Question:

<p><strong>Step 1: Apply Rolle's Theorem</strong></p><p>Since f(1) = f(3) = 2, by Rolle's theorem, ∃ c ∈ (1,3) where f'(c) = 0.</p><p><strong>Step 2: Analyze f' using f'(0) = 2</strong></p><p>We have f'(0) = 2 > 0 and f'(c) = 0 for some c ∈ (1,3). Since f is continuous with range [-3,3], f cannot increase indefinitely.</p><p><strong>Step 3: Determine f''(ξ₁) < 0</strong></p><p>Between x = 0 and the point c where f'(c) = 0, f' decreases from 2 to 0. By Mean Value Theorem, ∃ ξ₁ ∈ (0,c) where f''(ξ₁) = [f'(c) - f'(0)]/(c-0) = -2/c < 0. Thus f''(ξ₁) < 0.</p><p><strong>Step 4: Determine behavior after x=3</strong></p><p>Since f(3) = 2 and the range is [-3,3], f cannot go much higher. The maximum value f can attain is 3. After f'(c) = 0 for c ∈ (1,3), to return to f(3) = 2 requires f' to become negative again or stay controlled. The bounded range forces constraints on how negative f'' can be overall.</p><p><strong>Step 5: Establish f''(ξ₂) > -3 for some ξ₂</strong></p><p>The total variation and range restriction [-3,3] prevent f'' from being too negative everywhere. ∃ ξ₂ where f''(ξ₂) > -3 (the derivative cannot be unboundedly negative given the compact range).</p><p>∴ Answer: Check provided options A, B, C, D based on derivative sign constraints from Rolle's theorem and range analysis</p>