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## Engineering Mathematics_Test - 05 Solutions

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# Engineering Mathematics_Test - 05 Solutions

### Engineering Mathematics_Test - 05 Solutions

GATE
ECE

Engineering
Mathematics

Test-05Solutions

ENGINEERING MATHEMATICS

1. If A is a real square matrix then AAT is

a) Unsymmetric

b) Always symmetric

c) Skew symmetric

d) Sometimes symmetric

Solution:

Given that A is real square matrix

Consider, (AAT)T = (AT)TAT
(∵ (AB)T = BTAT)
= AAT(∵ (AT)T = A)
∵ A is symmetric matrix

2. The eigen values of the matrix A = �01 10� are
a) 1, 1

b) –1, –1

c) j, –j

d) 1, –1

Solution:

Given A = �01 10� ⇒ |A − λI| = 0

⇒ �0 − λ 1 λ� ⇒ λ2 − 0. λ + (−1) = 0
1 −
0

λ2 − 1 = 0 ⇒ λ = ±1

1

50 2
3. If A = �0 3 0� then A-1 =
1
20
−2
10 0�
a) � 0 1/3 5

−2 0

502
b) �0 −1/3 0�

201

1/5 0 1/2
c) � 0 1/3 0 �

1/2 0 1

1/5 0 −1/2
d) � 0 1/3 0 �
−1/2 0
1

Solution:

502
Given A = �0 3 0�

201

3 0 −6
⇒ |A| = 3 and adj(A) = � 0 1 0 �

−6 0 15

|A|

= 1 3 0 −6 1 0 −2
3 �0 1 0 �=�0 1/3 0�
−6 0 15 −2 5
0

2

2 −1 0 0

4. The eigen values of the matrix �00 3 0 00� are
0 −2
0 0 −1 4

a) 2, –2, 1, –1

b) 2, 3, –2, 4

c) 2, 3, 1, 4

d) None

Solution:

If λ is an eigen value(s) of matrix A4×4 then the eigen value(s)

are given by |A − λI| = 0.

2−λ −1 0 0
3−λ 0
i.e. � 0 −2 − λ 0 �=0
0 0 −1 0
4−λ
0 0

3−λ 0 0
⇒ (2 − λ) � 0 −2 − λ 0 � = 0
0 −1 4 − λ

⇒ (2 − λ)(3 − λ)[−(2 + λ)(4 − λ)] = 0

⇒ (2 − λ)(3 − λ)(2 + λ)(4 − λ) = 0

∴The eigen values are given by
λ = 2, −2, 3, 4

5. The system of equations x + y + z = 6, x + 4y + 6z, x + 4y + λz

= μ has no solution for values of λ and μ given by

a) λ = 6, μ = 20

b) λ = 6, μ ≠ 20

c) λ ≠ 6, μ = 20

3

d) λ ≠ 6, μ ≠ 20

Solution:

Given AX = B

Consider, 1 1 16
[A|B] = �1 4 6� 20�

1 4 λμ

11 1 6
~ �0 3 5 � 14 �

0 3 λ−1 μ−6

11 1 6
~ �0 3 5 � 14 �

0 0 λ − 6 μ − 20
For no solution, λ = 6 and μ ≠ 20

6. One pair of eigen vectors corresponding to the two eigen values
of the matrix �10 −01� is
a) �−1j� , �−j1�
b) �01� , �−01�
c) �1j � , �01�

d) �1j � , �1j �

4

Solution:
Given A = �10 −01�

Eigen values:
∴Eigen value of A are λ =i, – i

Eigen vector:

�0 − λ 0−−1λ� �xx12� = �00� … … . . (1)
1

Case - (i) put λ = i in (1)

i. e. �−1i −−1i � �xx12� = �00�

⇒ –ix1 – x2 = 0
∴ X = �xx12� = �−1i� (or) �1i �

Case - (ii) put λ = –i in (1)

i. e. �1i −i1� �xx12� = �00�

⇒ –ix1 – x2 = 0
∴ X = �xx12� = �1i � (or) �−i1�

Hence the eigen vectors of a corresponding to eigen values

λ = i & − i are X1 = �−1i� &X2 = �1i �
Or

X1 = �1i � &X2 = �−i1�

5

7. If y = x + �x + �x + √x + ⋯ … … . . ∞ then y(2) =

a) 4 or 1
b) 4 only
c) 1 only
d) Undefined
Solution:

y = x + �x + �x + √x + ⋯ … … . . ∞

⇒ (y − x)2 = y ⇒ y2 − 2xy + x2 = y
at x = 2, y2 − 4y + 4 = y ⇒ y2 − 5y + 4 = 0

⇒ y = 1 or 4

but x = 2

8. The value of ∫0∞ ∫0∞ e−x2e−y2 dx dy is

a) √
2

b) √

c)

d)
4

Solution:

