Engineering-Mathematics

Question 1

If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X + 2)2] equals _________.

A
54
B
55
C
56
D
57
       Engineering-Mathematics       Probability       GATE 2017 [Set-2]       Video-Explanation
Question 1 Explanation: 
In Poisson distribution:
Mean = Variance
E(X) = E(X2) - (E(X))2 = 5
E(X2) = 5 + (E(X))2 = 5 + 25 = 30
So, E[(X + 2)2] = E[X2 + 4 + 4X]
= E(X2) + 4 + 4E(X)
= 30 + 4 + 4 × 5
= 54
Question 2

If the ordinary generating function of a sequence is , then a3 - a0 is equal to ________.

A
15
B
16
C
17
D
18
       Engineering-Mathematics       Combinatorics       GATE 2017 [Set-2]       Video-Explanation
Question 2 Explanation: 
Question 3
 
A
1
B
Limit does not exist
C
53/12
D
108/7
       Engineering-Mathematics       Calculus       GATE 2019       Video-Explanation
Question 3 Explanation: 
Question 4

Let G be an arbitrary group. Consider the following relations on G:

    • R
1
    • : ∀a,b ∈ G, aR
1
    • b if and only if ∃g ∈ G such that a = g
-1
    • bg
 
    • R
2
    • : ∀a,b ∈ G, aR
2
    • b if and only if a = b
-1

Which of the above is/are equivalence relation/relations?

A
R2 only
B
R1 and R2
C
Neither R1 and R2
D
R1 only
       Engineering-Mathematics       Set-Theory       GATE 2019       Video-Explanation
Question 4 Explanation: 
A relation between the elements of a set is symmetric, reflexive and transitive then such relation is called as equivalence relation.
Consider Statement R1:
Reflexive:
aR1a
⇒ a = g-1ag
Left multiply both sides by g
⇒ ga = gg-1ag
Right multiply both sides by g-1
⇒ gag-1 = gg-1agg-1
⇒ gag-1 = a [∴ The relation is reflexive]
Symmetric:
If aR1b, then ∃g ∈ G such that gag-1 = b then a = g-1bg, which is Correct.
⇒ So, given relation is symmetric.
Transitive:
The given relation is Transitive.
So, the given relation R1 is equivalence.
R2:
The given relation is not reflexive.
So, which is not equivalence relation.
Such that a ≠ a-1.
So, only R1 is true.
Question 5

Let X be a square matrix. Consider the following two statements on X.

    • I. X is invertible.
 
    II. Determinant of X is non-zero.

Which one of the following is TRUE?

A
I implies II; II does not imply I.
B
II implies I; I does not imply II.
C
I and II are equivalent statements.
D
I does not imply II; II does not imply I.
       Engineering-Mathematics       Linear-Algebra       GATE 2019       Video-Explanation
Question 5 Explanation: 
X is invertible means, that X is non-singular matrix.
That means we can also say that determinant of X is non-zero.
Question 6

Let G be an undirected complete graph on n vertices, where n > 2. Then, the number of different Hamiltonian cycles in G is equal to

A
n!
B
1
C
(n-1)!
D
       Engineering-Mathematics       Graph-Theory       GATE 2019       Video-Explanation
Question 6 Explanation: 
A Hamiltonian cycle is a closed loop on a graph where every node (vertex) is visited exactly once.
The total number of hamiltonian cycles in a complete graph are
(n-1)!/2, where n is number of vertices.
Question 7

Let U = {1,2,...,n}. Let A = {(x,X)|x ∈ X, X ⊆ U}. Consider the following two statements on |A|.

Which of the above statements is/are TRUE?

A
Only II
B
Only I
C
Neither I nor II
D
Both I and II
       Engineering-Mathematics       Set-Theory       GATE 2019       Video-Explanation
Question 7 Explanation: 
Let us consider U = {1, 2}
and given A = {(x, X), x∈X and X⊆U}
Possible sets for U = {Φ, {1}, {2}, {1, 2}}
if x=1 then no. of possible sets = 2
x=2 then no. of possible sets = 2
⇒ No. of possible sets for A = (no. of sets at x=1) + (no. of sets at x=2) = 2 + 2 = 4
Consider statement (i) & (ii) and put n=2

Statement (i) is true

Statement (i) and (ii) both are true.
Answer: (C)
Video Explanation
Question 8

Suppose Y is distributed uniformly in the open interval (1,6). The probability that the polynomial 3x2 + 6xY + 3Y + 6 has only real roots is (rounded off to 1 decimal place) _____.

A
0.3
B
0.9
C
0.1
D
0.8
       Engineering-Mathematics       Probability       GATE 2019       Video-Explanation
Question 8 Explanation: 
Given polynomial equation is
3x2 + 6xY + 3Y + 6
= 3x2 + (6Y)x + (3Y + 6)
which is in the form: ax2 + bx + c
For real roots: b2 - 4ac ≥ 0
⇒ (6Y)2 - 4(3)(3Y + 6) ≥ 0
⇒ 36Y2 - 36Y - 72 ≥ 0
⇒ Y2 - Y - 2 ≥ 0
⇒ (Y+1)(Y-2) ≥ 0
Y = -1 (or) 2
The given interval is (1,6).
So, we need to consider the range (2,6).
The probability = (1/(6-1)) * (6-2) = 1/5 * 4 = 0.8
Question 9

Consider the first order predicate formula φ:

    ∀x[(∀z z|x ⇒ ((z = x) ∨ (z = 1))) ⇒ ∃w (w > x) ∧ (∀z z|w ⇒ ((w = z) ∨ (z = 1)))]

Here 'a|b' denotes that 'a divides b', where a and b are integers. Consider the following sets:

    S1.  {1, 2, 3, ..., 100}
    S2.  Set of all positive integers
    S3. Set of all integers

Which of the above sets satisfy φ?

A
S1 and S3
B
S1, S2 and S3
C
S2 and S3
D
S1 and S2
       Engineering-Mathematics       Propositional-Logic       GATE 2019
Question 9 Explanation: 
The first order logic gives the meaning that if z is a prime number then there exists another prime number in the set which is larger than it.
One of the case:
If -7 is a number which is prime (either divided by -7 or 1 only). then there exists some number like -3 which is larger than -7 also satisfy the property (either divided by -3 or 1 only).
So, S3 is correct
It's true for all integers too.
Question 10

Consider the following matrix:

The absolute value of the product of Eigen values of R is ______.

A
12
B
17
C
10
D
8
       Engineering-Mathematics       Linear-Algebra       GATE 2019       Video-Explanation
Question 10 Explanation: 
Question 11

The largest eigenvalue of A is ________

A
3
B
4
C
5
D
6
       Engineering-Mathematics       Linear-Algebra       GATE 2018
Question 11 Explanation: 

Correction in Explanation:

⇒ (1 - λ)(2 - λ) - 2 = 0
⇒ λ2 - 3λ=0
λ = 0, 3
So maximum is 3.
Question 12

Two people, P and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equi-probable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is __________.

A
0.021
B
0.022
C
0.023
D
0.024
       Engineering-Mathematics       Probability       GATE 2018       Video-Explanation
Question 12 Explanation: 
When two identical dice are rolled
⇾ A person wins who gets lower number compared to other person.
⇾ There could be “tie”, if they get same number.
Favorable cases = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Probability (tie) = 6/36 (when two dice are thrown, sample space = 6 × 6 = 36)
= 1/6
“Find the probability that one of them wins in the third attempt"
⇾ Which means, first & second time it should be tie and third time it should not be tie
⇾ P (tie) * P (tie) * P (not tie)
⇒ 1/6* 1/6 * (1 - 1/6)
⇒ (5/36×6)
= 0.138/6
= 0.023
Question 13

The chromatic number of the following graph is _______.

A
1
B
2
C
3
D
4
       Engineering-Mathematics       Graph-Theory       GATE 2018       Video-Explanation
Question 13 Explanation: 
Chromatic number of the following graph is “3”
Question 14

Let G be a finite group on 84 elements. The size of a largest possible proper subgroup of G is _________.

A
41
B
42
C
43
D
44
       Engineering-Mathematics       Set-Theory       GATE 2018       Video-Explanation
Question 14 Explanation: 
Lagranges Theorem:
For any group ‘G’ with order ‘n’, every subgroup ‘H’ has order ‘k’ such that ‘n’ is divisible by ‘k’.
Solution:
Given order n = 84
Then the order of subgroups = {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84}
As per the proper subgroup definition, it should be “42”.
Question 15

Which one of the following is a closed form expression for the generating function of the sequence {an}, where an = 2n+3 for all n = 0, 1, 2, …?

A
3/(1-x)2
B
3x/(1-x)2
C
2-x/(1-x)2
D
3-x/(1-x)2
       Engineering-Mathematics       Combinatorics       GATE 2018       Video-Explanation
Question 15 Explanation: 
Question 16
A
0.289
B
0.298
C
0.28
D
0.29
       Engineering-Mathematics       Calculus       GATE 2018       Video-Explanation
Question 16 Explanation: 
Question 17

Assume that multiplying a matrix G1 of dimension p×q with another matrix G2 of dimension q×r requires pqr scalar multiplications. Computing the product of n matrices G1G2G3…Gn can be done by parenthesizing in different ways. Define GiGi+1 as an explicitly computed pair for a given parenthesization if they are directly multiplied. For example, in the matrix multiplication chain G1G2G3G4G5G6 using parenthesization(G1(G2G3))(G4(G5G6)), G2G3 and G5G6 are the only explicitly computed pairs.

Consider a matrix multiplication chain F1F2F3F4F5, where matrices F1, F2, F3, F4 and F5 are of dimensions 2×25, 25×3, 3×16, 16×1 and 1×1000, respectively. In the parenthesization of F1F2F3F4F5 that minimizes the total number of scalar multiplications, the explicitly computed pairs is/ are

A
F1F2 and F3F4 only
B
F2F3 only
C
F3F4 only
D
F1F2 and F4F5 only
       Engineering-Mathematics       Linear-Algebra       GATE 2018       Video-Explanation
Question 17 Explanation: 
→ As per above information, the total number of scalar multiplications are 2173.
→ Optimal Parenthesization is:
((F1(F2(F3 F4)))F5)
→ But according to the problem statement we are only considering F3, F4 explicitly computed pairs.
Question 18

Consider the first-order logic sentence

φ ≡ ∃s∃t∃u∀v∀w∀x∀y ψ(s,t,u,v,w,x,y)

where ψ(s,t,u,v,w,x,y) is a quantifier-free first-order logic formula using only predicate symbols, and possibly equality, but no function symbols. Suppose φ has a model with a universe containing 7 elements.

Which one of the following statements is necessarily true?

A
There exists at least one model of φ with universe of size less than or equal to 3.
B
There exists no model of φ with universe of size less than or equal to 3.
C
There exists no model of φ with universe of size greater than 7.
D
Every model of φ has a universe of size equal to 7.
       Engineering-Mathematics       Propositional-Logic       GATE 2018
Question 18 Explanation: 
φ = ∃s∃t∃u∀v∀w∀x∀y ψ (s,t,u,v,w,x,y)
"∃" there exists quantifier decides whether a sentence belong to the model or not.
i.e., ~∃ will make it not belong to the model. (1) We have ‘7’ elements in the universe, So max. size of universe in a model = ‘7’
(2) There are three '∃' quantifiers, which makes that a model have atleast “3” elements. So, min. size of universe in model = ‘7’.
(A) is False because: (2)
(B) is true
(C) is false because of (1)
(D) is false, because these all models with size {3 to 7} not only ‘7’.
Question 19

Consider Guwahati (G) and Delhi (D) whose temperatures can be classified as high (H), medium (M) and low (L). Let P(HG) denote the probability that Guwahati has high temperature. Similarly, P(MG) and P(LG) denotes the probability of Guwahati having medium and low temperatures respectively. Similarly, we use P(HD), P(MD) and P(LD) for Delhi.

The following table gives the conditional probabilities for Delhi’s temperature given Guwahati’s temperature.

Consider the first row in the table above. The first entry denotes that if Guwahati has high temperature (HG) then the probability of Delhi also having a high temperature (HD) is 0.40; i.e., P(HD ∣ HG) = 0.40. Similarly, the next two entries are P(MD ∣ HG) = 0.48 and P(LD ∣ HG) = 0.12. Similarly for the other rows.

If it is known that P(HG) = 0.2, P(MG) = 0.5, and P(LG) = 0.3, then the probability (correct to two decimal places) that Guwahati has high temperature given that Delhi has high temperature is _______ .

A
0.60
B
0.61
C
0.62
D
0.63
       Engineering-Mathematics       Linear-Algebra       GATE 2018       Video-Explanation
Question 19 Explanation: 

The first entry denotes that if Guwahati has high temperature (HG ) then the probability that Delhi also having a high temperature (HD ) is 0.40.
P (HD / HG ) = 0.40
We need to find out the probability that Guwahati has high temperature.
Given that Delhi has high temperature (P(HG / HD )).

P (HD / HG ) = P(HG ∩ HD ) / P(HD )
= 0.2×0.4 / 0.2×0.4+0.5×0.1+0.3×0.01
= 0.60
Question 20

Let N be the set of natural numbers. Consider the following sets,

    • P: Set of Rational numbers (positive and negative)
 
    • Q: Set of functions from {0, 1} to N
 
    • R: Set of functions from N to {0, 1}
 
    • S: Set of finite subsets of N
 

Which of the above sets are countable?

A
Q and S only
B
P and S only
C
P and R only
D
P, Q and S only
       Engineering-Mathematics       Linear-Algebra       GATE 2018       Video-Explanation
Question 20 Explanation: 
Set of rational numbers are countable. It is proved by various methods in literature.
Set of functions from {0,1} to N is countable as it has one to one correspondence to N.
Set of functions from N to {0,1} is uncountable, as it has one to one correspondence to set of real numbers between (0 and 1).
Set of finite subsets of N is countable.
Question 21

Consider a matrix P whose only eigenvectors are the multiples of

 

Consider the following statements.

    • (I)
P
    •  does not have an inverse
 
    • (II)
P
    •  has a repeated eigenvalue
 
    • (III)
P
     cannot be diagonalized

Which one of the following options is correct?

A
Only I and III are necessarily true
B
Only II is necessarily true
C
Only I and II are necessarily true
D
Only II and III are necessarily true
       Engineering-Mathematics       Linear-Algebra       GATE 2018       Video-Explanation
Question 21 Explanation: 

Though the multiple of a vector represents same vector, and each eigen vector has distinct eigen value, we can conclude that ‘p’ has repeated eigen value.
If the unique eigen value corresponds to an eigen vector e, but the repeated eigen value corresponds to an entire plane, then the matrix can be diagonalized, using ‘e’ together with any two vectors that lie in plane.
But, if all eigen values are repeated, then the matrix cannot be diagonalized unless it is already diagonal.
So (III) holds correct.
A diagonal matrix can have inverse, So (I) is false.
Then (II) and (III) are necessarily True.
Question 22

Let G be a graph with 100! vertices, with each vertex labeled by a distinct permutation of the numbers 1, 2, …, 100. There is an edge between vertices u and v if and only if the label of u can be obtained by swapping two adjacent numbers in the label of v. Let y denote the degree of a vertex in G, and z denote the number of connected components in G.

Then, y + 10z = ___________.

A
109
B
110
C
111
D
112
       Engineering-Mathematics       Graph-Theory       GATE 2018       Video-Explanation
Question 22 Explanation: 
G is a graph with 100! vertices. Label of each vertex obtains from distinct permutation of numbers “1, 2, … 100”.
There exists edge between two vertices iff label of ‘u’ is obtained by swapping two adjacent numbers in label of ‘v’.
Example:
12 & 21, 23 & 34
The sets of the swapping numbers be (1, 2) (2, 3) (3, 4) … (99).
The no. of such sets are 99 i.e., no. of edges = 99.
As this is regular, each vertex has ‘99’ edges correspond to it.
So degree of each vertex = 99 = y.
As the vertices are connected together, the number of components formed = 1 = z
y + 102 = 99 + 10(1) = 109
Question 23

The statement (¬p) ⇒ (¬q) is logically equivalent to which of the statements below?

    • I. p ⇒ q
 
    • II. q ⇒ p
 
    • III. (¬q) ∨ p
 
    IV. (¬p) ∨ q
A
I only
B
I and IV only
C
II only
D
II and III only
       Engineering-Mathematics       Propositional-Logic       GATE 2017 [Set-1]       Video-Explanation
Question 23 Explanation: 
Method 1:
Construct Truth tables:
~p ⇒ ~q


II, III are equivalent to (~p) ⇒ (~q)
Method 2:
(I) p⇒q ≡ ~p∨q
(II) q⇒p ≡ ~q∨p
(III) (~q) ∨ p ≡ ~q∨p
(IV) (~p) ∨ p ≡ ~p∨q
Also, from question:
(~p) ⇒ (~q)
≡ p∨~q
So, (II) & (III) are equivalent to the statement given in question.
Question 24

Consider the first-order logic sentence F: ∀x(∃yR(x,y)). Assuming non-empty logical domains, which of the sentences below are implied by F?

    • I. ∃y(∃xR(x,y))
 
    • II. ∃y(∀xR(x,y))
 
    • III. ∀y(∃xR(x,y))
 
    IV. ¬∃x(∀y¬R(x,y))
A
IV only
B
I and IV only
C
II only
D
II and III only
       Engineering-Mathematics       Prepositional-Logic       GATE 2017 [Set-1]       Video-Explanation
Question 24 Explanation: 
Lets convert the given first order logic sentence into some english sentence.
F: ∀x(∃yR(x,y)) (given)
: For all girls there exist a boyfriend
(x for girls and y for boys)
I: ∃y(∃xR(x,y))
: There exist some boys who have girlfriends.
(Subset of statement F, so True)
II: ∃y(∀xR(x,y))
: There exists some boys for which all the girls are girlfriend. (False)
III: ∀y(∃xR(x,y))
: For all boys exists a girlfriend. (False)
IV: ~∃x(∀y~R(x,y))
= ∀x(~∀y~R(x,y))
= ∀x(∃yR(x,y)) (∵ ~∀y=∃y, ~∃x=∀x)
(True)
Question 25

Let c1, cn be scalars not all zero. Such that the following expression holds:

where ai is column vectors in Rn. Consider the set of linear equations.

Ax = B.

where A = [a1.......an] and

Then, Set of equations has

A
a unique solution at x = Jn where Jn denotes a n-dimensional vector of all 1
B
no solution
C
infinitely many solutions
D
finitely many solutions
       Engineering-Mathematics       Linear-Algebra       GATE 2017 [Set-1]       Video-Explanation
Question 25 Explanation: 
ai is a column vector
AX = B

As given that
and c1&cn ≠ 0
means c0a0 + c1a1 + ...cnan = 0, represents that a0, a1... an are linearly dependent.
So rank of 'A' = 0, (so if ‘B’ rank is = 0 infinite solution, ‘B’ rank>0 no solution) ⇾(1)
Another condition given here is, 'Σai = b',
so for c1c2...cn = {1,1,...1} set, it is having value 'b',
so there exists a solution if AX = b →(2)
From (1)&(2), we can conclude that AX = B has infinitely many solutions.
Question 26

Let T be a binary search tree with 15 nodes. The minimum and maximum possible heights of T are:

Note: The height of a tree with a single node is 0.
A
4 and 15 respectively
B
3 and 14 respectively
C
4 and 14 respectively
D
3 and 15 respectively
       Engineering-Mathematics       Binary-Trees       GATE 2017 [Set-1]       Video-Explanation
Question 26 Explanation: 
T be a Binary Search Tree with 15 nodes.
The height of a tree with single node is 0.
Minimum possible height is when it is a complete binary tree.

Maximum possible height is when it is a skewed tree left/right.

So the minimum and maximum possible heights of T are: 3 and 14 respectively.
Question 27

Let X be a Gaussian random variable with mean 0 and variance σ2. Let Y = max(X, 0) where max(a,b) is the maximum of a and b. The median of Y is __________.

A
0
B
1
C
2
D
3
       Engineering-Mathematics       Probability       GATE 2017 [Set-1]       Video-Explanation
Question 27 Explanation: 
Given that, Y = max(X,0),which means, Y is 0 for negative values of X and Y is X for positive values of X.
Median is a point, where the probability of getting less than median is 1/2 and probability of getting greater than median is 1/2.
From the given details, we can simply conclude that, median is 0. (0 lies exactly between positive and negative values)
Question 28
The value of
A
is 0
B
is -1
C
is 1
D
does not exist
       Engineering-Mathematics       Calculus       GATE 2017 [Set-1]       Video-Explanation
Question 28 Explanation: 

If "x=1" is substituted we get 0/0 form, so apply L-Hospital rule

Substitute x=1
⇒ (7(1)6-10(1)4)/(3(1)2-6(1)) = (7-10)/(3-6) = (-3)/(-3) = 1
Question 29

Let p, q and r be prepositions and the expression (p → q) → r be a contradiction. Then, the expression (r → p) → q is

A
a tautology
B
a contradiction
C
always TRUE when p is FALSE
D
always TRUE when q is TRUE
       Engineering-Mathematics       Prepositional-Logic       GATE 2017 [Set-1]       Video-Explanation
Question 29 Explanation: 
Given that (p→q)→r is a contradiction.
So r = F and (p→q) = T.
We have to evaluate the expression
(r→p)→q
Since r = F, (r→p) = T (As F→p, is always true)
The final expression is T→q and this is true when q is true, hence option D.
Question 30

Let u and v be two vectors in R2 whose Euclidean norms satisfy ||u||=2||v||. What is the value of α such that w = u + αv bisects the angle between u and v?

A
2
B
1/2
C
1
D
-1/2
       Engineering-Mathematics       Linear-Algebra       GATE 2017 [Set-1]       Video-Explanation
Question 30 Explanation: 

Let u, v be vectors in R2, increasing at a point, with an angle θ.
A vector bisecting the angle should split θ into θ/2, θ/2
Means ‘w’ should have the same angle with u, v and it should be half of the angle between u and v.
Assume that the angle between u, v be 2θ (thus angle between u,w = θ and v,w = θ)
Cosθ = (u∙w)/(∥u∥ ∥w∥) ⇾(1)
Cosθ = (v∙w)/(∥v∥ ∥w∥) ⇾(2)
(1)/(2) ⇒ 1/1 = ((u∙w)/(∥u∥ ∥w∥))/((v∙w)/(∥v∥ ∥w∥)) ⇒ 1 = ((u∙w)/(∥u∥))/((v∙w)/(∥v∥))
⇒ (u∙w)/(v∙w) = (∥u∥)/(∥v∥) which is given that ∥u∥ = 2 ∥v∥
⇒ (u∙w)/(v∙w) = (2∥v∥)/(∥v∥) = 2 ⇾(3)
Given ∥u∥ = 2∥v∥
u∙v = ∥u∥ ∥v∥Cosθ
=2∙∥v∥2 Cosθ
w = u+αv
(u∙w)/(v∙w) = 2
(u∙(u+αv))/(v∙(u+αv)) = 2
(u∙u+α∙u∙v)/(u∙v+α∙v∙v) = 2a∙a = ∥a∥2
4∥v∥2+α∙2∙∥v∥2 Cosθ = 2(2∥v∥2 Cosθ+α∙∥v∥2)
4+2αCosθ = 2(2Cosθ+α)
4+2αCosθ = 4Cosθ+2α ⇒ Cosθ(u-v)+2α-4 = 0
4-2α = Cosθ(4-2α)
(4-2α)(Cosθ-1) = 0
4-2α = 0
Question 31

Let A be m×n real valued square symmetric matrix of rank 2 with expression given below.

Consider the following statements

    • (i) One eigenvalue must be in [-5, 5].
 
    • (ii) The eigenvalue with the largest magnitude
 
    must be strictly greater than 5.

Which of the above statements about engenvalues of A is/are necessarily CORRECT?

