返回特征解读

2018 年高考数学（全国卷 3文科）

exams_raw/普通高考/2018/2018全国3文(云南,广西,贵州,西藏,四川).pdf · HS-MATH-1024-v2.1-solution-aware

23 题0 个小问/题组
1

一、选择题 · 组合数学

已知集合 A={x∣x−1⩾0}\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=13}{\mid }\htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=16}{-} \htmlData{tutor-start=17,tutor-end=18}{1} \htmlData{tutor-start=19,tutor-end=29}{\geqslant }\htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=32}{\}}, B={0,1,2}\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}{0}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=15}{\}}, 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=9,tutor-end=10}{=} ( ) (A) {0}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=5}{\}} (B) {1}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=5}{\}} (C) {1,2}\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}{2}\htmlData{tutor-start=6,tutor-end=8}{\}} (D) {0,1,2}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=11}{\}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 1 题

解题过程

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

2

一、选择题 · 数学竞赛/待细分

(1+i)(2−i)=\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{+} \mathrm{\htmlData{tutor-start=13,tutor-end=14}{i}}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{2} \htmlData{tutor-start=19,tutor-end=20}{-} \mathrm{\htmlData{tutor-start=29,tutor-end=30}{i}}\htmlData{tutor-start=31,tutor-end=32}{)} \htmlData{tutor-start=33,tutor-end=34}{=} ( ) (A) −3−i\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3} \htmlData{tutor-start=3,tutor-end=4}{-} \mathrm{\htmlData{tutor-start=13,tutor-end=14}{i}} (B) −3+i\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3} \htmlData{tutor-start=3,tutor-end=4}{+} \mathrm{\htmlData{tutor-start=13,tutor-end=14}{i}} (C) 3−i\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=3}{-} \mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}} (D) 3+i\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=3}{+} \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 3文科第 2 题

解题过程

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

3

一、选择题 · 数学竞赛/待细分

中国古建筑借助榫卯将木构件连接起来, 构件的凸出部分叫榫头, 凹进部分叫卯眼, 图中木构件右边的小长方体是榫头. 若如图摆放的木构件与某一带卯眼的木构件咬合成长方体, 则咬合时带卯眼的木构件的俯视图可以是 ( ) (A) (B) (C) (D)

原卷图示 1
原卷图示 1原卷第 1 页 · qwen3.7-plus_layout_detection · 需复核

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 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

若 sin⁡α=13\sin \htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{=} \dfrac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{3}}, 则 cos⁡2α=\cos \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=13}{\alpha }\htmlData{tutor-start=13,tutor-end=14}{=} ( ) (A) 89\dfrac{\htmlData{tutor-start=7,tutor-end=8}{8}}{\htmlData{tutor-start=10,tutor-end=11}{9}} (B) 79\dfrac{\htmlData{tutor-start=7,tutor-end=8}{7}}{\htmlData{tutor-start=10,tutor-end=11}{9}} (C) −79\htmlData{tutor-start=0,tutor-end=1}{-}\dfrac{\htmlData{tutor-start=8,tutor-end=9}{7}}{\htmlData{tutor-start=11,tutor-end=12}{9}} (D) −89\htmlData{tutor-start=0,tutor-end=1}{-}\dfrac{\htmlData{tutor-start=8,tutor-end=9}{8}}{\htmlData{tutor-start=11,tutor-end=12}{9}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 4 题

解题过程

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

5

一、选择题 · 数学竞赛/待细分

若某群体中的成员只用现金支付的概率为 0.45\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{5}, 既用现金支付也用非现金支付的概率为 0.15\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{5}, 则不用现金支付的概率为 ( ) (A) 0.3\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{3} (B) 0.4\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{4} (C) 0.6\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{6} (D) 0.7\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{7}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 5 题

解题过程

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

6

一、选择题 · 代数

函数 f(x)=tan⁡x1+tan⁡2x\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}{=} \dfrac{\tan \htmlData{tutor-start=19,tutor-end=20}{x}}{\htmlData{tutor-start=22,tutor-end=23}{1} \htmlData{tutor-start=24,tutor-end=25}{+} \tan^{\htmlData{tutor-start=32,tutor-end=33}{2}} \htmlData{tutor-start=35,tutor-end=36}{x}} 的最小正周期为 ( ) (A) π4\dfrac{\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{4}} (B) π2\dfrac{\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{2}} (C) π\htmlData{tutor-start=0,tutor-end=3}{\pi} (D) 2π\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\pi}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 6 题

