Question

Let $f\left(x\right)$ be a real valued function, then the domain of $f\left(x\right)=\sqrt{x-\sqrt{1-x^2}}$ is

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

Answer 1

The correct option is D $[\frac{1}{\sqrt{2}},1]$

For the function $f\left(x\right)=\sqrt{x-\sqrt{1-x^2}}$ to be a real valued function,
$x-\sqrt{1-x^2}\ge 0,1-x^2\ge 0x\in [-1,1]...\left(1\right)\Rightarrow x\ge \sqrt{1-x^2},x\in [-1,1]\Rightarrow x\ge 0(\because x\ge \sqrt{1-x^2}\ge 0)...\left(2\right)\Rightarrow x^2\ge 1-x^2,x\in [-1,1]\Rightarrow x^2\ge \frac{1}{2},x\in [-1,1]\Rightarrow x\in (-\infty ,-\frac{1}{\sqrt{2}}]\cup [\frac{1}{\sqrt{2}},\infty )...\left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$
$x\in [\frac{1}{\sqrt{2}},1]$