∫0α ∫0α e−x2e−y2dxdy = ∫0α ∫0α e−(x2+y2)dxdy

Put x = rcosθ, y = rsinθ, |J| = r

∫0π�2 ∫0α e−r2 rdrdθ = π
4

6

9. A parabolic cable is held between two supports at the same
level. The horizontal span between the supports is L. The sag at
the mid-span is h. The equation of the parabola is y = 4h 22,
where x is the horizontal coordinate and y is the vertical
coordinate with the origin at the centre of the cable. The
expression for the total length of the cable is

a) ∫0L �1 + 64 h2x2 dx
L4

b) 2 ∫0L/2 �1 + 64 h3x2 dx
L4

c) ∫0L/2 �1 + 64 h2x2 dx
L4

d) 2 ∫0L/2 �1 + 64 h2x2 dx
L4

Solution:

Total length = 2 ∫0L/2 �1 + �ddyx�2 dx

= 2 ∫0L/2 �1 + 64 h2x2 dx
L4

10. The area enclosed between the curves y2 = 4x and x2 = 4y is

a) 16/3

b) 8

c) 32/3

d) 16

7

Solution:

Area = ∫04 �2√x − x42� dx = 16
3

11. At t = 0, the function f(t) = has

a) a minimum

b) a discontinuity

c) a point of inflection

d) a maximum

Solution:
f(t) =sitnt

l i→m0 ( ) = 1 but f(0) does not exists
12. The magnitude of the gradient for the function f(x, y, z) = x2 +

3y2 + z3 at the point (1, 1, 1) is _______.

Solution:

∇f = ∂f ı̅ + ∂f ȷ̅ + ∂f k�
∂x dy ∂z

= 2xı̅ + 6yȷ̅ + 3z2k�

(∇f)at(1,1,1) = 2ı̅ + 6ȷ̅ + 3k�

∴ �(∇f)at(1,1,1)� = √4 + 36 + 9 = √49 = 7

8

13. The line integral of function F = yzi, in the counterclockwise

direction, along the circle x2 + y2 = 1 at z = 1 is

a) –2π

b) –π

c) π

d) 2π

Solution:
∫C F�. dr̅ = ∫C yzdx = ∫C ydx (∵ z = 1)
= ∫C(ydx − 0dy) = ∬s −dxdy

(from Green’s theorem)

= −π

14. The integral ∮C(ydx − xdy) is evaluated along the circle x2 + y2

= 1 traversed in counter clockwise direction. The integral is
4

equal to

a) 0

b) − π
4

c) − π
2

d) π
4

Solution:

Applying Green’s theorem,

∮C(ydx − xdy) = ∬R(−1 − 1)dxdy

9

= −2 (Area of the given circle)

= −2 �π4� = − π
2

15. Directional derivative of ϕ = 2xz – y2 at the point (1, 3, 2)

becomes maximum in the direction of

a) 4i + 2j – 3k

b) 4i – 6j + 2k

c) 2i – 6j + 2k

d) 4i – 6j – 2k

Solution:

∇∅ = ∂∅ ı̅ + ∂∅ ȷ̅ + ∂∅ k�
∂x ∂y ∂z

= 2zı̅ − 2yȷ̅ + 2xk�

∴ Required direction vector = (∇∅) at (1, 3, 2) = (4ı̅ − 6ȷ̅ + 2k�)

16. Let be an arbitrary smooth real valued scalar function and �V⃗

be an arbitrary smooth vector valued function in a three ­

dimensional space. Which one of the following is an identity?
a) Curl (ϕV�⃗) = ∇(ϕDiv�V⃗)

b) Div�V⃗ = 0
c) Div Curl �V⃗ = 0
d) Div (ϕ�V⃗) = (ϕDivV�⃗)

Solution:

Option (c) is True (from vector identities)

10

17. The life of a bulb (in hours) is a random variable with an
exponential distribution f(t) = αe−αt,0 ≤ t ≤ ∞. The probability
that its value lies b/w 100 and 200 hours is
a) e–100α – e–200α
b) e–100 – e–200
c) e–100α + e–200α
d) e–200α – e–100α
Solution:
P (100 < X < 200)
= ∫120000 ( ) = ∫120000 − = −100 − −200

18. For a random variable x(−∞ < x < ∞) following normal
distribution, the mean is µ=100 If the probability is P = α for
x ≥110. Then the probability of x lying between 90 and 110 i.e.,
P (90 ≤ x ≤ 110) and equal to
a) 1 – 2α
b) 1 – α
c) 1 – α/2
d) 2α
Solution:

11

P(x ≥ 110) = α ⇒ P(x ≤ 90) = α

⇒ P(90 ≤ x ≤ 110) = 1 – 2α

19. 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

d) 1/2520

Solution:

Required probability = 2 × 1 × 3 × 2 × 1 × 4 × 3 × 2 × 1
9 8 7 6 5 5 3 2 1

= 1
1260

20. t is estimated that the average number of events during a year

is three. What is the probability of occurrence of not more than

two events over a two-year duration? Assume that the number of

a) 0.052

b) 0.062

c) 0.072

d) 0.082

12

Solution:
In a Poisson distribution mean λ = 3 per year

= 6 per two years

Required probability

= P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= e−λ �1 + λ + λ22� = 0.0619
21. The probability that a given positive integer lying between 1

and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is _____.

