Question:
Let $f(x)=\begin{cases}x^{p}cos\frac{1}{x},x\ne0\\ 0,x=0\end{cases}$ Then $f(x)$ is differentiable at $x=0$ if}
Let $f(x)=\begin{cases}x^{p}cos\frac{1}{x},x\ne0\\ 0,x=0\end{cases}$ Then $f(x)$ is differentiable at $x=0$ if}
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Continuity requires $p > 0$. Differentiability requires one extra power, so $p > 1$.
Updated On: Jul 3, 2026
- $P>0$
- $P=0$
- $P>1$
- $P<1$
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The Correct Option is C
Solution and Explanation
Step 1: Concept Differentiability at $x=0$ requires $\lim_{h\to0} \frac{f(h)-f(0)}{h}$ to exist and be finite.
Step 2: Meaning Substituting the function: $\lim_{h\to0} \frac{h^p \cos(1/h) - 0}{h} = \lim_{h\to0} h^{p-1} \cos(1/h)$.
Step 3: Analysis For this limit to be 0 (and exist), the power of $h$ must be positive. Therefore, $p - 1 > 0$.
Step 4: Conclusion Solving the inequality gives $p > 1$. Final Answer: (C)
Step 2: Meaning Substituting the function: $\lim_{h\to0} \frac{h^p \cos(1/h) - 0}{h} = \lim_{h\to0} h^{p-1} \cos(1/h)$.
Step 3: Analysis For this limit to be 0 (and exist), the power of $h$ must be positive. Therefore, $p - 1 > 0$.
Step 4: Conclusion Solving the inequality gives $p > 1$. Final Answer: (C)
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