The range of ${sin}^{- 1} (2^{x})$ is

(\frac{- π}{2}, \frac{π}{2}]

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(0, \frac{π}{2}]

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(0, π)

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(0, π]

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Solution

The correct option is B $(0, \frac{π}{2}]$
$We know that 2^{x} > 0 {sin}^{- 1} (2^{x}) in order to be defined, we should have 2^{x} \leq 1 . Therefore, 0 < 2^{x} \leq 1 \Rightarrow 0 < {sin}^{- 1} (2^{x}) \leq \frac{π}{2}$

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