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Suppose $\quad f:[-2,2] \rightarrow R \quad$ is defined by
$f(x)=\left\{\begin{array}{ccc}-1, & \text { for } & -2 \leq x \leq 0 \\x-1, & \text { for } & 0 \leq x \leq 2 \\\{x \in[-2,2]: x \leq 0 & \text { and } & f(|x|)=x\end{array}\right.$
is equal tos
Options:
$f(x)=\left\{\begin{array}{ccc}-1, & \text { for } & -2 \leq x \leq 0 \\x-1, & \text { for } & 0 \leq x \leq 2 \\\{x \in[-2,2]: x \leq 0 & \text { and } & f(|x|)=x\end{array}\right.$
is equal tos
Solution:
1100 Upvotes
Verified Answer
The correct answer is:
$\left\{-\frac{1}{2}\right\}$
We have,
$f(x)=\left\{\begin{array}{c}-1, \text { for }-2 \leq x \leq 0 \\x-1, \text { for } 0 \leq x \leq 2\end{array}\right.$
Now, take $x=-\frac{1}{2}$
$\therefore \quad f\left(\left|-\frac{1}{2}\right|\right)=f\left(\frac{1}{2}\right)=\frac{1}{2}-1=-\frac{1}{2}$
Hence, $\quad f(|x|)=x$
$\therefore \quad$ Domain of $f(x)=\left\{-\frac{1}{2}\right\}$
$f(x)=\left\{\begin{array}{c}-1, \text { for }-2 \leq x \leq 0 \\x-1, \text { for } 0 \leq x \leq 2\end{array}\right.$
Now, take $x=-\frac{1}{2}$
$\therefore \quad f\left(\left|-\frac{1}{2}\right|\right)=f\left(\frac{1}{2}\right)=\frac{1}{2}-1=-\frac{1}{2}$
Hence, $\quad f(|x|)=x$
$\therefore \quad$ Domain of $f(x)=\left\{-\frac{1}{2}\right\}$
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