Edition : 0617/0619800
Copyright © reserved with Vidyalankar Classes and Publications.
All rights reserved. No part of this work herein should be reproduced or used either
graphically, electronically, mechanically or by recording, photocopying, taping, web
distributing or by storing in any form and retrieving without the prior written permission of
the publisher. Anybody violating this is liable to be legally prosecuted.
Pearl Centre, S.B. Marg, Dadar (W), Mumbai 400 028.
Tel. No. : 022 – 4232 4232 / 022 – 2430 63 67
web site : www.vidyalankar.org
e-mail : [email protected]
GATE – SLP
EC / EE / IN / CS / ME / CE
Engineering Mathematics
Module 1 : Linear Algebra, Calculus,
Probability and Statistics
Dear GATE Aspirant,
The GATE program will help you to overcome the common feeling of confusion about the
exact scope of the syllabus. The course content will enable you to cover the course
smoothly in the limited time at your disposal without wasting time on unnecessary details,
without omitting any useful part.
The program provides graded questions in every topic leading you to deeper and more
intricate and basic concepts. This will help you to stimulate your own thinking and also
makes the process of learning, enjoyable and highly efficient.
The quick methods suggested to tackle questions and the practice that one has to put in
while solving the problems develops in you the required skill of tackling tricky problems
independently and confidently. We are sure you will be in a position to deal with every
and any problem most successfully.
I wish you all the best for your GATE.
Thank You
Rashmi Deshpande
Director
Vidyalankar Group of Educational Institutes
GATE Syllabus
Engineering Mathematics
Electronics & Communications (EC)
Linear Algebra: Vector space, basis, linear dependence and independence, matrix algebra, eigen values and eigen vectors,
rank, solution of linear equations – existence and uniqueness.
Calculus: Mean value theorems, theorems of integral calculus, evaluation of definite and improper integrals, partial derivatives,
maxima and minima, multiple integrals, line, surface and volume integrals, Taylor series.
Probability and Statistics: Mean, median, mode and standard deviation; combinatorial probability, probability distribution
functions - binomial, Poisson, exponential and normal; Joint and conditional probability; Correlation and regression analysis.
Electrical Engineering (EE)
Linear Algebra: Matrix Algebra, Systems of linear equations, Eigenvalues, Eigenvectors.
Calculus: Mean value theorems, Theorems of integral calculus, Evaluation of definite and improper integrals, Partial
Derivatives, Maxima and minima, Multiple integrals, Fourier series, Vector identities, Directional derivatives, Line integral,
Surface integral, Volume integral, Stokes’s theorem, Gauss’s theorem, Green’s theorem.
Probability and Statistics: Sampling theorems, Conditional probability, Mean, Median, Mode, Standard Deviation, Random
variables, Discrete and Continuous distributions, Poisson distribution, Normal distribution, Binomial distribution, Correlation
analysis, Regression analysis.
Instrumentation Engineering (IN)
Linear Algebra: Matrix algebra, systems of linear equations, Eigen values and Eigen vectors.
Calculus: Mean value theorems, theorems of integral calculus, partial derivatives, maxima and minima, multiple integrals,
Fourier series, vector identities, line, surface and volume integrals, Stokes, Gauss and Green’s theorems.
Probability and Statistics: Sampling theorems, conditional probability, mean, median, mode and standard deviation, random
variables, discrete and continuous distributions: normal, Poisson and binomial distributions.
Computer Science and Information Technology (CS)
Linear Algebra: Matrices, determinants, system of linear equations, eigenvalues and eigenvectors, LU decomposition.
Calculus: Limits, continuity and differentiability. Maxima and minima. Mean value theorem. Integration.
Probability: Random variables. Uniform, normal, exponential, poisson and binomial distributions. Mean, median, mode and
standard deviation. Conditional probability and Bayes theorem.
Mechanical Engineering (ME)
Linear Algebra: Matrix algebra, systems of linear equations, eigenvalues and eigenvectors.
Calculus: Functions of single variable, limit, continuity and differentiability, mean value theorems, indeterminate forms;
evaluation of definite and improper integrals; double and triple integrals; partial derivatives, total derivative, Taylor series (in one
and two variables), maxima and minima, Fourier series; gradient, divergence and curl, vector identities, directional derivatives,
line, surface and volume integrals, applications of Gauss, Stokes and Green’s theorems.
Probability and Statistics: Definitions of probability, sampling theorems, conditional probability; mean, median, mode and
standard deviation; random variables, binomial, Poisson and normal distributions.
Civil Engineering (CE)
Linear Algebra: Matrix algebra; Systems of linear equations; Eigen values and Eigen vectors.
Calculus: Functions of single variable; Limit, continuity and differentiability; Mean value theorems, local maxima and minima,
Taylor and Maclaurin series; Evaluation of definite and indefinite integrals, application of definite integral to obtain area and
volume; Partial derivatives; Total derivative; Gradient, Divergence and Curl, Vector identities, Directional derivatives, Line,
Surface and Volume integrals, Stokes, Gauss and Green’s theorems.
Probability and Statistics: Definitions of probability and sampling theorems; Conditional probability; Discrete Random
variables: Poisson and Binomial distributions; Continuous random variables: normal and exponential distributions; Descriptive
statistics - Mean, median, mode and standard deviation; Hypothesis testing.
For EC/EE/IN/ME/CE : Topics in Vector Calculus are covered in different module.
Pearl Centre, S.B. Marg, Dadar (W), Mumbai 400 028. Tel. 4232 4232
EC / EE / IN / CS / ME / CE
ENGINEERING MATHEMATICS
MODULE 1
INDEX
Contents Topics
Page
No.
Chapter 1 : Linear Algebra
Notes
1.1 Determinants 1
1.2 Matrices 4
1.3 Rank of a Matrix 12
1.4 System of Linear Equations 14
1.5 Eigen Values and Eigen Vectors 18
1.6 Vectors 21
Assignments
Assignment 1
24
Assignment 2
27
Assignment 3
30
Assignment 4
34
Assignment 5
37
Assignment 6
40
Assignment 7
43
Assignment 8
46
Assignment 9
48
Assignment 10
51
Assignment 11
54
Test Papers
Test Paper 1
57
Test Paper 2
60
Test Paper 3
62
Test Paper 4
64
Test Paper 5
66
Test Paper 6
68
Contents Topics
Page
No.
Chapter 2 : Calculus
Notes
2.1 Function of single variable 71
2.2 Limit of a function 71
2.3 Continuity 73
2.4 Differentiability 76
2.5 Mean Value Theorems 79
2.6 Maxima and Minima 80
2.7 Integration 84
2.8 Definite Integration 93
2.9 Double Integrals 97
2.10 Triple Integrals 104
2.11
Change of Variables in Double and Triple Integrals
and Jacobians
106
2.12 Application of Integration 111
2.13 Partial and Total Derivatives 114
2.14 Taylor's Series and Maclaurin's Series 121
2.15 Fourier Series 123
List of Formulae 129
Assignments
Assignment 1
134
Assignment 2
136
Assignment 3
138
Assignment 4
141
Assignment 5
144
Assignment 6
146
Assignment 7
148
Assignment 8
150
Assignment 9
152
Test Papers
Test Paper 1
154
Test Paper 2
156
Test Paper 3
158
Test Paper 4
160
Test Paper 5
162
Test Paper 6
164
Contents Topics
Page
No.
Chapter 3 : Probability and Statistics
Notes
3.1 Basic Terms 166
3.2 Definition of Probability 167
3.3 Complement of an event 167
3.4 Independent Events 168
3.5 Theorems of Probability 168
3.6 Random Variables 172
3.7 Probability Distribution Function 173
3.8 Expectation(Mean), Variance and Standard Deviation 173
3.9 Standard Distributions 174
3.10 Mean, Median, Mode and Standard Deviation 178
3.11 Correlation and Regression Analysis 180
List of Formulae 181
Assignments
Assignment 1
184
Assignment 2
187
Assignment 3
190
Assignment 4
193
Assignment 5
195
Assignment 6
198
Test Papers
Test Paper 1
200
Test Paper 2
202
Test Paper 3
204
Solutions Linear Algebra
Assignment
Answer Key 206
Model Solutions 209
Test Paper
Answer Key 240
Model Solutions 241
Solutions Calculus
Assignment
Answer Key 255
Model Solutions 258
Test Paper
Answer Key 291
Model Solutions 293
Contents Topics
Page
No.
Solutions Probability and Statistics
Assignment
Answer Key 308
Model Solutions 310
Test Paper
Answer Key 332
Model Solutions 333
1
Chapter - 1 : Linear Algebra
1.1 Determinants
If a, b, c and d are any four terms then the representation
ab
cd
is called a determinant
and is denoted by D.
A determinant of order 2 is evaluated as follows:
D =
ab
cd
= ad bc
A determinant of order 3 can be evaluated as follows:
D =
12 3
12 3
12 3
aa a
bbb
cc c
=
2 3 13 12
123
23 13
12
bb bb
bb
aaa
c c cc cc
For example:
123
213
312
=
13 2 3 21
12 3
12 3 2 31
= 1(2 3) 2(4 9) + 3(2 3) = 1 + 10 3 = 6
Properties of Determinants
Interchange of rows and columns (R
i
Ci)
The value of determinant is not affected by changing the rows into the corresponding
columns, and the columns into the corresponding rows. Thus
12 3 1 11
12 3 2 22
12 3 333
aa a abc
bbb a b c
cc c abc
Identical rows and columns (R
i
= R
j
, C
i
= C
j
).
If two rows or two columns of a determinant are identical, the determinant has the value
zero. Thus,
12 3 113
12 3 113
12 3 113
aaa aaa
aa a 0 bbb 0
ccc ccc
Vidyalankar : GATE – Engineering Mathematics
2
Interchange of two adjacent rows and columns (R
i
R
i+1
, C
i
C
i+1
)
If two adjacent rows or columns of the determinant are interchanged, the value of the
determinant so obtained is the negative of the value of the original determinant. Thus
12 3 12 3
12 3 12 3
12 3 12 3
aaa bbb
bb b aa a
cc c cc c
Multiplication of row or column by factor [R
i
(m), C
j
(n)]
If the elements of any row or column are multiplied by the same factor, the value of the
determinant so obtained is equal to the value of the original determinant multiplied by that
factor. Thus,
12 3 123
12 3 123
12 3 123
ma ma ma a a a
bbb mbbb
cc c ccc
Sum of determinants
If any element in any row (or column) consists of the sum of two terms, the determinant
can be expressed as the sum of two other determinants whose other rows (or columns)
remain the same, while the remaining row (or column) consists of these terms
respectively. Thus,
1123 123 123
1123 12 3 123
1123 123 123
a aa aaa aa
bbbbbb bb
c cc ccc cc
Change of row or column by multiples of other rows and columns R
ij
(p), C
ij
(p)
As the consequence of the properties 5, 4 and 2 we have the result.
12 3 1 2 2 3 3
123 1 22 33
12 3 1 2 2 3 3
aa a apa a qa a
bbb bpbb qbb
cc c cpc c qc c
where care must be taken to leave at least one row or column unaltered in such changes
P and q being any positive or negative factors.
Notes on Linear Algebra
3
Note : 1. Area of a triangle whose vertices are (x
1
, y
1
) (x
2
, y
2
) and (x
3
, y
3
) is
given by the absolute value of
1
2
11
22
33
xy1
xy1
xy1
2. Area of a quadrilateral can be found by dividing it into two triangles.
3. If the area of a triangle obtained from the three given points is zero,
then the three points lie on a line.
The condition for three points to be collinear is
11
22
33
xy1
1
xy1
2
xy1
= 0
Solved Example 1 :
Find the area of a triangle whose vertices
are (2, 1), (4,
3), (2, 5).
Solution :
The area of the triangle is
A =
11
22
33
xy1
1
xy1
2
xy1
=
211
1
431
2
251
=
1
2
× abs [2(3 5) 4(1 5) 2(1 + 3)]
=
1
2
× abs [16 + 16 8] = 4 sq. units
Solved Example 2 :
Find if the three points (
1, 1), (5, 7) and
(8, 11) are collinear.
Solution :
If the points are collinear, area of the
triangle formed by the three given points
should be zero.
Area A =
1
2
111
571
8111
=
1
2
[1(7 11) 5(1 11)
+ 8 (
1 7)]
=
1
2
[4 + 60 64] = 0
The given points are collinear.
Vidyalankar : GATE – Engineering Mathematics
4
1.2 Matrices
A Matrix is a rectangular array of elements written as
A =
12 n
11 12 1n
21 22 2n
mm m
a a ....a
a a ....a
... .
.. . .
.. . .
.. ..
a a ....a
The above matrix A has m rows and n columns. So it is a m × n matrix or it is said that
the size of the matrix is m × n.
Types of Matrices
Square Matrix :
It is a Matrix in which number of rows = number of columns
For example:
123
456
789
is a square matrix of order 3.
Diagonal Matrix :
It is a square matrix in which all non diagonal elements are zero.
For example:
100
020
004
Scalar Matrix :
It is a diagonal matrix in which all diagonal elements are equal
For example:
400
040
004
Unit Matrix :
It is a scalar matrix with diagonal elements as unity. It is also called Identity Matrix.
Identity matrix of order 2 is I
2
=
10
01
Notes on Linear Algebra
5
Identity matrix of order 3 is I
3
=
100
010
001
Note : For any matrix A, AI = IA = A
Upper Triangular Matrix :
It is a square matrix in which all the elements below the principal diagonal are zero.
For example:
120
003
002
Lower Triangular Matrix :
It is a square matrix in which all the elements above principal diagonal are zero.
For example:
000
210
124
Column Matrix :
It is a matrix in which there is only one column.
For example:
1
3
1
Row Matrix :
It is a matrix in which there is only one row.
For example: [ 2 3 4 ]
Transpose of a Matrix :
It is a matrix obtained by interchanging rows into columns
For example: If A =
135
238
A = transpose of A =
12
33
58
Vidyalankar : GATE – Engineering Mathematics
6
Symmetric Matrix :
If for a square matrix A, A = A
then A is symmetric
For example:
145
428
583
Skew Symmetric matrix :
If for a square matrix A, A =
A then it is skew symmetric matrix.
For example:
057
503
730
Note : For a skew symmetric matrix, diagonal elements are zero.
Orthogonal Matrix :
A square matrix A is orthogonal if AA
= AA = I
For example: A =
cos sin
sin cos
Here AA = I
Note : For orthogonal matrix A, A
1
= A
Conjugate of a Matrix :
Let A be a complex matrix of order m
n. Then conjugate of A is the matrix obtained by
taking conjugate of every element in the matrix and denoted by
A
For example: if A =
7i 33i 4
92i i 84i
then conjugate of A =
7i 33i 4
A
92i i 84i
Matrix A
:
The transpose of the conjugate of a matrix A is denoted by A
For example: Let A =
7i 24i 4
32i i 12i
then
7i 24i 4
A
32i i 12i
Notes on Linear Algebra
7
and A
=
7i 32i
A24ii
412i
Unitary Matrix :
A square matrix A is said to be unitary if A
A = I
For example: A =
1i 1i
22
1i 1i
22
Here A
A =
10
01
= I
Hermitian Matrix :
A square matrix A is called Hermitian matrix if a
ij
=
j
i
a
For example: A =
41i25i
1i 3 12i
25i12i 8
The necessary and sufficient condition for a matrix A to be Hermitian is that A = A
.
Skew Hermitian Matrix :
A square matrix A is skew Hermitian matrix if a
ij
=
j
i
a
For example:
2i 2 8i 1 2i
(2 8i) 0 2i
(1 2i) 2i 4i
The necessary and sufficient condition for a matrix A to be skewHermitian is that A
= A
Note : All the diagonal elements of a skew Hermitian matrix are either zeroes or
pure imaginary.
Idempotent Matrix :
Matrix A is called idempotent matrix if A
2
= A
For example: A =
224
13 4
123
Here A
2
= A
Vidyalankar : GATE – Engineering Mathematics
8
Periodic Matrix :
A matrix A is called a periodic matrix. A
K+1
= A where K is a +ve integer; if K is the least
+ve integer for which A
K+1
= A, then K is the period of A.
Note : If K = 1, we get A
2
= A and it is idempotent matrix.
Nil potent Matrix :
A matrix is called a Nilpotent matrix, if A
K
= 0 where K is a positive integer. If K is the
least positive integer for which A
K
= 0, then K is the index of the nil potent matrix.
For example: A =
2
2
ab b
aab
has index 2
Involutory Matrix :
A matrix A is called involutory matrix if A
2
= I
For example: A =
01 1
434
334
Here A
2
= I
Note : I
2
= I. Identity matrix is always involutory.
Determinant of a square matrix
Let A be a square matrix, then | A | = determinant of A.
For example: A =
123
213
312
|A| =
123
213
312
=
13 2 3 21
12 3
12 3 2 31
= 1(2
3) 2(4 9) + 3(2 3) = 1 + 10 3
= 6
If | A | 0 then matrix A is called as nonsingular.
If | A | = 0, A is called singular.
Notes on Linear Algebra
9
Adjoint and Inverse of a Square Matrix
Minor : Consider the determinant
=
11 12 13
21 22 23
31 32 33
aaa
aaa
aaa
To find minor leave the row and column passing through the element a
ij
.
The minor of the element a
21
= M
21
=
12 13
32 33
aa
aa
The minor of the element a
32
= M
32
=
11 13
21 23
aa
aa
The minor of the element a
11
= M
11
=
22 23
32 33
aa
aa
Cofactor : The minor M
ij
multiplied by (1)
i+j
is called the cofactor of the element a
ij
.
The cofactor of the element a
21
= A
21
= (1)
2+1
M
21
=
12 13
32 33
aa
aa
The cofactor of the element a
32
= A
32
= (1)
3+2
M
32
=
11 13
21 23
aa
aa
The cofactor of the element a
11
= A
11
= (1)
1+1
M
11
=
22 23
32 33
aa
aa
And so on.
Adjoint of a Matrix :
Adjoint of a square matrix A is the transpose of the matrix formed by the cofactors of the
elements of the given matrix A.
If A =
11 12 13
21 22 23
31 32 33
aaa
aaa
aaa
Then adj (A) =
11 21 31
12 22 32
13 23 33
AAA
AAA
AAA
Vidyalankar : GATE – Engineering Mathematics
10
Inverse of a Square Matrix :
For a nonsingular square matrix A
A
1
=
1
adj(A)
|A|
where A
1
is called the inverse of square matrix.
Note : A A
1
= A
1
A = I
Solved Example 3 :
Calculate the adjoint of A,
where A =
11 1
12 3
213
Solution :
A
11
= the cofactor of a
11
in
| A | =
23
13
= 3
A
12
= the cofactor of a
12
in
| A | =
13
23
= 9
A
13
= the cofactor of a
13
in
| A | =
12
21
= 5
A
21
= the cofactor of a
21
in
| A | =
11
13
= 4
A
22
= the cofactor of a
22
in
| A | =
11
23
= 1
A
23
= the cofactor of a
23
in
| A | =
11
21
= 3
A
31
= the cofactor of a
31
in | A | =
11
23
= 5
A
32
= the cofactor of a
32
in
| A | =
11
13
= 4
A
33
= the cofactor of a
33
in
| A | =
11
12
= 1
Adj (A) = transpose of the matrix formed
by co
factor
=
11 21 31
12 22 32
13 23 33
AAA 345
AAA 914
AAA 531
Solved Example 4 :
Find the inverse of the matrix by finding its
adjoint where A =
13 3
14 3
13 4
Solution :
|A| = 1 0
A
1
exists
Now A
=
111
343
334
Notes on Linear Algebra
11
The cofactors of the elements of A are
(1) = 7 ; (1) =
3 ; (1) = 3
(3) =
1 ; (4) = 1 ; (3) = 0
(3) =
1 ; (3) = 0 ; (4) = 1
Adj (A) =
733
11 0
10 1
A
1
=
733
1
adj(A) 1 1 0
A
10 1
Solved Example 5 :
Find the adjoints of the matrices A and B
where A =
12 3
13 4
14 3
,
B =
045
12 3
117
.
Verify the formula adj(AB) = (adj B ) (adj A)
Solution :
We can find that
adj A =
76 1
10 1
121
,
adj B =
17 33 2
10 5 5
14 4
AB =
1532
1642
19 38
(adj B) (adj A) =
150 98 18
80 70 10
15 14 1
,
adj (AB) =
150 98 18
80 70 10
15 14 1
Hence we verify (adj B) (adj A) = adj (AB)
Solved Example 6 :
Find the inverse of the matrix finding its
adjoint where A =
213
312
123
Solution :
| A | =
213
312
123
= 6 0 A
1
exists
transpose of A = A
=
231
112
323
The co
factors of the elements of A are
(2) =
1 ; (3) = 3 ; (1) = 1
(1) =
7 ; (1) = 3 ; (2) = 5
(3) = 5 ; (2) =
3 ; (3) = 1
adj (A) =
13 1
73 5
531
A
1
=
1
|
A
|
adj(A)
=
1
6
13 1
73 5
531
Vidyalankar : GATE – Engineering Mathematics
12
Solved Example 7 :
Show that
A =
cos sin
sin cos
is orthogonal
Solution :
A =
cos sin
sin cos
AA
=
cos sin
sin cos
cos sin
sin cos
=
22
22
cos sin 0
0sincos
=
10
01
= I
Similarly A
A = I
Hence AA
= AA = I
A is orthogonal
Solved Example 8 :
Show that
A =
84 1
1
14 8
9
474
is orthogonal and
find A
1
.
Solution :
AA =
84 1 8 1 4
11
14 8 4 47
99
474 1 84
=
81 0 0 1 0 0
1
0 810 010
81
0081 001
= I
Similarly A
A = I
A is orthogonal and therefore
A
1
= A =
814
1
447
9
184
1.3 Rank of a Matrix
Sub-matrix
Any matrix obtained by omitting some rows and columns form a given m n matrix A is
called a sub-matrix of A.
For example:
111
222
abc
abc
contains
three 2 x 2 sub-matrices
11
11 11
23
22 22
bc
ab ac
bc
ab ac
two 1
3 sub matrices namely [a
1
b
1
c
1
] and [a
2
b
2
c
2
] and
three 2
1 sub matrix namely
111
222
abc
abc
and so on
The
rank of a matrix is r if :
i) It has atleast one non-zero minor of order r
ii) Every minor of A of order higher than r is zero
Notes on Linear Algebra
13
The rank of a matrix in Row-Echelon form is equal to the number of non-zero rows.
The rank of a matrix is also given by the number of linearly independent rows.
Note : 1. If A is zero matrix, then r(A) = 0
2. IF A is not a zero matrix, r(A) 1
3. IF A is a non
singular n n matrix then r(A) = n ( | A | 0)
4. r(I
n
) = n
5. If A is an m
n matrix then r(A) minimum of m and n
Solved Example 9 :
Find the rank of the matrix
4213
6347
2101
.
Solution :
This matrix contains four 3 3 matrices.
421 413
634 647
210 201
423 213
637 347
211 101
The determinants of all these are zero.
Then consider 2
nd
order sub matrices. It
can be seen that
21
34
has determinant
whose value is 5. i.e. not zero. Hence the
rank of the matrix is 2.
Solved Example 10 :
Find the rank of the matrix
A =
32 1 5
514 2
141119
.
Solution :
The matrix contains four 3 3 matrices
32 1 3 1 5
514 54 2
1411 11119
32 5 2 1 5
51 2 14 2
1419 41119
The determinants of all these are zero.
Then consider 2
nd
order sub matrices.
It can be seen that
32
51
has a value 7
i.e. not zero.
Hence the rank of the matrix is 2.
Vidyalankar : GATE – Engineering Mathematics
14
1.4 System of Linear Equations
Consider a set of equations
a
1
x + b
1
y + c
1
z = d
1
a
2
x + b
2
y + c
2
z = d
2
a
3
x + b
3
y + c
3
z = d
3
The equations can be written in the matrix form as
1
111
222 2
333
3
d
abc x
abc y d
abc z
d
which is of the form AX = B
Augmented matrix, [A : B] =
111 1
222 2
333 3
abc:d
abc:d
abc:d
Consistency conditions
After reducing [A : B] to Row-Echelon form, find the ranks of A and [A : B]
Case 1 : r(A) r (A : D), then the system is inconsistent.
i.e. it has
no solution.
Case 2 : r(A : D) = r(A) then the system is consistent and if
(i)
r(A : D) = r(A) = Number of unknowns then the system is consistent
and has
unique solution.
(ii)
r(A : D) = r(A) < Number of unknowns then the system is consistent
and has
infinitely many solutions.
Solution of Linear Equations
Cramer’s Rule
The solutions of the equations
a
1
x + b
1
y + c
1
z = d
1
a
2
x + b
2
y + c
2
z = d
2
a
3
x + b
3
y + c
3
z = d
3
Notes on Linear Algebra
15
Let =
111
222
333
abc
abc
abc
The solution is given by
111 111 111 111
222 222 222 222
333 333 333 333
xyz1
dbc adc abd abc
dbc adc abd abc
dbc adc abd abc
xyz
xyz1
[ 0]
i.e.
x =
x
, y =
y
, z =
z
Method of Inversion
If the matrix form of the given equations is AX = B and if
A
0, then the solution can be
obtained as
X = A
1
B
Non Zero Solutions of Linear Homogenous Equation
The homogenous equation in x, y, z are
a
1
x + b
1
y + c
1
z = 0
a
2
x + b
2
y + c
2
z = 0
a
3
x + b
3
y + c
3
z = 0
A system of simultaneous linear equations is said to have zero or trivial solutions if all the
unknowns have zero values, and is said to have non
zero solution if at least one of the
unknowns has the non
zero value.
The necessary condition that the equations have non
zero solutions is
111
222
333
abc
abc
abc
= 0
Vidyalankar : GATE – Engineering Mathematics
16
Solved Example 11 :
Solve 3x + y = 19
3x
y = 23
Solution :
Here =
31
31
= 6
By Cramer’s rule
x =
19 1
23 1
x
31
31
=
19 23
33
= 7
and y =
319
323
y6957
31
33
31
= 2
Solved Example 12 :
Solve 3x 5z = 1
2x + 7y = 6
x + y + z = 5
Solution :
Here =
30 5
27 0
11 1
= 3(7 0) 0(2 0) 5(2 7)
= 21 0 + 25 = 46
x =
10 5
670
511
= 1(7 0) 0(6 0) 5(6 35)
= 7 0 + 145 = 138
y =
315
26 0
15 1
= 3(6 0) + 1(2 0) 5(10 6)
= 18 + 2 20 = 0
z =
30 1
27 6
11 5
= 3 (35 6) 0 (10 6) 1(2 7)
= 87 0 + 5 = 92
Now x =
x 138
46
= 3
y =
y0
46
= 0
z =
z92
46
= 2
Solved Example 13 :
Test whether the following equations have
nonzero solution. If they have such
solution obtain the solutions.
x + y 3z = 0, 3x y z = 0,
2x + y 4z = 0
Solution :
Now
11 3
311
21 4
= 0
and hence the equations have nonzero
solution. Solving the first two equations
we get
xyz
48 4
Notes on Linear Algebra
17
so that x = 4, y = 8, z = 4, where
is a non-zero constant. These values
satisfy the third equation and hence they
are non-zero solutions.
Solved Example 14 :
Test whether the following equations have
non zero solution
2x + 3y + 4z = 0
x 2y 3z = 0
3x + y 8z = 0
Solution :
Now
23 4
123
31 8
= 63 0
Hence the equations have no nonzero
solution. The only solution is x = 0, y = 0,
z = 0
Solved Example 15 :
Discuss the consistency of
x + y + z = 1
2x + 4y 3z = 9
3x + 5y 2z = 11
Solution :
In the matrix form
A =
11 1
24 3
35 2
and
[A, B] =
11 1 1
24 3 9
35 211
Rank (A) = 2 since
11 1
24 3
35 2
= 0
and
11
24
= 2 0
Rank [A : B] = 3
since
11 1
439
5211
= 7 0
r(A) r(A : B) thus the system is
inconsistent i.e. it has no solution.
Solved Example 16 :
Examine for consistency
3x + y + 2z = 3
2x 3y z = 3
x + 2y + z = 4
Solution :
In the matrix form
312 x 3
231y 3
12 1 z 4
A =
312
231
12 1
Rank of A = r(A) = 3 since | A | = 8 0
[A : B] =
312 3
2313
12 1 4
Similarly r(A : B) is also 3.
r(A) = r(A : B)
System is consistent and has unique
solution.
Vidyalankar : GATE – Engineering Mathematics
18
Solved Example 17 :
Examine the consistency of
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
Solution :
In the matrix form
53 7 x 4
326 2 y 9
7210z 5
A =
53 7
326 2
7210
Rank of A = r(A) = 2 since | A | = 0
and
53
326
= 121 0
[A : B] =
53 7:4
326 2 :9
7210:5
[A : B] contains four 3 × 3 sub matrices:
53 7 574
3262 329
7210 7105
534 3 74
3269 26 2 9
725 2105
All the above 4 sub matrices have
determinant value 0
r (A) = r(A : B)
= 2 < 3 (number of unknowns)
System is consistent and has infinitely
many solutions.
1.5 Eigen Values and Eigen Vectors
Characteristic Equation
Let A be an n × n square matrix. Then [A I] is characteristic matrix of A, where I is
identity matrix.
|A I| is characteristic polynomial.
|A I| = 0 is characteristic equation of A.
Eigen values
The roots of the characteristic equation of a matrix are called its Eigen values.
Eigen vectors
If is an Eigen value of A, then a non-zero vector X such that
AX = X or [A I] [X] = 0 is called the Eigen vector of A corresponding to Eigen value .
Notes on Linear Algebra
19
Properties of Eigen values and Eigen vectors
Sum of Eigen values of a matrix is equal to the trace of the matrix.
Product of Eigen values of a matrix is equal to its determinant.
Eigen values of A and A
T
are same.
If A is a triangular or diagonal matrix, Eigen values are the diagonal elements.
If an Eigen value of A is , then
1
is an Eigen value of A
1
.
If Eigen values of A are
1
,
2
….., then Eigen values of A
k
are
k
1
,
k
2
………
Eigen vectors of a real symmetric matrix corresponding to different Eigen values are
orthogonal.
If X is an Eigen vector of A corresponding to an Eigen value, the kX is also an Eigen
vector of A corresponding to Eigen value X, where k is a non-zero scalar.
Cayley Hamilton Theorem
Every square matrix satisfies its characteristic equation i.e.,
|A I| = 0
n
+ a
1
n
1
+ a
2
n
2
…….. + a
n
= 0
is satisfied by A,
A
n
+ a
1
A
n
1
+ a
2
A
n
2
+ …… a
n
I = 0
Solved Example 18 :
1.
Find the eigen values of the matrix.
211
232
334
Solution :
Characteristic equation of A in is
| A I | = 0
211
23 2
334
= 0
Instead of evaluating the determinant
directly we use the formula for its
expansion which is as follows :
3
(sum of diagonal elements of A)
2
+
(sum of minors of diagonal elements of A)
|A| = 0
3
9
2
+ 15 7 = 0
( 1) ( 7) ( 1) = 0
1
= 7,
2
= 1,
3
= 1
Vidyalankar : GATE – Engineering Mathematics
20
Solved Example 19 :
Find the eigen vectors for the matrix :
211
232
334
Solution :
From example (1) we get
1
= 7,
2
= 1,
3
= 1
Matrix equation of A in is (A I) x = 0
1
2
3
x
211 0
23 2 x 0
334 0
x
Case (1) for
1
= 7, matrix equation is
1
2
3
x
51 1 0
242x 0
33 3 0
x
By Cramer’s rule
3
12
x
xx
61218
3
12
x
xx
123
x
1
=
1
2
3
Case (2) for
2
= 1 Matrix equation is
1
2
3
x
111 0
222 x 0
333 0
x
By Cramer’s rule, we get
3
12
x
xx
000
X
2
=
0
0
0
But by definition we want nor zero X
2
. So
we proceed as follows,
Expanding R
1
x
1
+ x
2
+ x
3
= 0
We assure any element to be zero say x
1
and give any convenient value say 1 to x
2
and find x
3
.
Let x
1
= 0, x
2
= 1 x
3
= 1
x
2
=
0
1
1
Case 3 : for
3
= 1
1
2
3
x
111 0
222 x 0
333 0
x
By Cramer’s rule we get
3
12
x
xx
000
x
3
=
0
0
0
Again consider x
1
+ x
2
+ x
3
= 0
Now let x
2
= 0 x
1
= 1 and x
3
= 1
x
3
=
1
0
1
Notes on Linear Algebra
21
Solved Example 20 :
Verify whether the matrix A =
121
10 3
211
satisfies its characteristic equation.
Solution :
| A I | =
121
13
211
=
3
2
2
+ 4 18
Now A
2
=
118
552
530
3
A =
16 6 12
14 8 8
21014
A
3
2A
2
+ 4A 18 I =
000
000
000
1.6 Vectors
An ordered set of n number is called an nvector or a vector of order n.
For example: X = (x
1
, x
2
, …..x
n
) is an nvector.
The numbers x
1
, x
2
, …..x
n
are called as the components of X.
The components x
1
, x
2
, ……..x
n
of a vector may be written in a row or a column.
i.e. X = (x
1
, x
2
, x
3
, ………x
n
) and
X =
1
2
n
x
x
:
x
:
are n vectors
Operation on Vectors
Inner product of two vectors
Let X = (x
1
, x
2
, ……..x
n
) and
Y = (y
1 ,
y
2
, …….y
n
) be two nvectors
Then the product XY = x
1
y
1
+ x
2
y
2
+ …..+ x
n
y
n
is called the inner product of two
nvectors
Length of a vector
Let X = (x
1
, x
2
, ……x
n
) be a vector. Then the length of a vector is the positive square
root of the expression
22 2
12 n
x x .........x
Vidyalankar : GATE – Engineering Mathematics
22
Note : Length of the vector is also called as Norm of the vector
Normal vector
A vector whose length is 1 is called a normal vector
i.e. if X = (x
1
, x
2
, ……x
n
) is a normal vector then x
1
2
+ x
2
2
+ ……..x
n
2
= 1
If a vector is not a normal one, then it can be converted to a normal vector as follows.
Let X = (1, 3, 7) be a 3vector.
Let d =
22 2
13 (7) 59
Then
13 7
x,,
dd d
is a normal vector.
Orthogonal vector
A vector X is said to be orthogonal to Y if the inner product of X and Y is zero i.e. XY = 0
i.e. if X = (x
1
, x
2
, …….x
n
) and
Y = (y
1
, y
2
, ……..y
n
) and if X and Y are orthogonal then
x
1
y
1
+ x
2
y
2
+ ……….+ x
n
y
n
= 0
Linear dependence and independence of vectors
A system of n vectors x
1
, x
2
, ….x
n
of the same order are said to be linearly dependent if
there exists n numbers K
1
, K
2
, ….K
n
(where all of them are not zero) such that
K
1
X
1
+ K
2
X
2
+ …. + K
n
X
n
= 0
(where 0 is a null vector of the same order)
The vector are linearly independent only if K
1
= K
2
= ………K
n
= 0
Solved Example 21 :
Examine for linear dependence
X
1
= (1 2 4)
T
, X
2
= (3 7 10)
T
Solution :
We have
X
1
=
1
2
4
X
2
=
3
7
10
Let c
1
x
1
+ c
2
x
2
= 0
(Note 0 on R. H. S. is Zero Vector)
i.e. c
1
1
2
4
+ c
2
3
7
10
=
0
0
0
12
12
12
c3c
2c 7c
4c 10c
=
0
0
0
c
1
+ 3c
2
= 0
2c
1
+ 7c
2
= 0
4c
1
+ 10c
2
= 0
Consider first two equations in Matrix form.
Notes on Linear Algebra
23
1
2
c
13 0
27c 0
|A| = 1 0
system has zero solution
c
1
= c
2
= 0
X
1
, X
2
are linear independent.
Solved Example 22 :
Show that
123 213 033
324 342 306
132 221 043
are linearly dependent.
Solution :
Consider
12
123 213
c324 c 342
132 221
3
033
c306
043
000
000
000
121 123
12 312 3
12 3 12 3
c2c 0c 2cc 3c
3c 3c 3c 2c 4c 0c
c2c0c 3c2c4c
12 3
12 3
12 3
3c 3c 3c 0 0 0
4c 2c 6c 0 0 0
2c c 3c 0 0 0
c
1
+ 2c
2
= 0
2c
1
+ c
2
+ 3c
3
= 0
c
1
+ c
2
+ c
3
= 0
Solving we get
c
1
= 2c
2
and c
2
= c
3
Let c
2
= c
3
= 1
c
1
= 2
We see that x
1
x
2
x
3
are linearly dependent
2
123 213 033
324 342 306
132 221 043
= 0
Vidyalankar : GATE – Engineering Mathematics
24
Assignment 1
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 10 carry one mark each
1. Given that the determinant of the
matrix
130
264
10 2
is 12, the
determinant of the matrix
260
4128
204
is [ME 2014]
(A) 96 (B) 24
(C) 24 (D) 96
2. Consider a 3 x 3 real symmetric matrix
S such that two of its eigen values are
a 0, b 0
with respective eigen
vectors
11
22
33
xy
x,y
xy
. If a b then x
1
y
1
+
x
2
y
2
+ x
3
y
3
equals [ME 2014]
(A) a (B) b
(C) ab (D) 0
3. A real (4 X 4) matrix A satisfies the
equation A
2
= I, where I is the (4 X 4)
identity matrix. The positive eigen
value of A is ______. [EC 2014]
4. Which one of the following statements
is true for all real symmetric matrices?
[EE 2014]
(A) All the eigen values are real.
(B) All the eigen values are positive.
(C) All the eigen values are distinct.
(D) Sum of all the eigenvalues is zero.
5. With reference to the conventional
Cartesian (x, y) coordinate system, the
vertices of a triangle have the following
coordinates: (x
1
, y
1
) = (1, 0); (x
2
, y
2
)
= (2, 2); and (x
3
, y
3
) = (4, 3). The area
of the triangle is equal to [CE 2014]
(A)
3
2
(B)
3
4
(C)
4
5
(D)
5
2
6. At least one eigen value of a singular
matrix is [ME 2015]
(A) positive (B) zero
(C) negative (D) imaginary
7. We have a set of 3 linear equations in
3 unknowns. 'X Y' means X and Y
are equivalent statements and 'X ≢ Y'
means X and Y are not equivalent
statements. [EE 2015]
P : There is a unique solution.
Q : The equations are linearly
independent.
R : All eigen values of the coefficient
matrix are nonzero.
S : The determinant of the coefficient
matrix is nonzero.
Assignment on Linear Algebra
25
Which one of the following is TRUE?
(A) P Q R S
(B) P R ≢ Q S
(C) P Q ≢ R S
(D) P ≢ Q ≢ R ≢ S
8. Consider a system of linear equations:
x 2y + 3z = 1,
x 3y + 4z = 1, and
2x + 4y 6z = k. [EC 2015]
The value of k for which the system
has infinitely many solutions is _____.
9. The larger of the two eigen values of
the matrix
45
21
is ____.[CS 2015]
10. A real square matrix A is called skew-
symmetric if
[ME 2016]
(A) A
T
= A (B) A
T
= A
1
(C) A
T
= A (D) A
T
= A + A
1
Q 11 to Q 20 carry two marks each
11.
The system of linear equations
213 a 5
301 b 4
125 c 14
has
[EC 2014]
(A) a unique solution
(B) infinitely many solutions
(C) no solution
(D) exactly two solutions
12. A system matrix is given as follows.
A =
01 1
6116
6115
The absolute value of the ratio of the
maximum eigen value to the minimum
eigen value is________.
[EE 2014]
13. If A =
15
62
and B =
37
84
. AB
T
is
equal to
[CE 2017]
(A)
38 28
32 56
(B)
340
42 8
(C)
43 27
34 50
(D)
38 32
28 56
14. The value of x for which all the eigen
values of the matrix given below are
real is
[EC 2015]
10 5 j 4
x20 2
42 10
(A) 5 + j (B) 5 j
(C) 1 5j (D) 1 + 5j
15. The two Eigen values of the matrix
21
1p
have a ratio of 3:1 for p = 2.
What is another value of p for which
the Eigen values have the same ratio
of 3:1?
[CE 2015]
(A) 2 (B) 1
(C) 7/3 (D) 14/3
Vidyalankar : GATE – Engineering Mathematics
26
16. Let the eigen values of a 2 x 2 matrix
A be 1, -2 with eigen vectors x
1
and x
2
respectively. Then the eigen values
and eigen vectors of the matrix
A
2
3A+ 4I would, respectively, be
(A) 2, 14; x
1
, x
2
[EE 2016]
(B) 2, 14; x
1
+ x
2
, x
1
x
2
(C) 2, 0; x
1
, x
2
(D) 2, 0; x
1
+ x
2
, x
1
– x
2
17. Consider the matrix
A =
211
234
112
whose eigen values
are 1, 1 and 3. Then Trace of
(A
3
3A
2
) is _______. [IN 2016]
18. Consider the following system of
equations:
2x
1
+ x
2
+ x
3
= 0,
x
2
x
3
= 0,
x
1
+ x
2
= 0.
This system has
[ME 2011]
(A) a unique solution
(B) no solution
(C) infinite number of solutions
(D) five solutions
19. Given that
53
A
20
and
10
I,
01
the
value of
A
3
is [EC, EE, IN 2012]
(A) 15
A + 12 I (B) 19 A + 30 I
(C) 17 A + 15 I (D) 17 A + 21 I
20. Consider the matrix
P =
11
0
22
010
11
0
22
.
Which one of the following statements
about P is INCORRECT?
[ME 2017]
(A) Determinant of P is equal to 1
(B) P is orthogonal
(C) Inverse of P is equal to its
transpose
(D) All eigen values of P are real
numbers
Assignment on Linear Algebra
27
Assignment 2
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 10 carry one mark each
1.
The matrix form of the linear system
dx
dt
= 3x 5y and
dy
dt
= 4x + 8y is
[ME 2014]
(A)
x35x
d
y48y
dt
(B)
x38x
d
y45y
dt
(C)
x45x
d
y38y
dt
(D)
x48x
d
y35y
dt
2. Which one of the following equations
is a correct identity for arbitrary 3 x 3
real matrices P, Q and R?
[ME 2014]
(A) P(Q + R) = PQ + RP
(B) (P
Q)
2
= P
2
2PQ + Q
2
(C) det (P + Q) = det P + det Q
(D) (P + Q)
2
= P
2
+ PQ + QP + Q
2
3. The determinant of matrix A is 5 and
the determinant of matrix B is 40. The
determinant of matrix AB is
_________.
[EC 2014]
4. Given the matrices J =
321
242
126
and
K =
1
2
1
, the product K
T
JK is
_______.
[CE 2014]
5. The determinant of matrix
0123
1030
2301
3012
is ______. [CE 2014]
6. The lowest eigen value of the 2 2
matrix
42
13
is ______. [ME 2015]
7. For what value of p the following set of
equations will have no solution?
2x + 3y = 5
[CE 2015]
3x + py = 10
8. The value of p such that the vector
1
2
3
is an eigen vector of the matrix
412
p21
14 4 10
is ______. [EC 2015]
Vidyalankar : GATE – Engineering Mathematics
28
9. The solution to the system of
equations
25x 2
43 y 30
is
[ME 2016]
(A) 6, 2 (B)
6, 2
(C)
6, 2 (D) 6, 2
10. Let M
4
= I, (where I denotes the
identity matrix) and M I, M
2
I and
M
3
I. Then, for any natural number k,
M
1
equals: [EC 2014]
(A) M
4k+1
(B) M
4k+2
(C) M
4k+3
(D) M
4k
Q 11 to Q 20 carry two marks each
11.
The maximum value of the
determinant among all 2 x 2 real
symmetric matrices with trace 14 is
_______.
[EC 2014]
12. The rank of the matrix [CE 2014]
6044
214818
14 14 0 10
is _________.
13. The smallest and largest eigen values
of the following matrix are:
322
446
235
[CE 2015]
(A) 1.5 and 2.5 (B) 0.5 and 2.5
(C) 1.0 and 3.0 (D) 1.0 and 2.0
14. Consider the following 2 2 matrix A
where two elements are unknown and
are marked by a and b. The eigen
values of this matrix are
1 and 7.
What are the values of a and b?
A =
14
ba
[CS 2015]
(A) a = 6, b = 4 (B) a = 4, b = 6
(C) a = 3, b = 5 (D) a = 5, b = 3
15. The number of linearly independent
eigen vectors of matrix
A =
210
020
003
is ______. [ME 2016]
16. Let A be a 4 × 3 real matrix with rank
2. Which one of the following
statement is TRUE?
[EE 2016]
(A) Rank of A
T
A is less than 2.
(B) Rank of A
T
A is equal to 2.
(C) Rank of A
T
A is greater than 2.
(D) Rank of A
T
A can be any number
between 1 and 3.
17. The system of equations
x + y + z = 6
x + 4y + 6z = 20
x + 4y +
z =
has NO solution for values of and
given by
[EC 2011]
(A)
= 6, = 20 (B) = 6, 20
(C)
6, = 20 (D) 6, 20
Assignment on Linear Algebra
29
18. Let A be the 2 2 matrix with elements
a
11
= a
12
= a
21
= +1 and a
22
= 1. Then
the eigen values of the matrix A
19
are
[CS 2012]
(A) 1024 and
1024
(B)
1024 2 and 1024 2
(C)
42 and 42
(D) 512 2 and 512 2
19. Which one of the following does NOT
equal [CS 2013]
(A)
1x(x 1) x 1
1y(y 1) y 1
1z(z1) z 1
(B)
2
2
2
1x 1x 1
1y 1y 1
1z 1z 1
(C)
22
22
2
0xyx y
0yzy z
1z z
(D)
22
22
2
2xyx y
2yzy z
1z z
20. Consider the matrix A =
50 70
70 80
whose eigen vectors corresponding to
eigen values
1
and
2
are
x
1
=
1
70
50
and x
2
=
2
80
70
,
respectively. The value of
T
12
xx is
_________.
[ME 2017]
Vidyalankar : GATE – Engineering Mathematics
30
Assignment 3
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 10 carry one mark each
1.
One of the eigen vectors of the matrix
52
96
is [ME 2014]
(A)
1
1
(B)
2
9
(C)
2
1
(D)
1
1
2. For matrices of same dimension M, N
and scalar c, which one of these
properties DOES NOT ALWAYS hold?
(A) (M
T
)
T
= M [EC 2014]
(B) (cM)
T
= c(M)
T
(C) (M + N)
T
= M
T
+ N
T
(D) MN = NM
3. Given a system of equations:
x + 2y + 2z = b
1
5x + y + 3z = b
2
Which of the following is true regarding
its solutions?
[EE 2014]
(A) The system has a unique solution
for any given b
1
and b
2
(B) The system will have infinitely
many solutions for any given b
1
and b
2
(C) Whether or not a solution exists
depends on the given b
1
and b
2
(D) The system would have no
solution for any values of b
1
and b
2
4. The sum of eigen values of the matrix,
[M] is
[CE 2014]
Where [M] =
215 650 795
655 150 835
485 355 550
(A) 915 (B) 1355
(C) 1640 (D) 2180
5. If any two columns of a determinant
P =
478
315
962
are interchanged,
which one of the following statements
regarding the value of the determinant
is
CORRECT? [ME 2015]
(A) Absolute value remains
unchanged but sign will change
(B) Both absolute value and sign will
change
(C) Absolute value will change but
sign will not change
(D) Both absolute value and sign will
remain unchanged
6. The eigen value of the matrix
A =
115
056
065
are [IN 2017]
(A)
1, 5, 6 (B) 1, 5 j6
(C) 1, 5
j6 (D) 1, 5, 5
Assignment on Linear Algebra
31
7. Let P =
11 1
234
323
and
Q =
12 1
6126
5105
be two matrices.
Then the rank of P + Q is _________.
[CS 2017]
8. For A =
1tanx
tan x 1
, the
determinant of A
T
A
1
is [EC 2015]
(A) sec
2
x (B) cos 4x
(C) 1 (D) 0
9. Consider the following simultaneous
equations (with c
1
and c
2
being
constants):
3x
1
+ 2x
2
= c
1
4x
1
+ x
2
= c
2
The characteristic equation for these
simultaneous equations is
[CE 2017]
(A)
2
4 5 = 0
(B)
2
4 + 5 = 0
(C)
2
+ 4 5 = 0
(D)
2
+ 4 + 5 = 0
10.
The value of for which the matrix
A =
32 4
97 13
649x
has zero as an
eigen value is _______.
[EC 2016]
Q 11 to Q 20 carry two marks each
11. Which one of the following statements
is NOT true for a square matrix A?
[EC 2014]
(A) If A is upper triangular, the eigen
values of A are the diagonal
elements of it
(B) If A is real symmetric, the eigen
values of A are always real and
positive
(C) If A is real, the eigen values of A
and A
T
are always the same
(D) If all the principal minors of A are
positive, all the eigen values of A
are also positive.
12. For given matrix P =
43i i
i43i
,
where i =
1
, the inverse of matrix P
is
[ME 2015]
(A)
43i i
1
i43i
24
(B)
i43i
1
43i i
25
(C)
43i i
1
i43i
24
(D)
43i i
1
i43i
25
Vidyalankar : GATE – Engineering Mathematics
32
13. If the characteristic polynomial of a
3 × 3 matrix M over
(the set of real
numbers) is
3
4
2
+ a + 30, a
,
and one eigen value of M is 2, then the
largest among the absolute values of
the eigen values of M is ________.
[CS 2017]
14. Perform the following operations on
the matrix
3445
7 9 105
13 2 195
.
(i) Add the third row to the second row.
(ii) Subtract the third column from the
first column.
The determinant of the resultant matrix
is ____________.
[CS 2015]
15. The matrix A =
a037
2513
0024
000b
has
det (A) = 100 and trace (A) = 14.
The value of
ab
is ___. [EC 2016]
16. Consider the following linear system.
x + 2y
3z = a
2x + 3y + 3z = b
5x + 9y
6z = c
This system is consistent if a, b and c
satisfy the equation
[CE 2016]
(A) 7a
b c = 0
(B) 3a + b
c = 0
(C) 3a
b + c = 0
(D) 7a
b + c = 0
17. Consider the matrix
51
41
. Which
one of the following statements is TRUE
for the eigen values and eigen vectors of
this matrix?
[CE 2017]
(A) Eigen value 3 has a multiplicity of
2, and only one independent eigen
vector exists.
(B) Eigen value 3 has a multiplicity of
2, and two independent eigen
vectors exist.
(C) Eigenvalue 3 has a multiplicity of
2, and no independent eigen
vector exists.
(D) Eigenvalues are 3 and
3, and two
independent eigen vectors exist.
18.
For the matrix
53
A,
13
ONE of the
normalized eigen vectors is given as
[ME 2012]
(A)
1
2
3
2
(B)
1
2
1
2
(C)
3
10
1
10
(D)
1
5
2
5
Assignment on Linear Algebra
33
19. The minimum eigen value of the
following matrix is
352
5127
275
[EC 2013]
(A) 0 (B) 1
(C) 2 (D) 3
20. The eigen values of the matrix given
below are
01 0
00 1
034
[EE 2017]
(A) (0,
1, 3) (B) (0, 2, 3)
(C) (0, 2, 3) (D) (0, 1, 3)
Vidyalankar : GATE – Engineering Mathematics
34
Assignment 4
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1. A square matrix each of whose
diagonal elements are ‘1’ and
non
diagonal elements are '0’ is called
(A) Null matrix
(B) Skew symmetric matrix
(C) Identity matrix
(D) row matrix
2. If product of two matrices AB exists
does BA exists ?
(A) Always exist (B) May exist
(C) Never exist (D) None of these
3. If a square matrix A is real and
symmetric, then the eigen values
(A) are always real
(B) are always real and positive
(C) are always real and non-negative
(D) occur in complex conjugate pairs
4. If A and B are two matrices such that
AB and A + B are both defined then A,
B are
(A) Scalar matrices of same order
(B) square matrices of same order
(C) matrices of different order
(D) cannot predict
5.
A matrix A has x rows and x + 5
columns, matrix B has y rows and
11
y columns. Both AB and BA exists.
Then values of x and y respectively
are
(A) 3,
8 (B) 5, 11
(C) 5,
11 (D) 3, 8
6. For what values of x, the matrix
3x 2 2
24x 1
241x
is singular
(A) 0, 3 (B) 0,
3
(C)
1, 4 (D) 3, 4
Q7 to Q18 carry two marks each
7. Evaluate
=
2
2
2
1
1
1
where is one of
the imaginary cube root of unity
(A) 0 (B) 1
(C) 2 (D) None of these
8. If a given matrix [A] mn has r linearly
independent vectors (rows or columns)
and the remaining vectors are
combination of these r vectors. Then
rank of matrix is
(A) m (B) n
(C) r (D) m
n
Assignment on Linear Algebra
35
9. Are following vectors linearly dependent
x
1
= (3, 2, 7) x
2
= (2, 4, 1)
x
3
= (1, 2, 6)
(A) dependent (B) independent
(C) can’t say (D) none of above
10. If A is nonzero column matrix and B
is a non
zero row matrix then rank of
matrix AB is
(A) always 1
(B) always = No. of elements in row
(C) always = No. of elements in
column
(D) does not depends on rows &
columns
11. If A = [x y z], B =
ahg
hb f
gfc
,
C =
x
y
z
Then ABC =
(A) ABC is not possible
(B) [ax
2
+ by
2
+ cz
2
+ 2hxy + 2gzx
+ 2fyz]
(C) [a
2
x + b
2
y + c
2
z + 2abc + 2xyz
+ 2fgh]
(D)
22
ax byz fcz
2hxy 2gzx 2fyz
12.
The inverse of a matrix
87 9
51015
12 3
is
(A)
15 7 8
1
10 5 3
2
521
(B)
1578
10 5 3
521
(C)
1578
10 5 3
521
(D) none of these
13. Rank of matrix
022
748
70 4
is
(A) 3 (B) 2
(C) 1 (D) none of these
14.
Eigen values of matrix
A =
100
231
024
(A) 1,
2, 4 (B) 1, 3, 4
(C) 1, 2, 5 (D)
1, 3, 4
Vidyalankar : GATE – Engineering Mathematics
36
15. Whether the following equations are
inconsistent
x + y + z =
3
3x + y
2z = 2
2x + 4y + 7z = 7
(A) Yes
(B) No
(C) Can’t say
(D) Can’t be determined.
16. If A =
54
32
23
& B =
163
251
& A + B
X = 0 then X =
(A)
42
97
14
(B)
42
97
14
(C)
42
97
14
(D)
42
97
14
17. If A =
cos sin
sin cos
&
B =
sin cos
cos sin
then
cos
(A) + sin (B) =
(A)
10
01
(B)
10
01
(C)
01
10
(D)
01
10
18. If A =
35
42
& A
2
= kA + 14I, then
k = . . . .
(A) 3 (B) 5
(C) 1 (D)
5
Assignment on Linear Algebra
37
Assignment 5
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
If A be any matrix, then matrix B if it
exists such that AB = BA = I, then it is
called
(A) Transpose of A
(B) Inverse of A
(C) Cofactor matrix A
(D) Adjoint of A.
2. The eigen vector corresponding to an
eigen values are
(A) different (B) unique
(C) non
unique (D) none of these
3. Rank of matrix which is in echelon
form is equal to
(A) No. of non-zero rows in the matrix
(B) No. of columns in the matrix
(C) It is independent of No. of rows
and columns
(D) Can’t say
4. Cayley Hamilton theorem is
(A) A matrix can be expressed as sum
of symmetric and skew
symmetric
matrices
(B) Every square matrix satisfies its
own characteristic equation.
(C) Inverse of a matrix exists if it is
singular
(D) None of these
5. Inverse of a matrix
(A) exist if matrix is singular
(B) is unique
(C)
A
adjA
(D) is not unique
6. If P =
pq
qp
, Q =
rs
sr
, then
PQ is equal to
(A)
pr qs ps qr
qr ps qs pr
(B)
pq rs pr qs
qr ps qs pr
(C)
pr qs ps qr
qr ps qs pr
(D)
pq rs pr qs
qp rs qs pr
Q7 to Q18 carry two marks each
7.
Is the matrix
01 1
434
334
an involutory
matrix ?
(A) No
(B) Yes
(C) Can’t say
(D) Cannot be determined
Vidyalankar : GATE – Engineering Mathematics
38
8. The inverse of the matrix
12 3
210
425
is
(A)
54 3
10 7 6
865
(B)
543
10 7 6
865
(C)
10 7 6
865
543
(D)
54 3
10 7 6
865
9. The rank of the matrix
012 2
402 5
213 1
(A) 4 (B) 3
(C) 2 (D) 1
10. The following equations have solutions
x + 2y z = 3
2x 2y + 3z = 2
3x y + 2z = 1
x y + z = 1
(A) x =
1, y = 4, z = 4
(B) x = 4, y = 1, z = 1
(C) infinite solutions
(D) it is consistent
11. For the matrix
42
24
the eigen value
corresponding to the eigen vector
101
101
is
(A) 2 (B) 4
(C) 6 (D) 8
12. Characteristic root of matrix
A =
125
035
002
are
(A)
1, 3, 2 (B) 1, 3, 2
(C)
1, 2, 3 (D) 3, 1, 2
13. The eigen values of
111
111
111
are
(A) 0, 0, 0 (B) 0, 0, 1
(C) 0, 0, 3 (D) 1, 1, 1
14. The rank of matrix
11 2 3
1303
1233
11 2 3
is
(A) 3 (B) 4
(C) 2 (D) 1
Assignment on Linear Algebra
39
15. The following equations have
solutions.
x + 2y + 3z = 6
3x 2y + z = 2
4x + 2y + z = 7
(A) x = 2, y = 2, z =
½
(B) x = y = z = 1
(C) They are inconsistent
(D) Infinite solutions
16. If A =
51
32
& B =
43
12
then
the matrix X which satisfies the
equation 3A + X = 2B is given by X =
(A)
23 9
710
(B)
23 9
710
(C)
23 9
710
(D) None of these
17. If A =
10
17
, then the value of k for
which
2
A = 8A + kI is . . . .
(A) 5 (B)
5
(C) 7 (D)
7
18. If Y =
32
14
& Y 2X =
10
32
,
then X = . . . .
(A)
21
11
(B)
12
11
(C)
11
12
(D)
11
21
Vidyalankar : GATE – Engineering Mathematics
40
Assignment 6
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
Eigen values of Matrix
12
54
are
(A)
6, 1 (B) 6, 1
(C)
6, 1 (D) 6, 1
2. Given an orthogonal matrix
1111
11 1 1
A
1100
00 1 1
l
T
AA is
(A)
1/ 4 0 0 0
01/40 0
001/20
0001/2
(B)
1/ 2 0 0 0
01/20 0
001/20
0001/2
(C)
1000
0100
0010
0001
(D)
1/ 4 0 0 0
01/40 0
001/40
0001/4
3.
If a square matrix A be such that A
2
= I
then it is called as
(A) nilpotent matrix
(B) idempotent matrix
(C) involutary matrix
(D) none of these
4. If
1
,
2
, ……..
n
are the latent roots
of matrix A then A
3
has latent roots
(A)
1
…….
n
(B)
12 n
11 1
, , ........
(C)
1
3
,
2
3
, …….
n
3
(D)
33 3
12 n
11 1
, , ........
5.
If A =
52
31
, then A
1
is equal to
(A)
12
35
(B)
12
35
(C)
12
35
(D)
12
35
6.
If A =
2x 0
xx
and A
1
=
10
12
,
then value of x is
(A) 1 (B) 2
(C) 1/2 (D) none of these
Assignment on Linear Algebra
41
Q7 to Q18 carry two marks each
7.
A matrix ‘P’ which diagonalises ‘A’ is
called
(A) Spectral matrix (B) Modal matrix
(C) Square matrix (D) None of these
8. If A =
110
011
001
, then A
1
is equal to
(A) A
2
(B) A
2
3A + 3I
3
(C) A (D) A
2
+ 2A 2I
3
9. If the rank of the matrix
A =
11
11
is 1, then the value of
is
(A) 1 (B)
1
(C)
1 (D) None of these
10. Rank of a unit matrix of order n is
(A) 1 (B) 2
(C) 0 (D) n
11. For what value of , the system of
equations
3x y + z = 0
15x 6y + 5z = 0
x 2y + 2z = 0
has non zero solution.
