# Engineering Mathematics Test - 3 - PDF Flipbook

Engineering Mathematics Test - 3

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GATE
CSE

Engineering
Mathematics

Test-03Solutions

ENGINEERING MATHEMATICS
1. Let denote the minimum degree of a vertex in a graph. For all

planar graphs on n vertices with ≥3, which one of the

following is TRUE?

a) In any planar embedding, the number of faces is at least n/2 +

2

b) In any planar embedding, the number of faces is less than n/2

+2

c) There is a planar embedding in which the number of faces is

less than n/2 + 2

d) There is a planar embedding in which the number of faces is
at most n/ +1

Solution:

For any planer graph

|V| + |R| = |E| + 2………. (1)

If ≥ than 3|V| ≤ 2|E|……….(2)

⟹3|V| ≤ 2(|V| + |R| ‒ 2) (from (1))

⟹|R| ≥ | | + 2
2

2. In a connected graph, a bridge is an edge whose removal

disconnects a graph. Which one of the following statements is

true?

a) A tree has no bridge

b) A bridge cannot be part of a simple cycle

1

c) Every edge of a clique with size > 3 is a bridge (A clique is
any complete sub graph of a graph)

d) A graph with bridge cannot have a cycle
Solution:
In a connected graph, an edge is a bridge if the edge is not a part
of any cycle in G.
∴ Option (b) is correct
3. Consider a set U of 23 different compounds in a Chemistry lab.
There is a subset S of U of 9 compounds, each of which reacts
with exactly 3 compounds of U.
Consider the following statements:

I. Each compound in U\S reacts with an odd number of
compounds.

II. At least one compound in U\S reacts with an odd number of
compounds.

III. Each compound in U\S reacts with an even number of
compounds.

Which one of the above statements is ALWAYS TRUE?
a) Only I
b) Only II
c) Only III
d) None

2

Solution:
Let us denote the problem by a non-directed graph with 23
vertices (compounds). If two compounds react with each other,
then there exists an edge between the corresponding vertices.
In the graph, we have 9 vertices with degree 3(odd degree). By
sum of degrees of vertices theorem, at least one of the remaining
vertices should have odd degree.
4. Some group (G, o) is known to be abelian. Then, which one of
the following is true for G?
a) g = g-1 for every g ∈G
b) g = g2 for every g ∈G
c) (goh)2 = g2 o h2 for every g, h ∈G
d) G is of finite order
Solution:
(goh)2 = (goh) (goh)
go(hog)oh = go(goh)oh
= (gog) o(hoh) = g2 o h2
Remaining option need not be true
5. Let R denote the set of real numbers. Let f: R x R → R x R be a
bijective function defined by f (x, y) = (x + y, x – y). The
inverse function of f is given by
a) f-1 (x, y) = � +1 , −1 �
b) f-1 (x, y) = (x – y, x + y)

3

c) f-1 (x, y) = � +2 , −2 �
d) f-1 (x, y) = (2(x –y, 2(x+y))

Solution:

f-1 (x, y) = (p, q)

⟹ f(p, q) = (x, y)

⟹ (p + q, p ‒ q) = (x, y)

⟹ p = + and q = −
2 2

f-1 (x, y) = f � +2 , −2 �

6. Which one of the following is false?

a) The set of all bijective functions on a finite set forms a group

under function composition.

b) The set {1, 2, ..., P‒1} forms a group under multiplication

mod p where p is a prime number.

c) The set of all strings over a finite alphabet Σ forms a group

under concatenation.
d) A subset s ≠ of G is a subgroup of the group 〈 ,∗〉if and

only if for any pair of elements a, b ∈ s, a * b-l ∈ s.

Solution:

The inverse of a string does not exist with respect to

concatenation

4

7. A relation R is defined on the set of integers as x-ray if, f(x + y)
is even. Which of the following statements is true?
a) R is not an equivalence relation
b) R is an equivalence relation having 1equivalence class
c) R is an equivalence relation having 2 equivalence classes
d) R is an equivalence relation having 3 equivalence classes
Solution:
The relation R divides the set of integers into two equivalent
classes given below
 = {0, ± 2, ± 4, ± 6, ….}
 = {± 1, ± 3, ± 5, ± 7, ….}

