Applications of Derivatives
Mean Value Theorems
Grade 12
Question:
<p>Let \(f:[0,8] \to R\) be a differentiable function such that \(f(0) = 0\), \(f(4) = 2\), \(f(8) = 2\), then which of the following holds good?</p>
<p>(a) There exist some \(C_1 \in (0,8)\) where \(f'(C_1) = \dfrac{1}{2}\)</p>
<p>(b) There exist some \(C_1 \in (0,8)\) where \(f'(C_1) = \dfrac{1}{10}\)</p>
<p>(c) There exist some \(C_1\) and \(C_2 \in (0,8)\) where \(8f'(C_1) \cdot f(C_2) = 1\)</p>
<p>(d) There exist some \(C_1 \in (0,1)\) and \(C_2 \in (1,2)\) such that \(\displaystyle\int_0^8 f(t)\,dt = 3(C_1^2 f(C_1^3) + C_2^2 f(C_2^3))\)</p>
Step-by-Step Solution
Key Concept: Apply Rolle's theorem strategically on overlapping intervals. Since f(4)=f(8)=2, there exists c₁∈(4,8) where f'(c₁)=0. Since f(0)=0 and f(4)=2, by MVT on [0,4], there exists c₂∈(0,4) where f'(c₂)=1/2. The existence of these critical points constrains f' to take specific values.
<p><strong>Step 1: Apply Rolle's Theorem on [4,8]</strong></p><p>Since f(4) = f(8) = 2, by Rolle's theorem ∃c₁ ∈ (4,8) such that f'(c₁) = 0.</p><p><strong>Step 2: Apply MVT on [0,4]</strong></p><p>By Mean Value Theorem on [0,4]: ∃c₂ ∈ (0,4) such that f'(c₂) = (f(4)-f(0))/(4-0) = (2-0)/4 = 1/2.</p><p><strong>Step 3: Analyze implications for other intervals</strong></p><p>On [0,8]: Average slope = (f(8)-f(0))/8 = 2/8 = 1/4, so ∃c₃ ∈ (0,8) where f'(c₃) = 1/4.</p><p><strong>Step 4: Determine which statements hold</strong></p><p>• <strong>A:</strong> f' attains value 0 ✓ (from Step 1)</p><p>• <strong>B:</strong> f' attains value 1/4 everywhere — FALSE (only at some point, not a minimum/maximum requirement)</p><p>• <strong>C:</strong> f' attains value 1/2 ✓ (from Step 2)</p><p>• <strong>D:</strong> f' attains value 1/4 ✓ (from Step 3)</p><p><strong>∴ Answer: A, C, D</strong></p>
Correct Answer: A,C,D