- 已知函数$f(x) = x(\ln x - ax)$（$a \in R$）有两个极值点$x_1, x_2$（$x_1 < x_2$），则
$A. 0 < a < \frac{1}{2}$ $B. 1 < x_2 < \frac{1}{2a}$ $C. x_2 - x_1 > \frac{1}{2a} - 1$ $D. f(x_1) < 0, f(x_2) > -\frac{1}{2}$
解：对于A选项
$f'(x) = \ln x - ax + x(\frac{1}{x} - a) = \ln x - 2ax + 1$
$\therefore \ln x - 2ax + 1 = 0$有二根$x_1, x_2$（$x_1 < x_2$）
$\therefore 2a = \frac{\ln x + 1}{x}$有二根$x_1, x_2$（$x_1 < x_2$）
设$g(x) = \frac{\ln x + 1}{x}$，则$g'(x) = -\frac{\ln x}{x^2}$
$\because x > 1$时，$g'(x) < 0$，$g(x)$单减
$0 < x < 1$时，$g'(x) > 0$，$g(x)$单增，如右图所示
$\therefore g(x)_{极大值} = g(1) = 1$
$\therefore 0 < 2a < 1$
$\therefore 0 < a < \frac{1}{2}$，故A正确
对于B选项
$g(\frac{1}{2a}) = \frac{\ln\frac{1}{2a}}{\frac{1}{2a}} = 2a(1 - \ln 2a)$
$\because 0 < a < \frac{1}{2}$
$\therefore \ln 2a < 0$
$\therefore g(\frac{1}{2a}) > 2a = g(x_2)$
$\because g(x)$在$(1, +\infty)$上单减
$\therefore x_2 > \frac{1}{2a}$，故B错误
对于C选项
$\because x_1 < 1, x_2 > \frac{1}{2a}$
$\therefore x_2 - x_1 > \frac{1}{2a} - 1$，故C正确
对于D选项
由上面解答可知$0 < x_1 < 1 < \frac{1}{2a} < x_2$
$f(x)$在$(0, x_1)$和$(x_2, +\infty)$单减，$(x_1, x_2)$上单增
$\therefore f(x_1) < f(1) = -a < 0$
$f(x_2) > f(\frac{1}{2a}) = \frac{1}{2a}(-\ln 2a - \frac{1}{2}) > -\frac{1}{2}$