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

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

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

For All real value of a

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

answer is A.

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Detailed Solution

$- \frac{π}{2} \leq \sin^{- 1} x \leq \frac{π}{2} \Rightarrow 2 \sin^{- 1} a \leq \frac{π}{2} \Rightarrow \sin^{- 1} a \leq \frac{π}{4} ∴ a \leq \frac{1}{\sqrt{2}}$

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