- 若曲线$f(x) = \frac{x}{e^x}$有三条过点$(0, a)$的切线，则实数$a$的取值范围为
$A.\left( 0, \frac{1}{e} \right)$ $B.\left( 0, \frac{4}{e^2} \right)$ $C.\left( 0, \frac{1}{e^2} \right)$ $D.\left( 0, \frac{4}{e} \right)$
解：设切点$(x_0, \frac{x_0}{e^{x_0}})$，$f'(x) = \frac{1 - x}{e^x}$
切线方程为$y - \frac{x_0}{e^{x_0}} = \frac{1 - x_0}{e^{x_0}}(x - x_0)$
代入$(0, a)$可得$a = \frac{x_0^2}{e^{x_0}}$
设$g(x) = \frac{x^2}{e^x}$，$g'(x) = \frac{-x(x - 2)}{e^x}$
$\therefore 0 < x < 2$时$g'(x) > 0$，$g(x)$单增
$x < 0$或$x > 2$时$g'(x) < 0$，$g(x)$单减
$\therefore g(x)_{极小值} = g(0) = 0$，$g(x)_{极大值} = g(2) = \frac{4}{e^2}$
$\because$过$(0, a)$可以作三条切线
$\therefore 0 < a < \frac{4}{e^2}$
- 已知双曲线$C: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$（$a > 0, b > 0$）的左、右焦点分别为$F_1(-c, 0)$，$F_2(c, 0)$，点$M$是双曲线$C$上一点，点$N\left( \frac{1}{3}c, 0 \right)$，且$\angle F_1MN = 30°$，$\overrightarrow{MN} \cdot \overrightarrow{MF_2} = 0$，则双曲线$C$的离心率为
$A. \frac{\sqrt{21}}{3}$ $B. \frac{2\sqrt{6}}{3}$ $C. \sqrt{3}$ $D. \frac{2\sqrt{13}}{3}$
解：过点$F_1$作$F_1A // MF_2$交$MN$延长线于点$A$
$\therefore \frac{F_1A}{MF_2} = \frac{F_1N}{NF_2} = \frac{c + \frac{1}{3}c}{c - \frac{1}{3}c} = 2$
$\therefore F_1A = 2MF_2$
设$MF_2 = x$，则$F_1A = 2x$，$F_1M = 2a + x$
$\because$在$\triangle F_1AM$中，$\angle F_1AN = \angle NMF_2 = 90°$，$\angle F_1MN = 30°$
$\therefore \frac{2x}{2a + x} = \frac{1}{2}$
$\therefore x = \frac{2a}{3}$
$\therefore$在$\triangle F_1MF_2$中$(2c)^2 = \left( \frac{8a}{3} \right)^2 + \left( \frac{2a}{3} \right)^2 - 2 \cdot \frac{8a}{3} \cdot \frac{2a}{3} \cdot \cos 120°$
$\therefore e = \frac{c}{a} = \frac{\sqrt{21}}{3}$