Engineering Mathematics Test - 4 - PDF Flipbook

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GATE
EEE

Engineering
Mathematics

Test-04Solutions

ENGINEERING MATHEMATICS

1. In the Gauss-elimination for a solving system of linear algebraic

equations, triangularization leads to

a) Diagonal matrix

b) Lower triangular matrix

c) Upper triangular matrix

d) Singular matrix

Answer: (c)

Solution:

In the Gauss-elimination method, the coefficient matrix of a

given system reduces to upper-triangular matrix.

2. If A and B are two matrices and AB exists then BA exists,

a) Only if A has as many rows as B has columns

b) Only if both A and B are square matrices

c) Only if A and B are skew matrices

d) Only if both A and B are symmetric

Answer: (a)

Solution:

If Am×n and Bp×q are two matrices and m = q, p = n then both

AB and BA exist.

14 87

3. The rank of the matrix �04 0 3 01� is
2 3

3 12 24 2

a) 3

b) 1

1

c) 2

d) 4

Answer: (d)

Solution:

14 87

Given A = �04 0 3 01�
2 3

3 12 24 2

R3 → R3 − 4R1; R4 → R4 − 3R1

14 8 7

�00 0 3 −027�
−14 −29
0 0 0 −19

R2 ↔ R3

14 8 7
−14 −29 −027�
�00 0 3
0 0 0 −19

∴ ρ(A) = 4

4. For what values of ‘a’ if any will the following system of

equations in x, y, and z have a solution?

2x + 3y = 4, x + y + z = 4, x + 2y – z = a

a) any real number

b) 0

c) 1

d) there is no such value

Answer: (b)

2

Solution:
1 1 14

Consider [A|B] = �2 3 0 � 4�
1 2 −1 a

R2 → R2 − 2R1; R3 → R3 − R1
11 1 4

~ �0 1 −2� −4 �
0 1 −2 a − 4

R3 → R3 − R2
11 1 4

~ �0 1 −2� −4�
00 0 a

If a = 0 then the system will have solution
5. The following system of equations x + y + z = 3, x + 2y + 3z =

4, x + 4y + kz = 6 will not have a unique solution for k equal to
a) 0
b) 5
c) 6
d) 7
Answer: (d)
Solution:

Given AX = B
111x 3

⇒ �1 2 3� �y� = �4�
14kz 6
1 1 13

[A|B] = �1 2 3� 4�
1 4 k6

3

11 1 3
~ �0 1 2 � 1�

0 3 k−1 3
11 1 3
~ �0 1 2 � 1�
0 0 k−7 0
If k – 7 ≠ 0 then the system will have unique solution
∴ For k = 7 the system will not have unique solution.
6. The Eigen values of a 2 × 2 matrix X are –2 and –3. The Eigen
values of matrix (X + I)-1(X + 5I) are
a) –3, –4
b) –1, –2
c) –1, –3
d) –2, –4
Answer: (c)
Solution:
Eigen values of X are –2, –3
Eigen values of I are 1, 1
Eigen values of X + I are –2 + 1, -3 + 1 i.e., –1, –2
Given (X + I)-1(X + 5I) = (X + I)-1(X + I + 4I)
= (X + I)-1(X + I) + 4(X + I)-1
= I + 4(X + I)-1
∴The eigen values of I + 4(X + I)-1 are –3, –1

4

7. xli→mπ4 Sin x2−�xπ4−π4� =
a) 0

b) 1/2

c) 1

d) 2

Answer: (d)

Solution:

Let x − π = t, lti→m0 Sin 2t = 2
4 t

8. The value of the integral ∫−11 1 dx is
x2

a) 2

b) Does not exists

c) -2

d) ∞

Answer: (d)

Solution:

∫−11 1 dx = ∫−01 1 dx + ∫01 1 dx = ∞
x2 x2 x2

Improper of second kind

9. If S = ∫1∞ x−3dx then S has the value

a) − 1
3

b) 1
4

c) 1
2

d) 1

5

Answer: (c)

Solution:

∫1∞ x−3dx = x−−22�1∞ = − 1 [0 − 1] = 1
2 2

10. Consider the function y = x2 – 6x + 9. The maximum value of

y obtained when x varies over the interval 2 to 5 is

a) 1

b) 3

c) 4

d) 9

Answer: (c)

Solution:

y = x2 – 6x + 9

y1 = 2x – 6 = 0

⇒ x = 3 is a stationary point

at x = 3, we have y11 = 2
∴f(x) has a minimum at x = 3

Maximum value of y occurs at the end points of the interval.