A
Both (I) and (II)
B
(I) only
C
(II) only
D
Neither (I) nor (II)
       Engineering-Mathematics       Linear-Algebra       GATE 2017 [Set-1]       Video-Explanation
Question 31 Explanation: 
Let

be a real valued, rank = 2 matrix.

a2+b2+c2+d2 = 50
Square values are of order 0, 1, 4, 9, 16, 25, 36, …
So consider (0, 0, 5, 5) then Sum of this square = 0+0+25+25=50
To get rank 2, the 2×2 matrix can be

The eigen values are,
|A-λI| = 0 (The characteristic equation)

-λ(-λ)-25 = 0
λ2-25 = 0

So, the eigen values are within [-5, 5], Statement I is correct.
The eigen values with largest magnitude must be strictly greater than 5: False.
So, only Statement I is correct.
Question 32

The number of integers between 1 and 500 (both inclusive) that are divisible by 3 or 5 or 7 is _________.

A
271
B
272
C
273
D
274
       Engineering-Mathematics       Combinatorics       GATE 2017 [Set-1]       Video-Explanation
Question 32 Explanation: 

Let A = number divisible by 3
B = numbers divisible by 5
C = number divisible by 7
We need to find “The number of integers between 1 and 500 that are divisible by 3 or 5 or 7" i.e., |A∪B∪C|
We know,
|A∪B∪C| = |A|+|B|+C-|A∩B|-|A∩C|-|B∩C|+|A∩B|
|A| = number of integers divisible by 3
[500/3 = 166.6 ≈ 166 = 166]
|B| = 100
[500/5 = 100]
|C| = 71
[500/7 = 71.42]
|A∩B| = number of integers divisible by both 3 and 5 we need to compute with LCM (15)
i.e.,⌊500/15⌋ ≈ 33
|A∩B| = 33
|A∩C| = 500/LCM(3,7) 500/21 = 23.8 ≈ 28
|B∩C| = 500/LCM(5,3) = 500/35 = 14.48 ≈ 14
|A∩B∩C| = 500/LCM(3,5,7) = 500/163 = 4.76 ≈ 4
|A∪B∪C| = |A|+|B|+|C|-|A∩B|-|A∩C|-|B∩C|+|A∩B∩C|
= 166+100+71-33-28-14+4
= 271
Question 33

If f(x) = Rsin(πx/2) + S, f'(1/2) = √2 and , then the constants R and S are, respectively.

A
B
C
D
       Engineering-Mathematics       Calculus       GATE 2017 [Set-2]       Video-Explanation
Question 33 Explanation: 

Question 34

Let p, q, r denote the statements “It is raining”, “It is cold”, and “It is pleasant”, respectively. Then the statement “It is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold” is represented by

A
(¬p ∧ r) ∧ (¬r → (p ∧ q))
B
(¬p ∧ r) ∧ ((p ∧ q) → ¬r)
C
(¬p ∧ r) ∨ ((p ∧ q) → ¬r)
D
(¬p ∧ r) ∨ (r → (p ∧ q))
       Engineering-Mathematics       Prepositional-Logic       GATE 2017 [Set-2]       Video-Explanation
Question 34 Explanation: 
p: It is raining
q: It is cold
r: It is pleasant
“If it is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold.”
We can divide the statement into two parts with “Conjunction”.

i.e., ¬r→(p∧q) ⇾(2)
From (1) & (2), the given statement can be represented as
Question 35

Consider the set X = {a,b,c,d e} under the partial ordering

    R = {(a,a),(a,b),(a,c),(a,d),(a,e),(b,b),(b,c),(b,e),(c,c),(c,e),(d,d),(d,e),(e,e)}.

The Hasse diagram of the partial order (X,R) is shown below.

The minimum number of ordered pairs that need to be added to R to make (X,R) a lattice is _________.

A
0
B
1
C
2
D
3
       Engineering-Mathematics       Set-Theory       GATE 2017 [Set-2]       Video-Explanation
Question 35 Explanation: 
R = {(a,a), (a,b), (a,c), (a,d), (a,e), (b,b), (b,c), (b,e), (c,c), (c,e), (d,d), (d,e), (e,e)}
As per the definition of lattice, each pair should have GLB, LUB.
The given ‘R’ has GLB, LUB for each and every pair.
So, no need to add extra pair.
Thus no. of required pair such that Hasse diagram to become lattice is “0”.
Question 36

Then the rank of P+Q is _________.

A
2
B
3
C
4
D
5
       Engineering-Mathematics       Matrices       GATE 2017 [Set-2]       Video-Explanation
Question 36 Explanation: 


R2→R2+R1

The number of non-zero rows of P + Q = 2,
So Rank = 2
Note: “Rank” is the number of independent vectors.
Method-1:
Each vector is a row in matrix.
Echelon form of a matrix has no. of zeroes increasing each rows.
The total non-zero rows left give the rank.
Method-2:
Find determinant of matrix, for 3×5, if determinant is ‘0’, the max rank can be 2.
If determinant of any n×n is non-zero, then rows proceed with (n-1)×(n-1).
Question 37

G is an undirected graph with n vertices and 25 edges such that each vertex of G has degree at least 3. Then the maximum possible value of n is ___________.

A
16
B
17
C
18
D
19
       Engineering-Mathematics       Graph-Theory       GATE 2017 [Set-2]       Video-Explanation
Question 37 Explanation: 
An undirected graph ‘G’ has ‘n’ vertices & 25 edges.
Degree of each vertex ≥ 3

|v| = 2|E|
The relation between max and min degree of graph are
m ≤ 2|E| / |v| ≤ M
Given minimum degree = 3
So, 3 ≤2 |E| / |v|
3|v| ≤ 2|E|
3(n) ≤ 2(25)
n ≤ 50/3
n ≤ 16.6
(n = 16)
Question 38

P and Q are considering to apply for job. The probability that p applies for job is 1/4. The probability that P applies for job given that Q applies for the job 1/2 and The probability that Q applies for job given that P applies for the job 1/3.The probability that P does not apply for job given that Q does not apply for the job

A
B
C
D
       Engineering-Mathematics       Probability       GATE 2017 [Set-2]       Video-Explanation
Question 38 Explanation: 
Probability that ‘P’ applies for a job = 1/4 = P(p) ⇾ (1)
Probability that ‘P’ applies for the job given that Q applies for the job = P(p/q) = 1/2 ⇾ (2)
Probability that ‘Q’ applies for the job, given that ‘P’ applies for the job = P(p/q) = 1/3 ⇾ (3)
Bayes Theorem:
(P(A/B) = (P(B/A)∙P(A))/P(B) ; P(A/B) = P(A∩B)/P(B))
⇒ P(p/q) = (P(q/p)∙P(p))/p(q)
⇒ 1/2 = (1/3×1/4)/p(q)
p(q) = 1/12×2 = 1/(6) (P(q) = 1/6) ⇾ (4)
From Bayes,
P(p/q) = (P(p∩q))/(P(q))
1/2 = P(p∩q)/(1⁄6)
(p(p∩q) = 1/12)
We need to find out the “probability that ‘P’ does not apply for the job given that q does not apply for the job = P(p'/q')
From Bayes theorem,
P(p'/q') = (P(p'∩q'))/P(q') ⇾ (5)
We know,
p(A∩B) = P(A) + P(B) - P(A∪B)
also (P(A'∩B') = 1 - P(A∪B))
P(p'∩q') = 1 - P(p∪q)
= 1 - (P(p) + P(q) - P(p∩q))
= 1 - (P(p) + P(q) - P(p) ∙ P(q))
= 1 - (1/4 + 1/6 - 1/12)
= 1 - (10/24 - 2/24)
= 1 - (8/24)
= 2/3
(P(p'∩q') = 2/3) ⇾ (6)
Substitute in (5),
P(p'⁄q') = (2⁄3)/(1-P(q)) = (2⁄3)/(1-1/6) = (2⁄3)/(5⁄6) = 4/5
(P(p'/q') = 4/5)
Question 39

For any discrete random variable X, with probability mass function P(X=j)=pj, pj≥0, j∈{0, ..., N} and , define the polynomial function . For a certain discrete random variable Y, there exists a scalar β∈[0,1] such that gy(Z)=(1-β+βz)N. The expectation of Y is

 
A
Nβ(1 - β)
B
C
N(1 - β)
D
Not expressible in terms of N and β alone
       Engineering-Mathematics       Probability       GATE 2017 [Set-2]       Video-Explanation
Question 39 Explanation: 
For a discrete random variable X,
Given gy (z) = (1 - β + βz)N ⇾ it is a binomial distribution like (x+y)n
Expectation (i.e., mean) of a binomial distribution will be np.
The polynomial function ,
given
Mean of Binomial distribution of b(xj,n,p)=
The probability Mass function,

Given:
Question 40

If the characteristic polynomial of a 3 × 3 matrix M over R (the set of real numbers) is λ3 - 4λ2 + aλ + 30, a ∈ ℝ, and one eigenvalue of M is 2, then the largest among the absolute values of the eigenvalues of M is ________.

A
5
B
6
C
7
D
8
       Engineering-Mathematics       Linear-Algebra       GATE 2017 [Set-2]       Video-Explanation
Question 40 Explanation: 
For a 3 × 3 matrix ‘M’, the characteristic equation |A – λI| is
λ3 - 4λ2 + aλ + 30 = 0 ⇾ (1)
One eigen value is ‘2’, so substitute it
23 - 4(2)2 + a(2) + 30 = 0
8 - 16 + 2a + 30 = 0
2a = -22
a = -11
Substitute in (1),
λ3 - 4λ2 - 11 + 30 = 0

So, (1) can be written as
(λ - 2)(λ2 - 2λ - 15) = 0
(λ - 2)(λ2 - 5λ + 3λ - 15) = 0
(λ - 2)(λ - 3)(λ - 5) = 0
λ = 2, 3, 5
Max λ=5
Question 41

Let p,q,r,s represent the following propositions.

    • p: x ∈ {8,9,10,11,12}
 
    • q: x is a composite number
 
    • r: x is a perfect square
 
    s: x is a prime number

The integer x≥2 which satisfies ¬((p ⇒ q) ∧ (¬r ∨ ¬s))  is _________.

A
11
B
12
C
13
D
14
       Engineering-Mathematics       Prepositional-Logic       GATE 2016 [Set-1]       Video-Explanation
Question 41 Explanation: 
Given,
~((p→q) ∧ (~r ∨ ~S))
⇒ first simplify the given statement by converging them to ∧, ∨
⇒ [~(p→q) ∨ (~(~r ∨ ~s)]
Demorgan’s law:
⇒ [~(~p ∨ q) ∨ (r ∧ s)]
∵ p→q ≡ ~p ∨ q
⇒ [(p ∧ ~q) ∨ (r ∧ s)]
p ∧ ~q is {8,9,10,11,12} ∧ {not a composite number} i.e. {11}
r ∧ s is {perfect square} ∧ {prime} i.e. no answer
So, the one and only answer is 11.
Question 42

Let an be the number of n-bit strings that do NOT contain two consecutive 1s. Which one of the following is the recurrence relation for an?

A
an = a(n-1) + 2a(n-2)
B
an = a(n-1) + a(n-2)
C
an = 2a(n-1) + a(n-2)
D
an = 2a(n-1) + 2a(n-2)
       Engineering-Mathematics       Combinatorics       GATE 2016 [Set-1]       Video-Explanation
Question 42 Explanation: 
an = number of n-bit strings, that do not have two consecutive 1’s.
If n=1, we have {0,1}
# Occurrences = 2
If n=2, we have {00,01,10}
# Occurrences = 3
If n=3, we have {000,001,010,100,101}
# Occurrences = 5
It is evident that a3 = a1 + a2
Similarly, an = an-1 + an-2
Question 43
A
4
B
3
C
2
D
1
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Question 43 Explanation: 
Question 44

A probability density function on the interval [a,1] is given by 1/x2 and outside this interval the value of the function is zero. The value of a is _________.

A
0.7
B
0.6
C
0.5
D
0.8
       Engineering-Mathematics       Probability       GATE 2016 [Set-1]       Video-Explanation
Question 44 Explanation: 
The property of probability density function is area under curve = 1
or

where (a, b) is internal and f(x) is probability density function.
Given,
f(x) = 1/x2 , a≤x≤1
The area under curve,

- 1 + 1/a = 1
1/a = 2
a = 0.5
Question 45

Two eigenvalues of a 3 × 3 real matrix P are (2 + √-1) and 3. The determinant of P is __________.

A
18
B
15
C
17
D
16
       Engineering-Mathematics       Linear-Algebra       GATE 2016 [Set-1]       Video-Explanation
Question 45 Explanation: 
If an eigen value of a matrix is a complex number, then there will be other eigen value, which is conjugate of the complex eigen value.
So, For the given 3×3 matrix there would be 3 eigen values.
Given eigen values are : 2+i and 3.
So the third eigen value should be 2-i.
As per the theorems, the determinant of the matrix is the product of the eigen values.
So the determinant is (2+i)*(2-i)*3 = 15.
Question 46

The coefficient of x12 in (x3 + x4 + x5 + x6 + ...)3 is _________.

A
10
B
11
C
12
D
13
       Engineering-Mathematics       Combinatorics       GATE 2016 [Set-1]       Video-Explanation
Question 46 Explanation: 
Co-efficient of x12 in (x3 + x4 + x5 + x6+...)3
⇒ [x3(1 + x + x2 + x3 + ...)]3
= x9(1 + x + x2 + x3 + ...)3
First Reduction:
As x9 is out of the series, we need to find the co-efficient of x3 in (1 + x + x2 + ⋯)3

Here, m=3, k=3, the coefficient

= 5C3 = 5!/2!3! = 10
Question 47

Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1. Let a99 = K × 104. The value of K is ___________.

A
198
B
199
C
200
D
201
       Engineering-Mathematics       Combinatorics       GATE 2016 [Set-1]       Video-Explanation
Question 47 Explanation: 
an = 6n2 + 2n + a(n-1)
Replace a(n-1)

⇒ an = 6n2 + 2n + 6(n-1)2 + 2(n-1) + 6(n-2)2 + 2(n-2) + ⋯ a1
Given that a1 = 8, replace it
⇒ an = 6n2 + 2n + 6(n-1)2 + 2(n-1) + 6(n-2)2 + 2(n-2) + ⋯8
= 6n2 + 2n + 6(n-1)2 + 2(n-1) + 6(n-2)2 + 2(n-2) + ⋯ + 6(1)2 + 2(1)

= 6(n2 + (n-1)2 + (n-2)2 + ⋯ + 22 + 12) + 2(n + (n-1) + ⋯1)
Sum of n2 = (n(n+1)(2n+1))/6
Sum of n = (n(n+1))/2
= 6 × (n(n+1)(2n+1))/6 + 2×(n(n+1))/2
= n(n+1)[1+2n+1]
= n(n+1)[2n+2]
= 2n(n+1)2
Given a99 = k×104
a99 = 2(99)(100)2 = 198 × 104
∴k = 198
Question 48

A function f:N+ → N+, defined on the set of positive integers N+, satisfies the following properties:

    • f(n) = f(n/2)           if n is even
 
    f(n) = f(n+5)           if n is odd

Let R = {i|∃j: f(j)=i} be the set of distinct values that f takes. The maximum possible size of R is __________.

A
2
B
3
C
4
D
5
       Engineering-Mathematics       Set-Theory       GATE 2016 [Set-1]       Video-Explanation
Question 48 Explanation: 
f(n)= f(n⁄2)          if n is even
f(n)= f(n+5)        if n is odd

We can observe that

and f(5) = f(10) = f(15) = f(20)
Observe that f(11) = f(8)
f(12) = f(6) = f(3)
f(13) = f(9) = f(14) = f(7) = f(12) = f(6) = f(3)
f(14) = f(9) = f(12) = f(6) = f(3)
f(16) = f(8) = f(4) = f(2) = f(1) [repeating]
So, we can conclude that
‘R’ can have size only ‘two’ [one: multiple of 5’s, other: other than 5 multiples]
Question 49

Consider the following experiment.

Step1.
    • Flip a fair coin twice.
Step2.
    • If the outcomes are (TAILS, HEADS) then output
Y
    • and stop.
Step3.
    • If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output
N
    • and stop.
Step4.
    If the outcomes are (TAILS, TAILS), then go to Step 1.

The probability that the output of the experiment is Y is (up to two decimal places) ________.

A
0.33
B
0.34
C
0.35
D
0.36
       Engineering-Mathematics       Probability       GATE 2016 [Set-1]       Video-Explanation
Question 49 Explanation: 
If a coin is flipped twice, the possible outcomes {HH, HT, TH, TT}
Stop conditions:
If outcome = TH then Stop [output 4] --------------- (1)
else
outcome = HH/ HT then Stop [output N] -------------- (2)
We get ‘y’ when we have (1) i.e., ‘TH’ is output.
(1) can be preceded by ‘TT’ also, as ‘TT’ will reset (1) again
Probability of getting y = TH + (TT)(TH) + (TT)(TT)(TH) + …
= 1/2 × 1/2 + 1/2 × 1/2 × 1/2 × 1/2 + ...
= (1/4)/(1-1/4)
= 1/3
= 0.33
Question 50

Consider the following expressions:

    • (i) false
 
    • (ii) Q
 
    • (iii) true
 
    • (iv) P ∨ Q
 
    (v) ¬Q ∨ P

The number of expressions given above that are logically implied by P ∧ (P ⇒ Q) is _________.

A
4
B
5
C
6
D
7
       Engineering-Mathematics       Propositional-Logic       GATE 2016 [Set-2]       Video-Explanation
Question 50 Explanation: 
The expression is logically implied by P ∧ (P → Q) means
(P ∧ (P → Q))→ expression is a tautology. So we have to find
How many tautological formulas are there for the given inputs.
(P ∧ (P → Q)) → True is always tautology
(P ∧ (P → Q)) → False is not a tautology
(P ∧ (P → Q)) → Q is a tautology
(P ∧ (P → Q)) → ¬Q ∨ P is a tautology
(P ∧ (P → Q)) → P ∨ Q is a tautology
So there are 4 expressions logically implied by (P ∧ (P → Q))
Question 51

Let f(x) be a polynomial and g(x) = f'(x) be its derivative. If the degree of (f(x) + f(-x)) is 10, then the degree of (g(x) - g(-x)) is __________.

A
9
B
10
C
11
D
12
       Engineering-Mathematics       Calculus       GATE 2016 [Set-2]       Video-Explanation
Question 51 Explanation: 
If the degree of a polynomial is ‘n’ then the derivative of that function have (n – 1) degree.
It is given that f(x) + f(-x) degree is 10.
It means f(x) is a polynomial of degree 10.
Then obviously the degree of g(x) which is f’(x) will be 9.
Question 52

The minimum number of colours that is sufficient to vertex-colour any planar graph is ________.

A
4
B
5
C
6
D
7
       Engineering-Mathematics       Graph-Theory       GATE 2016 [Set-2]       Video-Explanation
Question 52 Explanation: 
The 4-colour theorem of the planar graph describes that any planar can atmost be colored with 4 colors.
Here it is asked about the sufficient number of colors, so with the worst case of 4 colors we can color any planar graph.
Question 53

Consider the systems, each consisting of m linear equations in n variables.

    • I. If m < n, then all such systems have a solution
 
    • II. If m > n, then none of these systems has a solution
 
    III. If m = n, then there exists a system which has a solution

Which one of the following is CORRECT?

A
I, II and III are true
B
Only II and III are true
C
Only III is true
D
None of them is true
       Engineering-Mathematics       Linear-Algebra       GATE 2016 [Set-2]       Video-Explanation
Question 53 Explanation: 
i) If m In AX = B,
If R(A) ≠ R(A|B)
then there will be no solution.
ii) False, because if R(A) = R(A|B),
then there will be solution possible.
iii) True, if R(A) = R(A|B),
then there exists a solution.
Question 54

Suppose that a shop has an equal number of LED bulbs of two different types. The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is of Type 2 is 0.4. The probability that an LED bulb chosen uniformly at random lasts more than 100 hours is _________.

A
0.55
B
0.56
C
0.57
D
0.58
       Engineering-Mathematics       Probability       GATE 2016 [Set-2]       Video-Explanation
Question 54 Explanation: 

The bulbs of Type 1, Type 2 are same in number.
So, the probability to choose a type is 1/2.
The probability to choose quadrant ‘A’ in diagram is
P(last more than 100 hours/ type1) = 1/2 × 0.7
P(last more than 100 hours/ type2) = 1/2 × 0.4
Total probability = 1/2 × 0.7 + 1/2 × 0.4 = 0.55
Question 55

Suppose that the eigenvalues of matrix A are 1, 2, 4. The determinant of (A-1)T is _________.

A
0.125
B
0.126
C
0.127
D
0.128
       Engineering-Mathematics       Linear-Algebra       GATE 2016 [Set-2]       Video-Explanation
Question 55 Explanation: 
Determinant of a matrix is product of the eigen values.
Given that eigen values are 1, 2, 4.
So, its determinant is 1*2*4 = 8
The determinant of (A-1)T = 1/ AT = 1/|A| = 1/8 = 0.125
Question 56

A binary relation R on ℕ × ℕ is defined as follows: (a,b)R(c,d) if a≤c or b≤d. Consider the following propositions:

    • P:
R
    • is reflexive
 
    • Q:
R
    is transitive

Which one of the following statements is TRUE?

A
Both P and Q are true.
B
P is true and Q is false.
C
P is false and Q is true.
D
Both P and Q are false.
       Engineering-Mathematics       Set-Theory       GATE 2016 [Set-2]       Video-Explanation
Question 56 Explanation: 
For every a,b ∈ N,
a≤c ∨ b≤d
Let a≤a ∨ b≤b is true for all a,b ∈ N
So there exists (a,a) ∀ a∈N.
It is Reflexive relation.
Consider an example
c = (a,b)R(c,d) and (c,d)R(e,f) then (a,b)R(e,f)
This does not hold for any (a>e) or (b>f)
eg:
(2,2)R(1,2) as 2≤2
(1,2)R(1,1) as 1≤1
but (2,2) R (1,1) is False
So, Not transitive.
Question 57

Which one of the following well-formed formulae in predicate calculus is NOT valid?

A
(∀x p(x) ⇒ ∀x q(x)) ⇒ (∃x ¬p(x) ∨ ∀x q(x))
B
(∃x p(x) ∨ ∃x q(x)) ⇒ ∃x (p(x) ∨ q(x))
C
∃x (p(x) ∧ q(x)) ⇒ (∃x p(x) ∧ ∃x q(x))
D
∀x (p(x) ∨ q(x)) ⇒ (∀x p(x) ∨ ∀x q(x))
       Engineering-Mathematics       Propositional-Logic       GATE 2016 [Set-2]       Video-Explanation
Question 57 Explanation: 
For the formulae to be valid there should not be implication like T → F.
But in option (D), we can generate T → F.
Hence, not valid.
Question 58

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
       Engineering-Mathematics       Set-Theory       GATE 2016 [Set-2]       Video-Explanation
Question 58 Explanation: 
There are set of ‘23’ different compounds.
U = 23
∃S ∋ (S⊂U)
Each component in ‘S’ reacts with exactly ‘3’ compounds of U,

If a component ‘a’ reacts with ‘b’, then it is obvious that ‘b’ also reacts with ‘a’.
It’s a kind of symmetric relation.>br> If we connect the react able compounds, it will be an undirected graph.
The sum of degree of vertices = 9 × 3 = 27
But, in the graph of ‘23’ vertices the sum of degree of vertices should be even because
(di = degree of vertex i.e., = no. of edges)
But ‘27’ is not an even number.
To make it an even, one odd number should be added.
So, there exists atleast one compound in U/S reacts with an odd number of compounds.
Question 59

The value of the expression 1399(mod 17), in the range 0 to 16, is ________.

