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2020 年高考数学（全国卷 3文科）

exams_raw/普通高考/2020/2020全国3文(云南,广西,贵州,西藏,四川).pdf · HS-MATH-1024-v2.1-solution-aware

23 题0 个小问/题组
1

一、选择题 · 组合数学

已知集合 A={1,2,3,5,7,11}\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}{1}\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}{3}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{7}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=18}{\}}，B={x∣3<x<15}\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}{x}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{<}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{<}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{5}\htmlData{tutor-start=12,tutor-end=14}{\}}，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} 中元素的个数为 ( ) (A) 2 (B) 3 (C) 4 (D) 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 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 zˉ(1+i)=1−i\bar{\htmlData{tutor-start=5,tutor-end=6}{z}}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{+}\mathrm{\htmlData{tutor-start=18,tutor-end=19}{i}}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{-}\mathrm{\htmlData{tutor-start=32,tutor-end=33}{i}}，则 z=\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=} ( ) (A) 1−i\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}} (B) 1+i\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}} (C) −i\htmlData{tutor-start=0,tutor-end=1}{-}\mathrm{\htmlData{tutor-start=9,tutor-end=10}{i}} (D) i\mathrm{\htmlData{tutor-start=8,tutor-end=9}{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 · 数学竞赛/待细分

设一组样本数据 x1,x2,⋯,xn\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \cdots\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{x}_{\htmlData{tutor-start=25,tutor-end=26}{n}} 的方差为 0.01\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{1}，则数据 10x1,10x2,⋯,10xn\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{x}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{x}_{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{,} \cdots\htmlData{tutor-start=24,tutor-end=25}{,} \htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{x}_{\htmlData{tutor-start=31,tutor-end=32}{n}} 的方差为 ( ) (A) 0.01 (B) 0.1 (C) 1 (D) 10

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

Logistic 模型是常用数学模型之一，可应用于流行病学领域。有学者根据公布数据建立了某地区新冠肺炎累计确诊病例数 I(t)\htmlData{tutor-start=0,tutor-end=1}{I}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{)} (t\htmlData{tutor-start=0,tutor-end=1}{t} 的单位：天) 的 Logistic 模型：I(t)=K1+e−0.23(t−53)\htmlData{tutor-start=0,tutor-end=1}{I}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\frac{\htmlData{tutor-start=11,tutor-end=12}{K}}{\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{+}\mathrm{\htmlData{tutor-start=24,tutor-end=25}{e}}^{\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=31}{.}\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{3}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{t}\htmlData{tutor-start=35,tutor-end=36}{-}\htmlData{tutor-start=36,tutor-end=37}{5}\htmlData{tutor-start=37,tutor-end=38}{3}\htmlData{tutor-start=38,tutor-end=39}{)}}}，其中 K\htmlData{tutor-start=0,tutor-end=1}{K} 为最大确诊病例数。当 I(t∗)=0.95KI(t^*)=0.95K 时，标志着已初步遏制疫情，则 t∗t^* 约为 (ln⁡19≈3\ln \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{9} \htmlData{tutor-start=7,tutor-end=15}{\approx }\htmlData{tutor-start=15,tutor-end=16}{3}) ( ) (A) 60 (B) 63 (C) 66 (D) 69

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

已知 sin⁡θ+sin⁡(θ+π3)=1\sin\htmlData{tutor-start=4,tutor-end=10}{\theta}\htmlData{tutor-start=10,tutor-end=11}{+}\sin\left(\htmlData{tutor-start=21,tutor-end=27}{\theta}\htmlData{tutor-start=27,tutor-end=28}{+}\frac{\htmlData{tutor-start=34,tutor-end=37}{\pi}}{\htmlData{tutor-start=39,tutor-end=40}{3}}\right)\htmlData{tutor-start=48,tutor-end=49}{=}\htmlData{tutor-start=49,tutor-end=50}{1}，则 sin⁡(θ+π6)=\sin\left(\htmlData{tutor-start=10,tutor-end=16}{\theta}\htmlData{tutor-start=16,tutor-end=17}{+}\frac{\htmlData{tutor-start=23,tutor-end=26}{\pi}}{\htmlData{tutor-start=28,tutor-end=29}{6}}\right)\htmlData{tutor-start=37,tutor-end=38}{=} ( ) (A) 12\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}} (B) 33\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{3}} (C) 23\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{3}} (D) 22\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}

