<p>The minimum value of the expression \(\dfrac{\sin^3\alpha + 6\sin^2\alpha + \sin\alpha + 2\cos^2\alpha - 8}{\sin\alpha - 1}\) is equal to:</p>
Step-by-Step Solution
Key Concept: Substitute t = sin α where t ∈ [-1,1], then factor the numerator as a cubic in t and simplify by canceling (t-1) from numerator and denominator, reducing to a quadratic whose minimum can be found using calculus.
<p><strong>Step 1:</strong> Let t = sin α where t ∈ [-1, 1) and t ≠ 1. Use cos²α = 1 - sin²α = 1 - t².</p><p>Numerator becomes: t³ + 6t² + t + 2(1 - t²) - 8 = t³ + 6t² + t + 2 - 2t² - 8 = t³ + 4t² + t - 6</p><p><strong>Step 2:</strong> Factor the numerator t³ + 4t² + t - 6. Testing t = 1: 1 + 4 + 1 - 6 = 0, so (t - 1) is a factor.</p><p>Polynomial division: t³ + 4t² + t - 6 = (t - 1)(t² + 5t + 6) = (t - 1)(t + 2)(t + 3)</p><p><strong>Step 3:</strong> The expression becomes: $\frac{(t-1)(t+2)(t+3)}{t-1} = (t+2)(t+3)$ for t ≠ 1</p><p><strong>Step 4:</strong> Expand: f(t) = t² + 5t + 6 for t ∈ [-1, 1)</p><p><strong>Step 5:</strong> Find minimum: f'(t) = 2t + 5 = 0 ⟹ t = -5/2 (outside domain). Since f'(t) > 0 on [-1, 1), f is increasing on this interval.</p><p><strong>Step 6:</strong> Minimum occurs at t = -1: f(-1) = 1 - 5 + 6 = 2</p><p><strong>Step 7:</strong> As t → 1⁻, f(t) → 18 (not attained)</p><p>∴ Answer: B (Minimum value = 2)</p>
Correct Answer: B