Question

A function \( f \) is defined by \( f(x) = 2 + (x-1)^{2/3 \) on \( [0,2] \). Which of the following statements is incorrect?}

• A. \( f \) is not derivable in \( (0,2) \).
• B. \( f \) is continuous in \( [0,2] \).
• C. \( f(0) = f(2) \).
• D. Rolle’s theorem is applicable on \( [0,2] \).

Hint:

For Rolle’s theorem questions: \begin{itemize} \item Always check differentiability inside interval. \item Fractional powers like \( (x-a)^{2/3} \) cause cusps. \end{itemize}

Answer 1

The Correct Option is D

Concept: Rolle’s Theorem requires: \begin{itemize} \item Continuity on \( [a,b] \) \item Differentiability on \( (a,b) \) \item \( f(a) = f(b) \) \end{itemize} Step 1: {\color{red}Check continuity.} \[ f(x) = 2 + (x-1)^{2/3} \] This is continuous everywhere ⇒ continuous on \( [0,2] \). Step 2: {\color{red}Check endpoint values.} \[ f(0) = 2 + (-1)^{2/3} = 3 \] \[ f(2) = 2 + (1)^{2/3} = 3 \] So \( f(0)=f(2) \). Step 3: {\color{red}Check differentiability.} Derivative: \[ f'(x) = \frac{2}{3}(x-1)^{-1/3} \] At \( x=1 \), derivative is undefined (infinite slope). So function is not differentiable in \( (0,2) \). Step 4: {\color{red}Conclusion.} Since differentiability fails, Rolle’s theorem is not applicable. Thus statement (D) is incorrect.