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

Answer

To solve the problem, we need to analyze the given equation:

\[ |a-1| + |b-1| = |a| + |b| = |a+1| + |b+1| \]

Given that \(a\) and \(b\) are distinct real numbers, we will break down the absolute values based on different cases.

### Step 1: Analyze the first equation

Starting with the first part of the equation:

\[ |a-1| + |b-1| = |a| + |b| \]

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

**Case 1:** \(a \geq 1\) and \(b \geq 1\)

In this case, we have:

\[ |a-1| = a-1, \quad |b-1| = b-1, \quad |a| = a, \quad |b| = b \]

Thus, the equation becomes:

\[ (a-1) + (b-1) = a + b \]

This simplifies to:

\[ a + b - 2 = a + b \implies -2 = 0 \quad \text{(not valid)} \]

**Case 2:** \(a < 1\) and \(b < 1\)

Here, we have:

\[ |a-1| = 1-a, \quad |b-1| = 1-b, \quad |a| = -a, \quad |b| = -b \]

Thus, the equation becomes:

\[ (1-a) + (1-b) = -a - b \]

This simplifies to:

\[ 2 - (a + b) = -a - b \implies 2 = 0 \quad \text{(not valid)} \]

**Case 3:** \(a \geq 1\) and \(b < 1\)

In this case:

\[ |a-1| = a-1, \quad |b-1| = 1-b, \quad |a| = a, \quad |b| = -b \]

Thus, the equation becomes:

\[ (a-1) + (1-b) = a - b \]

This simplifies to:

\[ a - b = a - b \quad \text{(valid)} \]

### Step 2: Analyze the second equation

Now we analyze the second part of the equation:

\[ |a| + |b| = |a+1| + |b+1| \]

Using the same cases as before:

**Case 1:** \(a \geq 1\) and \(b \geq 1\)

We have:

\[ |a| = a, \quad |b| = b, \quad |a+1| = a+1, \quad |b+1| = b+1 \]

Thus, the equation becomes:

\[ a + b = (a + 1) + (b + 1) \implies a + b = a + b + 2 \quad \text{(not valid)} \]

**Case 2:** \(a < 1\) and \(b < 1\)

We have:

\[ |a| = -a, \quad |b| = -b, \quad |a+1| = 1+a, \quad |b+1| = 1+b \]

Thus, the equation becomes:

\[ -a - b = (1 + a) + (1 + b) \implies -a - b = 2 + a + b \quad \text{(not valid)} \]

**Case 3:** \(a \geq 1\) and \(b < 1\)

We have:

\[ |a| = a, \quad |b| = -b, \quad |a+1| = a + 1, \quad |b+1| = 1 + b \]

Thus, the equation becomes:

\[ a - b = (a + 1) + (1 + b) \implies a - b = a + 2 + b \]

This simplifies to:

\[ -a - b = 2 \quad \text{(valid)} \]

### Step 3: Solve for minimum value of \(|a-b|\)

From the valid conditions, we have \(a \geq 1\) and \(b < 1\). Let's set \(a = 1\) and \(b = -1\) to find the minimum value of \(|a-b|\):

\[ |a-b| = |1 - (-1)| = |1 + 1| = |2| = 2 \]

### Conclusion

The minimum value of \(|a-b|\) is:

\[ \boxed{2} \]