If a,b respectively be the number of solutions and sum of solutions of `|(2x)/(x-1)|-|x| = (x^(2))/(|x-1|)`, then

Options

A. a=3
B. b=1
C. b=2
D. a=2

Answer

To solve the equation \( \left| \frac{2x}{x-1} \right| - |x| = \frac{x^2}{|x-1|} \), we will analyze the expression step by step.

### Step 1: Rewrite the equation
We start with the equation:
\[ \left| \frac{2x}{x-1} \right| - |x| = \frac{x^2}{|x-1|} \]

### Step 2: Identify cases based on the absolute values
The absolute values in the equation suggest that we need to consider different cases based on the values of \( x \). The critical points are \( x = 0 \) and \( x = 1 \). We will analyze three cases:
1. Case 1: \( x < 0 \)
2. Case 2: \( 0 \leq x < 1 \)
3. Case 3: \( x \geq 1 \)

### Step 3: Case 1: \( x < 0 \)
In this case, \( |x| = -x \) and \( |x-1| = -(x-1) = 1-x \). Thus, the equation becomes:
\[ \left| \frac{2x}{x-1} \right| = -\frac{2x}{x-1} \quad \text{(since } x < 0\text{)} \]
So we rewrite the equation:
\[ -\frac{2x}{x-1} + x = \frac{x^2}{1-x} \]
Multiplying through by \( (1-x) \) to eliminate the denominator:
\[ -2x + x(1-x) = x^2 \]
This simplifies to:
\[ -x^2 - 2x = 0 \]
Factoring gives:
\[ x(2+x) = 0 \]
Thus, \( x = 0 \) or \( x = -2 \). Since we are in the case \( x < 0 \), we accept \( x = -2 \).

### Step 4: Case 2: \( 0 \leq x < 1 \)
Here, \( |x| = x \) and \( |x-1| = 1-x \). The equation becomes:
\[ \frac{2x}{1-x} - x = \frac{x^2}{1-x} \]
Multiplying through by \( 1-x \):
\[ 2x - x(1-x) = x^2 \]
This simplifies to:
\[ 2x - x + x^2 = x^2 \]
This leads to:
\[ x = 0 \]
This case gives us the solution \( x = 0 \).

### Step 5: Case 3: \( x \geq 1 \)
In this case, \( |x| = x \) and \( |x-1| = x-1 \). The equation becomes:
\[ \frac{2x}{x-1} - x = \frac{x^2}{x-1} \]
Multiplying through by \( x-1 \):
\[ 2x - x(x-1) = x^2 \]
This simplifies to:
\[ 2x - x^2 + x = x^2 \]
Rearranging gives:
\[ 2x = 2x^2 \]
Factoring gives:
\[ x(2x - 2) = 0 \]
Thus, \( x = 0 \) or \( x = 1 \). Since we are in the case \( x \geq 1 \), we accept \( x = 1 \).

### Step 6: Summary of Solutions
From all three cases, we have found the solutions:
1. \( x = -2 \) from Case 1
2. \( x = 0 \) from Case 2
3. \( x = 1 \) from Case 3

### Step 7: Count the number of solutions and their sum
The number of solutions \( a = 3 \) (i.e., \( -2, 0, 1 \)). The sum of the solutions \( b = -2 + 0 + 1 = -1 \).

### Final Answer
Thus, the values of \( a \) and \( b \) are:
- \( a = 3 \)
- \( b = -1 \)