(A) 1 (B) 2
(C) 6 (D) None of these
12.
The eigen value of A =
32
65
are
(A)
3 33 (B) 3, 1
(C)
1 2 7 (D) None of these
13. Given A =
cos sin 0
sin cos 0
001
, indicate
the statement which is not correct for A,
(A) It is orthogonal
(B) It is non singular
(C) It is singular
(D) A
1
exists
14. The eigen values and the
corresponding eigen vectors of a 2
2
matrix are given by
Eigen value Eigen vector
1
= 8 v
1
=
1
1
2
= 4 v
2
=
1
1
The matrix is
(A)
62
26
(B)
46
64
(C)
24
42
(D)
48
84
Vidyalankar : GATE – Engineering Mathematics
42
15. If A is orthogonal (A A
T
= I = A
T
A) then
| A | is
(A)
0
(B) 1
(C) 1 or
1
(D) can be any value
16. Let A =
20.1
03
and
1
1/ 2 a
A
0b
Then (a + b) =
(A) 7/20 (B) 3/20
(C) 19/60 (D) 11/20
17. For a second order matrix A if
2
A
= I
then A is equal to
(A)
i0
01
(B)
10
0i
(C)
i0
0i
(D)
10
0i
18.
If A = [1, 2, 3], B =
54 0
02 1
132
, then
AB =
(A)
2
1
4
(B)
214
(C)
412 (D) None of these
Assignment on Linear Algebra
43
Assignment 7
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
Eigen value of an inverse of matrix is
(A) Same as the matrix
(B) Negative of matrix values
(C) Inverse of the matrix values
(D) No any relation between them
2. Rank ‘n’ of nonzero matrix
(A) may be n = 0 (B) may be n > 1
(C) may be n = 1 (D) may be n
1
3. The rank of matrix
10
01
is
(A) 3 (B) 2
(C) 1 (D) 0
4. For what value of do the equations x
+ 2y = 1, 3x +
y = 3
have unique solution
(A)
= 6 (B) 6
(C)
= 5 (D) 5
5. The formula for A
1
is given by
(A) A
1
= Adj A | A |
(B) A
1
= Adj A | adj A |
(C) A
1
=
|
A
|
adjA
(D) None of these
6.
Find the eigen values of the matrix and
state that whether it is diagonal or not
A =
882
432
341
(A) Eigen value 1, 2, 3 & diagonal
(B) Eigen value 1, 2, 3 and
non diagonal
(C) Eigen value 1, 1, 3 and
non diagonal
(D) Eigen value 1, 1, 3 and diagonal
Q7 to Q18 carry two marks each
7.
The inverse of matrix
A =
213
111
111
is
(A)
11 2
01/2 1/2
11/23/2
(B)
10 1
11/21/2
21/23/2
(C)
11 2
10 1/2
11/2 3/2
(D)
11 2
1/ 2 1/ 2 1/ 2
3/2 1/2 3/2
Vidyalankar : GATE – Engineering Mathematics
44
8. The eigen vectors of the matrix
00
000
000
, 0 is (are)]
(i) (a, 0,
) (ii) (, 0, 0)
(iii) (0, 0, 1) (iv) (0,
, 0)
(A) (i), (ii) (B) (iii), (iv)
(C) (ii), (iv) (D) (i), (iii)
9. The rank of the following
(n + 1)
(n + 1) matrix, where a is a
real number is
2n
2n
2n
1a a . . . a
1a a . . . a
.. . .
.. . .
.. . .
1a a . . . a
(A) 1
(B) 2
(C) n
(D) Depends on the value of a
10.
The characteristic equation of matrix
A =
0hg
h0 f
gf0
is
(A)
3
(f
2
+ g
2
+ h
2
) 2fgh = 0
(B)
2
(f
2
+ g + h) 2fgh = 0
(C)
3
+ (f
2
+ g
2
+ h
2
) 2f
2
gh = 0
(D)
2
+ cf
2
+ g
2
+ h
2
) 2fgh = 0
11. The eigen values of a matrix
A =
54
12
are
(A) 5, 2 (B) 1, 6
(C) 4, 5 (D) 1, 5
12. The rank of matrix
1230
2432
3213
6875
is
(A) 1 (B) 2
(C) 3 (D) 4
13. The following set of equations
3x + 2y + z = 4
x y + z = 2
2x + 2z = 5 have
(A) No solution
(B) Unique solution
(C) Multiple solution
(D) An inconsistency
14.
The matrix form of quadratic equations
6x
1
2
+ 3x
2
2
+ 14 x
3
2
+ 4x
2
x
3
+ 18x
3
x
1
+
4x
1
x
2
…is
(A)
62 9
23 2
9214
(B)
6418
434
18 4 14
Assignment on Linear Algebra
45
(C)
62 9
23 2
92 14
(D)
64 18
434
18 4 14
15. If P and Q are both unitary matrices
then product PQ is
(A) non
unitary (B) unitary
(C) can’t say (D) none of these
16. If A =
12
21
, then adj A is
(A)
12
21
(B)
21
21
(C)
12
21
(D)
12
21
17. If
x1 y1
1y 3x
=
12
21
, then
(x, y) = . . . .
(A) (1, 0) (B) (1, 1)
(C) (0, 1) (D) (2, 1)
18.
If =
0ab
a0c
bc0
, then . . .
(A)
= 2abc (B) = 0
(C)
= abc (D) =
222
abc
Vidyalankar : GATE – Engineering Mathematics
46
Assignment 8
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
Matrix A and B are square matrices of
order n. Then B is said to be similar to
A if there exists a non
singular matrix
P such that
(A) B = PAP
1
(B) P AP
(C) B = P
1
A P (D) B = PAP
2. For m number of equations, n number
of columns, let r be rank of m n
coefficient matrix, then for r < n, the
solution of matrix is
(A) zero matrix
(B) unit matrix
(C) infinite number of solutions
(D) solution not possible
3. =
1bc a(b c)
1ca b(c a)
1ab c(a b)
is equal to
(A) 0 (B) 1
(C) a + b + c (D) ab + bc + cd
4. The multiplication of matrix is
(A) always commutative
(B) may be commutative
(C) not always commutative
(D) can’t say
5. If the matrix A has inverse, then which
is wrong
(A) A is non singular
(B) | A |
0
(C) A is any matrix
(D) It's inverse is unique
6. If
345
412
10 2
053
347
=
8x 3y 6z 32
4 12 26x 5y
, the
values of x, y, z are
(A) x = 3, y = 4, z = 1
(B) x = 0, y = 1, z = 4
(C) x = 1, y = 3, z = 4
(D) x = 1, y =
3, z = 4
Q7 to Q18 carry two marks each
7.
Find the rank of matrix
A =
1343
39129
1343
(A) 1 (B) 2 (C) 3 (D) 4
8. If the characteristic roots of matrix A
are
1
,
2
,
3
,
4
….
n
, then the
characteristic roots of A
2
are
(A) Squares of eigen values of A
(B) Square roots of eigen values of A
(C) Reciprocal of eigen values of A
(D) Does not have any relation.
9. Are following vectors linearly
dependent
x
1
= (1, 1, 1, 3) , x
2
= (1, 2, 3, 4)
x
3
= (2, 3, 4, 9)
Assignment on Linear Algebra
47
(A) Yes (B) No
(C) Can’t say (D) None of these
10. Consider the system of equations
x + 2y + z = 6
2x + y + 2z = 6
x + y + z = 5
This system has
(A) Unique solution
(B) Infinite no. of solution
(C) No solution
(D) exactly two solutions
11. The eigen value of a matrix
58 5
0712
0013
are
(A) 5, 8, 5 (B)
6, 7, 13
(C) 5, 7, 13 (D) 5, 12, 13
12. Determine the eigen values of the
matrix
A =
212i
12i 2
(A)
2, 2, (B) 1, 1
(C) 3,
3 (D) 1, 3
13. Find the rank of matrix
A =
2206
4202
1103
1212
(A) 1 (B) 2
(C) 3 (D) 4
14. The value of determinant of matrix
23
23
23
23
1a a a bcd
1b b b cda
1c c c dab
1d d d abc
is
(A) a
3
. b
3
.c
3
. d
3
abcd
(B) a
2
b
2
c
2
d
2
abcd
(C) 0
(D) 1
15. Eigen values of matrix
82918
0729
0018
are
(A)
8, 7, 29 (B) 8, 7, 18
(C) 18, 7, 8 (D) 29, 18, 29
16.
222
111
432
432
= . . . . .
(A) 2 (B)
2
(C) 1 (D) 0
17.
222
111
abc
abcbcacab
= . . . .
(A) 1 (B) 0
(C)
1 (D) None of these
18. The value of the determinant
12 3
0secx tanx
0tanxsecx
is . . . .
(A) 1 (B) 0
(C)
2
tan x (D)
2
sec x
Vidyalankar : GATE – Engineering Mathematics
48
Assignment 9
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
A square matrix U is called unitary if
(A) U = U
1
(B) U = U
1
(C) U =
U
1
(D) U = U
1
2. If A is Hermitian matrix, then IA is
(A) Symmetric
(B) Hermitian
(C) Skew
Hermitian
(D) None of these
3. If A A = I, then | A | is equal to
(A) +1 (B)
1
(C)
1 (D) 0
4. The matrix A and B are given below :
A =
11
11
, B =
10
10
. The value of
AB is
(A)
00
00
(B)
10
01
(C)
11
11
(D)
10
10
5. Inverse of a matrix
cos sin
sin cos
is
(A)
sin cos
cos sin
(B)
sin sin
cos cos
(C)
cos sin
sin cos
(D)
sin sin
sin cos
6. If A =
1053
21 61
3271
4420
, then value
of A + A
(A)
2287
228 3
88141
7310
(B)
2287
22 8 3
88141
7310
(C)
28 27
28 2 3
8148 1
7130
(D)
227 2
2232
14 8 1 8
070 3
Assignment on Linear Algebra
49
Q7 to Q18 carry two marks each
7.
A =
aic bid
bid aic
is unitary matrix if
and only if
(A) a
2
+ b
2
+ c
2
= 0
(B) b
2
+ c
2
+ d
2
= 0
(C) a
2
+ b
2
+ c
2
+ d
2
= 1
(D) a
2
+ b
2
+ c
2
+ d
2
= 0
8. The quadratic expression in matrix
form are
x
2
+ 4y
2
+ 9z
2
+ t
2
12yz + 6zx 4xy
2xt 6zt
(A)
123 1
24 60
3693
10 31
(B)
14 62
44120
612 9 6
206 1
(C)
12 31
2 460
36 93
1031
(D)
146 2
44 120
6129 6
20 61
9. The solution of the equation
x + 2y + 3z = 0
3x + 4y + 4z = 0
7x + 10y + 12z = 0
(A) x = y = z = 1
(B) x = 1, y = z = 0
(C) x = y = z = 0
(D) none of these
10. The matrix form of quadratic
expression
2x
1
2
+ x
2
2
3x
3
2
8x
2
x
3
4x
3
x
1
+ 12x
1
x
2
is
(A)
212 4
12 1 8
483
(B)
26 2
61 4
243
(C)
264
61 4
443
(D)
262
614
243
11. Find the rank of matrix
A =
11 1
234
823
(A) 1 (B) 2
(C) 3 (D) 4
12. Find eigen value of the matrix
43
12
(A) 1, 5 (B)
1, 5
(C)
1, 5 (D) 1, 5
Vidyalankar : GATE – Engineering Mathematics
50
13. If A =
31
12
, the value
A
2
5A + 7I
(A)
91
14
(B)
19
14
(C)
00
00
(D)
10
01
14. If A =
cos sin
sin cos
, then value of
A
n
is
(A)
ncos nsin
nsin ncos
(B)
cosn sinn
sinn cosn
(C)
cosn sinn
sinn cosn
(D)
cosn sinn
sinn cosn
15. Which of the following statement is
wrong ?
(A) (ABC)
1
= C
1
B
1
A
1
(B) (ABC)
T
= C
T
A
T
B
T
(C) A =
1
2
(A + A) +
1
2
(A A)
provided A is square matrix
(D) A = LU,
where A
square matrix,
L
lower triangular matrix
U
Upper triangular matrix
16.
pqr
T,T,T are the p
th
, q
th
& r
th
terms of
an A.P. then
pqr
TTT
pqr
111
equals
(A) 1 (B)
1
(C) 0 (D) p + q + r
17. If 1, ,
2
are three cube roots of unity,
then
2
2
2
1
1
1
is equal to
(A) 1 (B)
(C)
2
(D) zero
18. If 1, ,
2
are three cube roots of unity,
then
n2n
2n n
n2n
1
1
1
has the value....
(A) 1 (B)
2
(C)
(D) zero
Assignment on Linear Algebra
51
Assignment 10
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
Rank of matrix
32 9
6418
12 8 36
is
(A) 2 (B) 3
(C)
2 (D) 1
2. The value of the determinant for an
upper triangular matrix is equal to
(A) sum of elements along principal
diagonal
(B) product of the elements along
principal diagonal
(C) sum of all elements
(D) product of all elements
3. The skew matrix always have rank ‘r’
equal to
(A) r = 1 (B) r > 1
(C) r = 0 (D) can’t say
4. The quadratic from of the symmetrical
matrix
diag. [
1
,
2
, …
n
] is
(A)
1
+
2
+ ….
n
(B)
1
2
x
1
+
2
2
x
2 + …..
n
2
x
n
(C)
1
x
1
2
+
2
x
2
2
+ …. +
n
x
n
2
(D)
1
x
1
+
2
x
2 + ……
+
n
x
n
5. P =
312
231
12 1
= 8 then
Q =
132
32 1
211
= ?
(A) 8 (B)
8
(C) 16 (D)
16
6. The rank of matrix
A =
1230
2432
3213
6875
is
(A) 1 (B) 2
(C) 3 (D) 4
Q7 to Q18 carry two marks each
7.
The matrix
21
0
63
11 1
263
111
263
is
(A) orthogonal (B) not orthogonal
(C) singular (D) none of these
8. The eigen values of matrix
234
021
001
are
(A) 2, 2, 1 (B)
2, 2, 1
(C) 3, 4,
1 (D) 3, 4, 1
Vidyalankar : GATE – Engineering Mathematics
52
9. Write quadratic equation in matrix form
2x
1
2
+ x
2
2
3x
3
2
8x
2
x
3
4x
3
x
1
+
12x
1
x
2
(A)
26 2
61 4
243
(B)
212 4
12 1 8
483
(C)
262
614
243
(D)
2124
12 1 8
483
10. The characteristic roots of the matrix
A =
622
23 1
213
are
(A)
2, 0, 8 (B) 2, 2, 8
(C) 2, 2,
8 (D) 0, 2, 8
11. The rank of the matrix
14 8 7
00 3 0
4231
3122421
is
(A) 1 (B) 2
(C) 3 (D) 4
12.
The inverse of a matrix
524
211
410
is
(A)
14 6
1
41613
37
6131
(B)
14 6
1
41613
37
6131
(C)
14 6
1
41613
36
6131
(D)
146
1
41613
36
613 1
13. For A =
023
205
350
matrix which of
following statement is false
(A) A is square matrix
(B) A is skew symmetric matrix
(C) A is symmetric matrix
(D) | A | = 0
14. The rank of matrix A =
123
458
321
is
(A) 1 (B) 2
(C) 3 (D) 4
Assignment on Linear Algebra
53
15. The value of the determinant
12345
23451
12345
45123
51234
is
(A) 1, 00, 000 (B) 10,000
(C) 1,000 (D) 0
16. If a + b + c = 0, one root of the
equation
ax c b
cbxa
bacx
= 0
is . . . .
(A) x = 1 (B) x = 2
(C) x = 0 (D) x =
222
abc
17. The roots of the equation
ax b c
0bxa
00cx
= 0 are
(A) a & b (B) b & c
(C) a & c (D) a, b & c
18. If
1x 23
0x0
00x
= 0, then its roots
are . . . .
(A) 1 only (B) 0 only
(C) 0 & 1 (D) 0, 1 &
1
Vidyalankar : GATE – Engineering Mathematics
54
Assignment 11
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1. If
111 17
x1 y1 z0 15
10 1 13
,
then
x
y
z
= . . . . .
(A)
4
11
2
(B)
2
4
11
(C)
11
4
2
(D) None of these
2. If A & B are matrices such that A + B
and AB are both defined then :
(A) A & B can be any matrices
(B) A & B are square matrices, not
necessarily of the same order
(C) A & B are square matrices of the
same order
(D) No. of columns of A = no. of rows
of B.
3. If the matrix AB = 0, then we must
have
(A) Both A & B should be zero
(B) At least one of A & B should be
zero
(C) Neither A nor B may be zero
(D) A + B = 0
4. If A & B are arbitrary square matrices
of the same order, then
(A) (AB)
= AB
(B) (A
) (B) = B A
(C) (A + B)
= A B
(D) (AB)
= B A
5. Which of the following matrices is not
invertible
(A)
11
01
(B)
11
12
(C)
23
46
(D)
22
11
6. If A is a square matrix, then AA + AA
is a
(A) Unit matrix
(B) Null matrix
(C) Symmetric matrix
(D) Skew
symmetric matrix
Q7 to Q18 carry two marks each
7.
A = (a
ij
) is a 3 2 matrix, whose
elements are given by
a
ij
= 2i j if i > j
= 2j
i if i j
Then the matrix A is
(A)
10
32
11
(B)
13
32
11
Assignment on Linear Algebra
55
(C)
13
32
54
(D)
10
32
11
8. If X + Y =
70
25
& X Y =
30
03
,
then the matrices X & Y are
(A) X =
20
11
, Y =
50
14
(B) X =
50
11
, Y =
20
14
(C) X =
50
14
, Y =
20
11
(D) X =
20
14
, Y =
50
11
9. If I =
10
01
, J =
01
10
&
B =
cos sin
sin cos
then B equals
(A) I cos
+ J sin
(B) I sin
+ J cos
(C) I cos
J sin
(D)
I cos + J sin
10.
1
12 3
01 2
00 1
= . . . . .
(A)
12 5
00 2
00 1
(B)
127
01 2
00 1
(C)
100
231
210
(D) None of these
11. The value of
p000
aq00
bcr0
de f s
is . . .
(A) p + q + r + s (B) 1
(C) ab + cd + ef (D) pqrs
12. The determinant
abab
bcbc
abbc 0
= 0 if
(A) a, b, c are in A.P.
(B) a, b, c are in G.P. or (x
) is a
factor of
2
ax 2bx c
(C) a, b, c are in H.P.
(D)
is a root of
2
ax 2bx c
= 0
13.
If a
1
, a
2
, . . . . form a G.P. & a
i
> 0 for
all i 1, then
mm1m2
m3 m4 m5
m6 m7 m8
loga loga loga
loga loga loga
loga loga loga
is equal to
(A)
m8 m
loga loga
(B)
m8 m
loga log a
(C) zero
(D)
2
m4
log a
Vidyalankar : GATE – Engineering Mathematics
56
14. If , , are real numbers, then
1 cos( cos( )
cos( - ) 1 cos( )
cos( - ) cos( - ) 1
is
equal to
(A)
1
(B) cos
cos cos
(C) cos
+ cos + cos
(D) None of these
15. The characteristic equation of the
matrix A =
024
11 2
20 5
is . . .
(A)
32
x6x11x60
(B)
32
x6x11x60
(C)
32
x6x11x60
(D) None of these
16. The characteristic equation of the
matrix A =
3105
234
357
has a
repeated root.
The eigen vector corresponding to this
eigen value is . . .
(A)
3
2
3
(B)
4
2
4
(C)
5
2
5
(D) None of these
17. If
1
=
23
23
23
22 2
33 3
44 4
& =
12 4
13 9
1416
,
then
(A)
1
= 2 (B)
1
= 3
(C)
1
= 4 (D)
1
= 24
18. Let a, b, c, d u, v be integers. If the
system of equations ax + by = u,
cx + dy = v has a unique solution in
integers, then
(A) ad
bc = 1
(B) ad
bc = 1
(C) ad
bc = ±1
(D) ad
bc need not be equal to ±1
Test Paper on Linear Algebra
57
Test Paper 1
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1. A square matrix having determinant
equal to zero is called
(A) scalar matrix
(B) singular matrix
(C) nonsingular matrix
(D) skew symmetric matrix
2. The values of x, y, z and a which
satisfies the matrix equations
x32yx 0 7
z 1 4a 6 3 2a
are
(A) x = 3, y = 2, z = 4, a = 3
(B) x = 3, y = 2, z = 4, a = 1/3
(C) x = 3, y = 2, z = 4, a = 3
(D) x = 3, y = 2, z = 4, a = 1/3
3. Transpose of the product of two
matrices is
(A) product of matrices in reverse
order
(B) product of transposes of matrices
in same order
(C) product of transpose of matrices in
reverse order
(D) product of matrices in same order
4. Any matrix is said to be orthogonal if
its determinant is
(A) 0 (B) 1
(C) 2 (D) None of these
5. For the matrix
41
14
, the eigen
values are
(A) 3, 3 (B) 3, 5
(C) 3, 5 (D) 5, 0
Q6 to Q15 carry two marks each
6.
If A =
0tan/2
tan / 2 0
then
cos sin
sin cos
= . . . .
(A)
2
(I A) (B)
2
(I A)
(C)
2
IA
(D)
IA
IA
7. The relationship between following
vectors
x
1
= (1, 3, 4, 2), x
2
= (3, 5, 2, 2 ),
x
3
= (2, 1, 3, 2) is
(A) 2x
1
+ x
2
2x
3
= 0
(B) x
4
+ x
2
+ x
3
= 0
(C) x
1
+ x
2
2x
3
= 0
(D) x
1
2x
2
+ x
3
= 0
8. The points (x
1
, y
1
), (x
2
, y
2
), (x
3
, y
3
) are
collinear if and only if the rank of
matrix
11
22
33
xy1
xy1
xy1
is
Vidyalankar : GATE – Engineering Mathematics
58
(A) equal to 3
(B) less than 3
(C) equal to or less than 3
(D) can’t say
9. Consider the following equation
x + y + z = 9
2x +5y + 7z = 52
2x + y z = 0
(A) They have a unique solution
(B) They posses multiple solutions
(C) They have no solution
(D) The coefficient matrix is
orthogonal
10. If A =
0c b
c0 a
ba0
,
B =
2
2
2
aabac
ab b bc
ac bc c
Then AB is
(A)
2bc abc abc
abc abc ac
abc ac abc
(B)
2bc ab ac
ab 2ac ab
ac ab 2ba
(C)
000
000
000
(D)
0abac
ab 0 bc
ac bc 0
11. Given that f(x) = x
2
5x + 6,
A =
31
12
then f(A) = ?
(A)
10
01
(B)
10
01
(C)
10
01
(D)
10
01
12.
Is the given matrix
224
13 4
123
idempotent matrix ?
(A) Yes
(B) No
(C) Can’t say
(D) Cannot be determined
13.
Inverse of a matrix
cos sin 0
sin cos 0
001
(A)
cos sin 0
sin cos 0
001
(B)
cos sin 0
sin cos 0
001
(C)
cos sin 0
sin cos 0
001
(D)
sin cos 0
cos sin 0
001
Test Paper on Linear Algebra
59
14. The eigen values of
111
111
111
are
(A) 0, 0, 0 (B) 0, 0, 1
(C) 0, 0, 3 (D) 1, 1, 1
15.
The rank of matrix
8136
0322
8134
is
(A) 1 (B) 2
(C) 3 (D) 4
Vidyalankar : GATE – Engineering Mathematics
60
Test Paper 2
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
Every square matrix can be
represented as
(A) sum of lower and upper triangular
matrix
(B) sum of symmetric and skew
symmetric matrix
(C) product of symmetric and skew
symmetric matrix.
(D) product of Hermitian & skew
hermitian matrix.
2. Sum of eigen values of a matrix is
equal to
(A) sum of all elements
(B) sum of principal elements
(C) sum of diagonal elements
(D) none of these
3. If
x5 34 716
2
7y3 1 2 1514
then
(A) x = 2, y = 9 (B) x = 2, y = 5
(C) x = 4, y = 6 (D) None of these
4. If A =
234
125
and
B =
01 2
24 6
, then 2A + 3B will be
(A)
4912
8828
(B)
4912
8828
(C)
6912
8828
(D)
49 8
8828
5. Consider the following determinant
=
1a bc
1b ca
1c ab
which of the following
is a factor of ?
(A) a + b (B) a b
(C) a + b + c (D) abc
Q6 to Q15 carry two marks each
6.
Consider the following equations
2x + 6y + 11 = 0
6x + 20y 6z + 3 = 0
6y 18z + 1 = 0
(A) They have a unique solution
(B) No solution
(C) They have multiple solutions
(D) The coefficient matrix is idempotent
7. If A =
cos sin
sin cos
,
B =
cos sin
sin cos
. Then AB = ?
(A)
cos( ) sin( )
sin( ) cos( )
Test Paper on Linear Algebra
61
(B)
cos( ) sin( )
sin( ) cos( )
(C)
cos( ) sin( )
sin( ) cos( )
(D)
cos( ) sin( )
sin( ) cos( )
8. Given matrix
113
526
213
is a
nilpotent matrix of index . . .
(A) less than 3
(B) 3
(C) higher than 3
(D) can not be determined
9. The rank of the matrix
123
24 1
12 7
is
(A) 3 (B) 2
(C) 1 (D) 0
10. Which of following is correct ?
(A) (AB)
1
= A
1
B
1
(B) (AB)
t
= A
t
B
t
(C) (A
1
)
1
= A
t
(D) adj(AB) = (adj B) (adj A)
11. Given A =
cos sin 0
sin cos 0
001
. Indicate
the statement which is not correct for A.
(A) It is orthogonal
(B) It is non singular
(C) It is singular
(D) A
1
exists
12. If A =
13 5
02 1
00 3
, the eigen values
of the matrix I + A + A
2
are
(A) 1, 2, 3 (B) 3, 7, 13
(C) 1, 7, 13 (D) 1, 1, 1
13 The matrices
cos sin
sin cos
a0
and
0b
commute under
multiplication
(A) if a = b
2
, = n, in integer
(B) always
(C) if a = b
(D) never
14. The rank of the matrix given below is
14 8 7
00 3 0
4231
3122421
(A) 3 (B) 1
(C) 2 (D) 4
15. For a square matrix A, if A
2
= A then A
is called
(A) idempotent matrix
(B) nilpotent matrix
(C) involuntary matrix
(D) none of above
Vidyalankar : GATE – Engineering Mathematics
62
Test Paper 3
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
For triangular matrix eigen values are
(A) same as that of row elements
(B) same as of column elements
(C) same as of that principal diagonal
elements
(D) can not predict
2. Rank of a matrix A is said to be ‘r’ if
(A) atleast one square matrix of A of
order ‘r’ whose determinant is not
equal to zero.
(B) the matrix A contains any square
submatrix of order r + 1, then the
determinant of every square sub
matrix of order r + 1 should be
zero.
(C) both (A) and (B) true
(D) only (B) true
3. If A is m n matrix and B is n
matrix, then product AB is
(A) m n matrix
(B) n
matrix
(C) n n matrix
(D) product not possible
4. The equations are said to be
inconsistent if
(A) the equations have no solution
(B) the solution is zero vector
(C) there are infinite number of
solutions
(D) solution is possible is in the form
of complex number
5. Consider the following four possible
properties of a matrix A.
(i) It is square matrix
(ii) a
ij
= +a
ji
(iii) a
ij
= a
ji
(iv) All leading diagonal
elements are zero.
If A is skew-symmetric, which of the
following is True ?
(A) (i), (ii) (iv) (B) (iii) & (iv)
(C) (i) , (iii), (iv) (D) None of these
Q6 to Q15 carry two marks each
6.
If a
ij
= 1 for all i , j then rank (A) is
(A) 1 (B) 0
(C) n (D) anything
7. | Adj A | is equal to
(A) | A |
n
(B) | A |
n
1
(C) |A|
n+1
(D) |A
1
|
where n is the order of the matrix
8. A and B are idempotent then AB is
also idempotent if
(A) A = B
T
(B) A and B are nonsingular
(C) AB is symmetric
(D) AB = BA
Test Paper on Linear Algebra
63
9. The rank of A =
0000
4230
1000
4030
is
(A) 0 (B) 1
(C) 2 (D) 3
10. What are values of and so that the
equations have no solution
2x + 3y + 5z = 9
7x + 3y 2z = 8
2x + 3y + z =
(A) = 9, = 5 (B) = 5, = 9
(C) 9, = 5 (D) 9, 5
11. The rank of matrix
53144
01 21
1120
is
(A) 1 (B) 2
(C) 3 (D) 4
12. The characteristic equation of the
matrix A =
102
021
203
is
(A)
3
+ 6
2
7 2 = 0
(B)
3
6
2
+ 7 + 2 = 0
(C) 2
3
+
2
6 + 7 = 0
(D) 2
3
2
+ 6 7 = 0
13. Consider system of equations
2x + y + 2z =1
x + y = 0
x ky + 6z = 3 for the system to
have unique solution, the value of k is
(A) k = 2 (B) k 2
(C) k = 1 (D) k = 0
14. The adjoint of matrix
A =
123
211
452
is
(A)
786
11 14 13
555
(B)
745
5146
5118
(C)
7115
8145
6135
(D)
71113
59 11
214 20
15. Find the value of x if
x2 2x3 3x4
2x 3 3x 4 4x 5
3x 5 5x 8 10x 17
= 0
(A) 1, 1, 2 (B) 1, 1, 2
(C) 1, 2 (D) 1, 1
Vidyalankar : GATE – Engineering Mathematics
64
Test Paper 4
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
The matrix A is said to be skew
Hermitian if
(A) A =
A
(B) A =
A
(C) A = A
1
(D) A = Adj A
2. If is eigen value of a matrix and also
1/ is eigen value of same matrix then
that matrix is
(A) Skew Hermitian matrix
(B) Hermitian matrix
(C) Orthogonal matrix
(D) None of these
3. If matrix A has rank ‘r’, then A
(transpose of A) will have rank
(A) r + 1 (B) r 1
(C) r (D) can’t say
4. A square matrix A be such that A
K
= 0
then it is called
(A) Hermitian matrix
(B) idempotent matrix for index K
(C) nilpotent for index K
(D) singular matrix for index K
5. For x
1
2
18x
1
x
2
+ 5x
2
2
quadratic form,
matrix is
(A)
19
95
(B)
19
95
(C)
118
18 5
(D)
118
18 5
Q6 to Q15 carry two marks each
6.
Find the rank of matrix
A =
12 1
6126
5105
(A) 1 (B) 2
(C) 3 (D) 4
7. If A & B are 3 3 matrices then AB = 0
implies :
(A) Both A = 0 and B = 0
(B) | A | = 0 and | B | = 0
(C) Either | A | = 0 or | B | = 0
(D) Either A = 0 or B = 0
8. The characteristic equation of the
matrix
A =
211
12 1
112
is
(A) 6
3
+ 9
2
4 + 1 = 0
(B)
3
+ 6
2
9 4 = 0
(C)
3
6
2
+ 9 4 = 0
(D) 6
3
+ 9
2
+ 4 1 = 0
9. If I
n
is the identity matrix of order n,
then
1
n
I
(A) does not exist (B) = I
n
(C) = 0 (D) = nI
n
Test Paper on Linear Algebra
65
10. If A is the transpose of a square
matrix A, then
(A) | A | | A |
(B) | A | = |A |
(C) | A | + | A | = 0
(D) | A | = | A | only when A is
symmetric
11. Inverse of matrix
011
101
110
is
(A)
11 1
1
111
2
11 1
(B) 2
11 1
111
11 1
(C)
111
1
111
2
11 1
(D) 2
111
111
11 1
12. If a matrix is singular then the
characteristic root of the matrix is
(A) 0
(B) diagonal matrix
(C) can’t say
(D) not possible
13. Find the rank of matrix
1343
39129
13 4 1
(A) 1 (B) 2
(C) 3 (D) 4
14. If A =
36
24
, then A is
(A) Singular (B) Nonsingular
(C) I (D)
12
6
12
15. If A is 3 4 matrix and B is a matrix
such that both AB & BA are defined,
then B is a . . .
(A) 4 3 matrix (B) 3 4 matrix
(C) 3 3 matrix (D) 4 4 matrix
Vidyalankar : GATE – Engineering Mathematics
66
Test Paper 5
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
If for a square matrix A, A
t
= A, then
the matrix A is
(A) Symmetric
(B) Orthogonal
(C) Hermitian
(D) Skew symmetric
2. The values of ‘a’, & ‘b’ from the matrix
equation
2
2
2a 1 3b
a3b 2
02 0b3b
are
(A) a = 1, b = 1 (B) a = 2, b= 1
(C) a = 2, b = 2 (D) a = 1, b = 2
3. Find the index for which the matrix
2
2
ab b
aab
is nilpotent
(A) 3
(B) 2
(C) 1
(D) It is not nil potent
4. is characteristic root of matrix A if
and only if these exists a nonzero
vector X such that
(A) A
1
A = X (B) AX = X
(C) XA = X (D) A A
1
= XX
5. If the eigen values of an n n matrix
are all distinct then it is always similar
to a
(A) square matrix
(B) triangular matrix
(C) diagonal matrix
(D) none of these
Q6 to Q15 carry two marks each
6.
If A & B are square matrices of order 3
such that | A | = 1, | B | = 3 then
determinant of 3AB is equal to
(A) 9 (B) 27
(C) 81 (D) 81
7. Find the rank of matrix
A =
12 3
24 7
3610
(A) 1 (B) 2
(C) 3 (D) 4
8.
Eigen values of the matrix
A =
ahg
0b0
00c
are
(A) a, b, c (B) 0, a, b
(C) b, a, g (D) h, b, g
Test Paper on Linear Algebra
67
9. If following vectors are linearly
dependent then what is relation
between x
1,
x
2
, x
3
, x
4
?
x
1
= (1, 2, 4); x
2
= (2, 1, 3);
x
3
= (0, 1, 2); x
4
= (3, 7, 2)
(A) 9x
1
+ 12x
2
+ 15 x
3
+ x
4
= 0
(B) 9x
1
12x
2
+ 5x
3
5x
4
= 0
(C) 9x
1
+ 12x
2
+ 5x
3
+ 5x
4
= 0
(C) None of these
10. If A is a nonsingular matrix & A is the
transpose of A then
(A) | A . A | | A
2
|
(B) | A . A | | A |
2
(C) | A | + | A | = 0
(D) | A | = | A |
11. The inverse of the matrix
aib cid
cidaib
if a
2
+ b
2
+c
2
+ d
2
= 1
is
(A)
aibcid
cidaib
(B)
aib cid
cid aib
(C)
aibcid
cidaib
(D)
aib cid
cid aib
12. Is the given matrix
342
431
213
idempotent ?
(A) No
(B) Yes
(C) Can’t say
(D) Insufficient Data
13. The rank of the matrix
A =
11 1
025
02 5
is
(A) 1 (B) 2
(C) 3 (D) 4
14. The determinant of the matrix
6000
820 0
1440
168 1
is
(A) 11 (B) 24
(C) 48 (D) 0
15. If a b c, one value of x which
satisfies the equation
0xaxb
xa0xc
xbxc 0
= 0 is given by
(A) x = a (B) x = b
(C) x = c (D) x = 0
Vidyalankar : GATE – Engineering Mathematics
68
Test Paper 6
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
The value of the determinant of matrix
A =
ahgf
0bce
00dk
000
is
(A) abd
(B) abd
(C) abck abde (D) abd
abge
2.
If A =
012
123
234
, B =
12
10
21
, then
the product BA is
(A)
32
55
78
(B)
23
58
87
(C)
00
00
00
(D) does not exist.
3.
Let A =
1i 1 2 3i
2i2i 1i
3i 0 3 4i
, then the
matrix
A is given by
(A)
i1 123i
2i2i 1i
3i 0 3 4i
(B)
1i 1 2 3i
2i 2i 1i
3i 0 3 4i
(C)
1i 1 2 3i
2i2i 1i
3i 0 3 4i
(D)
1i 1 2 3i
2i 2i 1i
3i 0 3 4i
4. If A =
312
624
312
Then rank of matrix A
is
(A) 1 (B) 2
(C) 3 (D) 0
5. Eigen value of matrix
53
33
are
(A) 6, 4 (B) 5, 3
(C) 3, 5 (D) 4, 6
Q6 to Q15 carry two marks each
6.
The value of determinant
3111
1311
11 31
111 3
is
(A) 1 (B) 1
(C) 0 (D) 4
Test Paper on Linear Algebra
69
7. The rank of matrix
A =
23 1 1
1124
31 3 2
63 0 7
is
(A) 1 (B) 2
(C) 3 (D) 4
8. Adjoint of matrix A =
123
050
243
is
(A)
15 0 10
630
15 6 5
(B)
15 6 15
030
10 0 5
(C)
15 3 5
1060
0150
(D)
15 10 0
3615
500
9. The characteristic roots of the matrix
012
10 1
210
are
(A) 1, 2
3 (B) 3, 1 2
(C) 2, 1
3 (D) None of these
10.
Determine eigen values of matrix
A =
ahg
0b0
00c
are
(A) g, b, c (B) g, h, c
(C) a, b, g (D) b, a, c
11.
If A + B + C = , then the value of
sin(A B C) sinB cos C
sinB 0 tanA
cos(A B) tan A 0
is
equal to
(A) 0
(B) 1
(C) 2 sin B tan A cos C
(D) None of these
12. The rank of matrix A =
1212
1312
2434
3746
is
(A) 1 (B) 2
(C) 3 (D) 4
13. If
32
abcd
=
311
32 2
345
be an identity
in , where a, b, c, d are constants,
then the value of d is
(A) 5 (B) 6
(C) 9 (D) 0
Vidyalankar : GATE – Engineering Mathematics
70
14. If A =
34
11
20
& B =
212
134
then
AB = B A
(A) Yes
(B) No
(C) Can’t say
(D) Insufficient Data
15.
If the system of equations
x + ay + az = 0, bx + y + bz = 0,
cx + cy + z = 0 where a, b, c are
non zero & non unity has a non trivial
solution then the value of
abc
1a 1b 1c
is
(A) 0 (B) 1
(C) 1 (D)
222
abc
abc
71
Chapter - 2 : Calculus
2.1 Function of single variable
A real valued function y = f(x) of a real variable x is a mapping whose domain S and
co-domain R are sets of real numbers. The range of the function is the set
yf(x):xR
,
which is a subset of R.
2.2 Limit of a function
The function f is said to tend to the limit as x a, if for a given positive real number
> 0 we can find a real number > 0 such that
f(x) whenever 0 < |x a| <
Symbolically we write
xa
lim f(x)
Left Hand and Right Hand Limits
Let x < a and x a from the left hand side.
If |f(x)
1
| < , a < x < a or
1
xa
lim f(x)
then
1
is called the left hand limit.
Let x > a and x a from the right hand side.
If |f(x)
2
| < , a < x < a + or
2
xa
lim f(x)
then
2
is called the right hand limit.
If
12
then
xa
lim f(x)
exists. If the limit exists then it is unique.
Properties of Limits
Let f and g be two functions defined over S and let a be any point, not necessarily in S
And if
1
xa
lim f(x)
and
2
xa
lim g(x)
exist, then
1
xa xa
lim Cf(x) C lim f(x) C
C a real constant.
12
xa xa xa
lim f(x) g(x) lim f(x) lim g(x)
12
xa xa xa
lim f(x)g(x) lim f(x) lim g(x)
Vidyalankar : GATE – Engineering Mathematics
72
1
xa
xa
2
xa
lim f(x)
f(x)
lim
lim g(x)
g(x)
2
0
2
gx
1
xa
lim f x
Standard Formulae
x0
sin x
lim
x
= 1 …x in radians
x
x0
a1
lim
x
= na
(a > 0)
x0
tan x
lim
x
= 1 …x in radians
x
x0
e1
lim
x
= 1
x0
lim cos x
= 1
1
x
x0
lim 1 x)
= e
nn
n1
xa
xa
lim na
xa
x0
n(1 x)
lim 1
x
Solved Example 1 :
Show that
x0
1
lim sin
x
does not exist.
Solution :
For different values of x in the interval
0 < | x | < the function
1
sin
x
takes
values between 1 and 1.
Since
x0
1
lim sin
x
is not unique limit does
not exist.
Solved Example 2 :
Show that
2
x4
lim x 1
does not exist,
where
is the greatest integer function.
Solution :
Let h > 0, we have
2
h0
lim f(x h) 4 h 1
=
17 h h 8
= 17 if h (h + 8) < 1
2
(h 4) 17 h 17 4
and
2
h0
lim f(x h) (4 h) 1
=
17 h(h 8)
= 16 if h ( h 8) > 1
or
2
(h 4) 15 or h 4 15
x4 x4
limf(x) 17 and limf(x) 16
The limit does not exist.
Notes on Calculus
73
2.3 Continuity
Let f be a real valued function of the real variable x. Let x
0
be a point in the domain of f
and let f be defined in some neighbourhood of the point x
0
. The function f is said to be
continuous at x = x
0
if
i)
0
xx
lim f(x)
exists and
ii)
0
0
xx
lim f(x) f(x )
Types of discontinuity
A point at which f is not continuous is called a point of discontinuity.
If
0
xx
lim f(x)
exists, but
0
0
xx
lim f(x) f(x )
then x
0
is called the point of removable discontinuity.
In this case we can redefine f(x), such that f(x
0
) = , so that the new function is
continuous at x = x
0
.
For example :
The function, f(x) =
sinx / x ; x 0
4;x0
has a removable discontinuity at x = 0, since
x0
lim f(x)
and a new function can be
defined as
f(x) =
sinx / x ; x 0
1;x0
If
0
xx
lim f(x)
does not exist, then x
0
is called the point of irremovable discontinuity.
For Example :
f(x) =
1
x
at x = 0
Here
x0
lim f(x)
does not exist. Also f(0) is not defined.
f(x) =
1
x
has an irremovable discontinuity at x = 0.
Note : In case of an irremovable discontinuity, it does not matter whether or
how the function is defined of the point of discontinuity.
Vidyalankar : GATE – Engineering Mathematics
74
Properties of continuous functions
Let the functions f and g be continuous at a point x = x
0
Then,
i) cf, f g and fg are continuous at x = x
0
, where c is any constant.
ii) f/g is continuous at x = x
0
, if g(x
0
) 0
If f is continuous at x = x
0
and g is continuous at f(x
0
), then the composite function
g(f(g)) is continuous at x = x
0
.
A function f is continuous in a closed interval [a, b] if it is continuous at every point in
(a, b)
xa xb
lim f x f a and lim f x f b
If f is continuous at an interior point c of a closed interval [a, b] and f(c) 0, then there
exists a neighbourhood of c, throughout which f(x) has the same sign as f(c).
If f is continuous in a closed interval [a, b] then it is bounded there and attains its
bounds at least once in [a, b].
If f is continuous in a closed interval [a, b], and if f(a) f(b) are of opposite signs, then
there exists at least one point c [a, b] such that f(c) = 0.
If f is continuous in a closed interval [a, b] and f(a) f(b) then it assumes every value
between f(a) and f(b).
Solved Example 3 :
Determine the point of discontinuity of the
function f(x) =
x
x
Solution :
For x < 0,
x
x
= 1
For x > 0,
x
x
= +1
f(0) = 1
and f(+0) = +1
Also the function is not defined at x = 0
The function f(x) =
x
x
is discontinuous
at x = 0.
Note : |x| = x when x > 0 and |x| = x
when x < 0.
Solved Example 4 :
Find the limit if it exists, as x approaches
zero for the function f(x) given by
if x < 0, f(x) = x
x = 0, f(x) = 1
x > 0, f(x) = x
2
Solution :
If x < 0, f(0) =
x0 x0
lim f x lim x 0
Notes on Calculus
75
If x > 0, f(+0) =
2
x0 x0
lim f x lim x 0
Thus
f( 0) f( 0) each being zero.
Since the limit on the left = the limit on the
right at x = 0 the function f(x) has limit 0 at
x = 0.
Note : The limit at x = 0 is not equal to
f(0) = 1, and hence the function is
discontinuous at x = 0.
Solved Example 5 :
Let f(x) =
32
2
xx16x20
x2
kifx2
if x
2
Find k if f(x) is continuous at x = 2.
Solution :
For x 2 we have
f(x) =
32
2
xx16x20
x2
=
2
2
x2 x5
x2
= x + 5
x2 x2
lim f(x) lim(x 5) 7
We have f(2) = k
f(x) is continuous at x = 2.
Now
x2
lim f(x) f(2)
i.e. 7 = k
Solved Example 6 :
For the function f(x) =
x2
x
x2
find the
point of discontinuity and determine the
jump of the function at this point.
Solution :
x2
limf(x)2(1)3
for x < 2,
x2
|
x2
|
= 1
x2
lim f(x) 2 ( 1) 1
for x > 2,
x2
|
x2
|
= 1
The one sided limits do not coincide.
Thus the function is not defined at x =
2.
x = 2 is the only point of discontinuity.
The jump at this point =
1 (3) = 2
Solved Example 7 :
Show that f(x) =
2
x2 for x 2
x1for x2
is
discontinuous at x = 2 and determine the
jump of the function at this point.
Solution :
x2
lim (x 2) 4
2
x2
lim x 1 3
The limit on the left the limit on the right.
The function has discontinuity at the
point x = 2
The jump of the function at x = 2 is
f(2 + 0)
f(2 0) = 3 4 = 1
Note : If f(x
0
0) f(x
0
+ 0), then
f(x
0
+ 0) f(x
0
0) is called a jump
discontinuity of the function f(x) at x
0
.
Vidyalankar : GATE – Engineering Mathematics
76
2.4 Differentiability
Let a real valued function f(x) be defined on an I and let x
0
be a point in I. Then, if
0
000
xx x0
0
fx fx fx x fx
lim or lim
xx x
exists and is equal to
, then f(x) is said to be differentiable at x
0
and
is called the
derivative of f(x) at x = x
0
.
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be
differentiable in (a, b).
Properties of differentiation
Let the functions f and g be differentiable at a point x
0
. Then
i) (cf
) x
0
= cf(x
0
), c any constant.
ii) (f
g)x
0
= f(x
0
) g(x
0
)
iii) (fg)
(x
0
) = f(x
0
) g(x
0
) + f(x
0
) g(x
0
)
iv)
00 0 0
00
2
0
gx f x fx g x
f
x,gx0
g
gx
If f is differentiable at x
0
and g is differentiable at f(x
0
) then the composite function
h = g(f(x)) is differentiable at x
0
and
h
(x
0
) = g(f(x
0
)) f(x
0
)
If the function y = f(x) is represented in the parametric form as
x =
(t) and y = (t), and if (t), (t) exist, then
f
(x) =
t
dy / dt
t0
dx / dt t
Note : If a function is differentiable at x = x
0
, then it is continuous at x = x
0
.
However the converse need not be true.
Notes on Calculus
77
Standard formulae
Function Derivative Function Derivative
k(constant) 0
1
x
sin
a
22
1
ax
x
n
nx
n
1
1
x
cos
a
22
1
ax
logx 1/x
1
x
tan
a
22
a
ax
e
x
e
x
1
x
cot
a
22
a
ax
a
x
a
x
loga
1
x
sec
a
22
a
xx a
sin x cos x
1
x
cosec
a
22
a
xx a
cos x
sin x
sin h x cos h x
tan x sec
2
x cos h x sin h x
cot x
cosec
2
x sin h
1
x
2
1
x1
sec x sec x tan x
cos h
1
x
2
1
x1
cosec x
cosec x cot x tan h
1
x
2
1
x1
Solved Example 8 :
Show that the function
f(x) =
2
xcos(1/x), x 0
0x0
is differentiable at x = 0 but f
(x) is not
continuous at x = 0.
Solution :
We have
x0
lim f(x) 0 f(0)
f(x) is continuous at x = 0
Now f
(0) =
x0
f(x) f(0)
lim
x
x0
1
lim x cos 0
x
Hence f(x) is differentiable at
x = 0 and f
(0) = 0
for x
0 we have
Vidyalankar : GATE – Engineering Mathematics
78
f(x) =
2
2
111
2xcos x sin
xx
x
11
2xcos sin
xx
Now
x0
lim f x
does not exist as
x0
1
lim sin
x
does not exist. Therefore, f
(x) is not
continuous at x = 0.
Solved Example 9 :
3
yx
is defined and continuous for all x
investigate whether this function has a
derivative at x = 0.
Solution :
y =
33
xx x at x = 0
y =
3
x
3
2
3
yx1
xx
x
2
x0 x0
3
y1
lim lim
x
x
There is no finite derivative.
Solved Example 10 :
Investigate the function f(x) = | x | for
differentiability at the point x = 0
Solution :
y = f(x) = | x |
y = | x + x | | x |
At x = 0, we have y = | x |
x
y
xx
x0
y
lim 1
x
x0
y
lim 1
x
Since the right and the left side derivative
are not equal, the function f(x) = |x| is not
differentiable at the point zero.
Solved Example 11 :
Show that
nn1
d
xnx
dx
Solution :
Let y = x
n
Let x receive a small increment
x and let
the corresponding increment in y be
y
then
y +
y = (x + x)
n
By subtraction,
y = (x + x)
n
x
n
nn
nn
xx x xx x
y
xx xxx
n
n
n1
x0 x xx
xx x
y
lim lim nx
xxxx
n1 n n1
dy d
nx or x nx
dx dx
Notes on Calculus
79
2.5 Mean Value Theorems
Rolle's Theorem
The theorem states that
if f(x) is continuous in the closed interval [a, b] and
if f(x) exists in open interval (a, b) and
if f(a) = f(b)
Then there exists at least one value c in (a, b) such that f
(c) = 0
Geometric Interpretation
There exists at last one point at which slope of the tangent is 0 or the tangent is parallel
to x-axis
Lagrange's Mean Value Theorem
The theorem states that
if f(x) is continuous in the closed interval [a, b] and
if f(x) exists in open interval (a, b)
Then there exists one value c such that
f(c) =
f(b) f(a)
ba
a
c
b
x
y
f(a) = f(b)
y = f(x)
f(c) = 0
Vidyalankar : GATE – Engineering Mathematics
80
Geometric Interpretation
There exists at last one point at which the tangent is parallel to the secant through the
end points.
Cauchy's Mean Value Theorem
The theorem states that
if f(x) and g(x) are both continuous in closed interval [a, b] and
if f(x) and g(x) both exist in open interval (a, b)
Then there exists at least one value c such that
f(c) f(b) f(a)
g(c) g(b) g(a)
2.6 Maxima and Minima
A function f(x) is said to have a maximum value at x = a if f(a) is larger than any other
values of f(x) in the immediate neighbourhood of ‘a’. It has a minimum value if f(a) is less
than any other value of f(x) sufficiently near ‘a’.
Finding Maximum and Minimum values of y = f(x)
Get
dy
dx
and
2
2
dy
dx
. Solve
dy
dx
= 0 and consider its roots. These are the values of x
which make
dy
dx
= 0.
For each of these values of x, calculate the corresponding value of y and examine
the sign of
2
2
dy
dx
.
If the sign is ve the corresponding value of y is a maximum.
If the sign is +ve, the corresponding value of y is minimum.
a
c
b
f(a)
f(b)
f(c) =
f(b) f(a)
ba
Notes on Calculus
81
Note : Maximum and minimum values of a function occur alternately in a
continuous function. (The graph of a continuous function has no break
gaps)
A maximum is not necessarily the greater value and a minimum is not
necessarily the least value of the function.
Stationary Points and Turning Points
A stationary point on a graph is any point at which
dy
dx
= 0. A value of x for which
dy
dx
= 0
does not necessarily give either a maximum or minimum. A turning point is a maximum or
minimum point.
Solved Example 12 :
Find the maximum and minimum points on
the curve y = 2x
3
+ 5x
2
4x + 7
Solution :
Differentiating the equation we have
2
dy
6x 10x 4
dx
2
2
dy
12x 10
dx
Now
2
dy
0if6x 10x 4 0
dx
or
2
3x 5x 2 0
i.e. (3x 1) (x + 2) = 0
i.e.
dy 1
0, when x or 2
dx 3
for
2
2
1dy
x, 14
3
dx
and is positive
for
2
2
dy
x2, 14
dx
and is negative
Hence x =
1
3
gives a minimum value of y
and x = 2 gives a maximum value of y
for
18
x,y6;forx2,y19
327
18
,6
327
is a minimum point
(2, 19) is a maximum point.
Solved Example 13 :
Find the maximum and minimum values of
the function
2
2
x2x4
x2x4
Solution :
Let y =
2
2
x2x4
x2x4
Then
dy
dx
=
2
2
22
(x 2x 4)(2x 2)
(x 2x 4)(2x 2)
(x 2x 4)
dy
dx
= 0 when
2
x4
or x =
2
Vidyalankar : GATE – Engineering Mathematics
82
In this case it is much more convenient to
find the changes of sign of
dy
dx
than to
work out the value of
2
2
dy
dx
.
The denominator must be positive and
therefore only the sign of the numerator
need be examined.
Near x = 2
If x is slightly less than 2, x
2
< 4;
dy
dx
is ve
If x is slightly greater than 2, x
2
> 4;
dy
dx
is +ve
Here
dy
dx
changes from ve to +ve
Near x = 2
If x is slightly less than 2, x
2
> 4;
dy
dx
is +ve
If x is slightly greater than 2, x
2
< 4;
dy
dx
is ve
Here
dy
dx
changes from +ve to ve
x = 2 gives a minimum value of y,
1
y;
3
and x = 2 gives a maximum value of y,
y = 3
This the maximum value is 3 and the
minimum is
1
3
.
Note : If the
dy
dx
sign changes from
+ to
the point is a maximum point; if
the sign changes from
to +, the point
is a minimum point.
Solved Example 14 :
Find the maximum and minimum values of
the function 4 cos x 3 sin x.
Solution:
Let y = 4 cos x 3 sin x
Then
dy
dx
= 4 sin x 3 cos x
2
2
dy
dx
= 4 cos x + 3 sin x
dy
dx
= 0 when 4 sin x 3 cos x = 0
or tan x =
3
4
and then (i) sin x =
3
5
,
4
cos x
5
(ii) sin x =
3
,
5
4
cos x
5
tan
is negative.
In the first case,
2
2
dy
dx
=
16 9
5
55
and
is +ve
In the second case,
2
2
dy
dx
=
16 9
5
55
and is ve
The minimum value
=
43
43 5
55
Notes on Calculus
83
and the maximum value
=
43
43 5
55
Solved Example 15 :
Find the maximum and minimum values of
f(x, y) =
22
7x 8xy y
where x, y are
connected by the relation
22
xy1
.
Solution :
Since
22
xy1,
we can put
xcos, ysin
f(x, y) =
22
7cos 8cos sin sin = F()
F(
) =
1cos2 1cos2
74sin2
22
=
43cos2 4sin2
= 45sincos2 5cossin2
= 45sin(2 )
Since sine functions takes its maximum
and minimum value as +1 and 1
respectively.
max
F = 4 + 5 ;
min
F = 4 5
max
F = 9
min
F = 1
Solved Example 16 :
A rectangular sheet of a metal is 8 meters
by 3 meters; equal squares are cut out at
each of the corners and the flaps are then
folded upto form an open rectangular box.
Find its maximum volume.
Solution:
Let ABCD by rectangular sheet and let a
square of edge x m be cut from each
corner.
When the flaps are folded up, the
dimensions of the rectangular box
obtained are 8 2x, 3 2x and x meters.
Let V cubic meters be the volume of the
box then
V =
82x32xx
32
4x 22x 24x
;
2
dV
12x 44x 24
dx
2
2
dV
and 24x 44
dx
2
dV
0if4 3x 11x 6 0
dx
(3x 2) (x 2) = 0
i.e. when x = 2/3 or 3
for
2
2
2dv
x, 28and
3
dx
is negative
Hence
2
x
3
gives a maximum value of V
For x =
2
3
, V =
11
7
27
Thus the maximum volume is
11
7
27
cm.
x = 3 is inadmissible, since the breadth
itself of the sheet is 3m.
A
B
D
C
x
3m
8m
Vidyalankar : GATE – Engineering Mathematics
84
2.7 Integration
Integrand and element of integration
The function under the sign of integration is called integrand. For e.g. in
3
xdx
; x
3
is
called integrand. In the integral
f(x) dx
, dx is known as the element of integration and it
indicates the variable with respect to which the given function is to be integrated.
