CHAPTER 1
ENGINEERING MATHEMATICS
YEAR 2012
MCQ 1.1
The area enclosed between the straight line y = x and the parabola y = x2
in the x -y plane is
(B) 1/4
(A) 1/6
(C) 1/3
MCQ 1.2
ONE MARK
(D) 1/2
Consider the function f (x) = x in the interval − 1 # x # 1. At the point
x = 0 , f (x) is
(A) continuous and differentiable
(B) non-continuous and differentiable
(C) continuous and non-differentiable
(D) neither continuous nor differentiable
MCQ 1.3
MCQ 1.4
x is
lim b 1 − cos
2
l
x"0
x
(A) 1/4
(B) 1/2
(C) 1
(D) 2
At x = 0, the function f (x) = x3 + 1 has
(A) a maximum value
(B) a minimum value
(C) a singularity
MCQ 1.5
(D) a point of inflection
For the spherical surface x2 + y2 + z2 = 1, the unit outward normal vector at
the point c 1 , 1 , 0 m is given by
2 2
1
1
(B) 1 i − 1 j
(A)
i+
j
2
2
2
2
1
1
(C) k
(D)
i+
j+ 1 k
3
3
3
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PAGE 2
ENGINEERING MATHEMATICS
CHAP 1
YEAR 2012
MCQ 1.6
MCQ 1.7
MCQ 1.8
(A) f (t) = sin t
1
is given by
s (s + 1)
(B) f (t) = e−t sin t
(C) f (t) = e−t
(D) f (t) = 1 − e−t
The inverse Laplace transform of the function F (s) =
5 3
For the matrix A = >
, ONE of the normalized eigen vectors given as
1 3H
J 1 N
J 1N
K
O
K 2O
2O
O
(A) K
(B) K
K −1 O
KK 3 OO
K
O
2
L
P
L 2P
J 3 N
J 1 N
K
O
K
O
10 O
5O
(C) K
(D) K
K −1 O
K 2 O
K
O
K
O
10
L
P
L 5P
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
(A) 1/20
(B) 1/12
(C) 3/10
MCQ 1.9
TWO MARKS
(D) 1/2
Consider the differential equation x2 (d 2 y/dx 2) + x (dy/dx) − 4y = 0 with the
boundary conditions of y (0) = 0 and y (1) = 1. The complete solution of the
differential equation is
(B) sin a πx k
(A) x2
2
(C) ex sin a πx k
2
(D) e−x sin a πx k
2
MCQ 1.10
x + 2y + z = 4
2x + y + 2z = 5
x−y+z = 1
The system of algebraic equations given above has
(A) a unique solution of x = 1, y = 1 and z = 1.
(B) only the two solutions of (x = 1, y = 1, z = 1) and (x = 2, y = 1, z = 0)
(C) infinite number of solutions
(D) no feasible solution
YEAR 2011
MCQ 1.11
ONE MARK
A series expansion for the function sin θ is
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CHAP 1
ENGINEERING MATHEMATICS
(A) 1 − θ + θ − ...
2! 4!
(B) θ − θ + θ − ...
3! 5!
(C) 1 + θ + θ + θ + ...
2! 3!
What is lim sin θ equal to ?
θ"0 θ
(A) θ
(D) θ + θ + θ + ...
3! 5!
(B) sin θ
(C) 0
(D) 1
2
4
2
MCQ 1.12
MCQ 1.13
3
3
5
5
(D) complex
The product of two complex numbers 1 + i and 2 − 5i is
(A) 7 − 3i
(B) 3 − 4i
(C) − 3 − 4i
MCQ 1.15
3
Eigen values of a real symmetric matrix are always
(A) positive
(B) negative
(C) real
MCQ 1.14
PAGE 3
(D) 7 + 3i
If f (x) is an even function and a is a positive real number, then
equals
(A) 0
(B) a
#−a f (x) dx
a
(D) 2 # f (x) dx
a
(C) 2a
0
YEAR 2011
MCQ 1.16
TWO MARKS
1 dx , when evaluated by using Simpson’s 1/3 rule on two
x
equal sub-intervals each of length 1, equals
(A) 1.000
(B) 1.098
The integral
#1
3
(C) 1.111
MCQ 1.17
dy
Consider the differential equation
= (1 + y2) x . The general solution with
dx
constant c is
2
(B) y = tan2 a x + c k
(A) y = tan x + tan c
2
2
(C) y = tan2 a x k + c
2
MCQ 1.18
(D) 1.120
2
(D) y = tan b x + c l
2
An unbiased coin is tossed five times. The outcome of each toss is either a
head or a tail. The probability of getting at least one head is
(B) 13
(A) 1
32
32
(C) 16
32
(D) 31
32
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PAGE 4
MCQ 1.19
ENGINEERING MATHEMATICS
CHAP 1
Consider the following system of equations
2x1 + x2 + x 3 = 0
x2 − x 3 = 0
x1 + x 2 = 0
This system has
(A) a unique solution
(B) no solution
(C) infinite number of solutions
(D) five solutions
YEAR 2010
MCQ 1.20
ONE MARK
The parabolic arc y = x , 1 # x # 2 is revolved around the x -axis. The
volume of the solid of revolution is
(A) π/4
(B) π/2
(C) 3π/4
(D) 3π/2
MCQ 1.21
d 3f
f d 2f
= 0 , is a
3 + 2
dη2
dη
(A) second order nonlinear ordinary differential equation
The Blasius equation,
(B) third order nonlinear ordinary differential equation
(C) third order linear ordinary differential equation
(D) mixed order nonlinear ordinary differential equation
MCQ 1.22
The value of the integral
(A) − π
#− 33 1 dx
+ x2
(C) π/2
MCQ 1.23
(B) − π/2
(D) π
The modulus of the complex number b 3 + 4i l is
1 − 2i
(A) 5
(B) 5
(C) 1/ 5
MCQ 1.24
is
(D) 1/5
The function y = 2 − 3x
(A) is continuous 6x ! R and differentiable 6x ! R
(B) is continuous 6x ! R and differentiable 6x ! R except at x = 3/2
(C) is continuous 6x ! R and differentiable 6x ! R except at x = 2/3
(D) is continuous 6x ! R except x = 3 and differentiable 6x ! R
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 5
YEAR 2010
MCQ 1.25
TWO MARKS
2 2
H is
One of the eigen vectors of the matrix A = >
1 3
2
(A) > H
−1
2
(B) > H
1
4
(C) > H
1
MCQ 1.26
MCQ 1.27
1
(D) > H
−1
The Laplace transform of a function f (t) is 2 1
. The function f (t) is
s (s + 1)
(A) t − 1 + e−t
(B) t + 1 + e−t
(C) − 1 + e−t
(D) 2t + et
A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box
at random one at a time without replacement. The probability of drawing 2
washers first followed by 3 nuts and subsequently the 4 bolts is
(A) 2/315
(B) 1/630
(C) 1/1260
MCQ 1.28
(D) 1/2520
Torque exerted on a flywheel over a cycle is listed in the table. Flywheel
energy (in J per unit cycle) using Simpson’s rule is
Angle (Degree)
0
60c
120c
180c
240c
300c
360c
Torque (N-m)
0
1066
− 323
0
323
− 355
0
(A) 542
(B) 993
(C) 1444
(D) 1986
YEAR 2009
MCQ 1.29
3/5 4/5
H, the transpose of the matrix is equal to the
For a matrix 6M @ = >
x 3/5
T
−1
inverse of the matrix, 6M @ = 6M @ . The value of x is given by
(B) − 3
(A) − 4
5
5
(C) 3
5
MCQ 1.30
ONE MARK
(D) 4
5
The divergence of the vector field 3xzi + 2xyj − yz2 k at a point (1, 1, 1) is
equal to
(A) 7
(B) 4
(C) 3
(D) 0
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PAGE 6
MCQ 1.31
ENGINEERING MATHEMATICS
The inverse Laplace transform of 1/ (s2 + s) is
(B) 1 − et
(A) 1 + et
(C) 1 − e−t
MCQ 1.32
CHAP 1
(D) 1 + e−t
If three coins are tossed simultaneously, the probability of getting at least
one head is
(A) 1/8
(B) 3/8
(C) 1/2
(D) 7/8
YEAR 2009
MCQ 1.33
An analytic function of a complex variable z = x + iy is expressed as
f (z) = u (x, y) + iv (x, y) where i = − 1 . If u = xy , the expression for v
should be
(x + y) 2
x2 − y2
(B)
(A)
+k
+k
2
2
(C)
MCQ 1.34
MCQ 1.35
MCQ 1.36
TWO MARKS
y2 − x2
+k
2
The solution of x
(D)
(x − y) 2
+k
2
dy
+ y = x 4 with the condition y (1) = 6 is
5
dx
4
(A) y = x + 1
5 x
4
(B) y = 4x + 4
5
5x
4
(C) y = x + 1
5
5
(D) y = x + 1
5
A path AB in the form of one quarter of a circle of unit radius is shown
in the figure. Integration of (x + y) 2 on path AB traversed in a counterclockwise sense is
(A) π − 1
2
(B) π + 1
2
(C) π
2
(D) 1
The distance between the origin and the point nearest to it on the surface
z2 = 1 + xy is
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CHAP 1
ENGINEERING MATHEMATICS
(A) 1
(C)
MCQ 1.37
3
2
(B)
(D) 2
3
The area enclosed between the curves y2 = 4x and x2 = 4y is
(B) 8
(A) 16
3
(C) 32
3
MCQ 1.38
PAGE 7
(D) 16
The standard deviation of a uniformly distributed random variable between
0 and 1 is
(B) 1
(A) 1
12
3
(C) 5
(D) 7
12
12
YEAR 2008
MCQ 1.39
ONE MARK
In the Taylor series expansion of ex about x = 2 , the coefficient of (x − 2) 4 is
(A) 1/4 !
(B) 2 4 /4!
(C) e2 /4!
MCQ 1.40
MCQ 1.41
MCQ 1.42
Given that xp + 3x = 0 , and x (0) = 1, xo(0) = 0 , what is x (1) ?
(A) − 0.99
(B) − 0.16
(C) 0.16
(D) 0.99
1/3
The value of lim x − 2
x " 8 (x − 8)
(A) 1
16
(B) 1
12
(C) 1
8
(D) 1
4
A coin is tossed 4 times. What is the probability of getting heads exactly 3
times ?
(B) 3
(A) 1
8
4
(C) 1
2
MCQ 1.43
(D) e 4 /4!
R1
S
The matrix S3
SS1
T
other two eigen
(A) p
(D) 3
4
2 4VW
0 6W has one eigen value equal to 3. The sum of the
1 pWW
valueX is
(B) p − 1
(C) p − 2
(D) p − 3
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PAGE 8
MCQ 1.44
ENGINEERING MATHEMATICS
CHAP 1
The divergence of the vector field (x − y) i + (y − x) j + (x + y + z) k is
(A) 0
(B) 1
(C) 2
(D) 3
YEAR 2008
MCQ 1.45
TWO MARKS
Consider the shaded triangular region P shown in the figure. What is
## xydxdy ?
P
MCQ 1.46
(A) 1
6
(B) 2
9
(C) 7
16
(D) 1
The directional derivative of the scalar function f (x, y, z) = x2 + 2y2 + z at
the point
P = (1, 1, 2) in the direction of the vector a = 3i − 4j is
MCQ 1.47
(A) − 4
(B) − 2
(C) − 1
(D) 1
For what value of a, if any will the following system of equation in x, y and z
have a solution ?
2x + 3y = 4
x+y+z = 4
3x + 2y − z = a
MCQ 1.48
MCQ 1.49
(A) Any real number
(B) 0
(C) 1
(D) There is no such value
Which of the following integrals is unbounded ?
π/4
(A)
#0 tan xdx
(B)
#0 3 x2 1+ 1 dx
(C)
#0 3xe−x dx
(D)
#0
The integral
# f (z) dz
1
1 dx
1−x
evaluated around the unit circle on the complex plane
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CHAP 1
MCQ 1.50
ENGINEERING MATHEMATICS
for f (z) = cos z is
z
(A) 2πi
(B) 4πi
(C) − 2πi
(D) 0
The length of the curve y = 2 x3/2 between x = 0 and x = 1 is
3
(A) 0.27
(B) 0.67
(C) 1
MCQ 1.51
(D) 1.22
1
1
1 2
The eigen vector of the matrix >
are written in the form > H and > H.
H
a
b
0 2
What is a + b ?
(A) 0
(B) 1
2
(C) 1
MCQ 1.52
Let f = yx . What is
(A) 0
(C) 1
MCQ 1.53
PAGE 9
(D) 2
22 f
at x = 2, y = 1 ?
2x2y
(B) ln 2
(D) 1
ln 2
It is given that y m + 2yl + y = 0, y (0) = 0, y (1) = 0 . What is y (0.5) ?
