Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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) $\sqrt{\left( 2-x \right)}$

Last updated date: 21st Jul 2024
Total views: 349.5k
Views today: 6.49k
Verified
349.5k+ views
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 ${{f}^{-1}}\left( x \right)$ .

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