If `f(x) = x^(a) log x and f(0) = 0 ` then the value of `alpha` for which Rolle's theorem can be applied in [0,1] is

Options

A. -2
B. -1
C. 0
D. `1/2`

Answer

To solve the problem, we need to determine the value of \( \alpha \) for which Rolle's theorem can be applied to the function \( f(x) = x^{\alpha} \log x \) on the interval \([0, 1]\), given that \( f(0) = 0 \).

### Step-by-Step Solution:

1. **Understanding the Conditions for Rolle's Theorem**:
- For Rolle's theorem to be applicable, the function \( f(x) \) must be continuous on the closed interval \([0, 1]\) and differentiable on the open interval \((0, 1)\).
- Additionally, we need \( f(0) = f(1) \).

2. **Evaluate \( f(0) \)**:
- We know that \( f(0) = 0 \) is given.

3. **Finding \( f(1) \)**:
- Calculate \( f(1) \):
\[ f(1) = 1^{\alpha} \log(1) = 1 \cdot 0 = 0 \]
- Thus, \( f(1) = 0 \).

4. **Condition for Continuity at \( x = 0 \)**:
- We need to check the continuity of \( f(x) \) at \( x = 0 \). This requires:
\[ \lim_{x \to 0^+} f(x) = f(0) \]
- Therefore, we need to evaluate:
\[ \lim_{x \to 0^+} x^{\alpha} \log x \]

5. **Rewriting the Limit**:
- We can rewrite the limit as:
\[ \lim_{x \to 0^+} \frac{\log x}{x^{-\alpha}} \]
- This limit is in the form \( \frac{-\infty}{\infty} \) as \( x \to 0^+ \).

6. **Applying L'Hôpital's Rule**:
- We apply L'Hôpital's Rule:
\[ \lim_{x \to 0^+} \frac{\log x}{x^{-\alpha}} = \lim_{x \to 0^+} \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(x^{-\alpha})} \]
- The derivatives are:
- \( \frac{d}{dx}(\log x) = \frac{1}{x} \)
- \( \frac{d}{dx}(x^{-\alpha}) = -\alpha x^{-\alpha - 1} \)

7. **Substituting the Derivatives**:
- Therefore, we have:
\[ \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\alpha x^{-\alpha - 1}} = \lim_{x \to 0^+} \frac{-x^{\alpha + 1}}{\alpha} \]

8. **Evaluating the Limit**:
- As \( x \to 0^+ \), \( -x^{\alpha + 1} \) approaches \( 0 \) if \( \alpha + 1 > 0 \) (i.e., \( \alpha > -1 \)).
- Thus, for the limit to exist and equal \( 0 \), we need \( \alpha > 0 \).

9. **Conclusion**:
- Therefore, the value of \( \alpha \) for which Rolle's theorem can be applied in the interval \([0, 1]\) is:
\[ \alpha > 0 \]