Question

If Rolle's Theorem is applicable for the function:

\[ f(x) = \begin{cases} x^p \log x, & x \neq 0 \\ 0, & x = 0 \end{cases} \]

on the interval \([0,1]\), then a possible value of \( p \) is:

• A. \( -2 \)
• B. \( -1 \)
• C. \( 0 \)
• D. \( 1 \)

Hint:

For Rolle's Theorem, always check continuity, differentiability, and \( f(a) = f(b) \). If these hold, use \( f'(c) = 0 \) for some \( c \) in \( (a,b) \).

Answer 1

The Correct Option is D

Step 1: Conditions for Rolle's Theorem 
Rolle's Theorem states that if a function \( f(x) \) is: 1. Continuous on the closed interval \([0,1]\), 2. Differentiable on the open interval \((0,1)\), 3. \( f(0) = f(1) \), then there exists \( c \in (0,1) \) such that \( f'(c) = 0 \). 

Step 2: Checking Continuity 
For \( f(x) \) to be continuous at \( x = 0 \), we check: \[ \lim_{x \to 0^+} f(x) = f(0). \] \[ \lim_{x \to 0^+} x^p \log x = 0. \] Using the standard limit result, \[ \lim_{x \to 0^+} x^p \log x = 0, \quad \text{if } p>0. \] Since \( f(0) = 0 \), the function is continuous at \( x = 0 \) if \( p>0 \). 

Step 3: Checking Differentiability 
Differentiate \( f(x) \) for \( x \neq 0 \): \[ f'(x) = p x^{p-1} \log x + x^{p-1}. \] For \( f(x) \) to be differentiable at \( x = 0 \), the right-hand derivative should exist: \[ \lim_{x \to 0^+} p x^{p-1} \log x + x^{p-1}. \] Using limits, \( f(x) \) is differentiable at \( x = 0 \) if \( p>1 \). 

 

Step 4: Verifying \( f(0) = f(1) \) 
\[ f(1) = 1^p \log 1 = 0. \] Since \( f(0) = 0 \), the condition \( f(0) = f(1) \) holds. 

Step 5: Conclusion 
For Rolle's theorem to hold, we need \( p = 1 \) to satisfy all conditions. Thus, the correct answer is: \[ \mathbf{1}. \]