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:
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Previous year questions with solutions · 2024 – 2026
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 20 | 19 | 1.1 | 4.21% ▼ 23.1% vs last year |
| 2025 | 26 | 19 | 1.4 | 5.47% ▲ 17.3% vs last year |
| 2024 | 28 | 20 | 1.4 | 4.67% |
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:
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 :
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Let f(x)= 2x+2+1622x+1+2x+4+32. Then the value of 8 (f ( 115 )+f ( 215 )+ +f ( 5915 ) ) is equal to
The function f:(- , ) (- ,1), defined by f(x)= 2x-2-x2x+2-x is :
Let R=\(1,2),(2,3),(3,3)\ be a relation defined on the set \1,2,3,4\. Then the minimum number of elements, needed to be added in R so that R becomes an equivalence relation, is:
Let X=R . Define a relation R on X as : (a1,b1)R(a2,b2) b1=b2 In the light of the above statements, choose the correct answer from the options given below :
Let f(x) be a real differentiable function such that f(0)=1 and f(x+y)=f(x)f'(y)+f'(x)f(y) for all x,y . Then n=1100 ef(n) is equal to :
The number of non-empty equivalence relations on the set \1,2,3\ is :
Let A=\1,2,3,4\ and B=\1,4,9,16\. Then the number of many-one functions f:A B such that 1 f(A) is equal to :
Let [t] be the greatest integer less than or equal to t. Let A be the set of all prime factors of 2310 and f : A Z be the function f(x) = [ 2 (x2 + [ x35 ] ) ]. The number of one-to-one functions from A to the range of f is
Let f(x) = cases -a & if -a x 0 \ x + a & if 0 < x a cases where a > 0 and g(x) = (f(|x|) - |f(x)|)/2. Then the function g : [-a, a] [-a, a] is
Let A = \2, 3, 6, 8, 9, 11\ and B = \1, 4, 5, 10, 15\. Let R be a relation on A B defined by (a, b)R(c, d) if and only if 3ad - 7bc is an even integer. Then the relation R is
Let the relations R1 and R2 on the set X = \1, 2, 3, , 20\ be given by R1 = \(x, y) : 2x - 3y = 2\ and R2 = \(x, y) : -5x + 4y = 0\. If M and N be the minimum number of elements required to be added in R1 and R2, respectively, in order to make the relations symmetric, then M + N equals
The function f(x) = x2 + 2x - 15x2 - 4x + 9, x R is
Let A = \1, 2, 3, 4, 5\. Let R be a relation on A defined by xRy if and only if 4x 5y. Let m be the number of elements in R and n be the minimum number of elements from A A that are required to be added to R to make it a symmetric relation. Then m + n is equal to :
Let A = \1, 3, 7, 9, 11\ and B = \2, 4, 5, 7, 8, 10, 12\. Then the total number of one-one maps f : A B, such that f(1) + f(3) = 14, is :
Let f, g : R R be defined as : f(x) = |x - 1| and g(x) = cases ex, & x 0 \ x + 1, & x 0. cases Then the function f(g(x)) is
Let a relation R on N N be defined as: (x1, y1) R (x2, y2) if and only if x1 x2 or y1 y2. Consider the two statements: Then which one of the following is true?
Let f:R and g:R be defined as f(x)= cases e x, & x>0\ e-x, & x 0 cases and g(x)= casesx, & x 0\ ex, & x<0 cases. Then, gof :R is :
Consider the relations R1 and R2 defined as a R1 b a2+b2=1 for all a, b R and (a, b) R2(c, d) a+d=b+c for all (a, b),(c, d) N N. Then :
If f(x)= 4x+36x-4, x 23 and (f f)(x)=g(x), where g: R- \ 23 \ R- \ 23 \, then (g g g)(4) is equal to
Let f: R-\0\ R be a function satisfying f ( xy )= f(x)f(y) for all x,y,f(y) 0. If f'(1)=2024, then
Let R be a relation on Z Z defined by (a, b)R(c, d) if and only if ad-bc is divisible by 5. Then R is
If f(x)= cases2+2x, & -1 x<0\ 1- x3, & 0 x 3 cases; g(x)= cases-x, & -3 x 0\ x, & 0<x 1 cases, then range of (f g)(x) is
If R is the smallest equivalence relation on the set \1,2,3,4\ such that \(1,2),(1,3)\ , then the number of elements in R is .
Let S=\1,2,3, ,10\. Suppose M is the set of all the subsets of S, then the relation R=\(A, B):A B ; A, B M\ is :
The function f:N-\1\ ; defined by f(n)= the highest prime factor of n, is :
Let f:R- \ -12 \ and g:R- \ -52 \ be defined as f(x)= 2x+32x+1 and g(x)= |x|+12x+5. Then, the domain of the function fog is :