Download KENDRIYA VIDYALAYA SANGATHAN RAIPUR REGION STUDY

KENDRIYA VIDYALAYA
SANGATHAN
RAIPUR REGION
STUDY MODULES CLASS-XII
MATHEMATICS
2012-13
Minimum area of XII Maths to get at least 60%
Topic
Weightage
1
1.
Equivalence relation, Binary operation and Composition of two functions
4 Marks & 1Mark
2.
Inverse Trigonometric functions
4 Marks
3.
Matrices (Order of a matrix, Symmetric and skew symmetric matrix &
Transpose of a matrix)
1 Mark
4.
Express a matrix as the sum of a symmetric and skew symmetric matrix or a
problem based on transpose of a matrix
4 Marks
5.
Determinant (Simplification by using properties)
4 Marks
6.
Solution of a system of linear equations by using matrix method
6 Marks
7.
To check the continuity of a given function at a given point
4 Marks
8.
To find the derivative of a function based on the model chain rule, Second
derivative, (variable)variable+(variable)variabl
4 Marks
9.
(a)To find the interval in which a function is Increasing or decreasing ,To find
the rate of change of quantities, To find the equation of tangent and normal
(b) Maxima and Minima (To indicate flow chart in summary)
Integration by substitution, using partial fractions, integration of the
types
4 Marks
4 Marks
12.
Definite integral based on the properties
,
Integration as the limit of a sum (To indicate flow chart in summary)
13.
Area bounded by two curves(To indicate flow chart in summary)
6 Marks
14.
Differential equations (Variable Separable, homogeneous, Linear Differential
equation) (To indicate flow chart for linear differential equation).
6 Marks
15.
To find shortest distance between two lines / coordinate of foot
perpendicular from origin to a plane/ Image of a point in a plane
(To indicate flow chart in summary)
Linear programming problem (To indicate flow chart in summary)
4 Marks
A problem based on Dot & Cross product of vectors ( To find a unit vector
perpendicular to two given vectors, Area of triangle and parallelogram if
represent the adjacent sides, Condition of perpendicularity and
parellelity, based on properties of dot and cross product of vectors)
Probability Bayes theorem application (Indicate formula and flow chart)
4Marks & 1Mark
10
11.
16.
17.
18.
RELATIONS AND FUNCTIONS
2
6 Marks
4 Marks
6 Marks
6 Marks
6Marks
SCHEMATIC DIAGRAM
S Topic
N
1
Relations and
functions
Syllabus
Concept
Degree of
Importance
Types of relations: reflexive
symmetric transitive and
equivalence relations. One
to one and onto functions,
composite functions inverse
of a function Binary
operations.
1 Types of relations
***
2 One to one and onto
**
3 Inverse of a function
***
4 Composition of a
function
5.Binary operation
***
***
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, a1 , a2  A (iii) Transitive, if a1 , a2   R and
a2 , a3   R implies that
a1 , a3   R, a1 , a2 , 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}. Shoe 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
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.
3
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}
1 in each case



is an equivalence relation. Find the set of all element related to
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 triangle 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?
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 element in A
related to the right angle triangle T with sides 3,4,and 5?
Concept :One-one (injective),Onto (surjective) and One-one
onto
(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
4
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
x
' x  R is one
1 x

Show that the function f : R  {x  R : 1 ‹ x ‹1} defined by f ( x) 

one and onto function.
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 injective
 
 
Consider a function f : 0,   R given by f x   sin x and g : 0,   R given by
 2
 2
.Show that f and g are one one ,but f +g is not one-one.
 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
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
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.
5
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 .


4x
 4
Let f : R     R be a function defined as f ( x) 
. The inverse of f is the
3x  4
 3
 4
map g : Range f  R    given by
 3
LEVEL – II
Consider f : R  4,  given by f ( x)  x 2  3 . 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 invertible

with f 1 ( 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 .
6
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*7,20*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
ab
* b=a + ab
(iv)a * b = (a-b)2
(v)a * b =
4
2
(vi)a * b = ab
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) a * b  1a, b  R
( a  b)
a, b  R
2
Let A  N  N and * be the binary operation on A defined by
a, b  c, d   a  b, b  d  .Show that * is commutative and associative. Find the identity
(b) a * b 



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 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].Do o distributive over *? Justify
your answer.
7

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 .
if a  b  6
a  b
Define a binary operation * on the set {0,1,2,3,4,5} as a * b  
a  b  6 if a  b  6
Show that zero is the identity for this operation and each element a  0 of the set of the
invertible with 6-a being the inverse of a
INVERSE TRIGONOMETRIC FUNCTIONS
Introduction
Topic
Concepts
Degree of Importance
Inverse Trigonometric
Functions
(i) Principal Value
Branch Table
(ii) Properties of Inverse
Trigonometric Functions
**
***
Reference from
NCERT Book Vol.1
Ex: 2.1 Q.N. 11, 12, 13
Ex: 2.2 Q.N. 11, 19, 20
Ex 2.2 Q.N. 7, 13, 15
Misc. Ex. Q.N. 7, 8, 10,
11, 12
Principal Value Branch Table
Functions
Domain
Principal Value Branches
y = sin-1 x
[-1, 1]
  
 2 , 2 
y = cos-1 x
[-1, 1]
0, 
y = cosec-1 x
R – (-1, 1)
y = sec-1 x
R – (-1, 1)
y = tan-1 x
R
y = cot-1 x
R
Properties of Inverse Trigonometric Functions:
For suitable Values of domain, we have:
8
  
 2 , 2   0
0,     
2
  
 , 
 2 2
0, 
1.
(a) y = sin-1 x  x = sin y
2.
(a) sin (sin-1 x) = x
3.
(a) sin-1   = cosec-1 x
(b) cos-1   = sec-1 x
(c) tan-1   = cot-1 x
4.
5.
(a) cos-1 (-x) =  - cos-1 x
(a) sin-1 (-x) = - sin-1 x
(b) cot-1 (-x) =  - cot-1 x
(b) tan-1 (-x) = - tan-1 x
(c) sec-1 (-x) =  - sec-1 x
(c) cosec-1 (-x) = - cosec-1 x
6.
(a) sin-1x + cos-1x =
(b) tan-1x + cot-1x =
(c) cosec-1x + sec-1x =
7.
(a) tan
8.
2 tan 1 x  sin 1
9.
(a) sin-1x + sin-1y = sin-1 x 1  y 2  y 1  x 2
1
x
1
1
x

2
x  tan 1 y  tan 1
x y
1  xy
1
x

2
(b) tan
1
x  tan 1 y  tan 1
2
2x
2x
1 1  x

cos
 tan 1
2
2
1 x
1 x
1 x2
(b) sin-1x - sin-1y = sin-1
10.
(b) x = sin y  y = sin-1 x
(b) sin-1 (sin x) = x

x
1 y2  y 1 x2

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 
2  2
 3 

1 
1 
sin 1  sin
  sin sin   
  sin  sin

5 
5 
5  5


 

3.
If tan 1 3  cot 1 x  , find x
2
9
x y
1  xy

2
Solution:
tan 1 3  cot 1 x 
cot 1 x 

2
2
 tan 1 3
= cot
4.

1
tan-1x + cot-1x =
3x 3
 12 
1  3 
1  56 
  sin    sin  
 13 
5
 65 
Prove that: cos 1 
Solution:
12
 12 
cos 1    x  cos x 
13
 13 
2
5
 12 
sin x  1  cos x  1    
13
 13 
2
3
3
and sin 1    y  sin y 
5
5
2
4
3
cos y  1  sin y  1    
5
5
2
sin (x + y) = sin x cos y + cos x sin y
5.
 5  4   12  3  56
       
 13  5   13  5  65
 56 
=  x  y  sin 1  
 65 
 12 
3
 56 
cos 1    sin 1    sin 1  
 13 
5
 65 
 1  sin x  1  sin x  x
 
Prove that cot 1 
  , x   0, 
 4
 1  sin x  1  sin x  2
Solution:
10

2
6.
 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 x 
 cot 1 


1  sin x  1  sin x 
 1  sin x  1  sin x
1  sin x  1  sin x  2 1  sin x 1  sin x 
 cot 1 

1  sin x   1  sin x 


 2  2 1  sin 2 x 
1 1  cos x 
 cot 1 
  cot 

2 sin x
 sin x 


x 

2 cos 2

2   cot 1  cot x   x
 cot 1 
x
x
2 2

 2 sin cos 
2
2

1
1
1
1 
Prove the following: tan 1    tan 1    tan 1    tan 1    .
 3
5
7
8 4
Solution:
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 
11
7.

 x 1 
1  x  1 
  tan 
  , then find the value of x.
 x2
 x2 4
If tan 1 
Solution:

 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 
tan 1 
  tan 1  tan 

 x2
 x2
x 1 

 1

1 
1
x

2

  tan 1 
 tan

1 x 1 
 2x  3 


x2 

x 1
1
1
1


 2x 2 1  0  x 2   x  
x  2 2x  3
2
2
PRACTICE PROBLEMES:
Level-1
1.
Write the Principal value of the following:
(i) tan-1  3 
2.
3.
4.
5.
(ii) sin-1   1 
(iii) cos-1   1 
 2

Evaluate: cot tan 1 a  cot 1 a.
Prove: 3sin-1x = sin-1(3x-4x3)
Find x if sec-1(2) + cosec-1x = /2
Solve tan-12x + tan-13x = /4
Level-2
1.
Write the principal value of the following:
2 
2 

