返回特征解读

2018 年高考数学（北京卷文科）

exams_raw/普通高考/2018/2018北京文.pdf · HS-MATH-1024-v2.1-solution-aware

20 题0 个小问/题组
1

一、选择题 · 组合数学

已知集合 A={x∣∣x∣<2}\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=11}{\mid }\htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{|}\htmlData{tutor-start=14,tutor-end=15}{<}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=18}{\}}，B={−2,0,1,2}\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=17}{\}}，则 A∩B=\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{=} ( ) (A) {0,1}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=8}{\}} (B) {−1,0,1}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=12}{\}} (C) {−2,0,1,2}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=15}{\}} (D) {−1,0,1,2}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=15}{\}}

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

2

一、选择题 · 数学竞赛/待细分

在复平面内，复数 11−i\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{-}\mathrm{\htmlData{tutor-start=19,tutor-end=20}{i}}} 的共轭复数对应的点位于 ( ) (A) 第一象限 (B) 第二象限 (C) 第三象限 (D) 第四象限

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

3

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 s\htmlData{tutor-start=0,tutor-end=1}{s} 值为 ( ) (A) 12\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}} (B) 56\frac{\htmlData{tutor-start=6,tutor-end=7}{5}}{\htmlData{tutor-start=9,tutor-end=10}{6}} (C) 76\frac{\htmlData{tutor-start=6,tutor-end=7}{7}}{\htmlData{tutor-start=9,tutor-end=10}{6}} (D) 712\frac{\htmlData{tutor-start=6,tutor-end=7}{7}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{2}}

原卷图示 1
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：北京卷文科第 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。

4

一、选择题 · 代数

设 a,b,c,d\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{c}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{d} 是非零实数，则“ad=bc\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{d}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{c}”是“a,b,c,d\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{c}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{d} 成等比数列”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：北京卷文科第 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。

5

一、选择题 · 数学竞赛/待细分

“十二平均律”是通用的音律体系，明代朱载堉最早用数学方法计算出半音比例，为这个理论的发展做出了重要贡献。十二平均律将一个纯八度音程分成十二份，依次得到十三个单音，从第二个单音起，每一个单音的频率与它的前一个单音的频率的比都等于 212\sqrt[\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{2}]{\htmlData{tutor-start=10,tutor-end=11}{2}}。若第一个单音的频率为 f\htmlData{tutor-start=0,tutor-end=1}{f}，则第八个单音的频率为 ( ) (A) 23f\sqrt[\htmlData{tutor-start=6,tutor-end=7}{3}]{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{f} (B) 223f\sqrt[\htmlData{tutor-start=6,tutor-end=7}{3}]{\htmlData{tutor-start=9,tutor-end=10}{2}^{\htmlData{tutor-start=12,tutor-end=13}{2}}} \htmlData{tutor-start=16,tutor-end=17}{f} (C) 2512f\sqrt[\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{2}]{\htmlData{tutor-start=10,tutor-end=11}{2}^{\htmlData{tutor-start=13,tutor-end=14}{5}}} \htmlData{tutor-start=17,tutor-end=18}{f} (D) 2712f\sqrt[\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{2}]{\htmlData{tutor-start=10,tutor-end=11}{2}^{\htmlData{tutor-start=13,tutor-end=14}{7}}} \htmlData{tutor-start=17,tutor-end=18}{f}

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

6

一、选择题 · 平面几何

某四棱锥的三视图如图所示，在此四棱锥的侧面中，直角三角形的个数为 ( ) (A) 1 (B) 2 (C) 3 (D) 4

原卷图示 1
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：北京卷文科第 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。

