2019 年高考数学（北京卷理科）

exams_raw/普通高考/2019/2019北京理.pdf · HS-MATH-1024-v2.1-solution-aware

20 题0 个小问/题组

一、选择题 · 数学竞赛/待细分

已知复数 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{+}\mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=6}{\cdot}\bar{\htmlData{tutor-start=11,tutor-end=12}{z}}\htmlData{tutor-start=13,tutor-end=14}{=}$ （） (A) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}$ (B) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{5}}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{3}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{5}$

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 1 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

一、选择题 · 数学竞赛/待细分

执行如图所示的程序框图，输出的 $\htmlData{tutor-start=0,tutor-end=1}{s}$ 值为（） (A) $\htmlData{tutor-start=0,tutor-end=1}{1}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{2}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{3}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{4}$

**原卷图示 1**原卷第 1 页 · qwen3.7-plus_layout_detection · 需复核

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 2 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

一、选择题 · 平面几何

已知直线 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的参数方程为 $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{t}\htmlData{tutor-start=20,tutor-end=21}{,} \\ \htmlData{tutor-start=25,tutor-end=26}{y}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{4}\htmlData{tutor-start=30,tutor-end=31}{t}\htmlData{tutor-start=31,tutor-end=32}{,} \end{cases}$ （ $\htmlData{tutor-start=0,tutor-end=1}{t}$ 为参数），则点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}$ 到直线 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的距离是（） (A) $\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{5}}$ (B) $\dfrac{\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{5}}$ (C) $\dfrac{\htmlData{tutor-start=7,tutor-end=8}{4}}{\htmlData{tutor-start=10,tutor-end=11}{5}}$ (D) $\dfrac{\htmlData{tutor-start=7,tutor-end=8}{6}}{\htmlData{tutor-start=10,tutor-end=11}{5}}$

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 3 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

一、选择题 · 平面几何

已知椭圆 $\dfrac{\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}}{\htmlData{tutor-start=14,tutor-end=15}{a}^{\htmlData{tutor-start=17,tutor-end=18}{2}}}\htmlData{tutor-start=20,tutor-end=21}{+}\dfrac{\htmlData{tutor-start=28,tutor-end=29}{y}^{\htmlData{tutor-start=31,tutor-end=32}{2}}}{\htmlData{tutor-start=35,tutor-end=36}{b}^{\htmlData{tutor-start=38,tutor-end=39}{2}}}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{1}$ （ $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{>}\htmlData{tutor-start=4,tutor-end=5}{0}$ ）的离心率为 $\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}}$ ，则（） (A) $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{b}^{\htmlData{tutor-start=10,tutor-end=11}{2}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{b}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{b}$

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 4 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

一、选择题 · 数学竞赛/待细分

若 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$ 满足 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=13}{\leqslant }\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{y}$ ，且 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=11}{\geqslant }\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{y}$ 的最大值为（） (A) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{7}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{1}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{5}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{7}$

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 5 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

一、选择题 · 数学竞赛/待细分

在天文学中，天体的明暗程度可以用星等或亮度来描述。两颗星的星等与亮度满足 $\htmlData{tutor-start=0,tutor-end=1}{m}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{m}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{=}\dfrac{\htmlData{tutor-start=19,tutor-end=20}{5}}{\htmlData{tutor-start=22,tutor-end=23}{2}}\lg\dfrac{\htmlData{tutor-start=34,tutor-end=35}{E}_{\htmlData{tutor-start=37,tutor-end=38}{1}}}{\htmlData{tutor-start=41,tutor-end=42}{E}_{\htmlData{tutor-start=44,tutor-end=45}{2}}}$ ，其中星等为 $\htmlData{tutor-start=0,tutor-end=1}{m}_{\htmlData{tutor-start=3,tutor-end=4}{k}}$ 的星的亮度为 $\htmlData{tutor-start=0,tutor-end=1}{E}_{\htmlData{tutor-start=3,tutor-end=4}{k}}$ （ $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}$ ）。已知太阳的星等是 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{6}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{7}$ ，天狼星的星等是 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{.}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{5}$ ，则太阳与天狼星的亮度的比值为（） (A) $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}^{\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{1}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{.}\htmlData{tutor-start=3,tutor-end=4}{1}$ (C) $\lg \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{1}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}^{\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{.}\htmlData{tutor-start=8,tutor-end=9}{1}}$

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 6 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

