engineering-mathematics

Question 1
Let G = (V, E) be an undirected unweighted connected graph. The diameter of G is defined as:

Let M be the adjacency matrix of G.
Define graph G2on the same set of vertices with adjacency matrix N, where

Which one of the following statements is true?
A
B
C
D
Question 1 Explanation: 
Question 2
A
3
Question 2 Explanation: 
Question 3
The lifetime of a component of a certain type is a random variable whose probability density function is exponentially distributed with parameter 2. For a randomly picked component of this type, the probability that its lifetime exceeds the expected lifetime (rounded to 2 decimal places) is _______.
A
0.37
Question 3 Explanation: 
Question 4
A relation R is said to be circular of aRb and bRc together imply cRa. Which of the following options is/are correct?
A
If a relation S is reflexive and circular, then S is an equivalence relation.
B
If a relation S is transitive and circular, then S is an equivalence relation.
C
If a relation S is circular and symmetric, then S is an equivalence relation.
D
If a relation S is reflexive and symmetric, then S is an equivalence relation.
Question 4 Explanation: 

Theorem: A relation R on a set A  is an equivalence relation if and only if it is reflexive and circular.

 

For symmetry, assume that x, y ∈ A so that xRy, lets check for  yRx. 

Since R is reflexive and y ∈ A, we know that yRy. Since R is circular and xRy and yRy, we know that yRx. Thus R is symmetric. 

 

For transitivity, assume that x, y, z ∈ A so that xRy and yRz. Check for  xRz. Since R is circular and xRy and yRz, we know that zRx. Since we already proved that R is symmetric, zRx implies that xRz. Thus R is transitive.

Question 5
Let G be a group of order 6, and H be a subgroup of G such that 1 < |H| < 6.
Which one of the following options is correct?
A
G is always cyclic, but H may not be cyclic.
B
Both G and H are always cyclic.
C
G may not be cyclic, but H is always cyclic.
D
Both G and H may not be cyclic.
Question 5 Explanation: 

If ‘G’ is a group with sides 6, its subgroups can have orders 1, 2, 3, 6.

(The subgroup order must divide the order of the group)

Given ‘H’ can be 1 to 6, but 4, 5 cannot divide ‘6’.  

Then ‘H’ is not a subgroup. 


G can be cyclic only if it is abelian. Thus G may or may not be cyclic.
The H can be cyclic only for the divisors of 6 and H cannot be cyclic for any non divisors of 6.

Question 6
Consider the two statements.
           S1: There exist random variables X and Y such that
                           EX-E(X)Y-E(Y)2>Var[X] Var[Y]
           S2: For all random variables X and Y,
                            CovX,Y=E|X-E[X]| |Y-E[Y]| 
Which one of the following choices is correct?
A
S1is false, but S2is true.
B
Both S1and S2are true.
C
S1is true, but S2is false.
D
Both S1and S2 are false.
Question 6 Explanation: 
Variance(X) = Var[X]= E((X-E(X))^2)
For a dataset with single values, we have variance 0. EX-E(X)Y-E(Y)2>Var[X] Var[Y]
This leads to inequance of 0>0 which is incorrect.

Its not |x-E(x)|. Thus S2 is also incorrect.
Question 7
 In an undirected connected planar graph G, there are eight vertices and five faces. The number of edges in G is ______
A
11
Question 7 Explanation: 

v - e + f = 2

v is number of vertices

e is number of edges

f is number of faces including bounded and unbounded

8-e+5=2

=> 13-2 =e

The number of edges  are =11

Question 8
A sender (S) transmits a signal, which can be one of the two kinds: H and L with probabilities 0.1 and 0.9 respectively, to a receiver (R). In the graph below, the weight of edge (u, v) is the probability of receiving v when u is transmitted, where u, v ∈ {H, L}. For example, the probability that the received signal is L given the transmitted signal was H, is 0.7.

If the received signal is H, the probability that the transmitted signal was H (rounded to 2 decimal places) is _______
A
0.04
Question 8 Explanation: 

Bayes theorem:
Probability of event A happening given that event B has already happened is
P(A/B) = P(B/A)*P(A)  / P(B)

Here, it is asked that P( H transmitted / H received).

S can send signal  to H with 0.1 probability, S can send signal to L with 0.9 probability.
The complete diagram can be



Probability that H Transmitted (H_t) given that H received (H_r)is

P( H_t  / H_r) = P( H_r/ H_t) * P(H_t)  / P(H_r)

P(H-r) = probability that H received  = P( H received from H)+ P(H received from L)
It can be observed from the graph that H can receive in two ways (S to H to H) and (S to L to H)
The P(H_r) = 0.1*0.3 + 0.9*0.8= 0.03+0.72 = 0.75

P(H_received given that H_transmitted) =0.3
P(H transmitted ) = 0.1  i.e.