7

一、选择题 · 平面几何

在平面直角坐标系中，AB^,CD^,EF^,GH^\widehat{\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{B}}\htmlData{tutor-start=12,tutor-end=13}{,} \widehat{\htmlData{tutor-start=23,tutor-end=24}{C}\htmlData{tutor-start=24,tutor-end=25}{D}}\htmlData{tutor-start=26,tutor-end=27}{,} \widehat{\htmlData{tutor-start=37,tutor-end=38}{E}\htmlData{tutor-start=38,tutor-end=39}{F}}\htmlData{tutor-start=40,tutor-end=41}{,} \widehat{\htmlData{tutor-start=51,tutor-end=52}{G}\htmlData{tutor-start=52,tutor-end=53}{H}} 是圆 x2+y2=1\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1} 上的四段弧（如图），点 P\htmlData{tutor-start=0,tutor-end=1}{P} 在其中一段上，角 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 以 Ox\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{x} 为始边，OP\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{P} 为终边。若 tan⁡α<cos⁡α<sin⁡α\tan \htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{<}\cos \htmlData{tutor-start=17,tutor-end=23}{\alpha}\htmlData{tutor-start=23,tutor-end=24}{<}\sin \htmlData{tutor-start=29,tutor-end=35}{\alpha}，则 P\htmlData{tutor-start=0,tutor-end=1}{P} 所在的圆弧是 ( ) (A) AB^\widehat{\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{B}} (B) CD^\widehat{\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{D}} (C) EF^\widehat{\htmlData{tutor-start=9,tutor-end=10}{E}\htmlData{tutor-start=10,tutor-end=11}{F}} (D) GH^\widehat{\htmlData{tutor-start=9,tutor-end=10}{G}\htmlData{tutor-start=10,tutor-end=11}{H}}

原卷图示 1
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：北京卷文科第 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。

8

一、选择题 · 组合数学

设集合 A={(x,y)∣x−y⩾1,ax+y>4,x−ay⩽2}\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=16}{\mid }\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{y} \htmlData{tutor-start=20,tutor-end=30}{\geqslant }\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{,} \htmlData{tutor-start=33,tutor-end=34}{a} \htmlData{tutor-start=35,tutor-end=36}{x}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{y}\htmlData{tutor-start=38,tutor-end=39}{>}\htmlData{tutor-start=39,tutor-end=40}{4}\htmlData{tutor-start=40,tutor-end=41}{,} \htmlData{tutor-start=42,tutor-end=43}{x}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{a} \htmlData{tutor-start=46,tutor-end=47}{y} \htmlData{tutor-start=48,tutor-end=58}{\leqslant }\htmlData{tutor-start=58,tutor-end=59}{2}\htmlData{tutor-start=59,tutor-end=61}{\}}，则 ( ) (A) 对任意实数 a,(2,1)∈A\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=12}{\in }\htmlData{tutor-start=12,tutor-end=13}{A} (B) 对任意实数 a,(2,1)∉A\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)} \notin \htmlData{tutor-start=15,tutor-end=16}{A} (C) 当且仅当 a<0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0} 时，(2,1)∉A\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=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)} \notin \htmlData{tutor-start=13,tutor-end=14}{A} (D) 当且仅当 a⩽32\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\frac{\htmlData{tutor-start=18,tutor-end=19}{3}}{\htmlData{tutor-start=21,tutor-end=22}{2}} 时，(2,1)∉A\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=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)} \notin \htmlData{tutor-start=13,tutor-end=14}{A}

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

9

二、填空题 · 数学竞赛/待细分

设向量 a=(1,0),b=(−1,m)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{=}\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}{0}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{,} \boldsymbol{\htmlData{tutor-start=34,tutor-end=35}{b}}\htmlData{tutor-start=36,tutor-end=37}{=}\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{,} \htmlData{tutor-start=42,tutor-end=43}{m}\htmlData{tutor-start=43,tutor-end=44}{)}。若 a⊥(ma−b)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=20}{\perp}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{m} \boldsymbol{\htmlData{tutor-start=35,tutor-end=36}{a}}\htmlData{tutor-start=37,tutor-end=38}{-}\boldsymbol{\htmlData{tutor-start=50,tutor-end=51}{b}}\htmlData{tutor-start=52,tutor-end=53}{)}，则 m=\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{=} \_\_\_\_\_\_。

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

10

二、填空题 · 平面几何

已知直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 过点 (1,0)\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}{)} 且垂直于 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴。若 l\htmlData{tutor-start=0,tutor-end=1}{l} 被抛物线 y2=4ax\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4} \htmlData{tutor-start=8,tutor-end=9}{a} \htmlData{tutor-start=10,tutor-end=11}{x} 截得的线段长为 4，则抛物线的焦点坐标为 \_\_\_\_\_\_。

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

11

二、填空题 · 数学竞赛/待细分

能说明“若 a>b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{b}，则 1a<1b\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}}\htmlData{tutor-start=11,tutor-end=12}{<}\frac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{b}}”为假命题的一组 a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b} 的值依次为 \_\_\_\_\_\_。

