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2016 年高考数学（全国卷 3文科）

exams_raw/普通高考/2016/2016全国3文(云南,广西,贵州).pdf · HS-MATH-1024-v2.1-solution-aware

24 题0 个小问/题组
1

一、选择题 · 组合数学

设集合 A={0,2,4,6,8,10}\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}{0}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{6}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{8}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=23}{\}}，B={4,8}\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}{4}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{8}\htmlData{tutor-start=8,tutor-end=10}{\}}，则 ∁AB=\htmlData{tutor-start=0,tutor-end=11}{\complement}_{\htmlData{tutor-start=13,tutor-end=14}{A}} \htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{=} （ ） (A) {4,8}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{8}\htmlData{tutor-start=6,tutor-end=8}{\}} (B) {0,2,6}\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}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=11}{\}} (C) {0,2,6,10}\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}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=15}{\}} (D) {0,2,4,6,8,10}\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}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{6}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{8}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=21}{\}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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

若 z=4+3i\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{3}\text{\htmlData{tutor-start=11,tutor-end=12}{i}}，则 zˉ∣z∣=\frac{\bar{\htmlData{tutor-start=11,tutor-end=12}{z}}}{\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{z}\htmlData{tutor-start=17,tutor-end=18}{|}}\htmlData{tutor-start=19,tutor-end=20}{=} （ ） (A) 1\htmlData{tutor-start=0,tutor-end=1}{1} (B) −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} (C) 45+35i\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{+}\frac{\htmlData{tutor-start=18,tutor-end=19}{3}}{\htmlData{tutor-start=21,tutor-end=22}{5}}\text{\htmlData{tutor-start=29,tutor-end=30}{i}} (D) 45−35i\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{-}\frac{\htmlData{tutor-start=18,tutor-end=19}{3}}{\htmlData{tutor-start=21,tutor-end=22}{5}}\text{\htmlData{tutor-start=29,tutor-end=30}{i}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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 · 平面几何

已知向量 BA→=(12,32)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{A}}\htmlData{tutor-start=19,tutor-end=20}{=}\left(\frac{\htmlData{tutor-start=32,tutor-end=33}{1}}{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{,} \frac{\sqrt{\htmlData{tutor-start=51,tutor-end=52}{3}}}{\htmlData{tutor-start=55,tutor-end=56}{2}}\right)，BC→=(32,12)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{C}}\htmlData{tutor-start=19,tutor-end=20}{=}\left(\frac{\sqrt{\htmlData{tutor-start=38,tutor-end=39}{3}}}{\htmlData{tutor-start=42,tutor-end=43}{2}}\htmlData{tutor-start=44,tutor-end=45}{,} \frac{\htmlData{tutor-start=52,tutor-end=53}{1}}{\htmlData{tutor-start=55,tutor-end=56}{2}}\right)，则 ∠ABC=\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{=} （ ） (A) 30∘30^\circ (B) 45∘45^\circ (C) 60∘60^\circ (D) 120∘120^\circ

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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 · 数学竞赛/待细分

某旅游城市为向游客介绍本地的气温情况，绘制了一年中月平均最高气温和平均最低气温的雷达图. 图中 A\htmlData{tutor-start=0,tutor-end=1}{A} 点表示十月的平均最高气温约为 15∘C15^\circ\text{C}，B\htmlData{tutor-start=0,tutor-end=1}{B} 点表示四月的平均最低气温约为 5∘C5^\circ\text{C}. 下面叙述不正确的是 （ ） (A) 各月的平均最低气温都在 0∘C0^\circ\text{C} 以上 (B) 七月的平均温差比一月的平均温差大 (C) 三月和十一月的平均最高气温基本相同 (D) 平均气温高于 20∘C20^\circ\text{C} 的月份有 5 个

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：全国卷 3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

小敏打开计算机时，忘记了开机密码的前两位，只记得第一位是 M,I,N\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{I}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{N} 中的一个字母，第二位是 1,2,3,4,5\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{5} 中的一个数字，则小敏输入一次密码能够成功开机的概率是 （ ） (A) 815\frac{\htmlData{tutor-start=6,tutor-end=7}{8}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{5}} (B) 18\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{8}} (C) 115\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{5}} (D) 130\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{0}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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 · 数学竞赛/待细分

