《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. 在△ABC中，角A，B，C所对的边分别为a，b，c，已知b = 4，B = 30°，C = 45°.
(1)求c的值；
(2)求△ABC的面积S.

答案： 解：
(1)依题意，由正弦定理$\frac{b}{\sin B} = \frac{c}{\sin C}$，得$c = \frac{b}{\sin B}\times\sin C = \frac{4\times\frac{\sqrt{2}}{2}}{\frac{1}{2}} = 4\sqrt{2}$
(2)在$\triangle ABC$中，$A + B + C = 180^{\circ}$$\therefore \sin A = \sin[180^{\circ} - (B + C)] = \sin(B + C) = \sin(30^{\circ} + 45^{\circ})$$= \sin30^{\circ}\cos45^{\circ} + \cos30^{\circ}\sin45^{\circ} = \frac{1}{2}\times\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2} = \frac{\sqrt{2} + \sqrt{6}}{4}$$\therefore S_{\triangle ABC} = \frac{1}{2}bc\sin A = \frac{1}{2}\times4\times4\sqrt{2}\times\frac{\sqrt{2} + \sqrt{6}}{4} = 4 + 4\sqrt{3}$

23. 已知各项均为正数的数列{aₙ}的前n项和为Sₙ，满足S₁ ＞ 1，且6Sₙ = (aₙ + 1)(aₙ + 2).
(1)求数列{aₙ}的通项公式；
(2)设数列{bₙ}满足$bₙ = \frac{1}{(aₙ + 1)(aₙ + 4)}$，记Tₙ为{bₙ}的前n项和，求Tₙ.

答案： 解：
(1)$a_1 = S_1 = \frac{1}{6}(a_1 + 1)(a_1 + 2)$，解得$a_1 = 1$或$a_1 = 2$，由$a_1 = S_1 ＞ 1$，可得$a_1 = 2$由$a_{n + 1} = S_{n + 1} - S_n = \frac{1}{6}(a_{n + 1} + 1)(a_{n + 1} + 2) - \frac{1}{6}(a_n + 1)(a_n + 2)$得$(a_{n + 1} + a_n)(a_{n + 1} - a_n - 3) = 0$即$a_{n + 1} + a_n = 0$或$a_{n + 1} - a_n - 3 = 0$数列$\{a_n\}$各项均为正数，由$a_{n + 1} + a_n = 0$，得$a_{n + 1} = -a_n$不成立，所以$a_{n + 1} - a_n - 3 = 0$，即$a_{n + 1} - a_n = 3$所以数列$\{a_n\}$是首项为2，公差为3的等差数列，数列$\{a_n\}$的通项公式$a_n = 3n - 1(n\in\mathbf{N}_+)$
(2)由$b_n = \frac{1}{(a_n + 1)(a_n + 4)} = \frac{1}{(3n - 1 + 1)(3n - 1 + 4)} = \frac{1}{3n(3n + 3)}$，所以$b_n = \frac{1}{9}\times\frac{1}{n(n + 1)} = \frac{1}{9}(\frac{1}{n} - \frac{1}{n + 1})$$T_n = b_1 + b_2 + b_3 + \cdots + b_n = \frac{1}{9}[(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \cdots + (\frac{1}{n} - \frac{1}{n + 1})]$$= \frac{1}{9}(1 - \frac{1}{n + 1}) = \frac{n}{9(n + 1)}(n\in\mathbf{N}_+)$

24. 已知曲线上任意一点M到两个定点F₁(-$\sqrt{3}$，0)和F₂($\sqrt{3}$，0)的距离之和为4.
(1)求此曲线的方程；
(2)设过点P(0，-2)的直线l与曲线交于A，B两点，且$\overrightarrow{OA}\cdot\overrightarrow{OB}=0$，求直线l的方程.

答案：
(1)根据椭圆的定义，可知动点M的轨迹为椭圆，其中$a = 2$，$c = \sqrt{3}$，则$b = \sqrt{a^2 - c^2} = 1$，所以动点M的轨迹方程为$\frac{x^2}{4} + y^2 = 1$
(2)当直线的斜率不存在时，不满足题意；当直线的斜率存在时，设直线l的方程为$y = kx - 2$，设$A(x_1, y_1)$，$B(x_2, y_2)$，$\because \overrightarrow{OA}\cdot\overrightarrow{OB} = 0$，$\therefore x_1x_2 + y_1y_2 = 0$$\because y_1 = kx_1 - 2$，$y_2 = kx_2 - 2$，$\therefore y_1y_2 = (kx_1 - 2)\cdot(kx_2 - 2) = k^2x_1x_2 - 2k(x_1 + x_2) + 4$$\therefore (1 + k^2)x_1x_2 - 2k(x_1 + x_2) + 4 = 0$ …①由方程组$\begin{cases}\frac{x^2}{4} + y^2 = 1\\y = kx - 2\end{cases}$得$(1 + 4k^2)x^2 - 16kx + 12 = 0$，则$\triangle = (-16k)^2 - 4\times12\times(1 + 4k^2) ＞ 0$即：$k ＜ -\frac{\sqrt{3}}{2}$或$k ＞ \frac{\sqrt{3}}{2}$，则$x_1 + x_2 = \frac{16k}{1 + 4k^2}$，$x_1\cdot x_2 = \frac{12}{1 + 4k^2}$，代入①，得$(1 + k^2)\cdot\frac{12}{1 + 4k^2} - 2k\cdot\frac{16k}{1 + 4k^2} + 4 = 0$即$k^2 = 4$，解得，$k = 2$或$k = -2$所以，直线l的方程为$y = 2x - 2$或$y = -2x - 2$