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2017 年高考数学（全国卷 2理科）

exams_raw/普通高考/2017/2017全国2理(甘肃,青海,内蒙古,黑龙江,吉林,辽宁,海南,宁夏,新疆,陕西,重庆).pdf · HS-MATH-1024-v2.1-solution-aware

23 题0 个小问/题组
1

一、选择题 · 数学竞赛/待细分

3+i1+i=()\frac{\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{+}\mathrm{\htmlData{tutor-start=16,tutor-end=17}{i}}}{\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{+}\mathrm{\htmlData{tutor-start=30,tutor-end=31}{i}}}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{(}\quad\htmlData{tutor-start=40,tutor-end=41}{)} (A) 1+2i\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}} (B) 1−2i\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}} (C) 2+i\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{+}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}} (D) 2−i\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{-}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 组合数学

设集合 A={1,2,4}\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=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=13}{\}}, B={x∣x2−4x+m=0}\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=6,tutor-end=11}{\mid }\htmlData{tutor-start=11,tutor-end=12}{x}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{4}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{m}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=25}{\}}. 若 A∩B={1}\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=11}{\{}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=14}{\}}, 则 B=()\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\quad\htmlData{tutor-start=8,tutor-end=9}{)} (A) {1,−3}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=9}{\}} (B) {1,0}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=8}{\}} (C) {1,3}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=8}{\}} (D) {1,5}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=8}{\}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

我国古代数学名著《算法统宗》中有如下问题：“远望巍巍塔七层，红光点点倍加增，共灯三百八十一，请问尖头几盏灯？”意思是：一座 7 层塔共挂了 381 盏灯，且相邻两层中的下一层灯数是上一层灯数的 2 倍，则塔的顶层共有灯 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) 1 盏 (B) 3 盏 (C) 5 盏 (D) 9 盏

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 平面几何

如图，网格纸上小正方形的边长为 1，粗实线画出的是某几何体的三视图，该几何体由一平面将一圆柱截去一部分后所得，则该几何体的体积为 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) 90π\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=5}{\pi} (B) 63π\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=5}{\pi} (C) 42π\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=5}{\pi} (D) 36π\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=5}{\pi}

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 2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 · 数学竞赛/待细分

设 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+3y−3⩽0,2x−3y+3⩾0,y+3⩾0,\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}{3}\htmlData{tutor-start=18,tutor-end=19}{y}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{3} \htmlData{tutor-start=22,tutor-end=32}{\leqslant }\htmlData{tutor-start=32,tutor-end=33}{0}\htmlData{tutor-start=33,tutor-end=34}{,} \\ \htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{x}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{3}\htmlData{tutor-start=42,tutor-end=43}{y}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{3} \htmlData{tutor-start=46,tutor-end=56}{\geqslant }\htmlData{tutor-start=56,tutor-end=57}{0}\htmlData{tutor-start=57,tutor-end=58}{,} \\ \htmlData{tutor-start=62,tutor-end=63}{y}\htmlData{tutor-start=63,tutor-end=64}{+}\htmlData{tutor-start=64,tutor-end=65}{3} \htmlData{tutor-start=66,tutor-end=76}{\geqslant }\htmlData{tutor-start=76,tutor-end=77}{0}\htmlData{tutor-start=77,tutor-end=78}{,} \end{cases} 则 z=2x+y\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}{y} 的最小值是 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) −15\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{5} (B) −9\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{9} (C) 1 (D) 9

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 数学竞赛/待细分

安排 3 名志愿者完成 4 项工作，每人至少完成 1 项，每项工作由 1 人完成，则不同的安排方式共有 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) 12 种 (B) 18 种 (C) 24 种 (D) 36 种

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 数学竞赛/待细分

甲、乙、丙、丁四位同学一起去向老师询问成语竞赛的成绩。老师说：你们四人中有 2 位优秀，2 位良好，我现在给甲看乙、丙的成绩，给乙看丙的成绩，给丁看甲的成绩。看后甲对大家说：我还是不知道我的成绩。根据以上信息，则 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) 乙可以知道四人的成绩 (B) 丁可以知道四人的成绩 (C) 乙、丁可以知道对方的成绩 (D) 乙、丁可以知道自己的成绩

