《2026年一本密卷高考数学》

14. 已知$\triangle ABC$的内角$A,B,C$对边分别为$a,b,c,BC$边上的高为$h,h = b + c - a$,则$\sin A$的最小值为$\frac{24}{25}$.

答案：
14.$\frac{24}{25}$ 在$\triangle ABC$中，$b + c = a + h$，$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc} = \frac{(b + c)^{2} - a^{2} - 2bc}{2bc} = \frac{(a + h)^{2} - a^{2}}{2bc} - 1 = \frac{h^{2} + 2ah}{2bc} - 1$，即$1 + \cos A = \frac{h^{2} + 2ah}{2bc}$；又$\frac{1}{2}bc\sin A = \frac{1}{2}ah$，即$\sin A = \frac{ah}{bc}$，又$\sin A ＞ 0$；故$\frac{1 + \cos A}{\sin A} = \frac{h^{2} + 2ah}{2ah} = 1 + \frac{h}{2a} = \frac{2\cos^{2}\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}} = \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{1}{\tan\frac{A}{2}} ＞ 0$。
如图，在$\triangle ABC$中，过$B$作$CB$的垂线$EB$，且使$EB = 2h$，则$AE = AB = c$。

$\because b + c = a + h \geqslant |CE|$，即$(a + h)^{2} \geqslant a^{2} + (2h)^{2} = a^{2} + 4h^{2}$，可得$0 ＜ \frac{h}{2a} \leqslant \frac{1}{3}$，$\therefore 1 ＜ 1 + \frac{h}{2a} \leqslant \frac{4}{3}$，即$1 ＜ \frac{1}{\tan\frac{A}{2}} \leqslant \frac{4}{3}$，$\therefore\frac{3}{4} \leqslant \tan\frac{A}{2} ＜ 1$，$\sin A = \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin^{2}\frac{A}{2} + \cos^{2}\frac{A}{2}} = \frac{2\tan\frac{A}{2}}{1 + \tan^{2}\frac{A}{2}}$。设$\tan\frac{A}{2} = t$，$t \in [\frac{3}{4},1)$，$t + \frac{1}{t}$在区间$[\frac{3}{4},1)$单调递减，$\therefore t + \frac{1}{t} \in (2,\frac{25}{12}]$，即$\frac{1}{t + \frac{1}{t}} \in [\frac{12}{25},\frac{1}{2})$，$\therefore\frac{24}{25} \leqslant \sin A ＜ 1$，当且仅当$\frac{h}{2a} = \frac{1}{3}$时，即$A,C,E$三点共线时等号成立。
验证：如下图中$\triangle ABC$，若$b = c = \frac{5}{6}a$，$h = \frac{2}{3}a$时，满足$b + c - a = h$，此时$\cos\frac{A}{2} = \frac{h}{b} = \frac{\frac{2}{3}a}{\frac{5}{6}a} = \frac{4}{5}$，$\sin\frac{A}{2} = \frac{3}{5}$，故存在这样的$\triangle ABC$，使得$\sin A = 2\sin\frac{A}{2}\cos\frac{A}{2} = \frac{24}{25}$成立。

因此$\sin A$的最小值为$\frac{24}{25}$。

15. (13分)已知数列$\{a_{n}\}$为等差数列,且满足$a_{2n}=2a_{n}+1(n\in\mathbf{N}^{*})$.
(1)若$a_{1}=1$,求数列$\{\frac{1}{a_{n}a_{n + 1}}\}$的前$n$项和$S_{n}$;
(2)若数列$\{b_{n}\}$满足$2b_{1}+b_{2}=b_{1}b_{2}$,且数列$\{a_{n}· b_{n}\}$的前$n$项和$T_{n}=(3n - 4)· 2^{n + 1}+8$,求数列$\{b_{n}\}$的通项公式.

