<p>If the function \(f(x) = 2x^3 - 9ax^2 + 12a^2x + 1\), where \(a > 0\), attains its maximum and minimum at \(p\) and \(q\), respectively, such that \(p^2 = q\), then \(a\) equals</p>
Step-by-Step Solution
Key Concept: For a cubic with local extrema at p and q, the derivative f'(x) = 0 gives two critical points. Using Vieta's formulas on f'(x) and the constraint p² = q connects the critical points algebraically, allowing us to solve for a.
<p><strong>Step 1:</strong> Find the derivative: f'(x) = 6x² - 18ax + 12a²</p><p><strong>Step 2:</strong> At extrema, f'(x) = 0: 6x² - 18ax + 12a² = 0, or x² - 3ax + 2a² = 0</p><p><strong>Step 3:</strong> By Vieta's formulas for critical points p and q: p + q = 3a and pq = 2a²</p><p><strong>Step 4:</strong> Use the constraint p² = q. Substitute q = p² into Vieta's sum: p + p² = 3a</p><p><strong>Step 5:</strong> Substitute q = p² into Vieta's product: p · p² = 2a², so p³ = 2a²</p><p><strong>Step 6:</strong> From p³ = 2a², we get a² = p³/2, so a = (p³/2)^(1/2). From p + p² = 3a, substitute: p + p² = 3(p³/2)^(1/2)</p><p><strong>Step 7:</strong> Let p = 2t. Then 2t + 4t² = 3(4t³)^(1/2) = 6t^(3/2). Simplifying: t(2 + 4t) = 6t^(3/2), so 2 + 4t = 6t^(1/2). Squaring and solving yields t = 1, hence p = 2</p><p><strong>Step 8:</strong> From p³ = 2a²: 8 = 2a², so a² = 4, thus a = 2 (since a > 0)</p><p>∴ Answer: C</p>
Correct Answer: C