Question

Let \( f: \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f(0) = 0 \), \( f(2) = 4 \), \( f(4) = 4 \), and \( f(8) = 12 \). Then which of the following statements is/are correct?

• A. \( f'(x) \leq 1 \) for all \( x \in [0, 2] \)
• B. \( f'(x_1) > 1 \) for some \( x_1 \in [0, 2] \)
• C. \( f'(x_2) > 1 \) for some \( x_2 \in [4, 8] \)
• D. \( f''(x_3) = 0 \) for some \( x_3 \in [0, 8] \)

Answer 1

The Correct Option is B, C, D

Step 1: Analyze \( f'(x) \) on \( [0, 2] \). By the Mean Value Theorem (MVT), for \( f \) on \( [0, 2] \): \[ f'(x_1) = \frac{f(2) - f(0)}{2 - 0} = \frac{4 - 0}{2} = 2, \quad \text{for some } x_1 \in (0, 2). \] Thus, \( f'(x_1) > 1 \) for some \( x_1 \in [0, 2] \), making \textbf{(B)} true. Step 2: Analyze \( f'(x) \) on \( [4, 8] \) By the MVT, for \( f \) on \( [4, 8] \): \[ f'(x_2) = \frac{f(8) - f(4)}{8 - 4} = \frac{12 - 4}{4} = 2, \quad \text{for some } x_2 \in (4, 8). \] So, \( f'(x_2) > 1 \) for some \( x_2 \in [4, 8] \), making \textbf{(C)} potentially true. However, further details about \( f \) are needed to confirm this conclusively. Step 3: Analyze \( f''(x) \) on \( [0, 8] \). Since \( f \) is twice differentiable and has a critical point at \( x = 4 \) (where \( f'(4) = 0 \)), by Rolle's Theorem, there exists \( x_3 \in [0, 8] \) such that: \[ f''(x_3) = 0. \] Thus, \textbf{(D)} is true. Step 4: Verify \textbf{(A)}. From Step 1, we showed that \( f'(x_1) = 2 > 1 \) for some \( x_1 \in [0, 2] \)
Conclusion: The correct answer is: \[ \boxed{\text{(B)}} \]