Functions Question 669
Question: The function $ f(x),=,|x|+|x-1| $ is
[RPET 1996; Kurukshetra CEE 2002]
Options:
A) Continuous at $ x=1, $ but not differentiable at $ x=1 $
B) Both continuous and differentiable at $ x=1 $
C) Not continuous at $ x=1 $
D) Not differentiable at $ x=1 $
Show Answer
Answer:
Correct Answer: A
Solution:
We have, $ f(x)=|x|+|x-1| $ $ = \begin{cases} -2x+1, & x<0 & {} \\ x-x+1, & 0\le x<1 & = \\ x+x-1, & x\ge 1 & {} \\ \end{cases} . \begin{cases} -2x+1, & x<0 \\ 1 & 0\le x<1 \\ 2x-1, & x\ge 1 \\ \end{cases} . $ Clearly, $ \underset{x\to {0^{-}}}{\mathop{\lim }},f(x)=1,\underset{x\to {0^{+}}}{\mathop{\lim }},f(x)=1,\underset{x\to {1^{-}}}{\mathop{\lim }},f(x)=1 $ and $ \underset{x\to {1^{+}}}{\mathop{\lim }},f(x)=1 $ . So, $ f(x) $ is continuous at $ x=0,1. $ Now $ f’(x)={ \begin{array}{*{35}{l}} -2,x<0 \\ 0,,0\le x<1 \\ 2,,x\ge 1 \\ \end{cases} . $ Here x = 0, $ f’({0^{+}})=0 $ while $ f’({0^{-}})=-2 $ and at x = 1, $ f’({1^{+}})=2 $ while $ f’({1^{-}})=0 $ Thus, $ f(x) $ is not differentiable at x = 0 and 1.
BETA