一、选择题 · 数学竞赛/待细分

设点 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{C}$ 不共线，则" $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}}$ 与 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{C}}$ 的夹角为锐角"是" $\htmlData{tutor-start=0,tutor-end=1}{|}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{B}}\htmlData{tutor-start=20,tutor-end=21}{+}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{A}\htmlData{tutor-start=38,tutor-end=39}{C}}\htmlData{tutor-start=40,tutor-end=41}{|}\htmlData{tutor-start=41,tutor-end=42}{>}\htmlData{tutor-start=42,tutor-end=43}{|}\overrightarrow{\htmlData{tutor-start=59,tutor-end=60}{B}\htmlData{tutor-start=60,tutor-end=61}{C}}\htmlData{tutor-start=62,tutor-end=63}{|}$ "的（） (A) 充分而不必要条件 (B) 必要而不充分条件 (C) 充分必要条件 (D) 既不充分也不必要条件

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 7 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

一、选择题 · 数论

数学中有许多形状优美、寓意美好的曲线，曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:}\htmlData{tutor-start=2,tutor-end=3}{x}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{|}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{|}\htmlData{tutor-start=19,tutor-end=20}{y}$ 就是其中之一（如图）。给出下列三个结论： ①曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 恰好经过 $\htmlData{tutor-start=0,tutor-end=1}{6}$ 个整点（即横、纵坐标均为整数的点）； ②曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 上任意一点到原点的距离都不超过 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}$ ； ③曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 所围成的"心形"区域的面积小于 $\htmlData{tutor-start=0,tutor-end=1}{3}$ 。其中，所有正确结论的序号是（） (A) ① (B) ② (C) ①② (D) ①②③

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 8 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

二、填空题 · 代数

函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\sin^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{x}$ 的最小正周期是______。

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 9 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

二、填空题 · 代数

设等差数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的前 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项和为 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ ，若 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{3}$ ， $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{5}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{0}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{5}}\htmlData{tutor-start=5,tutor-end=6}{=}$ ______， $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ 的最小值为______。

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 10 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

二、填空题 · 数学竞赛/待细分

某几何体是由一个正方体去掉一个四棱柱所得，其三视图如图所示。如果网格纸上小正方形的边长为 $\htmlData{tutor-start=0,tutor-end=1}{1}$ ，那么该几何体的体积为______。

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 11 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

二、填空题 · 平面几何

已知 $\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{m}$ 是平面 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 外的两条不同直线。给出下列三个论断： ① $\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=7}{\perp }\htmlData{tutor-start=7,tutor-end=8}{m}$ ； ② $\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=10}{\parallel}\htmlData{tutor-start=10,tutor-end=16}{\alpha}$ ； ③ $\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=6}{\perp}\htmlData{tutor-start=6,tutor-end=12}{\alpha}$ 。以其中的两个论断作为条件，余下的一个论断作为结论，写出一个正确的命题：______。

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 12 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

二、填空题 · 代数

设函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\mathrm{\htmlData{tutor-start=13,tutor-end=14}{e}}^{\htmlData{tutor-start=17,tutor-end=18}{x}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{a}\mathrm{\htmlData{tutor-start=29,tutor-end=30}{e}}^{\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{x}}$ （ $\htmlData{tutor-start=0,tutor-end=1}{a}$ 为常数）。若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 为奇函数，则 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}$ ；若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 是 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{R}}$ 上的增函数，则 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的取值范围是。

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 13 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

二、填空题 · 数学竞赛/待细分

李明自主创业，在网上经营一家水果店，销售的水果中有草莓、京白梨、西瓜、桃，价格依次为 $\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{0}$ 元/盒、 $\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{5}$ 元/盒、 $\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{0}$ 元/盒、 $\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{0}$ 元/盒。为增加销量，李明对这四种水果进行促销：一次购买水果的总价达到 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{0}$ 元，顾客就少付 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 元。每笔订单顾客网上支付成功后，李明会得到支付款的 $\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=4}{\%}$ 。 ①当 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{0}$ 时，顾客一次购买草莓和西瓜各 $\htmlData{tutor-start=0,tutor-end=1}{1}$ 盒，需要支付元； ②在促销活动中，为保证李明每笔订单得到的金额均不低于促销前总价的七折，则 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的最大值为。

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 14 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

三、解答题 · 平面几何

在 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 中， $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$ ， $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{2}$ ， $\cos \htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{-}\dfrac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}}$ 。 (1) 求 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{c}$ 的值； (2) 求 $\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{)}$ 的值。

