If $y = {sin}^{- 1} [2 x]$ , where $[x]$ is the greatest integer function, then $x$ belongs to

[\frac{1}{2}, \frac{1}{2}]

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[- \frac{1}{2}, 1]

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[- \frac{1}{2}, 0]

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[- \frac{1}{2}, 1)

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Solution

The correct option is D $[- \frac{1}{2}, 1)$
Given

$y = {sin}^{- 1} [2 x]$

The range of ${sin}^{- 1} x$ is $[- 1, 1]$

$⟹ - 1 \leq [2 x] \leq 1$

$⟹ - 1 \leq 2 x < 2$

$⟹ - \frac{1}{2} \leq x < 1$

$⟹ x \in [- \frac{1}{2}, 1)$

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