Question

Given $f\left(x\right)=\frac{1}{\sqrt{|x|-x}}$ and $g\left(x\right)=\frac{1}{\sqrt{x-|x|}}$. Then

• A. $D\left(f\right)\ne \varphi ,D\left(g\right)=\varphi$
• B. $D\left(f\right)\ne \varphi ,D\left(g\right)\ne \varphi$
• C. $D\left(f\right)=\varphi ,D\left(g\right)=\varphi$
• D. $D\left(f\right)=\varphi ,D\left(g\right)\ne \varphi$

Answer 1

The correct option is A $D\left(f\right)\ne \varphi ,D\left(g\right)=\varphi$

$f\left(x\right)=\frac{1}{\sqrt{|x|-x}}$
For $f$ to be defined,
$|x|-x>0$
When $x<0$
$-x-x>0$
$\Rightarrow -2x>0\Rightarrow x\in (-\infty ,0)$
and when $x\ge 0$
$x-x>0$
$\Rightarrow 0>0$, which is not possible.
$\therefore D\left(f\right)=(-\infty ,0)$

$g\left(x\right)=\frac{1}{\sqrt{x-|x|}}$
For $g$ to be defined,
$x-|x|>0$
When $x<0$
$x-(-x)>0$
$\Rightarrow 2x>0$
$\Rightarrow x>0$, which is not possible as $x<0$
and when $x\ge 0$
$x-x>0$
$\Rightarrow 0>0$, which is not possible.
$\therefore D\left(g\right)=\varphi$