解题过程

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

7

一、选择题 · 代数

下列函数中, 其图象与函数 y=ln⁡x\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \ln \htmlData{tutor-start=8,tutor-end=9}{x} 的图象关于直线 x=1\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1} 对称的是 ( ) (A) y=ln⁡(1−x)\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \ln\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{1} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)} (B) y=ln⁡(2−x)\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \ln\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)} (C) y=ln⁡(1+x)\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \ln\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{1} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)} (D) y=ln⁡(2+x)\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \ln\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 7 题

解题过程

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

8

一、选择题 · 平面几何

直线 x+y+2=0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{y} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{2} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{0} 分别与 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴, y\htmlData{tutor-start=0,tutor-end=1}{y} 轴交于 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B} 两点, 点 P\htmlData{tutor-start=0,tutor-end=1}{P} 在圆 (x−2)2+y2=2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=4}{-} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{)}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{y}^{\htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{2} 上, 则 △ABP\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}{P} 面积的取值范围是 ( ) (A) [2,6]\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}{6}\htmlData{tutor-start=5,tutor-end=6}{]} (B) [4,8]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{8}\htmlData{tutor-start=5,tutor-end=6}{]} (C) [2,32]\htmlData{tutor-start=0,tutor-end=1}{[}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{3}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{]} (D) [22,32]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{2}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{3}\sqrt{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{]}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 8 题

解题过程

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

9

一、选择题 · 代数

函数 y=−x4+x2+2\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}^{\htmlData{tutor-start=8,tutor-end=9}{4}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{2} 的图象大致为 ( ) (A) (B) (C) (D)

原卷图示 1
原卷图示 1原卷第 1 页 · qwen3.7-plus_layout_detection · 需复核

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 9 题

解题过程

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

10

一、选择题 · 数学竞赛/待细分

已知双曲线 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=24,tutor-end=25}{-} \dfrac{\htmlData{tutor-start=33,tutor-end=34}{y}^{\htmlData{tutor-start=36,tutor-end=37}{2}}}{\htmlData{tutor-start=40,tutor-end=41}{b}^{\htmlData{tutor-start=43,tutor-end=44}{2}}} \htmlData{tutor-start=47,tutor-end=48}{=} \htmlData{tutor-start=49,tutor-end=50}{1} (a>0,b>0\htmlData{tutor-start=0,tutor-end=1}{a} \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=7,tutor-end=8}{b} \htmlData{tutor-start=9,tutor-end=10}{>} \htmlData{tutor-start=11,tutor-end=12}{0}) 的离心率为 2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}, 则点 (4,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)} 到 C\htmlData{tutor-start=0,tutor-end=1}{C} 的渐近线的距离为 ( ) (A) 2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}} (B) 2\htmlData{tutor-start=0,tutor-end=1}{2} (C) 322\dfrac{\htmlData{tutor-start=7,tutor-end=8}{3}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{2}}}{\htmlData{tutor-start=18,tutor-end=19}{2}} (D) 22\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 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

△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}, 若 △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} 的面积为 a2+b2−c24\dfrac{\htmlData{tutor-start=7,tutor-end=8}{a}^{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{b}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{c}^{\htmlData{tutor-start=26,tutor-end=27}{2}}}{\htmlData{tutor-start=30,tutor-end=31}{4}}, 则 C=\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{=} ( ) (A) π2\dfrac{\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{2}} (B) π3\dfrac{\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{3}} (C) π4\dfrac{\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{4}} (D) π6\dfrac{\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{6}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 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

设 A,B,C,D\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{D} 是同一个半径为 4\htmlData{tutor-start=0,tutor-end=1}{4} 的球的球面上四点, △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} 为等边三角形且其面积为 93\htmlData{tutor-start=0,tutor-end=1}{9}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{3}}, 则三棱锥 D−ABC\htmlData{tutor-start=0,tutor-end=1}{D} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{A}\htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{C} 体积的最大值为 ( ) (A) 123\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{2}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}} (B) 183\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{8}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}} (C) 243\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{4}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}} (D) 543\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{4}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 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

