Question

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

• A. $R$
• B. $[-2,2]$
• C. $[-1,1]$
• D. $\varphi$

Answer 1

The correct option is D $\varphi$

$f\left(x\right)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1-x}}$

For $f\left(x\right)$ to be defined

$\frac{x-2}{x+2}\ge \notin \frac{1-x}{1+x}\ge 0\notin x\ne -2,-1$

(1) $\frac{x-2}{x+2}\ge 0$ (refer image)

Note:

Domain of $\frac{x-2}{x+2}÷0$ is $D_1$

$\notin$domain of

$\frac{1-x}{1+x}\le 0$ is $D_2$ here

$D_1\in (-\infty ,-2)\cup [2,\infty )$

(2) $\frac{1-x}{1+x}\ge 0$

$\Rightarrow \frac{x-1}{x+1}\le 0$

$\therefore$ Domain of $x,D=D_1\cap D_2=\varphi$