Limits, Continuity & Differentiability
Continuity and Differentiability
Grade 12
Question:
<p>Given that <span>\(f(x) \geq 0\)</span> and continuous <span>\(\forall x \in \mathbb{R}\)</span>, and <span>\(A = \int_{\pi/4}^{\beta} f(x)\, dx = \left(\beta \sin\beta + \dfrac{\pi}{4}\cos\beta + \sqrt{2}\right)\beta\)</span>, <span>\(\beta > \dfrac{\pi}{4}\)</span>. Find <span>\(f\left(\dfrac{\pi}{2}\right)\)</span>.</p>
<p>\(1 + \dfrac{\pi}{4} + \sqrt{2}\)</p>
<p>\(1 - \dfrac{\pi}{4} - \sqrt{2}\)</p>
<p>\(1 - \dfrac{\pi}{4} + \sqrt{2}\)</p>
<p>\(\dfrac{\pi}{4} + \sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Differentiate the integral equation with respect to β using Leibniz rule to extract f(β), then evaluate at β = π/2 by ensuring continuity conditions are satisfied.
<p><strong>Step 1:</strong> Apply Leibniz rule to differentiate both sides with respect to β:</p><p>Left side: dA/dβ = f(β)</p><p><strong>Step 2:</strong> Differentiate the right side using product rule:</p><p>RHS = d/dβ[(β sin β + (π/4)cos β + √2)β]</p><p>= (sin β + β cos β - (π/4)sin β)·β + (β sin β + (π/4)cos β + √2)·1</p><p>= β sin β + β² cos β - (πβ/4)sin β + β sin β + (π/4)cos β + √2</p><p>= 2β sin β + β² cos β - (πβ/4)sin β + (π/4)cos β + √2</p><p><strong>Step 3:</strong> Therefore:</p><p>f(β) = 2β sin β + β² cos β - (πβ/4)sin β + (π/4)cos β + √2</p><p><strong>Step 4:</strong> Evaluate at β = π/2:</p><p>f(π/2) = 2(π/2)sin(π/2) + (π/2)² cos(π/2) - (π·π/2)/4·sin(π/2) + (π/4)cos(π/2) + √2</p><p>= 2(π/2)(1) + (π²/4)(0) - (π²/4)(1) + (π/4)(0) + √2</p><p>= π + 0 - π²/4 + 0 + √2</p><p>= π - π²/4 + √2 = <strong>(4π - π² + 4√2)/4</strong></p><p>∴ Answer: C</p>
Correct Answer: C