若 tan⁡θ=−13\tan\htmlData{tutor-start=4,tutor-end=11}{\theta }\htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{-}\frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{3}}，则 cos⁡2θ=\cos \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=13}{\theta }\htmlData{tutor-start=13,tutor-end=14}{=} （ ） (A) −45\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{4}}{\htmlData{tutor-start=10,tutor-end=11}{5}} (B) −15\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{5}} (C) 15\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{5}} (D) 45\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{5}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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 · 数学竞赛/待细分

已知 a=243\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}^{\frac{\htmlData{tutor-start=11,tutor-end=12}{4}}{\htmlData{tutor-start=14,tutor-end=15}{3}}}，b=323\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}^{\frac{\htmlData{tutor-start=11,tutor-end=12}{2}}{\htmlData{tutor-start=14,tutor-end=15}{3}}}，c=2513\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{5}^{\frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{3}}}，则 （ ） (A) b<a<c\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{c} (B) a<b<c\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{c} (C) b<c<a\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{c} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{a} (D) c<a<b\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{b}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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 · 数学竞赛/待细分

执行下图的程序框图，如果输入的 a=4,b=6\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{6}，那么输出的 n=\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=} （ ） (A) 3 (B) 4 (C) 5 (D) 6

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：全国卷 3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

一、选择题 · 平面几何

在 △ABC\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，B=π4\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=11}{\pi}}{\htmlData{tutor-start=13,tutor-end=14}{4}}，BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} 边上的高等于 13BC\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}，则 sin⁡A=\sin \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{=} （ ） (A) 310\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0}} (B) 1010\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{0}}}{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{0}} (C) 55\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{5}}}{\htmlData{tutor-start=16,tutor-end=17}{5}} (D) 31010\frac{\htmlData{tutor-start=6,tutor-end=7}{3}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{0}}}{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{0}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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

一、选择题 · 数学竞赛/待细分

如图，网格纸上小正方形的边长为 1，粗实线画出的是某多面体的三视图，则该多面体的表面积为 （ ） (A) 18+365\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{6}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{5}} (B) 54+185\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{8}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{5}} (C) 90 (D) 81

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：全国卷 3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

一、选择题 · 数学竞赛/待细分

在封闭的直三棱柱 ABC−A1B1C1\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}{-}\htmlData{tutor-start=4,tutor-end=5}{A}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{B}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{C}_{\htmlData{tutor-start=17,tutor-end=18}{1}} 内有一个体积为 V\htmlData{tutor-start=0,tutor-end=1}{V} 的球，若 AB⊥BC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{C}，AB=6,BC=8,AA1=3\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{6}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{B}\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{8}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{A}_{\htmlData{tutor-start=16,tutor-end=17}{1}}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{3}，则 V\htmlData{tutor-start=0,tutor-end=1}{V} 的最大值是 （ ） (A) 4π\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=4}{\pi} (B) 9π2\frac{\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{2}} (C) 6π\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=4}{\pi} (D) 32π3\frac{\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=11}{\pi}}{\htmlData{tutor-start=13,tutor-end=14}{3}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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

