Question
Domain of definition of the function $$f\left( x \right) = \sqrt {{{\sin }^{ - 1}}\left( {2x} \right) + \frac{\pi }{6}} $$ for real valued $$x,$$ is
A.
$$\left[ { - \frac{1}{4},\frac{1}{2}} \right]$$
B.
$$\left[ { - \frac{1}{2},\frac{1}{2}} \right]$$
C.
$$\left( { - \frac{1}{2},\frac{1}{2}} \right)$$
D.
$$\left[ { - \frac{1}{4},\frac{1}{4}} \right]$$
Answer :
$$\left[ { - \frac{1}{4},\frac{1}{2}} \right]$$
Solution :
For $$f\left( x \right) = \sqrt {{{\sin }^{ - 1}}\left( {2x} \right) + \frac{\pi }{6}} $$ to be defined and real if $$si{n^{ - 1}}2x + \frac{\pi }{6} \geqslant 0$$
$$ \Rightarrow {\sin ^{ - 1}}2x \geqslant - \frac{\pi }{6}\,......\left( 1 \right)$$
But we know that
$$ - \frac{\pi }{2} \leqslant {\sin ^{ - 1}}2x \leqslant \frac{\pi }{2}\,......\left( 2 \right)$$
Combining (1) and (2), we get
$$\eqalign{
& - \frac{\pi }{6} \leqslant {\sin ^{ - 1}}2x \leqslant \frac{\pi }{2} \cr
& \Rightarrow \sin \left( { - \frac{\pi }{6}} \right) \leqslant 2x \leqslant \sin \left( {\frac{\pi }{2}} \right) \Rightarrow - \frac{1}{2} \leqslant 2x \leqslant 1 \cr
& \Rightarrow - \frac{1}{4} \leqslant x \leqslant \frac{1}{2}\therefore {D_f} = \left[ { - \frac{1}{4},\frac{1}{2}} \right] \cr} $$