1. With K as a constant, the possible solution for the
first order differential equation dy/dx =
e
-3x
is
a. - 1/3 e
-3x + K (Ans)
b. - 1/3 e
3x + K
c. -3e
-3x + K
d. -3e
-x + K
Ans : dy/dx = e
-3x
dy = e
-3x dx
Taking integration on both sides,
y = e
-3x
/-3 + K = - 1/3 e-3x + K
2. Roots of the algebraic equation x
3 + x
2 + x + 1 = 0 are
a. +1, +j - j
b. +1, -1,+1
c. 0,0,0
d. -1, +j, -j (Ans)
Ans : x
3 + x
2 + x + 1 = 0
(x
2 + 1) (x +1) = 0
x
2 + 1 = 0 ⇒ x = ± j
x + 1 = 0 ⇒ x = -1
3. The Fourier series expansion f(t) = a0 + ∞Σn = 1 an cos nωt + bn sin nωt of the periodic
signal shown below will contain the following non-zero terms
f (t)
t
0
a. a0 and bn, n = 1, 3, 5, ... ∞
b. a0, an and bn, n = 1, 2, 3, ... ∞
c. a0 and an, n = 1, 2, 3, ... ∞
d. a0 and an, n = 1, 3, 5, ... ∞ (Ans)
Ans : The given signal satisfies even symmetry so bn = 0 and it also satisfies half wave
symmetry so, it will contain only odd harmonics.
4. The value of the quantity P, where P = ∫1
0 xe
x dx, is equal to
a. 0
b. 1 (Ans)
c. e
d. 1/e
Ans : P = ∫1
0 xe
x dx
= [x ∫e
x dx]1
0 - ∫1
0 1.e
x dx
= e - e + e
0 = 1
5. Divergence of the three-dimensional radial vector field →r is —
a. 3 (Ans)
b. 1/r
c. Îi + Ĵ + ^ k
d. 3( Î + Ĵ + ^ k )
7. The trace and determinant of a 2 * 2 matrix are known to be - 2 and - 35 respectively. Its eigen values are
a. - 30 and - 5
b. - 37 and - 1
c. - 7 and 5 (Ans)
d. 17.5 and - 2
Ans : Given, trace of 2 * 2 matrix = -2 and determinant = -35
Let A = [a11 a12]
a21 a22 2 * 2
then, trace of matrix
A = Addition of principal diagonal elements
- 2 = a11 + a22 ....(i)
and ⃒A⃒ =a11 . a22 - a12 . a21 = - 35 .....(ii)
Also the eigen values of [A] = Trace of matrix A
λ 1+ λ 2 = - 2 ........(iii)
So, option (c) will satisfy Eq. (iii).
Hence, eigen values are - 7 and 5.
8. X is a uniformly distributed random variable that takes values between O and 1. The value of E{X 3} will be
a. 0
b. 1/8
c. 1/4 (Ans)
d. 1/2
Ans :
Since, X is a uniformly distributed random variable between (0,1) = (a, b)
Then, probability density density function
f (x) = 1/(b - a) ⇒ 1/(1 - 0) ⇒ 1
So, f (x) = {1, 0 < x < 1
0, other value of x
So, E(x
3) = ∫1
x = 0 x
3 . f (x) dx
= ∫1
0 x
3 . 1 dx
= [x
4 /4] 1
0 = 1/4 (1 - 0)
= 1/4
9. The characteristic equation of a (3 * 3) matrix P is defined as a (λ) = ⃒λ| - P⃒ = λ
3 + λ
2 +
2λ + I = 0
If I denotes identity matrix, then the inverse of matrix P will be
a. (P
2 + P + 2I)a. (P
2 + P + 2I)
b. (P
2 + P + I)
c. - (P
2 + P + I)
d. - (P
2 + P + 2I) (Ans)
Ans : Given, the characteristic equation is
λ
3 + λ
2 + 2λ + I = 0 ..........(i)
We know that, By Caylay - Hamilton theorem Every square matrix satisfies its
characteristic equation.
