The trigonometric equation ^ -1 x=2 ^ -1 a has a solution for

$\sin ^{-1} x=2 \sin ^{-1} a$ $-\frac{\pi}{2} \leq \sin ^{-1} \mathrm{x} \leq \frac{\pi}{2} ; \therefore-\frac{\pi}{2} \leq 2 \sin ^{-1} \mathrm{a} \leq \frac{\pi}{2} \quad-\frac{\pi}{4} \leq \sin ^{-1} \mathrm{a} \leq \frac{\pi}{4}$ or $\frac{-1}{\sqrt{2}} \leq \mathrm{a} \leq \frac{1}{\sqrt{2}}$ $\therefore|\mathrm{a}| \leq \frac{1}{\sqrt{2}}\left(\right.$ As $\frac{1}{\sqrt{2}}>\frac{1}{2}$ ). Out of given four option no one is absolutely correct but (c) could be taken into consideration. $\rightarrow|\mathrm{a}| \leq \frac{1}{\sqrt{2}}$ is correct, if $\mathrm{a} < \frac{1}{\sqrt{2}}$ is taken as correct then it domain satisfy for $\mathrm{a}=\frac{1}{\sqrt{3}}$ but equation is satisfied. $\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>\frac{1}{2}$