Question:

<p><strong>Step 1:</strong> Differentiate the integral condition with respect to a.</p><p>Given: $\int_0^a f(g(x))\,dx = \frac{1}{2} - \frac{e^{-2a}}{2}$</p><p>Differentiating both sides: $f(g(a)) = e^{-2a}$</p><p>Therefore: $f(g(x)) = e^{-2x}$ for all real $x$.</p><p><strong>Step 2:</strong> Analyze the given functional equation.</p><p>Given: $xg(f(x))f'(g(x))g'(x) = f(g(x))g'(f(x))f'(x)$</p><p>Substitute $f(g(x)) = e^{-2x}$:</p><p>$xg(f(x))f'(g(x))g'(x) = e^{-2x}g'(f(x))f'(x)$</p><p><strong>Step 3:</strong> Differentiate $f(g(x)) = e^{-2x}$ to find a relation for $f'$.</p><p>$\frac{d}{dx}[f(g(x))] = -2e^{-2x}$</p><p>$f'(g(x))g'(x) = -2e^{-2x}$</p><p><strong>Step 4:</strong> Substitute this into the functional equation.</p><p>$xg(f(x)) \cdot (-2e^{-2x}) = e^{-2x}g'(f(x))f'(x)$</p><p>$-2xg(f(x))e^{-2x} = e^{-2x}g'(f(x))f'(x)$</p><p>Dividing by $e^{-2x}$ (nonzero):</p><p>$-2xg(f(x)) = g'(f(x))f'(x)$</p><p><strong>Step 5:</strong> Recognize that the right side is $\frac{d}{dx}[g(f(x))]$.</p><p>$\frac{d}{dx}[g(f(x))] = -2xg(f(x))$</p><p><strong>Step 6:</strong> Solve this differential equation.</p><p>Let $h(x) = g(f(x))$. Then: $h'(x) = -2xh(x)$</p><p>$\frac{dh}{h} = -2x\,dx$</p><p>$\ln h = -x^2 + C$</p><p>$h(x) = Ke^{-x^2}$</p><p><strong>Step 7:</strong> Use the initial condition $g(f(0)) = 1$.</p><p>$h(0) = g(f(0)) = 1$</p><p>$Ke^{0} = 1 \Rightarrow K = 1$</p><p>Therefore: $g(f(x)) = e^{-x^2}$</p><p><strong>Step 8:</strong> Find $g(f(4))$.</p><p>$g(f(4)) = e^{-16}$</p><p>Given that $g(f(4)) = e^{-\lambda}$ where $\lambda \in \mathbb{N}$:</p><p>$e^{-\lambda} = e^{-16}$</p><p>$\lambda = 16$</p><p><strong>∴ Answer: 16</strong></p>