A) p = 0
B) p > 0
C) p < 0
D) No value of p
Correct Answer: B
Solution :
[b] Given function is defined as: \[f(x)=\left\{ \begin{matrix} {{x}^{p}}\cos \left( \frac{1}{x} \right) & x\ne 0 \\ 0, & x=0 \\ \end{matrix} \right.\] For continuity: \[LHS:\underset{x\to 0}{\mathop{\lim }}\,f(x)=RHS\,\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)\] \[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,{{x}^{p}}\cos \left( \frac{1}{x} \right)=0\] \[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,{{x}^{p}}\cos \left( \frac{1}{x} \right)=0\] \[\cos \left( \frac{1}{x} \right)\] is always a finite quantity if \[x\to 0\] \[\Rightarrow {{x}^{p}}=0\] Which is possible only if \[p>0\].
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