1 
cos 1  cos
  sin  sin

3 
3 


2.
Write in the simplest form:
 1  x 2  1
tan 1 
x 0
,
x


3.
Prove that sin-1(8/17) + sin-1(3/5) = tan-1(77/36).
4.
Prove that 2 tan-1(1/2)+ tan-1(1/7) = tan-1(31/17)
5.
Solve for x
2
2
1 2 x
1 1  x
tan
 cot

2
2x
3
1 x
12
2
Level-3
1.
2.
3.
4.
5.
 1 x  1 x   1
1
Prove that : tan 1 
   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 = .
 x y 
x
 
Prove that: tan 1  tan 1 
y
 x y 4

 1  x 
 .
Write in the simplest form: cos 2 tan 1 


 1  x 
Study Module for ALGEBRA
MATRICES & DETERMINANTS
SCHEMATIC DIAGRAM
Matrices &
Determinants
(i)Introduction
(ii) Some solved problems
**
***
(iii) Order, Addition,
Multiplication and transpose of
matrices
***
(iv) Cofactors & Adjoint of a
matrix
(v)Inverse of a matrix &
applications
**
***
(vi)To find difference between
│A│, │adj A │,
│ kA│, │A.adjA│
*
(vii) Properties of Determinants
**
INTRODUCTION:
13
Definitions
Q 1,2,3.4,5,6 (Given below in the
booklet)
NCERT Text Book XII Ed. 2007
Ex 3.1 –Q.No 4,6,10
Ex 3.2 –Q.No 7,9,17,18
Ex 3.3 –Q.No 5;6,10
Theorem 4 (page 91), Q 6 (page
100)
Ex 4.4 –Q.No 5
Ex 4.5 –Q.No 11,12,13,17,18
Ex 4.6 –Q.No 15,16
Example – 32 ,33
MiscEx 4–Q.No,5,6,8,12,15
Ex 4.1 –Q.No 3,4,7,8
Ex 4.2 – Q No. 15
Ex 4.5 – Q No. 17
Ex 4.2–Q.No 7,8 11,12,13;14
Example –16,18,28
MATRIX: If mn elements can be arrange 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
order.
Addition and subtraction of matrices: Two matrices A and B can be add or subtract if they are of the
same order i.e. if A and B are two matrices of order
then A B is also a matrix of order
.
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
matrix and B be an
matrix then the
product of matrices A and B is another matrix of order
.
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
.
Properties of transpose of the Matrices: For any matrices A and B of suitable orders, we have
(i)
Symmetric Matrix: A square matrix M is said to be symmetric if
e.g.
Note: there will be symmetry about the principal diagonal in Symmetric Matrix.
Skew symmetric Matrix: A square matrix M is said to be symmetric if
e.g.
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
Let A
be an Square Matrix of order2
defined by
= ay-bx
Determinant of a matrix of order 3
order 3
then the determinant of A is denoted by
: Let us consider the determinant of a square matrix of
=
14
and
Expansion along first row
We can expand the determinant with respect to any row or any column.
Minors and cofactors:
Minor of an element of a determinant is the determinant obtained by
deleting its ith row and jth column in which element lies. Minor of an elemen
denoted by
.
Cofactors: cofactors of an element
by
and is defined by
is the minor of .
Adjoint of a Matrix: Let A =
is
where
be a Matrix of order
Then adj(A) =
Again let A =
be a Matrix of order
Then adj(A) =
Inverse of a Matrix: Inverse of a Square Matrix A is defined as
Note: If A be a given Square Matrix of order n then
(i)
A(adj(A) = adj(A)A=
where I is the Identity Matrix of order n.
(ii)
(iii)
A square Matrix A is said to be singular and non-singular according as
=
)
IMPORTANT SOLVED PROBLEMS
15
Q1. If A=
Verify that (AB)’=B’A’
Solution: - We have
If A=
Then
AB =
Now
A’ =
=
, B’=
B’A’ =
=
= (AB)’
Clearly (AB)’ = B’A’
Q2. If
then find the value of x and y.
Sol. Given
or
So 2x – y = 10 and 3x + y = 5
On solving we get x = 3 and y = -4
Q3. If F(x) =
Sol. Given F(x) =
prove that F(x) F(y) = F(x+Y)
so F(y) =
16
Hence F(x) .F(y) =
=
Hence F(x) F(y) = F(x+Y)
Q4. Express the given Matrix as the sum of a symmetric and skew symmetric matrix
A=
Sol. Here
P=
Now
so
P=
is a symmetric Matrix.
Also let Q = =
Hence Q is an Skew Symmetric Matrix.
Now P + Q=
Thus A is represented as the sum of a symmetric and skew symmetric matrix.
Q5. Using the property of determinant prove that
Sol.Applying
17
L.H.S =
Taking common a + b + c from first Row we get
L.H.S =(a + b+ c)
Now applying
L.H.S =(a + b+ c)
Expanding along first Row L.H.S = (a+b+c)
= 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
A=
Now
adj(A) =
Hence
So
Hence x = 3 , y = -2 , z = 1
18
Flow chart:
Step 1.Write the given system of equation in the form of A X = B
Step2. Find
Step3. Find adj(A)
Step4. Find
Step5. Find
Step6 Find the value of x , y and z
ASSIGNMENTS
(i). Order, Addition, Multiplication and transpose of matrices:
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 =
3.
If A =
4. If A =
, B=
and B =
|i – 3j |
then find A –2 B.
,
, write the order of AB and BA.
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
19
LEVEL III
1.
If
, then find the value of A2 –3A + 2I
A =
2. Express the matrix A as the sum of a symmetric and a skew symmetric matrix, where:
A=
3. If A =
=
(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 =
Verify A(adjA) = (adjA) A =
LEVEL II
I if
1. 1. A =
2.
1. 2. A =
2. (iii) Inverse of a Matrix & Applications
LEVEL I
1.
If A =
, write
-1
A in terms of A
2.
If A is square matrix
2
satisfying A = I, then what is the inverse of A ?
3.
For what value of k , the
matrix A =
is not invertible ?
20
4.
LEVEL II
, show that A2 –5A – 14I = 0. Hence find A-1
1.
If A =
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 and for
swhich A
=
has exactly two distinct solutions.
LEVEL III
1
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
2. Using matrices, solve the following system of equations:
a. x + 2y - 3z = - 4
2x + 3y + 2z = 2
3x - 3y – 4z = 11
b. 4x + 3y + 2z = 60
x + 2y + 3z = 45
6x + 2y + 3z = 70
3. Find the product AB, where A =
,B=
and use it to
solve the equations x – y = 3, 2x + 3y + 4z = 17, y + 2z = 7
4. Using matrices, solve the following system of equations:
-
+
=4
+
-
= 0
+
+
= 2
5. Using elementary transformations, find the inverse of the matrix
(iv)To Find The Difference Between
21
LEVEL I
1.
Evaluate
2.
What is the value of
3.
If A is nonsingular matrix of order 3 and
4.
For what valve of a,
, where I is identity matrix of order 3?
= 3, then find
is a singular matrix?
LEVEL II
1. If A is a square matrix of order 3 such that
2. If A is a nonsingular matrix of order 3 and
= 64, find
= 7, then find
LEVEL III
and
3
1.
If A =
= 125, then find a.
2.
A square matrix A, of order 3, has
= 5, find
(v).Properties of Determinants
LEVEL I
1.
2.
1. Find positive valve of x if
=
2. Evaluate
LEVEL II
1. Using properties of determinants, prove the following :
bc
a
a
b
ca
b  4abc
c
c
ab
2.
1  a 2  b2
2ab
 2b
2
2
2ab
1 a  b
2a
 1  a 2  b2
2b
 2a
1 a 2  b2

22

3
3.
= (1 + pxyz)(x - y)(y - z) (z - x)
LEVEL III
1. Using properties of determinants, solve the following for x :
a.
2.
= 0
b.
= 0
c.
= 0
If a, b, c, are positive and unequal, show that the following determinant is negative:
=
3.
a2 1
ab
ac
2
ab
b 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 b  c
ca c  a  0
ab a  b
23
b 2  bc c 2  bc
 bc
6.
a 2  ac
 ac
c 2  ac  (ab  bc  ca ) 3
a 2  ab b 2  ab
 ab
7.
= 2abc( a + b + c)3
8.
bc ca a b
ca a b bc  0
a b bc ca
If a, b, c are real numbers, and
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.
1.
2.
+
(ii). Cofactors & Adjoint of a matrix
LEVEL I
1. 46
2.
(iii) Inverse of a Matrix & Applications
LEVEL I
-1 = -
1. A
A
-1 =
2. A
A
3. k = 17
24
4. 2  2, 3  3
LEVEL II
1.
3. 0
1. x = 1, y = 2, z = 3.
LEVEL III
2. x = 3, ,y = -2, z = 1. 3. AB = 6I, x =
4. x = ½, y = -1, z = 1.
5.
, y = - 1, z =
(iv). To Find The Difference Between
LEVEL I
1.
2. 27
3. 24
LEVEL II
1. 8
2. 49
LEVEL III
1. a = 3
2. 125
(v). Properties of Determinants
LEVEL I
+ +
LEVEL II
2. [Hint: Apply C1  –bC3 and C2  aC3]
1. x = 4
2.
+
LEVEL III
1a. 4
1b.
1c.
25
4.
CONTINUITY AND DIFFERENTIABILITY
Concept :- Continuity
Suppose f is a real function on a subset of the real number and let c be a point in the
domain of f.
then f is continuous at c if lim f  x   f c 
x c
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 x  0
For what value of  is the function defined by f x   
if x  0
4 x  1
continuous at x  0 ? What about continuity at x  1 ?
LEVEL-III
3
 1  sin x

if x 
 3 cos 2 x
2



Let f  x   
If f x  be a continuous function at x 
, find a
a
if x 
2
2

 b1  sin x  if x  
   2 x 2
2
and b.
 3ax  b if x  1

if x  1 is continuous at x = 1,find the value of a and
If the function f x    11
5ax  2b if x  1

b.
Find the relation ship 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
Concept :- Differentiation, Implicit function, logarithmic functions , Parametric
forms ,
26
Second order derivatives and lagrange\s Mean value Theorems
log m  n   log m  log n
m
log    log m  log n
n
log m n  n log m
log a b 
log b b
log b a
log a a  1
PRACTICE PROBLEMS
x  3x
LEVEL-I
 4
w.r.t x.
3x  4 x  5
2