答案： 15.解：
(1)当$n = 1$时，由$a_{2n} = 2a_{n} + 1$，则$a_{2} = 2a_{1} + 1$，由$a_{1} = 1$，则$a_{2} = 3$，所以等差数列$\{ a_{n}\}$的公差为$a_{2} - a_{1} = 2$，所以$a_{n} = 1 + (n - 1) · 2 = 2n - 1$，$\frac{1}{a_{n}a_{n + 1}} = \frac{1}{(2n - 1)(2n + 1)} = \frac{1}{2}(\frac{1}{2n - 1} - \frac{1}{2n + 1})$，故数列$\{ a_{n}\}$的前$n$项和$S_{n} = \frac{1}{2}(1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + ·s + \frac{1}{2n - 1} - \frac{1}{2n + 1}) = \frac{1}{2}(1 - \frac{1}{2n + 1}) = \frac{n}{2n + 1}$。
(2)当$n = 1$时，$a_{1}b_{1} = T_{1} = (3 - 4) × 2^{2} + 8 = 4$，可得$b_{1} = \frac{4}{a_{1}}$，当$n \geqslant 2$时，$a_{n}b_{n} = T_{n} - T_{n - 1} = [(3n - 4) · 2^{n + 1} + 8] - [(3n - 7) · 2^{n} + 8] = 3n · 2^{n + 1} - 2^{n + 3} - 3n · 2^{n} + 7 · 2^{n} = 2^{n}(6n - 8 - 3n + 7) = 2^{n}(3n - 1)$，将$n = 1$代入上式，则$a_{1}b_{1} = 2 × (3 - 1) = 4 = T_{1}$，综上所述，$a_{n}b_{n} = 2^{n}(3n - 1)$，$n \in \mathbf{N}^{*}$。$a_{2}b_{2} = 2^{2} × (3 × 2 - 1) = 20$，可得$b_{2} = \frac{20}{a_{2}}$，又因为$a_{2} = 2a_{1} + 1$，则$b_{2} = \frac{20}{2a_{1} + 1}$，由方程$2b_{1} + b_{2} = b_{1}b_{2}$，可得$\frac{2}{20}(2a_{1} + 1) + \frac{1}{4}a_{1} = 1$，解得$a_{1} = 2$，由$a_{2} = 2a_{1} + 1 = 5$，则等差数列$\{ a_{n}\}$的公差为$3$，所以$a_{n} = 3n - 1$，由$a_{n}b_{n} = 2^{n}(3n - 1)$，$n \in \mathbf{N}^{*}$，则$b_{n} = 2^{n}$。

16. (15分)某商场进行抽奖活动,设置摸奖箱内有红球$1$个,白球$2$个,黑球$3$个,小球除颜色外没有任何区别.规定:摸到红球记$1$分,摸到白球记$0$分,摸到黑球记$-1$分.抽奖人摸$3$个球为一次抽奖,总分记为$X$,若$X\geqslant 0$,则获奖.
方案一:从中一次摸$1$个球,记录分数后不放回.
方案二:从中一次摸$1$个球,记录分数后放回.
(1)若甲顾客按照方案一摸球记分,求甲顾客获奖的概率;
(2)若乙顾客按照方案一摸球记分,求第二次摸到红球条件下,乙顾客获奖的概率;
(3)若丙顾客按照方案二摸球记分,求$X$的分布列和数学期望.

答案： 16.解：
(1)若$X \geqslant 0$，则甲顾客摸到$1$个红球$2$个白球、或者是$1$个红球$1$个白球$1$个黑球，所以$P(X \geqslant 0) = \frac{C_{2}^{2}C_{1}^{1} + C_{2}^{1}C_{3}^{1}C_{3}^{1}}{C_{6}^{3}} = \frac{7}{20}$。
(2)记事件$N$：乙顾客按照方案一摸球获奖，由
(1)可知$P(N) = \frac{7}{20}$，记事件$M$：乙顾客第二次摸到红球，则$P(MN) = \frac{A_{2}^{2} + C_{2}^{1}C_{3}^{1}C_{3}^{1}}{A_{6}^{3}} = \frac{7}{60}$，$P(M) = \frac{A_{2}^{3}}{A_{6}^{3}} = \frac{20}{120} = \frac{1}{6}$，所以$P(N|M) = \frac{P(MN)}{P(M)} = \frac{\frac{7}{60}}{\frac{1}{6}} = \frac{7}{10}$。
(3)摸到$1$次红球的概率为$\frac{1}{6}$，摸到$1$次白球的概率为$\frac{1}{3}$，摸到$1$次黑球的概率为$\frac{1}{2}$，则$X$的可能取值有$- 3、 - 2、 - 1、0、1、2、3$，$P(X = - 3) = (\frac{1}{2})^{3} = \frac{1}{8}$，$P(X = - 2) = C_{3}^{1} × \frac{1}{3} × (\frac{1}{2})^{2} = \frac{1}{4}$，$P(X = - 1) = C_{3}^{1} × \frac{1}{6} × (\frac{1}{2})^{2} + C_{3}^{1} × \frac{1}{2} × (\frac{1}{3})^{2} = \frac{7}{24}$，$P(X = 0) = (\frac{1}{3})^{3} + A_{3}^{3} × \frac{1}{6} × \frac{1}{3} × \frac{1}{2} = \frac{11}{54}$，$P(X = 1) = C_{3}^{1} × \frac{1}{6} × (\frac{1}{3})^{2} + C_{3}^{1} × \frac{1}{2} × (\frac{1}{6})^{2} = \frac{7}{72}$，$P(X = 2) = C_{3}^{2} × (\frac{1}{6})^{2} × \frac{1}{3} = \frac{1}{36}$，$P(X = 3) = (\frac{1}{6})^{3} = \frac{1}{216}$，随机变量$X$的分布列如下表所示：
|$X$|$- 3$|$- 2$|$- 1$|$0$|$1$|$2$|$3$|
|--|--|--|--|--|--|--|--|
|$P$|$\frac{1}{8}$|$\frac{1}{4}$|$\frac{7}{24}$|$\frac{11}{54}$|$\frac{7}{72}$|$\frac{1}{36}$|$\frac{1}{216}$|
故$E(X) = - 3 × \frac{1}{8} - 2 × \frac{1}{4} - 1 × \frac{7}{24} + 0 × \frac{11}{54} + 1 × \frac{7}{72} + 2 × \frac{1}{36} + 3 × \frac{1}{216} = - 1$。