一、选择题 · 平面几何

已知 O\htmlData{tutor-start=0,tutor-end=1}{O} 为坐标原点，F\htmlData{tutor-start=0,tutor-end=1}{F} 是椭圆 C:x2a2+y2b2=1 (a>b>0)\htmlData{tutor-start=0,tutor-end=1}{C}\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=54,tutor-end=55}{>} \htmlData{tutor-start=56,tutor-end=57}{b} \htmlData{tutor-start=58,tutor-end=59}{>} \htmlData{tutor-start=60,tutor-end=61}{0}\htmlData{tutor-start=61,tutor-end=62}{)} 的左焦点，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} 分别为 C\htmlData{tutor-start=0,tutor-end=1}{C} 的左、右顶点，P\htmlData{tutor-start=0,tutor-end=1}{P} 为 C\htmlData{tutor-start=0,tutor-end=1}{C} 上一点，且 PF⊥x\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{F} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{x} 轴. 过点 A\htmlData{tutor-start=0,tutor-end=1}{A} 的直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与线段 PF\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{F} 交于点 M\htmlData{tutor-start=0,tutor-end=1}{M}，与 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴交于点 E\htmlData{tutor-start=0,tutor-end=1}{E}. 若直线 BM\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{M} 经过 OE\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{E} 的中点，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的离心率为 （ ） (A) 13\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}} (B) 12\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}} (C) 23\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{3}} (D) 34\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{4}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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} 满足约束条件 {2x−y+1⩾0,x−2y−1⩽0,x⩽1,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{y}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{1} \htmlData{tutor-start=21,tutor-end=31}{\geqslant }\htmlData{tutor-start=31,tutor-end=32}{0}\htmlData{tutor-start=32,tutor-end=33}{,} \\ \htmlData{tutor-start=37,tutor-end=38}{x}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=41}{y}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{1} \htmlData{tutor-start=44,tutor-end=54}{\leqslant }\htmlData{tutor-start=54,tutor-end=55}{0}\htmlData{tutor-start=55,tutor-end=56}{,} \\ \htmlData{tutor-start=60,tutor-end=61}{x} \htmlData{tutor-start=62,tutor-end=72}{\leqslant }\htmlData{tutor-start=72,tutor-end=73}{1}\htmlData{tutor-start=73,tutor-end=74}{,} \end{cases} 则 z=2x+3y−5\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}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{y}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{5} 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：全国卷 3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 · 代数

函数 y=sin⁡x−3cos⁡x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sin \htmlData{tutor-start=7,tutor-end=8}{x} \htmlData{tutor-start=9,tutor-end=10}{-} \sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}\cos \htmlData{tutor-start=24,tutor-end=25}{x} 的图象可由函数 y=2sin⁡x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\sin \htmlData{tutor-start=8,tutor-end=9}{x} 的图象至少向右平移______个单位长度得到.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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

二、填空题 · 平面几何

已知直线 l:x−3y+6=0\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{3}}\htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{6}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{0} 与圆 x2+y2=12\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}\htmlData{tutor-start=13,tutor-end=14}{2} 交于 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} 两点，过 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} 分别作 l\htmlData{tutor-start=0,tutor-end=1}{l} 的垂线与 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴交于 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} 两点，则 ∣CD∣=\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}______.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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

二、填空题 · 代数

已知 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}{)} 为偶函数，当 x⩽0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{0} 时，f(x)=e−x−1−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}{=}\htmlData{tutor-start=5,tutor-end=6}{e}^{\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{x}，则曲线 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 (1,2)\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=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)} 处的切线方程是______.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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 · 代数

已知各项都为正数的数列 {an}\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}{\}} 满足 a1=1,an2−(2an+1−1)an−2an+1=0\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{n}}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{n}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{1}}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{a}_{\htmlData{tutor-start=34,tutor-end=35}{n}}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{2}\htmlData{tutor-start=38,tutor-end=39}{a}_{\htmlData{tutor-start=41,tutor-end=42}{n}\htmlData{tutor-start=42,tutor-end=43}{+}\htmlData{tutor-start=43,tutor-end=44}{1}}\htmlData{tutor-start=45,tutor-end=46}{=}\htmlData{tutor-start=46,tutor-end=47}{0}. (1) 求 a2,a3\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=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{3}}; (2) 求 {an}\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：全国卷 3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

