If `f(x)={(x^(alpha)logx , x > 0),(0, x=0):}` and Rolle's theorem is applicable to `f(x)` for `x in [0, 1]` then `alpha` may equal to (A) -2 (B) -1 (C) 0 (D) `1/2`

Options

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

Answer

To determine the value of \(\alpha\) for which Rolle's theorem is applicable to the function

\[ f(x) = \begin{cases} x^{\alpha} \log x & \text{if } x > 0 \\ 0 & \text{if } x = 0 \end{cases} \]

on the interval \([0, 1]\), we need to check the conditions of Rolle's theorem: continuity on the closed interval \([0, 1]\) and differentiability on the open interval \((0, 1)\).

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

To check continuity at \(x = 0\), we need to evaluate:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^{\alpha} \log x \]

We want this limit to equal \(f(0)\), which is \(0\).

### Step 2: Evaluate the Limit

We rewrite the limit:

\[ \lim_{x \to 0^+} x^{\alpha} \log x = \lim_{x \to 0^+} \frac{\log x}{x^{-\alpha}} \]

This limit is of the form \(\frac{-\infty}{\infty}\) when \(\alpha > 0\). We can apply L'Hôpital's rule, which states that we can differentiate the numerator and denominator:

\[ \text{Numerator: } \frac{d}{dx}(\log x) = \frac{1}{x} \]
\[ \text{Denominator: } \frac{d}{dx}(x^{-\alpha}) = -\alpha x^{-\alpha - 1} \]

Thus, we have:

\[ \lim_{x \to 0^+} \frac{\log x}{x^{-\alpha}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\alpha x^{-\alpha - 1}} = \lim_{x \to 0^+} \frac{-x^{\alpha}}{\alpha} \]

### Step 3: Evaluate the New Limit

Now, we need to evaluate:

\[ \lim_{x \to 0^+} \frac{-x^{\alpha}}{\alpha} \]

This limit approaches \(0\) if \(\alpha > 0\). Therefore, for continuity at \(x = 0\), we require \(\alpha > 0\).

### Step 4: Check Differentiability on \((0, 1)\)

Next, we need to check if \(f(x)\) is differentiable on \((0, 1)\). The function \(f(x) = x^{\alpha} \log x\) is differentiable for \(x > 0\) as long as \(\alpha\) is a real number.

### Step 5: Determine Valid Values for \(\alpha\)

Now we check the options given:

- **Option A**: \(\alpha = -2\) → Not valid (fails continuity).
- **Option B**: \(\alpha = -1\) → Not valid (fails continuity).
- **Option C**: \(\alpha = 0\) → Not valid (fails continuity).
- **Option D**: \(\alpha = \frac{1}{2}\) → Valid (satisfies continuity).

### Conclusion

Thus, the only value of \(\alpha\) that satisfies the conditions of Rolle's theorem for the function \(f(x)\) on the interval \([0, 1]\) is

\[ \boxed{\frac{1}{2}} \]