Relations and Functions (Class 12)

Let A = \-2, -1, 0, 1, 2, 3\. Let R be a relation on A defined by xRy if and only if y = \x, 1\. Let l be the number of elements in R. Let m and n be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then l + m + n is equal to

1. 11
2. 12
3. 14
4. 13

Let f : R R be a twice differentiable function such that ( x y)(f(2x + 2y) - f(2x - 2y)) = ( x y)(f(2x + 2y) + f(2x - 2y)), for all x, y R. If f'(0) = 12, then the value of 24f'' ( 5 3 ) is :

1. 2
2. 3
3. -3
4. -2

Let A be the set of all functions f : Z Z and R be a relation on A such that R = \(f, g) : f(0) = g(1) and f(1) = g(0)\. Then R is :

1. Symmetric and transitive but not reflexive
2. Symmetric but neither reflexive nor transitive
3. Transitive but neither reflexive nor symmetric
4. Reflexive but neither symmetric nor transitive

Define a relation R on the interval [0, 2 ) by x R y if and only if 2 x - 2 y = 1. Then R is :

1. both reflexive and symmetric but not transitive
2. both reflexive and transitive but not symmetric
3. reflexive but neither symmetric nor transitive
4. an equivalence relation

If f(x) = 2x2x + 2, x R, then k=181 f ( k82 ) is equal to

1. 82
2. 81 2
3. 41
4. 812

The relation R = \(x, y) : x, y Z and x + y is even\ is:

1. reflexive and transitive but not symmetric
2. reflexive and symmetric but not transitive
3. an equivalence relation
4. symmetric and transitive but not reflexive

Let f: R - \0\ (- , 1) be a polynomial of degree 2, satisfying f(x)f ( 1x ) = f(x) + f ( 1x ). If f(K) = -2K, then the sum of squares of all possible values of K is:

1. 9
2. 1
3. 6
4. 7

Let f : [0, 3] A be defined by f(x) = 2x3 - 15x2 + 36x + 7 and g : [0, ) B be defined by g(x) = x2025x2025 + 1. If both the functions are onto and S = \x Z; x A or x B\, then n(S) is equal to :

1. 29
2. 31
3. 30
4. 36

Let f:R-\0\ be a function such that f(x)-6f ( 1x )= 353x- 52. If x 0 ( 1 x+f(x) )= ; , , then +2 is equal to

1. 6
2. 5
3. 3
4. 4

Let f(x)= 2x+2+1622x+1+2x+4+32. Then the value of 8 (f ( 115 )+f ( 215 )+ +f ( 5915 ) ) is equal to

1. 108
2. 92
3. 118
4. 102