Engineering Mathematics Test - 1 - PDF Flipbook
Engineering Mathematics Test - 1
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GATE
MECHANICAL
Engineering
Mathematics
Test-01Solutions
ENGINEERING MATHEMATICS
1. A set of linear equations is represented by the matrix equations Ax
= b. The necessary condition for the existence of a solution for this
system is
a) A must be invertible
b) b must be linearly dependent on the columns of A
c) b must be linearly independent on the rows of A
d) none
Answer: (b)
Solution:
If the system is inconsistent then ρ(A) = ρ(Ab) number of linearly
independent columns of A.
∴ The column b must be linearly dependent on the columns of A.
2. If A is any n × n matrix and k is a scalar then |kA| = α|A| where α
is
a) Kn
b) nk
c) kn
d) k
n
Answer: (c)
Solution:
By the properties of determinant of matrices,
we have |kAn×n| = kn|An×n|
∴ α = kn where k is a scalar.
1
123
3. The rank of matrix A = �3 4 5� is
468
a) 0
b) 1
c) 2
d) 3
Answer: (c)
Solution:
123
Given A = �3 4 5�
468
R2 → R2 − 3R1; R3 → R3 − 4R1
12 3
~ �0 −2 −4�
0 −2 −4
R3 → R3 − R2
12 3
~ �0 −2 −4�
00 0
∴ ρ(A) = 2.
4. The necessary condition to diagonalise a matrix is that
a) All its Eigen values should be distinct
b) Its Eigen vectors should be independent
c) Its Eigen values should be real
d) The matrix is non-singular
Answer: (b)
2
Solution:
By a theorem, A square matrix A of order n is diagonalizable if and
only if it has n linearly independent Eigen vectors.
5. The product [P] [Q]T of the following two matrices [P] and [Q]
where [P] = �24 35�, [Q] = �49 28� is
a) �5362 2464�
b) �4246 5362�
c) �6315 4222�
d) �3242 5466�
Answer: (a)
Solution:
Given [P] = �24 53� and [Q] = �49 82�
[P][Q]T = �24 53� �84 92�
= �3562 4246�
6. Eigen values of the following matrix are �−41 −41�
a) 3, –5
b) –3, 5
c) –3, –5
d) 3, 5
Answer: (a)
3
Solution:
Given A = �−41 −41�
|A − λI| = 0
⇒ �−14− λ 4 λ� = 0
−1 −
⇒ λ2 + 2λ − 15 = 0
⇒ λ = 3, −5
7. A system of equations represented by AX = 0 where X is a column
vector of unknown and A is a matrix containing coefficients has a
non-trivial solution when A is.
a) Non-singular
b) Singular
c) Symmetric
d) Hermitian
Answer: (b)
Solution:
Given system is homogeneous system
AX = 0. This homogeneous system will have trivial and non-trivial
solution.
If the coefficient matrix A is non-singular in AX = 0 then AX = 0
will have trivial solution.
If the coefficient matrix A is singular matrix in AX = 0 then AX =
0 will have nontrivial solution.
Hence option (b) is correct.
4
8. The integration of ∫ logx dx has the value
a) (x log x – 1)
b) log x – x
c) x (log x – 1)
d) None of the above
Answer: (c)
Solution:
Using integration by parts
∫ logx dx = x logx − ∫ 1 x. dx
x
= x logx − x
9. θli→m0 sinθmθ, where m is an integer, is one of the following:
a) m
b) mπ
c) mθ
d) 1
Answer: (a)
Solution:
Standard limit formula, xli→m0 sin mx = m
x
10. Limit of the function f(x) = 1−a4 as x → ∞ is given by
x4
a) 1
b) e−a4
c) ∞
d) 0
5
Answer: (d)
Solution:
xl→im∞ 1−a4 = 0
x4
11. Limit of the following series as x approaches π is
2
f(x) = x − x3 + x5 − x7 + − − − − −
3! 5! 7!
a) 2π
3
b) π
2
c) π
3
d) 1
Answer: (d)
Solution:
f(x) = x − x3 + x5 − ⋯ … … ..
3! 5!
