Question

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

• A. $(-\infty ,-2]\cup [2,\infty )$
• B. $[-1,1]$
• C. $\varphi$
• D. None of these

Answer 1

The correct option is C

$\varphi$

$f\left(x\right)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1+x}}$
For f(x) to be defined,
$x+2\ne 0\Rightarrow x\ne -2\dots \dots \left(1\right)$
And $1+x\ne 0$
$\Rightarrow x\ne -1\dots \dots \left(2\right)$
Also, $\frac{x-2}{x+2}\ge 0$
$\Rightarrow \frac{(x-2)(x+2)}{(x+2{)}^2}\ge 0\Rightarrow (x-2)(x+2)\ge 0\Rightarrow xϵ(-\infty ,-2)\cup [2,\infty ]\dots \dots \left(3\right)$
And, $\frac{1-x}{1+x}\ge 0$
$\Rightarrow \frac{(1-x)(1+x)}{(1+x{)}^2}\ge 0\Rightarrow (1-x)(1+x)\ge 0\Rightarrow xϵ(-\infty ,-1)\cup [1,\infty )\dots \dots \left(4\right)$
From (1), (2), (3) and (4), we get,
$xϵ\varphi$
Thus, $dom\left(f\right(x\left)\right)=\varphi$