Question

The number of real values of $x$ which satisfy the equation $|\frac{x}{x-1}|+|x|=\frac{x}{|x-1|}$ is

• A. $2$
• B. $1$
• C. $infinite$
• D. $zero$

Answer 1

Answer: (B)

Given $\left|\frac{x}{x-1}\right|+\left|x\right|=\frac{x}{|x-1|}$
$\left(i\right)whenx>1$,
$\frac{x}{x-1}+x=\frac{x}{x-1}$
$\Rightarrow x=0$, does not exist
$\left(ii\right)$ When $0\le x<1$
$\frac{x}{1-x}+x=\frac{x}{1-x}$
$\Rightarrow x=0$
$\left(iii\right)$ when $-\infty <x<0$
$\therefore \frac{-x}{-\left(x-1\right)}-x=\frac{x}{-\left(x-1\right)}$
$\Rightarrow \frac{2x}{\left(x-1\right)}-x=0$
$\Rightarrow x\left[\frac{2-x+1}{x-1}\right]=0$
$x=0,x=3$, does not exist
Hence, only one solution exist.