《2025年小题狂做高中数学选择性必修第三册人教版》

1. [2025 湖北七市州期末]某校将开展三项不同的社会实践活动，现招募了 5 名学生志愿者参与.要求每个活动至少安排 1 名志愿者，至多安排 2 名志愿者，已知学生甲和乙是好朋友，须一起参与同一个活动，那么不同人员的分配方案种数共有
(A)

A.18
B.24
C.30
D.36

答案： 1. A. $C_{3}^{1}A_{3}^{3}=3 \times 3 \times 2 \times 1=18$.

2. [2024 辽宁大连模拟]$\mathrm{C}_{6}^{0}-\frac{\mathrm{C}_{6}^{1}}{3}+\frac{\mathrm{C}_{6}^{2}}{3^{2}}-\frac{\mathrm{C}_{6}^{3}}{3^{3}}+\frac{\mathrm{C}_{6}^{4}}{3^{4}}-\frac{\mathrm{C}_{6}^{5}}{3^{5}}+\frac{\mathrm{C}_{6}^{6}}{3^{6}}=$
(B)

A.$-\frac{64}{729}$
B.$\frac{64}{729}$
C.$-\frac{1}{729}$
D.$\frac{1}{729}$

答案： 2. B. $\frac{C_{6}^{0}}{3^{0}} - \frac{C_{6}^{1}}{3^{1}} + \frac{C_{6}^{2}}{3^{2}} - \frac{C_{6}^{3}}{3^{3}} + \frac{C_{6}^{4}}{3^{4}} - \frac{C_{6}^{5}}{3^{5}} + \frac{C_{6}^{6}}{3^{6}} = C_{6}^{0}(-1)^{0}×(\frac{1}{3})^{0} + C_{6}^{1}(-1)^{1}×(\frac{1}{3})^{1} + C_{6}^{2}(-1)^{2}×(\frac{1}{3})^{2} + C_{6}^{3}(-1)^{3}×(\frac{1}{3})^{3} + C_{6}^{4}(-1)^{4}×(\frac{1}{3})^{4} + C_{6}^{5}(-1)^{5}×(\frac{1}{3})^{5} + C_{6}^{6}(-1)^{6}×(\frac{1}{3})^{6} = ( - 1 + \frac{1}{3})^{6} = (-\frac{2}{3})^{6} = \frac{64}{729}$

3. [多选题][2025 浙江浙南期末]已知数列$\{ a_{n} \}$满足$a_{1}=1,a_{1}+\frac{a_{2}}{2}+·s+\frac{a_{n}}{n}=\frac{na_{n + 1}}{2(n + 1)}$，
令$b_{n}=\frac{1}{1012}(\frac{a_{n}}{n}-1)$,则
(ABD)

A.$a_{10}=100$
B.数列$\{ b_{n} \}$是等差数列
C.$b_{2 024}$为整数
D.数列$\{ b_{n} \}$的前 2 025 项和为 2 025

答案： 3. ABD. 因为$a_{1} + \frac{a_{2}}{2} + ·s + \frac{a_{n}}{n} = \frac{na_{n + 1}}{2(n + 1)}$ ①，所以当$n\geq2$时，$a_{1} + \frac{a_{2}}{2} + ·s + \frac{a_{n - 1}}{n - 1} = \frac{(n - 1)a_{n}}{2n}$ ②，① - ②得：
$\frac{a_{n}}{n} = \frac{na_{n + 1}}{2(n + 1)} - \frac{(n - 1)a_{n}}{2n} = \frac{n^{2}a_{n + 1} + 2n - (n^{2} - 1)a_{n}}{2n(n + 1)}$，化简得$n^{2}a_{n + 1} = (n + 1)^{2}a_{n}$，所以$\frac{a_{n + 1}}{(n + 1)^{2}} = \frac{a_{n}}{n^{2}}$，所以数列$\{\frac{a_{n}}{n^{2}}\}$是首项为$\frac{a_{1}}{1^{2}} = 1$的常数列，所以$\frac{a_{n}}{n^{2}} = 1$，所以$a_{n} = n^{2}$，$b_{n} = \frac{1}{1012}(\frac{a_{n}}{n} - 1) = \frac{n - 1}{1012}$。
对于A，$a_{10} = 10^{2} = 100$，故A正确；
对于B，因为$b_{n} = \frac{n - 1}{1012}$，所以数列$\{b_{n}\}$是首项为$b_{1} = \frac{1 - 1}{1012} = 0$，公差为$\frac{1}{1012}$的等差数列，故B正确；
对于C，$b_{2024} = \frac{2024 - 1}{1012} = \frac{2023}{1012}$，不是整数，故C错误；
对于D，因为数列$\{b_{n}\}$是首项为$0$，公差为$\frac{1}{1012}$的等差数列，所以其前$2025$项和为$S_{2025} = 2025×0 + \frac{2025×2024}{2}×\frac{1}{1012} = 2025$，故D正确。

