Question

If $f\left(x\right)={\left(x+1\right)}^2-1,x\ge -1$. Then,

Statement-I: The set $\left\{x:f\left(x\right)=f^{-1}\left(x\right)\right\}=\left\{0,-1\right\}$.

Statement-II: $f$ is a bijection.

• A. Statement-I is correct and Statement-II is correct; Statement-II is a correct explanation for Statement-I.
• B. Statement-I is correct and Statement-II is correct; Statement-II is not a correct explanation for Statement-I.
• C. Statement-I is correct and Statement-II is incorrect;
• D. Statement-I is incorrect and Statement-II is incorrect;

Answer 1

The correct option is C

Statement-I is correct and Statement-II is incorrect;

Explanation for the correct option:

Step 1: Verify the statement-I

The given function is $f\left(x\right)={\left(x+1\right)}^2-1,x\ge -1$.

To determine the inverse of the given function, let us assume that $y={\left(x+1\right)}^2-1$.

Now, switch the places of $x$ and $y$.

$$\begin{gathered} x={\left(y+1\right)}^2-1 \\ \Rightarrow {\left(y+1\right)}^2=x+1 \\ \Rightarrow \left(y+1\right)=\sqrt{x+1} \\ \Rightarrow y=\sqrt{x+1}-1 \end{gathered}$$

Thus, $f^{-1}\left(x\right)=\sqrt{x+1}-1$.

So, for $f\left(x\right)=f^{-1}\left(x\right)$.

$$\begin{gathered} \Rightarrow {\left(x+1\right)}^2-1=\sqrt{x+1}-1 \\ \Rightarrow {\left(x+1\right)}^2=\sqrt{x+1} \\ \Rightarrow {\left[{\left(x+1\right)}^2\right]}^2={\left[\sqrt{x+1}\right]}^2 \\ \Rightarrow {\left(x+1\right)}^4=x+1 \\ \Rightarrow {\left(x+1\right)}^4-\left(x+1\right)=0 \\ \Rightarrow \left(x+1\right)\left[{\left(x+1\right)}^3-1\right]=0 \end{gathered}$$

Thus, either $\left(x+1\right)=0$ or $\left[{\left(x+1\right)}^3-1\right]=0$.

Thus, either $x=-1$ or ${\left(x+1\right)}^3=1$.

Hence, $x=-1,0$.

Therefore, it is true that the set $\left\{x:f\left(x\right)=f^{-1}\left(x\right)\right\}=\left\{0,-1\right\}$.

Step 2: Verify the statement-II

The given function is $f\left(x\right)={\left(x+1\right)}^2-1,x\ge -1$.

$$\begin{gathered} x\ge -1 \\ \Rightarrow {\left(x+1\right)}^2\ge 0 \\ \Rightarrow {\left(x+1\right)}^2-1\ge -1 \\ \Rightarrow f\left(x\right)\ge -1 \end{gathered}$$

Thus, the Range of the function is $[-1,\infty )$.

As the co-domain of the function is not given, then it can be considered as $\mathrm{ℝ}$.

So, the range of the function is not equal to the co-domain of the function.

Thus the function is not an onto function.

Therefore, the function is not a bijection.

So, Statement-I is correct and Statement-II is incorrect;

Hence, option(C) is the correct option.