<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 calculus (derivative = 0) combined with the constraint that cos²x + sin²x = 1 to find critical points. Alternatively, recognize this as a weighted sum problem solvable by AM-GM or Cauchy-Schwarz with careful algebraic manipulation.
<p><strong>Step 1:</strong> Let f(x) = sec²x + 4csc²x. Differentiate with respect to x:</p><p>f'(x) = 2sec²x·tan x - 8csc²x·cot x</p><p><strong>Step 2:</strong> Set f'(x) = 0:</p><p>2sec²x·tan x = 8csc²x·cot x</p><p>2·(sin x/cos³x) = 8·(cos x/sin³x)</p><p>2sin⁴x = 8cos⁴x</p><p>sin⁴x = 4cos⁴x</p><p>sin²x = 2cos²x (taking positive root)</p><p><strong>Step 3:</strong> Using sin²x + cos²x = 1:</p><p>2cos²x + cos²x = 1</p><p>3cos²x = 1 → 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>∴ Minimum value = <strong>9</strong> (Answer: C)</p>
Correct Answer: C