17. (15分)已知抛物线$C:y^{2}=4x$.
(1)倾斜角为$\frac{\pi}{3}$的直线$l$过$C$的焦点,且与$C$交于$M,N$两点,求$|MN|$;
(2)设$P$是$C$上一点,$A,B$是$C$的准线上两个不同的点,且圆$x^{2}+y^{2}=1$是$\triangle PAB$的内切圆.
①若$|AB|=2\sqrt{5}$,求点$P$的横坐标;
②求$\triangle PAB$面积的最小值.

答案：
17.解：
(1)抛物线C的焦点坐标为$(1,0)$，直线的方程为$y = \sqrt{3}(x - 1)$。设点$M、N$的横坐标为$x_{M}、x_{N}$。由$\begin{cases} y = \sqrt{3}(x - 1) \\ y^{2} = 4x \end{cases}$消$y$得$3x^{2} - 10x + 3 = 0$，于是$x_{M} + x_{N} = \frac{10}{3}$，故$|MN| = x_{M} + x_{N} + 2 = \frac{16}{3}$。
(2)①设$P(x_{0},y_{0})$，于是有$y_{0}^{2} = 4x_{0}$，抛物线C的准线方程为$x = - 1$，设$A( - 1,y_{1})、B( - 1,y_{2})$，过$P$的直线$PA,PB$的方程可设为$y - y_{0} = k(x - x_{0})$，由题意，两直线均与圆$x^{2} + y^{2} = 1$相切，故$\frac{| - kx_{0} + y_{0}|}{\sqrt{k^{2} + 1}} = 1$，整理得$k^{2}(x_{0}^{2} - 1) - 2kx_{0}y_{0} + y_{0}^{2} - 1 = 0$，设直线$PA、PB$的斜率为$k_{1}、k_{2}$，于是$|AB| = |y_{1} - y_{2}| = |(x_{0} + 1)(k_{1} - k_{2})| = |x_{0} + 1|\frac{2\sqrt{x_{0}^{2} + y_{0}^{2} - 1}}{|x_{0}^{2} - 1|} = 2\sqrt{5}$，将$y_{0}^{2} = 4x_{0}$代入上式，化简得$2x_{0}^{2} - 7x_{0} + 3 = 0$，解得$x_{0} = 3$或$x_{0} = \frac{1}{2}$(舍)，故点$P$的横坐标为$3$。

②由①，$|AB| = |y_{1} - y_{2}| = |x_{0} + 1|\frac{2\sqrt{x_{0}^{2} + y_{0}^{2} - 1}}{|x_{0}^{2} - 1|}$，点$P$到$AB$的距离$d = |x_{0} + 1|$，故$\triangle PAB$的面积$S = \frac{1}{2}|AB|d = \frac{|x_{0} + 1|\sqrt{x_{0}^{2} + y_{0}^{2} - 1}}{|x_{0} - 1|}$，不妨令$t = x_{0} - 1 ＞ 0$，于是$S = \sqrt{(t + \frac{4}{t} + 4)(t + \frac{4}{t} + 6)} = \sqrt{(t + \frac{4}{t} + 5)^{2} - 1}$，当且仅当$t = 2$，即$x_{0} = 3$时，$\triangle PAB$的面积取到最小值，最小值为$4\sqrt{5}$。