1．若$a = \frac{1011}{506}\text{e}^{\frac{1009}{1014}}$，$b = \frac{1009}{507}\text{e}^{\frac{1011}{1012}}$，$c = \frac{2016}{507}$，$\text{e}$是自然对数的底数，则（ ）
A．$c > a > b$ B．$b > a > c$ C．$c > b > a$ D．$a > b > c$
解：设$f(x) = \text{e}^{x} - \text{e}x$，$x \in (1, + \infty)$，则$f'(x) = \text{e}^{x} - \text{e} > 0$在$(1, + \infty)$上恒成立，所以$f(x)$在$(1, + \infty)$上单调递增，则$f(x) > f(1) = 0$，所以$\text{e}^{x} > \text{e}x$，
所以$b = \frac{1009}{507}\text{e}^{\frac{1011}{1012}} > \frac{1009}{507} \times \text{e} \times \frac{1011}{1012} > \frac{1009}{507} \times 2 = \frac{2018}{507} > \frac{2016}{507} = c$，
设$h(x) = \frac{2x}{\text{e}^{x}},x \in (0,1)$，则$h'(x) = \frac{2(1 - x)}{\text{e}^{x}} > 0$，所以$h(x)$在$(0,1)$上单调递增，则$h(x) > h(0) = 0$，
又$0 < \frac{1009}{1014} < \frac{1009 + 2}{1014 - 2} = \frac{1011}{1012} < 1$，则$h\left( \frac{1009}{1014} \right) < h\left( \frac{1011}{1012} \right)$，即$\frac{\frac{1009}{507}}{\text{e}^{\frac{1009}{1014}}} < \frac{\frac{1011}{506}}{\text{e}^{\frac{1011}{1012}}}$，则$\frac{1009}{507}\text{e}^{\frac{1011}{1012}} < \frac{1011}{506}\text{e}^{\frac{1009}{1014}}$，故$b < a$，综上可得：$a > b > c$.