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2019 年高考数学（全国卷 2文科）

exams_raw/普通高考/2019/2019全国2文(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 · 组合数学

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

设 z=i(2+i)\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{+}\mathrm{\htmlData{tutor-start=23,tutor-end=24}{i}}\htmlData{tutor-start=25,tutor-end=26}{)}，则 zˉ=\bar{\htmlData{tutor-start=5,tutor-end=6}{z}}\htmlData{tutor-start=7,tutor-end=8}{=} （ ） (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}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{2}\mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}} (C) 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}} (D) −1−2i\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\mathrm{\htmlData{tutor-start=12,tutor-end=13}{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

已知向量 a=(2,3)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{)}，b=(3,2)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{)}，则 ∣a−b∣=\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{-}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}\htmlData{tutor-start=31,tutor-end=32}{=} （ ） (A) 2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}} (B) 2\htmlData{tutor-start=0,tutor-end=1}{2} (C) 52\htmlData{tutor-start=0,tutor-end=1}{5}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}} (D) 50\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{0}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

生物实验室有 5 只兔子，其中只有 3 只测量过某项指标，若从这 5 只兔子中随机取出 3 只，则恰有 2 只测量过该指标的概率为 （ ） (A) 23\dfrac{\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{3}} (B) 35\dfrac{\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{5}} (C) 25\dfrac{\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{5}} (D) 15\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{5}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

在“一带一路”知识测验后，甲、乙、丙三人对成绩进行预测. 甲：我的成绩比乙高. 乙：丙的成绩比我和甲的都高. 丙：我的成绩比乙高. 成绩公布后，三人成绩互不相同且只有一个人预测正确，那么三人按成绩由高到低的次序为 （ ） (A) 甲、乙、丙 (B) 乙、甲、丙 (C) 丙、乙、甲 (D) 甲、丙、乙

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 代数

设 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 为奇函数，且当 x⩾0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{0} 时，f(x)=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}{=}\mathrm{\htmlData{tutor-start=13,tutor-end=14}{e}}^{\htmlData{tutor-start=17,tutor-end=18}{x}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}，则当 x<0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0} 时，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}{)}\htmlData{tutor-start=4,tutor-end=5}{=} （ ） (A) e−x−1\mathrm{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{x}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1} (B) e−x+1\mathrm{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{x}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{1} (C) −e−x−1\htmlData{tutor-start=0,tutor-end=1}{-}\mathrm{\htmlData{tutor-start=9,tutor-end=10}{e}}^{\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{x}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1} (D) −e−x+1\htmlData{tutor-start=0,tutor-end=1}{-}\mathrm{\htmlData{tutor-start=9,tutor-end=10}{e}}^{\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{x}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{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 α,β\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=13}{\beta} 为两个平面，则 α//β\htmlData{tutor-start=0,tutor-end=7}{\alpha }\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{/} \htmlData{tutor-start=10,tutor-end=15}{\beta} 的充要条件是 （ ） (A) α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 内有无数条直线与 β\htmlData{tutor-start=0,tutor-end=5}{\beta} 平行 (B) α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 内有两条相交直线与 β\htmlData{tutor-start=0,tutor-end=5}{\beta} 平行 (C) α,β\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=13}{\beta} 平行于同一条直线 (D) α,β\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=13}{\beta} 垂直于同一平面

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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 · 代数

若 x1=π4\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\dfrac{\htmlData{tutor-start=13,tutor-end=16}{\pi}}{\htmlData{tutor-start=18,tutor-end=19}{4}}，x2=3π4\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\dfrac{\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=17}{\pi}}{\htmlData{tutor-start=19,tutor-end=20}{4}} 是函数 f(x)=sin⁡ωx\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=17}{\omega }\htmlData{tutor-start=17,tutor-end=18}{x} (ω>0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=7}{\omega}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{)} 两个相邻的极值点，则 ω=\htmlData{tutor-start=0,tutor-end=6}{\omega}\htmlData{tutor-start=6,tutor-end=7}{=} （ ） (A) 2\htmlData{tutor-start=0,tutor-end=1}{2} (B) 32\dfrac{\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{2}} (C) 1\htmlData{tutor-start=0,tutor-end=1}{1} (D) 12\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{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文科第 8 题

解题过程

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

9

一、选择题 · 平面几何

若抛物线 y2=2px\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x} (p>0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{>}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)} 的焦点是椭圆 x23p+y2p=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}{3}\htmlData{tutor-start=15,tutor-end=16}{p}}\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}{p}}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{1} 的一个焦点，则 p=\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=} （ ） (A) 2\htmlData{tutor-start=0,tutor-end=1}{2} (B) 3\htmlData{tutor-start=0,tutor-end=1}{3} (C) 4\htmlData{tutor-start=0,tutor-end=1}{4} (D) 8\htmlData{tutor-start=0,tutor-end=1}{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