(A) 0
(B) 0.37
(C) 0.62
(D) 1.13
YEAR 2007
MCQ 1.54
The minimum value of function y = x2 in the interval [1, 5] is
(A) 0
(B) 1
(C) 25
MCQ 1.55
(D) undefined
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
pairs
MCQ 1.56
ONE MARK
(D) occur in complex conjugate
If ϕ (x, y) and ψ (x, y) are functions with continuous second derivatives, then
ϕ (x, y) + iψ (x, y) can be expressed as an analytic function of x + iψ (i = − 1)
, when
2ϕ
2ϕ
2ψ 2ϕ 2ψ
2ψ 2ϕ 2ψ
(B)
(A)
=− ,
=
=− ,
=
2x
2x 2y 2y
2y
2x 2x 2y
(C)
22ϕ 22ϕ 22 ψ 22 ψ
+
=
+
=1
2x2 2y2 2x2 2y2
(D)
2ϕ 2ϕ 2ψ 2ψ
+
=
+
=0
2x 2y 2x 2y
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PAGE 10
MCQ 1.57
ENGINEERING MATHEMATICS
The partial differential equation
CHAP 1
22 ϕ 22 ϕ 2ϕ 2ϕ
+
+
+
= 0 has
2x2 2y2 2x 2y
(A) degree 1 order 2
(B) degree 1 order 1
(C) degree 2 order 1
(D) degree 2 order 2
YEAR 2007
MCQ 1.58
If y = x + x +
(A) 4 or 1
TWO MARKS
x + x + ......3 , then y ^2 h =
(B) 4 only
(C) 1 only
MCQ 1.59
The area of a triangle formed by the tips of vectors a, b and c is
(A) 1 (a − b) : (a − c)
2
MCQ 1.60
(D) − 2 # x # 2
If F (s) is the Laplace transform of function f (t), then Laplace transform of
#0
MCQ 1.62
(B) 1 (a − b) # (a − c)
2
(C) 1 a # b # c
(D) 1 (a # b) : c
2
2
dy
The solution of
= y2 with initial value y (0) = 1 bounded in the interval
dx
(A) − 3 # x # 3
(B) − 3 # x # 1
(C) x < 1, x > 1
MCQ 1.61
(D) undefined
t
f (τ) dτ is
(A) 1 F (s)
s
(B) 1 F (s) − f (0)
s
(C) sF (s) − f (0)
(D) # F (s) d s
A calculator has accuracy up to 8 digits after decimal place. The value of
2π
#0 sin xdx
when evaluated using the calculator by trapezoidal method with 8 equal
intervals, to 5 significant digits is
(A) 0.00000
(B) 1.0000
(C) 0.00500
MCQ 1.63
(D) 0.00025
Let X and Y be two independent random variables. Which one of the
relations between expectation (E), variance (Var) and covariance (Cov)
given below is FALSE ?
(A) E (XY ) = E (X ) E (Y )
(B) Cοv (X, Y ) = 0
(C) Var (X + Y ) = Var (X ) + Var (Y )
(D) E (X 2 Y 2) = (E (X )) 2 (E (Y )) 2
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 11
2
ex − b1 + x + x l
2
lim
=
3
x"0
x
MCQ 1.64
MCQ 1.65
(A) 0
(B) 1/6
(C) 1/3
(D) 1
2 1
H is
The number of linearly independent eigen vectors of >
0 2
(A) 0
(B) 1
(C) 2
(D) infinite
YEAR 2006
MCQ 1.66
ONE MARK
Match the items in column I and II.
Column I
P. Gauss-Seidel method
Column II
1. Interpolation
Q. Forward Newton-Gauss method 2. Non-linear differential equations
MCQ 1.67
R. Runge-Kutta method
3. Numerical integration
S. Trapezoidal Rule
4. Linear algebraic equations
(A) P-1, Q-4, R-3, S-2
(B) P-1, Q-4, R-2, S-3
(C) P-1. Q-3, R-2, S-4
(D) P-4, Q-1, R-2, S-3
The solution of the differential equation
(A) (1 + x) e+x
(C) (1 − x) e+x
MCQ 1.68
2
2
2
dy
+ 2xy = e−x with y (0) = 1 is
dx
2
(B) (1 + x) e−x
(D) (1 − x) e−x
2
Let x denote a real number. Find out the INCORRECT statement.
(A) S = {x : x > 3} represents the set of all real numbers greater than 3
(B) S = {x : x2 < 0} represents the empty set.
(C) S = {x : x ! A and x ! B} represents the union of set A and set B .
(D) S = {x : a < x < b} represents the set of all real numbers between a and
b, where a and b are real numbers.
MCQ 1.69
A box contains 20 defective items and 80 non-defective items. If two items
are selected at random without replacement, what will be the probability
that both items are defective ?
(B) 1
(A) 1
5
25
(C) 20
99
(D) 19
495
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PAGE 12
ENGINEERING MATHEMATICS
CHAP 1
YEAR 2006
MCQ 1.70
TWO MARKS
3 2
H are 5 and 1. What are the eigen
Eigen values of a matrix S = >
2 3
values of the matrix S 2 = SS ?
(A) 1 and 25
(B) 6 and 4
(C) 5 and 1
MCQ 1.71
(D) 2 and 10
Equation of the line normal to function f (x) = (x − 8) 2/3 + 1 at P (0, 5) is
(A) y = 3x − 5
(B) y = 3x + 5
(C) 3y = x + 15
MCQ 1.72
Assuming i =
(D) 3y = x − 15
− 1 and t is a real number,
3 + i1
2
2
(A)
(B)
MCQ 1.74
π/3
eit dt is
3 − i1
2
2
(D) 1 + i c1 − 3 m
2
2
(C) 1 + i 3
2
2
MCQ 1.73
#0
2
If f (x) = 2x2 − 7x + 3 , then lim f (x) will be
x"3
5x − 12x − 9
(A) − 1/3
(B) 5/18
(C) 0
(D) 2/5
Match the items in column I and II.
Column I
Column II
P.
Singular matrix
1.
Determinant is not defined
Q.
Non-square matrix
2.
Determinant is always one
R.
Real symmetric
3.
Determinant is zero
S.
Orthogonal matrix
4.
Eigenvalues are always real
5.
Eigenvalues are not defined
(A) P-3, Q-1, R-4, S-2
(B) P-2, Q-3, R-4, S-1
(C) P-3, Q-2, R-5, S-4
(D) P-3, Q-4, R-2, S-1
MCQ 1.75
For
d 2y
dy
2x
2 + 4 dx + 3y = 3e , the particular integral is
dx
(A) 1 e2x
15
(B) 1 e2x
5
(C) 3e2x
(D) C1 e−x + C2 e−3x
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CHAP 1
MCQ 1.76
MCQ 1.77
ENGINEERING MATHEMATICS
PAGE 13
Multiplication of matrices E and F is G . matrices E and G are
R1 0 0V
Rcos θ − sin θ 0V
S
W
S
W
E = S sin θ
cos θ 0W and G = S0 1 0W
SS0 0 1WW
SS 0
0 1WW
T
X
T
X
What is the matrix F ?
Rcos θ − sin θ 0V
R cos θ cos θ 0V
W
W
S
S
(B) S− cos θ sin θ 0W
(A) S sin θ
cos θ 0W
SS 0
SS
0 1WW
0
0 1WW
TR
VX
RT cos θ sin θ 0XV
W
S
S sin θ − cos θ 0W
(C) S− sin θ cos θ 0W
(D) Scos θ
sin θ 0W
W
SS
S
S 0
0
0 1W
0 1WW
X
X
T
T
Consider the continuous random variable with probability density function
f (t) = 1 + t for − 1 # t # 0
= 1 − t for 0 # t # 1
The standard deviation of the random variable is
(B) 1
(A) 1
3
6
(C) 1
3
(D) 1
6
YEAR 2005
MCQ 1.78
ONE MARK
Stokes theorem connects
(A) a line integral and a surface integral
(B) a surface integral and a volume integral
(C) a line integral and a volume integral
(D) gradient of a function and its surface integral
MCQ 1.79
A lot has 10% defective items. Ten items are chosen randomly from this lot.
The probability that exactly 2 of the chosen items are defective is
(A) 0.0036
(B) 0.1937
(C) 0.2234
(D) 0.3874
MCQ 1.80
#−a (sin6 x + sin7 x) dx
a
(A) 2 # sin6 x dx
0
a
is equal to
(C) 2 # (sin6 x + sin7 x) dx
a
(B) 2 # sin7 x dx
a
0
(D) zero
0
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MCQ 1.81
ENGINEERING MATHEMATICS
A is a 3 # 4 real matrix and Ax = b is an inconsistent system of equations.
The highest possible rank of A is
(A) 1
(B) 2
(C) 3
MCQ 1.82
CHAP 1
(D) 4
Changing the order of the integration in the double integral I =
leads to I =
#r #p
s
q
MCQ 1.84
f (x, y) dxdy What is q ?
(A) 4y
(B) 16 y2
(C) x
(D) 8
TWO MARKS
(D) 100 units
By a change of variable x (u, v) = uv, y (u, v) = v/u is double integral, the
integrand f (x, y) changes to f (uv, v/u) φ (u, v). Then, φ (u, v) is
(A) 2v/u
(B) 2uv
(C) v2
MCQ 1.86
f (x, y) dydx
Which one of the following is an eigen vector of the matrix
R
V
S5 0 0 0W
S0 5 0 0W
S0 0 2 1W
S
W
S0 0 3 1W
R V
R V
S 1W
S0W
T
X
S− 2W
S0W
(A) S W
(B) S W
S 0W
S1W
S 0W
S0W
TR VX
TR X V
1
S W
S 1W
S 0W
S− 1W
(C) S W
(D) S W
S 0W
S 2W
S− 2W
S 1W
T X
T X
With a 1 unit change in b, what is the change in x in the solution of the
system of equations x + y = 2, 1.01x + 0.99y = b ?
(A) zero
(B) 2 units
(C) 50 units
MCQ 1.85
2
4
YEAR 2005
MCQ 1.83
8
#0 #x
(D) 1
The right circular cone of largest volume that can be enclosed by a sphere
of 1 m radius has a height of
(A) 1/3 m
(B) 2/3 m
(C) 2 2 m
3
(D) 4/3 m
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CHAP 1
MCQ 1.87
ENGINEERING MATHEMATICS
2 ln (x)
dy
and y (1) = 0 , then what is y (e) ?
+ 2xy =
x
dx
(A) e
(B) 1
If x2
(C) 1/e
MCQ 1.88
PAGE 15
(D) 1/e2
The line integral # V :dr of the vector V :(r) = 2xyzi + x2 zj + x2 yk from the
origin to the point P (1, 1, 1)
(A) is 1
(B) is zero
(C) is – 1
(D) cannot be determined without specifying the path
MCQ 1.89
Starting from x 0 = 1, one step of Newton-Raphson method in solving the
equation x3 + 3x − 7 = 0 gives the next value (x1) as
(B) x1 = 1.406
(A) x1 = 0.5
(C) x1 = 1.5
MCQ 1.90
(D) x1 = 2
A single die is thrown twice. What is the probability that the sum is neither
8 nor 9 ?
(A) 1/9
(B) 5/36
(C) 1/4
(D) 3/4
• Common Data For Q. 91 and 92
MCQ 1.91
The complete solution of the ordinary differential equation
d 2y
dy
+ p + qy = 0 is y = c1 e−x + c2 e−3x
dx
dx2
Then p and q are
(B) p = 3, q = 4
(A) p = 3, q = 3
(C) p = 4, q = 3
MCQ 1.92
(D) p = 4, q = 4
Which of the following is a solution of the differential equation
d 2y
dy
2 + p dx + (q + 1) y = 0
dx
−3x
(A) e
(B) xe−x
(C) xe−2x
(D) x2 e−2x
YEAR 2004
MCQ 1.93
If x = a (θ + sin θ) and y = a (1 − cos θ), then
ONE MARK
dy
will be equal to
dx
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MCQ 1.94
MCQ 1.95
ENGINEERING MATHEMATICS
CHAP 1
(A) sin b θ l
2
(B) cos b θ l
2
(C) tan b θ l
2
(D) cot b θ l
2
The angle between two unit-magnitude coplanar vectors P (0.866, 0.500, 0)
and Q (0.259, 0.966, 0) will be
(A) 0c
(B) 30c
(C) 45c
(D) 60c
R1
S
The sum of the eigen values of the matrix given below is S1
SS3
T
(A) 5
(B) 7
(C) 9
TWO MARKS
From a pack of regular playing cards, two cards are drawn at random.
What is the probability that both cards will be Kings, if first card in NOT
replaced ?
(B) 1
(A) 1
26
52
(C) 1
169
MCQ 1.97
(D) 1
221
0, for t < a
A delayed unit step function is defined as U (t − a) = *
Its Laplace
1
,
for
t
$
a
transform is
−as
(B) e
(A) ae−as
s
as
(C) e
s
MCQ 1.98
3VW
1W
1WW
X
(D) 18
YEAR 2004
MCQ 1.96
2
5
1
as
(D) e
a
The values of a function f (x) are tabulated below
x
f (x)
0
1
1
2
2
1
3
10
Using Newton’s forward difference formula, the cubic polynomial that can
be fitted to the above data, is
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 17
(A) 2x3 + 7x2 − 6x + 2
(B) 2x3 − 7x2 + 6x − 2
(C) x3 − 7x2 − 6x2 + 1
(D) 2x3 − 7x2 + 6x + 1
The volume of an object expressed in spherical co-ordinates is given by
MCQ 1.99
V=
MCQ 1.100
2π
#0 #0
π/3
#0
1
r2 sin φdrdφdθ
The value of the integral is
(A) π
3
(B) π
6
(C) 2π
3
(D) π
4
For which value of x will the matrix given below become singular ?