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 · 平面几何

在平面内，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} 是两个定点，C\htmlData{tutor-start=0,tutor-end=1}{C} 是动点。若 AC→⋅BC→=1\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{C}} \htmlData{tutor-start=20,tutor-end=26}{\cdot }\overrightarrow{\htmlData{tutor-start=42,tutor-end=43}{B}\htmlData{tutor-start=43,tutor-end=44}{C}}\htmlData{tutor-start=45,tutor-end=46}{=}\htmlData{tutor-start=46,tutor-end=47}{1}，则点 C\htmlData{tutor-start=0,tutor-end=1}{C} 的轨迹为 ( ) (A) 圆 (B) 椭圆 (C) 抛物线 (D) 直线

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 · 平面几何

设 O\htmlData{tutor-start=0,tutor-end=1}{O} 为坐标原点，直线 x=2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 与抛物线 C:y2=2px\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}{p}\htmlData{tutor-start=11,tutor-end=12}{x} (p>0\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}) 交于 D,E\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{E} 两点。若 OD⊥OE\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{O}\htmlData{tutor-start=10,tutor-end=11}{E}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的焦点坐标为 ( ) (A) (14,0)\left(\frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{4}}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{0}\right) (B) (12,0)\left(\frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{0}\right) (C) (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=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)} (D) (2,0)\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}{0}\htmlData{tutor-start=5,tutor-end=6}{)}

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 · 平面几何

点 (0,−1)\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}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)} 到直线 y=k(x+1)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)} 距离的最大值为 ( ) (A) 1 (B) 2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}} (C) 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}} (D) 2

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 · 数学竞赛/待细分

下图为某几何体的三视图，则该几何体的表面积是 ( ) (A) 6+42\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{4}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{2}} (B) 4+42\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{4}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{2}} (C) 6+23\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{3}} (D) 4+23\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{3}}

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 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

设 a=log⁡32,b=log⁡53,c=23\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\log_{\htmlData{tutor-start=8,tutor-end=9}{3}} \htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{b}\htmlData{tutor-start=15,tutor-end=16}{=}\log_{\htmlData{tutor-start=22,tutor-end=23}{5}} \htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{,} \htmlData{tutor-start=28,tutor-end=29}{c}\htmlData{tutor-start=29,tutor-end=30}{=}\frac{\htmlData{tutor-start=36,tutor-end=37}{2}}{\htmlData{tutor-start=39,tutor-end=40}{3}}，则 ( ) (A) a<c<b\htmlData{tutor-start=0,tutor-end=1}{a}\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}{b} (B) a<b<c\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} (C) b<c<a\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}{a} (D) c<a<b\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=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：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 · 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，cos⁡C=23,AC=4,BC=3\cos \htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{=}\frac{\htmlData{tutor-start=13,tutor-end=14}{2}}{\htmlData{tutor-start=16,tutor-end=17}{3}}\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=21}{A}\htmlData{tutor-start=21,tutor-end=22}{C}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{4}\htmlData{tutor-start=24,tutor-end=25}{,} \htmlData{tutor-start=26,tutor-end=27}{B}\htmlData{tutor-start=27,tutor-end=28}{C}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{3}，则 tan⁡B=\tan \htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{=} ( ) (A) 5\sqrt{\htmlData{tutor-start=6,tutor-end=7}{5}} (B) 25\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{5}} (C) 45\htmlData{tutor-start=0,tutor-end=1}{4}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{5}} (D) 85\htmlData{tutor-start=0,tutor-end=1}{8}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{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 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

已知函数 f(x)=sin⁡x+1sin⁡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=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{+}\frac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\sin \htmlData{tutor-start=26,tutor-end=27}{x}}，则 ( ) (A) 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}{)} 的最小值为 2 (B) 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}{)} 的图象关于 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴对称 (C) 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}{)} 的图象关于直线 x=π\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pi} 对称 (D) 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}{)} 的图象关于直线 x=π2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=11}{\pi}}{\htmlData{tutor-start=13,tutor-end=14}{2}} 对称

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 {x+y⩾0,2x−y⩾0,x⩽1,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{y} \htmlData{tutor-start=18,tutor-end=28}{\geqslant }\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{,} \\ \htmlData{tutor-start=34,tutor-end=35}{2}\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=39,tutor-end=49}{\geqslant }\htmlData{tutor-start=49,tutor-end=50}{0}\htmlData{tutor-start=50,tutor-end=51}{,} \\ \htmlData{tutor-start=55,tutor-end=56}{x} \htmlData{tutor-start=57,tutor-end=67}{\leqslant }\htmlData{tutor-start=67,tutor-end=68}{1}\htmlData{tutor-start=68,tutor-end=69}{,} \end{cases} 则 z=3x+2y\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{y} 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：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 · 数学竞赛/待细分