Differentiate

Find

d2y
Find 2 , if y  x 3  tan x .
dx


2
dy
of the function:- x y  y x  1 .
dx
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
2x
3x
 5  6y  0 .
If y  3e  2e , prove that
2
dx
dx
LEVEL – II
x

1

 1 
1

Differentiate the function :-  x    x  x  .
x


Differentiate the function :-

Differentiate the function :- x x cos x 

Find




x cos x x  x sin x  x .
1
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

If y  3e 2 x  2e 3 x , prove that
d2y
dy
 5  6y  0 .
2
dx
dx
LEVEL –III
27
dy
, if y x  x y  x x  a b .
dx

Find

If x  a sin t , y  a cos

If y  tan 1 x ,show that
1

1
t
, show that

x
2
2
dy
y
 .
dx
x
 1 y2  2 x x 2  1y1  2
2
3

If
x  a 2   y  b2
  dy  2  2
1    
  dx  
2
 c , for some c>0 prove that 
is a
d2y
dx 2
constant independent of a and b


dy cos 2 a  y 

If cos y  x cosa  y  with cos a  1 ,prove that
dx
sin a
dy
1

If x 1  y  y 1  x  0 ,for, -1< x <1 , Prove that
dx
1  x 2

d2y
dy
 a2 y  0
If y  e
,1  x  1 ,Show that 1  x  2  x
dx
dx
cos x
sin x
Differentiate the function :- x  sin x  .

If y  sin 1 x , show that 1  x 2

Find

If x 

cos1 x
2

dy
of the function:dx
sin 3 t
cos 2t
,y 
cos 3 t
cos 2t
d
2
dx
y
2
x
dy
 0.
dx
yx  xy .
. Find
x  3x 2  4
dy
dx

Differentiate

For a positive constant a find

d2y
If x  acos t  t sin t  and y  asin t  t cos t  , find
dx 2
3x 2  4 x  5
w.r.t x.
1
t
dy
 1
,where y  a t , x   t  
dx
 t
 If y  500e 7 x  600e 7 x ,show that
d2y
 49 y
dx 2
28
a
Concept :- Rate of change of quantities ,increasing or decreasing
Rate of change of quantity:- If two varia bles x and y are varying with respect to
another variable
dy
dy dt
dx
t if x  f t  and y  f t  then by chain rule

, if
0
dx dy
dt
dt
Increasing or Decreasing function:- Let I be a open interval contained in the
domain of a real valued function f . Then f is said to be
(i) Increasing on I if x1  x2 in I  f x1   f x2 , x1 , x2  I
(ii) Strictly increasing on I if x1  x2 in I  f x1   f x2 , x1 , x2  I
(iii)Decreasing on I if x1  x2 in I  f x1   f x2 , x1 , x2  I
(iv) Strictly decreasing on I if x1  x2 in I  f x1   f x2 , x1 , x2  I
Theorem ;- Let f be continuous on a, b and differentiable on the open interval a, b
then
(i)f is increasing in a, b if f / x   0 for each x  a, b
(ii) 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
PRACTICE PROBLEMS







LEVEL – I
Show that the function f given by f x   x 3  3x 2  4 x, x  R is strictly increasing on R
An edge of a variable cube is increasing at the rate of 3cm/s. How fast is the volume of
the cube increasing when the edge is 10 cm long?
 
Find the interval in which the function given by f  x   sin 3x, x  0,  is (a) increasing
 2
(b)decreasing
LEVEL-II
Find the interval in which the function f given by f x   sin x  cos x,0  x  2 is strictly
increasing or strictly decreasing.
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 remain spherical, has a variable diameter 2 x  1 find the rate
2
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 ycoordinate is changing 8 times as fast as the x-coordinate.
29
LEVEL – III



Find the interval in which the function f given by f  x  
4 sin x  2 x  x cos x
is (i)
2  cos x
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 meter 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 interval in which the function f given by f x   x 3  3 , x  0 is (i) increasing
x
(ii) decreasing
Concept :Tangents and Normal , Approximation
Tangents and Normal:- The slope of the tangent to the curve
point x0 , y 0  is
given by
y  f x  at the
dy 
dx   x0 , y0 

 1
The slope of the normal to the curve y  f x  at the point x0 , y 0  is given by
dy 

dx   x0 , y0 
The equation of a tangent at x0 , y 0  to the curve y  f x  is given by
y  y 0  f / x0 x  x0 
The equation of a tangent at x0 , y 0  to the curve y  f x  is given by
1
x  x0 
y  y0  /
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 .
 dy 
(ii)The differential of y, denoted by dy , is defined by dy  f /  x0 dx or dy   x .
 dx 
PRACTICE PROBLEMS
LEVEL -I
 Find the slope of the tangent to the curve y  3x 2  4 x at x=4
30
 Use differential to approximate
36.6
 .Find the slope of the normal to the curve x  a cos 3  , y  a sin 3  , at ,  

4
LEVEL –II
 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.
 If the radius of the sphere is measured as 9m with an error of 0.03m, then find the
approximate error
in calculating its surface area.
LEVEL-III
 Find the equation of the normal to the curve y  x 3  2 x  6 which are parallel to the
line
 Prove that the curves x  y 2 and xy  k cut at right angle  if 8k 3  1
 Find the equation of the tangent to the curve y  3 x  2 which is parallel to the line
4x  2 y  5  0

Concept : Maxima or minima
First derivative test : Let f be a function defined on an open interval I. Let f be
continuous at a critical point c in I. Then
1.If f / ( x ) changes sign from positive to negative as x increases through c, i.e., if f / x   0 at
every point sufficiently close to and to the left of c, and f / ( x )  0 at every point sufficiently
close to and to the right of c is a point of local maxima.
2. If f / ( x ) changes sing from negative to positive as x increases through c, i.e., if f / ( x )  0 at
every point sufficiently close to and to the left of c, and f / ( x )  0 at every point sufficiently
close to and to the right of c, then c is a point of local minima.
3. If f / ( x ) does not change sign as x increases through c, then c is neither a point of local
maxima nor a point of local minima. Infact , such a point is called point of inflection.
SECOND DERIVATIVE TEST :-Let f be a function defined on an interval I and c I. let f be
twice differentiable at c. Then
(1) x  c is a point of local maxima if f  (c) = 0 and f  (c) < 0 the value f (c) is local
maximum value of f.
(2) x = c is a point of local minima if f (c ) = 0 and f  (c) > 0 in this case ,f (c) is local
minimum value of f.
(3) The test fails if f ( c )  0 and f ( c )  0 .In case go back to first derivative test.
31
PRACTICE PROBLEMS
LEVEL-I
 Show that the function given by f  x  
log x
has maximum at x = c
x
 Prove that the radius of the right circular cylinder of greatest curved surface area which
can be
inscribed in a given cone is half of that of the cone

Show that the height of the cylinder of maximum volume that can be inscribed in a
sphere of radius
2R
R is
.Also find the maximum volume.
3
LEVEL – II
 A tank with rectangular base and rectangular sides, open at the top is to be constructed
so that its
depth is 2 m and volume is 8m 3
base and Rs45
. If building of tank cost Rs 70 per sq meters for the
per sq metre for sides .What is the cost of least expensive tank?
 The sum of the perimeter of a circle and square is k, where k is some costant. Prove
that the sum of
their areas is least when the side of square is double the radius of the circle.
 Show that semi-vertical angle of the cone of the maximum volume and of given slant
height is
tan 1 2
LEVEL – III
 Find the maximum area of an isosceles triangle inscribed in the ellipse
2
x
y2

 1 with its vertex at
a 2 b2
one end of the major axis
 A point on the hypotenuse of a triangle is at distance a and b from the sides of the
triangle .Show that
3
2 2
 23

the minimum length of the hypotenuse is  a  b 3  .