A
4
B
5
C
6
D
7
       Engineering-Mathematics       Modular-Arithmetic       GATE 2016 [Set-2]       Video-Explanation
Question 59 Explanation: 
Fermat’s theorem,
a(p-1) ≡ 1 mod p (p is prime)
From given question,
p = 17
a(17-1) ≡ 1 mod 17
a16 ≡ 1 mod 17
1316 ≡ 1 mod 17
Given:
1399 mod 17

133 mod 17
2197 mod 17
4
Question 60

In the LU decomposition of the matrix ,if the diagonal elements of U are both 1, then the lower diagonal entry l22 of L is

A
5
B
6
C
7
D
8
       Engineering-Mathematics       Linear-Algebra       GATE 2015 [Set-1]
Question 60 Explanation: 
A = LU

l11 = 2 -----(1)
l11u12 = 2
u12 = 2/2
u12 = 1 ----- (2)
l21 = 4 ----- (3)
l21u12+l22 = 9
l22 = 9 - l21u12 = 9 - 4 × 1 = 5
Question 61

If g(x) = 1 - x and h(x) , then is:

A
h(x)/g(x)
B
-1/x
C
g(x)/h(x)
D
x/(1-x)2
       Engineering-Mathematics       Calculus       GATE 2015 [Set-1]
Question 61 Explanation: 
g(x)= 1 – x, h(x)=x/x-1 -------- (2)
Replace x by h(x) in (1), replacing x by g(x) in (2),
g(h(x))=1-h(x)=1-x/x-1=-1/x-1
h(g(x))=g(x)/g(x)-1=1-x/-x
⇒ g(h(x))/h(g(x))=x/(x-1)(1-x)=(x/x-1)/1-x=h(x)/g(x)
Question 62

Suppose L = {p, q, r, s, t} is a lattice represented by the following Hasse diagram:

For any x, y ∈ L, not necessarily distinct, x ∨ y and x ∧ y are join and meet of x, y respectively. Let L3 = {(x,y,z): x, y, z ∈ L} be the set of all ordered triplets of the elements of L. Let pr be the probability that an element (x,y,z) ∈ L3 chosen equiprobably satisfies x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z). Then

A
pr = 0
B
pr = 1
C
0 < pr ≤ 1/5
D
1/5 < pr < 1
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-1]
Question 62 Explanation: 
Total number of elements i.e., ordered triplets (x,y,z) in L3 are 5×5×5 i.e., 125 n(s) = 125
Let A be the event that an element (x,y,z)∈ L3 satisfies x ∨(y∧z) = (x∨y) ∧ (x∨z) Since q∨(r∧s) = q∨p = q
and (q∨r)∧(q∨s) = t∧t = t q∨(r∧s) ≠ (q∨r)∧(q∨s)
Therefore, (x,y,z) = (q,r,s),(q,s,r),(r,q,s),(r,s,q),(s,r,q),(s,q,r)
i.e., 3! = 6 elements will not satisfy distributive law and all other (x,y,z) choices satisfy distributive law
n(A) = 125-6 = 119
∴ required probability is 119/125
⇒ 1/5
Question 63

Consider the following 2 × 2 matrix A where two elements are unknown and are marked by a and b. The eigenvalues of this matrix are –1 and 7. What are the values of a and b?

 
A
a=6, b=4
B
a=4, b=6
C
a=3, b=5
D
a=5, b=3
       Engineering-Mathematics       Linear-Algebra       GATE 2015 [Set-1]
Question 63 Explanation: 
Given λ1=-1 and λ2=7 are eigen values of A
By properties,

⇒ 6=1+a and -7=a-4b
⇒ a=5 ⇒ -7=5-4b
⇒ b=3
Question 64

Let G = (V, E) be a simple undirected graph, and s be a particular vertex in it called the source.  For x ∈ V, let d(x)denote the shortest distance in G from s to x. A breadth first search (BFS) is performed starting at s. Let T be the resultant BFS tree. If (u, v) is an edge of G that is not in T, then which one of the following CANNOT be the value of d(u- d(v) ?   

A
-1
B
0
C
1
D
2
       Engineering-Mathematics       BFS       GATE 2015 [Set-1]
Question 64 Explanation: 
In an undirected graph if (u, v) be the edge then (u, v) also the edge. So shortest path that can be obtained by the (u, v) of (u, v).
Then the difference between the d(u) and d(v) is not more than '1'.
In the option 'D' the difference is given as '2' it is not possible in the undirected graph.
Question 65
A
-1
B
-2
C
-3
D
-4
       Engineering-Mathematics       Calculus       GATE 2015 [Set-1]
Question 65 Explanation: 
Question 66

Let G be a connected planar graph with 10 vertices. If the number of edges on each face is three, then the number of edges in G is _______________.

A
24
B
25
C
26
D
27
       Engineering-Mathematics       Graph-Theory       GATE 2015 [Set-1]
Question 66 Explanation: 
By Euler’s formula,
|V| + |R| = |E| + 2 ------(1) where |V|, |E|, |R| are respectively number of vertices, edges and faces (regions)
Given |V| = 10 ------(2) and number of edges on each face is three
∴3|R| = 2|E| ⇒ |R| = 2/3|E| ------(3)
Substituting (2), (3) in (1), we get
10 + 2/3|E| = |E| + 2 ⇒ |E|/3 = 8 ⇒ |E| = 24
Question 67

Consider the operations f(X, Y, Z) = X'YZ + XY' + Y'Z'  and  g(X′, Y, Z) = X′YZ + X′YZ′ + XY Which one of the following is correct?

A
Both {f} and {g} are functionally complete
B
Only {f} is functionally complete
C
Only {g} is functionally complete
D
Neither {f} nor {g} is functionally complete
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-1]
Question 67 Explanation: 
A function is functionally complete if (OR, NOT) or (AND, NOT) are implemented by it.
f(X,X,X) = X'XX'+XX'+X'X'
= 0+0+X'
= X'
Similarly, f(Y,Y,Y) = Y' and f(X,Z,Z) = Z'
f(Y',Y',Z') = (Y')'Y'Z'+Y'(Y')'+(Y')'(Z')'
= YY'Z'+Y'Y+YZ
= 0+0+YZ
= YZ
We have derived NOT and AND. So f(X,Y,Z) is functionally complete.
g(X,Y,Z) = X'YZ+X'YZ'+XY
g(X,X,X) = X'XX+X'XZ'+XX
= 0+0+X
= X
Similarly, g(Y,Y,Y) = Y and g(Z,Z,Z) = Z
NOT is not derived. Hence, g is not functionally complete.
Question 68
A
0.99
B
1.00
C
2.00
D
3.00
       Engineering-Mathematics       Calculus       GATE 2015 [Set-1]
Question 68 Explanation: 

= 2-1/1(2)+3-2/2(3)+4-3/3(4)+…+100-99/99(100)
= 1/1-1/2+1/2-1/3+1/3…+1/98-1/99+1/99-1/100
= 1-1/100
= 99/100
= 0.99
Question 69

Let R be the relation on the set of positive integers such that aRb if and only if a and b are distinct and have a common divisor other than 1. Which one of the following statements about R is true?

A
R is symmetric and reflexive but not transitive
B
R is reflexive but not symmetric and not transitive
C
R is transitive but not reflexive and not symmetric
D
R is symmetric but not reflexive and not transitive
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-2]
Question 69 Explanation: 
Reflexive:
In aRb, 'a' and 'b' are distinct. So it can never be reflexive.
Symmetric:
In aRb, if 'a' and 'b' have common divisor other than 1, then bRa, i.e., 'b' and 'a' also will have common divisor other than 1. So, yes symmetric.
Transitive:
Take (3, 6) and (6, 2) elements of R. For transitivity (3, 2) must be the element of R, but 3 and 2 don't have a common divisor. So not transitive.
Question 70

Consider the following statements:

    S1: If a candidate is known to be corrupt, then he will not be elected.
    S2: If a candidate is kind, he will be elected.

Which one of the following statements follows from S1 and S2 as per sound inference rules of logic?

A
If a person is known to corrupt, he is kind
B
If a person is not known to be corrupt, he is not kind
C
If a person is kind, he is not known to be corrupt
D
If a person is not kind, he is not known to be corrupt
       Engineering-Mathematics       Prepositional-Logic       GATE 2015 [Set-2]
Question 70 Explanation: 
Let p: candidate known to be corrupt
q: candidate will be elected
r: candidate is kind
then S1 = p→~q
= q→~p (conrapositive rule)
and S2: r→q ⇒ r→~p (transitive rule)
i.e., If a person is kind, he is not known to be corrupt. ∴ Option is C
Question 71

The larger of the two eigenvalues of the matrix is _________.

A
6
B
7
C
8
D
9
       Engineering-Mathematics       Linear-Algebra       GATE 2015 [Set-2]
Question 71 Explanation: 
Characteristic equation is

⇒ λ2 - 5λ - 6 = 0 ⇒ (λ - 6)(λ + 1) = 0 ⇒ λ = 6, -1
∴ Larger eigen value is 6.
Question 72

The cardinality of the power set of {0, 1, 2, … 10} is _________.

A
2046
B
2047
C
2048
D
2049
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-2]
Question 72 Explanation: 
Cardinality of the power set of {0, 1, 2, … , 10} is 211 i.e., 2048.
Question 73

The number of divisors of 2100 is ______.

A
36
B
37
C
38
D
39
       Engineering-Mathematics       Combinatorics       GATE 2015 [Set-2]
Question 73 Explanation: 
Let N = 2100
= 22+3×52×7 (i.e., product of primes)
Then the number of divisions of 2100 is
(2+1)∙(1+1)∙(2+1)∙(1+1) i.e., (3)(2)(3)(2) i.e., 36.
Question 74

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 bridges
B
A bridge cannot be part of a simple cycle
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 bridges cannot have a cycle
       Engineering-Mathematics       Graph-Theory       GATE 2015 [Set-2]
Question 74 Explanation: 
Since, every edge in a tree is bridge
∴ (A) is false
Since, every edge in a complete graph kn(n≥3) is not a bridge ⇒
(C) is false
Let us consider the following graph G:

This graph has a bridge i.e., edge ‘e’ and a cycle of length ‘3’
∴ (D) is false
Since, in a cycle every edge is not a bridge
∴ (B) is true
Question 75

Consider six memory partitions of sizes 200 KB, 400 KB, 600 KB, 500 KB, 300 KB and 250 KB, where KB refers to kilobyte. These partitions need to be allotted to four processes of sizes 357 KB, 210KB, 468 KB and 491 KB in that order. If the best fit algorithm is used, which partitions are NOT allotted to any process?

A
200KBand 300 KB
B
200KBand 250 KB
C
250KBand 300 KB
D
300KBand 400 KB
       Engineering-Mathematics       Memory-Management       GATE 2015 [Set-2]
Question 75 Explanation: 

Since Best fit algorithm is used. So, process of size,
357KB will occupy 400KB
210KB will occupy 250KB
468KB will occupy 500KB
491KB will occupy 600KB
So, partitions 200KB and 300KB are NOT alloted to any process.
Question 76

The number of onto function (surjective function) from set X = {1, 2, 3, 4} to set Y = {a, b, c} is ______.

A
36
B
37
C
38
D
39
       Engineering-Mathematics       Relations-and-Functions       GATE 2015 [Set-2]
Question 76 Explanation: 
Number of onto function from set X to set Y with |X| = m, |Y| = n is

m = 4, n = 3 ⇒ number of onto function is
Question 77

Perform the following operations on the matrix

    (i) add the third row to the second row
    (ii) Subtract the third column from the first column

The determinant of the resultant matrix is ________.

A
0
B
1
C
2
D
3
       Engineering-Mathematics       Linear-Algebra       GATE 2015 [Set-2]
Question 77 Explanation: 

Method 2: Determinant is unaltered by the operations (i) and (ii)
∴ Determinant of the resultant matrix = Determinant of the given matrix

(Since C1, C3 are proportional i.e., C3 = 15C1)
Question 78

Which one of the following well formed formulae is a tautology?

A
∀x ∃y R(x,y) ↔ ∃y ∀x R(x,y)
B
(∀x [∃y R(x,y) → S(x,y)]) → ∀x∃y S(x,y)
C
[∀x ∃y (P(x,y) → R(x,y)] ↔ [∀x ∃y (¬ P(x,y)∨R(x,y)]
D
∀x ∀y P(x,y) → ∀x ∀y P(y,x)
       Engineering-Mathematics       Propositional-Logic       GATE 2015 [Set-2]
Question 78 Explanation: 
Since P→R = ¬P∨R
[∀x ∃y (P(x,y) → R(x,y)] ↔ [∀x ∃y (¬ P(x,y)∨R(x,y)] is a tautology.
Question 79

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, or, 1 mod 4
       Engineering-Mathematics       Graph-Theory       GATE 2015 [Set-2]
Question 79 Explanation: 
An n vertex self-complementary graph has exactly half number of edges of the complete graph i.e., n(n-1)/4 edges. Since n(n – 1) must be divisible by 4, n must be congruent to 0 or 1 module 4.
Question 80

The secant method is used to find the root of an equation f(x) = 0. It is started from two distinct estimates xa and xb for the root. It is an iterative procedure involving linear interpolation to a root. The iteration stops if f(xb) is very small and then xb is the solution. The procedure is given below. Observe that there is an expression which is missing and is marked by? Which is the suitable expression that is to be put in place of? So that it follows all steps of the secant method?

Secant

Initialize: xa, xb, ε, N     // ε = convergence indicator
fb = f(xb) i = 0
while (i < N and |fb| > ε) do
   i = i + 1                 // update counter
   xt = ?                    // missing expression for
                             // intermediate value
   xa = xb                   // reset xa
   xb = xt                   // reset xb
   fb = f(xb)                // function value at new xb
end while
if |fb| > ε
  then // loop is terminated with i = N
  write “Non-convergence”
else
  write “return xb”
end if 
A
xb – (fb–f(xa))fb /(xb–xa)
B
xa – (fa–f(xa))fa /(xb–xa)
C
xb – (xb–xa)fb /(fb–f(xa))
D
xa – (xb–xa) fa /(fb–f(xa))
       Engineering-Mathematics       Secant-Method       GATE 2015 [Set-2]
Question 80 Explanation: 
The solution for this question is direct formula of secant method. In numerical analysis, the secant method is a root-finding algorithm that uses a succession of roots of secant lines to better approximate a root of a function f. The secant method can be thought of as a finite difference approximation of Newton's method. However, the method was developed independently of Newton's method, and predates it by over 3,000 years.

The first two iterations of the secant method. The red curve shows the function f and the blue lines are the secants. For this particular case, the secant method will not converge.
Question 81

Let f(x) = x -(1/3) and A denote the area of the region bounded bu f(x) and the X-axis, when x varies from -1 to 1. Which of the following statements is/are TRUE?

    I) f is continuous in [-1,1]
    II) f is not bounded in [-1,1]
    III) A is nonzero and finite
A
II only
B
III only
C
II and III only
D
I, II and III
       Engineering-Mathematics       Calculus       GATE 2015 [Set-2]
Question 81 Explanation: 
Since f(0)→∞
∴ f is not bounced in [-1, 1] and hence f is not continuous in [-1, 1].

∴ Statement II & III are true.
Question 82

Let X and Y denote the sets containing 2 and 20 distinct objects respectively and F denote the set of all possible functions defined from X to Y. Let f be randomly chosen from F. The probability of f being one-to-one is ______.

A
0.95
B
0.96
C
0.97
D
0.98
       Engineering-Mathematics       Relations-and-Functions       GATE 2015 [Set-2]
Question 82 Explanation: 
|X| = 2, |Y| = 20
Number of functions from X to Y is 202 i.e., 400 and number of one-one functions from X to Y is 20P2 i.e., 20×19 = 380
∴ Probability of a function f being one-one is 380/400 i.e., 0.95
Question 83

In a room there are only two types of people, namely Type 1 and Type 2. Type 1 people always tell the truth and Type 2 people always lie. You give a fair coin to a person in that room, without knowing which type he is from and tell him to toss it and hide the result from you till you ask for it. Upon asking, the person replies the following:

“The result of the toss is head if and only if I am telling the truth.”

Which of the following options is correct?

A
The result is head
B
The result is tail
C
If the person is of Type 2, then the result is tail
D
If the person is of Type 1, then the result is tail
       Engineering-Mathematics       Prepositional-Logic       GATE 2015 [Set-3]
Question 83 Explanation: 
We do not know the type of person from whom those words are coming from and so we can have two cases.
Case 1:
The person who speaks truth. This definitely implies that result of toss is Head.
Case 2:
The person who lies. In this the reality will be the negation of the statement.
The negation of (x⇔y) is exactly one of x or y holds. "The result of the toss is head if and only if I am telling the truth". So here two possibilities are there,
→ It is head and lie spoken.
→ It is not head and truth spoken.
Clearly, the second one cannot speaks the truth. So finally it is head.
Hence, option (A).
Question 84

Suppose U is the power set of the set S = {1, 2, 3, 4, 5, 6}. For any T ∈ U, let |T| denote the number of element in T and T' denote the complement of T. For any T, R ∈ U, let TR be the set of all elements in T which are not in R. Which one of the following is true?

A
∀X ∈ U (|X| = |X'|)
B
∃X ∈ U ∃Y ∈ U (|X| = 5, |Y| = 5 and X ∩ Y = ∅)
C
∀X ∈ U ∀Y ∈ U (|X| = 2, |Y| = 3 and X \ Y = ∅)
D
∀X ∈ U ∀Y ∈ U (X \ Y = Y' \ X')
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-3]
Question 84 Explanation: 
As X and Y are elements of U, so X and Y are subsets of S.
(A) False. Consider X = {1,2}. Therefore, X' = {3,4,5,6}, |X| = 2 and |X'| = 4.
(B) False. Because for any two possible subsets of S with 5 elements should have atleast 4 elements in common. Hence X∩Y cannot be null.
(C) False. Consider X = {1,4}, Y= {1,2,3} then X\Y = {4} which is not null.
(D) True. Take any possible cases.
Question 85

The number of 4 digit numbers having their digits in non-decreasing order (from left to right) constructed by using the digits belonging to the set {1, 2, 3} is _____.

A
15
B
16
C
17
D
18
       Engineering-Mathematics       Combinatorics       GATE 2015 [Set-3]
Question 85 Explanation: 
Just try to write all the four digit numbers containing the digits 1, 2 and 3 only, in decreasing order.
1 1 1 1
1 1 1 2
1 1 1 3
1 1 2 2
1 1 2 3
1 1 3 3
1 2 2 2
1 2 2 3
1 2 3 3
1 2 3 3
1 3 3 3
2 2 2 2
2 2 2 3
2 2 3 3
2 3 3 3
3 3 3 3
Hence, total 15 4-digit no. are possible.
Question 86

In the given matrix, one of the eigenvalues is 1. the eigenvectors corresponding to the eigenvalue 1 are

⎡ 1 -1  2 ⎤
⎢ 0  1  0 ⎥
⎣ 1  2  1 ⎦
A
{α(4,2,1) | α≠0, α∈R}
B
{α(-4,2,1) | α≠0, α∈R}
C
{α(2,0,1) | α≠0, α∈R}
D
{α(-2,0,1) | α≠0, α∈R}
       Engineering-Mathematics       Linear-Algebra       GATE 2015 [Set-3]
Question 86 Explanation: 
X be an eigen vector corresponding to eigen value λ =1, then
AX = λX ⇒ (A - I)X = 0

⇒ -y+2z = 0 and x+2y = 0
⇒ y = 2z and x/(-2) = y
∴ x/(-2) = y = 2z ⇒ x/(-4) = y/2 = z/1 = α(say)

∴ Eigen vectors are {α(-4,2,1 | α≠0, α∈R}
Question 87

Consider a machine with a byte addressable main memory of 220 bytes, block size of 16 bytes and a direct mapped cache having 212 cache lines. Let the addresses of two consecutive bytes in main memory be (E201F)16 and (E2020)16. What are the tag and cache line address (in hex) for main memory address (E201F)16?

A
E, 201
B
F, 201
C
E, E20
D
2, 01F
       Engineering-Mathematics       Cache       GATE 2015 [Set-3]
Question 87 Explanation: 
Block size is of 16 bytes, so we need 4 offset bits. So the lowest 4 bit is offset bits.
No. of cache lines in cache is 212 bytes which needs 12 bits. So next lower 12 bits are line indexing bits.
And the remaining top 4 bits are tag bits (out of 20). So answer is (A).
Question 88

If for non-zero x, where a≠b then is

A
B
C
D
       Engineering-Mathematics       Calculus       GATE 2015 [Set-3]
Question 88 Explanation: 
Given,
af(x) + bf(1/x) = 1/x - 25 ------ (1)
Put x = 1/x,
af(1/x) + bf(x) = x - 25 ----- (2)
Multiply equation (1) with 'a' and Multiply equation (2) with 'b', then
abf(1/x) + a2 = a/x - 25a ----- (3)
abf(1/x) + b2 = bk - 25b ----- (4)
Subtract (3) - (4), we get
(a2 - b2) f(x) = a/x- 25a - bx + 25b
f(x) = 1/(a2 - b2) (a/x - 25a - bx +25b)
Now from equation,

Hence option (A) is the answer.
Question 89

The velocity v (in kilometer/minute) of a motorbike which starts from rest, is given at fixed intervals of time t(in minutes) as follows:

t   2  4  6  8  10  12  14  16 18 20
v  10  18 25 29 32  20  11  5  2  0

The approximate distance (in kilometers) rounded to two places of decimals covered in 20 minutes using Simpson’s 1/3rd rule is _________.

A
309.33
B
309.34
C
309.35
D
309.36
       Engineering-Mathematics       Simpson\'s-1/3rd-rule       GATE 2015 [Set-3]
Question 89 Explanation: 
Let ‘S’ be the distance covered in 20 minutes, then by simpson’s 1/3rd rule,
∵v = velocity
= 2/3[(0+0)+4(10+25+32+11+2)+2(18+29+20+5)]
= 309.33 km
(Here length of each of the subinterval is h = 2)
Question 90

If the following system has non-trivial solution,

    px + qy + rz = 0
    qx + ry + pz = 0
    rx + py + qz = 0

then which one of the following options is True?

A
p-q+r = 0 or p = q = -r
B
p+q-r = 0 or p = -q = r
C
p+q+r = 0 or p = q = r
D
p-q+r = 0 or p = -q = -r
       Engineering-Mathematics       Linear-Algebra       GATE 2015 [Set-3]
Question 90 Explanation: 
For non-trivial solution the value of determinant should be zero.
Question 91

Let R be a relation on the set of ordered pairs of positive integers such that ((p,q),(r,s)) ∈ R if and only if p - s = q - r. Which one of the following is true about R?

A
Both reflexive and symmetric
B
Reflexive but not symmetric
C
Not reflexive but symmetric
D
Neither reflexive nor symmetric
       Engineering-Mathematics       Set-Theory       GATE 2015 [Set-3]
Question 91 Explanation: 
Since p-q ≠ q-p
∴(p,q) R (p,q)
⇒ R is not reflexive.
Let (p,q) R (r,s) then p-s = q-r
⇒ r-q = s-p
⇒ (r,s) R (p,q)
⇒ R is symmetric.
Question 92

Let G = (V,E) be a directed graph where V is the set of vertices and E the set of edges. Then which one of the following graphs has the same strongly connected components as G?

A
G1=(V,E1) where E1={(u,v)|(u,v)∉E}
B
G2=(V,E2 )where E2={(u,v)│(u,v)∈E}
C
G3=(V,E3) where E3={(u,v)|there is a path of length≤2 from u to v in E}
D
G4=(V4,E) where V4 is the set of vertices in G which are not isolated
       Engineering-Mathematics       Graph-Theory       GATE 2014 [Set-1]
Question 92 Explanation: 
G(V, E) is a directed graph.
→ It strongly connected.
(A) G1=(V,E1) where E1={(u,v)|(u,v)∉E}
If (u, v) does not belong to the edge set ‘E’, then it indicates there are no edges. So, it is not connected.
(B) G2=(V,E2 )where E2={(u,v)│(u,v)∈E}
Given that ‘G’ is directed graph, i.e., it has path from each vertex to every other vertex.
Though direction is changed from (u, v) to (v, u), it is still connected component same as ‘G’.
(C) G3=(V,E3) where E3={(u,v)|there is a path of length≤2 from u to v in E}
This can also be true.
eg:

Both from each vertex to other vertex is also exists. So it is also strongly connected graph.
(D) G4=(V4,E) where V4 is the set of vertices in G which are not isolated.
If ‘G’ has same ‘x’ no. of isolated vertices, one strongly connected component
then no. of SCC = x + 1
G4 contain only ‘1’ component, which is not same as G.
Question 93

The value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a 4-by-4 symmetric positive definite matrix is ________.