已知向量 a=(1,2)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=16}{=} \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}{2}\htmlData{tutor-start=22,tutor-end=23}{)}, b=(2,−2)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}} \htmlData{tutor-start=15,tutor-end=16}{=} \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}{-}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{)}, c=(1,λ)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{c}} \htmlData{tutor-start=15,tutor-end=16}{=} \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=28}{\lambda}\htmlData{tutor-start=28,tutor-end=29}{)}. 若 c∥(2a+b)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{c}} \htmlData{tutor-start=15,tutor-end=25}{\parallel }\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{2}\boldsymbol{\htmlData{tutor-start=39,tutor-end=40}{a}} \htmlData{tutor-start=42,tutor-end=43}{+} \boldsymbol{\htmlData{tutor-start=56,tutor-end=57}{b}}\htmlData{tutor-start=58,tutor-end=59}{)}, 则 λ=\htmlData{tutor-start=0,tutor-end=8}{\lambda }\htmlData{tutor-start=8,tutor-end=9}{=} ______.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 13 题

解题过程

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

14

二、填空题 · 数学竞赛/待细分

某公司有大量客户, 且不同龄段客户对其服务的评价有较大差异. 为了解客户的评价, 该公司准备进行抽样调查, 可供选择的抽样方法有简单随机抽样、分层抽样和系统抽样, 则最合适的抽样方法是______.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 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 · 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+y+3⩾0,x−2y+4⩾0,x−2⩽0,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{x} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{y} \htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{3} \htmlData{tutor-start=25,tutor-end=35}{\geqslant }\htmlData{tutor-start=35,tutor-end=36}{0}\htmlData{tutor-start=36,tutor-end=37}{,} \\ \htmlData{tutor-start=41,tutor-end=42}{x} \htmlData{tutor-start=43,tutor-end=44}{-} \htmlData{tutor-start=45,tutor-end=46}{2}\htmlData{tutor-start=46,tutor-end=47}{y} \htmlData{tutor-start=48,tutor-end=49}{+} \htmlData{tutor-start=50,tutor-end=51}{4} \htmlData{tutor-start=52,tutor-end=62}{\geqslant }\htmlData{tutor-start=62,tutor-end=63}{0}\htmlData{tutor-start=63,tutor-end=64}{,} \\ \htmlData{tutor-start=68,tutor-end=69}{x} \htmlData{tutor-start=70,tutor-end=71}{-} \htmlData{tutor-start=72,tutor-end=73}{2} \htmlData{tutor-start=74,tutor-end=84}{\leqslant }\htmlData{tutor-start=84,tutor-end=85}{0}\htmlData{tutor-start=85,tutor-end=86}{,} \end{cases} 则 z=x+13y\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{+} \dfrac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{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 3文科第 15 题

解题过程

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

16

二、填空题 · 代数

已知函数 f(x)=ln⁡(1+x2−x)+1\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \ln\htmlData{tutor-start=10,tutor-end=11}{(}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{1} \htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{x}^{\htmlData{tutor-start=24,tutor-end=25}{2}}} \htmlData{tutor-start=28,tutor-end=29}{-} \htmlData{tutor-start=30,tutor-end=31}{x}\htmlData{tutor-start=31,tutor-end=32}{)} \htmlData{tutor-start=33,tutor-end=34}{+} \htmlData{tutor-start=35,tutor-end=36}{1}, f(a)=4\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{4}, 则 f(−a)=\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} ______.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 16 题

解题过程

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

17

三、解答题 · 代数

等比数列 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 中, a1=1\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}{1}, a5=4a3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{5}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{3}}. (1) 求 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 的通项公式; (2) 记 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}} 为 {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. 若 Sm=63\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{m}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=10}{3}, 求 m\htmlData{tutor-start=0,tutor-end=1}{m}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 17 题

