Put g (x) = t
𝒅𝒙
3. Integration of the types ∫ 𝒂𝒙𝟐+𝒃𝒙+𝒄 ,
∫
𝒅𝒙
√𝒂𝒙𝟐 +𝒃𝒙+𝒄
, and ∫(𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄)𝒅𝒙
31
In this three forms change 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 in the form A2 + X2 , X2 – A2, or A2 – X2
Where X is of the form x + k and A is a constant ( by completing square method)
Then integral can be finding by using any of the special integral formulae
𝒑𝒙+𝒒
Integration of the types ∫ 𝒂𝒙𝟐 +𝒃𝒙+𝒄 𝒅𝒙,
4.
∫√
𝒑𝒙+𝒒
𝒂𝒙𝟐 +𝒃𝒙+𝒄
𝒅𝒙 and
∫(𝒑𝒙 + 𝒒)√𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 dx
In this three forms split the linear px + q = 𝝀
𝒅
𝒅𝒙
(𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 ) + 𝝁
Then divide the integral into two integrals
The first integral can be obtain by method of substitution and the second integral
by completing square method as explained in 3
i,e., to evaluate ∫
𝒑𝒙+𝒒
√𝒂𝒙𝟐 +𝒃𝒙+𝒄
𝒅𝒙 = ∫
𝝀 (𝟐𝒂𝒙+𝒃 )+ 𝝁
𝒂𝒙𝟐 + 𝒃𝒙+𝒄
𝟐𝒂𝒙+𝒃
dx
𝒅𝒙
= 𝝀 ∫ 𝒂𝒙𝟐 + 𝒃𝒙+𝒄 𝒅𝒙 + 𝝁 ∫ 𝒂𝒙𝟐 + 𝒃𝒙+𝒄
↓ ↓
Find by substitution method + by completing square method
5. Integration of rational functions
In the case of rational function, if the degree of the numerator is equal or greater
than degree of the denominator, then first divide the numerator by denominator and
write it as
𝑵𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓
𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓
= 𝑸𝒖𝒐𝒕𝒊𝒆𝒏𝒕 +
𝑹𝒆𝒎𝒂𝒊𝒏𝒅𝒆𝒓
𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒓
, then integrate
6. Integration by partial fractions
Integration by partial fraction is applicable for rational functions. There first we
must check that degree of the numerator is less than degree of the denominator, if
not, divide the numerator by denominator and write as
𝑵𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓
𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓
= 𝑸𝒖𝒐𝒕𝒊𝒆𝒏𝒕 +
𝑹𝒆𝒎𝒂𝒊𝒏𝒅𝒆𝒓
𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒓
and proceed for
𝑹𝒆𝒎𝒂𝒊𝒏𝒅𝒆𝒓
partial fraction of 𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒓
Sl.
No.
1
Form of the rational functions
𝒑𝒙 + 𝒒
(𝒙 − 𝒂)(𝒙 − 𝒃 )
Form of the partial fractions
𝑨
𝑩
+
𝒙−𝒂 𝒙−𝒃
32
2
𝒑𝒙 + 𝒒
(𝒙 − 𝒂)𝟐
𝑨
𝑩
+
𝒙 − 𝒂 (𝒙 − 𝒂)𝟐
3.
𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓
(𝒙 − 𝒂 )(𝒙 − 𝒃 )𝒙 − 𝒄 )
𝑨
𝑩
𝑪
+
+
𝒙−𝒂 𝒙−𝒃 𝒙−𝒄
4
𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓
(𝒙 − 𝒂 )𝟐 (𝒙 − 𝒃 )
𝑨
𝑩
𝑪
+
+
𝒙 − 𝒂 (𝒙 − 𝒂 )𝟐 𝒙 − 𝒃
𝑨
𝑩𝒙 + 𝑪
+ 𝟐
𝒙 − 𝒂 𝒙 + 𝒃𝒙 + 𝒄
𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓
(𝒙 − 𝒂 )(𝒙𝟐 + 𝒃𝒙 + 𝒄 )
Where x2 + bx + c cannot be
factorized further
5
DEFINITE INTEGRATION
Working Rule for different types of definite integrals
1. Problems in which integral can be found by direct use of standard formula or by
transformation method
Working Rule
(i) Find the indefinite integral without constant c
(ii) Then put the upper limit b in the place of x and lower limit a in the place of x and
subtract the second value from the first. This will be the required definite integral.
2. Problems in which definite integral can be found by substitution method
Working Rule
When definite integral is to be found by substitution then change the lower and upper
limits of integration. If substitution is z = φ(x) and lower limit integration is a and
upper limit is b Then new lower and upper limits will be φ(a) and φ(b) respectively.
Properties of Definite integrals
𝒃
𝒃
1.
∫𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒕)𝒅𝒕
𝟐.
∫𝒂 𝒇(𝒙)𝒅𝒙 = − ∫𝒃 𝒇(𝒙)𝒅𝒙.
3.
∫𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒙)𝒅𝒙 + ∫𝒄 𝒇(𝒙)𝒅𝒙,
4.
∫𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒂 + 𝒃 − 𝒙)𝒅𝒙
5.
∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
6.
∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒙)𝒅𝒙 + ∫𝟎 𝒇(𝟐𝒂 − 𝒙)𝒅𝒙
7.
if 𝒇(𝟐𝒂 − 𝒙) = 𝒇(𝒙) and
∫𝟎 𝒇(𝒙)𝒅𝒙 = 𝟐 ∫𝟎 𝒇(𝒙)𝒅𝒙 ,
= 0,
if 𝒇(𝟐𝒂 − 𝒙) = −𝒇(𝒙)
𝒂
𝒂
(i)
∫−𝒂 𝒇(𝒙)𝒅𝒙 = 𝟐 ∫𝟎 𝒇(𝒙)𝒅𝒙, if 𝒇 is an even function, i.e., if 𝒇(-x) = 𝒇(𝒙)
8.
𝒃
𝒂
𝒃
𝒄
𝒃
𝒃
𝒂
𝒂
𝟐𝒂
𝟐𝒂
𝒂
In particular, ∫𝒂 𝒇(𝒙)𝒅𝒙 = 0
𝒃
𝒂
a<c<b
𝒂
𝒂
33
(ii)
𝒂
∫−𝒂 𝒇(𝒙)𝒅𝒙 = 𝟎,
if 𝒇 is an odd function, i.e., if 𝒇(−x) = −𝒇(𝒙)
Problem based on property
𝒃
𝒄
𝒃
∫𝒂 𝒇(𝒙)𝒅𝒙 = ∫𝒂 𝒇(𝒙)𝒅𝒙 + ∫𝒄 𝒇(𝒙)𝒅𝒙,
a<c<b
Working Rule
This property should be used if the integrand is different in different parts of the
interval [a,b] in which function is to be integrand. This property should also be used
when the integrand (function which is to be integrated) is under modulus sign or is
discontinuous at some points in interval [a,b]. In case integrand contains modulus then
equate the functions whose modulus occur to zero and from this find those values of x
which lie between lower and upper limits of definite integration and then use the
property.
Problem based on property
𝒂
𝒂
∫𝟎 𝒇(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
Working Rule
Let
𝒂
I = ∫𝟎 𝒇(𝒙)𝒅𝒙
𝒂
I = ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
