Definite Integration
Leibniz Rule and Limits
Grade 12
Question:
<p>Let \( f(x) \) be a continuous function \( \forall\, x \in \mathbb{R} \) 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 \( a, k \in \mathbb{N} \), then the value of \( k^a \) is equal to:</p>
<p>(a) 4</p>
<p>(b) 16</p>
<p>(c) 64</p>
<p>(d) 256</p>
Step-by-Step Solution
Key Concept: Use L'Hôpital's rule on the limit of an integral expression, then apply Leibniz rule for differentiation under the integral sign. The key is recognizing that the numerator and denominator both approach 0, requiring differentiation of the upper limit.
<p><strong>Step 1: Identify the indeterminate form</strong></p><p>As x → π/4: numerator → ∫₂² f(t)dt = 0 and denominator → 0. This is 0/0 form, so apply L'Hôpital's rule.</p><p><strong>Step 2: Differentiate numerator using Leibniz rule</strong></p><p>d/dx[∫₂^(sec²x) f(t)dt] = f(sec²x)·d/dx(sec²x) = f(sec²x)·2sec²x·tanx</p><p><strong>Step 3: Differentiate denominator</strong></p><p>d/dx[x² - π²/16] = 2x</p><p><strong>Step 4: Apply L'Hôpital's limit</strong></p><p>lim(x→π/4) [f(sec²x)·2sec²x·tanx]/(2x) = k/π·f(a)</p><p><strong>Step 5: Evaluate at x = π/4</strong></p><p>At x = π/4: sec²(π/4) = 2, tan(π/4) = 1</p><p>[f(2)·2·2·1]/(2·π/4) = k/π·f(a)</p><p>[4f(2)]/(π/2) = k/π·f(a)</p><p>8f(2)/π = k/π·f(a)</p><p><strong>Step 6: Compare coefficients</strong></p><p>This gives us: 8f(2) = k·f(a)</p><p>Since this holds for arbitrary continuous f, we have: a = 2 and k = 8</p><p><strong>Step 7: Calculate k^a</strong></p><p>k^a = 8² = 64</p><p>∴ Answer: <strong>64</strong></p>
Correct Answer: C