If a,b `in` R distinct numbers satisfying |a-1| + |b-1| = |a| + |b| = |a+1| + |b+1|, Then the minimum value of |a-b| is :

Options

A. 3
B. 0
C. 1
D. 2

Explanation

To solve the problem, we need to analyze the given conditions involving the absolute values of \(a\) and \(b\). The conditions are:

1. \(|a - 1| + |b - 1| = |a| + |b|\)
2. \(|a| + |b| = |a + 1| + |b + 1|\)

We will break down the solution step by step.

### Step 1: Analyze the first condition
The first condition is \(|a - 1| + |b - 1| = |a| + |b|\).

We can consider different cases based on the values of \(a\) and \(b\) relative to 1.

#### Case 1: Both \(a\) and \(b\) are greater than or equal to 1.
In this case:
\[ |a - 1| = a - 1 \quad \text{and} \quad |b - 1| = b - 1 \]
Thus, the equation becomes:
\[ (a - 1) + (b - 1) = a + b \]
This simplifies to:
\[ a + b - 2 = a + b \quad \Rightarrow \quad -2 = 0 \quad \text{(not possible)} \]

#### Case 2: Both \(a\) and \(b\) are less than 1.
In this case:
\[ |a - 1| = 1 - a \quad \text{and} \quad |b - 1| = 1 - b \]
Thus, the equation becomes:
\[ (1 - a) + (1 - b) = -a - b \]
This simplifies to:
\[ 2 - a - b = -a - b \quad \Rightarrow \quad 2 = 0 \quad \text{(not possible)} \]

#### Case 3: One is less than 1 and the other is greater than or equal to 1.
Assume \(a < 1\) and \(b \geq 1\):
\[ |a - 1| = 1 - a \quad \text{and} \quad |b - 1| = b - 1 \]
Thus, the equation becomes:
\[ (1 - a) + (b - 1) = -a + b \]
This simplifies to:
\[ 1 - a + b - 1 = -a + b \quad \Rightarrow \quad 0 = 0 \quad \text{(always true)} \]

### Step 2: Analyze the second condition
The second condition is \(|a| + |b| = |a + 1| + |b + 1|\).

Using the same cases:

#### Case 1: Both \(a\) and \(b\) are greater than or equal to 0.
In this case:
\[ |a| = a \quad \text{and} \quad |b| = b \]
Thus, the equation becomes:
\[ a + b = (a + 1) + (b + 1) \quad \Rightarrow \quad a + b = a + b + 2 \quad \Rightarrow \quad 0 = 2 \quad \text{(not possible)} \]

#### Case 2: Both \(a\) and \(b\) are less than 0.
In this case:
\[ |a| = -a \quad \text{and} \quad |b| = -b \]
Thus, the equation becomes:
\[ -a - b = (-a + 1) + (-b + 1) \quad \Rightarrow \quad -a - b = -a - b + 2 \quad \Rightarrow \quad 0 = 2 \quad \text{(not possible)} \]

#### Case 3: One is negative and the other is positive.
Assume \(a < 0\) and \(b \geq 0\):
\[ |a| = -a \quad \text{and} \quad |b| = b \]
Thus, the equation becomes:
\[ -a + b = (-a + 1) + (b + 1) \quad \Rightarrow \quad -a + b = -a + b + 2 \quad \Rightarrow \quad 0 = 2 \quad \text{(not possible)} \]

### Conclusion
From our analysis, we find that the only valid case is when one of the numbers is less than 1 and the other is greater than or equal to 1.

To find the minimum value of \(|a - b|\), we can set \(a = 0\) and \(b = 2\) (or vice versa), which gives us:
\[ |a - b| = |0 - 2| = 2 \]

Thus, the minimum value of \(|a - b|\) is \(\boxed{2}\).