4. [多选题][2025 江苏宿迁期末]在平行六面体$ABCD-A_{1}B_{1}C_{1}D_{1}$中，各棱长均为 6，
$\angle A_{1}AB=\angle A_{1}AD=\angle DAB=\frac{\pi}{3}$,则下列结论正确的有
(BC)

A.$AC_{1}=6\sqrt{3}$
B.四边形$A_{1}BCD_{1}$为正方形
C.$AA_{1}$与平面$ABCD$所成角的余弦值为$\frac{\sqrt{3}}{3}$
D.四边形$CC_{1}D_{1}D$内存在点$P$，使得直线$AP$与$BD_{1}$所成角为$\frac{\pi}{6}$

答案：
4. BC. 对于A，因为$\overrightarrow{AC_{1}} = \overrightarrow{AA_{1}} + \overrightarrow{AD} + \overrightarrow{AB}$，所以$|\overrightarrow{AC_{1}}|^{2} = (\overrightarrow{AA_{1}} + \overrightarrow{AD} + \overrightarrow{AB})^{2} = |\overrightarrow{AA_{1}}|^{2} + |\overrightarrow{AD}|^{2} + |\overrightarrow{AB}|^{2} + 2\overrightarrow{AA_{1}}·\overrightarrow{AD} + 2\overrightarrow{AD}·\overrightarrow{AB} + 2\overrightarrow{AA_{1}}·\overrightarrow{AB}$，因为$|\overrightarrow{AA_{1}}|^{2} = |\overrightarrow{AD}|^{2} = |\overrightarrow{AB}|^{2} = 36$，$\overrightarrow{AA_{1}}·\overrightarrow{AD} = |\overrightarrow{AA_{1}}||\overrightarrow{AD}|\cos\angle A_{1}AD = 6×6×\frac{1}{2} = 18$，同理，$\overrightarrow{AD}·\overrightarrow{AB} = 18$，$\overrightarrow{AA_{1}}·\overrightarrow{AB} = 18$，所以$|\overrightarrow{AC_{1}}|^{2} = 36 + 36 + 36 + 2×18 + 2×18 + 2×18 = 216$，所以$|\overrightarrow{AC_{1}}| = \sqrt{216} = 6\sqrt{6}$，故A错误；
对于B，因为$\overrightarrow{BC_{1}}·\overrightarrow{BA_{1}} = \overrightarrow{AD}·(\overrightarrow{AA_{1}} - \overrightarrow{AB}) = \overrightarrow{AD}·\overrightarrow{AA_{1}} - \overrightarrow{AD}·\overrightarrow{AB} = 18 - 18 = 0$，所以$\overrightarrow{BC_{1}}\perp\overrightarrow{BA_{1}}$，又$|\overrightarrow{AA_{1}}| = |\overrightarrow{AB}| = 6$，$\angle A_{1}AB = \frac{\pi}{3}$，所以$\triangle A_{1}AB$为等边三角形，即$|\overrightarrow{BA_{1}}| = 6$，因为$\overrightarrow{BC_{1}}//\overrightarrow{A_{1}D_{1}}$，$|\overrightarrow{BC_{1}}| = |\overrightarrow{BA_{1}}| = |\overrightarrow{A_{1}D_{1}}| = 6$，$\overrightarrow{BA_{1}}\perp\overrightarrow{BC_{1}}$，所以平行四边形$A_{1}BCD_{1}$为正方形，故B正确；
对于C，设平面$A_{1}BCD_{1}$的法向量为$\overrightarrow{n}$且$\overrightarrow{n} = a\overrightarrow{AB} + b\overrightarrow{AD} + c\overrightarrow{AA_{1}}$，则$\begin{cases}\overrightarrow{n}·\overrightarrow{AD} = a\overrightarrow{AB}·\overrightarrow{AD} + b\overrightarrow{AD}^{2} + c\overrightarrow{AA_{1}}·\overrightarrow{AD} = 0\\\overrightarrow{n}·\overrightarrow{AB} = a\overrightarrow{AB}^{2} + b\overrightarrow{BA}·\overrightarrow{AD} + c\overrightarrow{AA_{1}}·\overrightarrow{AB} = 0\end{cases}$，即$\begin{cases}18a + 36b + 18c = 0\\36a + 18b + 18c = 0\end{cases}$，取$a = - 1$，则$b = - 1$，$c = 3$，所以$\overrightarrow{n} = -\overrightarrow{AB} - \overrightarrow{AD} + 3\overrightarrow{AA_{1}}$；则$|\overrightarrow{n}| = \sqrt{(-\overrightarrow{AB} - \overrightarrow{AD} + 3\overrightarrow{AA_{1}})^{2}} = \sqrt{11×36 - 12×6×6×\frac{1}{2} + 2×6×6×\frac{1}{2}} = 6\sqrt{6}$，设$\overrightarrow{AA_{1}}$与平面$A_{1}BCD_{1}$所成角为$\alpha$，则$\sin\alpha = |\cos\langle\overrightarrow{n},\overrightarrow{AA_{1}}\rangle| = \frac{|\overrightarrow{n}·\overrightarrow{AA_{1}}|}{|\overrightarrow{n}|·|\overrightarrow{AA_{1}}|} = \frac{|(-\overrightarrow{AB} - \overrightarrow{AD} + 3\overrightarrow{AA_{1}})·\overrightarrow{AA_{1}}|}{|\overrightarrow{AA_{1}}|·|-\overrightarrow{AB} - \overrightarrow{AD} + 3\overrightarrow{AA_{1}}|} = \frac{72}{6×6\sqrt{6}} = \frac{\sqrt{6}}{3}$，故C正确；
对于D，因为$P$是在四边形$DD_{1}C_{1}C$内的动点，所以设$\overrightarrow{AP} = \overrightarrow{AD} + m\overrightarrow{DD_{1}} + n\overrightarrow{DC}$，其中$0\lt m\lt1$，$0\lt n\lt1$，故$\overrightarrow{AP} = \overrightarrow{AD} + m\overrightarrow{AA_{1}} + n\overrightarrow{AB}$，故$|\overrightarrow{AP}| = \sqrt{(\overrightarrow{AD} + m\overrightarrow{AA_{1}} + n\overrightarrow{AB})^{2}} = \sqrt{36(1 + m^{2} + n^{2}) + 36(m + n) + 36mn}$，$|\overrightarrow{BD_{1}}| = \sqrt{(\overrightarrow{AD} + \overrightarrow{AA_{1}} - \overrightarrow{AB})^{2}} = \sqrt{3×36 - 36 - 36 + 36} = 6\sqrt{2}$，而$\overrightarrow{AP}·\overrightarrow{BD_{1}} = (\overrightarrow{AD} + m\overrightarrow{AA_{1}} + n\overrightarrow{AB})·(\overrightarrow{AD} + \overrightarrow{AA_{1}} - \overrightarrow{AB}) = 36 + 18 - 18 + 18m + 36m - 18m + 18n + 18n - 36n = 36 + 36m$，假设存在这样的点$P$，则$\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} = \frac{|\overrightarrow{AP}·\overrightarrow{BD_{1}}|}{|\overrightarrow{AP}|·|\overrightarrow{BD_{1}}|} = \frac{36 + 36m}{6\sqrt{2}×\sqrt{36(1 + m^{2} + n^{2}) + 36(m + n) + 36mn}}$，整理得$m^{2} + (3n - 1)m + 3n^{2} + 3n + 1 = 0$，此时$\Delta = (3n - 1)^{2} - 12n^{2} - 12n - 4 = - 3n^{2} - 18n - 3\lt0$，该方程无解，故四边形$CC_{1}DD_{1}$内不存在点$P$，使得直线$\overrightarrow{AP}$与$\overrightarrow{BD_{1}}$所成角为$\frac{\pi}{6}$。

