返回特征解读

2020 年高考数学（全国卷 2文科）

exams_raw/普通高考/2020/2020全国2文(甘肃,青海,内蒙古,黑龙江,吉林,辽宁.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∣<3,x∈Z}\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}{|}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{<} \htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=28}{\in }\mathbf{\htmlData{tutor-start=36,tutor-end=37}{Z}}\htmlData{tutor-start=38,tutor-end=40}{\}}, B={x∣∣x∣>1,x∈Z}\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}{x} \htmlData{tutor-start=8,tutor-end=13}{\mid }\htmlData{tutor-start=13,tutor-end=14}{|}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{>} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=28}{\in }\mathbf{\htmlData{tutor-start=36,tutor-end=37}{Z}}\htmlData{tutor-start=38,tutor-end=40}{\}}, 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) ∅\htmlData{tutor-start=0,tutor-end=11}{\varnothing} (B) {−3,−2,2,3}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=16}{\}} (C) {−2,0,2}\htmlData{tutor-start=0,tutor-end=2}{\{}\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=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=12}{\}} (D) {−2,2}\htmlData{tutor-start=0,tutor-end=2}{\{}\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=6,tutor-end=7}{2}\htmlData{tutor-start=7,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 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

(1−i)4=\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}^{\htmlData{tutor-start=16,tutor-end=17}{4}} \htmlData{tutor-start=19,tutor-end=20}{=} ( ) (A) −4\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{4} (B) 4\htmlData{tutor-start=0,tutor-end=1}{4} (C) −4i\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{4}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}} (D) 4i\htmlData{tutor-start=0,tutor-end=1}{4}\mathrm{\htmlData{tutor-start=9,tutor-end=10}{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

如图，将钢琴上的12个键依次记为 a1,a2,⋯,a12\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=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \cdots\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{a}_{\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{2}}. 设 1⩽i<j<k⩽12\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{i} \htmlData{tutor-start=14,tutor-end=15}{<} \htmlData{tutor-start=16,tutor-end=17}{j} \htmlData{tutor-start=18,tutor-end=19}{<} \htmlData{tutor-start=20,tutor-end=21}{k} \htmlData{tutor-start=22,tutor-end=32}{\leqslant }\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{2}. 若 k−j=3\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{j}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3} 且 j−i=4\htmlData{tutor-start=0,tutor-end=1}{j}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{i}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{4}, 则称 ai,aj,ak\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{j}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{k}} 为原位大三和弦；若 k−j=4\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{j}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{4} 且 j−i=3\htmlData{tutor-start=0,tutor-end=1}{j}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{i}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3}, 则称 ai,aj,ak\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{j}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{k}} 为原位小三和弦. 用这12个键可以构成的原位大三和弦与原位小三和弦的个数之和为 ( ) (A) 5 (B) 8 (C) 10 (D) 15

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 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

在新冠肺炎疫情防控期间，某超市开通网上销售业务，每天能完成1200份订单的配货，由于订单量大幅增加，导致订单积压. 为解决困难，许多志愿者踊跃报名参加配货工作. 已知该超市某日积压500份订单未配货，预计第二天的新订单超过1600份的概率为0.05，志愿者每人每天能完成50份订单的配货，为使第二天完成积压订单及当日订单的配货的概率不小于0.95，则至少需要志愿者 ( ) (A) 10名 (B) 18名 (C) 24名 (D) 32名

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

已知单位向量 a,b\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{,} \boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}} 的夹角为 60∘60^\circ, 则在下列向量中，与 b\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}} 垂直的是 ( ) (A) a+2b\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{2}\boldsymbol{\htmlData{tutor-start=30,tutor-end=31}{b}} (B) 2a+b\htmlData{tutor-start=0,tutor-end=1}{2}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}} \htmlData{tutor-start=16,tutor-end=17}{+} \boldsymbol{\htmlData{tutor-start=30,tutor-end=31}{b}} (C) a−2b\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=16}{-} \htmlData{tutor-start=17,tutor-end=18}{2}\boldsymbol{\htmlData{tutor-start=30,tutor-end=31}{b}} (D) 2a−b\htmlData{tutor-start=0,tutor-end=1}{2}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}} \htmlData{tutor-start=16,tutor-end=17}{-} \boldsymbol{\htmlData{tutor-start=30,tutor-end=31}{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