A
0
B
1
C
2
D
3
       Engineering-Mathematics       Linear-Algebra       GATE 2014 [Set-1]
Question 93 Explanation: 
For real symmetric matrix, the eigen values are orthogonal to each other. So their dot product will be zero.
The finite dimensional spectral theorem says that any symmetric matrix whose entries are real can be diagonalized by an orthogonal matrix.
Question 94

Let the function

where and f(θ) denote the derivative of f with respect to θ. Which of the following is/are TRUE?

    (I) There exists such that such that f(θ)=0.
    (II) There exists such that such that f(θ)≠0.

A
I only
B
II only
C
Both I and II
D
Neither I nor II
       Engineering-Mathematics       Calculus       GATE 2014 [Set-1]
Question 94 Explanation: 
Rolle’s theorem:
Rolle’s theorem states that for any continuous, differentiable function that has two equal values at two distinct points, the function must have a point on the function where the first derivative is zero. The technical way to state this is: if f is continuous and differentiable on a closed interval [a,b] and if f(a) = f(b), then f has a minimum of one value c in the open interval [a, b] so that f'(c) = 0.
We can observe that, sin, cos are continuous, but, Tan is not continuous at π/2. As the mentioned interval does not contain π/2, we can conclude that it is continuous.
As per Rolls theorem both statement 1 and statement 2 are true.
Question 95

The function f(x) = x sinx satisfies the following equation: f''(x) + f(x) + tcosx = 0. The value of t is __________.

A
-2
B
-3
C
-4
D
-5
       Engineering-Mathematics       Calculus       GATE 2014 [Set-1]
Question 95 Explanation: 
f(x) = x Sinx
f ’(x) = x(Sinx)’ + Sin(x)(x’)
= xCosx + Sinx ---------①
f ’’(x) = x (Cosx)’ + Cos (x)’+ Cos x
= -x Sinx + 2Cosx -----------②
Given: f ’’(x) + f(x) + t Cosx = 0
Replace ① & ②,
-xSinx + 2Cosx + xSinx + tCosx = 0
2Cosx + tCosx = 0
t = -2
Question 96

A function f(x) is continuous in the interval [0,2]. It is known that f(0) = f(2) = -1 and f(1) = 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)
       Engineering-Mathematics       Calculus       GATE 2014 [Set-1]
Question 96 Explanation: 
Consider this function as sum of two functions as, g(y) = f(y) -f(y+1)
Since function f is continuous in [0, 2], therefore g would be continuous in [0, 1]
g(0) = -2, g(1) = 2
Since g is continuous and goes from negative to positive value in [0,1]. Therefore at some point g would be 0 in (0,1).
g=0 ⇒ f(y) = f(y+1) for some y in (0,1).
Apply similar logic to option D, Let g(y) = f(y) + f(2 - y)
Since function f is continuous in [0, 2], therefore g would be continuous in [0, 1] (sum of two continuous functions is continuous)
g(0) = -2, g(1) = 2
Since g is continuous and goes from negative to positive value in [0, 1]. Therefore at some point g would be 0 in (0, 1).
There exists y in the interval (0, 1) such that:
g=0 ⇒ f(y) = -f(2 – y)
Both A, D are answers.
Question 97

Four fair six-sided dice are rolled. The probability that the sum of the results being 22 is X/1296. The value of X is ___________.

A
10
B
11
C
12
D
13
       Engineering-Mathematics       Probability       GATE 2014 [Set-1]
Question 97 Explanation: 
Each dice can result from {1, 2, 3, 4, 5, 6}
To get ‘22’ as Sum of four outcomes
x1 + x2 + x3 + x4 = 22
The maximum Sum = 6+6+6+6 = 24 which is near to 22
So, keeping three 6’s, 6+6+6+x = 22
x = 4 combination① = 6 6 6 4
Keeping two 6’s, 6+6+x1+x2 = 22
x1+x2 = 10 possible x’s (5, 5) only
combination② = 6 6 5 5
No. of permutation with 6664 = 4!/ 3! = 4
“ “ “ 6655 = 4!/ 2!2! = 6
Total = 4+6 = 10 ways out of 6×6×6×6 = 1296
Pnb (22) = 10/1296 ⇒ x = 10
Question 98

A pennant is a sequence of numbers, each number being 1 or 2. An n-pennant is a sequence of numbers with sum equal to n. For example, (1,1,2) is a 4-pennant. The set of all possible 1-pennants is {(1)}, the set of all possible 2-pennants is {(2), (1,1)}and the set of all 3-pennants is {(2,1), (1,1,1), (1,2)}. Note that the pennant (1,2) is not the same as the pennant (2,1). The number of 10-pennants is ________.

A
89
B
90
C
91
D
92
       Engineering-Mathematics       Combinatorics       GATE 2014 [Set-1]
Question 98 Explanation: 
No twos: 1111111111 ⇒ 1 pennant
Single two: 211111111 ⇒ 9!/8!1! = 9 pennants
Two twos: 22111111 ⇒ 8!/6!2! = 28
Three twos: 2221111 ⇒ 7!/3!4! = 35
Four twos: 222211 ⇒ 6!/4!2! = 15
Five twos: 22222 ⇒ 1
Total = 89 pennants.
Question 99

Let S denote the set of all functions  f:{0,1}→ {0,1}. Denote by N the number of functions from S to the set {0,1}. The value of log2log2N is ______.

A
16
B
17
C
18
D
19
       Engineering-Mathematics       Relations-and-Functions       GATE 2014 [Set-1]
Question 99 Explanation: 
The number of functions from A to B where size of A = |A| and size of B = |B| is |B||A|
{0,1}4 = {0,1}×{0,1}×{0,1}×{0,1} = 16
|S| = 216
N = 2|S|
loglogN=loglog2|S| = log |S| = log216 = 16
Question 100

Consider an undirected graph where self-loops are not allowed. The vertex set of G is {(i,j): 1 ≤ i ≤ 12, 1 ≤ j ≤ 12}. There is an edge between (a,b) and (c,d) if |a - c| ≤ 1 and |b - d| ≤ 1. The number of edges in this graph is __________.

A
506
B
507
C
508
D
509
       Engineering-Mathematics       Graph-Theory       GATE 2014 [Set-1]
Question 100 Explanation: 
The total number of vertices in the graph is 12*12 = 144. The vertices are allowed to connect in both horizontal and vertical directions which are separated by at most 1 distance.
If we observe the graph, it looks like a 12 by 12 grid. Each corner vertex has a degree of 3 and we have 4 corner vertices. 40 external vertices of degree 5 and remaining 100 vertices of degree 8.
From Handshaking theorem, sum of the degrees of the vertices is equal to the 2*number of edges in the graph.
⇒ (4*3) + (40*5) + (100*8) = 2*E
⇒ 1012 = 2*E
⇒ E = 506
Question 101

An ordered -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, 1, 1, 1, 0)
       Engineering-Mathematics       Graph-Theory       GATE 2014 [Set-1]
Question 101 Explanation: 
This can be checked by Havel-hakimi theorem:
A) (1, 1, 1, 1, 1, 1)

Yes, it is a graph.
We will see that option (C) is not graphic.
Question 102

Which one of the following propositional logic formulas is TRUE when exactly two of p, q, and r are TRUE?

A
((p↔q)∧r)∨(p∧q∧∼r)
B
(∼(p↔q )∧r)∨(p∧q∧∼r)
C
((p→q)∧r)∨(p∧q∧∼r)
D
(∼(p↔q)∧r)∧(p∧q∧∼r)
       Engineering-Mathematics       Prepositional-Logic       GATE 2014 [Set-1]
Question 102 Explanation: 
Method 1: construct the tables for all options and check with T, T, F combinations.
Method2: directly check with one of {TTF, TFT, FTT} options.
As there are two T’s in each option, replace them and check with the third value.
Eg: Place p=q= T
(∼(p↔q)∧r)∨(p∧q∧∼r)
=(∼(T↔T)∧r)∨(T∧T∧∼r)
=(∼(T)∧r)∨(T∧∼r)
=(F∧r)∨(T∧∼r)
=(F)∨(∼r)
=∼r
This is true for r=F.
Similarly with p=r=T and q=F.
q=r=T and p=F
Option B is the answer.
Question 103

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 =_____________.

A
11.90
B
11.91
C
11.92
D
11.93
       Engineering-Mathematics       Probability       GATE 2014 [Set-2]
Question 103 Explanation: 
There are 10 systems.
Out of which ‘4’ are chosen at random without replaced and created.
If at least three of them are working then system is deemed functional
i.e., there should be only ‘one’ non-working system in set of ‘4’.
It is possible with combination
W – W – W – N,
W – W – N – W,
W – N – W – W,
N – W – W – W.
For W – W – W – N, the probability = (choosing working out of 10) × (choosing working out of 9) × (choosing working out of 8) × (choosing non-working out of 7)
= 4/10×3/9×2/8×1/7
where 4/10 ⇒ 4 working out of 10
3/9 ⇒ 3 working are remaining out of ‘9’ as ‘1’ is already taken
For ‘4’ Sum combinations
Total probability = 4×[4/10×3/9×2/8×1/7] = 600/5040
We need 100p ⇒ 100×600/5040 = 11.90
Question 104

Each of the nine words in the sentence "The quick brown fox jumps over the lazy dog" is written on a separate piece of paper. These nine pieces of paper are kept in a box. One of the pieces is drawn at random from the box. The expected length of the word drawn is _____________. (The answer should be rounded to one decimal place.)

A
3.9
B
4.0
C
4.1
D
4.2
       Engineering-Mathematics       Probability       GATE 2014 [Set-2]
Question 104 Explanation: 
"The quick brown fox jumps over the lazy dog"
There are ‘9’ words in this sentence.
No. of characters in each word
The (3)
quick (5)
brown (5)
fox (3)
jumps (5)
over (4)
the (3)
lazy (4)
dog (3)
Each word has equal probability.
So expected length = 3×1/9+5×1/9+5×1/9+3×1/9+5×1/9+ 4×1/9+3×1/9+4×1/9+3×1/9
= 35/9
= 3.9
Question 105

The maximum number of edges in a bipartite graph on 12 vertices is ______.

A
36
B
37
C
38
D
39
       Engineering-Mathematics       Graph-Theory       GATE 2014 [Set-2]
Question 105 Explanation: 
Max. no. of edges possible in bipartite graph, only if vertices are equal in each set i.e., 12/2 = 6 in each set.

Total no. of edges = 6×6 = 36
Question 106

If the matrix A is such that

then the determinant of A is equal to

A
0
B
1
C
2
D
3
       Engineering-Mathematics       Linear-Algebra       GATE 2014 [Set-2]
Question 106 Explanation: 
Question 107

A non-zero polynomial f(x) of degree 3 has roots at x = 1, x = 2 and x = 3. Which one of the following must be TRUE?

A
f(0)f(4) < 0
B
f(0)f(4) > 0
C
f(0) + f(4) > 0
D
f(0) + f(4) < 0
       Engineering-Mathematics       Calculus       GATE 2014 [Set-2]
Question 107 Explanation: 
Roots of polynomial of degree ‘3’ are 1, 2, 3
Polynomial will be
f(x) = (x-1)(x-2)(x-3)
f(0) = -1 × -2 × -3 = -6
f(4) = 3 × 2 × 1 = 6
f(0) ∙ f(4) = - 36
f(0) + f(4) = 6 - 6 = 0
Option (A) is correct.
Question 108

In the Newton-Raphson method, an initial guess of x0 = 2 is made and the sequence x0, x1, x2 … is obtained for the function

0.75x3 – 2x2 – 2x + 4 = 0 

Consider the statements

(I) x3 = 0.
(II) The method converges to a solution in a finite number of iterations.

Which of the following is TRUE?

A
Only I
B
Only II
C
Both I and II
D
Neither I nor II
       Engineering-Mathematics       Newton-Raphson-Method       GATE 2014 [Set-2]
Question 108 Explanation: 
Note: Numerical methods are not in GATE CS syllabus.
Question 109

The product of the non-zero eigenvalues of the matrix

1 0 0 0 1
0 1 1 1 0
0 1 1 1 0
0 1 1 1 0
1 0 0 0 1

is ______.

A
6
B
7
C
8
D
9
       Engineering-Mathematics       Linear-Algebra       GATE 2014 [Set-2]
Question 109 Explanation: 
Characteristic equation for matrix ‘A’ is
AX = λX

x1 + x5 = λx1 ---------- (1)
x1 + x5 = λx5 ---------- (2)
(1) + (2) ⇒ 2(x1 + x5) = λ(x1 + x5) ⇒ λ1 = 2
x2 + x3 + x4 = λ∙x2 -------- (4)
x2 + x3 + x4 = λ∙x3 -------- (5)
x2 + x3 + x4 = λ∙x4 -------- (6)
(4)+(5)+(6) = 3(x2 + x3 + x4) = λ(x2 + x3 + x4 ) ⇒ λ2 = 3
Product = λ1 × λ2 = 2×3 = 6
Question 110

The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is ______.

A
0.259 to 0.261
B
0.260 to 0.262
C
0.261 to 0.263
D
0.262 to 0.264
       Engineering-Mathematics       Probability       GATE 2014 [Set-2]
Question 110 Explanation: 
Let A = divisible by 2, B = divisible by 3 and C = divisible by 5, then
n(A)=50, n(B)=33, n(C)=20
n(A∩B)=16, n(B∩C)=6, n(A∩C)=10
n(A∩B∩C)=3
P(A∪B∪C) = P(A)+P(B)+P(C)-P(A∩B)-P(B∩C) -P(A∩C)+P(A∩B∩C) = 74/100
∴ Required probability is P(A∩B∩C) = 1-P(A∪B∪C) = 0.26
Question 111

The number of distinct positive integral factors of 2014 is _________.

A
0.26
B
0.27
C
8
D
0.29
       Engineering-Mathematics       Combinatorics       GATE 2014 [Set-2]
Question 111 Explanation: 
First lets find prime factorization of 2014
= 2' × 19' × 53'
Now number of distinct integral factors of 2014 will be,
(1+1)×(1+1)×(1+1) = 2×2×2 = 8
Question 112

Consider the following relation on subsets of the set S of integers between 1 and 2014. For two distinct subsets U and V of S we say U < V if the minimum element in the symmetric difference of the two sets is in U. Consider the following two statements:

S1: There is a subset of S that is larger than every other subset.
S2: There is a subset of S that is smaller than every other subset.

Which one of the following is CORRECT?

A
Both S1 and S2 are true
B
S1 is true and S2 is false
C
S2 is true and S1 is false
D
Neither S1 nor S2 is true
       Engineering-Mathematics       Set-Theory       GATE 2014 [Set-2]
Question 112 Explanation: 
Given: S = {1, 2, 3, …, 2014}
U⊂S, V⊂S
Let U = {1, 2, 3}
V = {2, 3, 4}
Symmetric difference:
(U – V) ∪ (V – U) = {1} ∪ {4} = {1, 4}
The minimum element in the symmetric difference is 1 and 1∈U.
S1: Let S = Universal set = {1, 2, … 2014}
This universal set is larger than every other subset.
S2: Null set is smaller than every other set.
Let U = { }, V = {1}
Symmetric difference = ({ } – {1}) ∪ ({1} – { }) = { } ∪ {1} = {1}
So, U < V because { } ∈ U.
Question 113

A cycle on n vertices is isomorphic to its complement. The value of n is _____.

A
5
B
6
C
7
D
8
       Engineering-Mathematics       Graph-Theory       GATE 2014 [Set-2]
Question 113 Explanation: 
Complement of a graph means, we need to remove the existing edges from the graph and place the new edges which were not earlier among the actual graph.
In a cycle of n vertices, each vertex is connected to other two vertices. So each vertex degree is 2.
When we complement it, each vertex will be connected to remaining n-3 vertices ( one is self and two other vertices in actual graph).
As per given question,
n-3 =2
n=5
Cycle of 5 vertices is

Complement of the above graph1 is

Graph1 and Graph2 are complement each other.
So, the value of n is 5.
Question 114

Which one of the following Boolean expressions is NOT a tautology?

A
((a ⟶ b) ∧ (b ⟶ c)) ⟶ (a ⟶ c)
B
(a ⟷ c) ⟶ (∽ b ⟶ (a ∧ c))
C
(a ∧ b ∧ c) ⟶ (c ∨ a)
D
a ⟶ (b ⟶ a)
       Engineering-Mathematics       Prepositional-Logic       GATE 2014 [Set-2]
Question 114 Explanation: 
A:
((a → b) ∧ (b → c)) → (a → c)
If (a → b) is false with a = T, b = F,
then (F ∧ (b → c)) → (a → c)
F → (a → c)
which is True for any (a → c)
This is tautology.
B:
(a ⟷ c) ⟶ (∽b ⟶ (a ∧ c))
For (a ⟷ c) be True and
∽b → (a ∧ c) should be False
Let a = c = F
(F → F) → (∽b (F ∩ F))
T → (∽b → F)
This is False for b = F
So, this is not True.
C:
(a ∧ b ∧ c) ⟶ (c ∨ a)
(c ∨ a) is False only for a = c = F
if (c ∨ a) is False
(F ∧ b ∧ F) → F
F → F which is Tautology
True always.
D:
a ⟶ (b ⟶ a)
a ⟶ (~b ∨ a)
(~a ∨ a) ∨ ~b = T ∨ ~b = T which is tautology
Question 115

Consider the following statements:

P: Good mobile phones are not cheap
Q: Cheap mobile phones are not good
L: P implies Q
M: Q implies P
N: P is equivalent to Q

Which one of the following about L, M, and N is CORRECT?

A
Only L is TRUE.
B
Only M is TRUE.
C
Only N is TRUE.
D
L, M and N are TRUE.
       Engineering-Mathematics       Prepositional-Logic       GATE 2014 [Set-3]
Question 115 Explanation: 
In the given statements observe that "not cheap" & cheap, "good & not good" are used.
So, given statement can be sub divided such that we can utilize the negation of this atomic statements.
Suppose, X is Good mobile and Y is cheap then
P: (Good(x) → ~cheap(x)) → (~good(x) ∨ ~cheap(x))
Q: cheap(x) → ¬good(x) ⟺ ((¬cheap(x) ∨ good(x)) ⟺ ¬good(x) ∨ ¬cheap(x))
All these are contra positive.
All L, M, N are true.
Question 116

Let X and Y be finite sets and f: X→Y be a function. Which one of the following statements is TRUE?

A
For any subsets A and B of X, |f(A ∪ B)| = |f(A)|+|f(B)|
B
For any subsets A and B of X, f(A ∩ B) = f(A) ∩ f(B)
C
For any subsets A and B of X, |f(A ∩ B)| = min{ |f(A)|,f|(B)|}
D
For any subsets S and T of Y, f -1 (S ∩ T) = f -1 (S) ∩ f -1 (T)
       Engineering-Mathematics       Set-Theory       GATE 2014 [Set-3]
Question 116 Explanation: 
The function f: x→y.
We need to consider subsets of 'x', which are A & B (A, B can have common elements are exclusive).
Similarly S, T are subsets of 'y'.

To be a function, each element should be mapped with only one element.
(a) |f(A∪B)| = |f(A)|+|f(B)|
|{a,b,c}|∪|{c,d,e}| = |{a,b,c}| + |{c,d,e}|
|{a,b,c,d,e}| = 3+3
5 = 6 FALSE
(d) To get inverse, the function should be one-one & onto.
The above diagram fulfills it. So we can proceed with inverse.
f-1 (S∩T ) = f-1 (S)∩f-1 (T)
f-1 (c) = f-1 ({a,b,c})∩f-1 ({c,d,e})
2 = {1,2,3}∩{2,4,5}
2 = 2 TRUE
Question 117

Let G be a group with 15 elements. Let L be a subgroup of G. It is known that L ≠ G and that the size of L is at least 4. The size of L  is __________.

A
5
B
6
C
7
D
8
       Engineering-Mathematics       Set-Theory       GATE 2014 [Set-3]
Question 117 Explanation: 
Lagrange's theorem, in the mathematics of group theory, states that for any finite group G, the order (number of elements) of every subgroup H of G divides the order of G.
So, 15 is divided by {1, 3, 5, 15}.
As minimum is 4 and total is 15, we eliminate 1,3,15.
Answer is 5.
Question 118

Which one of the following statements is TRUE about every n × n matrix with only real eigenvalues?

A
If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative.
B
If the trace of the matrix is positive, all its eigenvalues are positive.
C
If the determinant of the matrix is positive, all its eigenvalues are positive.
D
If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive.
       Engineering-Mathematics       Linear-Algebra       GATE 2014 [Set-3]
Question 118 Explanation: 
The sum of the n eigenvalues of A is the same as the trace of A (that is, the sum of the diagonal elements of A).
• The product of the n eigenvalues of A is the same as the determinant of A. •
A: Yes, for sum to be negative there should be atleast one negative number.
B: There can be one small negative number and remaining positive, where sum is positive.
C: Product of two negative numbers is positive. So, there no need of all positive eigen values.
D: There is no need for all eigen values to be positive, as product of two negative numbers is positive.
Question 119

If V1 and V2 are 4-dimensional subspace of a 6-dimensional vector space V, then the smallest possible dimension of V1∩V2   is ______.

A
2
B
3
C
4
D
5
       Engineering-Mathematics       Set-Theory       GATE 2014 [Set-3]
Question 119 Explanation: 
In a 6 dimensional vector space, sub space of 4 dimensional subspace V1, V2 are provided. Then the V1∩V2?
For eg: a two dimensional vector space have x, y axis. For dimensional vector space, it have x, y, z axis.
In the same manner, 6 dimensional vector space has x, y, z, p, q, r (assume).
Any subspace of it, with 4 dimensional subspace consists any 4 of the above. Then their intersection will be atmost 2.
[{x,y,z,p} ∩ {r,q,p,z}] = #2
V1 ∩ V2 = V1 + V2 - V1 ∪ V2 = 4 + 4 + (-6) = 2
Question 120

If , then the value of k is equal to ________.

A
4
B
5
C
6
D
7
       Engineering-Mathematics       Calculus       GATE 2014 [Set-3]
Question 120 Explanation: 
The graph x.Sinx from 0 to 2π is

We have |xSinx|,

We can observe that it is positive from 0 to π and negative in π to 2π.
To get positive value from π to 2π we put ‘-‘ sign in the (π, 2π)
Question 121

With respect to the numerical evaluation of the definite integral ,  where a  and b are given, which of the following statements is/are TRUE?

    I) The value of obtained using the trapezoidal rule is always greater than or equal to the exact value of the definite integral.
    II) The value of obtained using the Simpson’s rule is always equal to the exact value of the definite integral.

A
I only
B
II only
C
Both I and II
D
Neither I nor II
       Engineering-Mathematics       Calculus       GATE 2014 [Set-3]
Question 121 Explanation: 
Note: Numerical methods are out of syllabus for the GATE -CS.
Question 122

The value of the integral given below is

A
-2π
B
π
C
D
       Engineering-Mathematics       Calculus       GATE 2014 [Set-3]
Question 122 Explanation: 
Question 123

Let S be a sample space and two mutually exclusive events A and B be such that A∪B = S. If P(∙) denotes the probability of the event, the maximum value of P(A)P(B) is __________.

A
0.25
B
0.26
C
0.27
D
0.28
       Engineering-Mathematics       Probability       GATE 2014 [Set-3]
Question 123 Explanation: 
We know that
P(A∪B) = P(A) + P(B) + P(A∩B) = 1 →①
But, as A and B are mutually exclusive events
P(A∩B) = 0
∴ P(A∪B) = P(A) + P(B) = 1 →②
Arithmetic mean of two numbers ≥ Geometric mean of those two numbers
(P(A)+P(B))/2 ≥ √(P(A)∙P(B))
1/2 ≥ √(P(A)∙P(B)) (∵from ②)
Squaring on both sides
1/4 ≥ P(A)∙P(B)
P(A)∙P(B) ≤ 1/4
∴ Maximum value of P(A)P(B) = 1/4 = 0.25
Question 124

Consider the set of all functions f: {0,1, … ,2014} → {0,1, … ,2014} such that f(f(i)) = i, for all 0 ≤ i ≤ 2014. Consider the following statements:

    P. For each such function it must be the case that for every i, f(i) = i.
    Q. For each such function it must be the case that for some i, f(i) = i.
    R. Each such function must be onto.