下图是我国 2008 年至 2014 年生活垃圾无害化处理量（单位：亿吨）的折线图. 注：年份代码 1-7 分别对应年份 2008-2014 (1) 由折线图看出，可用线性回归模型拟合 y\htmlData{tutor-start=0,tutor-end=1}{y} 与 t\htmlData{tutor-start=0,tutor-end=1}{t} 的关系，请用相关系数加以说明； (2) 建立 y\htmlData{tutor-start=0,tutor-end=1}{y} 关于 t\htmlData{tutor-start=0,tutor-end=1}{t} 的回归方程（系数精确到 0.01），预测 2016 年我国生活垃圾无害化处理量。 参考数据：∑i=17yi=9.32\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{7}} \htmlData{tutor-start=15,tutor-end=16}{y}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{9}\htmlData{tutor-start=24,tutor-end=25}{.}\htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{2}，∑i=17tiyi=40.17\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{7}} \htmlData{tutor-start=15,tutor-end=16}{t}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{y}_{\htmlData{tutor-start=24,tutor-end=25}{i}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{4}\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{.}\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{7}，∑i=17(yi−yˉ)2=0.55\sqrt{\sum_{\htmlData{tutor-start=12,tutor-end=13}{i}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}}^{\htmlData{tutor-start=18,tutor-end=19}{7}} \htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{y}_{\htmlData{tutor-start=25,tutor-end=26}{i}} \htmlData{tutor-start=28,tutor-end=29}{-} \bar{\htmlData{tutor-start=35,tutor-end=36}{y}}\htmlData{tutor-start=37,tutor-end=38}{)}^{\htmlData{tutor-start=40,tutor-end=41}{2}}} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{.}\htmlData{tutor-start=48,tutor-end=49}{5}\htmlData{tutor-start=49,tutor-end=50}{5}，7≈2.646\sqrt{\htmlData{tutor-start=6,tutor-end=7}{7}} \htmlData{tutor-start=9,tutor-end=17}{\approx }\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{.}\htmlData{tutor-start=19,tutor-end=20}{6}\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{6}。 参考公式：相关系数 r=∑i=1n(ti−tˉ)(yi−yˉ)∑i=1n(ti−tˉ)2∑i=1n(yi−yˉ)2\htmlData{tutor-start=0,tutor-end=1}{r} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\sum_{\htmlData{tutor-start=16,tutor-end=17}{i}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{1}}^{\htmlData{tutor-start=22,tutor-end=23}{n}} \htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{t}_{\htmlData{tutor-start=29,tutor-end=30}{i}} \htmlData{tutor-start=32,tutor-end=33}{-} \bar{\htmlData{tutor-start=39,tutor-end=40}{t}}\htmlData{tutor-start=41,tutor-end=42}{)}\htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{y}_{\htmlData{tutor-start=46,tutor-end=47}{i}} \htmlData{tutor-start=49,tutor-end=50}{-} \bar{\htmlData{tutor-start=56,tutor-end=57}{y}}\htmlData{tutor-start=58,tutor-end=59}{)}}{\sqrt{\sum_{\htmlData{tutor-start=73,tutor-end=74}{i}\htmlData{tutor-start=74,tutor-end=75}{=}\htmlData{tutor-start=75,tutor-end=76}{1}}^{\htmlData{tutor-start=79,tutor-end=80}{n}} \htmlData{tutor-start=82,tutor-end=83}{(}\htmlData{tutor-start=83,tutor-end=84}{t}_{\htmlData{tutor-start=86,tutor-end=87}{i}} \htmlData{tutor-start=89,tutor-end=90}{-} \bar{\htmlData{tutor-start=96,tutor-end=97}{t}}\htmlData{tutor-start=98,tutor-end=99}{)}^{\htmlData{tutor-start=101,tutor-end=102}{2}} \sum_{\htmlData{tutor-start=110,tutor-end=111}{i}\htmlData{tutor-start=111,tutor-end=112}{=}\htmlData{tutor-start=112,tutor-end=113}{1}}^{\htmlData{tutor-start=116,tutor-end=117}{n}} \htmlData{tutor-start=119,tutor-end=120}{(}\htmlData{tutor-start=120,tutor-end=121}{y}_{\htmlData{tutor-start=123,tutor-end=124}{i}} \htmlData{tutor-start=126,tutor-end=127}{-} \bar{\htmlData{tutor-start=133,tutor-end=134}{y}}\htmlData{tutor-start=135,tutor-end=136}{)}^{\htmlData{tutor-start=138,tutor-end=139}{2}}}}。 回归方程 y^=a^+b^t\hat{\htmlData{tutor-start=5,tutor-end=6}{y}} \htmlData{tutor-start=8,tutor-end=9}{=} \hat{\htmlData{tutor-start=15,tutor-end=16}{a}} \htmlData{tutor-start=18,tutor-end=19}{+} \hat{\htmlData{tutor-start=25,tutor-end=26}{b}}\htmlData{tutor-start=27,tutor-end=28}{t} 中斜率和截距的最小二乘估计公式分别为： b^=∑i=1n(ti−tˉ)(yi−yˉ)∑i=1n(ti−tˉ)2\hat{\htmlData{tutor-start=5,tutor-end=6}{b}} \htmlData{tutor-start=8,tutor-end=9}{=} \frac{\sum_{\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{1}}^{\htmlData{tutor-start=28,tutor-end=29}{n}} \htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{t}_{\htmlData{tutor-start=35,tutor-end=36}{i}} \htmlData{tutor-start=38,tutor-end=39}{-} \bar{\htmlData{tutor-start=45,tutor-end=46}{t}}\htmlData{tutor-start=47,tutor-end=48}{)}\htmlData{tutor-start=48,tutor-end=49}{(}\htmlData{tutor-start=49,tutor-end=50}{y}_{\htmlData{tutor-start=52,tutor-end=53}{i}} \htmlData{tutor-start=55,tutor-end=56}{-} \bar{\htmlData{tutor-start=62,tutor-end=63}{y}}\htmlData{tutor-start=64,tutor-end=65}{)}}{\sum_{\htmlData{tutor-start=73,tutor-end=74}{i}\htmlData{tutor-start=74,tutor-end=75}{=}\htmlData{tutor-start=75,tutor-end=76}{1}}^{\htmlData{tutor-start=79,tutor-end=80}{n}} \htmlData{tutor-start=82,tutor-end=83}{(}\htmlData{tutor-start=83,tutor-end=84}{t}_{\htmlData{tutor-start=86,tutor-end=87}{i}} \htmlData{tutor-start=89,tutor-end=90}{-} \bar{\htmlData{tutor-start=96,tutor-end=97}{t}}\htmlData{tutor-start=98,tutor-end=99}{)}^{\htmlData{tutor-start=101,tutor-end=102}{2}}}，a^=yˉ−b^tˉ\hat{\htmlData{tutor-start=5,tutor-end=6}{a}} \htmlData{tutor-start=8,tutor-end=9}{=} \bar{\htmlData{tutor-start=15,tutor-end=16}{y}} \htmlData{tutor-start=18,tutor-end=19}{-} \hat{\htmlData{tutor-start=25,tutor-end=26}{b}}\bar{\htmlData{tutor-start=32,tutor-end=33}{t}}。

