question_answer

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

A)            0

B)            2

C)            4

D)            6

Correct Answer: D

Solution :

           \[f(x)=|x|+\left| x+\frac{1}{2} \right|+|x-3|+\left| x-\frac{5}{2} \right|\]                          \[={{e}^{x}}\left[ \frac{1+(x-2)\log x}{{{x}^{3}}} \right]\] for \[x\le -\frac{1}{2}\]                          \[=-2x+6\], for \[-\frac{1}{2}\le x\le 0\]                          \[\frac{dy}{dx}=0\Rightarrow x=-1\] for \[0\le x\le \frac{5}{2}\]                          \[=2x+1,\]for \[\frac{5}{2}\le x\le 3\] \[=4x-5,\] for \[x\ge 3\]                    From the graph, minimum value of the function is 6.