$The domain of the function$ $f (x) = \sqrt{x - \sqrt{1 - x^{2}}} is$

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$(- \infty, - \frac{1}{2}] \cup [\frac{1}{\sqrt{2}}, + \infty)$

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

$[- 1, 1]$

$[- 1, - \frac{1}{\sqrt{2}}] \cup [\frac{1}{\sqrt{2}}, 1]$

answer is D.

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Detailed Solution

$\begin{array}{l} For f (x) to be defined, we must have \\ x - \sqrt{1 - x^{2}} \geq 0 or x \geq \sqrt{1 - x^{2}} > 0 \\ ∴ x^{2} \geq 1 - x^{2} or x^{2} \geq \frac{1}{2} \\ Also, 1 - x^{2} \geq 0 or x^{2} \leq 1 \\ Now, x^{2} \geq \frac{1}{2} \Rightarrow (x - \frac{1}{\sqrt{2}}) (x + \frac{1}{\sqrt{2}}) \geq 0 \\ \Rightarrow x \leq - \frac{1}{\sqrt{2}} or x \geq \frac{1}{\sqrt{2}} \\ Also, x^{2} \leq 1 \Rightarrow (x - 1) (x + 1) \leq 0 \\ Thus, x > 0, x^{2} \geq \frac{1}{2} and x^{2} \leq 1 \\ \Rightarrow x \in [\frac{1}{\sqrt{2}}, 1] \end{array}$