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

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In all value of $α$

$|α| \leq \frac{1}{\sqrt{2}}$

$|α| \geq \frac{1}{\sqrt{2}}$

$\frac{- 1}{2} |α| < \frac{1}{\sqrt{2}}$

answer is B.

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Detailed Solution

$\sin^{- 1} x = 2 \sin^{- 1} α -π/2≤ \sin^{- 1} x ≤π/2 -π/2≤ 2 \sin^{- 1} α ≤π/2 - π / 4 \leq \sin^{- 1} α \leq π / 4 \Rightarrow - \frac{1}{\sqrt{2}} \leq α \leq \frac{1}{\sqrt{2}}$

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