Differentiation Question 285
Question: The value of $ \frac{d}{dx}[|x-1|+|x-5|] $ at $ x=3 $ is
[MP PET 2000]
Options:
A) - 2
B) 0
C) 2
D) 4
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=|x-1|+|x-5| $
$ f(x)= \begin{cases} & -(x-1)-(x-5),x<1 \\ & (x-1)-(x-5),1<x<5 \\ & x-1+x-5,x>5 \\ \end{cases} . $
$ f(x)= \begin{cases} 6-2x, & x<1 \\ 4, & 1<x<5 \\ 2x-6, & x>5 \\ \end{cases} . $
$ \because $ $ x=3\in (1,5) $ , \ For x = 3, $ f(x)=4,\ \ f’(x)=0 $ .
BETA
