If `f(x)={{:(x^(p+1)cos.(1)/(x)":", x ne 0),(0":", x=0):}` then at x = 0 the function f(x) is

Options

A. continuous if `p gt -1` and differentiable if `p gt 0`
B. Continuous if `p gt0` and differentiable if `p gt -1`
C. Continuous and differetiable if `p gt -1`
D. None of these

Answer

To determine the behavior of the function \( f(x) \) at \( x = 0 \), we need to analyze its continuity and differentiability based on the given definition of the function.

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} x^{p+1} \cos\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

### Step 2: Check continuity at \( x = 0 \)
For \( f(x) \) to be continuous at \( x = 0 \), the following condition must hold:
\[ \lim_{x \to 0} f(x) = f(0) \]
Since \( f(0) = 0 \), we need to find \( \lim_{x \to 0} f(x) \).

#### Step 2.1: Calculate the limit as \( x \to 0 \)
We will evaluate the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x^{p+1} \cos\left(\frac{1}{x}\right) \]
The term \( \cos\left(\frac{1}{x}\right) \) oscillates between -1 and 1 as \( x \to 0 \). Therefore, we can bound the limit:
\[ -x^{p+1} \leq x^{p+1} \cos\left(\frac{1}{x}\right) \leq x^{p+1} \]
Now, we need to consider the behavior of \( x^{p+1} \) as \( x \to 0 \):
- If \( p + 1 > 0 \) (i.e., \( p > -1 \)), then \( x^{p+1} \to 0 \).
- If \( p + 1 \leq 0 \) (i.e., \( p \leq -1 \)), then \( x^{p+1} \) does not approach 0.

Thus, for continuity at \( x = 0 \):
\[ \lim_{x \to 0} f(x) = 0 \quad \text{if } p > -1 \]

### Step 3: Check differentiability at \( x = 0 \)
For \( f(x) \) to be differentiable at \( x = 0 \), the following must hold:
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h} \]
Substituting \( f(h) \):
\[ \lim_{h \to 0} \frac{h^{p+1} \cos\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h^{p} \cos\left(\frac{1}{h}\right) \]
Again, since \( \cos\left(\frac{1}{h}\right) \) oscillates between -1 and 1, we can bound this limit:
\[ -h^{p} \leq h^{p} \cos\left(\frac{1}{h}\right) \leq h^{p} \]
Now, we consider the behavior of \( h^{p} \) as \( h \to 0 \):
- If \( p > 0 \), then \( h^{p} \to 0 \).
- If \( p \leq 0 \), then \( h^{p} \) does not approach 0.

Thus, for differentiability at \( x = 0 \):
\[ \lim_{h \to 0} h^{p} \cos\left(\frac{1}{h}\right) = 0 \quad \text{if } p > 0 \]

### Conclusion
- The function \( f(x) \) is continuous at \( x = 0 \) if \( p > -1 \).
- The function \( f(x) \) is differentiable at \( x = 0 \) if \( p > 0 \).

### Final Answer
The function \( f(x) \) is continuous at \( x = 0 \) if \( p > -1 \) and differentiable at \( x = 0 \) if \( p > 0 \).