Limits, Continuity & Differentiability
Mean Value Theorems
Grade 12
Question:
<p>Assume that \(f\) is continuous on \([a, b]\), \(a > 0\) and differentiable on \((a, b)\). If \(\dfrac{f(a)}{a} = \dfrac{f(b)}{b}\), then there exists \(x_0 \in (a, b)\) such that:</p>
<p>\(x_0 f'(x_0) = f(x_0)\)</p>
<p>\(f'(x_0) + x_0 f(x_0) = 0\)</p>
<p>\(x_0 f'(x_0) + f(x_0) = 0\)</p>
<p>\(f'(x_0) = x_0^2 f(x_0)\)</p>
Step-by-Step Solution
Key Concept: Apply Rolle's theorem to the auxiliary function g(x) = f(x)/x, which will have equal values at endpoints, guaranteeing a zero derivative at some interior point x₀.
<p><strong>Step 1:</strong> Construct the auxiliary function g(x) = f(x)/x on [a,b]. Since f is continuous on [a,b] and a > 0, g is continuous on [a,b].</p><p><strong>Step 2:</strong> Since f is differentiable on (a,b), g is differentiable on (a,b) with g'(x) = [f'(x)·x - f(x)]/x².</p><p><strong>Step 3:</strong> Observe that g(a) = f(a)/a = f(b)/b = g(b), so the boundary condition for Rolle's theorem is satisfied.</p><p><strong>Step 4:</strong> By Rolle's theorem, there exists x₀ ∈ (a,b) such that g'(x₀) = 0.</p><p><strong>Step 5:</strong> Therefore: [f'(x₀)·x₀ - f(x₀)]/x₀² = 0, which gives f'(x₀)·x₀ - f(x₀) = 0.</p><p>∴ Answer: C (there exists x₀ ∈ (a,b) such that <strong>f'(x₀) = f(x₀)/x₀</strong> or equivalently <strong>x₀f'(x₀) = f(x₀)</strong>)</p>
Correct Answer: C