32
 Show that the height of the cylinder of greatest volume which can be inscribed in a
right circular
cone of height h and semi vertical angle α is one third that of the cone and the greatest
volume of
4
cylinder is
h 3 tan 2 
27
 Prove that the volume of largest cone that can be inscribed in a sphere of radius R is
8
of the
27
volume of the sphere

Show that semi-vertical angle of right circular cone of given surface area and
maximum volume is
1
sin 1  
 3
 An apache helicopter of enemy is flying along the curve given by y  x 2  7 .A soldier
at (3,7), want
to shoot down the helicopter when it is nearest to him. find the nearest distance.
INTEGRATION
INTRODUCTION
IF f(x) is derivative of function g(x), then g(x) is known as antiderivative or integral of f(x)
i.e.,
(g(x)) = f(x)
STANDARD SET OF FORMULAS
1.
2.
3.
=
x+ c
= log |x| + C
dx
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
dx
= sin -1 x + c
33
16.
17.
18.
19.
INTEGRALS OF LINEAR FUNCTIONS
1.
2.
3.
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.
+c
2.
3.
4.
5.
+ C
6.
7.
8.
dx
=
9.
dx
=
+
INTEGRATION BY PARTS
1.
OR
The integral of product of two functions = (first function ) x integral of the second function
– integral of ( differential coefficient of the first function ) x (integral of the second
function)
We can choose first and second function according to I L A T E where I
inverse
trigonometric function
2.
dx
= ex f (x) + C
34
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 k , (ii cos k ,(iii) tan k , (iv) cot k , (v) sec k ,(vi) cosec k ,(vii) sec2 k ,
(viii) cosec2 k , (ix) sec k tan k (x) cosec k cot k
For this use the following formulae whichever applicable
(i)
(ii)
(iii)
(iv)
(v)
x
=
x–1
(vi) cot2 x
(Vii) 2sin x sin y = cos (x – y ) – cos ( x + y)
(vii) 2 cos x cos y = 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
= cosec2x – 1
2. Integration by substitution
Integration of the algebraic function which are of the form
(a) P (x) . (ax + b ) n
Or
where P (x) is a polynomial of x and n is a positive rational
number
Substitution Put ax + b = t
(b) When the integrand is the product of two functions and one of them is a function g (x)
and the other is k g’ (x), where k is a constant then
Put g (x) = t
3.
Integration of the types
,
, and
In this three forms change
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 find by using any of the special integral formulae
4.
Integration of the types
,
and
dx
In this three forms split the linear px +q =
Then divide the integral into two integrals
The first integral can be find out by method of substitution and the second integral by
completing square method as explained in 3
I,e., to evaluate
=
dx
35
=
Find by substitution method + by completing square method
Integration of rational functions
5.
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
, then integrate
6.
Integration by partial fractions
Integration by partial fraction is applicable for rational functions. There first we must
check thatdegree of the numerator is less than degree of the denominator, if not, divide the
numerator by denominator and write as
and
proceed for partial fraction of
Sl.
No.
Form of the rational functions
Form of the rational functions
1
2
3.
4
5
Where x2 + bx + c cannot be factorized
further
36
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
1.
. In particular,
3.
0
+
4.
5.
6.
+
7.
, if
=0 ,
8.
=
if
(i)
(ii)
, if
, if
and
=
is an even function, i.e., if (-x) =
is an odd function, i.e., if ( x) =
Problem based on property
+
,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.
37
Problem based on property
Working Rule
Let
I=
Then
I=
(1) + (2) =>
2I =
+
I=
This property should be used when
Problem based on property
0, if
+
becomes an integral function of x.
is an odd function and
2
, if
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/ even.
PROBLEM BAESD ON LIMIT OF SUM
Working rule
=
=
Where nh = b – a
The following results are used for evaluating questions based on limit of sum.
(i)
1 + 2 + 3 + ….. + (n-1) =
(ii)
+
+
+ …. +
=
(iii)
+
+
+ ….. +
=
(iv)
a + ar + …… +
=
(r 1)
IMPORTANT SOLVED PROBLEMS
Evaluate the following integrals
1.
Solution
put 1+log x = t
=
=
38
=
2.
Solution
Put
=
=
=
=
=
3.
+C=
dx
Solution
Put tan x = t 2then
sec2x dx = 2t dt =>
dx =
=
dx =
=
(by dividing nr and dr by t2 )
=
=
=
( 1st integral put
=
)
2nd integral put
=
=
=
=
39
= uthen
4.
Solution
5x + 3 = A (2x + 4 ) + B
=> A=
=
=
=
=
dx
=5
5.
Solution
I =
-------------------(1)
Again I =
Using the property
I=
-------------------- (2)
Adding (1) and (2) we get
2I=
=
=
=
I
6.
) dx =
=
Evaluate
using limit of sum
Solution
Comparing the given integral with
a = 1, b = 4 f (x ) = x2 – x
and
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
=
40
=
=
=
=
=
=
=
+
=
PRACTICE PROBLEMS
LEVEL I
Evaluate the following integrals
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
LEVEL II
Evaluate the following integrals
1.
2.
3.
4.
5.
6.
41
7.
9.
10.
11.
12.
13.
14.
15.
as a limit of sum
LEVEL III
Evaluate the following integrals
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
as a limit of sum
42
APPLICATION OF INTEGATION
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
Area =
=
(ii)
2.
Area bounded by the curve y = f(x), the y axis and between abscassas at y = c and y
= d is given by
Area =
Where
=
y=
43
3. 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)
4.
Area bounded by two curves
and
for all
and between the ordinates at
, such that
is given by
Area
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 formula
(b) If the required area is included between two abscissa y = c and = d then use the
Formula
Finding the area included between two curves
WORKING RULE
1. Draw the graph of the given curves.
2. Mark the region whose area is to be determined.
3. Find whether the area is bounded between two given curves and two ordinates or
between the two given curves and two abscissa.
(a) If the required area is bounded between 2 ordinates , use the formula
44
(b) If the required area is included between two abscissa use the formula
SOME IMPORTANT POINTS TO BE KEPT IN MIND FOR SKETCHING
THE GRAPH
1.
= 4ax is a parabola with vertex at origin,
symmetric to X axis and right of origin
2.
= - 4ax is a parabola with vertex at origin,
symmetric to X axis and left of origin
3.
= 4ay is a parabola with vertex at origin,
symmetric to y axis and above origin
4.
= - 4ay is a parabola with vertex at origin,
symmetric to y axis and below origin
5.
+
= 1 is an ellipse symmetric to both axis,
7.
Cut x axis at ( a,0) and y axis at (0,
is a circle symmetric to both the axes
with centre at origin and radius r
is a circle with centre at (h,k) and radius r.
8.
ax +by + c = 0 representing a straight line
6.
IMPORTANT SOLVED PROBLEMS
1. Calculate the area of the region bounded by the parabola y =x2 and x = y2
Solution
45
Parabola y 2 = x is symmetrical to X- axis and x2 = y is symmetrical to y axis
Solving both the equations we get the point of intersection of the two curves as (0,0) and
(1,1)
Required area =
=
=
= sq. units
2. Find the area of the region ; { (x, y ) /
Solution
}
Curves y2 =4x, parabola symmetric to x axis and the circle
at (0,0 ) and radius
Sketch both the curves and shaded the area
46
circle with centre
The point of intersection of y2 =4x and
are the points (0, 0 )
Required area = 2 x Area of OBALO
= 2x [ area of OBLO + area of BLAB]
= 2
dx
= 2
=
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 shaded the area
Area enclosed =
=
) dx
=
=
4.
= 8 sq.units
Find the area of the region { (x,y):
Solution
Sketch the region whose area is to be found out.
47
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
=
=
=
+
sq.units
5. Find the area cut off from the parabola 4y = 3 x2 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
=
dx
= 27 sq.unts
PRACTICE PROBLEMS
LEVEL I
1. Find the areaof 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
48
4. Sketch the graph y = | x – 5 | . Evaluate
What does this value of the integral
represent on the graph
5. 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) : x2 + y2
2. Find the of the circle x2 + y2 = a2
3. Find the area of the region { (x , y) :
4. Sketch the region common to the circle x2 + y2 = 4 and the parabla y2 = 4 x. Also findthe
area of the region by integration
5. Find the area of the ellipse
=1
6. Find the area of the smaller region bounded by the ellipse
= 1 and the straight
line
7. Find the area enclosed by the curve x = 3 cos t, y = 2 sin t.
8. Find thee are abounded by the lines x +2y = 2, y – x = 1 and 2x + y = 7
9. Find the area of the region bounded by the parabola y ¾ x2 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
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. Find the area bounded by the curves y = 6x – x2 and y = x2 – 2x
5. Make a rough sketch of the region given below and find its area using integration
5.
Using the method of integration , find the area bounded by the curve |x| + |y| = 1
VALUE BASED QUESTIONS
1. A field is in the form of parabola x2 = 4y. A farmer has planted trees in the exterior to
the region bounded by the parabola and y = |x| and the remaining part for
playing games for children. Find the area of the ground where the students are
playing games. What are the importance of games in students life
2. Evaluate
dx. What is integrity in real life
49
3. The area between x = y2 and x = 4 is divided into equal parts by the line x = a, find the
value of a. A man constructed a house in one of the area and planted trees in the
other area. Find the area where he constructed house? What is the importance of
plantation of trees?
4. Evaluate
. What is the limit of integrity in real life?
5. A farmer has a plot in the shape of a circle x2+ y2 = 4. He divides his property among
his son and his daughter in such a way that for son he gives the area interior to
the parabola y2= x and for daughter the area interior to the parabola y2 = -x. How
much area the son got? Have both of them got equal share? What is the value for
that?
DIFFERENTIAL EQUATIONS
1.
INTRODUCTION
Problems based on the order and degree of the differential equations
Working rule
(a) In order to find the order of a differential equations, 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.
2.
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.
3.
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
=
50
i.e., in which
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.
In order to solve the equation
the form
Write down this equation in
, then the solution will be
where C is
an arbitrary constant.
4.
Problems based on solution of differential equations which are
homogeneous.
Working rule
(a)
Write down the given differential equation in the form
(b)
If
(c)
In order to solve, put
variable
=
then differential equation is homogeneous.
and . At the end put
so that
and then separate the
in place of .
IMPORTANT SOLVED PROBLEMS
1.
Solve the differential equation (x + y ) dy + ( x – y ) dx = 0
Solution
(x + y ) dy + ( x – y ) dx = 0
Put y = vx