= sinx
xli→mπ4 sinx = 1
12. xli→m0 ex−�1+x+x22� =
x3
a) 0
b) 1
6
c) 1
3
d) 1
Answer: (b)
6
Solution:
Use expansion of ex and apply L hospital’s rule
= xli→m0 ex−�1+x+x22�
x3
= xli→m0 �x33! +x44! +−−−∞�
x3
= 1
6
13. Given a vector field F�, the divergence theorem states that
a) ∫S F�. �d��s = ∫V ∇. F� dv
b) ∫S F�. �d��s = ∫V ∇� × F� dv
c) ∫S F� × �d��s = ∫V ∇. F� dv
d) ∫S F� × �d��s = ∫V ∇. F� dv
Answer: (a)
14. The divergence of the vector field (x − y)ı̅ + (y − x)ȷ̅ + (x + y +
z)k� is
a) 0
b) 1
c) 2
d) 3
Answer: (d)
Solution:
V = (x – y)i + (y – x) j + (x + y + z) k
div V = 1 + 1 + 1 = 3
7
15. Match the following.
P. Stoke's Theorem
Q. Gauss's Theorem
R. Divergence Theorem
S. Cauchy's Integral Theorem
1. ∯ D. ds = Q
2. ∮ f(z)dz = 0
3. ∭(∇. A)dv = ∯ A. ds
4. ∬(∇ × A). ds = ∮ A. dl
Codes:
P QRS
a) 2 1 4 3
b) 4 1 3 2
c) 4 3 1 2
d) 3 4 2 1
Answer: (b)
Solution:
Stokes theorem: ∮ A. dl = ∬(∇ × A). ds
Gauss’s theorem: ∯s D. ds = Q
Divergence theorem: ∯s A. ds = ∭(∇. A)dv
Cauchy’s integral theorem: ∮C f(Z) dz = 0
From the above theorems, option (b) is true.
8
16. Value of the integral ∮c xydy − y2dx, c is the square cut from the
first quadrant by the line x = 1 and y = 1 will be (Use Green's
theorem to change the line integral into double integral)
a) 1/2
b) 1
c) 3/2
d) 5/3
Answer: (c)
Solution:
∮C xy dy – y2dx
∮C M dx – N dy = ∫ ∫R �∂∂Nx − ∂∂My � dx dy
= ∫x1=0 ∫y1=0(y + 2y)dxdy
= 3
2
17. The differential equation y" + (x3 Sin x)5 y1 + y = Cos x3 is
a) Homogeneous
b) Non-linear
c) 2nd order linear
d) non-homogeneous with constant coefficients
Answer: (c)
9
Solution:
The given differential equation is a 2nd order linear, non-
homogeneous differential equation with variable coefficients.
18. For d2y + 4 dy + 3y = 3e2x the particular integral is
dx2 dx
a) 1 e2x
15
b) 1 e2x
5
c) 3e2x
d) C1e−x + C2e−3x
Answer: (b)
Solution:
Given (D2 + 4D + 3)y = 3 2
P.I = 3e2x
D2+4D+3
= 3e2x
4+8+3
= 3e2x
15
= e2x
5
19. Which of the following is a solution to the differential equation
d x(t) + 3x(t) = 0, x(0) = 2?
