Question:

Step 1: Rewrite the functional equation. Let $g(x) = e^{-x}f(x)$. Then: $$f(xy) = e^{xy-x-y}(e^y f(x) + e^x f(y))$$ $$e^{-xy}f(xy) = e^{-x}f(x) \cdot e^{-y} \cdot e^y + e^{-y}f(y) \cdot e^{-x} \cdot e^x$$ Wait, let $g(x) = f(x)e^{-x}$. Then: $$g(xy) = e^{-xy}f(xy) = e^{-xy} \cdot e^{xy-x-y}(e^y f(x) + e^x f(y))$$ $$= e^{-x-y}(e^y f(x) + e^x f(y)) = e^{-x}f(x) + e^{-y}f(y) = g(x) + g(y)$$ Step 2: So $g(xy) = g(x) + g(y)$, which is the logarithmic functional equation. Thus $g(x) = c\ln x$, giving $f(x) = ce^x \ln x$. Step 3: Apply $f'(1) = e$: $f'(x) = ce^x \ln x + ce^x/x$. At $x=1$: $f'(1) = c(e \cdot 0 + e) = ce = e \Rightarrow c = 1$. So $f(x) = e^x \ln x$. Step 4: Check option (a): $\displaystyle\lim_{x\to e} \dfrac{f(x)-e^x}{x-e} = \lim_{x\to e} \dfrac{e^x\ln x - e^x}{x-e} = \lim_{x\to e} \dfrac{e^x(\ln x - 1)}{x-e}$. Since $\ln e = 1$, this is $0/0$ form. Apply L'Hôpital: $\lim_{x\to e} e^x(\ln x - 1) + e^x/x = e^e(0) + e^e/e = e^{e-1}$. So $\left[e^{e-1}\right] = e^{(e-1)}$. ✓ Step 5: Check option (b): $f(x) = xe^x$ means $e^x \ln x = xe^x \Rightarrow \ln x = x$. This has no solution since $\ln x < x$ for all $x > 0$ (with equality nowhere). So number of roots is 0, not 2. ✗ Step 6: Check option (c): $\displaystyle\int_1^e e^x \ln x\, dx$. Since $\ln x < 1$ for $x \in (1, e)$, we have $e^x \ln x < e^x$, so $\int_1^e e^x \ln x\, dx < \int_1^e e^x\, dx = e^e - e < e^e(e-1)$. ✓ Step 7: Check option (d): $f'(x) = e^x \ln x + e^x/x = e^x(\ln x + 1/x)$. For $x > 0$, $e^x > 0$. At $x=1$: $\ln 1 + 1 = 1 > 0$. For $x \in (0,1)$: $\ln x < 0$ but $1/x > 1$, so $\ln x + 1/x > 0$ (since $1/x$ dominates). For $x > 1$: both terms positive. So $f'(x) > 0$ for all $x > 0$, meaning $f$ is strictly increasing. ✓