Question

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

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

Answer 1

The correct option is A

$\left[-1,-\frac{1}{\sqrt{2}}\right]\cup \left[\frac{1}{\sqrt{2}},1\right]$

Find the domain of the function $f\left(x\right)$:

Given function $f\left(x\right)=\sqrt{\left(x-\sqrt{\left(1-x^2\right)}\right)}$has two domains involved in it.

Both are square root functions and it is defined to be positive values always. Hence the functions must be $\ge 0$.

Solving function 1: $\sqrt{\left(x-\sqrt{\left(1-x^2\right)}\right)}$

$$\begin{gathered} \Rightarrow x-\sqrt{\left(1-x^2\right)}\ge 0 \\ \Rightarrow x\ge \sqrt{\left(1-x^2\right)} \\ \Rightarrow x^2\ge 1-x^2 \\ \Rightarrow 2x^2-1\ge 0 \\ \Rightarrow 2x^2\ge 1 \\ \Rightarrow x^2\ge \frac{1}{2} \\ \Rightarrow x\ge \pm \frac{1}{\sqrt{2}} \\ \therefore x\in (-\infty ,-\frac{1}{\sqrt{2}}]\cup [\frac{1}{\sqrt{2}},\infty ) \end{gathered}$$

Solving function 2: $\sqrt{\left(1-x^2\right)}$

$$\begin{gathered} \Rightarrow 1-x^2\ge 0 \\ \Rightarrow x^2-1\le 0 \\ \Rightarrow x^2\le 1 \\ \Rightarrow x\le \pm 1 \\ \therefore x\in \left[-1,1\right] \end{gathered}$$

The final domain is determined on basis of the intersection point of the above functions.

Hence, the domain of the given function is $\left[-1,-\frac{1}{\sqrt{2}}\right]\cup \left[\frac{1}{\sqrt{2}},1\right]$.

Therefore, the correct answer is option (A).