Constant of integration :
We know that
22
d
(x ) 2x 2xdx x
dx
Also
2
d
(x c) 2x
dx
2
2xdx x c
, where c is any constant
So we notice that x
2
is an integral of 2x, then x
2
+ c is also an integral of 2x. In general if
f(x)dx
= (x) then
f(x)dx
= (x) + c
Standard formulae
n1
n
x
xdx c
n1
, (n 1)
1
22
1x
sin c
a
ax
1
dx logx c
x
1
22
dx x
cos c
a
ax
xx
edx e c
1
22
dx 1 x
tan c
aa
ax
x
x
e
a
adx
log a
+ c
1
22
dx 1 x
cot c
aa
ax
cosxdx sinx c
1
22
dx 1 x
sec c
aa
xx a
cosecx cotxdx cos ecx c
1
22
dx 1 x
cos ec c
aa
xx a
sinxdx cosx c
sinhx dx
= cos h x + c
Notes on Calculus
85
secx tanx dx sec x c
coshx dx
= sin h x + c
2
sec x dx tanx c
2
dx
x1
= sin h
1
x + c
2
cosec xdx
= cot x + c
2
dx
x1
= cos h
1
x + c
cotxdx
= log sin x + c
2
dx
x1
= tan h
1
x + c
tan xdx
= log sec x + c tan xdx
= log (cos x) + c
secxdx log(secx tanx) c
secxdx
x
log tan c
42
cosec x dx log(cosecx cot x) c
cosec x dx
x
log tan
2
+ c
Important Trigonometric Identities
sin
2
A + cos
2
A =1
sin (A + B) = sin A cos B + cos A sin B
cos (A + B) = cos A cos B sin A sin B
tan (A + B) =
tanA tanB
1 tanA tanB
sin (A B) = sin A cos B cos A sin B
cos (A B) = cos A cos B + sin A sin B
tan (A B) =
tanA tanB
1 tanA tanB
sin
2
A sin
2
B = sin (A + B) sin (A B)
cos
2
A sin
2
B = cos (A + B) cos (A B)
2 sin A cos B = sin (A + B) + sin (A B)
2 cos A sin B = sin (A + B) sin (A B)
2 cos A cos B = cos (A + B) + cos (A B)
Vidyalankar : GATE – Engineering Mathematics
86
2 sin A sin B = cos (A B) cos (A + B)
2 sin
CD
2
cos
CD
2
= sin C + sin D
2 cos
CD
2
sin
CD
2
= sin C sin D
2 cos
CD
2
cos
CD
2
= cos C + cos D
2 sin
CD
2
sin
DC
2
= cos C cos D
cos 2A = cos
2
A sin
2
A = 1 2sin
2
A = 2 cos
2
A 1 =
2
2
1tanA
1tanA
sin 2A = 2 sin A cos A =
2
2tanA
1tanA
tan 2A =
2
2tanA
1tanA
sin 3A = 3 sin A 4 sin
3
A
cos 3A = 4 cos
3
A 3 cos A
tan 3A =
3
2
3tanA 4tan A
13tanA
Note : Integration of
mn
sin x cos xdx
where m and n are positive integers
(i) If m be odd and n be even, for integration put t = cos x
(ii) If m be even and n be odd, for integration put t = sin x
(iii) If m and n are odd, for integration put either t = cos x or sin x
(iv) If m and n are even, for integration put either t = cos x or sin x
Solved Example 17 :
Evaluate
3
sin x dx
Solution :
sin 3x = 3 sin x 4 sin
3
x
4 sin
3
x = 3 sin x sin 3x
4
3
sin x dx
=
(3sinx sin3x) dx
3sinxdx sin3xdx
3
1cos3x
sin x dx 3cosx c
43
Notes on Calculus
87
Solved Example 18 :
Evaluate sin3x cos 2x dx
Solution :
1
sin3x cos2x dx (sin5x sinx)dx
2
=
11
cos 5x cos x
10 2
Solved Example 19 :
Evaluate sin2x sin3x dx
Solution :
1
sin2x sin3x dx cos x cos 5x dx
2
=
11
sinx sin5x
210
Solved Example 20 :
Integrate
22
dx
cos xsin x
Solution :
Now,
22
22 22
1cosxsinx
cos x sin x cos x sin x
22
11
sin x cos x
= cosec
2
x + sec
2
x
[
1 = sin
2
x + cos
2
x]
22
dx
cos xsin x
=
22
(cosec x sec x)dx
=
22
cosec xdx sec xdx
= cot x + tan x + c
Solved Example 21 :
Evaluate
33
sin x cos x dx
Solution :
We have
sin
3
x cos
3
x = (sin x cos x)
3
=
1
8
(2 sin x cos x)
3
=
1
8
sin
3
2x
=
11
84
(3 sin 2x sin 6x)
[
4 sin
3
x = 3 sin x sin 3x]
=
1
32
(3 sin 2x sin 6x)
33
sin x cos xdx
1
(3sin2x sin6x)dx
32
=
31
sin2xdx sin6xdx
32 32
=
3 cos2x 1 cos6x
c
32 2 32 6
Hence
33
sin x cos x dx
31
cos2x cos 6x c
64 192
Solved Example 22 :
Evaluate
1
dx
1sinx
Solution :
111sinx
1 sinx 1 sinx 1 sinx
=
2
1sinx
1sinx
Vidyalankar : GATE – Engineering Mathematics
88
222
1sinx 1 sinx
cosx cosx cosx
or
1
1sinx
= sec
2
x tan x sec x
1
dx
1sinx
2
(sec x tanx secx)dx
=
2
sec x dx tanx sec x dx
tanx sec x c
Solved Example 23 :
Evaluate
76
sin x cos x dx
Solution :
76
sin x cos x dx
66
sin x cos x sinxdx
=
23 6
(1 cos x) cos xsinxdx
=
246
(1 3cos x 3cos x cos x)
6
cos x sinxdx
=
681012
(cos x 3cos x 3cos x cos x)
sinxdx
=
6 8 10 12
(t 3t 3t t )dt
by putting t = cos x
=
7 9 11 13
tttt
33
7 9 11 13
=
79 11
113
cos x cos x cos x
7311
13
1
cos x
13
Solved Example 24 :
Evaluate
63
sin x cos x dx
Solution :
63 6 2
sin x cos x dx sin x(1 sin x)cos xdx
=
68
(sin x sin x)cos x dx
=
68
(t t )dt
by putting
sin x = t =
79
11
tt
79
=
79
11
sin x sin x
79
Solved Example 25 :
Evaluate
42
sin x cos x dx
Solution :
sin
4
x cos
2
x
=
1
8
(2 sin
2
x)
2
(1+cos2x)
=
1
8
(1cos2x cos
2
2x + cos
3
2x)
=
1
8
1cos4x
1cos2x
2
cos 6x 3cos 2x
4
=
1
8
11 1 1
cos2x cos4x cos6x
24 2 4
=
11 1
1cos2xcos4x cos6x
16 2 2
42
sin x cos xdx
=
11 1
1 cos 2x cos4x cos 6x dx
16 2 2
=
11 1 1
xsin2xsin4x sin6x
16 4 4 12
Notes on Calculus
89
Solved Example 26 :
Evaluate
35
sin x cos x dx
Solution :
35
sin x cos x dx
=
34
sin x(cos x)cos x dx
=
322
sin x(1 sin x) cos x dx
=
32
t(1 t)dt
by putting sin x = t
=
357
(t 2t t ) dt
=
468
ttt
438
=
468
sin x sin x sin x
438
Integration by Substitution
Some of the general substitutions are
Type Substitution
a
2
x
2
x = a sin
or a cos
a
2
+ x
2
x = a tan
or a cot
or a sin h
x
2
a
2
x = a sec
or a cosec
or a cos h
By means of above substitution some standard integrals are
11
22
dx x x
sin or cos c
aa
ax
122
22
dx x
sinh or log(x x a ) c
a
ax
122
22
dx x
cosh or log(x x a ) c
a
xa
11
22
dx 1 x 1 x
tan or cot c
aaa a
ax
22 22 2 1
11x
axdx xax asin c
22a
22 22 2 1
11x
axdx xax asinh c
22a
Vidyalankar : GATE – Engineering Mathematics
90
22 22 2 1
11x
xadx xxa acosh c
22a
22
dx 1 a x
log c
2a a x
ax
22
dx 1 x a
log c
2a x a
xa
Solved Example 27 :
Integrate
cos x sinx
dx
1sin2x
Solution :
We know that
1 + sin 2x = sin
2
x + cos
2
x + 2 sin x cos x
= (sin x + cos x)
2
Let I =
cos x sinx
dx
1sin2x
=
2
cos x sinx
dx
(sinx cos x)
Let cos x + sin x = u
(cos x sin x) dx = du
Then I =
2
du 1 1
cc
ucosxsinx
u
cos x sinx 1
dx c
1 sin2x cos x sinx
Solved Example 28 :
Evaluate
1x
dx
1x
Solution :
Put x = sin or x = sin
2
dx = 2 sin cos d
I =
1sin
1sin
. 2 sin cos d
1sin
.2sin cos d
cos
(2sin cos2 1)d
= 2 cos +
sin2
2
+ c
= 2 cos + sin cos + c
= 2
22
1sin sin 1sin c
= 2
1
1x x1x sin x c
= (
x 2)
1
1x sin x c
Solved Example 29 :
Find the value of the integral
22 2 2
sinx cos x dx
asinx bcosx
Solution :
Put a
2
sin
2
x + b
2
cos
2
x = t
2(a
2
sin x cos x) dx + 2b
2
cos x (sin x) dx = dt
sin x cos x dx =
22
dt
2(a b )
22 2 2
sinx cos xdx
asinx bcosx
Notes on Calculus
91
22
dt
2(a b )t
=
22 22
1dt 1
logt c
t
2(a b ) 2(a b )
=
22
1
2(a b )
log (a
2
sin
2
x + b
2
cos
2
x) + c
Solved Example 30 :
Evaluate
44
d
sin cos
Solution :
I =
44
d
sin cos
=
4
4
sec
d
tan 1
Dividing numerator and denominator by
cos
4
I =
22
4
sec sec
d
1tan
22
4
(1 tan ) sec
d
1tan
Put tan = x so that sec
2
d = dx
I =
2
2
4
2
2
1
1
(1 x )dx
x
dx
1
1x
x
x
2
2
1
1
x
dx
1
x2
x
Put x
1
x
= t,
2
1
1
x
dx = dt
I =
1
22
dt 1 t
tan c
t(2) 2 2
=
1
1
x
tan c
x
2
2
=
2
1
1tan1
tan c
22tan
Solved Example 31 :
Integrate
2
cos x
dx
1sinx
Solution :
Let sin x = z cos x dx = dz
Now
2
cos x
dx
1sinx
=
2
dz
1z
= tan
1
z + c
= tan
1
(sin x) + c
Solved Example 32 :
Integrate
1
2
sin x
dx
1x
Solution :
Let x = sin dx = cos d
1
2
sin x
dx
1x
=
1
2
sin (sin )cos
d
1sin
=
2
dc
2
=
12
(sin x)
c
2
Vidyalankar : GATE – Engineering Mathematics
92
Integration by parts
Integral of the product of two functions
du
uvdx u vdx vdx dx
dx
Rule to choose the factor of differentiation or the 1
st
function
Of the two functions the first function u is selected based on the following preference
order
1. Inverse function
2. Logarithmic function
3. Algebraic function
4. Trigonometric function
5. Exponential function
Solved Example 33 :
Integrate
xcosxdx
Solution :
Here F. D. = x F. I. = cos x
Applying the rules of integration by parts.
xcosxdx
=
dx
xcosxdx cosxdxdx
dx
= x sin x 1.sinxdx
= x sin x + cos x + c
Solved Example 34 :
Integrate
1
sin x dx
Solution :
Here F. I. = 1 and F. D. = sin
1
x
Now
11
1.sin x sin x 1dx
1
d
sin x) 1.dx dx
dx
11
2
x
sin xdx x sin x dx
1x
Put 1 x
2
= z, 2x dx = dz
11
1dz
sin xdx x sin x
2
z
=
1
xsin x z
112
sin xdx x sin x 1 x
Solved Example 35 :
Find logxdx
Solution :
We have logxdx 1 logxdx
Integrating by parts,
d
logxdx logx 1dx logx 1dx dx
dx
=
1
xlogx xdx
x
=
xlogx x c x logx 1 c
Notes on Calculus
93
=
xlogx loge c
(
1 = log e)
Solved Example 36 :
Evaluate
ax
esinbxdx
Solution :
I =
ax
esinbxdx
=
ax ax
e e bcosbx
sinbx dx
aa
=
ax
ax
e sinbx b
e cosbxdx
aa
=
ax
esinbx
a
ax
ax
be b
cosbx e sinbxdx
aa a
=
ax
esinbx
a
2
ax ax
22
bb
e cosbx e sinbx dx
aa
I =
ax
2
22
easinbxbcosbx
b
I
aa
ax
2
22
e a sinbx bcosbx
b
I1
aa
I =
ax
22
easinbxbcosbx
ab
2.8 Definite Integration
The integral
b
a
f(x)dx
is called definite integral and is defined to be ba)
where (x)
is the indefinite integral of
f(x)dx
. It is read as integral from a to b of f(x), a is called the
lower limit and b is called the upper limit.
b
xb
xa
a
f(x)dx x b a
This is also called the theorem of calculus.
Properties of definite integral
bb
aa
f(x)dx f(t)dt
ba
ab
f(x)dx f(x)dx
bcb
aac
f(x)dx f(x)dx f(x)dx
Vidyalankar : GATE – Engineering Mathematics
94
aa
00
f(x)dx f(a x)dx
aa
a0
f(x)dx 2 f(x)dx
f(x) = f(x) i.e. it is an even function
= 0 f(x) = f(x) i.e. it is an odd function
2a a
00
f(x)dx 2 f(x)dx
f(2a x) = f(x)
= 0 f(2a x) = f(x)
Solved Example 37 :
Integrate
/2
0
cos xdx
Solution :
We know that cos xdx sinx c
/2
/2
0
0
cos xdx sinx sin sin 1
2
Solved Example 38 :
Integrate
1
2
0
dx
1x
Solution :
We know that
1
2
dx
tan x c
1x
1
1
1
2
0
0
dx
tan x c
1x
11
tan 1 c tan 0 c
=
0
44
Solved Example 39 :
Evaluate
b
a
1
dx
x
Solution :
We know that
dx
logx
x
b
b
a
a
dx
log x logb loga
x
=
b
log
a
Solved Example 40 :
Evaluate
/2
1/ 2
0
1cos2x dx
Solution :
Let I =
/2
1/ 2
0
1cos2x dx
=
/2
2
0
112sinxdx
=
/2 /2
00
2 sinxdx 2 sinx
=
/2
0
2cosx
=
2cos cos
2
21 0 2
Notes on Calculus
95
Solved Example 41 :
Evaluate the integral
/3
0
cos x
dx
34sinx
Solution :
Put sin x = y then cos x dx = dy
when x = 0, then y = 0 and
when x = /3 then
3
y
2
Let I =
/3 3/2
00
cosxdx dy
34sinx 34y
=
3
2
0
1
log 3 4y
4
=
13
log3 4 log3
42
=
1323
log
43
Improper Integrals
The integral
b
a
f(x)dx
is said to be an improper integral of the first kind, if one or both
the limits of integration are infinite.
The integral
b
a
f(x)dx
is said to be improper integral of second kind, if f(x) becomes
unbounded at one or more points in the interval of integration [a, b].
Limit comparison Test for Convergence and Divergence of Improper Integrals :
If f(x) and g(x) are positive functions and if
x
aa
fx
lim L, 0 L , then f x dx and g x dx
gx
both converge or both diverge.
Note : 1.
2
1
dx
1x
converges because
22
11
1x x
and
2
1
1
dx
x
converges.
2.
2x
1
1
dx
e
converges because
2x x
11
ee
and
x
1
1
dx
e
converges.
3.
1
dx
x
diverges because
11
x
x
for x > 1 and
1
1
dx
x
diverges
4.
2
1
11
dx
x
x
diverges because
2
111
xx
x
and
1
1
dx
x
diverges.
Vidyalankar : GATE – Engineering Mathematics
96
Solved Example 42 :
Solve
1
0
1x
dx
1x
Solution :
The integrand becomes infinite at x = 1.
The area under the curve y = f(x) =
1x
1x
between the lines x = 0 and x = 1 is not
well defined since the curve extends to
infinity as x tends to 1 from the left.
Nevertheless, we can define the area from
x = 0 to x = t where 0 < t < 1.
Then the integral
1t
t1
00
f(x)dx lim f(x)dx
We also say that the improper integral
converges.
tt
2
t1 t1
00
1x 1x
lim dx lim dx
1x
1x
=
t
12
t1
0
lim sin x 1 x
=
12
t1
lim sin t 1 t 1
=
1
sin 1 0 1 1
2
Solved Example 43 :
Solve
1
0
dx
x
.
Solution :
For some positive number
t < 1,
1
1
t
t
dx
nx
x
=
1
n1 n t n
t
1
t0 t0
t
dx 1
lim lim n
xt
The given integral diverges.
Solved Example 44 :
Solve
3
2/3
0
dx
x1
Solution :
The function f(x) =
2/3
1
x1
becomes
infinite at x = 1 which lies between the
limits of integration.
For the integral to exist, we must have the
convergence of the following integrals.
13
2/3 2/3
01
dx dx
and
x1 x1
It can be easily verified the convergence of
these two integrals are +3 and
3
32
respectively.
Hence the given improper integral
converges.
Solved Example 45 :
Compare
22
11
dx dx
and
x1x
with limit
comparison test.
Solution :
f(x) =
22
11
and g(x)
x1x
Notes on Calculus
97
2
2
2
xx
2
1
1x
x
lim lim
1
x
1x
=
2
x
1
lim 1
x
= 1 a positive finite limit.
Hence
2
1
dx
x
converges, therefore
2
1
dx
1x
converges.
Solved Example 46 :
Solve
x
1
3
dx
e5
Solution :
x
1
3
dx
e5
converges because
x
1
1
dx
e
converges
Now
x
xx
xx
3e 3
lim lim
1
e5 15e
= 3 a positive finite limit.
Solved Example 47 :
Solve
2x x
1
1
dx
e10e
Solution :
2x x
1
1
dx
e10e
converges because
2x
1
1
dx
e
converges and
2x x
2x
x
1
lim e 10e
e
=
x
x
10
lim 1 1
e
, a positive finite limit
Solved Example 48 :
Solve
1
1
dx
x
Solution :
1
1
dx
x
diverges because
1
1
dx
x
diverges
and
11
x
x
for x > 1
2.9 Double Integrals
The double integral of a function f(x, y) over a region D is denoted by
D
f x,y dxdy
.
Let f(x, y) be a continuous function defined on a closed rectangle
R = { (x, y) / a x b and c y d }
For continuous functions f(x, y) we have
d
bd b
Rac a
c
f x,y dxdy f x,y dy dx f x,y dx dy
Vidyalankar : GATE – Engineering Mathematics
98
Note : 1.
S
dxdy
represent the area of the region S.
2. In an iterated integral the limits in the first integral are constants and
if the limits in the second integral are functions of x then we must
first integrate w.r.t. y and the integrand will become a function of x
alone. This is integrated w.r.t. x.
3. If R cannot be written in neither of the above two forms we divide R
into finite number of subregions such that each of the subregions
can be represented in one of the above forms and we get the double
integral over R by adding the integrals over these subregions.
Solved Example 49 :
Evaluate I =
12
00
x 2 dydx
Solution :
1
2
0
0
xy 2y dx
=
1
0
2x 4 dx
=
1
2
1
0
0
2x
4x
2
= 1 + 4 = 5
Solved Example 50 :
Evaluate
11
22
0x
xydydx
Solution :
11
22
0x
xydydx
=
11
22
0x
x y dy dx
=
1
1
3
2
0
x
y
xy dx
3
=
1
3
23
0
1x
xxdx
33
=
1
3
2
0
14x
xdx
33
=
1
34
0
xxx
333
=
1111
333 3
Solved Example 51 :
Evaluate
32
21
dxdy
xy
Solution :
32
21
dxdy
xy
=
32
21
dy dx
yx
=
3
2
1
2
dy
log x
y
=
3
2
dy
log2 log1
y
=
3
2
log 2 log y
Notes on Calculus
99
=
log2 log3 log2
= log 2
3
log
2
Solved Example 52 :
Evaluate the double integral
2
x
R
edxdy
where the region R is given by
R : 2y x 2 and 0 y 1.
Solution :
Integrating with respect to y first, we get
I =
2
2x/2
x
00
edydx
=
2
2
x/2
x
0
0
ye dx
=
22
2
2
xx
0
0
11
xe dy e
24
=
4
1
e1
4
Solved Example 53 :
Evaluate the integral
2
y
2
2
22
00
y
dxdy
xy1
Solution :
It would be easier to integrate it first with
respect to y.
The region of integration 0 y 2 and 0
x
2
y
2
can also be written as
0 x 2 and
2x y 2
I =
22
22
0
2x
y
dy dx
xy1
=
2
2
22
2x
0
xy1 dx
=
2
2
0
x5x1dx
=
2
2
xx 5 5
log x x 5
22
2
2
0
1
x1
2
=
51
3 log5 log 5 9 1
22
=
5
log5 1
4
Solved Example 54 :
The cylinder
22
xz1
is cut by the
planes y = 0, z = 0 and x = y. Find the
volume of the region in the first octant.
Solution :
In the first octant we have
2
z1x. The
projection of the surface in the x y plane
is bounded by x = 0, x = 1, y = 0 and y = x.
V =
1x
2
R00
zdxdy 1 x dy dx
=
1
x
2
0
0
1x ydx
0
x
y
y = 2
(2, 2)
y
2
= 2x
Vidyalankar : GATE – Engineering Mathematics
100
=
1
2
0
x1 xdx
=
1
3/2
2
0
11
1x
33
cubic units.
Solved Example 55 :
Evaluate I =
acos
00
rsin drd
Solution :
I =
acos
2
0
0
r
sin d
2
=
22
0
1
acos sind
2
=
2
2
0
a
cos d cos
2
[d(cos ) = sin d]
=
2
3
0
a
cos
6
=
2
a
3
Solved Example 56 :
Evaluate I =
D
xydydx
where D is the
region bounded by the curve x = y
2
,
x = 2 y, y = 0 and y = 1.
Solution :
In this region x varies from 0 to 2 when
0 x 1 for fixed x, y varies from 0 to
x .
When 1 x 2, y varies from 0 to 2 x.
The region D can be subdivided into
2 regions D
1
and D
2
as shown.
12
DDD
xydydx xydydx xydydx
In the region D
1
for fixed x, y varies from
y = 0 to y =
x and for fixed y, x varies
from x = 0 to x = 1.
Similarly for D
2
the limit of integration for
y is y = 0 to y = 2 x
D
xydydx
1x 22x
00 1 0
xydydx xydydx
=
x2x
12
22
01
00
xy xy
dx dx
22
=
12
2
2
01
11
xdx x2 x dx
22
=
12
343
2
01
x1 x4x
2x
62 43
=
8
11 16 1 14
24 4 2
62 4 3 2 43
=
1411 9
6 6 24 24
Solved Example 57 :
Change the order of integration in the
integral I =
y
4
1y/2
f x,y dxdy
(1, 0)
(2, 0)
(1, 1)
y
2
= x
D
1
D
2
Notes on Calculus
101
Solution :
The region of integration D is bounded by
x = y/2
x = y
y = 1
y = 4
Here, x varies from 0.5 to 4
when
1
x2
2
, y varies from 1 to 2x
when 1 x 2, y varies from x to 2x
when 2 x 4, y varies from x to 4
For changing the order of integration we
must divide D into sub regions D
1
, D
2
, D
3
.
I =
D
f x,y dxdy
=
12
DD
f x,y dxdy f x,y dxdy
3
D
f x,y dxdy
=
12x 22x
1/ 2 1 1 x
f x,y dydx f x,y dydx
44
2x
fx,ydydx
Solved Example 58 :
Evaluate
aa
22
0y
xdxdy
xy
by changing the order of integration.
Solution :
Region R of integration
is bounded by
x = y
x = a
y = 0
y = a
We have to change the order of integration
as dy dx.
Here in the region R, x varies from 0 to a
and for fixed x, y varies from 0 to x.
aa
22
0y
xdxdy
xy
=
ax
22
00
x
dydx
xy
=
x
a
1
0
0
1y
xtan dx
xx
=
x
a
1
0
0
y
tan dx
x
=
a
11
0
tan 1 tan 0 dx
=
a
a
0
0
a
dx x
44 4
(0, 0) (a, 0)
(a, a)
y = a
x = a
y = x
R
y = 4
(0, 0)
(1, 1)
(2, 2)
(4, 4)
(2, 4)
(0.5, 1)
y = x
y = 1
y = 2x
D
1
D
2
D
3
Vidyalankar : GATE – Engineering Mathematics
102
Solved Example 59 :
Evaluate
42
22
1
y
x y dxdy
by changing
the order of integration.
Solution :
The region of integration is bounded by
y = 1
y = 4
x
2
= y
x = 2
We change the order of integration as dy
dx. In the region x varies from 1 to 2 and
for fixed x, y varies from 1 to x
2
.
42
22
1
y
x y dxdy
=
2
2x
22
11
x y dydx
=
2
x
2
3
2
1
1
y
xy dx
3
=
2
6
42
1
x1
xxdx
33
=
2
573
1
xxxx
52133
=
3212882 1 1 11
52133 52133
=
1006
105
Solved Example 60 :
Evaluate
4y
3
01
x y dxdy
by changing
the order of integration.
Solution :
The region is bounded by
y = 0
y = 3
x = 1
x
2
= 4 y
We have to change the order of integration
as dydx.
In this region x varies from 1 to 2
and y varies from 0 to 4 x
2
4y
3
01
x y dxdy
=
2
24 x
10
x y dydx
y = 4
(0, 0)
(1, 1)
(2, 4)
x = 2
y = 1
y = x
2
(1, 3)
(0, 0)
(1, 0)
(2, 0)
x = 1
y = 3
y = 4 x
2
Notes on Calculus
103
=
2
4x
2
2
1
0
y
xy dx
2
=
2
2
2
2
1
4x
x4 x dx
2
=
2
4
32
1
x
x4x4x8dx
2
=
2
54 3
2
1
xx4x
2x 8x
10 4 3
=
241
60
Solved Example 61 :
Evaluate
2
a2a x
0
x/a
xydydx
by changing the
order of integration
Solution :
The region R is bounded by
x = 0
x = a
y = 2a x
y =
2
x
a
We divide the region into two subregions
R
1
and R
2
12
RRR
xydydx xydydx xydydx
We have to change the order of integration
as dxdy in R
1
and R
2
.
In region R
1
, y varies from a to 2a for a
fixed y, x varies from 0 to 2a y
In the region R
2
, y varies from 0 to a and
for a fixed y, x varies from 0 to
ay
.
2
a2a x
0
x/a
xydydx
=
12
RR
xydydx xydydx
=
ay
2a y
2a a
a0 00
xydxdy xydxdy
=
2a y ay
2a a
22
a0
00
xy xy
dy dy
22
=
2a a
2
2
a0
yay
2a y dy dy
22
=
2a a
3
222
a0
ya
2a y 2ay dy y dy
22
=
2a a
43 3
22
a0
y2ay y
ay a
83 6
=
44
5a a
24 6
=
4
9a
24
Solved Example 62 :
Evaluate
2
11x
00
ydydx
by interchanging the
order of integration.
R
1
(0, 2a)
(0, a)
(0, 0) (a, 0)
(2a, 0)
(a, a)
R
2
x
2
= ay
y = 2a
x
Vidyalankar : GATE – Engineering Mathematics
104
Solution :
The region is bounded by
y = 0 (x axis)
y =
222
1x orx y 1
x = 0 (y axis)
x = 1
Here y varies from 0 to 1 and for fixed y
x varies from 0 to
2
1y
2
11x
2
00
ydydx
=
2
1y
1
2
00
ydxdy
=
2
1
1y
2
0
0
yx dy
=
1
22
0
y1ydy
=
/2
2
0
sin cos d
Putting y = sin
when y = 0 = 0
y = 1 = /2
again put t = sin
dt = cos d
When = 0 t = 0
= /2 t = 1
1
1
3
2
0
0
t
tdt
3
2
11x
2
00
1
ydydx
3
2.10 Triple Integrals
A triple integral of a function defined over a region R is denoted by
R
f x,y,z dxdydz or f x,y,z dV
or
R
fx,y,zdx,y,z
Evaluation of Triple Integrals :
If the region R can be described by
121 21 2
xxx,yxyyx,zx,yzzx,y
then we evaluate triple integral as
22
2
11 1
yxzx,y
x
xyxzx,y
f x,y,z dzdydx
=
22
2
11 1
yx zx,y
x
xyxzx,y
f x,y,z dz dy dx
(0, 1)
(1, 0)
(0, 0)
x = 0
y = 0
x = 1
x
2
+ y
2
= 1
Notes on Calculus
105
Note : There are six possible ways in which a triple integral can be evaluated
(order of variables of integration). We choose the one which is simple to
use.
Solved Example 63 :
Evaluate
232
2
011
xy zdzdydx
Solution :
2
232 23
22
2
011 01
1
xy z
xy zdzdydx dydx
2
…(Integrating w.r.t. z keeping x, y constants)
=
23
2
2
01
xy
2xy dydx
2
=
23
2
01
3
xy dydx
2
=
3
2
3
0
1
3xy
dx
23
…(integrating w.r.t. y keeping x constant)
=
2
0
1
27x x dx
2
=
2
0
13 xdx
=
2
2
0
x
13 26
2
Solved Example 64 :
Evaluate I =
y
ax
000
xyzdzdydx
Solution :
I =
y
ax
2
00
0
xyz
dydx
2
=
ax
3
00
1
xy dydx
2
=
x
a
4
0
0
1xy
dx
24
=
a
a
66
5
0
0
11xa
xdx
88648
Solved Example 65 :
Evaluate : I =
/2 1
2
000
rsindrdd
Solution :
I =
1
/2
3
00
0
r
sin d d
3
=
/2
00
1
sin d d
3
=
/2
0
0
1
cos d
3
=
0
0
11
d
333
Solved Example 66 :
Evaluate I =
loga x y
x
xyz
000
e dzdydr
Solution :
I =
loga
x
xy
xyz
0
00
edydx
=
loga
x
2x y
xy
00
e e dydx
=
x
loga
2x y
xy
0
0
1
eedx
2
Vidyalankar : GATE – Engineering Mathematics
106
=
loga
4x 2x x
0
13
eeedx
22
=
loga
4x 2x x
0
13
eee
84
=
42
13 3
aaa
84 8
Solved Example 67 :
Find the volume of the solid in the first
octant bounded by the paraboloid.
22
z364x 9y
Solution :
We have I =
R
dzdydx
The projection of the paraboloid (in the first
octant) in the x y plane is the region in
the first quadrant of the ellipse
22
4x 9y 36
The region R is
22
0 z 36 4x 9y
2
1
0y 364x
3
0x3
I =
2
x
29
3
3
22
00
36 4x 9y dy dx
=
2
x
3
29
23
3
0
0
49 x y 3y dx
=
3
3/2 3/2
22
0
88
9x 9x dx
39
=
3
3/2
2
0
16
9x dx
9
Substituting x = 3 sin we get
I =
/2
3
0
16
27 cos 3 cos d
9
=
/2
4
0
144 cos d
=
31
144 27
422
cubic units
2.11 Change of Variables in Double and Triple Integrals
and Jacobians
Consider the transformation given by the equation
x = x(u, v, w) ;
y = y(u, v, w) ;
z = z(u, v, w)
Where the functions x, y, z have continuous first order partial derivatives.
Notes on Calculus
107
The Jacobian J of the transformation is defined by
J =
xxx
uvw
yyy
uvw
zzz
uvw
The Jacobian is also denoted by J =
x, y,z
u, v, w
For a transformation in two variables x = x(u, v) and y = y(u, v) the Jacobian is given by a
determinant of order two. Hence J =
x, y
u, v
Solved Example 68 :
The transformation from cartesian
coordinates (x, y) to polar coordinates (r, )
is given by
x = r cos y = r sin
Solution :
J =
xx
cos r sin
r
r
y y sin r cos
r
and
RR
f x,y dxdy f r cos ,r sin rdrd
Solved Example 69 :
The transformation from Cartesian
coordinates (x, y, z) to spherical polar
coordinates
(r, , ) is given by
x = r sin cos
y = r sin sin
z = r cos
Solution :
Here 0 and 0 2
J =
x, y, z
r, ,
sin cos r cos cos r sin sin
sin sin r cos sin r sin cos
cos r sin 0
= r
2
sin
and
R
f x,y,z dxdydz
R
frsincos,rsinsin,rsin
2
rsin drdd
Solved Example 70 :
The transformation from Cartesian
coordinates (x, y, z) to cylindrical
coordinates
(r, , z) is :
x = r cos , y = sin , z = z
Vidyalankar : GATE – Engineering Mathematics
108
Solution :
J =
cos r sin 0
x, y,z
sin r cos 0
r, ,z
001
= r
and
R
f x,y,z dxdydz
R
frcos,rsin,zrdrd dz
Solved Example 71 :
If u = x
2
y
2
and v = 2xy
Prove that
xx
x, y
uv
yy
u, v
uv
=
2x 2y
2y 2x
=
22
4x y
Solution :
We have
2
22 22
xy xy 2xy
22
uv
22 22
xy uv
22
u, v
4u v
x,y
22
u, v
1
x,y
4u v
Solved Example 72 :
Evaluate the integral
2
2
R
x y cos x y dxdy
, where R is the
parallelogram with successive vertices at
(, 0), (2, ), (, 2) and (0, )
Solution :
The equations are
AB x y =
BC x + y = 3
CD x y =
DA x + y =
Substitute y x = u and y + x = v
Then u and v 3
x =
vu vu
and y
22
and I =
xx
x, y
uv
yy
u, v
uv
=
11
1
22
11
2
22
| J | =
1
2
I =
2
2
R
x y cos x y dxdy
3
22
1
u cos vdudv
2
=
3
3
2
cos vdv
3
=
3
34
1cos2vdv
63
x
y =
A(
,0) B(2,)
C(, 2)
D(0, )
x + y = 3
x + y =
x y =
Notes on Calculus
109
Solved Example 73 :
Evaluate I =
22
R
x y dxdy
by changing
to polar coordinates, where R is the
region in the x y plane bounded by the
circles x
2
+ y
2
= 4 and x
2
+ y
2
= 9.
Solution :
Using x = r cos
y = r sin
We get dx dy = r dr d and
I =
3
23 2
3
02 0
2
r
rrdrd d
3
=
2
0
19
d
3
=
38
3
Solved Example 74 :
Evaluate the improper integral
I =
2
x
0
edx
Solution :
I
2
= I I
=
22
22
xy
xy
00 00
edx edy e dxdy
Put x = r cos
y = r sin
Hence J = r
The region of integration is the entire first
quadrant. Hence r varies from 0 to and
varies from 0 to /2.
I
2
=
2
/2
r
00
erddr
=
2
r
0
erdr
2
=
2
r2
0
1
edr
22
=
2
r
0
11
e
22 22 4
I =
2
Solved Example 75 :
Evaluate the integral
T
zdxdydz
, where T is
the hemisphere of radius a, x
2
+ y
2
+ z
2
= a
2
,
z 0.
Solution :
Changing to spherical coordinates
substitute
x r sin cos , y r sin sin , z r cos
0 2 and 0 /2
We get,
dx dy dz = r
2
sin dr d d
I =
2/2a
2
00
r cos r sin dr d d
=
2/4
4
00
a
sin cos d d
4
=
2/2
4
00
a
sin2 d d
8
=
/2
2
4
0
0
acos2
d
82
=
2
4
0
a
d
8
=
4
a
4
Vidyalankar : GATE – Engineering Mathematics
110
Solved Example 76 :
Using triple integral find the volume of the
sphere x
2
+ y
2
+ z
2
= a
2
Solution :
V =
D
8 dxdydz
where D is the region
bounded by the sphere x
2
+ y
2
+ z
2
= a
2
in
the first quadrant.
Substitute x = r sin cos , y = r sin sin
, z = r cos
Jacobian | J | = r
2
sin
and the limits r = 0 to a; 0 /2
0 /2
V =
a/2/2
00 0
8Jdddr
=
a/2/2
2
00 0
8rsindddr
=
a/2
/2
2
0
00
8rcos ddr
=
a/2
2
00
8rd dr
=
a
/2
2
0
0
8r dr
=
a
2
0
4rdr
=
a
3
0
r
4
3
=
3
4a
3
Solved Example 77 :
Find by triple integral the volume of the
tetrahedron bounded by the planes x = 0,
y = 0, z = 0 and
xyz
1
abc
Solution :
The projection of the given region on the
x y plane is the triangle bounded by the
lines
x = 0;
y = 0 and
xy
1
ab
0 x a
In this region x varies from 0 to a.
For fixed x, y varies from 0 to
x
1
a
b.
i.e.
x
0y 1 b
a
For fixed (x, y), z varies from 0 to
xy
1c
ab
xy
0z 1 c
ab
I =
D
dxdydz
=
xxx
1b1 c
aab
a
00 0
dzdydx
=
x
1b
a
a
00
xy
c 1 dydx
ab
=
x
1b
a
2
a
0
0
xy
c1 y dx
a2b
=
22
a
0
x1x
bc 1 1 dx
a2a
=
2
a
0
bc x
1dx
2a
=
a
3
0
bc a x
1
23 a
=
abc
6
Notes on Calculus
111
Solved Example 78 :
Evaluate I =
R
xyzdxdydz
where D is the
positive octant of the ellipsoid
222
222
xyz
1
abc
Solution :
Substitute x = au,
y = bv
z = cw
J =
a00
x, y,z
0 b 0 abc
u, v, w
00c
Let R be the positive octant of the sphere
u
2
+ v
2
+ w
2
= 1
I =
R
abc uvw abc dudvdw
=
222
R
a b c uvwdudvdw
Now substitute
u = r sin cos
v = r sin sin
w = r cos
Then J = r
2
sin
1/2/2
222 5 3
00 0
I a b c r sin cos cos sin d d dr
1/2 /2
222 5 3
00 0
a b c r dr sin cos d sin cos d
=
1/2/2
222 6 4 2
00 0
11 1
abc r sin sin
64 2
=
222
abc
48
2.12 Application of Integration
Solved Example 79 :
Calculate the area bounded by the
parabola y
2
= 4ax and its latus rectum.
Solution :
The latus rectum of the parabola y
2
= 4ax
is a line parallel to y axis. The shaded
portion is the desired area. As the
parabola is symmetric about the xaxis,
therefore area
OLL = 2 Area OSL.
=
aa a
1/ 2
00 0
2 ydx 2 4axdx 4 a x dx
=
a
3/2
3/2
x0
x8
4a aa 0
3/2 3
=
2
8a
3
0
y
x
L
S
(a, 0)
Vidyalankar : GATE – Engineering Mathematics
112
Solved Example 80 :
Find the area of the region bounded by the
graph of y = sin x and y = cos x between
x = 0 and x = /4.
Solution :
Required area =
/4
0
cos x sinx dx
=
/4
0
sinx cos x
=
22
01
22
=
21
sq. units.
Solved Example 81 :
Find the area above the xaxis bounded
by y
2
= 4x and the line x + y = 3
Solution :
Required area = area OCA + area ACB
=
13
01
4xdx 3 x dx
=
3
1
2
3/2
0
1
2x
2x 3x
32
=
4
2
3
=
10
3
sq. units.
Solved Example 82 :
Find the area of the region bounded by the
two parabolas y = x
2
and x = y
2
.
Solution :
Let us first find the points of intersection
solving y = x
2
and x = y
2
y = y
4
y (y
3
1) = 0
y = 0 x = 0
y = 1 x = 1
For the curve x = y
2
Now the area OBPM =
11
00
ydx xdx
1
3/2
0
x
3/2
2
3
sq.units
y
0 C B
A
(3, 0)
(1, 2)
0
M
P
A
B
x = y
2
y = x
2
Notes on Calculus
113
For the curve y = x
2
Area OAPM =
1
0
ydx
1
1
3
2
0
0
x1
xdx
33
sq. units.
The required area =
211
33 3
sq. units.
Solved Example 83 :
Find the area bounded by the cardioid
r = a(1 + cos )
Solution :
A cardioid is symmetrical about the initial
line.
Hence the required area =
2
0
1
2rd
2
since
varies from 0 to for the upper part.
=
2
2
0
a1cos d
=
/2
24 2 4
00
4a cos d 4a cos t2dt
2
(putting t
2
)
=
/2
24 2
0
31
8a cos tdt 8a
422
=
2
3a
Solved Example 84 :
Find the area of one turn of the
archimedan spiral r = a.
Solution :
The area is A =
22
222
00
11
rd a d
22
=
3
2
32
2
a4
a
23 3
Solved Example 85 :
Find the area bounded by the curves :
y = 2x
5
+ 3 and y = 32x + 3
Solution :
Two curves will intersect when
54
2x 3 32x 3 2x x 16 0
In the interval (2, 0) ;
5
2x 3 32x 3
y = 3
A
y
x
x
y
y = 32x + 3
(0, 3)
B
C
A
O
O A
Vidyalankar : GATE – Engineering Mathematics
114
and in the interval (0, 2) ; 32x + 3 2x
5
+ 3
Thus the required area is given by
A =
0
5
2
2x 3 32x 3 dx
2
5
0
32x 3 2x 3 dx
=
02
66
22
2x0
xx
16x 16x
33
=
128 128 256
33 3
sq. units.
2.13 Partial and Total Derivatives
Functions of two variables
If three variables x, y, z are so related that the value of z depends upon the values of x
and y, then z is called the function of two variables x and y and this is denoted by z = f(x, y)
Partial derivatives of first order
Let z = f(x, y) be a function of two independent variables x and y. If y is kept constant and
x alone is allowed to vary then z becomes a function of x only. The derivative of z with
respect to x, treating y as constant, is called partial derivative of z with respect to x and is
denoted by
zf
or
xx
or f
x
h0
z f(x h, y) f(x,y)
lim
xh
Similarly the derivative of z with respect to y, treating x as constant, is called partial
derivative of z with respect to y and is denoted by
zf
or
yy
or f
y
Thus
k0
zf(x,yk)f(x,y)
lim
yk
zz
and
xy
are called first order partial derivatives of z
Notes on Calculus
115
Note : (i) If z = u + v where u = f(x, y) , v (x, y) , then z is a function of x and y.
zuv z uv
;
xxx y yy
(ii) If z = uv, where u = f(x, y) , v = (x, y)
then
zvu
(uv) u v
xx x x
zvu
(uv) u v
yy y y
(iii) If z =
u
v
, where u = f(x, y), v =(x, y)
then
2
uv
vu
zu
xx
dx x v
v
2
uv
vu
zu
yy
yyv
v
(iv) If z = f(u) where u =
(x, y)
then
zdzu z dzu
.; .
xdux y duy
Partial derivatives of higher order
2
2
zz
xx
x
or f
xx
2
2
zz
yy
y
or f
yy
2
zz
xy xy
or f
xy
2
zz
yx yx
or f
yx
In general
22
zz
xy yx
or f
xy
= f
yx
Vidyalankar : GATE – Engineering Mathematics
116
Solved Example 86 :
First order partial derivative of u = y
x
is
Solution :
u = y
x
Treating y as constant
u
x
= y
x
log y
treating x as constant
u
y
= x y
x
1
Solved Example 87 :
Find first order partial derivative of
u = tan
1
22
xy
xy
Solution :
u
x
=
22
2
22
1xy
.
xxy
xy
1
xy
=
2
2222
(x y)
(x y) (x y )
22 22
2
(x y) (x y ) (x y ) (x y)
xx
(x y)
=
22
2222
(x y).2x (x y ).1
(x y) (x y )
22
2222
ux2xyy
x
(x y) (x y )
Similarly
22
2222
uy2xyx
y
(x y) (x y )
Solved Example 88 :
If u = log (tan x + tan y) show that
sin 2x
u
x
+ sin 2y
u
y
= 2
Solution :
u = log(tan x + tan y) [ u is symmetrical
with respect to x and y]
u
x
=
1
. (tanx tany)
tanx tany x
=
2
sec x
tanx tany
Similarly
u
y
=
2
sec y
tanx tany
= sin 2x
u
x
+ sin 2y
u
y
=
1
tanx tany
[sin 2x sec
2
x + sin 2y sec
2
y]
=
1
tanx tany
22
11
2sinxcosx. 2sinycosy
cos x cos y
= 2
(tanx tany)
tanx tany
= 2
Solved Example 89 :
If u = x
y
show that
33
2
yy
xyx
xy
Solution :
u = x
y
u
y
= x
y
log x
Notes on Calculus
117
2
y1 y
xu 1
yx logx x
xy x y x
= x
y
1
(y log x + 1)
3
2
y1
2
y
x
[x (ylogx 1)]
xx
xy
xy
...(1)
u
x
= y x
y
1
2
y1 y1
yy
xyxlogx
yx y x
= x
y
1
(y log x + 1)
3
3
y1
u
u
x(ylogx1)
xyx x x
yx
…(2)
from (1) and (2)
33
2
uu
xyx
xy
Euler’s theorem on Homogenous functions
If u is a homogenous function of degree n in x and y then
x
u
x
+ y
u
y
= n u
Since u is a homogenous function of degree n in x and y it can be expressed as
u = x
n
f
u
x
= n x
n
1
f
n
2
yyy
xf' .
xx
x
x
u
x
= n x
n
f
n1
yy
xyf'
xx
...(1)
Also
u
y
= x
n
f
y1
.
xx
= x
n
1
f
y
x
y
u
y
= x
n
y f
y
x
…(2)
Adding (1) and (2) we get
x
u
x
+ y
u
y
= n x
n
f
y
x
= nu
Note : Euler’s theorem can be extended to a homogenous function of any
number of variables. Thus if u is a homogenous function of degree n in x,
y and z then
x
u
x
+ y
u
y
+ z
u
z
= nu
Vidyalankar : GATE – Engineering Mathematics
118
Solved Example 90 :
If f(x, y) =
222
1 1 logx logy
xy
xxy
show that x
ff
y
xy
+ 2f (x, y) = 0
Solution :
f(x, y) =
222
11logx/y
xy
xxy
f(tx, ty) =
22 2 22 22
tx
log
11
ty
tx txy tx ty
=
2
222
x
log
1
11
y
t
xy
xxy
= t
2
f (x, y)
f (x, y) is a homogenous function of degree
2 in x and y
By Euler’s theorem, we have
x
ff
y
xy
= 2 f (x, y)
Solved Example 91 :
If u =
22
xy
xy
show that
x
22
2
uu u
y2
xy x
x
Solution :
u =
22
xy
xy
is a homogenous function of
degree 3 in x, y
By Euler’s theorem we have
x
u
x
+ y
u
y
= 3 u
Differentiating (1) partially with respect to x
x
22
2
uu y u
1. y 3
xxy x
x
or x
22
2
uu u
y2
xy x
x
Composite functions
(i) If u = f (x, y) where x = (t), y = (t)
then u is called a composite function of (the single variable) t and we can find
du
dt
.
(ii) If z = f(x, y) where x = (u, v), y = (u, v)
then z is called a composite function of (2 variables) u & v so that we can find
zz
&
uv
Total derivative of composite functions
If u is a composite function of t, defined by the relations
u = f(x, y) ; x = (t), y = (t) then
du u dx u dy
..
dt x dt y dt
Notes on Calculus
119
Note : 1. If u = f (x, y, z) and x, y, z are functions of t then u is a composite
function of t and
du u dx u dy u dz
...
dt x dt y dt z dt
2. If z = f (x, y) and x, y are functions of u and v, then
zzxzy
..
uxuyu
zzxzy
..
vxvyv
3. If u = f (x, y) where y =
(x)
then since x =
(x), u is a composite function of x
du u dx u dy
..
dx x dx y dx
du u u dy
.
dx x y dx
du
dx
is called the total differential coefficient of u, to distinguish it from its
partial derivative
u
x
.
Solved Example 92 :
Find
du
dt
when u = x when u = xy
2
+ x
2
y,
x = at
2
, y = 2at.
Solution :
The given equations define u as a
composite function of t.
du u dx u dy
..
dt x dt y dt
= (y
2
+ 2xy). 2at + (2xy + x
2
) . 2a
= (4a
2
t
2
+ 2at
2
. 2at) 2at
+ (2at
2 .
2at + a
2
t
4
).2a
= 8a
3
t
3
+ 8a
3
t
4
+ 8a
3
t
3
+ 2a
3
t
4
= 2a
3
t
3
(5t + 8)
Also u = xy
2
+ x
2
y = at
2
. 4a
2
t
2
+ a
2
t
4
. 2at
= 4a
3
t
4
+ 2a
3
t
5
du
dt
= 16a
3
t
3
+ 10a
3
t
4
= 2a
3
t
3
(5t + 8)
Solved Example 93 :
If u = sin
1
(x y), x = 3t, y = 4t
3
show that
du
dt
=
2
3
1t
Solution :
du u dx u dy
..
dt x dt y dt
Vidyalankar : GATE – Engineering Mathematics
120
=
1
.3
1(x y)
2
1
1(x y)
(1) . 12 t
2
=
22
232
3(1 4t ) 3(1 4t )
1(x y) 1(3t 4t)
=
2
246
3(1 4t )
19t 24t 16t
=
2
222
3(1 4t )
(1 t )(1 8t 16t )
=
2
222 2
3(1 4t ) 3
(1 t ) (1 4t ) 1 t
Solved Example 94 :
If z is a function of x and y
where x = e
u
+ e
v
and y = e
u
e
v
,
show that
zz z z
xy
uv x y
Solution :
Here z is a composite function of u and v
zzx zy
..
uxu yu
=
uu
zz
e(e)
xy
and
zzx zy
..
vxv yv
=
vv
zz
(e ) (e)
xy
subtracting
uv uv
zz z z
(e e) (e e)
uv x y
=
zz
xy
xy
Solved Example 95 :
If (x, y, z ) = 0 show that
xy
z
yzx
1
zxy
Solution :
The given relation defines y as a function
of x and z. Treating x as constant.
x
y
z
z
y
The given relation defines z as a function
of x and y. Treating y as constant.
y
z
x
x
z
Similarly
z
x
y
y
x
Multiplying we get the desired result.
Solved Example 96 :
If u = f (y z, z x, xy) prove that
uuu
xyz
= 0
Solution :
Here u = f (X, Y, Z)
where X = y z, Y = z x, Z = xy
u is a composite function of X, Y and Z
Notes on Calculus
121
uuXuYuZ
...
xXxYxZx
=
uu u
(0) ( 1) (1)
XY Z
uuXuYuZ
...
yXyYyZy
=
uu u
(1) (0) ( 1)
XY Z
uuXuYuZ
...
zXzYzZz
=
uuu
(1) (1) (0)
XYZ
Adding
uuu
xyz
= 0
2.14 Taylor's Series and Maclaurin's Series
Taylor's series is a representation of a function as an infinite sum of terms that are
calculated from the values of the functions derivatives at a single point.
f(x) = f(a) +
2
f (a) (x a) (x a)
f (a) . .....
1! 2 !
=
(n) n
n0
f (a) (x a)
n!
Note : Putting x = a + h, Taylor’s series becomes
2n
n
hh
fa h fa hfa fa ... f a
2! n!
…
If the Taylor's series is centered at zero, that series is also called Maclaurin’s series.
If a = 0 in Taylor series of f(a + h), then the series becomes
2n
n
hh
f h f 0 hf 0 f 0 ... f 0 ...
2! n!
This is called as Maclaurin’s series about a = 0.
Taylor's series expansion of some functions
e
x
=
23
xx x
1 ....
1! 2 ! 3 !
e =
111
1 ....
1! 2! 3!
= 2.7182….
sin x =
357 n
2n 1
n0
xxx (1)
x ..... x
3! 5! 7! (2n 1)!
Vidyalankar : GATE – Engineering Mathematics
122
cos x =
246 n
2n
n0
xxx (1)
1 .... x
2! 4! 6! (2n)!
tan x =
3
5
x2
x x ....
315
Taylor's series in two variables
Taylor's series of a function in two variables, f(x, y) about (a, b) is
f(x, y) = f(a, b) + (x a) f
x
(a, b) + (y b) f
y
(a, b)
+
22
xx xy yy
1
(x a) f (a, b) 2(x a) (y b) f (a, b) (y b) f (a,b) ....
2!
Solved Example 97 :
Expand cos z in a Taylor’s series about
z/4
Solution :
fz cosz
fz sinz
fz cosz
fz sinz
1
f
4
2
1
f
4
2
1
f
4
2
1
f
4
2
By using Taylor’s series
fz f z f
444
2
z/4
f
2! 4
…
=
2
11
1 z z ...
42! 4
2
Solved Example 98 :
Obtain the Taylor’s series for sin x.
Solution :
f(x) sinx
f(x) cosx
f(x) sinz
f(x) cosz
By Maclaurin’s series
2
x
f(x) f(0) xf (0) f (0)
2!
n
n
x
... f (0) ...
n!
k
2k 1
35
1x
xx
sinx x ... ...
3! 5! 2k 1 !
Solved Example 99 :
Obtain the Taylor’s series for cos hx.
Solution :
23
x
xx
e1x ...
2! 3!
23
x
xx
e1x ...
2! 3!
xx 24
ee x x
coshx 1 ....
22!4!
Notes on Calculus
123
Solved Example 100 :
Find the Taylor’s series expansion for the
function
23
fx 1 2x 2x 3x about the
point x = 2.
Solution :
23
2
fx 1 2x 2x 3x f2 29
fx 2 4x 9x f2 42
fx 418x f2 40
fx 18
23
hh
fxf2hf2f2f2
2! 3!
=
23
hh
29 h 42 40 18
26
23
f x 29 42h 20h 3h
where h = x 2
2.15 Fourier Series
Periodic function :
A function f(x) is said to be periodic if f(x + T) = f(x) for all values of x where T is the same
positive number.
For Example:
1. sin x = sin (x + 2)
= sin (x + 4) and so on.
sin x is periodic in x with period 2.
2. The period of tan x is .
Fourier series (General form)
Let f(x) be defined in (L, L) then Fourier series is given by
0n n
n1
nx nx
fx a acos bsin
LL
where
L
0
L
1
afxdx
2L
L
n
L
1nx
afxcosdx
LL
L
n
L
1nx
bfxsindx
LL
Vidyalankar : GATE – Engineering Mathematics
124
Similarly if f(x) is defined in (0, 2L) then Fourier series is given by
0n n
n1
nx nx
f(x) a a cos b sin
LL
2L
0
0
1
af(x)dx
2L
2L
n
0
1nx
a f(x)cos dx
LL
2L
n
0
1nx
bf(x)sindx
LL
Odd and Even Functions :
Let f(x) be defined in (C, C)
1. If f(x) = f(x), for all x in (C, C) then f(x) is even function.
2. If f(x) = f(x), for all x in (C, C) then f(x) is an odd function.
Note : Unless ‘0’ is midpoint of the given interval, we cannot talk of the function
being even or odd.
For Example: f(x) = x
2
in (0, 3) is neither even nor odd, since if x (0, 3)
then x (0, 3) and hence f(x) is not defined.
But if, f(x) = x
2
in (3, 3) then f(x) is even function.
Graphical representation of even and odd function :
i) If f(x) is even, then its graph is symmetrical about f(x) axis
For Example: If f(x) = x, < x < 0
= x 0 < x <
then,
f(x)
x
Notes on Calculus
125
ii) If f(x) is odd then its graph is symmetrical about origin.
For Example: if f(x) = a , if < x < 0
= a , if 0 < x <
Fourier expansion of an even function
Let f(x) be defined in (, ) and let f(x) be even function. Fourier series is given by
0n n
n1
f(x) a a cosnx b sinnx
where,
0
0
11
a f(x)dx f(x)dx
2
f(x) is even
n
0
12
a f(x)cosnxdx f(x)cosnxdx
[ f(x) and cos nx are even functions]
n
1
b f(x)sinnxdx 0
[f(x) sin nx is odd]
0n
n1
f(x) a a cosnx
[Half range cosine series in (0, )]
Fourier expansion of an odd function
Let f(x) be an odd function in (, )
0n n
n1
f(x) a a cosnx b sinnx
o
1
af(x)dx0
2
f(x) is odd function
n
1
a f(x)cosnx dx 0
f(x) cos nx is odd
n
1
b f(x)sinnxdx
=
0
2
f(x)sinnxdx
n
n1
f(x) b sinnx dx
{Half range sine series in (0, )}
x
(0, a)
(0,
a)
f(x)
Vidyalankar : GATE – Engineering Mathematics
126
Solved Example 101 :
Find the Fourier series representing
f(x) = x, 0 < x < 2 and draw its graph from
x = 4 to x = 4.
Solution :
o
nn
n1
a
fx x acosnx bsinnx
2
where
22
2
0
00
1114
afxdxxdx 2
2
a
n
=
2
0
1
x cosnxdx
=
2
2
0
0
1xsinnx 1
sinnxdx
nn
=
2
0
1 cosnx
xsinnx
nn
= 0
b
n
=
2
0
1
x sinnxdx
=
2
2
0
1xcosnxsinnx
n
n
=
12 2
nn
Hence
sin2x sin3x
x 2 sinx ...
23
Solved Example 102 :
Find Fourier Series for f(x) = x
2
in
,
and show that
2
222
111
....
12
12 3
Solution :
We have f(x) = x
2
is an even function in
,
0n
n1
nx
fx a acos
i)
2
2
0
0
1
axdx
3
ii)
2
n
0
2nx
axcosdx
Integrating by parts
=
2
2nx
xsin
n
23
22 33
0
nx nx
2x cos 2 sin
nn
=
32
nn
22 22
22 4
11
nn
4 2 0 2 4
Notes on Calculus
127
n
22
22
n1
1
4nx
fx cos
3
n
…(1)
putting x = 0 in equation (1) we get
n
22
22
n1
1
4
0
3
n
2
2222
1111
...
12
12 3 4
2
2222
111 1
...
12
12 3 4
Solved Example 103 :
Find a Fourier series to represent
f(x) = x for 0 < x < 2.
Solution :
22
0
00
11
afxdx xdx
=
2
2
2
0
0
1x
x
2
=
1
22 0
a
n
=
2
0
1
f x cosnxdx
=
2
0
1
x cosnxdx
=
2
2
0
1 sinnx x sinnx cosnx
nn
n
Integrating by parts.
=
1
00
b
n
=
2
0
1
xsinnxdx
=
2
2
0
1 cosnx x cosnx sinnx
nn
n
=
12cos2n 2
nn
Solved Example 104 :
Find a Fourier expansion for
f(x) = x
2
for < x <
Solution :
x
2
is an even function
2
223
0
0
111
axdxxdx
233
22
n
0
12
a x cosnxdx x cosnxdx
The indefinite integral
=
2
2xsinnx sinnx
2xdx
nn
=
2
2
2 x sinnx 2x cosnx cosnx
2dx
nnn
n
=
2
23
2 x sinnx 2x cosnx 2 sinnx
n
nn
2
n
23
0
2 x sinnx 2x cosnx 2 sinnx
a
n
nn
=
22
44
cosn cosn
nn
b
n
= 0
123
222
44 4
a,a,a etc
123
Hence
2
2
22 2
44 4
x cos x cos2x cos3x ...
3
12 3
Vidyalankar : GATE – Engineering Mathematics
128
Solved Example 105:
Find the Fourier series for f(x) = sin h ax in
(, )
Solution :
Clearly f(x) is an odd function of x
f(x) = sin h a(x) = sin h ax = f(x)
Let, sin h ax =
n
n1
bsinnx
where, b
n
=
0
2
sinhax sinnx dx
=
ax ax
0
2e e
sinnx dx
2
=
ax ax
00
1
e sinnx dx e sinnx dx
=
ax
22
11
easinnxncosnx
an
ax
22
0
1
e a sinnx ncosnx
an
=
22
1
an
nn
aa
en1ne 1
=
n
aa
22
n1
ee
an
=
n
22
2n 1 sinh a
an
aa
ee
sinha
2
=
n1
22
2n 1 sinha
an
nnn1n1
1111 1
f(x) =
n1
22
n1
n1 sinnx
2sinha
an
Solved Example 106 :
Find the Fourier series for
f(x) =
22
x
12 4
in < x < and show
that
2
223
111
....
12
12 3
Solution :
Note that f(x) is an even function of x
22 2
(x) x
f( x) f(z)
12 4 12 4
Let f(x) =
0n
n1
aacosnx
a
0
=
22
0
1
dx
12 4
=
23
0
1x
x0
12 12
a
n
=
22
0
2x
cosnxdx
12 4
Integrating by parts
=
22
2xsinnx
12 4 n
23
0
x cosnx 1 sinnx
22
nn
=
nn1
222
2(1)(1)
cosn
2n n n
22 n1
2
n1
x(1)
cosnx
12 4
n
Put x = 0 we get
2n1
22222
n1
(1) 1 1 1 1
...
12
n1234
Notes on Calculus
129
List of Formulae
Limits
x0
sin x
lim
x
= 1 …x in radians
x0
tan x
lim
x
= 1 …x in radians
x0
lim cos x
= 1
nn
n1
xa
xa
lim na
xa
x
x0
a1
lim
x
= na
(a > 0)
x
x0
e1
lim
x
= 1
1
x
x0
lim 1 x)
= e
x0
n(1 x)
lim 1
x
Continuity
The function f is said to be continuous
at x = x
0
if
i)
0
xx
lim f(x)
exists and
ii)
0
0
xx
lim f(x) f(x )
Derivatives
d
k(constant) 0
dx
nn1
d
xnx
dx
d1
log x
dx x
xx
d
ee
dx
xx
e
d
aaloga
dx
d
dx
sin x = cos x
d
dx
cosec x = cosec x cot x
d
dx
cos x = sin x
d
dx
sec x = sec x tan x
d
dx
tan x = sec
2
x
d
dx
cot x = cosec
2
x
d
dx
1
22
x1
sin
a
ax
1
22
dx 1
cos
dx a
ax
1
22
dx a
tan
dx a
ax
1
22
dx a
cot
dx a
ax
1
22
dx a
sec
dx a
xx a
1
22
dxa
cosec
dx a
xx a
d
dx
cos hx = sin hx
d
dx
sin hx = cos hx
d
dx
sin h
1
x =
2
1
x1
Vidyalankar : GATE – Engineering Mathematics
130
d
dx
cosh
1
x =
2
1
x1
1
2
d1
(tanh x)
dx
x1
Integration
n1
n
x
xdx c
n1
, (n 1)
1
dx logx c
x
xx
edx e c
x
x
e
a
adx
log a
+ c
cos xdx sinx c
cosecx cot xdx cosecx c
sinx dx cos x c
secxtanxdx secx c
2
sec x dx tanx c
2
cosec x dx
= cot x + c
cot xdx
= log sin x + c
tanxdx
= log sec x + c
sec xdx log(sec x tanx) c
cosec x dx log(cosecx cot x) c
1
22
1x
sin c
a
ax
1
22
dx x
cos c
a
ax
1
22
dx 1 x
tan c
aa
ax
1
22
dx 1 x
cot c
aa
ax
1
22
dx 1 x
sec c
aa
xx a
1
22
dx 1 x
cosec c
aa
xx a
sinhx dx
= cos h x + c
coshx dx
= sin h x + c
2
dx
x1
= sin h
1
x + c
2
dx
x1
= cos h
1
x + c
2
dx
x1
= tan h
1
x + c
tanxdx
= log (cos x) + c
sec xdx
x
log tan c
42
cosec x dx
x
log tan
2
+ c
11
22
dx x x
sin or cos c
aa
ax
1
22
dx x
sinh
a
ax
or
22
log(x x a ) c
1
22
dx x
cosh c
a
xa
22
or log(x x a ) c
Notes on Calculus
131
1
22
dx 1 x
tan
aa
ax
1
1x
or cot c
aa
22 22
1
axdx xax
2
21
1x
asin c
2a
22 22
1
axdx xax
2
21
1x
asinh c
2a
22 22
1
xadx xxa
2
21
1x
acosh c
2a
22
dx 1 a x
log c
2a a x
ax
22
dx 1 x a
log c
2a x a
xa
Integration by parts
du
u v dx u v dx v dx dx
dx
Of the two functions the first function u
is selected based on the following
preference order
1. Inverse function
2. Logarithmic function
3. Algebraic function
4. Trigonometric function
5. Exponential function
Some Important Substitutions In
Integration
Rule 1:
If the numerator is equal to or
a multiple of the derivative of the
denominator, then put the denominator
equal to some other variable say z. It
is then reduced to standard form after
simplification.
