答案
\(1-5\) \(CADCA\)
\(6-8\) \(DBD\)
\(9\) \(ABD\)
\(10\) \(AC\)
\(11\) \(ACD\)
\(12.\) \({(\frac{\sqrt{6}}{3},\frac{\sqrt{6}}{3}),(-\frac{\sqrt{6}}{3},-\frac{\sqrt{6}}{3})}\)
\(13.\) 奇 \(\pi\)
\(14.\) \(4 + \frac{2\sqrt{6}}{3}\)
直线 \(l : \frac{x}{a} + \frac{y}{b} = 1\) 过点 \((a,0),(0,b)\)，
\((0,0)\) 到直线 \(l\) 的距离为 $d = \frac{|ab|}{\sqrt{a^{2} + b^{2}}} $
\(\because\) 直线 \(l\) 与圆有交点，\(\therefore\) \(d < 1\)，
化简得 \(\frac{1}{a^2} + \frac{1}{b^2} \ge 1\)
故选 \(D\)
设 \(CD = x\)，\(AC = y\)
\(\because\) \(BD = 1\)，\(AB = 3\)，\(CD^2 + BD^2 = BC^2\)，\(BC^2 + AC^2 = AB^2\)
\(\therefore\) \(x^2 + 1^2 + y^2 = 3^2\)，即 \(x^2 + y^2 = 8\)
\(\therefore\) \(V = \frac{1}{3} \times S_{\bigtriangleup BCD} \times AC = \frac{1}{3} \times ( \frac{1}{2} \times x \times 1 ) \times y = \frac{xy}{6} \le \frac{x^2 + y^2}{12} = \frac{2}{3}\)
故选 \(D\)
\(f'(x) = \begin{cases} e^x (x\ge 0)\\-e^{x} (x > 0)\end{cases}\)
\(\because x_1 < 0, x_2 > 0\) 且两直线相切
\(\therefore f'(x_1) \times f'(x_2) = -1\)
\(\therefore -e^{x_1} \cdot e^{x_2} = -1\)
\(\therefore e^{x_1 + x_2} = 1\)
\(\therefore x_1 + x_2 = 0\)
\(\frac{|AM|}{|BN|} = \frac{|x_1|\cdot \sqrt({f'(x_1)^{2} + 1} }{|x_2|\sqrt({f'(x_2)^{2} + 1}} = \frac{|x_1|\cdot \sqrt{(-e^{x_1})^2 + 1} }{|x_2| \sqrt{(e^{x_2})^2 + 1}} = \frac{|-x_2|\cdot \sqrt{(-e^{-x_2})^2 + 1} }{|x_2| \sqrt{(e^{x_2})^2 + 1}} = \frac{1}{e^{x_2}}\)
\(\because x_2 \in (0, +\infty )\)
\(\therefore e^{x_2} \in (1,\infty)\)
\(\therefore \frac{|AM|}{|BN|} \in (0,1)\)
故选 \(B\)
\(\because AB = 2\)
\(\therefore O_1A = O_1B = \sqrt{2}, O_2A = O_2B = 2\)
\(\therefore O_2E = \sqrt{3}\)
\(\therefore OA^2 = OO_1^2 + O_1A^2 = O_2E^2 + O_1A^2 = 5\)
\(\therefore r = OA = \sqrt{5}\)
\(\therefore S = 4\pi r^2 = 20\pi\)
故选 \(D\)
\(A.\) \(f(3) = cos6, \frac{3\pi}{2} < 6 < 2\pi , f(1) = cos2, \frac{\pi}{2} < 6 < \pi\)
\(\because 2\pi - 6 < 2 - \frac{\pi}{2}\)
\(\therefore cos6 > cos2\)，即 \(f(3) > f(1)\)
\(B.\) \(\because lnx > 1 - \frac{1}{x} > sin(1 - \frac{1}{x})\)
\(\therefore ln1.1 > sin(1 - \frac{1}{1.1}) = sin\frac{1}{11}\)
\(\because 1 > 2ln1.1 > 2sin\frac{1}{11} > 0\)
\(\therefore cos(2sin\frac{1}{11}) > cos(2ln1.1)\)，即 \(f(sin\frac{1}{11}) > f(ln1.1)\)
\(C.\) \(\because \frac{\pi}{2} < 2ln3 < \pi, \frac{3\pi}{2} < 2e < 2\pi\)
\(\because 2e - \frac{3\pi}{2} > 2ln3 - \frac{\pi}{2}\)
\(\therefore cos2e > |cos(2ln3)|\)，即 \(f(2e) > |f(ln3)|\)
\(D.\) \(f'(x) = -2sin2x\)
\(\because \frac{3\pi}{2} < 2e < \frac{11\pi}{6}\)
\(\therefore sin(\frac{3\pi}{2}) < sin2e < sinn(\frac{11\pi}{6})\)，即 \(-1 < sin2e < -\frac{1}{2}\)
\(\therefore f'(2e) = -2sin2e > 1 > cos2e = f(2e)\)
底面放三个小球，上面再放一个小球时体积最小
把四个小球的球心连接，可以得到一个棱长为 \(2\) 的小正四面体
可以求出这个小正四面体的中心到各个面的距离为 \(\frac{\sqrt{6}}{6}\)
\(\because\) 小正四面体的中心和正四面体容器的中心重合
\(\therefore\) 正四面体容器的中心到各个面的距离是 \(\frac{\sqrt{6}}{6} + 1\) (小球的半径为 \(1\)) ,
$\because $ 正四面体的中心到各个面的距离是正四面体的高的 \(\frac{1}{4}\)
\(\therefore\) 正四面体容器的高的最小值为 \(4 + \frac{2\sqrt{6}}{3}\)
故答案为 \(4 + \frac{2\sqrt{6}}{3}\)
标签：because,frac,正四面体,卷选,2024,therefore,sqrt,pi,联考 From： https://www.cnblogs.com/HT-walnut/p/18015235