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Question
If f : [−2, 2] → R is defined by \[f\left( x \right) = \begin{cases}- 1, & \text{ for } - 2 \leq x \leq 0 \\ x - 1, & \text{ for } 0 \leq x \leq 2\end{cases}\] , then
{x ∈ [−2, 2] : x ≤ 0 and f (|x|) = x} =
Options
(a) {−1}
(b) {0}
(c) \[\left\{ - \frac{1}{2} \right\}\]
(d) ϕ
Solution
(c) \[\left\{ - \frac{1}{2} \right\}\]
Given:
\[f\left( x \right) = \begin{cases}- 1, & \text { for } - 2 \leq x \leq 0 \\ x - 1, &\text{ for } 0 \leq x \leq 2\end{cases}\]We know, \[\left| x \right| \geq 0\]
If \[x \leq 0\] , then \[\left| x \right| = - x\] ...(2)
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