Which one of the following is CORRECT?

A
P, Q and R are true
B
Only Q and R are true
C
Only P and Q are true
D
Only R is true
       Engineering-Mathematics       Set-Theory       GATE 2014 [Set-3]
Question 124 Explanation: 
f: {0,1,…,2014} → {0,1,…,2014} and f(f(i)) = i

So f(i)should be resulting only {0, 1, …2014}
So, every element in range has a result value to domain. This is onto. (Option R is correct)
We have ‘2015’ elements in domain.
So atleast one element can have f(i) = i,
so option ‘Q’ is also True.
∴ Q, R are correct.
Question 125

There are two elements x, y in a group (G,∗) such that every element in the group can be written as a product of some number of x's and y's in some order. It is known that

  x * x = y * y = x * y * x = y * x * y * x = e   

where e is the identity element. The maximum number of elements in such a group is ________.

A
4
B
5
C
6
D
7
       Engineering-Mathematics       Graph-Theory       GATE 2014 [Set-3]
Question 125 Explanation: 
We know
a*a-1 = e

1. x*x = e So x-1 is x ⇒ x is element of Group
2. y*y = e So y-1 = y ⇒ y is element of Group

4. (y*x)*(y*x) = x*y*y*x = x*x*e = e So (y*x)-1 = (y*x)
In ③, ④
x*y, y*x has same inverse, there should be unique inverse for each element.
x*y = y*x (even with cumulative law, we can conclude)
So {x, y, e, x*y} are element of Group.
Question 126

If G is a forest with n vertices and k connected components, how many edges does G have?

A
⌊n/k⌋
B
⌈n/k⌉
C
n–k
D
n-k+1
       Engineering-Mathematics       Graph-Theory       GATE 2014 [Set-3]
Question 126 Explanation: 
Suppose, if each vertex is a component, then k=n, then there will not be any edges among them Replace the same in options
Option 1, 2 will give answer 1. (i.e. one edge among them),
Option 3: n-k = 0 edges.
Option 4: n-k+1 = 1 edge, which is false.
Question 127

Let δ denote the minimum degree of a vertex in a graph. For all planar graphs on n vertices with δ ≥ 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/(δ+1)
       Engineering-Mathematics       Set-Theory       GATE 2014 [Set-3]
Question 127 Explanation: 
Euler’s formula:
v – e + f = 2 →①
Point ① degree of each vertex is minimum ‘3’.

3×n ≥ 2e
e ≤ 3n/2
From ① :
n-3n/2+f = 2 ⇒
Question 128

The CORRECT formula for the sentence, “not all rainy days are cold” is

A
∀d (Rainy(d) ∧∼Cold(d))
B
∀d (∼Rainy(d) → Cold(d))
C
∃d (∼Rainy(d) → Cold(d))
D
∃d (Rainy(d) ∧∼Cold(d))
       Engineering-Mathematics       Prepositional-Logic       GATE 2014 [Set-3]
Question 128 Explanation: 
Not all rainy days are cold
= ∼[∀rainy days are cold]
= ∼[∀ days (rainy days ⇒ cold days]
= ∃ days[∼(cold days ∨ ∼rainy days)]
= ∃ days[rainy days ∧ ∼cold days]
Question 129

A binary operation ⊕ on a set of integers is defined as x ⊕ y = x+ y2. Which one of the following statements is TRUE about ⊕?

A
Commutative but not associative
B
Both commutative and associative
C
Associative but not commutative
D
Neither commutative nor associative
       Engineering-Mathematics       Set-Theory       GATE 2013
Question 129 Explanation: 
Cumulative property:
A binary relation on a set S is called cumulative if a*b = b*a ∀ x,y∈S.
Associative property:
A binary relation on set is called associative if (a*b)*c = a*(b*c) ∀ x,y∈S.
Given x⊕y = x2 + y2 --------(1)
Replace x, y in (1)
y⊕x = y2 + x2 which is same as (1), so this is cumulative
(x⊕y)⊕z = (x2 + y2) ⊕ z
= (x2 + y2) + z2
= x2 + y2 + z2 + 2x2y2 ----------(2)
x⊕(y ⊕ z) = x ⊕ (y2 + z2)
= x2 + (y2 + z2)2
= x2 + y2 + z2 + 2y2z2 ----------- (3)
(2) & (3) are not same so this is not associative.
Question 130

Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?

A
8/(2e3)
B
9/(2e3)
C
17/(2e3)
D
26/(2e3)
       Engineering-Mathematics       Probability       GATE 2013
Question 130 Explanation: 
The formula for the Poisson probability mean function
P(x : λ) = (e λx)/x! for x = 0,1,2….
‘λ’ is the average number (mean)
Given that mean = λ = 3
The probability of observing fewer than three cars is
P(zero car) + P(one car) + P(two cars)
= (e-3 30)/0!+(e-3 31)/1!+(e-3 32)/2!
= e-3+e-3∙3+(e-3)∙9)/2
= (17e-3<)/2
= 17/(2e3 )
Question 131

Which one of the following does NOT equal to

 
A
B
C
D
       Engineering-Mathematics       Linear-Algebra       GATE 2013
Question 131 Explanation: 

Try to derive options from the given matrix.
Observe that col 2 + col 3 will reuse x(x+1) term
C2 → C1 + C2


Question 132

Which one of the following functions is continuous at x = 3?

A
B
C
D
       Engineering-Mathematics       Number-System-and-Calculus       GATE 2013
Question 132 Explanation: 
Option A:
At x = 3, f(x) = 2
LHL (3), f(3-) = (x+3 )/3 = (3+3)/3 = 2
RHL (3), f(3+) = x-1 = 3-1 = 2
∴ f(x) is continuous.
Option B:
At x = 3, f(x) = 4
For x ≠ 3, f(x) = f(3+) = f(3-) = 8-x = 8-3 = 5
This is not continuous.
Option C:
At x ≤ 3, f(x) = x+3 = 3+3 = 6
At RHL(3), f(3+) = 4
This is not continuous.
Option D:
f(x) at x = 3 is not defined.
There is a break at x = 3, so this is not continuous.
Question 133

Function f is known at the following points:

 

The value of computed using the trapezoidal rule is

A
8.983
B
9.003
C
9.017
D
9.045
       Engineering-Mathematics       Number-System-and-Calculus       GATE 2013
Question 133 Explanation: 
Note: Numerical methods are not in GATE syllabus now.
Question 134

What is the logical translation of the following statement?

  "None of my friends are perfect."
A
∃x(F(x)∧¬P(x))
B
∃x(¬F(x)∧P(x))
C
∃x(¬F(x)∧¬P(x))
D
¬∃x(F(x)∧P(x))
       Engineering-Mathematics       Prepositional-Logic       GATE 2013
Question 134 Explanation: 
Let F(x) = x is my friend
P(x) = x is perfect
The meaning of ∃x(P(x)∧F(x)) is atleast one person who is my friend and perfect.
The negation of ∃x(P(x)∧F(x)) is “This is not the case that atlease one person who is my friend and perfect”.
So ~∃x(P(x)∧F(x)) is none of my friends are perfect.
Question 135

The following code segment is executed on a processor which allows only register operands in its instructions. Each instruction can have atmost two source operands and one destination operand. Assume that all variables are dead after this code segment.

c = a + b;
   d = c * a;
   e = c + a;
   x = c * c;
   if (x > a) {
      y = a * a;
   }
   else {
     d = d * d;
     e = e * e;
  }

Suppose the instruction set architecture of the processor has only two registers. The only allowed compiler optimization is code motion, which moves statements from one place to another while preserving correctness. What is the minimum number of spills to memory in the compiled code?

A
0
B
1
C
2
D
3
       Engineering-Mathematics       Prepositional-Logic       GATE 2013
Question 135 Explanation: 
After applying the code motion optimization the statement d = c*a; and e = c+a; can be moved down to else block as d and c are not used anywhere before that and also value of a and c is not changing.

In the above code total number of spills to memory is 1.
Question 136

Consider the following logical inferences.

    I1: If it rains then the cricket match will not be played.
    The cricket match was played.
    Inference: There was no rain.
    I2: If it rains then the cricket match will not be played.
    It did not rain.
    Inference: The cricket match was played.

Which of the following is TRUE?

A
Both I1 and I2 are correct inferences
B
I1 is correct but I2 is not a correct inference
C
I1 is not correct but I2 is a correct inference
D
Both I1 and I2 are not correct inferences
       Engineering-Mathematics       Propositional-Logic       GATE 2012
Question 136 Explanation: 
I1: If it rains then the cricket match will not be played.
The cricket match was played.
Let p = it rains
q = playing cricket/ match played
If (it rains) then (the match will not be played)
p ⇒ (∼q)
Inference: There was no rain. (i.e., p = F)
So for any F ⇒ (∼q) is true.
So this inference is valid.
I2: If it rains then the cricket match will not be played.
It did not rain.
p ⇒ (∼q)
Inference: The cricket match was played.
q = T
p ⇒ (∼q)
p ⇒ (∼T)
p ⇒ F
This is false for p = T, so this is not true.
Question 137

Consider the function f(x) = sin(x) in the interval x ∈ [π/4, 7π/4]. The number and location(s) of the local minima of this function are

A
One, at π/2
B
One, at 3π/2
C
Two, at π/2 and 3π/2
D
Two, at π/4 and 3π/2
       Engineering-Mathematics       Calculus       GATE 2012
Question 137 Explanation: 
f(x) = sin x
f’(x) = cos x

[just consider the given interval (π/4, 7π/4)]
f'(x) = 0 at π/2, 3π/2
To get local minima f ’’(x) > 0, f ’’(x) = - sin x
f ’’(x) at π/2, 3π/2
f ’’(x) = -1< 0 local maxima
f ’’ (3π/2) = 1 > 0 this is local minima
In the interval [π/4, π/2] the f(x) is increasing, so f(x) at π/4 is also a local minima.
So there are two local minima for f(x) at π/4, 3π/2.
Question 138

Let A be the 2×2 matrix with elements a11 = a12 = a21 = +1 and a22 = -1. Then the eigenvalues of the matrix A19 are

A
1024 and -1024
B
1024√2 and -1024√2
C
4√2 and -4√2
D
512√2 and -512√2
       Engineering-Mathematics       Linear-Algebra       GATE 2012
Question 138 Explanation: 
a11 = a12 = a21 = 1, a22 = -1
The 2×2 matrix =
Cayley Hamilton theorem:
If matrix A has ‘λ’ as eigen value, An has eigen value as λn.
Eigen value of
|A-λI| = 0

-(1-λ)(1+λ)-1 = 0
-(1-λ2 )-1 = 0
-1 = 1-λ2
λ2 = 2
λ = ±√2
A19 has (√2)19 = 29×√2 (or) (-√2)19 = -512√2
= 512√2
Question 139

What is the correct translation of the following statement into mathematical logic?

“Some real numbers are rational”

A
∃x (real(x) ∨ rational(x))
B
∀x (real(x) → rational(x))
C
∃x (real(x) ∧ rational(x))
D
∃x (rational(x) → real(x))
       Engineering-Mathematics       Propositional-Logic       GATE 2012
Question 139 Explanation: 

∃x (real(x) ∧ rational(x))
(A) ∃x(real(x) ∨ rational(x))
means There exists some number, which are either real or rational.
(B) ∀x (real(x)→rational(x))
If a number is real then it is rational.
(D) ∃x (rational(x)→real(x))
There exists a number such that if it is rational then it is real.
Question 140

Let G be a simple undirected planar graph on 10 vertices with 15 edges. If G is a connected graph, then the number of bounded faces in any embedding of G on the plane is equal to

A
3
B
4
C
5
D
6
       Engineering-Mathematics       Graph-Theory       GATE 2012
Question 140 Explanation: 
If the graph is planar, then we have to consider Euler’s formula
v-e+f = 2
Given 10 vertices & 15 edges
10-15+f = 2
f = 2+15-10
f = 7
There will be an unbounded face always. So, number of faces = 6.
Question 141

Which of the following graphs is isomorphic to

A
B
C
D
       Engineering-Mathematics       Graph-Theory       GATE 2012
Question 141 Explanation: 
Original graph:

(A) 3 cycle graph not in original one.

(B) Correct 5 cycles & max degree is 4.
(C) Original graph doesn’t have a degree of 3.

(D) 4 cycles not in original one.
Question 142

The bisection method is applied to compute a zero of the function f(x) = x4 - x3 - x2 - 4 in the interval [1,9]. The method converges to a solution after ________ iterations.

A
1
B
3
C
5
D
7
       Engineering-Mathematics       Numerical-Methods       GATE 2012
Question 142 Explanation: 
Note: Numerical methods are not present in the GATE CS syllabus anymore.
Question 143

Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?

A
10/21
B
5/12
C
2/3
D
1/6
       Engineering-Mathematics       Probability       GATE 2012
Question 143 Explanation: 
The value on the die for first time rolling = {1, 2, 3}
The value on second time can be {1, 2, 3, 4, 5, 6}
So the Sum can be

We have Sample space = 36
The no. of events where (Sum = atleast 6) = {6, 7, 6, 7, 8, 6, 7, 8, 9}
So the probability atleast ‘6’ while getting {1, 2, 3} in first time = 9/36 → ①
If we get ‘6’ in the first time itself, then we do not go for rolling die again.
So, its probability = 1/6
Total probability = 1/6 + 9/36 = 1/6 + 1/4 = 10/24 = 5/12
Question 144

How many onto (or surjective) functions are there from an n-element (n ≥ 2) set to a 2-element set?

A
2n
B
2n-1
C
2n-2
D
2(2n– 2)
       Engineering-Mathematics       Functions       GATE 2012
Question 144 Explanation: 

Onto function is possible if m ≥ n. So, no. of onto functions possible is,
nm - nC1 (n-1)m + nC2 (n-2)m + .......
Here in Question,
m = n, n = 2
So, the final answer will be,
= 2n - 2C1 (2-1)n + 2C2 (2-2)n
= 2n - 2 × 1 + 0
= 2n - 2
Question 145

Let G be a complete undirected graph on 6 vertices. If vertices of G are labeled, then the number of distinct cycles of length 4 in G is equal to

A
15
B
30
C
45
D
360
       Engineering-Mathematics       Graph-Theory       GATE 2012
Question 145 Explanation: 
Complete graph means there exists an edge between every pair of vertices.
It is asked to find the distinct cycle of length 4. As it is complete graph, if we chose any two vertices, there will be an edge.
So, to get a cycle of length 4 (means selecting the 4 edges which can form a cycle) we can select any four vertices.
The number of such selection of 4 vertices from 6 vertices is 6C4 => 15.
From each set of 4 vertices, suppose a set {a, b, c, d} we can have cycles like
a-b-c-d
a-b-d-c
a-c-b-d
a-c-d-b
a-d-b-c
a-d-c-b (Total 6, which is equal to number of cyclic permutations (n-1)! )
As they are labelled you can observe, a-b-c-d and a-d-c-b are same, in different directions.
So, we get only three combinations from the above 6.
So, total number of distinct cycles of length 4 will be 15*3 = 45.
If it is asked about just number of cycles then 15*6 = 90
Question 146

If the difference between the expectation of the square of a random variable (E[X2]) and the square of the expectation of the random variable (E[X])2 is denoted by R, then

A
R = 0
B
R < 0
C
R ≥ 0
D
R > 0
       Engineering-Mathematics       Probability       GATE 2011
Question 146 Explanation: 
We know that difference of E(X2) and E(X))2 is nothing but variance or V(X) which is always greater than or equal to zero.
So the answer will be R≥0.
Question 147

Consider the matrix as given below.

Which one of the following provides the CORRECT values of eigenvalues of the matrix?

A
1, 4, 3
B
3, 7, 3
C
7, 3, 2
D
1, 2, 3
       Engineering-Mathematics       Linear-Algebra       GATE 2011
Question 147 Explanation: 
Given matrix is upper triangular matrix and its diagonal elements are its eigen values = 1, 4, 3
Question 148

Which one of the following options is CORRECT given three positive integers x,y and z, and a predicate

P(x) = ¬(x=1)∧∀y(∃z(x=y*z) ⇒ (y=x)∨(y=1))
A
P(x) being true means that x is a prime number
B
P(x) being true means that x is a number other than 1
C
P(x) is always true irrespective of the value of x
D
P(x) being true means that x has exactly two factors other than 1 and x
       Engineering-Mathematics       Propositional-Logic       GATE 2011
Question 148 Explanation: 
Statement: x is not equal to 1 and if there exists some z for all y such that product of y and z is x, then y is either the number itself or 1.
This is the definition of prime numbers.
Question 149

Given i=√-1, what will be the evaluation of the definite integral

A
0
B
2
C
-i
D
i
       Engineering-Mathematics       Calculus       GATE 2011
Question 149 Explanation: 
We know that,
Question 150

Consider a finite sequence of random values X = [x1, x2, …, xn]. Let μx be the mean and σx be the standard deviation of X. Let another finite sequence Y of equal length be derived from this as yi = a * xi + b, where a and b are positive constants. Let μy be the mean and σy be the standard deviation of this sequence. Which one of the following statements is INCORRECT?

A
Index position of mode of X in X is the same as the index position of mode of Y in Y.
B
Index position of median of X in X is the same as the index position of median of Y in Y.
C
μy = aμx + b
D
σy = aσx + b
       Engineering-Mathematics       Probability-and-Statistics       GATE 2011
Question 150 Explanation: 
σy is standard deviation then
y)2 is variance so,
yi = a * xi + b
y)2 = a2 x)2
⇒ σy = a σx
Hence option (D) is incorrect.
Question 151

A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second.

A
1/5
B
4/25
C
1/4
D
2/5
       Engineering-Mathematics       Probability       GATE 2011
Question 151 Explanation: 
The possible events are
(2,1) (3,2) (4,3) (5,4).
So only 4 possibilities are there and sample space will be,
5C1 × 4C1 = 20
So probability = 4/20 = 1/5
Question 152

Let G = (V,E) be a graph. Define ξ(G) = Σd id x d, where id is the number of vertices of degree d in G. If S and T are two different trees with ξ(S) = ξ(T),then

A
|S| = 2|T|
B
|S| = |T| - 1
C
|S| = |T|
D
|S| = |T| + 1
       Engineering-Mathematics       Graph-Theory       GATE 2010
Question 152 Explanation: 

id= no. of vertices of degree ‘d’ in ‘G’
Eg:

No. of vertices with degree ‘2’ = 3
ξ(G') = 3 × 2 = '6' i.e., sum of degrees
By Handshaking Theorem,
The sum of degrees would be equal to twice the no. of edges
|V| = 2|E|
It is given that ξ(G) = ξ(S) then
Sum of degrees of vertices in G is equal to sum of degrees of vertices in S
i.e., 2*(no. of edges in G) = 2*no. of edges in S no. of edges in G = no. of edges in S
Eg:

ξ(G) = (2 × 2) + (2 × 3) = 4 + 6 = 10

ξ(S) = 2 × 5 = 10
You can observe that, though no. of vertices are different, but still no. of edges are same.
Question 153

Newton-Raphson method is used to compute a root of the equation x2 - 13 = 0 with 3.5 as the initial value. The approximation after one iteration is

A
3.575
B
3.676
C
3.667
D
3.607
       Engineering-Mathematics       Newton-Raphson-Method       GATE 2010
Question 153 Explanation: 
Note: Out of syllabus.
Question 154

What is the possible number of reflexive relations on a set of 5 elements?

A
210
B
215
C
220
D
225
       Engineering-Mathematics       Sets-And Relation       GATE 2010
Question 154 Explanation: 
Let set = ‘A’ with ‘n’ elements,
Definition of Reflexive relation:
A relation ‘R’ is reflexive if it contains xRx ∀ x∈A
A relation with all diagonal elements, it can contain any combination of non-diagonal elements.
Eg:
A={1, 2, 3}


So for a relation to be reflexive, it should contain all diagonal elements. In addition to them, we can have possible combination of (n2-n)non-diagonal elements (i.e., 2n2-n)
Ex:
{(1,1)(2,2)(3,3)} ----- ‘0’ non-diagonal element
{(1,1)(2,2)(3,3)(1,2)} ----- ‘1’ non-diagonal element
{(1,1)(2,2)(3,3)(1,2)(1,3)} “
___________ “
___________ “
{(1,1)(2,2)(3,3)(1,2)(1,3)(2,1)(2,3)(3,1)(3,2)} (n2-n) diagonal elements
____________________
Total: 2n2-n
For the given question n = 5.
The number of reflexive relations = 2(25-5) = 220
Question 155

Consider the set S = {1, ω, ω2}, where ω and ω2 are cube roots of unity. If * denotes the multiplication operation, the structure (S,*) forms

A
A group
B
A ring
C
An integral domain
D
A field
       Engineering-Mathematics       Sets-And Relation       GATE 2010
Question 155 Explanation: 
A Group is an algebraic structure which satisfies
1) Closure
2) Associativity
3) Have Identity element
4) Invertible
Over ‘*’ operation the S = {1, ω, ω2} satisfies the above properties.
The identity element is ‘1’ and inverse of 1 is 1, inverse of ‘w’ is 'w2' and inverse of 'w2' is 'w'.
Question 156

What is the value of

A
0
B
e-2
C
e-1/2
D
1
       Engineering-Mathematics       Calculus       GATE 2010
Question 156 Explanation: 
Question 157

Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty?

A
pq + (1 - p)(1 - q)
B
(1 - q)p
C
(1 - p)q
D
pq
       Engineering-Mathematics       Probability       GATE 2010
Question 157 Explanation: 
The probability of the computer being declared faulty is,
= Probability of testing process gives the correct result × Probability that computer is faulty + Probability of testing process giving incorrect result × Probability that computer is not faulty
= p × q + (1 - p) (1 - q)
Question 158

What is the probability that divisor of 1099 is a multiple of 1096?

A
1/625
B
4/625
C
12/625
D
16/625
       Engineering-Mathematics       Probability       GATE 2010
Question 158 Explanation: 
Probability that divisor of 1099 is a multiple of 1096
We can write 1099 as 1096×103
So, (1099)/(1096) to be a whole number, [1096×103/1096] ➝ (1)
We can observe that every divisor of 103 is a multiple of 1096
So number of divisor of 103 to be found first
⇒ 103 = (5×2)3 = 23×53
No. of divisors = (3 + 1) (3 + 1) = 16
Total number of divisor of 1099 are 1099 = 299×599 = 100×100 = 10000
Probability that divisor of 1099 is a multiple of 1096 is
⇒ 16/10,000
Question 159

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 can not 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
       Engineering-Mathematics       Graph-Theory       GATE 2010
Question 159 Explanation: 
Havel Hakimi theorem:
⇾ Arrange the degree of vertices in descending order
eg. d1, d2, d3... dn
⇾ Discard d1, subtrack ‘1’ from the next 'd1' degrees
eg:
⇒ 1 1 0 1
⇾ We should not get any negative value if its negative, this is not valid sequence
⇾ Repeat it till we get ‘0’ sequence
I. 7, 6, 5, 4, 4, 3, 2, 1
➡️5, 4, 3, 3, 2, 1, 0
➡️3, 2, 2, 1, 0, 0
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
II. 6, 6, 6, 6, 3, 3, 2, 2
➡️5, 5, 5, 2, 2, 1, 2
put them in descending order
➡️5, 5, 5, 2, 2, 2, 1
➡️4, 4, 1, 1, 1, 1
➡️3, 0, 0, 0, 1 (descending order)
➡️3, 1, 0, 0, 0
➡️0, -1, -1, 0
[This is not valid]
III. 7, 6, 6, 4, 4, 3, 2, 2
➡️5, 5, 3, 3, 2, 1, 1
➡️4, 2, 2, 1, 0, 1
➡️4, 2, 2, 1, 1, 0 (descending order)
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
IV. 8, 7, 7, 6, 4, 2, 1, 1
There is a degree ‘8’, but there are only ‘8’ vertices.
A vertex cannot have edge to itself in a simple graph. This is not valid sequence.
Question 160

Consider the following matrix . If the eigenvalues 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
       Engineering-Mathematics       Linear-Algebra       GATE 2010
Question 160 Explanation: 

Trace = {Sum of diagonal elements of matrix}

Here given that eigen values are 4, 8
Sum = 8 + 4 = 12
Trace = 2 + y
⇒ 2 + y = 12
y = 10

Determinant = |2y - 3x|
Product of eigen values = 8 × 4 = 32
2y - 3x = 32
(y = 10)
20 - 3x = 32
-12 = 3x
x = -4
∴ x = -4, y = 10
Question 161

Suppose the predicate F(x, y, t) is used to represent the statement that person x can fool person y at time t. Which one of the statements below expresses best the meaning of the formula ∀x∃y∃t(¬F(x,y,t))?

