If f : [−2, 2]
Question:

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) $\phi$

Solution:

(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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