y(2) = 1 , y(5) = 4

∴ The Maximum value of y = 4

11. 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

d) 1.22

6

Answer: (d)

Solution:

y = 2 x3�2
3

Length = ∫01 �1 + �ddyx�2 dx

= ∫01 √1 + xdx

= 2 �2√2 − 1�
3

= 1.22

12. If a� and b� are two arbitrary vectors with Magnitudes a and b

respectively, �a� × b��2 will be equal to

a) a2b2 − �a�. b��2

b) ab − a�. b�

c) a2b2 + �a�. b��2

d) ab + a�. b�

Answer: (a)

Solution:

�a� × b��2 = |a�|2�b��2sin2�a�, b��

= a2b2 �1 − cos2�a�, b���

= a2b2 �1 − �a�. b��2/a2b2�

= a2b2 − �a�. b��2

7

13. The direction of vector A is radially outward from the origin,
with |A| = Krn where r2 = x2 + y2 + z2 and K is constant. The
value of n for which ∇. A = 0 is

a) –2

b) 2

c) 1

d) 0

Answer: (a)

Solution:

|A| = Krn r�
r

Consider ∇. A� = 0

⇒ K∇. (rn−1r̅) = 0

⇒ K(n + 2)rn−1 = 0

⇒ n = −2

14. For the parallelogram OPQR shown in the sketch. �O��P� = aı̂ +

bȷ̂ and �O��R� = cı̂ + dȷ.̂ The area of the parallelogram is

Q

R

P

O

a) ad – bc
b) ac + bd
c) ad + bc
d) ab – cd

8

Answer: (a)

Solution:
Area = �a� × b�� = (ai + bj) × (ci + dj)
�O��P� × �O��R� = (ai + bj) × (ci + dj)

i jk
= �a b 0�

cd0
O���P� × �O��R� = i(0) + j(0) + k�(ad − bc)

|�O��P� × �O��R�| = ad − bc

15. Let ∇.(f V) = x2y + y2z + z2x, where f and V are scalar and

vector fields respectively. If V = yi + zj + xk, then V·(∇f) is

a) x2y + y2z + z2x

b) 2xy + 2yz + 2zx

c) x + y + z

d) 0

Answer: (a)

Solution:

∇.(f V) = (∇f).V +f (∇.V)
divV� = ∇. V� = 0

∴ ∇.(f V) = (∇f).V

= V. (∇f)
∴ V�. (∇f) = x2y + y2z + z2x (as per the given data

9

16. 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

Answer: (b)

Solution:

Refer the definition of regression model in the material book.

17. The probability that there are 53 Sundays in a randomly chosen

leap year is

a) 1
7

b) 1
14

c) 1
28

d) 2
7

Answer: (d)

Solution:

Leap year = 366 days = (52 × 7) + 2 days

Sample space = {Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu,

Thu-Fri, Fri-Sat, Sat-Sun}

Required probability = 2
7

10

18. X is uniformly distributed random variable that take values
between 0 and 1. The value of E(X3) will be
a) 0
b) 1/8

c) 1/4

d) 1/2

Answer: (c)

Solution:

For uniform distribution, f(x) = 1
b−a

For a < x < b

Here f(x) = 1 = 1
1−0

E(x3) = ∫1 x3f(x)dx = 1/4

19. A coin is tossed 4 times. What is the probability of getting

heads exactly 3 times?
a) 1/4
b) 3/8

c) 1/2
d) 3/4

Answer: (a)

Solution:

n = 4, p = ½, q = ½

P(x = 3) = 4c3 (1/2)3(1/2) = 4 = 1/4
16

(Binomial Distribution)

11

20. A fair coin is tossed independently four times. The probability

of the event "The number of times heads show up is more than

the number of times tails show up" is

a) 1/16

b) 1/8

c) 1/4

d) 5/16

Answer: (d)

Solution:

Let X be the random variable which denote number of heads.

Given n = 4

Required probability = P(X = 3) + P(X = 4)

= 4c3 (1/2)3(1/2) + 4c4 (1/2)4 = 5/16

21. The following differential equation has 3 d2y + 4 �ddyx�3 + y2 +
dx2

2=x

a) degree = 2, order = 1

b) degree = l, order =2

c) degree = 4, order = 3

d) degree = 2, order = 3

Answer: (b)

Solution:

The order and degree of the differential equation are 2 and 1.

