《2025年热搜题高中数学必修第二册人教版》

7. [2024·南京秦淮中学期末]如图所示，在平面直角坐标系$Oxy$中，已知四边形$OABC$是等腰梯形，$OA// BC$，$A(6,0)$，$C(1,\sqrt{3})$，点$M$满足$\overrightarrow{OM} = \frac{1}{2}\overrightarrow{OA}$，点$P$在线段$BC$上运动（包括端点）。
(1)求$\angle OCM$的余弦值。
(2)是否存在实数$\lambda$，使$(\overrightarrow{OA} - \lambda\overrightarrow{OP})\perp\overrightarrow{CM}$成立？若存在，求出实数$\lambda$的取值范围；若不存在，请说明理由。

答案： 7.
(1)由题意可得$\overrightarrow{CO} = (-1,-\sqrt{3})$，$\overrightarrow{CM} = \overrightarrow{OM} + \overrightarrow{CO} = \frac{1}{2}\overrightarrow{OA} + \overrightarrow{CO} = (3,0) + (-1,-\sqrt{3}) = (2,-\sqrt{3})$，故$\cos \angle OCM = \cos \langle \overrightarrow{CO},\overrightarrow{CM} \rangle = \frac{\overrightarrow{CO} · \overrightarrow{CM}}{|\overrightarrow{CO}||\overrightarrow{CM}|} = \frac{\sqrt{7}}{14}$.
(2)设$P(t,\sqrt{3})$，其中$1 \leq t \leq 5$，$\lambda\overrightarrow{OP} = (\lambda t,\sqrt{3}\lambda)$，$\overrightarrow{OA} - \lambda\overrightarrow{OP} = (6 - \lambda t,-\sqrt{3}\lambda)$，$\overrightarrow{CM} = (2,-\sqrt{3})$，若$(\overrightarrow{OA} - \lambda\overrightarrow{OP}) \perp \overrightarrow{CM}$，则$(\overrightarrow{OA} - \lambda\overrightarrow{OP}) · \overrightarrow{CM} = 0$，即$12 - 2\lambda t + 3\lambda = 0$，可得$(2t - 3)\lambda = 12$.若$t = \frac{3}{2}$，则$\lambda$不存在；若$t \neq \frac{3}{2}$，则$\lambda = \frac{12}{2t - 3}$.因为$t \in [1,\frac{3}{2}) \cup (\frac{3}{2},5]$，所以$-1 \leq 2t - 3 \leq 7$，且$2t - 3 \neq 0$，所以$\lambda \in (-\infty,-12] \cup [\frac{12}{7},+\infty)$，所以实数$\lambda$的取值范围是$(-\infty,-12] \cup [\frac{12}{7},+\infty)$.

8. [2024·太原五中月考]已知向量$a = (4\sin\alpha,1 - \cos\alpha)$，$b = (1,-2)$，若$a· b = -2$，则$\frac{\sin\alpha\cos\alpha}{2\sin^2\alpha - \cos^2\alpha}=$(A)

A.1
B.$-1$
C.$-\frac{2}{7}$
D.$-\frac{1}{2}$

答案： 8.A【解析】由$a · b = -2$，得$4\sin \alpha - 2(1 - \cos \alpha) = -2$，整理得$\tan \alpha = -\frac{1}{2}$，所以$\frac{\sin \alpha \cos \alpha}{2\sin^{2} \alpha - \cos^{2} \alpha} = \frac{\tan \alpha}{2\tan^{2} \alpha - 1} = \frac{-\frac{1}{2}}{2 × (-\frac{1}{2})^{2} - 1} = 1$. 故选A.