设双曲线 C:x2a2−y2b2=1\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=22,tutor-end=23}{-}\frac{\htmlData{tutor-start=29,tutor-end=30}{y}^{\htmlData{tutor-start=32,tutor-end=33}{2}}}{\htmlData{tutor-start=36,tutor-end=37}{b}^{\htmlData{tutor-start=39,tutor-end=40}{2}}}\htmlData{tutor-start=42,tutor-end=43}{=}\htmlData{tutor-start=43,tutor-end=44}{1} (a>0,b>0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,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}{b}\htmlData{tutor-start=6,tutor-end=7}{>}\htmlData{tutor-start=7,tutor-end=8}{0}) 的一条渐近线为 y=2x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{x}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的离心率为______。

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 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 · 代数

设函数 f(x)=exx+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=4,tutor-end=5}{=}\frac{\mathrm{\htmlData{tutor-start=19,tutor-end=20}{e}}^{\htmlData{tutor-start=23,tutor-end=24}{x}}}{\htmlData{tutor-start=27,tutor-end=28}{x}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{a}}。若 f′(1)=e4\htmlData{tutor-start=0,tutor-end=1}{f}'\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=5,tutor-end=6}{=}\frac{\mathrm{\htmlData{tutor-start=20,tutor-end=21}{e}}}{\htmlData{tutor-start=24,tutor-end=25}{4}}，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：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 · 平面几何

已知圆锥的底面半径为 1，母线长为 3，则该圆锥内半径最大的球的体积为______。

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 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

设等比数列 {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+a2=4,a3−a1=8\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}{a}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{4}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{1}}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{8}。 (1) 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； (2) 记 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}} 为数列 {log⁡3an}\htmlData{tutor-start=0,tutor-end=2}{\{}\log_{\htmlData{tutor-start=8,tutor-end=9}{3}} \htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{n}}\htmlData{tutor-start=16,tutor-end=18}{\}} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和。若 Sm+Sm+1=Sm+3\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{m}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{S}_{\htmlData{tutor-start=9,tutor-end=10}{m}\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}{S}_{\htmlData{tutor-start=17,tutor-end=18}{m}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{3}}，求 m\htmlData{tutor-start=0,tutor-end=1}{m}。

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

某学生兴趣小组随机调查了某市 100 天中每天的空气质量等级和当天到某公园锻炼的人次，整理数据得到下表（单位：天）： (表格略) (1) 分别估计该市一天的空气质量等级为 1, 2, 3, 4 的概率； (2) 求一天中到该公园锻炼的平均人次的估计值（同一组中的数据用该组区间的中点值为代表）； (3) 若某天的空气质量等级为 1 或 2，则称这天“空气质量好”；若某天的空气质量等级为 3 或 4，则称这天“空气质量不好”。根据所给数据，完成下面的 2×2\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{2} 列联表，并根据列联表，判断是否有 95% 的把握认为一天中到该公园锻炼的人次与该市当天的空气质量有关？ (列联表略) 附：K2=n(ad−bc)2(a+b)(c+d)(a+c)(b+d)\htmlData{tutor-start=0,tutor-end=1}{K}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{n}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{a}\htmlData{tutor-start=15,tutor-end=16}{d}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{b}\htmlData{tutor-start=18,tutor-end=19}{c}\htmlData{tutor-start=19,tutor-end=20}{)}^{\htmlData{tutor-start=22,tutor-end=23}{2}}}{\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{a}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{b}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{c}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{d}\htmlData{tutor-start=35,tutor-end=36}{)}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{a}\htmlData{tutor-start=38,tutor-end=39}{+}\htmlData{tutor-start=39,tutor-end=40}{c}\htmlData{tutor-start=40,tutor-end=41}{)}\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{b}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{d}\htmlData{tutor-start=45,tutor-end=46}{)}}， (临界值表略)

原卷图示 1
原卷图示 1原卷第 1 页 · qwen3.7-plus_layout_detection · 需复核
原卷图示 2
原卷图示 2原卷第 1 页 · qwen3.7-plus_layout_detection · 需复核
原卷题面及图示 3
原卷题面及图示 3原卷第 1 页 · question_region_fallback · 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，在长方体 ABCD−A1B1C1D1\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=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{A}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{B}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{C}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{D}_{\htmlData{tutor-start=23,tutor-end=24}{1}} 中，点 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} 分别在棱 DD1,BB1\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{D}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{B}_{\htmlData{tutor-start=12,tutor-end=13}{1}} 上，且 2DE=ED1,BF=2FB1\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{E}\htmlData{tutor-start=5,tutor-end=6}{D}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{F}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{F}\htmlData{tutor-start=17,tutor-end=18}{B}_{\htmlData{tutor-start=20,tutor-end=21}{1}}. 证明： (1) 当 AB=BC\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}{B}\htmlData{tutor-start=4,tutor-end=5}{C} 时，EF⊥AC\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{C}; (2) 点 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 在平面 AEF\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.

