Engineering Mathematics Test - 2 - PDF Flipbook
Engineering Mathematics Test - 2
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GATE
CSE
Engineering
Mathematics
Test-02Solutions
ENGINEERING MATHEMATICS
1. Let G1 = (V, E1) and G2 = (V, E2) be connected graphs on the
same vertex set V with more than two vertices. If G1∩G2 = (V,
E1 ∩ E2) is not a connected graph, then the graph G1∪G2 = (V,
E1 ∪ E2)
a) cannot have a cut vertex
b) must have a cycle
c) must have a cut-edge (bridge)
d) has chromatic number strictly greater than those of G1 and G2
Answer: (b)
Solution:
From the given data, we can conclude that G1 and G2 are not
same; Suppose G1 is not having any cycle. The G1 is a spanning
tree of G.
∴ G1 ∪ G2 must containatleast one cycle.
2. Which one of the following graphs is NOT planar?
a) G1
b) G2
c) G3
d) G4
Answer: (a)
1
Solution:
G1 ≅ K3,
By kurastowski’s theorem, G1 is not planer the other 3 graph
can be redrawing without cross overs.
3. Consider the undirected graph G defined as follows. The
vertices of G are bit strings of length n. We have an edge
between vertex u and vertex v if and only if u and v differ in
exactly one bit position (in other words, v can be obtained from
u by flipping a single bit). The ratio of the chromatic number of
G to the diameter of G is
a) 1/2n-1
b) 1/n
c) 2/n
d) 3/n
Answer: (c)
Solution:
G can be represented by n-cube in n-cube, all the cycles are of
even length
∴G is a bipartite graph
⟹ chromatic of G is a distance between two vertices with
differs in all n-bit position.
Diameter of G = n.
2
4. Which of the following statements is true for every planar graph
on n vertices?
a) The graph is connected
b) The graph is Eulerian
c) The graph has a vertex-cover of size at most 3n/4
d) The graph has an independent set of size at least n/3
Answer: (c)
Solution:
(a) A planer graph need not be connected
(b) A planer graph Euler path may or may not exists.
For example, K4 is a planer graph, but Euler path does not
exist.
(d) K4 is a planner graph in which maximum independent set
has only one vertex
∴ Option (d) is false
5. G is a simple undirected graph. Some vertices of G are of odd
degree. Add a node v to G and make it adjacent to each odd
degree vertex of G. The resultant graph is sure to be
a) Regular
b) complete
c) Hamiltonian
d) Euler
Answer: (d)
c
3
Solution:
In a simple undirected graph, the number of vertices with odd
degree is even.
If we add, a new vertex to all the vertices of odd degree, then
degree of every vertex in the graph is even
∴ The graph is a Euler graph
6. The degree sequence of a simple graph is the sequence of the
degrees of the nodes in the graph in decreasing order. Which of
the following sequences cannot be the degree sequence of any
graph?
I. 7, 6, 5, 4, 4, 3, 2, 1
II. 6, 6, 6, 6, 3, 3, 2, 2
III. 7, 6, 6, 4, 4, 3, 2, 2
IV. 8, 7, 7, 6, 4, 2, 1, 1
a) I and II
b) III and IV
c) IV only
d) II and IV
Answer: (d)
Solution:
Applying Havel - Hakimi result we can show that I and III can
be represented by a simple non directed graph, and sequence II
cannot be represented by a simple graph. Sequence IV cannot be
represented by a simple graph, because in a simple graph of
order 8, degree of vertex cannot be 8.
4
7. Let G be a complete undirected graph on 6 vertices. If vertices
of G are labelled, then the number of distinct cycles of length 4
in G is equal to
a) 15
b) 30
c) 90
d) 360
Answer: (a)
Solution:
In a complete undirected graph on ‘6’ vertices, any 4 vertices
from a cycle of length 4
∴ The number of distic cycle of length 4 in G = C(6, 4) = 15
8. An ordered n-tuple (d1, d2, ...., dn) with d1 ≥ d2 ≥ …. ≥ dn is
called graphic if there exists a simple undirected graph with n
vertices having degrees d1, d2, ...., dn. respectively. Which of the
following 6- tuples is NOT graphic?
a) (1, 1, 1, 1, 1, 1)
b) (2, 2, 2, 2, 2, 2)
c) (3, 3, 3, 1, 0, 0)
d) (3, 2, l, l, 1, 0)
Answer: (c)
Solution:
In option a, a sample graph with degree sequence (1, 1, 1, 1, 1,
1) is shown below.
5
In option (b, a sample graph with degree sequence (2, 2, 2, 2, 2,
2) is shown below.