R 8 x 0V
S
W
= S 4 0 2W
SS12 6 0WW
T
X
(A) 4
(B) 6
(C) 8
(D) 12
YEAR 2003
MCQ 1.101
MCQ 1.102
ONE MARK
2
lim sin x is equal to
x
x"0
(A) 0
(B) 3
(C) 1
(D) − 1
The accuracy of Simpson’s rule quadrature for a step size h is
(A) O (h2)
(B) O (h3)
(C) O (h 4)
MCQ 1.103
(D) O (h5)
4 1
For the matrix >
the eigen values are
1 4H
(B) − 3 and − 5
(A) 3 and − 3
(C) 3 and 5
(D) 5 and 0
YEAR 2003
MCQ 1.104
TWO MARKS
Consider the system of simultaneous equations
x + 2y + z = 6
2x + y + 2z = 6
x+y+z = 5
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PAGE 18
ENGINEERING MATHEMATICS
CHAP 1
This system has
(A) unique solution
(B) infinite number of solutions
(C) no solution
(D) exactly two solutions
MCQ 1.105
The area enclosed between the parabola y = x2 and the straight line y = x is
(A) 1/8
(B) 1/6
(C) 1/3
MCQ 1.106
(D) 1/2
The solution of the differential equation
(A) y =
1
x+c
dy
+ y2 = 0 is
dx
3
(B) y = − x + c
3
(C) cex
linear
MCQ 1.107
(D) unsolvable as equation is non-
The vector field is F = xi − yj (where i and j are unit vector) is
(A) divergence free, but not irrotational
(B) irrotational, but not divergence free
(C) divergence free and irrotational
(D) neither divergence free nor irrational
MCQ 1.108
Laplace transform of the function sin ωt is
(B) 2 ω 2
(A) 2 s 2
s +ω
s +ω
(C) 2 s 2
(D) 2 ω 2
s −ω
s −ω
MCQ 1.109
A box contains 5 black and 5 red balls. Two balls are randomly picked one
after another form the box, without replacement. The probability for balls
being red is
(A) 1/90
(B) 1/2
(C) 19/90
(D) 2/9
YEAR 2002
MCQ 1.110
ONE MARK
Two dice are thrown. What is the probability that the sum of the numbers
on the two dice is eight?
(B) 5
(A) 5
36
18
(C) 1
(D) 1
3
4
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CHAP 1
ENGINEERING MATHEMATICS
MCQ 1.111
Which of the following functions is not differentiable in the domain [− 1, 1] ?
(B) f (x) = x − 1
(A) f (x) = x2
(C) f (x) = 2
MCQ 1.112
PAGE 19
(D) f (x) = maximum ( x, − x )
A regression model is used to express a variable Y as a function of another
variable X .This implies that
(A) there is a causal relationship between Y and X
(B) a value of X may be used to estimate a value of Y
(C) values of X exactly determine values of Y
(D) there is no causal relationship between Y and X
YEAR 2002
MCQ 1.113
TWO MARKS
The following set of equations has
3x + 2y + z = 4
x−y+z = 2
MCQ 1.114
− 2x + 2z = 5
(A) no solution
(B) a unique solution
(C) multiple solutions
(D) an inconsistency
The function f (x, y) = 2x2 + 2xy − y3 has
(A) only one stationary point at (0, 0)
(B) two stationary points at (0, 0) and b 1 , − 1 l
6 3
(C) two stationary points at (0, 0) and (1, − 1)
(D) no stationary point
MCQ 1.115
Manish has to travel from A to D changing buses at stops B and C enroute.
The maximum waiting time at either stop can be 8 min each but any time
of waiting up to 8 min is equally, likely at both places. He can afford up
to 13 min of total waiting time if he is to arrive at D on time. What is the
probability that Manish will arrive late at D ?
(B) 13
(A) 8
13
64
(C) 119
128
(D) 9
128
YEAR 2001
MCQ 1.116
ONE MARK
The divergence of vector i = xi + yj + zk is
(A) i + j + k
(B) 3
(C) 0
(D) 1
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PAGE 20
ENGINEERING MATHEMATICS
MCQ 1.117
MCQ 1.118
Consider the system of equations given below
x+y = 2
2x + 2y = 5
This system has
(A) one solution
(B) no solution
(C) infinite solutions
(D) four solutions
What is the derivative of f (x) = x at x = 0 ?
(A) 1
(B) − 1
(C) 0
MCQ 1.119
CHAP 1
(D) Does not exist
The Gauss divergence theorem relates certain
(A) surface integrals to volume integrals
(B) surface integrals to line integrals
(C) vector quantities to other vector quantities
(D) line integrals to volume integrals
YEAR 2001
TWO MARKS
3
MCQ 1.120
MCQ 1.121
The minimum point of the function f (x) = b x l − x is at
3
(A) x = 1
(B) x =− 1
(C) x = 0
(D) x = 1
3
The rank of a 3 # 3 matrix C (= AB), found by multiplying a non-zero
column matrix A of size 3 # 1 and a non-zero row matrix B of size 1 # 3 , is
(A) 0
(B) 1
(C) 2
MCQ 1.122
(D) 3
An unbiased coin is tossed three times. The probability that the head turns
up in exactly two cases is
(B) 1
(A) 1
9
8
(C) 2
3
(D) 3
8
**********
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 21
SOLUTION
SOL 1.1
Option (A) is correct.
For
y = x straight line and
y = x 2 parabola, curve is as given. The shaded
region is the area, which is bounded by the both curves (common area).
We solve given equation as follows to gett the intersection points :
In y = x2 putting y = x we have x = x2 or
x2 − x = 0 & x (x − 1) = 0 & x = 0, 1
Then from y = x , for
x = 0 & y = 0 and x = 1 & y = 1
Curve y = x2 and y = x intersects at point (0, 0) and (1, 1)
So, the area bounded by both the curves is
y = x2
x=1
A=
# # dydx
x=0
3
y=x
x=1
=
y = x2
# dx # dy
x=0
y=x
x=1
=
# dx6y @
x2
x
x=0
2 1
= :x − x D = 1 − 1 =− 1 = 1 unit2
3
2 0
3 2
6
6
SOL 1.2
x=1
# (x
=
Area is never negative
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− x) dx
x=0
Option (C) is correct.
Given f (x) = x (in − 1 # x # 1)
For this function the plot is as given below.
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PAGE 22
ENGINEERING MATHEMATICS
CHAP 1
At x = 0 , function is continuous but not differentiable because.
For
x > 0 and x < 0
f l (x) = 1 and f l (x) =− 1
lim f l(x) = 1 and lim f l(x) =− 1
x " 0+
x " 0−
R.H.S lim = 1 and L.H.S lim =− 1
Therefore it is not differentiable.
SOL 1.3
Option (B) is correct.
Let
y = lim
x"0
(1 − cos x)
x2
It forms : 0 D condition. Hence by L-Hospital rule
0
d
y = lim dx
x"0
(1 − cos x)
= lim sin x
2
d
x " 0 2x
dx (x )
Still these gives : 0 D condition, so again applying L-Hospital rule
0
y = lim
x"0
SOL 1.4
(sin x)
= lim cos x = cos 0 = 1
2
2
2
x"0
2 # dxd (x)
Option (D) is correct.
We have
f (x) = x3 + 1
f l(x) = 3x2 + 0
Putting f l(x) equal to zero
f l(x) = 0
Now
At x = 0,
SOL 1.5
d
dx
3x2 + 0 = 0 & x = 0
f ll(x) = 6x
fll(0) = 6 # 0 = 0
Hence x = 0 is the point of inflection.
Option (A) is correct.
Given :
x2 + y2 + z2 = 1
This is a equation of sphere with radius r = 1
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 23
The unit normal vector at point c 1 , 1 , 0 m is OA
2 2
Hence
OA = c 1 − 0 m i + c 1 − 0 m j + (0 − 0) k = 1 i + 1 j
2
2
2
2
SOL 1.6
Option (D) is correct.
First using the partial fraction :
A (s + 1) + Bs
1
=A+ B =
s (s + 1) s s + 1
s (s + 1)
(A + B) s
1
A
+
=
s (s + 1)
s (s + 1)
s (s + 1)
Comparing the coefficients both the sides,
(A + B) = 0 and A = 1, B =− 1
1
So
=1− 1
s (s + 1) s s + 1
F (s) =
F (t) = L−1 [F (s)]
= L−1 ; 1 E = L−1 :1 − 1 D = L−1 :1D − L−1 : 1 D
s
s+1
s s+1
s (s + 1)
= 1 − e−t
SOL 1.7
Option (B) is correct.
5
A =>
1
For finding eigen values, we write
Given
3
3H
the characteristic equation as
A − λI = 0
5−λ
3
=0
1 3−λ
(5 − λ) (3 − λ) − 3 = 0
λ2 − 8λ + 12 = 0 & λ = 2, 6
Now from characteristic equation for eigen vector.
&
For λ = 2
&
6A − λI @"x , = 60@
5−2
3 X1
0
=> H
>
H
>
H
1 3 − 2 X2
0
3 3 X1
0
>1 1H>X H = >0H
2
X1 + X 2 = 0
So
& X1 =− X2
1
eigen vector = * 4
−1
Magnitude of eigen vector =
(1) 2 + (1) 2 =
2
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PAGE 24
ENGINEERING MATHEMATICS
CHAP 1
R 1 V
W
S
2W
S
Normalized eigen vector = S
−1W
W
S
2
X
T
SOL 1.8
Option (D) is correct.
Given :
No. of Red balls = 4
No. of Black ball = 6
3 balls are selected randomly one after another, without replacement.
1 red and 2 black balls are will be selected as following
Manners
Probability for these sequence
R B B
4
6
5=1
10 # 9 # 8 6
B R B
6
4
5=1
10 # 9 # 8 6
B B R
6
5
4=1
10 # 9 # 8 6
Hence Total probability of selecting 1 red and 2 black ball is
P =1+1+1 = 3=1
6 6 6 6 2
SOL 1.9
Option (A) is correct.
d2y
dy
We have
x2 2 + x − 4y = 0
dx
dx
z
Let x = e then
z = log x
dz = 1
x
dx
dy
dy
dy
So, we get
= b lb dz l = 1
x dz
dx
dz dx
dy
x
= Dy
dx
Again
...(1)
where d = D
dz
d 2y
d dy
d 1 dy
− 1 dy + 1 d dy dz
b l
2 = dx b dx l = dx b x dz l =
x2 dz x dz dz dx
dx
dy
d 2y
d 2 y dy
= −21 + 1 2 dz = 12 c 2 − m
dz
x dz x dz dx
x dz
x2 d 2 y
= (D2 − D) y = D (D − 1) y
dx2
Now substitute in equation (i)
[D (D − 1) + D − 4] y = 0
(D2 − 4) y = 0 & D = ! 2
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 25
y = C1 x2 + C2 x−2
y (0) = 0 , equation (ii) gives
0 = C1 # 0 + C 2
C2 = 0
And from y (1) = 1, equation (ii) gives
1 = C1 + C 2
So the required solution is
From the given limits
...(ii)
C1 = 1
Substitute C1 & C2 in equation (ii), the required solution be
y = x2
SOL 1.10
Option (C) is correct.
For given equation matrix form is as follows
R4V
R1 2 1V
W
S W
S
A = S2 1 2W, B = S5W
SS1WW
SS1 − 1 1WW
X
T X
The augmented matrix Tis
R1 2 1 : 4V
W
S
R2 " R2 − 2R1, R 3 " R 3 − R1
8A : BB = S2 1 2 : 5W
SS1 − 1 1 : 1WW
TR1 2 1 : X4V
W
S
R 3 " R3 − R2
+ S0 − 3 0 : − 3W
SS0 − 3 0 : − 3WW
RT1 2 1 : 4VX
W
S
+ S0 − 3 0 : − 3W
R2 " R2 / − 3
SS0 0 0 : 0WW
RT1 2 1 : 4V X
S
W
+ S0 1 0 : 1W
SS0 0 0 : 0WW
T
X
This gives rank of A, ρ (A) = 2 and Rank of 8A : BB = ρ 8A : BB = 2
Which is less than the number of unknowns (3)
ρ 6A@ = ρ 8A : BB = 2 < 3
Hence, this gives infinite No. of solutions.
SOL 1.11
Option (B) is correct.
3
5
7
sin θ = θ − θ + θ − θ + ......
3
5
7
SOL 1.12
Option (D) is correct.
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PAGE 26
ENGINEERING MATHEMATICS
CHAP 1
y = lim sin θ
θ"0 θ
Let
d
(sin θ)
= lim dθ d
= lim cos θ
1
θ"0
θ"0
(
θ
)
dθ
= cos 0 =1
1
SOL 1.13
Applying L-Hospital rule
Option (C) is correct
Let a square matrix
x y
A =>
y xH
We know that the characteristic equation for the eigen values is given by
A − λI = 0
x−λ
y
=0
y x−λ
(x − λ) 2 − y2 = 0
(x − λ) 2 = y2
x − λ =! y & λ = x ! y
So, eigen values are real if matrix is real and symmetric.
SOL 1.14
Option (A) is correct.
Let, z1 = (1 + i), z2 = (2 − 5i)
z = z1 # z2 = (1 + i) (2 − 5i)
= 2 − 5i + 2i − 5i2 = 2 − 3i + 5 = 7 − 3i
SOL 1.15
Option (D) is correct.
For a function, whose limits bounded between − a to a and a is a positive
real number. The solution is given by
#−a f (x) dx
a
SOL 1.16
i 2 =− 1
2 # f (x) dx ;
f (x) is even
0
f (x) is odd
a
=*
0
;
Option (C) is correct.
1 dx
x
From this function we get a = 1, b = 3 and n = 3 − 1 = 2
Let,
So,
f (x) =
#1
3
h =b−a = 3−1 = 1
n
2
We make the table from the given function y = f (x) = 1 as follows :
x
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CHAP 1
ENGINEERING MATHEMATICS
x
f (x) = y = 1
x
x=1
y1 = 1 = 1
1
x=2
y2 = 1 = 0.5
2
x=3
y 3 = 1 = 0.333
3
PAGE 27
Applying the Simpson’s 1/3 rd formula
3
#1 x1 dx = h3 6(y1 + y3) + 4y2@ = 13 6(1 + 0.333) + 4 # 0.5@
= 1 [1.333 + 2] = 3.333 = 1.111
3
3
SOL 1.17
SOL 1.18
Option (D) is correct.
dy
Given :
= (1 + y2) x
dx
dy
= xdx
(1 + y2)
Integrating both the sides, we get
dy
= # xdx
# 1+
y2
2
tan−1 y = x + c &
2
2
y = tan b x + c l
2
Option (D) is correct.
