A) \[0<p\le 1\]
B) \[1\le p<\infty \]
C) \[-\infty <p<0\]
D) \[p=0\]
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 |
| \[\therefore \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,f(x)=f(0)=0\] |
| \[\Rightarrow \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,{{x}^{P}}\sin \frac{1}{x}=0\Rightarrow p>0\] |
| \[f(x)\] is differentiable at \[x=0,\]if \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{f(x)-f(0)}{x-0}\]exists |
| \[\Rightarrow \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{x}^{P}}\sin \frac{1}{x}-0}{x-0}\] exists |
| \[\Rightarrow \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,{{x}^{P-1}}\sin \frac{1}{x}\] exists |
| \[\Rightarrow \,\,p-1>0\] or \[p>1\] |
| \[\therefore \] for \[0<p\le 1,\] \[f(x)\] is a continuous function at \[x=0\]but not differentiable. |
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