Question
The domain of the function $$f\left( x \right) = {\sin ^{ - 1}}\left\{ {{{\log }_2}\left( {\frac{1}{2}{x^2}} \right)} \right\}{\text{ is}}$$
A.
$$\left[ { - 2, - 1} \right) \cup \left[ {1,2} \right]$$
B.
$$\left( { - 2, - 1} \right] \cup \left[ {1,2} \right]$$
C.
$$\left[ { - 2, - 1} \right] \cup \left[ {1,2} \right]$$
D.
$$\left( { - 2, - 1} \right) \cup \left( {1,2} \right)$$
Answer :
$$\left[ { - 2, - 1} \right] \cup \left[ {1,2} \right]$$
Solution :
For $$f\left( x \right)$$ to be defined, we must have
$$\eqalign{
& - 1 \leqslant {\log _2}\left( {\frac{1}{2}{x^2}} \right) \leqslant 1 \cr
& \Rightarrow {2^{ - 1}} \leqslant \frac{1}{2}{x^2} \leqslant {2^1} \cr
& \Rightarrow 1 \leqslant {x^2} \leqslant 4\,\,\,.....\left( 1 \right) \cr
& {\text{Now}},1 \leqslant {x^2} \cr
& \Rightarrow {x^2} - 1 \geqslant 0{\text{ i}}{\text{.e}}{\text{. }}\left( {x - 1} \right)\left( {x + 1} \right) \geqslant 0 \cr
& \Rightarrow x \leqslant - 1{\text{ or }}x \geqslant 1\,\,\,.....\left( 2 \right) \cr
& {\text{Also}},\,\,{x^2} \leqslant 4 \cr
& \Rightarrow {x^2} - 4 \leqslant 0{\text{ i}}{\text{.e}}{\text{. }}\left( {x - 2} \right)\left( {x + 2} \right) \leqslant 0 \cr
& \Rightarrow - 2 \leqslant x \leqslant 2\,\,\,\,\,\,.....\left( 3 \right) \cr} $$
Form (2) and (3), we get the domain of
$$\eqalign{
& f = \left\{ {\left( { - \infty , - 1} \right] \cup \left[ {1,\infty } \right)} \right\} \cap \left[ { - 2,2} \right] \cr
& = \left[ { - 2, - 1} \right] \cup \left[ {1,2} \right] \cr} $$