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：全国卷 3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

解答题 · 数学竞赛/待细分

如图，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，PA⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=8}{\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}，AD∥BC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=13}{\parallel }\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{C}，AB=AD=AC=3\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\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}{A}\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{3}，PA=BC=4\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}{B}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}，M\htmlData{tutor-start=0,tutor-end=1}{M} 为线段 AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} 上一点，AM=2MD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{M}\htmlData{tutor-start=5,tutor-end=6}{D}，N\htmlData{tutor-start=0,tutor-end=1}{N} 为 PC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C} 的中点。 (1) 证明：MN∥\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N} \htmlData{tutor-start=3,tutor-end=12}{\parallel} 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}； (2) 求四面体 N−BCM\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{M} 的体积。

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：全国卷 3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

解答题 · 平面几何

已知抛物线 C:y2=2x\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{x} 的焦点为 F\htmlData{tutor-start=0,tutor-end=1}{F}，平行于 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的两条直线 l1,l2\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{l}_{\htmlData{tutor-start=10,tutor-end=11}{2}} 分别交 C\htmlData{tutor-start=0,tutor-end=1}{C} 于 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} 两点，交 C\htmlData{tutor-start=0,tutor-end=1}{C} 的准线于 P,Q\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{Q} 两点。 (1) 若 F\htmlData{tutor-start=0,tutor-end=1}{F} 在线段 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 上，R\htmlData{tutor-start=0,tutor-end=1}{R} 是 PQ\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{Q} 的中点，证明 AR∥FQ\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{R} \htmlData{tutor-start=3,tutor-end=13}{\parallel }\htmlData{tutor-start=13,tutor-end=14}{F}\htmlData{tutor-start=14,tutor-end=15}{Q}； (2) 若 △PQF\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{Q}\htmlData{tutor-start=12,tutor-end=13}{F} 的面积是 △ABF\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}{F} 的面积的两倍，求 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 中点的轨迹方程。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3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。

