Question

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

• A. $2$
• B. $1$
• C. Infinite
• D. Zero

Answer 1

The correct option is B

$1$

Explanation for correct option

The given equation is,

$\left|\frac{x}{x-1}\right|+\left|x\right|=\frac{x}{\left|x-1\right|}$

$\left(1\right)$ When $x>1$

$$\begin{gathered} \frac{x}{x-1}+x=\frac{x}{x-1} \\ x=0 \end{gathered}$$

So, $x=0$, doesn't exist.

$\left(2\right)$ When $0\le x\le 1$

$$\begin{gathered} \frac{x}{1-x}+x=\frac{x}{1-x} \\ x=0 \end{gathered}$$

$\left(3\right)$ When $-\infty <x<0$

$$\begin{gathered} \frac{-x}{-\left(x-1\right)}-x=\frac{x}{-\left(x-1\right)} \\ \frac{2x}{x-1}-x=0 \\ \frac{2x-x^2+x}{x-1}=0 \\ x\left(3-x\right)=0 \end{gathered}$$

$x=0,x=3,$ does not exist.

Hence, the correct option is $\left(\mathrm{B}\right)$.