$The domain of f (x) = {sin}^{- 1} (2 x^{2} - 3], where [.] denotes the greatest integer function, is$

(- \sqrt{\frac{5}{2}}, - 1] \cup [1, \sqrt{\frac{5}{2}})

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(- \sqrt{\frac{3}{2}}, - 1] \cup (- \sqrt{\frac{5}{2}}, \sqrt{\frac{5}{2}})

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(- \sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})

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(- \sqrt{\frac{5}{2}}, \sqrt{\frac{5}{2}})

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Solution

The correct option is A $(- \sqrt{\frac{5}{2}}, - 1] \cup [1, \sqrt{\frac{5}{2}})$
$Given function, f (x) = {sin}^{- 1} (2 x^{2} - 3], where (.] denotes the greatest integer function . We must have - 1 \leq (2 x^{2} - 3] \leq 1 \Rightarrow - 1 \leq 2 x^{2} - 3 < 2 \Rightarrow 2 \leq 2 x^{2} < 5 \Rightarrow 1 \leq x^{2} < \frac{5}{2} or x \in (- \sqrt{\frac{5}{2}}, - 1] \cup [1, \sqrt{\frac{5}{2}})$

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