Question

The solution set of $\left|\frac{3x}{x-3}\right|+|x|=\frac{x^2}{|x-3|}$ is

• A. $(3,\infty )$
• B. $[3,\infty )$
• C. $\left\{0\right\}\cup (3,\infty )$
• D. $\left\{0\right\}\cup [3,\infty )$

Answer 1

The correct option is C $\left\{0\right\}\cup (3,\infty )$

$\left|\frac{3x}{x-3}\right|+|x|=\frac{x^2}{|x-3|}$
Let $a=\frac{3x}{x-3}$
and $b=x$
Given equation can be written as
$|a|+|b|=|a+b|$
$\therefore a$ and $b$ should have same sign,
i.e., $ab\ge 0$
$\Rightarrow \frac{3x^2}{x-3}\ge 0$
Since $x^2\ge 0\forall x\in R$
$\therefore x-3>0$
$\Rightarrow x>3$
So, the solution set is $\left\{0\right\}\cup (3,\infty )$