5. [2024 福建福州一中期末]过点$(2,0)$作曲线$f(x)=x^{3}$的切线$l$,则直线$l$的方程可能为$y = 0$或$27x - y - 54 = 0$.

答案： 5. $y = 0$或$27x - y - 54 = 0$。由$f(x) = x^{3}$，得$f^\prime(x) = 3x^{2}$。设切点坐标为$(x_{0},x_{0}^{3})$，则$k = f^\prime(x_{0}) = 3x_{0}^{2}$，所以$k = f^\prime(x_{0}) = 3x_{0}^{2} = \frac{x_{0}^{3} - 0}{x_{0} - 2}$，解得$x_{0} = 0$或$x_{0} = 3$。当$x_{0} = 0$时，切线方程为$y = 0$；当$x_{0} = 3$时，切点为$(3,27)$，斜率$k = 27$，故切线方程为$y - 27 = 27(x - 3)$，整理得$27x - y - 54 = 0$。

6. [2025 河南郑州期中]设$P$是双曲线$\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$上一点，$M,N$分别是圆$(x - 5)^{2}+y^{2}=\frac{1}{4}$和$(x + 5)^{2}+y^{2}=\frac{9}{4}$上的点，则$\vert PM\vert-\vert PN\vert$的最大值为8.

答案：
6. 8. 圆$(x + 5)^{2} + y^{2} = \frac{9}{4}$的圆心$F_{1}(-5,0)$，半径$r_{1} = \frac{3}{2}$，圆$(x - 5)^{2} + y^{2} = \frac{1}{4}$的圆心$F_{2}(5,0)$，半径$r_{2} = \frac{1}{2}$，双曲线$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$的实半轴长$a = 3$，半焦距$c = 5$，则$F_{1}$，$F_{2}$为其左、右焦点。如图所示，对任一确定的点$P$，有$|PM|_{max} = |PF_{2}| + r_{2}$，$|PN|_{min} = |PF_{1}| - r_{1}$。要使$|PM| - |PN|$取最大值，点$P$必在双曲线左支上，所以$(|PM| - |PN|)_{max} = |PF_{2}| + r_{2} - (|PF_{1}| - r_{1}) = 2a + r_{1} + r_{2} = 8$。