Then
𝒂
𝒂
2I = ∫𝟎 𝒇(𝒙)𝒅𝒙 + ∫𝟎 𝒇(𝒂 − 𝒙)𝒅𝒙
(1) + (2) =>
𝟏
𝒂
I = 𝟐 ∫𝟎 {𝒇(𝒙) + 𝒇(𝒂 − 𝒙)}𝒅𝒙
This property should be used when 𝒇(𝒙) + 𝒇(𝒂 − 𝒙) becomes an integral function of x.
Problem based on property
𝒂
∫−𝒂 𝒇(𝒙)𝒅𝒙 =0, if 𝒇(𝒙) is an odd function
and
𝒂
𝒂
∫−𝒂 𝒇(𝒙)𝒅𝒙 = 2∫𝟎 𝒇(𝒙)𝒅𝒙, if 𝒇(𝒙) is an even function.
Working Rule
This property should be used only when limits are equal and opposite and the
function which is to be integrated is either odd or even.
PROBLEM BAESD ON LIMIT OF SUM
Working rule
𝒃
∫𝒂 𝒇 (𝒙)𝒅𝒙 = 𝐥𝐢𝐦 𝒉 { 𝒇 (𝒂) + 𝒇 (𝒂 + 𝒉 ) + … . +𝒇 ( 𝒂 + (𝒏 − 𝟏 )𝒉 )}
𝒉→𝟎
Where nh = b – a
The following results are used for evaluating questions based on limit of sum.
n 1
n(n  1)
(i)
1 + 2 + 3 + ….. + (n-1) =  k 
2
k 1
(ii)
𝟏𝟐 + 𝟐𝟐 + 𝟑𝟐 + …. + (𝒏 − 𝟏)𝟐 = ∑
(iii)
𝟏𝟑 + 𝟐𝟑 + 𝟑𝟑 + ….. + (𝒏 − 𝟏)𝟑 = [
(𝒏−𝟏)𝒏(𝟐𝒏−𝟏)
𝟔
(𝒏−𝟏)𝒏 𝟐
𝟐
]
34
(iv)
𝒂[𝒓𝒏 − 𝟏]
a + ar + …… + 𝒂𝒓𝒏−𝟏 =
𝒓−𝟏
(r≠1)
IMPORTANT SOLVED PROBLEMS
Evaluate the following integrals
1.
∫
( 𝟏+𝒍𝒐𝒈 𝑿 )𝟐
𝑿
𝒅𝒙
Solution
𝟏
put 1+log x = t
( 𝟏+𝒍𝒐𝒈 𝑿 )𝟐
∫
𝑿
𝒅𝒙 =
∫ 𝒕𝟐 𝒅𝒕
𝒕𝟑
=
∫√
𝒆𝒙
𝟓−𝟒𝒆𝒙 −𝒆𝟐𝒙
+ 𝒄
𝟑
(𝟏+𝒍𝒐𝒈𝒙 )𝟑
=
2.
𝒅𝒙 = 𝒅𝒕
𝒙
𝟑
+𝒄
𝒅𝒙
Solution
Put 𝒆𝒙 = 𝒕 𝒕𝒉𝒆𝒏 𝒆𝒙 𝒅𝒙 = 𝒅𝒕
𝒆𝒙
𝒅𝒕
∫
𝒅𝒙 = ∫
√𝟓 − 𝟒𝒆𝒙 − 𝒆𝟐𝒙
√𝟓 − 𝟒𝒕 − 𝒕𝟐
𝒅𝒕
=∫
√− 𝟐
=∫
(𝒕 + 𝟒𝒕−𝟓 )
𝒅𝒕
√− ( 𝒕𝟐 + 𝟒𝒕+𝟒−𝟒−𝟓 )
𝒅𝒕
=∫
=∫
√− { ( 𝒕+𝟐 )𝟐 −𝟗 }
𝒅𝒕
√𝟑𝟐 −( 𝒕+𝟐 )𝟐
= 𝐬𝐢𝐧−𝟏
3.
𝒕+𝟐
𝟑
𝒆𝒙 +𝟐
+ C = 𝐬𝐢𝐧−𝟏 (
𝟑
)+ 𝑪
∫ √𝒕𝒂𝒏𝒙 dx
Solution
Put tan x = t 2then
∫ √𝒕𝒂𝒏𝒙 dx =∫ 𝒕
𝟐𝒕 𝒅𝒕
sec2x dx = 2t dt => dx =
𝟏+ 𝒕𝟒
𝟐𝒕𝟐
𝟐𝒕 𝒅𝒕
= ∫ 𝟏+ 𝒕𝟒 𝒅𝒕
𝟏+ 𝒕𝟒
𝟐
𝒅𝒕 (by dividing nr and dr by t2 )
= ∫𝟏
𝟐
𝟐+ 𝒕
𝒕
𝟏
= ∫
𝟏
(𝟏+ 𝟐 ) + (𝟏− 𝟐 )
𝒕
𝒕
𝟏
𝒕𝟐 + 𝟐
𝒕
𝒅𝒕
𝟏
= ∫
𝟏
𝟏+ 𝟐
𝒕
𝒅𝒕
𝟏
𝒕𝟐 + 𝟐
𝒕
+∫
𝟏
= ∫
𝒅𝒕 + ∫
(𝒕− ) + 𝟐
𝒅𝒖
𝟏
𝒕𝟐 + 𝟐
𝒕
𝒅𝒕
𝟏
𝟏+ 𝟐
𝒕
𝟏 𝟐
𝒕
𝟏− 𝟐
𝒕
𝟏− 𝟐
𝒕
𝟏 𝟐
𝒕
𝒅𝒕
(𝒕+ ) − 𝟐
𝒅𝒗
𝟏
= ∫ 𝒖𝟐 + 𝟐 + ∫ 𝒗𝟐 − 𝟐( 1st integral put 𝒕 − 𝒕 = u
then(𝟏 +
𝟏
𝒕𝟐
) 𝒅𝒕 = 𝒅𝒖
35
𝟏
2nd integral put𝒕 + 𝒕 = 𝒗 𝒕𝒉𝒆𝒏 ( 𝟏 −
𝟏
=
√𝟐
𝟏
=
=
=
∫√
4.
√𝟐
𝒕𝟐
) 𝒅𝒕 = 𝒅𝒗
𝟏
𝒗− √𝟐
𝐭𝐚𝐧−𝟏 ( ) + 𝟐√𝟐 𝒍𝒐𝒈 | 𝒗+√𝟐 | + 𝑪
√𝟐
𝐭𝐚𝐧
√𝟐
𝟏
𝒖
𝟏
−𝟏
𝒕−
(
𝟏
𝒕
√𝟐
𝟏
) + 𝟐√𝟐 𝒍𝒐𝒈 |
𝒕𝟐 − 𝟏
𝐭𝐚𝐧−𝟏 (
𝟏
𝒕
𝟏
𝒕+ +√𝟐
𝒕
𝒕+ − √𝟐
| +𝑪
𝒕𝟐 + 𝟏− √𝟐 𝒕
𝟏
) + 𝟐√𝟐 𝒍𝒐𝒈 |𝒕𝟐 + 𝟏+
𝒕
√𝟐
𝟏
𝒕𝒂𝒏𝒙−
𝟏
𝐭𝐚𝐧−𝟏 ( 𝒕𝒂𝒏𝒙)
√𝟐
√𝟐
𝟓𝒙+𝟑
| +𝑪
√𝟐 𝒕
𝒕𝒂𝒏𝒙+ 𝟏− √𝟐𝒕𝒂𝒏𝒙
𝟏
+ 𝟐√𝟐 𝒍𝒐𝒈 |𝒕𝒂𝒏𝒙+ 𝟏−
| +𝑪
√𝟐𝒕𝒂𝒏𝒙
𝒅𝒙
𝒙𝟐 + 𝟒𝒙+ 𝟏𝟎
Solution
𝟓
= > A = 𝟐 𝒂𝒏𝒅 𝑩 = −𝟕
5x + 3 = A (2x + 4 ) + B
∫
𝟓
𝟓𝒙 + 𝟑
√𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎
𝒅𝒙 = ∫
𝟓
(𝟐𝒙+𝟒)
∫ √ 𝟐𝟐
=
𝒙 + 𝟒𝒙+𝟏𝟎
(𝟐𝒙+𝟒)
𝟓
=
𝟐
𝟐
√𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎
𝒅𝒙 + ∫
√
∫√
𝒙𝟐 + 𝟒𝒙+𝟏𝟎
𝟓
𝒅𝒕
𝟓
=
𝟐
𝟕
𝒙𝟐 + 𝟒𝒙+𝟏𝟎
𝟏
𝒅𝒙 + 𝟕 ∫
√
+𝟕 ∫
∫
𝟐 √𝒕
√
=
(𝟐𝒙 + 𝟒) − 𝟕
𝒙𝟐 + 𝟒𝒙+𝟒−𝟒+𝟏𝟎
𝟏
𝒙+𝟐 )𝟐 +𝟔
𝒅𝒙
𝒙𝟐 + 𝟒𝒙+𝟏𝟎
𝟏
× 𝟐√ 𝒕 + 𝟕 ∫
√(
𝒅𝒙
𝒅𝒙
𝒅𝒙
dx
= 5 √𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎 + 𝟕 𝒍𝒐𝒈 | 𝒙 + 𝟐 + √𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎| + 𝑪
𝝅
∫𝟎
5.
𝒙 𝒕𝒂𝒏 𝒙
𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙
𝒅𝒙
Solution
I
𝝅 𝒙 𝒕𝒂𝒏 𝒙
=∫𝟎 𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙 𝒅𝒙 -------------------(1)
𝝅 (𝝅−𝒙) 𝒕𝒂𝒏(𝝅−𝒙)
𝒂
Again I = ∫𝟎 𝒔𝒆𝒄 (𝝅−𝒙)+𝒕𝒂𝒏 (𝝅−𝒙) 𝒅𝒙 Using the property ∫𝟎 𝒇
𝝅 (𝝅−𝒙) 𝒕𝒂𝒏 𝒙
I = ∫𝟎 𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙 𝒅𝒙
-------------------- (2)
Adding (1) and (2) we get
𝝅
2 I = ∫𝟎
=
=
𝒙 𝒕𝒂𝒏 𝒙
𝝅 (𝝅−𝒙) 𝒕𝒂𝒏 𝒙
𝒅𝒙 + ∫𝟎
𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙
𝝅
𝒕𝒂𝒏𝒙
𝝅 ∫𝟎 𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙 𝒅𝒙
𝟐
𝝅
𝝅 ∫𝟎 (𝒕𝒂𝒏
𝒙 𝒔𝒆𝒄 𝒙 − 𝒕𝒂𝒏
𝒅𝒙
𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙
𝝅 𝒕𝒂𝒏𝒙 (𝒔𝒆𝒄 𝒙−𝒕𝒂𝒏 𝒙 )
= 𝝅 ∫𝟎 𝒔𝒆𝒄𝟐 𝒙− 𝒕𝒂𝒏𝟐 𝒙 𝒅𝒙
𝝅
𝒙) dx = 𝝅 ∫𝟎 (𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄 𝒙 −
𝝅
𝟎
= 𝝅⌈𝒔𝒆𝒄 𝒙 − 𝒕𝒂𝒏 𝒙 + 𝒙⌉
I
6.
𝝅
= 𝝅 ( 𝟐 − 𝟏)
𝟒
Evaluate ∫𝟏 (𝒙𝟐 − 𝒙) 𝒅𝒙 using limit of sum
Solution
𝒃
Comparing the given integral with ∫𝒂 𝒇 (𝒙)𝒅𝒙
a = 1, b = 4 f (x ) = x2 – x
∴ 𝒏𝒉 = 𝟒 − 𝟏 = 𝟑and
𝒂
(𝒙)𝒅𝒙 = ∫𝟎 𝒇(𝒂 − 𝒙 )𝒅𝒙
𝒔𝒆𝒄𝟐 𝒙 + 𝟏)𝒅𝒙
36
f ( a+ (n-1)h )= f ( 1 + (n – 1 ) h )
= ( 1 + (n – 1 ) h ) 2 - [1 + (n – 1 ) h]
= 1 + 2 (n-1 )h + (n – 1 ) 2 h2 -1 – (n – 1 ) h
= (n – 1 ) h + (n – 1 ) 2 h2
𝒃
= 𝐥𝐢𝐦 𝒉 ∑𝒏−𝟏
𝒇 ( 𝒂 + (𝒏 − 𝟏 )𝒉 )
𝟎
∫𝒂 𝒇 (𝒙)𝒅𝒙
𝒉 →𝟎
𝟒
𝟐 𝟐
∫𝟏 (𝒙𝟐 − 𝒙) 𝒅𝒙 = 𝐥𝐢𝐦𝒉{ ∑𝒏−𝟏
𝟎 (𝒏 − 𝟏 ) 𝒉 + (𝒏 − 𝟏) 𝒉 }
𝒉→𝟎
= 𝐥𝐢𝐦𝒉{∑(𝒏 − 𝟏 )𝒉 + ∑(𝒏 − 𝟏)𝟐 𝒉𝟐 }
𝒉→𝟎
= 𝐥𝐢𝐦𝒉{ 𝒉 ∑(𝒏 − 𝟏 ) + 𝒉𝟐 ∑(𝒏 − 𝟏)𝟐 }
𝒉→𝟎
= 𝐥𝐢𝐦 𝒉𝟐
𝒏 ( 𝒏−𝟏 )
𝟐
𝒉→𝟎
𝒏𝒉 ( 𝒏𝒉−𝒉 )
= 𝐥𝐢𝐦
+
𝒉 →𝟎
= 𝐥𝐢𝐦
𝟗
𝒉→𝟎
𝟓𝟒
=𝟐 +
=
𝟐
𝟑 ( 𝟑−𝒉 )
𝟐
+
+ 𝒉𝟑
𝒏 ( 𝒏−𝟏 )( 𝟐𝒏−𝟏 )
𝟔
𝒏𝒉 ( 𝒏𝒉−𝒉 )(𝟐𝒏𝒉−𝒉 )
𝟔
𝟑 ( 𝟑−𝒉 )( 𝟔−𝒉 )
𝟔
𝟔
𝟐𝟕
𝟐
PRACTICE PROBLEMS
LEVEL I
Evaluate the following integrals
1.
∫(𝟐𝒙 − 𝟑 𝒄𝒐𝒔𝒙 + 𝒆𝒙 ) 𝒅𝒙
2.
3.
4.
5.
∫
𝒙𝟒 +𝟓𝒙𝟐 + 𝟑
√𝒙
𝟑
𝒅𝒙
∫ 𝒔𝒊𝒏 𝒙 𝒅𝒙
∫ 𝒔𝒊𝒏 𝟑𝒙 𝒄𝒐𝒔 𝟓𝒙 𝒅𝒙
∫ 𝒔𝒊𝒏𝟑 𝒙𝒄𝒐𝒔𝟓 𝒙 𝒅𝒙
𝒅𝒙
6.
∫ 𝒙𝟐 + 𝟒𝒙+𝟏
7.
∫√
8.
∫𝟎𝟒 𝒕𝒂𝒏𝟐 𝒙 𝒅𝒙
9.
∫ 𝒙𝟐 𝒆𝒙 𝒅𝒙
10.
∫𝟎
𝒅𝒙
𝒙𝟐 + 𝟓𝒙+𝟖
𝝅
𝟏
𝒙
𝒙𝟐 + 𝟏
𝒅𝒙
37
LEVEL II
Evaluate the following integrals
𝒔𝒊𝒏 𝒙
1.
∫ 𝒔𝒊𝒏 (𝒙−𝒂 ) 𝒅𝒙
2.
∫ ( 𝟏−𝒙 )( 𝟏+ 𝒙𝟐 ) 𝒅𝒙
3.
∫ √( 𝒙−𝟓 )(𝒙−𝟑 ) 𝒅𝒙
4.
∫ 𝒆𝒙 (𝟏−𝒄𝒐𝒔 𝟒𝒙) 𝒅𝒙
5.
∫ 𝒙𝟐 𝐭𝐚𝐧−𝟏 𝒙 𝒅𝒙
6.
∫ 𝒄𝒐𝒔𝟐 𝒙+𝒔𝒊𝒏 𝟐𝒙
7.
∫ ( 𝟏+ 𝒙𝟐 )( 𝟑+ 𝒙𝟐 ) 𝒅𝒙
9.
∫ 𝐭𝐚𝐧−𝟏 √𝟏+𝒙 𝒅𝒙
10.
11.
12.
13.
𝟐
𝒙+𝟐
𝒔𝒊𝒏 𝟒𝒙−𝟒
𝒅𝒙
𝟐𝒙
𝟏−𝒙
𝝅
∫𝟎
𝒙𝒕𝒂𝒏 𝒙
𝒅𝒙
𝒔𝒆𝒄 𝒙 𝒄𝒐𝒔𝒆𝒄 𝒙
𝝅
𝒙
∫𝟎 𝟏+𝒔𝒊𝒏𝒙 𝒅𝒙
𝟑
∫−𝟏 | 𝒙𝟑 − 𝒙 | 𝒅𝒙
𝟑
√𝒙
∫𝟏 √𝒙+ √𝟓−𝒙 𝒅𝒙
𝝅
14.
∫𝟎𝟒 𝒍𝒐𝒈 ( 𝟏 + 𝒕𝒂𝒏 𝒙 )𝒅𝒙
15.
∫𝟎 (𝒙𝟐 + 𝟐) 𝒅𝒙 as a limit of sum
𝟐