Integrating both sides we get
=-
51
2.
Solve
Solution
= -
Integrating both sides w.r.t x
= ( put t = 1 – x2 and put u = 1 – y2 )
=
+C
3.
Solve the differential equation
Solution
The diff.eqn is in the form
Where P =
and Q = 2x2
I.F =
Multiplying both sides of diff.eqn by I.F we get
Integrating both sides w.r.t.x we get
Y
4.
= x3
+ Cx
Solve the diff. eqn
Solution
Put y = vx =>

= v – tan v

 Cot v dv = Integrating both sides w.r.t.x ,we get
52
Log sin v = - log x + log C
Log x sin
x sin
=C
5.
, find the equation of the curve passing through the points
(1, -1)
Solution


 Integrating both sides weget
 Y –2 log (y + 2 ) = x +2 log x + C
The curve is pasing through the the point (1,-1)
- 1 –2 log 1 = 1 + 2 log 1 + C => C = -2
The equation of the line is
6.
Solve the differential equation
Solution

Integrating both sides w.r.t x we get
-
PRACTICE PROBLEMS
LEVEL I
53
1.
Find the order and degree of the following differential equation.
Show that
3.
is a solution of the differential equation.
5.
Form the differential equation corresponding to
eliminating a.
Form the differential equation representing the family of curves
where
are arbitrary constants.
Solve the differential equation.
6.
Solve the differential equation.
4.
by
,
LEVEL II
Solve the following differential equations.
1.
2.
3.
=0
4.
5.
3
.
6.
7.
8.
9.
10.
The population of a city increases at a rate proportional to the number of inhabitants
present at any time t. If the population of the city was 200000 in 1990 and 250000 in
2000, what will be the population in 2010?
LEVEL III
Solve the following differential equations
1.
y–x
2.
( 1 + sin2x ) dy + ( 1 + y2 ) cos x dx = 0, given that x= , y = 0
3.
Show that the general solution of the differential equation
=a
Given by ( x + y + 1 ) = A ( 1 – x – y – 2xy ), where A is a parameter
54
= 0 is
Prove that x2 – y2 = c ( x2 + y2 )2 is the general solution of differential equation
( x3 – 3xy2 ) dx = ( y3 – 3x2 y) dy where c is a parameter
4.
5.
( x3 + x2 + x +1)
6.
2x2 + x.
7.
8.
In a college hostel accommodating 1000 students, one of the hostellers came in
carrying a flu virus, and the hostel was isolated. If the rate at which the virus spreads is
assumed to be proportional to the product of the number N of infected students and the
number of non-infected students, and if infected students are 50 after 4 days then show that
more than 95% of the hostellers will be infected after 10 days
VALUE BASED QUESTIONS
1.
2.
3.
In a college hostel accommodating 1000 students, one of the hostellers came in
carrying a flu virus, and the hostel was isolated. If the rate at which the virus spreads
is assumed to be proportional to the product of the number N of infected students and
the number of non-infected students, and if infected students are 50 after 4 days then
show that more than 95% of the hostellers will be infected after 10 days. If Shyam
was the first student to be infected what precautions he should have initiated to avoid
this situation.
The population of a city increases at a rate proportional to the number of inhabitants
present at any time t. If the population of the city was 200000 in 1990 and 250000 in
2000, what will be the population in 2010? What is the main awareness to be given
among the people to control population?
If the interest is compounded continuously at 6% per annum, how much worth Rs
1000 will be after 10 years? How long will it take to double Rs 1000? (Given e0.6 =
1.822) What is the importance of savings among the students?
VECTORS
SUMMARY
55
1. Position vector of a point P(x,y,z) is given as
2.
3.
4.
5.
6.
7.
8.
9.
(= ) =x +yj+zk,and its
magnitude by √x2+y2+z2.
The scalar components of a vector are its direction ratios, and represent its
projections along the respective axes.
The magnitude(r),direction ratios (a,b,c) and direction cosines (l,m,n) of any
vector are related as: l=a/r ,m= b/r,n=c/r.
The vector sum of the three sides of a triangle taken in order is 0 .
The vector sum of two coinitial vectors is given by the diagonal of the
parallelogram whose adjacents sides are the given vectors.
The multiplication of a given vector by a scalar α, changes the magnitude of
the by the multiple │α│, and keeps the direction same (or makes it
opposite) according as the value of α is positive (or negative).
For a given vector a ,the vector ậ =a/│a │gives the unit vector in the
direction of a.
The position vector of a point R dividing a line segment joing the points P
and Q whose position vectors are a and b respectively ,in the ratio m : n
(i) internally , is given by na +mb /m+n .
(ii) Externally, is given by mb –na /m-n .
The scalar product of two given vectors a and b having angle ϴ
between them is defined as a.b= ǀaǀǀbǀcos .
Also, when a.b is given, the angle ‘
between the vectors a and b may be
determined by
Cos = a.b
ǀaǀǀbǀ
10.
If
is the angle between the two vectors a and b, then their cross
product is given as
a×b=ǀaǀǀbǀsin
56
Where n is a unit vector perpendicular to the plane containing a and b. such
that a, b, n form right handed system of co-ordinate axes.
11.
If we have two vectors a and b, given in component form as
a=a1 i+ a2j+a3k and b=b1i+b2j+b3k and λ any scalar, then
a+b=(a1+b1)i+(a2+b2)j+(a3+b3)k;
λa=(λa1)i + (λa2 )j+ (λa3)k;
a.b = a1b1+a2b2+a3b3;
And a×b = i
j k
a1 b1 c1
a2 b 2 c2
FLOW CHART
(a) To find shortest distance between two skew lines
1
Find a̅1 ,b̅1 ,a̅2 ,b̅2
2
3
4
Find a̅2 - a̅1
Find b̅1 x b̅2
Find | b̅1 x b̅2 |
5
(b̅1 x b̅2). (a̅2 - a̅1 )
6
d=| (b̅1 x b̅2). (a̅2 - a̅1 )/ | b̅1 x b̅2 ||
57
(b)
To find Coordinates of foot of perpendicular from origin to
the plane
1. Direction ratio of any line perpendicular to the given plane
2. Equation of the line through origin and perpendicular to the
given plane
3. Coordinates of any point(P) on the line
4. Find the value of t, by putting the coordinates of P in the
plane
5. Find required coordinates of foot of perpendicular
58
(c) To find coordinates of image of a point in the plane
1. Direction ratios of any line (PP’) perpendicular to the given
plane.
2. Equation of the line PP’ through a given point (P) and
perpendicular the given plane.
3. Coordinates of any point on the plane (say P’)
4. Coordinates of midpoint (M) of PP’
5. Find the value of r ,as M will satisfy the plane
6. Then find coordinates of P’ Images of point P
59
Important Solved Problems
1. Find vector in the direction of vector a̅=î-2ĵ that has magnitude 7
units
Sol: The unit vector in the direction of given vector a̅ is
â= a̅/|a̅|= (1/√5)(î-2ĵ) = 1/√5 î -2/√5ĵ
So, the vector having magnitude equal to 7 and in the direction
of a̅ is
7â = 7(1/√5 î -2/√5ĵ) = 7/√5 î -14/√5ĵ
60
2. Show that the points A ,B and C with position vectors, a̅= 3î-4ĵ-4k̂ ,
b̅ = 2î-ĵ-k̂ and c̅ = î-3ĵ-5k,̂ respectively form the vertices of a right
angled triangle.
Sol:
there O̅A̅ =3î-4ĵ-4k̂
O̅B̅=2î-ĵ-k̂
O̅C̅= î-3ĵ-5k̂
A̅B̅=O̅B̅-O̅A̅
=(2î-ĵ-k̂)-( 3î-4ĵ-4k̂)
= -î+3ĵ+5k̂
|A̅B̅|2=(-1)2+32+52 =35
B̅C̅=O̅C̅-O̅B̅
=( î-3ĵ-5k̂)-( 2î-ĵ-k̂)
= -î-2ĵ-6k̂
|B̅C̅|2=(-1)2+(-2)2+(-6)2 = 41
C̅A̅=O̅A̅-O̅C̅
=3 î-4ĵ-4k̂-( î-3ĵ-5k̂)
=2î-ĵ+k̂
|CA|2= 22+(-1)2+12=6
SINCE 35 +6 =41
SO |A̅B̅|2 +|C̅A̅|2=|B̅C̅|2
So, the points A,B,C form the vertices of a right angled triangle .
3.Three vectors a̅, b̅ and c̅ satisfy the condition a̅+ b̅+ c̅=0. Evaluate
μ= a̅.b̅ + b̅.c̅ + c̅.a̅, if │ a̅│=1,│ b̅│=4,│ c̅│=2
Sol: since, a̅+ b̅+ c̅=0
a̅.( a̅+ b̅+ c̅)=0
a̅.a̅ + a̅.b̅+ a̅.c̅=0
a̅.b̅+ a̅.c̅ = -│ a̅│2 = -1
---- (1)
Again, b̅.( a̅+ b̅+ c̅)=0
 a̅.b̅ + b̅.c̅= -│ b̅│2= -16
---- (2)
similarily, a̅.c̅+ b̅.c̅= -4
---- (3)
adding (1), (2) and (3), we have,
61
2(a̅.b̅ + b̅.c̅ + c̅.a̅) = -21
 2μ = -21
 i.e μ=-21/2.
4.Find a unit vector perpendicular to each of the vector (a̅+ b̅) and a̅- b̅ ,
where a̅= î +ĵ+k̂ , b̅= î +2ĵ+3k̂
Sol: We have a̅+ b̅ = 2î +3ĵ+4k̂ and a̅- b̅ = -ĵ-2k̂
A vector which is perpendicular to both a̅+ b̅ and a̅- b̅ is given by
(a̅+ b̅)x (a̅- b̅)= î
ĵ
k̂ = -2î +4ĵ-2k̂ = c̅ (say)
2 3 4
0 -1 -2
│ c̅│=
=
=2
Required unit vector is
c̅/│ c̅│= -1/ î + 2/ ĵ - 1/ k̂
5) Find the area of a parallelogram whose adjacent sides are given by the vectors a =3Î+Ĵ+4k and
b=Î-Ĵ+k.
Solution—a x b= i j
k
3 1 4 =5Î+Ĵ-4k
1 -1 1
|a x b|=√25+1+16=√42
∴ area of parallelogram =√42
6) let a= Î+4Ĵ+2k, b= 3Î-2Ĵ+7k and c= 2Î-Ĵ+4k. find a vector d which is perpendicular to both a
and b, and c x d=18.
Solution given a= Î+4Ĵ+2k, b= 3Î-2Ĵ+7k, c= 2Î-Ĵ+4k vector d is perpendicular to both a and b
i.e d is parallel to vector a x b .
∴a x b= I j k
1 4 2
= 32 Î-Ĵ-14 k
62
3 -2 7
Let d=μ(32 Î-Ĵ-14k)
Also c x d=18
(2 Î-Ĵ-4k) xμ( 32 Î-Ĵ-14k) =18
 μ( 64+1-56) =18
 9μ= 18
 μ =2
d =2( 32 Î-Ĵ-14k) = 64 Î-2Ĵ-28k
Practice problems
Level 1
1) Find a unit vector in the direction of the vector a =2Î+ Ĵ +2k .
2) Find the position vector of the mid points A(5Î+3 Ĵ) and B(3Î- Ĵ).
3) If
=√3 , |b|=2 and a. b=√6.then find the angle between a and b.
4) Find the projection of the vector Î+3 Ĵ +7k on the vector 7Î –j Ĵ +8k.
5) Find the area of triangle having the points A(1,1,1 ),B( 1,2,3)and C(2,3,1) as its vertices.
Level 2
1.
Find a vector of magnitude 5 units, perpendicular to each of the vectors ( + ),(
- )where
=î+ Ĵ
+k and =î+2 Ĵ +3k .
2. Write the position vector of a point R which divides the line joining the points P and Q where