dt
a) x(t) = 3e-t
b) x(t) = 2e-3t
c) x(t) = − 3 2
2
d) x(t) = 3t2
Answer: (b)
10
Solution:
Given d x(t) + 3x(t) = 0 …… (1)
dt
And x(0) = 2 …….. (2)
⇒ dx = –3 dt
x
⇒ log x = –3k + c
x = e−3tk …….. (3)
using (2), (3) becomes
2 = e0k
⸫k=2
Hence x(t) = 2e−3t is a solution of (1)
20. The solution of the initial value problem dy = −2xy; y(0) = 2 is
dx
a) 1 + e−x2
b) 2e−x2
c) 1 + ex2
d) 2ex2
Answer: (b)
Solution:
dy = –2xy
dx
dy = –2x dx
dx
⇒ log y = –x2 + C
⸫ y = e−x2+C ……….. (1)
2 = eC (⸫ y(0) = 2)
⇒ C = log 2 ……………. (2)
11
⸫ using (2) in (1)
y = e−x2+log 2 = 2 e−x2
21. Consider the differential equation dx = 10 − 0.2x with initial
dt
condition x(0) = 1. The response x(t) for t > 0 is
a) 2 − e−0.2t
b) 2 − e0.2t
c) 50 − 49e−0.2t
d) 50 − 49e0.2t
Answer: (c)
Solution:
dx + (0.2) x = 10 [Linear differential equation]
dt
I.F = e∫(0.2)dt = e0.2t
The solution is x(t) × e0.2t = ∫ 10 e0.2t dt
⇒ x(t) = 50 + Ce−0.2t ……… (1)
x(0) = 1 ⇒ 1 = 50 + Ce−0 ⇒ C = –49
The solution is x(t) = 50 – 49e−0.2t
22. E1 and E2 are events in a probability space satisfying the following
constraints P(E1) = P(E2); P (E1 U E2) = 1; E1 & E2 are independent
then P(E1) =
a) 0
b) 1
4
c) 1
2
d) 1
12
Answer: (d)
Solution:
Let P(E1) = P(E2) = p
P(E1 ∪ E2) = p(E1) + P(E2) – P(E1 ∩ E2)
1 = p + p – p2 (⸫ E1 & E2 are indep.)
⇒p=1
23. If a fair coin is tossed 4 times, what is the probability that two
heads and two tails will result?
a) 3
8
b) 1
2
c) 5
8
d) 3
4
Answer: (a)
Solution:
Using binomial distribution
Required probability = P (X = 2) = 4C2 × �21�2 × �12�2 = 3
8
24. If P and Q are two random events, then which of the following is
true?
a) Independence of P and Q implies that probability (P ∩ Q) = 0
b) Probability (P ∩ Q) ≥ probability (P) + probability (Q)
c) If P and Q are mutually exclusive, then they must be
independent
d) Probability (P ∩ Q) probability (P)
13
Answer: (d)
Solution:
Verifying each options
Probability of p ∩ q ≤ min {P(p), P(q)}
25. If calls arrive at a telephone exchange such that the time of arrival
of any call is independent of the time of arrival of earlier or future
calls, the probability distribution function of the total number of
calls in a fixed time interval will be
a) Poisson
b) Gaussian
c) Exponential
d) Gamma
Answer: (a)
Solution:
The number of phone calls in a fixed time interval is discrete
random variable.
⸫ Option (a) follows discrete probability distribution and
remaining options (b), (c) & (d) are continuous probability
distribution.
26. 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
a) 0.001
b) 0.057
c) 0.107
14
d) 0.3
Answer: (b)
Solution:
Let p = 0.1, q = 0.9 & n = 10
Using Binomial Distribution,
The required probability = 10C3(0.1)3(0.9)7 = 0.057
27. The real part of the complex number z = x + iy is given by
a) Re(z) = z − z∗
b) Re(z) = z−z∗
2
c) Re(z) = z+z∗
2
d) Re(z) = z + z∗
Answer: (c)
Solution:
Given z = x + iy
⇒ z = z* = x – iy
⸫ Re (z) = z + z∗
2
28. ii, where i = √−1 is given by
a) 0
b) e−π/2
c) π
2
d) 1
Answer: (b)
15
Solution:
i = √−1 = x
let y = Xx ⇒ log y = x log x
⇒ log y = i log i = i log iπ
e2
⇒ log y = i × i π = - π
2 2
⇒ y = −π
e2
29. Let z3 = z�, where z is a complex number not equal to zero. Then z
is a solution of
a) z2 = 1
b) z3 = 1
c) z4 = 1
d) z9 = 1
Answer: (a)
Solution:
Given Z3 = Z
⇒ Z3 Z = ZZ = Z2
⇒ Z2 = 1
30. The formula used to compute an approximation for the second
derivative of a function fat a point x0 is
a) f(x0+h) + f(x0−h)
2
b) f(x0+h) − f(x0−h)
2h
c) f(x0+h) + 2f(x0) + f(x0−h)
h2
16
d) f(x0+h) − 2f(x0) + f(x0−h)
h2
Answer: (d)
Solution:
The finite difference approximation for the 2nd derivative of a
function f at a point x0 is f 11(x) = f(x0+h)) − 2 f(x0) + f(x0−h)
h2
17
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