Question

Let $A=x\in R:x\le 1$ and $f:A\to A$ be defined as $f\left(x\right)=x(2-x)$, then $f^{-1}\left(x\right)$ is

• A. $1+\sqrt{1-x}$
• B. $1-\sqrt{1-x}$
• C. $\sqrt{1-x}$
• D. $1\pm \sqrt{1-x}$

Answer 1

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

Solution -

$A=x\in R:x\le 1$

$f:A\to A$

that means

$x\le 1$ & $y\le 1$

and $f^{-1}\left(x\right)$

also $x\le 1$ & $y\le 1$

$f\left(x\right)=x(2-x)$

$y=x(2-x)$

$x^2+y-2x=0$

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

Put $x=y,y=x$ to get $y=f^{-i}\left(x\right)$

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

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

If $y\le 1$ in $f^{-1}\left(x\right)$ then

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

B is correct.