2018 年高考数学（全国卷 2文科）

exams_raw/普通高考/2018/2018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

一、选择题 · 数学竞赛/待细分

$\mathrm{\htmlData{tutor-start=8,tutor-end=9}{i}}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{3}\mathrm{\htmlData{tutor-start=22,tutor-end=23}{i}}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{=}$ ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{2}\mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}$

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 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。

一、选择题 · 组合数学

已知集合 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{7}\htmlData{tutor-start=14,tutor-end=16}{\}}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{B}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=22}{\{}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{,} \htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{,} \htmlData{tutor-start=28,tutor-end=29}{4}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{5}\htmlData{tutor-start=32,tutor-end=34}{\}}$ , 则 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{=}$ ( ) (A) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=5}{\}}$ (B) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=5}{\}}$ (C) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=8}{\}}$ (D) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{7}\htmlData{tutor-start=18,tutor-end=20}{\}}$

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。

一、选择题 · 代数

函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\frac{\mathrm{\htmlData{tutor-start=19,tutor-end=20}{e}}^{\htmlData{tutor-start=23,tutor-end=24}{x}}\htmlData{tutor-start=25,tutor-end=26}{-}\mathrm{\htmlData{tutor-start=34,tutor-end=35}{e}}^{\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{x}}}{\htmlData{tutor-start=43,tutor-end=44}{x}^{\htmlData{tutor-start=46,tutor-end=47}{2}}}$ 的图象大致为 ( )

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

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

**原卷图示 3**原卷第 1 页 · qwen3.7-plus_layout_detection · 需复核

**原卷图示 4**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。

一、选择题 · 数学竞赛/待细分

已知向量 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{,} \boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}$ 满足 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{,} \boldsymbol{\htmlData{tutor-start=32,tutor-end=33}{a}} \htmlData{tutor-start=35,tutor-end=41}{\cdot }\boldsymbol{\htmlData{tutor-start=53,tutor-end=54}{b}}\htmlData{tutor-start=55,tutor-end=56}{=}\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{1}$ , 则 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{2}\boldsymbol{\htmlData{tutor-start=35,tutor-end=36}{a}}\htmlData{tutor-start=37,tutor-end=38}{-}\boldsymbol{\htmlData{tutor-start=50,tutor-end=51}{b}}\htmlData{tutor-start=52,tutor-end=53}{)}\htmlData{tutor-start=53,tutor-end=54}{=}$ ( ) (A) 4 (B) 3 (C) 2 (D) 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 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。

一、选择题 · 数学竞赛/待细分

从 2 名男同学和 3 名女同学中任选 2 人参加社区服务, 则选中的 2 人都是女同学的概率为 ( ) (A) 0.6 (B) 0.5 (C) 0.4 (D) 0.3

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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。

一、选择题 · 数学竞赛/待细分

双曲线 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}\htmlData{tutor-start=19,tutor-end=20}{-}\frac{\htmlData{tutor-start=26,tutor-end=27}{y}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}{\htmlData{tutor-start=33,tutor-end=34}{b}^{\htmlData{tutor-start=36,tutor-end=37}{2}}}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{1} \htmlData{tutor-start=42,tutor-end=44}{\ }\htmlData{tutor-start=44,tutor-end=45}{(}\htmlData{tutor-start=45,tutor-end=46}{a}\htmlData{tutor-start=46,tutor-end=47}{>}\htmlData{tutor-start=47,tutor-end=48}{0}\htmlData{tutor-start=48,tutor-end=49}{,} \htmlData{tutor-start=50,tutor-end=51}{b}\htmlData{tutor-start=51,tutor-end=52}{>}\htmlData{tutor-start=52,tutor-end=53}{0}\htmlData{tutor-start=53,tutor-end=54}{)}$ 的离心率为 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}$ , 则其渐近线方程为 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=6}{\pm }\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{x}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=6}{\pm }\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{x}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=6}{\pm }\frac{\sqrt{\htmlData{tutor-start=18,tutor-end=19}{2}}}{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{x}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=6}{\pm }\frac{\sqrt{\htmlData{tutor-start=18,tutor-end=19}{3}}}{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{x}$

