Applications Of Derivatives Question 297
Question: The minimum value of $ |x|+|x+\frac{1}{2}|+|x-3|+|x-\frac{5}{2}| $ is
Options:
A) 0
B) 2
C) 4
D) 6
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)=|x|+| x+\frac{1}{2} |+|x-3|+| x-\frac{5}{2} | $
$ =e^{x}[ \frac{1+(x-2)\log x}{x^{3}} ] $ 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 $
minimum value of the function is 6.
BETA
