2018 年高考数学（全国卷 2理科）

exams_raw/普通高考/2018/2018全国2理(甘肃,青海,内蒙古,黑龙江,吉林,辽宁,海南.宁夏,新疆,陕西,重庆).pdf · HS-MATH-1024-v2.1-solution-aware

23 题0 个小问/题组

一、选择题 · 数学竞赛/待细分

$\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\mathrm{\htmlData{tutor-start=17,tutor-end=18}{i}}}{\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{2}\mathrm{\htmlData{tutor-start=32,tutor-end=33}{i}}}\htmlData{tutor-start=35,tutor-end=36}{=}$ ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{4}}{\htmlData{tutor-start=10,tutor-end=11}{5}}\htmlData{tutor-start=12,tutor-end=13}{-}\frac{\htmlData{tutor-start=19,tutor-end=20}{3}}{\htmlData{tutor-start=22,tutor-end=23}{5}}\mathrm{\htmlData{tutor-start=32,tutor-end=33}{i}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{4}}{\htmlData{tutor-start=10,tutor-end=11}{5}}\htmlData{tutor-start=12,tutor-end=13}{+}\frac{\htmlData{tutor-start=19,tutor-end=20}{3}}{\htmlData{tutor-start=22,tutor-end=23}{5}}\mathrm{\htmlData{tutor-start=32,tutor-end=33}{i}}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{5}}\htmlData{tutor-start=12,tutor-end=13}{-}\frac{\htmlData{tutor-start=19,tutor-end=20}{4}}{\htmlData{tutor-start=22,tutor-end=23}{5}}\mathrm{\htmlData{tutor-start=32,tutor-end=33}{i}}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{5}}\htmlData{tutor-start=12,tutor-end=13}{+}\frac{\htmlData{tutor-start=19,tutor-end=20}{4}}{\htmlData{tutor-start=22,tutor-end=23}{5}}\mathrm{\htmlData{tutor-start=32,tutor-end=33}{i}}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 1 题

解题过程

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

一、选择题 · 组合数学

已知集合 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{y}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=14}{\mid }\htmlData{tutor-start=14,tutor-end=15}{x}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{y}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=35}{\leqslant }\htmlData{tutor-start=35,tutor-end=36}{3}\htmlData{tutor-start=36,tutor-end=37}{,} \htmlData{tutor-start=38,tutor-end=39}{x}\htmlData{tutor-start=39,tutor-end=42}{\in}\mathbf{\htmlData{tutor-start=50,tutor-end=51}{Z}}\htmlData{tutor-start=52,tutor-end=53}{,} \htmlData{tutor-start=54,tutor-end=55}{y}\htmlData{tutor-start=55,tutor-end=58}{\in}\mathbf{\htmlData{tutor-start=66,tutor-end=67}{Z}}\htmlData{tutor-start=68,tutor-end=70}{\}}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 中元素的个数为 ( ) (A) 9 (B) 8 (C) 5 (D) 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理科第 2 题

解题过程

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

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 ( ) (A) (B) (C) (D)

**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理科第 3 题

解题过程

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

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 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{,} \boldsymbol{\htmlData{tutor-start=32,tutor-end=33}{a}}\htmlData{tutor-start=34,tutor-end=39}{\cdot}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{b}}\htmlData{tutor-start=53,tutor-end=54}{=}\htmlData{tutor-start=54,tutor-end=55}{-}\htmlData{tutor-start=55,tutor-end=56}{1}$ ，则 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=19}{\cdot}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{2}\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{a}}\htmlData{tutor-start=35,tutor-end=36}{-}\boldsymbol{\htmlData{tutor-start=48,tutor-end=49}{b}}\htmlData{tutor-start=50,tutor-end=51}{)}\htmlData{tutor-start=51,tutor-end=52}{=}$ ( ) (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理科第 4 题

解题过程

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

一、选择题 · 数学竞赛/待细分

双曲线 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}\htmlData{tutor-start=19,tutor-end=20}{-}\frac{\htmlData{tutor-start=26,tutor-end=27}{y}^{\htmlData{tutor-start=29,tutor-end=30}{2}}}{\htmlData{tutor-start=33,tutor-end=34}{b}^{\htmlData{tutor-start=36,tutor-end=37}{2}}}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{1}\htmlData{tutor-start=41,tutor-end=43}{\ }\htmlData{tutor-start=43,tutor-end=44}{(}\htmlData{tutor-start=44,tutor-end=45}{a}\htmlData{tutor-start=45,tutor-end=46}{>}\htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{,} \htmlData{tutor-start=49,tutor-end=50}{b}\htmlData{tutor-start=50,tutor-end=51}{>}\htmlData{tutor-start=51,tutor-end=52}{0}\htmlData{tutor-start=52,tutor-end=53}{)}$ 的离心率为 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}$ ，则其渐近线方程为 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pm}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{x}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pm}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{3}}\htmlData{tutor-start=13,tutor-end=14}{x}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pm}\frac{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{2}}}{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{x}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pm}\frac{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}}{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{x}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 5 题

解题过程

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

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， $\cos\frac{\htmlData{tutor-start=10,tutor-end=11}{C}}{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{=}\frac{\sqrt{\htmlData{tutor-start=28,tutor-end=29}{5}}}{\htmlData{tutor-start=32,tutor-end=33}{5}}\htmlData{tutor-start=34,tutor-end=35}{,} \htmlData{tutor-start=36,tutor-end=37}{B}\htmlData{tutor-start=37,tutor-end=38}{C}\htmlData{tutor-start=38,tutor-end=39}{=}\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{,} \htmlData{tutor-start=42,tutor-end=43}{A}\htmlData{tutor-start=43,tutor-end=44}{C}\htmlData{tutor-start=44,tutor-end=45}{=}\htmlData{tutor-start=45,tutor-end=46}{5}$ ，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理科第 6 题

