JEE Main · 2009 · 12 Apr 2009 · Offline · Sets Relations and Functions

Let

f (x) = (x + 1)^{2} - 1, x \geq - 1

Statement-1: The set

{x : f (x) = f^{- 1} (x)} = {0, - 1}

Statement-2:

f

is a bijection.

(A)Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(B)Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
(C)Statement-1 is true, Statement-2 is false✓ Correct
(D)Statement-1 is false, Statement-2 is true

Answer: (C)

Solution

No codomain is specified for

f (x) = (x + 1)^{2} - 1

, so it cannot be guaranteed onto and need not be a bijection, which makes the reason statement false. For the first statement, since

f

is increasing on

x \geq - 1

any solution of

f (x) = f^{- 1} (x)

must satisfy

f (x) = x

; this gives

x^{2} + 2 x = x

, that is

x^{2} + x = 0

, whose roots

0

and

- 1

both lie in the domain, so the solution set is indeed

{0, - 1}

and the first statement is true.

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option