Question

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

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

Answer 1

Answer: (D)

For $f(x)$ to be defined, we must have
$x-\sqrt{1-x^2}\ge 0$ or $x\ge \sqrt{1-x^2}>0$
$\therefore x^2\ge 1-x^2$ or $x^2\ge \frac{1}{2}$
Also, $1-x^2\ge 0$ or $x^2\le 1$
Now, $x^2\ge \frac{1}{2}\Rightarrow \left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right)\ge 0$
$\Rightarrow x\le -\frac{1}{\sqrt{2}}$ or $x\ge \frac{1}{\sqrt{2}}$
Also, $x^2\le 1\Rightarrow (x-1)(x+1)\le 0$
$\Rightarrow -1\le x\le 1$
Thus, $x>0,x^2\ge \frac{1}{2}$ and $x^2\le 1$
$\Rightarrow x\in \left[\frac{1}{\sqrt{2}},1\right]$