记 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}} 为等比数列 {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}{\}} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和. 若 a5−a3=12\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}{a}_{\htmlData{tutor-start=11,tutor-end=12}{3}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{2}, a6−a4=24\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{6}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{4}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{4}, 则 Snan=\frac{\htmlData{tutor-start=6,tutor-end=7}{S}_{\htmlData{tutor-start=9,tutor-end=10}{n}}}{\htmlData{tutor-start=13,tutor-end=14}{a}_{\htmlData{tutor-start=16,tutor-end=17}{n}}} \htmlData{tutor-start=20,tutor-end=21}{=} ( ) (A) 2n−1\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{1} (B) 2−21−n\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{2}^{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{n}} (C) 2−2n−1\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{2}^{\htmlData{tutor-start=7,tutor-end=8}{n}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}} (D) 21−n−1\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{n}} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{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 · 数学竞赛/待细分

执行下面的程序框图，若输入的 k=0,a=0\htmlData{tutor-start=0,tutor-end=1}{k}\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}{a}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{0}, 则输出的 k\htmlData{tutor-start=0,tutor-end=1}{k} 为 ( ) (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

若过点 (2,1)\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}{1}\htmlData{tutor-start=5,tutor-end=6}{)} 的圆与两坐标轴都相切，则圆心到直线 2x−y−3=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=4}{-} \htmlData{tutor-start=5,tutor-end=6}{y} \htmlData{tutor-start=7,tutor-end=8}{-} \htmlData{tutor-start=9,tutor-end=10}{3} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{0} 的距离为 ( ) (A) 55\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{5}}}{\htmlData{tutor-start=16,tutor-end=17}{5}} (B) 255\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{5}}}{\htmlData{tutor-start=17,tutor-end=18}{5}} (C) 355\frac{\htmlData{tutor-start=6,tutor-end=7}{3}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{5}}}{\htmlData{tutor-start=17,tutor-end=18}{5}} (D) 455\frac{\htmlData{tutor-start=6,tutor-end=7}{4}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{5}}}{\htmlData{tutor-start=17,tutor-end=18}{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 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

设 O\htmlData{tutor-start=0,tutor-end=1}{O} 为坐标原点，直线 x=a\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a} 与双曲线 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=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=53,tutor-end=54}{>}\htmlData{tutor-start=54,tutor-end=55}{0}\htmlData{tutor-start=55,tutor-end=56}{,} \htmlData{tutor-start=57,tutor-end=58}{b}\htmlData{tutor-start=58,tutor-end=59}{>}\htmlData{tutor-start=59,tutor-end=60}{0}\htmlData{tutor-start=60,tutor-end=61}{)} 的两条渐近线分别交于 D,E\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{E} 两点. 若 △ODE\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{O}\htmlData{tutor-start=11,tutor-end=12}{D}\htmlData{tutor-start=12,tutor-end=13}{E} 的面积为8，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的焦距的最小值为 ( ) (A) 4 (B) 8 (C) 16 (D) 32

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 · 代数

设函数 f(x)=x3−1x3\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}{x}^{\htmlData{tutor-start=10,tutor-end=11}{3}} \htmlData{tutor-start=13,tutor-end=14}{-} \frac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{x}^{\htmlData{tutor-start=27,tutor-end=28}{3}}}, 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}{)} ( ) (A) 是奇函数，且在 (0,+∞)\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=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} 单调递增 (B) 是奇函数，且在 (0,+∞)\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=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} 单调递减 (C) 是偶函数，且在 (0,+∞)\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=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} 单调递增 (D) 是偶函数，且在 (0,+∞)\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=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} 单调递减

