The trigonometric equation $\sin^{- 1} x = 2 \sin^{- 1} (a)$ has a solution for

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$| a | \leq \frac{1}{\sqrt{2}}$

$| a | < \frac{1}{2}$

all real values of a

$| a | \geq \frac{1}{\sqrt{2}}$

answer is A.

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Detailed Solution

$\frac{- π}{2} \leq 2 \sin^{- 1} (a) \leq \frac{π}{2} \Rightarrow \frac{- π}{4} \leq \sin^{- 1} (a) \leq \frac{π}{4} \Rightarrow \frac{- 1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}} \Rightarrow |a| \leq \frac{1}{\sqrt{2}}$

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