Then, from Eq. (i), putting (λ = P),
P
3 + P
2 + 2P + I = 0
Operating (P
-1) on both sides,
(P
-1 P
3) + (P
-1 P
2) + 2(P
-1 P) + (P
-1 I) = (0.P
-1)
⇒ P
2 + P + 2I + P
-1 = 0
⇒ P
-1 = - (P
2 + P + 2I)
10. If the rank of a (5 * 6) matrix Q is 4, then which one of the following statements is correct?
a. Q will have four linearly independent rows and four linearly independent columns (Ans)
b. Q will have four linearly independent rows and five linearly independent columns
c. QQT will be invertible
d. QT Q will be invertible
12. In the matrix equation px = q, which of the following is a necessary condition for the existence of at least one solution for the unknown vector x ?
a. Augmented matrix [pq] must have the same rank as matrix p (Ans)
b. Vector q must have only non - zero elements
c. Matrix p must be singular
d. Matrix p must be square
Ans : The matrix equation px = q will have solutions if it is consistent.
So, for consistency of matrix equation is Rank of [pq] = Rank of [p]
where, [pq] → Augmented matrix
13. 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) (Ans)
Ans : Given, P and Q are two random events, then
(a) Independence of P and Q, Prob (P ⌒ Q) = Prob (P) . Prob (Q)
If two events P and Q are mutually disjoint, then Prob (P ⌒ Q) = 0
So, option (a) is incorrect.
(b) For two events P and Q,
Prob (P ∪Q) = Prob (P) + Prob (Q) - Prob (P ⌒ Q)
Here, Prob (P ∪Q) ≤ Prob (P) + Prob (Q)
So, option (b) is incorrect.
(c) There is no relation between mutually exclusive and independent for two random variables
P and Q.
So, option (c) is incorrect.
(d) And Prob (P ⌒ Q) ≤ Prob (P) which is true for every events.
So, option (d) is correct.
14. If S = ∫∞1 x
-3 dx, then S has the value
a. -1/3
b. 1/4
c. 1/2 (Ans)
d. 1
Ans : S = ∫∞1 x
-3 dx
= [x
-2 /-2]∞1 = - 1/2[1/x
2]∞1 = 1/2 [1/∞ - 1/1] = - 1/2(0-1)
= 1/2
15. The solution of the first order differential equation x '(t) = -3x(t), x(0) = x0 is
a. x (t) = x0e
-3t (Ans)
b. x (t) = x0e
-3
c. x (t) = x0e
-1/3
d. x (t) = x0e
-t
Ans : x '(t) = -3x (t) ⇒ dx/dt = -3x
⇒ ∫dx/x = -∫3 dt (on integrating)
⇒ log x = -3t + log c
⇒ x = ce
-3t .........(i)
But at (t = 0) and (x = x0),
⇒ x0 = ce
0 ⇒ (c = x0)
Then, from Eq. (i), x = x0 e
-3t
or x (t) = x0 e
-3t
16. The two vectors [1,1,1] and [1,a,a
2], where a = (- 1/2 + j √3/2) are
a. orthonormal
b. orthogonal (Ans)
c. parallel
d. collinear
Ans : These two vectors is orthogonal because the dot product of these two vectors is zero.
17. The matrix [A] = 2 1 is decomposed into a product of a lower triangular matrix [L]
and an upper triangular matrix [U].
4 -1
The properly decomposed [L] and [U] matrices respectively are
a. 1 0 and 1 1
4 -1 0 -2
b. 2 0 and 1 1
4 -1 0 1
c. 1 0 and 2 1
4 1 0 -1
d. 2 0 1 0.5 (Ans)
4 -3 and 0 1
Ans : [A] = [L] [U]
= 2 0 1 0.5 = 2+0 1+0
4 -3 0 1 4+0 2-3
= 2 1
4 -1
18. The function f (x) = 2x - x
2 + 3 has
a. a maximum at x = 1 and a minimum at x = 5
b. a maximum at x = 1 and a minimum at x = -5
c. only a maximum at x = 1 (Ans)
d. only a minimum at x = 1
Ans : f (x) = 2x - x
2 + 3
f' (x) = 2 - 2x = 0
∴ x = 1
at x = 1, f" (x) = - 2 < 0 ⇒ f" (x) < 0
So, f (x) having only a maximum at x = 1.