The probability of getting head p = 1
2
And the probability of getting tail q = 1 − 1 = 1
2 2
The probability of getting at least one head is
5
0
P (x $ 1) = 1 − 5C 0 (p) 5 (q) 0 = 1 − 1 # b 1 l b 1 l
2 2
= 1 − 15 = 31
32
2
SOL 1.19
Option (C) is correct.
Given system of equations are,
2x1 + x2 + x 3 = 0
x2 − x 3 = 0
x1 + x 2 = 0
Adding the equation (i) and (ii) we have
...(i)
...(ii)
...(iii)
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ENGINEERING MATHEMATICS
CHAP 1
2x1 + 2x2 = 0
...(iv)
x1 + x 2 = 0
We see that the equation (iii) and (iv) is same and they will meet at infinite
points. Hence this system of equations have infinite number of solutions.
SOL 1.20
Option (D) is correct.
The volume of a solid generated by revolution about x -axis bounded by the
function f (x) and limits between a to b is given by
V =
Given
Therefore,
SOL 1.21
#a
y =
V =
b
πy2 dx
x and a = 1, b = 2
#1
2
2 2
2
π ( x ) 2 dx = π # xdx = π :x D = π : 4 − 1 D = 3π
2 1
2 2
2
1
Option (B) is correct.
d 3f
f d 2f
=0
+
dη3 2 dη2
Order is determined by the order of the highest derivation present in it. So,
It is third order equation but it is a nonlinear equation because in linear
equation, the product of f with d 2 f/dη2 is not allow.
Therefore, it is a third order non-linear ordinary differential equation.
Given:
SOL 1.22
Option (D) is correct.
Let
I =
#− 33 1 dx
+ x2
= 6tan−1 x @3
= [tan−1 (+ 3) − tan−1 (− 3)]
−3
= π − a− π k = π
2
2
SOL 1.23
tan−1 (− θ) =− tan−1 (θ)
Option (B) is correct.
z = 3 + 4i
1 − 2i
Divide and multiply z by the conjugate of (1 − 2i) to convert it in the form
of a + bi we have
(3 + 4i) (1 + 2i)
z = 3 + 4i # 1 + 2i =
1 − 2i
1 + 2i
(1) 2 − (2i) 2
Let,
2
= 3 + 10i +2 8i = 3 + 10i − 8
1 − (− 4)
1 − 4i
= − 5 + 10i =− 1 + 2i
5
z =
(− 1) 2 + (2) 2 =
5
a + ib =
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a2 + b2
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CHAP 1
SOL 1.24
ENGINEERING MATHEMATICS
PAGE 29
Option (C) is correct.
Z
if x < 2
]2 − 3x
3
]
]
y = f (x) = [0
if x = 2
3
]
]]− (2 − 3x) if x > 2
3
Checking the continuity\of the function.
At x = 2 ,
Lf (x) = lim f b 2 − h l = lim 2 − 3 b 2 − h l
3
3
3
h"0
h"0
and
= lim 2 − 2 + 3h = 0
h"0
Rf (x) = lim f b 2 + h l = lim 3 b 2 + h l − 2
3
3
h"0
h"0
= lim 2 + 3h − 2 = 0
h"0
Since
L lim f (x) = R lim f (x)
h"0
h"0
So, function is continuous 6 x ! R
Now checking the differentiability :
f ^ 23 − h h − f ^ 23 h
2 − 3 ^ 23 − h h − 0
Lf l (x) = lim
= lim
h"0
h"0
−h
−h
= lim 2 − 2 + 3h = lim 3h =− 3
h"0
h"0 −h
−h
and
Since
SOL 1.25
f ^ 23 + h h − f ^ 23 h
h"0
h
Rf l (x) = lim
3 ^ 23 + h h − 2 − 0
= lim
= lim 2 + 3h − 2 = 3
h"0
h"0
h
h
Lf lb 2 l ! Rf lb 2 l, f (x) is not differentiable at x = 2 .
3
3
3
Option (A) is correct.
2 2
H
A =>
1 3
And λ1 and λ2 are the eigen values of the matrix A.
The characteristic equation is written as
Let,
A − λI = 0
2 2
1 0
>
H − λ>
H =0
1 3
0 1
2−λ
2
=0
1 3−λ
...(i)
(2 − λ) (3 − λ) − 2 = 0
λ2 − 5λ + 4 = 0 & λ = 1 & 4
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PAGE 30
ENGINEERING MATHEMATICS
CHAP 1
Putting λ = 1 in equation (i),
0
2−1
2 x1
=> H
>
H
>
H
0
1 3 − 1 x2
x1
where > H is eigen vector
x2
2 x1
0
=> H
H
>
H
2 x2
0
1
>1
x1 + 2x2 = 0 or x1 + 2x2 = 0
Let
x2 = K
Then
x1 + 2K = 0 & x1 =− 2K
So, the eigen vector is
− 2K
−2
> K H or > 1H
2
Since option A> H is in the same ratio of x1 and x2 . Therefore option (A)
−1
is an eigen vector.
SOL 1.26
Option (A) is correct.
f (t) is the inverse Laplace
So,
f (t) = L − 1 ; 2 1
s (s + 1)E
1
= A + B2 + C
s
s+1
s
s2 (s + 1)
As (1 + s) + B (s + 1) + Cs2
s2 (s + 1)
s2 (A + C) + s (A + B) + B
=
s2 (s + 1)
Compare the coefficients of s2, s and constant terms and we get
A + C = 0 ; A + B = 0 and B = 1
Solving above equation, we get A =− 1, B = 1 and C = 1
Thus
f (t) = L − 1 :− 1 + 12 + 1 D
s s
s+1
=
=− 1 + t + e−t = t − 1 + e−t
SOL 1.27
L − 1 : 1 D = e−at
s+a
Option (C) is correct.
The box contains :
Number of washers = 2
Number of nuts = 3
Number of bolts = 4
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 31
Total objects = 2 + 3 + 4 = 9
First two washers are drawn from the box which contain 9 items. So the
probability of drawing 2 washers is,
2
7!2!
n
P1 = 9C2 == 1 =
Cn = 1
= 2 = 1
C2
9 # 8 # 7! 9 # 8 36
9!
7! 2!
After this box contains only 7 objects and then 3 nuts drawn from it. So the
probability of drawing 3 nuts from the remaining objects is,
3
4! 3!
P2 = 7C 3 = 1 =
= 1
C3
7!
7 # 6 # 5 # 4! 35
4! 3!
After this box contain only 4 objects, probability of drawing 4 bolts from
the box,
4
P3 = 4C 4 = 1 = 1
C4 1
Therefore the required probability is,
P = P1 P2 P3 = 1 # 1 # 1 = 1
36
35
1260
SOL 1.28
Option (B) is correct.
Given :
h = 60c − 0 = 60c
h = 60 # π = π = 1.047 radians
180
3
From the table, we have
y 0 = 0 , y1 = 1066 , y2 =− 323 , y 3 = 0 , y 4 = 323 , y5 =− 355 and y6 = 0
From the Simpson’s 1/3rd rule the flywheel Energy is,
E = h 6(y 0 + y6) + 4 (y1 + y 3 + y5) + 2 (y2 + y 4)@
3
Substitute the values, we get
E = 1.047 6(0 + 0) + 4 (1066 + 0 − 355) + 2 (− 323 + 323)@
3
= 1.047 64 # 711 + 2 (0)@ = 993 Nm rad (Joules/cycle)
3
SOL 1.29
Option (A) is correct.
Given :
And
3
5
M =>
x
4
5
3
5
H
[M]T = [M] −1
Τ
−1
We know that when 6A@ = 6A@ then it is called orthogonal matrix.
6M @T = I
6M @
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PAGE 32
ENGINEERING MATHEMATICS
CHAP 1
6M @T 6M @ = I
Substitute the values of M and M T , we get
>
3
5
4
5
x
3
5
H.>x
3
5
4
5
3
5
1 0
H = >0 1H
3
4
3 V
b 5 # 5 l + 5 xW
W = >1 0H
4
4
3
3 W
0 1
b 5 # 5 l + b 5 # 5 lW
X
2
12
3
9
1 0
25 + 5 x
25 + x
=>
>12 + 3 x
H
0 1H
1
25
5
Comparing both sides a12 element,
12 + 3 x = 0 " x =− 12
5 =− 4
25 5
25 # 3
5
R 3
S b # 3 l + x2
5
S 5
S 4 # 3 + 3x
Sb 5
5l 5
T
SOL 1.30
Option (C) is correct.
Let,
V = 3xzi + 2xyj − yz2 k
We know divergence vector field of V is given by (4: V)
So,
4: V = c 2 i + 2 j + 2 k m : ^3xzi + 2xyj − yz2 k h
2x
2y
2z
4: V = 3z + 2x − 2yz
At point P (1, 1, 1)
(4: V) P (1, 1, 1) = 3 # 1 + 2 # 1 − 2 # 1 # 1 = 3
SOL 1.31
Option (C) is correct.
f (s) = L − 1 ; 2 1 E
s +s
1
First, take the function 2
and break it by the partial fraction,
s +s
Solve by
1 =
1
1
1
= −
* 1 =A+ B 4
s (s + 1) s (s + 1)
s2 + s
(s + 1) s s + 1
So,
L − 1 c 2 1 m = L − 1 ;1 − 1 E = L − 1 :1D − L − 1 : 1 D = 1 − e−t
s
s+1
s (s + 1)
s +s
Let
SOL 1.32
Option (D) is correct.
Total number of cases = 23 = 8
& Possible cases when coins are tossed simultaneously.
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CHAP 1
ENGINEERING MATHEMATICS
H
H
H
T
H
T
T
T
H
H
T
H
T
H
T
T
PAGE 33
H
T
H
H
T
T
H
T
From these cases we can see that out of total 8 cases 7 cases contain at least
one head. So, the probability of come at least one head is = 7
8
SOL 1.33
Option (C) is correct.
Given :
z = x + iy is a analytic function
f (z) = u (x, y) + iv (x, y)
u = xy
Analytic function satisfies the Cauchy-Riemann equation.
2u = 2v and 2u =−2v
2x
2y
2y
2x
So from equation (i),
2u = y
2x
&
..(i)
2v = y
2y
2u = x &
2v =− x
2y
2x
Let v (x, y) be the conjugate function of u (x, y)
dv = 2v dx + 2v dy = (− x) dx + (y) dy
2x
2y
Integrating both the sides,
# dv
=− # xdx +
# ydy
2
y2
v =− x + + k = 1 (y2 − x2) + k
2
2
2
SOL 1.34
Option (A) is correct.
dy
Given
x + y = x4
dx
dy
+ 1 y = x3
dx b x l
...(i)
dy
It is a single order differential equation. Compare this with
+ Py = Q
dx
and we get
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PAGE 34
ENGINEERING MATHEMATICS
P =1
x
CHAP 1
Q = x3
Its solution will be
y (I.F.) =
# Q (I.F.) dx + C
I.F. = e # Pdx = e
Complete solution is given by,
yx =
# x1 dx
= e log x = x
# x3 # xdx + C
e
=
# x4 dx + C
5
= x +C
5
...(ii)
and y (1) = 6 at x = 1 & y = 6 From equation (ii),
5
5
6 1 = 1+C & C = 6−1 = 1
5
5#
5 5
Then, from equation (ii), we get
5
4
yx = x + 1 & y = x + 1
5
5 x
SOL 1.35
Option (B) is correct.
The equation of circle with unit radius and centre at origin is given by,
x2 + y2 = 1
Finding the integration of (x + y) 2 on path AB traversed in counter-clockwise
sense So using the polar form
Let: x = cos θ , y = sin θ , and r = 1
So put the value of x and y and limits in first quadrant between 0 to π/2 .
Hence,
I =
#0
π/2
=
#0
π/2
=
#0
π/2
(cos θ + sin θ) 2 dθ
(cos2 θ + sin2 θ + 2 sin θ cos θ) dθ
(1 + sin 2θ) dθ
Integrating above equation, we get
π/2
= :θ − cos 2θ D = ;a π − cos π k − b 0 − cos 0 lE
2 0
2
2
2
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 35
= b π + 1 l − b− 1 l = π + 1
2 2
2
2
SOL 1.36
Option (A) is correct.
The given equation of surface is
...(i)
z2 = 1 + xy
Let P (x, y, z) be the nearest point on the surface (i), then distance from the
origin is
d =
(x − 0) 2 + (y − 0) 2 + (z − 0) 2
d 2 = x2 + y2 + z2
z2 = d 2 − x2 − y2
From equation (i) and (ii), we get
...(ii)
d 2 − x2 − y2 = 1 + xy
d 2 = x2 + y2 + xy + 1
Let
...(iii)
f (x, y) = d 2 = x2 + y2 + xy + 1
The f (x, y) be the maximum or minimum according to d 2 maximum or
minimum.
Differentiating equation (iii) w.r.t x and y respectively, we get
2f
2f
= 2y + x
= 2x + y or
2x
2y
2f
2f
Applying maxima minima principle and putting
and
equal to zero,
2x
2y
2f
2f
= 2x + y = 0 or
= 2y + x = 0
2x
2y
Solving these equations, we get x = 0 , y = 0
So, x = y = 0 is only one stationary point.
22 f
Now
p = 2 =2
2x
22 f
q =
=1
2x2y
22 f
=2
2y2
or
pr − q2 = 4 − 1 = 3 > 0 and r is positive.
