If $f:[-2,2] \rightarrow \mathrm{R}$ is defined by $f(x)=\left\{\begin{aligned}-1, & \text { for }-2 \leq x \leq 0 \\ x-1, & \text { for } 0 \leq x \leq 2 \end{aligned}\right.$, then
$\{x \in[-2,2]: x \leq 0$ and $f(|x|)=x\}=$
(a) {−1}
(b) {0}
(c) $\left\{-\frac{1}{2}\right\}$
(d) $\phi$
(c) $\left\{-\frac{1}{2}\right\}$
Given:
$f(x)=\left\{\begin{aligned}-1, & \text { for }-2 \leq x \leq 0 \\ x-1, & \text { for } 0 \leq x \leq 2 \end{aligned}\right.$
We know,
$|x| \geq 0$
$\Rightarrow f(|x|)=|x|-1$ ...(1)
Also,
If $x \leq 0$, then $|x|=-x$
∴ {x ∈ [−2, 2]: x ≤ 0 and f (|x|) = x}
$=\{x:|x|-1=x\} \quad[$ Using $(1)]$
$=\{x:-x-1=x\} \quad[\operatorname{Using}(2)]$
$=\left\{x: 2 x=\frac{-1}{2}\right\}$
$=\left\{x: x=\frac{-1}{2}\right\}$
$=\left\{\frac{-1}{2}\right\}$
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