# ISRO CS 2017

• Last Updated : 27 Oct, 2021

 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 {n,1} {n, n} {n2, 1} {1, n2}

Question 1-Explanation:
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 Reflexive, not symmetric, transitive Not reflexive, not antisymmetric, transitive Reflexive, antisymmetric, transitive Not reflexive, not antisymmetric, not transitive

Question 2-Explanation:
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 1 1/21 0 0.5

Question 3-Explanation:
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 α = -6, β = -1/2 α = 2, β = -1/2 α = 2, β = 1/2 α = -6, β =1/2

Question 4-Explanation:
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 [-1, 0] [-1, 0) [-∞, 0] [-∞,1]

Question 5-Explanation:
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 tautology contradiction contingency absurdity

Question 6-Explanation:
(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 ¬∃(x) [ T(x) ⋀ ¬P(x) ] ¬∃(x) [ T(x) ⋁ ¬P(x) ] ¬∃(x) [ ¬T(x) ⋀ ¬P(x) ] ¬∃(x) [ T(x) ⋀ P(x) ]

Question 7-Explanation:
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 3 8 4 2

Question 8-Explanation:
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 injective and surjective surjective but not injective injective but not surjective neither injective nor surjective

Question 9-Explanation:
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 log(base a) 2 = x and log(base a) 8 = y, what is the value of log(base a) 16? 2x + y 2x - y x + y 4x

Question 10-Explanation:

We know that log(base a) 2 = x and log(base a) 8 = y.
We need to find the value of log(base a) 16.
Let's express 16 as a power of 2: {16 = 2^4}
Now, we can rewrite log(base a) 16 using the properties of logarithms:
log(base a) 16 = log(base a) (2^4)
Using the power rule of logarithms, we can bring the exponent out front:
log(base a) 16 = 4 * log(base a) 2
Substituting the given values, we have: log(base a) 16 = 4x