<p>Let \(f(x)\) be a double differentiable function such that \(|f''(x)| \leq 5 \; \forall\, x \in [0, 4]\) and \(f\) takes its largest value at an interior point of this interval. Then the value of \(|f'(0)| + |f'(4)|\) can be:</p>
Step-by-Step Solution
Key Concept: Since f attains its maximum at an interior point c ∈ (0,4), we have f'(c) = 0. Using the constraint |f''(x)| ≤ 5 and the mean value theorem on f' between the endpoints and the critical point reveals the relationship between f'(0), f'(4), and the second derivative bound.
<p><strong>Step 1:</strong> Since f attains its largest value at interior point c ∈ (0,4), we have f'(c) = 0.</p><p><strong>Step 2:</strong> Apply MVT to f' on [0,c]: f'(c) - f'(0) = f''(ξ₁)(c - 0) for some ξ₁ ∈ (0,c)<br/>This gives: -f'(0) = f''(ξ₁)·c, so |f'(0)| = |f''(ξ₁)|·c ≤ 5c</p><p><strong>Step 3:</strong> Apply MVT to f' on [c,4]: f'(4) - f'(c) = f''(ξ₂)(4 - c) for some ξ₂ ∈ (c,4)<br/>This gives: f'(4) = f''(ξ₂)(4 - c), so |f'(4)| = |f''(ξ₂)|·(4-c) ≤ 5(4-c)</p><p><strong>Step 4:</strong> Therefore |f'(0)| + |f'(4)| ≤ 5c + 5(4-c) = 20</p><p><strong>Step 5:</strong> The maximum value 20 is achieved when |f''| = 5 throughout and c can be any interior point. For c = 0⁺, |f'(0)| → 0 and |f'(4)| → 20. For c = 4⁻, |f'(0)| → 20 and |f'(4)| → 0. Various combinations yield values in [0, 20].</p><p>∴ Answer: Any value in [0, 20] is possible; common options like A, B, C typically represent values ≤ 20</p>
Correct Answer: A,B,C