16、若函数$f(x) = x\ln x - \frac{1}{2}ax^2$存在极大值点$x_0$，且$2f(x_0) > e^2$，则实数$a$的取值范围__
解：$f(x) = x\ln x - \frac{1}{2}ax^2$，则$f'(x) = \ln x + 1 - ax$
$\therefore \ln x_0 + 1 - ax_0 = 0$
$\therefore a = \frac{\ln x_0 + 1}{x_0}$
设$g(x) = \frac{\ln x + 1}{x}$，则$g'(x) = \frac{-\ln x}{x^2}$
$\therefore 0 < x < 1$时$g(x)$单增，$x > 1$时$g(x)$单减，如右图所示
$\because x_0$为$f(x)$极大值点
$\therefore x < x_0$时$g(x) > a$，$x > x_0$时$g(x) < a$
$\therefore 0 < a < 1$
$\because 2f(x_0) > e^2$
$\therefore 2(x_0\ln x_0 - \frac{1}{2}ax_0^2) > e^2$
$\therefore 2(x_0\ln x_0 - \frac{\ln x_0 + 1}{x_0} \cdot x_0^2) > e^2$
$\therefore x_0\ln x_0 - x_0 > e^2$
设$h(x) = x\ln x - x$，则$h'(x) = \ln x$
$\therefore 0 < x < 1$时$h(x)$单减，$x > 1$时$h(x)$单增
$\because h(e^2) = e^2\ln e^2 - e^2 = e^2$
$\therefore x_0 > e^2$
$\therefore a = g(x_0) < g(e^2) = \frac{\ln e^2 + 1}{e^2} = \frac{3}{e^2}$
$\therefore 0 < a < \frac{3}{e^2}$