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2020 年高考数学（全国卷 3理科）

exams_raw/普通高考/2020/2020全国3理(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={(x,y)∣x,y∈N∗,y⩾x}A = \{(x, y) \mid x, y \in \mathbf{N}^*, y \geqslant x\}, B={(x,y)∣x+y=8}\htmlData{tutor-start=0,tutor-end=1}{B} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{y}\htmlData{tutor-start=11,tutor-end=12}{)} \htmlData{tutor-start=13,tutor-end=18}{\mid }\htmlData{tutor-start=18,tutor-end=19}{x} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{y} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{8}\htmlData{tutor-start=27,tutor-end=29}{\}}, 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) 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理科第 1 题

解题过程

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

2

一、选择题 · 数学竞赛/待细分

复数 11−3i\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{3}\mathrm{\htmlData{tutor-start=20,tutor-end=21}{i}}} 的虚部是 ( ) (A) −310\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{0}} (B) −110\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{0}} (C) 110\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0}} (D) 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}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 2 题

解题过程

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

3

一、选择题 · 数学竞赛/待细分

在一组样本数据中, 1, 2, 3, 4 出现的频率分别为 p1,p2,p3,p4\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{p}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{p}_{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{p}_{\htmlData{tutor-start=24,tutor-end=25}{4}}, 且 ∑i=14pi=1\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}{4}} \htmlData{tutor-start=15,tutor-end=16}{p}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{1}, 则下面四种情形中, 对应样本的标准差最大的一组是 ( ) (A) p1=p4=0.1,p2=p3=0.4\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{p}_{\htmlData{tutor-start=11,tutor-end=12}{4}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{.}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{p}_{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{p}_{\htmlData{tutor-start=32,tutor-end=33}{3}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{.}\htmlData{tutor-start=39,tutor-end=40}{4} (B) p1=p4=0.4,p2=p3=0.1\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{p}_{\htmlData{tutor-start=11,tutor-end=12}{4}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{.}\htmlData{tutor-start=18,tutor-end=19}{4}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{p}_{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{p}_{\htmlData{tutor-start=32,tutor-end=33}{3}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{.}\htmlData{tutor-start=39,tutor-end=40}{1} (C) p1=p4=0.2,p2=p3=0.3\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{p}_{\htmlData{tutor-start=11,tutor-end=12}{4}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{.}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{p}_{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{p}_{\htmlData{tutor-start=32,tutor-end=33}{3}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{.}\htmlData{tutor-start=39,tutor-end=40}{3} (D) p1=p4=0.3,p2=p3=0.2\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{p}_{\htmlData{tutor-start=11,tutor-end=12}{4}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{.}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{p}_{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{p}_{\htmlData{tutor-start=32,tutor-end=33}{3}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{.}\htmlData{tutor-start=39,tutor-end=40}{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理科第 3 题

解题过程

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

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=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{K}}{\htmlData{tutor-start=16,tutor-end=17}{1} \htmlData{tutor-start=18,tutor-end=19}{+} \mathrm{\htmlData{tutor-start=28,tutor-end=29}{e}}^{\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{0}\htmlData{tutor-start=34,tutor-end=35}{.}\htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=37}{3}\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{t}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{5}\htmlData{tutor-start=41,tutor-end=42}{3}\htmlData{tutor-start=42,tutor-end=43}{)}}}, 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理科第 4 题

解题过程

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

5

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=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{p}\htmlData{tutor-start=13,tutor-end=14}{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)\htmlData{tutor-start=0,tutor-end=1}{(}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{4}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{)} (B) (12,0)\htmlData{tutor-start=0,tutor-end=1}{(}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{)} (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理科第 5 题