position vectors are i+2 Ĵ -k and – î + Ĵ +k respectively in the ratio 2:1 externally .
3. If =2i+2 Ĵ +3k , =-i+2 Ĵ +k and =3i+ Ĵ are such that +λ is perpendicular to , find the value of λ.
4. Find λ ,if(2Î+ 6Ĵ+14k) x(Î- λĴ+7k)=
5. Find the angle between two vector
and
if | |=3, | |=4 and | x |=6
Level-3
1. If a line make α, ß, with the x - axis, y-axis and z-axis respectively, then find the value of Sin^2 α + Sin^2 ß
+Sin^2
→→
→
→→
→
2. If a = 5î- ĵ - 7k^ and b =î – ĵ + μk^, find μ such that a + b and a - b are orthogonal.
→
→
3. If a unit vector a makes angle π/4 with î, π/3 with ĵ and an acute angle θ with k^ , the find component of a
and angle θ. →
→
→
→
63
→→
→
4.
5.
Vectors a and b are such that | a | = √3 , | b | = (2/3) and a x →
unit vector. Write the angle between a
→
→
→
→ → b is a →
→
and b.
If a = î + ĵ + k^ and b = ĵ - k^, find a vector c such that a x c = b and a .c = 3.
Value based questions
→→
→
→
1. Find | a x b |, if a = 2î + ĵ +3k^ and b =3î + 5 ĵ – 2k^. we should be firm and determined about
in life. (Hint: Some value as you perceive is a possible answer) Why?
2. Find the area of a triangle having the points A(1,1,1), B(1,2,3) and C(2,3,1,) as its vertices. What is the
difference between finding its area and a triangle in which all points are in a single plane?
3D GEOMETRY
1.
2.
3.
4.
5.
6.
7.
INTRODUCTION
Summary
Direction cosines of a line are the cosines of the angles made by
the line with the positive direction of the coordinate axes.
If l ,m ,n are the direction cosines of a line, then l2 + m2 +n2 = 1
Direction cosines of a line joining two points P(x1, y1 ,z1) and Q(x2,
y2, z2) are x2-x1/PQ, y2-y1/PQ, z2-z1/PQ
Where PQ=
(x2-x1)2 +( y2-y1)2 +( z2-z1)2
Direction ratios of a line are the
numbers which are proportional to the direction cosines of a line
Skew lines are lines in space which
are neither parallel nor intersecting. They lie in different planes.
Angles between skew lines is the
angle between two intersecting lines drawn from any point
(preferably through the origin) parallel to each of the skew lines.
If l1, m1, n1 and l2 , m2, n2, are the
direction cosines of two lines;
And is the acute angle between the two lines; Then
cos⍬=│l1l2 + m1m2 + n1n2│
64
8.
Vector equation of a line that
passes through the given point whose position vector is and
parallel to a given vector
=
9.
is
+λ
Equation of a line through a point
(x1, y1, z1) and having directional cosines l, m, n is
x-x1/l = y-y1/m = z-z1/n
10. The vector equation of a line which passes through two points
whose position vectors are
and
is =
+ λ( -
)
11.Cartesian equation of a line that passes through two points (x1, y1,
z1) and (x2, y2, z2) is
x-x1/x2-x1 = y-y1/y2-y1 = z-z1/z2-z1
12If ⍬ is the acute angle between = a̅1 + λ b̅1 and = a̅2 + λb̅2 , then ,
cos⍬ = │ b̅1. b̅2│/││ b̅1││ b̅2││
13.If x-x1 /l1 = y-y1/m1 = z-z1/n1 and x-x2/l2 = y-y2/m2 = z-z2/n2
are the equations of two lines, then the acute angle between the
two lines is given by cos⍬ = │l1 l2 + m1 m2 + n1 n2│.
14. Shortest distance between two skew lines is the line segment
perpendicular
to both the lines .
15.Shortest distance between = a̅1 + λ b̅1 and = a̅2 + μb̅2 is
│( b̅1 x b̅2).( a̅2 - a̅1)│/ │( b̅1 x b̅2)│
Distance between parallel lines = a̅1 + λ b̅1 and = a̅2 + μb̅2 ,
16.
is
x(a̅2 - a̅1)│ / ││ ││
65
17. In the vector form ,equation of a plane which is at a distance d
from the origin and is the unit vector normal to the plane
through the origin is
.
=d
18. Equation of a plane which is at a distance of d from the origin and
the direction cosine of the normal to the plane as l, m, n is
lx+my+nz=d
19. The equation of a plane through a point whose position vector is
a̅ and perpendicular to the vector
is ( - a̅).
=0
20. Equation of a plane perpendicular to a given line with direction
ratios A, B, C and passing through a given point (x1, y1, z1) is
A(x-x1) + B(y-y1) + C(z-z1) = 0
21. Equation of a plane passing through three non collinear points
(x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is
x-x1
y2-y1
z2-z1
x2-x1
y2-y1
z2-z1
x3-x1
y3-y1
z3-z1
=0
22. Vector equation of a plane that contains three non collinear points having
position vectors
and is
23. Equation of a plane that cuts the co-ordinates axes at (a, 0, 0), (0,b, 0) and
(0,0,c) is
+ + =1
24. Vector equation of a plane that passes through the intersection of planes
2= d1 and
2=d2 is . ( 1+ λ 2)=d1+λd2, where λ is any nonzero constant.
25. Vector equation of a plane that passes through the intersection of two given
planes A1x+B1y+C1z+D1=0 and A2x+B2y+C2z+D2=0 is
(A1x+B1y+C1z+D1) + λ (A2x+B2y+C2z+D2) =0.
26. Two lines = 1+λ 1and = 2+µ 2 are co-planar if ( 2- 1).( 1 × 2)=0.
66
27. Two lines x-x1 = y-y1 =z-z1 and x-x2 = y-y2 = z-z2 are coplanar if
a1
b1
c1
a2
b2
c2
x2-x1
y2-y1
z2-z1
a1
b1
c1
a2
b2
c2
=0.
28. In the vector form, if is the angle between the two planes . 1=d1 and . 2=d2
then =cos-1 1. 2 ǀ 1 ǀǀ 2 ǀ.
29. The angle between the line = +λ and the plane . =d is
Sin = . /ǀ ǀǀ ǀ
30. The angle
given by
cos
=
between the planes A1x+B1y+C1z+D1=0 and A2x+B2y+C2z+D2=0 is
A1A2+B1B2+C1C2
A12+B12+C12
A22+B22+C22
31. The distance of a point whose position vector is a from the plane . =d is
ǀd- . ǀ.
32. The distance from a point (x1, y1, z1) to the plane Ax+By+Cz+D=0 is
Ax1+By1+Cz1+D
A2+B2+C2
67
IMPORTANT SOLVED PROBLEMS
Ques 1: Find the shortest distance between the lines l1 and l2 whose
vectors equations are
= i+j+λ (2i-j+k)
=2i+j-k+μ (3i-5j+2k)
Sol: Here
=i+j, 1=2i-j+k
=2i+j-k,
-
=
│ 1x
=3i-5j+2k
=i-k and 1 X
= (2i-j+k) x (3i-5j+2k)
i
j
k
2
-1
1
3
-5
2
│=√9+1+49
=√59
( 1x
).(
-
) = 3-0+7
=10
Shortest distance is
68
( 1x
).(
│ 1x
-
)
│
= 10/√59
Ques 2: Find the coordinate of the foot of the perpendicular drawn from
the origin to the plane
2x-3y+4z-6=0.
Sol. : Given plane is 2x-3y+4z-6=0
………………………….(1)
Direction ratio of OP is 2,-3, 4.
Equation of the line OP is
X-0/2 = y-0/-3 = z-0/4
Any point on this line is (2t,-3t, 4t).
Let it be P.
P (2t,-3t, 4t) lies in the plane 2x-3y+4z-6=0
2(2t)-3(-3t) +4(4t)-6=0
T= 6/29
∴ Co-ordinates of P are
(2X6/29, -3X6/29, 4X6/29)
= (12/29, -18/29, 24/29)
Ques. 3: Find the co-ordinates of the image of the point (1,3,4) in the
plane 2x-y+z+3=0.
Sol: Let P′ be the image of the point P (1, 3, 4) in the plane 2x-y+z+3=0.
Since PP′ is perpendicular to 2x-y+z+3=0
∴ Direction ratio of PP′ is
2,-1, 1.
Equation of the line PP′ is
X-1/2=y-3/-3=z-4/1=r (say)
Co-ordinates of the point on this line be (2r+1,-r+3,r+4)
Let it be P′
But M is the mid-point of PP′
∴ Co-ordinates of the point M are
(Zr+1+1/2,-r+3+3/2, r+4+4/2)=(r+1,-r/2+3, r/2+4).
Since M (r+1,-r/2+3, r/2+4) lies on the plane 2x-y+z+3=0
69
∴ 2(r+1)-(-r/2+3)+(r/2+4)+3=0
2r+2+r/2-3+r/2+4+3=0
Or, 3r+6=0
Or, r=-2
∴ Image of the point P is (2(-2) +1,-(-2) +3,-2+4))
= (-3, 5, 2).
Ques 4: find the equation of the plane passingthrough the line of
intersection of the planes .(î+j+k)=1 and (2i+3j- k)+4=0 and parallel
to x-axis
Solⁿ: given planes are,
x+y+z-1=0 ……(1)
and 2x+3y-z+4=0 …….(2)
equation of planes through the intersection of (1) and (2) is
x+y+z-1+λ(2x+3y-z+4)=0
or (1+2λ)x+ (1+3λ)y+(1-λ)z-1+4λ=0 ……….(3)
plane (3) is parallel to x-axis
1+2λ=0
or λ= -1/2
equation of plane is
0.x+(-1/2)y+3/2z-1-2=0
i.e. y-3z+6 =0
or .(j-3k)+6=0
Ques 5: Find the equation of the plane that contains the print(1,-1,2) and is
perpendicular to each of the planes
2x+3y-2z = 5 and x+2y-3z=8
Solⁿ: The equation of the plane containing the given point is
A(x-1)+ B(y-1) +C(z-2)=0 ………………(1)
Since, (1) is perpendicular to 2x+3y-2z=5
2A+3B-2C=0 …………………(2)
Since, (1) is perpendicular to x+2y-3z=8
A+2B-3C=0 ………………(3)
Solving equation (2)&(3),
A=-5C and B=+4C
Now, putting these values in (1),
We get,
-5C(x-1) + 4C(y+1)+C(z-2)=0
5x– 4y–z =7
70
Ques 6: find the co-ordinates of the point where the line through (5,1,6)
and (3,4,1) crosses the yz- plane.
Solⁿ: Equation of yz-plane is x=0 ……………(1)
The equation of the line joining (5,1,6) and (3,4,1) i
=
=
or
=
=
= t (say) …………….(2)
Any point on this line is P( 5-2t, 1+3t, 6-5t)
Since, this lies on plane (1)
5-2t=0
t=5/2
the required point of intersection of plane (1) and line (2) is
(0, 17/2, -13/2)
Practice Problems
Level-1
1.
2.
3.
4.
5.
Write the direction –cosines of the line joining the points (1, 0, 0) and (0, 1, 1).
Write the vector equation of the line (x-5)/3=(y-4)/7 = (6-z)/2.
Find the equation of plane with intercepts 2, 3, 4 on the x, y, z axis respectively.
Find the angle between the lines whose direction ratios are (1, 1, 2) and (√3-1, -√3-1, 4).
→
Find the distance of a point (2, 5, -3) from the plane r. (6Î - 3ĵ +2k^) = 4.
Level-2
1. Write the direction-cosines of a line parallel to the line (3-x)/ (3) = (y-2)/ (-2) = (z-2)/ (6).
2. Show that the lines (x-1))/2=(y-2)/3 = (z-3)/4 and (x-4)/5=(y-1)/2=z intersect. Find their point
of intersection.
3. Find the shortest distance between the lines
→
r = (î + 2ĵ + 3k^) + ƛ (î-3 ĵ+2k^)
→
r = (4+2μ) î + (5+3μ) ĵ + (6+ μ) k^.
4. Find the value of P, such that the lines (x/1) = (y/3) = (z/2) and (x/ -3) = (y/5) = (z/2) are
perpendicular to each other.
5. Find the equation of the plane which is perpendicular to the plane 5x+3y+6z+8=0 and which
contains the line of intersection of the planes x+2y+3z-4=0 and 2x+y-z+5=0.