A
Everyone can fool some person at some time
B
No one can fool everyone all the time
C
Everyone cannot fool some person all the time
D
No one can fool some person at some time
       Engineering-Mathematics       Propositional-Logic       GATE 2010
Question 161 Explanation: 
F(x,y,t) ⇒ Person 'x' can fool person 'y' at time 't'.
For better understanding propagate negation sign outward by applying Demorgan's law.
∀x∃y∃t(¬F(x, y, t)) ≡ ¬∃x∀y∀t(F(x,y,t))
Now converting ¬∃x∀y∀t(F(x,y,t)) to English is simple.
¬∃x∀y∀t(F(x,y,t)) ⇒ There does not exist a person who can fool everyone all the time.
Which means "No one can fool everyone all the time".
Hence, Option (B) is correct.
Question 162

Which one of the following in NOT necessarily a property of a Group?

A
Commutativity
B
Associativity
C
Existence of inverse for every element
D
Existence of identity
       Engineering-Mathematics       Sets-And Relation       GATE 2009
Question 162 Explanation: 
A Group should satisfy Closure, Associative, should have identity element and each element has inverse.
So, commutativity is not required.
Question 163

What is the chromatic number of an n-vertex simple connected graph which does not contain any odd length cycle? Assume n ≥ 2.

A
2
B
3
C
n-1
D
n
       Engineering-Mathematics       Graph-Theory       GATE 2009
Question 163 Explanation: 
If n≥ 2 and there are no odd length cycles, then it is a bipartite graph. A bipartite graph has the chromatic number 2.
Eg: Consider a square, which has 4 edges. It can be represented as bipartite ,with chromatic number 2.
Question 164

Which one of the following is TRUE for any simple connected undirected graph with more than 2 vertices?

A
No two vertices have the same degree.
B
At least two vertices have the same degree.
C
At least three vertices have the same degree.
D
All vertices have the same degree.
       Engineering-Mathematics       Graph-Theory       GATE 2009
Question 164 Explanation: 
Method 1:
If all vertices have different degrees, then the degree sequence will be {1,2,3,....n-1}, it will not have ‘n’( A simple graph will not have edge to itself, so it can have edges with all other (n-1) vertices). Degree sequence has only (n-1) numbers, but we have ‘n’ vertices. So, by Pigeonhole principle there are two vertices which has same degree.
Method 2:
A) Consider a triangle, all vertices has same degree, so it is false
C) Consider a square with one diagonal, there are less than three vertices with same degree, so it is false
D) Consider a square with one diagonal, vertices have different degrees. So, it is false.
We can conclude that option B is correct.
Question 165

Consider the binary relation R = {(x, y), (x, z), (z, x), (z, y)} on the set {x, y, z}. Which one of the following is TRUE?

A
R is symmetric but NOT antisymmetric
B
R is NOT symmetric but antisymmetric
C
R is both symmetric and antisymmetric
D
R is neither symmetric nor antisymmetric
       Engineering-Mathematics       Sets-And Relation       GATE 2009
Question 165 Explanation: 
Symmetric Relation: A relation R on a set A is called symmetric if (b,a) € R holds when (a,b) € R.
Antisymmetric Relation: A relation R on a set A is called antisymmetric if (a,b)€ R and (b,a) € R then a = b is called antisymmetric.
In the given relation R, for (x,y) there is no (y,x). So, this is not Symmetric. (x,z) is in R also (z,x) is in R, but as per antisymmetric relation, x should be equal to z, where this fails.
So, R is neither Symmetric nor Antisymmetric.
Question 166

An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the

If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closest to the probability that the face value exceeds 3?

A
0.453
B
0.468
C
0.485
D
0.492
       Engineering-Mathematics       Probability       GATE 2009
Question 166 Explanation: 
P(0) = Probability of getting odd no. face.
P(e) = Probability of getting even no. face.
It is given that,
P(0) = 0.9 P(e) ----- (I)
Also we know that,
P(0) + P(e) = 1 ----- (II)
Solving equation (I) and (II) we get,
P(e) = 0.52
Also even no. can be 2 or 4 or 6.
And given in question that P(2) = P(4) = P(6).
So, 3 × P(2) = 0.52
P(2) = 0.175
So, P(2) = P(4) = P(6) = 0.175
Also in question it is given that,
P(e/>3) = 0.75
P(even no. greater than 3)/ P(no. greater than 3) = 0.75
P(4,6)/P(>3) = 0.75
(0.175×2)/P(>3) = 0.75
P(>3) = 0.35/0.75 = 0.467
Question 167

For the composition table of a cyclic group shown below

Which one of the following choices is correct?

A
a, b are generators
B
b, c are generators
C
c, d are generators
D
d, a are generators
       Engineering-Mathematics       Sets-And-Relation       GATE 2009
Question 167 Explanation: 
Each element of set can be written as a power of g in multiplicative notation, or as a multiple of g in additive notation. This element g is called a generator of the group.
We can observe that, a is an identity element. ( a *x = x ). An identity element cannot be a generator, as it cannot produce any other element ( always a*a*... = a).
Also, b*b =a, so it also cannot produce all other elements ( always b*b*... =a , where a is identify element).
c,d are able to produce other elements like { c*c =b, c*(c*c) = c*b= d, c*(c*(c*c))) = c*(c*b)= c*d=a. }. Similar with d.
Question 168

Which one of the following is the most appropriate logical formula to represent the statement?

      “Gold and silver ornaments are precious”.

The following notations are used:

  
       G(x): x is a gold ornament
       S(x): x is a silver ornament
       P(x): x is precious
A
∀x(P(x) → (G(x) ∧ S(x)))
B
∀x((G(x) ∧ S(x)) → P(x))
C
∃x((G(x) ∧ S(x)) → P(x)
D
∀x((G(x) ∨ S(x)) → P(x))
       Engineering-Mathematics       Propositional-Logic       GATE 2009
Question 168 Explanation: 
Interpreting the options will lead to
(A) for all ornaments, if it is precious then they should be gold and silver.
But, given statement does not says that, “ only gold and silver are precious “ . So this is wrong.
(B) For all ornaments, which contains gold and silver are precious.

Which is only the shaded region in the venn diagrams. But, it misses p,r regions. So, this is wrong option.
C) Some ornaments, which are gold and silver are precious. It is false, because all gold or silver ornaments are precious.
D) For all ornaments, Any ornament which is gold or silver is precious. Which is true.
Question 169

The binary operation □ is defined as follows

Which one of the following is equivalent to P ∨ Q ?

A
¬Q□¬P
B
P□¬Q
C
¬P□Q
D
¬P□¬Q
       Engineering-Mathematics       Propositional-Logic       GATE 2009
Question 169 Explanation: 
The options are simple to draw the truth table then go with the corresponding options.

P∨Q = P□️Q
So, option B is correct.
Question 170
is equivalent to
A
0
B
1
C
ln 2
D
1/2 ln 2
       Engineering-Mathematics       Calculus       GATE 2009
Question 170 Explanation: 
Question 171

Consider the following well-formed formulae:

    I. ¬∀x(P(x))
    II. ¬∃(P(x))
    III. ¬∃(¬P(x))
    IV. ∃x(¬(P(x))

Which of the above are equivalent?

A
I and III
B
I and IV
C
II and III
D
II and IV
       Engineering-Mathematics       Propositional-Logic       GATE 2009
Question 171 Explanation: 
I) ¬∀x(P(x)) = ∃x(¬P(x)) [Demorgan's Law]
II ) ¬∃x(P(x))= ∀x(~P(x))
III) ¬∃x(¬P(x)) = ∀x(P(x))
Question 172
equals
A
1
B
-1
C
D
-∞
       Engineering-Mathematics       Calculus       GATE 2008
Question 172 Explanation: 
Question 173

If P, Q, R are subsets of the universal set U, then (P∩Q∩R) ∪ (Pc∩Q∩R) ∪ Q∪ Rc is

A
Qc ∪ Rc
B
P ∪ Qc ∪ Rc
C
Pc ∪ Qc ∪ Rc
D
U
       Engineering-Mathematics       Sets-And-Relation       GATE 2008
Question 173 Explanation: 
Given,
(P∩Q∩R)∪(Pc∩Q∩R)∪Qc∪Rc
It can be written as the p.q.r + p'.q.r +q' + r'
=> (p+p').q.r + q' + r'
=> q.r + (q'+r')
=> q.r + q' + r' = 1 i.e., U
Question 174

The following system of equations

    x1 + x2 + 2x3  = 1
    x1 + 2x2 + 3x3 = 2
    x1 + 4x2 + ax3 = 4

has a unique solution. The only possible value(s) for a is/are

A
0
B
either 0 or 1
C
one of 0, 1 or -1
D
any real number
       Engineering-Mathematics       Linear-Algebra       GATE 2008
Question 174 Explanation: 
The conjugate matrix [A|B] is

When a-5 = 0, then rank(A) = rank[A|B]<3,
So infinite number of solutions.
But, it is given that the given system has unique solution i.e., rank(A) = rank[A|B] = 3 will be retain only if a-5 ≠ 0.
Question 175

The minimum number of equal length subintervals needed to approximate to an accuracy of atleast using the trapezoidal rule is

A
1000e
B
1000
C
100e
D
100
       Engineering-Mathematics       Trapezidal Rule       GATE 2008
Question 175 Explanation: 
Note: Out of syllabus.
Question 176

The Newton-Raphson iteration can be used to compute the

A
square of R
B
reciprocal of R
C
square root of R
D
logarithm of R
       Engineering-Mathematics       Newton-Raphson-Method       GATE 2008
Question 176 Explanation: 
Note: Out of syllabus.
Question 177

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
       Engineering-Mathematics       Graph-Theory       GATE 2008
Question 177 Explanation: 
Lets do with elimination method,
(A) Consider the following disconnected graph which is planar.

So false.
(B) A graph is Eulerian if all vertices have even degree but a planar graph can have vertices with odd degree.
So false.
(D) Consider K4 graph. It has independent set size 1 which is less than 4/3.

So false.
Hence, option (C) is correct.
Question 178

Let and , where k is a positive integer. Then

A
P = Q - k
B
P = Q + k
C
P = Q
D
P = Q + 2 k
       Engineering-Mathematics       Permutation-and-Combination       GATE 2008
Question 178 Explanation: 

P = 1+3+5+7+...+(2k-1)
= (2-1)+(4-1)+(6-1)+(8-1)+...+(2k-1)
= (2+4+6+8+...+2k)+(-1+-1+-1+k times)
= Q-(1+1+...+k times)
= Q-k
Question 179

A point on a curve is said to be an extremum if it is a local minimum or a local maximum. The number of distinct extrema for the curve 3x4 - 16x3 + 24x2 + 37 is

A
0
B
1
C
2
D
3
       Engineering-Mathematics       Calculus       GATE 2008
Question 179 Explanation: 
f(x) = 3x4 - 16x3 + 24x2 + 37
f’(x) = 12x3 + 48x2 + 48x = 0
12x(x2 - 4x + 4) = 0
x=0; (x-2)2 = 0
x=2
f’’(x) = 36x2 - 96x + 48
f ”(0) = 48
f ”(2) = 36(4) - 96(2) + 48
= 144 - 192 + 48
= 0
At x=2, we can’t apply the second derivative test.
f’(1) = 12; f’(3) = 36, on either side of 2 there is no sign change then this is neither minimum or maximum.
Finally, we have only one Extremum i.e., x=0.
Question 180

Aishwarya studies either computer science or mathematics everyday. If she studies computer science on a day, then the probability that she studies mathematics the next day is 0.6. If she studies mathematics on a day, then the probability that she studies computer science the next day is 0.4. Given that Aishwarya studies computer science on Monday, what is the probability that she studies computer science on Wednesday?

A
0.24
B
0.36
C
0.4
D
0.6
       Engineering-Mathematics       Probability       GATE 2008
Question 180 Explanation: 
→ Aishwarya studied CS on Monday. Then we have two possibilities to she study computer science on wednesday.
(i) She study Mathematics on Tuesday and computer science on wednesday.
⇒ 0.6×0.4
⇒ 0.24
(ii) She study computer science on Tuesday and computer science on wednesday.
⇒ 0.4×0.4
⇒ 0.16
→ The probability that she study computer science on wednesday is
0.24+0.16 = 0.40
Question 181

How many of the following matrices have an eigenvalue 1?

 
A
one
B
two
C
three
D
four
       Engineering-Mathematics       Linear-Algebra       GATE 2008
Question 181 Explanation: 




Answer: We have only one matrix with eigen value 1.
Question 182

Let X be a random variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean -1 and variance unknown. If P(X ≤ -1) = P(Y ≥ 2), the standard deviation of Y is

A
3
B
2
C
√2
D
1
       Engineering-Mathematics       Probability       GATE 2008
Question 182 Explanation: 
P(X ≤ -1) = P(Y ≥ 2)
We can compare their values using standard normal distributions.

The above equation satisfies when σy will be equal to 3.
Question 183

Let fsa and pda be two predicates such that fsa(x) means x is a finite state automaton, and pda(y) means that y is a pushdown automaton. Let equivalent be another predicate such that equivalent (a, b) means a and b are equivalent. Which of the following first order logic statements represents the following:

Each finite state automaton has an equivalent pushdown automaton

A
(∀x fsa(x)) ⇒ (∃y pda(y) ∧ equivalent(x,y))
B
∼∀y(∃x fsa(x) ⇒ pda(y) ∧ equivalent(x,y))
C
∀x ∃y(fsa(x) ∧ pda(y) ∧ equivalent(x,y))
D
∀x ∃y(fsa(y)∧ pda(x) ∧ equivalent(x,y))
       Engineering-Mathematics       Propositional-Logic       GATE 2008
Question 183 Explanation: 
Go through the options.
Option A:
If everything is a FSA. Then there exists an equivalent PDA for everything.
Option B:
Not for the case Y, if there exists a FSA then it can have equivalent PDA.
Option C:
Everything is a PDA and consists equivalent PDA.
Option D:
Everything is a PDA and has exist an equivalent FSA. In option A we are getting the equivalent of a and b.
So answer is option A.
Question 184

P and Q are two propositions. Which of the following logical expressions are equivalent?

    I. P∨∼Q
    II. ∼(∼P∧Q)
    III. (P∧Q)∨(P∧∼Q)∨(∼P∧∼Q)
    IV. (P∧Q)∨(P∧∼Q)∨(∼P∧Q)
A
Only I and II
B
Only I, II and III
C
Only I, II and IV
D
All of I, II, III and IV
       Engineering-Mathematics       Propositional-Logic       GATE 2008
Question 184 Explanation: 
I. P∨∼Q (✔️)
II. ∼(∼P∧Q)⇒(P∨∼Q)≡I (✔️)
III. (P×Q)∨(P×∼Q)∨(∼P×∼Q)
P∧(Q∨∼Q)∨(∼P∧∼Q)
P∨(∼P×∼Q)
(P∨∼P)×(P∨∼Q)
(P∨∼Q)≡I=II (✔️)
IV. (P×Q)∨(P∧∼Q)∨(∼P×Q)
P×(Q∨∼Q)∨(∼P∧Q)
P∨(∼P×Q)
(P∨∼P)×(P∨Q)
(P∨Q)≠I (❌)
So I≡II≡III (✔️)
Question 185

Consider the following two statements about the function f(x)=|x|

P. f(x) is continuous for all real values of x
Q. f(x) is differentiable for all real values of x 

Which of the following is TRUE?

A
P is true and Q is false.
B
P is false and Q is true.
C
Both P and Q are true.
D
Both P and Q are false.
       Engineering-Mathematics       Calculus       GATE 2007
Question 185 Explanation: 
f(x) = |x|
→ f(x) is continuous for all real values of x

For every value of x, there is corresponding value of f(x).
For x is positive, f(x) is also positive
x is negative, f(x) is positive.
So, f(x) is continuous for all real values of x.
→ f(x) is not differentiable for all real values of x. For x<0, derivative is negative
x>0, derivative is positive.
Here, left derivative and right derivatives are not equal.
Question 186

Let S be a set of n elements. The number of ordered pairs in the largest and the smallest equivalence relations on S are:

A
n and n
B
n2 and n
C
n2 and 0
D
n and 1
       Engineering-Mathematics       Sets-And-Relation       GATE 2007
Question 186 Explanation: 
A relation is said to be equivalent relation if it satisfies,
→ Reflexive
→ Symmetric
→ Transitive
Let a set S be,
S = {1, 2, 3}
Now, the smallest relation which is equivalence relation is,
S×S = {(1,1), (2,2), (3,3)}
= 3
= n (for set of n elements)
And, the largest relation which is equivalence relation is,
S×S = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}
= 9
= 32
= n2 (for set of n elements)
Question 187

Let G be the non-planar graph with the minimum possible number of edges. Then G has

A
9 edges and 5 vertices
B
9 edges and 6 vertices
C
10 edges and 5 vertices
D
10 edges and 6 vertices
       Engineering-Mathematics       Graph-Theory       GATE 2007
Question 187 Explanation: 
Using Euler’s formula we know that,
if n ≥ 3 then e ≤ 3n-6 (for planarity)
where n = no. of vertices
e = no. of edges
Now lets check the options.
A) e=9, n=5
9 ≤ 3(5) - 6
9 ≤ 15 - 6
9 ≤ 9
Yes, it is planar.
B) e=9, n=6
9 ≤ 3(6) - 6
9 ≤ 18 - 6 9 ≤ 12 Yes, it is planar.
iii) e=10, n=5
10 ≤ 3(5) - 6
10 ≤ 15 - 6
10 ≤ 9 No, it is not planar.
So, option C is non-planar graph.
iv) e=10, n=6
10 ≤ 3(6) - 6
10 ≤ 18 - 6
10 ≤ 12
Yes, it is planar.
Question 188

How many different non-isomorphic Abelian groups of order 4 are there?

A
2
B
3
C
4
D
5
       Engineering-Mathematics       Sets-And-Relation       GATE 2007
Question 188 Explanation: 
In this problem first of all we find the exponents of prime nos. that can be taken out from given number.
4 = 22
So, prime no. is 2 and power of 2 is 2. So exponent value 2 is considered now.
Now the no. of ways we can divide 2 into sets will be the answer.
So division can be done as,
{1,1}, {0,2}
in two ways. Hence, answer is 2.
Question 189

Let Graph(x) be a predicate which denotes that x is a graph. Let Connected(x) be a predicate which denotes that x is connected. Which of the following first order logic sentences DOES NOT represent the statement: “Not every graph is connected”?

A
¬∀x (Graph (x) ⇒ Connected (x))
B
¬∃x (Graph (x) ∧ ¬Connected (x))
C
¬∀x (¬Graph (x) ∨ Connected (x))
D
∀x (Graph (x) ⇒ ¬Connected (x))
       Engineering-Mathematics       Propositional-Logic       GATE 2007
Question 189 Explanation: 
Option (A) and (C) are same, because in option (C) the given expression is
Given expression is
¬∀x(¬Graph(x) ∨ Connected(x)
which can be rewritten as,
¬∀x(Graph(x) ⇒ Connected(x)
which is equivalent to option (A)
(∵ ¬p∨q ≡ p→q)
So, option (A) and (C) cannot be the answer.
Coming to option (B), the given expression is,
∃x (Graph (x) ∧ ¬Connected (x))
"There exist some graph which is not connected", which is equivalent in saying that "Not every graph is connected".
Coming to option (D),
For all x graph is not connected, which is not correct.
Hence, option (D) is the answer.
Question 190

Which of the following graphs has an Eulerian circuit?

A
Any k-regular graph where k is an even number.
B
A complete graph on 90 vertices.
C
The complement of a cycle on 25 vertices.
D
None of the above.
       Engineering-Mathematics       Graph-Theory       GATE 2007
Question 190 Explanation: 
Two necessary condition for the existence of Eulerian circuits is
→ all vertices in the graph have an "even degree".
→ And the graph must be corrected.
Now in option (C) it is saying that the complement of a cycle on 25 vertices without complement the degree of each vertex is 2.
Now since there are 25 vertices, so maximum degree of each vertex will be 24 and so in complement of cycle each vertex degree will be 24 - 2 = 22.
There is a theorem which says "G be a graph with n vertices and if every vertex has a degree of atleast n-1/2 then G is connected."
So we can say that complement of cycle with 25 vertices fulfills both the conditions, and hence is Eulerian circuit.
Question 191

Suppose we uniformly and randomly select a permutation from the 20! Permutations of 1, 2, 3,….., 20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?

A
1/2
B
1/10
C
9!/20!
D
None of these
       Engineering-Mathematics       Probability       GATE 2007
Question 191 Explanation: 
1, 2, 3, 4, …….20
→ Total no. of possible even number = 10
→ Here we are not considering odd number.
→ The probability that 2 appears at an earlier position than any other even number is =1/10
Question 192

Let A be a 4 x 4 matrix with eigenvalues -5, -2, 1, 4. Which of the following is an eigenvalue of

[A  I]
[I  A]

where I is the 4 x 4 identity matrix?

A
-5
B
-7
C
2
D
1
       Engineering-Mathematics       Linear-Algebra       GATE 2007
Question 192 Explanation: 
Eigen value of A[4×4] matrix is -5, -2, 1, 4.

|(A-λI)2-I| = 0 [a2-b2 = (a+b)(a-b)]
|(A-λI+I)(A-λI-I) = 0
|(A-(λ-I)I)(A-(λ+I)I| = 0
Let us assume:
λ-1=k & λ +1=k
λ =k+1 λ =k-1
⇓ ⇓
for k=-5; λ=-4 λ =-6
k=-2; λ=-1 λ =-3
k=1; λ=2 λ = 0
k=4; λ=5 λ = 3
So, λ = -4,-1,2,5,-6,-3,0,3
Check with the option
Option C = 2
Question 193

Consider the set S = {a,b,c,d}. Consider the following 4 partitions π1, π2, π3, π4 on Let p be the partial order on the set of partitions S' = {π1, π2, π3, π4} defined as follows: πi p πj if and only if πi refines πj. The poset diagram for (S', p) is:

A
B
C
D
       Engineering-Mathematics       Sets-And Relation       GATE 2007
Question 193 Explanation: 
π4 = refines every partition. So it has to be bottom of poset diagram.
And, neither π2 refines π3, nor π3 refines π2.
Here, only π1 refined by every set, so it has to be at the top.
Finally, option C satisfies all the property.
Question 194

Consider the set of (column) vectors defined by X = {x ∈ R3| x1+x2+x3 = 0, where xT = [x1,x2,x3]T}. Which of the following is TRUE?

