8、已知$x^{\frac{1}{x^2}} - a = 0$在$(0, +\infty)$上有两个不相等的实数根，则实数$a$的取值范围（ ）
$A.(0, \frac{1}{2e}]$ $B.(0, \frac{1}{2e})$ $C.[1, e^{\frac{1}{2e}}]$ $D.(1, e^{\frac{1}{2e}})$
解：$\because x^{\frac{1}{x^2}} - a = 0$
$\therefore x^{\frac{1}{x^2}} = a$
$\therefore \frac{\ln x}{x^2} = \ln a$
设$f(x) = \frac{\ln x}{x^2}$，则$f'(x) = \frac{1 - 2\ln x}{x^3}$
$\therefore 0 < x < \sqrt{e}$时$f(x)$单增，$x > \sqrt{e}$时$f(x)$单减
$\therefore f(x)_{极大值} = f(\sqrt{e}) = \frac{1}{2e}$
$\because x^{\frac{1}{x^2}} - a = 0$在$(0, +\infty)$上有两个不相等的实数根
$\therefore 0 < \ln a < \frac{1}{2e}$
$\therefore 1 < a < e^{\frac{1}{2e}}$