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

执行如图的程序框图，如果输入的 a=−1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}，则输出的 S=()\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\quad\htmlData{tutor-start=8,tutor-end=9}{)} (A) 2 (B) 3 (C) 4 (D) 5

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 2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 · 平面几何

若双曲线 C:x2a2−y2b2=1 (a>0,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=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} \htmlData{tutor-start=45,tutor-end=47}{\ }\htmlData{tutor-start=47,tutor-end=48}{(}\htmlData{tutor-start=48,tutor-end=49}{a}\htmlData{tutor-start=49,tutor-end=50}{>}\htmlData{tutor-start=50,tutor-end=51}{0}\htmlData{tutor-start=51,tutor-end=52}{,} \htmlData{tutor-start=53,tutor-end=54}{b}\htmlData{tutor-start=54,tutor-end=55}{>}\htmlData{tutor-start=55,tutor-end=56}{0}\htmlData{tutor-start=56,tutor-end=57}{)} 的一条渐近线被圆 (x−2)2+y2=4\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{y}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{4} 所截得的弦长为 2，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的离心率为 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) 2 (B) 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}} (C) 2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}} (D) 233\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{3}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

已知直三棱柱 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}} 中，∠ABC=120∘,AB=2,BC=CC1=1\angle ABC=120^\circ, AB=2, BC=CC_{1}=1，则异面直线 AB1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 与 BC1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 所成角的余弦值为 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) 32\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{2}} (B) 155\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}{5}} (C) 105\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}{5}} (D) 33\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{3}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 代数

若 x=−2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2} 是函数 f(x)=(x2+ax−1)ex−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=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}\mathrm{\htmlData{tutor-start=25,tutor-end=26}{e}}^{\htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{1}} 的极值点，则 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}{)} 的极小值为 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} (B) −2e−3\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{e}}^{\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{3}} (C) 5e−3\htmlData{tutor-start=0,tutor-end=1}{5}\mathrm{\htmlData{tutor-start=9,tutor-end=10}{e}}^{\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{3}} (D) 1

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 平面几何

已知 △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} 是边长为 2 的等边三角形，P\htmlData{tutor-start=0,tutor-end=1}{P} 为平面 ABC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C} 内一点，则 PA→⋅(PB→+PC→)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{P}\htmlData{tutor-start=17,tutor-end=18}{A}} \htmlData{tutor-start=20,tutor-end=26}{\cdot }\htmlData{tutor-start=26,tutor-end=27}{(}\overrightarrow{\htmlData{tutor-start=43,tutor-end=44}{P}\htmlData{tutor-start=44,tutor-end=45}{B}}\htmlData{tutor-start=46,tutor-end=47}{+}\overrightarrow{\htmlData{tutor-start=63,tutor-end=64}{P}\htmlData{tutor-start=64,tutor-end=65}{C}}\htmlData{tutor-start=66,tutor-end=67}{)} 的最小值是 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) −2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2} (B) −32\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{2}} (C) −43\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{4}}{\htmlData{tutor-start=10,tutor-end=11}{3}} (D) −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

一批产品的二等品率为 0.02，从这批产品中每次随机取一件，有放回地抽取 100 次，X\htmlData{tutor-start=0,tutor-end=1}{X} 表示抽到的二等品件数，则 DX=\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{X}\htmlData{tutor-start=2,tutor-end=3}{=} \underline{\hspace{2cm}}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 代数

函数 f(x)=sin⁡2x+3cos⁡x−34(x∈[0,π2])\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\sin^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\sqrt{\htmlData{tutor-start=22,tutor-end=23}{3}}\cos \htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{-}\frac{\htmlData{tutor-start=37,tutor-end=38}{3}}{\htmlData{tutor-start=40,tutor-end=41}{4}} \left(\htmlData{tutor-start=49,tutor-end=50}{x} \htmlData{tutor-start=51,tutor-end=55}{\in }\left[\htmlData{tutor-start=61,tutor-end=62}{0}\htmlData{tutor-start=62,tutor-end=63}{,} \frac{\htmlData{tutor-start=70,tutor-end=73}{\pi}}{\htmlData{tutor-start=75,tutor-end=76}{2}}\right]\right) 的最大值是 \underline{\hspace{2cm}}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 代数

