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 \ \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)=0\] 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 \[\underset{x\to 0}{\mathop{\lim }}\,{{x}^{p-1}}\sin \left( \frac{1}{x} \right)\] 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.
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