The equation $|\frac{x}{x-1}|+|x|=\frac{x^2}{|x-1|}$ has

exactly one solution exactly two solutions at most two solutions infinite number of solutions

infinite number of solutions

We have, $|\frac{x}{x-1}|+|x|=\frac{x^2}{|x-1|}$ $⇒|\frac{x}{x-1}|+|x|=|\frac{x^2}{x-1}|$ $⇒|\frac{x}{x-1}|+|x|=|\frac{x^2-x+x}{x-1}|$ $⇒|\frac{x}{x-1}|+|x|=|x+\frac{x}{x-1}|$ $⇒\frac{x}{x-1}×x≥0$   $[∵ |a|+|b|=|a+b|\, if\, ab ≥ 0]$ $⇒\frac{x^2}{x-1}≥0⇒x-1>0⇒x∈(1,∞)$