A
{[1,-1,0]T, [1,0,-1]T} is a basis for the subspace X.
B
{[1,-1,0]T, [1,0,-1]T} is a linearly independent set, but it does not span X and therefore is not a basis of X.
C
X is not a subspace of R3
D
None of the above
       Engineering-Mathematics       Linear-Algebra       GATE 2007
Question 194 Explanation: 
Since, [1,-1,0]T and [1,0,-1]T are linearly independent set and spans X. So is the basis for subspace X.
Question 195

Consider the series obtained from the Newton-Raphson method. The series converges to

A
1.5
B
√2
C
1.6
D
1.4
       Engineering-Mathematics       Newton-Raphson-Method       GATE 2007
Question 195 Explanation: 
Given series is
xn+1 = xn/2+9/8xn ⟶ (I); x0 = 0.5
Equation based on Newton-Raphson is
xn+1 = xn-f(xn)/f'(xn) ⟶ (II)
Equate I and II
xn-f(xn)/f'(xn) = xn/2+9/8xn
xn-f(xn)/f'(xn) = xn-xn/2+9/8xn
xn-f(xn)/f'(xn) = xn-(4xn2-9)/8xn
So, f(x) = 4xn2-9
4x2-9 = 0
4x2 = 9
x2 = 9/4
x = ±3/2
x=±1.5
Question 196

Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at (i,j) then it can move to either (i+1,j) or (i,j+1).

How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0,0)?

A
B
220
C
210
D
None of the above
       Engineering-Mathematics       Combinatorics       GATE 2007
Question 196 Explanation: 
Lets name one unit up movement as 'u' and one unit right movement as 'r'.
So now we have 10 u's and 10 r's, i.e.,
uuuuuuuuuurrrrrrrrrr
So, finally the no. of arrangements of above sequences is,
Question 197

Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at (i,j) then it can move to either (i+1,j) or (i,j+1).

Suppose that the robot is not allowed to traverse the line segment from (4,4) to (5,4). With this constraint, how many distinct paths are there for the robot to reach (10,10) starting from (0,0)?

A
29
B
219
C
D
       Engineering-Mathematics       Combinatorics       GATE 2007
Question 197 Explanation: 
So to solve this lets find the no. of paths possible if line segment from (4,4) to (5,4) is taken. And then we will subtract it from the total no. of paths possible.
So, no. of paths possible if line segment from (4,4) to (5,4) is taken is,
= paths possible from (0,0) to (4,4) * paths possible from (5,4) to (10,10)
= {uuuurrrr} * {uuuuuurrrrr}

Hence, the final answer is
Question 198

Consider the polynomial p(x) = a0 + a1x + a2x2 + a3x3, where ai ≠ 0, ∀i. The minimum number of multiplications needed to evaluate p on an input x is:

A
3
B
4
C
6
D
9
       Engineering-Mathematics       Numerical-Methods       GATE 2006
Question 198 Explanation: 
Given polynomial equation is
p(x) = a0 + a1x + a2x2 + a3x3 where ai≠0
This can be written as
p(x) = a0 +x( a1 + a2x + a3x2)=a0+(a1+(a2+a3x)x)x
Total no. of multiplications required is 3
i.e., a3x = K.....(i)
(a2+K)x = M..... (ii)
(a1+M)x=N...... (iii)
Question 199

Let X, Y, Z be sets of sizes x, y and z respectively. Let W = X×Y and E be the set of all subsets of W. The number of functions from Z to E is:

A
Z2xy
B
Z×2xy
C
Z2x+y
D
2xyz
       Engineering-Mathematics       Sets-And-Functions       GATE 2006
Question 199 Explanation: 
A set consists of n elements then no. of possible subsets are 2n.
A set ‘P’ consists of m elements and ‘Q’ consists of n elements then total number of function from P to Q is mn.
⇒ E be the no. of subsets of W = 2|w| = 2|xxy| = 2xy
No. of function from Z to E is = (2xy)z = (2xy)z = 2xyz
Question 200

The set {1, 2, 3, 5, 7, 8, 9} under multiplication modulo 10 is not a group. Given below are four plausible reasons. Which one of them is false?

A
It is not closed
B
2 does not have an inverse
C
3 does not have an inverse
D
8 does not have an inverse
       Engineering-Mathematics       Sets-And Relation       GATE 2006
Question 200 Explanation: 
The given set is x = {1,2,3,5,7,8,9}
Option A:
It is not closed under multiplication. After multiplication modulo (10) we get ‘0’. The ‘0’ is not present in the set.
(2*5)%10 ⇒ 10%10 = 0
Option B:
2 does not have an inverse such as
(2*x)%10 ≠ 1
Option C:
3 have an inverse such that
(3*7)%10 = 1
Option D:
8 does not have an inverse such that
(8*x)%10 ≠ 1
Question 201

We are given a set X = {x1, .... xn} where xi = 2i. A sample S ⊆ X is drawn by selecting each xi independently with probability pi = 1/2. The expected value of the smallest number in sample S is:

A
1/n
B
2
C
√n
D
n
       Engineering-Mathematics       Probability       GATE 2006
Question 201 Explanation: 
The smallest element in sample S would be xi for which i is the smallest.
The given probability Pi is for selection of each item independently with probability 1/2.
Now, Probability for x1 to be smallest in S = 1/2
Now, Probability for x2 to be smallest in S = Probability of x1 not being in S × Probability of x2 being in S
= 1/2 × 1/2
Similarly, Probability xi to be smallest = (1/2)i
Now the Expected value is
Question 202

For each element in a set of size 2n, an unbiased coin is tossed. The 2n coin tosses are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is:

A
B
C
D
       Engineering-Mathematics       Probability       GATE 2006
Question 202 Explanation: 
Total number of coins = 2n
No. of elements selected = n
Probability of getting head = ½
Probability of n heads out of 2n coin tosses is
2nCn*(1/2)n*(1/2)n = 2nCn/4n
Question 203

Let E, F and G be finite sets.

Let X = (E∩F) - (F∩G) and Y = (E - (E∩G)) - (E-F).

Which one of the following is true?

A
X ⊂ Y
B
X ⊃ Y
C
X = Y
D
X - Y ≠ ∅ and Y - X ≠ ∅
       Engineering-Mathematics       Sets-And-Relation       GATE 2006
Question 203 Explanation: 
Let us consider
E = {(1,1), (1,2), (1,3), (2,2), (2,3), (3,3), (3,1)}
F = {(1,1), (2,2), (3,3)}
G = {(1,3), (2,1), (2,3), (3,1)}
X = (E∩F) - (F∩G)
= {(1,1), (2,2), (3,3) - ∅}
= {(1,1), (2,2), (3,3)} (✔️)
Y = (E - (E∩G) - (E - F))
= (E - {(1,3), (2,3), (3,1)} - {(1,2), (1,3), (2,3), (3,1)})
= {(1,1), (1,2), (1,3), (2,2), (2,3), (3,3), (3,1)} - {(1,3), (2,2), (2,1)} - (1,2), (1,3), (2,3), (3,1)}
= {(1,1), (1,2), (2,2), (3,3)} - {(1,2), (1,3), (2,3), (3,1)}
= {(1,1), (2,2), (3,3)} (✔️)
X = Y

X = (E∩F) - (F∩G) = {2,5} - {5} = {2}
Y = (E - (E∩G) - (E - F))
= {(1,2,4,5) - (4,5) - (1,4)}
= {(1,2) - (1,4)}
= {2}
X = Y
Question 204

F is an n×n real matrix. b is an n×1 real vector. Suppose there are two n×1 vectors, u and v such that, u≠v and Fu=b, Fv=b. Which one of the following statements is false?

A
Determinant of F is zero
B
There are an infinite number of solutions to Fx=b
C
There is an x≠0 such that Fx=0
D
F must have two identical rows
       Engineering-Mathematics       Linear-Algebra       GATE 2006
Question 204 Explanation: 
⇾ Fu = b, Fv = b
Fu = Fv
Fu - Fv = 0
F(u - v) = 0
Given u ≠ v
F = 0 (i.e., Singular matrix, so determinant is zero)
Option A is true.
⇾ Fx = b; where F is singular
It can have no solution (or) infinitely many solutions.
Option B is true.
⇾ x ≠ 0 such that Fx = 0 is True because F is singular matrix (“stated by singular matrix rules). Option C is true.
⇾ F can two identical columns and rows.
Option D is false.
Question 205

Given a set of elements N = {1, 2, ..., n} and two arbitrary subsets A⊆N and B⊆N, how many of the n! permutations π from N to N satisfy min(π(A)) = min(π(B)), where min(S) is the smallest integer in the set of integers S, and π(S) is the set of integers obtained by applying permutation π to each element of S?

A
(n-|A ∪ B|) |A| |B|
B
(|A|2+|B|2)n2
C
n!(|A∩B|/|A∪B|)
D
       Engineering-Mathematics       Combinatorics       GATE 2006
Question 205 Explanation: 
Given a set of elements N = {1, 2, 3, ...N}
Two arbitrary subsets A⊆N and B⊆N.
Out of n! permutations π from N to N, to satisfy
min(π(A)) = min (π(B))
*) π(S) is the set of integers obtained by applying permutation π to each element of S.
If min(π(A)) =min (π(B)), say y = π(x) is the common minimum.
Since the permutation π is a 1-to-1 mapping of N,
x ∈ A∩B
∴ A∩B cannot be empty.
⇒ y = π(x)
= π(A∩B) is the minimum of π(A∪B) is the minimum of π(A) and π(B) are to be same.
You can think like
*) If the minimum of π(A) and π(B) are same [min π(A)] = min [π(B)]
then min(π(A∩B)) = min(π(A∪B))
∴ Total number is given by n! |A∩B|/|A∪B|
*) Finally
Considering all possible permutations, the fraction of them that meet this condition |π(A∩B)| / |π(A∪B)|
[The probability of single permutation].
Ex: N = {1, 2, 3, 4} A = {1, 3} B = {1, 2, 4}

Since π is one to one mapping
|π(A∩B)| = |A∩B|
∴ π(A) = {1, 2}
π(B) = {1, 4, 3}
π(A∩B) = {1}
π(A∪B) = {1, 2, 3, 4}
4! × 1/4 = 6
Question 206

Let S = {1,2,3,....,m}, m > 3. Let x1, x2,....xn be the subsets of S each of size 3. Define a function f from S to the set of natural numbers as, f(i) is the number of sets of Xj that contain the element i. That is f(i) = |{j|i ∈ Xj|}|.

Then is

A
3m
B
3n
C
2m+1
D
2n+1
       Engineering-Mathematics       Sets And Functions       GATE 2006
Question 206 Explanation: 
Question 207

Which one of the first order predicate calculus statements given below correctly express the following English statement?

     Tigers and lions attack if they are hungry or threatened.
A
∀x [(tiger(x) ∧ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]
B
∀x [(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) ∧ attacks(x)}]
C
∀x [(tiger(x) ∨ lion(x)) → {(attacks(x) → (hungry (x)) ∨ threatened (x))}]
D
∀x [(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]
       Engineering-Mathematics       Propositional-Logic       GATE 2006
Question 207 Explanation: 
Tigers and lions attack if they are hungry (or) threatened.
Here we have two cases.
i) If Tiger is hungry (or) threaten that will attack.
ii) If Lion is hungry (or) threaten that will attack.
If Tiger is hungry (or) threaten then both lion and tiger will not attack only Tiger will attack and vice-versa.
Then answer is
∀x[(tiger(x) ∨ lion(x)) → {(hungry(x) ∨ threatened(x)) → attacks(x)}]
Note: Don’t confuse with the statement Tiger and Lion.
Question 208

Consider the following propositional statements:

    P1 : ((A ∧ B) → C)) ≡ ((A → C) ∧ (B → C))
    P2 : ((A ∨ B) → C)) ≡ ((A → C) ∨ (B → C))

Which one of the following is true?

A
P1 is a tautology, but not P2
B
P2 is a tautology, but not P1
C
P1 and P2 are both tautologies
D
Both P1 and P2 are not tautologies
       Engineering-Mathematics       Propositional-Logic       GATE 2006
Question 208 Explanation: 
It’s better to draw truth table such that

Both P1 and P2 are not Tautologies.
Question 209

A logical binary relation ⊙,is defined as follows:

Let ~ be the unary negation (NOT) operator, with higher precedence than ⊙. Which one of the following is equivalent to A∧B ?

A
(~A⊙B)
B
~(A⊙~B)
C
~(~A⊙~B)
D
~(~A⊙B)
       Engineering-Mathematics       Sets-And-Relation       GATE 2006
Question 209 Explanation: 
Question 210

The 2n vertices of a graph G corresponds to all subsets of a set of size n, for n ≥ 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements.

The number of vertices of degree zero in G is:

A
1
B
n
C
n+1
D
2n
       Engineering-Mathematics       Graph-Theory       GATE 2006
Question 210 Explanation: 
No. of vertices with degree zero is
= no. of subsets with size less than or equal to 1
= n+1, because in question it is given that the two vertices are connected if and only if the corresponding sets intersect in exactly two elements.
Question 211

The 2n vertices of a graph G corresponds to all subsets of a set of size n, for n ≥ 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements.

The maximum degree of a vertex in G is:

A
B
2n-2
C
2n-3 × 3
D
2n-1
       Engineering-Mathematics       Graph-Theory       GATE 2006
Question 211 Explanation: 
The degree of each subset with k elements will be
(k(c))2 2n-k
∴ We need to find 'k' value such that, the value will be maximum.[k should be an integer].
If you differentiate (k(c))2 2n-k w.r.t. k and equal to 0.
You will get k = 2/(loge)2 which is not an integer.
So you can see it like

∴ The maximum degree 3⋅2n-3 at k=3 or k=4.
Question 212

The 2n vertices of a graph G corresponds to all subsets of a set of size n, for n ≥ 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements.

The number of connected components in G is:

A
n
B
n+2
C
2n/2
D
2n / n
       Engineering-Mathematics       Graph-Theory       GATE 2006
Question 212 Explanation: 
Not connected nodes is n+1.
While other nodes are connected so that total number of connected components is (n+1)+1
(here we are adding 1 because it is connected corresponding remaining vertices)
= n+2
Question 213

Let A, B and C be non-empty sets and let X = (A - B) - C and Y = (A - C) - (B - C). Which one of the following is TRUE?

A
X = Y
B
X ⊂ Y
C
Y ⊂ X
D
None of these
       Engineering-Mathematics       Sets-And Relation       GATE 2005
Question 213 Explanation: 
Consider, A = {1, 2, 3, 4, 5, 6}
B = {1, 3, 4, 5}
C = {2, 4, 5, 6}
X = (A - B) - C
X = {2, 6} - {2, 4, 5, 6}
= ∅
Y = (A - C) - (B - C)
= {1, 3} - { 1, 3}
= ∅
X = Y
X = (A - B) - C
= (1, 5) - (5, 7, 4, 3)
= (1)
Y = (A - C) - (B - C)
= (1, 4) - (2, 4)
= (1)
X = Y
Question 214

Let G be a simple connected planar graph with 13 vertices and 19 edges. Then, the number of faces in the planar embedding of the graph is:

A
6
B
8
C
9
D
13
       Engineering-Mathematics       Graph-Theory       GATE 2005
Question 214 Explanation: 
No. of faces in a planar embedding of a graph is
F = E - V + 2 [From Euler's formula i.e., F + V - E = 2]
F = 19 - 13 +2
F = 8
Question 215

Let G be a simple graph with 20 vertices and 100 edges. The size of the minimum vertex cover of G is 8. Then, the size of the maximum independent set of G is

A
12
B
8
C
Less than 8
D
More than 12
       Engineering-Mathematics       Graph-Theory       GATE 2005
Question 215 Explanation: 
No. of vertices = 20
Edges = 100
Minimum cover of vertex G is = 8
Maximum Independent set of G = No. of vertices - Minimum cover of vertex G
= 20 - 8
= 12
Question 216

Let f(x) be the continuous probability density function of a random variable X. The probability that a < X ≤ b, is:

A
f(b - a)
B
f(b) - f(a)
C
D
       Engineering-Mathematics       Probability       GATE 2005
Question 216 Explanation: 
f(x) be the continuous probability density function of random variable X.
Then the probablity be area of the corresponding curve i.e.,
Question 217

The set {1, 2, 4, 7, 8, 11, 13, 14} is a group under multiplication modulo 15. The inverses of 4 and 7 are respectively:

A
3 and 13
B
2 and 11
C
4 and 13
D
8 and 14
       Engineering-Mathematics       Sets-And Relation       GATE 2005
Question 217 Explanation: 
Let say,
Inverse of 4 = m; Inverse of 7 = n
(4×m)%15=1; (7*n)%15=1
Option A: m=3 n=13
12%15≠1 (✖️) 91%15=1 (✔️)
Option B: m=2 n=11
8%15≠1 (✖️) 11%15≠1 (✖️)
Option C: m=4 n=13
16%15=1(✔️) 91%15=1 (✔️)
Option D: m=8 n=14
120%15≠1(✖️) 98%15≠1(✖️)
Question 218

Let P, Q and R be three atomic prepositional assertions. Let X denote (P ∨ Q) → R and Y denote (P → R) ∨ (Q → R). Which one of the following is a tautology?

A
X ≡ Y
B
X → Y
C
Y → X
D
¬Y → X
       Engineering-Mathematics       Propositional-Logic       GATE 2005
Question 218 Explanation: 
X: (P∨Q) → R
⇒ ∼(P∨Q) ∨ R
⇒ (∼P∧∼Q) ∨ R
⇒ (∼P∨R) × (∼Q∨R)
⇒ (P→R) ∧ (Q→R)
Option B: X→Y
[(P→R) × (Q→R)] → [(P→R) ∨ (Q→R)]
∼[(P→R) × (Q→R) ∨ (P→R) ∨ (Q→R)]
[∼(P→R) ∨ ∼(Q→R)] ∨ [(P→R) ∨ (Q→R)]
[∼(P→R) ∨ (P→R)] ∨ [∼(P→R) ∨ (Q→R)] ∨ [(Q→R) ∨ (P→R)] ∨ [∼(Q→R) ∨ (Q→R)]
T ∨ [∼(P→R) ∨ (Q→R)] ∨ [(Q→R) ∨ (P→R)] V T
T (✔️)
Question 219

What is the first order predicate calculus statement equivalent to the following?

   Every teacher is liked by some student  
A
∀(x) [teacher(x) → ∃ (y) [student(y) → likes (y, x)]]
B
∀(x) [teacher(x) → ∃ (y) [student(y) ∧ likes (y, x)]]
C
∃(y) ∀(x) [teacher(x) → [student(y) ∧ likes (y, x)]]
D
∀(x) [teacher(x) ∧ ∃ (y)[student(y) → likes (y, x)]]
       Engineering-Mathematics       Propositional-Logic       GATE 2005
Question 219 Explanation: 
Option A: If x is a teacher, then there exist a y such that if y is a student , then y likes x.
Option B: If x is a teacher, then there exists some y, who is a student and like x. (✔️)
Option C: There exists a student who likes all teachers.
Option D: If x is a teacher and then there exists some y, if y is a student then y likes x.
Question 220

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 an equivalence relation.
       Engineering-Mathematics       Sets-And-Relation       GATE 2005
Question 220 Explanation: 
R∪S might not be transitive.
Let (a,b) present in R and (b,c) present in S and (a,c) is not present in either of them. Then R∪S will contain (a,b) and (b,c) but not (a,c) and hence not transitive.
And equivalence relation must satisfy 3 property:
(i) Reflexive
(ii) Symmetric
(iii) Transitive
But as we have seen that for R∪S, Transitivity is not satisfied.
Question 221

Let f: B → C and g: A → B be two functions and let h = f∘g. Given that h is an onto function. Which one of the following is TRUE?

A
f and g should both be onto functions
B
f should be onto but g need not be onto
C
g should be onto but f need not be onto
D
both f and g need not be onto
       Engineering-Mathematics       Sets And Functions       GATE 2005
Question 221 Explanation: 
Given:
f: B→C and g: A→B are two functions.
h = f∘g = f(g(x))
→ If his onto function, that means for every value in C, there must be value in A.
→ Here, we are mapping C to A using B, that means for every value in C there is a value in B then f is onto function.
→ But g may (or) may not be the onto function i.e., so values in B which may doesn't map with A.
Question 222

What is the minimum number of ordered pairs of non-negative numbers that should be chosen to ensure that there are two pairs (a,b) and (c,d) in the chosen set such that

    a ≡ c mod 3 and b ≡ d mod 5  
A
4
B
6
C
16
D
24
       Engineering-Mathematics       Sets-And-Relation       GATE 2005
Question 222 Explanation: 
a = cmod3
That means a = 0,1,2 ⇒ |3|
b = dmod5
That means b = 0,1,2,3,4 ⇒ |4|
→ Total no. of order pairs = 3 * 5 = 15
→ Ordered pair (c,d) has 1 combination.
Then total no. of combinations = 15+1 = 16
Question 223

Consider the set H of all 3 × 3 matrices of the type

|a   f   e|
|0   b   d|
|0   0   c|

where a, b, c, d, e and f are real numbers and abc ≠ 0. Under the matrix multiplication operation, the set H is

A
a group
B
a monoid but not a group
C
a semi group but not a monoid
D
neither a group nor a semi group
       Engineering-Mathematics       Sets-And-Relation       GATE 2005
Question 223 Explanation: 
abc != 0 & Identity matrix is identity then it is non-singular and inverse is also defined.
The algebraic structure is a group because the given matrix can have inverse and the inverse is non-singular.
Question 224

Which one of the following graphs is NOT planar?

A
G1
B
G2
C
G3
D
G4
       Engineering-Mathematics       Graph-Theory       GATE 2005
Question 224 Explanation: 
G2 can also be drawn as

which is planar
G3 can also be drawn as

which is planar
G4 can also be drawn as

which is planar
But G1 cannot be drawn as planar graph.
Hence, option (A) is the answer.
Question 225

Consider the following system of equations in three real variables xl, x2 and x3

2xl - x2 + 3x3 = 1 
3xl- 2x2 + 5x3 = 2 
-xl + 4x2 + x3 = 3 

This system of equations has

A
no solution
B
a unique solution
C
more than one but a finite number of solutions
D
an infinite number of solutions
       Engineering-Mathematics       Linear-Algebra       GATE 2005
Question 225 Explanation: 

2(-2 - 20) +1(3 + 5) + 3(12 - 2)
= -44 + 8 + 30
= -6 ≠ 0
→ |A| ≠ 0, we have Unique Solution.
Question 226

What are the eigenvalues of the following 2 × 2 matrix?

|2   -1|
|-4   5| 
A
-1 and 1
B
1 and 6
C
2 and 5
D
4 and -1
       Engineering-Mathematics       Linear-Algebra       GATE 2005
Question 226 Explanation: 

|A| = (2 - λ)(5 - λ) - (4) = 0
10 - 7λ+ λ2 - 4 = 0
λ2 - 7λ + 6 = 0
λ2 - 6λ - λ + 6 = 0
(λ - 6) -1(λ - 6) = 0
λ = 1 (or) 6
Question 227

Let , where |x|<1. What is g(i)?

A
i
B
i+1
C
2i
D
2i
       Engineering-Mathematics       Combinatorics       GATE 2005
Question 227 Explanation: 

Put g(i) = i+1

S = 1 + 2x + 3x2 + 4x3 + .....
Sx = 1x + 2x2 + 3x3 + 4x4 + ......
S - Sx = 1 + x + x2 + x3 + .....
[Sum of infinite series in GP with ratio < 1 is a/1-r]
S - Sx = 1/(1-x)
S(1-x) = 1/(1-x)
S = 1/(1-x)2
Question 228

Box P has 2 red balls and 3 blue balls and box Q has 3 red balls and 1 blue ball. A ball is selected as follows:

(i) Select a box
(ii) Choose a ball from the selected box such that each ball in
     the box is equally likely to be chosen. The probabilities of
     selecting boxes P and Q are (1/3) and (2/3), respectively.

