Question
The trigonometric equation $${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$$ has a solution for
A.
$$\left| a \right| \geqslant \frac{1}{{\sqrt 2 }}$$
B.
$$\frac{1}{2} < \left| a \right| < \frac{1}{{\sqrt 2 }}$$
C.
$$\left| a \right| < \frac{1}{2}$$
D.
None of these
Answer :
None of these
Solution :
$$\eqalign{
& {\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a \cr
& - \frac{\pi }{2} \leqslant {\sin ^{ - 1}}x \leqslant \frac{\pi }{2}; \cr
& \therefore \,\, - \frac{\pi }{2} \leqslant 2{\sin ^{ - 1}}a \leqslant \frac{\pi }{2} \cr
& - \frac{\pi }{4} \leqslant {\sin ^{ - 1}}a \leqslant \frac{\pi }{4}\,\,{\text{or }}\frac{{ - 1}}{{\sqrt 2 }} \leqslant a \leqslant \frac{1}{{\sqrt 2 }} \cr
& \therefore \,\,\left| a \right| \leqslant \frac{1}{{\sqrt 2 }} \cr} $$