Question

The solution set of $\left|\frac{x}{x-1}\right|+|x|=\frac{x^2}{|x-1|}$ is

• A. $x\ge 0$
• B. $x>0$
• C. $x\epsilon (1,\infty )$
• D. $x\epsilon (1,\infty )\cup \left\{0\right\}$

Answer 1

The correct option is D $x\epsilon (1,\infty )\cup \left\{0\right\}$

We know if $|a|+|b|=|a+b|$
$\Rightarrow ab\ge 0$
Thus for, $\frac{x^2}{x-1}\ge 0$
$\Rightarrow x=0$ or $x>1$
$\Rightarrow x\in \left\{0\right\}\cup (1,\infty )$