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 · 代数

已知函数 f(x)=x3−kx+k2\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}{x}^{\htmlData{tutor-start=8,tutor-end=9}{3}}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{k}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{k}^{\htmlData{tutor-start=17,tutor-end=18}{2}}. (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) 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}{)} 有三个零点，求 k\htmlData{tutor-start=0,tutor-end=1}{k} 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 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 · 平面几何

已知椭圆 C:x225+y2m2=1 (0<m<5)\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}{2}\htmlData{tutor-start=17,tutor-end=18}{5}} \htmlData{tutor-start=20,tutor-end=21}{+} \frac{\htmlData{tutor-start=28,tutor-end=29}{y}^{\htmlData{tutor-start=31,tutor-end=32}{2}}}{\htmlData{tutor-start=35,tutor-end=36}{m}^{\htmlData{tutor-start=38,tutor-end=39}{2}}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{1} \htmlData{tutor-start=46,tutor-end=48}{\ }\htmlData{tutor-start=48,tutor-end=49}{(}\htmlData{tutor-start=49,tutor-end=50}{0}\htmlData{tutor-start=50,tutor-end=51}{<}\htmlData{tutor-start=51,tutor-end=52}{m}\htmlData{tutor-start=52,tutor-end=53}{<}\htmlData{tutor-start=53,tutor-end=54}{5}\htmlData{tutor-start=54,tutor-end=55}{)} 的离心率为 154\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{5}}}{\htmlData{tutor-start=17,tutor-end=18}{4}}，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. (1) 求 C\htmlData{tutor-start=0,tutor-end=1}{C} 的方程; (2) 若点 P\htmlData{tutor-start=0,tutor-end=1}{P} 在 C\htmlData{tutor-start=0,tutor-end=1}{C} 上，点 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 在直线 x=6\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{6} 上，且 ∣BP∣=∣BQ∣,BP⊥BQ\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{P}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{B}\htmlData{tutor-start=7,tutor-end=8}{Q}\htmlData{tutor-start=8,tutor-end=9}{|}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{P} \htmlData{tutor-start=14,tutor-end=20}{\perp }\htmlData{tutor-start=20,tutor-end=21}{B}\htmlData{tutor-start=21,tutor-end=22}{Q}，求 △APQ\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{P}\htmlData{tutor-start=12,tutor-end=13}{Q} 的面积.

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

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 C\htmlData{tutor-start=0,tutor-end=1}{C} 的参数方程为 {x=2−t−t2,y=2−3t+t2,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{t}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{t}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{,} \\ \htmlData{tutor-start=30,tutor-end=31}{y}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{3}\htmlData{tutor-start=35,tutor-end=36}{t}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{t}^{\htmlData{tutor-start=40,tutor-end=41}{2}}\htmlData{tutor-start=42,tutor-end=43}{,} \end{cases} (t\htmlData{tutor-start=0,tutor-end=1}{t} 为参数且 t≠1\htmlData{tutor-start=0,tutor-end=1}{t} \neq \htmlData{tutor-start=7,tutor-end=8}{1})，C\htmlData{tutor-start=0,tutor-end=1}{C} 与坐标轴交于 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. (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}{|}; (2) 以坐标原点为极点，x\htmlData{tutor-start=0,tutor-end=1}{x} 轴正半轴为极轴建立极坐标系，求直线 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 的极坐标方程.

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 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

设 a,b,c∈R,a+b+c=0,abc=1\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=8,tutor-end=12}{\in }\mathbf{\htmlData{tutor-start=20,tutor-end=21}{R}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{a}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{b}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{c}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{,} \htmlData{tutor-start=33,tutor-end=34}{a}\htmlData{tutor-start=34,tutor-end=35}{b}\htmlData{tutor-start=35,tutor-end=36}{c}\htmlData{tutor-start=36,tutor-end=37}{=}\htmlData{tutor-start=37,tutor-end=38}{1}. (1) 证明：ab+bc+ca<0\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}{b}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{c}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{<}\htmlData{tutor-start=9,tutor-end=10}{0}; (2) 用 max⁡{a,b,c}\max\htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{c}\htmlData{tutor-start=13,tutor-end=15}{\}} 表示 a,b,c\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} 中的最大值，证明：max⁡{a,b,c}⩾43\max\htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{c}\htmlData{tutor-start=13,tutor-end=15}{\}} \htmlData{tutor-start=16,tutor-end=26}{\geqslant }\sqrt[\htmlData{tutor-start=32,tutor-end=33}{3}]{\htmlData{tutor-start=35,tutor-end=36}{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 23 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。