Differentiation Question 308
Question: The differential coefficient of the function $ |x-1|+|x-3| $ at the point $ x=2 $ is
[RPET 2002; Pb. CET 2000, 04]
Options:
A) - 2
B) 0
C) 2
D) Undefined
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=|x-1|+|x-3| $
$ f(x)= \begin{cases} -(x-1)-(x-3), & x<1 \\ (x-1)-(x-3), & x>1 \\ (x-1)-(x-3), & x<3 \\ (x-1)+(x-3), & x>3 \\ \end{cases} . $
$ = \begin{cases} 4-2x, & x<1 \\ 2, & 1<x<3 \\ 2x-4, & x>3 \\ \end{cases} . $
At $ x=2 $ , $ f(x)= $ 2. Hence $ {f}’(x)=0 $ .
BETA