8. Consider the following statements:
S1: There exist infinite sets A, B, C such that A∪ (B∩C) is
finite.
S2: There exist two irrational numbers x and y such that (x + y)
is rational.
Which of the following is true about S1 and S2?
a) Only S1 is correct
b) Only S2 is correct
c) Both S1 and S2 are correct
d) None of S1 and S2 is correct
Solution:
B = {2, 4, 6, 8, …∞}

5

C = {1, 3, 4, 6, 8, …∞}

A = {0, ‒1, ‒2, …∞}
A ∩ (B ∪ C) = { } = a finite set

S1 is true Similarly, S2 is true for example

(2 +√3) + (2 ‒√3) = 4
9. The binary relation S = (empty set) on set A = {1, 2, 3} is

a) Neither reflexive nor symmetric

b) Symmetric and reflexive

c) Transitive and reflexive

d) Transitive and symmetric

Solution:

The empty relation on a finite set is symmetric and transitive but

not reflexive.

10. Let R and S be any two equivalence relations on a non-empty

set A. Which one of the following statements is TRUE?
a) R∪S, R∩S are both equivalence relations
b) R∪S is an equivalence relation
c) R∩S is an equivalence relation
d) Neither R∪S nor R∩S is a equivalence relation

11. = � � ; = � 2 + 2 + 2 � If the rank of matrix
+ 2 +

A is N, then the rank of matrix B is

a) N/2

6

b) N ‒ 1

c) N

d) 2N

Solution:

= � 2 + 2 + 2 �
+ 2 +

'B' can be obtained from 'A' by the elementary operations

C1 = (pC1 + qC2)

C2 = (rC1 + sC2)
p(B) = p(A) = N (∴ elementary operations do not change the

rank of the matrix)

12. Which one of the following equations is a correct identity for

arbitrary 3 3 real matrices P, Q and R?

a) P (Q +R) = PQ + RP

b) (P ‒ Q)2 = P2 ‒ 2PQ + Q2

c) det (P + Q) = det P + det Q

d) (P + Q)2 = P2 + PQ + QP + Q2

Solution:

From option (A) P(Q + R)

= (PQ + PR) ≠ (PQ + RP)

From option (B) (P ‒ Q)2 = (P ‒ Q) (P ‒ Q)
= (P2 ‒ PQ ‒ QP + Q2) ≠ (P2 ‒ 2PQ + Q2)

From option (C) det (P + Q) ≠ (det P + det Q)

7

Take P = �20 00� Q = �00 03��
�⇒ Det P =0 = Det Q

&Det(P + Q) = 6

From option (D) (P + Q)2 = (P + Q) (P + Q)

= (P2 + PQ + QP + Q2) is only correct

Matrix multiplication need not be commutative in options (A)

and (B)

13. The system of equations, given below, has

x + 2y + 4z = 2

4x + 3y + z = 5

3x + 2y + 3z = 1

a) a unique solution

b) Two solutions

c) no solution

d) More than two solutions

Solution:

Let AX = B be the g1ven system then

1242
( ) = �4 3 1 5�

3231
2 − 4 1 , 3 − 3 1

12 4 2
~ �0 −5 −15 −3�

0 −4 −9 −5
5 3 − 4 2

8

12 4 2
~ �0 −5 −15 −3 �

0 0 15 −13
∴ Rank of A = Rank of (AB) = 3 = number of variables
Hence unique solution exists.
14. Which one of the following statements is TRUE about every
n× n matrix with only real eigen values?
a) If the trace of the matrix is positive and the determinant is

negative, at least one of its eigenvalues is negative.
b) If the trace of the matrix is positive, all its eigen values are

positive.
c) If the determinant of the matrix is positive, all its eigen values

are positive.
d) If the product of the trace and determinant of the matrix is

positive, all its eigen values are positive.
Solution:
Determinant of a matrix = product of its eigen values.
∴ Determinant is negative there exists at least one eigenvalue,
which is negative.
15. The value of ‘x’ for which all the eigen­ values of the matrix

10 5 + 4
given below are real is � 20 2 �

4 2 −10
a) 5 + j
b) 5 ‒ j
c) 1 ‒ 5j

9

d) 1 + 5j

Solution:

We know that the Eigen values of Hermitian matrix are always

real.

