A function f is defined as follows :
`f(x) = x^(p) cos(1//x), x ne 0, f(0)=0`
What conditions whould be imposed on p so that
(i) f may be continuous at x=0
(ii) f may have a differential coefficient at x=0

Answer

To solve the problem, we need to analyze the function \( f(x) = x^p \cos\left(\frac{1}{x}\right) \) for \( x \neq 0 \) and \( f(0) = 0 \). We will determine the conditions on \( p \) for two scenarios: (i) continuity at \( x = 0 \) and (ii) differentiability at \( x = 0 \).

### Step 1: Continuity at \( x = 0 \)

For \( f \) to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0} f(x) = f(0) \]

Since \( f(0) = 0 \), we need to find:

\[ \lim_{x \to 0} x^p \cos\left(\frac{1}{x}\right) \]

The cosine function oscillates between -1 and 1. Therefore, we can write:

\[ -x^p \leq x^p \cos\left(\frac{1}{x}\right) \leq x^p \]

As \( x \to 0 \), both \( -x^p \) and \( x^p \) approach 0 if \( p > 0 \). By the Squeeze Theorem, we conclude:

\[ \lim_{x \to 0} x^p \cos\left(\frac{1}{x}\right) = 0 \]

Thus, for continuity at \( x = 0 \), we require:

\[ p > 0 \]

### Step 2: Differentiability at \( x = 0 \)

For \( f \) to be differentiable at \( x = 0 \), we need to check if the derivative \( f'(0) \) exists. The derivative is defined as:

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h} \]

Substituting \( f(h) \):

\[ f'(0) = \lim_{h \to 0} \frac{h^p \cos\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h^{p-1} \cos\left(\frac{1}{h}\right) \]

Again, using the oscillatory nature of the cosine function:

\[ -h^{p-1} \leq h^{p-1} \cos\left(\frac{1}{h}\right) \leq h^{p-1} \]

For the limit to exist and equal 0, we need \( p - 1 > 0 \) or \( p > 1 \). Therefore, for differentiability at \( x = 0 \), we require:

\[ p > 1 \]

### Final Conditions

Combining our results, we find:

1. For continuity at \( x = 0 \): \( p > 0 \)
2. For differentiability at \( x = 0 \): \( p > 1 \)

Thus, the final condition that must be imposed on \( p \) is:

\[ p > 1 \]