So,
f (x, y) = d 2 is minimum at (0, 0).
Hence minimum value of d 2 at (0, 0).
r =
d 2 = x2 + y2 + xy + 1 = 1
d = 1 or f (x, y) = 1
So, the nearest point is
&
z2 = 1 + xy = 1 + 0
z =! 1
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PAGE 36
SOL 1.37
ENGINEERING MATHEMATICS
CHAP 1
Option (A) is correct.
Given : y2 = 4x and x2 = 4y draw the curves from the given equations,
The shaded area shows the common area. Now finding the intersection
points of the curves.
y2 = 4x = 4 4y = 8 y
Squaring both sides
x=
4y From second curve
y 4 = 8 # 8 # y & y (y3 − 64) = 0
y =4 & 0
Similarly put y = 0 in curve x2 = 4y
x2 = 4 # 0 = 0 & x = 0
And Put
y =4
x2 = 4 # 4 = 16 x = 4
So,
x = 4, 0
Therefore the intersection points of the curves are (0, 0) and (4, 4).
So the enclosed area is given by
A=
#x
x2
(y1 − y2) dx
1
Put y1 and y2 from the equation of curves y2 = 4x and x2 = 4y
2
4
A = # b 4x − x l dx
4
0
4
x2
1
b 2 x − 4 l dx = 2 # x dx − 4
0
Integrating the equation, we get
=
#0
4
#0
4
x2 dx
3 4
4
A = 2 :2 x3/2D − 1 :x D
3
4 3 0
0
3
= 4 # 43/2 − 1 # 4 = 4 # 8 − 16 = 16
3
3
3
3
3
4
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CHAP 1
SOL 1.38
ENGINEERING MATHEMATICS
PAGE 37
Option (A) is correct.
The cumulative distribution function
Z0,
x#a
]
]x − a
,
a<x<b
f (x) = [
b−a
]]
0,
x$b
\
and density function
1 , a#x#b
f (x) = *b − a
0,
a > x, x > b
Mean
E (x) =
/ xf (x) = a +2 b
b
x=a
Variance = x2 f (x) − x 2 = x2 f (x) − 6xf (x)@2
Substitute the value of f (x)
Variance =
b
/
x=a
==
2
b
x2
/
1 dx −
1
) x b − a dx 3
b−a
x=a
b
b
2
x3
− >) x
3
G
3 (b − a) a
2 (b − a) aH
2
3
3
(b2 − a2) 2
= b −a −
3 (b − a) 4 (b − a) 2
=
(b − a) (b2 + ab + a2) (b + a) 2 (b − a) 2
−
3 (b − a)
4 (b − a) 2
=
4 (b2 + ab + a2) + 3 (a + b) 2
(b − a) 2
=
12
12
Standard deviation =
Given : b = 1, a = 0
Variance =
So, standard deviation = 1 − 0 =
12
SOL 1.39
(b − a) 2
(b − a)
=
12
12
1
12
Option (C) is correct.
Taylor’s series expansion of f (x) is given by,
(x − a)
(x − a) 2
(x − a) 3
f l (a) +
f m (a) +
f lll (a) + ....
1
2
3
f mm (a)
Then from this expansion the coefficient of (x − a) 4 is
4
Given
a =2
f (x) = ex
f l (x) = ex
f m (x) = ex
f (x) = f (a) +
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PAGE 38
ENGINEERING MATHEMATICS
CHAP 1
f n (x) = ex
f mm (x) = ex
2
Hence, for a = 2 the coefficient of (x − a) 4 is e
4
SOL 1.40
Option (D) is correct.
Given :
xp + 3x = 0 and x (0) = 1
D= d
dt
(D2 + 3) x = 0
The auxiliary Equation is written as
m2 + 3 = 0
m =! 3 i = 0 !
Here the roots are imaginary
m1 = 0 and m2 =
Solution is given by
3i
3
x = em t (A cos m2 t + B sin m2 t)
1
= e0 [A cos 3 t + B sin 3 t]
= [A cos 3 t + B sin 3 t]
Given :
x (0) = 1 at t = 0 , x = 1
Substituting in equation (i),
...(i)
1 = [A cos 3 (0) + B sin 3 (0)]= A + 0
A =1
Differentiateing equation (i) w.r.t. t ,
xo = 3 [− A sin 3 t + B cos 3 t]
Given
xo(0) = 0 at t = 0 , xo = 0
Substituting in equation (ii), we get
...(ii)
0 = 3 [− A sin 0 + B cos 0]
B =0
Substituting A & B in equation (i)
x = cos 3 t
x (1) = cos 3 = 0.99
SOL 1.41
Option (B) is correct.
Let
1/3
f (x) = lim x − 2
x " 8 (x − 8)
= lim
1
3
x"8
x−2/3
1
0 form
0
Applying L-Hospital rule
Substitute the limits, we get
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 39
f (x) = 1 (8) −2/3 = 1 (23) −2/3 = 1 = 1
3
3
4 # 3 12
SOL 1.42
Option (A) is correct.
In a coin probability of getting Head
p = 1 = No. of Possible cases
2
No. of Total cases
Probability of getting tail
q = 1−1 = 1
2 2
So the probability of getting Heads exactly three times, when coin is tossed
4 times is
3
1
P = 4C 3 (p) 3 (q) 1 = 4C 3 b 1 l b 1 l
2 2
= 4#1 #1 = 1
8
2 4
SOL 1.43
Option (C) is correct.
R1 2 4V
W
S
Let,
A = S3 0 6W
SS1 1 pWW
Let the eigen values of this matrix areT λ1, λ2 &Xλ3
Here one values is given so let λ1 = 3
We know that
Sum of eigen values of matrix= Sum of the diagonal element of matrix A
λ1 + λ2 + λ3 = 1 + 0 + p
λ2 + λ3 = 1 + p − λ1 = 1 + p − 3 = p − 2
SOL 1.44
Option (D) is correct.
We know that the divergence is defined as 4:V
Let
And
So,
V = (x − y) i + (y − x) j + (x + y + z) k
4 = c 2 i + 2 j + 2 km
2x
2y
2z
4:V = c 2 i + 2 j + 2 k m : 6(x − y) i + (y − x) j + (x + y + z) k @
2x
2y
2z
= 2 (x − y) + 2 (y − x) + 2 (x + y + z)
2x
2y
2z
= 1+1+1 = 3
SOL 1.45
Option (A) is correct.
Given :
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PAGE 40
ENGINEERING MATHEMATICS
CHAP 1
The equation of line in intercept form is given by
x +y =1
2 1
x +y =1
a b
x + 2y = 2 & x = 2 (1 − y)
The limit of x is between 0 to x = 2 (1 − y) and y is 0 to 1,
Now
##p xydxdy
=
y=1
2 (1 − y)
y=1
4 (1 − y)
− 0E dy
2
#y = 0 #x = 0
=
#y = 0
=
#y = 0
y=1
y;
xydxdy =
y=1
#y = 0
x2 2 (1 − y) ydy
:2D
0
2
2y (1 + y2 − 2y) dy =
y=1
#y = 0
2 (y + y3 − 2y2) dy
Again Integrating and substituting the limits, we get
##p xydxdy
SOL 1.46
y 2 y 4 2y 3 1
= 2; + −
= 2 :1 + 1 − 2 − 0D
2
3 E0
2 4 3
4
= 2:6 + 3 − 8D = 2 = 1
12
12 6
Option (B) is correct.
Direction derivative of a function f along a vector P is given by
a =grad f : a
a
2f
2f
2f
where
grad f = c
i+
j+
k
2x
2y
2z m
f (x, y, z) = x2 + 2y2 + z , a = 3i − 4j
3i − 4j
a = grad (x2 + 2y2 + z) :
(3) 2 + (− 4) 2
(3i − 4j)
6x − 16y
=
= (2xi + 4yj + k) :
5
25
At point P (1, 1, 2) the direction derivative is
a = 6 # 1 − 16 # 1 =− 10 =− 2
5
5
SOL 1.47
Option (B) is correct.
Given :
2x + 3y = 4
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 41
x+y+z = 4
x + 2y − z = a
It is a set of non-homogenous equation, so the augmented matrix of this
system is
R
V
S2 3 0 : 4W
6A : B@ = S1 1 1 : 4W
SS1 2 − 1 : aWW
TR
X V
4W
S2 3 0 :
S
4W
R 3 " R 3 + R2 , R2 " 2R2 − R1
+ 0 −1 2 :
SS2 3 0 : 4 + aWW
TR
V X
S2 3 0 : 4W
+ S0 − 1 2 : 4W
R 3 " R 3 − R1
SS0 0 0 : aWW
X
T
So, for a unique solution of the system of equations, it must have the condition
ρ [A: B] = ρ [A]
So, when putting a = 0
We get
ρ [A: B] = ρ [A]
SOL 1.48
Option (D) is correct.
Here we check all the four options for unbounded condition.
π/4
(A)
#0 tan xdx = 8log sec x B0π/4 = 9log sec π4 − log sec 0 C
= log 2 − log 1 = log 2
(B)
(C)
Let
#0 3 x2 1+ 1 dx
#0 3xe−x dx
I =
= 6tan−1 x @3
= tan−1 3 − tan−1 (0) = π − 0 = π
0
2
2
3
#0 xe
3 −x
#e
dx = x
0
= 6− xe @ +
−x 3
0
dx −
# :dxd (x) # e
−x
dx D dx
0
3
#e
3
−x
−x
= 6− e−x (x + 1)@3
dx = 6− xe−x − e−x@3
0
0
0
=− [0 − 1] = 1
1
1
(D)
#0 1 −1 x dx =− #0 x −1 1 dx =−6log (x − 1)@10 −6log 0 − log (− 1)@
Both log 0 and log (–1) undefined so it is unbounded.
SOL 1.49
Option (A) is correct.
Let
I=
# f (z) dz
and f (z) = cos z
z
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PAGE 42
ENGINEERING MATHEMATICS
CHAP 1
cos z dz
...(i)
z−0
Given that z = 1 for unit circle. From the Cauchy Integral formula
...(ii)
# zf−(z)a dz = 2πi f (a)
Compare equation (i) and (ii), we can say that,
I =
Then
# cosz z dz = #
a = 0 and f (z) = cos z
Or,
f (a) = f (0) = cos 0 = 1
Now from equation (ii) we get
# zf−(z)0 dz = 2πi # 1 = 2πi
SOL 1.50
a=0
Option (D) is correct.
y = 2 x3/2
3
Given
...(i)
#x
We know that the length of curve is given by
1
x2
)
dy 2
b dx l + 1 3 dx
...(ii)
Differentiate equation(i) w.r.t. x
3
dy
= 2 # 3 x 2 − 1 = x1/2 = x
3
2
dx
dy
Substitute the limit x1 = 0 to x2 = 1 and
in equation (ii), we get
dx
L =
#0
1
_ ( x ) 2 + 1 i dx =
#0
1
x + 1 dx
1
= :2 (x + 1) 3/2D = 1.22
3
0
SOL 1.51
Option (B) is correct.
1 2
λ1 and λ2 is the eigen values of the matrix.
A =>
0 2H
For eigen values characteristic matrix is,
Let
A − λI = 0
1
>0
2
1
− λ>
H
2
0
0
=0
1H
(1 − λ)
2
=0
0 (2 − λ)
...(i)
(1 − λ) (2 − λ) = 0 & λ = 1 & 2
So, Eigen vector corresponding to the λ = 1 is,
0 2 1
>0 1H>a H = 0
2a + a = 0 & a = 0
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 43
Again for λ = 2
−1 2 1
> 0 0H>b H = 0
− 1 + 2b = 0
Then sum of
SOL 1.52
a &b & a + b = 0 + 1 = 1
2 2
Option (C) is correct.
Given
f (x, y) = yx
First partially differentiate the function w.r.t. y
2f
= xyx − 1
2y
Again differentiate. it w.r.t. x
22 f
= yx − 1 (1) + x ^yx − 1 log y h = yx − 1 ^x log y + 1h
2x2y
At :
SOL 1.53
b=1
2
x = 2, y = 1
22 f
= (1) 2 − 1 (2 log 1 + 1) = 1 (2 # 0 + 1) = 1
2x2y
Option (A) is correct.
Given :
y m + 2yl + y = 0
2
(D + 2D + 1) y = 0
The auxiliary equation is
where D = d/dx
m2 + 2m + 1 = 0
(m + 1) 2 = 0 , m =− 1, − 1
The roots of auxiliary equation are equal and hence the general solution of
the given differential equation is,
..(i)
y = (C1 + C2 x) em x = (C1 + C2 x) e−x
Given y (0) = 0 at x = 0, & y = 0
Substitute in equation (i), we get
0 = (C1 + C2 # 0) e−0
0 = C1 # 1 & C1 = 0
Again y (1) = 0 , at x = 1 & y = 0
Substitute in equation (i), we get
0 = [C1 + C2 # (1)] e−1 = [C1 + C2] 1
e
1
C1 + C 2 = 0 & C 2 = 0
Substitute C1 and C2 in equation (i), we get
y = (0 + 0x) e−x = 0
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PAGE 44
ENGINEERING MATHEMATICS
And
SOL 1.54
SOL 1.55
CHAP 1
y (0.5) = 0
Option (B) is correct.
Given :
y = x2
and interval [1, 5]
At
x=1 &y =1
And at
x=5
y = (5) 2 = 25
Here the interval is bounded between 1 and 5
So, the minimum value at this interval is 1.
...(i)
Option (A) is correct
Let square matrix
x y
A =>
y xH
The characteristic equation for the eigen values is given by
A − λI = 0
x−λ
y
=0
y x−λ
(x − λ) 2 − y2 = 0
(x − λ) 2 = y2
x − λ =! y
λ = x!y
So, eigen values are real if matrix is real and symmetric.