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。

一、选择题 · 平面几何

在 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 中, $\cos \frac{\htmlData{tutor-start=11,tutor-end=12}{C}}{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{=}\frac{\sqrt{\htmlData{tutor-start=29,tutor-end=30}{5}}}{\htmlData{tutor-start=33,tutor-end=34}{5}}\htmlData{tutor-start=35,tutor-end=36}{,} \htmlData{tutor-start=37,tutor-end=38}{B}\htmlData{tutor-start=38,tutor-end=39}{C}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{1}\htmlData{tutor-start=41,tutor-end=42}{,} \htmlData{tutor-start=43,tutor-end=44}{A}\htmlData{tutor-start=44,tutor-end=45}{C}\htmlData{tutor-start=45,tutor-end=46}{=}\htmlData{tutor-start=46,tutor-end=47}{5}$ , 则 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}$ ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{4}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}$ (B) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{0}}$ (C) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{9}}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{5}}$

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 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。

一、选择题 · 数学竞赛/待细分

为计算 $\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{-}\frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{3}}\htmlData{tutor-start=27,tutor-end=28}{-}\frac{\htmlData{tutor-start=34,tutor-end=35}{1}}{\htmlData{tutor-start=37,tutor-end=38}{4}}\htmlData{tutor-start=39,tutor-end=40}{+}\cdots\htmlData{tutor-start=46,tutor-end=47}{+}\frac{\htmlData{tutor-start=53,tutor-end=54}{1}}{\htmlData{tutor-start=56,tutor-end=57}{9}\htmlData{tutor-start=57,tutor-end=58}{9}}\htmlData{tutor-start=59,tutor-end=60}{-}\frac{\htmlData{tutor-start=66,tutor-end=67}{1}}{\htmlData{tutor-start=69,tutor-end=70}{1}\htmlData{tutor-start=70,tutor-end=71}{0}\htmlData{tutor-start=71,tutor-end=72}{0}}$ , 设计了如图的程序框图, 则在空白框中应填入 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{i}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{i}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{2}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{i}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{3}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{i}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{4}$

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。

一、选择题 · 平面几何

在正方体 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{A}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{B}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{C}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{D}_{\htmlData{tutor-start=23,tutor-end=24}{1}}$ 中, $\htmlData{tutor-start=0,tutor-end=1}{E}$ 为棱 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 的中点, 则异面直线 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}$ 所成角的正切值为 ( ) (A) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$ (B) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$ (C) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{5}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$ (D) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{7}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$

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 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。

一、选择题 · 代数

若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\cos \htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{-}\sin \htmlData{tutor-start=17,tutor-end=18}{x}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{]}$ 是减函数, 则 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的最大值是 ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=9}{\pi}}{\htmlData{tutor-start=11,tutor-end=12}{4}}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=9}{\pi}}{\htmlData{tutor-start=11,tutor-end=12}{2}}$ (C) $\frac{\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{4}}$ (D) $\htmlData{tutor-start=0,tutor-end=3}{\pi}$

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。

一、选择题 · 平面几何

已知 $\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{F}_{\htmlData{tutor-start=10,tutor-end=11}{2}}$ 是椭圆 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的两个焦点, $\htmlData{tutor-start=0,tutor-end=1}{P}$ 是 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 上的一点, 若 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=13}{\perp }\htmlData{tutor-start=13,tutor-end=14}{P}\htmlData{tutor-start=14,tutor-end=15}{F}_{\htmlData{tutor-start=17,tutor-end=18}{2}}$ , 且 $\angle PF_{2}F_{1}=60^\circ$ , 则 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的离心率为 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\frac{\sqrt{\htmlData{tutor-start=14,tutor-end=15}{3}}}{\htmlData{tutor-start=18,tutor-end=19}{2}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{-}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}$ (C) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}}$ (D) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}$