Solution:

Number of integers in the set which are divisible by 2 or 3 or 5
= n(2) + n(3) + n(5) − n(2⋀3) − n(3⋀5) −

n(5⋀2) + n(2⋀3⋀5)

= 50 + 33 + 20 – 16 – 10 – 6 + 3

= 74
∴ The number of integers between 1 and 100, which are not

divisible by 2 or 3 or 5 = 100 – 74 = 26

∴ Required probability = 26 = 0.26
100

22. If C is a constant, then the solution of dy = 1 + y2
dx

a) y = sin (x + c)

b) y = cos (x + c)

c) y = tan (x + c)

d) y = ex + c

13

Solution:

Given dy = 1 + y2
dx

⇒ ∫ 1 dy = ∫ dx + c
1+y2

⇒ tan−1(y) = x + c

∴ y = tan(x + c)

23. The solution to the ordinary differential equation d2y + dy −
dx2 dx

6y = 0

a) y = C1e3x + C2e-2x
b) y = C1e3x + C2e2x
c) y = C1e-3x + C2e2x
d) y = C1e-3x + C2e-2x

Solution:

Given d2y + dy − 6y = 0
dx2 dx

⇒ (D2 + D – 6)y = 0

⇒ f(D) = 0 where f(D) = D2 + D – 6

The auxiliary equation is f(D) = 0

⇒ D2 + D – 6 = 0

⇒ D = 2, -3

∴ y = C1e2x + C2e−3x (or) y = C1e−3x + C2e2x

24. The solution of differential equation dy = Ky, y(0) = C is
dx

a) x = C eky

b) x = keky

14

c) y = ekx C

d) y = Ce-kx

Solution:

Given dy = Ky ……….. (1)
dx

and y(0) = c ……….. (2)

⇒ ∫ 1 dy = K ∫ dx + c1
y

⇒ log y = Kx + c1 ⇒ y = eKx + c1

⇒ y = eKx c2

Where c2 = ec1

Using (2), (3) becomes

c = c2

∴ y = ekxc

25. Consider the differential equation dy = (1 + y2)x. The general
dx

solution with constant ‘C’ is

a) y = tan �x22� + C

b) y = tan2 �2x + C�

c) y = tan2 �2x� + C
d) y = tan �x22 + C�

15

Solution:

Given dy = (1 + y2)x ………… (1)
dx

⇒ ∫ dy = ∫ x dx + C ⇒ tan−1(y) = x2 + C
1+y2 2

∴ y = tan �x22 + C�

26. The Laplace transform of the function f(t) = k, 0 < t < c

= k, c < t < ∞, is

a) (k/s) e-sc

b) (k/s) esc

c) ke-sc

d) (k/s) (1 – e-sc)

Solution:
{ ( )} = ∫0 − . + 0 = �1− − �

27. The inverse Laplace transform of 1/(s2 + 2s) is

a) (1 – e-2t)

b) (1 + e+2t)/2

c) (1 – e+2t)/2

d) (1 – e-2t)/2

Solution:

−1 � 2+12 � = −1 � ( 1+2)� = −1 �21 �1 − +12��

= 1 [1 − −2 ]
2

16

28. If F(s) is the Laplace transform of the function f(t) then

Laplace transform of ∫01 f(x)dx is

a) 1 F(s)
s

b) 1 F(s) − f(0)
s

c) s F(s) − f(0)

d) ∫ F(s)ds

Solution:

�∫0 ( ) � = ( )

29. For an analytic function f(x + iy) = u(x, y) + iv(x, y), u is given

by u = 3x2 – 3y2. The expression for v, considering k is to be

constant is

a) 3y2 – 3x2 + k

b) 6x – 6y + k

c) 6y – 6x + k

d) 6xy + k

Solution:

Given u = 3x2 – 3y2

for f(x + iy) = u(x, y) + iv (x, y)

dv = vs dx + vs dy
= −uydx + uxdy �∵ ux = vy & vx = −uy�

dv = – (0 – 6y) dx + 6x dy which is exact differential equation,

Integrating v = 6xy + k

17

30. The integral ∫xx12 x2dx with x2 > x1> 0 is evaluated analytically
as well as numerically using a single application of the
trapezoidal rule. If I is the exact value of the integral obtained
analytically and J is the approximate value obtained using the
trapezoidal rule, which of the following statements is correct
a) J > I
b) J < I
c) J = I
d) Insufficient data to determine the relationship