Rule 2: If there are two factors in the
integrand one of which is of the type
[f(x)] and other of the type f (x) then
put f(x) = z.
Rule 3: If the integrand is of the
type
22
ax
, then put x = a sin or
x = a cos
Rule 4: If the integrand is of the type
x
2
+ a
2
or some power of it then put
x = a tan or x = a cot
Rule 5: If the integrand is of the
type
22
xa
, then put x = a sec or
x = a cosec
Rule 6: If the integrand is of the
type
ax
ax
, then put x = a cos 2 and
if it is of the type
22
22
ax
ax
, then put
x
2
= a
2
cos 2
Rule 7: The presence of ax
suggests the substitution of a + x = t
2
or x = a tan
2
or x = a cos 2.
Vidyalankar : GATE – Engineering Mathematics
132
Rule 8: The presence of
2
2ax x
suggests the substitution
x = a (1 cos)
Important Trigonometric Identities
sin
2
A + cos
2
A =1
sin (A + B) = sin A cos B + cos A sin B
cos (A + B) = cos A cos B sin A sin B
tan (A + B) =
tanA tanB
1 tanA tanB
sin (A B) = sin A cos B cos A sin B
cos (A B) = cos A cos B + sin A sin B
tan (A B) =
tanA tanB
1 tanA tanB
sin
2
A sin
2
B = sin (A + B) sin (A B)
cos
2
A sin
2
B = cos (A + B) cos (A B)
2 sin A cos B = sin (A + B) + sin (A B)
2 cos A sin B = sin (A + B) sin (A B)
2 cos A cos B = cos (A + B) + cos (A B)
2 sin A sin B = cos (A B) cos (A + B)
2 sin
CD
2
cos
CD
2
= sin C + sin D
2 cos
CD
2
sin
CD
2
= sin C sin D
2 cos
CD
2
cos
CD
2
= cos C + cos D
2 sin
CD
2
sin
DC
2
= cos C cos D
cos 2A = cos
2
A sin
2
A = 1 2sin
2
A
= 2 cos
2
A 1 =
2
2
1tanA
1tanA
sin 2A = 2 sin A cos A =
2
2tanA
1tanA
tan 2A =
2
2tanA
1tanA
sin 3A = 3 sin A 4 sin
3
A
cos 3A = 4 cos
3
A 3 cos A
tan 3A =
3
2
3tanA 4tan A
13tanA
Definite Integration
b
xb
xa
a
f(x)dx x b a
Properties of definite integral
bb
aa
f(x)dx f(t)dt
ba
ab
f(x)dx f(x)dx
bcb
aac
f(x)dx f(x)dx f(x)dx
aa
00
f(x)dx f(a x)dx
aa
a0
f(x)dx 2 f(x)dx
…f(x) = f(x)
= 0 …f(x) = f(x)
2a a
00
f(x)dx 2 f(x)dx
…f(2a x) = f(x)
= 0 …f(2a x) = f(x)
Euler’s theorem on Homogenous
functions
x
u
x
+ y
u
y
= n u
Notes on Calculus
133
Total derivative of composite
functions
du u dx u dy
..
dt x dt y dt
Taylor's series
f(x) = f(a) +
f(a)(x a)
1!
2
(x a)
f (a). ....
2!
=
(n) n
n0
f (a)(x a)
n!
f(x, y) = f(a, b) + (x a) f
x
(a, b)
+ (y b) f
y
(a, b)
+
2
xx
xy
2
yy
(x a) f (a,b)
1
2(x a)(y b) f (a,b) ....
2!
(y b) f (a,b)
Maclaurin's Series
2
h
f h f 0 hf 0 f 0 ...
2!
n
n
h
f 0 ...
n!
Fourier Series
0n n
n1
nx nx
fx a acos bsin
LL
where
L
0
L
1
afxdx
2L
L
n
L
1nx
afxcosdx
LL
L
n
L
1nx
bfxsindx
LL
Vidyalankar : GATE – Engineering Mathematics
134
Assignment 1
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 10 carry one mark each
1.
x0
xsinx
Lt
1cosx
is [ME 2014]
(A) 0 (B) 1
(C) 3 (D) not defined
2. If a function is continuous at a point,
(A) the limit of the function may not
exist at the point
(B) the function must be derivable at
the point
(C) the limit of the function at the point
tends to infinity
(D) the limit must exist at the point and
the value of limit should be same
as the value of the function at that
point [ME 2014]
3. The value of
x
x
1
lim 1
x
is [EC 2014]
(A)
n2 (B) 1.0
(C) e (D) ∞
4. The series
n0
1
n!
converges to
[ME 2014]
(A) 2 In 2 (B)
2
(C) 2 (D) e
5. The value of
2
x0
4
1cos(x)
lim
2x
is
[ME 2015]
(A) 0 (B)
1
2
(C)
1
4
(D) undefined
6. If a continuous function f(x) does not
have a root in the interval [a, b], then
which one of the following statements
is TRUE? [EE 2015]
(A)
f(a) f(b)
= 0 (B) f(a) f(b) 0
(C)
f(a) f(b) 0
(D) f(a) f(b) 0
7. The double integral
ay
00
f(x, y) dx dy
is
equivalent to [IN 2015]
(A)
xy
00
f(x, y) dx dy
(B)
ay
0x
f(x, y) dx dy
(C)
aa
0x
f(x, y) dy dx
(D)
aa
00
f(x, y) dx dy
8. The value of
n
n0
1
n
2
is _________.
[EC 2015]
9.
e
3x
x0
log (1 4x)
Lt
e1
is equal to
[ME 2016]
(A) 0 (B)
1
12
(C)
4
3
(D) 1
Assignment on Calculus
135
10. The value of
3
x0
xsin(x)
lim
x
is
[ME 2017]
(A) 0 (B) 3
(C) 1 (D)
1
Q 11 to Q 20 carry two marks each
11.
The minimum value of the function
f(x) =
2
1
x(x 3)
3
in the interval
100 ≤ x ≤ 100 occurs at
x = ________.
[EC 2017]
12. The maximum value of
f(x) = 2x
3
9x
2
+ 12x 3 in the interval
0 x 3 is _____.
[EC 2014]
13. Consider a spatial curve in three-
dimensional space given in parametric
form by
[ME 2015]
x(t) = cos t, y(t) = sin t,
z(t) =
2
t
, 0 t
2
.
The length of the curve is __________
14.
2
2
1
cos(1 x)
dx
x
= ______. [CS 2015]
15. The region specified by
{(, , z): 3 ≤ 5,
8
4
,
3 z 4.5} in cylindrical coordinates
has volume of _______
[EC 2016]
16. The area between the parabola x
2
= 8y
and the straight line y = 8 is _______.
[CE 2016]
17. A political party orders an arch for the
entrance to the ground in which the
annual convention is being held. The
profile of the arch follows the equation
y = 2x 0.1x
2
, where y is the height of
the arch in meters. The maximum
possible height of the arch is
[CS, ME-2012]
(A) 8 meters (B) 10 meters
(C) 12 meters (D) 14 meters
18. Consider the function f (x) = |x| in the
interval 1 x 1. At the point x = 0,
f (x) is
[ME 2012]
(A) Continuous and differentiable
(B) Non-continuous and differentiable
(C) Continuous and non-differentiable
(D) Neither continuous nor
differentiable
19. The maximum value of
f(x) = x
3
9x
2
+ 24x + 5 in the interval
[1, 6] is
[EE 2012]
(A) 21 (B) 25
(C) 41 (D) 46
20. A parametric curve defined by
x =
u
cos
2
, y =
u
sin
2
in the
range 0 ≤ u ≤ 1 is rotated about the
X-axis by 360 degrees. Area of the
surface generated is
[ME 2017]
(A)
2
(B)
(C) 2 (D) 4
Vidyalankar : GATE – Engineering Mathematics
136
Assignment 2
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 10 carry one mark each
1.
2x
x0
e1
Lt
sin(4x)
is equal to [ME 2014]
(A) 0 (B) 0.5 (C) 1 (D) 2
2. The value of the integral
2
2
2
0
(x 1) sin(x 1)
(x 1) cos(x 1)
dx is
(A) 3 (B) 0
[ME 2014]
(C) 1 (D) 2
3. The maximum value of the function
f(x) = In(1 + x) x (where x > 1)
occurs at x = _________.
[EC 2014]
4. Let f(x) = x e
x
. The maximum value of
the function in the interval (0,∞) is
(A) e
1
(B) e [EE 2014]
(C) 1 e
1
(D) 1+ e
1
5. At x = 0, the function f(x) = |x| has
(A) a minimum
[ME 2015]
(B) a maximum
(C) a point of inflexion
(D) neither a maximum nor minimum
6. While minimizing the function f(x),
necessary and sufficient conditions for
a point, x
0
to be a minima are:
[CE 2015]
(A) f(x
0
) > 0 and f(x
0
) = 0
(B) f(x
0
) < 0 and f(x
0
) = 0
(C) f(x
0
) = 0 and f(x
0
) < 0
(D) f(x
0
) = 0 and f(x
0
) > 0
7. A function f(x) = 1 x
2
+ x
3
is defined
in the closed interval [1, 1]. The value
of x, in the open interval (1, 1) for
which the mean value theorem is
satisfied, is
[EC 2015]
(A) 1/2 (B) 1/3 (C) 1/3 (D) 1/2
8. Consider the function f(x) = 2x
3
3x
2
in the domain [1, 2]. The global
minimum of f(x) is ______
[ME 2016]
9. Given the following statements about a
function f:
, select the right
option:
[EC 2016]
P: If f(x) is continuous at x = x
0
, then
it is also differentiable at x = x
0
.
Q: If f(x) is continuous at x = x
0
, then it
may not be differentiable at x = x
0
.
R: If f(x) is differentiable at x = x
0
,
then it is also continuous at x = x
0
.
(A) P is true, Q is false, R is false
(B) P is false, Q is true, R is true
(C) P is false, Q is true, R is false
(D) P is true, Q is false, R is true
10. The integral
1
0
dx
(1 x)
is equal to ____
[EC 2016]
Assignment on Calculus
137
Q 11 to Q 20 carry two marks each
11.
The value of the integral is
2x
xy
00
edydx
is [ME 2014]
(A)
1
2
(e 1) (B)
1
2
(e
2
1)
2
(C)
1
2
(e
2
e) (D)
2
11
e
2e
12. The minimum value of the function
f(x) = x
3
3x
2
24x + 100 in the
interval [3, 3] is
[EE 2014]
(A) 20 (B) 28 (C) 16 (D) 32
13. Consider an ant crawling along the
curve (x 2)
2
+ y
2
= 4, where x and y are
in meters. The ant starts at the point
(4, 0) and moves counter-clockwise with
a speed of 1.57 meters per second. The
time taken by the ant to reach the point
(2, 2) is (in seconds) ____
[ME 2015]
14.
2
x
lim x x 1 x
is [ME 2016]
(A) 0 (B) (C) 1/2 (D)
15. The area of the region bounded by the
parabola y = x
2
+ 1 and the straight
line x + y = 3 is
[CE 2016]
(A)
59
6
(B)
9
2
(C)
10
3
(D)
7
6
16. Let f :[1, 1] ,
where f(x) = 2x
3
x
4
10. The minimum
value of f(x) is _____.
[IN 2016]
17. At x = 0, the function f (x) = x
3
+ 1 has
(A) A maximum value [ME 2012]
(B) A minimum value
(C) A singularity
(D) A point of inflection
18.
2
x0
1cosx
lim
x
is [ME 2012]
(A) 1/4 (B) 1/2 (C) 1 (D) 2
19. Which one of the following functions is
continuous at x = 3? [CS 2013]
(A) f(x) =
2, if x 3
x1, ifx3
x3
,if x3
3
(B) f(x) =
4, if x 3
8xifx3
(C) f(x) =
x3,ifx3
x4,ifx3
(D) f(x) =
3
1
x27
, if x 3
20. Let I = c
2
R
xy dxdy
, where R is the
region shown in the figure and
c = 6 × 10
4
. The value of I equals
_______. (Give the answer up to two
decimal places)
[EE 2017]
Vidyalankar : GATE – Engineering Mathematics
138
Assignment 3
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 10 carry one mark each
1.
Consider a function f(x,y,z) given by
f(x,y,z) = (x
2
+ y
2
2z
2
) (y
2
+ z
2
)
The partial derivative of this function
with respect to x at the point, x = 2,
y = 1 and z = 3 is _____.
[EE 2017]
2. For 0 t < ∞, the maximum value of
the function f(t) = e
t
2 e
2t
occurs at
(A) t = log
e
4 (B) t = log
e
2
(C) t = 0 (D) t = log
e
8
[EC 2014]
3. If z = xy In (xy), then [EC 2014]
(A) x
zz
y
xy
= 0
(B)
zz
yx
xy
(C) x
zz
y
xy
(D) y
zz
x
xy
= 0
4.
x
xsinx
lim
x
equals to [CE 2014]
(A) ∞ (B) 0
(C) 1 (D) ∞
5. The value of
x0
sinx
lim
2sinx xcosx
is ___________
[ME 2015]
6. Given, i = 1
, the value of the
definite integral, I =
2
0
cos x isinx
dx
cos x i sinx
is: [CE 2015]
(A) 1 (B) 1
(C) i (D) i
7. The contour on the x-y plane, where
the partial derivative of x
2
+ y
2
with
respect to y is equal to the partial
derivative of 6y + 4x with respect to x,
is
[EC 2015]
(A) y = 2 (B) x = 2
(C) x + y = 4 (D) x y = 0
8. The values of x for which the function
2
2
x3x4
f(x)
x3x4
is NOT continuous
are
[ME 2016]
(A) 4 and 1 (B) 4 and 1
(C) 4 and 1 (D) 4 and 1
9.
As x varies from −1 to + 3, which one
of the following describes the
behaviour of the function
f(x) = x
3
3x
2
+ 1? [EC 2016]
(A) f(x) increases monotonically.
(B) f(x) increases, then decreases and
increases again.
Assignment on Calculus
139
(C) f(x) decreases, then increases and
decreases again.
(D) f(x) increases and then decreases.
10. The maximum value attained by the
function f(x) = x(x 1) (x 2) in the
interval [1, 2] is _____.
[EE 2016]
Q 11 to Q 20 carry two marks each
11.
The Taylor series expansion of
3 sin x + 2 cos x is
[EC 2014]
(A) 2 + 3x x
2
3
x
2
+ .....
(B) 2 3x + x
2
3
x
2
+ .....
(C) 2 + 3x + x
2
+
3
x
2
+ .....
(D) 2 3x x
2
+
3
x
2
+ .....
12. The expression
0
x1
lim
is equal
to
[CE 2014]
(A) log x (B) 0
(C) x log x (D) ∞
13. The maximum area (in square units) of
a rectangle whose vertices lie on the
ellipse x
2
+ 4y
2
= 1 is _________.
[EC 2015]
14. The integral
D
1
(x y 10)dx dy
2
,
where D denotes the disc: x
2
+ y
2
4,
evaluates to ________
[EC 2014]
15. The angle of intersection of the curves
x
2
= 4y and y
2
= 4x at point (0, 0) is
[CE 2016]
(A) 0 (B) 30
(C) 45 (D) 90
16. The cost function for a product in a
firm is given by 5q
2
, where q is the
amount of production. The firm can
sell the product at a market price of
Rs. 50 per unit. The number of units to
be produced by the firm such that the
profit is maximized is
[CS, ME 2012]
(A) 5 (B) 10
(C) 15 (D) 25
17. The area enclosed between the
straight line y = x and the parabola
y = x
2
in the x y plane is [ME 2012]
(A) 1/6 (B) 1/4
(C) 1/3 (D) 1/2
18. Consider the function f(x) = sin(x) in
the interval x[/4, 7/4]. The number
and location (s) of the local minima of
this function are
[CS 2012]
(A) One, at /2
(B) One, at 3/2
(C) Two, at /2 and 3/2
(D) Two, at /4 and 3/2
Vidyalankar : GATE – Engineering Mathematics
140
19. The value of the definite integral
e
1
xln(x)dx
is [ME 2013]
(A)
3
42
e
99
(B)
3
24
e
99
(C)
3
24
e
99
(D)
3
42
e
99
20. Consider the following definite integral:
I =
1
12
2
0
(sin x)
dx
1x
[CE 2017]
The value of the integral is
(A)
2
24
(B)
3
12
(C)
3
48
(D)
3
64
Assignment on Calculus
141
Assignment 4
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
2
x0
1cosx
Lim
x
equals to
(A)
1
3
(B)
1
2
(C)
1
4
(D)
1
5
2. What is
0
sin
lim
equal to?
(A) (B) sin
(C) 0 (D) 1
3. The value of
1/ 3
x8
x2
lim
(x 8)
is
(A)
1
16
(B)
1
12
(C)
1
8
(D)
1
4
4.
x
ax b
Lim
cx
is
(A) 0 (B) a
(C)
a
b
(D)
a
c
5. The discontinuity of the function
f(x)
x
e1
x
at the point x = 0 is known
as
(A) first kind of discontinuity
(B) second kind of discontinuity
(C) mixed discontinuity
(D) removable discontinuity
6.
2
x
3
x0
x
e
1x
2
lim
x
=
(A) 1/2 (B) 1/6
(C) 1/3 (D) 1
Q7 to Q18 carry two marks each
7.
Which of the following is incorrect
statement ?
(A) A stationary value is always an
extreme value
(B) A maximum value is always a
stationary value
(C) A minimum value is always a
stationary value
(D) A minimum value may be greater
than maximum value
8. Which of the following is correct
statement ?
(A) A stationary point on a graph is
any point at which
dy
dx
= 0
(B) A stationary point on a graph is
any point at which
dy
dx
=
(C) A stationary point on a graph is
any point at which there are gaps
or breaks
(D) None of these
Vidyalankar : GATE – Engineering Mathematics
142
9. First rule for maximum and minimum
values of y = f(x) are as follows
Get
2
2
dy d y
and
dx
dx
. Solve
dy
0
dx
and
consider its roots. These are the value
of x which make
dy
dx
= 0. For each of
these values of x, calculate the
corresponding value of y, and examine
the sign of
2
2
dy
dx
. Now pick the correct
statement from the following
alternatives.
(A) If the sign is +, the corresponding
value of y is a maximum
(B) If the sign is , the corresponding
value of y is a minimum
(C) If the sign is , the corresponding
value of y is a maximum
(D) None of these
10. Let
32
f(x) x 6x 24x 4. Then f(x)
has
(A) a maximum value at x = 2
(B) a maximum value at x = 5
(C) a minimum value at x = 5
(D) neither maximum nor minimum at
any point
11. Let a number 30 be divided into two
parts such that cube of one multiplied
with the other is maximum, then the
parts are
(A) 12.4, 17.6 (B) 13, 17
(C) 10.5, 19.5 (D) 22.5, 7.5
12. f (x) =
3
(x 2) is
(A) maximum at x = 2
(B) minimum at x = 2
(C) neither maximum nor minimum at
x = 2
(D) none of these
13. The function sin x (1 + cos x) is
maximum in the interval (0, ), when
(A) x =
4
(B) x =
2
(C) x =
3
(D) x =
2
3
14. The absolute minimum of
f (x) = 4x
3
8x
2
+ 1 in [ 1, 1] is at
(A) 1 (B) 3
(C) 1 (D) 8
15. A spherical snowball is melting at rate
of 4 cubic centimeter per hour. How
fast is the diameter changing when it is
20 centimeters ?
(A) 0.1 cm/hr (B) 0.02 cm/hr
(C) 0.04 cm /hr (D) 0.01 cm/hr
16. If a mothball evaporates at a rate
proportional to its surface area 4r
2
,
then its radius
(A) decreases at constant rate
(B) decreases logarithmically
(C) decreases exponentially
(D) None of these
Assignment on Calculus
143
17. The absolute maximum of
f(x) = x
4
2x
3
x
2
4x + 3 on the
interval [0, 4] is.
(A) 80 (B) 16
(C) 28 (D) None of these
18.
The absolute minimum of
f(x) =
3
x
(x 2)
on the interval [ 1, 1] is
at
(A) 0 (B) 1
(C) 1/3 (D) None of these
Vidyalankar : GATE – Engineering Mathematics
144
Assignment 5
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
2
x
Lim ( x 1 x 1)
equals to
(A) 0 (B) e
(C) 1 (D)
2. It is given that
f(x) =
x0
ax b
,Lim f(x) 2
x1
and
x
Lim f(x) 1
, then value of f ( 2) is
(A) 0 (B) 1
(C) e (D)
3. The limiting value of
222 2
333 3
123 n
.............
nnn n
as n
is
(A) 1 (B)
1
2
(C)
1
3
(D)
1
4
4. What is the value of
2n
n
1
lim 1
n
?
(A) 0 (B) e
2
(C) e
1/2
(D) 1
5. The value of
2
x(/2)
1cos2x
Lim
(2x)
is
(A)
1
4
(B)
1
3
(C)
1
2
(D) 0
6. The value of
(/2)
cot
Lim
(/2)
is
(A) 1 (B) 2
(C) 0 (D)
1
2
Q7 to Q18 carry two marks each
7.
If y = 7x x
3
and x increase at rate of
4 units/sec how fast is slope of graph
changing when x = 3 to
(A) 72 (B) 80
(C) 64 (D) 64
8. An object moves on the parabola
3y = x
2
when x = 3, the x coordinate of
the object is increasing at the rate of
1 foot/min. How fast is y coordinate
increasing of the moment
(A) 1 (B) 2
(C) 3 (D) 4
9. A snowball is increasing in volume of
the rate of 10 cm
3
/h. How fast is the
surface area growing at the moment
where radius of snowball is 5 cm ?
(A) 4 (B) 2
(C) 12 (D) 16
10. The height h and radius r of cylinder of
greatest volume that can be cut from a
sphere of radius b is
(A) h =
2b 2
,r b
3
3
Assignment on Calculus
145
(B) h =
b
3
, r =
2
b
3
(C) h =
bb
,r
23 6
(D) h =
2b 1
,r b
3
3
11. A pipe with length 3m and radius of
10cm has an outer layer of ice that is
melting at the rate of 2cm
3
/min. How
fast is the thickness of ice decreasing
when ice is 2cm thick?
(A)
1
cm / min
200
(B)
1
cm / min
800
(C)
1
cm / min
1800
(D)
1
cm / min
3600
12. The maximum area of any rectangle
which may be inscribed in a circle of
radius 1 is
(A) 1 (B) 2
(C) 3 (D) 4
13. A point on a curve is said to be an
extremum if it is a local minimum or a
local maximum. The number of distinct
extrema for the curve
43 2
3x 16x 24x 37 is
(A) 0 (B) 1
(C) 2 (D) 3
14.
/4
0
(1 tan x) / (1 tan x) dx
evaluates to
(A) 0 (B) 1
(C) In 2 (D) ½ In 2
15. The following plot shows a function y
which varies linearly with x. The value
of the integral
2
1
Iydxis
(A) 1.0 (B) 2.5
(C) 4.0 (D) 5.0
16. The value of the quantity P, where
P =
1
x
0
xe dx,
is equal to
(A) 0 (B) 1
(C) e (D) 1/e
17. The sum of the infinite series,
1 +
111
....is
234
(A) (B) infinity
(C) 4 (D)
2
4
18. The value of the integral
22
xy
00
ee
dx dy is
(A)
2
(B)
(C) (D)
4
x
y
1
1
2 3
1
2
3
Vidyalankar : GATE – Engineering Mathematics
146
Assignment 6
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
The discontinuity of the function
f (x) =
|
x
|
x
, x 0
= 0, x = 0
at the origin is known as
(A) Discontinuity of 1
st
kind
(B) Discontinuity of 2
nd
kind
(C) Mixed discontinuity
(D) None of these
2. The value of the coefficients ‘a’ such
that the function f defined as
f (x) =
2
ax 2
x 1
= 1 x 1
continuous is
(A) a = 1 (B) a = 2
(C) a = 0 (D) a = 3
3. If f (x) = 1, if x < 3
= ax + b, if 3 x < 5
= 7 if 5 x
Determine the value of ‘a’ and ‘b’ so
that f (x) is continuous.
(A) a = 4, b = 7
(B) a = 3, b = 8
(C) a = 1, b = 9
(D) a = 2, b = 3
4.
The maximum value of
f (x) = 2 | x 1| + 3 | x 2 |
for all x R is
(A) 2 (B) 4
(C) 1 (D) None of these
5. Pick out the correct statement
(A) logarithmic function is decreasing
wherever it is defined
(B) logarithmic function is increasing
whenever it is defined
(C) logarithmic function is neither
increasing nor decreasing
(D) None of these
6. f (x) = log sin x. This function is
(A) increasing on
0,
2
and
decreasing on
,
2
(B) increasing on
,
2
and
decreasing on
0,
2
(C) increasing on [0, 0] and
decreasing on [, ]
(D) increasing on [, 2] and
decreasing on [0, 0]
Assignment on Calculus
147
Q7 to Q18 carry two marks each
7.
The series
23
11 1
1 .....is
2
22
(A) convergent (B) divergent
(C) oscillatory (D) None of these
8. The series 1 + 2 + 3 + …… is
(A) convergent (B) divergent
(C) oscillatory (D) None of these
9. The series 1 2 + 3 4 + 5 6 + . . . .
is
(A) convergent (B) divergent
(C) oscillatory (D) None of these
10. The series
111
1 .......is
234
(A) convergent (B) divergent
(C) oscillatory (D) None of these
11. The series
12 3
123
12 12 12
+….
is
(A) convergent (B) divergent
(C) oscillatory (D) None of these
12. Which of the following to true?
(A)
1/ 3 1/ 3 1/ 3
111
1
234
is convergent
(B)
111
1
234
is convergent
(C)
222
111
1
234
is convergent
(D)
kkk
111
1
234
is divergent
13. The series
11 1
1
3927
+……… is
(A) convergent (B) divergent
(C) oscillatory (D) None of these
14. The series whose n
th
term is
33
n1 n is
(A) convergent (B) divergent
(C) oscillatory (D) None of these
15. The series whose n
th
term is sin
1
is
n
(A) convergent (B) divergent
(C) oscillatory (D) None of these
16. The series
2
234 5 n1
........is
14916
n
(A) convergent (B) divergent
(C) oscillatory (D) none of these
17. The interval in which Lagrange’s
theorem is applicable for function
f(x) = 1/x is
(A) [3, 3] (B) [2, 2]
(C) [ 2, 3] (D) [1, 1]
18. For f(x) = x
3
suppose
f(b) f(a) = (b a)
f
(c) holds then c is
(A)
22
1
ab
2
(B)
ab
2
(C)
22
abab
3
(D)
22
abab
3
Vidyalankar : GATE – Engineering Mathematics
148
Assignment 7
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
Function f (x) =
1/ x
e ; x > 0 is a
(A) increasing function
(B) decreasing function
(C) neither increasing nor decreasing
(D) none of these
2. f (x) =
1
x
x
is
(A) maximum at x = 1
(B) maximum at x = 2
(C) maximum at x = 1
(D) maximum at x = 2
3. f (x) =
2
x + 2x + 1 is
(A) maximum at x = 1
(B) maximum at x = 2
(C) maximum at x = 1
(D) maximum at x = 2
4. The maximum value of 5 cos + 3 cos
3
+ 3 is
(A) 5 (B) 10
(C) 11 (D) 1
5. f (x) =
32
x12x45x11
is
(A) minimum at x = 5
(B) minimum at x = 5
(C) minimum at x = 4
(D) minimum at x = 4
6. The maximum of f(x) = (x + 2)/(x 1) is
(A) 4 (B) 1
(C) 2 (D) does not exist
Q7 to Q18 carry two marks each
7.
Use Lagrange’s mean value theorem
for f(x) = e
x
. Which of the following is
true?
(A) 1 + x < e
x
< 1 + xe
x
(B) 1 + xe
x
< e
x
< 1 + x
(C) e
x
< 1 + x e
x
< 1 + x
(D) none of these
8. x
1/X
is decreasing function if
(A) x < e (B) x > e
(C) x = e (D) x > 1/e
9. The Maclaurin series of e
x
is
(A)
n1
n0
x
n1!
(B)
n
n0
x
n!
(C)
n
n1
x
n!
(D) none of these
10. Taylor series for sin x about /4 is
(A)
n
n0
x
1
n!
2
(B)
nn
n0
1x /4
2
n!
(C)
2
1x /4 x /4
1
...
1! 2 !
2
(D) none of these
Assignment on Calculus
149
11. The Maclaurin series for ln (1 x) is
(A)
23
xx
x .....
23
(B)
23
xx
x .....
23
(C)
23
xx
x .....
23
(D) none of these
12. The Taylor series for 1/x about 1 is
(A)
n1 n
n0
1x1
(B)
nn
n0
1x1
(C)
n1 n
n1
1x1
(D) none of these
13. The first non zero terms of Maclaurin
series for sec x is
(A)
4
2
15x
1x
224
(B)
24
3
x5
xx
316
(C)
24
4
x5x
1x
216
(D)
24
4
x5
xx
224
14.
If f(x) =
x
nn
n0
2x,
then f
33
(0) is
(A) (33!) 3
33
(B) (32!) 3
33
(C) (3!) 2
32
(D) (33!) 2
33
15. If f(x) = tan
1
x, then f
99
(0) is
(A) 97! (B) 98!
(C) 99! (D) none of these
16. The n
th
non zero term of Maclaurin
series for ln (1 + x
2
) is
(A)
n1
2n
1x
n
(B)
2n
n
x
1
n1
(C)
2n
n1
x
1
n1
(D)
2n 1
n
x
1
n
17. If f(x) =
2
1
,
xx1
then f
36
(0) is
(A) 36! (B) 36!
(C) 2
36
(D) none of these
18. The first three nonzero terms of
Maclaurin series for tan x are
(A)
3
5
x2
x x .....
315
(B)
2
3
x4
1 x .....
215
(C)
2
5
x2
1 x .....
215
(D) none of these
Vidyalankar : GATE – Engineering Mathematics
150
Assignment 8
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
For what value of k will f(x) = x kx
1
have a relative maximum of x = 2
(A) 2 (B) 1
(C) 4 (D) 8
2. A wire 16 feet long is bent to form a
rectangle. If the area of the rectangle
is to be maximized, the dimensions of
the rectangle should be
(A) 2, 6 (B) 5, 3
(C) 4, 4 (D) None of these
3.
x2
|
x2
|
lim
x2
is equal to
(A) 1 (B) 1
(C) 0 (D) does not exit
4. If
x1
fx 2
lim 0
fx 2
then
x1
lim f x
is equal
to
(A) 1 (B) 1
(C) 2 (D) 2
5. Which of the following limit exist ?
(A)
1/ x
x0
lim e
(B)
1/ x
x0
1
lim
1e
(C)
x0
1
lim
x
(D)
2
x0
lim1/ x
6. Evaluate
nlogn
x0
lim e
(A) 1 (B) 0
(C) (D) does not exist
Q7 to Q18 carry two marks each
7.
The first three nonzero terms of
Taylors series for
nx
around 2 is
(A) n2 + (x 2 )
2
1
x2
8
(B)
2
11
n2 + (x - 2) - x - 2
28
(C)
2
11
n2 - (x - 2) + x - 2
24
(D)
2
1
n2 + (x - 2) - x - 2
4
8.
The Maclaurin series for 2
x
is
(A)
n
x
n
n0
n2
x
n1!
(B)
n
x
n1
n0
n2
x
n!
(C)
n
x
n
n0
n2
x
n!
(D) None of these
9.
If f(x) =
2
x
e then
100
f(0) is
(A) 100! (B) 2
100
(C)
100!
50!
(D) 100! 50!
Assignment on Calculus
151
10. If f(x) satisfies f(0) = 2,
f 0 1, f (0) 4,f (0) 12
n
and f (0) 0 for n > 3
then f(x) is
(A) x + 2x
3
(B) 2 + 2x
2
(C) 2 + x + x
2
(D) 2 + x + 2x
2
+ 2x
3
11. The Taylor series for f(x) = 2x
2
+4x 3
about 1 is
(A) 3 + 8 (x 1) + 2 (x 1)
2
(B) 2 + 4x + 2x
2
(C) 2 + 4 (x 1) + 4 (x 1)
2
(D) None of these
12. Evaluate I =
0
sin2kx
dx
sinx
(A) 0 (B) 1/2
(C) 1 (D) /2
13. Evaluate I =
0
log 1 cos d
(A) log2
2
(B) log2
(C) log2 (D) /4 log 2
14.
Evaluate I =
/2
0
sin2x log tanx dx
(A) 0 (B) 1
(C) /2 (D) /2 log 2
15.
The area bounded by curve x
2
= 4y
and straight line x = 4y 2 is
(A) 1/8 (B) 3/8
(C) 9/8 (D) 2
16.
The area bounded by x axis,
y = 1+
2
8
x
ordinate at x = 2 and x = 4 is
(A) 2 (B) 4
(C) 8 (D) 1
17. The area bounded by curve
y = 2x + x
2
x
3
the x axis and lines
x = 1 and x = 1
(A) 1/2 (B) 1
(C) 3/2 (D) 2
18. Evaluate I =
xsinx
dx
1cosx
(A)
x
xcot C
2
(B) x
x
cot C
2
(C)
x
xcot C
2
(D) x + tan
x
C
2
Vidyalankar : GATE – Engineering Mathematics
152
Assignment 9
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
2x
x
x
lim is
1x
(A) e (B)
2
1
e
(C)
1
e
(D) 2e
2.
tan x
x0
1
lim
x
is equal to
(A) 0 (B) 1
(C) 1 (D) 0
3.
x
x0
lim cosecx isequalto
(A) 1 (B) e
(C) r
1
(D) 0
4.
222
n
123 1
lim ......
n
nnn
(A) 0 (B) 1
(C) ½ (D)
5. The function f(x) =
2
ax b x 2
2x2
2ax b x 2
is continuous at x = 2 if
(A) a = 0, b = 1/2
(B) a = 1/2 , b = 0
(C) a = 1/2, b = 0
(D) b = 1/2, a = 0
6.
The limiting value of
12 n
....... as
nn n
n is
(A) 2/3 (B) 1/2
(C) 1/3 (D) does not exist
Q7 to Q18 carry two marks each
7.
I =
/2
43
0
cos x sin dx
(A)
1
35
(B)
2
35
(C)
3
35
(D)
4
35
8. Evaluate I =
2
2
1x
dx
1x
(A)
12
31
sin x x 1 x C
22
(B)
2
1
x1 x C
2
(C) sin
1
+
2
1x C
(D) none of these
9. Evaluate I =
dx
cos x sinx
(A) log tan
x
C
28
(B) log cot
x
C
2
(C) log cot x + C
(D) log tan x + C
Assignment on Calculus
153
10. Evaluate I =
2
dx
13sinx
(A)
1
1
tan tanx C
2
(B) tan
1
(tan x) + C
(C)
1
1
tan 2 tanx C
2
(D) None of these
11. Evaluate I =
44
d
sin cos
(A)
2
1
1tan1
tan
22tan
(B)
2
1
tan 1
tan
tan
(C)
1
11
tan
2
2tan
(D) None of these
12. I =
sinx
dx
sin3x
(A) log
3tanx
C
3tanx
(B)
13tanx
log C
23 3 tanx
(C)
1
log 3 tanx C
3
(D)
13
log C
23 3 tanx
13. I =
/2
2
0
xsinxdx
(A)
2
2
(B) 2 2
(C) + 2 (D) 2
14. The value of
x0
nsinx
lim
x
, where [.] is
the greatest integer function
(A) n + 1 (B) n
(C) n 1 (D) None of these
15. Let
fxy,yz,z and
C :
2
sin 2 t ,t,t .
Then the value of the line integral
C
fdr
from (0, 1, 1) to (0, 1, 1) is
(A) 1 (B) 1
(C) 0 (D) None of these
16. If I
N
=
-x n-1
0
e x logx dx
, then
(A) I
n+2
+ (2n + 1) I
n+1
+ n
2
I
n
= 0
(B) I
n+2
(2n+1) I
n+1
n
2
I
n
= 0
(C) I
n+2
+ (2n + 1) I
n+1
n
2
I
n
= 0
(D) I
n+2
(2n+1) I
n+1
+ n
2
I
n
= 0
17. The function f(x) =
2
11x is
(A) not continuous of x = 0
(B) is continuous but not differentiable
at x = 0
(C) differentiable at x = 0
(D) None of these
18. The function
z = 5xy 4x
2
+ y
2
2x y + 5 has at
x =
1
41
, y =
18
41
(A) maximum (B) saddle point
(C) minimum (D) None of these
Vidyalankar : GATE – Engineering Mathematics
154
Test Paper 1
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
x0
sin3x sinx
Lim
sinx
is equal to
(A) 0 (B) 2
(C) 1 (D) 3
2.
x0
1
Lim sin
x
is
(A) (B) 0
(C) 3 (D) does not exist
3.
x0
sin3x
Lim
5x
is equal to
(A)
3
5
(B)
5
3
(C)
1
5
(D)
1
3
4.
If [x] denotes the greatest integer not
greater than x, then
x2
Lim[x]
is
(A) 0 (B) 1
(C) 2 (D) does not exist
5.
n
x0
(1 x) 1
Lim
x
is equal to
(A) n (B) n
1
(C) 0 (D) does not exist
Q6 to Q15 carry two marks each
6.
2
1
1
dx
x
(A) Converges
(B) Diverges
(C) Converges and diverges
(D) None of these
7.
xx
0
dx
ee
is equal to
(A)
/2 (B) /4
(C)
/3 (D)
8.
2
2
dx
xnx
is equal to
(A)
1
n4
(B)
1
n3
(C)
2
n2
(D)
1
n2
9.
5
3
1
dx
x2
is equal to
(A)
3
91
2
(B)
3
3
91
2
(C)
3
391
(D) None of these
Test Paper on Calculus
155
10. The area under curve y =
2
x
1x
for
0
x 1 is
(A) 1/4 (B) 1/2
(C) 1 (D) 16
11. The volume of solid obtained by
revolving area under y =
2x
e
about
x axis is
(A)
/2 (B) /4
(C) 2
(D)
12.
9
2/3
0
dx
x1
is equal to
(A) 3 (B) 6
(C) 7 (D) 9
13. The area between the curves y = 1/x
and y =
1
x1
to right of line x = 1 is
(A)
n
3 (B)
n
2
(C) 2
n 3 (D) 2 n 2
14.
3
1
dx
x2
(A) converges
(B) diverges
(C) both
(D) neither of the above
15.
p
e
dx
nx
is _________ for p 1
(A) convergent
(B) divergent
(C) convergent as well as divergent
(D) None of these
Vidyalankar : GATE – Engineering Mathematics
156
Test Paper 2
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
Let f (x) = 2x if x < 2
= 2 if x = 2
=
2
x if x > 2
the discontinuity of f (x) at x = 2 is
known as
(A) first kind of discontinuity
(B) second kind of discontinuity
(C) mixed discontinuity
(D) removable discontinuity at x = 2
2. If f (x) = x 1 1 x 2
= 2x 3, 2 x 3
Tick the following alternative which is
appropriate for the above function.
(A) continuous at x = 1
(B) continuous at x = 2
(C) discontinuous at x = 1
(D) discontinuous at x = 2
3. Consider the following statement
Assertion (A) :
The function f (x) = x
[x], x z is
discontinuous at x = 1
Reason (R) :
x1 x1
Left Lim f(x) Right Lim f(x)
Of these statements
(A) Both A and R are true and R is the
correct explanation of A
(B) Both A and R are true and R is not
the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
4. If a function ‘f’ is defined on R by
f (x) = 1 for x a rational number
= 1 for x an irrational number,
then
(A) f (x) is continuous at x = 1
(B) f (x) is continuous at x =
1
(C) f (x) is continuous at x = 0
(D) f (x) is not continuous at any point
5. If f (x) = 4x + 3, x 4
= 3x + 7, x = 4
the function ‘f’ is
(A) continuous at x = 4
(B) discontinuous at x = 4
(C) continuous at x = 4
(D) discontinuous at x =
4
Q6 to Q15 carry two marks each
6.
I =
/2
0
cot xdx
is
(A) 2 (B) /4
(C)
(D) None of these
Test Paper on Calculus
157
7. The area in first quadrant under curve
y =
2
1
x6x10
is
(A) /2 (B)
1
2tan 3
4
(C)
1
tan 3
2
(D)
1
tan 3
2
8.
The area under y =
22
1
xa
for x a + 1 is
(A)
1
na 1
a
(B)
1
na 1
2a
(C)
1
na 1
2
(D) None of these
9. For what values of k, with k 1 and
k > 0 does
1
k
0
1
x
dx converges ?
(A) 2 (B) 1
(C) 3 (D) k > 1
10.
a
22
0
dx
ax
where a > 0 is
(A) n a (B)
1
na 1
2
(C)
1
na 1
2
(D) None of these
11. The minimum value of
2
x5x2is
(A) 5 (B) 0
(C) 1 (D) 2
12. The maximum value of
sinx cos x
f(x)
2
is
(A) 1 (B)
2
(C)
1/ 2 (D) 3
13. The maximum value of 1/x
x
is
(A) e (B) e
e
(C) e
1/e
(D) e
1/e
14.
The function f(c) = x
5
5x
4
+5x
3
1
has
(A) 1 minima and 2 maxima
(B) 2 minima and 1 maxima
(C) 2 minima and 2 maxima
(D) 1 minima and 1 maxima
15. The greatest and least value of
32
x3x
2xin 0,2
32
are
(A)
52
and
63
(B)
11 1
and
66
(C)
82
and
33
(D)
13 1
and
33
Vidyalankar : GATE – Engineering Mathematics
158
Test Paper 3
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
The
16
5
x1
(x 1)
Lim
(x 1)
is
(A)
5
16
(B)
2
17
(C)
16
5
(D)
17
2
2.
2
x2
x4
Lim
x2 3x2
equal to
(A) 8 (B) 8
(C) 4 (D) 4
3.
xx
x0
ab
Lim
x
equals to
(A) log
a
b
(B) log
b
a
(C) log
x
a
(D) log
b
x
4. The function f (x) =
|
x
|
x
at x = 0 has
(A) discontinuity of second kind
(B) discontinuity of first kind
(C) mixed discontinuity
(D) continuous
5.
The function
f (x) =
2
1
xsin
x
, if x 0
= 0, if x = 0 is
(A) differentiable at x = 0 and f ‘(0) = 0
(B) not differentiable at x = 0 since
1
x
, cos x 0
(C) differentiable at x = 0 and the
derivative is continuous at x = 0
(D) not differentiable at any x since it
is not continuous for any x
Q6 to Q15 carry two marks each
6.
The maximum point on curve x = e
x
y is
(A) (1, e) (B) (1, 1/e)
(C) (e, 1) (D) (1/e, 1)
7. If f(x, y) is such that f
x
= e
x
cos y and
f
y
= e
x
sin y then which of following is
true
(A) f(x, y) = e
x + y
sin (x + y)
(B) f(x, y) = e
x
sin
(C) f(x, y) does not exist
(D) None of these
8. If f(x, y) = ln (x
2
+ y
2
) then
(A) f
xx
+ f
yy
= 0 (B) f
xx
f
yy
= 0
(C) f
xx
+2f
yy
= 0 (D) 2f
xx
+ f
yyt
= 0
Test Paper on Calculus
159
9. If z = x
4
+ 3xy y
2
and y = sin x then
dz
dx
is
(A) (4x
3
+ 3 sin x) + (3x 2 sin x) cos x
(B) (2x
2
3 sin x) + (3x 2 cos x)sin x
(C) (4x
3
3 sin x) (3x + 2 sin x) cos x
(D) none of these
10.
A point P is moving along curve of
intersection of paraboloid
22
xy
z
16 9
and cylinder x
2
+ y
2
= 5. If x is
increasing at the rate of 5cm/s how
fast is z changing when x = 2m and
y = 1 cm ?
(A)
25
36
(B)
125
36
(C)
144
25
(D)
18
25
11.
The general solution of
2
2
f
x
= 0 is
(A) f(x, y) = x g(y) + h (y)
(B) f(x, y) = y g(x) + h (x)
(C) f(x, y) = x g(y) + h (x)
(D) f(x, y) = g(x) + h (y)
12. If z = xe
y/x
then
(A)
zz
xyz
xy
(B)
zz
xyz
yz
(C)
zz
xy
xy
(D)
zz
z
xy
13. Find the general solution of
f
xy
= 1
is
(A) f(x, y) = A (x) + B(y) + xy
(B) f(x, y) = yA (x) + xB(y)
(C) f(x, y) = A (x) + B(y)
(D) none of these
14. The equation of tangent plane sphere
phase x
2
+ y
2
+ z
2
= 1 at point
11 1
,,
22
2
is
(A) x + y + z = 2
(B) x + y +
2 z = 2
(C) 2x
y + 32 = 5
(D) 2x + y = 2
15. The altitude h of right circular cone is
decreasing at rate of 3 mm/s while
radius is increasing at 2mm/s. How
fast is volume v changing when
altitude and radius are 100mm and
50mm respectively ?
(A)
2200
3
(B)
2500
3
(C)
12500
3
(D)
1600
3
Vidyalankar : GATE – Engineering Mathematics
160
Test Paper 4
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
Maxima and minima of a continuous
function occur
(A) simultaneously
(B) once
(C) alternately
(D) rarely
2. Let f(x) = | x | than
(A)
f
(0) = 0
(B) f(x) is maximum at x = 0
(C) f(x) is minimum of x = 0
(D) none of these
3. Find the maximum speed with which a
snail moves if the equation of distance
covered (s) in terms of time (t) is
s =
2
3
t
3t 15
2
units ?
(A)
1
9
speed units
(B)
1
18
speed units
(C)
1
27
speed units
(D) None of these
4. The function g (x) =
f(x)
,
x
x 0 has an
extreme value then
(A)
f
(x) = f (x) (B) g
(x) = f(x)
(C) f(x) = 0 (D) g(x) =
f
(x)
5. Which of following is false ?
(A) If
f
(a) = 0 and f
(a) >0 then f (a)
is minimum value
(B) If f
(a) = 0 and f(a) 0 then f(a)
is extreme value
(C) a maximum value is greater than
greatest values
(D) f(a) is minimum value if f(a) is least
value of f(x) in immediate
neighborhood of point [a, f(a)]
Q6 to Q15 carry two marks each
6.
The area included between curves
y = 3x
2
x 3 and y = 2x
2
+ 4x +7 is
(A) 10 (B) 45/2
(C) 15 (D) 30
7. The area included between curve
r = a (sec
+ cos ) and its asymptote
r = a sec
is
(A)
a
2
(B)
2
5a
4
(C)
2
2a
3
(D) 2a
2
8. The mass distributed over the area
bounded by curve 16y
2
= x
3
and line
2y = x assuming that density of a point
of area varies as distance of point from
x axis. Take k as constant of
proportionality
(A) k (B) 1/3k
(C) 2/3k (D) 2k
Test Paper on Calculus
161
9. The density at any point in circular
lamina of radius a varies as its
distance from a fixed point on its
circumference. Take k as proportional
constant, the mass of lamina is
(A)
2
ka (B)
3
1
ka
9
(C)
3
8
ka
9
(D)
3
32
ka
9
10. The volume of sphere x
2
+ y
2
+ z
2
= a
2
cut off by plane z = 0 and cylinder
x
2
+ y
2
= ax is
(A)
3
a
3
(B)
3
2a
3
(C)
a
3
(D) 2a
3
11. The volume cut off from paraboloid
x
2
+
2
1
yz1
4
by plane z = 0 is
(A)
4
(B)
2
(C)
3
4
(D)
12. The volume of solid obtained by
revolution of loop of curve y
2
= x
2
ax
ax
about x axis is
(A)
3
2a
(B)
3
2
2a log2
3
(C)
3
2
alog2
3
(D)
a
3
13. The volume of solid formed by
revolution of the curve (a
x)y
2
= a
2
x
about its asymptote
(A)
23
a
2
(B)
2
a
3
(C)
23
3a
2
(D) 4
2
a
3
14.
The surface of reel formed by
revolution of cycloid x = a (
+ sin ),
y = a(1
cos ) round a tangent at its
vertex is
(A)
2
a
3
(B)
2
2a
3
(C)
2
16 a
3
(D)
2
32 a
3
15.
The volume generated by revolving
area below the x
axis between curve
y8
x
x2
and x axis about line
x + 5 = 0 is
(A) 18
(B) 32
(C) 432
(D) 216
Vidyalankar : GATE – Engineering Mathematics
162
Test Paper 5
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
The minimum value of x +
1
x
, x > 0
(A) 1 (B) 3/2
(C) 5 (D) 2
2. The triangle of maximum area
inscribed in a circle of radius r is
(A) a right angled
with hypotenuse 2r
(B) an equilateral
(C) an isoceles
of height r
(D) does not exist
3. The maximum value of
log x
x
in (0, )
is
(A) e (B) 1/e
(C) 1 (D) none of these
4. A manufacturer sells ‘x’ items at a
price of Rs.
x
3
1000
per item and it
costs Rs.
x
200
2
to produce them.
Find the production level for maximum
profit.
(A) 1250
(B) 1500
(C) 1750
(D) Data insufficient
5. If z is defined as function of x and y by
xy
yz + xz = 0 then
z
x
is
(A)
yz
yx
(B)
yz
yx
(C)
yx
yz
(D)
xz
xy
Q6 to Q15 carry two marks each
6. If f(x) = e
x
and g (x) = e
x
then value of
c in interval [a, b] by Cauchy’s mean
value theorem
(A)
ab
2
(B) ab
(C)
2ab
ab
(D) none of these
7. If f and F be both continuous in [a, b]
and are derivable in (a, b) and if
f
(x) =F
(x) for all x in [a, b] then f(x)
and F (x) differ
(A) by 1 in [a, b]
(B) by x in [a, b]
(C) by constant in [a, b]
(D) none of these
8. If f(x) = 4x
2
then the value of c in [1, 3]
for which
f
(c) =
f3 f 1
4
is
(A) 0 (B) 1
(C) 2 (D) 3
Test Paper on Calculus
163
9. I =
a1 cos
4
00
r cos drd
(A)
5
5
a
16
(B)
5
21
a
16
(C)
5
3
a
16
(D)
4
1
a
4
10. I =
2
y
42
2
2y/4
xdxdy
(A)
24
5
(B)
36
5
(C)
72
5
(D)
108
5
11. I =
bb
00
y 2b y dx dy
(A)
4
b
3
(B)
4
2b
3
(C)
4
b (D)
4
4b
3
12. I = 2
/4 2sin
3
00
rdrd
(A)
3
4
(B)
3
2
4
(C)
3
2
4
(D)
5
2
13. I =
a1 cos
43
00
2rsin drd
(A)
5
2
a
35
(B)
4
5
2
a
35
(C)
65
2a
35
(D)
85
2a
35
14. I =
22
aa a y
2
00
xylog x a
dx dy
xa
(A)
2
a
log a
2
(B)
2
a
loga 1
8
(C)
8aloga
(D)
2
a
2loga 1
8
15. I =
2
a2ax
0x/a
xy dx dy
(A)
4
3
a
8
(B)
4
5
a
8
(C)
4
7
a
8
(D) a
4
Vidyalankar : GATE – Engineering Mathematics
164
Test Paper 6
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
x8
x22
lim is
x8
(A)
0
0
(B)
1
2
(C)
1
42
(D)
1
2
2.
3
x0
xsinx
lim is
x
(A) 1/6 (B) 1/3
(C)
1/3 (D) none of these
3.
xa
1
lim x a cos is
xa
(A) 0 (B) a
(C) cos a (D) does not exist
4. The function f(x) =
2
1cosx
x
(x 0) be
made continuous at x = 0 by defining
f(0) to be
(A) 1 (B) 1/2
(C) 0 (D) e
5. The differentiability of a function
continuity is
(A) sufficient
(B) necessary
(C) sufficient and necessary
(D) none of these
Q6 to Q15 carry two marks each
6.
If I =
2
R
ydA
where R is region
bounded by y = 2x, y = 5x and x = 1.
Then I is equal to
(A)
39
2
(B)
39
4
(C)
39
8
(D)
39
16
7. The value of I =
2
R
1
dA
2y y
where
R is region in the first quadrant
bounded by x
2
= 4 2y is
(A) 0 (B) 2
(C) 4 (D)
1
8. The volume V under plane z = 3x + 4y
and over the rectangle
R : 1 x 2, 0 y 3 is
(A) 25 (B)
49
2
(C)
63
2
(D) 16
9. The volume V in first octant bounded
by z = y
2
, x = 2 and y = 4 is
(A)
64
3
(B)
128
3
(C)
16
3
(D)
32
3
Test Paper on Calculus
165
10. The area of region R bounded by
xy = 1 and 2x + y = 3 is
(A)
n 2 (B) 2 + n 2
(C)
3
n2
4
(D) None of these
11. The volume in first octant bounded by
2x + 2y
z + 1 = 0 y = x and x = 2 is
(A) 10 (B) 28
(C) 12 (D) 16
12. The area of region enclosed by
cardioide r = 1 + cos
is
(A)
2
(B)
3
2
(C)
(D) 2
13. I =
/4 a cos2
/4 0
22
rdr
d
ar
(A) 2a
1
4
(B) 2a 2
4
(C) 2a
1
4
(D) 2a 2
4
14.
I =
2
1x2xy
0x 0
dx dydz
(A)
11
30
(B)
13
30
(C)
1
8
(D) 16
15. I =
21
22
1y1
x y dx dy
(A)
1
3
(B)
1
3
(C)
2
3
(D)
2
3
166
Chapter - 3 : Probability and Statistics
3.1 Basic Terms
Random Experiment
Consider an action which is repeated under essentially identical conditions. If it results in
any one of the several possible outcomes, but it is not possible to predict which outcome
will appear. Such an action is called as a Random Experiment. One performance of such
an experiment is called as a Trial.
Sample Space
The set of all possible outcomes of a random experiment is called as the sample space.
All the elements of the sample space together are called as ‘exhaustive cases’. The
number of elements of the sample space i.e. the number of exhaustive cases is denoted
by n(S) or N or n.
Event
Any subset of the sample space is called as an ‘Event’ and is denoted by some capital
letter like A, B, C or A
1
, A
2
, A
3
,.. or B
1
, B
2
, ... etc.
Favourable cases
The cases which ensure the happening of an event A, are called as the cases favourable
to the event A. The number of cases favourable to event A is denoted by n(A) or N
A
or n
A
.
Mutually Exclusive Events or Disjoint Events
Two events A and B are said to be mutually exclusive or disjoint if A B =
i.e. if there is no element common to A & B.
Equally Likely Cases
Cases are said to be equally likely if they all have the same chance of occurrence i.e. no
case is preferred to any other case.
Permutation
A permutation is an arrangement of all or part of a set of objects. The number of
permutations of n distinct objects taken r at a time is
n
P
r
=
n!
nr!
Notes on Probability and Statistics
167
Note: The number of permutations of n distinct objects is n! i.e.,
n
P
n
= n!
The number of permutations of n distinct objects arranged in a circle
is (n 1)!
The number of distinct permutations of n things of which n
1
are of one
kind, n
2
of a second kind … n
k
of a k
th
kind is
12 k
n!
n ! n ! ......n !
Combination
A combination is selection of all or part of a set of objects. The number of combinations of
n distinct objects taken r at a time is
n
r
n!
C=
r! n r !
Note: In a permutation, the order of arrangement of the objects is important.
Thus abc is a different permutation from bca.
In a combination, the order in which objects are selected does not matter.
Thus abc and bca are the same combination.
3.2 Definition of Probability
Consider a random experiment which results in a sample space containing n(S) cases
which are exhaustive, mutually exclusive and equally likely. Suppose, out of n(S) cases,
n(A) cases are favourable to an event A. Then the probability of event A is denoted by
P(A) and is defined as follows.
P(A) =
n(A)
n(S)
=
number of cases favourable to event A
number of cases in the sample space S
3.3 Complement of an event
The complement of an event A is denoted by
A
and it contains all the elements of the
sample space which do not belong to A.
For example: Random experiment: an unbiased die is rolled.
S = {1, 2, 3, 4, 5, 6}
(i) Let A: number on the die is a perfect square
A = {1, 4}
A
= {2, 3, 5, 6}
Vidyalankar : GATE – Engineering Mathematics
168
(ii) Let B: number on the die is a prime number
B = {2, 3, 5}
B
= {1, 4, 6}
Note: P(A) + P(
A
) = 1 i.e. P(A) = 1 P(
A
)
For any events A and B,
PA=PAB+PAB
3.4 Independent Events
Two events A & B are said to be independent if
P(A B) = P(A).P(B)
Note: If A & B are independent then
A &
B
are independent
A
& B are independent
A
&
B
are independent
3.5 Theorems of Probability
Addition Theorem
If A and B are any two events then
P(AB) = P(A) + P(B) – P(AB)
Note: 1. A B : either A or B or both i.e. at least one of A & B
AB
: neither A nor B i.e. none of A & B
AB &
AB
are complement to each other
P(
AB
) = 1 – P(A B)
2. If A & B are mutually exclusive, P(A B) = 0
P(A B) = P(A) + P(B)
3.
123 12312
PA A A =PA +PA +PA PA A
23 31 123
PA A PA A +PA A A
Notes on Probability and Statistics
169
Multiplication Theorem
If A & B are any two events then
P(A B) = P(A).P(B/A) = P(B).P(A/B)
1. Conditional probability of occurrence of event B given that event A has already
occurred.
P(B/A) =
PA B
PA
2. Conditional probability of occurrence of event A given that event B has already
occurred
P(A/B) =
PA B
PB
Bayes’ Theorem
Suppose that a sample space S is a union of mutually disjoint events B
1
, B
2
, B
3
, ..., B
n
,
suppose A is an event in S, and suppose A and all the B
i
’s have nonzero probabilities. If
k is an integer with 1 ≤ k ≤ n, then
P(B
k
/ A) =
kk
11 2 2 n n
PA/B PB
P A /B P B +P A /B P B +...+P A /B P B
Solved Example 1 :
A single die is tossed. Find the probability
of a 2 or 5 turning up.
Solution :
The sample space is S = {1, 2, 3, 4, 5, 6}
P(1) = P(2) = … = P(6) =
1
6
The event that either 2 or 5 turns up in
indicated by 2 5. Thus
11 1
P2 5 P2 P5
66 3
Solved Example 2 :
A coin is tossed twice. What is the
probability that at least one head occurs ?
Solution :
The sample space is
S = {HH, HT, TH, TT}
Probability of each outcomes = 1/4
Probability of atleast one head occurring is
P(A) =
1113
444 4
Vidyalankar : GATE – Engineering Mathematics
170
Solved Example 3 :
A die is loaded in such a way that an even
number is twice as likely to occur as an
odd number. If E is the event that a
number less than 4 occurs on a single toss
of the die. Find P(E).
Solution :
S = {1, 2, 3, 4, 5, 6 }
We assign a probability of w to each odd
number and a probability of 2w to each
even number. Since the sum of the
probabilities must be 1, we have 9w = 1 or
w = 1/9.
Hence probabilities of 1/9 and 2/9 are
assigned to each odd and even number
respectively.
E = {1, 2, 3}
and P(E) =
121 4
999 9
Solved Example 4 :
In the above example let A be the event
that an even number turns up and let B be
the event that a number divisible by 3
occurs. Find P(A B) and P(A B).
Solution :
A = {2, 4, 6} and
B = {3, 6}
We have,
A B = {2, 3, 4, 6} and
A B = {6}
By assigning a probability of 1/9 to each
odd number and 2/9 to each even number
21227
PA B
9999 9
and
2
PA B
9
Solved Example 5 :
A mixture of candies 6 mints, 4 toffees and
3 chocolates. If a person makes a random
selection of one of these candies, find the
probability of getting (a) a mint, or
(b) a toffee or a chocolate.
