25~26高三上·河北沧州沧县中学一模·第8题

C

已知$x_0^2\text{e}^{x_0} + \text{e}^4\ln x_0 = 4\text{e}^4$，则$\ln x_0 + x_0$的值为（   ）
A.$1$      B.$2$
C.$4$      D.$6$

【答案】C

【解析】由$x_0^2\text{e}^{x_0} + \text{e}^4\ln x_0 = 4\text{e}^4$，
得$x_0^2\text{e}^{x_0} = 4\text{e}^4 – \text{e}^4\ln x_0$，
即$x_0^2\text{e}^{x_0} = \text{e}^4\ln \text{e}^4 – \text{e}^4\ln x_0 = \text{e}^4\ln \frac{\text{e}^4}{x_0}$，
显然$x_0 > 0$，即$x_0\text{e}^{x_0} = \frac{\text{e}^4}{x_0}\ln \frac{\text{e}^4}{x_0} = \ln \frac{\text{e}^4}{x_0} \cdot \text{e}^{\ln \frac{\text{e}^4}{x_0}}$，
因为$x_0 > 0$，所以$x_0\text{e}^{x_0} > 0$，$\text{e}^{\ln \frac{\text{e}^4}{x_0}} > 0$，所以$\ln \frac{\text{e}^4}{x_0} > 0$，
令$f(x) = x\text{e}^x, x > 0$，则$f(x_0) = f\left(\ln \frac{\text{e}^4}{x_0}\right)$，
由于$f'(x) = (x\text{e}^x)’ = \text{e}^x + x\text{e}^x = (1 + x)\text{e}^x$，且$x > 0$，即得$f'(x) > 0$，
故$f(x) = x\text{e}^x$在$(0, +\infty)$上单调递增，
所以$x_0 = \ln \frac{\text{e}^4}{x_0}$，
即$x_0 = \ln \text{e}^4 – \ln x_0$，即$x_0 + \ln x_0 = \ln \text{e}^4 = 4$．

故选：C．