Question

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

• A. 3
• B. 0
• C. 1
• D. 2

Answer 1

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.

$\therefore |a-1|+b-1=a-b=a+1+|b+1|$

Equating $1^{st}$ two.

$|a-1|+b-1=a-b$

For this to be true, $a-1\le ,0\Rightarrow a\le 1$

Equality last two

$a-b=a+1+|b+1|$

For this to be true, $b+1\ge 0\Rightarrow b\ge -1$

$\therefore (a,b)\in R-(-1,1)$

$\therefore |a-b{|}_{mn}=|1-(-1)|=2$