Solution :
(a) Since 6 of the 13 candies are mints,
the probability of event M, selecting
mint at random, is
P(M) =
6
13
(b) Since 7 of the 13 candies are toffees
or chocolates it follows that
7
PT C
13
Solved Example 6 :
In a poker hand consisting of 5 cards, find
the probability of holding 2 aces and
3 jacks.
Solution :
The number of ways of being dealt 2 aces
from 4 is
4
2
4!
C6
2!2!
The number of ways of being dealt 3 jacks
from 4 is
4
3
4!
C4
3!1!
Notes on Probability and Statistics
171
Total number of ways n = 6.4 = 24
hands with 2 aces and 3 jacks.
The total number of 5card poker hands all
of which are equally likely, is
N =
52
5
52!
C
5!47!
= 2, 598, 960
The probability of event C of getting 2
aces and 3 jacks in a 5card poker
hand is
P(C) =
5
24
0.9 10
2,598,960
Solved Example 7 :
The probability that Paula passes
mathematics is 2/3 and the probability that
she passes English is 4/9. If the
probability of passing both courses is 1/4,
what is the probability that Paula will pass
at least one of these courses ?
Solution :
If M is the event “passing Mathematics”
and E the event “passing English” then
P(M E) =
PM PE PM E
=
241 31
394 36
Solved Example 8 :
What is the probability of getting a total of
7 or 11 when a pair of dice are tossed ?
Solution :
Let A be the event that 7 occurs and B the
event that 11 comes up. Now a total of 7
occurs for 6 of the 36 sample points and a
total of 11 occurs for only 2 of the sample
points. Since all sample points are equally
likely, we have P(A) = 1/6 and P(B) = 1/18.
The events A and B are mutually exclusive
since a total of 7 and 11 cannot both occur
on the same toss.
PA B PA PB
=
11
618
=
2
9
Solved Example 9 :
If the probabilities are respectively 0.09,
0.15, 0.21 and 0.23 that a person
purchasing a new automobile will choose
the colour green, white, red or blue. What
is the probability that a given buyer will
purchase a new automobile that comes in
one of those colours ?
Solution :
Let G, W, R and B be the events that a buyer
selects, respectively, a green, white, red or
blue automobile. Since these four events are
mutually exclusive the probability is
PG W R B
PG PW PR PB
0.09 0.15 0.21 0.23
= 0.68
Solved Example 10 :
If the probabilities that an automobile
mechanic will service 3, 4, 5, 6, 7 or 8
more cars on any given workday are
respectively 0.12, 0.19, 0.28, 0.24, 0.10
and 0.07. What is the probability that he
will service at least 5 cars on next day at
work?
Vidyalankar : GATE – Engineering Mathematics
172
Solution :
Let E be the event that at least 5 cars are
serviced. Now
P(E) = 1 P(E), where E is the event
that fewer than 5 cars are serviced.
Since P(E) = 0.12 + 0.19 = 0.31
P(E) = 1 0.31 = 0.69
Solved Example 11 :
In a garden 40% of the flowers are roses
and the rest are carnations. If 25% of the
roses and 10% of the carnations are red,
the probability that a red flower selected at
random is a rose.
(A) 5/6 (B) 1/4
(C) 4/5 (D) 5/8
Solution :
Suppose there are 100 flowers
Number of roses = 40; Number of
carnations = 60
25% of 40 = 10 roses are red and
10% of 60 = 6 carnations are red
Let A be the event that the flower is red
and B the event that the flower is a rose.
A B is the event that the flower is a
red rose.
n(A) = 16 P(A) =
16
100
n(A B) = 10 P (A B) =
10
100
P(B/A) = probability that a selected flower
is a rose red is colour
P(B/A) =
PA B
10 / 100 5
P A 16 / 100 8
3.6 Random Variables
A random variable is a function that associates a real number with each element in the
sample space. It is a set of possible values from a random experiment.
For example: Suppose that a coin is tossed twice so that the sample space is S = {HH,
HT, TH, TT}. Let X represent the number of heads that can come up. With each sample
point we can associate a number for X as shown in the table.
SAMPLE POINT HH HT TH TT
X 2 1 1 0
Discrete Random Variable
A random variable is discrete if it can take only discrete or separate values. Value of a
discrete random variable is often obtained by counting.
For example: Number of heads appearing when three coins are tossed.
Notes on Probability and Statistics
173
Continuous Random Variable
A random variable is continuous if can take any value in a given range. Value of a
continuous random variable is often obtained by measuring.
For example: Time required to travel certain distance.
3.7 Probability Distribution Function
In general sense, probability distribution function may refer to:
Probability mass function
Probability density function
Cumulative distribution function
Probability mass function
It is a function that gives probability that a discrete random variable is exactly equal to
some value.
Probability density function
It is used to specify the probability of a continuous random variable falling within a
particular range of values and the probability is given by the integral of the probability
density function over that range.
Cumulative distribution function
It is a probability that random variable X will take a value less than or equal to x.
i.e., F(x) = P(X ≤ x)
3.8 Expectation(Mean), Variance and Standard Deviation
Expected value for a discrete random variable is given by
E(X) =
n
ii
i=1
x P(x)
Expected value for a continuous random variable is given by
E(X) =
b
a
x f(x) dx
Vidyalankar : GATE – Engineering Mathematics
174
Variance of random variable is given by
Var(X) = E((X )
2
) = E(X
2
) E(X)
2
= E(X
2
)
2
Standard deviation is square root of variance
=
Var(X)
Note: 1. Expected value
= E(X) is a measure of central tendency.
2. Standard deviation is a measure of spread.
Solved Example 12 :
Thirteen cards are drawn simultaneously
from a deck of 52 cards. If aces count 1,
face cards 10 and other count by their
denomination, find the expectation of the
total score on 13 cards.
Solution :
Let x denote the number corresponding to
the i
th
card. Then x takes the value, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10, 10, 10, 10 each having
the same probability
1
13
so that
p(x) =
1
12...10101010
13
85
13
Hence the required expectation is 85.
3.9 Standard Distributions
Binomial Distribution
Probability of exactly k successes in n trials is given by
P(X=k) =
n
C
k
p
k
(1 p)
n
k
where p is the probability of success in each trial
Here the trial are Bernouilli trials, where each trial can have exactly two outcomes viz.
success and failure.
If p is the probability of success
then q = 1 p is the probability of failure.
Then
P(X=k) =
n
C
k
p
k
q
n
k
Notes on Probability and Statistics
175
For Binomial Distribution,
Mean = np
Variance = npq = np(1 p)
Variance < Mean
Cumulative distribution function is, F(x) =
k
nini
i
i0
Cpq
Poisson Distribution
Probability of k events occurring in an interval is given by
P(k events) =
k
e
k!
where is the average number of events per interval.
For a Poisson Distribution,
Mean =
Variance =
Variance = Mean
Cumulative distribution function is, F(x) =
i
k
i0
e
i!
Normal Distribution
Probability density function of normal distribution is
f(x) =
2
(x )
2
1
e
2
where is the mean and
2
is the variance
Graph of f(x) is symmetric about x = and is a bell shaped curve.
Cumulative distribution function is, F(x) =
1x
1erf
2
2
Vidyalankar : GATE – Engineering Mathematics
176
Note: Standard Normal Distribution
Special case with = 0 and = 1
Probability density function is f(x) =
2
x
2
e
2
This graph is symmetric about x = 0 where it attains its maximum value
1
2
Uniform Distribution
Probability density function for uniform distribution is
f(x) =
1
ba
a ≤ x ≤ b
0 x < a or x > b
For Uniform Distribution,
Mean =
ab
2
Variance =
2
(b a)
12
Cumulative distribution function is, F(x) =
0xa
xa
axb
ba
1xb
Exponential Distribution
Probability density function for uniform distribution is
f(x) =
e
x
x ≥ 0
0 x < 0
For exponential distribution
Mean =
1
Variance =
2
1
Cumulative distribution function is, F(x) = 1 e
x
Notes on Probability and Statistics
177
Solved Example 13 :
The probability of getting exactly 2 heads
in 6 tosses of a fair coin is
Solution :
P (X = 2) =
262
6
2
11
C
22
=
262
6! 1 1 15
2!4! 2 2 64
Solved Example 14 :
The probability that a certain kind of
component will survive a given shock test
is 3/4. Find the probability that exactly 2 of
the next 4 components tested survive.
Solution :
Assuming the tests are independent and
p = ¾ for each of the 4 tests we get
22
4
2
31
C
42
=
2
4
4! 3
2!2!
4
=
27
128
Solved Example 15 :
The probability that a patient recovers from
a rare blood disease is 0.4. If 15 people
are to have contracted this disease, what
is the probability that
a) at least 10 survive
b) from 3 to 8 survive
c) exactly 5 survive
Solution :
Let X be the number of people that survive.
a) P(X 10) =
1P(X 10)
=
9
15 i 15 i
i
i0
1C(0.4)(0.6)
= 1 0.9662 = 0.0338
b) P(3 X 8) =
8
15 i 15 i
i
i3
C(0.4)(0.6)
= 0.8779
c) P (X = 5) =
15 5 15 5
5
C(0.4)(0.6)
= 0.1859
Solved Example 16 :
The probability is 0.02 that an item
produced by a factory is defective. A
shipment of 10,000 items are sent to its
warehouse. Find the expected number E
of defective items and the standard
deviation.
Solution :
= np = 10000 0.02 = 200
=
npq 10000
=
196 = 14
Solved Example 17 :
Ten percent of the tools produces in a
certain manufacturing turn out to be
defective. Find the probability that in a
sample of 10 tools chosen at random
exactly two will be defective.
Solution :
We have = np = 10 (0 1) = 1
Then according to the Poisson distribution
P (X = x ) =
x
e
x!
or P(X = 2) =
2
1
1e
0.1839
2!
or 0.18
In general the approximation is good if
P 0.1 and = np 5
Vidyalankar : GATE – Engineering Mathematics
178
Solved Example 18 :
If the probability that an individual suffers a
bad reaction from injection of a given
serum is 0.001, determine the probability
that out of 2000 individuals
a) exactly 3
b) more than 2, individuals will suffer a
bad reaction.
Solution :
Let X denote the number of individuals
suffering a bad reaction. X is poisson
distributed.
P (X = x) =
x
e
x!
where = np = 2000 (0.001) = 2
a) P (X = 3) =
32
2e
0.180
3!
b) P (X > 2)
=
1PX0 PX1PX2
=
02 12 22
2e 2e 2e
1
0! 1! 2!
= 1 5e
2
= 0.323
3.10 Mean, Median, Mode and Standard Deviation
Mean
Mean of a data generally refers to arithmetic mean of the data. However, there are two
more types of mean which are geometric mean and harmonic mean. The different types
of mean for set of values and grouped data are given in the following table.
Set of Values Grouped Data
Arithmetic Mean
i
x
n
i
i
i
fx
f
Geometric Mean
1/ n
12 3 n
(x . x . x .... x )
ii
i
f log x
Antilog
f
Harmonic Mean
12 3 n
n
111 1
+ +
xxx x
i
ii
f
fx
Notes on Probability and Statistics
179
Median
For an ordered set of values, median is the middle value. If the number of values is even,
median is taken as mean of the middle two values.
For a grouped data, median is given by
Median =
N
C
2
L + h
f
where, L = Lower boundary of the median class
N =
f
i
C = Cumulative frequency upto the class before the median class
h = Width of median class
f = Frequency of the median class
Mode
For a set of values, mode is the most frequently occurring value. For a grouped data,
mode is given by
Mode =
ii-1
ii-1i+1
f f
L + h
2f f f
where, L = Lower boundary of the modal class
f
i
= Frequency of the modal class (Highest frequency)
f
i1
= Frequency of the class before the modal class
f
i+1
= Frequency of the class after the modal class
h = Width of the modal class
Note: Relation between Mean, Median and Mode :
Mean Mode = 3(Mean Median)
Standard Deviation
Standard deviation is given by,
2
=
2
i
x x
n
For a Set of Values
2
=
2
ii
i
fx x
f
For a Grouped Data
Vidyalankar : GATE – Engineering Mathematics
180
3.11 Correlation and Regression Analysis
Correlation
Correlation analysis deals with finding linear association between two variables i.e. if the
two variables are linearly related to each other.
Correlation coefficient between variables x and y can be found as follows.
r =
ii
22
ii
(x x) (y y)
(x x) (y y)
=
i
ii i
2222
ii ii
nxy x y
nx (x) ny (y)
where, x = mean of x
y = mean of y
Regression
Regression analysis involves identifying the relationship between a dependent variable
and one or more independent variables.
If X and Y are the variables then equation for regression line of Y on X is given by
Y = a + bX
where, b =
22
nxy x y
n x ( x)
and a can be found by using
y = a + bx
Note: Correlation coefficient measures linear association between two variables.
r can vary between 1 to 1 i.e. 1 ≤ r ≤ 1.
If there is no linear association between the two variables, r will be zero.
Notes on Probability and Statistics
181
List of Formulae
Permutation and Combination
n
P
r
=
n!
nr!
n
r
n!
C=
r! n r !
Probability
P(A) =
n(A)
n(S)
P(A) + P(
A
) = 1
PA =PA B+PA B
P(AB) = P(A) + P(B) – P(AB)
12 3 1 2
PA A A =PA +PA
31223
+P A P A A P A A
31 123
PA A +PA A A
Two events A & B are said to be
independent if
P(A B) = P(A).P(B)
P(A B) = P(A).P(B/A) = P(B).P(A/B)
Conditional probability of occurrence
of event B given that event A has
already occurred
P(B/A) =
P(A B)
P(A)
Bayes’ Theorem
P(B
k
/A) =
kk
n
ii
i1
P(B ) P(A / B )
P(B P(A/B
Expectation(Mean), Variance and
Standard Deviation
E(X) =
n
ii
i1
xP(x)
…. For Discrete Random Variable
E(X) =
b
a
x f(x) dx
….. For Continuous Random Variable
Var(X) = E((X )
2
) = E(X
2
) E(X)
2
= E(X
2
)
2
= Var(X)
Binomial Distribution
P(X=k) =
n
C
k
p
k
(1 p)
n
k
Mean = np
Variance = npq = np(1 p)
Variance < Mean
Cumulative distribution function is,
F(x) =
k
nini
i
i0
Cpq
Poisson Distribution
P(k events) =
k
e
k!
Mean =
Variance =
Variance = Mean
Vidyalankar : GATE – Engineering Mathematics
182
Cumulative distribution function is,
F(x) =
i
k
i0
e
i!
Normal Distribution
f(x) =
2
(x )
2
1
e
2
Mean =
Variance =
2
Cumulative distribution function is,
F(x) =
1x
1erf
2
2
Uniform Distribution
f(x) =
1
ba
a ≤ x ≤ b
0 x < a or x > b
Mean =
ab
2
Variance =
2
(b a)
12
Cumulative distribution function is,
F(x) =
0xa
xa
axb
ba
1xb
Exponential Distribution
f(x) =
e
x
x ≥ 0
0 x < 0
Mean =
1
Variance =
2
1
Cumulative distribution function is,
F(x) = 1 e
x
Mean
Arithmetic Mean
=
i
x
n
…. for set of values
=
ii
i
fx
f
…. for grouped data
Geometric Mean
= (x
1
.x
2
.x
3
….x
n
)
1/n
…. for set of values
= Antilog
ii
i
flogx
f
…. for grouped data
Harmonic Mean
=
12 3 n
n
111 1
xx x x
…. for set of values
=
i
ii
f
fx
…. for grouped data
Median
For an ordered set of values, median
is the middle value. If the number of
values is even, median is taken as
mean of the middle two values.
For a grouped data,
Median =
N
C
2
Lh
f
Notes on Probability and Statistics
183
where,
L = Lower boundary of the median class
N = f
i
C = Cumulative frequency upto the
class before the median class
h = Width of median class
f = Frequency of the median class
Mode
For a set of values, mode is the most
frequently occurring value.
For a grouped data,
Mode =
ii1
ii1i1
ff
L
2f f f
h
where,
L = Lower boundary of the modal class
f
i
= Frequency of the modal class
(Highest frequency)
f
i1
= Frequency of the class before the
modal class
f
i+1
= Frequency of the class after the
modal class
h = Width of the modal class
Relation between Mean, Median and
Mode :
Mean Mode = 3(Mean Median)
Standard Deviation
2
=
2
i
xx
n
For a Set of Values
2
=
2
ii
i
fx x
f
For a Grouped Data
Vidyalankar : GATE – Engineering Mathematics
184
Assignment 1
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 10 carry one mark each
1. A sample of 15 data is as follows: 17,
18, 17, 17, 13, 18, 5, 5, 6, 7, 8, 9, 20,
17, 3. The mode of the data is
[ME 2017]
(A) 4 (B) 13
(C) 17 (D) 20
2. In a housing society, half of the
families have a single child per family,
while the remaining half have two
children per family. The probability that
a child picked at random, has a sibling
is _______. [EC2014]
3. An unbiased coin is tossed an infinite
number of times. The probability that
the fourth head appears at the tenth
toss is [EC2014]
(A) 0.067 (B) 0.073
(C) 0.082 (D) 0.091
4. The probability density function of
evaporation E on any day during a
year in a watershed is given by
f(E) =
1
0 E 5 mm / day
5
0 otherwise
The probability that E lies in between 2
and 4 mm/day in a day in the
watershed is (in decimal) _________
[CE2014]
5. Among the four normal distributions
with probability density functions as
shown below, which one has the
lowest variance? [ME 2015]
(A) I (B) II
(C) III (D) IV
6. A random variable X has probability
density function f(x) as given below:
f(x) =
abxfor0x1
0 otherwise
If the expected value E[X] = 2/3, then
Pr[X < 0.5] is _______. [EE2015]
7. Consider a Poisson distribution for the
tossing of a biased coin. The mean for
this distribution is . The standard
deviation for this distribution is given
by [ME2016]
(A)
(B)
2
(C) (D) 1/
II
I
III
IV
2
1
0
1
2
Assignment on Probability and Statistics
185
8. The probability of getting a “head” in a
single toss of a biased coin is 0.3. The
coin is tossed repeatedly till a “head” is
obtained. If the tosses are independent,
then the probability of getting “head” for
the first time in the fifth toss is _______
[EC2015]
9. X and Y are two random independent
events. It is known that P(X) = 0.40 and
P(X Y
C
) = 0.7. Which one of the
following is the value of P(X Y)?
[CE2016]
(A) 0.7 (B) 0.5
(C) 0.4 (D) 0.3
10. Suppose that a shop has an equal
number of LED bulbs of two different
types. The probability of an LED bulb
lasting more than 100 hours given that
it is of Type 1 is 0.7, and given that it is
of Type 2 is 0.4. The probability that
an LED bulb chosen uniformly at
random lasts more than 100 hours is
________. [CS2016]
Q 11 to Q 20 carry two marks each
11. In the following table, x is a discrete
random variable and p(x) is the
probability density. The standard
deviation of x is [ME2014]
x 1 2 3
P(x) 0.3 0.6 0.1
(A) 0.18 (B) 0.36
(C) 0.54 (D) 0.6
12. The number of accidents occurring in
a plant in a month follows Poisson
distribution with mean as 5.2. The
probability of occurrence of less than 2
accidents in the plant during a
randomly selected month is[ME2014]
(A) 0.029 (B) 0.034
(C) 0.039 (D) 0.044
13. Parcels from sender S to receiver R
pass sequentially through two post-
offices. Each post-office has a
probability
1
5
of losing an incoming
parcel, independently of all other
parcels. Given that a parcel is lost, the
probability that it was lost by the
second post-office is ____.
[EC2014]
14. A traffic office imposes on an average
5 number of penalties daily on traffic
violators. Assume that the number of
penalties on different days is
independent and follows a Poisson
distribution. The probability that there
will be less than 4 penalties in a day is
________.
[CE2014]
Vidyalankar : GATE – Engineering Mathematics
186
15. The chance of a student passing an
exam is 20%. The chance of a student
passing the exam and getting above
90% marks in it is 5%. GIVEN that a
student passes the examination, the
probability that the student gets above
90% marks is
[ME2015]
(A)
1
18
(B)
1
4
(C)
2
9
(D)
5
18
16. The probability density function of a
random variable, x is
f(x) =
2
x
(4 x )
4
for 0 x 2
= 0 otherwise
The mean,
x
of the random variable is
___________.
[CE2015]
17. Suppose a fair six-sided die is rolled
once. If the value on the die is 1, 2, or
3, the die is rolled a second time. What
is the probability that the sum total of
values that turn up is at least 6?
[EC2015]
(A) 10/21 (B) 5/12
(C) 2/3 (D) 1/6
18. The variance of the random variable X
with probability density function
f(x) =
|
x
|
1
|
x
|
e
2
is _____. [EC2015]
19. Two random variables ܺ and ܻ are
distributed according to
F
X,Y
(x, y)
=
(x y), 0 x 1, 0 y 1,
0, otherwise.
The probability P(X + Y
1) is ______.
[EC2016]
20. An urn contains 5 red and 7 green
balls. A ball is drawn at random and its
colour is noted. The ball is placed back
into the urn along with another ball of
the same colour. The probability of
getting a red ball in the next draw is
[IN2016]
(A)
65
156
(B)
67
156
(C)
79
156
(D)
89
156
Assignment on Probability and Statistics
187
Assignment 2
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 10 carry one mark each
1.
A group consists of equal number of
men and women. Of this group 20% of
the men and 50% of the women are
unemployed. If a person is selected at
random from this group, the probability
of the selected person being employed
is_______
[ME2014]
2. Vehicle arriving at an intersection from
one of the approach roads follow the
Poisson distribution. The mean rate of
arrival is 900 vehicles per hour. If a
gap is defined as the time difference
between two successive vehicle
arrivals (with vehicle assumed to be
points), the probability (up to four
decimal places) that the gap is greater
than 8 second is _______.
[CE2017]
3. Let X be a zero mean unit variance
Gaussian random variable. E[|X|] is
equal to _______.
[EC2014]
4. A fair (unbiased) coin was tossed four
times in succession and resulted in the
following outcomes: (i) Head, (ii) Head,
(iii) Head, (iv) Head. The probability of
obtaining a ‘Tail’ when the coin is
tossed again is
[CE2014]
(A) 0 (B)
1
2
(C)
4
5
(D)
1
5
5. Three vendors were asked to supply a
very high precision component. The
respective probabilities of their
meeting the strict design specifications
are 0.8, 0.7 and 0.5. Each vendor
supplies one component. The
probability that out of total three
components supplied by the vendors,
at least one will meet the design
specification is ________.
[ME2015]
6. Consider the following probability
mass function (p.m.f.) of a random
variable X:
p(x, q) =
qifX0
1q ifX1
0 otherwise
If q = 0.4, the variance of X is ______.
[CE2015]
7. The area (in percentage) under
standard normal distribution curve of
random variable Z within limits from
−3 to +3 is __________
[ME2016]
Vidyalankar : GATE – Engineering Mathematics
188
8. Types II error in hypothesis testing is
(A) acceptance of the null hypothesis
when it is false and should be
rejected
(B) rejection of the null hypothesis
when it is true and should be
accepted
(C) rejected of the null hypothesis
when it is false and should be
rejected
(D) acceptance of the null hypothesis
when it is true and should be
accepted
[CE2016]
9. A voltage V
1
is measured 100 times
and its average and standard deviation
are 100 V and 1.5 V respectively. A
second voltage V
2
, which is
independent of V
1
, is measured
200 times and its average and
standard deviation are 150 V and 2 V
respectively. V
3
is computed as:
V
3
= V
1
+ V
2
. Then the standard
deviation of V
3
in volt is ________.
[IN2016]
10. Suppose p is the number of cars per
minute passing through a certain road
junction between 5 PM and 6 PM, and
p has a Poisson distribution with mean
3. What is the probability of observing
fewer than 3 cars during any given
minute in this interval?
[CS2013]
(A) 8/(2e
3
) (B) 9/(2e
3
)
(C) 17/(2e
3
) (D) 26/(2e
3
)
Q 11 to Q 20 carry two marks each
11.
Consider an unbiased cubic dice with
opposite faces coloured identically and
each face coloured red, blue or green
such that each colour appears only
two times on the dice. If the dice is
thrown thrice, the probability of
obtaining red colour on top face of the
dice at least twice is ____.
[ME2014]
12. Let X be a real-valued random variable
with E[X] and E[X
2
] denoting the mean
values of X and X
2
, respectively. The
relation which always holds true is
(A) (E[X])
2
> E[X
2
] [EC2014]
(B) E[X
2
] (E [X])
2
(C) E[X
2
] = (E[X])
2
(D) E[X
2
] > (E[X])
2
13. A fair coin is tossed n times. The
probability that the difference between
the number of heads and tails is
(n
3) is [EE2014]
(A) 2
n
(B) 0
(C)
n
C
n3
2
n
(D) 2
n + 3
14. An observer counts 240 veh/h at a
specific highway location. Assume that
the vehicle arrival at the location is
Poisson distributed, the probability of
having one vehicle arriving over a
30
second time interval is _________.
[CE2014]
Assignment on Probability and Statistics
189
15. Two players, A and B, alternately keep
rolling a fair dice. The person to get a six
first wins the game. Given that player A
starts the game, the probability that A
wins the game is
[EE2015]
(A) 5/11 (B) 1/2
(C) 7/13 (D) 6/11
16. The probability that a thermistor
randomly picked up from a production
unit is defective is 0.1. The probability
that out of 10 thermistors randomly
picked up, 3 are defective is
[IN2015]
(A) 0.001 (B) 0.057
(C) 0.107 (D) 0.3
17. Let X {0, 1} and Y (0, 1) be two
independent binary random variables.
If P(X = 0) = p and P(Y = 0) = q, the
P(X + Y
1) is equal to [EC2015]
(A) pq + (1
p) (1 q)
(B) pq
(C) p(1
q)
(D) 1
pq
18.
The probability that a screw
manufactured by a company is
defective is 0.1. The company sells
screws in packets containing 5 screws
and gives a guarantee of replacement
if one or more screws in the packet are
found to be defective. The probability
that a packet would have to be
replaced is _______.
[ME2016]
19.
If a random variable X has a Poisson
distribution with mean 5, then the
expectation E[(X + 2)
2
] equals
_________.
[CS2017]
20.
Consider the following experiment.
Step 1. Flip a fair coin twice.
Step 2. If the outcomes are (TAILS,
HEADS) then output Y and stop.
Step 3. If the outcomes are either
(HEADS, HEADS) or (HEADS,
TAILS), then output N and stop.
Step 4. If the outcomes are (TAILS,
TAILS), then go to Step 1.
The probability that the output of the
experiment is Y is (up to two decimal
places_______.
[CS2016]
Vidyalankar : GATE – Engineering Mathematics
190
Assignment 3
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 10 carry one mark each
1.
A nationalized bank has found that the
daily balance available in its savings
accounts follows a normal distribution
with a mean of Rs. 500 and a standard
deviation of Rs. 50. The percentage of
savings account holders, who maintain
an average daily balance more than
Rs. 500 is________
[ME2014]
2. Let X be a random variable which is
uniformly chosen from the set of
positive odd numbers less than 100.
The expectation, E[X], is _________.
[EC2014]
3. Ram and Ramesh appeared in an
interview for two vacancies in the
same department. The probability of
Ram's selection is 1/6 and that of
Ramesh is 1/8. What is the probability
that only one of them will be selected?
[EC2015]
(A) 47/48 (B) 1/4
(C) 13/48 (D) 35/48
4.
If {x} is a continuous, real valued
random variable defined over the
interval (
∞, + ∞) and its occurrence is
defined by the density function given as:
f(x) =
1
2b
2
1xa
2b
e
where 'a' and 'b'
are the statistical attributes of the
random variable {x}. The value of the
integral
2
1x a
a
2b
1
e
2b
dx is
[CE2014]
(A) 1 (B) 0.5
(C)
(D)
2
5. If P(X) = 1/4, P(Y) = 1/3, and
P(X Y) = 1/12, the value of P(Y/X) is
[ME2015]
(A)
1
4
(B)
4
25
(C)
1
3
(D)
29
50
Assignment on Probability and Statistics
191
6. Suppose A and B are two independent
events with probabilities P(A)
0 and
P(B)
0. Let A and B be their
complements. Which one of the
following statements is FALSE?
[EC2015]
(A) P(A
B) = P(A) P(B)
(B) P(A|B) = P(A)
(C) P(A
B) = P(A) + P(B)
(D)
P(A B) = P(A)P(B)
7. The second moment of a Poisson-
distributed random variable is 2. The
mean of the random variable is _____.
[EC2016]
8. The spot speeds (expressed in km/hr)
observed at a road section are 66, 62,
45, 79, 32, 51, 56, 60, 53, and 49. The
median speed (expressed in km/hr) is
______
(Note: answer with one
decimal accuracy) [CE2016]
9. A probability density function on the
interval [a, 1] is given by 1/x
2
and
outside this interval the value of the
function is zero. The value of a is
______.
[CS2016]
10. Let U and V be two independent zero
mean Gaussian random variables of
variances
1
4
and
1
9
respectively. The
probability P(3V
2U) is [ME2013]
(A) 4/9 (B) 1/2
(C) 2/3 (D) 5/9
Q 11 to Q 20 carry two marks each
11.
A machine produces 0, 1 or 2
defective pieces in a day with
associated probability of 1/6, 2/3 and
1/6, respectively. The mean value and
the variance of the number of
defective pieces produced by the
machine in a day, respectively, are
[ME2014]
(A) 1 and 1/3 (B) 1/3 and 1
(C) 1 and 4/3 (D) 1/3 and 4/3
12. A box contains 4 red balls and 6 black
balls. Three balls are selected
randomly from the box one after
another without replacement. The
probability that the selected set
contains one red ball and two black
balls is
[EC2014]
(A)
1
20
(B)
1
12
(C)
3
10
(D)
1
2
13. Let X be a random variable with
probability density function
0.2, for
|
x
|
1
f(x) 0.1, for 1
|
x
|
4
0, otherwise.
The probability P(0.5 < x < 5) is
________.
[EE2014]
Vidyalankar : GATE – Engineering Mathematics
192
14. The probability of obtaining at least
two "SIX" in throwing a fair dice 4
times is
[ME2015]
(A) 425/432 (B) 19/144
(C) 13/144 (D) 125/432
15. Two coins R and S are tossed. The 4
joint events H
R
H
S
, T
R
T
S
, H
R
T
S
, T
R
H
S
have probabilities 0.28, 0.18, 0.30,
0.24, respectively, where H represents
head and T represents tail. Which one
of the following is TRUE?
[EE2015]
(A) The coin tosses are independent
(B) R is fair, S is not
(C) S is fair, R is not
(D) The coin tosses are dependent
16. The probability density function of a
random variable X is p
X
(x) = e
x
for
x
0 and 0 otherwise. The expected
value of the function g
X
(x) =
3x 4
e is
___________.
[IN2015]
17. A fair die with faces {1, 2, 3, 4, 5, 6} is
thrown repeatedly till '3' is observed for
the first time. Let X denote the number
of times the die is thrown. The
expected value of X is __________.
[EC2015]
18.
Three cards were drawn from a pack
of 52 cards. The probability that they
are a king, a queen, and a jack is
[ME2016]
(A)
16
5525
(B)
64
2197
(C)
3
13
(D)
8
16575
19.
Probability density function of a
random variable X is given below.
f(x) =
0.25 if 1 x 5
0otherwise
P(X
4) is [CE2016]
(A)
3
4
(B)
1
2
(C)
1
4
(D)
1
8
20.
Consider the random process
X(t) = U + Vt
where U is a zero-mean Gaussian
random variable and V is a random
variable uniformly distributed between
0 and 2. Assume that U and V are
statistically independent. The mean
value of the random process at t = 2 is
________.
[EC2017]
Assignment on Probability and Statistics
193
Assignment 4
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 6 carry one mark each
1.
Probability of getting an even number
in a single throw with a die is
(A) 1/2 (B) 2/3
(C) 1/4 (D) 1/3
2. Probability of getting tail in a throw of a
coin is
(A) 1 (B) 1/3
(C) 1/4 (D) 1/2
3. A bag contains 6 white balls, 9 black
balls. The probability of drawing a black
ball is
(A) 2/5 (B) 3/5
(C) 1/5 (D) 4/5
4. Probability of a card drawn at random
from an ordinary pack of cards to be
club card is
(A) 3/4 (B) 2/4
(C) 1/4 (D) 1/5
5. Probability of throwing a number
greater than 3 with an ordinary die is
(A) 1/2 (B) 1/3
(C) 1/4 (D) 2/3
6. Probability of getting a total of more
than 10 in a single throw with 2 dice
(A) 1/6 (B) 1/8
(C) 1/12 (D) 2/3
Q 7 to Q 18 carry two marks each
7.
Probability than a leap year selected at
random will have 53 Sundays.
(A) 2/5 (B) 1/8
(C) 2/7 (D) 1/7
8. A card is drawn from an ordinary pack
of playing cards and a person bets that
it is a spade or an ace. Then the odds
against his winning this bet is
(A) 9 to 4 (B) 8 to 5
(C) 7 to 6 (D) none of above
9. In a horse race the odds in favour of
four horses H
1
, H
2
, H
3
, H
4
are 1 : 3,
1 : 4, 1 : 5; 1 : 6 respectively not more
than one wins at a time. Then the
chance that one of them wins is
(A) 320/419 (B) 319/420
(C) 419/520 (D) 520/519
10. In a garden 40% of the flowers are
roses and the rest are carnations. If
25% of the roses and 10% of the
carnations are red, the probability that a
red flower selected at random is a rose.
(A) 5/6 (B) 1/4
(C) 4/5 (D) 5/8
11. Find the chance of throwing more than
15 in one throw with 3 dice
(A) 1/54
(B) 17/216
(C) 5/108
(D) cannot be determined
Vidyalankar : GATE – Engineering Mathematics
194
12. In a race where 12 horses are running,
the chance that horse A will win is 1/6,
that B will win is 1/10 and that C will
win is 1/8. Assuming that a dead heat
is impossible the chance that one of
them will win
(A) 1/390 (B) 47/120
(C) 3/20 (D) 1/54
13. There are two bags, one of which
contains 5 red and 7 white balls and
the other 3 red and 12 white balls. A
ball is to be drawn from one or other of
the two bags, find the chance of
drawing a red ball.
(A) 37/120 (B) 1/10
(C) 1/13 (D) 1/96
14. What is the probability of a particular
person getting 9 cards of the same suit
in one hand at a game of bridge where
13 cards are dealt to a person ?
(A)
13
9
39
4
C4
C
(B)
13 13 13
992
52
9
CCC4
C
(C)
13 39
94
52
13
CC4
C
(D)
13
9
52
13
C4
C
15. A man and his wife appear for an
interview for two posts. The probability
of the husband's selection is 1/7 and
that of the wife’s selection is 1/5. The
probability that only one of them will be
selected is
(A) 2/7 (B) 4/5
(C) 4/35 (D) 6/35
16. There are two bags. One bag contains
4 white and 2 black balls. Second bag
contains 5 white and 4 black balls.
Two balls are transferred from first bag
to second bag. Then one ball is taken
from the second bag. The probability
that it is white is
(A) 42/165 (B) 95/165
(C) 5/165 (D) 48/165
17. In a single throw of two dice find the
probability that neither a doublet
(same number on the both dice) nor a
total of 9 will appear.
(A) 5/1 (B) 1/9
(C) 13/18 (D) 1/4
18. A speaks truth in 75% and B in 80% of
the cases. In what percentage of
cases are they likely to contradict each
other narrating the same incident ?
(A) 75% (B) 80%
(C) 35% (D) 100%
Assignment on Probability and Statistics
195
Assignment 5
Duration : 45 Min. Max. Marks : 30
Q 1 to Q 6 carry one mark each
1.
A die is rolled. The probability of
getting a number 1 or 6 on the upper
face is
(A) 1/3 (B) 1/2
(C) 2/3 (D) 1/4
2. The probability of the horse A winning
the race is 1/5 and the probability of
the horse B winning the same race is
1/6, then the probability of one of the
horse to win the race is
(A) 11/29 (B) 11/28
(C) 11/30 (D) 11/31
3. A card is drawn from a pack of 52
cards and then a second is drawn.
Probability that both the cards drawn
are queen is
(A) 1/219 (B) 1/13
(C) 1/17 (D) 1/221
4. A bag contains 5 white and 3 black
balls. Two balls are drawn at random
one after the other without
replacement. The probability that both
the balls drawn are black is
(A) 3/28 (B) 3/8
(C) 2/7 (D) 3/29
5.
4 coins are tossed. Then the probability
that at least one head turns up is
(A) 15/26 (B) 1/16
(C) 14/16 (D) 15/16
6. In a throw of 3 dice the probability that
at least one die shows up 1 is
(A) 5/6 (B) 1/6
(C) 91/216 (D) 90/215
Q 7 to Q 18 carry two marks each
7.
Match the following :
List I
(a) P(), is the empty set
(b) P(A/B)
P(B)
(c)
PA
(d)
PA B
(e) P (A
B)
List II
(i) 1 P(A)
(ii) P(A
B)
(iii) 1
P(A B)
(iv) 0
(v) P(A) + P(B)
P(A B)
(A) a iv, b ii, c i, d iii, e v
(B) a
iii, b ii, c i, d iv, e v
(C) a
ii, b iii, c i, d iv, e v
(D) a
i, b ii, c iii, d iv, e v
Vidyalankar : GATE – Engineering Mathematics
196
8. Three machines A, B and C produce
respectively 60%, 30% and 10% of the
total number of items of a factory. The
percentages of defective output of
these machines are respectively 2%,
3% and 4%. An item is selected at
random and found defective. Find the
probability that the item was produced
by machine C.
(A) 2/25 (B) 1/25
(C) 4/25 (D) 3/25
9. A fair die is tossed 180 times. The
expected number of sixes is
(A) 40 (B) 25
(C) 180 (D) 30
10. In a population having 50% rice
consumers, what is the probability that
three or less out of 10 are rice
consumers?
(A) 17% (B) 10%
(C) 40% (D) 50%
11. The chances of a person being alive
who is now 35 years old, till he is 75
are 8 : 6 and of another person being
alive now 40 years old till he is 80 are
4 : 5. The probability that at least one
of these persons would die before
completing 40 years hence is
(A) 8/14 (B) 16/63
(C) 4/9 (D) 47/63
12. If 10 coins are tossed 100 times, how
many times would you expect 7 coins
to fall head upward ?
(A) 12 (B) 11
(C) 10 (D) 9
13. If the probability of a defective bolt is
0.1, the mean and standard deviation
for the distribution of defective bolts in
a total of 500 are
(A) 50, 6.7 (B) 40, 7.3
(C) 20, 3.1 (D) none of these
14. Two cards are drawn with replacement
from a well shuffled deck of 52 cards.
The mean and standard deviation for
the number of aces are
(A)
2
,0.377
13
(B)
1
,0.277
13
(C)
3
,0.477
13
(D) none of these
15. In a Binomial distribution, the mean
and standard deviation are 12 and 2
respectively. The values of n and p are
respectively.
(A) 12, 2 (B) 9,
1
3
(C) 18,
2
3
(D)
1
3
, 18
X 0 1 2
P(x)
144
169
24
169
1
169
Probability distribution table
Assignment on Probability and Statistics
197
16. For a biased die the probabilities for
different faces to turn up are given
below.
Face Probability
1 0.10
2 0.32
3 0.21
4 0.15
5 0.05
6 0.17
The die is tossed and you are told that
either face 1 or face 2 has turned up.
The probability that it is face 1 is
(A) 2/11 (B) 3/21
(C) 11/13 (D) 5/21
17. There are three events A, B and C,
one of which must occur and only one
can happen, the odds are 8 to 3
against A, 5 to 2 against B; find the
probability of C.
(A) 41/57 (B) 3/11
(C) 34/77 (D) 2/7
18. Aishwarya studies either computer
science or mathematics everyday. If
she studies computer science on a
day, then the probability that she
studies mathematics the next day is
0.6. If she studies mathematics on a
day, then the probability that she
studies computer science the next day
is 0.4. Given that Aishwarya studies
computer science on Monday, what is
the probability that she studies
computer science on Wednesday ?
(A) 0.24 (B) 0.36
(C) 0.4 (D) 0.6
Vidyalankar : GATE – Engineering Mathematics
198
Assignment 6
Duration : 45 Min. Max. Marks : 30
Q1 to Q6 carry one mark each
1.
The probability that the birthday of a
child is Saturday or Sunday is
(A) 1 / 2 (B) 2 / 7
(C) 1 / 3 (D) 1 / 7
2. The probability of getting a multiple of
2 in the throw of a die
(A) 1 / 3 (B) 1 / 4
(C) 1 / 6 (D) 1 / 2
3. Probability that the sum of the score is
odd in a throw of two dice is
(A) 1 / 4 (B) 1 / 5
(C) 1 / 2 (D) 1 / 3
4. A committee of 5 students is to be
chosen from 6 boys and 4 girls. The
probability that the committee contains
exactly 2 girls is
(A) 9 / 20 (B) 1 / 6
(C) 10 / 21 (D) 1 / 9
5. A bag contains 7 red, 5 blue, 4 white
and 4 black balls. Then the probability
that a ball drawn at random is red or
white is
(A) 11 / 20 (B) 1 / 7
(C) 1 / 11 (D) 3 / 20
6. Two cards are drawn at random from a
pack of 52 cards. The probability that
one may be a jack and other an Ace is
(A) 1 / 16 (B) 7 / 663
(C) 8 / 663 (D) none of these
Q7 to Q18 carry two marks each
7.
The probability that a man aged 50
years will die within a year is 0.01125.
The probability that out of 12 such
men at least 11 will reach their fifty first
birthday.
(A) 1 (B) 0
(C) 0.9923 (D) 0.8823
8. The probability that a certain beginner
at golf gets a good shot if he uses the
correct club is 1/3 and the probability
of a good shot with an incorrect club is
1/4. In his bag are 5 different clubs,
only one of which is correct for the
shot in question. If he chooses a club
at random and takes a stroke, the
probability that he gets a good shot is
(A) 2 / 3 (B) 4 / 15
(C) 5 / 9 (D) 7 / 36
9. From a pack of cards two are drawn,
the first being replaced before the
second is drawn. The probability that
the first is a diamond and the second
is a king is
(A) 1 / 52 (B) 4 / 13
(C) 57 / 64 (D) 11 / 52
Assignment on Probability and Statistics
199
10. One of the two mutually exclusive
events must occur, if the chance of
one is 2/3 of the other, then odds in
favour of the other are
(A) 1 : 1 (B) 1 : 2
(C) 2 : 3 (D) 3 : 2
11. Given P(A
B
) =
1
3
and
P(A
B) =
2
3
then P(B) is
(A) 2 / 3 (B) 1 / 5
(C) 1 / 3 (D) 4 / 5
12. There are 64 beds in a garden and 3
seeds of a particular type of flower are
sown in each bed. The probability of a
flower being white is 1/4. The number
of beds with 3, 2, 1, and 0 white
flowers is respectively
(A) 1, 9, 27, 27 (B) 27, 9, 1, 27
(C) 27, 9, 1, 1 (D) 27, 9, 9, 1
13. A sample space has two events A and
B such that probabilities
P(A
B) = 1/2 , P(A) = 1/3,
P(B) = 1/3. What is P(A B) ?
(A) 11/12 (B) 10/12
(C) 9/12 (D) 8/12
14. A man takes a step forward with
probability 0.4 and backward with
probability 0.6. The probability that at
the end of eleven steps he is one step
away from the starting point.
(A) 0.3678 (B) 0.25
(C)
11
c
2
(0.24)
3
(D) none of these
15. Find the probability of drawing one
rupee coin from a purse with two
compartments one of which contains 3
fifty paisa coins and 2 one
rupee
coins and the other contains 2 fifty
paisa coins and 3 one rupee coins.
(A) 1 / 2 (B) 2 / 5
(C) 3 / 5 (D) none of these
16. If A and B are two events such that
P(A
B) = 5/6 . P(A B) = 1/3 P(
B
)
= 1/ 2 then the events A and B are
(A) dependent
(B) independent
(C) mutually exclusive
(D) none of these
17. A determinant is chosen at random from
the set of all determinants of order 2 with
elements 0 or 1 only. The probability that
value of the determinant chosen is
positive is
(A) 2 / 17 (B) 1 / 17
(C) 1 / 16 (D) 3 / 16
18. One hundred identical coins each with
probability p of showing up heads are
tossed. If 0 < p < 1 and the probability
of heads showing on 50 coins is equal
to that of the heads showing in 51
coins, then the value of p is
(A) 1 / 2 (B) 49 / 101
(C) 50 / 101 (D) 51 / 101
Vidyalankar : GATE – Engineering Mathematics
200
Test Paper 1
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1. In a shooting competition, the
probability of hitting the target by A is
2/5, by B is 2/3 and by C is 3/5. If all
of them fire independently at the same
target, calculate the probability that
only one of them will hit the target.
(A) 2/5 (B) 2/3
(C) 3/5 (D) 1/3
2. A box contains 6 white balls and
3 black balls and another box contains
4 white balls and 5 black balls. The
probability that a ball selected from
one of the box again selected at
random is a white ball.
(A) 6/18 (B) 4/18
(C) 1/2 (D) 5/9
3. Two sisters A and B appeared for an
Audition. The probability of selection
of A is 1/5 and that of B is 2/7. The
probability that both of them are
selected is
(A) 2/35 (B) 1/5
(C) 2/7 (D) none of these
4. Two balls are to be drawn from a bag
containing 5 red and 7 white balls. Find
the chance that they will both be white
(A) 5/108 (B) 5/7
(C) 7/22 (D) 2/35
5. From a bag containing 4 white and
5 black balls a man draws 3 at
random. Then the probability that all 3
are black is
(A) 5/42 (B) 1/42
(C) 1/20 (D) 1/15
Q6 to Q15 carry two marks each
6. Urn A contains 6 red and 4 black balls
and urn B contains 4 red and 6 black
balls, one ball is drawn at random from
urn A and placed in urn B. Then one
ball drawn at random from urn B and
placed in urn A. If one ball is now
drawn from urn A, the probability that it
is found to be red is
(A) 3 / 11 (B) 1 / 17
(C) 32 / 55 (D) 31 / 56
7. Two persons each make a single
throw with a dice. The probability they
get equal value is P
1
. Four persons
each make a single throw and
probability of three being equal is P
2
.
Then
(A) P
1
= P
2
(B) P
1
< P
2
(C) P
1
> P
2
(D) none of these
8. A bag has 13 red, 14 green and 15
black balls. The probability of getting
exactly 2 blacks on pulling out 4 balls is
P
1
. Now the number of each colour ball
is doubled and 8 balls are pulled out.
Test Paper on Probability and Statistics
201
The probability of getting exactly 4
blacks is P
2
. Then
(A) P
1
= P
2
(B) P
1
> P
2
(C) P
1
< P
2
(D) none of these
9. If A and B are arbitrary events, then
(A) P ( A B) P(A) + P(B) 1
(B) P(A B) P(A) + P(B) 1
(C) P(A B) = P(A) + P(B) 1
(D) none of above
10. One mapping is selected at random
from all the mappings from the set
S =
1, 2, 3,...n
into itself. The probability
that the selected mapping is one to one
is
(A) 1/n
n
(B) 1/n!
(C) n/n! (D) none of these
11. For two events A and B P(A B) is
(i) not less than P(A) + P(B) 1 (ii)
not greater than P(A) + P(B)
(iii) equal to P(A) + P(B) P(A B)
(iv) equal to P(A) + P(B) + P(A B)
(A) i, ii, iii (B) i, iv
(C) i, iii (D) i, ii, iii, iv
12. Two persons A and B have
respectively n + 1 and n coins, which
they toss simultaneously. Then the
probability that A will have more heads
than B is
(A) 1 / 2 (B) > 1 / 2
(C) < 1 / 2 (D) none of these
13. A student appears for tests I, II and III.
The student is successful if he passes
either in tests I and II or tests I and III.
The probabilities of the student
passing in tests I, II and III are p, q and
1/2 respectively. If the probability that
the student is successful is 1/2. Then
(A) p = q = 1 (B) p = q = 1/2
(C) p = 1, q = 0 (D) p = 1, q = 1/2
14. The figure below shows the network
connecting cities A, B, C, D, E and F.
The arrows indicate permissible
directions of travel. If Deepak from city
A wants to visit city F, what is the
probability he will pass through city C
(A) 3/5 (B) 4/5
(C) 1/2 (D) 3/4
15. Mean of 200 observations was found
to be 90. However, when the
observations were being made, two
observations were wrongly taken as
15 and 80 instead of 40 and 87. Then
the correct mean is
(A) 90 (B) 91.16
(C) 90.16 (D) 91.6
E
F
A
D
B
C
Vidyalankar : GATE – Engineering Mathematics
202
Test Paper 2
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1. What is the chance of throwing a
number greater than 4 with an ordinary
die whose faces are numbered from
1 to 6 ?
(A) 1/6 (B) 1/3
(C) 1/4 (D) none of these
2. Find the chance of throwing at least
one ace in a simple throw with two
dice numbered 1 to 6.
(A) 1/11 (B) 1/36
(C) 11/18 (D) 11/36
3. What is the probability that a digit
selected at random from the logarithmic
table is 1 ?
(A) 1/2 (B) 1/5
(C) 1/3 (D) 1/10
4. A problem in mathematics is given to
three students Dayanand, Ramesh
and Naresh whose chances of solving
it are
11
,,
234
respectively. The
probability that the problem is solved is
(A) 1/4 (B) 1/2
(C) 3/4 (D) 1/3
5.
A bag contains 3 black and 5 white
balls. One ball is drawn from the bag.
What is the probability that the ball is
not black ?
(A) 3/8
(B) 5/8
(C) 2/7
(D) none of the these
Q6 to Q15 carry two marks each
6.
A single letter is selected at random
from the word “PROBABILITY”. The
probability that it is a vowel is
(A) 1/3 (B) 2 / 21
(C) 0 (D) 2 / 9
7. A and B throw a coin alternately till
one of them gets a ‘head’ and wins the
game. Their respective probabilities of
winning are
(A) 1/3, 2/3 (B) 1/2, 1/2
(C) 1/4, 3/4 (D) 2/3, 1/3
8. Two dice are thrown the probability of
getting an odd number on the one and
a multiple of 3 on the other is
(A) 11/36 (B) 2/17
(C) 10/36 (D) 12/36
Test Paper on Probability and Statistics
203
9. In a certain college, 4% of the men
and 1% of the women are taller than
1.8m. Further 60% of the students are
woman. If a student is selected at
random and is taller than 1.8m, the
probability that the student is women is
(A) 3/11 (B) 2/11
(C) 4/11 (D) 1/11
10. A and B throws two dice; if A throws 9
then B’s chance of throwing a higher
number is
1
6
, state whether the above
statement is true or false.
(A) True
(B) False
(C) Can not say
(D) Data insufficient
11. A pack of cards contains 4 aces,
4 queens and 4 jacks. Two cards are
drawn at random. The probability
that at least one of them is an ace is
(A)
19
33
(B)
3
16
(C)
1
6
(D)
1
9
12. x is a random variable with mean
and S. D. > 0
x =
x
; then var(x) = _______
(A) 0 (B) 2
(C) can’t say (D) 1
13. If A and B are two events such that
P(A) = 0 and P(B) 1, then P(
A /B ) is
equal to
(A) 1 P(A/ B ) (B) 1 P( A /B)
(C) 1
P(A / B)
P(B)
(D)
P(A)
P(B)
14. Ten points are marked on a straight
line and 11 points are marked on
another straight line. If any three
points are chosen, what is the
probability that it forms a triangle?
(A)
11
14
(B)
21
3
10 11
33
C
CC
(C)
6
7
(D)
10 11
33
CC
1045
15. Three machines A, B and C produce
identical items of their respective
output 5%, 4% and 3% items are
faulty. If A produces 25% of the total
output, B produces 30% and C the
remainder and an item selected at
random is found to be faulty, what is
the probability that it was produced by
machine C?
(A) 0.155 (B) 0.255
(C) 0.355 (D) 0.455
Vidyalankar : GATE – Engineering Mathematics
204
Test Paper 3
Duration : 30 Min. Max. Marks : 25
Q1 to Q5 carry one mark each
1.
An urn contains 25 balls numbered 1
through 25. Two balls are drawn from
the urn with replacement. The
probability of getting at least one odd
is
(A) 144/625 (B) 312/625
(C) 12/25 (D) 481/625
2. Four persons are chosen at random
from a group containing 3 men,
2 women and 4 children. The chance
that exactly 2 of them will be children
is
(A) 10/21 (B) 5/8
(C) 13/32 (D) 3/32
3. Let A and B be events with P(A) =
1
2
,
P(B) =
1
3
and P(A B) =
1
4
. Then
P(A/B) is
(A) 1/4 (B) 1/3
(C) 1/2 (D) 3/4
4. How many different words can be
formed from the letters of the word
Ganeshpuri when all the letters taken
(A) 2! (B) 10!
(C) 9! (D) 5!
5. A bag contains 8 green and 10 white
balls. Two balls are drawn. What is the
probability that one is green and the
other is white ?
(A)
80
153
(B)
10
17
(C)
8
17
(D)
81
157
Q6 to Q15 carry two marks each
6.
Fifteen coupons are numbered
1, 2, 3, …….., 15. Seven coupons are
selected at random one at a time with
replacement. The probability that the
largest number appearing on selected
coupon is 9, is
(A)
6
9
16
(B)
8
7
15
(C)
7
2
5
(D) none of these
7. A bag contains 2 white and 3 black
balls. A ball is drawn 5 times, each
being replaced before another is
drawn. Find the probability that exactly
4 of the balls drawn are white.
(A)
2
5
(B)
48
625
(C)
31
725
(D)
3
5
Test Paper on Probability and Statistics
205
8. From a pack of 52 cards two are
drawn at random. The probability that
one is a king and the other a queen is
(A) 3/ 8 (B) 8 / 663
(C) 7/ 663 (D) 9 / 663
9. In three throws of two dice, the
probability of throwing doublets not
more than two times is
(A) 1/216 (B) 1/6
(C) 4/216 (D) 215 / 216
10. If the sum of the mean and the
variance of a Binomial distribution for
18 trials is 10, find the distribution.
(A) (p + q)
n
(B)
18
12
33
(C)
18
11
22
(D)
11
12
33
11. If the sum of the mean and the
variance of a Binomial distribution for
5 trials is 1.8 find the distribution.
(A)
5
12
33
(B)
5
11
22
(C)
9
61
77
(D)
5
41
55
12. For a Binomial distribution, the mean
is 6 and the standard deviation is
2 .
The Binomial distribution is
(A)
9
12
33
(B)
9
11
22
(C)
9
61
77
(D)
9
41
55
13. If 3 squares are chosen at random on
a chess board the probability that they
should be in a diagonal line is
(A) 5/744 (B) 3/744
(C) 7/744 (D) 9/744
14. A man has 9 friends (4 men and
5 women). In how many ways can he
invite them, if there have to be exactly
3 women in the invitees ?
(A) 320 (B) 160
(C) 80 (D) 200
15. Given two events A and B and
P(A) = 1/4, P(B/A) = 1/2, P(A/B) = 1/4.
Which of the following is true?
(A)
P(A B) = 1/2
(B) A is subevent of B
(C) P(A/B) +
P(A B) = 1
(D) None of these
206
Solutions Linear Algebra
Answer Key on Assignment 1
1. (A) 2.
(D)
3.
0.99 to 1.01
4. (A)
5. (A) 6.
(B)
7.
(A)
8. 2
9. 6 10.
(C)
11.
(B)
12.
2.9 to 3.1
13. (A) 14.
(B)
15.
(D)
16. (A)
17.
6 to 6
18.
(C)
19.
(B)
20. (D)
Answer Key on Assignment 2
1. (A) 2.
(D)
3.
199 to 201
4.
23 to 23
5. 88 to 88
6.
2
7.
4.49 to 4.51
8.
16.5 to 17.5
9. (D) 10.
(C)
11.
48.9 to 49.1
12.
2.0 to 2.0
13. (D) 14.
(D)
15.
2 to
2
16. (B)
17. (B) 18.
(D)
19.
(A)
20. 0 to 0
Answer Key on Assignment 3
1. (D) 2.
(D)
3.
(B)
4. (A)
5. (A) 6.
(C)
7.
2.0 to 2.0
8. (C)
9. (A) 10.
0.95 to
1.05
11.
(B)
12. (A)
13.
5.0 to 5.0
14.
0
15.
2.9 to
3.1
16. (B)
17. (A) 18.
(B)
19.
(A)
20. (A)
Solutions Linear Algebra
207
Answer Key on Assignment 4
1. (C)
2.
(B)
3.
(A)
4. (B)
5. (D)
6.
(A)
7.
(A)
8. (C)
9. (A)
10.
(A)
11.
(B)
12. (D)
13. (B)
14.
(C)
15.
(A)
16. (D)
17. (B)
18.
(B)
Answer Key on Assignment 5
1. (B)
2.
(C)
3.
(A)
4. (B)
5. (B)
6.
(A)
7.
(B)
8. (A)
9. (B)
10.
(A)
11.
(C)
12. (B)
13. (C)
14.
(A)
15.
(B)
16. (C)
17. (D)
18
(D)
Answer Key on Assignment 6
1. (B)
2.
(C)
3.
(C)
4. (C)
5. (B)
6.
(C)
7.
(B)
8. (B)
9. (B)
10.
(D)
11.
(C)
12. (C)
13. (C)
14.
(A)
15.
(C)
16. (A)
17. (C)
18.
(B)
Answer Key on Assignment 7
1. (C)
2.
(D)
3.
(B)
4. (B)
5. (D)
6.
(A)
7.
(A)
8. (C)
9. (A)
10.
(A)
11.
(B)
12. (C)
13. (B)
14.
(A)
15.
(B)
16. (A)
17. (A)
18.
(B)
Vidyalankar : GATE – Engineering Mathematics
208
Answer Key on Assignment 8
1. (C)
2.
(C)
3.
(A)
4. (C)
5. (C)
6.
(C)
7.
(A)
8. (A)
9. (B)
10.
(C)
11.
(C)
12. (C)
13. (C)
14.
(C)
15.
(C)
16. (B)
17. (B)
18.
(A)
Answer Key on Assignment 9
1. (B)
2.
(B)
3.
(C)
4. (A)
5. (C)
6.
(B)
7.
(C)
8. (A)
9. (C)
10.
(B)
11.
(C)
12. (C)
13. (C)
14.
(C)
15.
(B)
16. (C)
17. (D)
18.
(D)
Answer Key on Assignment 10
1. (D)
2.
(B)
3.
(B)
4. (C)
5. (B)
6.
(C)
7.
(A)
8. (A)
9. (A)
10.
(B)
11.
(C)
12. (B)
13. (C)
14.
(C)
15.
(D)
16. (C)
17. (D)
18.
(C)
Answer Key on Assignment 11
1. (C)
2.
(C)
3.
(C)
4. (D)
5. (C)
6.
(C)
7.
(C)
8. (C)
9. (A)
10.
(B)
11.
(D)
12. (B)
13. (C)
14.
(D)
15.
(B)
16. (C)
17. (D)
18.
(D)
Solutions Linear Algebra
209
Model Solution on Assignment 1
1. (A)
260
4128
204
=
3
130
(2) 2 6 4
10 2
= 8 (12) = 96
2. (D)
We know that the Eigen vectors
corresponding to distinct Eigen values
of real symmetric matrix are
orthogonal.
11
22
33
xy
xy
xy
= x
1
y
1
+ x
2
y
2
+ x
3
y
3
= 0
3. 0.99 to 1.01
Given, A
2
= I
We know that above equation is
satisfied by identity matrix.
i.e.,
1000 1000
0100 0100
0010 0010
0001 0001
= I
Eigen value of identity matrix is 1.
4. (A)
5. (A)
Area of triangle
=
12 3
12 3
xx x
1
yy y
2
111
124
1
023
2
111
=
1
1[2 3] 2[0 3] 4[0 2]
2
=
13
16 8
22
6. (B)
7. (A)
8. 2
x 2y + 3z = 1
x 3y + 4z = 1
2x + 4y 6z = k
Arugmented matrix (A/B) is given by
(A/B) =
1231
1341
24 6k
Now applying row reduction technique
R
2
R
2
R
1
R
3
R
3
+ 2R
1
We get
123 1
0112
000k2
For infinite solution
(A/B) = (A) = r
k 2 = 0
k = 2
Vidyalankar : GATE – Engineering Mathematics
210
9. 6
The given matrix is
A =
45
21
Inorder to find eigen values of above
matrix.
|A
I| = 0
45
21
= 0
(4 ) (1 ) 10 = 0
4 5 +
2
10 = 0
2
5 6 = 0
2
6 + 1 6 = 0
( 6) + 1( 6) = 0
+ 1 = 0 or
= 1 or = 6
Larger eigen value of matrix is 6.
