<p><strong>250.</strong> If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 - x\sin 2\theta + 2\cos^2\theta = 0\), \(\theta \in R\) and the maximum value of \((2-\alpha)(2-\beta)\) is \((a + \sqrt{a})\), then \(a\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to express (2-α)(2-β) in terms of sum and product of roots, then maximize using the constraint that the discriminant must be non-negative for real roots.
<p><strong>Step 1:</strong> From the equation x² - x sin 2θ + 2cos²θ = 0, by Vieta's formulas:</p><ul><li>α + β = sin 2θ</li><li>αβ = 2cos²θ</li></ul><p><strong>Step 2:</strong> Expand (2-α)(2-β) = 4 - 2α - 2β + αβ = 4 - 2(α+β) + αβ</p><p>= 4 - 2sin 2θ + 2cos²θ</p><p><strong>Step 3:</strong> For real roots, discriminant Δ ≥ 0:</p><p>sin²2θ - 8cos²θ ≥ 0</p><p>Using sin 2θ = 2sin θ cos θ and cos²θ = (1+cos 2θ)/2:</p><p>4sin²θ cos²θ - 4(1+cos 2θ) ≥ 0</p><p><strong>Step 4:</strong> Let f(θ) = 4 - 2sin 2θ + 2cos²θ. Using cos²θ = (1+cos 2θ)/2:</p><p>f(θ) = 5 - 2sin 2θ + cos 2θ</p><p><strong>Step 5:</strong> Rewrite as f(θ) = 5 + √5 sin(2θ + φ) where √5 = √(4+1)</p><p>Maximum of sin(2θ + φ) = 1, giving max f(θ) = 5 + √5</p><p><strong>Step 6:</strong> Given maximum = a + √a, we have:</p><p>a + √a = 5 + √5</p><p>Comparing: a = 5</p><p>∴ Answer: B (a = 5)</p>
Correct Answer: B