Question

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

• 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

Answer: (D)

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

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