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Find the inverse of the quadratic function $f\left( x \right)=-{{x}^{2}}+2,x\ge 0$
(a) $\sqrt{\left( 2+x \right)}$
(b) $\left( 2-x \right)$
(c) $\sqrt{{{\left( 2+x \right)}^{2}}}$
(d) $\sqrt{\left( 2-x \right)}$

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Last updated date: 21st Jul 2024
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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 ${{f}^{-1}}\left( x \right)$ .
$\Rightarrow {{f}^{-1}}\left( x \right)=\sqrt{2-x}$