<p>Given limit = \(\displaystyle\lim_{x \to \frac{\pi}{4}} \dfrac{f(\sec^2 x) \cdot 2\sec^2 x \tan x}{2x} = \dfrac{8}{\pi} f(2)\). If \(k^a = 8^2\), find the value of \(k^a\).</p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's rule or substitution u = sec²x to convert the limit into a form involving f'(2), then match coefficients with the given limit value to deduce properties of f. The phrase 'f(2)' in the answer suggests f is differentiable at 2.
<p><strong>Step 1:</strong> Recognize that at x = π/4: sec²(π/4) = 2 and tan(π/4) = 1, so the numerator approaches f(2)·2·2·1 = 4f(2) and denominator approaches π/2. This suggests we need L'Hôpital's rule.</p><p><strong>Step 2:</strong> Note that d/dx[sec²x] = 2sec²x·tan x. So the numerator is f(sec²x)·(d/dx[sec²x]). By chain rule, d/dx[f(sec²x)] = f'(sec²x)·2sec²x·tan x.</p><p><strong>Step 3:</strong> Apply L'Hôpital's rule: lim(x→π/4) [d/dx(f(sec²x)·2sec²x·tan x)] / 2 = f'(2)·4 + f(2)·(2sec²x·tan x)'|ₓ₌π/₄. Simplifying the second derivative term and evaluating at x = π/4 gives f'(2)·4 = (8/π)f(2).</p><p><strong>Step 4:</strong> This yields f'(2) = (2/π)f(2). Given the constraint k^a = 8² = 64, and matching the coefficient 8/π in the original equation with our derivation:</p><p>∴ Answer: k^a = <strong>64</strong></p>
Correct Answer: C