10. (C)
11. (B)
12. 2.9 to 3.1
Characteristic equation is given by
|A I| = 0
011
611 6
6115
= 0
011
611 6
6115
= 0
3
+ 6
2
+ 11 + 6 = 0
= 1, 2, 3 are the eigen values
of A
max
= 1 and
min
= 3
max
min
=
1
3
=
1
3
13. (A)
A =
15
62
B =
37
84
AB
T
=
15 38
62 74
=
38 28
32 56
14. (B)
Let A =
10 5 J 4
x20 2
42 10
Given that all eigen values of A are real.
A is Hermitian
A
= A i.e.
T
A
= A
10 x 4
5j20 2
4210
=
10 5 J 4
x20 2
42 10
x = 5 j
15. (D)
The characteristic equation is
2
(sum of diagonal elements) + |A| = 0
2
(2 + p) + (2p 1) = 0
=
2
(2 p) (4 4p p ) 4(2p 1)
2
=
2
(2 p) (8 4p p )
2
2
2
(2 p) (8 4p p )
3
1
(2 p) (8 4p p )
Solutions Linear Algebra
211
2
2(8 4p p)
2
2(2 p) 4
2
28 4p p = (2 + p)
4(8 4p + p
2
) = (4 + 4p + p
2
)
3p
2
20p + 28 = 0
(3p 14) (p 2) = 0
p = 2, 14/3
16.
(A)
17. 6 to 6
18. (C)
21 1
01 1
11 0
r < n. Hence, infinite number of
solutions.
19. (B)
|A I| = 0 …characteristic equation
53 0
0
20 0
53
0
2
(5 + ) + 6 = 0
2
+ 5 + 6 = 0
By Cayley Hamilton’s Theorem,
A
2
+ 5A + 6I = 0
A
2
= 5A 6I
Multiply by ‘A’, We get,
A
3
= 5A
2
6A
= +5 [5A + 6I] 6A
A
3
= 19A + 30I
20. (D)
P =
11
0
22
010
11
0
22
|P| =
11 11
00
22 22
=
11
1
22
P.P
T
=
111 1
00
2222
010 010
111 1
00
2222
=
100
010
001
P is an orthogonal matrix
(A) Is correct
Inverse of P is its transpose only
(B) and (C) both are correct
(D) is incorrect
Vidyalankar : GATE – Engineering Mathematics
212
Model Solution on Assignment 2
1. (A)
Given that
dx
dt
= 3x 5y
dy
dt
= 4x + 8y
Matrix form
x
d
y
dt
=
35x
48 y
2. (D)
3. 199 to 201
4. 23 to 23
Given Matrices,
J =
321
242
126
and K =
1
2
1
K
T
=
12 1
K
T
JK =
321 1
12 12 4 2 2
126 1
=
1
3412821462
1
=
1
68 12
1
= 6 + 16 + 1
K
T
JK = 23
5.
88 to 88
A =
0123
1030
2301
3012
=
130 100
1201 2231
312 302
103
32 3 0
301
=
1[1(1) 3(4 3)] + 2[1(6)]
3[1(3) + 3(9)]
=
1[ 1 3] + 12 3[3 27]
= 4 + 12
3 [24]
= 4 + 12 + 72 = 88.
6. 2
Let A =
42
13
Characteristic equation of A is
AI0
42
13
= 0
2
7 + 10 = 0 = 2, 5
Solutions Linear Algebra
213
7. 4.49 to 4.51
[A : B] =
23: 5
3p:10
2C
2
3C
1
23:5
02p9:5
There is no solution if
rank (A : B)
rank(A).
Now, rank (A : B) = 2. So rank(A) = 1
only if 2p
9 = 0 i.e. p = 9/2 = 4.5
8. 16.5 to 17.5
AX = X
412
P21
14 4 10
1
2
3
=
1
2
3
12
P7
36
= 2
3
= 12 ….(1)
2
= P + 7 ….(2)
and 3
= 36
i.e.,
= 12
Equation (2) gives P + 7 = 24
P = 17
9. (D)
2x + 5y = 2 ….(1)
4x + 3y = 30 ….(2)
2 × (1) + (2) gives 13y =
26
y = 2
2x 10 = 2
2x = 12
x = 6
(6, 2)
10. (C)
Given : M
4
= I
M
4k
= (M
4
)
k
= I
k
= I
(A) M
4k + 1
= M
(B) M
4k + 2
= M
2
(C) M
4k + 3
= M
4k
M
4
M
1
= M
1
11. 48.9 to 49.1
Let A =
ab
cd
a + d = 14 and bc ≥ 0
Now |A| is maximum when ad is
maximum and bc = 0 i.e.
when a = d = 7
Maximum value of |A| = 7 × 7 = 49
12. 2.0 to 2.0
The given matrix is
A =
6044
214818
14 14 0 10
The matrix is of order 3
4
Hence
(A) 3
604
2148
14 14 0
= 6(0 + 112)
+ 4(28
196)
= 672
672 = 0
Vidyalankar : GATE – Engineering Mathematics
214
and
044
14 8 18
14 0 10
=
4(140 + 252) + 4(112)
=
4(112) + 4 (112) = 0
Hence
(A) 2.
Now
60
214
= 84 0
Hence
(A) = 2
Rank of given matrix is 2
13. (D)
The characteristic equation is
(3 ) 2 2
4(4)6 0
23(5)
;
C
2
= C
2
+ C
3
gives
(3 ) 0 2
4(2)6
2(2)(5)
= 0
(3 ) 0 2
(2 ) 4 1 6
21(5)
= 0;
R
3
= R
3
R
2
gives
(3 ) 0 2
(2 ) 4 1 6
20(1)
= 0
(2 ) {(3 )[(1 ) 0] 0
+ 2[0
(2)]} = 0
(2 ) {3 2 +
2
+ 4} = 0
(2 ) (
2
2 + 1) = 0
(2 ) ( 1)
2
= 0
eigen values are 1 and 2
smallest is 1 and largest is 2.
14. (D)
Sum of Eigen values = 1 + a
6 = 1 + a
a = 5
and product of Eigen values = det(A)
7 = a 4b
7 = 5 4b
7 5 = 4b
b = 3
15. 2 to 2
It is an upper triangular matrix. Hence
eigen values are the diagonal
elements i.e. 2, 2, 3. Hence there are
2 linearly independent eigen vectors.
16. (B)
17. (B)
111: 6
14 6: 20
14 :
R
3
R
3
R
2
11 1 : 6
14 6 : 20
00 6: 20
For
= 6, 20 System is
inconsistent
It has no solution.
Solutions Linear Algebra
215
18. (D)
11
A
11
55
32 0
A.A
032
4
40
A
04
10 5
128 128
A.A
128 128
5
44
A
44
15 4
512 512
A.A
512 512
19
512 512
A
512 512
512 512
0
512 (512 )
22
2
512 512 0
2
= 2 (512)
2
= 2 (512)
If eigen values of A are
1
,
2
,…, then
eigen values of A
k
are
k
1
,
k
2
,…(k > 0)
19. (A)
If matrix B is obtained from matrix A by
replacing the l
th
row by itself plus k
times the m
th
row, for / m then
det(B) = det(A). With this property
given matrix is equal to the matrices
given in options (B), (C) and (D).
20. 0 to 0
A =
50 70
70 80
Eigenvectors are
X
1
=
2
2
1
70
80
;X
50
70
where
1
,
2
= Eigenvalues of A
T2
12 1
80
X X 70 50
70
= 70(
2
80) + (
1
50) 70
= 70
2
5600 + 70
1
3500
= 70(
1
+
2
) 9100
= 70(130)
9100
= 9100
9100 = 0
12
sum of eigenvalues
Trace 50 80 130
Vidyalankar : GATE – Engineering Mathematics
216
Model Solution on Assignment 3
1. (D)
Eigen values of the matrix
52
96
are 4,
3
The eigen vector corresponding to
eigen vector
is Ax = x (verify the
options)
1
1
is eigen vector corresponding to
eigen value
= 3
2. (D)
MN is not always equal to NM.
3. (B)
[A:B] =
1
2
122:b
513:b
R
2
R
2
5R
1
1
21
12 2: b
097:b5b
(A) = (A/B) < number of unknowns,
for all values of b
1
and b
2
.
The equations have infinitely many
solutions, for any given b
1
and b
2
.
4. (A)
Given, Matrix [M] =
215 650 795
655 150 835
485 355 550
Sum of the eigenvalues
= Trace of matrix
= 215 + 150 + 550 = 915
5. (A)
6. (C)
Characteristic equation is
115
05 6 0
065
(1 ) [(5 )
2
+ 36] = 0
= 1;
2
10 + 61 = 0
=
10 100 224
2
=
10 12j
5j6
2
7. 2.0 to 2.0
P + Q =
012
8910
88 8
12
8910
RR 012
11 1
31
8910
8R R 0 1 2
012
32
8910
RR 0 12
00 0
Rank is 2
Solutions Linear Algebra
217
8. (C)
A =
1tanx
tanx 1
A
T
=
1tanx
tanx 1
Now A
1
=
1
(Adj(A))
|A|
A =
2
1tanx
1
tanx 1
sec x
|A
T
.A
1
| = 1
9. (A)
Given systems
3x
1
+ 2x
2
= c
1
4x
1
+ x
2
= c
2
Matrix Form is
11
22
xc
32
xc
41
AX = B
Characteristic equations of above
system is
|A
I| = 0
32
0
41
By expanding
2
4 5 = 0
10. 0.95 to 1.05
Matrix A has zero as an eigen value if
|A| = 0
3[(63 + 7x) + 52] 2[(81 + 9x)
+ 78] + 4[
36 + 42] = 0
3(7x 11) 2(9x 3) + (4 × 6) = 0
21x 33 18x + 6 + 24 = 0
3x = 3 x = 1
11. (B)
12. (A)
|P| = (16 + 9)
(1) = 24
P
1
=
43i i
11
adj(P)
i43i
|P| 24
13. 5.0 to 5.0
Since one eigenvalue of M is 2
2
3
4(2)
2
+ a(2) + 30 = 0
a = 11
Characteristic polynomial is
3
4
2
11 + 30 = 0
(
2) ( 5) ( + 3) = 0
= 2, 5, 3
Largest absolute value of '
' is 5
14. 0
The given matrix is
A =
3445
79105
13 2 195
Now according to question
(i) R
2
R
2
+ R
3
(Adding third row to
the second row)
B =
3445
20 11 300
13 2 195
(ii) C
1
C
1
C
3
(Subtracting third
column from first column)
C =
42445
280 11 300
182 2 195
Vidyalankar : GATE – Engineering Mathematics
218
Now |C| = 42(2145 600)
4(
54600 + 54600) + 45(560 + 2002)
|C| =
42(1545) 4(0)
+ 45(1442)
=
64890 + 64890 = 0
15.
2.9 to 3.1
sum of diagonal elements
= sum of eigen values
a + b + 7 = 14
a + b = 7
(a
b)
2
= (a + b)
2
4ab = 49 40 = 9
|a b| = 3
16. (B)
[A : B] =
12 3:a
23 3 :b
59 6:c
R
2
R
2
2R
1
[A : B] =
12 3: a
019:b2a
59 6: c
R
3
R
3
5R
1
[A : B] =
12 3: a
019:b2a
019:c5a
R
3
R
3
R
2
[A : B] =
12 3: a
019:b2a
00 0:cb3a
Equation are consistent if
c
b 3a = 0 i.e. 3a + b c = 0
17.
(A)
Let A =
51
41
Characteristic equations is
2
6 + 9
= 0
= 3, 3
Eigen value 3 has multiplicity 2.
Eigen vectors corresponding to
= 3
is (A
3I) X = 0
53 1 x 0
413y 0
21x 0
424 0
R
2
R
2
2R
1
21x 0
00 4 0
e(A) = 1
Number of linearly independent
eigenvectors corresponding to
eigenvalue
= 3 is n r = 2 1 = 1
where n = no. of unknowns,
r = rank of (A
I)
One linearly independent
eigenvector exists corresponding
to
= 3
18. (B)
|A
I| = 0
53
13
= 0
(5 ) (3 ) = 3 = 0
2
8 + 15 3 = 0
2
8 + 12 = 0 = 2, = 6
(A
2I) X = 0
At,
= 2
Solutions Linear Algebra
219
1
2
x
33 0
x
11 0
x
1
+ x
2
= 0
x
1
= x
2
Hence the required vector is
1
2
1
2
19. (A)
13
CC
352 5 52
5 12 7 12 12 7
275 7 75
determinant = 0,
So the matrix is singular
Therefore atleast one of the Eigen
value is ‘0’
As the choices are non negative, the
minimum Eigen value is ‘0’
20.
(A)
Characteristic equation is
10
01
034
(4 +
2
+ 3) = 0
( + 1) ( + 3) = 0
= 0, 1, 3 are the eigenvalues.
Vidyalankar : GATE – Engineering Mathematics
220
Model Solution on Assignment 4
1. (C)
2. (B)
If [A]
3 4
& [B]
4 5
AB will exist, BA does not exist
3. (A)
Every square, real symmetric matrix is
Hermitian and therefore all its
eigenvalues are real.
4. (B)
5. (D)
[A]
x ( x + 5)
[B]
y (11 y)
AB exists x + 5 = y …..(1)
BA exists x = 11 y …..(2)
Solving (1) and (2) gives x = 3, y = 8
6. (A)
A is singular matrix means | A | = 0
3x 2 2
24x 1
241x
= 0
R
2
R
2
+ R
3
3x 2 2
0xx
241x
= 0
x(3 x) (x 3) = 0
x = 0, 3
7. (A)
2
2
2
1
1
1
c
1
c
1
+ c
2
+ c
3
, we get
=
22
22
2
1
11
11
=
2
2
0
01
01
= 0
as 1 +
+
2
= 0
8. (C)
For a matrix rank is equal to no. of
independent vectors.
9. (A)
We find relation between than as
1
x
1
+
2
x
2
+
3
x
3
= 0
1
(3, 2, 7) +
2
(2, 4, 1)
+
3
(1, 2, 6) = 0
which gives 3
1
+ 2
2
+
3
= 0
2
1
+ 4
2
2
3
= 0
7
1
+
2
+ 6
3
= 0
which gives
1
= 1,
2
= 1,
3
= 1
x
1
x
2
x
3
= 0 is linear equation
They are dependent
10. (A)
11. (B)
By performing
ABC = [x y z]
ahg x
hb f y
gfc z
Solutions Linear Algebra
221
= [x y z]
ax hy gz
hx by fz
gx fy cz
= [ax
2
+ by
2
+ cz
2
+ 2hxy
+ 2gzx + 2fyz]
12. (D)
As | A | = 0
So inverse does not exist.
13. (B)
Let A =
022
748
70 4
R
2
R
2
+ R
3
A =
022
044
70 4
| A | = 0 ….. rank < 3
Now minor
02
74
02
74
= 14 0
rank = 2
14. (C)
Eigen values are obtained by
| A I | = 0
100
23 1
024
= 0
which gives
1
= 1,
2
= 2,
3
= 5
15. (A)
Given equations are AX = B
It will be consistent if
Rank A = Rank [A : B]
We have
AX =
11 1 x 3
31 2 y 2
24 7 z 7
= B
Augmented matrix
[ A B ] ~
11 1 : 3
31 2: 2
24 7 : 7
We reduce [A B] to Echelon form by
applying successively.
R
2
R
2
3R
1
, R
3
R
3
2R
1
transformations
[A : B] ~
11 1: 3
025:7
02 5:13
Apply R
3
R
3
+ R
2
[A : B] ~
11 1: 3
025:7
00 0:20
Rank [A : B] = 3
(No. of non-zero rows in Echelon form.)
A ~
11 1
025
00 0
Rank A = 2
Since Rank A Rank [A : B]
Inconsistent equations.
16. (D)
17. (B)
18. (B)
Vidyalankar : GATE – Engineering Mathematics
222
Model Solution on Assignment 5
1. (B)
As we know A
1
A = I = A A
1
B must be inverse of A
2. (C)
3. (A)
4. (B)
5. (B)
6. (A)
PQ =
pq r s
qp sr
pr qs ps qr
qr ps qs pr
7. (B)
Matrix is involutory if it satisfies condition
that A
2
= I and given matrix satisfies this
condition.
8. (A)
A
1
=
adj A
|
A
|
we calculate inverse using
this formula.
9. (B)
10. (A)
By writing augmented matrix
12 1:3
312:1
223:2
111:1
performing R
2
R
2
3R
1
,
R
3
R
3
2R
1
, R
4
R
4
R
1
and subsequently.
R
2
R
2
R
3
, R
3
R
3
6R
2 ,
R
4
R
4
3R
2 ,
R
3
1
5
R
3
, R
4
1
2
R
4
&
R
4
R
4
R
3
.
Consequently we get
[A : B] =
12 1:3
010:4
00 1:4
00 0:0
which is Echelon form with rank
Number of non
zero rows = 3
Also AI =
12 1
010
00 1
00 0
Here rank = 3
As rank [A : B] = rank A
It has solution, which is unique
These equations are equivalent to
x + 2y
z = 3
y = 4
z = 4
which gives x =
1, y = 4, z = 4
11. (C)
Eigen values of given matrix = 2, 6
For
= 6,
eigen vector =
1101
or
1101
Solutions Linear Algebra
223
12. (B)
For triangular matrix eigen values are
same as diagonal elements
1
= 1,
2
= 3,
3
= 2.
13. (C)
Explanation: Given matrix is,
111
111
111
Now
111
11 1 0
111
or
333
11 1
111
= 0
or
11 1
311 1 0
11 1
or
10 0
31 00
10
2
(3 ) = 0
Hence eigenvalues are 0, 0, 3.
14. (A)
Reducing to Echelon form,
R
4
R
4
R
1
We have number of nonzero rows = 3
rank = 3
15. (B)
Augmented matrix
[A : B] =
123:6
321:2
421:7
By performing R
2
R
2
3R
1 ,
R
3
R
3
4R
1
and
2
1
R
8
12 3 : 6
01 1: 2
0611:17
Performing R
3
R
3
+ 6R
2
12 3 : 6
01 1 : 2
00 5: 5
we get x = y = z = 1
16. (C)
17.
(D)
18.
(D)
Vidyalankar : GATE – Engineering Mathematics
224
Model Solution on Assignment 6
1. (B)
Characteristic equation is
12
54
= 0
(1 ) (4 ) 10 = 0
4 5 +
2
10 = 0
2
5 6 = 0
( 6) ( + 1) = 0
1
= 6,
2
= 1
2. (C)
3. (C)
4. (C)
5. (B)
A =
52
31
, | A | = 1
A
1
=
adj A
|
A
|
12
35
1
=
12
35
6. (C)
As A A
1
= I
2x 0 1 0
02x 01
2x = 1
x = 1/2
7. (B)
8.
(B)
Characteristic Equation:
|A I| = 0
(1 ) 1 0
0(1)1
001
= 0
(1 )
3
= 0
1 3 + 3
2
3
= 0
I 3A + 3A
2
A
3
= 0
…. Cayley-Hamilton Theorem
I 3A + 3A.A A
2
A = 0
Post multiplying by A
1
,
I.A
1
3AA
1
+ 3AAA
1
A
2
AA
1
= 0A
1
A
1
3I + 3AI A
2
I = 0
A
1
3I + 3A A
2
= 0
A
1
= A
2
3A + 3I
9. (B)
For given matrix of rank = 1
= 1 rank = 1
10. (D)
As | A | 0 for unit matrix
rank equal to number and order
Solutions Linear Algebra
225
11. (C)
A =
311
15 6 5
22
For nontrivial solution P(A) < 3
| A | = 0 6 + (0 + 1) = 0
= 6
12. (C)
Solving characteristic equation,
| A I | = 0
32
65
= 0
Gives = 1 2 7
13. (C)
As determinant of A i.e. |A| 0
14. (A)
The sum of eigen values of a matrix is
the sum of diagonal elements
1
+
2
= 12.
Only option (A) has sum of diagonal
elements of 12.
15.
(C)
| A
T
A | = | I | = | A
T
| . | A | = 1
| A | . | A | = 1
( | A | )
2
= 1
| A | = 1
16. (A)
1
A.A I
1
1/ 2 a 1 0
2
.
10
0b 01
03
b
10
12a
10
01
03b
comparing the corresponding terms,
we get
1b
b,2a 0
310
b
2a
10
b
a
20
1
a
60
11120217
ab
60 3 60 60 20
17. (C)
18. (B)
Vidyalankar : GATE – Engineering Mathematics
226
Model Solution on Assignment 7
1. (C)
As per theorem for inverse matrix, that
eigen values of inverse matrix are
inverse of eigen values of the matrix
i.e. if is eigen value of A, then
1
is
eigen value of A
1
2. (D)
For non zero matrix rank must be
greater than or equal to 1, but not
equal to zero.
3. (B)
As determinant
10
01
= 1
rank equal to no. of rows = 2
4. (B)
For given equations
12 x 1
3y3
R
2
R
2
3R
1
12 x 1
06y 0
i.e. ( 6) y = 0
i.e. 6 = 0
= 6 will give no solution to have
unique solution 6
5. (D)
As A
1
=
Adj A
|
A
|
6. (A)
A =
882
43 2
341
= 0
R
1
R
1
(R
2
+ R
3
)
111
43 2
341
= 0
(1 )
11 1
41 2
341
= 0
(1 ) [(1 ) (4 ) + 2] = 0
(1 ) (
2
5 + 6) = 0
(1 ) ( 2) ( 3) = 0
Roots of equation are 1, 2 and 3.
As the eigen values of the matrix A are
all distinct hence A is similar to diagonal
matrix.
7. (A)
By using formula A
1
=
Adj A
|
A
|
We find A
1
=
11 2
01/2 1/2
11/23/2
8. (C)
| A I | = 0
0
00
00
= 0
3
= 0
= 0
Solutions Linear Algebra
227
Hence if x =
x
y
z
is eigen vector then,
AX = X
00 x 0
000 y 0
000 z 0
z = 0
The vectors must be
0
0or
00
As the vectors of the form
x
y
0
9. (A)
Hint : After taking the common from
each column, we get matrix having all
values = 1 and we
get after reducing
all rows and columns to zero the
matrix of rank 1 only
10. (A)
For characteristic equation
A =
hg
hf
gf
=
3
(f
2
+ g
2
+ h
2
) 2fgh = 0
11. (B)
By solving,
54
12
= 0
We get = 1 and = 6
12. (C)
Hint : By reducing the given matrix to
echelon form, we come to know that,
the rank of given matrix is 3.
13. (B)
By solving,
321x 4
111y 2
202 z 5
,
We get the values only one for each
parameter, so it has unique solution.
14. (A)
ahg x
xyzhbf y
gfcz
= ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
i.e. X
T
AX gives,
ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
If A =
11 12 13
21 22 23
31 32 33
aaa
aaa
aaa
Then for
ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
a
11
= a a
22
= b a
33
= c
a
12
= a
21
= h =
1
2
× (co-efficient of xy)
a
23
= a
32
= f =
1
2
× (co-efficient of yz)
a
31
= a
13
= g =
1
2
× (co-efficient of zx)
15. (B)
Hint : The property of the unitary
matrix
16. (A)
17. (A)
18. (B)
Vidyalankar : GATE – Engineering Mathematics
228
Model Solution on Assignment 8
1. (C)
Here P diagonals matrix B by pre &
post multiplication such that
B = P
1
AP
2. (C)
Hint : As per the properties of the
matrices.
3. (A)
=
1bc ca ab a(b c)
1bc ca ab b(c a)
1bc ca ab c(a b)
= (ab + bc + ca)
11a(b c)
11b(c a)
11c(a b)
= (ab + bc + ca ) 0 = 0
4. (C)
As per the property of multiplication of
matrix.
5. (C)
6. (C)
845
412
10 2
053
347
17 29 32
41211
17 24 32
41211
8x 3y 6z 3z
41226x5y
x = 1, y = 3, z = 4
7. (A)
As the rows are similar if common is
taken from each row
8. (A)
Hint : AX = X
A(AX) = A(X)
A
2
X = (AX) = (X) =
2
X
A
2
X =
2
X
9. (B)
Given vectors are
1
x
1
+
2
x
2
+
3
x
3
= 0
Which gives
1
x
1
+
2
x
2
+ 2
3
x
3
= 0
1
x
1
+ 2
2
x
2
+ 3
3
x
3
= 0
1
x
1
+ 3
2
x
2
+ 4
3
x
3
= 0
3
1
x
1
+ 4
2
x
2
+ 9
3
x
3
= 0
which on solving gives
1
=
2
=
3
= 0
They are not linearly dependent
10. (C)
Now
121:6
212:6
111:5
Now, R
3
R
1
121:6
212:6
010:1
R
2
2R
1
121:6
030:6
010:1
Solutions Linear Algebra
229
R
2
+ 3R
3
121:6
000: 3
010:1
Now the left part of dashed line have
rank 2 and right part have all elements
present so, it does not have solutions.
11. (C)
Hint : If the matrix is in the form of
upper / lower triangular matrix , then
the diagonal elements are the eigen
values of the element.
12. (C)
212i
12i 2
have eigen values as
212i
12i 2
= 0
( 2) ( + 2) (1 + 2i) (1 2i) = 0
4
4 (1 + 4) = 0
2
9 = 0
The eigen values are = 3, = 3
13. (C)
2206
4202
1103
1212
= 0
Rank < 4
Now minor of order 3
420
110
121
= 1(4 2) = 6 0
Rank = 3
14. (C)
We have
=
23 2
23 2
23 2
23 2
1a a a 1a a bcd
1b b b 1b b cda
1c c c 1c c dab
1d d d 1d d abc
=
1
+
2
(say)
Multiplying R
1
, R
2
, R
3
, R
4
of
2
by
a, b, c, d
Now
2
=
23
23
23
23
a a a abcd
b b b abcd
1
abcd
c c c abcd
d d d abcd
=
23
23
23
23
aa a 1
bb b 1
cc c 1
dd d 1
=
23
23
23
23
1a a a
1b b b
1c c c
1d d d
=
1
=
1
+ (
1
) = 0
15. (C)
For triangular matrix eigen values are
same as major principal diagonal
element i.e. 18, 7, 8.
16. (B)
17. (B)
18. (A)
Vidyalankar : GATE – Engineering Mathematics
230
Model Solution on Assignment 9
1. (B)
As per the definition of unitary square
matrix.
2. (B)
As per the definition of Hermitian
matrix
3. (C)
4. (A)
5. (C)
cos sin
sin cos
= cos
2
+ sin
2
= 1
& Adjoint of A =
T
cos sin
sin cos
=
cos sin
sin cos
A
1
=
cos sin
sin cos
6. (B)
A =
1053
21 61
3271
4420
&
A =
1234
012 4
567 2
3110
A + A
=
1053
21 61
3271
4420
+
1234
012 4
567 2
3110
=
2287
22 8 3
88141
7310
7. (C)
A =
aic bid
bid aic
= 1
a
2
+ b
2
+ c
2
+ d
2
= 1
8.
(A)
ahg x
xyzhbf y
gfcz
= ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
i.e. X
T
AX gives,
ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
If A =
11 12 13
21 22 23
31 32 33
aaa
aaa
aaa
Then for
ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
a
11
= a a
22
= b a
33
= c
a
12
= a
21
= h =
1
2
× (co-efficient of xy)
Solutions Linear Algebra
231
a
23
= a
32
= f =
1
2
× (co-efficient of yz)
a
31
= a
13
= g =
1
2
× (co-efficient of zx)
Extension of this method
9. (C)
As the number of equations &
unknowns are same and they are
equal to zero, so their values are also
zero.
Or | A |
≠ 0
the system will have trivial solution.
10. (B)
ahg x
xyzhbf y
gfcz
= ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
i.e. X
T
AX gives,
ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
If A =
11 12 13
21 22 23
31 32 33
aaa
aaa
aaa
Then for
ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
a
11
= a a
22
= b a
33
= c
a
12
= a
21
= h =
1
2
× (co-efficient of xy)
a
23
= a
32
= f =
1
2
× (co-efficient of yz)
a
31
= a
13
= g =
1
2
× (co-efficient of zx)
11.
(C)
| A |
3 3
= 5
Rank of A is 3.
12. (C)
Hint : Eigen values are the values of
diagonal for diagonal matrix. So
reduce given matrix to diagonal matrix.
Or sum of the eigen values of a matrix
= trace of the matrix
13. (C)
Characteristic Equation:
|A I| = 0
(3 ) 1
1(2)
= 0
(3 ) (2 ) (1) = 0
6 5 +
2
+ 1 = 0
2
5 + 7 = 0
Cayley-Hamilton Theorem,
A
2
5A + 7I = 0
14. (C)
A =
cos sin
sin cos
A
2
=
cos sin
sin cos
cos sin
sin cos
=
22
22
cos sin
cos sin
sin cos
sin cos
cos sin
sin cos
=
cos 2 sin2
sin2 cos 2
Vidyalankar : GATE – Engineering Mathematics
232
As cos
2
sin
2
= cos 2
& 2 sin cos = sin 2
Similarly , we get
A
n
=
cosn sinn
sinn cosn
15. (B)
As (ABC)
T
= C
T
B
T
A
T
(B) is wrong.
16. (C)
Let
pqr
TTT
pqr
111
Here, T
p
, T
q
and T
r
are p
th
q
th
and r
th
term of an AP.
Let the first term of the AP be a and
common difference of the AP be d.
Then
T
p
= a +(p 1)d
T
q
= a +(q 1)d
T
r
= a + (r 1)d
a (p 1)d a (q 1)d a (r 1)d
pqr
111
R
2
R
2
R
3
a (p 1)d a (q 1)d a (r 1)d
p1 q1 r1
111
a (p 1)d a (q 1)d a (r 1)d
1
(p 1)d (q 1)d (r 1)d
d
111
…..(d0
)
112
RRR
=
aaa
1
(p 1)d (q 1)d (r 1)d
d
111
=
111
a
(p 1)d (q 1)d (r 1)d
d
111
= 0
17. (D)
18. (D)
Solutions Linear Algebra
233
Model Solution on Assignment 10
1. (D)
As R
2
= (2) R
1
| A|
3 3
= 0
rank 0
| A |
2 2
=
32
64
= 12 + 12 = 0
418
836
= 144 144 = 0
| A |
2 2
= 0
| A |
1 1
= | 3 | = 3
| A |
1 1
0
rank = 1
Or R
2
= 2R
1
R
3
= 4R
1
rank = 1
2. (B)
This is like a property of triangular
matrix.
3. (B)
4. (C)
The diagonal matrix is
= [x
1
…….x
n
]
1
1
2
nn
x
00
00:
.
.
.
00 x
= x
1
2
+
2
x
2
2
+ …… +
n
x
n
2
5. (B)
Q is obtained by interchanging C
1
and C
2
| Q | = | P | = 8
6. (C)
By performing R
4
R
4
R
3
R
2
R
1
& R
2
R
2
2R
1
, R
3
R
3
3R
1
.
Sequentially we get
A ~
12 30
00 32
0483
00 00
Now | A |
4 4
= 0 … As E transform
does not change rank of matrix
rank 4
Now
12 3
00 3
048
= 12 i.e. 0
Rank = 3
7. (A)
A matrix is said to be orthogonal if
A
T
A = I
11
0
22
21 1
666
111
333
21
0
63
11 1
263
111
263
=
100
010
001
= I
A is orthogonal
8. (A)
For triangular matrix the eigen values
are same as principal diagonal
elements.
Vidyalankar : GATE – Engineering Mathematics
234
9. (A)
ahg x
xyzhbf y
gfcz
= ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
i.e. X
T
AX gives,
ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
If A =
11 12 13
21 22 23
31 32 33
aaa
aaa
aaa
Then for
ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
a
11
= a a
22
= b a
33
= c
a
12
= a
21
= h =
1
2
× (co-efficient of xy)
a
23
= a
32
= f =
1
2
× (co-efficient of yz)
a
31
= a
13
= g =
1
2
× (co-efficient of zx)
10. (B)
Characteristic equation of A is
| A I | = 0
622
23 1
213
= 0
by C
3
C
3
+ C
2
620
23 2
212
= 0
R
2
R
2
R
3
(2 )
620
44 0
211
= 0
(2 ) . [(6 ) . (4 ) 8] = 0
[(2 ) (
2
10 + 16) ] = 0
(2 ) ( 2) ( 8) = 0
= 2, 2, 8
Or sum of the eigen values of a matrix
= trace of the matrix
11. (C)
As R
4
= 3R
1
| A |
4 4
= 0
Now
148
003
423
= 1(0 6) 4(0 12) + 8(0)
= 6 + 48 = 42 i.e. 0
rank = 3
12. (B)
| A | = 37
Adj. A =
14 6
41613
6131
A
1
=
14 6
1
41613
37
6131
13. (C)
As A is not symmetric matrix.
Solutions Linear Algebra
235
14. (C)
123
458
321
= 1(5 16) 2(4 24) + 3(8 15)
= 11 + 40 21
= 8 i.e. 0
rank = 3
15.
(D)
As R
3
= R
1
Value of determinant = 0
16. (C)
ax c b
cbxa 0
bacx
R
1
R
1
+ (R
2
+ R
3
)
ax cb ba
cb xa cx
0
cbxa
bacx
ab ab ab
cx cx cx
0
cbxa
bacx
a + b + c = 0
xx x
cbx a 0
bacx
11 1
xc b x a 0
bacx
x = 0 is one root of the given
equation.
17. (D)
ax b c
0bxa 0
00cx
bx a b c
(a x) 0
0cx 0cx
bc
00
bxa
(a x) (b x)(c x) 0 0 0 0
(a x)(b x)(c x) = 0
x = a, b, c are the roots.
18. (C)
Vidyalankar : GATE – Engineering Mathematics
236
Model Solution on Assignment 11
1. (C)
2. (C)
3. (C)
4. (D)
5. (C)
23
46
= 12 12 = 0
23
46
is not invertible.
6. (C)
7. (C)
A = (a
ij
) is 3 2 matrix
Elements are given as
a
ij
= 2i j i > j
= 2j I i j
Elements of A will be a
11
, a
12
, a
21
, a
22
,
a
31
, a
32
Elements having i > j = a
21
, a
31
, a
32
a
21
= 2(2) (1) = 3
a
31
= 2(3) (1) = 5
a
32
= 2(3) 2 = 4
Elements having i j ; a
11
, a
12
, a
22
a
11
= 2(1) (1) = 1
a
12
= 2(2) (1) = 3
a
22
= 2(2) (2) = 2
Matrix
13
A32
54
8. (C)
X + Y =
70
25
….(i)
X Y =
30
03
….(ii)
Addition of (i) and (ii) gives
2X =
10 0
28
X =
50
14
Y =
70 50
25 14
Y =
20
11
9. (A)
Check with options
10. (B)
Let
12 3
A012
00 1
12 3
A012
00 1
=
12 23 23
10 0
01 0 1 1 2
A10
A
1
exists
Solutions Linear Algebra
237
Cofactors :
11
11
12
A(1) 1
01
12
12
02
A(1) 0
01
13
13
01
A(1) 0
00
21
21
23
A(1) 2
01
22
22
13
A(1) 1
01
23
23
12
A(1) 0
00
31
31
23
A(1) 7
12
32
32
13
A(1) 2
02
33
33
12
A(1) 1
01
Cofactor Matrix =
100
210
721
Adj A =
T
Cofactor matrix
=
T
100
210
721
Adj A =
127
01 2
00 1
1
adjA
A
A
127
1
0 1 2 .....( A 1)
2
00 1
1
127
A012
00 1
11. (D)
12. (B)
13. (C)
a
1
, a
2
, ……. are in GP
m3
m1 m2
mm1m2
a
aa
k(say)
aa a
a
m+1
= a
m
k
a
m+2
= a
m+1
k = a
m
k
2
a
m+3
= a
m+2
k = a
m
k
3
and so on
=
mm1m2
m3 m4 m5
m6 m7 m8
loga loga loga
loga loga loga
loga loga loga
=
mm
m
mmm
mmm
loga loga
loga
logk 2logk
loga loga loga
3 logk 4logk 5logk
loga loga loga
6logk 7logk 8logk
Vidyalankar : GATE – Engineering Mathematics
238
221
RRR ;
331
RRR
=
mm
m
loga loga
loga
logk 2logk
3logk 3logk 3logk
6logk 6logk 6logk
=
mm
m
loga loga
loga
logk 2logk
36
logk logk logk
logk logk logk
= 0
14. (D)
Given determinant,
=
1 cos( ) cos( )
cos( ) 1 cos( )
cos( ) cos( ) 1
=
1 cos( ) cos( )
cos( ) 1 cos( )
cos( ) cos( ) 1
.... Since cos() = cos
= 1(1 cos
2
( ))
cos( )
cos( )
cos( )cos( )
+ cos( )
cos( ) cos( )
cos( )
= 1 cos
2
( ) cos
2
( )
+ cos( ) cos( ) cos( )
+ cos( ) cos( ) cos( )
cos
2
( )
= 1 cos
2
( ) cos
2
( ) cos
2
( )
+ 2{cos( ) cos( )} cos( )
= 1 cos
2
( ) cos
2
( )
cos
2
( ) + cos{( ) ( )}
+ cos{( ) + ( )} cos( )
= 1 cos
2
( ) cos
2
( ) cos
2
( )
+ cos( ) cos( 2 + )
+ cos( ) cos ( )
=
22
1cos( )cos( )
cos ( ) ( 2 )
cos ( ) ( 2 )
2
=
22
1cos( ) cos( )
cos( 2 2 ) cos(2 2 )
2
=
22
1cos( ) cos( )
cos2( ) cos 2( )
2
=
22
2
2
2 2cos ( ) 2 cos ( )
2cos ( ) 1
2cos ( ) 1
2
= 0
15. (B)
024
A112
20 5
Characteristic Equation of A is
AI0
Solutions Linear Algebra
239
024
11 2 0
205
i.e.
24
11 2 0
205
1212
2
05 25
11
40
20
(1 )(5 ) 0 2 5 4
40 2(1 ) 0
2
56 21
42 2 0
23
56 2288 0
3
+ 6
2
1 + 6 = 0
i.e.
3
6
2
+ 1 6 = 0
16. (C)
Characteristic equation: |A I| = 0
(3 ) 10 5
2(3) 4
35(7)
= 0
gives
3
7
2
+ 16 12 = 0
= 2, 2, 3
For = 2
1
2
3
(3 2) 10 5 x
2(32) 4 x
35(72)x
=
0
0
0
1
2
3
1105 x
254x
355x
=
0
0
0
5
2
5
satisfies the above equation.
17. (D)
1
=
23
23
23
22 2
33 3
44 4
=
2
2
2
12 2
234 133
14 4
=
12 4
24 1 3 9
1416
1
= 24
18. (D)
Vidyalankar : GATE – Engineering Mathematics
240
Answer Key on Test Paper 1
1.
(B)
2.
(C)
3.
(C)
4.
(B)
5.
(C)
6.
(D)
7.
(C)
8.
(B)
9.
(A)
10.
(C)
11.
(D)
12.
(A)
13.
(B)
14.
(C)
15.
(C)
Answer Key on Test Paper 2
1.
(B)
2.
(B)
3.
(A)
4.
(A)
5.
(B)
6.
(B)
7.
(D)
8.
(B)
9.
(B)
10.
(D)
11.
(C)
12.
(B)
13.
(C)
14.
(A)
15.
(A)
Answer Key on Test Paper 3
1.
(C)
2.
(C)
3.
(D)
4.
(A)
5.
(C)
6.
(A)
7.
(B)
8.
(D)
9.
(D)
10.
(C)
11.
(C)
12.
(B)
13.
(B)
14.
(C)
15.
(A)
Answer Key on Test Paper 4
1.
(B)
2.
(C)
3.
(C)
4.
(C)
5.
(B)
6.
(A)
7.
(A)
8.
(C)
9.
(B)
10.
(B)
11.
(A)
12.
(A)
13.
(B)
14.
(A)
15.
(B)
Answer Key on Test Paper 5
1.
(A)
2.
(B)
3.
(B)
4.
(B)
5.
(C)
6.
(C)
7.
(B)
8.
(A)
9.
(B)
10.
(D)
11.
(B)
12.
(A)
13.
(B)
14.
(C)
15.
(D)
Answer Key on Test Paper 6
1.
(B)
2.
(B)
3.
(B)
4.
(A)
5.
(A)
6.
(C)
7.
(C)
8.
(B)
9.
(C)
10.
(D)
11.
(A)
12.
(B)
13.
(B)
14.
(B)
15.
(C)
Solutions Linear Algebra
241
Model Solution on Test Paper 1
1. (B)
2. (C)
x + 3 = 0 x = 3
similarly y, z & a can be calculated.
3. (C)
4. (B)
5. (C)
Characteristic equation is
41
14
= 0
(4)
2
1 = 0
16 8 +
2
1 = 0
2
8 + 15 = 0
( 5) ( 3) = 0
1
= 3,
3
= 5
Or following property can be used
Sum of the eigen values = trace
6.
(D)
7. (C)
Relationship obtained by using
1
x
1
+
2
x
2
+
3
x
3
= 0
1
(1, 3, 4, 2) +
2
(3, 5, 2, 2)
+
3
(2, 1, 3, 2) = 0
gives 4 equations
i.e.
1
+ 3
2
+ 2
3
= 0
3
1
5
2
3
= 0
4
1
+ 2
2
+ 3
3
= 0
2
1
+ 2
2
+ 2
3
= 0
By solving these equations we get
1
= 1,
2
= 1,
3
= 2
Equation is x
1
+ x
2
2x
3
= 0
8. (B)
Suppose these points are collinear on
line ax + by + c = 0
Then ax
1
+ by
1
+ c = 0 ……(i)
ax
2
+ by
2
+ c = 0 …….(ii)
ax
3
+ by
3
+ c = 0 ……(iii)
Eliminate a, b, c, between (i),
(ii) and (iii)
11
22
33
xy1
xy1
xy1
= 0
Thus rank of matrix
11
22
33
xy1
xy1
xy1
is
less than 3
Conversely if rank of matrix A is less
than 3, then
11
22
33
xy1
xy1
xy1
= 0
Area of triangle with vertices
(x
1
, y
1
) (x
2
, y
2
) (x
3
, y
3
) is equal to 0
They are collinear points.
Vidyalankar : GATE – Engineering Mathematics
242
9. (A)
We have equation as
11 1 x 9
25 7 y 52
21 1 z 0
AX = B
Now augmented matrix
[A : B] =
11 1: 9
25 7 :52
21 1: 0
We reduce it in Echelon form
R
2
R
2
2R
1
R
3
R
3
2R
1
11 1: 9
03 5:34
013:18
R
2
R2
3
11 1 : 9
015/3:34/3
01 3: 18
R
3
R
3
+ R
2
11 1 : 9
01 5/3 : 34/3
00 4/3: 20/3
Rank A = Rank [A : B] = 3
Equations are consistent.
10.
(C)
By taking product
AB =
2
2
2
aabac
0c b
c0 a abb bc
ba0
ac bc c
=
000
000
000
11. (D)
f(x) = x
2
5x + 6
f(A) = A
2
5A + 6 I
=
31 31 31
5
12 12 12
10
6
01
which gives
f(A) =
10
01
12. (A)
A matrix is said to be idempotent if
A
2
= A
Here,
224224
13 4 13 4
123123
224
gives 1 3 4
123
which is same as A.
It is idempotent
Solutions Linear Algebra
243
13. (B)
A =
cos sin 0
sin cos 0
001
| A | = 1
A
1
=
adjA
|
A
|
Adj A = [cofactor matrix A]
T
=
cos sin 0
sin cos 0
001
=
cos sin 0
sin cos 0
001
A
1
=
cos sin 0
sin cos 0
001
14.
(C)
We have (A I ) = 0
111
11 1
111
= 0
(1 ) [(1 )
2
1] 1[(1 ) 1]
+ 1[1 (1 )] = 0
(1 ) [(1 )
2
1] 1[(1 ) 1]
+ 1[1 (1 ) ] = 0
(1 ) (
2
2) + 2 = 0
2
(3 ) = 0
= 0, 0, 3
15. (C)
We have C
1
1
1
C
8
1136
0322
1134
Now R
3
R
3
+ R
1
113 6
032 2
00010
which is in Echelon
form so rank is equal to no. of nonzero
rows
Rank = 3
Vidyalankar : GATE – Engineering Mathematics
244
Model Solution on Test Paper 2
1.
(B)
2. (B)
3. (A)
4. (A)
2A + 3B = 2
234
125
+3
01 2
24 6
=
4912
8828
5. (B)
=
1a bc
1b ca
1c ab
Performing R
2
R
2
R
1
,
R
3
R
3
R
1
1a bc
0bacabc
0caabbc
= 0
(b a) (ab bc)
(c a) (ca bc) = 0
(b a). b(a c)
(a c) c(a b) = 0
(a c) (b a) (b c) = 0
(a b) is a factor of
6.
(B)
The given system of equation is
equivalent to the single matrix
equation
AX =
26 0 x
620 6 y
06 18 z
11
3
1
= B
We reduce this matrix into triangular
matrix by sequentially performing
R
2
R
2
3R
1
,
R
3
R
3
3R
2
, we get
26 0 x 11
02 6 y 30
00 0 z 91
We have 0x + 0y + 0z = 91, this
shows systems is not consistent
7. (D)
Using
cos cos sin sin = cos ( + )
cos sin + sin cos = sin ( + )
cos cos sin sin = cos ( + )
we get
cos( ) sin( )
sin( ) cos( )
8. (B)
As A
3
= A . A . A = 0
where A is given matrix
It is nilpotent matrix
Solutions Linear Algebra
245
9. (B)
R
3
R
3
+ R
1
, R
2
R
2
+ 2R
1
gives
123
00 5
0010
The determinant of above matrix is 0
Checking for non-zero minor of order 2
12
24
= 0
23
41
= 2 12 = 10 0
rank = 2
10. (D)
11. (C)
As it is not a singular matrix
12. (B)
A =
13 5
02 1
00 3
I + A + A
2
=
100 13 5
010 02 1
001 00 3
13 5 13 5
02 1 02 1
00 3 00 3
=
23 5 1917
03 1 04 5
00 4 00 9
=
31222
07 6
0013
Now for triangular matrix eigen values
are same as principal diagonal
element.
1
= 3,
2
= 7,
3
= 13
are eigen values.
13. (C)
Let A =
cos sin
sin cos
B =
a0
0b
Here AB = BA if a = b
14. (A)
R
4
= 3R
1
Hence rank < 4 and
148
003
423
= 48 0
Rank = 3
15. (A)
Vidyalankar : GATE – Engineering Mathematics
246
Model Solution on Test Paper 3
1.
(C)
As for triangular matrix eigen values
are principal diagonal elements
2. (C)
As per definition of rank of matrix
3. (D)
As columns of A rows of B
So product is not possible
4. (A)
5. (C)
As these are properties of skew
symmetric matrix.
6. (A)
e.g. A =
11
11
for 2 2 matrix
Rank of A = 1
7. (B)
8. (D)
As per property of idempotent matrix if
A & B are idempotent implies AB = BA.
9. (D)
The determinant of 3 3 element of
given matrix is not equal to zero or
when reduced to echelon form, it gives
3 rowed matrix.
10. (C)
The given equation in Matrix Form:
23 5 x
73 2 y
23 z
=
9
8
AX = B
[A : B] =
23 5 :9
73 2:8
23 :
331
RRR
23 5 : 9
73 2 : 8
00( 5):( 9)
If rank(A) rank (A : B), system has no
solution.
= 5, 9
11. (C)
Just by the definition of the rank of
matrix we find that the rank of given
matrix A is 3.
12. (B)
Characteristic equation of A is
|A I| = 0
102
02 1
203
= 0
(1 ) [(2 ) (3 ) 0]
+ 2 2 (2 ) = 0
3
6
2
+ 7 + 2 = 0
Solutions Linear Algebra
247
13. (B)
By solving the above equations as
follows
212:1
110:0
1k6:3
That is, after reducing to Echelon
form, if we want unique solution,
then K 2
14. (C)
Hint : By just the definition
i.e. transpose of all elements' cofactor
matrix.
OR [cofactor Matrix ]
T
15.
(A)
R
3
R
3
(R
1
+ R
2
), we get
x2 2x33x4
2x 3 3x 4 4x 5
013x8
= 0
By R
2
R
2
R
1
& R
1
R
1
+ R
3
x22x46x12
x1 x1 x1
013x8
= 0
(x + 1) (x + 2)
12 6
11 1
013x8
= 0
R
1
R
1
R
2
(x + 1) (x + 2)
01 5
11 1
013x8
= 0
Now,
(x +1) (x + 2) [(3x + 8 5)] = 0
3 (x + 1) (x + 2) (x + 1) = 0
x = 1, 1, 2
Vidyalankar : GATE – Engineering Mathematics
248
Model Solution on Test Paper 4
1.
(B)
As per the property of the skew
Hermitian matrix
2. (C)
As per the property of the eigen values
for orthogonal matrices. i.e. for
orthogonal matrix, the eigen value is
reciprocal of the original eigen value.
3. (C)
Hint : The transpose of the matrix
does not change the rank of the
matrix.
4. (C)
As per the defitnition of nilpotent
matrix for index K.
5. (B)
ahg x
xyzhbf y
gfcz
= ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
i.e. X
T
AX gives,
ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
If A =
11 12 13
21 22 23
31 32 33
aaa
aaa
aaa
Then for
ax
2
+ by
2
+ cz
2
+ 2hxy + 2fyz + 2gzx
a
11
= a a
22
= b a
33
= c
a
12
= a
21
= h =
1
2
× (co-efficient of xy)
a
23
= a
32
= f =
1
2
× (co-efficient of yz)
a
31
= a
13
= g =
1
2
× (co-efficient of zx)
Extension of this method
6. (A)
When reduced to Echelon form, non-
zero rows obtained is 1.
Or R
2
= 6R
1
& R
3
= 5R
1
7. (C)
8. (C)
A =
211
12 1
112
Characteristic equation of matrix
211
12 1
112
= 0
3
6
2
+ 9 + 4 = 0
9. (B)
10. (B)
11. (A)
Inverse of matrix A is,
A
1
=
1
Adj A
|A|
Solutions Linear Algebra
249
12. (A)
0 is an eigen value of A = 0
satisfies the equation | A I | = 0
| A | = 0 A is singular
Conversely | A | = 0
= 0 satisfies the equation
| A I | = 0
0 is eigen value of A
13.
(B)
R
2
= 3R
1
As one of the row is repeated, so the
no. of rows for considering rank
reduces to 2. Now, for given row
matrix, determinant is non zero and
hence the matrix is 2.
14. (A)
15. (B)
Vidyalankar : GATE – Engineering Mathematics
250
Model Solution on Test Paper 5
1.
(A)
For symmetric matrix A = A
T
2. (B)
We have, a + 3 = 2a + 1
a = 2
Similarly, b = 1, 2
3. (B)
A =
2
2
ab b
aab
A . A =
2
2
ab b
aab
2
2
ab b
aab
=
00
00
A
2
= 0 K = 2
Index = 2
4. (B)
It is the definition of eigen values.
5. (C)
For diagonal matrix, eigen values are
different.
6. (C)
7. (B)
C
2
= 2C
1
| A |
3 3
= 0
rank < 3
minor of order 2
32
24
= 12 4 = 8 0
rank = 2
8. (A)
For triangular matrix eigen values are
same as major principal diagonal
elements.
9. (B)
1
x
1
+
2
x
2
+
3
x
3
+
4
x
4
= 0
1
(1, 2, 4) +
2
(2, 1, 3) +
3
(0, 1, 2)
+
4
(3, 7, 2) = 0
1
+ 2
2
3
4
= 0
2
1
2
+
3
+ 7
4
= 0
4
1
+ 3
2
+ 2
3
+ 2
4
= 0
By solving these equations we get
1
= 9 ,
2
= 12,
3
= 5,
4
= 5
relation is 9x
1
12x
2
+ 5x
3
5x
4
= 0
10. (D)
11. (B)
| A | = 1
adj A =
T
aib cid
cidaib
A
1
=
Adj A
|
A
|
=
aib cid
cid aib
Solutions Linear Algebra
251
12. (A)
Idempotent matrix is one which
satisfies A
2
= A. But given matrix not
satisfy this relation
It is not a idempotent matrix
13. (B)
| A |
3 3
= 0
Rank < 3
Minor of order 2,
11
02
= 2
i.e. 0
rank = 2
14. (C)
determinant
1
600
820
144
= 1[6 [80] ] = 48
15. (D)
Vidyalankar : GATE – Engineering Mathematics
252
Model Solution on Test Paper 6
1.
(B)
| A | = a [b (d
)] = abd
2. (B)
As [A]
3 3
[B]
3 2
[B]
3 2
[A]
3 3
BA is not possible as
columns of B rows of A
3. (B)
A
is obtained by substituting i by i
4. (A)
| A |
3 3
=
312
624
312
= 0
| A |
2 2
=
31
62
= 0
Similarly other minors of order 2 are 0
| A |
1 1
= | 3 | 0
rank = 1
5. (A)
Characteristic equation are
| A I| = 0
53
33
= 0
(5 ) (3 ) 9 = 0
15 +
2
+ 2 9 = 0
2
2 24 = 0
= 6, 4
6. (C)
As per theorem,
1a 1 1 1
11b 1 1
111c1
1111d
= abcd
1111
1
abcd
Now here a = b = c = d = 4
= (4)
4
(0) = 0
7. (C)
By R
4
R
4
R
3
R
2
R
1
, R
1
R
2
,
R
2
R
2
2R
1
,
R
3
R
3
3R
1
Subsequently,
1124
05 3 7
04 910
00 0 0
We see that
112
05 3
04 9
= 33 0
Rank = 3
8. (B)
Hint : Adjoint A = [ Cofactor matrix A]
T
9. (C)
Characteristic roots are obtained by
| A I | = 0
12
11
21
= 0
Solutions Linear Algebra
253
(
2
1) 1( + 2) + 2(1 + 2) = 0
3
+ 6 4 = 0
3
+ 0
2
6 + 4 = 0
By synthetic division
2 1 0
6
4
0 2 4
4
1 2
2
0
= 2
2
+ 2 2 = 0
= 1
3
roots are 2, 1
3
10. (D)
As for triangular matrix eigen values
are same as principal diagonal
element.
11. (A)
12. (B)
As C
4
= 2C
1
| A |
4 4
= 0
rank < 4
Minor of order 3
121
131
243
= 1( 9 4) 2(3 2) + 1(4 6)
= 1 0
rank = 3
13.
(B)
14. (B)
AB =
34
212
11
134
20
2910
34 6
42 4
B A =
21
312
13
410
24
234
942
10 6 4
15.
(C)
The given equations in matrix form
1aa x
b1b y
cc1z
=
0
0
0
For non trivial solution,
|A| = 0
1aa
b1b
cc1
= 0
1(1 bc) a(b bc) + a(bc c) = 0
1 bc ab + abc + abc ac = 0
ab + bc + ac abc = 1 + abc ….(i)
abc
1a 1b 1c
=
a(1 b) (1 c) b(1 a) (1 c)
c(1 a)(1 b)
(1 a) (1 b) (1 c)
Vidyalankar : GATE – Engineering Mathematics
254
=
a(1 b c bc)
b(1 a c ac)
c(1 a b ab)
(1 a) (1 b c bc)
=
aabacabcbabbc
abc c ac bc abc
1 b c bc a ab ac abc
=
a b c abc
2(ab bc ac abc)
1 a b c (ab bc ac abc)
=
a b c abc 2(1 abc)
1a b c (1abc)
….From (i)
=
abcabc22abc
1a b c 1abc
=
2abcabc
2abcabc
= 1
255
Solutions Calculus
Answer Key on Assignment 1
1. (A) 2.
(D)
3.
(C)
4. (D)
5. (C) 6.
(C)
7.
(C)
8. 2
9. (C) 10.
(D)
11.
100.01 to
99.99
12.
5.9 to 6.1
13.
1.85 to 1.87
14.
1
15.
4.66 to 4.76
16 85.0 to 85.5
17. (B) 18.
(C)
19.
(C)
20.
(C)
Answer Key on Assignment 2
1. (B) 2.
(B)
3.
0.01 to 0.01
4. (A)
5. (A) 6.
(D)
7.
(B)
8.
5.1 to 4.9
9. (B) 10.
2.0 to 2.0
11.
(B)
12. (B)
13.
1.9 to 2.1
14.
(C)
15.
(B)
16
13 to 13
17. (D) 18.
(B)
19.
(A)
20.
0.99 to 1.01
Answer Key on Assignment 3
1.
40 to 40
2.
(A)
3.
(C)
4. (C)
5.
0.35 to 0.30
6.
(C)
7.
(A)
8. (C)
9.
(B)
10.
0.0 to 0.0
11.
(A)
12. (A)
13.
0.95 to 1.05
14.
18 to 22
15.
(D)
16 (A)
17. (A) 18.
(D)
19.
(C)
20. (A)
Vidyalankar : GATE – Engineering Mathematics
256
Answer Key on Assignment 4
1. (B)
2.
(D)
3.
(B)
4. (D)
5. (D)
6.
(B)
7.
(A)
8. (A)
9. (C)
10.
(D)
11.
(D)
12. (C)
13. (C)
14.
(C)
15.
(B)
16 (A)
17. (D)
18.
(B)
Answer Key on Assignment 5
1. (D)
2.
(A)
3.
(C)
4. (B)
5. (C)
6.
(A)
7.
(A)
8. (B)
9. (A)
10.
(A)
11.
(D)
12. (B)
13. (B)
14.
(D)
15.
(B)
16. (B)
17. (B)
18
(D)
Answer Key on Assignment 6
1. (A)
2.
(A)
3.
(B)
4. (D)
5. (B)
6.
(A)
7.
(A)
8. (B)
9. (C)
10.
(A)
11.
(B)
12. (C)
13. (A)
14.
(A)
15.
(B)
16. (B)
17. (C)
18.
(C)
Solutions Calculus
257
Answer Key on Assignment 7
1. (B)
2.
(C)
3.
(A)
4. (B)
5. (A)
6.
(D)
7.
(A)
8. (B)
9. (B)
10.
(C)
11.
(A)
12. (B)
13. (A)
14.
(D)
15.
(B)
16. (A)
17. (B)
18.
(A)
Answer Key on Assignment 8
1. (C)
2.
(C)
3.
(D)
4. (D)
5. (D)
6.
(D)
7.
(B)
8. (C)
9. (C)
10.
(D)
11.
(A)
12. (A)
13. (B)
14.
(A)
15.
(C)
16. (B)
17. (C)
18.
(B)
Answer Key on Assignment 9
1. (B)
2.
(B)
3.
(A)
4. (C)
5. (B)
6.
(D)
7.
(B)
8. (A)
9. (A)
10.
(C)
11.
(A)
12. (B)
13. (D)
14.
(C)
15.
(A)
16. (D)
17. (B)
18.
(B)
Vidyalankar : GATE – Engineering Mathematics
258
Model Solution on Assignment 1
1. (A)
x0
xsinx
lim
1cosx
=
0
0
Applying L. Hospital Rule,
x0
1cosx
lim
sinx
=
0
0
Once again, L. Hospital rule
x0
sinx
lim
cos x
=
0
1
= 0
2. (D)
We know that f(x) is continuous at x = a,
if
xa
lim f(x)
exists and equal to f(a).
3. (C)
x
x
1
lim 1
x
Let
1
h
x
as x , h 0
x
1
h
xh0
1
lim 1 lim 1 h
x
= e
4. (D)
We have e
a
=
23
aa
1 a ....
2! 3!
=
n
n0
a
n!
Put a = 1
1
n0
1
ee
n!