19. At t = 0, the function f (t) = sin t/ t has
a. a minimum
b. a discontinuity
c. a point of inflection
d. a maximum (Ans)
Ans : Given, f (t) = sin t/ t ;
At t = 0, first we will check continuity of the function.
LHL f (0 - h) = lim sin (0 - h)
h → 0 (0 - h)
= lim - sin h [ ∵ sin (-θ) = - sin θ]
h → 0 -h
= 1 [ ∵ lim sin θ/θ = 1]
h → 0
RHL f (0 + h) = lim sin (0 + h)/(0 + h)
h → 0
= lim sin h/h = 1
h → 0and f(0) = 1
since, LHL = RHL = f(0)
So, the function is continuous at t = 0.
Now, we check the function is maximum or minimum.
f '(t) = 1/t cos t -1/ t2 sin t
and f "(t) = 1/t sin t - 1/t2 cos t - 1/t2 cos t + 2 sin t/ t3
= - sin t /t - 2 cos t/ t2 + 2 sin t/t3
For max or min value of f (x),
f '(x) = 0
cos t /t - sin t/ t2 = 0 ⇒ tan t/t = 1
Now, lim f "(t) = - lim sin t/t
t → 0 t → 0
+ lim (2 sin t - 2 t cos t)/t3 [∴(0/0) form]
t → 0
= - 1 + lim (2 cos t - 2 cos t + 2 t sin t)/3 t2 [∵ Using L hospital's rule]
t → 0
= - 1 + lim 2 t sin t/3 t
t → 0
= - 1 + 2/3 lim sin t/t
t → 0
= -1 + 2/3 * 1 = - 1/3 < 0 (Maxima)
So, function f (t) is maximum at t = 0.
20. A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is
a. 1/3
b. 3/7
c. 1/2 (Ans)
d. 4/7
Ans : Probability (IInd ball is red/lst ball is white)
= P (IInd is red and Ist is white)
P (Ist ball is white)
= P (Ist ball is white and IInd ball is red)
P (Ist ball is white)
(Probability of Ist ball is white)
= x (Probability of II nd ball is red)
(Probability of Ist ball is white)
= Probability of IInd ball is red
= 3/6 = 1/2
e
-3x
is
a. - 1/3 e
-3x + K (Ans)
b. - 1/3 e
3x + K
c. -3e
-3x + K
d. -3e
-x + K
Ans : dy/dx = e
-3x
dy = e
-3x dx
Taking integration on both sides,
y = e
-3x
/-3 + K = - 1/3 e-3x + K
2. Roots of the algebraic equation x
3 + x
2 + x + 1 = 0 are
a. +1, +j - j
b. +1, -1,+1
c. 0,0,0
d. -1, +j, -j (Ans)
Ans : x
3 + x
2 + x + 1 = 0
(x
2 + 1) (x +1) = 0
x
2 + 1 = 0 ⇒ x = ± j
x + 1 = 0 ⇒ x = -1
3. The Fourier series expansion f(t) = a0 + ∞Σn = 1 an cos nωt + bn sin nωt of the periodic
signal shown below will contain the following non-zero terms
f (t)
t
0
a. a0 and bn, n = 1, 3, 5, ... ∞
b. a0, an and bn, n = 1, 2, 3, ... ∞
c. a0 and an, n = 1, 2, 3, ... ∞
d. a0 and an, n = 1, 3, 5, ... ∞ (Ans)
Ans : The given signal satisfies even symmetry so bn = 0 and it also satisfies half wave
symmetry so, it will contain only odd harmonics.
4. The value of the quantity P, where P = ∫1
0 xe
x dx, is equal to
a. 0
b. 1 (Ans)
c. e
d. 1/e
Ans : P = ∫1
0 xe
x dx
= [x ∫e
x dx]1
0 - ∫1
0 1.e
x dx
= e - e + e
0 = 1
5. Divergence of the three-dimensional radial vector field →r is —
a. 3 (Ans)
b. 1/r
c. Îi + Ĵ + ^ k
d. 3( Î + Ĵ + ^ k )
7. The trace and determinant of a 2 * 2 matrix are known to be - 2 and - 35 respectively. Its eigen values are
a. - 30 and - 5
b. - 37 and - 1
c. - 7 and 5 (Ans)
d. 17.5 and - 2
Ans : Given, trace of 2 * 2 matrix = -2 and determinant = -35
Let A = [a11 a12]
a21 a22 2 * 2
then, trace of matrix
A = Addition of principal diagonal elements
- 2 = a11 + a22 ....(i)
and ⃒A⃒ =a11 . a22 - a12 . a21 = - 35 .....(ii)
Also the eigen values of [A] = Trace of matrix A
λ 1+ λ 2 = - 2 ........(iii)
So, option (c) will satisfy Eq. (iii).