 5Sinx  3Cosx 
dx
Sinx  Cosx 
0
2
16.
 
LEVEL III
Evaluate the following integrals
𝒔𝒊𝒏 𝒙
1.
∫ 𝒔𝒊𝒏𝟒𝒙 𝒅𝒙
2.
∫(√𝒕𝒂𝒏 𝒙 + √𝒄𝒐𝒕 𝒙 ) 𝒅𝒙
3.
∫ 𝒔𝒊𝒏𝟒 𝒙+ 𝒄𝒐𝒔𝟒 𝒙 𝒅𝒙
4.
∫ √ 𝟏+√𝒙 𝒅𝒙
5.
6.
∫ 𝒔𝒊𝒏 (𝒍𝒐𝒈 𝒙 )𝒅𝒙
∫ 𝒍𝒐𝒈 ( 𝒙 + √𝒙𝟐 + 𝒂𝟐 ) 𝒅𝒙
7
∫𝟎 𝒙 𝒍𝒐𝒈 (𝒔𝒊𝒏 𝒙 )𝒅𝒙
8.
∫𝟏 (𝟐𝒙𝟐 + 𝒙 + 𝟕 ) 𝒅𝒙 as a limit of sum
𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙
𝟏− √𝒙
𝝅
𝟐
37
38
xdx
9.
 1  x tan x
10
x4
 ( x  1)( x 2  1) dx

1 
11.  log log x  
dx
log x 2 

38
39
APPLICATION OF INTEGRATION
INTRODUCTION
Area under Simple Curves
(i)
Area bounded by the curve y = f(x), the x-axis and between the ordinates at x = a and x = b
is given by
𝒃
𝒃
Area =∫𝒂 𝒚 𝒅𝒙 = ∫𝒂 𝒇(𝒙) 𝒅𝒙
(ii)
Area bounded by the curve y = f(x), the y axis and between abscissas at y = c and
y = d is given by
𝒅
𝒅
Area = ∫𝒄 𝒙 𝒅𝒙 = ∫𝒄 𝒈(𝒚)𝒅𝒚
Where
y = 𝒇(𝒙) => 𝑥 = 𝑔(𝑦)
Note: If area lies below x-axis or to left side of y-axis, then it is negative and in such a
case we like its absolute value. (Numerical value)
39
40
4.
Area bounded by two curves 𝒚 = 𝒇(𝒙) and 𝒚 = 𝒈(𝒙), such that 𝟎 ≤ 𝒈(𝒙) ≤ 𝒇(𝒙) for
all 𝒙 ∈ [𝒂, 𝒃] and between the ordinates at 𝒙 = 𝒂, 𝒙 = 𝒃 is given by
𝒃
Area = ∫𝒂 {𝒇(𝒙) − 𝒈(𝒙)} 𝒅𝒙
Finding the area enclosed between a curve, X- axis and two ordinates or a curve , Y- axis
and two abscissa
WORKING RULE
1. Draw the rough sketch of the given curve
2. Find whether the required area is included between two ordinate or two abscissa
3. (a) If the required area is included between two ordinates x = a and x= b then use the
𝒃
formula ∫𝒂 𝒚 𝒅𝒙
(b) If the required area is included between two abscissas y = c and y = d then use the
𝒅
formula ∫𝒄 𝒙 𝒅𝒚
Finding the area included between two curves
WORKING RULE
1. Draw the graph of the given curves.
2. Obtain the point of intersections of the curves.
3. Mark the region whose area is to be determined.
4. Find whether the area is bounded between two given curves and two ordinates or between
the two given curves and two abscissas.
(a) If the required area is bounded between 2 ordinates x = a and x = b, then use the
𝒃
Formula: ∫𝒂 [𝒇 (𝒙) − 𝒈 (𝒙)]𝒅𝒙
(b) If the required area is included between two abscissas y = c and y = d, then use the
𝒅
Formula: ∫𝒄 [𝒇(𝒚) − 𝒈(𝒚)] 𝒅𝒚
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41
SOME IMPORTANT POINTS TO BE KEPT IN MIND FOR SKETCHING THE GRAPH
1.
𝒚𝟐 = 4ax is a parabola with vertex at origin,
symmetric to X axis and right of origin
2.
𝒚𝟐 = - 4ax is a parabola with vertex at origin,
symmetric to X axis and left of origin
3.
𝒙𝟐 = 4ay is a parabola with vertex at origin,
symmetric to y axis and above origin
4.
𝒙𝟐 = - 4ay is a parabola with vertex at origin,
symmetric to y axis and below origin
5.
𝒙𝟐
𝒂𝟐
𝒚𝟐 = - 4ax
y2 = 4ax
𝒙𝟐 = 4ay
𝒙𝟐 = - 4ay
𝒚𝟐
+ 𝒃𝟐 = 1 is an ellipse symmetric to both axis,
Cut x axis at ( ± a, 0) and y axis at (0, ± 𝒃)
𝒙𝟐
𝒂𝟐
6.
7.
8.
9.
𝒙𝟐 + 𝒚𝟐 = 𝒓𝟐 is a circle symmetric to both the axes
with centre at origin and radius r
𝒚𝟐
+ 𝒃𝟐 = 1
𝒙 𝟐 + 𝒚𝟐 = 𝒓 𝟐
(𝒙 − 𝒉)𝟐 + (𝒚 − 𝒌)𝟐 = 𝒓𝟐 is a circle with centre at (h, k) and radius r.
ax +by + c = 0 representing a straight line
Graph of 𝒚 = |𝒙|
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42
10.
Graph of 𝒚 = |𝒙 − 𝟑|
y-axis
(0 , 3)
O
( 3, 0)
x-axis
IMPORTANT SOLVED PROBLEMS
1. Calculate the area of the region bounded by the parabola y = x2 and x = y2
Solution
Parabola y 2 = x (𝒐𝒓 𝒚 = √𝒙) is symmetrical to x- axis and x2 = y is symmetrical to y - axis
Solving both the equations we get the point of intersections of the two curves as (0, 0) and (1,
1)
𝟏
Required area = ∫𝟎 (√𝒙 − 𝒙𝟐 ) 𝒅𝒙
𝟑
= [
𝒙𝟐
𝟐
𝟑
𝟐
−
=𝟑 −
𝟏
𝟑
𝒙𝟑
𝟑
]
𝟏
𝟎
𝟏
= 𝟑 sq. units
2. Find the area of the region ; { (x, y ) : 𝒚𝟐 ≤ 𝟒𝒙 , 𝟒𝒙𝟐 + 𝟒𝒚𝟐 ≤ 𝟗}
Solution:
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43
𝟑 𝟐
Curves y2 = 4x, parabola symmetric to x axis and the curve 𝒙𝟐 + 𝒚𝟐 = (𝟐) is a circle with
centre at ( 0, 0 ) and radius
𝟑
𝟐
Sketch both the curves and shaded the area
The point of intersection of y2 = 4x and 𝒙𝟐 + 𝒚𝟐 =
𝟗
𝟏
𝟏
are the points (𝟐 , √𝟐) and (𝟐 , −√𝟐)
𝟒
𝟑
𝟏
Thus the co-ordinates of the points are O(0, 0), A(𝟐 , 𝟎) and B(𝟐 , √𝟐)
Required area = 2 × Area of (OBALO)
= 2×[ area of (OBLO) + area of (BLAB)]
𝟏
𝟑
𝟗
= 2 (∫𝟎𝟐 𝟐√𝒙 + ∫𝟏𝟐 √𝟒 − 𝒙𝟐 ) dx
𝟐
𝟑
= 2{(𝟐
𝒙𝟐
𝟑
𝟐
𝟏
𝒙
𝟗
+ [ 𝟐 √𝟒 − 𝒙𝟐 +
)𝟐
𝟎
𝟑
𝟐𝒙
𝐬𝐢𝐧−𝟏 𝟑 ] 𝟐𝟏}
𝟖
𝟗
𝟐
=
𝟗𝝅
𝟖
−
𝟗
−𝟏 𝟏
𝐬𝐢𝐧
(𝟑)
𝟒
+
√𝟐
𝟔
sq. units
3. Find the area bounded between the lines y = 2x + 1, y = 3x + 1, x = 4 using integration
Solution:
Draw the rough sketch and shade the region bounded by the given lnes
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44
𝟒
Area enclosed = ∫𝟎 (𝒇 (𝒙) − 𝒈(𝒙))𝒅𝒙
𝟒
= ∫𝟎 (𝟑𝒙 + 𝟏 − 𝟐𝒙 − 𝟏) dx
𝟒
4.
= ∫𝟎 𝒙 𝒅𝒙
𝒙𝟐 𝟒
= [ 𝟐 ] = 8 sq.units
𝟎
Find the area of the region { (x, y): 𝟎 ≤ 𝒚 ≤ 𝒙𝟐 + 𝟏, 𝟎 ≤ 𝒚 ≤ 𝒙 + 𝟏, 𝟎 ≤ 𝒙 ≤ 𝟐}
Solution
Sketch the region whose area is to be found out.
The point of intersection of y = x2 +1 and y = x+1 are he points (0, 1) and (1, 2)
The required area = area of the region (OPQRSTO)
= area of the region OTQPO + area of the region TSRQT
𝟏
𝟐
= ∫𝟎 (𝒙𝟐 + 𝟏 )𝒅𝒙 + ∫𝟏 (𝒙 + 𝟏 ) 𝒅𝒙
𝒙𝟑
𝒙𝟐
𝟏
𝟐
= [ 𝟑 + 𝒙]
+ [ 𝟐 + 𝒙]
𝟎
𝟏
𝟐𝟑
= 𝟔 sq.units
5. Find the area cut off from the parabola 4y = 3x2 by the line 2y = 3x + 12
Solution
Given 4y = 3x2 and 3x – 2y +12 = 0
Solve both the equation we get the point of
intersecction of both the curves
(-2, 3) and (4, 12)
Required area = area of AOBA
𝟒 𝟑𝒙+𝟏𝟐
= ∫−𝟐[
𝟐
−
𝟑𝒙𝟐
𝟒
] dx
= 27 sq.unts
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45
PRACTICE PROBLEMS
LEVEL I
1. Find the area of the region bounded by the parabola y2 = 4ax, its axis and two ordinates x = 4
and x = 9
2. Find the area bounded by the parabola x2 = y, y axis and the line y =1
3. Find the area bounded by the curve y = 4x – x2, x axis and the ordinates x = 1 and x = 3
4. Find the area of the region bounded by the curve y2 = 4x and the line x = 3
LEVEL II
1. Find the area of the region {(x, y) : x + y ≤ 4, 𝑥 + 𝑦 ≥ 2}
2. Find the of the circle x2 + y2 = a2
3. Find the area of the region {(x , y) : 𝑦 2 ≤ 6 , 𝑥 2 + 𝑦 2 ≤ 16}
4. Sketch the region common to the circle x2 + y2 = 4 and the parabola y2 = 4x. Also find the area
of the region by integration.
2
𝑥2
5. Find the area of the ellipse 𝑎2 +
𝑦2
𝑏2
2
=1
6. Find the area of the smaller region bounded by the ellipse
𝑦
2
𝑥2
9
+
𝑦2
4
𝑥
= 1 and the straight line 3 +
=1
7. Find the area enclosed by the curve x = 3cost, y = 2 sint.
8. Find the area bounded by the lines x +2y = 2, y – x = 1 and 2x + y = 7
3
9. Find the area of the region bounded by the parabola 𝑦 = 4 𝑥 2 and the line 3x – 2y +12 =0
10. Using integration, find the area of the triangle ABC with vertices A (-1, 0 ), B (1 ,3 ) and C
(3,2)
LEVEL III
1. Using integration find the area of the following region
{(𝑥, 𝑦 ): |𝑥 + 2 | ≤ 𝑦 ≤
√20 − 𝑥 2 }
2. Sketch the region enclosed between the circles x2 + y2 = 1 and x2 + (y-1)2 = 1. Also find
the area of the region using integration
3. Find the area of the region lying above the x axis and included between the circle
x2 + y2 = 8x and the parabola y2 = 4x
4. Using the method of integration, find the area bounded by the curve |x| + |y| = 1
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46
1.
2.
3.
DIFFERENTIAL EQUATIONS
INTRODUCTION
Problems based on the order and degree of the differential equations
Working rule
(a) In order to find the order of a differential equation, see the highest derivative in the
given differential equation. Write down the order of this highest order derivatives.
(b) In order to find the degree of a differential equation write down the power of the
highest order derivative after making the derivatives occurring in the given differential
equation free from radicals and fractions.
Problems based on formation of differential equation.
Working rule
(a) Write the given equation.
(b) Differentiate the given equation w.r.t. independent variable of x as many times as the
number of arbitrary constants.
(c) Eliminate the arbitrary constants from given equation and the equations obtained by
differentiation.
Problems based on solution of differential equation in which variables are separable.
Working rule
This differential equation can be solved by the variable separable method which can be put
in the form
𝑑𝑦
𝑑𝑥
= 𝑓(𝑥). 𝑔(𝑦).
𝑑𝑦
i.e., in which 𝑑𝑥 can be expressed as the product of two functions, one of which is a
function of x only and the other a function of y only.
𝑑𝑦
In order to solve the equation 𝑑𝑥 = 𝑓(𝑥). 𝑔(𝑦). Write down this equation in the
𝑑𝑦
𝑑𝑦
form 𝑔(𝑦) = 𝑓(𝑥). 𝑑𝑥, then the solution will be ∫ 𝑔(𝑦) = ∫ 𝑓(𝑥)𝑑𝑥 + 𝑐, where C is an
4.
5.
arbitrary constant.
Problems based on solution of differential equations which are homogeneous.
Working rule
𝑑𝑦
(a)
Write down the given differential equation in the form 𝑑𝑥 = 𝑓(𝑥, 𝑦)
(b)
If 𝑓(𝑘𝑥, 𝑘𝑦) = 𝑓(𝑥, 𝑦). then differential equation is homogeneous.
(c)
In order to solve, put 𝑦 = 𝑣𝑥, so that 𝑑𝑥 = 𝑣 + 𝑥
(d)
variables 𝑥 and 𝑣.
Now solve the obtained differential equation by the variable separable
𝑦
method. At the end put 𝑥 in place of 𝑣.
𝑑𝑦
𝑑𝑣
𝑑𝑥
and then separate the
Working Rule for Linear Differential Equation of first degree:
(a)
Type 1.
𝑑𝑦
𝑑𝑥
+ 𝑃𝑦 = 𝑄 , where P and Q are constants or function of x only
General solution is 𝑦. 𝐼𝐹 = ∫(𝑄. 𝐼𝐹)𝑑𝑥 + 𝐶 , where 𝐼𝐹 = 𝑒 ∫ 𝑃𝑑𝑥 .
46
47
𝑑𝑥
Type 2.𝑑𝑦 + 𝑃𝑥 = 𝑄 , where P and Q are constants or function of y only
(b)
General solution is𝑥. 𝐼𝐹 = ∫(𝑄. 𝐼𝐹)𝑑𝑦 + 𝐶 , where 𝐼𝐹 = 𝑒 ∫ 𝑃𝑑𝑦 .
Note: Particular solution can be obtained after getting the value of parameter C by
substituting the given initial values of the variables.
1.
IMPORTANT SOLVED PROBLEMS
Solve the differential equation (x + y )dy + ( x – y ) dx = 0
Solution
(x + y )dy + ( x – y ) dx = 0
𝑑𝑦
𝑑𝑥
=
𝑦−𝑥
𝑥+𝑦
Put y = vx
𝑑𝑦
𝑑𝑥
=𝑣+𝑥
𝑑𝑣
∴ 𝑣+𝑥
 𝑣+𝑥
 𝑥
 𝑥