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 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。

一、选择题 · 代数

已知 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 是定义域为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=17}{\infty}\htmlData{tutor-start=17,tutor-end=18}{)}$ 的奇函数, 满足 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{)}$ . 若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}$ , 则 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{f}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{+}\cdots\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{f}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{5}\htmlData{tutor-start=25,tutor-end=26}{0}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{=}$ ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{5}\htmlData{tutor-start=2,tutor-end=3}{0}$ (B) 0 (C) 2 (D) 50

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。

二、填空题 · 数学竞赛/待细分

曲线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\ln \htmlData{tutor-start=7,tutor-end=8}{x}$ 在点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$ 处的切线方程为______.

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 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。

二、填空题 · 数学竞赛/待细分

若变量 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y}$ 满足约束条件 $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{y}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{5} \htmlData{tutor-start=21,tutor-end=31}{\geqslant }\htmlData{tutor-start=31,tutor-end=32}{0}\htmlData{tutor-start=32,tutor-end=33}{,} \\ \htmlData{tutor-start=37,tutor-end=38}{x}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=41}{y}\htmlData{tutor-start=41,tutor-end=42}{+}\htmlData{tutor-start=42,tutor-end=43}{3} \htmlData{tutor-start=44,tutor-end=54}{\geqslant }\htmlData{tutor-start=54,tutor-end=55}{0}\htmlData{tutor-start=55,tutor-end=56}{,} \\ \htmlData{tutor-start=60,tutor-end=61}{x}\htmlData{tutor-start=61,tutor-end=62}{-}\htmlData{tutor-start=62,tutor-end=63}{5} \htmlData{tutor-start=64,tutor-end=74}{\leqslant }\htmlData{tutor-start=74,tutor-end=75}{0}\htmlData{tutor-start=75,tutor-end=76}{,} \end{cases}$ 则 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{y}$ 的最大值为______.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 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。

二、填空题 · 数学竞赛/待细分

已知 $\tan \left(\htmlData{tutor-start=11,tutor-end=17}{\alpha}\htmlData{tutor-start=17,tutor-end=18}{-}\frac{\htmlData{tutor-start=24,tutor-end=25}{5}\htmlData{tutor-start=25,tutor-end=28}{\pi}}{\htmlData{tutor-start=30,tutor-end=31}{4}}\right)\htmlData{tutor-start=39,tutor-end=40}{=}\frac{\htmlData{tutor-start=46,tutor-end=47}{1}}{\htmlData{tutor-start=49,tutor-end=50}{5}}$ , 则 $\tan \htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}$ ______.

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。

二、填空题 · 平面几何

已知圆锥的顶点为 $\htmlData{tutor-start=0,tutor-end=1}{S}$ , 母线 $\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{S}\htmlData{tutor-start=5,tutor-end=6}{B}$ 互相垂直, $\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{A}$ 与圆锥底面所成角为 $30^\circ$ , 若 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{S}\htmlData{tutor-start=11,tutor-end=12}{A}\htmlData{tutor-start=12,tutor-end=13}{B}$ 的面积为 8, 则该圆锥的体积为______.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 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。

三、解答题 · 代数

记 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ 为等差数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的前 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项和, 已知 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{7}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{S}_{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{5}$ . (1) 求 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的通项公式; (2) 求 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ , 并求 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ 的最小值.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2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。