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 · 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} 是面积为 934\frac{\htmlData{tutor-start=6,tutor-end=7}{9}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{4}} 的等边三角形，且其顶点都在球 O\htmlData{tutor-start=0,tutor-end=1}{O} 的球面上. 若球 O\htmlData{tutor-start=0,tutor-end=1}{O} 的表面积为 16π\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=5}{\pi}, 则 O\htmlData{tutor-start=0,tutor-end=1}{O} 到平面 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} 的距离为 ( ) (A) 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}} (B) 32\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{2}} (C) 1 (D) 32\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 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

一、选择题 · 数学竞赛/待细分

若 2x−2y<3−x−3−y\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{x}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{2}^{\htmlData{tutor-start=11,tutor-end=12}{y}} \htmlData{tutor-start=14,tutor-end=15}{<} \htmlData{tutor-start=16,tutor-end=17}{3}^{\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{x}} \htmlData{tutor-start=23,tutor-end=24}{-} \htmlData{tutor-start=25,tutor-end=26}{3}^{\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{y}}, 则 ( ) (A) ln⁡(y−x+1)>0\ln\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{x}\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=11,tutor-end=12}{>} \htmlData{tutor-start=13,tutor-end=14}{0} (B) ln⁡(y−x+1)<0\ln\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{x}\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=11,tutor-end=12}{<} \htmlData{tutor-start=13,tutor-end=14}{0} (C) ln⁡∣x−y∣>0\ln\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{y}\htmlData{tutor-start=7,tutor-end=8}{|} \htmlData{tutor-start=9,tutor-end=10}{>} \htmlData{tutor-start=11,tutor-end=12}{0} (D) ln⁡∣x−y∣<0\ln\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{y}\htmlData{tutor-start=7,tutor-end=8}{|} \htmlData{tutor-start=9,tutor-end=10}{<} \htmlData{tutor-start=11,tutor-end=12}{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 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 sin⁡x=−23\sin \htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\frac{\htmlData{tutor-start=16,tutor-end=17}{2}}{\htmlData{tutor-start=19,tutor-end=20}{3}}, 则 cos⁡2x=\cos \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{x} \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 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 · 代数

记 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}} 为等差数列 {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. 若 a1=−2\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}{-}\htmlData{tutor-start=9,tutor-end=10}{2}, a2+a6=2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{6}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{2}, 则 S10=\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}} \htmlData{tutor-start=7,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 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y} 满足约束条件 {x+y⩾−1,x−y⩾−1,2x−y⩽1,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{y} \htmlData{tutor-start=18,tutor-end=28}{\geqslant }\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{,} \\ \htmlData{tutor-start=35,tutor-end=36}{x}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{y} \htmlData{tutor-start=39,tutor-end=49}{\geqslant }\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}\htmlData{tutor-start=51,tutor-end=52}{,} \\ \htmlData{tutor-start=56,tutor-end=57}{2}\htmlData{tutor-start=57,tutor-end=58}{x}\htmlData{tutor-start=58,tutor-end=59}{-}\htmlData{tutor-start=59,tutor-end=60}{y} \htmlData{tutor-start=61,tutor-end=71}{\leqslant }\htmlData{tutor-start=71,tutor-end=72}{1}\htmlData{tutor-start=72,tutor-end=73}{,} \end{cases} 则 z=x+2y\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}{+} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{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 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

