Question:

Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ f(0) = 4, \, f(1) = -2, \, f(2) = 8, \, f(3) = 2. \] Then, which of the following is/are TRUE?

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Rolle’s Theorem guarantees that if a function is continuous on a closed interval and differentiable on the open interval, and if the function takes the same value at the endpoints, then there is at least one point in the open interval where the derivative is zero.
Updated On: Sep 6, 2025
  • \( |f'(x)|<5 \) for all \( x \in [0, 1] \).
  • \( |f'(x_1)| \geq 5 \) for some \( x_1 \in [0, 1] \).
  • \( f'(x_2) = 0 \) for some \( x_2 \in [0, 3] \).
  • \( f''(x_3) = 0 \) for some \( x_3 \in [0, 3] \).
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The Correct Option is C

Solution and Explanation

Step 1: Apply Rolle’s Theorem.
We know that \( f(0) = 4 \) and \( f(1) = -2 \), so the function changes sign over the interval \( [0, 1] \). By Rolle’s Theorem, since \( f(x) \) is continuous and differentiable on the closed interval \( [0, 1] \), and \( f(0) = f(1) \), there must exist some \( c_1 \in (0, 1) \) such that \( f'(c_1) = 0 \). Step 2: Conclusion.
From Rolle's Theorem and the function's values, we conclude that \( f'(x) = 0 \) for some \( x \in [0, 3] \), so option (C) is true. Final Answer: \[ \boxed{f'(x_2) = 0 \text{ for some } x_2 \in [0, 3].} \]
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