Applications of Derivatives
Local Maxima and Minima
Grade 12
Question:
<p>Let \(f\) be a differentiable function satisfying \(x^2 f'(x) + 2xf(x) = e^x\) and \(f(2) = \dfrac{e^2}{4}\), then:</p>
<p>(a) \(f(x)\) has no local maxima and no local minima.</p>
<p>(b) \(f(x)\) has both local maxima and local minima.</p>
<p>(c) \(f(x)\) has local maxima but no local minima.</p>
<p>(d) \(f(x)\) has no local maxima but local minima.</p>
Step-by-Step Solution
Key Concept: Recognize that the left side x²f'(x) + 2xf(x) is the derivative of x²f(x) using the product rule. This transforms the differential equation into d/dx[x²f(x)] = eˣ, which can be integrated directly.
<p><strong>Step 1:</strong> Recognize the pattern on the left side.</p><p>Notice that x²f'(x) + 2xf(x) = d/dx[x²f(x)] (by product rule: d/dx[x²f(x)] = x²f'(x) + f(x)·2x)</p><p><strong>Step 2:</strong> Rewrite the equation.</p><p>d/dx[x²f(x)] = eˣ</p><p><strong>Step 3:</strong> Integrate both sides.</p><p>x²f(x) = ∫eˣ dx = eˣ + C</p><p><strong>Step 4:</strong> Apply the initial condition f(2) = e²/4.</p><p>4 · (e²/4) = e² + C</p><p>e² = e² + C</p><p>C = 0</p><p><strong>Step 5:</strong> Solve for f(x).</p><p>x²f(x) = eˣ</p><p>f(x) = eˣ/x²</p><p>∴ Answer: A</p>
Correct Answer: A