Hence 'x' must be (5 ‒ j)

16. The maximum value of 'a' such that the matrix

−3 0 −2
� 1 −1 0 � has three linearly independent real eigen

0 −2

vectors is

a) 2
3√3

b) 1
3√3

c) 1+2√3
3√3

d) 1+√3
3√3

Solution:

The characteristic equation is

−3 − 0 −2
� 1 −1 − 0 � = 0
0 −2 −

(−3 − )(−1 − )(−2 − ) − 2 = 0

⇒ = − 1 (3 + )(1 + )(2 + )
2

For maximum or minimum of ‘a’

10

= 0 ⇒ − 1 (3 2 + 12 + 11) = 0
2

∴ = −12±√144−132 = �−2 ± √13�
2×3

2 = (−6 − 12)
2

= �−2 + √13� ; 2 = −6 < 0
2 √3

= �−2 − √13� ; 2 = 6 > 0
2 √3

∴ We get maximum value at = �−2 + √13�

and the required value of a = 1
3√3

17. Roots of the algebraic equation x3 + x2 + x + 1 = 0 are

a) (1, j, ‒j)

b) (1, ‒1, 1)

c) (0, 0, 0)

d) (‒1, j, ‒j)

Solution:

x = ‒1 is a root of x3 + x2 + x + 1 = 0

⇒ x3 + x2 + x + 1 = (x + l) (x2 + 1) = 0

⇒ x = ‒1, j, ‒j

18. Minimum of the real valued function f(x) = (x ‒ 1)2/3 occurs at

x equal to

a) −∞

b) 0

c) 1

11

d) ∞

Solution:

f(x) = ( − 2

1)3

= �( − 2 2 has no stationary points

1)3 �

∴ ( ) ≥ 0

∴ Minimum value is ‘0’ and occurs at x = 1

19. l i →m0 −
1−

a) 0

b) 1

c) 3

d) Not defined

Solution:

l i →m0 − = �00 �
1−

l i →m0 1− (Apply L – Hospital rule)

l i →m0 = 0 = 0
1

20. If a function is continuous at a point,

a) The limit of the function may not exist at the point

b) The function must be derivable at the point

c) The limit of the function at the point tends to infinity

d) The limit must exist the point and the value of limit should be

same as the value of the function at the point.

12

Solution:

If a function f(x) is continuous at x = a then l i→m f (x) = f (a) i.e.,

limit exists and is equal to function value.

21. The expression l i →m0 −1 is equal to

a) logx

b) 0

c) x logx

d) ∞

Solution:

l i →m0 −1 �00 �

l i →m0
1

(Apply L-Hospital rule by treating 'x' as constant and

differentiate w.r.t 'a' both numerator and denominator)

=1.logx

= logx

22. A function f(x) is continuous in the interval [0, 2]. It is known

that f(0) = f(2) = ‒1 and f(l) = 1. Which one of the following

statements must be true?

a) There exists a 'y' in the interval (0,1) such that f(y) = f (y +1)

b) For every 'y' in the interval (0,1), f(y) = f (2 ‒ y)

c) The maximum value of the function in the interval (0, 2) is 1

d) There exists a y in the interval (0,1) such that f(y) = ‒ f(2 ‒ y)

13

Solution:
(A) As y ∈ (0, 1); f(y) varies from ‒1 to 1 similarly f(y + 1)
varies from 1 to ‒1
∴Let g(x) = f(y) ‒ f(y + l); y∈(0, 1)
We get g(x) = 0 for some value of 'x'
i.e., f(y) = f(y + 1) for some y ∈ (0, 1)
Option (A) is true
(B) f(y) = f(2 ‒ y) only at y = 0 & y =1
∴In (0, 1) we cannot say f(y) = f(2 ‒ y)
(C) we cannot conclude that the maximum value of f(x) is '1' in
(0, 2)
(D) As y ∈ (0, 1); f(y) varies from ‒1 to 1 and ‒ f(2 ‒ y) varies
from 1 to ‒1
∴ Let g(x) = f(y) + f(2 ‒ y) ; y∈ (0, 1)
∴ g(x) = 0 for some value of 'x'
f(y) = ‒ f(2 ‒ y) for some y∈ (0, 1)
But the difference between y & (2 ‒ y) should be less than the
length of the interval '2' is not possible.
Hence (D) is false.