SOL 1.56
Option (B) is correct.
The Cauchy-Reimann equation, the necessary condition for a function f (z)
to be analytic is
2ϕ 2ψ
=
2y
2x
2ϕ
2ϕ 2ϕ 2ψ 2ψ
2ψ
when
,
,
,
exist.
=−
2x
2y
2x 2y 2y 2x
SOL 1.57
Option (A) is correct.
2 2 ϕ 2 2 ϕ 2ϕ 2ϕ
Given :
+ 2 +
+
=0
2x
2y
2x 2
2y
Order is determined by the order of the highest derivative present in it.
Degree is determined by the degree of the highest order derivative present in
it after the differential equation is cleared of radicals and fractions.
So, degree = 1 and order = 2
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CHAP 1
SOL 1.58
ENGINEERING MATHEMATICS
PAGE 45
Option (B) is correct.
Given
y = x+
x+
y−x = x+
Squaring both the sides,
x + x + .......3
...(i)
x + x + ....3
(y − x) 2 = x + x + x + ......3
(y − x) 2 = y
y2 + x2 − 2xy = y
We have to find y (2), put x = 2 in equation (ii),
From equation (i)
...(ii)
y2 + 4 − 4y = y
y2 − 5y + 4 = 0
(y − 4) (y − 1) = 0
y = 1, 4
From Equation (i) we see that
For y (2)
Therefore,
SOL 1.59
y = 2+
2+
2 + 2 + .....3 > 2
y =4
Option (B) is correct.
Vector area of TABC ,
A = 1 BC # BA = 1 (c − b) # (a − b)
2
2
= 1 [c # a − c # b − b # a + b # b]
2
= 1 [c # a + b # c + a # b]
2
b # b = 0 and c # b =− (b # c)
= 1 [(a − b) # (a − c)]
2
SOL 1.60
Option (C) is correct.
dy
dy
Given :
= y2 or 2 = dx
dx
y
Integrating both the sides
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ENGINEERING MATHEMATICS
# dy
y2
=
CHAP 1
# dx
−1 = x + C
y
Given y (0) = 1 at
x=0 &y=1
Put in equation (i) for the value of C
− 1 = 0 + C &C =− 1
1
From equation (i),
−1 = x − 1
y
y =− 1
x−1
For this value of y , x − 1 ! 0 or x ! 1
And
x < 1 or x > 1
SOL 1.61
...(i)
Option (A) is correct.
φ (t) =
Let
# f (t) dt and φ (0) = 0 then φl (t) = f (t)
t
0
We know the formula of Laplace transforms of φl (t) is
L 6φl (t)@ = sL 6φ ^ t h@ − φ (0) = sL 6φ (t)@
L 6φ (t)@ = 1 L 6φl (t)@
s
Substitute the values of φ (t) and φl (t), we get
t
L ; f (t) dtE = 1 L 6f (t)@
s
0
φ (0) = 0
#
L;
or
# f (t) dtE = s1 F (s)
t
0
SOL 1.62
Option (A) is correct.
From the Trapezoidal Method
b
#a f (x) dx = h2 6f (x0) + 2f (x1) + 2f (x2) .....2f (xn − 1) + f (xn)@
Interval h = 2π − 0 = π
8
4
Find
2π
#0 sin xdx
...(i)
Here f (x) = sin x
Table for the interval of π/4 is as follows
Angle θ
0
f (x) = sin x 0
π
4
π
2
0.707 1
3π
4
π
0.707 0
5π
4
3π
2
− 0.707 − 1
7π
4
− 0.707 0
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 47
Now from equation(i),
2π
#0 sin xdx = π8 [0 + 2 (0.707 + 1 + 0.707 + 0 − 0.707 − 1 − 0.0707 + 0)]
= π #0 = 0
8
SOL 1.63
Option (D) is correct.
The X and Y be two independent random variables.
So,
E (XY) = E (X) E (Y)
& covariance is defined as
Cov (X, Y) = E (XY) − E (X) E (Y)
= E (X) E (Y) − E (X) E (Y)
(i)
From eqn. (i)
=0
For two independent random variables
Var (X + Y) = Var (X) + Var (Y)
and
E (X 2 Y 2) = E (X 2) E (Y 2)
So, option (D) is incorrect.
SOL 1.64
Option (B) is correct.
2
Let,
ex − b1 + x + x l
2
f (x) = lim
3
x"0
x
x
e − (1 + x)
= lim
x"0
3x2
x
= lim e − 1
x " 0 6x
0 form
0
0 form
0
0 form
0
x
0
= lim e = e = 1
6
6
x"0 6
SOL 1.65
Option (B) is correct.
2 1
H
A =>
0 2
Let λ is the eigen value of the given matrix then characteristic matrix is
1 0
A − λI = 0
H = Identity matrix
Here I = >
0 1
2−λ
1
=0
0 2−λ
Let,
(2 − λ) 2 = 0
λ = 2, 2
So, only one eigen vector.
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PAGE 48
SOL 1.66
ENGINEERING MATHEMATICS
CHAP 1
Option (D) is correct.
Column I
P. Gauss-Seidel method
4. Linear algebraic equation
Q. Forward Newton-Gauss method
1. Interpolation
R. Runge-Kutta method
2. Non-linear differential equation
S. Trapezoidal Rule
3. Numerical integration
So, correct pairs are, P-4, Q-1, R-2, S-3
SOL 1.67
Option (B) is correct.
2
dy
Given :
+ 2xy = e−x and y (0) = 1
dx
It is the first order linear differential equation so its solution is
y (I.F.) =
So,
# Q (I.F.) dx + C
I. F . = e
# Pdx
=e
# 2xdx
x2
= e2 # xdx = e2 # 2 = ex
The complete solution is,
2
yex =
# e−x # ex dx + C
=
# dx + C = x + C
2
2
compare with
dy
+ P (y) = Q
dx
2
y = x +x2 c
e
Given
y (0) = 1
At
x =0 &y=1
Substitute in equation (i), we get
1 =C &C=1
1
2
Then
y = x +x2 1 = (x + 1) e−x
e
...(i)
SOL 1.68
Option (C) is correct.
The incorrect statement is, S = {x : x ! A and x ! B} represents the union
of setA and set B .
The above symbol (!) denotes intersection of set A and set B . Therefore
this statement is incorrect.
SOL 1.69
Option (D) is correct.
Total number of items = 100
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 49
Number of defective items = 20
Number of Non-defective items = 80
Then the probability that both items are defective, when 2 items are selected
at random is,
20!
20 # 19
80
20
C
C
18
!
2
!
2
0
2
=
P = 100
=
= 19
100 # 99
495
C2
100!
2
98!2!
Alternate Method :
Here two items are selected without replacement.
Probability of first item being defective is
P1 = 20 = 1
100 5
After drawing one defective item from box, there are 19 defective items in
the 99 remaining items.
Probability that second item is defective,
P2 = 19
899
then probability that both are defective
P = P1 # P2 = 1 # 19 = 19
5
99 495
SOL 1.70
Option (A) is correct.
3 2
H
S =>
2 3
Eigen values of this matrix is 5 and 1. We can say λ1 = 1 λ2 = 5
Then the eigen value of the matrix
Given :
S 2 = S S is λ12 , λ22
Because. if λ1, λ2, λ3 .... are the eigen values of A, then eigen value of Am are
λ1m, λm2 , λm3 ....
Hence matrix S 2 has eigen values (1) 2 and (5) 2 & 1 and 25
SOL 1.71
Option (B) is correct.
Given
f (x) = (x − 8) 2/3 + 1
The equation of line normal to the function is
(y − y1) = m2 (x − x1)
Slope of tangent at point (0, 5) is
m1 = f l (x) = :2 (x − 8) −1/3D
3
(0, 5)
1
m1 = f l (x) = 2 (− 8) −1/3 =− 2 (23) − 3 =− 1
3
3
3
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PAGE 50
ENGINEERING MATHEMATICS
CHAP 1
We know the slope of two perpendicular curves is − 1.
m1 m2 =− 1
m2 =− 1 = − 1 = 3
m1 − 1/3
The equation of line, from equation (i) is
(y − 5) = 3 (x − 0)
y = 3x + 5
SOL 1.72
Option (A) is correct.
Let
it π/3
iπ/3
0
eit dt = :e D & e − e
i 0
i
i
π
= 1 6e 3 i − 1@ = 1 9cos π + i sin π − 1C
i
3
3
i
f (x) =
#0
π/3
= 1 ;1 + i 3 − 1E = 1 ;− 1 + 3 iE
i 2
2
i 2
2
= 1 # i ;− 1 + 3 iE =− i ;− 1 + 3 iE
2
2
i
i 2
2
i2 =− 1
= i ; 1 − 3 iE = 1 i − 3 i 2 = 3 + 1 i
2
2
2
2
2
2
SOL 1.73
Option (B) is correct.
Given
2
f (x) = 2x2 − 7x + 3
5x − 12x − 9
Then
2
lim f (x) = lim 2x2 − 7x + 3
x"3
x " 3 5x − 12x − 9
Applying L – Hospital rule
= lim 4x − 7
x " 3 10x − 12
Substitute the limit, we get
lim f (x) = 4 # 3 − 7 = 12 − 7 = 5
10 # 3 − 12 30 − 12 18
x"3
SOL 1.74
Option (A) is correct.
(P) Singular Matrix " Determinant is zero A = 0
(Q) Non-square matrix " An m # n matrix for which m ! n , is called nonsquare matrix. Its determinant is not defined
(R) Real Symmetric Matrix " Eigen values are always real.
(S) Orthogonal Matrix " A square matrix A is said to be orthogonal if
AAT = I
Its determinant is always one.
SOL 1.75
Option (B) is correct.
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CHAP 1
ENGINEERING MATHEMATICS
Given :
PAGE 51
d 2y
dy
2x
2 + 4 dx + 3y = 3e
dx
d =D
dx
[D2 + 4D + 3] y = 3e2x
The auxiliary Equation is,
Then
SOL 1.76
m2 + 4m + 3 = 0 & m =− 1, − 3
C.F. = C1 e−x + C2 e−3x
2x
3e2x
P.I. = 2 3e
=
D + 4D + 3 (D + 1) (D + 3)
2x
2x
3e2x
=
= 3e = e
5
(2 + 1) (2 + 3) 3 # 5
Option (C) is correct.
Given
EF = G
Rcos θ − sin θ
S
cos θ
S sin θ
SS 0
0
T
We know that the
matrix
SOL 1.77
Put D = 2
where G = I = Identity matrix
R1
0VW
S
0W # F = S0
SS0
1WW
X
T
multiplication
0
1
0
of
0VW
0W
1WW
X
a matrix and its inverse be a identity
AA−1 = I
So, we can say that F is the inverse matrix of E
[adj.E]
F = E −1 =
E
Rcos θ − (sin θ) 0VT
R cos θ sin θ 0V
W
S
W
S
adjE = S sin θ
cos θ 0W = S− sin θ cos θ 0W
SS 0
SS
0 1WW
0
0 1WW
X
X
T
T
E = 6cos θ # (cos θ − 0)@ − 8^− sin θh # ^sin θ − 0hB + 0
= cos2 θ + sin2 θ = 1
R cos θ sin θ 0V
W
adj
.
E
6
@ S
Hence,
F =
= S− sin θ cos θ 0W
E
SS
0
0 1WW
X
T
Option (B) is correct.
The probability density function is,
1+t
for − 1 # t # 0
f (t) = )
1−t
for 0 # t # 1
For standard deviation first we have to find the mean and variance of the
function.
Mean (t ) =
=
#−13t f (t) dt
0
=
0
#−1 t (1 + t) dt + #0
#−1 (t + t2) dt + #0
1
1
t (1 − t) dt
(t − t2) dt
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PAGE 52
ENGINEERING MATHEMATICS
CHAP 1
2
3 0
2
3 1
= :t + t D + :t − t D = :− 1 + 1 D + :1 − 1 D = 0
2 3 −1
2 3 0
2 3
2 3
And
variance ^σ2h =
#− 33 (t − t ) 2 f (t) dt
t=0
0
=
#−1 t2 (1 + t) dt + #0
=
#−1 (t2 + t3) dt + #0
0
1
1
t2 (1 − t) dt
(t2 − t3) dt
3
4 0
3
4 1
= :t + t D + :t − t D
3 4 −1
3 4 0
=−:− 1 + 1 D + :1 − 1 − 0D = 1 + 1 = 1
3 4
3 4
12 12 6
Now, standard deviation
(σ2) s = 1 = 1
6
6
SOL 1.78
Option (A) is correct.
The Stokes theorem is,
#C F : dr
=
##S (4 # F) : ndS
=
##S (Curl F) : dS
Here we can see that the line integral # F : dr and surface integral
C
## (Curl F) : ds is related to the stokes theorem.
S
SOL 1.79
Option (B) is correct.
Let,
P = defective items
Q = non-defective items
10% items are defective, then probability of defective items
P = 0.1
Probability of non-defective item
Q = 1 − 0.1 = 0.9
The Probability that exactly 2 of the chosen items are defective is
= 10 C2 (P) 2 (Q) 8 = 10! (0.1) 2 (0.9) 8
8! 2!
= 45 # (0.1) 2 # (0.9) 8 = 0.1937
SOL 1.80
Option (A) is correct.
