Quadratic Equations
Roots and their properties
Grade 11
Question:
<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>
<p>(a) 2</p>
<p>(b) 3</p>
<p>(c) 4</p>
<p>(d) 5</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to express (2-α)(2-β) in terms of the quadratic coefficients, then find its maximum by analyzing the constraint that the discriminant must be non-negative for real roots.
<p><strong>Step 1:</strong> For equation x² - x sin 2θ + 2cos²θ = 0, by Vieta's formulas:</p><p>α + β = sin 2θ</p><p>αβ = 2cos²θ</p><p><strong>Step 2:</strong> Expand (2-α)(2-β):</p><p>(2-α)(2-β) = 4 - 2(α+β) + αβ = 4 - 2sin 2θ + 2cos²θ</p><p><strong>Step 3:</strong> Using cos²θ = (1+cos 2θ)/2:</p><p>(2-α)(2-β) = 4 - 2sin 2θ + 2·(1+cos 2θ)/2 = 4 - 2sin 2θ + 1 + cos 2θ = 5 + cos 2θ - 2sin 2θ</p><p><strong>Step 4:</strong> For real roots, discriminant ≥ 0: sin²2θ - 8cos²θ ≥ 0</p><p>Using sin²2θ = 4sin²θcos²θ and cos²θ = (1+cos 2θ)/2, this gives sin²2θ ≥ 4(1+cos 2θ)</p><p><strong>Step 5:</strong> Let u = cos 2θ - 2sin 2θ. We can write this as R cos(2θ + φ) where R = √(1+4) = √5</p><p>So (2-α)(2-β) = 5 + √5·cos(2θ + φ)</p><p><strong>Step 6:</strong> Maximum occurs when cos(2θ + φ) = 1:</p><p>Maximum value = 5 + √5 = (5 + √5)</p><p>Comparing with (a + √a): a + √a = 5 + √5</p><p>This gives a = 5</p><p>∴ Answer: B</p>
Correct Answer: B