9. [2024·枣庄三中期中]已知向量$a = (\cos\frac{3}{2}x,\sin\frac{3}{2}x)$，$b = (\cos\frac{x}{2},\sin\frac{x}{2})$，且$x\in[0,\frac{\pi}{2}]$，$f(x) = a· b - 2\lambda|a - b|$（$\lambda$为常数）。
(1)求$a· b$及$|a - b|$；
(2)若$f(x)$的最大值是$\frac{3}{2}$，求实数$\lambda$的值。

答案： 9.
(1)$a · b = \cos \frac{3}{2}x \cos \frac{x}{2} + \sin \frac{3}{2}x \sin \frac{x}{2} = \cos x$，$|a - b| = \sqrt{(\cos \frac{3}{2}x - \cos \frac{x}{2})^{2} + (\sin \frac{3}{2}x - \sin \frac{x}{2})^{2}} = \sqrt{2 - 2\cos x} = 2|\sin \frac{x}{2}|$，因为$x \in [0,\frac{\pi}{2}]$，所以$\sin \frac{x}{2} ＞ 0$，所以$|a - b| = 2\sin \frac{x}{2}$.
(2)$f(x) = \cos x - 4\lambda \sin \frac{x}{2} = -2(\sin \frac{x}{2} + \lambda)^{2} + 2\lambda^{2} + 1$.因为$x \in [0,\frac{\pi}{2}]$，所以$0 \leq \sin \frac{x}{2} \leq \frac{\sqrt{2}}{2}$.
①若$\lambda ＞ 0$，当且仅当$\sin \frac{x}{2} = 0$时，$f(x)$取得最大值1，这与已知矛盾；
②若$-\frac{\sqrt{2}}{2} \leq \lambda \leq 0$，当且仅当$\sin \frac{x}{2} = -\lambda$时，$f(x)$取得最大值$2\lambda^{2} + 1$，由已知得$2\lambda^{2} + 1 = \frac{3}{2}$，解得$\lambda = -\frac{1}{2}$；
③若$\lambda ＜ -\frac{\sqrt{2}}{2}$，当且仅当$\sin \frac{x}{2} = \frac{\sqrt{2}}{2}$时，$f(x)$取得最大值$-2\sqrt{2}\lambda$，由已知得$-2\sqrt{2}\lambda = \frac{3}{2}$，解得$\lambda = -\frac{3\sqrt{2}}{8}$，这与$\lambda ＜ -\frac{\sqrt{2}}{2}$矛盾.综上所述，$\lambda = -\frac{1}{2}$.

10. [2024·杭州二中月考]已知$\theta\in[0,\pi)$，向量$a = (\cos\theta,\sin\theta)$，$b = (1,0)$，$O$为坐标原点，$P_1$，$P_2$，$P_3$是坐标平面上的三点，且$\overrightarrow{OP_2} = 2[\overrightarrow{OP_1} - (a·\overrightarrow{OP_1})· a]$，$\overrightarrow{OP_3} = 2[\overrightarrow{OP_2} - (b·\overrightarrow{OP_2})· b]$。
(1)若$\theta = \frac{\pi}{2}$，$P_1$的坐标为$(20,21)$，求$\overrightarrow{OP_3}$；
(2)若$\theta = \frac{2\pi}{3}$，$|\overrightarrow{OP_1}| = 6$，求$|\overrightarrow{OP_3}|$的最大值；
(3)若存在$\alpha\in[0,\pi)$，使得当$\overrightarrow{OP_1} = (\cos\alpha,\sin\alpha)$时，$\triangle P_1P_2P_3$为等边三角形，求$\theta$的所有可能值。

