Mathematics - Functions Question with Solution | TestHub

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 $x^2\le 1$ $\Rightarrow x\in \left[\frac{1}{\sqrt{2}},1\right]$