Given that a ball selected in the above process is a red ball, the probability that it came from the box P is

A
4/19
B
5/19
C
2/9
D
19/30
       Engineering-Mathematics       Probability       GATE 2005
Question 228 Explanation: 
P → 2 red, 3 blue
Q → 3 red, 1 blue
The probability of selecting a red ball is
(1/3)(2/5) + (2/3)(3/4)
2/15 + 1/2 = 19/30
The probability of selecting a red ball from P
(1/3) * (2/5) = 2/15
→ The colour of ball is selected is to be red and that is taken from the box P.
⇒ Probability of selecting a red ball from P/Probability of selecting a red ball
⇒ (2/15)/(19/30)
⇒ 4/19
Question 229

A random bit string of length n is constructed by tossing a fair coin n times and setting a bit to 0 or 1 depending on outcomes head and tail, respectively. The probability that two such randomly generated strings are not identical is:

A
1/2n
B
1 - 1/n
C
1/n!
D
1-(1/2n)
       Engineering-Mathematics       Probability       GATE 2005
Question 229 Explanation: 
Total combinations of strings that can be generated are 2n. We will get one such string in the first experiment. So, favourable cases for the second string are 2n - 1, so that it doesn't match with the previous generated string.
Hence Probability = (2n - 1) /2n = 1 - 1/2n
Question 230

Identify the correct translation into logical notation of the following assertion.

      "Some boys in the class are taller than all the girls"  

Note: taller(x,y) is true if x is taller than y.

A
(∃x) (boy(x) → (∀y) (girl(y) ∧ taller(x,y)))
B
(∃x) (boy(x) ∧ (∀y) (girl(y) ∧ taller(x,y)))
C
(∃x) (boy(x) → (∀y) (girl(y) → taller(x,y)))
D
(∃x) (boy(x) ∧ (∀y) (girl(y) → taller(x,y)))
       Engineering-Mathematics       Propositional-Logic       GATE 2004
Question 230 Explanation: 
Don't confuse with '∧' and '→'
'∧' → predicts statements are always true, no matter the value of x.
'→' → predicts there is no need of left predicate to be true always, but whenever it becomes true, then right predicate must be true.
Option D:
There exists a some boys who are taller than of all girls y.
Question 231

Consider the binary relation:

       S = {(x,y)|y = x+1 and x,y ∈ {0,1,2, ...}}  

The reflexive transitive closure of S is

A
{(x, y)|y > x and x, y ∈ {0, 1, 2, ... }}
B
{(x, y)|y ≥ x and x, y ∈ {0, 1, 2, ... }}
C
{(x, y)|y < x and x, y ∈ {0, 1, 2, ... }}
D
{(x, y)|y ≤ x and x, y ∈ {0, 1, 2, ... }}
       Engineering-Mathematics       Sets-And-Relation       GATE 2004
Question 231 Explanation: 
Transitive means that x is related to greater value y and reflexive means x related to y.
Answer is option B.
{(x, y)|y ≥ x and x, y ∈ {0, 1, 2, ... }}
Question 232

If a fair coin is tossed four times. What is the probability that two heads and two tails will result?

A
3/8
B
1/2
C
5/8
D
3/4
       Engineering-Mathematics       Probability       GATE 2004
Question 232 Explanation: 
A fair coin is tossed 4 times.
Then total number of possibilities = 24 = 16
No. of possibilities getting 2 heads and 2 tails is
HHTT, HTHT, TTHH, THTH, THHT, HTTH = 6
Probability of getting 2 heads and 2 tails is
= No. of possibilities/Total no. of possibilities = 6/16 = 3/8
Question 233

The number of different n × n symmetric matrices with each element being either 0 or 1 is: (Note: power(2,x) is same as 2x)

A
power (2,n)
B
power (2,n2)
C
power (2, (n2 + n)/2)
D
power (2, (n2 - n)/2)
       Engineering-Mathematics       Linear-Algebra       GATE 2004
Question 233 Explanation: 
If a matrix is symmetric then
A[i][j] = A[j][i]
So, we have only two choices, they are either upper triangular elements (or) lower triangular elements.
No. of such elements are
n + (n-1) + (n-2) + ... + 1
n(n+1)/2
We have two choices, thus we have
2(n(n+1)/2) = 2((n2+n)/2) choices
i.e., Power (2, (n2+n)/2).
Question 234

Let A, B, C, D be n × n matrices, each with non-­zero determinant. If ABCD = 1, then B-1 is

A
D-1C-1A-1
B
CDA
C
ADC
D
Does not necessarily exist
       Engineering-Mathematics       Linear-Algebra       GATE 2004
Question 234 Explanation: 
ABCD are n × n matrices with non-zero determinant.
ABCD = I
Pre multiply A-1 on both sides
A-1ABCD = A-1⋅I
BCD = A-1
Pre multiply B-1 on both sides
B-1BCD = B-1A-1
CD = B-1A-1
Post multiply A on both sides
CDA = B-1A-1⋅A
∴ CDA = B-1(I)
∴ CDA = B-1
Question 235

The following propositional statement is (P → (Q v R)) → ((P v Q) → R)

A
satisfiable but not valid
B
valid
C
a contradiction
D
None of the above
       Engineering-Mathematics       Propositional-Logic       GATE 2004
Question 235 Explanation: 
(P→(Q∨R)) → ((P∨Q)→R)
If P=T; Q=T; R=T
(P→(T∨T)) → ((T∨T)→R)
(P→T) → (T→R)
(T→T) → (T→T)
T→T
T(Satisfiable)
Question 236

How many solutions does the following system of linear equations have?

  -x + 5y = -1
   x - y = 2
   x + 3y = 3 
A
infinitely many
B
two distinct solutions
C
unique
D
none
       Engineering-Mathematics       Linear-Algebra       GATE 2004
Question 236 Explanation: 

rank = r(A) = r(A|B) = 2
rank = total no. of variables
Hence, unique solution.
Question 237

The following is the incomplete operation table a 4-element group.

The last row of the table is

A
c a e b
B
c b a e
C
c b e a
D
c e a b
       Engineering-Mathematics       Sets-And-Relation       GATE 2004
Question 237 Explanation: 

The last row is c e a b.
Question 238

The inclusion of which of the following sets into

S = {{1,2}, {1,2,3}, {1,3,5}, (1,2,4), (1,2,3,4,5}}

is necessary and sufficient to make S a complete lattice under the partial order defined by set containment?

A
{1}
B
{1}, {2, 3}
C
{1}, {1, 3}
D
{1}, {1, 3}, (1, 2, 3, 4}, {1, 2, 3, 5}
       Engineering-Mathematics       Sets-And-Relation       GATE 2004
Question 238 Explanation: 
A lattice is complete if every subset of partial ordered set is to be a supremum and infimum element.
For the set {1, 2, 3, 4, 5} there is no supremum element i.e., {1}.
Then clearly we need to add {1}, then it is to be a lattice.
Question 239

An examination paper has 150 multiple-choice questions of one mark each, with each question having four choices. Each incorrect answer fetches -0.25 mark. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained by all these students is:

A
0
B
2550
C
7525
D
9375
       Engineering-Mathematics       Probability       GATE 2004
Question 239 Explanation: 
Probability of choosing a correct answer = 1/4
Probability of selecting a wrong answer = 3/4
For correct answer +1, for wrong answer-0.25;
Expected marks for each question = (1/4) × 1 + (3/4) -(0.25)
= 1/4 + (-3/16)
= 4-3/16
= 1/16
= 0.0625
Expected marks for 150 questions = 150 × 0.625 = 9.375
The sum total of expected marks obtained by 1000 students is = 1000×9.375 = 9375
Question 240

Mala has a colouring book in which each English letter is drawn two times. She wants to paint each of these 52 prints with one of k colours, such that the colour-pairs used to colour any two letters are different. Both prints of a letter can also be coloured with the same colour. What is the minimum value of k that satisfies this requirement?

A
9
B
8
C
7
D
6
       Engineering-Mathematics       Combinatorics       GATE 2004
Question 240 Explanation: 
No. of letters from A-Z is = 26
Each is printed twice the no. of letters = 26×2 = 52
If Mala has k colours, she can have k pairs of same colours.
She also can have kC2 different pairs in which each pair is having different colours.
So total no. of pairs that can be coloured = k+kC2
k+kC2 ≥ 26
k+k(k-1)/2 ≥ 26
k(k+1)/2 ≥ 26
k(k+1) ≥ 52
k(k+1) ≥ 7*8
k≥7
Question 241

In an M×N matrix such that all non-zero entries are covered in a rows and b columns. Then the maximum number of non-zero entries, such that no two are on the same row or column, is

A
≤ a+b
B
≤ max(a, b)
C
≤ min(M-a, N-b)
D
≤ min(a, b)
       Engineering-Mathematics       Linear-Algebra       GATE 2004
Question 241 Explanation: 
Entry will be a member of same row and same column.
→ Such that a row can have maximum of a elements and no row has separate element and for b also same.
→ By combining the both, it should be ≤ (a,b).
Question 242

The minimum number of colours required to colour the following graph, such that no two adjacent vertices are assigned the same colour, is

A
2
B
3
C
4
D
5
       Engineering-Mathematics       Graph-Theory       GATE 2004
Question 242 Explanation: 

→ a, b, c, d = 4
→ The minimum no. of colours required to colour a graph = 4 (no two adjacent vertices have same colours)
Question 243

Two n bit binary strings, S1 and S2, are chosen randomly with uniform probability. The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to d is

A
nCd /2n
B
nCd / 2d
C
d/2n
D
1/2d
       Engineering-Mathematics       Probability       GATE 2004
Question 243 Explanation: 
n binary bits with difference 'd' then no. of favourable cases = nCd
Total no. of cases where n positions have any binary bit = 2n
The probability of 'd' bits differ = nCd / 2n
Question 244

How many graphs on n labeled vertices exist which have at least (n2 - 3n)/2 edges?

A
B
C
D
       Engineering-Mathematics       Graph-Theory       GATE 2004
Question 244 Explanation: 
No. of atleast edges = (n2-3n)/2 = e
Maximum no. of vertices = n(n-1)/2 = v
No. of graphs with minimum b edges is
= C(v,e) + C(v,e+1) + C(v,e+2) + ... + C(v,v)
= C((v,v-e) + C(v,v-(e+1)) + C(v,v-(e+2)) + ... + C(v,0)
= C(a,n) + C(a,n-1) + C(a,n-2) + ... + C(a,0) (since a-b=n)
= C(n(n-1)/2,n) + C(n(n-1)/2,n-1) + ... + C(n(n-1)/2,0)
Question 245

A point is randomly selected with uniform probability in the X-Y plane within the rectangle with corners at (0,0), (1,0), (1,2) and (0,2). If p is the length of the position vector of the point, the expected value of p2 is

A
2/3
B
1
C
4/3
D
5/3
       Engineering-Mathematics       Probability       GATE 2004
Question 245 Explanation: 

Above diagram shows the scenario of our question.
The length p of our position vector (x,y) is
p = √x2 + y2
p2 = x2 + y2
E(p2) = E(x2 + y2) = E(x2) + E(y2)
Now we need to calculate the probability density function of X and Y.
Since distribution is uniform,
X goes from 0 to 1, so PDF(x) = 1/1-0 = 1
Y goes from 0 to 2, so PDF(y) = 1/2-0 = 1/2
Now we evaluate,

E(p2) = E(x2) + E(y2) = 5/3
Question 246

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 U G2 = (V, E1 U 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
       Engineering-Mathematics       Graph-Theory       GATE 2004
Question 246 Explanation: 
Lets try to take counter example for each of them,
(A)

False, since in G1∪G2 'C' is a cut vertex.
(B) True, for all conditions.
(C)

False. G1∪G2 has no bridge.
D)

False. G1∪G2, G1, G2 all the three graphs have chromatic number of 2.
Question 247

Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, the values of P(A|B) and P(B|A) respectively are

A
1/4, 1/2
B
1/2, 1/4
C
1/2, 1
D
1, 1/2
       Engineering-Mathematics       Probability       GATE 2003
Question 247 Explanation: 
P(A)=1, P(B)=1/2
P(A/B) = P(A∩B)/P(B)
= P(A)⋅P(B)/P(B) (consider P(A), P(B) are two independent events)
= 1
P(B/A) = P(B∩A)/P(A)
= P(B)⋅P(A)/P(A)
= 1/2
Question 248

n couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gatherings possible at the party is

A
B
C
D
       Engineering-Mathematics       Combinatorics       GATE 2003
Question 248 Explanation: 
The possibilities to attend party is
i) Both husband and wife comes
ii) Only wife comes
iii) Both are not come
The no. of different gatherings possible at party is
= 3 * 3 * 3 * 3 * ... n times
= 3n
Question 249

Consider the set ∑* of all strings over the alphabet ∑ = {0, 1}. ∑* with the concatenation operator for strings

A
does not form a group
B
forms a non-commutative group
C
does not have a right identity element
D
forms a group if the empty string is removed from Σ*
       Engineering-Mathematics       Sets-And-Relation       GATE 2003
Question 249 Explanation: 
In the concatenation '∊' is the identity element. And given one is not a group because no element has inverse element.
→ To perform concatenation with the given set can result a Monoid and it follows the property of closure, associativity and consists of identity element.
Question 250

Let G be an arbitrary graph with n nodes and k components. If a vertex is removed from G, the number of components in the resultant graph must necessarily lie between

A
k and n
B
k – 1 and k + 1
C
k – 1 and n – 1
D
k + 1 and n – k
       Engineering-Mathematics       Graph-Theory       GATE 2003
Question 250 Explanation: 
While a vertex is removed from a graph then that can be itself be forms a new component. The minimum number of components is k-1.
If a vertex is removed then it results that all the components are also be disconnected. So removal can create (n-1) components.
Question 251

Let (S, ≤) be a partial order with two minimal elements a and b, and a maximum element c. Let P: S → {True, False} be a predicate defined on S. Suppose that P(a) = True, P(b) = False and P(x) ⇒ P(y) for all x, y ∈ S satisfying x ≤ y, where ⇒ stands for logical implication. Which of the following statements CANNOT be true?

A
P(x) = True for all x ∈ S such that x ≠ b
B
P(x) = False for all x ∈ S such that x ≠ a and x ≠ c
C
P(x) = False for all x ∈ S such that b ≤ x and x ≠ c
D
P(x) = False for all x ∈ S such that a ≤ x and b ≤ x
       Engineering-Mathematics       Sets-And Relation       GATE 2003
Question 251 Explanation: 
c is the maximum element.
a or b the minimal element in set.
P(a) = True for all x ∈ S such that a ≤ x and b ≤ x.
Option D is False.
Question 252

Which of the following is a valid first order formula? (Here α and β are first order formulae with x as their only free variable)

A
((∀x)[α] ⇒ (∀x)[β]) ⇒ (∀x)[α⇒β]
B
(∀x)[α] ⇒ (∃x)[α ∧ β]
C
((∀x)[α ∨ β] ⇒ (∃x)[α] ⇒ (∀x)[α]
D
(∀x)[α ⇒ β] ⇒ ((∀x)[α] ⇒ (∀x)[β])
       Engineering-Mathematics       Propositional-Logic       GATE 2003
Question 252 Explanation: 
Option D is valid.
Here, α, β are holding values of x. Then and RHS saying that α holding the value of x and β is holding value of x.
Then LHS ⇒ RHS.
Question 253

Consider the following formula a and its two interpretations I1 and I2

α: (∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Qyy]] ⇒ (∀x)[¬Px]
I1: Domain: the set of natural numbers
    Px ≡ 'x is a prime number'
    Qxy ≡ 'y divides x'
I2: same as I1 except that Px = 'x is a composite number'.

Which of the following statements is true?

A
I1 satisfies α, I2 does not
B
I2 satisfies α, I1 does not
C
Neither I2 nor I1 satisfies α
D
Both I1 and I2 satisfy α
       Engineering-Mathematics       Propositional-Logic       GATE 2003
Question 253 Explanation: 
Given that:
(∀x)[Px ⇔ (∀y)[Qxy ⇔ ¬Qyy]] ⇒(∀x)[¬Px]
Qyy is always true, because y divide y, then ¬Qyy is false.
∀x[(P(x) ⇔ ∀y [Qxy ⇔ False]]
∀y [Qxy ⇔ False] can be written as ∀y[¬axy]
⇒(∀x)[P(x) ⇔ ∀y[¬Qxy]]
Here, ¬Qxy says that y doesnot divides x, which is not always be true.
For example, if x=y then it is false then ∀y[¬Qxy] is not true for all values of y.
⇒(∀x)[P(x) ⇔ False]
⇒(∀x)[¬P(x) = RHS]
LHS = RHS
⇒ Irrespective of x, whether x is prime of composite number I1 and I2 satisfies α.
Question 254

m identical balls are to be placed in n distinct bags. You are given that m ≥ kn, where, k is a natural number ≥1. In how many ways can the balls be placed in the bags if each bag must contain at least k balls?

A
B
C
D
       Engineering-Mathematics       Combinatorics       GATE 2003
Question 254 Explanation: 
Since we want at least k balls in each bag, so first we put kn balls into bags, k balls in each bag. Now we are left with m - kn balls, and we have to put them into n bags such that each bag may receive 0 or more balls. So applying theorem 2 of stars and bars with m - nk stars and n bars, we get number of ways to be

. So option (B) is correct.
Question 255

How many perfect matching are there in a complete graph of 6 vertices ?

A
15
B
24
C
30
D
60
       Engineering-Mathematics       Graph-Theory       GATE 2003
Question 255 Explanation: 
We have formula to find no. of perfect matching in complete graphs of 2n vertices,
(2n)!/n!×2n
Given, 2n = 6 ⇒ n = 3
So, finally, 6!/3!×23 = 15
Question 256

Let f : A → B be an injective (one-to-one) function. Define g: 2A→2B as: g(C) = {f(x)|x ∈ C}, for all subsets C of A.
Define h: 2B → 2A as: h(D) = {x|x ∈ A, f(x) ∈ D}, for all subsets D of B.

Which of the following statements is always true?

A
g(h(D)) ⊆ D
B
g(h(D)) ⊇ D
C
g(h(D)) ∩ D = ɸ
D
g(h(D)) ∩ (B—D) ≠ ɸ
       Engineering-Mathematics       Sets and Functions       GATE 2003
Question 256 Explanation: 
f: A→B be an injective (one-to-one)
→ g: 2A→2B be also one to one function and g(C) = f(x)|x∈C}, for all subsets C of A.
The range of this function is n(2A).
→ h: 2B→2A it is not a one to one function and given h(D) = {x|x∈A, f(x)∈D}, for all subsets D of B.
The range of this function is also n(2A).
→ The function g(h(D)) also have the range n(2A) that implies n(A)≤n(B), i.e., n(2A) is less than n(2B).
Then this result is g(h(D)) ⊆ D.
Question 257

Consider the set {a, b, c} with binary operators + and × defined as follows:

+	a	b	c		×	a	b	c
a	b	a	c		a	a	b	c
b	a	b	c		b	b	c	a
c	a	c	b		c	c	c	b 

For example, a + c = c, c + a = a, c × b = c and b × c = a. Given the following set of equations:

(a × x) + (a × y) = c
(b × x) + (c × y) = c 

The number of solution(s) (i.e., pair(s) (x, y)) that satisfy the equations is:

A
0
B
1
C
2
D
3
       Engineering-Mathematics       Sets-And-Relation       GATE 2003
Question 257 Explanation: 
Total possible values = 32 = 9 i.e., (a,a), (b,b), (c,c), (a,b), (a,c), (b,a), (b,c), (c,a), (c,b)
In those (x, y) = (b,c) & (c,b) are the possible solution for the corresponding equations.
(x, y) = (b,c) ⇒ (a*b)+(a*c) ⇒ (b*b)+(c*c)
⇒ (b) + (c) ⇒ c + b
⇒ c (✔️) ⇒ c (✔️)
(x,y) = (c,b) ⇒ (a*c)+(a*b) ⇒ (b*c)+(c*b)
⇒ c+b ⇒ a+c
⇒ c (✔️) ⇒ c (✔️)
Question 258

Let ∑ = (a, b, c, d, e) be an alphabet. We define an encoding scheme as follows:

g(a) = 3, g(b) = 5, g(c) = 7, g(d) = 9, g(e) = 11.

Let pi denote the i-th prime number (p1=2).

    For a non-empty string s=a1 ... an, where each ai∈Σ, define
    For a non-empty sequence 〈sj...sn〉 of strings from ∑+, define

Which of the following numbers is the encoding h of a non-empty sequence of strings?

A
273757
B
283858
C
293959
D
210510710
       Engineering-Mathematics       Sets-And-Relation       GATE 2003
Question 258 Explanation: 
The answers can have only three possibilities of sequences i.e., "a", "a", "a", and there is no multiplies of 7 and 9. So eliminates option A & C.
And f(S) = 23 = 8
So answer is 283858.
Question 259

A graph G = (V,E) satisfies |E|≤ 3|V|-6. The min-degree of G is defined as . Therefore, min-degree of G cannot be

A
3
B
4
C
5
D
6
       Engineering-Mathematics       Graph-Theory       GATE 2003
Question 259 Explanation: 
The minimum degree of G = minv∈V {degree(v)}
|E| ≤ 3|v| - 6
Based on handshaking lemma, the minimum degree is (min×|v|)/2
⇒ (min×|v|)/2 ≤ 3|v| - 6
Checking the options lets take min=6
(6×|v|)/2 ≤ 3|v| - 6
0 ≤ -6 (Not satisfied)
And which is inconsistent.
Question 260

Consider the following system of linear equation

Notice that the second and the third columns of the coefficient matrix are linearly dependent. For how many values of α, does this system of equations have infinitely many solutions?

A
0
B
1
C
2
D
infinitely many
       Engineering-Mathematics       Linear-Algebra       GATE 2003
Question 260 Explanation: 

This is in the form AX = B

⇒ R(AB) < n [If we want infinitely many solution]
then -1+5α = 0
5α = 1
α = 1/5 There is only one value of α. System can have infinitely many solutions.
Question 261

A piecewise linear function f(x) is plotted using thick solid lines in the figure below (the plot is drawn to scale).

If we use the Newton-Raphson method to find the roots of f(x) = 0 using x0, x1 and x2 respectively as initial guesses, the roots obtained would be

A
1.3, 0.6, and 0.6 respectively
B
0.6, 0.6, and 1.3 respectively
C
1.3, 1.3, and 0.6 respectively
D
1.3, 0.6, and 1.3 respectively
       Engineering-Mathematics       Newton-Raphson-Method       GATE 2003
Question 261 Explanation: 
Note: Out of syllabus.
Question 262

A program consists of two modules executed sequentially. Let f1(t) and f2(t) respectively denote the probability density functions of time taken to execute the two modules. The probability density function of the overall time taken to execute the program is given by:

A
f1(t) + f2(t)
B
C
D
max{f1(t), f2(t)}
       Engineering-Mathematics       Probability       GATE 2003
Question 262 Explanation: 
f1(t) and f2(t) are executed sequentially.
→ They representing the probability density functions of time taken to execute.
→ f1 can be executed in 'x' time.
f2 can be executed in 't-x' time.
→ The probability density function =
Question 263

The following resolution rule is used in logic programming.

Derive clause (P ∨ Q) from clauses (P ∨ R), (Q ∨ ¬R) 

Which of the following statements related to this rule is FALSE?

A
((P ∨ R) ∧ (Q ∨ ¬R)) ⇒ (P ∨ Q) is logically valid
B
(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R)) is logically valid
C
(P ∨ Q) is satisfiable if and only if (P∨R) ∧ (Q∨¬R) is satisfiable
D
(P ∨ Q) ⇒ FALSE if and only if both P and Q are unsatisfiable
       Engineering-Mathematics       Propositional-Logic       GATE 2003
Question 263 Explanation: 
(P ∨ Q) ⇒ ((P ∨ R) ∧ (Q ∨ ¬R))

It is may be True (or) False depending on values. So this is not valid.
Question 264

The rank of the matrix is

A
4
B
2
C
1
D
0
       Engineering-Mathematics       Linear-Algebra       GATE 2002
Question 264 Explanation: 
Number of non-zero rows is the rank of the matrix.
Question 265

The trapezoidal rule for integration give exact result when the integrand is a polynomial of degree:

A
0 but not 1
B
1 but not 0
C
0 or 1
D
2
       Engineering-Mathematics       Trapezoidal Rule       GATE 2002
Question 265 Explanation: 
If curve is of function of degree 1, then curve will be a straight line, and in that case also, all trapezoids will fit completely. So there is no error in degree 1.
Question 266

The minimum number of colours required to colour the vertices of a cycle with η nodes in such a way that no two adjacent nodes have the same colour is

A
2
B