Question

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

• A. $0$
• B. $2$
• C. $4$
• D. $6$

Answer 1

The correct option is D

$6$

Find the minimum value of the given expression

Given :$f\left(x\right)=\left|x\right|+\left|x+\frac{1}{2}\right|+\left|x-3\right|+\left|x-\frac{5}{2}\right|$

For, $x<-\frac{1}{2}$,

$$\begin{gathered} f\left(x\right)=-x+(-x-\frac{1}{2})+(-x+3)+\left(-x+\frac{5}{2}\right) \\ \Rightarrow f\left(x\right)=-4x+5 \end{gathered}$$

For, $-\frac{1}{2}\le x<0$,

$$\begin{gathered} f\left(x\right)=-x+x+\frac{1}{2}+3-x+\frac{5}{2}-x \\ \Rightarrow f\left(x\right)=6-2x \end{gathered}$$

For, $0\le x<\frac{5}{2}$,

$$\begin{gathered} f\left(x\right)=x+x+\frac{1}{2}+3-x+\frac{5}{2}-x \\ \Rightarrow f\left(x\right)=6 \end{gathered}$$

For, $\frac{5}{2}\le x\le 3$

$f\left(x\right)=2x+1$

Now plot the graph,

From the graph, the function have the minimum value of $6$.

Hence, option $\left(D\right)$ is the correct answer.