The domain of the function f(x)=x−1−x2 is

x−1−x2≥0⇒x≥1−x2>0 x2≥1−x2 1−x2≥0x2≥12 x2−1≤0x2−12≥0 (x+1)(x−1)≤0x2−12≥0 −1≤x≤1x+12 x−12≥0 x≤−12 or x≥12∴Req Domain is x∈12,1

The domain of the function f(x)=x−1−x2 is

x−1−x2≥0⇒x≥1−x2>0 x2≥1−x2 1−x2≥0x2≥12 x2−1≤0x2−12≥0 (x+1)(x−1)≤0x2−12≥0 −1≤x≤1x+12 x−12≥0 x≤−12 or x≥12∴Req Domain is x∈12,1

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

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

$[- 1, 1]$

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

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

answer is D.

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Detailed Solution

$x - \sqrt{1 - x^{2}} \geq 0 \Rightarrow x \geq \sqrt{1 - x^{2}} > 0 \begin{array}{ll} x^{2} \geq 1 - x^{2} 1 - x^{2} \geq 0 \\ x^{2} \geq \frac{1}{2} x^{2} - 1 \leq 0 \\ x^{2} - \frac{1}{2} \geq 0 (x + 1) (x - 1) \leq 0 \\ x^{2} - \frac{1}{2} \geq 0 - 1 \leq x \leq 1 \\ (x + \frac{1}{\sqrt{2}}) (x - \frac{1}{\sqrt{2}}) \geq 0 \end{array}$
$x \leq \frac{- 1}{\sqrt{2}} o r x \geq \frac{1}{\sqrt{2}}$
$∴$ Req Domain is $x \in [\frac{1}{\sqrt{2}}, 1]$