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Question

# If a, b ∈ R distinct numbers satisfying |a−1|+|b−1|=|a|+|b|=|a+1|+|b+1|, then the minimum value of |a−b| is

A

3

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B

0

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C

1

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D

2

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Solution

## The correct option is D 2 |a−1|+b−1=a−b=a+1+(b+1) By analyzing (a - 1) & (b - 1) needs to be of different sign so as cancel off the 1 & be be equal |a| + |b| 11 aly, for (a + 1) & (b + 1) too,they need to be of different sign. Therefore, a & b will be of different signs. So, let a > 0 & b < 0. ∴|a−1|+b−1=a−b=a+1+|b+1| Equating 1st two. |a−1|+b−1=a−b For this to be true, a−1≤,0⇒a≤1 Equality last two a−b=a+1+|b+1| For this to be true, b+1≥0⇒b≥−1 ∴(a,b)∈R−(−1,1) ∴|a−b|mn=|1−(−1)|=2

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