等差数列 {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}{\}} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和为 Sn,a3=3,S4=10\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{S}_{\htmlData{tutor-start=19,tutor-end=20}{4}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{0}，则 ∑k=1n1Sk=\sum_{\htmlData{tutor-start=6,tutor-end=7}{k}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \frac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{S}_{\htmlData{tutor-start=27,tutor-end=28}{k}}}\htmlData{tutor-start=30,tutor-end=31}{=} \underline{\hspace{2cm}}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 数学竞赛/待细分

已知 F\htmlData{tutor-start=0,tutor-end=1}{F} 是抛物线 C:y2=8x\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}{8}\htmlData{tutor-start=10,tutor-end=11}{x} 的焦点，M\htmlData{tutor-start=0,tutor-end=1}{M} 是 C\htmlData{tutor-start=0,tutor-end=1}{C} 上一点，FM\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{M} 的延长线交 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴于点 N\htmlData{tutor-start=0,tutor-end=1}{N}. 若 M\htmlData{tutor-start=0,tutor-end=1}{M} 为 FN\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{N} 的中点，则 ∣FN∣=\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=3}{N}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=} \underline{\hspace{2cm}}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 平面几何

△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} 的内角 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} 的对边分别为 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}，已知 sin⁡(A+C)=8sin⁡2B2\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{8}\sin^{\htmlData{tutor-start=17,tutor-end=18}{2}} \frac{\htmlData{tutor-start=26,tutor-end=27}{B}}{\htmlData{tutor-start=29,tutor-end=30}{2}}. (1) 求 cos⁡B\cos \htmlData{tutor-start=5,tutor-end=6}{B}; (2) 若 a+c=6,△ABC\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}{6}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=17}{\triangle }\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{B}\htmlData{tutor-start=19,tutor-end=20}{C} 面积为 2，求 b\htmlData{tutor-start=0,tutor-end=1}{b}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

海水养殖场进行某水产品的新、旧网箱养殖方法的产量对比，收获时各随机抽取了 100 个网箱，测量各箱水产品的产量（单位：kg），其频率分布直方图如图： (1) 设两种养殖方法的箱产量相互独立，记 A\htmlData{tutor-start=0,tutor-end=1}{A} 表示事件“旧养殖法的箱产量低于 50 kg，新养殖法的箱产量不低于 50 kg”，估计 A\htmlData{tutor-start=0,tutor-end=1}{A} 的概率； (2) 填写下面列联表，并根据列联表判断是否有 99% 的把握认为箱产量与养殖方法有关； \begin{tabular}{|c|c|c|} \hline & 箱产量 <50\htmlData{tutor-start=0,tutor-end=1}{<}\htmlData{tutor-start=1,tutor-end=2}{5}\htmlData{tutor-start=2,tutor-end=3}{0} kg & 箱产量 ⩾50\htmlData{tutor-start=0,tutor-end=10}{\geqslant }\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{0} kg \\ \hline 旧养殖法 & & \\ \hline 新养殖法 & & \\ \hline \end{tabular} (3) 根据箱产量的频率分布直方图，求新养殖法箱产量的中位数的估计值（精确到 0.01）. 附：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}{)} \begin{tabular}{|c|c|c|c|} \hline & 0.050 & 0.010 & 0.001 \\ \hline k\htmlData{tutor-start=0,tutor-end=1}{k} & 3.841 & 6.635 & 10.828 \\ \hline \end{tabular} 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}{)}}.

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 2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 · 平面几何

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，侧面 PAD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{D} 为等边三角形且垂直于底面 ABCD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}，AB=BC=12AD\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}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{D}，∠BAD=∠ABC=90∘\angle BAD = \angle ABC = 90^\circ，E\htmlData{tutor-start=0,tutor-end=1}{E} 是 PD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D} 的中点. (1) 证明：直线 CE//\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{/} 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) 点 M\htmlData{tutor-start=0,tutor-end=1}{M} 在棱 PC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C} 上，且直线 BM\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{M} 与底面 ABCD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D} 所成角为 45∘45^\circ，求二面角 M−AB−D\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{D} 的余弦值.

