Let `f (x)= [{:(x ^(2alpha+1)ln x ,,, x gt0),(0 ,,, x =0):}` If f (x) satisfies rolle's theorem in interval `[0,1],` then ` alpha` can be:

Options

A. `-1/2`
B. `-1/3`
C. `-1/4`
D. `-1`

Answer

To determine the values of \( \alpha \) for which the function \( f(x) \) satisfies Rolle's theorem in the interval \([0, 1]\), we will follow these steps:

### Step 1: Define the function
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} x^{2\alpha + 1} \ln x & \text{if } x > 0 \\ 0 & \text{if } x = 0 \end{cases} \]

### Step 2: Check continuity at \( x = 0 \)
For \( f(x) \) to satisfy Rolle's theorem, it must be continuous on the closed interval \([0, 1]\). This requires that:
\[ \lim_{x \to 0^+} f(x) = f(0) \]
Since \( f(0) = 0 \), we need to find:
\[ \lim_{x \to 0^+} x^{2\alpha + 1} \ln x \]

### Step 3: Evaluate the limit
To evaluate the limit, we rewrite it:
\[ \lim_{x \to 0^+} x^{2\alpha + 1} \ln x = \lim_{x \to 0^+} \frac{\ln x}{x^{-(2\alpha + 1)}} \]
This is an indeterminate form of type \( \frac{-\infty}{\infty} \).

### Step 4: Apply L'Hôpital's Rule
Using L'Hôpital's Rule, we differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}(\ln x) = \frac{1}{x} \]
\[ \text{Denominator: } \frac{d}{dx}(x^{-(2\alpha + 1)}) = -(2\alpha + 1)x^{-(2\alpha + 2)} \]
Thus, we have:
\[ \lim_{x \to 0^+} \frac{\ln x}{x^{-(2\alpha + 1)}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-(2\alpha + 1)x^{-(2\alpha + 2)}} \]
This simplifies to:
\[ \lim_{x \to 0^+} \frac{-x^{2\alpha + 2}}{2\alpha + 1} \]

### Step 5: Analyze the limit
For the limit to equal 0, we need \( 2\alpha + 2 > 0 \) or \( \alpha > -1 \). If \( \alpha = -1 \), the limit diverges.

### Step 6: Check differentiability
Next, we check if \( f(x) \) is differentiable in the interval \((0, 1)\). The function \( f(x) = x^{2\alpha + 1} \ln x \) is differentiable for \( x > 0 \).

### Step 7: Conclusion
From the analysis, we conclude:
- For \( f(x) \) to be continuous at \( x = 0 \), we require \( \alpha > -1 \).
- The function is differentiable in the interval \((0, 1)\).

Thus, the values of \( \alpha \) that satisfy Rolle's theorem in the interval \([0, 1]\) are:
\[ \alpha = -\frac{1}{3} \quad \text{and} \quad \alpha = -\frac{1}{4} \]