Limits, Continuity & Differentiability
Mean Value Theorem
Grade 12
Question:
<p>If \( f \) and \( g \) are differentiable functions in \([0, 1]\) satisfying \( f(0) = 2 = g(1) \), \( g(0) = 0 \) and \( f(1) = 6 \), then for some \( c \in [0, 1] \)</p>
<p>\( f'(c) = g'(c) \)</p>
<p>\( f'(c) = 2g'(c) \)</p>
<p>\( 2f'(c) = g'(c) \)</p>
<p>\( 2f'(c) = 3g'(c) \)</p>
Step-by-Step Solution
Key Concept: Apply Rolle's Theorem to the auxiliary function h(x) = f(x) - 2g(x), which satisfies h(0) = h(1) by the given boundary conditions. This guarantees that h'(c) = 0 for some c ∈ (0,1), yielding f'(c) = 2g'(c).
<p><strong>Step 1:</strong> Define an auxiliary function h(x) = f(x) - 2g(x) on [0,1].</p><p><strong>Step 2:</strong> Verify the boundary conditions:</p><p>h(0) = f(0) - 2g(0) = 2 - 2(0) = 2</p><p>h(1) = f(1) - 2g(1) = 6 - 2(2) = 6 - 4 = 2</p><p>So h(0) = h(1) = 2.</p><p><strong>Step 3:</strong> Since h(x) is continuous on [0,1] and differentiable on (0,1) with h(0) = h(1), by Rolle's Theorem there exists c ∈ (0,1) such that h'(c) = 0.</p><p><strong>Step 4:</strong> Computing the derivative:</p><p>h'(x) = f'(x) - 2g'(x)</p><p>h'(c) = 0 ⟹ f'(c) = 2g'(c)</p><p>∴ Answer: <strong>f'(c) = 2g'(c)</strong> for some c ∈ [0,1]</p>
Correct Answer: B