Functions Question 563
Question: Let $ f(x)= \begin{cases} & x^{p}\sin \frac{1}{x},x\ne 0 \\ & 0,x=0 \\ \end{cases} . $ then $ f(x) $ is continuous but not differential at $ x=0 $ if
[DCE 2005]
Options:
A) $ 0<p\le 1 $
B) $ 1\le p<\infty $
C) $ -\infty <p<0 $
D) p = 0
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=x^{p}\sin \frac{1}{x},x\ne 0 $ and $ f(x)=0,\ x=0 $ Since at $ x=0 $ , $ f(x) $ is a continuous function \ $ \underset{x\to 0}{\mathop{\lim }},f(x)=f(0)=0 $
Þ $ \underset{x\to 0}{\mathop{\lim }}x^{p}\sin \frac{1}{x}=0\Rightarrow p>0 $ . $ f(x) $ is differentiable at $ x=0 $ , if $ \underset{x\to 0}{\mathop{\lim }},\frac{f(x)-f(0)}{x-0} $ exists
Þ $ \underset{x\to 0}{\mathop{\lim }},\frac{x^{p}\sin \frac{1}{x}-0}{x-0} $ exists
Þ $ \underset{x\to 0}{\mathop{\lim }},{x^{p-1}}\sin \frac{1}{x} $ exists
Þ $ p-1>0 $ or $ p>1 $ If $ p\le 1 $ , then $ \underset{x\to 0}{\mathop{\lim }},{x^{p-1}}\sin ( \frac{1}{x} ) $ does not exist and at $ x=0 $ $ f(x) $ is not differentiable. \ for $ 0<p\le 1 $ f(x) is a continuous function at $ x=0 $ but not differentiable.
BETA