=
𝑥 + 𝑣𝑥
𝑣−1
=
𝑑𝑥
=
𝑑𝑥
𝑑𝑣
𝑣𝑥−𝑥
=
𝑑𝑥
𝑑𝑣
𝑑𝑣
𝑑𝑥
1+𝑣
𝑑𝑣
𝑑𝑥
𝑣+1
𝑣−1
− 𝑣
𝑣+1
−( 1+ 𝑣 2 )
1+𝑣
𝑑𝑥
𝑑𝑣 = −
1+ 𝑣 2
𝑥
Integrating both sides we get
1+𝑣
∫ 1+ 𝑣2 𝑑𝑣 = - ∫
𝑑𝑣
∫ 1+ 𝑣2 +
2
∫ 1+𝑣2 𝑑𝑣 = − log 𝑥 + 𝑐
1
tan−1 𝑣 + log|1 + 𝑣 2 | = − log 𝑥 + 𝑐
2
1
𝑦
2
1
tan−1 𝑥 +
𝑥
2𝑣
𝑦
tan−1 𝑥 +
2.
1
𝑑𝑥
2
𝑙𝑜𝑔 (1 +
𝑦2
)
𝑥2
2)
𝑙𝑜𝑔 (𝑥 2 + 𝑦
= −𝑙𝑜𝑔 𝑥 + 𝑐
= 𝑐
Solve 𝑥√1 − 𝑦 2 𝑑𝑥 + 𝑦 √1 − 𝑥 2 𝑑𝑦 = 0
Solution
𝑥√1 − 𝑦 2 𝑑𝑥
𝑥
√1− 𝑥 2
= - 𝑦 √1 − 𝑥 2 𝑑𝑦
𝑑𝑥 = −
𝑦
√1− 𝑦 2
Integrating both sides
𝑥
𝑦
∫ √1− 2 𝑑𝑥 = - ∫
𝑥
𝑑𝑦
√1− 𝑦 2
𝑑𝑦
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48
−1
2
𝑑𝑡
1
∫ √𝑡 =
𝑑𝑢
( put t = 1 – x2 and put u = 1 – y2 )
∫
2 √𝑢
− √𝑡 = √𝑢 + C
√1 − 𝑥 2 + √ 1 − 𝑦 2 = 𝐶
3.
Solve the differential equation
𝑑𝑦
𝑑𝑥
−
1
𝑥
. 𝑦 = 2𝑥 2
Solution
𝑑𝑦
The diff.eqn is in the form 𝑑𝑥 + 𝑃𝑦 = 𝑄
−1
Where P =
𝑥
and Q = 2x2
−1
= 𝑒 ∫ 𝑥 𝑑𝑥
I.F = 𝑒 ∫ 𝑃 𝑑𝑥
= 𝑒 −𝑙𝑜𝑔 𝑥
=
1
𝑥
Multiplying both sides of diff.eqn by I.F we get
1 𝑑𝑦
𝑑𝑥
1
−
𝑥 𝑑𝑥
𝑑
𝑥2
𝑦 = 2𝑥
1
(𝑦. 𝑥) = 2𝑥
Integrating both sides w.r.t.x we get
1
𝑦. = 𝑥 2 + 𝑐
𝑥
y = x3 + Cx
4.
𝑑𝑦
Solve the diff. equation 𝑥
𝑑𝑥
𝑦
= 𝑦 − 𝑥 tan 𝑥
Solution
𝑦
𝑦 − 𝑥 tan
𝑑𝑦
𝑥
=
𝑑𝑥
𝑥
𝑑𝑦
Put y = vx => 𝑑𝑥 = 𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
𝑑𝑣
𝑣𝑥 − 𝑥 𝑡𝑎𝑛𝑣
=
𝑑𝑥
𝑥
𝑑𝑣
 𝑣 + 𝑥 𝑑𝑥 = v – tan v
∴ 𝑣+𝑥
 𝑥
𝑑𝑣
𝑑𝑥
= −𝑡𝑎𝑛 𝑣
 cot v dv = -
𝑑𝑥
𝑥
Integrating both sides ,we get
∫ 𝑐𝑜𝑡 𝑣 𝑑𝑣
= −∫
𝑑𝑥
𝑥
log sin v = - log x + log C
𝑦
log[x. sin (𝑥 ) ] = 𝑙𝑜𝑔 𝑐
𝑦
x sin (𝑥 ) = C
5.
𝑥𝑦
𝑑𝑦
𝑑𝑥
= (𝑥 + 2 )(𝑦 + 2 ), find the equation of the curve passing through the points (1, -1)
Solution
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49
𝑥𝑦

𝑦
𝑦+2
𝑑𝑦 =
 (1 −
2
𝑥+2
𝑥
𝑑𝑦
= (𝑥 + 2 )(𝑦 + 2 )
𝑑𝑥
𝑑𝑥
2
) 𝑑𝑦 = (1 + 𝑥) 𝑑𝑥
𝑦+2
 Integrating both sides weget
 y – 2 log (y + 2) = x + 2 log x + C
The curve is passing through the the point (1, - 1)
- 1 –2 log 1 = 1 + 2 log 1 + C => C = -2
The equation of the line is 𝑦 − 𝑥 = 2 𝑙𝑜𝑔[x(y+2)] – 2
6.
𝑑𝑦
Solve the differential equation (𝑥 2 − 1) 𝑑𝑥 + 2𝑥𝑦 =
2
𝑥 2 −1
Solution
(𝑥 2 − 1)
𝑑𝑦
2𝑥
+ (𝑥 2 −1) . 𝑦 =
𝑑𝑥
2𝑥
I.F. = 𝑒 ∫𝑥2−1

