Question:

Step 1: Define a new function to simplify the problem. Let $u(x) = f(x) + g(x)$. We will find a differential equation for this combined function. Step 2: Find the derivative of $u(x)$. Taking the derivative of both sides: $$\frac{du}{dx} = f'(x) + g'(x)$$ Substituting the given expressions for $f'(x)$ and $g'(x)$: $$\frac{du}{dx} = \left(f^2(x) + g^2(x)\right) + \left(2f(x)g(x) + 1\right)$$ Rearranging: $$\frac{du}{dx} = f^2(x) + 2f(x)g(x) + g^2(x) + 1 = (f(x) + g(x))^2 + 1$$ Therefore: $$\frac{du}{dx} = u^2 + 1$$ Step 3: Separate variables and integrate. Rearranging the differential equation: $$\frac{du}{u^2 + 1} = dx$$ Integrating both sides: $$\int \frac{du}{u^2 + 1} = \int dx$$ $$\arctan(u) = x + C$$ Step 4: Find the constant of integration using initial conditions. At $x = 0$: $$u(0) = f(0) + g(0) = \frac{1}{5} + \frac{4}{5} = 1$$ Therefore: $$\arctan(1) = 0 + C$$ $$\frac{\pi}{4} = C$$ Step 5: Write the general solution and evaluate at $x = \frac{\pi}{12}$. The solution is: $$\arctan(f(x) + g(x)) = x + \frac{\pi}{4}$$ At $x = \frac{\pi}{12}$: $$\arctan\left(f\left(\frac{\pi}{12}\right) + g\left(\frac{\pi}{12}\right)\right) = \frac{\pi}{12} + \frac{\pi}{4}$$ $$\arctan\left(f\left(\frac{\pi}{12}\right) + g\left(\frac{\pi}{12}\right)\right) = \frac{\pi}{12} + \frac{3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$$ Step 6: Solve for the final answer. Taking the tangent of both sides: $$f\left(\frac{\pi}{12}\right) + g\left(\frac{\pi}{12}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$$ **Final Answer:** The value of $f\left(\frac{\pi}{12}\right) + g\left(\frac{\pi}{12}\right) = \sqrt{3}$, which corresponds to **Option 3**.