解题过程

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

18

三、解答题 · 数学竞赛/待细分

某工厂为提高生产效率, 开展技术创新活动, 提出了完成某项生产任务的两种新的生产方式. 为比较两种生产方式的效率, 选取 40\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{0} 名工人, 将他们随机分成两组, 每组 20\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0} 人, 第一组工人用第一种生产方式, 第二组工人用第二种生产方式. 根据工人完成生产任务的工作时间 (单位: min\mathrm{\htmlData{tutor-start=8,tutor-end=9}{m}\htmlData{tutor-start=9,tutor-end=10}{i}\htmlData{tutor-start=10,tutor-end=11}{n}}) 绘制了如图茎叶图: (1) 根据茎叶图判断哪种生产方式的效率更高? 并说明理由; (2) 求 40\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{0} 名工人完成生产任务所需时间的中位数 m\htmlData{tutor-start=0,tutor-end=1}{m}, 并将完成生产任务所需时间超过 m\htmlData{tutor-start=0,tutor-end=1}{m} 和不超过 m\htmlData{tutor-start=0,tutor-end=1}{m} 的工人数填入下面的列联表: | | 超过 m\htmlData{tutor-start=0,tutor-end=1}{m} | 不超过 m\htmlData{tutor-start=0,tutor-end=1}{m} | |---|---|---| | 第一种生产方式 | | | | 第二种生产方式 | | | (3) 根据 (2) 中的列联表, 能否有 99%\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{9}\htmlData{tutor-start=2,tutor-end=4}{\%} 的把握认为两种生产方式的效率有差异? 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}{=} \dfrac{\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{d} \htmlData{tutor-start=20,tutor-end=21}{-} \htmlData{tutor-start=22,tutor-end=23}{b}\htmlData{tutor-start=23,tutor-end=24}{c}\htmlData{tutor-start=24,tutor-end=25}{)}^{\htmlData{tutor-start=27,tutor-end=28}{2}}}{\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{a} \htmlData{tutor-start=34,tutor-end=35}{+} \htmlData{tutor-start=36,tutor-end=37}{b}\htmlData{tutor-start=37,tutor-end=38}{)}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{c} \htmlData{tutor-start=41,tutor-end=42}{+} \htmlData{tutor-start=43,tutor-end=44}{d}\htmlData{tutor-start=44,tutor-end=45}{)}\htmlData{tutor-start=45,tutor-end=46}{(}\htmlData{tutor-start=46,tutor-end=47}{a} \htmlData{tutor-start=48,tutor-end=49}{+} \htmlData{tutor-start=50,tutor-end=51}{c}\htmlData{tutor-start=51,tutor-end=52}{)}\htmlData{tutor-start=52,tutor-end=53}{(}\htmlData{tutor-start=53,tutor-end=54}{b} \htmlData{tutor-start=55,tutor-end=56}{+} \htmlData{tutor-start=57,tutor-end=58}{d}\htmlData{tutor-start=58,tutor-end=59}{)}}, | 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\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{0} | 0.010\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0} | 0.001\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{1} | |---|---|---|---| | k\htmlData{tutor-start=0,tutor-end=1}{k} | 3.841\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{8}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{1} | 6.635\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{6}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{5} | 10.828\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{.}\htmlData{tutor-start=3,tutor-end=4}{8}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{8} |

原卷图示 1
原卷图示 1原卷第 1 页 · qwen3.7-plus_layout_detection · 需复核

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 18 题

解题过程

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

19

解答题 · 平面几何

如图，矩形 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} 所在平面与半圆弧 CD^\widehat{\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{D}} 所在平面垂直，M\htmlData{tutor-start=0,tutor-end=1}{M} 是 CD^\widehat{\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{D}} 上异于 C,D\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{D} 的点. (1) 证明：平面 AMD⊥\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{D} \htmlData{tutor-start=4,tutor-end=9}{\perp} 平面 BMC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{C}； (2) 在线段 AM\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M} 上是否存在点 P\htmlData{tutor-start=0,tutor-end=1}{P}，使得 MC//\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{/} 平面 PBD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{D}？说明理由.

