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, $a b \geq 0$
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} \geq 0 \Rightarrow \frac{x^{2}}{x-1} \geq 0 $
$\Rightarrow x \in\{0\} \cup(1, \infty)$