Question

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

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

Answer 1

The correct option is B $1-\sqrt{1-x}$

We have, $f\left(x\right)=x(2-x)$

$y=x(2-x)$

$-y=x(x-2)$

$-y=x^2-2x$

$1-y=x^2-2x+1$

$1-y=(x-1{)}^2$

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

Now $f:(-\infty ,1]\to (-\infty ,1]$

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

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

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