解题过程

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

6

一、选择题 · 平面几何

已知向量 a,b\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{,} \boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}} 满足 ∣a∣=5,∣b∣=6,a⋅b=−6\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{5}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{|}\boldsymbol{\htmlData{tutor-start=35,tutor-end=36}{b}}\htmlData{tutor-start=37,tutor-end=38}{|} \htmlData{tutor-start=39,tutor-end=40}{=} \htmlData{tutor-start=41,tutor-end=42}{6}\htmlData{tutor-start=42,tutor-end=43}{,} \boldsymbol{\htmlData{tutor-start=56,tutor-end=57}{a}} \htmlData{tutor-start=59,tutor-end=65}{\cdot }\boldsymbol{\htmlData{tutor-start=77,tutor-end=78}{b}} \htmlData{tutor-start=80,tutor-end=81}{=} \htmlData{tutor-start=82,tutor-end=83}{-}\htmlData{tutor-start=83,tutor-end=84}{6}, 则 cos⁡⟨a,a+b⟩=\cos \htmlData{tutor-start=5,tutor-end=13}{\langle }\boldsymbol{\htmlData{tutor-start=25,tutor-end=26}{a}}\htmlData{tutor-start=27,tutor-end=28}{,} \boldsymbol{\htmlData{tutor-start=41,tutor-end=42}{a}} \htmlData{tutor-start=44,tutor-end=45}{+} \boldsymbol{\htmlData{tutor-start=58,tutor-end=59}{b}} \htmlData{tutor-start=61,tutor-end=69}{\rangle }\htmlData{tutor-start=69,tutor-end=70}{=} ( ) (A) −3135\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{5}} (B) −1935\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{9}}{\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{5}} (C) 1735\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{7}}{\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{5}} (D) 1935\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{9}}{\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{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理科第 6 题

解题过程

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

7

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=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{2}}{\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{A}\htmlData{tutor-start=23,tutor-end=24}{C} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{4}\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=31}{B}\htmlData{tutor-start=31,tutor-end=32}{C} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{3}, 则 cos⁡B=\cos \htmlData{tutor-start=5,tutor-end=6}{B} \htmlData{tutor-start=7,tutor-end=8}{=} ( ) (A) 19\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{9}} (B) 13\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}} (C) 12\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}} (D) 23\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{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理科第 7 题

解题过程

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

8

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=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{4}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{2}} (B) 4+42\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{4}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{2}} (C) 6+23\htmlData{tutor-start=0,tutor-end=1}{6} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{3}} (D) 4+23\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{3}}

原卷题面及图示 1
原卷题面及图示 1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理科第 8 题

解题过程

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

9

一、选择题 · 数学竞赛/待细分

已知 2tan⁡θ−tan⁡(θ+π4)=7\htmlData{tutor-start=0,tutor-end=1}{2}\tan\htmlData{tutor-start=5,tutor-end=12}{\theta }\htmlData{tutor-start=12,tutor-end=13}{-} \tan\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=26}{\theta }\htmlData{tutor-start=26,tutor-end=27}{+} \frac{\htmlData{tutor-start=34,tutor-end=37}{\pi}}{\htmlData{tutor-start=39,tutor-end=40}{4}}\htmlData{tutor-start=41,tutor-end=42}{)} \htmlData{tutor-start=43,tutor-end=44}{=} \htmlData{tutor-start=45,tutor-end=46}{7}, 则 tan⁡θ=\tan\htmlData{tutor-start=4,tutor-end=11}{\theta }\htmlData{tutor-start=11,tutor-end=12}{=} ( ) (A) −2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2} (B) −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} (C) 1 (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理科第 9 题

解题过程

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

10

一、选择题 · 平面几何

若直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与曲线 y=x\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{x}} 和 x2+y2=15\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{5}} 都相切, 则 l\htmlData{tutor-start=0,tutor-end=1}{l} 的方程为 ( ) (A) y=2x+1\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{1} (B) y=2x+12\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{+} \frac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}} (C) y=12x+1\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{x} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{1} (D) y=12x+12\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{x} \htmlData{tutor-start=17,tutor-end=18}{+} \frac{\htmlData{tutor-start=25,tutor-end=26}{1}}{\htmlData{tutor-start=28,tutor-end=29}{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理科第 10 题

解题过程

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

11

一、选择题 · 平面几何

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=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} (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}) 的左、右焦点分别为 F1,F2\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{F}_{\htmlData{tutor-start=10,tutor-end=11}{2}}, 离心率为 5\sqrt{\htmlData{tutor-start=6,tutor-end=7}{5}}. P\htmlData{tutor-start=0,tutor-end=1}{P} 是 C\htmlData{tutor-start=0,tutor-end=1}{C} 上一点, 且 F1P⊥F2P\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{P} \htmlData{tutor-start=7,tutor-end=13}{\perp }\htmlData{tutor-start=13,tutor-end=14}{F}_{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{P}. 若 △PF1F2\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{F}_{\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{F}_{\htmlData{tutor-start=19,tutor-end=20}{2}} 的面积为 4, 则 a=\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} ( ) (A) 1 (B) 2 (C) 4 (D) 8