In option (c), the degree sequence canot represent a simple non-
directed graph because, in a sample graph of order’6’ with 2
isolated vertices, if we have 3 verices with degree ‘3’, then a
vertex with degeree ‘1’ is not possible.
In option (d), a sample graph with degree sequence (3, 2, 1, 1, 1,
0) is shown below
9. If G is a forest with n vertices and k connected components,
how many edges does G have?
a) ⌊ / ⌋
b) ⌈ / ⌉
c) n – k
d) n – k + 1
Answer: (c)
6
Solution:
If ‘G’ is a simple graph with ‘n’ vertices and ‘k’ components,
then number of edges in the graph ≥ (n ‒ k).
A forest is a simple graph with minimum number of edges
because all the components in a forest are trees.
∴ number of edges in ‘G’ = (n ‒ k)
10. A graph is self-complementary if it is isomorphic to its
complement. For all self-complementary graphs on n vertices, n
is
a) A multiple of 4
b) Even
c) odd
d) Congruent to 0 mod 4,1mod 4.
Answer: (d)
Solution:
if G ≅ ̅ then n = 4k or 4k + 1 (k = 1, 2, 3….)
⟹ n is divisible by 4 or (n – 1) is divisible by 4.
⟹ n ≅ 0 mod 4 or n ≅ 1 mod 4
11. Let A, B, C, D be × matrices, each with non-zero
determinant. ABCD = I then B-1
a) D-1C-1A-1
b) CDA
c) ABC
d) Does not exist
Answer: (b)
7
Solution:
ABCD = I
⇒ ( )−1 = −1
D-1C-1B-1A-1 = I
D(D-1C-1B-1A-1)A = DIA
C-1B-1 = DA
C(C-1B-1) = CDA
∴ B-1 = CDA
3 4
12. For a matrix[ ] = �5 53�. Transpose of matrix is equal to the
5
inverse of the matrix, [M]T = [M]-1. The value of x if given by
a) − 4
5
b) − 3
5
c) 3
5
d) 4
5
Answer: (a)
Solution:
Given MT = M-1
⇒ =
3 43 = �01 01�
53� �54 3�
�5 55
5
8
Equating the corresponding elements on both sides, we have
= − 4
5
13. A real × matrix A = [aij] is defined as follows
�= 0 ,= ℎ , ∀ = The sum of all n eigen values of A is
a) ( +1)
2
b) ( −1)
2
c) ( +1)(2 +1)
2
d) 2
Answer: (a)
Solution:
= ⎡⎢⎢⎢001⋮ 0 0 ⋯ ⋯ 000⋮ ⎥⎥⎤⎥
1 0 ⋯ ⋯
0 3 ⋯ ⋯
⋮ ⋮ ⋯ ⋯
⎢⋮ ⋮ ⋮ ⋯ ⋯ ⋮⎥
⎣0 0 0 ⋯ ⋯ ⎦ ×
⇒Eigen values of A are 1, 2, 3, ---n
Sum of all Eigen values=1 + 2 + 3 ± – ∓n = ( +1)
2
14. Consider the following matrix A = � 2 3 �. If the eigen values
of A are 4 and 8 then
a) x = 4, y = 10
b) x = 5, y = 8
c) x = ‒3, y = 9
d) x = ‒4, y = 10
9
Answer: (d)
Solution:
Eigen values of A are 4 and 8, = �2 3 �
Sum of the Eigen values = Tr(A)
2+y=4+8
∴ = 10
Product of Eigen values = det (A)
2y – 3x = (4) (8) = 32
3x = 2y – 32
3x = 20 – 32 = –12
∴x=–4
15. Which one of the following statements is true for all real
symmetric matrices?
a) All the eigenvalues are real
b) All the eigen values are positive
c) All the eigen values are distinct
d) Sum of all the eigen values is zero
Answer: (a)
Solution:
By properties of eigen values, only (A) is true.
16. For the matrix A satisfying the equation given below, the eigen
123 123
values are [ ] �7 8 9� = �4 5 6�
456 789
a) (1, –j, j)
10
b) (1, 1, 0)
c) (1, 1, –1)
d) (1, 0, 0)
Answer: (c)
Solution:
The given matrix equation can be written as AB = C
The matrix ‘C’ can be obtained from ‘B’ by interchanging R2
and R3
The above elementary operation on 'B' is equivalent to
multiplying ‘B’ by the elementary matrix which can be obtained
from h by interchanging R2 and R3.