设有下列四个命题： p1\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}}: 两两相交且不过同一点的三条直线必在同一平面内； p2\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{2}}: 过空间中任意三点有且仅有一个平面； p3\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{3}}: 若空间两条直线不相交，则这两条直线平行； p4\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{4}}: 若直线 l⊂\htmlData{tutor-start=0,tutor-end=1}{l} \htmlData{tutor-start=2,tutor-end=9}{\subset} 平面 α\htmlData{tutor-start=0,tutor-end=6}{\alpha}, 直线 m⊥\htmlData{tutor-start=0,tutor-end=1}{m} \htmlData{tutor-start=2,tutor-end=7}{\perp} 平面 α\htmlData{tutor-start=0,tutor-end=6}{\alpha}, 则 m⊥l\htmlData{tutor-start=0,tutor-end=1}{m} \htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{l}. 则下述命题中所有真命题的序号是 ______. ① p1∧p4\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=13}{\wedge }\htmlData{tutor-start=13,tutor-end=14}{p}_{\htmlData{tutor-start=16,tutor-end=17}{4}}; ② p1∧p2\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=13}{\wedge }\htmlData{tutor-start=13,tutor-end=14}{p}_{\htmlData{tutor-start=16,tutor-end=17}{2}}; ③ ¬p2∨p3\htmlData{tutor-start=0,tutor-end=5}{\neg }\htmlData{tutor-start=5,tutor-end=6}{p}_{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=16}{\vee }\htmlData{tutor-start=16,tutor-end=17}{p}_{\htmlData{tutor-start=19,tutor-end=20}{3}}; ④ ¬p3∨¬p4\htmlData{tutor-start=0,tutor-end=5}{\neg }\htmlData{tutor-start=5,tutor-end=6}{p}_{\htmlData{tutor-start=8,tutor-end=9}{3}} \htmlData{tutor-start=11,tutor-end=16}{\vee }\htmlData{tutor-start=16,tutor-end=21}{\neg }\htmlData{tutor-start=21,tutor-end=22}{p}_{\htmlData{tutor-start=24,tutor-end=25}{4}}.

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 · 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 cos⁡2(π2+A)+cos⁡A=54\cos^{\htmlData{tutor-start=6,tutor-end=7}{2}}\left(\frac{\htmlData{tutor-start=20,tutor-end=23}{\pi}}{\htmlData{tutor-start=25,tutor-end=26}{2}} \htmlData{tutor-start=28,tutor-end=29}{+} \htmlData{tutor-start=30,tutor-end=31}{A}\right) \htmlData{tutor-start=39,tutor-end=40}{+} \cos \htmlData{tutor-start=46,tutor-end=47}{A} \htmlData{tutor-start=48,tutor-end=49}{=} \frac{\htmlData{tutor-start=56,tutor-end=57}{5}}{\htmlData{tutor-start=59,tutor-end=60}{4}}. (1) 求 A\htmlData{tutor-start=0,tutor-end=1}{A}; (2) 若 b−c=33a\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{c} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\sqrt{\htmlData{tutor-start=20,tutor-end=21}{3}}}{\htmlData{tutor-start=24,tutor-end=25}{3}}\htmlData{tutor-start=26,tutor-end=27}{a}, 证明：△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} 是直角三角形.

