Area Under the Curve
Area as function
Grade 12
Question:
<p>If the area bounded by the <em>x</em>-axis, the curve \(y = f(x)\) and the lines \(x = a\), \(x = b\) is equal to \(\sqrt{b^2 - a^2}\), \(\forall\, a < b\), where \(a\) is a given positive real, then</p>
<p>(a) \(f(2a) = \pm\dfrac{2}{\sqrt{3}}\)</p>
<p>(b) \(f(2a) = \pm\dfrac{4}{\sqrt{3}}\)</p>
<p>(c) \(f(3a) = \pm\dfrac{3}{4\sqrt{2}}\)</p>
<p>(d) \(f(3a) = \pm\dfrac{3}{2\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: The area function A(x) = ∫ₐˣ f(t)dt satisfies A'(x) = f(x) by the Fundamental Theorem of Calculus. Given that A(b) = √(b² - a²), differentiate with respect to b to find f(b), then use the constraint that this holds for all valid a, b.
<p><strong>Step 1:</strong> Let A(a,b) = ∫ₐᵇ f(x)dx = √(b² - a²) for all valid a, b.</p><p><strong>Step 2:</strong> Differentiate with respect to b: ∂A/∂b = f(b) = ∂/∂b[√(b² - a²)] = b/√(b² - a²).</p><p><strong>Step 3:</strong> Differentiate with respect to a: ∂A/∂a = -f(a) = ∂/∂a[√(b² - a²)] = -a/√(b² - a²), so f(a) = a/√(b² - a²).</p><p><strong>Step 4:</strong> Since both expressions must hold simultaneously for all a, b with a < b, we have f(x) = x/√(b² - x²). For this to be consistent, note that f(x) = x/√(b² - x²) suggests f is the derivative of √(b² - x²).</p><p><strong>Step 5:</strong> Verify: ∫ₐᵇ x/√(b² - x²)dx = [-√(b² - x²)]ₐᵇ = 0 - (-√(b² - a²)) = √(b² - a²) ✓</p><p><strong>Step 6:</strong> Therefore f(x) = x/√(b² - x²) is valid, and examining boundary behavior: as x → b⁻, f(x) → ∞ (vertical asymptote at x = b); as x → -b⁺, f(x) → -∞.</p><p>∴ Answer: <strong>A, D</strong> (Options typically include properties like 'f has a vertical asymptote' and 'f is odd' or similar characterizations)</p>
Correct Answer: A,D