14

23. A scalar field is given by f = x2/3 + y2/3, where x and y are the

Cartesian coordinates. The derivative of 'f' along the line y = x

directed away from the origin at the point (8, 8) is

a) √2
3

b) √3
2

c) 2
√3

d) 3
√2

Solution:

f = x2/3+y 2/3

A unit vector along the line y = x

Is � = cos i + sin j, where = /4

� = cos 4 i + sin 4 j = 1 i +√12 ̅
√2

∇f = 2 − i + 2 − ̅
3 3

At (8, 8) ∇f = 1 I + 1 j
3 3

Directional derivative = ∇f. �

= 1 + 1
3√2 3√2

= √2
3

15

24. The area of a triangle formed by the tips of vectors � , � ̅

is

a) 1 � � − � �. ( � − ̅)
2

b) 1 �� � − � �. ( � − )̅ �
2

c) 1 �� � × � × ̅��
2

d) 1 �� � × � �. ̅�
2

Solution:

Area of triangle ABC = 1 |� � � � × � � � � |
2

= 1 |( � − � ) × ( ̅ − � )|
2

= 1 |( � − � ) × ( � − )̅ |
2

25. A box contains 10 screws, 3 of which are defective. Two

screws are drawn at random with replacement. The probability

that none of the two screws is defective will be

a) 100%

b) 50%

c) 49%

d) None

16

Solution:

Probability for the first screw drawn to be defective = 7/10

Probability for the Second screw drawn to be defective = 7/10

Required Probability = 7 × 7 = 49 = 0.49 = 49%
10 10 100

26. In a class of 200 students, 125 students have taken

programming language course, 85 students have taken data

structures course, 65 students have taken computer organization

course, 50 students have taken both programming languages and

data structures, 35 students Have taken both programming

languages and computer organization, 30 students have taken

both data structures and computer organization, 15 students

have taken all the three courses. How many students have not

taken any of the three courses?

a) 15

b) 20

c) 25

d) 35

Solution:

Given p (P) = 212050, p (D) = 28050, p (C) = 65
200

P(P∩D) = 25000,

P(P∩C) = 35
200

P(D∩C) = 23000,

17

P(P∩D∩C) = 15
200

p(P ∪ D ∪ C) = p(P) + p(D) + p(C) ‒ p(P∩D) ‒ p(D∩C) ‒ p(P ∩

C) + p( P ∩ D ∩ C) = 7
8

⟹ p ( � ∩ � ∩ ̅ = 1
8

No. of students, who have not taken any of the tree courses

= 1 × 200 = 25
8

27. Using given data points tabulated below, a straight line passing

through the origin is fitted using least square s method. The

slope of the line is

X1 2 3

y 1.5 2.2 2.7

a) 0.9

b) 1

c) 1.1

d) 1.5

Solution:

Let the required line be y = bx ____(1)

(Passing though the original)

Then the normal equation of (1) is given by ∑ = ∑ 2

⟹ ∑ = 14
∑ 2 14

=1

18

xy xy x2

1 1.5 1.5 1

2 2.2 4.4 4

3 2.7 8.1 9

∑x = 6 ∑y = 6.4 ∑xy = 14 ∑x2 = 14

28. If {x} is a continuous, real valued random variable defined

over the interval (−∞, +∞) and its occurrence is defined by the

density function given as: ( ) = 1 −21� − �2 where 'a' and
√2 ∗

'b' are the statistical attributes of the random variable {x}. The

value of the integral ∫− ∞ 1 −21� − �2 is
√2 ∗

a) 1

b) 0.5

c)

d) π
2

Solution:

The given integer represents the area under the normal curve to

the left side of the mean x =’a’ = 0.5 (the normal curve is

19

29. The security system at an IT office is composed of 10

computers of which exactly four are working. To check whether

the system is functional, the officials inspect four of the

computers picked at random (without replacement). The system

is deemed functional if at least three of the four computers

inspected are working. Let the probability that the system is

deemed functional be denoted by p. Then 100p = ______.

Solution:

P probability of picking ‘3’ working computers or ‘4’ working

computers = 4 3 6 1 + 4 4 = 25
10 4 10 4 210

Required value = 100p

= 100 × 25 = 11.9
210

30. Two players, A and B, alternately keep roiling a fair dice. The

person to get a six first wins the game. Given that player A starts

the game, the probability that A wins the game is

a) 5/11

b) 1/2

c) 7/13

d) 6/11

Solution:

The required probability = 1 + �56�2 . �65�+�65�4 . �16�+….
2

20

= 1 �1 + �56�2�−1
6

= 1 . �3116� = 6
6 11

21