Question

Let $f(x)=(x+1)^2-1, x \geq-1$
Assertion (A) The set $\left\{x: f(x)=f^{-1}(x)\right\}=\{0,-1\}$.
Reason(R) $f$ is a bijection.

• A. Both A and R are correct, R is the correct explanation of A
• B. Both $A$ and $R$ are correct, $R$ is not the correct explanation of $A$
• C. $A$ is correct, $R$ is incorrect
• D. $R$ is correct, $A$ is incorrect

Answer 1

Answer: (C)

Given, $f(x)=(x+1)^2-1, x \geq-1$
$\Rightarrow f^{\prime}(x)=2(x+1) \geq 0$ for $x \geq-1$
$\Rightarrow f(x)$ is one-one.
Since, codomain of the given function is not given. Hence, it can be considered as $R$ the set of reals and consequently $f$ is not onto.
Hence, $f$ is not bijective.
Also, $f(x)=(x+1)^2-1 \geq-1$ for $x \geq-1$
$\Rightarrow R_f =[-1, \infty) \\ f^{-1}(x) =\sqrt{x+1}-1$
Clearly, $f(x)=f^{-1}(x)$ at $x=0$ and $x=-1$