LEVEL-3
1. Show that the four points (0,-1,-1), (4,5,1), (3,9,4) and (-4,4,4) are co-planar and find the
equation of the common plane.
2. Find the co-ordinates of the foot of the perpendicular and perpendicular distance of the
point (1, 3, 4) from the plane 2x-y+z+3=0. Find also, the image of the point in the plane.
3. Find the distance of the point P(6,5,9) from the plane determined by the points A(3,1,2), B(5,2,4) and C(-1,-1,6).
71
4. Find the equation of the plane passing through the point (1, 1, 1) and perpendicular to
the planes x+2y+3z-7=0 and 2x-3y+4z=0.
5. Find the image of the point (1,-2, 1) in the line (x-2)/3 =(y+1)/-1= (z+3)/2.
VALUE BASED QUESTIONS
1. Find the equation of the plane through the line of intersection of the planes x+y+z=1
and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0. Which will you choose if
you have to choose between good and truth?
2. Find the co-ordinates of the point where the line through the points A(3,4,1) and
B(5,1,6) crosses the xy-plane. Name any two values which we get from our friends.
Study Module of Linear programming problems
LINEAR PROGRAMMING
72
SCHEMATIC DIAGRAM
Topic
concept
Degree of Importance
Linear Programming
(i)Introduction
(ii)Some solved
problems
(iii) Diet Problem
**
***
iv) Manufacturing
Problem
***
Solved Ex. 8 Q. Nos
3,4,5,6,7 of Ex. 12.2
(v) Allocation Problem
**
Solved Ex.10 Q. Nos 4
& 10 Misc. Ex.
(vi) Transportation
Problem
*
Solved Example 7
Q. No 10 Ex.12.2,
Q. No 5 & 8 Misc. Ex.
vii) Miscellaneous
Problems
**
Solved Ex.11
Q. Nos 6 & 7 Misc. Ex.
***
References
From NCERT Book
Vol. II
Definitions
Q No. 1, 2, 3, 4 (Given
below in the booklet)
Q. Nos 1, 2 and 9 Ex.
12.2
Solved Ex. 9 Q. Nos 2
and 3 Misc. Ex.
Q. No 8 Ex. 12.2
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 non-negative 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.
Objective function: Linear function Z = ax + by, where a, b are constants, which has
to be maximised or minimized is called a linear objective function..
73
Objective function : Linear function Z = ax + by, where a, b are constants, which has to be
maximised or minimized is called a linear objective function.
Constraints: The linear inequalities or equations or restrictions on the variables of a
linear programming problem are called constraints. The conditions x 0, y
0 are
called non-negative restrictions.
Optimisation 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, 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.
The contents of one kg. food of each kind as given below:
74
Food
Vitamin A
Vitamin B
Vitamin C
Cost
X
1
2
3
16
Y
2
2
1
20
Minimum
Requirement
10
12
8
The above L.P.P. is given as
Minimum, Z = 16x + 20 y
subject to the constraints
x + 2y 10, 2x + 2y
3x + y
8, x, y
0
L1 : x + 2y = 10
A
x
y
10
0
12,
L2 : x + y = 6
B
0
5
C
x
y
6
0
L3 : 3x + y = 8
D
E
x
y
0
6
Corner points
A (10,0)
F (0, 8)
G (1, 5)
H (2,4)
2
2
Z = 16x + 20 y
160
160
116
112
Here the cost is minimum at H (2,4)
75
F
0
8
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
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.
Q2. An aero plane 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
76
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?
SOL. 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 = 1000x + 600y subject to the constraints x + y
L1 : x + y = 200
x
y
L2 : 4x + y = 0
A
B
0
200
200
0
C
x
y
0
0
D
50
200
L3 : x = 20
77
200 , x
20, y
4x or 4x – y
0
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. Oil Company has two depots A and B with capacities of 7000l respectively. The company is
to supply oil to three petrol pumps, D,E and F, whose requirements are 4500l, 3000l, and 3500l
respectively. The distances (in km) between the depots and the petrol pumps are given in the
following table:
78
Distance in (km.)
From/ To
D
E
F
A
7
6
3
B
3
4
2
Assuming that the transportation cost of 10 liters of oil is Rs. 1 per km, how should the delivery
be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Sol. Let Depot A transport x liters of petrol to petrol pump D, y liters to E and 7000 – (x + y)
liters to C.
D
4500l
A
B 4000l
E 3000l
7000l
F 3500 l
Also, let depot B transport (4500- x) liters of petrol to petrol pump to D, (3000 – y) liters to E
and x + y – 3500 to F.
Graphical representation of the problem is given below:
Here objective function is
C = 7x + 6y + 3 (7000 – x – y ) + 3 (4500 – x) + 4 (3000 - y) + 2 (x + y – 3500)
i.e. C = 3x + y
7000, 4500 – x
0, 3000 – y
0, x + y – 3500
79
0, x, y,
0
i.e., Minimize C = 3x + y + 39500 subject to the constraints x + y
4500, y 3000x, y 0
L1 : x + y = 7000
x
y
3500, x
L2 : x + y = 3500
A
B
7000
0
0
7000
L3 : x = 4500
7000, x + y
C
x
y
3500
0
D
0
3500
L4 : y = 3000
Q4.. 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
80
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:
Screw
Hand operated
Machine
6
Profit
A
Automatic
Machine
4
B
6
3
10
Time
available
240
240
Therefore, the above L.P.P. is given as
Maximise, P = 7x + 10 y subject to the constraints.
4x + 6y
240 ; 6x + 3y
i.e. 2x + 3y
240
120 : 2x + y 80, x, y 0
L1 ; 2x + 3y = 120
A
B
x
y
60
0
0
40
L2 : 2x + y = 80
C
D
x
y
40
0
0
80
81
7
Corner points
P = 7x + 10y
O (0,0)
0
C (40,0)
280
B (0,40)
400
E (30, 20)
410
(maximum)
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.
82
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 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?
83
(ii) Graphical method of solving LPP (bounded and unbounded
solutions)
LEVEL I
Solve the following Linear Programming Problems graphically:
1. Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
2. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
3. Minimise Z = 3x + 5y such that 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?
84
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
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.
2. An oil company requires 12,000, 20,000 and 15,000 barrels of high grade, medium grade and
low grade oil respectively. Refinery A produces 100, 300 and 200 barrels per day of high,
medium and low grade oil respectively whereas the Refinery B produces 200, 400 and 100
barrels per day respectively. If A costs Rs. 400 per day and B costs Rs. 300 per day to operate,
how many days should each be run to minimize the cost of requirement? How can you save
fuel in your daily routine?any two ways
Q3 A 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.
85
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?
(vi) Transportation Problem
LEVEL III
1. A medicine company has factories at two places A and B . From these places, supply is to be
made to each of its three agencies P, Q and R. The monthly requirement of these agencies are
respectively 40, 40 and 50 packets of the medicines, While the production capacity of the
factories at A and B are 60 and 70 packets are respectively. The transportation cost per packet
from these factories to the agencies are given:
Transportation cost per packet (in Rs.)
From
A
B
To
P
Q
R
5
4
3
4
2
5
How many packets from each factory be transported to each agency so that the cost of
transportation is minimum? Also find the minimum cost.
Q2.There are two factories located one at place P and the other at place Q. From these locations, a
certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly
requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production
capacity of the factories at P and Q are respectively 8 and 6 units. The cost of
transportation per unit is given below:
From / To
Cost in( Rs)
A
B
C
160
100
150
P
100
120
100
Q
How many units should be transported from each factory to each depot in order that
the transportation cost is minimum. What will be the minimum transportation cost?
Answers
(i) LPP and its Mathematical Formulation
LEVEL I
1. Z = 6x + 3y, 4x + y ≥ 80, x + 5y ≥115,
x, y ≥ 0
86
(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.
2. A should run for 60 days and B for 30 days.
Q3. 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.
87
(vi) Transportation Problem
LEVEL III
1. Minimum transportation cost is Rs. 400 when 10, 0 and 50 packets are transported from
factory at A and 30, 40 and 0 packets are transported from factory at B to the agencies at P, Q
and R respectively.
Q2. Minimum value of Z is 1550 at the point (0, 5).
The optimal transportation strategy will be to deliver 0, 5 and 3 units from
The factory at p and 5, 0 and 1 units from the factory at q to the depots at a, b and c
Respectively. Corresponding to this strategy, the transportation cost would be minimum,
I.e., rs 1550.
PROBABILITY:
INTRODUCTION:
Topic
Concepts
Degree of Importance
Probability
(i) Conditional Probability
***
(ii) Multiplication theorem on
probability
**
(iii) Independent events
***
(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
***
88
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
89
90
91
: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 writ 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
P(x) = pi
0
1/27
Mean number of successes =
x p
i
1
6/27
2
12/27
3
8/27
i
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
(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
Probability that exactly one of them will solve the problem
 
