$T h e d o m a i n o f f (x) = \sin^{- 1} [2 - 4 x^{2}] (w h e r e [.] i s G . I . F) i s$

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$[\frac{- \sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$

$[\frac{- 3}{2}, \frac{3}{2}]$

$[\frac{- \sqrt{3}}{2}, 0) \cup (0, \frac{\sqrt{3}}{2}]$

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answer is C.

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Detailed Solution

$\begin{array}{l} f (x) = \sin^{- 1} [2 - 4 x^{2}] \\ W e k n o w t h a t D o m a i n o f \sin^{- 1} x i s - 1 \leq x \leq 1 \\ ∴ - 1 \leq [2 - 4 x^{2}] \leq 1 \\ \Rightarrow - 1 \leq 2 - 4 x^{2} < 2 \\ l e t 2 - 4 x^{2} \geq - 1 \\ \Rightarrow 3 \geq 4 x^{2} \\ \Rightarrow {(2 x)}^{2} - {(\sqrt{3})}^{2} < 0 \\ \Rightarrow (2 x + \sqrt{3}) (2 x - \sqrt{3}) < 0 \end{array}$

$\begin{array}{l} x \in [\frac{- \sqrt{3}}{2}, \frac{\sqrt{3}}{2}] \\ l e t 2 - 4 x^{2} < 2 \\ \Rightarrow - 4 x^{2} < 0 \\ \Rightarrow x^{2} > 0 i s t r u e \forall x \in R e x c e p t x = 0 \\ D o m a i n = [\frac{- \sqrt{3}}{2}, 0) \cup (0, \frac{\sqrt{3}}{2}] \end{array}$