Hence, eigen values are - 7 and 5.
8. X is a uniformly distributed random variable that takes values between O and 1. The value of E{X 3} will be
a. 0
b. 1/8
c. 1/4 (Ans)
d. 1/2
Ans :
Since, X is a uniformly distributed random variable between (0,1) = (a, b)
Then, probability density density function
f (x) = 1/(b - a) ⇒ 1/(1 - 0) ⇒ 1
So, f (x) = {1, 0 < x < 1
0, other value of x
So, E(x
3) = ∫1
x = 0 x
3 . f (x) dx
= ∫1
0 x
3 . 1 dx
= [x
4 /4] 1
0 = 1/4 (1 - 0)
= 1/4
9. The characteristic equation of a (3 * 3) matrix P is defined as a (λ) = ⃒λ| - P⃒ = λ
3 + λ
2 +
2λ + I = 0
If I denotes identity matrix, then the inverse of matrix P will be
a. (P
2 + P + 2I)a. (P
2 + P + 2I)
b. (P
2 + P + I)
c. - (P
2 + P + I)
d. - (P
2 + P + 2I) (Ans)
Ans : Given, the characteristic equation is
λ
3 + λ
2 + 2λ + I = 0 ..........(i)
We know that, By Caylay - Hamilton theorem Every square matrix satisfies its
characteristic equation.
Then, from Eq. (i), putting (λ = P),
P
3 + P
2 + 2P + I = 0
Operating (P
-1) on both sides,
(P
-1 P
3) + (P
-1 P
2) + 2(P
-1 P) + (P
-1 I) = (0.P
-1)
⇒ P
2 + P + 2I + P
-1 = 0
⇒ P
-1 = - (P
2 + P + 2I)
10. If the rank of a (5 * 6) matrix Q is 4, then which one of the following statements is correct?
a. Q will have four linearly independent rows and four linearly independent columns (Ans)
b. Q will have four linearly independent rows and five linearly independent columns
c. QQT will be invertible
d. QT Q will be invertible
12. In the matrix equation px = q, which of the following is a necessary condition for the existence of at least one solution for the unknown vector x ?
a. Augmented matrix [pq] must have the same rank as matrix p (Ans)
b. Vector q must have only non - zero elements
c. Matrix p must be singular
d. Matrix p must be square
Ans : The matrix equation px = q will have solutions if it is consistent.
So, for consistency of matrix equation is Rank of [pq] = Rank of [p]
where, [pq] → Augmented matrix
13. 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) (Ans)
Ans : Given, P and Q are two random events, then
(a) Independence of P and Q, Prob (P ⌒ Q) = Prob (P) . Prob (Q)
If two events P and Q are mutually disjoint, then Prob (P ⌒ Q) = 0
So, option (a) is incorrect.
(b) For two events P and Q,
Prob (P ∪Q) = Prob (P) + Prob (Q) - Prob (P ⌒ Q)
Here, Prob (P ∪Q) ≤ Prob (P) + Prob (Q)
So, option (b) is incorrect.
(c) There is no relation between mutually exclusive and independent for two random variables
P and Q.
So, option (c) is incorrect.
(d) And Prob (P ⌒ Q) ≤ Prob (P) which is true for every events.
So, option (d) is correct.
