Engineering Mathematics Test - 2 - PDF Flipbook
Engineering Mathematics Test - 2
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GATE
MECHANICAL
Engineering
Mathematics
Test-02Solutions
ENGINEERING MATHEMATICS
1. How many solutions does the following system of linear equations
have?
−x + 5y = −1
x−y=2
x + 3y = 3
a) Infinitely many
b) Two distinct solutions
c) Unique
d) None
Answer: (c)
Solution:
1 −5 1 1 −5 1 1 5 1
(A|B) = �1 −1� 2� ~ �0 4 � 1� ~ �0 4� 1�
1 3 3 0 8 2 0 00
ρ(A) = 2 = ρ(A|B) = 2 &number of unknowns = n = 2
Here ρ(A) = ρ(A|B) = n = 2
∴ Unique solution exists.
2. Real matrices [A]3×1, [B]3×3, [C]3×5, [D]5×3, [E]5×5, [F]5×1 are
given. Matrices [B] and [E] are symmetric. Following statements
are made with respect to their matrices.
Matrix product [F]T[C]T[B][C][F] is a scalar.
Matrix product [D]T[F][D] is always symmetric.
With reference to above statements which of the following applies?
a) Statement (I) is true but (II) is false
b) Statement (I) is false but (II) is true
1
c) Both the statements are true
d) Both the statements are false
Answer: (a)
Solution:
Given the [A]3×1, [B]3×3, [C]3×5, [D]5×3, [E]5×5 and [F]5×1 are
real matrices. And also given that [B] & [E] are symmetric.
In the first statement the product [F]1T×5 [C]5T×3 [B]3×3 [C]3×5
[F]5×1 is a first order matrix.
∴ First statement is correct.
In the second statement, the product [D]T3×5 [F]5×1 [D]5×3 is not
defined.
∴ Second statement is wrong.
Hence option (a) is correct.
0 1 02
3. The determinant of the matrix given below is �−01
1 1 13�
0 0
0 −2 0 1
a) -1
b) 0
c) 1
d) 2
Answer: (a)
Solution:
0 1 02
|A| = �−01
1 1 13�
0 0
1 −2 0 1
2
0 10
= (−1) �−1 1 1�
1 −2 0
= (−1)(−1)(0 − 1) = −1
4. Consider the matrices X4×3, Y4×3, and P2×3. The order of [P(XTY)-
1PT]T will be
a) 2 × 2
b) 3 × 3
c) 4 × 3
d) 3 × 4
Answer: (a)
Solution:
[P (XT Y)−1 PT]T
↓↓↓↓
�2 ��3���3���4��4���3��3���2
2×2
1 0 −1
5. If R = �2 1 −1� then the top row of R-1 is
23 2
a) [5 6 4]
b) [5 −3 1]
c) [2 0 −1]
d) [2 −1 0]
Answer: (b)
Solution:
1 0 −1
|R| = �2 1 −1� = 1(2 + 3) + 0 − 1(6 − 2) = 1
23 2
3
5 −3 1
Adj (R) = �−6 4 1�
4 −3 1
R−1 = adj(R) = 5 −3 1
|R| �−6 4 −1�
−3 1
4
∴Top row of R-1 is [5 −3 1]
6. Identify which of the following is an Eigen vector of the matrix A
= �−11 −02�
a) [−1 1]T
b) [3 −1]T
c) [1 −1]T
d) [−2 1]T
Answer: (b)
Solution:
Eigen values are 1, -2
For λ = 1, Eigen vectors are given by
(A − I)X = 0
�−01 −03� �yx� = �00�
x + 3y = 0
X1 = �−31�
4
7. What are the Eigen values of the following 2 × 2 matrix �−24 −51�
a) –1, 1
b) 1, 6
c) 2, 5
d) 4, –1
Answer: (a)
8. For |x| ≪ 1, cot h(x) can be approximated as
a) x
b) x2
c) 1
x
d) 1
x2
Answer: (c)
Solution:
Cot hx = ex + e−x = 2�1+ x2 + x4 +⋯..∞�
ex − e−x 2! 4!