𝑑
𝑑𝑥
𝑑𝑥
𝑑𝑦
2
+ 2𝑥𝑦 = 2
𝑑𝑥
𝑥 −1
2
(𝑥 2 −1)2
= 𝑒 𝑙𝑜𝑔(𝑥
(𝑦 (𝑥 2 − 1)) =
2 −1)
= (𝑥 2 − 1)
2
𝑥 2 −1
Integrating both sides w.r.t x we get
2
𝑦 (𝑥 2 − 1) = ∫ 𝑥 2 −1 𝑑𝑥
𝑥−1
𝑦 (𝑥 2 − 1) = 𝑙𝑜𝑔 (𝑥+1) + 𝐶
PRACTICE PROBLEMS
LEVEL I
1.
Find the order and degree of the following differential equation.
𝑑3 𝑦
2
𝑑𝑦 3
(𝑑𝑥 3 ) − 𝑥 (𝑑𝑥 ) = 0
𝐵
𝑑2 𝑦
𝑑𝑦
is a solution of the differential equation: 𝑥 2 𝑑𝑥 2 + 𝑥 𝑑𝑥 − 𝑦 = 0
2.
Show that 𝑦 = 𝐴𝑥 +
3.
4.
5.
Form the differential equation corresponding to 𝑦 2 − 2𝑎𝑦 + 𝑥 2 = 𝑎2 by eliminating a.
Form the differential equation representing the family of curves 𝑦 = 𝑎 𝑠𝑖𝑛 (𝑥 + 𝑏), where
𝑎, 𝑏 are arbitrary constants.
Solve the differential equation: 𝑠𝑒𝑐 2 𝑥 𝑡𝑎𝑛 𝑦 𝑑𝑥 + 𝑠𝑒𝑐 2 𝑦 𝑡𝑎𝑛 𝑥 𝑑𝑦 = 0
6.
Solve the differential equation :
𝑥
𝑑𝑦
𝑑𝑥
=
49
1+ 𝑦 2
1+ 𝑥 2
50
LEVEL II
Solve the following differential equations.
𝑑𝑦
1.
𝑑𝑥
+
1
. 𝑦 = 2𝑥 2
𝑥
(𝑥 2 + 𝑥𝑦)𝑑𝑦 = (𝑥 2 + 𝑦 2 )𝑑𝑥
2.
𝑑𝑦
3.
𝑑𝑥
+
4𝑥
𝑥 2 +1
𝑑𝑦
𝑦+
1
(𝑥 2 +1)2
=0
+ 𝑐𝑜𝑠 𝑥. 𝑦 = 𝑐𝑜𝑠 𝑥. 𝑠𝑖𝑛2 𝑥
4.
𝑠𝑖𝑛 𝑥
5.
3𝑒 𝑥 𝑡𝑎𝑛 𝑦 𝑑𝑥 + (1 − 𝑒 2 )𝑠𝑒𝑐 2 𝑦 𝑑𝑦 = 0, 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝑦 =
6.
(𝑥 3 + 𝑦 3 )𝑑𝑦 − 𝑥 2 𝑦 𝑑𝑥 = 0
7.
𝑐𝑜𝑠 2 𝑥
𝑑𝑦
8.
𝑑𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑥
4
, 𝑤ℎ𝑒𝑛 𝑥 = 1.
+ 𝑦 = 𝑡𝑎𝑛 𝑥
+ 𝑦 = 𝑐𝑜𝑠 𝑥 − 𝑠𝑖𝑛 𝑥
𝑥 𝑙𝑜𝑔 𝑥
9.
𝜋
𝑑𝑦
+𝑦 =
𝑑𝑥
2
𝑥
𝑙𝑜𝑔 𝑥
LEVEL III
Solve the following differential equations
𝑑𝑦
𝑑𝑦
1.
y – x 𝑑𝑥 = a (𝑦 2 + 𝑥 2 𝑑𝑥 )
2.
(1 + sin2x) dy + (1 + y2 ) cos x dx = 0, given that x = 2 , y = 0
3.
(x3 + x2 + x +1)
𝑑𝑦
𝑑𝑥
𝜋
= 2x2 + x
VECTORS ALGEBRA
SUMMARY
⃗⃗⃗⃗⃗ (𝑟) = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂, and its magnitude
1. Position vector of a point P(x, y, z) is given as 𝑂𝑃
by√𝑥 2 + 𝑦 2 + 𝑧 2 .
2. The scalar components of a vector are its direction ratios, and represent its projections
along the respective axes.
3. The magnitude(r), direction ratios (𝑎, 𝑏, 𝑐) and direction cosines (𝑙, 𝑚, 𝑛 )of any vector
𝑎
𝑏
𝑐
are related as: 𝑙 = 𝑟 , 𝑚 = 𝑟 , 𝑛 = 𝑟
4. The vector sum of the three sides of a triangle taken in order is 0 .
5. The vector sum of two co initial vectors is given by the diagonal of the parallelogram
whose adjacent sides are the given vectors.
6. The multiplication of a given vector by a scalar α, changes the magnitude of the given
vector by the multiple │α│, and keeps the direction same (or makes it opposite) according
as the value of α is positive (or negative).
𝑎⃗
7. For a given vector 𝑎, the vector 𝑎̂ = |𝑎⃗| gives the unit vector in the direction of 𝑎.
8. The position vector of a point R dividing a line segment joining the points P and Q whose
position vectors are 𝑎 and 𝑏⃗ respectively, in the ratio m : n
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⃗ +𝑛𝑎⃗
𝑚𝑏
(i)
internally, is given by 𝑅⃗ =
(ii)
externally, is given by⃗⃗⃗𝑅 =
(iii)
if R is the mid point of PQ, then 𝑅⃗ =
𝑚+𝑛
⃗ −𝑛𝑎⃗
𝑚𝑏
𝑚−𝑛
.
.
⃗ +𝑎⃗
𝑏
2
.
9. The scalar product of two given vectors 𝑎 and𝑏⃗ having angle 𝜃 between them is defined as
⃗⃗⃗ = ǀ 𝑎ǀǀ𝑏⃗ǀcos 𝜃.
𝑎. 𝑏
Also, when 𝑎. 𝑏⃗ is given, the angle ‘𝜃′ between the vectors⃗⃗⃗𝑎 and ⃗⃗⃗
𝑏 may be determined by
⃗
𝑎⃗.𝑏
cos 𝜃 = ǀ 𝑎⃗ǀ ǀ𝑏⃗ǀ .
10. The vector product is given as⃗⃗⃗𝑎 × 𝑏⃗ = |𝑎||𝑏⃗| sin 𝜃 𝑛̂, where 𝑛̂ is a unit vector
perpendicular to the plane containing 𝑎
⃗⃗⃗ and 𝑏⃗ such that 𝑎, 𝑏⃗, 𝑛̂ form right handed system of
co-ordinate axes.
11. If we have two vectors 𝑎 and 𝑏⃗, given in component form as
𝑎 = 𝑎1 𝑖̂ + 𝑏1 𝑗̂ + 𝑐1 𝑘̂ and 𝑏⃗ = 𝑎2 𝑖̂ + 𝑏2 𝑗̂ + 𝑐2 𝑘̂
and λ any scalar, then
𝑎 + 𝑏⃗ = (𝑎1 + 𝑎2 )𝑖̂ + (𝑏1 + 𝑏2 )𝑗̂ + (𝑐1 + 𝑐2 )𝑘̂
λ𝑎 = λ𝑎1 𝑖̂ + λ𝑏1 𝑗̂ + λ𝑐1 𝑘̂
⃗⃗⃗ . 𝑏⃗= 𝑎1 𝑎2 + 𝑏1 𝑏2 + 𝑐1 𝑐2
𝑎
and
𝑖̂
⃗
⃗⃗⃗ × 𝑏 = |𝑎1
𝑎
𝑎2
𝑗̂
𝑏1
𝑏2
𝑘̂
𝑐1 |
𝑐2
Practice problems
LEVEL-1
1) Find the projection of 𝑖̂ - 𝑗̂ on 𝑖̂ + 𝑗̂ .
2) If |𝑎| =2, |𝑏⃗| = √3 and 𝑎.𝑏⃗ = √3. Find the angle between 𝑎 and 𝑏⃗ .
3) Find the value of λ when the projection of 𝑎 = λ𝑖̂ + 𝑗̂ + 4𝑘̂ on 𝑏⃗ = 2𝑖̂ + 6̂𝑗 + 3𝑘̂ is 4 units.
4) Find 𝑎 . ( 𝑏⃗ x 𝑐 ), if 𝑎 = 2𝑖̂ + 𝑗̂ + 3𝑘̂, 𝑏⃗ = −𝑖̂ + 2𝑗̂ + 𝑘̂ and 𝑐 = 3𝑖̂ + 𝑗̂ + 2𝑘̂
5) Show that the four points A, B, C and D with position vectors 4𝑖̂ + 5𝑗̂ + 𝑘̂, −𝑗̂ − 𝑘̂,
3𝑖̂ + 9𝑗̂ + 4𝑘̂ and 4 ( −𝑖̂ + 𝑗̂ + 𝑘̂ ) respectively are coplanar.
LEVEL-2
̂
1. Find the angle between the vectors 𝑖̂ -2𝑗̂ + 3𝑘 and 3𝑖̂ - 2𝑗̂ + 𝑘̂ .
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2. Find |𝑎 − 𝑏⃗| if |𝑎| =2,|𝑏⃗| =3 and 𝑎 .𝑏⃗ = 4.
3. If 𝑎 is a unit vector and (𝑥 - 𝑎 ) . (𝑥 + 𝑎) =15. Find |𝑥| .
4. If 𝑎 and 𝑏⃗ are two vectors such that |𝑎 + 𝑏⃗| =|𝑎| then prove that the vector 2𝑎 + 𝑏⃗ is
perpendicular to 𝑏⃗ .
5. If vectors 𝑎 =2𝑖̂ + 2𝑗̂ + 3𝑘̂,⃗⃗𝑏 = 𝑖̂ + ̂𝑗 + 𝑘̂ and 𝑐 = 3 ̂𝑖 + 𝑗̂ are such that 𝑎 + λ⃗⃗𝑏 is
perpendicular to⃗⃗𝑐. Find the value of λ.
6. If A and B be two points with position vectors 2𝑖̂ − 𝑗̂ + 2𝑘̂ and 𝑖̂ + 2𝑗̂ respectively. Find
the position vector of the point which divides AB in 1 : 2 internally.
LEVEL-3
1. Find the value of p so that 𝑎 = 2𝑖̂ + p𝑗̂ + 𝑘̂ and 𝑏⃗ = 𝑖̂ -2𝑗̂+ 3𝑘̂ are perpendicular to each
other.
2. Find |𝑎| if |𝑎|=2|𝑏⃗| and (𝑎 +𝑏⃗ ) . (𝑎 - 𝑏⃗ ) = 12.
3. If |𝑎 + 𝑏⃗| = 60, |𝑎 − 𝑏⃗| = 40 and |𝑏⃗| = 46. Find |𝑎| .
⃗⃗⃗ in the direction
4. If 𝑎 = (3𝑖̂ + 2𝑗̂ − 3𝑘̂) and⃗⃗𝑏 = (4𝑖̂ + 7𝑗̂ − 3𝑘̂). Find vector projection of 𝒂
of⃗⃗𝑏.
5. The two adjacent sides of a parallelogram are (2iˆ  4 ˆj  5kˆ) & (iˆ  2 j  3kˆ) .Find the unit
vectors parallel to its diagonals. Also find its area.
6. If (iˆ  ˆj  kˆ), (2iˆ  5 ˆj  3kˆ), (3iˆ  2 ˆj  2kˆ) & (iˆ  6 j  kˆ) are the position vectors of points
A, B, C & D respectively, then find the angle between AB & CD. Deduce that AB & CD
are parallel.
7. Find the value of 𝜆 such that the vectors (3iˆ  ˆj  5kˆ), (iˆ  2 ˆj  3kˆ) & (2iˆ  j  kˆ) are
coplanar.
8. The scalar product of the vector 𝑎 = 𝑖̂ + 𝑗̂ + 𝑘̂ with a unit vector along the sum of the
vectors 𝑏⃗ = 2𝑖̂ + 4𝑗̂ − 5𝑘̂ and 𝑐 = 𝜇𝑖̂ + 2𝑗̂ + 3𝑘̂ is equal to one .Find the value of 𝜇 and
hence find the unit vector along 𝑏⃗ + 𝑐
9. Find the value of 𝜆 if the points A( -1,4,-3 ), B(3,𝛌,-5), C(-3,8,-5) and D(-3,2,1) are
coplanar.
3D GEOMETRY
INTRODUCTION
Summary
1. Distance formula: Distance between two points A(𝑥1 , 𝑦1 , 𝑧1 ) and B (𝑥2 , 𝑦2 , 𝑧2 ) is
AB =√(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 + (𝑧2 − 𝑧1 )2
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2. Section formula: Coordinates of a point P, which divides the line segment joining two
given points A(𝑥1 , 𝑦1 , 𝑧1 ) and B(𝑥2 , 𝑦2 , 𝑧2 ) in the ratio m : n
𝑚𝑥2 +𝑛𝑥1 𝑚𝑦2 +𝑛𝑦1 𝑚𝑧2 +𝑛𝑧1
(i) internally, are P (
𝑚+𝑛
,
𝑚+𝑛
,
𝑚+𝑛
),
(ii) the coordinates of a point Q divides the line segment joining two given points in the
𝑚𝑥2 −𝑛𝑥1 𝑚𝑦2 −𝑛𝑦1 𝑚𝑧2 −𝑛𝑧1
,
,
)
𝑚−𝑛
𝑚−𝑛
𝑚−𝑛
𝑥 +𝑥 𝑦 +𝑦 𝑧 +𝑧
mid-point are R( 2 2 1 , 2 2 1 , 2 2 1 )
ratio m : n; externally are Q (
(iii)
the coordinates of
3. Direction cosines of a line :
(i)
(ii)
(iii)
The direction of a line OP is determined by the angles 𝛼, 𝛽, 𝛾 which makes with
OX, OY,OZ respectively. These angles are called the direction angles and their
cosines are called the direction cosines.
Direction cosines of a line are denoted by l, m, n; l = cos 𝛼, m = cos 𝛽, 𝑛 = cos 𝛾
Sum of the squares of direction cosines of a line is always 1.
l2 + m2 + n2 = 1
i.e cos2𝛼 + cos2𝛽 + cos2𝛾 =1
4. Direction ratio of a line :
(i) Numbers proportional to the direction cosines of a line
𝑙
𝑎
are called direction ratios of a line. If a, b, and c are, direction ratios of a line, then =
𝑚
𝑏
𝑛
𝑐
= .
(ii) If a, b, c are, direction ratios of a line, then the direction cosines are
± √𝑎2
𝑎
+𝑏2 +𝑐 2
, ± √𝑎2
𝑏
+𝑏2 +𝑐 2
, ± √𝑎2
𝑐
+𝑏2 +𝑐 2
(𝑖𝑖𝑖) Direction ratio of a line AB passing through the points A(x1, y1, z1) and
B (x2, y2, z2) are 𝑥2 − 𝑥1 , 𝑦2 − 𝑦1 , 𝑧2 − 𝑧1
5. STRAIGHT LINE:
(i)
Vector equation of a Line passing through a point 𝑎 and along the
⃗ , : ⃗⃗𝑟 = 𝑎 + 𝜇𝑏⃗,
direction 𝒃
(ii)
Cartesian equation of a Line:
𝑥−𝑥1
𝑎
=
𝑦−𝑦1
𝑏
=
𝑧−𝑧1
𝑐
. Where (x1, y1, z1) is the
given point and its direction ratios are a, b, c.
6. (i) Vector equation of a Line passing through two points, with position vectors 𝑎 𝑎𝑛𝑑 𝑏⃗
⃗ -𝑎)
𝑟 = 𝑎 + 𝝁(𝒃
𝑥−𝑥1
(ii) Cartesian equation of a Line: 𝑥
2 −𝑥1
𝑦−𝑦1
=𝑦
2 −𝑦1
𝑧−𝑧1
=𝑧
2 −𝑧1
, two points are (x1,y1) and (x2,y2).
⃗⃗⃗1 and 𝑟⃗⃗ = ⃗⃗⃗⃗
⃗⃗⃗⃗2,
7. ANGLE between two lines (i) Vector equations: 𝑟⃗⃗ = ⃗⃗⃗⃗
𝑎1 + 𝝀𝑏
𝑎2 + 𝝁𝑏
(ii) Cartesian equations: If lines are
cos 𝜃 =
𝑥−𝑥1
𝑎1
=
𝑦−𝑦1
𝑏1
=
𝑧−𝑧1
𝑐1
,
𝑥−𝑥2
𝑎2
=
𝑦−𝑦2
𝑏2
⃗⃗⃗⃗ . 𝑏2
⃗⃗⃗⃗
𝑏1
⃗⃗⃗⃗ |. |𝑏2
⃗⃗⃗⃗ |
|𝑏1
(iii) If two lines are perpendicular, then 𝑏⃗1. . 𝑏⃗2 = 0, i.e. 𝑎1 𝑎2 + 𝑏1 𝑏2 + 𝑐1 𝑐2 = 0
53
=
𝑧−𝑧2
𝑐2
