25~26高三下·山东德州一模·第8题

B

由题意可得$f(x_1) = x_1\mathrm{e}^{x_1} = m$，则$\ln m = \ln x_1 + x_1$.

由$g(x_2) = \mathrm{e}x_2(1 + \ln x_2) = m$，则$\ln m = 1 + \ln x_2 + \ln(1 + \ln x_2)$.

令$1 + \ln x_2 = t$，则$\ln m = t + \ln t$.

令$h(x) = x + \ln x$，可知函数$h(x)$在$(0, +\infty)$上单调递增，
所以当$\ln m = \ln x_1 + x_1 = t + \ln t$有唯一解，即$t = x_1$，即$1 + \ln x_2 = x_1$，可得$x_2 = \mathrm{e}^{x_1 – 1}$.

所以$\dfrac{\ln m}{x_1x_2} = \dfrac{x_1 + \ln x_1}{x_1\mathrm{e}^{x_1 – 1}}$.

令$u = x_1\mathrm{e}^{x_1}(u > 0)$，则$\ln u = x_1 + \ln x_1$，所以$\dfrac{x_1 + \ln x_1}{x_1\mathrm{e}^{x_1 – 1}} = \dfrac{\ln u}{\dfrac{u}{\mathrm{e}}} = \dfrac{\mathrm{e}\ln u}{u}$.

令$p(u) = \dfrac{\mathrm{e}\ln u}{u}\ (u > 0)$，则$p'(u) = \dfrac{\mathrm{e}(1 – \ln u)}{u^2}$.

令$p'(u) = 0$，即$1 – \ln u = 0$，解得$u = \mathrm{e}$.

当$0 < u < \mathrm{e}$时，$p'(u) > 0$，则$p(u)$在$(0, \mathrm{e})$上单调递增；
当$u > \mathrm{e}$时，$p'(u) < 0$，则$p(u)$在$(\mathrm{e}, +\infty)$上单调递减.

所以函数$p(u)$在$u = \mathrm{e}$处取得极大值，也是最大值，为$p(\mathrm{e}) = \dfrac{\mathrm{e}\ln \mathrm{e}}{\mathrm{e}} = 1$.

所以$\dfrac{\ln m}{x_1x_2}$的最大值为$1$.

故选：B.