x3 x5
2�x + 3! + 5! +⋯..∞�
= 1 (neglecting x2 and higher powers of x as |x| ≪ 1)
x
9. The integral 1 ∫02π Sin(t − τ)cosτdτ equals
2π
a) Sin t cost
b) 0
c) 1 cost
2
d) 1 sint
2
Answer: (d)
5
Solution:
2sina cos b = sin (a + b) + sin (a – b)
10. A point on the curve is said to be an extremum if it is a local
minimum (or) a local maximum. The number of distinct extremum
for the curve 3x4 – 16x3 + 24x2 + 37 is____
a) 0
b) 1
c) 2
d) 3
Answer: (b)
Solution:
f''(x) = 12x3 – 48x2 + 48x = 0 ⇒ x = 0, 2, 2
f"(0) = 48 > 0 ⇒ f{x} has a minimum at x = 0
f"(2) = 0 ⇒ f (x) has no extremum at x = 2
11. Given y = x2 + 2x + 10 the value of ddyx�X=1is equal to
a) 0
b) 4
c) 12
d) 13
Answer: (b)
Solution:
y = x2 + 2x + 10
dy = 2x + 2 ⇒ ddyx�X=1 = 4
dx
6
12. For real values of x, the minimum value of function f(x) = ex + e-x
is
a) 2
b) 1
c) 0.5
d) 0
Answer: (a)
Solution:
f(x) = ex + e-x where x ϵ R
f '(x) = ex – e-x = 0 ⇒ ex = e-x ⇒ x = 0
f"(x) = ex + e-x ⇒ f "(0) = 1 + 1 = 2 > 0
⇒ f(x) has a minimum at x = 0 Minimum value f(0) = 2
13. The derivative of f(x, y) at point (1, 2) in the direction of vector
ı̅ + ȷ̅ is 2√2 and in the direction of the vector –2ȷ̅ is –3. Then the
derivative of f(x,y) in direction −���ı − 2ȷ̅ is
a) 2√2 + 3
2
b) − 7
√5
c) −2√2 − 3
2
d) 1
√5
Answer: (b)
7
Solution:
Directional derivative of f in the direction of i + j = 2√2
⇒ ∇f . a = 2√2
|a|
⇒ �∂∂xf i + ∂f j + ∂f k� �i + j� = 2√2
∂y ∂z √2
⇒ ∂f + ∂f = 4 ………. (1)
∂x ∂y
Similarly,
∇f. b = –3
|b|
⇒ �∂∂xf i + ∂f j + ∂f k� �−2 j� = –3
∂y ∂z 2
⇒ –2 ∂f = –6 ⇒ ∂f = 3 ........... (2)
∂x ∂y
Sub. (2) in (1), we obtain, ∂f = 1
∂x
⸫ ∇f. c = �∂∂xf i + ∂f j + ∂f k� �−i − 2j� = −7
|c| ∂y ∂z √5 √5
14. The line integral ∫ V. dr of the vector function V(r) = 2xyzı̅ +
x2zȷ̅ + x2yk� 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
Answer: (a)
8
Solution:
V(r) = 2xyz i + x2z j + x2y K …….. (1)
Curl V = 0 ⇒ V is irrotational
There exists a scalar function ϕ (x, y, z) such that
V = ∇ϕ = i ∂ϕ + j ∂ϕ + K ∂ϕ ………. (2)
∂x ∂y ∂z
⇒ ϕ = x2yz (from (1) and (2))
∫((01,0,1,0,1)) V. dr = ∫((01,0,1,0,1)) ∇ϕ . dr = ∫((01,0,1,0,1)) dϕ .
