Limits, Continuity & Differentiability
Continuity
Grade None
Question:
<p>Let <span>\( f(x) \)</span> be a non-negative continuous function such that the area bounded by the curve <span>\( y = f(x) \)</span>, <span>\( x \)</span>-axis and the ordinates <span>\( x = \dfrac{\pi}{4} \)</span> and <span>\( x = \beta > \dfrac{\pi}{4} \)</span> is <span>\( \left( \beta \sin\beta + \dfrac{\pi}{4}\cos\beta + \sqrt{2}\beta \right) \)</span>. Then <span>\( f\!\left(\dfrac{\pi}{2}\right) \)</span> is</p>
<p>\( \left( \dfrac{\pi}{4} + \sqrt{2} - 1 \right) \)</p>
<p>\( \left( \dfrac{\pi}{4} - \sqrt{2} + 1 \right) \)</p>
<p>\( \left( 1 - \dfrac{\pi}{4} - \sqrt{2} \right) \)</p>
<p>\( \left( 1 - \dfrac{\pi}{4} + \sqrt{2} \right) \)</p>
Step-by-Step Solution
Key Concept: Use Leibniz rule for differentiation under the integral sign: if A(β) = ∫[a to β] f(x)dx, then dA/dβ = f(β). Differentiate the given area expression to recover f(β).
<p><strong>Step 1:</strong> Let the area function be:</p><p>A(β) = ∫[π/4 to β] f(x)dx = β sin β + (π/4)cos β + √2·β</p><p><strong>Step 2:</strong> By the Fundamental Theorem of Calculus (Leibniz Rule), differentiate both sides with respect to β:</p><p>f(β) = dA/dβ = d/dβ[β sin β + (π/4)cos β + √2·β]</p><p><strong>Step 3:</strong> Apply differentiation rules:</p><p>f(β) = (sin β + β cos β) + (π/4)(-sin β) + √2</p><p>f(β) = sin β + β cos β - (π/4)sin β + √2</p><p>f(β) = sin β(1 - π/4) + β cos β + √2</p><p><strong>Step 4:</strong> Evaluate at β = π/2:</p><p>f(π/2) = sin(π/2)(1 - π/4) + (π/2)cos(π/2) + √2</p><p>f(π/2) = 1·(1 - π/4) + (π/2)·0 + √2</p><p>f(π/2) = 1 - π/4 + √2</p><p>∴ Answer: D</p>
Correct Answer: D