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 11 题

解题过程

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

12

一、选择题 · 数学竞赛/待细分

已知 55<84,134<85\htmlData{tutor-start=0,tutor-end=1}{5}^{\htmlData{tutor-start=3,tutor-end=4}{5}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{8}^{\htmlData{tutor-start=11,tutor-end=12}{4}}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{4}} \htmlData{tutor-start=22,tutor-end=23}{<} \htmlData{tutor-start=24,tutor-end=25}{8}^{\htmlData{tutor-start=27,tutor-end=28}{5}}. 设 a=log⁡53,b=log⁡85,c=log⁡138\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \log_{\htmlData{tutor-start=10,tutor-end=11}{5}} \htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{b} \htmlData{tutor-start=18,tutor-end=19}{=} \log_{\htmlData{tutor-start=26,tutor-end=27}{8}} \htmlData{tutor-start=29,tutor-end=30}{5}\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{c} \htmlData{tutor-start=34,tutor-end=35}{=} \log_{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{3}} \htmlData{tutor-start=46,tutor-end=47}{8}, 则 ( ) (A) 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} (B) 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} (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理科第 12 题

解题过程

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

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=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{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理科第 13 题

解题过程

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

14

二、填空题 · 数学竞赛/待细分

(x2+2x)6\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{+} \frac{\htmlData{tutor-start=15,tutor-end=16}{2}}{\htmlData{tutor-start=18,tutor-end=19}{x}}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{6}} 的展开式中常数项是 \_\_\_\_\_\_. (用数字作答)

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 14 题

解题过程

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

15

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理科第 15 题

解题过程

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

16

二、填空题 · 平面几何

关于函数 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=5,tutor-end=6}{=} \sin \htmlData{tutor-start=12,tutor-end=13}{x} \htmlData{tutor-start=14,tutor-end=15}{+} \frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\sin \htmlData{tutor-start=30,tutor-end=31}{x}} 有如下四个命题: ① 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; ② 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}{)} 的图象关于原点对称; ③ 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=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=13}{\pi}}{\htmlData{tutor-start=15,tutor-end=16}{2}} 对称; ④ 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. 其中所有真命题的序号是 \_\_\_\_\_\_.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 16 题

解题过程

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

17

三、解答题 · 代数

设数列 {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=3,an+1=3an−4n\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{n}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{1}} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{3}\htmlData{tutor-start=22,tutor-end=23}{a}_{\htmlData{tutor-start=25,tutor-end=26}{n}} \htmlData{tutor-start=28,tutor-end=29}{-} \htmlData{tutor-start=30,tutor-end=31}{4}\htmlData{tutor-start=31,tutor-end=32}{n}. (1) 计算 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}}, 猜想 {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}{\}} 的通项公式并加以证明; (2) 求数列 {2nan}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{2}^{\htmlData{tutor-start=5,tutor-end=6}{n}} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=15}{\}} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 17 题

解题过程

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

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): | 空气质量等级 \ 锻炼人次 | [0,200]\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}{2}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{]} | (200,400]\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{]} | (400,600]\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{6}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{]} | | :--- | :---: | :---: | :---: | | 1 (优) | 2 | 16 | 25 | | 2 (良) | 5 | 10 | 12 | | 3 (轻度污染) | 6 | 7 | 8 | | 4 (中度污染) | 7 | 2 | 0 | (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? | | 人次 ⩽400\htmlData{tutor-start=0,tutor-end=10}{\leqslant }\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{0} | 人次 >400\htmlData{tutor-start=0,tutor-end=1}{>} \htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{0} | | :--- | :---: | :---: | | 空气质量好 | | | | 空气质量不好 | | | 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=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{n}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{a}\htmlData{tutor-start=17,tutor-end=18}{d}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{c}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}}}{\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{a}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{b}\htmlData{tutor-start=32,tutor-end=33}{)}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{c}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{d}\htmlData{tutor-start=37,tutor-end=38}{)}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{a}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{c}\htmlData{tutor-start=42,tutor-end=43}{)}\htmlData{tutor-start=43,tutor-end=44}{(}\htmlData{tutor-start=44,tutor-end=45}{b}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{d}\htmlData{tutor-start=47,tutor-end=48}{)}}, | P(K2⩾k)\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{K}^{\htmlData{tutor-start=5,tutor-end=6}{2}} \htmlData{tutor-start=8,tutor-end=18}{\geqslant }\htmlData{tutor-start=18,tutor-end=19}{k}\htmlData{tutor-start=19,tutor-end=20}{)} | 0.050 | 0.010 | 0.001 | | :--- | :---: | :---: | :---: | | k\htmlData{tutor-start=0,tutor-end=1}{k} | 3.841 | 6.635 | 10.828 |