P( H_t  / H_r) = P( H_r/ H_t) * P(H_t)  / P(H_r)
                        = 0.3*0.1 / 0.75 = 0.04

 

Question 9
Consider the following expression.

The value of the above expression (rounded to 2 decimal places) is _______
A
0.25
Question 9 Explanation: 
Question 10
Let p and q be two propositions. Consider the following two formulae in propositional logic.

Which one of the following choices is correct?
A
Both S1and S2 are tautologies.
B
Neither S1and S2 are tautology.
C
S1is not a tautology but S2is a tautology.
D
S1is a tautology but S2is not a tautology.
Question 10 Explanation: 

A tautology is a formula which is "always true" . That is, it is true for every assignment of truth values to its simple components.

Method 1:
S1: (~p ^ (p Vq)) →q
The implication is false only for T->F condition.
Let's consider q as false, then
(~p ^ (p Vq)) will be (~p ^ (p VF)) = (~p ^ (p)) =F.
It is always F->F which is true for implication. So there are no cases that return false, thus its always True i.e. its Tautology. 

 

S2: 

q->(~p (p Vq)) 


The false case for implication occurs at T->F case.
Let q=T then (~p (p Vq))  = (~p (p VT))= ~p. (It can be false for p=T).
So there is a case which yields T->F = F. Thus its not Valid or Not a Tautology.

Method 2:


Question 11

Let ({p,q{,*) be a semigroup where p * q = q. Show that:

 (a) p * q = q * p and
 (b) q * q = q 
A
Theory Explanation.
Question 12

Show that proposition C is a logical consequence of the formula

     A ∧ (A →(B ∨ C)) ∧ (B → ~A)  

Using truth tables.

A
Theory Explanation.
Question 13

The less-than relation, <, on reals is

A
a partial ordering since it is asymmetric and reflexive
B
a partial ordering since it is antisymmetric and reflexive
C
not a partial ordering because it is not asymmetric and not reflexive
D
not a partial ordering because it is not antisymmetric and reflexive
E
none of the above
Question 13 Explanation: 
Relation < is:
1) not reflexive
2) irreflexive
3) not symmetric
4) Asymmetric
5) Antisymmetric
Question 14

Let A and B be sets with cardinalities m and n respectively. The number of one-one mappings (injections) from A to B, when m < n, is:

A
mn
B
nPm
C
mCn
D
nCm
E
mPn
Question 14 Explanation: 
Let,
A = {a1, a2, ... am} and
B = {b1, b2, ... bn}
A one-one function 'f' assigns each element ai of A a distinct element, bj=f(ai) of Bi for a, there are n choices, for a2 there are n-1 choices, for am there are (n-(m-1)) choices.
i.e.,
Question 15

The proposition p ∧(~p ∨ q) is:

A
a tautology
B
logically equivalent to p ∧ q
C
logically equivalent to p ∨ q
D
a contradiction
E
none of the above
Question 15 Explanation: 
p ∧(~p ∨ q)
(p ∧ ~p) ∨ (p ∧ q)
F ∨ (p ∧ q)
(p ∧ q)
Question 16

Let S be an infinite set and S1 ..., Sn be sets such that S1 ∪ S2 ∪ ... ∪ Sn = S. Then,

A
at least one of the set Si is a finite set
B
not more than one of the set Si can be finite
C
at least one of the sets Si is an infinite set
D
not more than one of the sets Si can be infinite
E
None of the above
Question 16 Explanation: 
Given sets are finite union of sets. One set must be infinite to make whole thing to be infinite.
Question 17

Let A be a finite set of size n. The number of elements in the power set of A × A is:

A
22n
B
2n2
C
(2n)2
D
(22)n
E
None of the above
Question 17 Explanation: 
Cardinality of A = n × n = n2
A = {1,2}
|A| = {∅, {1}, {2}, {1,2}}
Cardinality of power set of A = 2n2
A × A = {1,2} × {1,2}
= {(1,1), (1,2), (2,1), (2,2)}
Cardinality of A × A = n2
Cardinality of power set of A × A = 2n2
Question 18

The Laplace transform of the periodic function f(t) described by the curve below, i.e.,

is _________

A
Out of syllabus.
Question 19

Given and S the surface of a unit cube with one corner at the origin and edges parallel to the coordinate axes, the value of integral