14. If S = ∫∞1 x
-3 dx, then S has the value
a. -1/3
b. 1/4
c. 1/2 (Ans)
d. 1
Ans : S = ∫∞1 x
-3 dx
= [x
-2 /-2]∞1 = - 1/2[1/x
2]∞1 = 1/2 [1/∞ - 1/1] = - 1/2(0-1)
= 1/2
15. The solution of the first order differential equation x '(t) = -3x(t), x(0) = x0 is
a. x (t) = x0e
-3t (Ans)
b. x (t) = x0e
-3
c. x (t) = x0e
-1/3
d. x (t) = x0e
-t
Ans : x '(t) = -3x (t) ⇒ dx/dt = -3x
⇒ ∫dx/x = -∫3 dt (on integrating)
⇒ log x = -3t + log c
⇒ x = ce
-3t .........(i)
But at (t = 0) and (x = x0),
⇒ x0 = ce
0 ⇒ (c = x0)
Then, from Eq. (i), x = x0 e
-3t
or x (t) = x0 e
-3t
16. The two vectors [1,1,1] and [1,a,a
2], where a = (- 1/2 + j √3/2) are
a. orthonormal
b. orthogonal (Ans)
c. parallel
d. collinear
Ans : These two vectors is orthogonal because the dot product of these two vectors is zero.
17. The matrix [A] = 2 1 is decomposed into a product of a lower triangular matrix [L]
and an upper triangular matrix [U].
4 -1
The properly decomposed [L] and [U] matrices respectively are
a. 1 0 and 1 1
4 -1 0 -2
b. 2 0 and 1 1
4 -1 0 1
c. 1 0 and 2 1
4 1 0 -1
d. 2 0 1 0.5 (Ans)
4 -3 and 0 1
Ans : [A] = [L] [U]
= 2 0 1 0.5 = 2+0 1+0
4 -3 0 1 4+0 2-3
= 2 1
4 -1
18. The function f (x) = 2x - x
2 + 3 has
a. a maximum at x = 1 and a minimum at x = 5
b. a maximum at x = 1 and a minimum at x = -5
c. only a maximum at x = 1 (Ans)
d. only a minimum at x = 1
Ans : f (x) = 2x - x
2 + 3
f' (x) = 2 - 2x = 0
∴ x = 1
at x = 1, f" (x) = - 2 < 0 ⇒ f" (x) < 0
So, f (x) having only a maximum at x = 1.
19. At t = 0, the function f (t) = sin t/ t has
a. a minimum
b. a discontinuity
c. a point of inflection
d. a maximum (Ans)
Ans : Given, f (t) = sin t/ t ;
At t = 0, first we will check continuity of the function.
LHL f (0 - h) = lim sin (0 - h)
h → 0 (0 - h)
= lim - sin h [ ∵ sin (-θ) = - sin θ]
h → 0 -h
= 1 [ ∵ lim sin θ/θ = 1]
h → 0
RHL f (0 + h) = lim sin (0 + h)/(0 + h)
h → 0
= lim sin h/h = 1
h → 0and f(0) = 1
since, LHL = RHL = f(0)
So, the function is continuous at t = 0.
Now, we check the function is maximum or minimum.
f '(t) = 1/t cos t -1/ t2 sin t
and f "(t) = 1/t sin t - 1/t2 cos t - 1/t2 cos t + 2 sin t/ t3
= - sin t /t - 2 cos t/ t2 + 2 sin t/t3
For max or min value of f (x),
f '(x) = 0
cos t /t - sin t/ t2 = 0 ⇒ tan t/t = 1
Now, lim f "(t) = - lim sin t/t
t → 0 t → 0
+ lim (2 sin t - 2 t cos t)/t3 [∴(0/0) form]
t → 0
= - 1 + lim (2 cos t - 2 cos t + 2 t sin t)/3 t2 [∵ Using L hospital's rule]
t → 0
= - 1 + lim 2 t sin t/3 t
t → 0
= - 1 + 2/3 lim sin t/t
t → 0
= -1 + 2/3 * 1 = - 1/3 < 0 (Maxima)
So, function f (t) is maximum at t = 0.
20. A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is
a. 1/3
b. 3/7
c. 1/2 (Ans)
d. 4/7
Ans : Probability (IInd ball is red/lst ball is white)
= P (IInd is red and Ist is white)
P (Ist ball is white)
= P (Ist ball is white and IInd ball is red)
P (Ist ball is white)
(Probability of Ist ball is white)
= x (Probability of II nd ball is red)
(Probability of Ist ball is white)
= Probability of IInd ball is red
= 3/6 = 1/2
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