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} =

 

विकल्प

• (a) {−1}

• (b) {0}

• (c) \[\left\{ - \frac{1}{2} \right\}\]

• (d) ϕ

   

Answer 1

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

Given:

\[f\left( x \right) = \begin{cases}- 1, &  \text { for } - 2 \leq x \leq 0 \\ x - 1, &\text{  for }  0 \leq x \leq 2\end{cases}\]We know,  \[\left| x \right| \geq 0\]

⇒  \[f\left( \left| x \right| \right) = \left| x \right| - 1\]      ...(1)

Also,
If  \[x \leq 0\] , then \[\left| x \right| = - x\]    ...(2)

 

∴ {x ∈ [−2, 2]: x ≤ 0 and f (|x|) = x} 

=\[\left\{ x: \left| x \right| - 1 = x \right\} [\text{ Using } (1)]\]

=\[\left\{ x: - x - 1 = x \right\} [\text{ Using }  (2)]\]

\[\left\{ x: 2x = \frac{- 1}{2} \right\}\]

 =\[\left\{ x: x = \frac{- 1}{2} \right\}\]

=\[\left\{ \frac{- 1}{2} \right\}\]