Question

The minimum value of $|x|+|x+\frac{1}{2}|+|x-3|+|x-\frac{5}{2}|$ is

• A. $2$
• B. $4$
• C. $6$
• D. $0$

Answer 1

Answer: (C)

$6$

For $x<-\frac{1}{2}$, the expression becomes $-x+-x-\frac{1}{2}-x+3-x+\frac{5}{2}$
$=-4x+5$; no minima exists $\to \left(1\right)$

(As $x$ approaches $-\frac{1}{2}$, values of the expression keeps on decreasing)

For $-\frac{1}{2}\le x<0$, the expression becomes $-x+x+\frac{1}{2}+3-x+\frac{5}{2}-x=6-2x\to \left(2\right)$ (again no minima exists)

For $0\le x<\frac{5}{2}$
$x+x+\frac{1}{2}-x+3-x+\frac{5}{2}=6\to \left(3\right)$

For $\frac{5}{2}\le x<3\Rightarrow 2x+1$

In the above interval, the minimum value is at the lower limit, that is $x=\frac{5}{2}$, where the expression has value $6\to \left(4\right)$
For $x\ge 3\Rightarrow 4x-5$.

In this interval, the minimum value of the expression is $7$, which occurs at $x=3\to \left(5\right)$

From (1), (2), (3), (4) and (5), the minimum value of the expression is $6$.