A function f is defined as follows
`f(x)=x^(p)cos((1)/(x)),xne0`
`f(0)=0`
What conditions should be imposed on p so that f may be continuous at x=0 ?

Options

A. p=0
B. `pgt0`
C. `plt0`
D. No value of p

Explanation

To determine the conditions on \( p \) for the function \( f(x) = x^p \cos\left(\frac{1}{x}\right) \) (for \( x \neq 0 \)) and \( f(0) = 0 \) to be continuous at \( x = 0 \), we need to analyze the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-step Solution:

1. **Definition of Continuity**:
A function \( f \) is continuous at a point \( c \) if:
\[ \lim_{x \to c} f(x) = f(c) \]
In this case, we need to check if:
\[ \lim_{x \to 0} f(x) = f(0) \]
Given \( f(0) = 0 \), we need to find:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^p \cos\left(\frac{1}{x}\right) \]

2. **Evaluating the Limit**:
We need to evaluate:
\[ \lim_{x \to 0} x^p \cos\left(\frac{1}{x}\right) \]
The function \( \cos\left(\frac{1}{x}\right) \) oscillates between -1 and 1 as \( x \) approaches 0. Therefore, we can say:
\[ -1 \leq \cos\left(\frac{1}{x}\right) \leq 1 \]
This implies:
\[ -x^p \leq x^p \cos\left(\frac{1}{x}\right) \leq x^p \]

3. **Applying the Squeeze Theorem**:
As \( x \to 0 \):
- If \( p > 0 \), then \( x^p \to 0 \).
- If \( p = 0 \), then \( x^p = 1 \) (which does not approach 0).
- If \( p < 0 \), then \( x^p \to \infty \) (which does not approach 0).

Therefore, if \( p > 0 \):
\[ \lim_{x \to 0} x^p \cos\left(\frac{1}{x}\right) = 0 \]
By the Squeeze Theorem, since both bounds approach 0.

4. **Conclusion**:
For the limit to equal \( f(0) = 0 \), we require:
\[ p > 0 \]

Thus, the condition that must be imposed on \( p \) for \( f \) to be continuous at \( x = 0 \) is:
\[ \boxed{p > 0} \]