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(iv) If two lines are parallel, then 𝑏⃗1 = 𝑡 𝑏⃗2 , where t is a scalar. OR 𝑏⃗1 × 𝑏⃗2 = 0, OR
𝑎1
𝑎2
𝑏
𝑐
= 𝑏1 = 𝑐1
2
2
(v) If 𝜃 𝑖𝑠 𝑡ℎ𝑒 angle between two lines with direction cosines, l1, m1, n1 and l2, m2, n2 then
𝑙
𝑚
𝑛
(b) if the lines are parallel, then 𝑙1 = 𝑚1 = 𝑛1
(a) cos 𝜃 = l1l2 + m1m2 + n1n2
2
2
2
(c) If the lines are perpendicular, then l1l2 + m1m2 + n1n2 = 0
8 Shortest distance between two skew- lines:
⃗⃗⃗1 , and : 𝑟⃗⃗ = ⃗⃗⃗⃗
⃗⃗⃗⃗2,
(i)
Vector equations: 𝑟⃗⃗ = ⃗⃗⃗⃗
𝑎1 + 𝝀𝑏
𝑎2 + 𝝁𝑏
SD = |
⃗⃗⃗⃗⃗ ×𝑏2
⃗⃗⃗⃗⃗ )
⃗⃗⃗⃗⃗ −𝑎1
⃗⃗⃗⃗⃗ ).(𝑏1
(𝑎2
|.
⃗⃗⃗⃗⃗ ×𝑏2
⃗⃗⃗⃗⃗ |
|𝑏1
If shortest distance is zero, then lines intersect and line intersects in space if they are
coplanar. Hence if above lines are coplanar
⃗⃗⃗⃗ × ⃗⃗⃗⃗
⃗⃗⃗⃗ − ⃗⃗⃗⃗
If (𝑎2
𝑎1). (𝑏1
𝑏2) = 0
(ii) Cartesian equations:
SD = |
𝑥−𝑥1
𝑎1
=
𝑦−𝑦1
𝑏1
𝑥2 −𝑥1
| 𝑎1
𝑎2
=
𝑧−𝑧1 𝑥−𝑥2
𝑐1
𝑦2 −𝑦1
𝑏1
𝑏2
,
𝑎2
=
𝑦−𝑦2
𝑏2
=
𝑧2 −𝑧1
𝑐1 |
𝑐2
√(𝑏1 𝑐2 −𝑏2 𝑐1 )2 + (𝑐1 𝑎2 − 𝑐2 𝑎1 )2 +(𝑎1 𝑏2 −𝑎2 𝑏1 )2
𝑧−𝑧2
𝑐2
|
If shortest distance is zero, then lines intersect and line intersects in space if they are
coplanar. Hence if above lines are coplanar
𝑥2 − 𝑥1 𝑦2 − 𝑦1 𝑧2 − 𝑧1
𝑏1
𝑐1 | = 0
| 𝑎1
𝑎2
𝑏2
𝑐2
9.Shortest distance between two parallel lines:
If two lines are parallel, then they are coplanar. Let the lines be: 𝑟 = ⃗⃗⃗⃗
𝑎1 + 𝝀𝑏⃗, and:
⃗𝑏 ×(𝑎2
⃗⃗⃗⃗⃗ −𝑎1
⃗⃗⃗⃗⃗ )
𝑟⃗⃗ = ⃗⃗⃗⃗
𝑎2 + 𝝁𝑏⃗, SD = |
|
⃗|
|𝑏
10.General equation of a plane in vector form :- It is given by 𝑟. 𝑛⃗ + 𝑑 = 0 , 𝑛⃗ is a
vector normal to plane.
11.General equation of a plane in Cartesian form :- 𝒂𝒙 + 𝒃𝒚 + 𝒄𝒛 + 𝒅 = 𝟎 , Where
a, b, c are direction ratios of normal to the plane.
12.General equation of a plane passing through a point :- if position vector of given
point is 𝑎 then equation is given by (𝑟 − 𝑎). 𝑛⃗ = 0, 𝑛⃗ is a vector perpendicular to the
plane.
13.General equation of a plane passing through a point :- if coordinates of point
are (𝑥, 𝑦, 𝑧) then equation is 𝑎(𝑥 − 𝑥1 ) + 𝑏(𝑦 − 𝑦1 ) + 𝑐(𝑧 − 𝑧1 ) = 0, a, b, c are
direction ratios of a line perpendicular to the plane.
14. Intercept form of equation of a plane:- General equation of a plane which cuts off intercepts
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55
𝑥
a, b and c on x-axis, y-axis, z-axis respectively is
𝑦
𝑧
+ 𝑏 + 𝑐 = 1.
𝑎
15. Equation of a plane in normal form:- 𝑟. 𝑛̂ = p,where 𝑛̂ is a unit vector along perpendicular
from origin and ‛p’ is distance of plane from origin.p is always positive.
16. Equation of a plane in normal form:- It is given by 𝑙𝑥 + 𝑚𝑦 + 𝑛𝑧 = 𝑝, where 𝑙, 𝑚, 𝑛 are
direction cosines of perpendicular from origin and ‛p’ is distance of plane from origin and p is
always positive.
17. Equation of a plane passing through three non-collinear points:- If 𝑎, 𝑏⃗, 𝑐 are the position
vectors of three non-collinear points, then equation of a plane through three points is given by
(𝑟 − 𝑎). {(𝑏⃗ − 𝑎) × (𝑐 − 𝑎)} = 0.
18. Equation of a plane passing through three non-collinear points (Cartesian system):If plane passing through points (𝒙𝟏 , 𝒚𝟏 , 𝒛𝟏 ), (𝒙𝟐 , 𝒚𝟐 , 𝒛𝟐 ) and (𝒙𝟑 , 𝒚𝟑 , 𝒛𝟑 ) then equation is:(𝑥 − 𝑥1 ) (𝑦 − 𝑦1)
(𝑧 − 𝑧1)
|(𝑥2 − 𝑥1 ) (𝑦2 − 𝑦1 ) (𝑧2 − 𝑧1 )| = 0
(𝑥3 − 𝑥1 ) (𝑦3 − 𝑦1 ) (𝑧3 − 𝑧1 )
⃗ . ⃗⃗⃗⃗
⃗ .𝒏
⃗⃗⃗⃗𝟐 + 𝒅𝟐 = 𝟎 then
19. If 𝜽 is the angle between two planes 𝒓
𝒏𝟏 + 𝒅𝟏 = 𝟎 and 𝒓
cos 𝜃 =
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗2
𝑛1 .𝑛
|𝑛
⃗⃗⃗⃗⃗1 ||𝑛
⃗⃗⃗⃗⃗2 |
(i) If planes are perpendicular, then ⃗⃗⃗⃗
𝑛1 . ⃗⃗⃗⃗
𝑛2 = 0
(ii) If planes are parallel, then ⃗⃗⃗⃗
𝑛1 × ⃗⃗⃗⃗
𝑛2 = 0 or ⃗⃗⃗⃗
𝑛1 = 𝝀𝑛
⃗⃗⃗⃗2 , 𝝀is a scalar.
20 If 𝜽 is angle between two planes
𝒂𝟏 𝒙 + 𝒃𝟏 𝒚 + 𝒄𝟏 𝒛 + 𝒅𝟏 = 𝟎 𝒂𝒏𝒅 𝒂𝟐 𝒙 + 𝒃𝟐 𝒚 + 𝒄𝟐 𝒛 + 𝒅𝟐 = 𝟎
then
𝐜𝐨𝐬 𝜽 =
𝒂𝟏 𝒂𝟐 +𝒃𝟏 𝒃𝟐 +𝒄𝟏 𝒄𝟐
√(𝒂𝟏 𝟐 +𝒃𝟏 𝟐 +𝒄𝟏 𝟐 )( 𝒂𝟐 𝟐 +𝒃𝟐 𝟐 +𝒄𝟐 𝟐 )
(i)
If planes are perpendicular ,then 𝑎1 𝑎2 + 𝑏1 𝑏2 + 𝑐1 𝑐2 = 0
(ii)
If planes are parallel , then
𝑎1
𝑎2
⃗ =𝒂
⃗ + 𝝀𝒎
⃗⃗⃗
21 If 𝜽 is angle between line 𝒓
𝑏
𝑐
= 𝑏1 = 𝑐1
2
2
⃗⃗ + 𝒅 = 𝟎, then 𝐬𝐢𝐧 𝜽 =
and the plane⃗⃗𝒓. 𝒏
⃗𝒎
⃗⃗ .𝒏
⃗
|𝒎
⃗⃗⃗ |.|𝒏
⃗|
(i) If line is parallel to plane, then 𝑚
⃗⃗ . 𝑛
⃗⃗⃗ = 0 and
(ii) If line is perpendicular to plane, then 𝑚
⃗⃗ × 𝑛⃗ = 0 or 𝑚
⃗⃗ = 𝑡𝑛⃗, t is a scalar.
𝒙−𝒙𝟏
𝒚−𝒚𝟏
𝒛−𝒛𝟏
22 If 𝜽 is angle between line 𝒂 = 𝒃 = 𝒄 and the plane 𝒂𝒙 + 𝒃𝒚 + 𝒄𝒛 + 𝒅 = 𝟎, then
𝟏
𝐬𝐢𝐧 𝜽 =
𝟏
𝟏
𝒂𝒂𝟏 + 𝒃𝒃𝟏 + 𝒄𝒄𝟏
√(𝒂𝟏 𝟐 + 𝒃𝟏 𝟐 + 𝒄𝟏 𝟐 )( 𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐 )
(i) If line is parallel to the plane, then 𝑎𝑎1 + 𝑏𝑏1 + 𝑐𝑐1 = 0
(ii) If line is perpendicular to the plane, then
𝑎
𝑎1
𝑏
𝑐
1
1
=𝑏 =𝑐
23. General equation of a plane parallel to the plane 𝑟. 𝑛⃗ + 𝑑 = 0 𝑖𝑠⃗⃗𝑟. 𝑛⃗ + 𝜆 = 0, where 𝜆 is a
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constant and can be calculated from given condition.
24 General equation of a plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + 𝜆 = 0,
where 𝜆 is a constant and can be calculated from given condition.
25 General equation of a plane (vector form) passing through the line of the intersection of planes
𝑟. ⃗⃗⃗⃗
𝑛1 + 𝑑1 = 0 and 𝑟. ⃗⃗⃗⃗
𝑛2 + 𝜆𝑑2 = 0 is 𝑟. (𝑛⃗1 + 𝜆𝑛⃗2 ) + (𝑑1 + 𝜆𝑑2 ) = 0 , where 𝜆 is a
constant and can be calculated from given condition.
26 General equation of a plane (Cartesian form) passing through the line of the intersection of
planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is
(a1x + b1y + c1z + d1) + 𝜆(a2x + b2y + c2z + d2) = 0, where 𝜆 is a constant and can be calculated
from given condition.
27 Distance of a plane (vector form) 𝑟. 𝑛⃗ + 𝑑 = 0, from a point with position vector 𝑎, is
⃗ +𝑑
𝑎⃗.𝑛
|.
|𝑛
⃗|
d=|
28 Distance of a plane (Cartesian form)ax + by + cz + d = 0, from a point (x1, y1, z1) is
𝑎𝑥1 +𝑏𝑦1 +𝑐𝑧1 +𝑑
𝑑= |
√𝑎2 +𝑏2 +𝑐 2
|.
FLOW CHART
(a)
To find shortest distance between two skew lines
1
Find 𝑎1, 𝑎2 , 𝑏⃗1 & 𝑏⃗2
2
Find 𝑎2 − 𝑎1
3
Find⃗⃗𝑏1 × 𝑏⃗2
5
Find |𝑏⃗1 × 𝑏⃗2 |
⃗⃗⃗1 × ⃗⃗⃗⃗
(𝑏
𝑏2 ). (𝑎
⃗⃗⃗⃗2 − ⃗⃗⃗⃗
𝑎1 )
6
Distance = |
4
⃗ 1 ×𝑏
⃗ 2 ).(𝑎⃗2 −𝑎⃗1 )
(𝑏
|
⃗ 1 ×𝑏
⃗ 2|
|𝑏
56
57
(b)
1.
2.
3.
4.
5.
To find Coordinates of foot of perpendicular from origin to the plane:
Direction ratio of any line perpendicular to the given plane.
Equation of the line through origin and perpendicular to the given plane.
Coordinates of any point (P) on the line.
Find the value of λ, by putting the coordinates of P in the plane.
Find required coordinates of foot of perpendicular
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58
(c)
To find coordinates of image of a point in the plane
1.
2.
Direction ratios of any line (PP’) perpendicular to the given plane.
3.
4.
5.
6.
Coordinates of any point on the plane (say P’)
Equation of the line PP’ through a given point (P) and perpendicular to the given
plane.
Coordinates of mid-point (M) of PP’
Find the value of r, as M will satisfy the plane
Then find coordinates of P’ Images of point P
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59
QUESTIONS ON 3-D.
LEVEL-1
4−𝑥
𝑦+3
𝑧+2
1. The equation of a line is given by 2 = 3 = 6 . Write the direction cosines of a line parallel
to given line.
2. Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to the
planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
3. Find the co-ordinates of the point where the line through (3, 4, 1) and (5, 1, 6) crosses the
xy-plane .
4. Find the shortest distance between the following pair of lines :
𝑥−1 𝑦−2 𝑧−3
𝑥−2 𝑦−4 𝑧−5
=
=
;
=
=
2
3
4
3
4
5
LEVEL-2
1. Find the Coordinates of foot of the perpendicular from origin to the plane 3x + 4y - 5z = 7
2. Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin is
1 1 1
1
 2 2  2
2
a b c
p
3. Find the distance of the point (−1, −5, −10) from the point of intersection of the line
𝑟 = (2𝑖̂ − 𝑗̂ + 2𝑘̂) + 𝜆(3 𝑖̂ + 4𝑗̂ + 2𝑘̂ ) and the plane 𝑟. (𝑖̂ − 𝑗̂ + 𝑘̂) = 5.
4. Find the distance between the point P(6, 5, 9) and the plane determined by the points
A(3, −1, 2), B(5, 2, 4) and C(−1, −1,6).
5. Find the equation of the plane passing through the points (3, 4, 1), (0, 1, 0) and is parallel to line
𝑥+3 𝑦−3 𝑧−2
=
=
2
7
5
6. Find the values of p so that the lines:
1−𝑥
3
=
7𝑦−14
2𝑝
=
𝑧−3
2
and
7−7𝑥
3𝑝
=
𝑦−5
1
=
6−𝑧
5
are at right
angles.
7. Find the shortest distance between the lines 𝑥 + 1 = 2𝑦 = −12𝑧 and 𝑥 = 𝑦 + 2 = 6𝑧 − 6
8. Find the shortest distance between the following pair of lines :
𝑥−1 𝑦+1
𝑥+1 𝑦−2
=
=𝑧 ;
=
; 𝑧=2
2
3
5
1
LEVEL-3
1. Find the equation of the plane which is perpendicular to the plane 5 x  3 y  6 z  8  0 and
which contains the line of intersection of the planes x  2 y  3z  4  0 and 2 x  y  z  5  0 .
2. Find the equation of a plane which is at a distance of 7 units from the origin and which is



