《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. 已知数列{aₙ}的前n项和为Sₙ = 2aₙ - 4.
(1)求数列{aₙ}的通项公式aₙ；
(2)设bₙ = $\frac{naₙ}{2^{n}}$，求数列{bₙ}的前n项和为Tₙ.

答案： 解：
(1)当 $n = 1$ 时，$a_1 = S_1 = 2a_1 - 4$，
$\therefore a_1 = 2a_1 - 4$ ，$a_1 = 4$
当 $n ＞ 1$ 时，$a_n = S_n - S_{n - 1} = (2a_n - 4) - (2a_{n - 1} - 4) = 2a_n - 2a_{n - 1}$
$\therefore a_n = 2a_{n - 1}$
$\therefore \frac{a_n}{a_{n - 1}} = 2$
$\therefore$ 数列 $\{a_n\}$ 是等比数列，公比 $q = 2$
$\therefore a_n = a_1q^{n - 1} = 4\times2^{n - 1} = 2^{n + 1}$
(2)$\because b_n = \frac{na_n}{2^{n}} = \frac{n\cdot2^{n + 1}}{2^{n}} = 2n$
$\therefore T_n = b_1 + b_2 + \cdots + b_n$
$= 2\times(1 + 2 + \cdots + n)$
$= 2\times\frac{n(1 + n)}{2}$
$= n^{2} + n$

23. △ABC中，角A，B，C所对的边分别为a，b，c. 已知a = 3，cosA = $\frac{\sqrt{6}}{3}$，B = A + $\frac{\pi}{2}$.
(1)求b的值；
(2)求△ABC的面积.

答案： 解：
(1)在 $\triangle ABC$ 中，
$\sin A = \sqrt{1 - \cos^{2}A} = \frac{\sqrt{3}}{3}$
$\because B = A + \frac{\pi}{2}$
$\therefore \sin B = \sin(A + \frac{\pi}{2}) = \cos A = \frac{\sqrt{6}}{3}$
由正弦定理可得，$b = \frac{a\sin B}{\sin A} = \frac{3\times\frac{\sqrt{6}}{3}}{\frac{\sqrt{3}}{3}} = 3\sqrt{2}$
(2)$\because B = A + \frac{\pi}{2}$
$\therefore \cos B = \cos(A + \frac{\pi}{2}) = -\sin A = -\frac{\sqrt{3}}{3}$
又 $\because A + B + C = \pi$
$\therefore C = \pi - (A + B)$
$\therefore \sin C = \sin[\pi - (A + B)]$
$= \sin(A + B)$
$= \sin A\cos B + \cos A\sin B$
$= \frac{\sqrt{3}}{3}\times(-\frac{\sqrt{3}}{3}) + \frac{\sqrt{6}}{3}\times\frac{\sqrt{6}}{3}$
$= \frac{1}{3}$
$\therefore S = \frac{1}{2}ab\sin C = \frac{1}{2}\times3\times3\sqrt{2}\times\frac{1}{3} = \frac{3\sqrt{2}}{2}$

24. 设椭圆C：$\frac{x^{2}}{a^{2}}$ + $\frac{y^{2}}{b^{2}}$ = 1(a ＞ b ＞ 0)过点(0，4)，离心率为$\frac{3}{5}$，过点(3，0)且斜率为$\frac{4}{5}$的直线与椭圆相交于A、B两点，椭圆左、右焦点分别为F₁、F₂.
(1)求椭圆C的方程；
(2)求△F₁AB的周长；
(3)求AB的垂直平分线方程.

答案： 解：
(1)将 $(0, 4)$ 代入 $C$ 的方程得 $\frac{16}{b^{2}} = 1$
$\therefore b = 4$，
$\because e = \frac{c}{a} = \frac{3}{5}$，$\therefore \frac{c^{2}}{a^{2}} = \frac{9}{25}$
$\therefore \frac{a^{2} - b^{2}}{a^{2}} = \frac{9}{25}$ 即 $1 - \frac{16}{a^{2}} = \frac{9}{25}$，
$\therefore a^{2} = 25$
$\therefore$ 椭圆 $C$ 的方程为 $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$
(2)$\because$ 直线过点 $(3, 0)$
$\therefore$ 直线过椭圆的右焦点 $F_2$
$\therefore \triangle F_1AB$ 的周长 $= AF_1 + AF_2 + BF_1 + BF_2 = 2a + 2a = 20$
(3)过点 $(3, 0)$ 且斜率为 $\frac{4}{5}$ 的直线方程为 $y = \frac{4}{5}(x - 3)$
设直线与椭圆 $C$ 的交点为 $A(x_1, y_1)$，$B(x_2, y_2)$将直线方程 $y = \frac{4}{5}(x - 3)$ 代入椭圆 $C$ 的方程得 $\frac{x^{2}}{25} + \frac{(x - 3)^{2}}{25} = 1$，即 $x^{2} - 3x - 8 = 0$
$\therefore x_1 + x_2 = 3$
设线段 $AB$ 的中点坐标为 $(x_0, y_0)$
$\therefore x_0 = \frac{x_1 + x_2}{2} = \frac{3}{2}$，$y_0 = \frac{4}{5}(x_0 - 3) = -\frac{6}{5}$
$\therefore$ 中点坐标为 $(\frac{3}{2}, -\frac{6}{5})$
$\because AB$ 的斜率为 $\frac{4}{5}$
$\therefore$ 垂直平分线的斜率为 $-\frac{5}{4}$
$\therefore$ 垂直平分线的方程为 $y + \frac{6}{5} = -\frac{5}{4}(x - \frac{3}{2})$
即：$y = -\frac{5}{4}x + \frac{27}{40}$