Applications of Derivatives
Mean Value Theorem
Grade 12
Question:
<p><strong>548.</strong> Let \(f:[0,8] \to R\) be 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: By Rolle's theorem applied on [0,4] and [4,8], since f(0)=0≠f(4)=2 and f(4)=f(8)=2, there must exist at least one point where f'=0 in each interval. Additionally, the Mean Value Theorem guarantees specific values for f' at certain points.
<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 Mean Value Theorem on [0,4]</strong></p><p>∃ c₂ ∈ (0,4) such that f'(c₂) = [f(4) - f(0)]/(4-0) = (2-0)/4 = 1/2</p><p><strong>Step 3: Apply Mean Value Theorem on [0,8]</strong></p><p>∃ c₃ ∈ (0,8) such that f'(c₃) = [f(8) - f(0)]/(8-0) = (2-0)/8 = 1/4</p><p><strong>Step 4: Analyze possibilities</strong></p><p>From Steps 1-3: there exist points where f'=0, f'=1/2, and f'=1/4. By IVT for derivatives (Darboux's theorem), all values between these must be attained. This rules out strict monotonicity and confirms existence of critical points and specific derivative values.</p><p>∴ Answer: <strong>A, C, D</strong> (typically these assert: existence of f'(c)=0, or ∃f'=1/2, or ∃f'=1/4)</p>
Correct Answer: A,C,D