Question

Least value of the function $f\left(x\right)=|x-a|+|x-b|+|x-c|+|x-d|$, where $a<b<c<d$ are fixed real numbers & $x$ takes arbitary values, is

• A. $a+d-b-c$
• B. $c+d-a-b$
• C. $b+c-a-d$
• D. $a+c-b-d$

Answer 1

The correct option is B $c+d-a-b$


It takes minimum at $x=b\text{and}c$ both.
So, $f\left(b\right)=b-a+0+c-b+d-b=(c+d-a-b)$