<p>Let \( f(x) = \dfrac{1}{\cos^2 x} + \dfrac{4}{\sin^2 x} \). The minimum value of \( f(x) \) for \( 0 < x < \dfrac{\pi}{2} \) is:</p>
Step-by-Step Solution
Key Concept: Use Cauchy-Schwarz inequality or calculus: express f(x) in terms of a single variable using substitution, then apply AM-GM or find critical points by setting f'(x) = 0.
<p><strong>Step 1:</strong> Let u = cos²x, so sin²x = 1 - u where u ∈ (0,1).</p><p>f(x) = 1/u + 4/(1-u)</p><p><strong>Step 2:</strong> Find critical points: f'(u) = -1/u² + 4/(1-u)² = 0</p><p>This gives: 4u² = (1-u)²</p><p>Taking positive root: 2u = 1-u → u = 1/3</p><p><strong>Step 3:</strong> So cos²x = 1/3, sin²x = 2/3</p><p><strong>Step 4:</strong> Substitute back:</p><p>f(x) = 1/(1/3) + 4/(2/3) = 3 + 6 = 9</p><p><strong>Verification:</strong> By Cauchy-Schwarz: (1/cos²x + 4/sin²x)(cos²x + sin²x) ≥ (1 + 2)² = 9</p><p>∴ Answer: <strong>9</strong></p>
Correct Answer: C