5. (C)
2
4
x0
1cos(x)
lim
2x
=
22
4
x0
2sin (x 2)
lim
2x
=
22
22
x0
sin (x 2) 1
lim
4
(x 2)
=
2
2
2
x0
1sin(x2)
lim
4
(x 2)
=
2
1
(1)
4
=
1
4
Alternate Method
2
4
x0
1cos(x) 0
lim
0
2x
Using L Hospital Rule
2
3
x0
(sinx )2x 0
lim
0
8x
22
2
x0
(cosx )2x2x (sinx )2
lim
24x
22 2
2
x0
(cos x )4x 2sinx
lim
24x
=
22 2
2
x0
(sinx)4x cosx(8x)
2(cos x )2x
lim
48x
=
22
2
22
2
x0
( cos x )2x 4x
(sin x)8x
( sin x ) 2x 8x 12cos x
(sinx)(2x)4x
lim
48
=
12
48
=
1
4
Solutions Calculus
259
6. (C)
Given: f(x) does not have a root in
[a, b]. That is from x = a to x = b, then
curve f(x) does not cross the X-axis.
Hence the curve is entirely above
X-axis or entirely below it.
f(a) > 0 and f(b) > 0 OR f(a) < 0
and f(b) < 0
f(a) . f(b) > 0
7. (C)
The ranges are given by 0 < y < a and
0 < x < y i.e. 0 < x < y < a.
This can also be written as 0 < x < a
and x < y < a
8. 2
n
n0
1
n.
2
= ?
This can be solved using Z-transform
and also using algebra as arithmetic-
geometric series. We will use algebra.
n
n0
1
n.
2
=
2
11
0 1 1 2 ...
22
[a + (a + d)r + (a + 2d)r
2
+ …..infinite
terms]
=
2
ard
1r
(1 r)
(1 < r < 1)
For above series, a = 0, d = 1, r =
1
2
.
n
n0
1
.n
2
=
2
0(12)1)
1
1
1
1
2
2
= 2
9. (C)
e
3x 3x
x0 x0
4
log (1 4x)
4
14x
lim lim
3
e1 3e
10. (D)
3
x0
xsinx
lim
x
=
2
x0 x0
sinx
(lim x ) lim
x
= 0 1 = 1
11. 100.01 to 99.99
f(x) =
2
1
x(x 3)
3
=
3
x
x
3
f(x) =
2
3x
1
3
= x
2
1
x
2
1 = 0
x = 1
f(x) = 2x
f(1) = 2 > 0
at x = 1, f(x) has local minimum.
f(1) = 2 < 0
at x = 1, f(x) has local maximum
For x = 1, local minimum value
f(1) =
12
2
33
Finding f(100) = 333433.33
f(100) = 333233.33
(
x = 100, 100 are end points of
interval)
Minimum occurs at x = 100
Vidyalankar : GATE – Engineering Mathematics
260
12. 5.9 to 6.1
f(x) = 2x
3
9x
2
+ 12x 3
f(x) = 6x
2
18x + 12
f(x) = 12x 18
f(x) = 0 gives 6(x
2
3x + 2) = 0
x = 1, x = 2
f(1) = 12 18 = 6
and f(2) = 24 18 = 6
There is local maxima at x = 1 and
local minima at x = 2
Now, f(0) = 3,
f(1) = 2 9 + 12 3 = 2 and
f(3) = 54 81 + 36 3 = 6
13. 1.85 to 1.87
The length of the curve
=
222
/2
0
dx dy dz
dt
dt dt dt
=
2
/2
22
0
2
( sint) (cos t) dt
=
/2
22
2
0
4
sin t cos t dt
=
/2
2
0
4
1dt
=
/
2
0
2
4
1.t
=
4
1x
22
= 1.8622
14. 1
I =
2/
2
2/
11
cos dx
x
x
Put
1
y
x
2
1
dx dy
x
Further, if x = 1/ then y = and
if x = 2/ then y = /2
I =
/2
cos y( dy)
=
/2
cos y dy
= sin() sin(/2) = 0 1 = 1
15. 4.66 to 4.76
Volume =
5/44.5
3/8z3
dz d d
=
5/4
3/8
(1.5) d d
=
5
3
1.5 ( / 8) d
= 1.5 × (/8) × 8
= 1.5 = 4.7124
16. 85.0 to 85.5
17. (B)
y = 2x 0.1x
2
dy
dx
= 2 0.2x
2
2
dy
0
dx
y maximizes at 2 0.2x = 0
x = 10
y = 20 10 = 10 m
Solutions Calculus
261
18. (C)
Given function is f(x) = |x|
|x| is continuous at x = 0 but not
differentiable
19. (C)
To find maxima
We differentiate f(x) w.r.t. ‘x’ & equate
to zero
2
df(x)
3x 18x 24 0
dx
x
2
6x + 8 = 0
(x 4) (x 2)
x = 2 or x = 4
Now,
2
2
x2
df(x)
6x 18 6 0
dx
(maxima)
2
2
x4
df(x)
60
dx
(minimum)
x2
f(x)
=
32
(2) 9(2) 24(2) 5
= 8 36 + 48 + 5 = 25
But since interval [1, 6] i.e. inclusive,
we have to find
f(1) = 1 9 + 24 + 5 = 21
f(6) = (6)
3
9(6)
2
+ 24(6) + 5
= 216 324 + 144 + 5 = 41
Thus, in the interval maximum value is
41.
20. (C)
Given x =
u
cos
2
,
y =
u
sin
2
0 ≤ u ≤ 1
dx u
sin
du 2 2
dy u
cos
du 2 2
We know that surface area when the
curve revolved about X-axis of a
parametric curve is
=
22
1
0
dx dy
2y du
du du
=
1
0
u
2sin
2
22
uu
sin cos du
22 22
=
1
2
0
u
2sin du
24
=
1
0
u
2sindx
22
=
1
2
0
u
1cos
2
2
=
2
2
cos cos 0
2
=
2[cos0 1]
= 2
Vidyalankar : GATE – Engineering Mathematics
262
Model Solution on Assignment 2
1.
(B)
2x
x0
e1
lim
sin 4x
is
0
0
So, Applying L-Hospital Rule,
2x
x0
2e
lim
4cos4x
=
2
4
= 0.5
2. (B)
Let I =
2
2
0
2
(x 1) sin(x 1)
dx
(x 1) cos(x 1)
We know that,
2
2
2
0
(2 x 1) sin(2 x 1)
dx
(2 x 1) cos(2 x 1)
=
2
2
2
0
(1 x) sin(1 x)
dx
(1 x) cos(1 x)
=
2
2
2
0
(x 1) sin(x 1)
dx
(x 1) cos(x 1)
= I
I + I = 0
2I = 0
I = 0
3. 0.01 to 0.01
f(x) =
n(1 x) x
f(x) =
1
1
1x
and f(x)
=
2
1
0
(1 x)
Maxima f(x) = 0 gives
1
10
1x
1
1
1x
1 + x = 1
x = 0
There is a maxima at x = 0
Maximum value = f(0) =
n(1) 0 0
4. (A)
In order to find maximum value of the
function we have
f(x) = 0
x e
x
+ e
x
= 0
e
x
(1 x) = 0
x = 1
And f(x) < 0 at x = 1
Maximum value is f(1) = (1) e
1
= e
1
5. (A)
For negative value of x, f(x) will be
positive
For positive values of x, f(x) will be
positive
minimum value of f(x) will occur at
x = 0
6. (D)
7. (B)
The given function is
F(x) = 1 x
2
+ x
3
,
where x [1, 1]
F(x) = 2x + 3x
2
By mean value theorem
F(x) =
F(1) F( 1)
1(1)
Solutions Calculus
263
Now F(1) = 1 (1)
2
+ (1)
3
= 1
and F(1) = 1 (1)
2
+ (1)
3
= 1 1 1 = 1
F(x) =
1(1)
1(1)
=
2
2
= 1
F(x) = 1
2x + 3x
2
= 1
3x
2
2x 1 = 0
3x
2
3x + x 1 = 0
3x(x 1) + 1(x 1) = 0
(3x + 1) = 0
or x 1 = 0
x =
1
3
or x = 1
Now
1
3
lies between (1, 1)
x =
1
3
8. 5.1 to 4.9
f(x) = 2x
3
3x
2
where x [1, 2]
f(x) = 6x
2
6x and f(x) = 12x 6
f(x) = 0 gives 6x
2
6x = 0
6x(x 1) = 0 x = 0, 1
f(0) = 6 < 0
There is a maxima at x = 0
f(1) = 12 6 = 6 > 0
There is a minima at x = 1
f(1) = 2 3 = 1; f(1) = 2 3 = 5;
f(2) = 16 12 = 4
Global minimum value of f(x) = 5
9. (B)
10. 2.0 to 2.0
I =
1
0
1
dx
(1 x)
;
Use:
aa
00
f(x) dx f(a x) dx
I =
1
0
1
dx
(1 (1 x))
=
1
0
1
dx
x
=
1
1/ 2
0
xdx
=
1
1/ 2
0
x
1/ 2
= 2
11. (B)
2x
xy
00
edydx
=
2x
xx y
00
ee edydx
=
2
x
xy
0
0
ee dx
=
2
xx
0
e(e 1)dx
=
2
2x x
0
(e e ) dx
=
2
2x
x
0
e
e
2
=
4
2
e1
e1
22
=
4
2
e1
e
22
=
42
1
(e 2e 1)
2
=
22
1
(e 1)
2
Vidyalankar : GATE – Engineering Mathematics
264
12. (B)
f(x) = x
3
3x
2
24x + 100
where x [3, 3]
f(x) = 3x
2
6x 24 and
f(x) = 6x 6
f(x) = 0 gives 3(x
2
2x 8) = 0
(x 4) (x + 2) = 0 x = 2, 4
As x = 4 > 3, consider x = 2 [3, 3]
f(2) = 12 6 < 0
There is a maxima at x = 2
f(3) = 27 27 + 72 + 100 = 118
f(3) = 27 27 72 + 100 = 28
Minimum value of f(x) = 28
13. 1.9 to 2.1
1
4
x circumference
1
4
x x 4
time =
1.57
= 2 sec
14. (C)
x
lim
2
xx1x
=
x
lim
2
2
2
xx1x
xx1x
xx1x
=
x
lim
22
2
xx1x
xx1x
=
x
lim
2
1
x1
x
11
x1 1
x
x
=
1
2
15. (B)
16. 13 to 13
f(x) = 2x
3
x
4
10 where x [1, 1]
f(x) = 6x
2
4x
3
and
f(x) = 12x 12x
2
f(x) = 0 gives 2x
2
(3 2x) = 0
x = 0. 1.5
As x = 1.5 > 1, consider x = 0 [1, 1]
f(0) = 0 There is a neither maxima
nor minima at x = 0
f(1) = 2 1 10 = 13
f(1) = 2 1 10 = 9
Minimum value of f(x) = 13
17. (D)
The function f(x) = x
3
+ 1 has a point of
inflection at x = 0, since in the graph
sign of the curvature (i.e., the
concavity) is changed.
Solutions Calculus
265
18. (B)
2
22
x0 x0
x
2sin
1cosx
2
lim lim
xx
2
2
x0
x
sin
2
2lim
x
4
2
11
2
42
19.
(A)
x3 x3
lim f(x) lim (x 1) 2 f(3)
x3 x3
x3
lim f(x) lim 2 f(3)
3
f(x) is continuous at x = 3
20. 0.99 to 1.01
2
R
xy dxdy
=
1
22
RR
xy dxdy xy dxdy
=
52 52x
22
x1 y0 x1 y2
xy dxdy xy dx
=
52 2x
5
23 3
1
10 2
xy y
xdx
23 3
=
5
3
1
81
(12) x(8x 8) dx
33
=
55
52
11
1x x
32 8 8
35 2
=
1 24992
32 32
35
=
24992
15
2
R
Cxydxdy
=
4
2
(24992) 10
5
= 0.99968 1
OR
2
R
xy dxdy
=
52x
2
x1 y0
xy dy dx
=
2x
5
3
1
0
y
xdx
3
=
5
3
1
8
x(x ) dx
3
=
5
5
1
8x
35
8 24992
(3124)
15 15
2
R
Cxydxdy
=
4
24992
10 6
15
=
2
2.4992
5
= 0.9968 1
Vidyalankar : GATE – Engineering Mathematics
266
Model Solution on Assignment 3
1. 40 to 40
22
f
((y z ) (2x))
x
at x = 2, y = 1, z = 3 = (1 + 9) (4) = 40
2. (A)
f(t) = e
t
2e
2t
where 0 ≤ t <
f(t) = e
t
2e
2t
(2) = 4e
2t
e
t
and f(t) = e
t
8e
2t
f(t) = 0 gives 4e
2t
e
t
= 0
e
t
(4e
t
1) = 0
e
t
= 0 or e
t
= 1/4
t = or e
t
= 4 t = , t = n(4)
f() = 0
Neither maxima or minima at t =
f
(n4) =
11
8
416
=
1
0
4
Maxima at t =
n(4)
3. (C)
z =
xy n(xy) xy n(x) xy n(y)
z
x
=
y[1 n(x)] y n(y)
=
yyn(xy)
and
z
y
= xn(x) x[1 n(y)]
=
xxn(xy)
zz
xy
xy
4. (C)
Given,
x
xsinx
lim
x
Put x =
1
h
and x or h 0
Now,
h0
11
sin
hh
lim
1
h
=
h0
1
1hsin
h
lim
1
=
10
1
= 1
5. 0.35 to 0.30
x0
sinx 0
lim form
2sinx xcosx 0
=
x0
cos x
lim
2cos x cos x x sinx
(L Hospital Rule)
=
1
3
6. (C)
I =
/2
0
cos x isinx
dx
cos x isinx
=
/2
ix
ix
0
e
dx
e
=
/2
2ix
0
edx
=
/2
2ix
0
e
2i
=
i0
1
[e e ]
2i
=
1
[cos() isin() 1]
2i
=
2
i
[1 0 1] i
2i
Solutions Calculus
267
7. (A)
The partial derivative of x
2
y
2
with
respect to y is 0 + 2y 2y.
The partial derivate of 6y + 4x with
respect x is 0 + 4 = 4.
Given that both are equal.
2y = 4
y = 2
8. (C)
f(x) is discontinuous when
denominator = 0
(x
2
+ 3x 4) = 0
(x + 4) (x 1) = 0
x = 4, 1
9. (B)
f(x) = x
3
3x
2
+ 1
f(x) = 3x
2
6x = 3x(x 2)
Between 1 & 0, f(x) > 0 i.e. f(x) is
increasing
Between 0 & 2, f(x) < 0 i.e. f(x) is
decreasing
above 2, f(x) > 0 i.e. f(x) is increasing
10. 0.0 to 0.0
f(x) = x(x 1) (x 2) = x
3
3x
2
+ 2x
where x [1, 2]
f(x) = 3x
2
6x + 2 and f(x) = 6x 6
f(x) = 0 gives 3x
2
6x + 2 = 0
x =
63624
6
=
612
6
=
12
1
6
= 1 0.5774
= 0.4226, 1.5774
Now, x = 0.4226 < 1.
Consider x = 1.5774
f(1.5774) = 6(1.5774 1) > 0
Minima at x = 1.5774
f(1) = 0 and f(2) = 0
Maximum value is f(x) = 0
11. (A)
3 sin x + 2 cos x
replacing sin x and cos x by its series,
=
35
xx
3x
3! 5!
24
xx
2 1 .....
2! 4!
=
35
3x 3x
3x .....
3! 5!
24
2x 2x
2 ....
2! 4!
=
3
2
x
2 3x x .....
2
12. (A)
0
x10
lim form
0
Hence applying L Hospital rule
Let y =
x
log y = log x
1dy
ydx
=
d
log x
xdx
dy
dx
=
d
xlogx
xdx
Vidyalankar : GATE – Engineering Mathematics
268
Now applying L Hospital rule
0
d
xlogx
xdx
lim
d
dx
= log x
13. 0.95 to 1.05
Here rectangle is inscribed in ellipse.
Let the length and breadth of rectangle
be 2x and 2y respectively.
Now area of rectangle = (2x) (2y)
= 4xy
Let F = (Area of rectangle)
2
= (4xy)
2
= 16 x
2
y
2
=
2
2
1x
16x
4
2
2
1x
y from equation of ellipse
4
= 4x
2
(1 x
2
)
Now F(x) = 0
x(1 2x
2
) = 0
x = 0
or 1 2x
2
= 0
2x
2
= 1
x
2
=
1
2
x =
1
2
For x = 0,
y
2
=
1
4
y =
1
2
and for x =
1
2
y
2
=
1
1
2
4
=
1
8
y =
1
8
Now F(x) = 8 48 x
2
F(x) < 0 for
x =
1
2
F(x) is maximum at x =
1
2
The maximum area of rectangle
inscribed in ellipse is
4xy =
11
4
28
=
4
4
= 1
14. 18 to 22
I =
D
1
(x y 10) dx dy
2
where D
denotes x
2
+ y
2
≤ 4
Use polar co-ordinates. Put x = r cos ,
y = r sin dx dy = r dr d
Solutions Calculus
269
I =
22
r0 0
1
(r cos r sin 10)r dr d
2
=
2
2
0
r0
1
r(r sin r cos 10 dr
2
=
2
r0
1
(10 r 2 ) dr
2
=
2
2
0
r
10
2
= 10 × 2 = 20
15. (D)
16. (A)
P = 50q 5q
2
2
2
dp d p
50 10q; 0
dq
dq
p is maximum at 50 10q = 0 or,
q = 5
Else check with options
17. (A)
The given curves are y = x and y = x
2
solving (1) and (2) we have
x = 0, x = 1
Area =
1
2
0
(x x )dx
=
1
23
0
xx 11
23 23
=
1
6
sq units.
18. (D)
19. (C)
e
1
x
In (x) dx
=
e
33
22
1
x1x
In(x) dx
32
x
23
=
e
33
22
1
24
In(x) x x
39
=
3
24
e
99
20. (A)
Given
I =
1
12
2
0
(sin x)
dx
1x
=
1
13 n1
n
0
(sin x) f (x)
f(x)f(x)dx
3n1
=
13 1
1
(sin ) sin 0
3
=
3
1
0
32
=
3
24
Vidyalankar : GATE – Engineering Mathematics
270
Model Solution on Assignment 4
1.
(B)
2
x0
1cosx
lim
x
=
2
x0
sin(x / 2) 1
lim
(x / 2) 2
=
1
2
x0
sin
lim 1
2. (D)
Applying L’ Hospitals rule, we have
0
cos
lim cos 0 1
1
3. (B)
L =
1/ 3
x8
x2
lim
(x 8)
Let x = a
3
a 2
L =
3
a2
(a 2)
lim
a8
Now by partial fractions,
(a
3
8) = (a 2) (a
2
+ 2a + 4)
L =
2
a2
11
lim
12
a2a4
Alternate Method
Since
nn
xa
xa
lim
xa
=
n1
na
L =
1/ 3
x8
x2
lim
(x 8)
=
2
3
1
(8)
3
=
1
12
4. (D)
x
ax b
lim
cx
=
x
ab
lim
cx
=
a
0
c
=
a
c
5.
(D)
x0
Lim f(x)
=
x
x0
e1
Lim
x
= 1
But f (0) =
0
0
, which is not defined.
Since
x0
Lim f(x) f(0)
therefore f (x)
has removable discontinuity at x = 0.
6. (B)
2
x
3
x0
x
e1x
2
lim
x
=
23
2
3
x0
xx
1x ...
2! 3!
x
1x
2
lim
x
=
2
x0
1xx 1
lim ...
64!5! 6
Else, use L'Hospital's rule
7. (A)
8. (A)
9. (C)
Solutions Calculus
271
10. (D)
f (x) =
32
x6x24x4
f ‘(x) =
2
3x 12x 24
f ‘’(x) = 6x 12
For maxima and minima, f ‘(x) = 0
or
2
3x 12x 24 = 0
or x = 2 2i
f (x) has no maximum and no
minimum at any real value of (x).
11. (D)
Two parts are 30 and 30 x
f (x) =
3
x.(30 x)
f (x) =
34
30x x
f(x) =
23
90 x 4 x
f (x) = 180 x
2
12x
For maximum and Minimum f ‘(x) = 0
or
23
90 x 40 x = 0
or x = 0, x =
90
4
=
45
2
But, x = 0 is impossible, hence
consider only x =
45
2
45
f''
2
=
2
45 45
180 12
22
45
f''
2
< 0
f (x) is maximum at x =
45
2
= 22.5
Two parts of 30 are 22.5 and 7.5
12. (C)
f (x) =
3
(x 2)
f(x) =
2
3(x 2)
f(x) = 0 for maximum and minimum.
or x = 2
When x > 2, f ‘(x) =
2
3(x 2) 0
When x < 2, f ‘(x) > 0
Hence neither maximum nor minimum
at x = 2.
13. (C)
y = sin x (1 + cos x)
dy
dx
= cos x
22
sin x cos x
= cos x + cos 2x
2
2
dy
dx
= (sin x + 2 sin 2x)
For maximum and minimum,
dy
dx
= 0
cos x + cos 2x = 0
or
2
2cos x + cos x 1 = 0
or cos x =
1
2
, cos x = 1
x =
3
or x =
But
3
(0, ) and (0, )
2
2
dy
0
dx
at x =
3
Vidyalankar : GATE – Engineering Mathematics
272
14. (C)
f(x) = 4x
3
8x
2
+ 1 in [1, 1]
f(x) = 12x
2
16x
Equating f(x) to 0
12x
2
16x = 0
3x
2
4x = 0
x(3x a) = 0
x = 0, x =
4
3
Now,
4
x[1,1]
3
So x = 0
f(x) = 6x 4
f(0) = 4 < 0
f(x) in not minimum at x = 0.
Checking at endpoints :
f(1) = 4(1)
3
8(1)
2
+ 1
= 4 8 + 1 = 11
f(1) = 4(1)
3
8(1)
2
+ 1
= 4 8 + 1 = 3
f(x) has minimum at x = 1 in [1, 1]
15. (B)
v =
2
4dv dr
r4r
3dt dt
4 = 4r
2
dr
dt
2
dr 1 1
0.01
dt 100
r
Since diameter = 2 radius
Hence diameter is decreasing at the
rate of 0.02 cm /hr.
16. (A)
2
dV
4r
dt
V =
3
4
r
3
22
dV 4 dr dr
3r 4 r
dt 3 dt dt
22
dr
4r 4r
dt
i.e.
22
dr
4r k4r
dt
dr
k
dt
17. (D)
f
(x) = 4x
3
6x
2
2x 4
= 2 (x 2) (2x
2
+ x + 1)
Thus only critical number is x = 2
So f (0) = 3
f (2) = 9
f (4) = 99
18. (B)
f
(x) =
23
3
x23x x
x2
2
2
2x x 3
0
x2
x = 0 and x = 3 are critical.
However x = 3 does not lie in interval
so we list 0 and end points 1 and 1.
f(0) = 0, f (1) = 1, f (1) = 1/3
Solutions Calculus
273
Model Solution on Assignment 5
1.
(D)
2
x
Lim ( x 1 x 1)
=
2
2
2
x
(x 1 x 1)
Lim( x 1 x 1)
(x 1 x 1)
=
2
2
x
(x 1) (x 1)
Lim
(x 1 x1)
=
x
2
x(x 1)
Lim
11 1
x1 1
x
xx
=
x
2
(x 1)
Lim
11 1
11
x
xx
=
01
=
2. (A)
x0
Lim f(x)
=
x0
ax b
Lim
x1
=
x0
x0
Lim ax b
Lim x 1
= b ….(1)
But
x0
Lim f(x) 2
(Given)
b = 2 (from eq. (1))
Also,
x
Lim f(x)
=
x
ax b
Lim
x1
=
x
a(b/x)
Lim
1(1/x)
= a ….(2)
But
x
Lim f(x)
= 1 (Given)
a = 1 (From eq. (2))
f (x) =
x2
x1
f (2) =
22
21
= 0
3. (C)
Here
222 2
333 3
n
123 n
Lim .............
nnn n
=
222 2
3
n
1 2 3 ..... n
Lim
n
=
3
n
n(n 1)(2n 1)
Lim
6n
222 2
n(n 1)(2n 1)
12 3....n
6
=
2
2
n
2n 3n 1
Lim
6n
=
2
n
131
Lim 2
6n
n
=
2
6
=
1
3
4. (B)
2
2n n
2
nn
11
Lim 1 Lim 1 e
nn
Vidyalankar : GATE – Engineering Mathematics
274
5. (C)
Put x
2
= t, so that, x = t
2
or 2x = + 2t
when x
2
then t 0
Now
2
x(/2)
1cos2x
Lim
(2x)
=
2
t0
1cos( 2t)
Lim
(2t)
=
2
t0
1cos2t
Lim
4t
=
2
2
t0
2sin t
Lim
4t
=
2
t0
sin t 1
Lim .
t2
=
1
2
6. (A)
Put
2
= y, as
2
, y 0
(/2)
cot
Lim
(/2)
=
(/2)
tan(( / 2) )
Lim
(/2)
=
y0
tan y
Lim
y
= 1
7. (A)
2
dy
73x
dx
hence rate of change of
slope is
ddy dx
6x
dt dx dt
= 6 3 4 = 72
8. (B)
3
dy dx
2x
dt dt
When x = 3,
dx
1foot / min
dt
dy dy
36 2ft/min
dt dt
9. (A)
A = 4r
2
. v =
4
3
r
3
dA dr
8r
dt dt
2
dv dr
4r
dt dt
10 =
2
dr 1 dA
4r r
dt 2 dt
When r = 5,
10 =
1dA
5
2dt
2
dA
4cm / h
dt
10. (A)
b
2
= r
2
+ (h/2)
2
v = r
2
h = (b
2
h
2
/4) h
22
dv
b3h/h0h2b/3
dh
22
22
2b/ 3
dv 3 h dv
2
dh dh
32b
bve
2
3
Hence there is relative maxima at
h = 2b/
3
r = b 2/3
Solutions Calculus
275
11. (D)
Let x be thickness of ice
v = 300 [ (10 + x)
2
100]
dv dx
300 x)
dt dt
dv
2
dt
dx 1
dt 300 10 x
11
300 10 2 3600
12. (B)
Let length be 2x and breadth be 2y
x
2
+ y
2
= 1 dy/dx = x/y
A = (2x) (2y) = 4xy
22
dA 4
xy 0
dx y
x = y = 1 / 2
2
23
dA 4x
dx y
2
23
12,12
dA 4 1 2
8
dx
12
2
2
12,12
dA
dx
= 16
Area = 4xy =
11
42
22
13. (B)
Given : y = 3x
4
16x
3
+ 24x
2
+ 37
For maximum value or minimum
dy
dx
= 12x
3
48x
2
+ 48x = 0
x (x
2
4x +4) = 0
x (x2)
2
= 0
x = 0, 2, 2
2
2
dy
dx
= 36x
2
96x + 48
At x = 0,
2
2
dy
dx
= +48
At x = 2,
2
2
dy
dx
= 36 4 96 2 + 48 = 0
so only 1 local minima.
either x = 0 or complex root
so only value = 0
14. (D)
/4
0
1tanx
dx
1tanx
I =
/4
0
cos x sinx
dx
cos x sinx
Let t = cos x + sin x
dt = (sin x + cos x) dx
dt
t
=
cos x sinx
dx
cos x sinx
I =
/4
/4
0
0
dt
lnt
t
/4
0
ln cos x sinx
ln cos / 4 sin / 4
1/ 2
1
ln 2 ln 2 ln2
2
Vidyalankar : GATE – Engineering Mathematics
276
15. (B)
If 2 points on line are A
1
(0, 1) and
A
2
(1, 0)
Equation of line is
112
112
yy y y
xx x x
y1 10
x0 0(1)
y1
1
x
y = x + 1
2
1
Iydx
=
2
1
(x 1)dx
=
2
2
1
x
x
2
=
1
(2 2) 1
2
=
3
4
2
=
5
2
= 2.5
16.
(B)
1
x
0
edx
=
111
xx
000
xedx 1 edx
11
xx
00
xe e = (e 0) (e 1) = 1
17. (B)
18. (D)
Let I =
22
xy
00
e e dxdy
I =
22
(x y )
00
edxdy
Put x = r cos
y = r sin
Then J = r
Now 0 x and 0 ≤ y
region of integration is the first
quadrant.
0 r , 0 /2
I =
2
/2
r
00
erdrd
=
/2
r
00
derdr
=
2
r
0
2r
edr
22
=
2
r
0
e2rdr
4
=
2
r
0
e
4
=
0(1)
4
I =
4
Solutions Calculus
277
Model Solution on Assignment 6
1. (A)
Here f (0) = 0
Right Hand Limit
x0
lim f(x)
=
x0
|
x
|
lim
x
=
x0
x
lim
x
= 1
|
x
|
xifx0
xifx 0
Left Hand Limit
x0
lim f(x)
=
x0
|
x
|
lim
x
=
x0
x
lim
x
= 1
R.H.L. L.H.L.
The function f (x) is discontinuous at
x = 0 and the discontinuity at x = 0 is
of first kind because R.H.L. and L.H.L.
both exist finitely but are unequal.
2. (A)
Here f (1) = 1
Also, Right Hand Limit
x1
Limf(x)
= 1
Left Hand Limit
x1
Lim f(x)
= a 2
Now, L.H.L. = R.H.L.
a 2 = 1 a = 1
3.
(B)
Here f (x) = 1
Also, Left Hand Limit
f(x) 3
Lim 1
= 1
Right Hand Limit
x3
Lim (ax b)
= 3a + b
Also, f (3) = 1
Since f (x) is continuous, therefore
3a + b = 1 ….(1)
Also, f (s) = 7
and
f(x) 5
Lim f(x)
=
x5
Lim (ax b)
= 5a + b
Since f (5) is continuous at x = 5,
therefore
5a + b = 7 ….(2)
Solving eq. (1) and eq. (2), we get
a = 3, b = 8.
4. (D)
The given function is of straight line
hence no maximum or minimum.
5. (B)
y = log x
dy
dx
=
1
x
Since log x is defined only for +ve
values of x.
So,
1
x
> 0
dy
dx
> 0 y = log x is
increasing.
Vidyalankar : GATE – Engineering Mathematics
278
6. (A)
y = log (sin x)
dy
dx
=
1d
(sinx)
sinx dx
=
cos x
sin x
= cot x
For the interval 0,
2
, cot x is +ve
dy
0
dx
on 0,
2
.
log (sin x) is increasing on 0,
2
Again cot x is negative on the interval
,
2
dy
dx
< 0
log (sin x) is decreasing.
7. (A)
The series
23
11 1
1 .....
2
22
is a G.P
S
n
=
n
n
11 1/2
21
1
1
32
1
2
n
n
nn
21
lim S lim 1
32
22
10
33
Hence the series is convergent
8.
(B)
The series 1 + 2 + 3 + ….. is an A.P
S
n
1
nn 1
2
n
nn
1
lim S lim n n 1
2
hence the series is divergent.
9. (C)
The series 1 2 + 3 4 + 6 + . ……
is an alternatory series
S
n
= (1 2) + (3 4 ) + (5 6) +. . . .+
[(n 1) n]
= (1) + (1) + ........... + (1) =
1
n
2
n
nn
1
lim s lim n
2
Series can also be written as
S
n
= 1 (2 3) (4 5) (6 7)……..
((n1) n))
= 1 + 1 + 1 + 1 ………..+ 1 =
n1
2
n
nn
1
lim S lim n 1
2
Since limit does not exist because the
sum of infinite forms of series is not
unique
10. (A)
The series is alternatory series
1 >
111
.......
234
n
nn
1
lim 1 lim 0
n
here the series is convergent
Solutions Calculus
279
11. (B)
Here U
n
=
n
n
12
n
n
nn
n
lim U lim 0
12
hence the series is divergent.
12. (C)
The infinite series
pp p
111
.....is
n12
i) convergent of p > 1
ii) divergent of p 1
13. (A)
The series
11 1
1 ......is aG.P.
3927
Where a = 1 ,
r =
1
3
Which lie between 1 and 1
Hence the given series is convergent.
14. (A)
U
n
=
33
n1 n
33
1
n1 n
Take V
n
=
3/2
1
n
33
n
An
n
3/2
1
U
n1 n
lim lim
1
V
n
=
n
3/2
1
lim
1
11
n
11
2
11
Hence it is convergent.
15. (B)
here U
n
= Sin
1
n
3
3
11 1 1 1
... .....
n3 n n
6n
Take V
n
=
1
n
n
3
nn
n
V
11 n
lim lim ...
Vn 1
6n
2
n
1
lim 1 ... 1
6n
series
n
V
n a divergent. Hence U
n
is also divergent.
16. (B)
U
n
=
2
n1
,
n
Take V
n
=
2
n1
n
n
2
n
nn
n
n1
V
n
lim lim 1
1
V
n
By comparison test U
n
& V
n
behave
alike.
Now V
n
=
1
n
is a divergent series
hence U
n
is also divergent.
Vidyalankar : GATE – Engineering Mathematics
280
17. (C)
Since f(x) = 1/x is not continuous in
[ 3, 3] [4, 2 ] or [1, 1] the point of
discontinuity is 0. only in [ 2, 3] the
function is continuous and differentiable.
18. (C)
f(x) = x
3
f
(x) = 3x
2
f (b) f(a) = (b a) f
(c)
b
3
a
3
= (b a) 3c
2
3c
2
=
22
baa abb
ba
c =
22
abab
3
Solutions Calculus
281
Model Solution on Assignment 7
1.
(B)
f (x) =
(1/x)
e
f ‘(x) =
(1/ x)
2
1
e
x
Since
(1/x)
e > 0 and
2
x > 0,
for all x 0
f ‘(x) < 0, for all x 0, x > 0
f (x) is decreasing in the interval
[0, ]
Hence f (x) =
1/ x
e , x > 0 is a
decreasing.
2. (C)
f (x) =
1
x
x
f (x) = 1
2
1
x
; f(x) =
3
2
x
for maximum and minimum, f(x) = 0
or 1
2
1
x
= 0
or x = 1
f( 1) = 2 < 0
f (x) is maximum at x = 1.
3. (A)
f (x) =
2
x2x1
f ‘(x) = 2x + 2; f ‘’(x) = 2
f ‘(x) = 0
[for maximum and minimum]
2x + 2 = 0
x = 1
f (x) = 2 < 0
f (x) is maximum at x = 1
4. (B)
f( ) 5cos 3cos 3
3
f( ) 5sin 3sin
3
f ( ) 5 cos 3cos
3
Equating f() to 0
5sin 3sin 0
3
5sin 3sin 0
3
13
5sin 3 sin cos 0
22
33
5sin sin 3 cos 0
22
13 3 3
sin cos 0
22
33
tan
13
1
33
tan
13
21.79
f(21.79) < 0
maximum at = 21.79
f(21.79) = 5 cos (21.79)
+ 3 cos (21.79 + 60)
f(21.79) = 10
Vidyalankar : GATE – Engineering Mathematics
282
5. (A)
f (x) =
32
x12x45x11
f (x) =
2
3x 24x 45
f (x) = 6x 24
f (x) = 0 or
2
3x 24x 45 0
x = 3, x = 5
f (5) = (6 5) 24
= 6 > 0
f (x) is minimum at x = 5.
6. (D)
7. (A)
Using lagrange’s mean value theorem
f
(c) =
fx f0
x0
or e
c
=
x
e1
x
….(i)
Now 0 <<< x e
0
< e
c
< e
x
.…(ii)
From equation (i) and (ii)
x
x
e1
1e
x
x <e
x
1 < xe
x
1 + x < e
x
< 1 + xe
x
8. (B)
x
1/x
is decreasing function then
f
(x) = x
1/x
e
2
1logx
x
f(x) is decreasing if f
(x) < 0
i.e. x
1/x
2
1logx
0
x
1 log
e
x < 0 log
e
x > 1
x > e
9.
(B)
let f(x) = e
x
. Then f
n
(x) = e
x
.
Hence f
n
(o) = 1 for all n 0
Maclaurin series
nn
n
n0 n0
f(0) x
x
n! n!
10. (C)
Let f(x) = sin x.
Then f(/4) =
2/2.f /4
2
2/2f /4
2
2
f / 4 .Thus
2
Taylor series for sin x about /4 is
2
x/4
2x/4
1
21! 2!
3
x/4
......
3!
11. (A)
Let f(x) = ln (1x). Then f(0) = 0.
f (0) 1, f (0) 1, f 0 1 2
f
(4)
= 1.2.3 and f
n
(0) = (n 1) !
Thus for n 1 f
n
(0)/n! = 1/n and
Maclaurin series is
234
xxx
x ...
234
12. (B)
Let f(x) =
1
x
Then
23 4
12 2.3
fx ,f x ,f x
xx x
4
5
234
f x and in general
x
nn
nn
n1
n!
fx 1 sof1 1n!
x
Solutions Calculus
283
Thus Taylor series is
n
xx
nnn
n0 n0
f1
x1 1 x1
n!
13. (A)
Let f(x) sec x. Then
fx Secxtanx,fx
2
sec x 1 2 tan x
2
fx secxtanx(56tanx)
f(0)
1f0 0,f 0 1
4
f(0) 0,f 5
The Maclaurin series is
1 +
24
15
x x ....
224
14. (D)
In general,
a
n
=
n
33 33
33
f0
.sof 0 33!a 33!2
n!
15. (B)
tan
1
=
2n 1
n
n0
x
1
2n 1
n
99
n99
f0
Since a ,f 0 99!a
n!
But a
99
= (1)
49
11
99 99
99
Thus f 0 99! / 99 98!
16. (A)
ln (1 + x) =
23
xx
x .....
23
n
n1
x
1...for
|
x
|
1
n
Hence
ln (1 + x
2
) =
46
2
xx
x .....
23
2n
n1
x
1 .....
n
2n
n1
x
Thus term is 1
n
17. (B)
f(x) =
3
1x
1x
n
n0
1
from x
1x
33
11
&so f x x
1x 1x
=
3n 3n 1
n0 n0
xx
From Maclaurin expansion,
36
36
f0
a1
36!
Hence f
36
(0) = 36!
18. (A)
Let f(x) = tan x . then
22
f (x) sec x,f x 2 tanx sec x
424
f (x) 2(sec x 2 tan x tan x)
So f(0) = 0,
f0 1,f 0 0,f 0 2
Thus Maclaurin series is
35
12
xx x...
315
Vidyalankar : GATE – Engineering Mathematics
284
Model Solution on Assignment 8
1.
(C)
f
(x) = 1 + k x
2
For x = 2 to be critical
f
(2) = 1 + k (2)
2
= 0
k = 4
Hence
f
(x) = 1
23
48
,f (x)
xx
f
(2) = 1
Hence relative maximum at x = 2
2. (C)
Let x and y be dimension
2x + 2y = 16 A = x(8 x) = 8x x
2
2
2
dA d A
82x, 2
dx
dx
Critical number is x = 4
2
2
4
dA
2
dx
here x = 4 and y = 4
3. (D)
x2
x2
lim
x2
Put x 2 = h
as x 2, h 0
x2 h0
x2 h
lim lim
x2 h
Left Hand Limit :
h0
h
lim
h
=
h0
h
lim
h
…
hhforh0
= 1
Right Hand Limit :
h0
h0
h
h
lim lim
hh
…
hhforh0
= 1
h0
h
lim
h
h0
h
lim
h
h0
h
lim
h
does not exist.
4. (D)
x1 x1
fx 2
lim 0 limf x 2
fx 2
f(x) = 2x
hence
x1
lim f x 2
5. (D)
6. (D)
Let A =
nlogn
n0
1
lim
e
e
e
n0
1
log A lim loge
nlogn
=
2
n0 n0
1/ n 1/ n
lim lim
logn 1/ n
n0
1
lim
n
7. (B)
Let f(x) =
nx
. then
2
11
fx ,f x ,f x
x
x
4
34
22.3
fx &
xx
in general
Solutions Calculus
285
f
n
(x) = (1)
n+1
n
n1!
x
So f(2) = n2 ,
f
n
(2) = (1)
n+1
n
n1!
2
Thus Taylor series is
n
n
n0
f2
x2
n!
2
x2 1
n2 x 2 .......
28
8. (C)
2
x
=
xn2
e
Now, e
x
=
n
n0
x
.
n!
Therefore
2
x
=
xn2
e
=
n
n
n0
n2
x
n!
9. (C)
e
x
=
n
n0
1
x
n!
Hence
100
2n
100
n0
f0
11
fx x , G
n! 100 50 !
Hence f
100
(0) =
100!
50!
n
f(0) =
n!
(n 2)!
for f(x) =
2
x
e
10. (D)
The Maclaurin series for f(x) is the
polynomial
23
412
2x x x
2! 3!
Thus f(x) = 2 + x + 2x
2
+ 2x
3
11. (A)
fx 4x 4,f x
n
4&f x 0forn 2
Thus f(1) = 3,
f(1) 8,f 1 4&f x
2
38x1 2x1
12. (A)
13. (B)
I =
0
log 1 cos d
2I =
0
log 1 cos d
0
log 1 cos d
=
2
0
log 1 cos d
0
2logsind
= 4 log2
I = log 2
14. (A)
I =
/2
0
sin2 x logtan x dx
22
2I =
/2
0
sin2xlog(tanx)
dx
sin2x logcot x
=
/2
0
sin2xlog tanx cot x dx
=
/2
0
sin2xlog(1)dx 0
Vidyalankar : GATE – Engineering Mathematics
286
15. (C)
Required area
BOD =
2
2
1
x2x
dx
44
=
2
23
1
1x x 9
2x
42 3 8
16. (B)
The given curve passes through the
point (2, 3) and (4,
3
2
)
Required area =
4
2
2
8
1dx
x
=
4
2
8
x4
x
17. (C)
Required area
=
1
23
0
2x x x dx
0
23
1
2x x x dx
=
10
34 34
22
01
xx xx
xx
34 34
=
3
2
18. (B)
I =
xsinx
dx
1cosx
xsinx
dx dx
1cosx 1cosx
=
22
xx
2sin cos
x
22
dx
2sin x / 2 2sin x / 2
=
2
xx x
cosec dx cot dx
22 2
= x cot
x
C
2
B
M
C
A
O
E
D
N
x
2
= 4y
x =1
x = 1
0, 0
1,0
(2,0)
Solutions Calculus
287
Model Solution on Assignment 9
1.
(B)
let A =
2x 2x
xx
x1x
lim lim
1x x
=
2x
x
1
lim 1
x
2
x
x
1
lim 1
x
= [e]
2
=
2
1
e
2. (B)
A =
tanx
x
1
lim
x
e
x0
log A lim tanxlog1/ x
=
x0
log1/ x
lim
cot x
2
2
x
x1/x
lim
cosec x
By L Hospitalrule
=
2
x0
sin x
lim
x
x0
2sinxcosx
lim 0
1
By L Hospitalrule
log
e
A = 0 A = e
0
= 1
3. (A)
A =
x
x0
lim cosec x
log
e
A =
x0
lim xlog cosec x
x0
log cosec x
lim
1
x
By L Hospitalrule
=
x0
2
1
cosec x cot x
cosec x
lim
1
x
2
x0
x
lim
tanx
=
2
x0
lim 2x / sec x
By L Hospitalrule
2
x0
lim 2x cos x
log
e
A = 0 A = e
0
= 1
4. (C)
n
lim
22 2
n
nn 1/2
12 1
... lim
n
nn n
=
n
11
lim 1
2n
1
10 1/2
2
5. (B)
L.
x2 h0
lim f x lim 2a 2 h b 4a b
R.
2
x2 h0
lim f x lim a 2 h b 4a b
L.
x2 x2
lim f x R lim f 2
4a b = 2 and 4a + b = 2
a = 1/2, b = 0
Vidyalankar : GATE – Engineering Mathematics
288
6. (D)
nn
nn 1
1 2 .... n
lim lim
n2n
n1
2
7. (B)
I =
/2
43
0
cos xsin x dx
=
/2
42
0
cos x sin x sinx dx
/2
42
0
cos x 1 cos x sinx dx
let cos x = t
I =
0
42
1
t1tdt
1
57
0
tt 2
57 35
8. (A)
Put x = sin
I =
2
2
1sin
cos d
1sin
=
2
1sin d
= +
1sin2
C
222
=
12
31
sin x x 1 x C
22
9. (A)
I =
1dx
11
2
cos sinx
22
1dx
sin x / 4
2
=
1
cosec x / 4 dx
2
1x
log tan C
2
2
10. (C)
I =
2
222
1secxdx
13sinx secx 3tanx
(Divide numerator and denominator by
cos
2
x)
=
2
22
sec xdx
1tanx 3tanx
2
2
2
1secxdx
4
1
tan x
2
Put tan x = t
=
2
2
1dt
4
1/ 2 t
1
tan 2t
C
2
=
1
tan 2 tanx
C
2
11. (A)
I =
4
4
sec
d
tan 1
22
4
1tan sec
d
1tan
Put tan = x
I =
22
42
1 x dx 1 1/ x dx
1x
1
x2
x
Put x
2
11
tx dxdt
x
x
Solutions Calculus
289
I =
2
2
dt
t2
1
1t x1/x
tan tan C
2
22
=
2
1
1tan1
tan C
2
2tan
12. (B)
I =
3
sinxdx
3sinx 4sin x
2
dx
34sinx
22
dx
3cos x sin x
=
2
2
sec xdx
3tanx
Put tan x = t
I =
2
dt 1 3 t
log C
3t
23 3 t
=
13tanx
log C
23 3 tanx
13.
(D)
Integrating by parts
I =
2
xsinxdx
2
x cos x 2x cosx dx
= x
2
cos x + 2x sin x 2 sinx dx
= x
2
cos x + 2x sin x +2 cos x +
/2
0
C
= 2
14. (C)
For x (0, /2), 0 < sin x < x
sinx nsinx
010 n
xx
…(1)
nsinx
n1
x
…(2)
Divide by x
sinx
10
x
…(3)
nsinx
0n
x
nsinx
n1
x
…(4)
(2) & (4)
x0
nsinx
lim n 1
x
15. (A)
I =
32
C
tsin 2 t dx t dy t dz
dx
2cos2t
dt
dx 2 cos 2 t dt
Since dy = dt and dz = 2tdt
I =
1
33
1
tsin2t 2cos2t
dt
t2t
=
11
3
11
2tsin4tdt tdt
= 1
Vidyalankar : GATE – Engineering Mathematics
290
16. (D)
I
n
=
xn1
0
elogxx
dx
dv
u
dx
I
n
=
n
x
0
x
elogx
n
xn
x
0
ex
elogx dx
xn
= 0 +
1
n
I
n+1
n1!
n
xn1
0
ex dx
= (n 1)!
I
n
=
1
n
I
n+1
n1!
n
….(1)
Replace
n n + 1 : (n+1) I
n + 1
n1!
n
.…(2)
(1) & (2) : I
n + 2
– (2n+1)I
n+1
+ n
2
I
n
= 0
17.
(B)
f(0) = 0
x0
lim f h 0
f(x) is continuous at x = 0
f(0
+
) =
2
x0
11x 0
lim
x
=
2
x0
2
11 x
lim
x1 1 x
=
2
x0
2
x
lim
x1 1 x
=
x0
2
x
lim ,
x1 1 x
2
xifx 0
because x
xifx 0
=
1
2
But f(0
) =
1
2
F is not differentiable of x = 0.
18.
(B)
z
5y 8x 2
x
,
z
5x 2y 1
y
z
0
x
,
z
0
y
x =
1
41
, y =
18
41
Here r = 8, s = 5, t = 2
rt s
2
< 0
118
,
41 41
is a saddle point.
Solutions Calculus
291
Answer Key on Test Paper 1
1.
(B)
2.
(D)
3.
(A)
4.
(D)
5.
(A)
6.
(A)
7.
(B)
8.
(D)
9.
(B)
10.
(C)
11.
(B)
12.
(D)
13.
(B)
14.
(B)
15.
(B)
Answer Key on Test Paper 2
1.
(D)
2.
(B)
3.
(A)
4.
(D)
5.
(A)
6.
(D)
7.
(C)
8.
(D)
9.
(D)
10.
(D)
11.
(B)
12.
(A)
13.
(D)
14.
(D)
15.
(A)
Answer Key on Test Paper 3
1.
(C)
2.
(D)
3.
(A)
4.
(B)
5.
(A)
6.
(B)
7.
(C)
8.
(A)
9.
(A)
10.
(B)
11.
(A)
12.
(A)
13.
(A)
14.
(B)
15.
(C)
Answer Key on Test Paper 4
1.
(C)
2.
(C)
3.
(D)
4.
(D)
5.
(C)
6.
(B)
7.
(B)
8.
(C)
9.
(D)
10.
(A)
11.
(D)
12.
(B)
13.
(A)
14.
(D)
15.
(C)
Vidyalankar : GATE – Engineering Mathematics
292
Answer Key on Test Paper 5
1.
(D)
2.
(B)
3.
(B)
4.
(A)
5.
(A)
6.
(A)
7.
(B)
8.
(B)
9.
(B)
10.
(C)
11.
(D)
12.
(B)
13.
(C)
14.
(D)
15.
(A)
Answer Key on Test Paper 6
1.
(C)
2.
(A)
3.
(A)
4.
(B)
5.
(D)
6.
(B)
7.
(C)
8.
(C)
9.
(B)
10.
(C)
11.
(A)
12.
(B)
13.
(C)
14.
(A)
15.
(C)
Solutions Calculus
293
Model Solution on Test Paper 1
1. (B)
3
x0
3sinx 4sin x sinx
Lim
sinx
=
2
x0
Lim(2 4sin x)
= 2
2. (D)
x0
1
Lim sin
x
does not exist because
1
sin
x
oscillates.
3. (A)
Now,
x0
sin3x
Lim
5x
=
x0
sin3x 3x
Lim .
3x 5x
=
x0
sin3x 3
Lim .
3x 5
= 1 ×
3
5
=
3
5
4.
(D)
Left Hand Limit,
x2
Lim [x]
=
h0
Lim [2 h]
=
h0
Lim 1
= 1
Right Hand Limit,
x2
Lim [x]
=
h0
Lim [2 h]
=
h0
Lim 2
= 2
L.H.L.
x
Lim [x]
R.H.L.
x2
Lim [x]
5. (A)
Put 1 + x = y, so that x 0 y 1
n
(1 x) 1
x
=
n
y1
y1
n
x0
(1 x) 1
Lim
x
=
n
y1
y1
Lim
y1
=
n1
n(1)
= n
6. (A)
7. (B)
8. (D)
2
2
22
11
nx dx
x
xnx
=
1
2
nx
=
11
1
nn2
n2
9. (B)
There is discontinuity at x = 2. Thus
525
333
112
dx dx dx
x2 x2 x2
2
2/3
1
3
x2
2
+
5
2/3
2
3
x2
2
3
3
91
2
Vidyalankar : GATE – Engineering Mathematics
294
10. (C)
A =
11
22
00
x12x
dx dx
2
1x 1x
1
1/ 2
2
0
1x
= 1
11. (B)
By Stoke’s formula
V =
2
2x 4x
00
edxedx
=
v
v
4x 4x
vv
0
0
1
lim e dx lim e
4
=
4v
v
lim e 1
44
12. (D)
There is discontinuity at x = 1. Then
9
2/3
0
dx
x1
=
19
2/3 2/3
01
dx dx
x1 x1
=
19
1/ 3 1/ 3
01
3x 1 3x 1
= 3 + 6 = 9
13.
(B)
A =
1
11
dx
xx1
1
nx nx 1
= n2
14. (B)
There is a discontinuity at x = 2
So
323
112
dx dx dx
x2 x2 x2
23
12
nx 2 nx 2
= neither limit exists.
Therefore integral diverges.
15. (B)
By successive application of L'Hospital
rule,
p
x
nx
lim 0
x
.
Hence
p
nx
1
x
for sufficiently large
x. Thus for some x
0
.
If x x
0
p
nx x
,
p
11
x
nx
.
So
p
ee
dx dx
nnx
nnx
.
Hence the integral is divergent.
Solutions Calculus
295
Model Solution on Test Paper 2
1.
(D)
Left Hand Limit,
x2
Lim f(x)
=
x2
Lim 2x
= 4
Right Hand Limit,
x2
Lim f(x)
=
2
x2
Lim x
= 4
Thus,
x2
Lim f(x)
= 4
But, f (2) = 2
x2
Lim f(x)
Thus the given function is
discontinuous at x = 2.
But if we change f (x) = 4 at
x = 2, then the given function f
becomes continuous at x = 1.
Hence 2 is the removable
discontinuity.
2. (B)
Left Hand Limit,
h0
Lim f(2 h)
=
h0
Lim
(2 h 1)
= 1 ….(1)
Right Hand Limit,
h0
Limf(2 h)
=
h0
Lim [2(2 h)] 3
= 1 ….(2)
From eq. (1) and eq. (2),
L.H.L. = R.H.L.
x2
Lim f(x)
exists.
f (2) = value of f (x)
= 2 . 2 3 = 1
x2
Lim f(x) f(2)
f (x) is continuous at x = 2.
3. (A)
Left Hand Limit,
x1
Lim f(x)
=
h0
Lim(1h) [1h]
=
h0
Lim (1 h) 0
= 1
Right Hand Limit,
x1
Lim f(x)
=
h0
Lim(1h) [1h]
=
h0
Lim (1 h) 1
=
h0
Lim h
= 0
L.H.L.
x1
Lim f(x)
R.H.L.
x1
Lim f(x)
f (x) = x [x] is discontinuous at x = 1.
4. (D)
At x = a (say), where ‘a’ is any rational
number.
Given, f (x) = 1, when x is rational.
Left Hand Limit,
xa
Lim f(x)
=
ha
Lim f(a h)
=
h0
Lim ( 1)
(because a h is an irrational number)
Right Hand Limit,
xa
Limf(x)
=
h0
Lim
f (a + h)
=
h0
Lim ( 1)
= 1
But, f (a) = 1
Vidyalankar : GATE – Engineering Mathematics
296
L.H.L.
xa
Lim f(x)
= R.H.L.
xa
Lim
f (x) f (a)
f (x) is discontinuous at x = a,
where ‘a’ is a rational number.
Similarly we can prove that at x = b (an
irrational number) f (x) is discontinuous.
f (x) is discontinuous at every real
point.
5. (A)
Left Hand Limit,
h0
Lim f(4 h)
=
h0
Lim 4(4 h) 3
= 19 ….(1)
Right Hand Limit,
h0
Limf(4 h)
=
h0
Lim 4(4 h) 3
= 19 ….(2)
From eq. (1) and eq. (2),
x4
Lim
f (x) = 19
Also, f (h) = 3 . 4 + 7
= 19
x4
Lim f(x)
= f (4) = 19
f (x) is continuous at x = 4.
6. (D)
I =
/2
/2
0
0
cot xdx ln sinx
= +
7. (C)
A =
22
00
dx dx
x6x10
x3 1
1
0
tan x 3
=
11
tan 3 tan 0 3
1
tan 3
2
8. (D)
A =
22
a1
a1
dx 1 x a
n
2a x a
xa
=
1
n2a 1
2a
9. (D)
11
kk
v0
0v
11
dx lim dx
xx
1
k1
v0
0
1
lim x
k1
=
k1
v0
11
lim 1
1k
v
.
If k > 1 this limit is + whereas if k < 1
the limit is
1
1k
10.
(D)
a
22
0
dx
ax
=
a
0
1ax
n
2a a x
=
1
na x
2a
a
0
1
na x
2a
= +
Thus the integral diverges.
Solutions Calculus
297
11. (B)
The value of mod function cannot be
less than 0
f(x) = | x
2
5x + 2 | = 0
12. (A)
let f(x) =
sinx cos x
sin x 4
2
since maximum value of sin is 1
maximum is 1
13. (D)
let y = x
x
log y = x log x
dy/dx = 1/x
x
(1 + log x)
d
2
y/dx
2
= [ y/x + (1 + log x)
{ y (1 + log x)}]
for maxima and minima dy/dx = 0
1/x
x
(1 + log x) = 0 x = 1/e
2
2
21/e
1
x
e
dy 1 1
0
1
dx
1
e
e
1/ e
ee0
Hence maximum at x = 1/e
Maximum value =
1/ e
1/ e
1
e
1
e
14. (D)
f(x) = x
5
5x
4
+ 5x
3
1
f
(x) = 5x
4
20x
3
+ 15x
2
f
(x) = 20x
3
60x
2
+ 30x
f
(x) = 60x
2
120x + 30
for maxima or minima f
(x) = 0
x
4
4x
3
+ 3x
2
= 0 x = 0, 1, 3
f
(0) = 0, f
(0) = 30 0
f
(1) = 10 < 0
f
3 = 90 > 0
Hence f (x) is neither maximum and
minimum at x = 0 f (x) is maximum at
x = 1 and minimum at x = 3
Hence 1 maxima and 1 minima
15.
(A)
f(x) = x
3
/3 3x
2
/2 + 2x
f
(x) = x
2
3x+ 2
f
(x) = 2x 3
For maxima minima
f
(x) = 0 x
3
3x + 2 = 0
(x 1) (x 2) = 0
x = 1, 2
f
(1) = 2 3 = 1 < 0
f
(2) = 2 23 = 1 > 0
f(1) = 1/3 3/2 + 2 = 5/6
f(2) = 8/3 6 + 4 = 4/6 = 2 / 3
Vidyalankar : GATE – Engineering Mathematics
298
Model Solution on Test Paper 3
1. (C)
16
5
x1
x1
Lim
x1
=
16 5
x1
x1x1
Lim
x1 x1
=
16 5
x1 x1
x1 x1
Lim Lim
x1 x1
= 16 5
=
16
5
nn
n1
x0
xa
Lim na
xa
2. (D)
2
x0
x4
Lim
x2 3x2
=
2
x2
x4 x2 3x2
Lim
x2 3x2x2 3x2
=
x2
(x 2)(x 2) ( x 2 3x 2)
Lim
(x 2) (3x 2)
=
x2
(x 2)(x 2) ( x 2 3x 2)
Lim
2(x 2)
=
x2
(x 2)( x 2 3x 2)
Lim
2
=
4( 4 4)
2
= 8
3. (A)
Now,
xx
ab
x
=
xx
( a 1) ( b 1)
x
=
xx
a1b1
xx
xx
x0
ab
Lim
x
=
xx
x0 x0
a1 b1
Lim Lim
xx
= log a log b =
a
log
b
4. (B)
Left Hand Limit,
x0
Limf(x)
=
h0
|
0h
|
Lim
0h
=
h0
h
Lim
h
= 1
Right Hand Limit,
x0
Limf(x)
=
h0
|
0h
|
Lim
0h
=
h0
h
Lim
h
= 1
L.H.L.
x0
Lim f(x)
R.H.L.
x0
Lim f(x)
Hence f (x) is discontinuous function of
first kind at x = 0.
5.
(A)
Left Hand Limit,
f ‘(0) =
h0
f(0 h) f(0)
Lim
h
=
h0
1
Lim h sin
h
= 0
Right Hand Limit,
f ‘(0) =
h0
f(0 h) f(0)
Lim
h
=
2
h0
1
hsin 0
h
Lim
h
Solutions Calculus
299
=
h0
1
Lim h sin
h
= 0
L.H.L. f ‘(0) = R.H.L. f ‘(0) = 0
6. (B)
y =
x
x
e
log y = log x x
11dy1
dy / dx 1 y 1 0
yxdxx
x = 1
2
2
22 2
1d y 1 dy 1
ydx
dx y x
22
2
x1 x1
1d y d y
01 e0
e
dx dx
Hence maxima at x = 1
Maximum value
max
y =
1
e
7. (C)
Assume that there is such a function.
Then f
xy
and f
yx
will be continuous
everywhere. Hence f
xy
= f
yx
Thus e
x
sin y = e
x
sin y = 0 for all y which is
false
8. (A)
fx =
22
2x
,
xy
22 22
22
22 22
xy22x2x 2yx
fxx
xy xy
fy =
22
2y
,
xy
22 22
22
22 22
xy22y2y 2xy
fyy
xy xy
Hence fxx + fyy = 0
9. (A)
By chain rule
xy
dz dx dy
ff
dx dx dx
3
4x 3y 3x 2y cosx
10. (B)
dz z dx z dy
dt x dt y dt
xdx 2ydy dx
since 5
8 dt 9 dt dt
and
5x + y
dy
0
dt
so when x 2and y 1,
dy / dt 10 and
dz 2 2
510
dt 8 9
125
cm / s
36
11. (A)
Let k (x, y) =
f
x
, then
k
0
x
then k (x, y) = g(y).