答案： 10.
(1)当$\theta = \frac{\pi}{2}$时，$a = (0,1)$.因为$P_1$的坐标为$(20,21)$，所以$\overrightarrow{OP_1} = (20,21)$，所以$\overrightarrow{OP_2} = 2[\overrightarrow{OP_1} - (a · \overrightarrow{OP_1}) · a] = 2[(20,21) - 21(0,1)] = (40,0)$，所以$\overrightarrow{OP_3} = 2[\overrightarrow{OP_2} - (b · \overrightarrow{OP_2}) · b] = 2[(40,0) - 40(1,0)] = (0,0)$.
(2)当$\theta = \frac{2\pi}{3}$时，$a = (-\frac{1}{2},\frac{\sqrt{3}}{2})$.因为$|\overrightarrow{OP_1}| = 6$，所以设$\overrightarrow{OP_1} = (6\cos \alpha,6\sin \alpha)$，则$\overrightarrow{OP_2} = 2[\overrightarrow{OP_1} - (a · \overrightarrow{OP_1}) · a] = 2[(6\cos \alpha,6\sin \alpha) - (3\sqrt{3}\sin \alpha - 3\cos \alpha) · (-\frac{1}{2},\frac{\sqrt{3}}{2})] = (3\sqrt{3}\sin \alpha + 9\cos \alpha,3\sin \alpha + 3\sqrt{3}\cos \alpha)$，所以$\overrightarrow{OP_3} = 2[\overrightarrow{OP_2} - (b · \overrightarrow{OP_2}) · b] = 2[(3\sqrt{3}\sin \alpha + 9\cos \alpha,3\sin \alpha + 3\sqrt{3}\cos \alpha) - (3\sqrt{3}\sin \alpha + 9\cos \alpha,3\sin \alpha + 3\sqrt{3}\cos \alpha) · (1,0)] = (0,6\sin \alpha + 6\sqrt{3}\cos \alpha)$，所以$|\overrightarrow{OP_3}| = |6\sin \alpha + 6\sqrt{3}\cos \alpha| = 12|\sin(\alpha + \frac{\pi}{3})|$，所以当$|\sin(\alpha + \frac{\pi}{3})| = 1$时，$|\overrightarrow{OP_3}|$取得最大值，最大值为12.
(3)因为$\overrightarrow{OP} = (\cos \alpha,\sin \alpha)$，所以$\overrightarrow{OP_2} = 2[\overrightarrow{OP} - (a · \overrightarrow{OP}) · a] = 2[(\cos \alpha,\sin \alpha) - (\cos \alpha \cos \theta + \sin \alpha \sin \theta)(\cos \theta,\sin \theta)] = 2(\sin \theta \sin(\theta - \alpha),\cos \theta \sin(\alpha - \theta))$，$\overrightarrow{OP_3} = 2[\overrightarrow{OP_2} - (b · \overrightarrow{OP_2}) · b] = 2[2(\sin \theta \sin(\theta - \alpha),\cos \theta \sin(\alpha - \theta)) - 2\sin \theta \sin(\theta - \alpha) · (1,0)] = 4(0,\cos \theta \sin(\alpha - \theta))$，$\overrightarrow{P_2P_3} = (2\sin \theta \sin(\theta - \alpha) - \cos \alpha,2\cos \theta \sin(\alpha - \theta) - \sin \alpha)$，所以$\overrightarrow{P_2P_3} = 2\sin(\theta - \alpha)(\sin \theta,\cos \theta)$.因为$\triangle P_1P_2P_3$为等边三角形，$\begin{cases} |\overrightarrow{P_2P_3}| = |\overrightarrow{P_2P_4}| = 2|\sin(\alpha - \theta)| = 1, \\ \frac{\overrightarrow{P_2P_3} · \overrightarrow{P_2P_4}}{|\overrightarrow{P_2P_3}||\overrightarrow{P_2P_4}|} = -\frac{1}{2}. \end{cases}$所以$|\sin(\alpha - \theta)| = \frac{1}{2}$，$(2\sin \theta \sin(\theta - \alpha) - \cos \alpha,2\cos \theta \sin(\alpha - \theta) - \sin \alpha) · (2\sin(\alpha - \theta) \sin \theta,2\sin(\alpha - \theta) \cos \theta) = -\frac{1}{2}$，整理得$\cos 2\alpha = -\frac{1}{2}$.又$2\alpha \in [0,2\pi)$，所以$\alpha = \frac{\pi}{3}$或$\alpha = \frac{2\pi}{3}$.当$\alpha = \frac{\pi}{3}$，$|\sin(\alpha - \theta)| = \frac{1}{2}$时，$\theta = \frac{\pi}{6}$或$\frac{\pi}{2}$；当$\alpha = \frac{2\pi}{3}$，$|\sin(\alpha - \theta)| = \frac{1}{2}$时，$\theta = \frac{5\pi}{6}$或$\frac{\pi}{2}$.综上所述，$\theta$的所有可能值为$\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6}$.