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
Answer: (a)
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
Answer: (b)
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
Answer: (b)
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
Answer: (c)
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))
Answer: (c)
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.
Answer: (c)
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
Answer: (c)
Solution:
The relation R divides the set of integers into two equivalent
classes given below
[0] = {0, ± 2, ± 4, ± 6, ….}
[1] = {± 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
Answer: (c)
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
Answer: (d)
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
Answer: (c)
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
Answer: (c)
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
Answer: (d)
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
Answer: (a)
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.
Answer: (a)
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
Answer: (b)
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
Answer: (b)
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)
Answer: (a)
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) ∞
Answer: (c)
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
Answer: (a)
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
Answer: (d)
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) ∞
Answer: (a)
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
Answer: (a)
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
Answer: (a)
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
Answer: (b)
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
Answer: (c)
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
Answer: (c)
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
Answer: (b)
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
Answer: (b)
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
symmetric about x = a).
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 = ______.
Answer: 11.9
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
Answer: (d)
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
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