25-26高三上·江西吉安期末·第14题

$(-\infty, -1]$

【题目】已知曲线$y = me^x(m \ge e)$与曲线$y = \ln x + n(n \in \mathbb{R})$有两条公切线，且它们的斜率之积为$1$，则实数$n$的取值范围为______.

【答案】$(-\infty, -1]$

【解析】设$f(x) = me^x$，$g(x) = \ln x + n$.

由题意得存在实数$x_1$，$x_2$使得$f(x)$在$x = x_1$处的切线和$g(x)$在$x = x_2$处的切线重合，
所以$f'(x_1) = g'(x_2) = \dfrac{f(x_1) – g(x_2)}{x_1 – x_2}$，
即$me^{x_1} = \dfrac{1}{x_2} = \dfrac{me^{x_1} – \ln x_2 – n}{x_1 – x_2}$.

由$\dfrac{1}{x_2} = \dfrac{me^{x_1} – \ln x_2 – n}{x_1 – x_2} = \dfrac{\dfrac{1}{x_2} – \ln x_2 – n}{x_1 – x_2}$$\Rightarrow x_1 – x_2 = 1 – x_2\ln x_2 – nx_2$，
即$x_1 = 1 – x_2\ln x_2 – (n – 1)x_2$.

又由$me^{x_1} = \dfrac{1}{x_2} \Rightarrow \ln m + x_1 = -\ln x_2$，
即$\ln m = -\ln x_2 – x_1 = -\ln x_2 – 1 + x_2\ln x_2 + (n – 1)x_2$.

令$h(x) = -\ln x – 1 + x\ln x + (n – 1)x$，则题目转化为$h(x) = \ln m$有两个不相等的实数根，且互为倒数.

设两根分别为$t$，$\dfrac{1}{t}$，
则由$h(t) = h\left(\dfrac{1}{t}\right)$得$-\ln t – 1 + t\ln t + (n – 1)t $$= -\ln \dfrac{1}{t} – 1 + \dfrac{1}{t}\ln \dfrac{1}{t} + (n – 1)\dfrac{1}{t}$，

化简得$\ln t = \dfrac{(n – 1)\left(\dfrac{1}{t} – t\right)}{t + \dfrac{1}{t} – 2} $$= \dfrac{(n – 1)(1 – t^2)}{t^2 + 1 – 2t} = (n – 1)\dfrac{1 + t}{1 – t}$.

所以$\ln m = (n – 1)\ln t – 1 + (n – 1)t $$= (n – 1)(-t – 1) – 1 + (n – 1)t = -n$，即$n = -\ln m$.

因为$m \ge e$，所以$n \le -1$，
故$n$的取值范围为$(-\infty, -1]$.

故答案为：$(-\infty, -1]$.