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 18 题

解题过程

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

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\htmlData{tutor-start=0,tutor-end=1}{E}，F\htmlData{tutor-start=0,tutor-end=1}{F} 分别在棱 DD1\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{D}_{\htmlData{tutor-start=4,tutor-end=5}{1}}，BB1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 上，且 2DE=ED1\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}}，BF=2FB1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{F}\htmlData{tutor-start=5,tutor-end=6}{B}_{\htmlData{tutor-start=8,tutor-end=9}{1}}。 (1) 证明：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； (2) 若 AB=2\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}{2}，AD=1\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}{1}，AA1=3\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{3}，求二面角 A−EF−A1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{F}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{A}_{\htmlData{tutor-start=8,tutor-end=9}{1}} 的正弦值。

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理科第 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

已知椭圆 C\htmlData{tutor-start=0,tutor-end=1}{C}：x225+y2m2=1\dfrac{\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}}{\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{5}}\htmlData{tutor-start=17,tutor-end=18}{+}\dfrac{\htmlData{tutor-start=25,tutor-end=26}{y}^{\htmlData{tutor-start=28,tutor-end=29}{2}}}{\htmlData{tutor-start=32,tutor-end=33}{m}^{\htmlData{tutor-start=35,tutor-end=36}{2}}}\htmlData{tutor-start=38,tutor-end=39}{=}\htmlData{tutor-start=39,tutor-end=40}{1}（0<m<5\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{5}）的离心率为 154\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{5}}}{\htmlData{tutor-start=18,tutor-end=19}{4}}，A\htmlData{tutor-start=0,tutor-end=1}{A}，B\htmlData{tutor-start=0,tutor-end=1}{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∣\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}{|}，BP⊥BQ\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{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理科第 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)=x3+bx+c\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}{b}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{c}，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 (12,f(12))\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{f}\left(\dfrac{\htmlData{tutor-start=33,tutor-end=34}{1}}{\htmlData{tutor-start=36,tutor-end=37}{2}}\right)\right) 处的切线与 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴垂直。 (1) 求 b\htmlData{tutor-start=0,tutor-end=1}{b}； (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}{)} 有一个绝对值不大于 1\htmlData{tutor-start=0,tutor-end=1}{1} 的零点，证明：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}{)} 所有零点的绝对值都不大于 1\htmlData{tutor-start=0,tutor-end=1}{1}。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 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}x=2-t-t^{2},\\\y=2-3t+t^{2},\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=6,tutor-end=7}{1}），C\htmlData{tutor-start=0,tutor-end=1}{C} 与坐标轴交于 A\htmlData{tutor-start=0,tutor-end=1}{A}，B\htmlData{tutor-start=0,tutor-end=1}{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理科第 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\htmlData{tutor-start=0,tutor-end=1}{a}，b\htmlData{tutor-start=0,tutor-end=1}{b}，c∈R\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=4}{\in}\mathbf{\htmlData{tutor-start=12,tutor-end=13}{R}}，a+b+c=0\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}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，abc=1\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}{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=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{,}\,\htmlData{tutor-start=14,tutor-end=15}{c}\htmlData{tutor-start=15,tutor-end=17}{\}} 表示 a\htmlData{tutor-start=0,tutor-end=1}{a}，b\htmlData{tutor-start=0,tutor-end=1}{b}，c\htmlData{tutor-start=0,tutor-end=1}{c} 中的最大值，证明：max⁡{a, b, c}⩾44\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=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=12}{,}\,\htmlData{tutor-start=14,tutor-end=15}{c}\htmlData{tutor-start=15,tutor-end=17}{\}}\htmlData{tutor-start=17,tutor-end=26}{\geqslant}\sqrt[\htmlData{tutor-start=32,tutor-end=33}{4}]{\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理科第 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。