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 · 数学竞赛/待细分

某沙漠地区经过治理，生态系统得到很大改善，野生动物数量有所增加. 为调查该地区某种野生动物的数量，将其分成面积相近的200个地块，从这些地块中用简单随机抽样的方法抽取20个作为样区，调查得到样本数据 (xi,yi) (i=1,2,⋯,20)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{i}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{y}_{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=17}{\ }\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{i}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{,} \cdots\htmlData{tutor-start=32,tutor-end=33}{,} \htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{0}\htmlData{tutor-start=36,tutor-end=37}{)}, 其中 xi\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 和 yi\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 分别表示第 i\htmlData{tutor-start=0,tutor-end=1}{i} 个样区的植物覆盖面积（单位：公顷）和这种野生动物的数量，并计算得 ∑i=120xi=60\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{x}_{\htmlData{tutor-start=19,tutor-end=20}{i}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{6}\htmlData{tutor-start=25,tutor-end=26}{0}, ∑i=120yi=1200\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{y}_{\htmlData{tutor-start=19,tutor-end=20}{i}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{0}, ∑i=120(xi−xˉ)2=80\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}_{\htmlData{tutor-start=20,tutor-end=21}{i}} \htmlData{tutor-start=23,tutor-end=24}{-} \bar{\htmlData{tutor-start=30,tutor-end=31}{x}}\htmlData{tutor-start=32,tutor-end=33}{)}^{\htmlData{tutor-start=35,tutor-end=36}{2}} \htmlData{tutor-start=38,tutor-end=39}{=} \htmlData{tutor-start=40,tutor-end=41}{8}\htmlData{tutor-start=41,tutor-end=42}{0}, ∑i=120(yi−yˉ)2=9000\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{y}_{\htmlData{tutor-start=20,tutor-end=21}{i}} \htmlData{tutor-start=23,tutor-end=24}{-} \bar{\htmlData{tutor-start=30,tutor-end=31}{y}}\htmlData{tutor-start=32,tutor-end=33}{)}^{\htmlData{tutor-start=35,tutor-end=36}{2}} \htmlData{tutor-start=38,tutor-end=39}{=} \htmlData{tutor-start=40,tutor-end=41}{9}\htmlData{tutor-start=41,tutor-end=42}{0}\htmlData{tutor-start=42,tutor-end=43}{0}\htmlData{tutor-start=43,tutor-end=44}{0}, ∑i=120(xi−xˉ)(yi−yˉ)=800\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}_{\htmlData{tutor-start=20,tutor-end=21}{i}} \htmlData{tutor-start=23,tutor-end=24}{-} \bar{\htmlData{tutor-start=30,tutor-end=31}{x}}\htmlData{tutor-start=32,tutor-end=33}{)}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{y}_{\htmlData{tutor-start=37,tutor-end=38}{i}} \htmlData{tutor-start=40,tutor-end=41}{-} \bar{\htmlData{tutor-start=47,tutor-end=48}{y}}\htmlData{tutor-start=49,tutor-end=50}{)} \htmlData{tutor-start=51,tutor-end=52}{=} \htmlData{tutor-start=53,tutor-end=54}{8}\htmlData{tutor-start=54,tutor-end=55}{0}\htmlData{tutor-start=55,tutor-end=56}{0}. (1) 求该地区这种野生动物数量的估计值（这种野生动物数量的估计值等于样区这种野生动物数量的平均数乘以地块数）； (2) 求样本 (xi,yi) (i=1,2,⋯,20)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{i}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{y}_{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=17}{\ }\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{i}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{,} \cdots\htmlData{tutor-start=32,tutor-end=33}{,} \htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{0}\htmlData{tutor-start=36,tutor-end=37}{)} 的相关系数（精确到0.01）； (3) 根据现有统计资料，各地块间植物覆盖面积差异很大. 为提高样本的代表性以获得该地区这种野生动物数量更准确的估计，请给出一种你认为更合理的抽样方法，并说明理由. 附：相关系数 r=∑i=1n(xi−xˉ)(yi−yˉ)∑i=1n(xi−xˉ)2∑i=1n(yi−yˉ)2\htmlData{tutor-start=0,tutor-end=1}{r} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\sum_{\htmlData{tutor-start=16,tutor-end=17}{i}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{1}}^{\htmlData{tutor-start=22,tutor-end=23}{n}} \htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{x}_{\htmlData{tutor-start=29,tutor-end=30}{i}} \htmlData{tutor-start=32,tutor-end=33}{-} \bar{\htmlData{tutor-start=39,tutor-end=40}{x}}\htmlData{tutor-start=41,tutor-end=42}{)}\htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{y}_{\htmlData{tutor-start=46,tutor-end=47}{i}} \htmlData{tutor-start=49,tutor-end=50}{-} \bar{\htmlData{tutor-start=56,tutor-end=57}{y}}\htmlData{tutor-start=58,tutor-end=59}{)}}{\sqrt{\sum_{\htmlData{tutor-start=73,tutor-end=74}{i}\htmlData{tutor-start=74,tutor-end=75}{=}\htmlData{tutor-start=75,tutor-end=76}{1}}^{\htmlData{tutor-start=79,tutor-end=80}{n}} \htmlData{tutor-start=82,tutor-end=83}{(}\htmlData{tutor-start=83,tutor-end=84}{x}_{\htmlData{tutor-start=86,tutor-end=87}{i}} \htmlData{tutor-start=89,tutor-end=90}{-} \bar{\htmlData{tutor-start=96,tutor-end=97}{x}}\htmlData{tutor-start=98,tutor-end=99}{)}^{\htmlData{tutor-start=101,tutor-end=102}{2}} \sum_{\htmlData{tutor-start=110,tutor-end=111}{i}\htmlData{tutor-start=111,tutor-end=112}{=}\htmlData{tutor-start=112,tutor-end=113}{1}}^{\htmlData{tutor-start=116,tutor-end=117}{n}} \htmlData{tutor-start=119,tutor-end=120}{(}\htmlData{tutor-start=120,tutor-end=121}{y}_{\htmlData{tutor-start=123,tutor-end=124}{i}} \htmlData{tutor-start=126,tutor-end=127}{-} \bar{\htmlData{tutor-start=133,tutor-end=134}{y}}\htmlData{tutor-start=135,tutor-end=136}{)}^{\htmlData{tutor-start=138,tutor-end=139}{2}}}}, 2≈1.414\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=17}{\approx }\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{.}\htmlData{tutor-start=19,tutor-end=20}{4}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{4}.

