MCQ (Previous Year Question) - Relations And Functions (Competition Level 2) - JEE MCQ

(a) Let g(x) = 1 + x – [x] & f(x) = . Then for all x, f(g(x)) is equal to [JEE 2001 (Scr.), each 1 mark]

(A) x (B) 1 (C) f(x) (D) g(x)

where [ * ] denotes the greatest integer function.

(b) If f : [1, ¥) → [2, ¥) is given by, f(x) = , then f^_1(x) equals.

(A)  (B)  (C)  (D) 1–

(c) The domain of definition of f(x) =  is

(A)  (B) (–2, ¥) (C) R {–1, –2, –3} (D) (–3, ¥) {–1, –2}

(d) Let E = {1, 2, 3, 4} & F = {1, 2}. Then the number of onto functions from E to F is

(A) 14 (B) 16 (C) 12 (D) 8

(e) Let f(x) = , x ¹ – 1. Then for what value of a is f(f(x)) = x ?

(A)  (B)  (C) 1 (D) – 1

(a) Suppose f(x) = (x + 1)² for x ³ – 1. If g(x) is the function whose graph is the reflection of the graph of f(x) with respect to the line y = x, then g(x) equals            [JEE. 2002 (Scr.), 3 + 3]

(A)  – 1, x ³ 0 (B)  (C)  (D) 

(b) Let function f : R → R be defined by f(x) = 2x + sinx for x Î R. Then f is

(A) one to one and onto (B) one to one but NOT onto

(C) onto but NOT one to one (D) neither one to one nor onto

 (a) Let f(x) =  defined from (0, ¥) → [0, ¥) then by f(x) is [JEE. 2003 (Scr.), 2+2]

(A) one - one but not onto (B) one- one and onto (C) Many one but not onto (D) Many one and onto

(b) Range of the function f(x) =  is

(A) [1, 2] (B) [1, ¥) (C)  (D) 

 Let f(x) = sinx + cosx, g (x) = x² – 1. Thus g(f(x)) is invertible for x Î       [JEE 2004 (Scr.), 1]

If the functions f(x) & g(x) are defined on R ® R such that then (f –g) (x) is                    [JEE 2005 (Scr.), 1]

Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to    [JEE 2010]

Let f(x) = x² and g(x) = sin x for all x Î R. Then the set of all x satisfying (f o g o g o f) (x) = g(g o g of) (x),

where (f o g) (x) = f(g(x)), is        [JEE 2011]

The function f : [0, 3] → [1, 29], defined by f(x) = 2x³ – 15x² + 36x + 1, is        [JEE 2012]

 Let f : (–1, 1)  →IR be such that f(cos 4q) =  for q Π. Then the value(s) for f is (are)