The trigonometric equation ${sin}^{- 1} x = 2 {sin}^{- 1} 2 a$ has a real solution, if

| a | > \frac{1}{\sqrt{2}}

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

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| a | > \frac{1}{2 \sqrt{2}}

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

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Solution

The correct option is D $| a | \leq \frac{1}{2 \sqrt{2}}$
We have that
$\frac{- π}{2} \leq {sin}^{- 1} x \leq \frac{π}{2}$

$\Rightarrow \frac{- π}{2} \leq 2 {sin}^{- 1} 2 a \leq \frac{π}{2}$

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

$\Rightarrow sin (\frac{- π}{4}) \leq 2 a \leq sin (\frac{π}{4})$

$\Rightarrow \frac{- 1}{\sqrt{2}} \leq 2 a \leq \frac{1}{\sqrt{2}}$

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

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