Question

Let $f\left(x\right)=(x+1{)}^2-1,x\ge -1$
Statement-1: The set $\{x:f(x)=f^{-1}(x\left)\right\}$ $=\{0,-1\}$.
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.
• D. Statement 1 is false, Statement 2 is true

Answer 1

The correct option is A Statement 1 is true, Statement 2 is true,Statement 2 is a correct explanation for statement 1

The co-domain of the given function is $[-1,\infty )$
Now, Let $y=(x+1{)}^2-1$
$\Rightarrow x=\sqrt{y+1}-1$
$\therefore f^{-1}\left(x\right)=\sqrt{y+1}-1$, which exists $\Rightarrow$ domain$=$ co-domain
Therefore, the function $f$ is one-one onto or bijection.
Hence, option 'A' is correct.