Let
f (x) =
=
#−a (sin6 x + sin7 x) dx
a
#−a sin6 xdx + #−a sin7 xdx
a
a
We know that
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CHAP 1
ENGINEERING MATHEMATICS
#−a f (x) dx
a
0
=* a
2 # f (x)
0
PAGE 53
when f (− x) =− f (x); odd function
when f (− x) = f (x); even function
Now, here sin6 x is an even function and sin7 x is an odd function. Then,
f (x) = 2 # sin6 xdx + 0 = 2 # sin6 xdx
a
a
0
0
SOL 1.81
Option (C) is correct.
We know, from the Echelon form the rank of any matrix is equal to the
Number of non zero rows.
Here order of matrix is 3 # 4 , then, we can say that the Highest possible
rank of this matrix is 3.
SOL 1.82
Option (A) is correct.
I =
Given
8
2
#0 #π/4 f (x, y) dydx
We can draw the graph from the limits of the integration, the limit of y is
from y = x to y = 2 . For x the limit is x = 0 to x = 8
4
Here we change the order of the integration. The limit of x is 0 to 8 but we
have to find the limits in the form of y then x = 0 to x = 4y and limit of y
is 0 to 2
So
8
2
#0 #x/4 f (x, y) dydx
=
2
#0 #0
4y
f (x, y) dxdy =
#r #p
s
q
f (x, y) dxdy
Comparing the limits and get
r = 0 , s = 2 , p = 0 , q = 4y
SOL 1.83
Option (A) is correct.
R
S5 0
S0 5
Let,
A =S
S0 0
S0 0
T
The characteristic equation for eigen values
V
0W
0W
1WW
1W
X
is given by,
0
0
2
3
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ENGINEERING MATHEMATICS
CHAP 1
A − λI = 0
5−λ
0
0
0
0 5−λ
0
0
A=
=0
0
0 2−λ
1
0
0
3 1−λ
Solving this, we get
(5 − λ) (5 − λ) [(2 − λ) (1 − λ) − 3] = 0
(5 − λ) 2 [2 − 3λ + λ2 − 3] = 0
(5 − λ) 2 (λ2 − 3λ − 1) = 0
So,
(5 − λ) 2 = 0 & λ = 5 , 5 and λ2 − 3λ − 1 = 0
− (− 3) ! 9 + 4
= 3 + 13 , 3 − 13
2
2
2
The eigen values are λ = 5 , 5, 3 + 13 , 3 − 13
2
2
R V
Sx1W
Sx2W
Let
X1 = S W
Sx 3W
Sx 4W
T X
be the eigen vector for the eigen value λ = 5
Then,
(A − λI ) X1 = 0
λ =
(A − 5I ) X1 = 0
VR V
0 0 0WSx1W
0 0 0WSx2W
=0
0 − 3 1WWSSx 3WW
0 3 − 4WSx 4W
XT X
or
− 3x 3 + x 4 = 0
3x 3 − 4x 4 = 0
This implies that x 3 = 0 , x 4 = 0
Let
x1 = k1 and x2 = k2
R V
Sk1W
Sk2W
So, eigen vector,
where k1 , k2 ε R
X1 = S W
S0W
S0W
T X
Option (C) is correct.
Given :
...(i)
x+y = 2
...(ii)
1.01x + 0.99y = b , db = 1 unit
We have to find the change in x in the solution of the system. So reduce y
R
S0
S0
S0
S
S0
T
SOL 1.84
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 55
From the equation (i) and (ii).
Multiply equation (i) by 0.99 and subtract from equation (ii)
1.01x + 0.99y − (0.99x + 0.99y) = b − 1.98
1.01x − 0.99x = b − 1.98
0.02x = b − 1.98
Differentiating both the sides, we get
0.02dx = db
dx = 1 = 50 unit
0.02
SOL 1.85
SOL 1.86
Option (A) is correct.
Given,
x (u, v) = uv
dx = v ,
du
And y (u, v) = v
u
2y
=− v2
2u
u
We know that,
R2x
S
2u
φ (u, v) = S2y
S
S2u
T
v
φ (u, v) = >− 2v
u
dx = u
dv
2y
=1
2v u
2x VW
2v W
2y W
2v W
X
u
1 H = v # 1 − u # − v = v + v = 2v
a u2 k
u u
u
u
u
Option (D) is correct.
Given : Radius of sphere r = 1
Let,
Radius of cone = R
Height of the cone = H
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PAGE 56
ENGINEERING MATHEMATICS
CHAP 1
Finding the relation between the volume and Height of the cone
From ΔOBD ,
OB 2 = OD 2 + BD 2
1 = (H − 1) 2 + R2 = H 2 + 1 − 2H + R2
R2 + H 2 − 2H = 0
R2 = 2H − H 2
Volume of the cone,
V = 1 πR 2 H
3
2
Substitute the value of R from equation (i), we get
V = 1 π (2H − H 2) H = 1 π (2H 2 − H 3)
3
3
Differentiate V w.r.t to H
dV = 1 π [4H − 3H 2]
3
dH
...(i)
d 2 V = 1 π [4 − 6H]
3
dH 2
For minimum and maximum value, using the principal of minima and
maxima.
Put dV = 0
dH
1 π [4H − 3H 2] = 0
3
H [4 − 3H] = 0 & H = 0 and H = 4
3
Again differentiate
d 2 V = 1 π 4 − 6 4 = 1 π [4 − 8] =− 4 π < 0 (Maxima)
# 3D 3
3 :
3
dH 2
2
(Minima)
And at H = 0 , d V2 = 1 π [4 − 0] = 4 π > 0
3
3
dH
So, for the largest volume of cone, the value of H should be 4/3
At H = 4 ,
3
SOL 1.87
Option (D) is correct.
2 ln (x)
dy
Given : x2
+ 2xy =
x
dx
2 ln (x)
dy 2y
+
=
x
dx
x3
Comparing this equation with the differential equation
dy
+ P (y) = Q we
dx
2 ln (x)
have P = 2 and Q =
x
x3
The integrating factor is,
2
I.F.= e # Pdx = e # x dx
e2 lnx = e lnx = x2
2
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 57
Complete solution is written as,
y (I.F.) =
y (x2) =
Integrating the value
Let,
# Q (I.F.) dx + C
x
2
# 2 ln
# x dx + C
x3
# ln x # x1 dx
I =
= 2 # ln x # 1 dx + C
x
(II)
Separately
# ln x # x1 dx
(I)
...(i)
(I)
...(ii)
(II)
= ln x # 1 dx −
x
= ln x ln x −
d (ln x)
1
# &dx
# # x dx 0 dx
# x1 # ln xdx
From equation(ii)
1 444 2
444 3
I
2I = (ln x) 2
(ln x) 2
2
Substitute the value from equation (iii) in equation (i),
2 (ln x) 2
+C
y (x2) =
2
I =
...(iii)
x2 y = (ln x) 2 + C
Given y (1) = 0 , means at x = 1 &y = 0
...(iv)
or
then
0 = (ln 1) 2 + C & C = 0
So from equation (iv), we get
Now at x = e ,
SOL 1.88
x2 y = (ln x) 2
(ln e) 2
y (e) =
= 12
2
e
e
Option (A) is correct.
Potential function of v = x2 yz at P (1, 1, 1) is = 12 # 1 # 1 = 1 and at origin
O (0, 0, 0) is 0.
Thus the integral of vector function from origin to the point (1, 1, 1) is
= 6x2 yz @ P − 6x2 yz @O
= 1−0 = 1
SOL 1.89
Option (C) is correct.
Let,
f (x) = x3 + 3x − 7
From the Newton Rapson’s method
f (xn)
xn + 1 = xn −
f l (xn)
...(i)
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PAGE 58
ENGINEERING MATHEMATICS
CHAP 1
We have to find the value of x1 , so put n = 0 in equation (i),
f (x 0)
x1 = x 0 −
f l (x 0)
= x3 + 3x − 7
= 13 + 3 # 1 − 7 = 1 + 3 − 7 =− 3
= 3x2 + 3
= 3 # (1) 2 + 3 = 6
(− 3)
x1 = 1 −
= 1 + 3 = 1 + 1 = 3 = 1.5
6
6
2 2
f (x)
f (x 0)
f l (x)
f l (x 0)
Then,
SOL 1.90
x0 = 1
Option (D) is correct.
We know a die has 6 faces and 6 numbers so the total number of ways
= 6 # 6 = 36
And total ways in which sum is either 8 or 9 is 9, i.e.
(2, 6), (3, 6) (3, 5) (4, 4) (4, 5) (5, 4) (5, 3) (6, 2) (6, 3)
Total number of tosses when both the 8 or 9 numbers are not come
= 36 − 9 = 27
Then probability of not coming sum 8 or 9 is, = 27 = 3
36
4
SOL 1.91
Option (C) is correct.
d 2y
dy
Given :
2 + p dx + qy = 0
dx
The solution of this equation is given by,
y = c1 emx + c2 enx
Here m & n are the roots of ordinary differential equation
Given solution is,
y = c1 e−x + c2 e−3x
Comparing equation (i) and (ii), we get m =− 1 and n =− 3
Sum of roots,
m + n =− p
− 1 − 3 =− p & p = 4
and product of roots, mn = q
(− 1) (− 3) = q & q = 3
SOL 1.92
...(i)
...(ii)
Option (C) is correct.
d 2y
dy
Given :
2 + p dx + (q + 1) y = 0
dx
[D2 + pD + (q + 1)] y = 0
From the previous question, put p = 4 and m = 3
[D2 + 4D + 4] y = 0
The auxilliary equation of equation (i) is written as
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...(i)
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 59
m2 + 4m + 4 = 0 & m =− 2, − 2
Here the roots of auxiliary equation are same then the solution is
Let c1 = 0
y = (c1 + c2 x) emx = xe−2x
e
o
c2 = 1
SOL 1.93
Option (C) is correct.
Given :
x = a (θ + sin θ), y = a (1 − cos θ)
First differentiate x w.r.t. θ,
dx = a [1 + cos θ]
dθ
And differentiate y w.r.t. θ
dy
= a [0 − (− sin θ)] = a sin θ
dθ
dy
dy
dθ = dy/dθ
We know,
=
dx
dθ # dx dx/dθ
dy
Substitute the values of
and dx
dθ
dθ
SOL 1.94
2 sin θ cos θ
dy
θ
1
sin
2
2
= a sin θ #
=
=
dx
2θ
a [1 + cos θ] 1 + cos θ
2 cos
2
θ
sin
2 = tan θ
=
cos θ + 1 = 2 cos2 θ
2
2
θ
cos
2
Option (C) is correct.
Given : P (0.866, 0.500, 0), so we can write
P = 0.866i + 0.5j + 0k
Q = (0.259, 0.966, 0), so we can write
Q = 0.259i + 0.966j + 0k
For the coplanar vectors
P : Q = P Q cos θ
cos θ =
P:Q
P Q
P : Q = (0.866i + 0.5j + 0k) : (0.259i + 0.966j + 0k)
= 0.866 # 0.259 + 0.5 # 0.966
So,
SOL 1.95
0.866 # 0.259 + 0.5 # 0.966
(0.866) 2 + (0.5) 2 + (0.259) 2 + (0.966) 2
0.70729
= 0.707
= 0.22429 + 0.483 =
0.99 # 1.001
0.99 # 1.001
θ = cos−1 (0.707) = 45c
cos θ =
Option (B) is correct.
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PAGE 60
ENGINEERING MATHEMATICS
CHAP 1
R1 2 3V
W
S
Let
A = S1 5 1W
SS3 1 1WW
X
T
We know that the sum of the eigen value of a matrix is equal to the sum of
the diagonal elements of the matrix
So, the sum of eigen values is,
1+5+1 = 7
SOL 1.96
Option (D) is correct.
Given : Total number of cards = 52 and two cards are drawn at random.
Number of kings in playing cards = 4
So the probability that both cards will be king is given by,
4
3
P = 52C1 # 51C1 = 4 # 3 = 1
52
51 221
C1
C1
SOL 1.97
n
Cr =
Option (B) is correct.
Given :
0,
U (t − a) = *
1,
for t < a
for t $ a
From the definition of Laplace Transform
L [F (t)] =
L 6U (t − a)@ =
=
#0 3e−st f (t) dt
#0 3e−st U (t − a) dt
#0
a −st
e
(0) +
#a 3e−st (1) dt = 0 + #a 3e−st dt
−as
−as
−st 3
L 6U (t − a)@ = :e D = 0 − :e D = e
−s
s
−s a
SOL 1.98
Option (D) is correct.
First we have to make the table from the given data
Take x 0 = 0 and h = 1
Then
P = x − x0 = x
h
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r n−r
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 61
From Newton’s forward Formula
P (P − 1) 2
P (P − 1) (P − 2) 3
f (x) = f (x 0) + P Δf (0) +
Δ f (0) +
Δ f (0)
1
2
3
x (x − 1) 2
x (x − 1) (x − 2) 3
= f (0) + xΔf (0) +
Δ f (0) +
Δ f (0)
2
6
x (x − 1)
x (x − 1) (x − 2)
= 1 + x (1) +
(− 2) +
(12)
2
6
= 1 + x − x (x − 1) + 2x (x − 1) (x − 2)
f (x) = 2x3 − 7x2 + 6x + 1
SOL 1.99
Option (A) is correct.
Given :
V =
π/3
2π
#0 #0 #0
1
r2 sin φdrdφdθ
First integrating the term of r , we get
V =
2π
#0 #0
π/3
r3 1 sin φdφdθ =
:3D
0
2π
#0 #0
π/3
1 sin φdφdθ
3
Integrating the term of φ, we have
#0
2π
=− 1
3
=− 1
3
#0
V =1
3
π/3
6− cos φ@0 dθ
π
1 2π 1
9cos 3 − cos 0C dθ =− 3 #0 :2 − 1D dθ
2π
2π
#0 b− 12 ldθ =− 13 # b− 12 l #0 dθ
2π
Now, integrating the term of θ, we have
V = 1 6θ@ 20π = 1 [2π − 0] = π
6
6
3
SOL 1.100
Option (A) is correct.
