Limits, Continuity & Differentiability
Differential equations / functional equations
Grade 12
Question:
<p><strong>196.</strong> Let <em>f</em> and <em>g</em> be defined such that \(f'(x) = f^2(x) + g^2(x)\) and \(g'(x) = 2f(x)g(x) + 1\). If \(f(0) = \dfrac{1}{5}\), \(g(0) = \dfrac{4}{5}\), then the value of \(f\!\left(\dfrac{\pi}{12}\right) + g\!\left(\dfrac{\pi}{12}\right)\) equals:</p>
<p>(a) 0</p>
<p>(b) \(\dfrac{\sqrt{3}}{2}\)</p>
<p>(c) \(\sqrt{3}\)</p>
<p>(d) \(\dfrac{1}{\sqrt{3}}\)</p>
Step-by-Step Solution
Key Concept: Recognize that f'(x) = f²(x) + g²(x) and g'(x) = 2f(x)g(x) + 1 suggest considering the derivative of (f + g)² or analyzing f² + g² as a derivative of a composite function involving inverse trigonometric identities.
<p><strong>Step 1:</strong> Analyze the given system:</p><p>f'(x) = f²(x) + g²(x)</p><p>g'(x) = 2f(x)g(x) + 1</p><p><strong>Step 2:</strong> Let h(x) = f(x) + g(x). Then:</p><p>h'(x) = f'(x) + g'(x) = [f²(x) + g²(x)] + [2f(x)g(x) + 1]</p><p>h'(x) = f²(x) + 2f(x)g(x) + g²(x) + 1 = [f(x) + g(x)]² + 1 = h²(x) + 1</p><p><strong>Step 3:</strong> This is the standard differential equation:</p><p>dh/dx = h² + 1 ⟹ dh/(h² + 1) = dx</p><p>∫dh/(h² + 1) = ∫dx ⟹ tan⁻¹(h) = x + C</p><p><strong>Step 4:</strong> Apply initial condition h(0) = f(0) + g(0) = 1/5 + 4/5 = 1:</p><p>tan⁻¹(1) = 0 + C ⟹ C = π/4</p><p><strong>Step 5:</strong> Therefore:</p><p>tan⁻¹(h(x)) = x + π/4</p><p>h(x) = tan(x + π/4)</p><p><strong>Step 6:</strong> Calculate h(π/12):</p><p>h(π/12) = tan(π/12 + π/4) = tan(π/12 + 3π/12) = tan(4π/12) = tan(π/3) = √3</p><p>∴ f(π/12) + g(π/12) = <strong>√3</strong></p>
Correct Answer: C