ISRO CS 2017

Question 1

Suppose A is a finite set with n elements. The number of elements and the rank of the largest equivalence relation on A are

• A. {n,1}
• B. {n, n}
• C. {n², 1}
• D. {1, n²}

Answer: (C)

ISRO CS 2017 Set Theory & Algebra
Discuss it

An Equivalence relation is always Reflexive, Symmetric and Transitive, so for a set of size 'n' elements the largest Equivalence relation will always contain n² elements whereas the smallest Equivalence relation on a set of 'n' elements contain n elements itself. The Rank of an Equivalence relation is equal to the number of induced Equivalence classes. Since we have maximum number of ordered pairs(which are reflexive, symmetric and transitive ) in largest Equivalence relation, its rank is always 1. So option C is correct.

Question 2

Consider the set of integers I. Let D denote “divides with an integer quotient” (e.g. 4D8 but 4D7). Then D is

• A. Reflexive, not symmetric, transitive
• B. Not reflexive, not antisymmetric, transitive
• C. Reflexive, antisymmetric, transitive
• D. Not reflexive, not antisymmetric, not transitive

Answer: (B)

ISRO CS 2017 Set Theory & Algebra Set Theory & Algebra
Discuss it

Reflexibility: for all x ∈ I, R(x,x) is reflexive but here R(0,0) is a violation as 0 belongs to the set of integers but does not satisfy this relation. Symmetric: for all x ∈ I, R(x,y)and R(y,x) is symmetric and clearly the above relation cannot be symmetric.Considering the example S={1,2}. For this to be symmetric (1,2)(2,1) both ordered pair should be present but it is also a violation as 2 can be divided by 1 but 1 cannot be divided by 2 to give an integer quotient. Antisymmetric: for all x ∈ I, R(x,y) and R(y,x) then x=y is antisymmetric. We can easily make a violation as R(-2,2) and R(2,-2) are not antisymmetric. Transitivity:for all x ∈ I , R(x,y), R(y,z) then R(x,z). This is always true for the above divides relation.So The relation is not-reflexive, not-antisymmetric but transitive.

Question 3

A bag contains 19 red balls and 19 black balls. Two balls are removed at a time repeatedly and discarded if they are of the same colour, but if they are different, black ball is discarded and red ball is returned to the bag. The probability that this process will terminate with one red ball is

• A. 1
• B. 1/21
• C. 0
• D. 0.5

Answer: (A)

ISRO CS 2017 Probability
Discuss it

Possible outcomes: RR, BB, RB and BR. If two same coloured balls appear both are discarded whereas if different coloured balls appear only black ball is discarded. Black balls will always be used up and 1 red ball will remain at the end regardless of the outcome of the experiment. The probability that this process will terminate with one red ball is always 1. Option (A) is correct.

Question 4

If x = -1 and x = 2 are extreme points of f(x) = α log |x| + β x² + x then

• A. α = -6, β = -1/2
• B. α = 2, β = -1/2
• C. α = 2, β = 1/2
• D. α = -6, β =1/2

Answer: (B)

ISRO CS 2017 Numerical Methods and Calculus
Discuss it

f(x) = α log |x| + β x² + x for extreme points f'(x)=0 f'(x)= α/x + 2βx + 1 = 0 for x= -1: -α -2β = -1 for x= 2: α/2 + 4β = -1 from here we can get the value of α=2 and β= -1/2

Question 5

Let f(x) = log|x| and g(x) = sin x . If A is the range of f(g(x)) and B is the range of g(f(x)) then A ∩ B is

• A. [-1, 0]
• B. [-1, 0)
• C. [-∞, 0]
• D. [-∞,1]

Answer: (A)

ISRO CS 2017 Numerical Methods and Calculus Set Theory & Algebra
Discuss it

Given, f(x) = log|x| and g(x) = sin(x) Now, f(g(x)) = log|g(x)| = log|sin(x)| So, A = range of log|sin(x)| = (-∞ ,0] And g(f(x)) = sin(f(x)) = sin(log|x|) So, B = range of sin(log|x|) = [-1, 1] Therefore, A ∩ B = [-1, 0] Note that -1 ≤ sin(x) ≤ 1 and -∞ ≤ log|x| ≤ ∞ So, option (A) is correct.

Question 6

The proposition (P⇒Q)⋀(Q⇒P) is a

• A. tautology
• B. contradiction
• C. contingency
• D. absurdity

Answer: (C)

ISRO CS 2017 Propositional and First Order Logic.
Discuss it

(P⇒Q)⋀(Q⇒P) = (¬P+Q)(¬Q+P) = (¬P¬Q + ¬PP + ¬QQ + PQ) = (¬P¬Q + PQ) Therefore it is a contingency.

Question 7

If T(x) denotes x is a trigonometric function, P(x) denotes x is a periodic function and C(x) denotes x is a continuous function then the statement “It is not the case that some trigonometric functions are not periodic” can be logically represented as

• A. ¬∃(x) [ T(x) ⋀ ¬P(x) ]
• B. ¬∃(x) [ T(x) ⋁ ¬P(x) ]
• C. ¬∃(x) [ ¬T(x) ⋀ ¬P(x) ]
• D. ¬∃(x) [ T(x) ⋀ P(x) ]

Answer: (A)

ISRO CS 2017 Propositional and First Order Logic.
Discuss it

some trigonometric functions are not periodic = ∃(x) [ T(x) ⋀ ¬P(x) ] And it's negation is = ¬∃(x) [ T(x) ⋀ ¬P(x) ] Which is equivalent to "It is not the case that some trigonometric functions are not periodic" is equivalent to "All trigonometric functions are periodic" can be expression as = ∀(x) [T(x) → P(x)] = ∀(x) [¬ T(x) ⋁ P(x)] = ∀(x) ¬ [ T(x) ⋀ ¬P(x)] = ¬∃(x) [ T(x) ⋀ ¬P(x) ] Option (A) is correct.

Question 8

The number of elements in the power set of { {1, 2}, {2, 1, 1}, {2, 1, 1, 2} } is

• A. 3
• B. 8
• C. 4
• D. 2

Answer: (D)

ISRO CS 2017 Set Theory & Algebra
Discuss it

This set has only one element because of no order and identical elements. So, total number of elements in power set of given set = 2ⁿ = 2¹ = 2 So, option (D) is correct.

Question 9

The function f: [0,3]→[1,29] defined by f(x) = 2x³ - 15x² + 36x + 1 is

• A. injective and surjective
• B. surjective but not injective
• C. injective but not surjective
• D. neither injective nor surjective

Answer: (B)

ISRO CS 2017 Set Theory & Algebra
Discuss it

Given, f(x) = 2x³ - 15x² + 36x + 1 ⇒ f′(x) = 6(x² − 5x + 6) = 6(x-2)(x-3) Since f(x) is non monotonic in ∈ [0,3] ⇒ f(x) is not one-one And, f(x) is increasing in x ∈[0,2) and decreasing in ∈(0,2] f(0) = 1, f(2) = 29 & f(3) = 28 Therefore Range of f(x) is [1,29] ⇒ f(x) is onto. So, option (B) is correct.

Question 10

If vectors and are perpendicular to each other, then value of λ is

• A. 2/5
• B. 2
• C. 3
• D. 5/2

Answer: (D)

ISRO CS 2017 Algebra Linear Algebra
Discuss it

Given, vectors and are perpendicular to each other, so So, option (D) is correct.