三、解答题 · 数学竞赛/待细分

如图是某地区 2000 年至 2016 年环境基础设施投资额 $\htmlData{tutor-start=0,tutor-end=1}{y}$ (单位: 亿元) 的折线图. 为了预测该地区 2018 年的环境基础设施投资额, 建立了 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 与时间变量 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 的两个线性回归模型. 根据 2000 年至 2016 年的数据 (时间变量 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 的值依次为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,} \cdots\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{7}$ ) 建立模型①: $\hat{\htmlData{tutor-start=5,tutor-end=6}{y}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{.}\htmlData{tutor-start=12,tutor-end=13}{4}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{.}\htmlData{tutor-start=17,tutor-end=18}{5}\htmlData{tutor-start=18,tutor-end=19}{t}$ ; 根据 2010 年至 2016 年的数据 (时间变量 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 的值依次为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,} \cdots\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{7}$ ) 建立模型②: $\hat{\htmlData{tutor-start=5,tutor-end=6}{y}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{9}\htmlData{tutor-start=9,tutor-end=10}{9}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{7}\htmlData{tutor-start=13,tutor-end=14}{.}\htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=16}{t}$ . (1) 分别利用这两个模型, 求该地区 2018 年的环境基础设施投资额的预测值; (2) 你认为用哪个模型得到的预测值更可靠? 并说明理由.

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。

解答题 · 数学竞赛/待细分

如图，在三棱锥 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}$ 中， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{2}}$ ， $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{P}\htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{P}\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{C}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{4}$ ， $\htmlData{tutor-start=0,tutor-end=1}{O}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}$ 的中点. (1) 证明： $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{O} \htmlData{tutor-start=3,tutor-end=8}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ ； (2) 若点 $\htmlData{tutor-start=0,tutor-end=1}{M}$ 在棱 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 上，且 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{M}\htmlData{tutor-start=5,tutor-end=6}{B}$ ，求点 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 到平面 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{M}$ 的距离.

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。

解答题 · 平面几何

设抛物线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{x}$ 的焦点为 $\htmlData{tutor-start=0,tutor-end=1}{F}$ ，过 $\htmlData{tutor-start=0,tutor-end=1}{F}$ 且斜率为 $\htmlData{tutor-start=0,tutor-end=1}{k}$ ( $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}$ ) 的直线 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 交于 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}$ 两点， $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{8}$ . (1) 求 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的方程； (2) 求过点 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}$ 且与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的准线相切的圆的方程.

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 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。

解答题 · 代数

已知函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\frac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{3}}\htmlData{tutor-start=16,tutor-end=17}{x}^{\htmlData{tutor-start=19,tutor-end=20}{3}}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{a}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{x}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{x}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{)}$ . (1) 若 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$ ，求 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的单调区间； (2) 证明： $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 只有一个零点.

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。

解答题 · 平面几何

在直角坐标系 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{y}$ 中，曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的参数方程为 $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{2}\cos\htmlData{tutor-start=21,tutor-end=27}{\theta}\htmlData{tutor-start=27,tutor-end=28}{,} \\ \htmlData{tutor-start=32,tutor-end=33}{y}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{4}\sin\htmlData{tutor-start=39,tutor-end=45}{\theta}\htmlData{tutor-start=45,tutor-end=46}{,} \end{cases}$ ( $\htmlData{tutor-start=0,tutor-end=6}{\theta}$ 为参数)，直线 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的参数方程为 $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{t}\cos\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{,} \\ \htmlData{tutor-start=34,tutor-end=35}{y}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{2}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{t}\sin\htmlData{tutor-start=43,tutor-end=50}{\alpha }\end{cases}$ ( $\htmlData{tutor-start=0,tutor-end=1}{t}$ 为参数). (1) 求 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的直角坐标方程； (2) 若曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 截直线 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 所得线段的中点坐标为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)}$ ，求 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的斜率.

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。

解答题 · 代数

设函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{|}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{|}$ . (1) 当 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$ 时，求不等式 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=15}{\geqslant }\htmlData{tutor-start=15,tutor-end=16}{0}$ 的解集； (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=15}{\leqslant }\htmlData{tutor-start=15,tutor-end=16}{1}$ ，求 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 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。