曲线 y=2sin⁡x+cos⁡x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\sin \htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\cos \htmlData{tutor-start=15,tutor-end=16}{x} 在点 (π,−1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=4}{\pi}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)} 处的切线方程为 （ ） (A) x−y−π−1=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=7}{\pi}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{0} (B) 2x−y−2π−1=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=9}{\pi}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0} (C) 2x+y−2π+1=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=9}{\pi}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0} (D) x+y−π+1=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=7}{\pi}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{=}\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 2文科第 10 题

解题过程

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

11

一、选择题 · 数学竞赛/待细分

已知 α∈(0,π2)\htmlData{tutor-start=0,tutor-end=7}{\alpha }\htmlData{tutor-start=7,tutor-end=11}{\in }\left(\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{,} \dfrac{\htmlData{tutor-start=27,tutor-end=30}{\pi}}{\htmlData{tutor-start=32,tutor-end=33}{2}}\right)，2sin⁡2α=cos⁡2α+1\htmlData{tutor-start=0,tutor-end=1}{2}\sin \htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=13}{\alpha}\htmlData{tutor-start=13,tutor-end=14}{=}\cos \htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=26}{\alpha}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{1}，则 sin⁡α=\sin \htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=} （ ） (A) 15\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{5}} (B) 55\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{5}}}{\htmlData{tutor-start=17,tutor-end=18}{5}} (C) 33\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{3}} (D) 255\dfrac{\htmlData{tutor-start=7,tutor-end=8}{2}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{5}}}{\htmlData{tutor-start=18,tutor-end=19}{5}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2文科第 11 题

解题过程

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

12

二、填空题 · 平面几何

设 F\htmlData{tutor-start=0,tutor-end=1}{F} 为双曲线 C:x2a2−y2b2=1\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \dfrac{\htmlData{tutor-start=10,tutor-end=11}{x}^{\htmlData{tutor-start=13,tutor-end=14}{2}}}{\htmlData{tutor-start=17,tutor-end=18}{a}^{\htmlData{tutor-start=20,tutor-end=21}{2}}}\htmlData{tutor-start=23,tutor-end=24}{-}\dfrac{\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=44,tutor-end=45}{=}\htmlData{tutor-start=45,tutor-end=46}{1} (a>0,b>0)\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}{0}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{)} 的右焦点，O\htmlData{tutor-start=0,tutor-end=1}{O} 为坐标原点，以 OF\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{F} 为直径的圆与圆 x2+y2=a2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{a}^{\htmlData{tutor-start=15,tutor-end=16}{2}} 交于 P,Q\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{Q} 两点. 若 ∣PQ∣=∣OF∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{Q}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{O}\htmlData{tutor-start=7,tutor-end=8}{F}\htmlData{tutor-start=8,tutor-end=9}{|}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的离心率为 （ ） (A) 2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}} (B) 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}} (C) 2\htmlData{tutor-start=0,tutor-end=1}{2} (D) 5\sqrt{\htmlData{tutor-start=6,tutor-end=7}{5}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

若变量 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−6⩾0,x+y−3⩽0,y−2⩽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}{6} \htmlData{tutor-start=22,tutor-end=32}{\geqslant }\htmlData{tutor-start=32,tutor-end=33}{0}\htmlData{tutor-start=33,tutor-end=34}{,} \\ \htmlData{tutor-start=38,tutor-end=39}{x}\htmlData{tutor-start=39,tutor-end=40}{+}\htmlData{tutor-start=40,tutor-end=41}{y}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{3} \htmlData{tutor-start=44,tutor-end=54}{\leqslant }\htmlData{tutor-start=54,tutor-end=55}{0}\htmlData{tutor-start=55,tutor-end=56}{,} \\ \htmlData{tutor-start=60,tutor-end=61}{y}\htmlData{tutor-start=61,tutor-end=62}{-}\htmlData{tutor-start=62,tutor-end=63}{2} \htmlData{tutor-start=64,tutor-end=74}{\leqslant }\htmlData{tutor-start=74,tutor-end=75}{0}\htmlData{tutor-start=75,tutor-end=76}{,} \end{cases} 则 z=3x−y\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}{y} 的最大值是______.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

我国高铁发展迅速，技术先进. 经统计，在经停某站的高铁列车中，有 10 个车次的正点率为 0.97，有 20 个车次的正点率为 0.98，有 10 个车次的正点率为 0.99，则经停该站高铁列车所有车次的平均正点率的估计值为______.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2文科第 14 题

解题过程

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

15

二、填空题 · 平面几何

△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 bsin⁡A+acos⁡B=0\htmlData{tutor-start=0,tutor-end=1}{b}\sin \htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{a}\cos \htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{0}，则 B=\htmlData{tutor-start=0,tutor-end=1}{B}\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 2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

中国有悠久的金石文化，印信是金石文化的代表之一. 印信的形状多为长方体、正方体或圆柱体，但南北朝时期的官员独孤信的印信形状是“半正多面体”（图 1）. 半正多面体是由两种或两种以上的正多边形围成的多面体. 半正多面体体现了数学的对称美. 图 2 是一个棱数为 48 的半正多面体，它的所有顶点都在同一个正方体的表面上，且此正方体的棱长为 1. 则该半正多面体共有______个面，其棱长为______.

