Functions Question 567
Question: If $ f(x)= \begin{cases} & ,e^{x};x\le 0 \\ & |1-x|;x>0 \\ \end{cases} . $ , then
[Roorkee 1995]
Options:
A) $ f(x) $ is differentiable at $ x=0 $
B) $ f(x) $ is continuous at $ x=0 $
C) $ f(x) $ is differentiable at $ x=1 $
D) $ f(x) $ is continuous at $ x=1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)= \begin{cases} & e^{x};,x\le 0 \\ & 1-x;0<x\le 1 \\ & x-1,;x>1 \\ \end{cases} . $ $ Rf’(0)=\underset{h\to 0}{\mathop{\lim }},\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }},\frac{1-h-1}{h}=-1 $ $ Lf’(0)=\underset{h\to 0}{\mathop{\lim }},\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }},\frac{{e^{-h}}-1}{-h}=1 $ So, it is not differentiable at $ x=0 $ . Similarly, it is not differentiable at $ x=1 $ . But it is continous at $ x=0 $ , 1.
BETA
