12、已知函数$f(x) = xe^x$与$g(x) = \ln x + (a + 2)x + 1$（$a \in R$）图像没有公共点，则实数$a$的取值范围是（ ）
$A.(-\infty, -3)$ $B.(-\infty, -1)$ $C.(-\infty, 0)$ $D.(-\infty, 1)$
解：设$h(x) = e^x - x - 1$，则$h'(x) = e^x - 1$
$\therefore x > 0$时$h(x)$单增，$x < 0$时$h(x)$单减
$\therefore h(x)_{极小值} = h(0) = 0$
$\therefore e^x \geq x + 1$
$\therefore e^{x + \ln x} \geq x + \ln x + 1$
$\therefore xe^x \geq x + \ln x + 1$
假设$f(x) = xe^x$与$g(x) = \ln x + (a + 2)x + 1$图像有交点，则$xe^x = \ln x + (a + 2)x + 1$有正解
$\therefore a + 2 = \frac{xe^x - (\ln x + 1)}{x}$有解
$\because xe^x - (\ln x + 1) \geq x$
$\therefore a + 2 \geq 1$
$\therefore a \geq -1$
$\because f(x) = xe^x$与$g(x) = \ln x + (a + 2)x + 1$图像无交点
$\therefore a < -1$，选B
解：假设$f(x) = xe^x$与$g(x) = \ln x + (a + 2)x + 1$图像有交点，则$xe^x = \ln x + (a + 2)x + 1$有正解
$\therefore a + 2 = \frac{xe^x - (\ln x + 1)}{x}$有解
设$h(x) = e^x - \frac{\ln x + 1}{x}$
$\therefore h'(x) = e^x - \frac{\frac{1}{x} \cdot x - (\ln x + 1)}{x^2} = \frac{x^2 e^x + \ln x}{x^2}$
设$t(x) = x^2 e^x + \ln x$，则$t'(x) = e^x(x^2 + 2x) + \frac{1}{x}$
$\therefore x > 0$时$t(x)$单增
若$x_0^2 e^{x_0} + \ln x_0 = 0$，则$x_0 e^{x_0} = -\frac{\ln x_0}{x_0} = \ln\frac{1}{x_0} \cdot \frac{1}{x_0}$
设$m(x) = xe^x$，则$m'(x) = e^x(x + 1)$
$\therefore x > 0$时，$m(x)$单增
$\therefore x_0 = \ln\frac{1}{x_0}$
$\therefore 0 < x < x_0$时$t(x) < 0$，$x > x_0$时$t(x) > 0$
$\therefore 0 < x < x_0$时$h'(x) < 0$，$x > x_0$时$h'(x) > 0$
$\therefore h(x)_{极小值} = h(x_0) = e^{x_0} - \frac{\ln x_0 + 1}{x_0} = \frac{1}{x_0} - \frac{-x_0 + 1}{x_0} = 1$
$\therefore a + 2 \geq 1$
$\therefore a \geq -1$
$\because f(x) = xe^x$与$g(x) = \ln x + (a + 2)x + 1$图像无交点
$\therefore a < -1$，选B