Question:

<p><strong>Step 1:</strong> Express $g(x)$ using the given condition: $g(x) = x + \int_0^1 e^{x+t} g(t)\, dt$</p><p><strong>Step 2:</strong> Factor out $e^x$ from the integral: $g(x) = x + e^x \int_0^1 e^t g(t)\, dt$. Let $K = \int_0^1 e^t g(t)\, dt$ (a constant).</p><p><strong>Step 3:</strong> So $g(x) = x + Ke^x$</p><p><strong>Step 4:</strong> Evaluate at $x=0$: $g(0) = 0 + Ke^0 = K$</p><p><strong>Step 5:</strong> Use the definition of $K$: $K = \int_0^1 e^t g(t)\, dt = \int_0^1 e^t(t + Ke^t)\, dt = \int_0^1 te^t\, dt + K\int_0^1 e^{2t}\, dt$</p><p><strong>Step 6:</strong> Calculate: $\int_0^1 te^t\, dt = [te^t - e^t]_0^1 = (e-e)-(0-1) = 1$ and $\int_0^1 e^{2t}\, dt = \frac{e^2-1}{2}$</p><p><strong>Step 7:</strong> Thus $K = 1 + K\cdot\frac{e^2-1}{2}$, giving $K(1 - \frac{e^2-1}{2}) = 1$, so $K\cdot\frac{3-e^2}{2} = 1$, thus $K = \frac{2}{3-e^2}$</p><p>∴ Answer: A (where $g(0) = \frac{2}{3-e^2}$ or equivalent form)</p>