(ii)
92


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) =
 
E x2
x p
i
0
188/221
1
32/221
2
1/221
i
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
How many times must a man toss a fair coin, so that the probability of having at least one
head is more than 80%.
Solution:
P(H) = ½
and
P(T) = ½
Let the coin be tossed n times
1
Required probability = 1 – P(all trials)
=1– n
2
If the given condition
1
80
4
1
2n  5
1– n 

1–  n

100
5 2
2
 
2
2
93

Least value of n is 3.
The fair coin has to be tossed 3 times for the desired situation.
Q.5
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 
(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 
 A  B   14  1
 A
Required probability = P   P
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.6
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
94
 A
1 28
PE1   P 

E
28 14
1 

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.7
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?
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).

Practice Problem
Level-1
1.
2.
3.
B
If P(A) = 0.3, P(B) = 0.2 find P   , if A and B are mutually exclusive events.
 A
Find the probability of drawing two white balls in succession from a bag containing 3 red
and 5 white balls, the ball first drawn is not replaced.
A coin is tossed thrice and all the 8 outcomes are equally likely:
E: the first throw results in head
95
4.
5.
6.
7.
8.
9.
10.
11.
12.
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.
A bag contains 6 reds and 5 blue balls and another bag contains 5 red and 8 blue balls. A
ball is drawn from the first bag and without noticing its colour is put in the second bag. A
ball is drawn from the second bag, find the probability that the ball drawn is blue in
colour.
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.
A purse contains two silver and 4 copper coins. A second purse contains 4 silver and 3
copper coins. If a coin is drawn at random from one of the two purses, what is the
probability that it is a silver coin
Two thirds of the students in a class are boys and the rest are girls. It is known that the
probability of a girl getting first class is 0.25 and that of a boy getting a first class 0.28.
Find the probability that a student chosen at random will get first class in the subject.
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
P(X)
2
0.3
3
0.4
4
0.3
Find the mean and variance of distribution.
Level-2
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?
A bag contains 5 white , 7 red and 3 black balls. If 3 balls are drawn one by one without
replacement, find what is the probability that none is red.
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
find the probability of drawing a one rupee coin from a purse with two compartments one
of which contains 3 fifty paise coins and 2 one rupee coins and other contains 2 fifty
paise coins and 3 one rupee coins.
Suppose five men out of 100 and twenty five women out of 1000 are good orator. An
orator is chosen at random. Find the probability that a male person is selected. Assume
that there are equal number of men and women.
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
96
7.
8.
9.
10.
picked up at random and found to be standard quality. Find the probability that it comes
from the second plant.
If a dice is thrown 5 times, what is the chance that an even number will appear exactly 3
times.
An experiment succeeds twice as often as it fails. Find the probability that in the next six
trials, there will be at least 4 successes.
A pair of dice is thrown 200 times. If getting a sum 9 is considered a success. Find the
mean and variance of the number of success.
Let X denote the number of colleges where you will apply after your results and P (X= x)
denotes your probability of getting admission in x number of colleges. It is given that

if x  0 or 1
 kx

P X  x   
2kx
if x  2

if x  3 or 4
k (5  x)

Where k is a positive constant
Find the mean and variance of the probability distribution.
Level-3
1.
2.
3.
4.
5.
6.
7.
8.
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.
A letter is known to have come either from LONDON or CLIFTON. On the envelop just
has two consecutive letters ON are visible what is the probability that the letter has come
from (i) LONDON (ii) CLIFTON
A laboratory blood test is 99% effective in detecting a certain disease when it is in fact,
present. However, the test also yields a false positive result for 0.5% of the healthy
person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will
imply he has the disease). If 0.1 percent of the population actually has the disease, what is
the probability that a person has the disease given that his test result is positive?
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 sum of mean and Variance of a binomial distribution for 5 trials be 1.8. Find the
probability distribution.
The mean and Variance of a binomial distribution are 4/3 and 8/9 respectively. Find
P(X  1).
97