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 · 平面几何

已知椭圆 C1:x2a2+y2b2=1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{:} \frac{\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}{\htmlData{tutor-start=20,tutor-end=21}{a}^{\htmlData{tutor-start=23,tutor-end=24}{2}}} \htmlData{tutor-start=27,tutor-end=28}{+} \frac{\htmlData{tutor-start=35,tutor-end=36}{y}^{\htmlData{tutor-start=38,tutor-end=39}{2}}}{\htmlData{tutor-start=42,tutor-end=43}{b}^{\htmlData{tutor-start=45,tutor-end=46}{2}}} \htmlData{tutor-start=49,tutor-end=50}{=} \htmlData{tutor-start=51,tutor-end=52}{1} (a>b>0\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{0}) 的右焦点 F\htmlData{tutor-start=0,tutor-end=1}{F} 与抛物线 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的焦点重合，C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的中心与 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的顶点重合. 过 F\htmlData{tutor-start=0,tutor-end=1}{F} 且与 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴垂直的直线交 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 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，交 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 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} 两点，且 ∣CD∣=43∣AB∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{4}}{\htmlData{tutor-start=16,tutor-end=17}{3}}\htmlData{tutor-start=18,tutor-end=19}{|}\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{B}\htmlData{tutor-start=21,tutor-end=22}{|}. (1) 求 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的离心率; (2) 若 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的四个顶点到 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的准线距离之和为 12，求 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 与 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的标准方程.

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 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

如图，已知三棱柱 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}} 的底面是正三角形，侧面 BB1C1C\htmlData{tutor-start=0,tutor-end=1}{B}\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}}\htmlData{tutor-start=11,tutor-end=12}{C} 是矩形，M,N\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{N} 分别为 BC,B1C1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{B}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{C}_{\htmlData{tutor-start=12,tutor-end=13}{1}} 的中点，P\htmlData{tutor-start=0,tutor-end=1}{P} 为 AM\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M} 上一点. 过 B1C1\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{1}} 和 P\htmlData{tutor-start=0,tutor-end=1}{P} 的平面交 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 于 E\htmlData{tutor-start=0,tutor-end=1}{E}，交 AC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} 于 F\htmlData{tutor-start=0,tutor-end=1}{F}. (1) 证明: AA1//MN\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{/} \htmlData{tutor-start=10,tutor-end=11}{M}\htmlData{tutor-start=11,tutor-end=12}{N}，且平面 A1AMN⊥\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{M}\htmlData{tutor-start=7,tutor-end=8}{N} \htmlData{tutor-start=9,tutor-end=14}{\perp} 平面 EB1C1F\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}}\htmlData{tutor-start=11,tutor-end=12}{F}; (2) 设 O\htmlData{tutor-start=0,tutor-end=1}{O} 为 △A1B1C1\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{B}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{C}_{\htmlData{tutor-start=23,tutor-end=24}{1}} 的中心，若 AO=AB=6\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{B} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{6}，AO//\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} \htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{/} 平面 EB1C1F\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}}\htmlData{tutor-start=11,tutor-end=12}{F}，且 ∠MPN=π3\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{M}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{N} \htmlData{tutor-start=11,tutor-end=12}{=} \frac{\htmlData{tutor-start=19,tutor-end=22}{\pi}}{\htmlData{tutor-start=24,tutor-end=25}{3}}，求四棱锥 B−EB1C1F\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{B}_{\htmlData{tutor-start=6,tutor-end=7}{1}}\htmlData{tutor-start=8,tutor-end=9}{C}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{F} 的体积.

