If the function f : [1, ¥) → [1, ¥) is defined by f(x) = 2x(x _ 1), then f_1(x) is [JEE 99, 2]
The domain of definition of the function, y(x) given by the equation, 2x + 2y = 2 is [JEE 2000(Scr.), 1]
Given X = {1, 2, 3, 4}, find all one–one, onto mappings, f : X → X such that, f(1) = 1, f(2) ¹ 2 and f(4) ¹ 4
(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 Î
. Then the value(s) for f
is (are)