《2024年高职高考全真模拟试卷辽海出版社高中数学全一册人教版》

22. 已知函数$f(x) = (\cos2x + \sin2x)(\cos2x - \sin2x)$.
(1)求函数$f(x)$的最小正周期；
(2)若$0 ＜ \alpha ＜ \frac{\pi}{2}$，$0 ＜ \beta ＜ \frac{\pi}{2}$，且$f(\frac{\alpha}{4}) = \frac{1}{3}$，$f(\frac{\beta}{4}) = \frac{2}{3}$，求$\sin(\alpha - \beta)$的值.

答案：
(1)$\because f(x)=(\cos2x+\sin2x)(\cos2x-\sin2x)=\cos^{2}2x-\sin^{2}2x=\cos4x$，$\therefore$函数$f(x)$的最小正周期为$T=\frac{2\pi}{4}=\frac{\pi}{2}$
(2)由
(1)得$f(x)=\cos4x$，$\because f(\frac{\alpha}{4})=\frac{1}{3}$，$f(\frac{\beta}{4})=\frac{2}{3}$，$\therefore\cos\alpha=\frac{1}{3}$，$\cos\beta=\frac{2}{3}$，$\because0＜\alpha＜\frac{\pi}{2}$，$0＜\beta＜\frac{\pi}{2}$，$\therefore\sin\alpha=\sqrt{1 - \cos^{2}\alpha}=\frac{2\sqrt{2}}{3}$，$\sin\beta=\sqrt{1 - \cos^{2}\beta}=\frac{\sqrt{5}}{3}$，$\therefore\sin(\alpha - \beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta=\frac{2\sqrt{2}}{3}\times\frac{2}{3}-\frac{1}{3}\times\frac{\sqrt{5}}{3}=\frac{4\sqrt{2}-\sqrt{5}}{9}$

23. 已知数列$\{a_{n}\}$的前$n$项和$S_{n} = n^{2}(n\in\mathbf{N}^{*})$，数列$\{b_{n}\}$为等比数列，且满足$b_{1} = a_{2}$，$3b_{2} = b_{3}$.
(1)求数列$\{a_{n}\}$以及数列$\{b_{n}\}$的通项公式；
(2)若$c_{n} = a_{n} + b_{n}$，求数列$\{c_{n}\}$的前$n$项和$T_{n}$.

答案：
(1)由已知$S_{n}=n^{2}$，得$a_{1}=S_{1}=1$，当$n\geq2$时，$a_{n}=S_{n}-S_{n - 1}=n^{2}-(n - 1)^{2}=2n - 1$，所以$a_{n}=2n - 1(n\in\mathbf{N}^{*})$，由已知$b_{1}=a_{2}=3$，设等比数列$\{b_{n}\}$的公比为$q$，由$3b_{2}=b_{3}$得$\frac{b_{3}}{b_{2}}=3$，即$q = 3$，故$b_{n}=3^{n}$
(2)由
(1)可知$c_{n}=a_{n}+b_{n}=2n - 1+3^{n}$，$T_{n}=c_{1}+c_{2}+c_{3}+\cdots+c_{n}=[1 + 3+5+\cdots+(2n - 1)]+[3 + 3^{2}+3^{3}+\cdots+3^{n}]=\frac{(1 + 2n - 1)n}{2}+\frac{3(1 - 3^{n})}{1 - 3}=n^{2}+\frac{3^{n + 1}}{2}-\frac{3}{2}$，所以$T_{n}=n^{2}+\frac{3^{n + 1}}{2}-\frac{3}{2}(n\in\mathbf{N}_{+})$

24. 已知双曲线中心在原点，焦点$F_{1}$，$F_{2}$在坐标轴上. $c = \sqrt{2}a$且过点$(3, -\sqrt{3})$.
(1)求双曲线方程；
(2)若点$M(3, m)$在双曲线上，求证：$MF_{1}\perp MF_{2}$.

答案：
(1)由$\frac{c}{a}=\sqrt{2}$得$a = b$，所以双曲线为等轴双曲线，故可设双曲线方程为$x^{2}-y^{2}=\lambda$，将点$(3,-\sqrt{3})$代入上式得$\lambda = 6$，则双曲线方程为$x^{2}-y^{2}=6$
(2)$\because M(3,m)$在双曲线上，$\therefore m^{2}=3$，又$\because a = b=\sqrt{6}$，$\therefore c^{2}=6 + 6 = 12$，则当$m=\sqrt{3}$时，$k_{MF_{1}}\cdot k_{MF_{2}}=\frac{\sqrt{3}}{3 - c}\cdot\frac{\sqrt{3}}{3 + c}=\frac{3}{9 - c^{2}}=-1$，$\therefore MF_{1}\perp MF_{2}$，当$m=-\sqrt{3}$时，由对称性仍有$MF_{1}\perp MF_{2}$