Question

Let \(f(x) = x^p \cos(1/x)\) when \(x \neq 0\) and \(f(x) = 0\), when \(x = 0\). Then \(f(x)\) will be differentiable at \(x = 0\), if

• A. \(p>0\)
• B. \(p>1\)
• C. \(0<p<1\)
• D. \(1/2<p<1\)

Hint:

For differentiable at 0, need \(p>1\); for continuous at 0, need \(p>0\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Differentiability at 0: \(\lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}\) must exist finitely.
Step 2: Detailed Explanation:
\(\lim_{x \to 0} \frac{x^p \cos(1/x)}{x} = \lim_{x \to 0} x^{p-1} \cos(1/x)\)
This limit exists finitely only if \(p - 1>0 \rightarrow p>1\).
Step 3: Final Answer:
\(p>1\).