R8 x
S
Let,
A =S 4 0
SS12 6
T
For singularity of the matrix A = 0
8 x 0
4 0 2 =0
12 6 0
0VW
2W
0WW
X
8 [0 − 2 # 6] − x [0 − 24] + 0 [24 − 0] = 0
8 # (− 12) + 24x = 0
− 96 + 24x = 0 & x = 96 = 4
24
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PAGE 62
SOL 1.101
ENGINEERING MATHEMATICS
CHAP 1
Option (A) is correct
2
2
f (x) = lim sin x = lim sin x # x
x
x
x
x"0
x"0
2
= lim b sin x l # x
x
x"0
Let,
lim sin x = 1
x"0 x
= (1) 2 # 0 = 0
Alternative :
2
Let
f (x) = lim sin x
x
x"0
f (x) = lim 2 sin x cos x
1
x"0
= lim sin 2x = sin 0 = 0
1
1
x"0
0
: 0 formD
Apply L-Hospital rule
SOL 1.102
Option (D) is correct.
Accuracy of Simpson’s rule quadrature is O (h5)
SOL 1.103
Option (C) is correct.
4 1
A =>
1 4H
The characteristic equation for the eigen value is given by,
Let,
A − λI = 0
1 0
I = Identity matrix >
0 1H
4 1
1 0
>1 4H − λ >0 1H = 0
4−λ
1
1
=0
4−λ
(4 − λ) (4 − λ) − 1 = 0
(4 − λ) 2 − 1 = 0
λ2 − 8λ + 15 = 0
Solving above equation, we get
λ = 5, 3
SOL 1.104
Option (C) is correct.
Given :
x + 2y + z = 6
2x + y + 2z = 6
x+y+z = 5
Comparing to Ax = B ,we get
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 63
R1 2 1V
R6V
W
S
S W
A = S2 1 2W, B = S6W
SS1 1 1WW
SS5WW
X
T X
T
Write the system of simultaneous equations in the form of Augmented
matrix,
R1 2 1 : 6V
W
S
R2 " R2 − 2R1 and R 3 " 2R 3 − R2
6A: B@ = S2 1 2 : 6W
SS1 1 1 : 5WW
TR1 2 1 : X6V
W
S
+ S0 − 3 0 : − 6W
R 3 " 3R 3 + R2
SS0 1 0 : 4WW
RT1 2 1 : 6VX
W
S
+ S0 − 3 0 : − 6W
SS0 0 0 : 6WW
X
T
It is a echelon form of matrix.
Since ρ 6A@ = 2 and ρ 5A: B? = 3
ρ [A] ! ρ [A: B ]
So, the system has no solution and system is inconsistent.
SOL 1.105
Option (B) is correct.
Given : y = x2 and y = x .
The shaded area shows the area, which is bounded by the both curves.
Solving given equation, we get the intersection points as,
In y = x2 putting y = x we have x = x2 or x2 − x = 0 which gives x = 0, 1
Then from y = x we can see that curve y = x2 and y = x intersects at point
(0, 0) and (1, 1). So, the area bounded by both the curves is
y = x2
x=1
A=
# #
x=0
dydx =
y=x
x=1
#
x=0
dx
y = x2
#
y=x
x=1
dy =
# dx6y @
x2
x
x=0
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PAGE 64
ENGINEERING MATHEMATICS
x=1
=
# (x
2
x=0
CHAP 1
3
2 1
− x) = :x − x D = 1 − 1 =− 1 = 1 unit2
3
2 0 3 2
6
6
Area is never negative
SOL 1.106
Option (A) is correct.
dy
+ y2 = 0
dx
dy
=− y2
dx
dy
− 2 = dx
y
Integrating both the sides, we have
dy
− # 2 = # dx
y
y−1 = x + c & y =
SOL 1.107
1
x+c
Option (C) is correct.
Given :
F = xi − yj
First Check divergency, for divergence,
Grade F = 4:F = ; 2 i + 2 j + 2 k E:6xi − yj @ = 1 − 1 = 0
2x
2y
2z
So we can say that F is divergence free.
Now checking the irrationalit;. For irritation the curl F = 0
Curl F = 4# F = ; 2 i + 2 j + 2 k E # [xi − yj]
2x
2y
2z
R
V
j
k W
S i
2
2
2
W = i [0 − 0] − j [0 − 0] + k [0 − 0] = 0
=S
S2x 2y 2z W
S x −y 0 W
T
X
So, vector field is irrotational. We can say that the vector field is divergence
free and irrotational.
SOL 1.108
Option (B) is correct.
Let
f (t) = sin ωt
From the definition of Laplace transformation
L [F (t)] =
=
#0 3e−st f (t) dt
#0 3e−st b e
iωt
=
#0 3e−st sin ωtdt
− e−iωt dt
l
2i
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 65
iωt
−iωt
3
sin ωt = e − e
= 1 # (e−st eiωt − e−st e−iωt) dt
2i
2i 0
3
= 1 # 6e(− s + iω) t − e− (s + iω) t@ dt
2i 0
Integrating above equation, we get
− (s + iω) t
(− s + iω) t
sin ωt = 1 = e
− e
2i − s + iω − (s + iω)G
0
3
− (s + iω) t
(− s + iω) t
= 1 =e
+e
2i − s + iω (s + iω)G
0
Substitute the limits, we get
−0
e0
sin ωt = 1 =0 + 0 − e
+ e
2i
(− s + iω) s + iω oG
=− 1 ; s + iω + iω − s E
2i (− s + iω) (s + iω)
2iω
=− 1 #
= −ω
= 2ω 2
2i
(iω) 2 − s 2 − ω2 − s 2
ω +s
Alternative :
From the definition of Laplace transformation
3
L [F (t)] =
#0 3e−st sin ωtdt
a =− s and
eat a sin bt − b cos bt
@
26
e
o
a +b
b=ω
3
−st
Then,
L [sin ωt] = ; 2e 2 ^− s sin ωt − ω cos ωt hE
s +ω
0
−3
−0
e
e
=; 2
(− s sin 3 − ω cos 3)E − ; 2
(− s sin 0 − ω cos 0)E
s + ω2
s + ω2
= 0 − 2 1 2 [0 − ω] =− 2 1 2 (− ω)
s +ω
s +ω
L [sin ωt] = 2 ω 2
s +ω
We know # eat sin btdt =
SOL 1.109
2
Option (D) is correct.
Given : black balls = 5, Red balls = 5, Total balls=10
Here, two balls are picked from the box randomly one after the other without
replacement. So the probability of both the balls are red is
5!
5!
5
# 3!2!
5
n
C
0
!
5
!
0 # C2
P =
= #
= 1 # 10 = 10 = 2 n Cr =
10
45
45 9
C2
10!
r n−r
3!2!
Alternate Method :
Given :
Black balls = 5 ,
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PAGE 66
ENGINEERING MATHEMATICS
CHAP 1
Red balls = 5
Total balls = 10
The probability of drawing a red bell,
P1 = 5 = 1
10 2
If ball is not replaced, then box contains 9 balls.
So, probability of drawing the next red ball from the box.
P2 = 4
9
Hence, probability for both the balls being red is,
P = P1 # P2 = 1 # 4 = 2
2
9 9
SOL 1.110
Option (A) is correct.
We know that a dice has 6 faces and 6 numbers so the total number of cases
(outcomes) = 6 # 6 = 36
And total ways in which sum of the numbers on the dices is eight,
(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)
So, the probability that the sum of the numbers eight is,
p = 5
36
SOL 1.111
Option (D) is correct.
We have to draw the graph on x -y axis from the given functions.
−x
f (x) = * 0
x
x #− 1
x=0
x$1
It clearly shows that f (x) is differential at x =− 1, x = 0 and x = 1,
i.e. in the domain [− 1, 1].
So, (a), (b) and (c) are differential and f (x) is maximum at (x, − x).
SOL 1.112
Option (B) is correct.
If the scatter diagram indicates some relationship between two variables X
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 67
and Y , then the dots of the scatter diagram will be concentrated round a
curve. This curve is called the curve of regression.
Regression analysis is used for estimating the unknown values of one variable
corresponding to the known value of another variable.
SOL 1.113
Option (B) is correct.
Given : 3x + 2y + z = 4
x−y+z = 2
− 2x + 2z = 5
The Augmented matrix of the given system of equation is
R 3 2 1 : 4V
W
S
6A : B@ = S 1 − 1 1 : 2W R 3 " R 3 + 2R2 , R2 " R2 − R1
SS− 2 0 2 : 5WW
X
T
R 3 2 1 : 4V
W
S
+ S− 2 − 3 0 : − 2W
SS 0 − 2 4 : 9WW
X of unknown)
Here ρ 6A : B@ = ρT6A@ = 3 = n (number
Then the system of equation has a unique solution.
SOL 1.114
Option (B) is correct.
Given :
f (x, y) = 2x2 + 2xy − y3
Partially differentiate this function w.r.t x and y ,
2f
2f
= 2x − 3y2
= 4x + 2y ,
2x
2y
For the stationary point of the function, put 2f/2x and 2f/2y equal to zero.
2f
...(i)
& 2x + y = 0
= 4x + 2y = 0
2x
2f
...(ii)
and
& 2x − 3y2 = 0
= 2x − 3y2 = 0
2y
From equation (i), y =− 2x substitute in equation (ii),
2x − 3 (− 2x) 2 = 0
2x − 3 # 4x2 = 0
6x2 − x = 0 & x = 0 , 1
6
From equation (i),
For x = 0 ,
y =− 2 # (0) = 0
and for x = 1 ,
y =− 2 # 1 =− 1
6
6
3
So, two stationary point at (0, 0) and b 1 , − 1 l
6
3
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PAGE 68
SOL 1.115
ENGINEERING MATHEMATICS
CHAP 1
Option (B) is correct.
Sample space = (1, 1), (1, 2) ... (1, 8)
(2, 1), (2, 2)
(3, 1), (3, 2)
h
h
(8, 1), (8, 2)
f
f
h
f
(2, 8)
(3, 8)
h
(8, 8)
Total number of sample space = 8 # 8 = 64
Now, the favourable cases when Manish will arrive late at D
= (6, 8), (8, 6)...(8, 8)
Total number of favourable cases = 13
Probability = Total number of favourable cases
Totol number of sample space
= 13
64
So,
SOL 1.116
Option (B) is correct.
Divergence is defined as d:r
where
r = xi + yj + zk
and
So,
d= 2 i+ 2 j+ 2 k
2x
2y
2z
d:r = c 2 i + 2 j + 2 k m:(xi + yj + zk)
2x
2y
2z
d:r = 1 + 1 + 1 = 3
SOL 1.117
Option (B) is correct.
Given :
x+y = 2
2x + 2y = 5
The Augmented matrix of the given system of equations is
1 1 : 2
6A : B@ = >2 2 : 5H
Applying row operation, R2 " R2 − 2R1
1 1 : 2
6A : B@ = >0 0 : 1H
ρ [A] = 1 ! ρ 6A : B@ = 2
So, the system has no solution.
SOL 1.118
Option (D) is correct.
Given :
f (x) = x
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CHAP 1
ENGINEERING MATHEMATICS
PAGE 69
x
if x > 0
f (x) = * 0
if x = 0
−x
if x < 0
f (0 − h) − f (0)
− (− h)
Lf l (x) = lim
= lim
− 0 =− 1
h"0
h"0
−h
−h
f (0 + h) − f (0)
Rf l (x) = lim
= lim h − 0 = 1
h"0
h"0
h
h
Since
Lfl (0) ! Rf l (0)
So, derivative of f (x) at x = 0 does not exist.
SOL 1.119
Option (A) is correct.
The surface integral of the normal component of a vector function F taken
around a closed surface S is equal to the integral of the divergence of F
taken over the volume V enclosed by the surface S .
Mathematically
## F:n dS
S
=
### div Fdv
V
So, Gauss divergence theorem relates surface integrals to volume integrals.
SOL 1.120
Option (A) is correct.
Given :
3
f (x) = x − x
3
f l (x) = x2 − 1
f m (x) = 2x
Using the principle of maxima – minima and put f l (x) = 0
x2 − 1 = 0 & x = ! 1
Hence at x =− 1,
(Maxima)
f m (x) =− 2 < 0
at x = 1,
(Minima)
f m (x) = 2 > 0
So, f (x) is minimum at x = 1
SOL 1.121
Option (B) is correct.
Ra V
S 1W
Let
A = Sb1W, B = 8a2 b2 c2B
SSc WW
1
T X
C = AB
Ra V
Ra a a b
1 2
S 1W
S1 2
Let
= Sb1W # 8a2 b2 c2B = Sb1 a2 b1 b2
SSc WW
SSc a c b
1 2
1
1 2
X
T matrix
T all the
The 3 # 3 minor of this
is zero and
zero. So the rank of this matrix is 1.
a1 c2VW
b1 c2W
c1 c2WW
2 # 2X minors are also
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PAGE 70
ENGINEERING MATHEMATICS
CHAP 1
ρ 6C @ = 1
SOL 1.122
Option (D) is correct.
In a coin probability of getting head p = 1 and probability of getting tail,
2
1
1
q = 1− =
2 2
When unbiased coin is tossed three times, then total possibilities are
H H H
H H T
H T H
T H H
H T T
T T H
T H T
T T T
From these cases, there are three cases, when head comes exactly two times.
So, the probability of getting head exactly two times, when coin is tossed 3
times is,
2
P = 3C2 (p) 2 (q) 1 = 3 # b 1 l # 1 = 3
2
2
8
**********
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