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2文科第 16 题

解题过程

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

17

三、解答题 · 数学竞赛/待细分

如图，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}\text{\htmlData{tutor-start=10,tutor-end=11}{-}}\htmlData{tutor-start=12,tutor-end=13}{A}_{\htmlData{tutor-start=15,tutor-end=16}{1}}\htmlData{tutor-start=17,tutor-end=18}{B}_{\htmlData{tutor-start=20,tutor-end=21}{1}}\htmlData{tutor-start=22,tutor-end=23}{C}_{\htmlData{tutor-start=25,tutor-end=26}{1}}\htmlData{tutor-start=27,tutor-end=28}{D}_{\htmlData{tutor-start=30,tutor-end=31}{1}} 的底面 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} 是正方形，点 E\htmlData{tutor-start=0,tutor-end=1}{E} 在棱 AA1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 上，BE⊥EC1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{E}\htmlData{tutor-start=10,tutor-end=11}{C}_{\htmlData{tutor-start=13,tutor-end=14}{1}}. (1) 证明：BE⊥\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=8}{\perp} 平面 EB1C1\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{C}_{\htmlData{tutor-start=9,tutor-end=10}{1}}； (2) 若 AE=A1E\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{A}_{\htmlData{tutor-start=6,tutor-end=7}{1}}\htmlData{tutor-start=8,tutor-end=9}{E}，AB=3\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{3}，求四棱锥 E-BB1C1C\htmlData{tutor-start=0,tutor-end=1}{E}\text{\htmlData{tutor-start=7,tutor-end=8}{-}}\htmlData{tutor-start=9,tutor-end=10}{B}\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}{C} 的体积.

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 · 代数

已知 {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}{\}} 是各项均为正数的等比数列，a1=2\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}{2}，a3=2a2+16\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{6}. (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) 设 bn=log⁡2an\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\log_{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{n}}，求数列 {bn}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{b}_{\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.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2文科第 18 题

解题过程

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

19

三、解答题 · 数学竞赛/待细分

某行业主管部门为了了解本行业中小企业的生产情况，随机调查了 100 个企业，得到这些企业第一季度相对于前一年第一季度产值增长率 y\htmlData{tutor-start=0,tutor-end=1}{y} 的频数分布表. | y\htmlData{tutor-start=0,tutor-end=1}{y} 的分组 | [−0.20,0)\htmlData{tutor-start=0,tutor-end=1}{[}\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=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{)} | [0,0.20)\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}{0}\htmlData{tutor-start=5,tutor-end=6}{.}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)} | [0.20,0.40)\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=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{.}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)} | [0.40,0.60)\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=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{.}\htmlData{tutor-start=9,tutor-end=10}{6}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)} | [0.60,0.80)\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=3,tutor-end=4}{6}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{.}\htmlData{tutor-start=9,tutor-end=10}{8}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)} | | :--- | :---: | :---: | :---: | :---: | :---: | | 企业数 | 2 | 24 | 53 | 14 | 7 | (1) 分别估计这类企业中产值增长率不低于 40% 的企业比例、产值负增长的企业比例； (2) 求这类企业产值增长率的平均数与标准差的估计值（同一组中的数据用该组区间的中点值为代表）.（精确到 0.01） 附：74≈8.602\sqrt{\htmlData{tutor-start=6,tutor-end=7}{7}\htmlData{tutor-start=7,tutor-end=8}{4}} \htmlData{tutor-start=10,tutor-end=18}{\approx }\htmlData{tutor-start=18,tutor-end=19}{8}\htmlData{tutor-start=19,tutor-end=20}{.}\htmlData{tutor-start=20,tutor-end=21}{6}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=23}{2}.

