Question

If f : [−2, 2] → R is defined by $f\left(x\right)=\left\{\begin{array}{ll}-1,& \mathrm{for}-2\le x\le 0\\ x-1,& \mathrm{for}0\le x\le 2\end{array}\right.$, then
{x ∈ [−2, 2] : x ≤ 0 and f (|x|) = x} =

(a) {−1}
(b) {0}
(c) $\left\{-\frac{1}{2}\right\}$
(d) ϕ

Answer 1

(c) $\left\{-\frac{1}{2}\right\}$

Given:
$f\left(x\right)=\left\{\begin{array}{ll}-1,& \mathrm{for}-2\le x\le 0\\ x-1,& \mathrm{for}0\le x\le 2\end{array}\right.$
We know,
$\left|x\right|\ge 0$
⇒ $f\left(\left|x\right|\right)=\left|x\right|-1$ ...(1)

Also,
If $x\le 0$, then $\left|x\right|=-x$ ...(2)

∴ {x ∈ [−2, 2]: x ≤ 0 and f (|x|) = x}
= $\left\{x:\left|x\right|-1=x\right\}[\mathrm{Using}(1\left)\right]$
= $\left\{x:-x-1=x\right\}[\mathrm{Using}(2\left)\right]$
= $\left\{x:2x=\frac{-1}{2}\right\}$
= $\left\{x:x=\frac{-1}{2}\right\}$
= $\left\{\frac{-1}{2}\right\}$