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 15 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

三、解答题 · 平面几何

如图，在四棱锥 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{D}$ 中， $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=7}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}$ ， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{D}$ ， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=12}{\parallel }\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{C}$ ， $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{D}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{2}$ ， $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{3}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D}$ 的中点，点 $\htmlData{tutor-start=0,tutor-end=1}{F}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}$ 上，且 $\dfrac{\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{F}}{\htmlData{tutor-start=11,tutor-end=12}{P}\htmlData{tutor-start=12,tutor-end=13}{C}}\htmlData{tutor-start=14,tutor-end=15}{=}\dfrac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{3}}$ 。 (1) 求证： $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=7}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{D}$ ； (2) 求二面角 $\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{E}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{P}$ 的余弦值； (3) 设点 $\htmlData{tutor-start=0,tutor-end=1}{G}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B}$ 上，且 $\dfrac{\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{G}}{\htmlData{tutor-start=11,tutor-end=12}{P}\htmlData{tutor-start=12,tutor-end=13}{B}}\htmlData{tutor-start=14,tutor-end=15}{=}\dfrac{\htmlData{tutor-start=22,tutor-end=23}{2}}{\htmlData{tutor-start=25,tutor-end=26}{3}}$ 。判断直线 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{G}$ 是否在平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{F}$ 内，说明理由。

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 16 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

三、解答题 · 数学竞赛/待细分

17. 改革开放以来，人们的支付方式发生了巨大转变. 近年来，移动支付已成为主要支付方式之一. 为了解某校学生上个月 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}$ 两种移动支付方式的使用情况，从全校学生中随机抽取了 100 人，发现样本中 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}$ 两种支付方式都不使用的有 5 人，样本中仅使用 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 和仅使用 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 的学生的支付金额分布情况如下： \begin{tabular}{|c|c|c|c|} \hline \diagbox{支付方式}{支付金额（元）} & $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{]}$ & $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{]}$ & 大于 2000 \\ \hline 仅使用 $\htmlData{tutor-start=0,tutor-end=1}{A}$ & 18 人 & 9 人 & 3 人 \\ \hline 仅使用 $\htmlData{tutor-start=0,tutor-end=1}{B}$ & 10 人 & 14 人 & 1 人 \\ \hline \end{tabular} (1) 从全校学生中随机抽取 1 人，估计该学生上个月 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}$ 两种支付方式都使用的概率； (2) 从样本仅使用 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 和仅使用 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 的学生中各随机抽取 1 人，以 $\htmlData{tutor-start=0,tutor-end=1}{X}$ 表示这 2 人中上个月支付金额大于 1000 元的人数，求 $\htmlData{tutor-start=0,tutor-end=1}{X}$ 的分布列和数学期望； (3) 已知上个月样本学生的支付方式在本月没有变化. 现从样本仅使用 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的学生中，随机抽查 3 人，发现他们本月的支付金额都大于 2000 元. 根据抽查结果，能否认为样本仅使用 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的学生中本月支付金额大于 2000 元的人数有变化？说明理由.

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 17 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

三、解答题 · 平面几何

18. 已知抛物线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{x}^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{p}\htmlData{tutor-start=14,tutor-end=15}{y}$ 经过点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}$ . (1) 求抛物线 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的方程及其准线方程； (2) 设 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 为原点，过抛物线 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的焦点作斜率不为 0 的直线 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 交抛物线 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 于两点 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{N}$ ，直线 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}$ 分别交直线 $\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{O}\htmlData{tutor-start=5,tutor-end=6}{N}$ 于点 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 和点 $\htmlData{tutor-start=0,tutor-end=1}{B}$ . 求证：以 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 为直径的圆经过 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 轴上的两个定点.