21

解答题 · 代数

设函数 f(x)=ln⁡x−x+1\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}{=} \ln \htmlData{tutor-start=11,tutor-end=12}{x} \htmlData{tutor-start=13,tutor-end=14}{-} \htmlData{tutor-start=15,tutor-end=16}{x} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{1}。 (1) 讨论 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}{)} 的单调性； (2) 证明当 x∈(1,+∞)\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}{1}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=17}{\infty}\htmlData{tutor-start=17,tutor-end=18}{)} 时，1<x−1ln⁡x<x\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{<} \frac{\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}}{\ln \htmlData{tutor-start=19,tutor-end=20}{x}} \htmlData{tutor-start=22,tutor-end=23}{<} \htmlData{tutor-start=24,tutor-end=25}{x}； (3) 设 c>1\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}，证明当 x∈(0,1)\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}{0}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)} 时，1+(c−1)x>cx\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{c}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{>} \htmlData{tutor-start=13,tutor-end=14}{c}^{\htmlData{tutor-start=16,tutor-end=17}{x}}。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 21 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。

22

解答题 · 平面几何

如图，⊙O\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O} 中 AB⌢\overset{\htmlData{tutor-start=9,tutor-end=15}{\frown}}{\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{B}} 的中点为 P\htmlData{tutor-start=0,tutor-end=1}{P}，弦 PC,PD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{P}\htmlData{tutor-start=5,tutor-end=6}{D} 分别交 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 于 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} 两点。 (1) 若 ∠PFB=2∠PCD\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=10}{B} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{P}\htmlData{tutor-start=22,tutor-end=23}{C}\htmlData{tutor-start=23,tutor-end=24}{D}，求 ∠PCD\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{D} 的大小； (2) 若 EC\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{C} 的垂直平分线与 FD\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{D} 的垂直平分线交于点 G\htmlData{tutor-start=0,tutor-end=1}{G}，证明 OG⊥CD\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{G} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{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：全国卷 3文科第 22 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。

23

解答题 · 数学竞赛/待细分

在直角坐标系 xOy\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{y} 中，曲线 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的参数方程为 {x=3cos⁡α,y=sin⁡α,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{=} \sqrt{\htmlData{tutor-start=24,tutor-end=25}{3}}\cos\htmlData{tutor-start=30,tutor-end=36}{\alpha}\htmlData{tutor-start=36,tutor-end=37}{,} \\ \htmlData{tutor-start=41,tutor-end=42}{y} \htmlData{tutor-start=43,tutor-end=44}{=} \sin\htmlData{tutor-start=49,tutor-end=55}{\alpha}\htmlData{tutor-start=55,tutor-end=56}{,} \end{cases} (α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 为参数)，以坐标原点为极点，以 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的正半轴为极轴，建立极坐标系，曲线 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的极坐标方程为 ρsin⁡(θ+π4)=22\htmlData{tutor-start=0,tutor-end=4}{\rho}\sin\left(\htmlData{tutor-start=14,tutor-end=21}{\theta }\htmlData{tutor-start=21,tutor-end=22}{+} \frac{\htmlData{tutor-start=29,tutor-end=32}{\pi}}{\htmlData{tutor-start=34,tutor-end=35}{4}}\right) \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{2}\sqrt{\htmlData{tutor-start=53,tutor-end=54}{2}}。 (1) 写出 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的普通方程和 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的直角坐标方程； (2) 设点 P\htmlData{tutor-start=0,tutor-end=1}{P} 在 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 上，点 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 在 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 上，求 ∣PQ∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{Q}\htmlData{tutor-start=3,tutor-end=4}{|} 的最小值及此时 P\htmlData{tutor-start=0,tutor-end=1}{P} 的直角坐标。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 23 题

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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。

24

解答题 · 代数

已知函数 f(x)=∣2x−a∣+a\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}{2}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{|} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{a}。 (1) 当 a=2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 时，求不等式 f(x)⩽6\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=15}{\leqslant }\htmlData{tutor-start=15,tutor-end=16}{6} 的解集； (2) 设函数 g(x)=∣2x−1∣\htmlData{tutor-start=0,tutor-end=1}{g}\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}{2}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{|}，当 x∈R\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\mathbf{\htmlData{tutor-start=14,tutor-end=15}{R}} 时，f(x)+g(x)⩾3\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}{g}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)} \htmlData{tutor-start=12,tutor-end=22}{\geqslant }\htmlData{tutor-start=22,tutor-end=23}{3}，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：全国卷 3文科第 24 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。