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
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.
Equating 1st two.
For this to be true, a−1≤,0⇒a≤1
Equality last two
For this to be true, b+1≥0⇒b≥−1