Question:

<p><strong>Step 1: Find sin⁻¹(2/√5) - sin⁻¹(1/√5)</strong></p><p>Let α = sin⁻¹(2/√5) and β = sin⁻¹(1/√5)</p><p>Then sin α = 2/√5, cos α = √(1 - 4/5) = 1/√5</p><p>And sin β = 1/√5, cos β = √(1 - 1/5) = 2/√5</p><p>Using sin(α - β) = sin α cos β - cos α sin β:</p><p>sin(α - β) = (2/√5)(2/√5) - (1/√5)(1/√5) = 4/5 - 1/5 = 3/5</p><p>So α - β = sin⁻¹(3/5)</p><p><strong>Note:</strong> The expression equals a constant sin⁻¹(3/5) ≈ 0.6435, not a minimum. Let's denote this as root r₁ = 3/5 (working with the sine value).</p><p><strong>Step 2: Find the limit lim(x→∞) [(x+1)(x+2) - x]/2</strong></p><p>Expand the numerator: (x+1)(x+2) - x = x² + 3x + 2 - x = x² + 2x + 2</p><p>The limit becomes: lim(x→∞) (x² + 2x + 2)/2</p><p>Dividing by highest power: lim(x→∞) (x²/2 + x + 1) = ∞</p><p><strong>Reinterpretation:</strong> If the problem intends [(x+1)(x+2) - x²]/2:</p><p>= (x² + 3x + 2 - x²)/2 = (3x + 2)/2 → ∞</p><p><strong>If the limit is:</strong> lim(x→∞) [√((x+1)(x+2)) - x]/2</p><p>Using conjugate: = lim(x→∞) [(x+1)(x+2) - x²]/[2(√((x+1)(x+2)) + x)]</p><p>= lim(x→∞) (3x + 2)/[2(√(x² + 3x + 2) + x)] = lim(x→∞) (3x + 2)/[2·2x] = 3/4</p><p>So r₂ = 3/4</p><p><strong>Step 3: Form quadratic with roots r₁ = 3/5 and r₂ = 3/4</strong></p><p>Sum of roots = 3/5 + 3/4 = 12/20 + 15/20 = 27/20</p><p>Product of roots = (3/5)(3/4) = 9/20</p><p>Quadratic: x² - (27/20)x + 9/20 = 0</p><p>Multiply by 20: 20x² - 27x + 9 = 0</p><p><strong>Checking option (b):</strong> 8x² - 14x + 3 = 0 dividing by 2: 4x² - 7x + 3/2 = 0</p><p>This has roots: x = [7 ± √(49-24)]/8 = [7 ± 5]/8, giving 3/2 and 1/4</p><p>Given the discrepancies between interpretation and provided options, and the ambiguity in problem statement:</p><p><strong>∴ Answer:</strong> Not specified</p>