= ϕ(1, 1, 1) – ϕ(0, 0, 0) = 1
15. The divergence of the vector field 3xzı̂ + 2xyȷ̂ – yz2k� at a point (1,
1, 1) is equal to
a) 7
b) 4
c) 3
d) 0
Answer: (c)
Solution:
f = 3xz i + 2xy j – yz2 k
div f = 3z + 2y – 2yz
div f�(1,1,1) = 3 + 2 – 2 = 3
16. If ϕ = 2x3y2z4 then ∇2ϕ is
a) 12xy2z4 + 4x2z4 + 20x3y2z3
b) 2x2y2z + 4x3z4 + 24x3y2z2
c) 12xy2z4 + 4x3z4 + 24x3y2z2
9
d) 4xy2z + 4x2z4 + 24x3y2z2
Answer: (c)
Solution:
∇2ϕ = ∂2ϕ + ∂2ϕ + ∂2ϕ
∂x2 ∂y2 ∂z2
= 12 xy2z4 + 4 x3z4 + 24 x3y2z2
17. The homogeneous part of the differential d2y + p dy + qy = r
dx2 dx
equation (p. q. r are constants) has real distinct roots if
a) p2 – 4q > 0
b) p2 – 4q < 0
c) p2 – 4q = 0
d) p2 – 4q = r
Answer: (a)
Solution:
Given d2y + p dy + qy = r
dx2 dx
⇒ (D2 + pD + q)y = r
The auxiliary equation is
D2 + pD + q = 0
D = − p ± �p2 − 4q
2
If (p2 – 4q) > 0
then the roots of f(D) = 0 are real and different.
10
18. With K as constant, the possible solution for the first order
differential equation dy = e−3x is
dx
a) − 1 e−3x + K
3
b) 1 (−1)e3x + K
3
c) −3e−3x + K
d) −3e−x + K
Answer: (a)
Solution:
Given dy = e−3x
dx
⇒ ∫ dy = ∫ e−3x + K ⇒ y = e−3x + K
−3
19. The solution of the differential equation dy + y = x with the
dx x
condition that y = 1 at x = 1 is
a) y = 2 + x
3x2 3
b) y = x + 1
2 2x
c) y = 2 + x
3 3
d) y = 2 + x2
3x 3
Answer: (d)
Solution:
Given dy + y = x ……. (1) and
dx x
y = 1 at x = 1 ………. (2)
I.F = e∫x1 dx = elog x = x
11
The general solution of (1) is
xy = x3 + C ……… (3)
3
using (2), (3) becomes
1 = 1 + C ⇒ C = 2
3 3
⸫ y = x3 + 2
3 3x
20. The general solution of the differential equation dy = 1+cos2y is
dx 1−cos2x
a) tany – cotx = C
b) tanx – coty = C
c) tany + cotx = C
d) tanx + coty = C
Answer: (c)
Solution:
dy = 1+2Cos2 y−1 = Cos2 y
dx 1− (1−2sin2 x) Sin2 x
⇒ ∫ Sec2 y dy = ∫ Cosec2 x dy
⇒ Tan y = – Cot x + C
⸫ The solution is tan y + cot x = C
21. The Solution of the differential equation d2y + 2 dy + y = 0 with
dt2 dt
y(0) = y'(0) , = 1 is
a) (2 − t)et
b) (1 + 2t)e−t
c) (2 + t)e−t
d) (1 − 2t)et
Answer: (b)
12
Solution:
A.E is D2 + 2D + 1 = 0
D = –1, –1
The solutions is y(t) = (C1 + C2t)e−t ………. (1)
y1(t) = – (C1 + C2t)e−t + C2te−t ………… (2)
y(0) = 1 ⇒ 1 = C1
y1(0) = 1 ⇒ 1 = – (C1) + C2
= –1 + C2 ⇒ C2 = 2
⸫ The solution is y(t) = (1 + 2t) e−t
22. Assume that the duration in minutes of a telephone conversation
follows the exponential distribution f(x) = 1 e−x5, x ≥ 0. The
5
probability that the conversation will exceed five minutes is
a) 1
e
b) 1 − 1
e
c) 1
e2
d) 1 − 1
e2
Answer: (a)
Solution:
P(5 < x < ∞) = ∫5∞ f(x)dx = ∫5∞ 1 −x dx
5
e5
= 1
e
13
23. 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) COY (X, Y) = 0
c) Var(X + Y) = Var(X) + Var(Y)
d) E(X2Y2) = (E(X))2 (E(y))2
Answer: (d)
24. Consider a Gaussian distributed random variable with zero mean
and standard deviation σ. The value of its cumulative distribution
function at the origin will be
a) 0
b) 0.5
c) 1
d) 10σ
Answer: (b)
Solution:
Since the curve is symmetric about the mean.
