Question

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

• A. $[-1,-\frac{1}{\sqrt{2}}]\cup [\frac{1}{\sqrt{2}},1]$
• B. $[-1,1]$
• C. $(-\infty ,-\frac{1}{2}]\cup [\frac{1}{\sqrt{2}},+\infty )$
• D. $[\frac{1}{\sqrt{2}},1]$

Answer 1

The correct option is A $[-1,-\frac{1}{\sqrt{2}}]\cup [\frac{1}{\sqrt{2}},1]$

$f\left(x\right)=\sqrt{x-\sqrt{1-x^2}}$

$x-\sqrt{1-x^2}\ge 0$ and $1-x^2\ge 0$

$\Rightarrow x\ge \sqrt{1-x^2}$ and $-1\le x\le 1$ ... (i)

Now, $x^2\ge (\sqrt{1-x^2}{)}^2$, for positive $x$

$x\le -\frac{1}{\sqrt{2}}$ or $x\ge \frac{1}{\sqrt{2}}$ ... (ii)

From (i) and (ii), we get

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

Hence, option 'D' is correct.