Question:
Let $A=\{x \in R: x \leq 1\}$ and $f: A \rightarrow A$ be defined as $f(x)=x(2-x)$. Then, $f^{-1}(x)$ is
(a) $1+\sqrt{1-x}$
(b) $1-\sqrt{1-x}$
(c) $\sqrt{1-x}$
(d) $1 \pm \sqrt{1-x}$
Solution:
Let y be the element in the codomain R such that
$f^{-1}(x)=y \ldots(1)$
$\Rightarrow f(y)=x$ and $\mathrm{y} \leq 1$
$\Rightarrow y(2-y)=x$
$\Rightarrow 2 y-y^{2}=x$
$\Rightarrow y^{2}-2 y+x=0$
$\Rightarrow y^{2}-2 y=-x$
$\Rightarrow y^{2}-2 y+1=1-x$
$\Rightarrow(y-1)^{2}=1-x$
$\Rightarrow y-1=\pm \sqrt{1-x}$
$\Rightarrow y=1 \pm \sqrt{1-x}$
$\Rightarrow y=1-\sqrt{1-x} \quad(\because \mathrm{y} \leq 1)$
The correct answer is (b).
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