Limits, Continuity & Differentiability
Mean Value Theorem
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 converts the given equality condition into a form where Rolle's theorem applies directly.
<p><strong>Step 1:</strong> Given that f is continuous on [a,b], differentiable on (a,b), and f(a)/a = f(b)/b.</p><p><strong>Step 2:</strong> Construct auxiliary function g(x) = f(x)/x, which is continuous on [a,b] and differentiable on (a,b).</p><p><strong>Step 3:</strong> Evaluate at endpoints: g(a) = f(a)/a and g(b) = f(b)/b. By hypothesis, g(a) = g(b).</p><p><strong>Step 4:</strong> By Rolle's Theorem, since g(a) = g(b), there exists x₀ ∈ (a,b) such that g'(x₀) = 0.</p><p><strong>Step 5:</strong> Calculate g'(x) = [f'(x)·x - f(x)·1]/x² = [xf'(x) - f(x)]/x².</p><p><strong>Step 6:</strong> Setting g'(x₀) = 0: xf'(x₀) - f(x₀) = 0, which gives xf'(x₀) = f(x₀) or equivalently f'(x₀) = f(x₀)/x₀.</p><p>∴ There exists x₀ ∈ (a,b) such that <strong>x₀f'(x₀) = f(x₀)</strong> (or f'(x₀)/f(x₀) = 1/x₀)</p>
Correct Answer: C