Let A={x ∈ R : x ≤1} and f : A→A be defined as f(x)
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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