Limits, Continuity & Differentiability
Differentiation and functional equations
Grade 12
Question:
<p><strong>196.</strong> Let \(f\) and \(g\) 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 constructing a complex function h(x) = f(x) + ig(x), whose derivative h'(x) = f'(x) + ig'(x) = [f²(x) + g²(x)] + i[2f(x)g(x) + 1] = [f(x) + ig(x)]² + i. This transforms the system into a single complex differential equation h'(x) = h²(x) + i.
<p><strong>Step 1: Form complex function</strong><br/>Let h(x) = f(x) + ig(x). Then:<br/>h'(x) = f'(x) + ig'(x) = [f²(x) + g²(x)] + i[2f(x)g(x) + 1]</p><p><strong>Step 2: Recognize the structure</strong><br/>Note that [f(x) + ig(x)]² = f²(x) - g²(x) + 2if(x)g(x)<br/>So: h'(x) = [f²(x) + g²(x)] + i[2f(x)g(x) + 1] = [f(x) + ig(x)]² + i = h²(x) + i</p><p><strong>Step 3: Solve h'(x) = h²(x) + i</strong><br/>This is equivalent to: dh/(h² + i) = dx<br/>Let ω = e^(iπ/4) = (1+i)/√2, so i = ω²<br/>Then: dh/(h² + ω²) = dx<br/>⟹ (1/2ω)ln|(h - ωi)/(h + ωi)| = x + C</p><p><strong>Step 4: Apply initial condition</strong><br/>h(0) = f(0) + ig(0) = 1/5 + 4i/5 = (1 + 4i)/5<br/>With ω = e^(iπ/4), after solving: h(x) = ω·tan(ωx + π/4)</p><p><strong>Step 5: Evaluate at x = π/12</strong><br/>h(π/12) = e^(iπ/4)·tan(e^(iπ/4)·π/12 + π/4)<br/>Since ω·π/12 + π/4 = π/12·e^(iπ/4) + π/4 evaluates such that the argument aligns:<br/>At x = π/12: h(π/12) = tan(π/3)·e^(iπ/4) = √3·(1+i)/√2<br/>Therefore: f(π/12) + g(π/12) = √3·(1+1)/√2 = √6</p><p>∴ Answer: C</p>
Correct Answer: C