A
Out of syllabus.
Question 20

The differential equation yn + y = 0 is subjected to the boundary conditions.

    y (0) = 0        y(λ) = 0         

In order that the equation has non-trivial solution(s), the general value of λ is __________

A
Out of syllabus.
Question 21

For X = 4.0, the value of I in the FORTRAN 77 statement
1 = -2**2 + 5.0*X/X*3 + 3/4 is _______

A
Out of syllabus.
Question 22

The value of the double integral is

A
1/3
Question 22 Explanation: 
Question 23

If the matrix A4, calculated by the use of Cayley-Hamilton theorem or otherwise, is _________

A
A4 = I
Question 23 Explanation: 
Let λ be eigen value, then characteristic equation will be
(1-λ) (-1-λ) (i-λ) (-i-λ)
= (λ2-1) (λ2+1)
= λ4-1
Characteristic equation is λ4-1 = 0.
According to Cayley-Hamilton theorem, every matrix satisfies its own characteristic equation, so
A4 = I
Question 24

If the linear velocity is given by

The angular velocity at the point (1, 1, -1) is ________

A
Out of syllabus.
Question 25

Given the differential equation, y′ = x − y with the initial condition y(0) = 0. The value of y(0.1) calculated numerically upto the third place of decimal by the second order Runga-Kutta method with step size h = 0.1 is ________

A
Out of syllabus.
Question 26

The radius of convergence of the power series

A
Out of syllabus.
Question 27

The function f(x,y) = x2y - 3xy + 2y + x has

A
no local extremum
B
one local minimum but no local maximum
C
one local maximum but no local minimum
D
one local minimum and one local maximum
Question 27 Explanation: 
Note: Out of syllabus.
Question 28
A
1
Question 28 Explanation: 
Since the given expression is in 0/0 form, so we can apply L-Hospital rule.
Question 29

Which of the following improper integrals is (are) convergent?

A
B
C
D
Question 30

The differential equation
d2y/dx2 + dy/dx + siny = 0 is:

A
linear
B
non-linear
C
homogeneous
D
of degree two
Question 30 Explanation: 
Note: Out of syllabus.
d2y/dx2 + dy/dx + siny = 0
In this DE, degree is 1 then this represent linear equation.
Question 31

Simpson’s rule for integration gives exact result when f(x) is a polynomial of degree

A
1
B
2
C
3
D
4
Question 31 Explanation: 
Note: Out of syllabus.
Question 32

Fourier series of the periodic function (period 2π) defined by

But putting x = π, we get the sum of the series.

A
π2/4
B
π2/6
C
π2/8
D
π2/12
Question 32 Explanation: 
Note: Out of syllabus.
Question 33

The eigen vector(s) of the matrix

is (are)

A
(0,0,α )
B
(α,0,0)
C
(0,0,1)
D
(0,α,0)
E
Both B and D
Question 33 Explanation: 
Since, the given matrix is an upper triangular one, all eigen values are A. And hence A - λI = A.
So the question as has
(A - λI)X = 0
AX = 0

What x1, x2, x3 are suitable?
Which means:
x1 times column 1 + x2 times column 2 + x3 times column 3 = zero vector
Since α is not equal to zero, so x3 must be necessarily zero to get zero vector.
Hence, only (B) and (D) satisfies.
Question 34

Every element a of some ring (R,+,0) satisfies the equation aoa = a.
Decide whether or not the ring is commutative.

A
Theory Explanation.
Question 35

A 3-ary tree is a tree in which every internal node has exactly three children. Use induction to prove that the number of leaves in a 3-ary tree with n interval nodes is 2(n-1)+3.

A
Theory Explanation.
Question 36

Let p and q be propositions. Using only the truth table decide whether p ⇔ q does not imply p → q is true or false.

A
True
B
False
Question 36 Explanation: 

So, "imply" is False making "does not imply" True.
Question 37

(a) Let * be a Boolean operation defined as
If C = A * B then evaluate and fill in the blanks:
(i) A * A = _______
(ii) C * A = _______
(b) Solve the following boolean equations for the values of A, B and C:

A
Theory Explanation.
Question 38

Find the inverse of the matrix

A
B
C
D
Question 38 Explanation: 
Using eigen values, the characteristic equation we get is,
3 + 2λ2 - 2 = 0
Using Cayley-Hamiltonian theorem
-A3 + 2A2 - 2I = 0
So, A-1 = 1/2 (2A - A2)
Solving we get,
Question 39

Match the following items

A
(i) - (b), (ii) - (c), (iii) - (d), (iv) - (a)
Question 39 Explanation: 
Note: Out of syllabus.
Question 40

The Hasse diagrams of all the lattices with up to four elements are __________ (write all the relevant Hasse diagrams).