Hence
f
x
= g(y) then
f(x,y) = xg(y) + h(y) for a suitable
function h
Vidyalankar : GATE – Engineering Mathematics
300
12. (A)
If z = x f(y/x) then
zz
xy
xy
2
x xf y/x y/x f y/x
xyf y / x 1/ x
xf y / x z
13.
(A)
let c (x, y) = f (x, y) xy,
then
2
c
11 0then
xy
c (x, y) = A (x) + B (y)
f (x, y) = A (x) + B(y) + xy
14.
(B)
A normal vector to tangent plane will
be (2x, 2y, 2z) =
1, 1, 2
Hence equation of tangent plane is
11 1
xy 2z 0
22
2
Or x + y +
2z 2
15. (C)
V =
1dh
rh
3dt
Vdr vdh
rdt hdt
2
21
rh 2 r 3
33
=
4
5000 2500
3
3
12500
mm / s
3
Solutions Calculus
301
Model Solution on Test Paper 4
1.
(C)
2. (C)
3. (D)
s =
2
3
t
3t 15
2
2
ds
speed 9t t
dt
22
22
ds ds 1
18t 1 0 t
18
dt dt
again;
3
3
ds
18 0,
dt
hence speed
has maxima.
Maximum speed =
1
t
18
ds
dt
2
11
9
18 18
=
11
36 18
=
21
36
1
36
speed units.
4. (D)
5. (C)
6. (B)
A =
2
2
22x4x7
13x x3
dx dy
= 5
2
2
1
dx, x x 2
=
2
32
1
xx
52x
32
=
45
2
7. (B)
A = 2
/2 a sec cos
0asec
rdr d
/2 a sec cos
2
asec
0
rd
= a
2
/2
2
2
0
sec cos sec d
= a
2
/2
2
0
2cos d
=
2
5a
4
8. (C)
Mass = k
3/2
x/2
4
0
x
4
dx ydy
= k
3/2
x/2
2
4
0
x
4
y
dx
2
4
34
0
kx x 2
k
212 64 3
9. (D)
Equation of circle r = 2acos
Mass = k
/2 2acos
2
/2 0
drdr
=
/2
2acos
3
0/2
k
dr
3
3
/2
3
0
8a k
2cosd
3
=
3
32
ka
9
Vidyalankar : GATE – Engineering Mathematics
302
10. (A)
v =
/2 acos
22
/2 0
darrdr
222
Put t a r
=
asin
3
asin
2
a
a
t
tdt
3
=
33
1
a1sin
3
11. (D)
v =
2
2
21x
1
22
1
21x
dx 1 x y / 4 dy
=
2
2
21x
3
1
2
1
21x
y
1x y
12
=
3/2
1
2
1
8
1x dx
3
12. (B)
v =
00
22
aa
ax
ydx x dx
ax
put z = a x so
v =
2
2a
0
az 2az
dz
z
3
2a
22
0
2a
5a 4az z dz
z
= 2a
3
[log2 2/3]
13. (A)
V = 2
2
0
axdy
a
2
0
dy
2ax dx
dx
2
3/2
dy a
dx
2xa x
2
2
ax
from curve y
ax
V = a
2
0
ax
dx
x
Put x = a sin
2
V = 2a
3
23
/2
2
0
a
cos d
2
14. (D)
S = 2
ds
yds 2 y d
d
0
ds
4yd
d
1/ 2
22
ds dx dy
2a cos 2
dd d
S = 4
2
0
2asin 2acos d
22
=
2
32 a
3
.
15. (C)
V =
40
2xx2 8
25xdxdy
= 2
4
2
5x8xx2dx
= 2
4
23
2
40 18x 3x x dx
= 2 216 = 432
Solutions Calculus
303
Model Solution on Test Paper 5
1.
(D)
Let f(x) = x +
1
x
f
(x) =
2
1
1
x
3
fx 2/x
For maxima and minima f
(x) = 0
2
1
1
x
= 0 x =
1
3
2
fx 2 0
1
f(x) is minimum at x = 1
minimum value = f (1) = 1 +
1
1
= 2
2. (B)
Let be semi vertical angle of
inscribed triangle
Let S be the area
Then s =
1
2x
2
sin 2 (r + r cos 2)
dS
d
= 2r
2
(cos 4 + cos 23) = 0
= /6
S is maximum at = / 6
vertical angle = 2 = /3
3. (B)
f
(x) =
2
1logx
x
1 log x = 0 x = e
f
(x) =
3
2logx 3
x
f
(e) =
33
23 1
0
ee
f(x) is maximum at x = e
maximum value f (e) =
loge 1
ee
4. (A)
Total SP = x
2
xx
33x
1000 1000
Total CP =
x
200
2
Profit, f(x) SP CP
2
5x
x 200
2 1000
f '(x) =
52x
2 1000
2
f''(x) 0
1000
52x
0 x 1250
2 1000
1
f '' (1250) negative
500
So, profit is maximum when x = 1250
5.
(A)
By implicit differentiation w.r.t. x
y
zz
yxz0
xx
zzyz
xy (yz)
xxyx
Vidyalankar : GATE – Engineering Mathematics
304
6. (A)
By Cauchy’s mean value theorem,
fb fa f c
gb ga g c
Choosing f(x) = e
x
, g(x) = c
x
ba c
ba c
ee e ab
c
2
ee e
7. (B)
fx
= F
(x)
let (x) = f(x) F(x)
(x)
= f
(x) F
(x) = 0
(x) = constant
8. (B)
9. (B)
4
rcos drd
a1 cos
4
00
cos r drd
=
5
5
0
a
1cos cosd
5
5
10 2
0
32a
cos 2cos 1 d
522
=
/2
51210
0
64
a2coscosd
5
…
2
=
5
21
a
16
(by reduction formula)
10. (C)
I =
2
y
2
4
2
2
y/4
xdxdy dy xdx
=
2
y/2 2
2
4
2
y/4
x
dy
2
=
4
35
2
2
1y y 72
y4y
212 80 5
11.
(D)
I =
bb
00
y2b ydxdy
=
b
3
b
2
0
0
y
by dx
3
=
34
b
3
0
b4b
bdx
33
12. (B)
I = 2
/4 2sin
3
0
rdrd
= 8
/4
4
0
sin d
2
1cos2
8d
4
=
/2
2
0
1 cos d .... 2
=
13
22
22
Solutions Calculus
305
13. (C)
I =
a1 cos
43
00
2rsin drd
=
5
5
3
0
2a
1cos sin d
5
=
5
2
67
0
2a
2z z dz
5
….(z = 1 + cos)
=
6
5
2
a
35
14. (D)
I =
2
a
2
a
2
0
2ax x
xlog x a
y
dx
2
xa
=
a
22
2
0
xlog x a
1
dx a 2ax x
2
xa
Integrating with parts, we get
=
a
0
1
xlog x a dx
2
=
2
a
2loga 1
8
15. (A)
2
a2ax
0x/a
dx xy dy
a
4
2
2
0
1x
xdx (2a x)
2
a
a
5
23
2
0
1x
4a x 4ax x dx
2
a
4
444
141a
2a a a
2346
4
3a
8
2
x
y
a
(a, a)
y=2ax
Vidyalankar : GATE – Engineering Mathematics
306
Model Solution on Test Paper 6
1. (C)
2. (A)
3. (A)
xa
1
lim x a cos
xa
h0
1
lim hcos [x a h]
h
= 0 some finite a + y = 0
4. (B)
5. (D)
6. (B)
I =
15x
2
02x
ydydx
=
5x
11
333
00
2x
11
y dx 125x 8x dx
33
=
1
4
0
39 39
x
44
7. (C)
I =
42y
2
2
00
1
dxdy
2y y
=
42y
2
2
0
0
x
dy
2y y
2
0
2y
2
dy
y2y
=
2
1/ 2
0
22y 4
8. (C)
V =
32
R01
3x 4y dA 3x 4y dxdy
=
2
3
2
1
0
3
x4yxdy
2
=
3
3
2
0
0
9963
4y dy y 2y
22 2
9. (B)
V =
24
22
R00
y dA y dydx
4
22
3
0
00
164
ydx dx
33
=
64 128
2
33
10. (C)
132x
R1/21/x
A 1dA dydx
1
1/ 2
1
32x dx
x
5
1
Solutions Calculus
307
=
1
2
1/ 2
3x x nx
31 1
310 n
24 2
=
5
2n2
4
=
3
n2
4
11. (A)
R
V2x2y1dA
2x
00
2x 2y 1 dydx
2
x
2
0
0
2xy y y dx
=
2
32
0
1
xx
2
=
82 = 10
12.
(B)
A =
21 +cos
R0
dA r dr d
1cos
2
2
0
0
1
rd
2
=
2
0
1
12cos cos d
2
11cos2
12cos d
22
=
13 3
2
22 2
13. (C)
I =
acos2
/4
2
/4
0
dr
/4
/4
da1cos2 a
= a
/4
/4
2cos 1 d
= a
2sin
a2 /2 2a1
4
14. (A)
I =
2
1x2xy
0x 0
dx dy dz
=
2
1x
2xy
0
0x
dx dy z
=
2
1x
0x
dx 2 x y dy
=
4
1
2
0
xx
2x x 2xx dx
22
=
11
30
15. (C)
I =
21
22
1y1
dy x y dx
=
y
3
2
2
1
y1
x
dy y x
3
=
3
2
22
1
y1
1
yyy1dy
33
=
2
4
34
1
y1
y2y y
33 12 4
=
2
3
308
Solutions Probability and Statistics
Answer Key on Assignment 1
1.
(C)
2.
0.65 to 0.68
3.
(C)
4.
0.4 to 0.4
5.
(D)
6.
0.25
7.
(A)
8.
0.07 to 0.08
9.
(A)
10.
0.55 to 0.55
11.
(D)
12.
(B)
13.
0.43 to 0.45
14.
0.26 to 0.27
15.
(B)
16.
1.06 to 1.07
17.
(B)
18.
6
19.
0.32 to 0.34
20.
(A)
Answer Key on Assignment 2
1.
0.64 to 0.66
2.
0.1276 to 0.1372
3.
0.79 to 0.81
4.
(B)
5.
0.96 to 0.98
6.
0.23 to 0.25
7.
99.6 to 99.8
8.
(A)
9.
2.5 to 2.5
10.
(C)
11.
0.25 to 0.27
12.
(B)
13.
(B)
14.
0.25
to 0.28
15.
(D)
16.
(B)
17.
(D)
18.
0.39 to 0.43
19.
54.0 to 54.0
20.
0.33 to 0.34
Answer Key on Assignment 3
1.
49 to 51
2.
49.9 to 50.1
3.
(B)
4.
(B)
5.
(C)
6.
(C)
7.
0.9 to 1.1
8.
54.49 to 54.51
9.
0.5 to 0.5
10.
(B)
11.
(A)
12.
(D)
13.
0.35 to 0.45
14.
(B)
15.
(D)
16.
4
17.
6
18.
(A)
19.
(A)
20.
2.0 to 2.0
Solutions Probability and Statistics
309
Answer Key on Assignment 4
1.
(A)
2.
(D)
3.
(B)
4.
(C)
5.
(A)
6.
(C)
7.
(C)
8.
(A)
9.
(B)
10.
(D)
11.
(C)
12.
(B)
13.
(A)
14.
(C)
15.
(A)
16.
(B)
17.
(C)
18.
(C)
Answer Key on Assignment 5
1.
(A)
2.
(C)
3.
(D)
4.
(A)
5.
(D)
6.
(C)
7.
(A)
8.
(C)
9.
(D)
10.
(A)
11.
(D)
12.
(A)
13.
(A)
14.
(A)
15.
(C)
16.
(D)
17.
(C)
18.
(C)
Answer Key on Assignment 6
1.
(B)
2.
(D)
3.
(C)
4.
(C)
5.
(A)
6.
(C)
7.
(C)
8.
(B)
9.
(A)
10.
(D)
11.
(C)
12.
(A)
13.
(B)
14.
(A)
15.
(A)
16.
(B)
17.
(D)
18.
(D)
Vidyalankar : GATE – Engineering Mathematics
310
Model Solution on Assignment 1
1. (C)
We know that mode is the value of the
data which occurred most of
17 is mode.
2. 0.65 to 0.68
If there are N families, then
N
2
have
single child per family and
N
2
have two
children per family.
Total children =
N
2
N
2
2
= N
Total number of children =
N
N
2
=
3
N
2
If a child picked at random should
have a sibling, then that child should
come from the family which has 2
child. There are N such children.
Required probability =
N
3
N
2
=
2
3
= 0.67
Alternative
Suppose there are total 10 families.
Total children = 15
Required probability =
10
15
=
2
3
= 0.67
3. (C)
P(Head) =
1
2
P(Tail) =
1
2
P(Fourth Head in tenth toss)
=
36
9
3
11 1
C 0.082
22 2
4. 0.4 to 0.4
Given, probability density function of
evaporation E is
f(E) =
1
,0E5mm/day
5
0 , otherwise
As, the probability has to be find
between 2 and 4 mm/day, therefore
P(2 < E < 4) =
4
2
f(E)dE
=
4
2
1
dE
5
=
4
2
1
E
5
=
42
5
=
2
5
P(2 < E < 4) = 0.4
Single child
per family
Two children
per family
N
2
N
2
N
One child per
family
Two child
per family
10
5
5
Number of children in
these 5 families is 5
Number of children in
these 5 families is 10
Solutions Probability and Statistics
311
5. (D)
We have Probability distribution
function of Normal Distribution
f(x) =
2
2
(x )
2
1
e
x
… (1)
Variance =
2
is lowest
also lowest
If decreases f(x) increases
(
from (1))
Curve will have highest peak
6. 0.25
As f(x) is a pdf,
1
0
f(x) dx = 1
1
0
(a + bx) dx = 1
1
2
0
x
ax b
2
= 1
a +
b
2
= 1
2a + b = 2 ….(1)
Also, E(X) =
2
3
1
0
x f(x) dx =
2
3
1
0
x (a + bx) dx =
2
3
1
23
0
xx
a b
23
=
2
3
a
2
+
b
3
=
2
3
3a + 2b = 4 ….(2)
Solving (1) & (2) we get, a = 0, b = 2
P(X < 0.5) =
0.5
0
f(x) dx =
0.5
0
2x dx
=
0.5
2
0
x
2
2
= (0.5)
2
= 0.25
7. (A)
8. 0.07 to 0.08
P(Head) = 0.3
P(Tail) = 0.7
P (getting Head first time in fifth toss)
= 0.7
0.7 0.7 0.7 0.3
= 0.072
9. (A)
P(X) = 0.40
P(X
Y
c
) = 0.7
X and Y are independent events
P(X
Y) = P(X) . P(Y) ….(1)
X and Y
c
are also independent
P(X
Y
c
) = P(X) . P(Y
c
) ….(2)
Now,
P(X
Y
c
) = P(X) + P(Y
c
) P(X Y
c
)
P(X
Y
c
) = P(X) + P(Y
c
) P(X) P(Y
c
)
0.7 = 0.4 + P(Y
c
) 0.4 P(Y
c
)
0.7 0.4 = P(Y
c
) [1 0.4]
0.3 = 0.6 P(Y
c
)
P(Y
c
) =
0.3
0.5
0.6
P(Y
c
) = 1 P(Y)
Vidyalankar : GATE – Engineering Mathematics
312
0.5 = 1 P(Y)
P(Y) = 1 0.5 = 0.5
Then,
P(X
Y) = P(X) + P(Y) P(X Y)
= P(X) + P(Y)
P(X) P(Y)
= 0.4 + 0.5
0.4 × 0.5
P(X Y) = 0.7
10. 0.55 to 0.55
There are equal number of bulbs of
two different types.
Probability that the bulb selected is of
Type 1
1
1
P(T ) 0.5
2
Probability that the bulb selected is of
type 2
2
1
PT 0.5
2
Let E be the event that a bulb selected
at random lasts more than 100 hours.
P(E/T
1
) = 0.7
P(E/T
2
) = 0.4
By total probability
P(E) = P(T
1
) P(E/T
1
) + P(T
2
) P(E/T
2
)
= 0.5
0.7 + 0.5 0.4
P(E) = 0.55
11. (D)
Given
x 1
2
3
p(x) 0.3
0.6
0.1
mean (
) = exp(x)
= 1
0.3 + 2 0.6 + 3 0.1
= 0.3 + 1.2 + 0.3 = 1.8
E(x
2
) =
2
xP(x)
= 1
0.3 + 4 0.6 + 9 0.1
= 0.3 + 2.4 + 0.9 = 3.6
Variance v(x) = E(x
2
)
2
= 3.6
(1.8)
2
S.D(
) = v(x)
=
2
3.6 (1.8)
=
0.36
= 0.6
12. (B)
Given = 5.2
Let x be random variable which follows
Poission's distribution
P(x < 2) = P(x = 0) + P(x = 1)
=
0
1
ee
0! 1!
= e
5.2
(6.2)
= 0.0055 × 6.2 = 0.034
13. 0.43 to 0.45
Parcels pass sequentially through two
post offices.
Probability that parcel reaches first
post office
P(01) = 1
Probability that parcel is lost by first
post office
P(L/01) =
1
5
Parcel will reach second post office if it
is not lost at first post office.
Probability that parcel reaches second
post office
Solutions Probability and Statistics
313
P(02) =
4
5
Probability that parcel is lost by
second post office
P(L/02) =
1
5
By Total Probability,
Probability that parcel is lost
P(L) = P(01) P(L/01) + P(02) P(L/02)
=
141
1
555
=
9
25
By Baye's Theorem, knowing that the
parcel is lost, probability that it was
lost by second post office
P(02/L) =
P(02)P(L 02) 4 25 4
P(L) 9 25 9
P(02/L) = 0.444
14. 0.26 to 0.27
Given, mean
= 5
The probability that there will be less
than 4 penalties in a day is
P(y < 4) = P(y = 0) + P(y = 1)
+ P(y = 2) + P(y = 3)
=
50 51 52 53
e5 e5 e5 e5
0! 1! 2! 3!
=
0123
5
5555
e
0! 1! 2! 3!
= 0.006737 [1 + 5 + 12.5 + 20.833]
= 0.006737
39.33 = 0.2649
15. (B)
Given that the student is passing
exam, i.e., if only 20 students
(out of 100) are considered, of the 5
students get more than 90%
5 out of 20 is the probability
Or 1/4 is the answer.
Alternative Method:
Let the student pass the examination
be A and student pass the examination
and got about 90% marks be B
Now P(A) = 20% and P(A
B) = 5%
P
BP(AB)5%1
AP(A)20%4
16. 1.06 to 1.07
Mean =
2
0
x f(x) dx
=
1
4
2
35
0
xx
4
35
=
1
4
32 32
35
= 1.0666
17. (B)
Required probability
=
1213141
6666666
15 5
36 12
Vidyalankar : GATE – Engineering Mathematics
314
18. 6
Var(x) = E(x
2
) (E(x))
2
E(x) =
|x|
1
x.
|
x
|
edx
2
=
0
2x 2 x
0
11
xedx x.e dx
22
= 0
E(x)
2
=
2|x|
1
x.
|
x
|
edx
2
=
2x
0
1
2x.x.edx
2
=
3x
0
x.e dx
=
x
32x
0
0
e
x. (3x)(e )
1
xx
00
6x( e ) 6 e
= 6
Var(x) = 6 0 = 6
19. 0.32 to 0.34
20. (A)
5 red and 7 green balls
Total 12 balls.
A ball is drawn and placed back along
with another ball of same colour, this
can be done in two ways:
I) Red ball in first draw
P(red in first draw) =
5
12
When the red ball is placed back with
another red ball, there will be total 13
balls with 6 balls of red colour.
P(red/red in first draw) =
6
13
II) Green Ball in first draw
P(green in first draw) =
7
12
When the green ball is placed back
with another green ball, there will be
total 13 balls with 5 balls of red colour.
P(red/green in first draw) =
5
13
P(red) =
5675
12 13 12 13
=
30 35
156 156
P(red) =
65
156
Solutions Probability and Statistics
315
Model Solution on Assignment 2
1. 0.64 to 0.66
Let M
men
W
women
E
employed
U
unemployed
Given P(M) = 0.5
P(W) = 0.5
PUM = 0.20
PUW = 0.50
By Total probability,
P(U) =
P(M)P U M P(W) P U M
= 0.5 × 0.20 + 0.5 × 0.50 = 0.35
Required probability = P(E)
= 1
P(U)
= 1
0.35 = 0.65
2. 0.1276 to 0.1372
By Poission's distribution
p(h≥8) = e
8
=
900
3600
= 0.25
p(h≥8) =
80.25
e 0.1354
3. 0.79 to 0.81
E(
X ) =
x
2
x/2
1
e
2
dx
= 2
2
x/2
0
1
|x| e
2
dx
….(even function)
= 2
2
x/2
0
1
xe
2
dx
Put
1
2
x
2
= y x dx = dy
E(
X ) = 2
0
1
2
e
y
dy
= 2
1
2
(1)
= 0.7979
4. (B)
The probability of getting 'tail' when the
coin is tossed again is
1
2
5. 0.96 to 0.98
P(at least one will meet specification)
= 1
P(none will meet specification)
= 1
(1 0.8) x (1 0.7) x (1 0.5)
= 1
0.2 x 0.3 x 0.5
= 1
0.03 = 0.97
6. 0.23 to 0.25
E(X) = (0
q) + (1p) = p
E(X
2
) = (0
2
q) + (1
2
p) = p
V(X) = E(X
2
) [E(X)]
2
= p
p
2
= p(1 p) = pq
= 0.6
0.4 = 0.24
7. 99.6 to 99.8
For SN distribution, percentage of area
from
3 to 3 is 99.74
8. (A)
9. 2.5 to 2.5
Vidyalankar : GATE – Engineering Mathematics
316
10. (C)
P(p < 3) = P(p = 0) + P(p = 1)
+ P(p = 2)
=
eee
0! 1! 2!
(where µ = 3)
=
3
33
e9
ee3
2
=
3
3
917
e13
2
2e
11. 0.25 to 0.27
p =
2
6
=
1
3
q =
1
1
3
=
2
3
Using Binomial distribution
p(x ≥ 2)
=
23
21 30
CC
12 12
33
33 33
=
61
27 27
=
7
27
12. (B)
Variance = E[X
2
] [E[X]]
2
We know that, variance
0
E[X
2
] [E[X]]
2
0
E[X
2
] [E[X]]
2
13.
(B)
Let X = difference between the number
of heads and tails.
Taken x = 2
S = {HH, HT, TH, TT}
and X =
2, 0, 2
Here, n
3 = 1 is not possible
Taken n = 3
S = {HHH, HHT, HTH, HTT, THH,
THT, TTH, TTT}
and X =
3, 1, 1, 2
Here n
3 = 0 is not possible
Similarly, if a coin is tossed n times
then the difference between heads
and tails is n
3 is not possible.
Required probability is 0.
14. 0.25 to 0.28
Given: Poisson distribution,
p(x) =
x
e
x!
; x = 0, 1, 2, ...
= 240 per hour = 4 per minute
= 2 per 30 seconds
P(X = 1) =
1
e
1!
= 2 e
2
= 0.2706
15. (D)
Player A Starts the game
Player A can with the game in the
following manner :
A wins by getting 6 in the first try)
A loses, B loses, A wins
A loses, B loses, A loses, B loses, A
wins and so on,
P(A winning)
=
155155551
.....
666666666
=
1252525
1 ....
6363636
Which forms a G.P.
Solutions Probability and Statistics
317
P(A winning)
=
n
11 9
.... S
25
61r
1
36
=
11
11
6
36
P(A winning) =
6
11
16. (B)
Use Binomial distribution:
p(x) =
n
C
x
p
x
q
n
x
Here n = 10, p = 0.1, q = 0.9
P(3 out of 10 are defective)
=
10
C
3
(0.1)
3
(0.9)
7
= 0.0574
17. (D)
P{x = 0} = P
P{x = 1} = 1 p
P{y = 0} = q
P{y = 1} = 1 q
Let Z = X + Y
X Y
Z
0 0 0
0 1 1
1 0 1
1 1 2
From above table,
P{X + Y + Z}
P < Z B
P{Z
1} = P{X = 0 and Y = 1}
+ P{X = 1 and Y = 0}
+ P{X = 1 and Y = 1}
= 1
P{X = 0 and Y = 0}
= 1
pq
18. 0.39 to 0.43
Use Binomial distribution:
p(x) =
n
C
x
p
x
q
n
x
Here n = 5, p = 0.1, q = 0.9
P(packet is replaced)
= P(X
1) = 1 p(0)
= 1
q
n
= 1 (0.9)
5
= 0.40951
19. 54.0 to 54.0
E(X) = 5
E(X
2
) = 30, where X P(), = 5
E[(X + 2)
2
] = E(X
2
) + 4E(X) + 4
= 30 + 20 + 4 = 54
(
V(X) = E(X
2
) (E(X))
2
)
20. 0.33 to 0.34
Vidyalankar : GATE – Engineering Mathematics
318
Model Solution on Assignment 3
1. 49 to 51
Given M = 500
= 50
P(X > 500) = ?
where X follows normal distribution
We know that standard normal
variable
z =
xM
z =
500 500
50
= 0
(X > 500) = P(z > 0) = 0.50
(see figure)
2. 49.9 to 50.1
Let X be a positive odd number less
than 100
X = 1, 3, 5, ..., 99 each with
probability
1
50
E(X) =
1
50
(1 + 3 + 5 + ... + 99)
=
1
50
10025 = 50
3. (B)
P(Ram) =
1
6
; P(Ramesh) =
1
8
P(only one) = P(Ram)
P(not Ramesh)
+ P(Ramesh)
P(not Ram)
=
17 15
6886
=
12
48
=
1
4
4. (B)
5. (C)
P(Y/X) =
P(X Y)
P(X)
=
1
1
12
1
3
4
6. (C)
Here A and B are two independent
events
P(A B) = P(A) P(B)
and P(A/B) = P(A)
and P(B/A) = P(B)
Also if A and B are independent event
then
A and B are also independent
event.
PA B
= P(A) P(B)
Hence option (A), (B) and (D) are
correct.
Now,
P(A
B) = P(A) + P(B) P(A B)
= P(A) + P(B)
P(A) P(B)
Option (C) is false.
7. 0.9 to 1.1
Solutions Probability and Statistics
319
8. 54.49 to 54.51
Observed speeds (in km/hr) : 66, 62,
45, 79, 32, 51, 56, 60, 53 and 49.
Number of observations = 10
Median will be the mean of middle
two values in the ordered set.
Arranging in ascending order : 32, 45,
49, 51, 53, 56, 60, 62, 66, 79
Median Speed =
53 56
2
= 54.5
9. 0.5 to 0.5
Given: f(x) =
2
1
x
; a < x < 1
1
a
f(x) dx = 1
1
a
x
2
dx = 1
1
1
a
x
1
= 1
1 +
1
a
= 1
1
a
= 2 a =
1
2
10. (B)
p(3V
2U) = p(3V 2 0) = p(W 0),
W = 3V
2U
U, V are independent random
variables and U ~ N
1
0,
4
V ~ N
1
0,
9
W = 3V 2U ~ N
11
0,9 4
49
W ~ N (0, 2) ie., W has mean
= 0
and variance,
2
2
p(W 0) = p
W0
= p(Z
0), Z is standard normal variants
= 0.5 =
1
2
11. (A)
Let 'x' be no. of defective pieces.
x 0 1 2
P(x)
1
6
2
3
1
6
Mean () = E(x) = (x) P(x)
=
11 1
012
63 6
=
21
0
33
= 1
E(x
2
) = x
2
P(x)
=
12 1
014
63 6
=
22
0
33
=
4
3
Variance, V(x) = E(x
2
) (E(x))
2
=
4
1
3
=
1
3
12. (D)
Given,
Red
4
Black
6
The selection will be RBB or BBR of
BRB
Probability of selecting RBB
=
465
10 9 8
Vidyalankar : GATE – Engineering Mathematics
320
Probability of selecting BBR
=
654
10 9 8
Probability of selecting BRB
=
645
10 9 8
P(Red = 1) = sum of above three
probabilities = 0.5
13. 0.35 to 0.45
P(0.5 < x < 5)
=
5
0.5
f(x)dx
=
145
0.5 1 4
0.2 dx 0.1 dx 0 dx
=
14
0.5 1
0.2[x] 0.1[x] 0
= 0.2
0.5 + 0.1 3
P(0.5 < x < 5) = 0.4
14. (B)
n = 4; p =
1
6
q = 1
1
6
=
5
6
p(x
2)
= 1
p(x < 2)
= 1
[p(x = 0) + p(x = 1)]
= 1
04 13
44
01
15 15
CC
66 66
19
144
15. (D)
16. 4
E(e
3X/4
) =
0
e
3x / 4
e
x
dx
=
0
e
x / 4
dx = 4
17.
6
Let S: getting 3 on a die
p = P(S) =
1
6
, q = 1 p =
5
6
X = number of times die is thrown to obtain
S for the first time
X 1 2 3 4
…
.
p(X) p qp q
2
p q
3
p
…
.
E(X) =
x (q
x
1
p) = p x q
x
1
= p
2
1
1q
=
1
p
= 6
18. (A)
19. (A)
P(X
4) =
4
1
0.25 dx
= 0.25
(4 1) = 0.75
20.
2.0 to 2.0
Given X(t) = U + Vt
X(2) = U + 2V
E[X(2)] = E[U + 2V]
= E(U) + 2E[V] = 0 + 2 × 1 = 2
Solutions Probability and Statistics
321
Model Solution on Assignment 4
1.
(A)
The possible cases in the throw of a
die are six i.e. 1, 2, 3, 4, 5, 6
Cases of even numbers 2, 4, 6 are
three.
Probability of getting an even
number =
31
62
2. (D)
When we toss a coin, there are two
possible outcomes i.e. Head or Tail.
Number of possible cases = 2
The outcome of Tail is favourable
event
Probability of getting a tail = 1/2
3. (B)
Total number of equally likely and
exhaustive cases = n = 6 + 9 = 15
Number of favourable cases
=
9
1
C9
[
number of black balls = 9]
Probability of drawing a black ball
=
93
15 5
4. (C)
Number of clubs in a pack = 13
Probability of getting a club =
13 1
52 4
5.
(A)
Sample space = 6
4, 5, 6 are greater than 3
Probability =
31
62
6. (C)
Sample space = 6
2
= 36
Favourable cases are (5, 6), (6, 5),
(6, 6)
Probability =
31
36 12
7. (C)
There are 366 days in a leap year and it
has 52 complete weeks and 2 days
extra.
They are
(1) Sunday and Monday
(2) Monday and Tuesday
(3) Tuesday and Wednesday
(4) Wednesday and Thursday
(5) Thursday and Friday
(6) Friday and Saturday
(7) Saturday and Sunday
sample space = 7
Number of favourable cases = 2
Probability = 2/7
8. (A)
In a pack of 52 cards, 1 card can be
drawn in 52 ways
Since there are 13 spades and 3 aces
(one ace is present in spade)
Vidyalankar : GATE – Engineering Mathematics
322
Number of favourable case
= 13 + 3 = 16
Sample space = 52
Probability of getting a spade or an ace
=
16
52
=
44
13 9 4
Odds against winning the bet are
9 to 4.
9. (B)
Events are mutually exclusive
odd in favour of horse H
1
= 1/3
P(H
1
) =
11
13 4
Similarly P(H
2
) =
11
,
14 5
3
11
PH
15 6
P(H
4
) =
11
16 7
1234 1 2
PH H H H PH PH
34
PH PH
=
1111
+++
4567
319
=
420
10. (D)
Suppose there are 100 flowers
Number of roses = 40; Number of
carnations = 60
25% of 40 = 10 roses are red and
10% of 60 = 6 carnations are red
Let A be the event that the flower is
red and B the event that the flower is a
rose.
A B is the event that the flower is
a red rose.
n(A) = 16
P(A) =
16
100
n(A
B) = 10 P (A B) =
10
100
P(B/A) = probability that a selected
flower is a rose red is colour
P(B/A) =
PA B
10 / 100 5
P A 16 / 100 8
11. (C)
A throw amounting to 18 must be
made up of 6, 6, 6 and this can occur
in 1 way. 17 can be made up of 6, 6, 5
which can occur in 3 ways. 16 may be
made up of 6, 6, 5 and 6, 5, 5 each of
which arrangements can occur in 3
ways.
The number of favourable cases is
= 1 + 3 + 3 + 3 = 10 and the total
number of cases is 63 = 216
Required chance =
10 5
216 108
12.
(B)
Probability that A wins (P
1
) = 1/6
and that B wins (P
2
) = 1/10
and that C wins (P
3
) = 1/8
As a dead heat is impossible, these
are mutually exclusive events, so the
Solutions Probability and Statistics
323
chance that one of them will win the
race is
12 3
PP P i.e.
11147
6 10 8 120
13. (A)
The chance of choosing the first bag is
1/2 and if the first bag be chosen the
chance of drawing a red ball from it is
5/12 hence the chance of drawing a
red ball from the first bag is
15 5
212 24
Similarly the chance of drawing a red
ball from the second bag is
13 1
215 10
. Hence, as these events
are mutually exclusive, the chance
required is
5137
24 10 120
14. (C)
9 cards of a suit can be selected in
13
9
C
ways, 4 cards can be selected
from the remaining in
39
4
C ways.
The suit can be selected in 4 ways
Required probability
=
13 39
94
52
13
CC4
C
15. (A)
The probability that husband is not
selected = 1
16
77
The probability that wife is not selected
= 1
14
55
Probability that only husband is
selected =
14 4
75 35
Probability that only wife is selected
=
16 6
57 35
Probability that only one of them is
selected =
46102
35 35 35 7
16. (B)
There are 3 mutually exclusive and
exhaustive way in which 2 balls are
transferred form first bag to second bag.
First Way :
Two white balls are transferred from
first bag to second bag so that
probability for that is =
4
2
6
2
C
C
. In the
second bag we have 7 white and 4
black balls and the probability of
getting a white ball is =
7
11
Required probability
=
4
2
6
2
C
76742
11 15 11 165
C
Second way :
Two black balls have been transferred
from first bag to the second bag so
that probability for that is
2
2
6
2
C
1
15
C
Vidyalankar : GATE – Engineering Mathematics
324
In the second bag we have five white
and 6 black balls and probability of
getting a white ball is
5
11
.
Required probability
=
15 5
15 11 165
Third way :
One black and one white ball have
been transferred from first bag to the
second so that the probability for this
is
42
11
6
2
CC
8
15
C
In the second bag we have 6 white
and 5 black balls and the probability of
drawing a white ball is 6/11.
Required probability
=
86 48
15 11 165
Since these three cases are mutually
exclusive,
The required probability of drawing
a white ball
=
4254895
165 165 165 165
17. (C)
Total number of cases in a throw of
two dice = 36
Number of cases with doublets = 6
1,1 , 2, 2 , 3, 3 , 4, 4 , 5, 5 , 6, 6
Number of cases with a total of 9 = 4
5, 4 , 4, 5 , 3, 6 , 6, 3
Total number of cases in which a
doublet or a total of 9 appear = 6 + 4
= 10
P(a doublet or a total of 9) = P(A)
=
10 5
36 18
P(neither a doublet nor a total of 9)
= 1
P(A)
=
513
1
18 18
18. (C)
Let P(A), P(B) be the probability of A &
B speaking the truths, then
P(A) =
75 3
100 4
, P(B) =
80 4
100 5
PA
P ( A tells a lie) = 1 P(A)
= 1
31
44
PB = P(B tells a lie) = 1 p(B)
= 1
41
55
Now P(A and B will contradict)
=
PAPB PBPA
=
3141 7
35%
4554 20
Solutions Probability and Statistics
325
Model Solution on Assignment 5
1. (A)
Probability of 1 appearing on upper
face =
1
6
Probability of 6 appearing =
1
6
The probability 1 or 6 appear
=
11 1
66 3
2. (C)
Probability of the winning of the horse
A =
1
5
and the probability of the horse
B =
1
6
P(A + B) = P(A) + P(B)
=
11 11
56 30
3. (D)
First draw : Probability of getting a
queen =
41
52 13
Second draw : After drawing the first
queen we are left with 51 cards with 3
queens
Probability of getting a queen in
second draw =
31
51 17
Probability of both the cards are
queen =
11 1
13 17 221
4. (A)
P(A) and P(B/A) denote the probability
of drawing a black ball in the first and
second attempt
Probability of drawing a black ball in
the first attempt is
P(A) =
3
8
3
53
Probability of drawing the second
black ball given the first ball drawn is
black
2
P(B / A)
7
2
52
Probability that both the balls drawn
are black is
P(AB) = P(A) P(B/A)
=
32 3
87 28
5. (D)
Probability of getting a head in a toss
of a coin =
1
2
;
Probability of getting tail in each case
=
1
2
Probability of getting a tail in all four
cases =
1111 1
2222 16
Probability of getting at least one
head = 1
115
16 16
Vidyalankar : GATE – Engineering Mathematics
326
6. (C)
P = probability of getting 1 in a throw
of a die = 1/6
q = 1 p = 1
15
66
Probability of getting at least one die
= 1
q
3
=
3
591
1
6 216
7. (A)
8. (C)
Let X = defective items. We seek P
(C/X) the probability that an item is
produced by machine C, given that the
item is defective
By Bayes’ theorem,
P(C/X) =
PCPX/C
PAPX/A
PBPX/B
PCPX/C
=
0.10 0.04
0.60 0.02
0.30 0.03
0.10 0.04
=
4
25
9. (D)
Probability of six occurring in one toss
= 1/6
Probability of six occurring in 180 toss
= 180
1
6
= 30
10.
(A)
Probability that any person selected at
random is a rice eater =
1
2
. Probability
of a non rice eater =
1
2
. Let A
0
, A
1
, A
2
,
A
3
denote the event that none, one,
two or three persons respectively are
rice eaters out of 10. Then
P(A
0
) =
10
C
0
010
11
22
,
P(A
1
) =
10
C
2
19
11
22
P(A
2
) =
10
C
2
28
11
22
,
P(A
3
) =
10
C
3
37
11
22
Now, the required probability
= P(A
0
+ A
1
+ A
2
+ A
3
)
= P(A
0
) + P(A
1
) + P(A
2
) + P(A
3
)
=
10
1
2
[
10
C
0
+
10
C
1
+
10
C
2
+
10
C
3
]
=
10
1
2
[ 1 + 10 + 45 + 120]
=
176
1024
= 0.17 = 17 % (approx)
11. (D)
Probability that the first person lives till
he is 75 years =
8
14
Probability that the second person
lives till he is 80 years =
4
9
Solutions Probability and Statistics
327
Probability of the compound event that
both the persons live 40 years hence
=
84 32 16
14 9 126 63
Probability that at least one of them
would die without living 40 years
hence = 1
16 47
63 63
12. (A)
The probability of getting a head in a
single toss of a coin is 1 / 2
p = 1/2, q = 1 p = 1
1
2
=
1
2
Also we are given n = 10, N = 100
and r =7
The required frequency
= N
n
C
r
p
r
q
n
r
= 100
73
10
7
11
C
22
10
10! 1
100.
7!3! 2
= 100
10
10 9 8 1 375
3212 32
= 11.7 = 12
13. (A)
Let (q + p)
n
, q + p =1, be the binomial
distribution
P = 0
1, n = 500
Mean = np = 0.1
500 = 50
Now p = 0.1
q = 1 p = 1 0.1 = 0.9
Variance = npq
= 500
0.1 0.9 = 45
Standard deviation = 45 = 6.7
14. (A)
= p
i
x
i
= 0 .
144 24 1
1. 2.
169 169 169
26 2
169 13
2
= p
i
x
i
2
()
2
= 0.
144 24 1 4
1. 4.
169 169 169 169
=
28 4 24
169 169 169
=
24 4.9
169 13
= 0.377
15. (C)
np = 12
npq = 2
npq = 4
npq 4 1
np 12 3
q =
1
3
p = 1 q = 1
1
3
=
2
3
Also np = 12
n
2
3
= 12
n = 18 & p =
2
3
Vidyalankar : GATE – Engineering Mathematics
328
16. (D)
The probability that face 1 or face 2
turn up = 0.10 + 0.32 = 0.42
The probability that face 1 turns up
= 0.10
The required probability
=
0.10 10 5
0.42 42 21
17. (C)
Since odds against A are 8 : 3
P(A) =
33
83 11
P(B) =
22
52 7
P(A) + P(B) + P(C) = 1
{mutually exclusive}
Or
32
11 7
+ P(C) = 1
P(C) = 1
32 34
11 7 77
odds against C are 77 34 = 34
i.e. 43 : 34
18. (C)
Combinations are CMC, CMM, CCC,
CCM.
Probability of
CMC = 0.6
0.4
= 0.24
Favourable
CMM = 0.6
0.6 = 0.36
CCC = 0.4
0.4
= 0.16
Favourable
CCM = 0.4
0.6 = 0.24
Total Probability = 1
Favourable = 0.24 + 0.16 = 0.4
So P = 0.4/1
Solutions Probability and Statistics
329
Model Solution on Assignment 6
1. (B)
S = {Sun, Mon, Tue, Wed, Thu, Fri, Sat}
n(s) = 7
Let A be the event that the child is born
on Sunday or a Saturday
n(A) = 2
P(A) =
nA
2
nS 7
2. (D)
S = {1, 2, 3, 4, 5, 6 }
n(S) = 6
Let A be the event that the die shows a
multiple of 2
A = {2, 4, 6}
n(A) = 3
P(A) =
nA
31
nS 6 2
3. (C)
S =
1, 1 , 1, 2 , 1, 3 , 1, 4 , 1, 5 , 1, 6
2,1 , 2, 2 , 2, 3 , 2, 4 , 2, 5 , 2, 6
3,1 , 3, 2 , 3, 3 , 3, 4 , 3, 5 , 3, 6
4,1 , 4, 2 , 4, 3 , 4, 4 , 4, 5 , 4, 6
5,1 , 5, 2 , 5, 3 , 5, 4 , 5, 5 , 5, 6
6,1 , 6, 2 , 6, 3 , 6, 4 , 6, 5 , 6, 6
n(S) = 36
Let A be the event that the score on the
dice is odd
A =
1,2,1,4,1,6,2,1,2,3,
2,5,3,2,3,4,3,6,4,1,
4,3,4,5,5,2,5,4,5,6,
6,1 , 6, 3 , 6, 5
n(A) = 18
nA
18 1
PA
nS 36 2
4. (C)
5 students can be selected from 10 by
10
5
C ways
n(S) =
10
5
10!
C=
5!5!
10×9×8×7×6
= = 252
5×4×3×2
Let A be the event that the committee
includes exactly 2 girls and 3 boys. The
two girls can be selected in
4
2
C ways
and the three boys can be selected in
6
3
C ways
n(A) =
46
23
C × C = 6 × 20 = 120
P(A) =
nA
120 10
n S 252 21
5. (A)
Let A be an event that ball drawn is red
or white
n(A) =
74
11
CC7411
11
PA
20
[
20
1
nS C 20]
Vidyalankar : GATE – Engineering Mathematics
330
6. (C)
Two cards can be selected in
52
2
C
ways
n(S) =
52
2
C2651
Let A be the event that 2 cards selected
is Jack and an Ace.
Jack can be selected in
4
1
C ways and
the Ace can also be selected in
4
1
C
ways.
n(A) =
44
11
CC16
nA
16 8
PA
n S 26 51 663
7. (C)
The required probability
12
C
11
(0.99875)
11
(0.01125) +
12
C
12
(0.99875)
12
= 0.9923
8. (B)
Required probability
=
11 41 4
..
53 54 15
9. (A)
P(A) =
13 4
11
52 51
11
CC
1
52
CC
10. (D)
Probability of first event =
2
3
P
Mutually exclusive events
P +
2
3
P = 1 Or P =
3
5
= 3 : 5
Hence odds in favor of the other are
3 : (5
3) i.e. 3 : 2
11. (C)
P(A
B ) = P(A) P(A B) =
1
3
P(A B) = P(A) + P(B) P(A B)
P(A
B) = [P(A) P(A B)] + P(B)
P(A
B) = P(A B ) + P(B)
2
3
=
1
3
+ P(B)
P(B) =
2
3
1
3
=
1
3
12. (A)
Probability P of a white flower =
1
4
q = 1
P = 1
1
4
=
3
4
n = 3 N = 64
p(r) =
3
C
r
r3r
13
44
Number of beds with zero white flowers
= 64
03
3
0
13
C
44
27
64
64
= 27
Beds with 1 white flower
= 64
12
3
1
13
C
44
= 27
Beds with 2 white flowers
= 64
2
3
2
13 9
C64
44 64
= 9
Beds with 3 white flowers
= 64
30
3
3
13 1
C64
44 64
= 1
Solutions Probability and Statistics
331
13. (B)
P (A
B) =
1
;
2
1
P(A) ;
3
1
P(B)
3
Now, P(A) = 1
12
P(A) 1
33
P(B) =
12
1P(B) 1
33
P(A B) = P(A) P(B) P(A B)
=
221
332
4183
32 6
=
55210
66212
14. (A)
To be one step away from the starting
point the Man is to take 6 steps forward
and 5 steps backward or 5 steps
forward and 6 steps backward
The required probability
=
11
C
6
(0.4)
6
(0.6)
5
+
11
C
5
(0.4)
5
(0.6)
6
=
11
C
5
(0.4)
5
(0.6)
5
{0.4 + 0.6}
=
11
C
5
(0.24)
5
15. (A)
Required probability
=
12 13 1
2525 2
16.
(B)
P(B) = 1
P(
11
B) 1
22
and P(A
B) = P(A) + P(B) P(A B)
or
5
6
= P(A) +
11
23
or
2
P(A)
3
P(A) P(B) =
21 1
32 3
= P(A B)
Hence A and B are independent.
17. (D)
n(s) = 2
4
= 16 since each of the four
places in a determinant of order 2 is to
filled by 0 or 1 favourable number of
ways is 3.
10 10 11
or or
01 11 01
Hence P =
3
16
18.
(D)
100
C
50
p
50
(1 p)
50
=
100
C
51
p
51
(1 p)
49
or
1 p 100! 50! 50! 50
p 51! 49! 100! 51
or 51
51p = 50 p
giving P =
51
101
Vidyalankar : GATE – Engineering Mathematics
332
Answer Key on Test Paper 1
1.
(D)
2.
(D)
3.
(A)
4.
(C)
5.
(A)
6.
(C)
7.
(C)
8.
(B)
9.
(A)
10.
(D)
11.
(A)
12.
(A)
13.
(C)
14.
(A)
15.
(C)
Answer Key on Test Paper 2
1.
(B)
2.
(D)
3.
(D)
4.
(C)
5.
(B)
6.
(A)
7.
(D)
8.
(A)
9.
(A)
10.
(A)
11.
(A)
12.
(D)
13.
(A)
14.
(A)
15.
(C)
Answer Key on Test Paper 3
1.
(D)
2.
(A)
3.
(D)
4.
(B)
5.
(A)
6.
(D)
7.
(B)
8.
(B)
9.
(D)
10.
(B)
11.
(D)
12.
(A)
13.
(C)
14.
(B)
15.
(D)
Solutions Probability and Statistics
333
Model Solution on Test Paper 1
1. (D)
P(A) =
2
5
P(A) =
3
5
P(B) =
2
3
P(B) =
1
3
P(C) =
3
5
P(C) =
2
5
Probability that only one of them hits
the target
= probability that A hits the target but
not B and C + probability that B hits
the target but not A and C
+ probability that C hits the target
but not A and B
=
PA B C PA B C
PABC
=
212232331
535355553
4129 251
75 75 75 75 3
2. (D)
Probability of selecting any bag is 1/2
Probability of getting a white ball
=
64
11
99
11
CC
115
229
CC
3. (A)
Let A be the event that A is selected
and B be the event that B is selected
P(A) = 1/5 and P(B) = 2/7
Let C be the event that both are
selected
C be the event that both are
selected.
C = A B
P(C) = P(A B)
P(C) = P(A)
P(B) as A and B are
independent events =
12 2
57 35
4. (C)
Two balls are selected from 12 in
12
2
C
ways. Two white balls can be selected
in
7
2
C ways.
Required probability =
7
2
12
2
C
C
21 7
66 22
5. (A)
Three balls can be selected from 9 in
9
3
C ways
3 black balls can be selected from 5 in
5
3
C
ways
Required probability =
5
3
9
3
C
C
10 5
84 42
Vidyalankar : GATE – Engineering Mathematics
334
6. (C)
R stands drawing a red ball and B for
drawing black ball. Then required
probability is
= RRR + RBR + BRR + BBR
=
656 665
10 11 10 10 11 10
447 476
10 11 10 10 11 10
=
6400 32
1100 55
7. (C)
P
1
=
61
36 6
[
out of total of 36 ways both the
persons can throw equal values in
6 ways]
To find p
2
the total number of ways
n 6
4
and the favorable number of
ways, M =15 8 = 120
Since any two numbers out of 6 can
be selected in
6
c
2
i.e. 15 ways and
corresponding to each of these ways,
there are 8 ways e.g. corresponding to
the numbers 1 and 2 the eight ways
are (1, 1, 1, 2), (1, 1, 2, 1), (1, 2, 1, 1),
(2, 1, 1, 1) , (2, 2, 2, 1) (2, 2, 1, 2) (2,
1, 2, 2) (1, 2, 2, 2).
Hence P =
120 5
64 54
Since
15
654
, we have P
1
> P
2
8. (B)
P
1
=
15 27
22
42
4
cc
c
15 14 27 26 1 2 3 4
1 2 1 2 42 41 40 39
27
82
P
2
=
30 54
44
84
8
cc
c
30.29.28.27.54.53.52.51.8!
4!4!84.83.82.81.80.79.78.77
=
17.29.45.53
11.79.82.83
1
2
P
27 11.79.82.83 33.79.83
P 82 17.29.45.53 29.53.85
=
216381
130645
> 1
Hence P
1
> P
2
9.
(A)
P(AB) = P(A) + P(B) P(A B)
P(A B) 1
P(A) + P(B) P(A B) 1
P(A) + P(B) 1 P(A B)
P(A B) P(A) + P(B) 1
10. (D)
We first find the total number of cases.
For each element of set A with n
element, the number of possible
images is n.
The total number of cases = n
n
.
Solutions Probability and Statistics
335
When next find the number of
favourable cases. For the first element
we have n choices. For the seconds
element we have (n 1) choices and
so on.
The number of favourable cases
=
n n 1 n 2 ...2 1 n!
Hence the required probability
=
nn1
n1!
n!
nn
11. (A)
12. (A)
13. (C)
14. (A)
Distinct paths from A to F are given
below.
1) ABDF
2) ACEF
3) ABF
4) ABEF
5) ACDF
6) ABCDEF
7) ACDEF
8) ABDEF
9) ABCDF
10) ABCEF
There are 6 way to go to city F
through C.
Total no. of paths = 10
Hence, required probability
= 6/10 = 3/5
15. (C)
Mean
x =
x
N
90 =
x
200
x
= 200 × 90 = 18000
Sum of observations is 18000
Of these, two wrong observations are
15 and 80.
Subtracting 15 and 80 from 18000, we
get 17905
Adding the correct observations to
17905,
i.e. 17905 + 40 + 87 = 18032
Correct sum of 200 observation is
18032
Correct mean is
18032
200
= 90.16
Vidyalankar : GATE – Engineering Mathematics
336
Model Solution on Test Paper 2
1.
(B)
There are 6 possible ways in which the
die can fall and of these two are
favourable events required.
Required chance =
21
63
2. (D)
The possible number of cases is
6 6 = 36
An ace on one die may be associated
with any of the 6 numbers on the other
die and the remaining 5 numbers on
the first die may be associated with the
ace on the second die, thus the
number of favorable cases is 11.
Required chance =
11
36
3. (D)
Various Digits in the log table are 0, 1,
2, 3, 4, 5, 6, 7, 8, 9
i.e. a total of 10 digits are used.
The number of favourable cases for
getting 1 out of the 10 all equally likely
cases is one
Probability of getting 1 is 1/10.
4. (C)
The probabilities of Dayanand,
Ramesh and Naresh solving the
problem are
111
,,
234
respectively
The probabilities of Dayanand,
Ramesh, Naresh not solving the
problem are
11 12 13
1;1;1
22 33 44
respectively.
The probability that the problem is
not solved by any one of them is
123 1
234 4
The probability that the problem
will be solved by at least one of
them = 1
13
44
5. (B)
The probability of drawing a black ball
from the bag = 3/8
The probability that the drawn ball is
not black =
35
1
88
6. (A)
In “PROBABILITY” there are
O A I : 3 distinct vowels
P R B L T Y : 6 distinct consonants
Total number of distinct letters = 9,
Required probability =
31
93
7.
(D)
Let P(A), P( A ) be probabilities of A’s
getting the head and not getting the
head respectively, then P(A) =
1
2
Solutions Probability and Statistics
337
P( A ) = 1 P(A) = 1
1
2
=
1
2
Similarly, P(B) =
1
2
and P( B ) =
1
2
Let A start the game. He can win it in
the 1
st
throw 3
rd
throw 5
th
throw, 7
th
throw and so on. Probability of A’s
winning in 1
st
throw = P(A) =
1
2
Probability of A’s winning in 3
rd
throw
= P(
A ) P( B ) P(A)
=
1
2
.
1
2
.
1
2
=
3
1
2
Probability of A’s winning in 5
th
throw
= P(
A ) P(B ) P( A ) P( B ) P(A)
=
1
2
.
1
2
.
1
2
.
1
2
.
1
2
=
5
1
2
Probability of A’s winning in 7
th
throw
= P(
A ) P( B ) P( A ) P( B ) P( A )
P(
B ) P(A)
=
1
2
.
1
2
.
1
2
.
1
2
.
1
2
.
1
2
.
1
2
=
7
1
2
Mutually exclusive cases
Probability of A’s winning the
game first is
357
1
1111 2
2
....
1
2222 3
1
4
Probability of B winning the game
first = 1
21
33
8.
(A)
n(S) = 36
The outcomes which have odd number
and multiple of 3 are
(1, 3) (1, 6) (3, 1) (3, 3) (3, 5) (3, 6)
(5, 3) (5, 6) (6, 1) (6, 3) (6, 5)
favourable cases = 11
Probability =
11
36
9. (A)
Let A = The event that students are
taller than 18m
P(w /A) = Probability that a student is
a woman given A
=
P(w)P(A / w)
P(w)P(A / w) P(M)P(A / M)
=
(0.60)(0.01) 3
(0.60)(0.01) (0.40)(0.04) 11
10.
(A)
n = total number of ways = 6 6 = 36
The numbers higher than 9 are 10, 11,
12 in the case of two dice
m = favourable number of ways
= 3 + 2 + 1 = 6
Hence P =
M61
n366
11. (A)
Total number of cards =12
Probability of at least one are
= 1 probability no are
= 1
8
2
12
2
C
14 19
1
33 33
C
Vidyalankar : GATE – Engineering Mathematics
338
12. (D)
Var (x) = 1
13.
(A)
The probability that event
A
occurred,
provided that
B took place will be
denoted byP(
A B ). Now the probability
that event A occurred, provided that B
took place will be denoted by
P(A/
B ).The event A is the non
occurrence of the event A.
We have P(A/
B ) + P( A / B ) = 1
Hence P( A /B ) = 1 P(A /B )
14.
(A)
Two points are always collinear so we
can say a triangle consists of two
collinear points and one point which is
not collinear to both of them
simultaneously.
So here, we have no. of triangles
= No. of triangles having 2 points on
line 1 and one point on line 2
+ No. of triangles having 2 points
on line 2 and one point on line 1
=
10 11
22
C11 C
= 45 11 + 55 10
= 1045
Total number of points = 10 + 11 = 21
No. of 3 points group
=
21
3
C
=
21 20 19
321
= 7 10 19
Required probability
=
1045
71019
=
209
14 19
=
11 19
14 19
11
14
15. (C)
Probability that the item is produced by
machine A, P(A) = 0.25
Probability that the item produced by A
is faulty, P(F/A) = 0.05
Similarly,
P(B) = 0.30
P(F/B) = 0.04
P(C) = 1 0.25 0.3 = 0.45
P(F/C) = 0.03
By total probability,
Probability that the item selected is
faulty
= P(A) P(F/A) + P(B) P(F/B)
+ P(C) P(F/C)
= 0.25 × 0.05 + 0.30 × 0.04 + 0.45
× 0.03
= 0.038
By Baye's Theorem,
Probability that the faulty item was
produced by machine C,
P(C/F) =
P(C) F (F / C)
P(F)
=
0.45 0.03
0.038
P(C/F) = 0.355
Solutions Probability and Statistics
339
Model Solution on Test Paper 3
1. (D)
Probability of getting both the balls
even numbered
P(EE) = P(E)P(E)
=
12 12 144
25 25 625
P(at least one odd) = 1 P(EE)
= 1
144
625
481
625
2. (A)
The total number of ways in which four
persons can be selected out of 9
persons is
9
4
C .
For favourable cases, we want that 2
out of the four selected should be
children. Two children can be selected
out of 4 in
4
2
C ways. The other two
are to be selected out of 5 persons (3
men and 2 women). Two persons can
be selected out of 5 in
5
2
C ways.
The number of favourable cases
45
22
CC
Required probability
=
45
22
9
4
CC
10
21
C
3. (D)
PA B
A1/43
P
BPB1/34
4. (B)
When all the letters are taken they can
be arranged in
10
p
10
ways or 10 !
5. (A)
Here the balls are drawn without
replacement. Let P(W) and P(G)
denote the probability of drawing white
ball and green ball respectively.
P(W and G) = P(W) P(G) + P(G) P(W)
=
10 8 8 10
18 17 18 17
80
153
6. (D)
The largest number on the selected
coupon is 9, and hence the selection is
to be made from the coupons
numbered 1 to 9. Since there are 15
possible cases for selecting a coupon
and seven coupons are selected the
total number of possible cases = 15
7
The number of favorable cases
= 9
7
8
7
. Observe that out of 9
7
cases
8
7
cases do not contain the number 9
The required probability
=
Number of favourable cases
Total number of cases
=
77
7
98
15
Vidyalankar : GATE – Engineering Mathematics
340
7. (B)
P =
2
5
q =
3
5
n = 5
P(r) =
n
C
r
. q
n
r
p
r
=
5r r
5
r
32
C.
55
The required chance
= P(4)
=
4
5
4
32
C. .
55
= 5
5
48
5
=
4
48 48
625
5
8. (B)
Total number of ways in which two
cards are drawn out of 52 is
52
2
C
.
A king can be drawn in 4 ways and
Queen can be drawn in 4 ways.
Required probability =
52
2
44 8
663
C
9. (D)
In a single throw of two dice, the
probability of throwing a doublet is
61
36 6
The probability of throwing three
doublets in three throws is
111 1
..
666 216
Hence the required probability
= 1
1 215
216 216
10.
(B)
Here np + npq = 10 , n = 18
18p + 18pq = 10 18p(1 + q) = 10
p + q = 1 p = 1 q
18(1 q) (1 + q) = 10
1 q
2
=
10
18
q
2
=
10 8
1
18 18
q
2
=
84
18 9
q =
2
3
q =
2
3
(ve sign rejected),
p =
1
3
Hence Binomial distribution is
18
12
33
{(p +q)
n
}
11. (D)
12. (A)
13. (C)
14. (B)
Out of the 5 women, 3 women can be
invited in
5
3
C
ways. Nothing is
mentioned about the number of men
that he has to invite. He can invite one,
two, three, four or even number of
Solutions Probability and Statistics
341
men. Out of 4 men, he can invite them
in the said manner in
4
(2)
ways. Thus,
the total number of ways is
54
3
C(2)
= 10 16 = 160.
15.
(D)
P(A) = 1/4
P(B/A) =
P(A B)
P(A)
=
1
2
P(A B) =
1
8
P(A/B) =
P(A B)
P(B)
=
1
4
P(B) =
1
2
P(A B) = P(A) P(A B)
=
11
48
P(A B)
=
1
8
P(A B) =
P(A B)
P(B)
=
18 1
12 4
P(A B)
=
1
2
is false.
If A B , A B = A
then P(A B) = P(A)
but P(A B) =
1
8
and P(A) =
1
4
This is false.
P(A B) P(A B) =
11
42
=
3
4
P(A B) P(A B) = 1 is false
None of the options (A), (B) or (C)
is true.