100
∴ = �0 0 1�
010
Now the eigenvalues of 'A' are obtained by solving
1 − 0 0
� 0 − 1 � = 0
0 1 −
⇒ (1 − )( 2 − 1) = 0
⇒ = 1,1, −1
17. The series ∑ =0 1 ( − 1)2 converges for
4
a) –2 < x < 2
b) 3 < x < 1
c) –l < x < 3
d) x < 3
Answer: (b)
11
Solution:
mth term, um = 1 ( − 1)2
4
⇒ +1 = ( −1)2 +12
4 +1
18. The function f(x) = 2x – x2 + 3
a) A maxima at x = 1 and a minima at x = 5
b) A maxima at x = 1 and a minima at x = -5
c) Only a maxima at x = 1
d) Only a minimal at x = 5
Answer: (c)
Solution:
f '(x) = 2 – 2x = 0 ⇒ x = 1
f "(x) = –2 < 0 f(x) has a maximum at x = 1.
19. At x = 0, the function f(x) = x3 + 1 has
a) a maximum value
b) a minimum value
c) a singularity
d) a point of inflection
Answer: (d)
Solution:
f (x) = x3 +1
f1(x) = 3x2 = 0 ⇒ x = 0
f 11(x) = 6x
f11(0) = 0
So, f(x) has a point of inflection at x = 0.
12
20. The Taylor series expansion of 3 sin x + 2cos x is
a) 2 + 3 − 2 − 3 + ⋯ … ..
2
b) 2 − 3 + 2 − 3 + ⋯ … ..
2
c) 2 + 3 + 2 + 3 + ⋯ … ..
2
d) 2 − 3 − 2 + 3 + ⋯ … ..
2
Answer: (a)
Solution:
3 sin x + 2cos x
= 3 �x − x3 + x5 − ⋯ … … . . � + 2 �x − x2 + x4 − ⋯ … . . �
3! 5! 2! 4!
= 2 + 3x − x2 − x3 + ⋯ … …
2
21. The maximum value of the function f(x) = ln(1+x) – x (where
x > –1) occurs at = ______.
Answer: 0
Solution:
f(x) = ln(1 + x) – x
f1(x) = 1 − 1
1+
f1(x) = 0 ⇒ x = 0
f11(x) = −1 is negative at x = 0
(1+ )2
∴f(x) is maximum at x = 0
13
22. Let f(x) = xe-x. The maximum value of the function in the
interval (0, ∞) is
a) e-1
b) e
c) 1 – e-1
d) 1 + e-1
Answer: (a)
Solution:
f '(x) = (‒xe-x + e-x)
f '(x) = 0 ⇒ x = 1
At x = 1, f "(x) < 0
∴ Maximum exists at x = 1 and is equal to f(1) = e-1
23. A vector �⃗ is given by �⃗ =x3y ⃗ ‒ x2y2 ⃗ ‒ x2yz ⃗ . Which
one of the following statements is TRUE?
a) �⃗ is solenoidal, but not irrotational
b) �⃗ is irrotational, but not solenoidal
c) �⃗ is neither solenoidal nor irrotational
d) �⃗ is both solenoidal and irrotational
Answer: (a)
Solution:
div �⃗ = 3x2y – 2x2y – x2y = 0
So, �⃗ is solenoidal vector?
14
̂ ̂ �
Curl �⃗ = � �
2 − 2
2 2
= {̂ – x2z – 0} {̂ –2xyz – 0}+ � ( – 2xy2 – x3)
Cut �⃗ ≠ 0�
So �⃗ = is NOT irrotational vector
24. Curl of vector V(x, y, z) = 2x2i + 3z2 j + y3k at x = y = z = 1 is
a) –3i
b) 3i
c) 3i – 4j
d) 3i – 6k
Answer: (a)
Solution:
Curl V = � �
2 2 3 2 2
= i(3y2 – 6z) – j(0) + k(0) = – 3i
25. Seven car accidents occurred in a week, what is the probability
that they all occurred on the same day?
a) 1
77
b) 1
76
c) 1
27
d) 7
27
Answer: (b)
15
Solution:
Probability that a car accident accrued on a particular day of the
work = 1
7
Probability that all 7 accidents occurred on a particular data of
the week = 1
77
Required probability = 7 × 1
77
= 1
76
26. A box contains 4 red balls and 6 black balls. Three balls are
selected randomly from the box one after another, without
replacement. The probability that the selected set contains one
red ball and two black balls is
a) 1/20
b) 1/12
c) 3/10
d) 1/2
Answer: (d)
Solution:
Given 4R and 6B
⟹ p[1R∩2B] = 4 1×6 2
10 3
= 4×15
120
= 1
2
16
27. Let X be a normal random variable with mean 1 and variance
4. The probability P{X
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