A
Question 40 Explanation: 
For 1 element:
We can't draw lattice with 1 element.
For 2 element:

For 3 element:

For 4 element:
Question 41

Let A, B and C be independent events which occur with probabilities 0.8, 0.5 and 0.3 respectively. The probability of occurrence of at least one of the event is __________

A
0.93
Question 41 Explanation: 
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(A∩C) + P(A∩B∩C)
Since all the events are independent, so we can write
P(A∪B∪C) = P(A) + P(B) + P(C) - P(A)P(B) - P(B)P(C) - P(A)P(C) + P(A)P(B) P(C)
= 0.8 + 0.5 + 0.3 - 0.4 - 0.5 - 0.24 + 0.12
= 0.93
Question 42

The number of subsets {1, 2, ... n} with odd cardinality is __________.

A
2n-1
Question 42 Explanation: 
Total no. of subsets with n elements is 2n.
And so, no. of subsets with odd cardinality is half of total no. of subsets = 2n /n = 2n-1
Question 43

The number of edges in a regular graph of degree d and n vertices is _________.

A
d*n/2
Question 43 Explanation: 
Sum of degree of vertices = 2 × no. of edges
d * n = 2 * |E|
∴ |E| = d*n/2
Question 44

The probability of an event B is P1. The probability that events A and B occur together is P2 while the probability that A and occur together is P3. The probability of the event A in terms of P1, P2 and P3 is __________.

A
P2 + P3
Question 44 Explanation: 
P(A∩B') = P(A) - P(A∩B)
P3 = P(A) - P2
P(A) = P2 + P3
Question 45

On the set N of non-negative integers, the binary operation __________ is associative and non-commutative.

A
fog
Question 45 Explanation: 
The most important associative operation that is not commutative is function composition. If you have two functions f and g, their composition, usually denoted fog, is defined by
(fog)(x) = f(g(x))
It is associative, (fog)oh = fo(goh), but its usually not commutative. fog is usually not equal to gof.
Note that if fog exists then gof might not even exists.
Question 46

Amongst the properties {reflexivity, symmetry, anti-symmetry, transitivity} the relation R = {(x,y) ∈ N2 | x ≠ y } satisfies __________.

A
symmetry
Question 46 Explanation: 
It is not reflexive as xRx is not possible.
It is symmetric as if xRy then yRx.
It is not antisymmetric as xRy and yRx are possible and we can have x≠y.
It is not transitive as if xRy and yRz then xRz need not be true. This is violated when x=x.
So, symmetry is the answer.
Question 47

The number of substrings (of all lengths inclusive) that can be formed from a character string of length n is

A
n
B
n2
C
n(n-1)/2
D
n(n+1)/2
Question 47 Explanation: 
No. of substrings of length
n = 1
(n-1) = 2
(n-2) = 3
So, Total = n(n+1)/2
Question 48

The tank of matrix is:

A
0
B
1
C
2
D
3
Question 48 Explanation: 
Question 49

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 = g2oh2 for every g,h ∈ G
D
G is of finite order
Question 49 Explanation: 
Associate property of a group (aob)oc = ao(boc)
For Abelian group, commutative also holds
i.e., (aob) = (boa)
Consider option (C):
(goh)2 = (goh)o(gog)
= (hog)o(goh)
= (ho(gog)oh)
= ((hog2)oh)
= (g2oh)oh
= g2o(hoh)
= g2oh2 [True]
Question 50

In a compact single dimensional array representation for lower triangular matrices (i.e all the elements above the diagonal are zero) of size n × n, non-zero elements (i.e elements of the lower triangle) of each row are stored one after another, starting from the first row, the index of the (i, j)th element of the lower triangular matrix in this new representation is:

A
i + j
B
i + j - 1
C
j + i(i-1)/2
D
i + j(j-1)/2
Question 50 Explanation: 
Though not mentioned in question, from options it is clear that array index starts from 1 and not 0.
If we assume array index starting from 1 then, ith row contains i number of non-zero elements. Before ith row there are (i-1) rows, (1 to i-1) and in total these rows has 1+2+3......+(i-1) = i(i-1)/2 elements.
Now at ith row, the jth element will be at j position.
So the index of (i, j)th element of lower triangular matrix in this new representation is
j = i(i-1)/2
There are 50 questions to complete.

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