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

12

二、填空题 · 数学竞赛/待细分

若双曲线 x2a2−y24=1(a>0)\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}\htmlData{tutor-start=19,tutor-end=20}{-}\frac{\htmlData{tutor-start=26,tutor-end=27}{y}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}{\htmlData{tutor-start=33,tutor-end=34}{4}}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{a}\htmlData{tutor-start=39,tutor-end=40}{>}\htmlData{tutor-start=40,tutor-end=41}{0}\htmlData{tutor-start=41,tutor-end=42}{)} 的离心率为 52\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{5}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}，则 a=\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=} \_\_\_\_\_\_。

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

13

二、填空题 · 数学竞赛/待细分

若 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y} 满足 x+1⩽y⩽2x\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=4,tutor-end=14}{\leqslant }\htmlData{tutor-start=14,tutor-end=15}{y} \htmlData{tutor-start=16,tutor-end=26}{\leqslant }\htmlData{tutor-start=26,tutor-end=27}{2} \htmlData{tutor-start=28,tutor-end=29}{x}，则 2y−x\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{x} 的最小值是 \_\_\_\_\_\_。

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

14

二、填空题 · 平面几何

若 △ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A} \htmlData{tutor-start=12,tutor-end=13}{B} \htmlData{tutor-start=14,tutor-end=15}{C} 的面积为 34(a2+c2−b2)\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{4}}\left(\htmlData{tutor-start=24,tutor-end=25}{a}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{c}^{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=36}{-}\htmlData{tutor-start=36,tutor-end=37}{b}^{\htmlData{tutor-start=39,tutor-end=40}{2}}\right)，且 ∠C\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{C} 为钝角，则 ∠B=\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{=} \_\_\_\_\_\_；ca\frac{\htmlData{tutor-start=6,tutor-end=7}{c}}{\htmlData{tutor-start=9,tutor-end=10}{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：北京卷文科第 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。

15

三、解答题 · 代数

设 {an}\left\{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\right\} 是等差数列，且 a1=ln⁡2,a2+a3=5ln⁡2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\ln \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{a}_{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{3}}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{5} \ln \htmlData{tutor-start=31,tutor-end=32}{2}。 (1) 求 {an}\left\{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\right\} 的通项公式； (2) 求 ea1+ea2+⋯+ean\mathrm{e}^{a_{1}}+\mathrm{e}^{a_{2}}+\cdots+\mathrm{e}^{a_{n}}。

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

16

三、解答题 · 代数

已知函数 f(x)=sin⁡2x+3sin⁡xcos⁡x\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=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{+}\sqrt{\htmlData{tutor-start=23,tutor-end=24}{3}} \sin \htmlData{tutor-start=31,tutor-end=32}{x} \cos \htmlData{tutor-start=38,tutor-end=39}{x}。 (1) 求 f(x)\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}{)} 的最小正周期； (2) 若 f(x)\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 [−π3,m]\left[\htmlData{tutor-start=6,tutor-end=7}{-}\frac{\htmlData{tutor-start=13,tutor-end=16}{\pi}}{\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{m}\right] 上的最大值为 32\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{2}}，求 m\htmlData{tutor-start=0,tutor-end=1}{m} 的最小值。

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 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。

17

三、解答题 · 数学竞赛/待细分

电影公司随机收集了电影的有关数据，经分类整理得到下表： | 电影类型 | 第一类 | 第二类 | 第三类 | 第四类 | 第五类 | 第六类 | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | | 电影部数 | 140 | 50 | 300 | 200 | 800 | 510 | | 好评率 | 0.4 | 0.2 | 0.15 | 0.25 | 0.2 | 0.1 | 好评率是指：一类电影中获得好评的部数与该类电影的部数的比值。 (1) 从电影公司收集的电影中随机选取 1 部，求这部电影是获得好评的第四类电影的概率； (2) 随机选取 1 部电影，估计这部电影没有获得好评的概率； (3) 电影公司为增加投资回报，拟改变投资策略，这将导致不同类型电影的好评率发生变化。假设表格中只有两类电影的好评率数据发生变化，那么哪类电影的好评率增加 0.1，哪类电影的好评率减少 0.1，使得获得好评的电影总部数与样本中的电影总部数的比值达到最大？（只需写出结论）

原卷图示 1
原卷图示 1原卷第 2 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：北京卷文科第 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。

