Question

Let $f:(–\infty ,1]\to (\infty ,1]$such that $f\left(x\right)=x\left(2-x\right)$. Then $f^{-1}\left(x\right)$ is

• A. $2+\sqrt{2-x}$
• B. $1-\sqrt{1-x}$
• C. $1+\sqrt{2-x}$
• D. None of these

Answer 1

The correct option is B

$1-\sqrt{1-x}$

Explanation for the correct option(s):

Determine the value of $f^{-1}\left(x\right)$:

Consider the given equation as: $f\left(x\right)=x\left(2-x\right)......\left(1\right)$

Let $f\left(x\right)=y$, substituting it in the Equation $\left(1\right)$

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

Adding one on both sides,

$$\begin{gathered} \Rightarrow 1-y=x(x-2)+1 \\ \Rightarrow 1-y=x^2-2x+1 \\ \Rightarrow 1-y={(x-1)}^2[\because {(x-1)}^2=x^2-2x+1] \\ \Rightarrow \pm \sqrt{1-y}=x-1 \\ \Rightarrow x=1\pm \sqrt{1-y} \end{gathered}$$

Since, We have, $f:(–\infty ,1]\to (\infty ,1]$if we take $x=1+\sqrt{1-y}$, the range is falsified, hence we consider,

$x=1-\sqrt{1-y}$

Therefore, $f^{-1}\left(x\right)=1-\sqrt{1-y}$

Hence, the correct answer is option B.