Explanation

(c)
Given, $\sin^{- 1}{x = 2\sin^{- 1}a}$
Since, $\ \ \ \ \ \ \ \ \ \ - \frac{\pi}{2} \leq \sin^{- 1}{x \leq \frac{\pi}{2}}$
$\Rightarrow \ \ \ \ \ \ \ - \frac{\pi}{2} \leq 2\sin^{- 1}{a \leq \frac{\pi}{2}}$
$\Rightarrow \ \ \ \ \ \ \ - \frac{\pi}{4} \leq \sin^{- 1}{a \leq \frac{\pi}{4}}$
$\Rightarrow \ \ \ \ \ \ \sin{\left( - \frac{\pi}{4} \right) \leq a \leq \sin\frac{\pi}{4}}$
$\Rightarrow \ \ \ \ \ \ \ - \frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a| \leq \frac{1}{\sqrt{2}}$