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
$x-\sqrt{1-x^{2}} \geq 0$ or $x \geq \sqrt{1-x^{2}}>0$
$\therefore x^{2} \geq 1-x^{2}$ or $x^{2} \geq \frac{1}{2}$
Also, $1-x^{2} \geq 0$ or $x^{2} \leq 1$
Now, $x^{2} \geq \frac{1}{2} \Rightarrow \left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right) \geq 0$
$\Rightarrow x \leq-\frac{1}{\sqrt{2}}$ or $x \geq \frac{1}{\sqrt{2}}$
Also, $x^{2} \leq 1 \Rightarrow (x-1)(x+1) \leq 0$
$\Rightarrow -1 \leq x \leq 1$
Thus, $x>0, x^{2} \geq \frac{1}{2}$ and $x^{2} \leq 1$
$\Rightarrow x \in\left[\frac{1}{\sqrt{2}}, 1\right]$