12

22. Solution of the differential equation 3y dy 2x = 0 represents a
dx

family of

a) ellipses

b) circles

c) parabolas

d) hyperbolas

Answer: (a)

Solution:

Given dy 3y + 2x = 0
dx

⇒ 3y dy + 2x dx = 0 ⇒ 3y2 + 2x2
2 2

⇒ x2 + y2 = C
1 (3/2)

The above equation represents a family of ellipses

23. The solution of x dy + y = x4 with condition y(1) = 6
dx 5

a) y = x4 + 1
5 x

b) y = 4x4 + 4
5 5x

c) y = x4 + 1
5

d) y = x5 + 1
5

Answer: (a)

Solution:

Given dy + 1 y = x3 → (1)
dx x

And y(1) = 6�5 → (2)

13

I.F = e∫x1dx = elogx = x

The solution is xy = x5 + c → (3)
5

Using (2), (3) becomes

6 = 1 + c ⇒c=1
5 5

∴ Solution is y = x4 + 1/x
5

24. The solution of the differential equation dy − y2 = 1 satisfying
dx

the condition y(0) = 1 is

a) y = ex2

b) y = √x

c) y = cot�x + π�4�
d) y = tan�x + π�4�

Answer: (d)

Solution:

Given dy − y2 = 1 ⇒ dy = 1 + y2 → (1)
dx dx

and y(0) = 1 → (2)

(1) ⇒ ∫ dy = ∫ dx + c
1+y2

⇒ tan−1(y) = x + c

Using (2), (1) becomes,

i.e., tan−1(1) = 0 + C

⇒ π = C
4

∴ Solution is y = tan �x + π4�

14

25. The order and degree of a differential equation d3y +
dx3

4��ddyx�3 + y2 are respectively.

a) 3 and 2

b) 2 and 3

c) 3 and 3

d) 3 and 1

Answer: (a)

Solution:

Given d3y + 4��ddyx�3 + y2 = 0
dx3

⇒ d3y = −4 ��ddyx�3 + 1
dx3
y2�2

⇒ � 3 3 �2 = 16 �ddyx�3 + 16 2

∴ Order = 3 and Degree = 2

26. Laplace transform of (a + bt)2 where 'a' and 'b' are constants is

given by:

a) (a + bs)2

b) 1/(a + bs)2

c) (a2/s) + (2ab / s2) + (2b2/s3)

d) (a2/s) + (2ab / s2 + (b2 / s3)

Answer: (c)

Solution:

L{(a + bt)2} = a2L{1} + 2abL{t} + b2L{t2} = a2 + 2ab + 2b2
s s2 s3

15

27. A delayed unit step function is defined as u(t − a) =

�01,, t < a . Its laplace transform is
t ≥ a

a) ae-as

b) e-as/s

c) eas/s

d) eas/a

Answer: (b)

Solution:

L{u(t − a)} = ∫0∞ e−stu(t − a)dt = e−as
s

28. In what range should Re(s) remain so that the Laplace

transform of the function e(a+2)t+5 exists?

a) Re (s) > a + 2

b) Re(s) > a + 7

c) Re (s) < 2

d) Re (s) > a + 5

Answer: (a)

Solution:

L�e(a+2)t+5� = e5. 1
s−(a+2)

Where s > (a + 2)

29. The value of the integral ∮C −3z+4 dz, when C is the circle
z2+4z+5

|z| = 1 is given by

a) 0

b) 1
10

16

c) 4
5

d) 1

Answer: (a)

Solution:

I = ∫C −3z+4 dz, where C is |Z| = 1
z2+4z+5

The singular points of the integrand are z = -2 + i, -2 –i

But, the given contour does not contain singular points

∴ By Cauchy’s Integral theorem, we have

I = ∫C −3z+4 dz = 2πi(0) = 0
z2+4z+5

30. The estimate of ∫01.5.5 obtained using Simpson’s rule with


three-point function evaluation exceeds the exact value by

a) 0.235

b) 0.068

c) 0.024

d) 0.012

Answer: (d)

Solution:

x 0.5 1 1.5

f(x) = 1 2 1 2
x 3

By Simpson’s rule

∫01.5.5 1 = ℎ ( 0 + 4 1 + 2)
3

= 1 (2 + 4 + 0.666) = 1.111
6

17

By direct integration,

∫01.5.5 1 = 3 = 1.0986


Estimated value – Exact value = 1.1111 – 1.0986 = 0.012

18