<p>Solve the differential equation with integrating factor. Given the solution passes through \((\pi/2, 8)\), find the minimum value of \(y\) where \(y = \dfrac{8(1-\cos x)}{\sin^2 x}\).</p>
Step-by-Step Solution
Key Concept: Recognize that y = 8(1-cosx)/sin²x can be rewritten using the identity 1-cosx = 2sin²(x/2) and sin²x = 4sin²(x/2)cos²(x/2), simplifying to y = 2tan²(x/2). Then find the minimum by calculus or recognize the function's behavior.
<p><strong>Step 1:</strong> Simplify using trigonometric identities.</p><p>Given: y = 8(1-cos x)/sin²x</p><p>Use 1 - cos x = 2sin²(x/2) and sin²x = 4sin²(x/2)cos²(x/2)</p><p>y = 8·2sin²(x/2) / [4sin²(x/2)cos²(x/2)] = 4/cos²(x/2) = 4sec²(x/2)</p><p><strong>Step 2:</strong> Verify the solution passes through (π/2, 8).</p><p>At x = π/2: y = 4sec²(π/4) = 4·(√2)² = 4·2 = 8 ✓</p><p><strong>Step 3:</strong> Find the minimum value.</p><p>Since y = 4sec²(x/2) and sec²(x/2) ≥ 1 for all x, with equality when cos(x/2) = ±1</p><p>The minimum value occurs when cos(x/2) = 1, i.e., x/2 = 0, giving x = 0</p><p>Minimum y = 4·1² = <strong>4</strong></p><p>∴ Answer: C</p>
Correct Answer: C