<p>Let \( f(x) \) be 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 applying the Mean Value Theorem to f' on [0,c] and [c,4] creates bounds on |f'(0)| and |f'(4)| that can be combined to limit their sum.
<p><strong>Step 1:</strong> Since f attains maximum at interior point c ∈ (0,4), we have f'(c) = 0.</p><p><strong>Step 2:</strong> Apply MVT to f' on [0,c]: There exists c₁ ∈ (0,c) such that f''(c₁) = [f'(c) - f'(0)]/(c - 0) = -f'(0)/c</p><p>Thus: |f'(0)| = c|f''(c₁)| ≤ 5c</p><p><strong>Step 3:</strong> Apply MVT to f' on [c,4]: There exists c₂ ∈ (c,4) such that f''(c₂) = [f'(4) - f'(c)]/(4 - c) = f'(4)/(4 - c)</p><p>Thus: |f'(4)| = (4 - c)|f''(c₂)| ≤ 5(4 - c)</p><p><strong>Step 4:</strong> Adding the inequalities: |f'(0)| + |f'(4)| ≤ 5c + 5(4 - c) = 20</p><p><strong>Step 5:</strong> The maximum value of 20 is achieved when c = 2 and |f''| achieves maximum value 5 on both intervals. Therefore |f'(0)| + |f'(4)| can take any value in [0, 20], making options A, B, C all possible.</p><p>∴ Answer: A,B,C</p>
Correct Answer: A,B,C