37．已知函数$f(x) = 2\text{e}x^{2} - a^{x}$（$a > 0$且$a \neq 1$）的极大值和极小值分别为$f\left( x_{1} \right)$，$f\left( x_{2} \right)$，且$x_{1} < x_{2}$，则$a$的取值范围是\\\\\\．
解：$f'(x) = 4\text{e}x - a^{x}\ln a$，所以方程$4\text{e}x - a^{x}\ln a = 0$的两个根为$x_{1}$，$x_{2}$，
即函数$y_{1} = 4\text{e}x$和$y_{2} = a^{x}\ln a$的图象有两个不同的交点，
因为$f(x)$的极大值和极小值分别为$f(x_{1})$，$f(x_{2})$，
故当$x \in (x_{1},x_{2})$时，$f'(x) < 0$，$y_{1}$的图象在$y_{2}$的下方，
当$x \in ( - \infty,x_{1})$、$(x_{2}, + \infty)$时，$f'(x) > 0$，$y_{1}$的图象在$y_{2}$的上方；
易知$0 < a < 1$，设过原点且与$y_{2}$图象相切的直线$l$斜率为$k$，则$k < 4\text{e}$，
设$l$与$y_{2} = a^{x}\ln a$切于点$(x_{0},a^{x_{0}}\ln a)$，而${y'}_{2} = a^{x}\ln^{2}a$，所以$k = a^{x_{0}}\ln^{2}a = \frac{a^{x_{0}}\ln a}{x_{0}}$，
解得$x_{0} = \frac{1}{\ln a}$，所以$k = a^{\frac{1}{\ln a}} \times \ln^{2}a < 4\text{e}$，
因为$a^{\frac{1}{\ln a}} = \text{e}$，即$\ln^{2}a < 4$，又$0 < a < 1$，所以$- 2 < \ln a < 0$，所以$a \in \left( \frac{1}{\text{e}^{2}},1 \right)$．