Question

The equation |x| + $|\frac{x}{x-1}|=\frac{x^2}{|x-1|}$ will be always true for $x$ , belonging to

• A. $[0,1)$
• B. $\{0\}\cup (1,\infty )$
• C. $(-1,1)$
• D. $(-\infty ,\infty )$

Answer 1

Answer: (B)

We have,
$|x|+\left|\frac{x}{x-1}\right|=\frac{x^2}{|x-1|}$
We know that,
$|a|+|b|=|a+b|$. Then, $ab\ge 0$
Here, $|x|+\left|\frac{x}{x-1}\right|=\left|x+\frac{x}{x-1}\right|$
$|x|+\left|\frac{x}{x-1}\right|=\left|\frac{x^2}{x-1}\right|=\frac{x^2}{|x-1|}$
$\therefore x\cdot \frac{x}{x-1}\ge 0\Rightarrow \frac{x^2}{x-1}\ge 0$
$\Rightarrow x\in \{0\}\cup (1,\infty )$