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文科第 19 题

解题过程

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

20

三、解答题 · 平面几何

已知 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}} 是椭圆 C:x2a2+y2b2=1 (a>b>0)\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \frac{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}} \htmlData{tutor-start=23,tutor-end=24}{+} \frac{\htmlData{tutor-start=31,tutor-end=32}{y}^{\htmlData{tutor-start=34,tutor-end=35}{2}}}{\htmlData{tutor-start=38,tutor-end=39}{b}^{\htmlData{tutor-start=41,tutor-end=42}{2}}} \htmlData{tutor-start=45,tutor-end=46}{=} \htmlData{tutor-start=47,tutor-end=48}{1} \htmlData{tutor-start=49,tutor-end=51}{\ }\htmlData{tutor-start=51,tutor-end=52}{(}\htmlData{tutor-start=52,tutor-end=53}{a} \htmlData{tutor-start=54,tutor-end=55}{>} \htmlData{tutor-start=56,tutor-end=57}{b} \htmlData{tutor-start=58,tutor-end=59}{>} \htmlData{tutor-start=60,tutor-end=61}{0}\htmlData{tutor-start=61,tutor-end=62}{)} 的两个焦点，P\htmlData{tutor-start=0,tutor-end=1}{P} 为 C\htmlData{tutor-start=0,tutor-end=1}{C} 上一点，O\htmlData{tutor-start=0,tutor-end=1}{O} 为坐标原点. (1) 若 △POF2\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{O}\htmlData{tutor-start=12,tutor-end=13}{F}_{\htmlData{tutor-start=15,tutor-end=16}{2}} 为等边三角形，求 C\htmlData{tutor-start=0,tutor-end=1}{C} 的离心率； (2) 如果存在点 P\htmlData{tutor-start=0,tutor-end=1}{P}，使得 PF1⊥PF2\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=13}{\perp }\htmlData{tutor-start=13,tutor-end=14}{P}\htmlData{tutor-start=14,tutor-end=15}{F}_{\htmlData{tutor-start=17,tutor-end=18}{2}}，且 △F1PF2\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{F}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{P}\htmlData{tutor-start=16,tutor-end=17}{F}_{\htmlData{tutor-start=19,tutor-end=20}{2}} 的面积等于 16，求 b\htmlData{tutor-start=0,tutor-end=1}{b} 的值和 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

三、解答题 · 代数

已知函数 f(x)=(x−1)ln⁡x−x−1\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \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}{)}\ln \htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{x} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{1}. 证明： (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}{)} 存在唯一的极值点； (2) 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=6}{=} \htmlData{tutor-start=7,tutor-end=8}{0} 有且仅有两个实根，且两个实根互为倒数.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

三、解答题 · 平面几何

在极坐标系中，O\htmlData{tutor-start=0,tutor-end=1}{O} 为极点，点 M(ρ0,θ0) (ρ0>0)\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=6}{\rho}_{\htmlData{tutor-start=8,tutor-end=9}{0}}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=18}{\theta}_{\htmlData{tutor-start=20,tutor-end=21}{0}}\htmlData{tutor-start=22,tutor-end=23}{)} \htmlData{tutor-start=24,tutor-end=26}{\ }\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=31}{\rho}_{\htmlData{tutor-start=33,tutor-end=34}{0}} \htmlData{tutor-start=36,tutor-end=37}{>} \htmlData{tutor-start=38,tutor-end=39}{0}\htmlData{tutor-start=39,tutor-end=40}{)} 在曲线 C:ρ=4sin⁡θ\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=8}{\rho }\htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{4}\sin\htmlData{tutor-start=15,tutor-end=21}{\theta} 上，直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 过点 A(4,0)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)} 且与 OM\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M} 垂直，垂足为 P\htmlData{tutor-start=0,tutor-end=1}{P}. (1) 当 θ0=π3\htmlData{tutor-start=0,tutor-end=6}{\theta}_{\htmlData{tutor-start=8,tutor-end=9}{0}} \htmlData{tutor-start=11,tutor-end=12}{=} \frac{\htmlData{tutor-start=19,tutor-end=22}{\pi}}{\htmlData{tutor-start=24,tutor-end=25}{3}} 时，求 ρ0\htmlData{tutor-start=0,tutor-end=4}{\rho}_{\htmlData{tutor-start=6,tutor-end=7}{0}} 及 l\htmlData{tutor-start=0,tutor-end=1}{l} 的极坐标方程； (2) 当 M\htmlData{tutor-start=0,tutor-end=1}{M} 在 C\htmlData{tutor-start=0,tutor-end=1}{C} 上运动且 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} 上时，求 P\htmlData{tutor-start=0,tutor-end=1}{P} 点轨迹的极坐标方程.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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

三、解答题 · 代数

已知 f(x)=∣x−a∣x+∣x−2∣(x−a)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{x} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{|}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{|}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{x}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{a}\htmlData{tutor-start=25,tutor-end=26}{)}. (1) 当 a=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 时，求不等式 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=6}{<} \htmlData{tutor-start=7,tutor-end=8}{0} 的解集； (2) 若 x∈(−∞,1]\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=14}{\infty}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{]} 时，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=6}{<} \htmlData{tutor-start=7,tutor-end=8}{0}，求 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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。