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 18 题

解题过程

该题已完成题面、原题号、PDF 来源锚定和题目级特征标注；答案、证明步骤与教材关联尚未录入。

三、解答题 · 代数

19. 已知函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{4}}\htmlData{tutor-start=18,tutor-end=19}{x}^{\htmlData{tutor-start=21,tutor-end=22}{3}} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{x}^{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{+} \htmlData{tutor-start=34,tutor-end=35}{x}$ . (1) 求曲线 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)}$ 的斜率为 1 的切线方程； (2) 当 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{[}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{]}$ 时，求证： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{6} \htmlData{tutor-start=6,tutor-end=16}{\leqslant }\htmlData{tutor-start=16,tutor-end=17}{f}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{)} \htmlData{tutor-start=21,tutor-end=31}{\leqslant }\htmlData{tutor-start=31,tutor-end=32}{x}$ ； (3) 设 $\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{f}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{)} \htmlData{tutor-start=13,tutor-end=14}{-} \htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{)}\htmlData{tutor-start=22,tutor-end=23}{|}$ ( $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=6}{\in }\mathbf{\htmlData{tutor-start=14,tutor-end=15}{R}}$ )，记 $\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 在区间 $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{]}$ 上的最大值为 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{)}$ . 当 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{)}$ 最小时，求 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的值.

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 19 题

解题过程

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三、解答题 · 代数

20. 已知数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 从中选取第 $\htmlData{tutor-start=0,tutor-end=1}{i}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 项、第 $\htmlData{tutor-start=0,tutor-end=1}{i}_{\htmlData{tutor-start=3,tutor-end=4}{2}}$ 项、 $\cdots$ 、第 $\htmlData{tutor-start=0,tutor-end=1}{i}_{\htmlData{tutor-start=3,tutor-end=4}{m}}$ 项 ( $\htmlData{tutor-start=0,tutor-end=1}{i}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{i}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{<} \cdots \htmlData{tutor-start=23,tutor-end=24}{<} \htmlData{tutor-start=25,tutor-end=26}{i}_{\htmlData{tutor-start=28,tutor-end=29}{m}}$ )，若 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}_{\htmlData{tutor-start=6,tutor-end=7}{1}}} \htmlData{tutor-start=10,tutor-end=11}{<} \htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{i}_{\htmlData{tutor-start=18,tutor-end=19}{2}}} \htmlData{tutor-start=22,tutor-end=23}{<} \cdots \htmlData{tutor-start=31,tutor-end=32}{<} \htmlData{tutor-start=33,tutor-end=34}{a}_{\htmlData{tutor-start=36,tutor-end=37}{i}_{\htmlData{tutor-start=39,tutor-end=40}{m}}}$ ，则称新数列 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}_{\htmlData{tutor-start=6,tutor-end=7}{1}}}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{i}_{\htmlData{tutor-start=17,tutor-end=18}{2}}}\htmlData{tutor-start=20,tutor-end=21}{,} \cdots\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{i}_{\htmlData{tutor-start=36,tutor-end=37}{m}}}$ 为 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的长度为 $\htmlData{tutor-start=0,tutor-end=1}{m}$ 的递增子列. 规定：数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的任意一项都是 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的长度为 1 的递增子列. (1) 写出数列 1, 8, 3, 7, 5, 6, 9 的一个长度为 4 的递增子列； (2) 已知数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的长度为 $\htmlData{tutor-start=0,tutor-end=1}{p}$ 的递增子列的末项的最小值为 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{m}_{\htmlData{tutor-start=6,tutor-end=7}{0}}}$ ，长度为 $\htmlData{tutor-start=0,tutor-end=1}{q}$ 的递增子列的末项的最小值为 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}_{\htmlData{tutor-start=6,tutor-end=7}{0}}}$ . 若 $\htmlData{tutor-start=0,tutor-end=1}{p} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{q}$ ，求证： $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{m}_{\htmlData{tutor-start=6,tutor-end=7}{0}}} \htmlData{tutor-start=10,tutor-end=11}{<} \htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{n}_{\htmlData{tutor-start=18,tutor-end=19}{0}}}$ ； (3) 设无穷数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的各项均为正整数，且任意两项均不相等. 若 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的长度为 $\htmlData{tutor-start=0,tutor-end=1}{s}$ 的递增子列末项的最小值为 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{s} \htmlData{tutor-start=3,tutor-end=4}{-} \htmlData{tutor-start=5,tutor-end=6}{1}$ ，且长度为 $\htmlData{tutor-start=0,tutor-end=1}{s}$ 末项为 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{s} \htmlData{tutor-start=3,tutor-end=4}{-} \htmlData{tutor-start=5,tutor-end=6}{1}$ 的递增子列恰有 $\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{s}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}}$ 个 ( $\htmlData{tutor-start=0,tutor-end=1}{s} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,} \cdots$ )，求数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的通项公式.

题解状态：标准答案与规范题解待补充

题目标签：北京卷理科第 20 题

解题过程

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