Step 1: Analyze the continuity of \( f(x) \) on \( [0, 2] \).
The function \( g(x) = (x - 1)^{2/3} = (\sqrt[3]{x - 1})^2 \) is continuous on \( [0, 2] \) as it is a composition of continuous functions.
Therefore, \( f(x) = 2 + g(x) \) is also continuous on \( [0, 2] \). Statement (B) is correct.
Step 2: Analyze the differentiability of \( f(x) \) on \( (0, 2) \).
The derivative of \( f(x) \) is \( f'(x) = \frac{2}{3}(x - 1)^{-1/3} = \frac{2}{3\sqrt[3]{x - 1}} \). This derivative is undefined at \( x = 1 \), which is in the interval \( (0, 2) \). Thus, \( f(x) \) is not differentiable in \( (0, 2) \). Statement (A) is correct.
Step 3: Check if \( f(0) = f(2) \).
\( f(0) = 2 + (0 - 1)^{2/3} = 2 + (-1)^{2/3} = 2 + 1 = 3 \)
\( f(2) = 2 + (2 - 1)^{2/3} = 2 + (1)^{2/3} = 2 + 1 = 3 \)
Since \( f(0) = f(2) = 3 \), statement (C) is correct.
Step 4: Check if Rolle's theorem is applicable on \( [0, 2] \).
Rolle's theorem requires three conditions to be met:
1. \( f \) is continuous on \( [0, 2] \). (Met)
2. \( f \) is differentiable on \( (0, 2) \). (Not met, as \( f \) is not differentiable at \( x = 1 \))
3. \( f(0) = f(2) \). (Met)
Since the second condition is not satisfied, Rolle's theorem is not applicable on \( [0, 2] \). Statement (D) is incorrect.