Question

Total number of real solutions of $\frac{x}{|x-1|}+|x|=\frac{x^2}{|x-1|}$ is equal to

• A. 4
• B. 2
• C. 8
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

The given equation is $\frac{x}{|x-1|}+|x|=\frac{x^2}{|x-1|}$.

Rearranging the terms, we get: $\frac{x^2}{|x-1|} - \frac{x}{|x-1|} = |x|$ $\frac{x^2-x}{|x-1|} = |x|$ $\frac{x(x-1)}{|x-1|} = |x|$

We analyze the term $\frac{x-1}{|x-1|}$. This term is equal to $1$ if $x-1 > 0$ (i.e., $x > 1$). This term is equal to $-1$ if $x-1 < 0$ (i.e., $x < 1$). The denominator $|x-1|$ cannot be zero, so $x \ne 1$.

The equation can be rewritten as: $x \cdot \left(\frac{x-1}{|x-1|}\right) = |x|$

Now we consider two cases:

Case 1: $x > 1$ In this case, $\frac{x-1}{|x-1|} = 1$. The equation becomes: $x \cdot (1) = |x|$ $x = |x|$ Since $x > 1$, $|x| = x$. So, the equation $x = x$ is true for all $x > 1$. Thus, all real numbers in the interval $(1, \infty)$ are solutions.

Case 2: $x < 1$ In this case, $\frac{x-1}{|x-1|} = -1$. The equation becomes: $x \cdot (-1) = |x|$ $-x = |x|$ The equation $|x| = -x$ is true if and only if $x \le 0$. We need to satisfy both conditions: $x < 1$ and $x \le 0$. The intersection of these conditions is $x \le 0$. Thus, all real numbers in the interval $(-\infty, 0]$ are solutions.

Combining the solutions from both cases, the set of all real solutions is $(-\infty, 0] \cup (1, \infty)$. This set contains infinitely many real numbers. Since the number of solutions is infinite, none of the options (A) 4, (B) 2, or (C) 8 are correct. Therefore, the correct option is (D) None of these.