<p>If \(x^2 - x\sin 2\theta + 2\cos^2\theta = (x-\alpha)(x-\beta)\), then the maximum value of \((2-\alpha)(2-\beta)\) is \(5 + \sqrt{a}\). Find \(a\).</p>
Step-by-Step Solution
Key Concept: Express (2-α)(2-β) using Vieta's formulas from the quadratic, then recognize it as a function of θ that must be maximized by analyzing its critical points.
<p><strong>Step 1:</strong> From the quadratic x² - x·sin2θ + 2cos²θ = (x-α)(x-β), by Vieta's formulas:</p><p>α + β = sin2θ</p><p>αβ = 2cos²θ</p><p><strong>Step 2:</strong> Calculate (2-α)(2-β):</p><p>(2-α)(2-β) = 4 - 2(α+β) + αβ = 4 - 2sin2θ + 2cos²θ</p><p><strong>Step 3:</strong> Use cos²θ = (1+cos2θ)/2:</p><p>(2-α)(2-β) = 4 - 2sin2θ + 2·(1+cos2θ)/2 = 4 - 2sin2θ + 1 + cos2θ = 5 + cos2θ - 2sin2θ</p><p><strong>Step 4:</strong> Express in form R·sin(2θ + φ). Let f = cos2θ - 2sin2θ:</p><p>f = √(1² + 2²)·sin(2θ + φ) = √5·sin(2θ + φ)</p><p>where tan φ corresponds to the phase shift.</p><p><strong>Step 5:</strong> Maximum value of √5·sin(2θ + φ) is √5:</p><p>Max[(2-α)(2-β)] = 5 + √5</p><p><strong>Step 6:</strong> Comparing with 5 + √a:</p><p>∴ a = 5</p>
Correct Answer: 5