Definite Integration
Limit as a Definite Integral / L'Hopital
Grade 12
Question:
<p>Let \( f(x) \) be a continuous function \( \forall\, x \in R \) such that \[ \lim_{x \to \pi/4} \frac{\displaystyle\int_{2}^{\sec^2 x} f(t)\,dt}{x^2 - \dfrac{\pi^2}{16}} = \frac{k}{\pi} f(a) \] where \( a, k \in N \), then the value of \( k^a \) is equal to:</p>
<p>4</p>
<p>16</p>
<p>64</p>
<p>256</p>
Step-by-Step Solution
Key Concept: Apply L'Hôpital's rule to the indeterminate form, then use Leibniz rule for differentiating integrals with variable limits. The denominator's derivative reveals the limit structure.
<p><strong>Step 1:</strong> Check the form at x = π/4.</p><p>Numerator at x = π/4: ∫₂^(sec²(π/4)) f(t)dt = ∫₂² f(t)dt = 0 (since sec(π/4) = √2, so sec²(π/4) = 2)</p><p>Denominator at x = π/4: (π/4)² - π²/16 = 0</p><p>This is 0/0 form, so apply L'Hôpital's rule.</p><p><strong>Step 2:</strong> Differentiate numerator using Leibniz rule:</p><p>d/dx[∫₂^(sec²x) f(t)dt] = f(sec²x)·d/dx(sec²x) = f(sec²x)·2sec²x·tanx</p><p><strong>Step 3:</strong> Differentiate denominator:</p><p>d/dx[x² - π²/16] = 2x</p><p><strong>Step 4:</strong> Apply L'Hôpital's limit:</p><p>lim(x→π/4) [f(sec²x)·2sec²x·tanx]/(2x)</p><p>At x = π/4: sec²(π/4) = 2, tan(π/4) = 1, x = π/4</p><p>= [f(2)·2(2)·1]/(2·π/4) = (4f(2))/(π/2) = 8f(2)/π</p><p><strong>Step 5:</strong> Compare with k/π·f(a):</p><p>8f(2)/π = k/π·f(a)</p><p>Therefore: k = 8, a = 2</p><p>∴ k^a = 8² = <strong>64</strong></p>
Correct Answer: C