- 已知实数$a > 0$，$e = 2.718\cdots$，对任意$x \in ( - 1, + \infty )$，不等式$e^{x} \geq ae\lbrack 2 + \ln(ax + a)\rbrack$恒成立，则实数$a$的取值范围是（ ）
A．$\left( 0,\frac{1}{e} \right]$ B．$\left( \frac{1}{e},1 \right)$ C．$\left( 0,\frac{2}{e} \right)$ D．$\left( \frac{2}{e},1 \right)$
解：原式可化为$\frac{e^{x}}{e^{\ln ae}} \geq 2 + \ln(ax + a)$
$\therefore e^{x - \ln ae} \geq 2 + \ln a + \ln(x + 1)$
$\therefore e^{x - \ln ae} - 1 \geq 1 + \ln a + \ln(x + 1) = \ln ae + \ln(x + 1)$
$\therefore e^{x - \ln ae} + x - \ln ae \geq \ln(x + 1) + x + 1 = e^{\ln(x + 1)} + \ln(x + 1)$
设$f(x) = e^{x} + x$，则$f'(x) = e^{x} + 1$
$\therefore f(x)$在$R$上单增
$\therefore x - \ln ae \geq \ln(x + 1)$
$\therefore \ln a \leq x - \ln(x + 1) - 1$
设$g(x) = x - \ln(x + 1) - 1$，则$g'(x) = 1 - \frac{1}{x + 1} = \frac{x}{x + 1}$
$\therefore - 1 < x < 0$时$g(x)$单减，$x > 0$时$g(x)$单增
$\therefore g(x)$极小值$= g(0) = - 1$
$\therefore \ln a \leq - 1 = \ln\frac{1}{e}$
$\therefore 0 < a \leq \frac{1}{e}$，故选 A