normal to the vector 3 i  5 j  6 k .
3. Find a unit vector perpendicular to the plane of the triangle ABC, where the coordinates of its
vertices are A (3, −1, 2), B (1, −1, −3) and C (4, −3, 1).
𝑥
𝑦−1
𝑧−2
4. Find the image of the point (1, 6, 3) in the line 1 = 2 = 3 .
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60
5. Find the image of the point (1,3,4) in the plane 𝑥 − 𝑦 + 𝑧 = 5.
6. Find the vector and the Cartesian equations of the plane passing through the intersection of the
planes 𝑟 . (𝑖̂ + 𝑗̂ + 𝑘̂) = 6 and 𝑟 . (2𝑖̂ + 3𝑗̂ + 4𝑘̂) = −5 and the point (1, 1, 1).
7. Find the shortest distance between the lines whose vector equations are:
𝑟 = (1 − 𝑡)𝑖̂ + (𝑡 − 2)𝑗̂ + (3 − 2𝑡)𝑘̂ 𝑎𝑛𝑑 𝑟 = (𝑠 + 1)𝑖̂ + (2𝑠 − 1)𝑗̂ − (2𝑠 + 1)𝑘̂
8. Find the distance of the point (1, −2, 3) from the plane 𝑥 − 𝑦 + 𝑧 = 5 measured ‖ to the
𝑥+1
𝑦+3
𝑧+1
line 2 = 3 = −6 .
Study Module of Linear programming problems
LINEAR PROGRAMMING
SCHEMATIC DIAGRAM
Topic
concept
Linear Programming
(i) Introduction
(ii )Some solved
problems
(iii) Diet Problem
Degree of
Importance
**
***
***
iv) Manufacturing
Problem
***
(v) Allocation
Problem
**
(vi) Transportation
Problem
*
vii) Miscellaneous
Problems
**
Introduction:
Linear programming problems: A Linear Programming Problem is one that is concerned with
finding the optimal value (maximum or minimum value) of a linear function (called objective
Function) of several variables (say x and y), subject to the conditions that the variables are nonnegative and satisfy a set of linear inequalities (called linear constraints). The term linear
implies that all the mathematical relations used in the problem are linear relations while the term
programming refers to the method of determining a particular plan of action.
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61
Objective function: Linear function Z = ax + by, where a, b are constants, which has to be
maximized or minimized is called a linear objective function.
Constraints: The linear inequalities or inequations or restrictions on the variables of a
linear programming problem are called constraints. The conditions x ≥ 0, y ≥0 are
called non-negative constraints or restrictions.
Optimization problem: A problem which seeks to maximise or minimise a linear
function (say of two variables x and y) subject to certain constraints as determined by
a set of linear inequalities is called an optimisation problem. Linear programming
problems are special type of optimisation problems.
Feasible region: The common region determined by all the constraints including
non- negative constraints x ≥ 0, y ≥0 of a linear programming problem is called the feasible
region (or solution region) for the problem.
Optimal (feasible) solution: Any point in the feasible region that gives the optimal
value (maximum or minimum) of the objective function is called an optimal solution.
IMPORTANT SOLVED PROBLEMS
Q1. A dietician wishes to mix together two kinds of foods X and Y in such a way that the mixture
contains at least 10 units of vitamin A, 12 units vitamin B and 8 units of vitamin C. The vitamin
contents on one kg. food is given below :
Food
Vitamin A
Vitamin B
Vitamin C
X
1
2
3
Y
2
2
1
One kg. of food X costs Rs. 16 and one kg. of food Y costs Rs. 20. Find the least cost of the
mixture which will produce a required diet?
Sol. Let x kg and y kg food of two kinds of foods X and Y to be mixed in a diet respectively.
The contents of one kg. food of each kind as given below:
Food
Vitamin A Vitamin B
Vitamin C
Cost
X
1
2
3
16
Y
Minimum
Requirement
2
2
1
20
10
12
8
The above L.P.P. is given as
Minimum, Z = 16x + 20 y
subject to the constraints
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62
x + 2y ≥10, 2x + 2y ≥ 12,
3x + y ≥ 8, x, y ≥ 0
L1 : x + 2y = 10
X
Y
L2 : x + y = 6
A
B
10
0
0
5
C
X
Y
6
0
L3 : 3x + y = 8
D
x
y
0
6
Corner points
A (10, 0)
F (0, 8)
G (1, 5)
H (2, 4)
E
F
2
2
0
8
Z = 16x + 20 y
160
160
116
112
Here the cost is minimum at H (2, 4)
Since the region is unbounded therefore Rs. 112 may be or may not be the minimum value of C.
For this draw of inequality
16x + 20y < 112 i.e. 4x + 5y -< 28
L : 4x + 5 y = 28
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63
x
y
7
0
2
4
Clearly open half plane has no common point with the feasible region so minimum value of Z is
Rs. 112.
Q 2. An aeroplane can carry a maximum of 200 passengers. A profit of Rs.1000 is made on each
executive class ticket and a profit of Rs.600 is made on each economy class ticket. The airline
reserves at least 20 seats for executive class. However at least 4 times as many passengers prefer
to travel by economy class than by the executive class. Determine how many tickets of each type
must be sold in order to maximize the profit for the airline. What is the maximum profit?
SOLUTION:- Let the number of executive class ticket = x
and the number of economy class tickets = y
Given, maximum capacity of passengers = 200
∴ x + y ≤ 200
Atleast 20 seats of executive class are reserved.
∴ x≥ 20
Also atleast 4x seats of economy class are reserved
∴ y≥4x
Therefore, above L.P.P. is given as
Maximum P = 1000 x + 600 y
subject to the constraints
x + y ≤ 200,
x ≥ 20,
y ≥ 4x or 4x – y ≤ 0
x ≥ 0, y ≥ 0
L1 : x + y = 200
X
Y
L2 : 4x + y = 0
A
B
0
200
200
0
X
C
D
0
50
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64
Y
0
200
L3 :x = 20
Corner Points
E (20, 80)
F ( 40, 160)
G (20, 180)
P = 1000x + 600y
68000
136000
128000
(Maximum)
∴ here profit is maximum at F (40, 160)
∴ 40 tickets of executive class and 160 tickets of economy class to sold to get maximum profit and
maximum profit is Rs. 136000.
Q3. A factory manufactures two types of screws, A and B; each type requiring the use of two
machines an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on
hand operated machines to manufacture a package of screws A, while it takes 6 minutes on
automatic and 3 minutes on the hand operated machines to manufacture a package of screws B.
Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package
of screws A at a profit Rs. 7 and screws B at a profit of Rs. 10. Assuming that he can sell all the
screws he manufactures, how many packages of each type should the factory owner produce in a
day in order to maximise his profit? Determine the maximum profit.
Sol. Let number of packages of screws A produced = x
And number of packages of screws B produced = y
The number of minutes for producing 1 unit of each item is given below:
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65
Screw
Automatic
Machine
4
6
240
A
B
Time
available
Hand operated
Machine
6
3
240
Therefore, the above L.P.P. is given as
Maximise, P = 7x + 10 y subject to the constraints.
4x + 6y ≤240 ; 6x + 3y ≤ 240
i.e. 2x + 3y ≤ 120 : 2x + y ≤80, x, y≥0
L1 ; 2x + 3y = 120
A
B
X
Y
60
0
Corner points
O (0,0)
C (40,0)
B (0,40)
E (30, 20)
0
40
L2 : 2x + y = 80
C
D
x
y
40
0
0
80
P = 7x + 10y
0
280
400
410
(maximum)
65
Profit
7
10
66
Here profit is maximum at E (30,20)
∴ Number of packages of screws A = 30
Number of packages of screws B = 20
Maximum profit = Rs. 410.
Flow Chart
Step 1.Write the given informations in the tabulated form.
Step2. Form the L.P.P model of the problem.
Step3. Draw all the constraints by converting them in to equations.
Now we solve the L.P.P. by CORNER POINT METHOD which has the following steps
Step 1. Find the feasible region of the linear programming problem bounded by all the
constraints and determine its corner points (vertices) either by inspection or by solving the two
equations of the lines intersecting at that point.
Step 2. Evaluate the objective function Z = ax + by at each corner point. Let M and m,
respectively denote the largest and smallest values of these points.
Step 3. (i) When the feasible region is bounded, M and m are the maximum and minimum values
of Z.
(ii) In case, the feasible region is unbounded, we have:
Step 4. (a) M is the maximum value of Z, if the open half plane determined by
ax+ by > M has no point in common with the feasible region. Otherwise, Z
has no maximum value.
(b) Similarly, m is the minimum value of Z, if the open half plane determined by ax + by <m has
no point in common with the feasible region. Otherwise, Z has no minimum value
ASSIGNMENTS
(i) LPP and its Mathematical Formulation
LEVEL I
1.A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g)
of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin
A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4
units of cholesterol and 3 units of vitamin A. The diet requires at least 240 units of calcium, at
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least 460 units of iron and at most 300 units of cholesterol. How many packets of each food
should be used to minimise the amount of vitamin A in the diet? What is the minimum amount
of vitamin A?
(ii) Graphical method of solving LPP (bounded and unbounded solutions)
LEVEL I
Solve the following Linear Programming Problems graphically:
1.Minimise Z = – 3x + 4 ysubject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
2.Maximise Z = 5x + 3ysubject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
3.Minimise Z = 3x + 5y suchthat x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
(iii) Diet Problem
LEVEL II
1.A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and
1,400 calories. Two foods X and Y are available at a cost of Rs. 4 and Rs. 3 per unit respectively.
One unit of the food X contains 200 units of vitamins, 1 unit of mineral and 40 calories, whereas
one unit of food Y contains 100 units of vitamins, 2 units of minerals and 40 calories. Find what
combination of X and Y should be used to have least cost? Also find the least cost.
2. A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture
contain at least 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2 units/kg of
vitamin A and 1 unit/kg of vitamin C.Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of
vitamin C. It costs Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’.
Formulate this problem as a linear programming problem to minimise the cost of such a mixture.
In what way a balanced and healthy diet is helpful in performing your day-to-day activities
(iv) Manufacturing Problem
LEVEL II
1.A company manufactures two articles A and B. There are two departments through which these
articles are processed: (i) assembly and (ii) finishing departments. The maximum capacity of the
assembly department is 60 hours a week and that of the finishing department is 48 hours a week.
The production of each article A requires 4 hours in assembly and 2 hours in finishing and that of
each unit of B requires 2 hours in assembly and 4 hours in finishing. If the profit is Rs. 6 for each
unit of A and Rs. 8 for each unit of B, find the number of units of A and B to be produced per
week in order to have maximum profit.
2. A company sells two different produces A and B. The two products are produced in a common
production process which has a total capacity of 500 man hours. It takes 5 hours to produce a unit
of A and 3 hours to produce a unit of B. The demand in the market shows that the maximum
number of units of A that can be sold is 70 and that for B is 125. Profit on each unit of A is Rs. 20
and that on B is Rs. 15. How many units of A and B should be produced to maximize the profit?
Solve it graphically. What safety measures should be taken while working in a factory?
Q3.A toy company manufactures two types of dolls, A and B. Market tests and available
resources have indicated that the combined production level should not exceed 1200 dolls per
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week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the
production level of dolls of type A can exceed three times the production of dolls of other type
by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on
dolls A and B, how many of each should be produced weekly in order to maximise the profit?
LEVEL III
1.A manufacture makes two types of cups, A and B. Three machines are required to manufacture
the cups and the time in minutes required by each is as given below:
Type of Cup
A
B
I
12
6
Machines
II
18
0
III
6
9
Each machine is available for a maximum period of 6 hours per day. If the profit on each cup A is
75 paise, and on B it is 50 paise, show that the 15 cups of type A and 30 cups of type B should be
manufactured per day to get the maximum profit.
(v) Allocation Problem
LEVEL II
1. Ramesh wants to invest at most Rs. 70,000 in Bonds A and B. According to the rules, he has to
invest at least Rs. 10,000 in Bond A and at least Rs. 30,000 in Bond B. If the rate of interest on
bond A is 8 % per annum and the rate of interest on bond B is 10 % per annum , how much
money should he invest to earn maximum yearly income ? Find also his maximum yearly income.
Q2A merchant plans to sell two types of personal computers – a desktop model and a portable
model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly
demand of computers will not exceed 250 units. Determine the number of units of each type of
computers which the merchant should stock to get maximum profit if he does not want to invest
more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is
Rs 5000.
LEVEL III
1. An aeroplane can carry a maximum of 250 passengers. A profit of Rs 500 is made on each
executive class ticket and a profit of Rs 350 is made on each economy class ticket. The airline
reserves at least 25 seats for executive class. However, at least 3 times as many passengers prefer
to travel by economy class than by the executive class. Determine how many tickets of each type
must be sold in order to maximize the profit for the airline. What is the maximum profit?
Answers
(i) LPP and its Mathematical Formulation
LEVEL I
1. Z = 6x + 3y, 4x + y ≥ 80, x + 5y ≥115, 3𝑥 + 2𝑦 ≤ 150x, y ≥0
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69
(ii) Graphical method of solving LPP (bounded and unbounded solutions)
1. Minimum Z = – 12 at (4, 0),
2. Maximum Z =
235
 20 45 
at  , 
19
 19 19 
3 1
3. Minimum Z = 7 at  , 
2 2
(iii) Diet Problem
LEVEL II
1. Least cost = Rs.110 at x = 5 and y = 30
2. Minimum cost = Rs.380 at x = 2 and y = 4
(iv) Manufacturing Problem
LEVEL II
1. Maximum profit is Rs. 120 when 12 units of A and 6 units of B are produced
2. For maximum profit, 25 units of product A and 125 units of product B are produced
and sold.
3. 800 dolls of type A and 400 dolls of type B; Maximum profit = Rs 16000
(v) Allocation Problem
LEVEL II
1. Maximum annual income = Rs. 6,200 on investment of Rs. 40,000 on Bond A and
Rs. 30,000 on Bond B.
Q2. 200 units of desktop model and 50 units of portable model;
Maximum profit =Rs 1150000.
LEVEL III
1.For maximum profit, 62 executive class tickets and 188 economy class ticket should be sold.
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PROBABILITY:
INTRODUCTION:
Topic
Concepts
Probability
Degree of Importance
(i) Conditional Probability
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(ii) Multiplication theorem on
probability
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(iii) Independent events
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(iv) Baye’s Theorem,
Partition of a sample space
and theorem of total
probability
(v) Random Variables &
Probability distribution Mean
& Variance of Random
Variables
(vi) Bernaulli’s trails and
Binomial distribution
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Reference from NCERT
Book Vol. II
Article 13.2 and 13.2.1
Solved Ex. 1 to 6
Ex, 13.1 Q.N.-1,5 to 15
Article 13.3
solved Ex. 8 & 9
Ex. 13.2 Q.N.-2,3,13,14, 16
Article 13.4
Solved Ex. 10 to 14
Ex 13.2 Q.N.-1,6,7,8,11
Article 13.5, 13.5.1, 13.5.2
Solved Ex. 15 to 21
Ex. 13.3 Q.N.-1 to 12
Misc. Ex. Q.N. 13 to 16
Articles 13.6, 13.6.1, 13.6.2
13.6.3
Solved Ex. 24 to 29
Ex 13.4 Q.No.-1,4 to 15
Articles 13.7, 13.7.1, 13.7.2
Solved Ex. 31 & 32
Ex. 13.5 Q.N.- 1 to 13
Concept Mapping:
𝑷(𝑬∩𝑭)
Conditional Probability: 𝑷(𝑬/𝑭) = 𝑷(𝑭) , P(F) ≠ 𝟎
Multiplication Theorem: 𝑃(𝐸 ∩ 𝐹) = 𝑃(𝐸). 𝑃(𝐹/𝐸)
If E and F are independent events then 𝑃(𝐸 ∩ 𝐹) = 𝑃(𝐸). 𝑃(𝐹)and vice versa
Baye’s Theorem: If E1, E2 and E3 are three events of sample space S and E1∪ E2∪ E3 = S
and pair wise disjoint sets i.e. 𝐸1 ∩ 𝐸2 = 𝐸2 ∩ 𝐸3 = 𝐸3 ∩ 𝐸1 = ∅ If A is any event with
non zero probability. Then
𝑷(𝑬𝟏 /𝑨) = 𝑷(𝑬
𝑷(𝑬𝟏 )𝑷(𝑨/𝑬𝟏 )
𝟏 )𝑷(𝑨/𝑬𝟏 )+𝑷(𝑬𝟐 )𝑷(𝑨/𝑬𝟐 )+𝑷(𝑬𝟑 )𝑷(𝑨/𝑬𝟑 )
Probability Distribution: The probability distribution of a random variable X is the system of
numbers
X
:
x1
x2
x3
xn
...
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P(X)
:
p1
p2
p3
pn
Where, pi > 0, ∑𝑛𝑖=1 𝑝𝑖 = 1, 𝑖 = 1, 2, … , 𝑛
Mean or Expectation of a random variable X i.e. E (X) = 𝜇 = ∑𝑛𝑖=1 𝑥𝑖 𝑝𝑖 , 𝑖 = 1,2, … , 𝑛
Variance of a random variable X: 𝑉𝑎𝑟(𝑋) = ∑𝑛𝑖=1 𝑥𝑖 2 𝑝(𝑥𝑖 ) − (∑𝑛𝑖=1 𝑥𝑖 𝑝(𝑥𝑖 )2
𝑂𝑟 𝜎 2 𝑥 = E(𝑋 2 ) − [E(X)]2
Standard Deviation: 𝜎𝑥 = √𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒
Binomial Distribution: B (n, p)
P(X = r) = nCr 𝑝𝑟 𝑞 𝑛−𝑟 , 𝑟 = 0, 1, … , 𝑛 and q = 1- p, where p is the probability of success.
…
Solved Examples:
1.
A biased die is twice as likely to show an even number as an odd number. The die is rolled
three times. If occurrence of an even number is considered a success, then write the
probability distribution of number of successes. Also find the mean number of successes.
Which human value is violated in this case?
Solution:
1
1
2
2
P (odd number) =
P (even number) =


