12、若存在$x_{0} \in \lbrack - 1,2\rbrack$使不等式$x_{0} + (e^{2} - 1)\ln a \geq \frac{2a}{e^{x_{0}}} + e^{x_{0}}x_{0} - 2$成立，则$a$的取值范围（ ）
A．$\left\lbrack \frac{1}{2e},e^{2} \right\rbrack$ B．$\left\lbrack \frac{1}{e^{2}},e^{2} \right\rbrack$ C．$\left\lbrack \frac{1}{e^{2}},e^{4} \right\rbrack$ D．$\left\lbrack \frac{1}{e},e^{4} \right\rbrack$
解：原式可化为$(e^{2} - 1)\ln a + x_{0} - e^{x_{0}}x_{0} \geq \frac{2a}{e^{x_{0}}} - 2$
$\therefore (e^{2} - 1)\ln(a - x_{0}) \geq 2e^{\ln a - x_{0}} - 2$
设$t = \ln a - x_{0}$，则$(e^{2} - 1)t \geq 2e^{t} - 2$
$\because - 1 \leq x_{0} \leq 2$
$\therefore \ln a - 2 \leq t \leq \ln a + 1$
$\therefore$存在$t \in \lbrack \ln a - 2,\ln a + 1\rbrack$使得$2e^{t} - \left( e^{2} - 1 \right)t - 2 \leq 0$
设$f(t) = 2e^{t} - \left( e^{2} - 1 \right)t - 2$，则$f(0) = f(2) = 0$，$f'(t) = 2e^{t} - \left( e^{2} - 1 \right)$
$\therefore t > \ln\frac{e^{2} - 1}{2}$时$f(t)$单增，$t < \ln\frac{e^{2} - 1}{2}$时$f(t)$单减
$f(t)$图像如右图所示
若不存在，使得$f(t) \leq 0$，则$\ln a + 1 < 0$或$\ln a - 2 > 2$
$\therefore a < \frac{1}{e}$或$a > e^{4}$
$\therefore \frac{1}{e} \leq a \leq e^{4}$