18

三、解答题 · 数学竞赛/待细分

如图，在四棱锥 P−ABCD\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，底面 ABCD\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} 为矩形，平面 PAD⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{D} \htmlData{tutor-start=4,tutor-end=9}{\perp} 平面 ABCD\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}，PA⊥PD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{P}\htmlData{tutor-start=10,tutor-end=11}{D}，PA=PD\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}{P}\htmlData{tutor-start=4,tutor-end=5}{D}，E,F\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{F} 分别为 AD,PB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{P}\htmlData{tutor-start=5,tutor-end=6}{B} 的中点。 (1) 求证：PE⊥BC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{C}； (2) 求证：平面 PAB⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B} \htmlData{tutor-start=4,tutor-end=9}{\perp} 平面 PCD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}； (3) 求证：EF//\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F} \htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{/} 平面 PCD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}；

原卷图示 1
原卷图示 1原卷第 2 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：北京卷文科第 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。

19

三、解答题 · 代数

设函数 f(x)=[ax2−(3a+1)x+3a+2]ex\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}{a}\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{-} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{a}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{)}\htmlData{tutor-start=23,tutor-end=24}{x} \htmlData{tutor-start=25,tutor-end=26}{+} \htmlData{tutor-start=27,tutor-end=28}{3}\htmlData{tutor-start=28,tutor-end=29}{a} \htmlData{tutor-start=30,tutor-end=31}{+} \htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{]}\mathrm{\htmlData{tutor-start=42,tutor-end=43}{e}}^{\htmlData{tutor-start=46,tutor-end=47}{x}}。 (1) 若曲线 y=f(x)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)} 在点 (2,f(2))\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}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{)} 处的切线斜率为 0，求 a\htmlData{tutor-start=0,tutor-end=1}{a}； (2) 若 f(x)\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 x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 处取得极小值，求 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围。

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 19 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。

20

三、解答题 · 平面几何

已知椭圆 M:x2a2+y2b2=1 (a>b>0)\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{:} \frac{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}} \htmlData{tutor-start=23,tutor-end=24}{+} \frac{\htmlData{tutor-start=31,tutor-end=32}{y}^{\htmlData{tutor-start=34,tutor-end=35}{2}}}{\htmlData{tutor-start=38,tutor-end=39}{b}^{\htmlData{tutor-start=41,tutor-end=42}{2}}} \htmlData{tutor-start=45,tutor-end=46}{=} \htmlData{tutor-start=47,tutor-end=48}{1} \htmlData{tutor-start=49,tutor-end=51}{\ }\htmlData{tutor-start=51,tutor-end=52}{(}\htmlData{tutor-start=52,tutor-end=53}{a}\htmlData{tutor-start=53,tutor-end=54}{>}\htmlData{tutor-start=54,tutor-end=55}{b}\htmlData{tutor-start=55,tutor-end=56}{>}\htmlData{tutor-start=56,tutor-end=57}{0}\htmlData{tutor-start=57,tutor-end=58}{)} 的离心率为 63\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{6}}}{\htmlData{tutor-start=16,tutor-end=17}{3}}，焦距为 22\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}，斜率为 k\htmlData{tutor-start=0,tutor-end=1}{k} 的直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与椭圆 M\htmlData{tutor-start=0,tutor-end=1}{M} 有两个不同的交点 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}。 (1) 求椭圆 M\htmlData{tutor-start=0,tutor-end=1}{M} 的方程； (2) 若 k=1\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，求 ∣AB∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|} 的最大值； (3) 设 P(−2,0)\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}，直线 PA\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A} 与椭圆 M\htmlData{tutor-start=0,tutor-end=1}{M} 的另一个交点为 C\htmlData{tutor-start=0,tutor-end=1}{C}，直线 PB\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B} 与椭圆 M\htmlData{tutor-start=0,tutor-end=1}{M} 的另一个交点为 D\htmlData{tutor-start=0,tutor-end=1}{D}。若 C,D\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{D} 和点 Q(−74,14)\htmlData{tutor-start=0,tutor-end=1}{Q}\left(\htmlData{tutor-start=7,tutor-end=8}{-}\frac{\htmlData{tutor-start=14,tutor-end=15}{7}}{\htmlData{tutor-start=17,tutor-end=18}{4}}\htmlData{tutor-start=19,tutor-end=20}{,} \frac{\htmlData{tutor-start=27,tutor-end=28}{1}}{\htmlData{tutor-start=30,tutor-end=31}{4}}\right) 共线，求 k\htmlData{tutor-start=0,tutor-end=1}{k}。

题解状态：标准答案与规范题解待补充

题目标签：北京卷文科第 20 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。