Question

The least value of $f\left(x\right)=|x-a|+|x-b|+|x-c|+|x-d|$, where $a<b<c<d$ are fixed real numbers, is

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

Answer 1

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


It takes minimum at $x=b\text{and}c$ both, so putting $x=b$, we get
$f\left(b\right)=b-a+0+c-b+d-b\Rightarrow f\left(b\right)=-a-b+c+d$