Q. Suppose $f:\lbrack - 2,\ 2\rbrack \rightarrow R$ is defined by $f(x) = \left\{ \begin{array}{r} - 1,\ for - 2 \leq x \leq 0 \\ x - 1\ for\ 0 \leq x \leq 2 \end{array} \right.\$, then $\{ x \in \lbrack - 2,\ 2\rbrack:x \leq 0\ and\ f(|x|) = x\}$ is equal to
A) $\{ - 1\}$
B) {0}
C) $\left\{ - \frac{1}{2} \right\}$
D) $\phi$
Explanation
(c)
Since, $x \in \lbrack - 2,\ 2\rbrack,\ \ \ \ x \leq 0$ and $f\left( |x| \right) = x$
For $- 2 \leq x \leq 0$
$f( - x) = x\ \ \Rightarrow \ \ \ \ \leq ( - x) - 1 = x\ \ \ \Rightarrow \ \ \ \ x = - \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.