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 nature of the function \( f(x) \) at \( x = 0 \), we will analyze its continuity and differentiability based on the given piecewise definition:

\[ f(x) = \begin{cases} x^{p+1} \cos\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

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

To check if \( f(x) \) is continuous at \( x = 0 \), we need to verify that:

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

Given that \( f(0) = 0 \), we need to find \( \lim_{x \to 0} f(x) \):

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

Since \( \cos\left(\frac{1}{x}\right) \) oscillates between -1 and 1, we can bound the limit:

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

### Step 2: Apply the Squeeze Theorem

As \( x \to 0 \):

- If \( p + 1 > 0 \), then \( x^{p+1} \to 0 \).
- If \( p + 1 = 0 \), then \( x^{p+1} \) approaches a constant (which is 1).
- If \( p + 1 < 0 \), then \( x^{p+1} \) approaches infinity.

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

\[ \lim_{x \to 0} f(x) = 0 \quad \text{if } p + 1 > 0 \quad (p > -1) \]

### Step 3: Conclusion on Continuity

Therefore, \( f(x) \) is continuous at \( x = 0 \) if \( p > -1 \).

### Step 4: Check for Differentiability at \( x = 0 \)

To check for differentiability at \( x = 0 \), we need to evaluate the left-hand derivative (LHD) and right-hand derivative (RHD):

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

This simplifies to:

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

Since \( \cos\left(-\frac{1}{h}\right) = \cos\left(\frac{1}{h}\right) \) oscillates between -1 and 1, we can apply the same bounding technique:

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

### Step 5: Apply the Squeeze Theorem for Differentiability

As \( h \to 0 \):

- If \( p > 0 \), then both bounds approach 0.
- If \( p = 0 \), the limit approaches 1.
- If \( p < 0 \), the limit does not exist.

Thus, for differentiability at \( x = 0 \):

\[ \text{LHD} = 0 \quad \text{if } p > 0 \]

### Step 6: Conclusion on Differentiability

Therefore, \( f(x) \) is differentiable at \( x = 0 \) if \( p > 0 \).

### Final Answer

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