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 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)=2ln⁡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}{2}\ln \htmlData{tutor-start=12,tutor-end=13}{x} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{1}. (1) 若 f(x)⩽2x+c\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=15}{\leqslant }\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{c}，求 c\htmlData{tutor-start=0,tutor-end=1}{c} 的取值范围; (2) 设 a>0\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0} 时，讨论函数 g(x)=f(x)−f(a)x−a\htmlData{tutor-start=0,tutor-end=1}{g}\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}{f}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{)} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{f}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{a}\htmlData{tutor-start=23,tutor-end=24}{)}}{\htmlData{tutor-start=26,tutor-end=27}{x} \htmlData{tutor-start=28,tutor-end=29}{-} \htmlData{tutor-start=30,tutor-end=31}{a}} 的单调性.

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

已知曲线 C1,C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{C}_{\htmlData{tutor-start=10,tutor-end=11}{2}} 的参数方程分别为 C1:{x=4cos⁡2θ,y=4sin⁡2θ,\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{:} \begin{cases} \htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{4}\cos^{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=40}{\theta}\htmlData{tutor-start=40,tutor-end=41}{,} \\ \htmlData{tutor-start=45,tutor-end=46}{y} \htmlData{tutor-start=47,tutor-end=48}{=} \htmlData{tutor-start=49,tutor-end=50}{4}\sin^{\htmlData{tutor-start=56,tutor-end=57}{2}}\htmlData{tutor-start=58,tutor-end=64}{\theta}\htmlData{tutor-start=64,tutor-end=65}{,} \end{cases} (θ\htmlData{tutor-start=0,tutor-end=6}{\theta} 为参数)，C2:{x=t+1t,y=t−1t,\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{:} \begin{cases} \htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{t} \htmlData{tutor-start=27,tutor-end=28}{+} \frac{\htmlData{tutor-start=35,tutor-end=36}{1}}{\htmlData{tutor-start=38,tutor-end=39}{t}}\htmlData{tutor-start=40,tutor-end=41}{,} \\ \htmlData{tutor-start=45,tutor-end=46}{y} \htmlData{tutor-start=47,tutor-end=48}{=} \htmlData{tutor-start=49,tutor-end=50}{t} \htmlData{tutor-start=51,tutor-end=52}{-} \frac{\htmlData{tutor-start=59,tutor-end=60}{1}}{\htmlData{tutor-start=62,tutor-end=63}{t}}\htmlData{tutor-start=64,tutor-end=65}{,} \end{cases} (t\htmlData{tutor-start=0,tutor-end=1}{t} 为参数). (1) 将 C1,C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{C}_{\htmlData{tutor-start=10,tutor-end=11}{2}} 的参数方程化为普通方程; (2) 以坐标原点为极点，x\htmlData{tutor-start=0,tutor-end=1}{x} 轴正半轴为极轴建立极坐标系. 设 C1,C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{C}_{\htmlData{tutor-start=10,tutor-end=11}{2}} 的交点为 P\htmlData{tutor-start=0,tutor-end=1}{P}，求圆心在极轴上，且经过极点和 P\htmlData{tutor-start=0,tutor-end=1}{P} 的圆的极坐标方程.

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

已知函数 f(x)=∣x−a2∣+∣x−2a+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=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{a}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{|} \htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{|}\htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{a} \htmlData{tutor-start=29,tutor-end=30}{+} \htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{|}. (1) 当 a=2\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2} 时，求不等式 f(x)⩾4\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}{4} 的解集; (2) 若 f(x)⩾4\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}{4}，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。