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Let f(x)=(x+1)^2-1, x ≥-1 Statement-1: The set x: f(x)=f^-1(x) =0,-1 Statement-2 : f is a bijection.

JEE Main 2009 — Mathematics Algebra

2009mcqmedium

Let f(x)=(x+1)2−1,x≥−1f(x)=(x+1)^2-1, x \geq-1f(x)=(x+1)2−1,x≥−1 Statement-1: The set {x:f(x)=f−1(x)}={0,−1}\left\{x: f(x)=f^{-1}(x)\right\}=\{0,-1\}{x:f(x)=f−1(x)}={0,−1} Statement-2 : f\mathrm{f}f is a bijection.

Official previous-year question

Held on 30 Apr 2009 · Verified 6 Jul 2026.

Options

  1. A

    Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

  2. B

    Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

  3. C

    Statement-1 is true, Statement-2 is false

  4. D

    Statement-1 is false, Statement-2 is true

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