Let `f(x) = {(x^(p) sin 1/x, x ge 0),(0, x =0):}` Then `f(x)` is continuous but not differentiable at x =0 if

Options

A. `p in [0,1]`
B. `p in [1, infty)`
C. `p in (-infty, 0)`
D. `p=0`

Answer

To determine the conditions under which the function \( f(x) \) is continuous but not differentiable at \( x = 0 \), we start with the given function:

\[ f(x) = \begin{cases} x^p \sin \frac{1}{x}, & x > 0 \\ 0, & x = 0 \end{cases} \]

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

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

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

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

\[ \lim_{x \to 0} x^p \sin \frac{1}{x} \]

### Step 2: Evaluate the Limit

Using the property that \( |\sin \frac{1}{x}| \leq 1 \), we can bound the limit:

\[ |x^p \sin \frac{1}{x}| \leq |x^p| \]

Thus, we have:

\[ \lim_{x \to 0} |x^p \sin \frac{1}{x}| \leq \lim_{x \to 0} |x^p| = 0 \quad \text{(for \( p > 0 \))} \]

This implies:

\[ \lim_{x \to 0} x^p \sin \frac{1}{x} = 0 \]

### Step 3: Condition for Continuity

For \( f(x) \) to be continuous at \( x = 0 \), we require \( p > 0 \).

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

Next, we check for differentiability at \( x = 0 \). The function \( f(x) \) is differentiable at \( x = 0 \) if the following limit exists:

\[ \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{x^p \sin \frac{1}{x}}{x} \]

This simplifies to:

\[ \lim_{x \to 0} x^{p-1} \sin \frac{1}{x} \]

### Step 5: Evaluate the Limit for Differentiability

Again, using the property of sine, we have:

\[ |x^{p-1} \sin \frac{1}{x}| \leq |x^{p-1}| \]

Thus, we need to analyze:

\[ \lim_{x \to 0} |x^{p-1}| \]

- If \( p - 1 > 0 \) (i.e., \( p > 1 \)), then \( \lim_{x \to 0} x^{p-1} = 0 \), and the limit exists.
- If \( p - 1 = 0 \) (i.e., \( p = 1 \)), then \( \lim_{x \to 0} x^{p-1} \) does not exist (it approaches 1).
- If \( p - 1 < 0 \) (i.e., \( p < 1 \)), then \( \lim_{x \to 0} x^{p-1} \) approaches infinity.

### Step 6: Conclusion

For \( f(x) \) to be continuous but not differentiable at \( x = 0 \), we need:

- \( p > 0 \) (for continuity)
- \( p < 1 \) (for non-differentiability)

Thus, the final condition is:

\[ p \in (0, 1) \]