Functions Question 128
Question: At the point $ x=1 $ , the given function $ f(x)= \begin{cases} & x^{3}-1;1<x<\infty \\ & x-1;-\infty <x\le 1 \\ \end{cases} . $ is
[Roorkee 1993]
Options:
A) Continuous and differentiable
B) Continuous and not differentiable
C) Discontinuous and differentiable
D) Discontinuous and not differentiable
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ Rf’(1)=\underset{h\to 0}{\mathop{\lim }},\frac{f(1+h)-f(1)}{h} $ $ =\underset{h\to 0}{\mathop{\lim }},\frac{{ {{(1+h)}^{3}}-1 }-0}{h}=3 $ $ Lf’(1)=\underset{h\to 0}{\mathop{\lim }},\frac{f(1-h)-f(1)}{-h} $ $ =\underset{h\to 0}{\mathop{\lim }},\frac{{ (1-h)-1 }-0}{-h}=1 $
$ \therefore ,Rf’(1)\ne Lf’(1) $
$ \Rightarrow f(x) $ is not differentiable at $ x=1. $ Now, $ f(1+0)=\underset{h\to 0}{\mathop{\lim }}f(1+h)=0 $ and $ f(1-0)=\underset{h\to 0}{\mathop{\lim }}f(1-h)=0 $
$ \therefore ,f(1+0)=f(1-0)=f(0) $
$ \Rightarrow f(x) $ is continuous at $ x=1. $ Hence at $ x=1,,f(x) $ is continuous and not differentiable.
BETA