原卷图示 1
原卷图示 1原卷第 2 页 · qwen3.7-plus_layout_detection · 需复核

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 19 题

解题过程

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

20

解答题 · 平面几何

已知斜率为 k\htmlData{tutor-start=0,tutor-end=1}{k} 的直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与椭圆 C:x24+y23=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}{4}} \htmlData{tutor-start=19,tutor-end=20}{+} \frac{\htmlData{tutor-start=27,tutor-end=28}{y}^{\htmlData{tutor-start=30,tutor-end=31}{2}}}{\htmlData{tutor-start=34,tutor-end=35}{3}} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{1} 交于 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B} 两点. 线段 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 的中点为 M(1,m)\htmlData{tutor-start=0,tutor-end=1}{M}\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}{m}\htmlData{tutor-start=6,tutor-end=7}{)} (m>0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{m} \htmlData{tutor-start=3,tutor-end=4}{>} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}. (1) 证明：k<−12\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{-}\frac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{2}}； (2) 设 F\htmlData{tutor-start=0,tutor-end=1}{F} 为 C\htmlData{tutor-start=0,tutor-end=1}{C} 的右焦点，P\htmlData{tutor-start=0,tutor-end=1}{P} 为 C\htmlData{tutor-start=0,tutor-end=1}{C} 上一点，且 FP→+FA→+FB→=0→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{F}\htmlData{tutor-start=17,tutor-end=18}{P}} \htmlData{tutor-start=20,tutor-end=21}{+} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{F}\htmlData{tutor-start=39,tutor-end=40}{A}} \htmlData{tutor-start=42,tutor-end=43}{+} \overrightarrow{\htmlData{tutor-start=60,tutor-end=61}{F}\htmlData{tutor-start=61,tutor-end=62}{B}} \htmlData{tutor-start=64,tutor-end=65}{=} \overrightarrow{\htmlData{tutor-start=82,tutor-end=83}{0}}. 证明：2∣FP→∣=∣FA→∣+∣FB→∣\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{|}\overrightarrow{\htmlData{tutor-start=18,tutor-end=19}{F}\htmlData{tutor-start=19,tutor-end=20}{P}}\htmlData{tutor-start=21,tutor-end=22}{|} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{|}\overrightarrow{\htmlData{tutor-start=42,tutor-end=43}{F}\htmlData{tutor-start=43,tutor-end=44}{A}}\htmlData{tutor-start=45,tutor-end=46}{|} \htmlData{tutor-start=47,tutor-end=48}{+} \htmlData{tutor-start=49,tutor-end=50}{|}\overrightarrow{\htmlData{tutor-start=66,tutor-end=67}{F}\htmlData{tutor-start=67,tutor-end=68}{B}}\htmlData{tutor-start=69,tutor-end=70}{|}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 20 题

解题过程

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

21

解答题 · 代数

已知函数 f(x)=ax2+x−1ex\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}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{x}^{\htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{1}}{\mathrm{\htmlData{tutor-start=37,tutor-end=38}{e}}^{\htmlData{tutor-start=41,tutor-end=42}{x}}}. (1) 求曲线 y=f(x)\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)} 在点 (0,−1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)} 处的切线方程； (2) 证明：当 a⩾1\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{1} 时，f(x)+e⩾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}{+} \mathrm{\htmlData{tutor-start=15,tutor-end=16}{e}} \htmlData{tutor-start=18,tutor-end=28}{\geqslant }\htmlData{tutor-start=28,tutor-end=29}{0}.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 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

在平面直角坐标系 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，⊙O\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O} 的参数方程为 {x=cos⁡θ,y=sin⁡θ,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{=} \cos \htmlData{tutor-start=23,tutor-end=29}{\theta}\htmlData{tutor-start=29,tutor-end=30}{,} \\ \htmlData{tutor-start=34,tutor-end=35}{y} \htmlData{tutor-start=36,tutor-end=37}{=} \sin \htmlData{tutor-start=43,tutor-end=49}{\theta}\htmlData{tutor-start=49,tutor-end=50}{,} \end{cases} (θ\htmlData{tutor-start=0,tutor-end=6}{\theta} 为参数)，过点 (0,−2)\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}{-}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{)} 且倾斜角为 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 的直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与 ⊙O\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O} 交于 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B} 两点. (1) 求 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 的取值范围； (2) 求 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 中点 P\htmlData{tutor-start=0,tutor-end=1}{P} 的轨迹的参数方程.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3文科第 22 题

解题过程

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

23

解答题 · 代数

设函数 f(x)=∣2x+1∣+∣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}{2}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{|} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{|}\htmlData{tutor-start=19,tutor-end=20}{x} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{|}. (1) 画出 y=f(x)\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)} 的图象； (2) 当 x∈[0,+∞)\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{[}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=17}{\infty}\htmlData{tutor-start=17,tutor-end=18}{)} 时，f(x)⩽ax+b\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=15}{\leqslant }\htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{b}，求 a+b\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{b} 的最小值.

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。