<p>If <em>α</em> and <em>β</em> be the roots of the equation \(x^2 + px - 1/(2p^2) = 0\), where \(p \in \mathbb{R}\). Then the minimum value of \(\alpha^4 + \beta^4\) is</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to express α+β and αβ, then build up to α⁴+β⁴ using the recurrence relation: α⁴+β⁴ = (α²+β²)² - 2(αβ)². Minimize the resulting expression in terms of p.
<p><strong>Step 1:</strong> By Vieta's formulas for x² + px - 1/(2p²) = 0:</p><p>α + β = -p and αβ = -1/(2p²)</p><p><strong>Step 2:</strong> Compute α² + β²:</p><p>α² + β² = (α+β)² - 2αβ = p² - 2(-1/(2p²)) = p² + 1/p²</p><p><strong>Step 3:</strong> Compute α⁴ + β⁴ using the identity:</p><p>α⁴ + β⁴ = (α²+β²)² - 2(αβ)² = (p² + 1/p²)² - 2(1/(2p²))²</p><p>= (p² + 1/p²)² - 1/(2p⁴)</p><p>= p⁴ + 2 + 1/p⁴ - 1/(2p⁴)</p><p>= p⁴ + 2 + (2-1)/(2p⁴) = p⁴ + 2 + 1/(2p⁴)</p><p><strong>Step 4:</strong> Let u = p⁴ where u > 0. Then f(u) = u + 2 + 1/(2u)</p><p>Minimize: f'(u) = 1 - 1/(2u²) = 0 ⟹ u² = 1/2 ⟹ u = 1/√2</p><p><strong>Step 5:</strong> Minimum value = 1/√2 + 2 + 1/(2·1/√2) = 1/√2 + 2 + √2/2 = √2 + 2 + √2/2 = 3√2/2 + 2 = (3√2 + 4)/2</p><p>∴ Answer: D</p>
Correct Answer: D