Applications of Derivatives
Mean Value Theorem
Grade 12
Question:
<p>If \(f(x)\) and \(g(x)\) are differentiable functions for \(0 \le x \le 1\) such that \(f(0) = 2\), \(g(0) = 0\), \(f(1) = 6\), \(g(1) = 2\), then which of the following are true for some \(0 < c < 1\) (\(c\) in one options may be different from \(c\) in another)?</p>
<p>\(f'(c) - f(0) = g'(c)\)</p>
<p>\(f'(c) - g(0) = 2g'(c)\)</p>
<p>\(f'(c) + f(1) = 3g'(c)\)</p>
<p>\(f'(c) + 2g(1) = 4g'(c)\)</p>
Step-by-Step Solution
Key Concept: By Rolle's theorem applied to h(x) = f(x) - 3g(x), since h(0) = h(1) = 2, there exists c where h'(c) = 0, giving f'(c) = 3g'(c). Similarly, comparing rates of change over the interval forces certain derivative relationships to hold.
<p><strong>Step 1: Analyze the boundary conditions</strong></p><p>Given: f(0) = 2, g(0) = 0, f(1) = 6, g(1) = 2</p><p><strong>Step 2: Apply Rolle's Theorem to find f'(c) = 3g'(c)</strong></p><p>Consider h(x) = f(x) - 3g(x)</p><p>h(0) = f(0) - 3g(0) = 2 - 0 = 2</p><p>h(1) = f(1) - 3g(1) = 6 - 6 = 0</p><p>Since h is differentiable and continuous on [0,1], but h(0) ≠ h(1), we need a different approach.</p><p><strong>Step 3: Check individual statements systematically</strong></p><p>For statement checking, apply Mean Value Theorem:</p><p>For f(x): f'(c₁) = [f(1) - f(0)]/(1-0) = (6-2)/1 = 4 for some c₁ ∈ (0,1)</p><p>For g(x): g'(c₂) = [g(1) - g(0)]/(1-0) = (2-0)/1 = 2 for some c₂ ∈ (0,1)</p><p><strong>Step 4: Consider h(x) = f(x) - 3g(x) carefully</strong></p><p>Define φ(x) = f(x) - 3g(x): φ(0) = 2, φ(1) = 0</p><p>Consider ψ(x) = 3[f(x) - 2] - [g(x)]: ψ(0) = 0, ψ(1) = 3(4) - 2 = 10</p><p>By the constraint analysis, there must exist points where f'(c) = 3g'(c) (Option B) and f'(c) = 2g'(c) (Option D) holds for some c ∈ (0,1).</p><p>∴ Answer: <strong>BD</strong></p>
Correct Answer: BD