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 2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

设 O\htmlData{tutor-start=0,tutor-end=1}{O} 为坐标原点，动点 M\htmlData{tutor-start=0,tutor-end=1}{M} 在椭圆 C:x22+y2=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}{2}} \htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{y}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{1} 上，过 M\htmlData{tutor-start=0,tutor-end=1}{M} 作 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的垂线，垂足为 N\htmlData{tutor-start=0,tutor-end=1}{N}，点 P\htmlData{tutor-start=0,tutor-end=1}{P} 满足 NP→=2NM→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{N}\htmlData{tutor-start=17,tutor-end=18}{P}} \htmlData{tutor-start=20,tutor-end=21}{=} \sqrt{\htmlData{tutor-start=28,tutor-end=29}{2}}\overrightarrow{\htmlData{tutor-start=46,tutor-end=47}{N}\htmlData{tutor-start=47,tutor-end=48}{M}}. (1) 求点 P\htmlData{tutor-start=0,tutor-end=1}{P} 的轨迹方程; (2) 设点 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 在直线 x=−3\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{3} 上，且 OP→⋅PQ→=1\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{O}\htmlData{tutor-start=17,tutor-end=18}{P}} \htmlData{tutor-start=20,tutor-end=26}{\cdot }\overrightarrow{\htmlData{tutor-start=42,tutor-end=43}{P}\htmlData{tutor-start=43,tutor-end=44}{Q}} \htmlData{tutor-start=46,tutor-end=47}{=} \htmlData{tutor-start=48,tutor-end=49}{1}. 证明：过点 P\htmlData{tutor-start=0,tutor-end=1}{P} 且垂直于 OQ\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{Q} 的直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 过 C\htmlData{tutor-start=0,tutor-end=1}{C} 的左焦点 F\htmlData{tutor-start=0,tutor-end=1}{F}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

已知函数 f(x)=ax2−ax−xln⁡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}{=} \htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{a}\htmlData{tutor-start=17,tutor-end=18}{x} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{x} \ln \htmlData{tutor-start=27,tutor-end=28}{x}，且 f(x)⩾0\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}{\geqslant }\htmlData{tutor-start=15,tutor-end=16}{0}. (1) 求 a\htmlData{tutor-start=0,tutor-end=1}{a}; (2) 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}{)} 存在唯一的极大值点 x0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}，且 e−2<f(x0)<2−2\htmlData{tutor-start=0,tutor-end=1}{e}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{x}_{\htmlData{tutor-start=14,tutor-end=15}{0}}\htmlData{tutor-start=16,tutor-end=17}{)} \htmlData{tutor-start=18,tutor-end=19}{<} \htmlData{tutor-start=20,tutor-end=21}{2}^{\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{2}}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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，以坐标原点为极点，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 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的极坐标方程为 ρcos⁡θ=4\htmlData{tutor-start=0,tutor-end=5}{\rho }\cos \htmlData{tutor-start=10,tutor-end=17}{\theta }\htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{4}. (1) M\htmlData{tutor-start=0,tutor-end=1}{M} 为曲线 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 上的动点，点 P\htmlData{tutor-start=0,tutor-end=1}{P} 在线段 OM\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M} 上，且满足 ∣OM∣⋅∣OP∣=16\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{M}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=11}{\cdot }\htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{O}\htmlData{tutor-start=13,tutor-end=14}{P}\htmlData{tutor-start=14,tutor-end=15}{|} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{6}，求点 P\htmlData{tutor-start=0,tutor-end=1}{P} 的轨迹 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 A\htmlData{tutor-start=0,tutor-end=1}{A} 的极坐标为 (2,π3)\left(\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{,} \frac{\htmlData{tutor-start=15,tutor-end=18}{\pi}}{\htmlData{tutor-start=20,tutor-end=21}{3}}\right)，点 B\htmlData{tutor-start=0,tutor-end=1}{B} 在曲线 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 上，求 △OAB\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{O}\htmlData{tutor-start=11,tutor-end=12}{A}\htmlData{tutor-start=12,tutor-end=13}{B} 面积的最大值.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

已知 a>0,b>0,a3+b3=2\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}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{b}^{\htmlData{tutor-start=19,tutor-end=20}{3}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{2}，证明： (1) (a+b)(a5+b5)⩾4\htmlData{tutor-start=0,tutor-end=1}{(}\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}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{5}}\htmlData{tutor-start=17,tutor-end=18}{)} \htmlData{tutor-start=19,tutor-end=29}{\geqslant }\htmlData{tutor-start=29,tutor-end=30}{4}; (2) a+b⩽2\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=4,tutor-end=14}{\leqslant }\htmlData{tutor-start=14,tutor-end=15}{2}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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。