1 2 3
1 2 3
Here occurrence of an even number is considered a success. Let the number of success is a
random variable x and can take values 0, 1, 2 or 3.
The probability distribution of number of successes is as below:
1 1 1 1
P(x = 0)
= P (no success)
= P (FFF)
=   
3 3 3 27
 2 1 1 6
P(x = 1)
= P (one success)
= P (SFF, FSF, FFS) = 3    
 3 3 3  27
 2 2 1  12
P (x = 2)
= P (two success)
= P (SSF, SFS, FSS) = 3    
 3 3 3  27
2 2 2 8
P (x = 3)
= P (three success) = P (SSS)
=    
 3 3 3  27
X = xi
0
1
P(x) = pi
1/27
6/27
Mean number of successes =  xi pi
2
12/27
3
8/27
1  
6  
12  
8  54

2
=  0    1     2     3   
27   27  
27   27  27

Having unbiased is violated in this case.
Q.2
Probabilities of solving a specific problem independently by A and B are 1/2 and 1/3
respectively. If both try to solve the problem independently, find the probability that (i) the
problem is solved (ii) exactly one of them solves the problem.
Solution:
P(A) = 1/2 = prob. that A will solve the problem
P(B) = 1/3 = prob. that B will solve the problem
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(i)
Probability that the problem is solved
=
1 - prob. that none of them solve the problem
 1  1 
1 2 2
1  P A .P B =
1  1  1    1     
=
 2  3 
2 3 3
 
(ii)
Probability that exactly one of them will solve the problem


P A B or A B .  P( A)  P( B)  P( A)  P( B)


1  1  1 1 1 2 1 1 3 1
  1     1             
2  3  2 3  2 3  2 3 6 2
Q.3
Two cards are drawn simultaneously without replacement from a well shuffled pack of 52
cards. Find the mean and variance of the number of aces.
Solution:
Let x denote the number of aces in a draw of two cards.
x is a random variable which can assume the values 0, 1 and 2
48  47
188
48C 2
P ( x  0)  P (no ace) 
 2 1 
52

51
221
52C 2
2 1
P ( x  1)  P (one ace and one non  ace)
4C  48C1 4  48  2 32
 1


52C 2
52  51
221
P ( x  2)  P (two aces) 
4C 2
43
1


52C 2 52  51 221
The probability distribution of x is
x or xi
P(x) = pi
Mean = E(x) =  xi pi
 
E x2
0
188/221
1
32/221
2
1/221
32  
1  2
 188  
= 0
  1 
  2

221   221  
221  13

32  
1  36
 188  
  xi2 pi   0 
  1 
  4

221   221  
221  221

2
36  2  400


Var ( x)  E x  E x  

221  13  2873
Q.4
A family has 2 children. Find the probability that both are boys, if it is known that
(i) at least one of the children is a boy
(ii) the elder child is a boy
Solution:
S  B1B2 , B1G2 , G1B2 , G1G2 
 
2
2
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(i)
at least one of the children is a boy
A = Both the children are boys B1B2 
B = At least one of the children is a boy  B1B2 , B1G2 , G1B2 
1

A  B
1
 A
Required probability = P   P
 4
3
P( B)
3
B
4
(ii)
The elder child is a boy
A = Both the children are boys = B1B2 
B = elder child is a boy =  B1B2 , B1G2 
 A  B   14  1
 A
Required probability = P   P
2
P( B)
2
B
4
Q.5
Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a
meditation and yoga course reduces the risk of heart attack by 30% and prescription of
certain drugs reduces his chances by 25%. At a time a patient can choose any one of the
two options with equal probabilities. It is given that after going through one of the two
options, the patient selected at random, suffers a heart attack. Find the probability that the
patient followed a course of meditation and yoga. In a student life state any one point how
yoga and meditation influence.
Solution:
Let E1 and E2 be events of selection of meditation and yoga and prescription of medicine
respectively.
Let A = event of having heart attack.
1
We have
PE1   PE 2  
2
 A 
30
28

P    40 
 40 % 
100
100

 E1  
 A
P
 E2
 
25
30

   40 
 40 % 
100
100

 
E 
Required probability = P  1 
 A
 A
1 28
PE1   P 

E1 
28 14

2
100




 A
 A  1  28  1  30 58 29
PE1   P   PE 2   P 
 E1 
 E 2  2 100 2 100
yoga and meditation improves our physical and mental health.
Q.6
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck
drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the
insured persons meets with an accident. What is the probability that he is a scooter driver?
Which mode of transport would you suggest to students and why?
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Solution:
Let E1, E2 and E3 are the events of selection of a scooter driver, car driver and truck driver
respectively.
Let A = event that the insured person meets with an accident.
2000 1
4000 1
6000 1
PE1  

P E 2  

P E 3  

12000 6
12000 3
12000 2
 A
 A
 A
P   0.01
P   0.03
P   0.15
 E1 
 E2 
 E3 
E 
Required probability = P  1 
 A
 A
1
PE1   P 
 0.01
E
 1
6


 A  1  0.01  1  0.03  1  0.15
 A
 A
PE1   P   PE 2   P   PE3   P 
3
2
 E1 
 E2 
 E3  6
0.01
0.01 1


0.01  0.06  0.45 0.52 52
Cycle should be suggested as it is good for (i) health (ii) no pollution (iii) saves energy (no
fuel).

1.
2.
3.
4.
5.
6.
7.
Practice Problem
Level-1
B
 
If P(A) = 0.3, P(B) = 0.2 find P   , if A and B are mutually exclusive events.
 A
A coin is tossed thrice and all the 8 outcomes are equally likely:
E: the first throw results in head
F: the last throw results in tail
Are the events independent?
Given P(A) = 1/4 , P(B) = 2/3 and P(AUB) = 3/4. Are the events independent?
If A and B are independent events, find P(B) if P(AUB) = 0.60 and P(A) = 0.35.
Two cards are drawn with replacement from a well shuffled pack of 52 cards. Find the
probability distribution of the number of spades.
4 defective apples are accidentally mixed with 16 good ones. Three apples are drawn at
random from the mixed lot. Find the probability distribution of the number of defective
apples.
A random variable X is specified by the following distribution:
X
2
3
4
P(X)
0.3
0.4
0.3
Find the mean and variance of distribution.
Level-2
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1.
2.
3.
1.
2.
3.
4.
5.
6.
A dice is thrown twice and sum of numbers appearing is observed to be 6. What is the
conditional probability that the number 4 has appeared at least ones?
The probability of A hitting a target is 3/7 and that of B hitting is 1/3. They both fire at
their target find the probability that:
(a) at least one of them will hit the target
(b) only one of them will hit the target
A company has two plants to manufacture bicycles. The first plants manufactures 60% of
the bicycle and second plant 40%. Out of that 80% of the bicycles are rated of standard
quality at the first plant and 90% of the standard quality at the second plant. A bicycle is
picked up at random and found to be standard quality. Find the probability that it comes
from the second plant.
Level-3
A class consists of 80 students, 25 of them are girls and 55 are boys. 10 of them rich and
remaining poor; 20 of them are fair complexioned. What is the probability that selecting a
fair complexioned rich girl.
Two integers are selected from integers 1 to 11. If the sum is even, find the probability that
both numbers are odd.
Given three identical boxes I, II and III, each containing two coins. In box I, both coins are
gold coins, in box II, both are silver coins and in the box III, there is one gold and one
silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold,
what is the probability that the other coin in the box is also of gold?
A coin is biased so that the head is 3 times as likely to occur as a tail. If the coin is tossed
twice, find the probability distribution of the number of tails.
The mean and Variance of a binomial distribution are 4/3 and 8/9 respectively. Find
P (X  1).
A card is drawn from a pack 52 cards is lost. From the remaining cards of the pack, two
cards are drawn and are found to be both hearts. Find the probability of the lost card being
a heart.
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