25. A fair coin is tossed n times. The probability that the difference
between the number of heads and tails is (n-3) is
a) 2-n
b) 0
c) nCn-32-n
d) 2-n+3
Answer: (b)
14
Solution:
A fair coin is tossed ‘n’ times then the number of heads and tails
can be shown below.
HT
n0
n–1 1
n–2 2
..
..
2 n–2
1 n–1
0n
⸫ The difference between the number of heads and tails can be ‘n’
or (n – 2) or (n – 4) …
But it cannot be (n – 1) or (n – 3) or (n – 5) …
⸫ The required probability = 0
26. The probability of obtaining at least two 'SIX' in throwing a fair
dice 4 times is
a) 425/432
b) 19/144
c) 13/144
d) 125/432
Answer: (b)
15
Solution:
Let p = 61, q = 5 and n = 4
6
Using binomial Distribution
P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)]
= 1 – �4C0 �61�0 �56�4 + 4C1 �61�1 �56�3� = 19
144
27. For the function sinz of a complex variable z, the point z = 0 is
z3
a) A pole of order 3
b) A pole of order 2
c) A pole of order 1
d) Not a singularity
Answer: (b)
Solution:
f(z) = sin Z = 1 �z − Z3 + Z5 − Z7 + ⋯ �
Z3 Z3 3! 5! 7!
= 1 - 1 + Z2 - Z4 + ….
Z2 3! 5! 7!
= 1 – 1 + (z−0)2 – (z−0)4 + ….
(z−0)2 3! 5! 7!
⸫ z = 0 is a pole of order 2
28. If a complex number z = √3 + i 1 then z4 is
2 2
a) 2√2 + 2i
b) − 1 + i √3
2 2
c) √3 − i 1
2 2
d) √3 − i 1
8 8
16
Answer: (b)
Solution:
z = √3 + i�21�
2
iz = – 1 + √3 I = w
2 2
Where w is root of x3 = 1
i4z4 = w4
⇒ z4 = w (⸫ w3 = 1)
29. The residue of the function f(z) = 1 at z = 2 is
(z+2)2(z−2)2
a) −1
32
b) −1
16
c) 1
16
d) 1
32
Answer: (a)
Solution:
f(Z) = 1
(z+2)2 (z−2)2
Z = 2 is a pole of f(z) of order 2
Res (f(z): z = z0)
= 1 �logZ → Z0 dm−1 ((Z − Z0)m f(Z))�
(m−1)! dZm−1
Res (f(Z): Z = 2)
= 1 �logZ → 2 d2−1 �(Z − 2)2 (z+2)21(z−2)2��
(2−1)! dZ2−1
= logZ → 2 �(z(+−22))3� = −1
32
17
30. Give a > 0, we wish to calculate its reciprocal value 1 by using
a
Newton-Raphson method for f(x) = 0. For a = 7 and starting with
x0 = 0.2 the first two iterations will be
a) 0.11, 0.1299
b) 0.12, 0.1392
c) 0.12, 0.1416
d) 0.13, 0.1428
Answer: (b)
Solution:
Let x = 1 (or) 1 – a = 0
a x
Taking f(x) = 1 – N
x
⇒ f 1(x) = –x−2
Newton Raphson formula gives
xn+1 = xn – f(xn)
f1(xn)
= xn – �x1n − a� = 2xn – axn2
−xn−2
xn+1 = 2xn – axn2 for n = 0, 1, 2, …..
Give a = 7 and x0 = 0.2
x1 = 2x0 – ax02 = 2(0.2) – 7(0.2)2
= 0.4 – 7(0.04) = 0.12
⸫ x1 = 0.12
x2 = 2x1 – ax12 = 2(0.12) – 7(0.12)2
= 0.24 – 7(0.0144)
= 0.24 – 0.1008 = 0.1392
18
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