Answer

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**Hint:**To find the inverse of $f\left( x \right)=-{{x}^{2}}+2$ , we have to replace $f\left( x \right)$ with y. Then, we have to solve for x. We will ignore the negative value since $x\ge 0$ . Now, we have to replace x with y. Finally, we have to replace y with ${{f}^{-1}}\left( x \right)$ .

**Complete step by step answer:**

We have to find the inverse of the quadratic function $f\left( x \right)=-{{x}^{2}}+2$ . Firstly, we have to replace $f\left( x \right)$ with y.

$\Rightarrow y=-{{x}^{2}}+2$

Let us solve for x. We have to take 2 to the LHS.

$\Rightarrow y-2=-{{x}^{2}}$

Now, we have to take the negative sign to the LHS.

$\begin{align}

& \Rightarrow -\left( y-2 \right)={{x}^{2}} \\

& \Rightarrow -y+2={{x}^{2}} \\

& \Rightarrow {{x}^{2}}=2-y \\

\end{align}$

Let us take square roots on both sides.

$\begin{align}

& \Rightarrow x=\pm \sqrt{2-y} \\

& \Rightarrow x=\sqrt{2-y},-\sqrt{2-y} \\

\end{align}$

We are given that $x\ge 0$ . Therefore, we will ignore the negative value.

$\Rightarrow x=\sqrt{2-y}$

Now, we have to replace x with y.

$\Rightarrow y=\sqrt{2-x}$

We have to replace y with ${{f}^{-1}}\left( x \right)$ .

$\Rightarrow {{f}^{-1}}\left( x \right)=\sqrt{2-x}$

Therefore, the inverse of $f\left( x \right)=-{{x}^{2}}+2$ is $\sqrt{2-x}$ .

**So, the correct answer is “Option d”.**

**Note:**Students must be thorough in solving algebraic equations and the rules involved in it. They should never miss to solve for x in step 2. If so, they have to solve for y in the second last step. After step 1, we will obtain

$\Rightarrow y=-{{x}^{2}}+2$

Then, we have to replace x with y.

$\Rightarrow x=-{{y}^{2}}+2,y\ge 0$

Now, we have to solve for y.

$\begin{align}

& \Rightarrow x-2=-{{y}^{2}} \\

& \Rightarrow -x+2={{y}^{2}} \\

& \Rightarrow {{y}^{2}}=2-x \\

\end{align}$

Let us take square roots on both sides.

$\Rightarrow y=\pm \sqrt{2-x}$

We have to ignore the negative value since $y\ge 0$ .

$\Rightarrow y=\sqrt{2-x}$

We have to replace y with ${{f}^{-1}}\left( x \right)$ .

$\Rightarrow {{f}^{-1}}\left( x \right)=\sqrt{2-x}$

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