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

\forall R

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

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

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

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Solution

The correct option is C $\frac{- 1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$
Given equation ${sin}^{- 1} x = 2 {sin}^{- 1} a$

$\Rightarrow - \frac{π}{2} \leq 2 {sin}^{- 1} a \leq \frac{π}{2} {∵ {sin}^{- 1} x = 2 {sin}^{- 1} a a n d - \frac{π}{2} \leq 2 {sin}^{- 1} x \leq \frac{π}{2}}$
$\Rightarrow - \frac{π}{4} \leq {sin}^{- 1} a \leq \frac{π}{4}$
$\Rightarrow sin (- \frac{π}{4}) \leq a \leq sin (\frac{π}{4})$
$\Rightarrow - \frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}}$

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Similar questions

{s i n}^{- 1} x = 3 {s i n}^{- 1} (a)

- \frac{1}{2} \leq a \leq \frac{1}{2}

\forall x \in [- 1, 1], {s i n}^{- 1} x \in [0, 2 π]

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s i n^{- 1} x = 2 s i n^{- 1}

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