解题过程

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

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理科第 7 题

解题过程

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

一、选择题 · 数论

我国数学家陈景润在哥德巴赫猜想的研究中取得了世界领先的成果。哥德巴赫猜想是“每个大于 2 的偶数可以表示为两个素数的和”，如 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{7}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{3}$ 。在不超过 30 的素数中，随机选取两个不同的数，其和等于 30 的概率是 ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{2}}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{4}}$ (C) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{5}}$ (D) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{8}}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 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

在长方体 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{A}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{B}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{C}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{D}_{\htmlData{tutor-start=23,tutor-end=24}{1}}$ 中， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{A}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{=}\sqrt{\htmlData{tutor-start=22,tutor-end=23}{3}}$ ，则异面直线 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 所成角的余弦值为 ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{5}}$ (B) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{5}}}{\htmlData{tutor-start=16,tutor-end=17}{6}}$ (C) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{5}}}{\htmlData{tutor-start=16,tutor-end=17}{5}}$ (D) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 9 题

解题过程

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

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 $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{]}$ 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理科第 10 题

解题过程

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

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理科第 11 题

解题过程

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

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 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \frac{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}}\htmlData{tutor-start=22,tutor-end=23}{+}\frac{\htmlData{tutor-start=29,tutor-end=30}{y}^{\htmlData{tutor-start=32,tutor-end=33}{2}}}{\htmlData{tutor-start=36,tutor-end=37}{b}^{\htmlData{tutor-start=39,tutor-end=40}{2}}}\htmlData{tutor-start=42,tutor-end=43}{=}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=46}{\ }\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{a}\htmlData{tutor-start=48,tutor-end=49}{>}\htmlData{tutor-start=49,tutor-end=50}{b}\htmlData{tutor-start=50,tutor-end=51}{>}\htmlData{tutor-start=51,tutor-end=52}{0}\htmlData{tutor-start=52,tutor-end=53}{)}$ 的左，右焦点， $\htmlData{tutor-start=0,tutor-end=1}{A}$ 是 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的左顶点，点 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 在过 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 且斜率为 $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{6}}$ 的直线上， $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{F}_{\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{F}_{\htmlData{tutor-start=19,tutor-end=20}{2}}$ 为等腰三角形， $\angle F_{1}F_{2}P=120^\circ$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的离心率为 ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{3}}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$ (C) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}}$ (D) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{4}}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 12 题

解题过程

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

二、填空题 · 数学竞赛/待细分

曲线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\ln\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)}$ 在点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}$ 处的切线方程为 \underline{\hspace{2cm}}。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 13 题

解题过程

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

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 $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{y}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{5}\htmlData{tutor-start=20,tutor-end=30}{\geqslant }\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{,} \\ \htmlData{tutor-start=36,tutor-end=37}{x}\htmlData{tutor-start=37,tutor-end=38}{-}\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{y}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{3}\htmlData{tutor-start=42,tutor-end=52}{\geqslant }\htmlData{tutor-start=52,tutor-end=53}{0}\htmlData{tutor-start=53,tutor-end=54}{,} \\ \htmlData{tutor-start=58,tutor-end=59}{x}\htmlData{tutor-start=59,tutor-end=60}{-}\htmlData{tutor-start=60,tutor-end=61}{5}\htmlData{tutor-start=61,tutor-end=71}{\leqslant }\htmlData{tutor-start=71,tutor-end=72}{0}\htmlData{tutor-start=72,tutor-end=73}{,} \end{cases}$ 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 \underline{\hspace{2cm}}。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 14 题

解题过程

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

二、填空题 · 数学竞赛/待细分

已知 $\sin\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{+}\cos\htmlData{tutor-start=15,tutor-end=20}{\beta}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{,} \cos\htmlData{tutor-start=28,tutor-end=34}{\alpha}\htmlData{tutor-start=34,tutor-end=35}{+}\sin\htmlData{tutor-start=39,tutor-end=44}{\beta}\htmlData{tutor-start=44,tutor-end=45}{=}\htmlData{tutor-start=45,tutor-end=46}{0}$ ，则 $\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=17}{\beta}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{=}$ \underline{\hspace{2cm}}。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 15 题

解题过程

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

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 $\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{S}\htmlData{tutor-start=5,tutor-end=6}{B}$ 所成角的余弦值为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{7}}{\htmlData{tutor-start=9,tutor-end=10}{8}}$ ， $\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{A}$ 与圆锥底面所成角为 $45^\circ$ ，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 $\htmlData{tutor-start=0,tutor-end=1}{5}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{5}}$ ，则该圆锥的侧面积为 \underline{\hspace{2cm}}。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 16 题

解题过程

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

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理科第 17 题

解题过程

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

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理科第 18 题

解题过程

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

解答题 · 平面几何

设抛物线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y}^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{x}$ 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 $\htmlData{tutor-start=0,tutor-end=1}{k}$ （ $\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}$ ）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， $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{8}$ . (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理科第 19 题

解题过程

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

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， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{C} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{2}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{2}}$ ， $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{P}\htmlData{tutor-start=6,tutor-end=7}{B} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{C} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{A}\htmlData{tutor-start=16,tutor-end=17}{C} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{4}$ ， $\htmlData{tutor-start=0,tutor-end=1}{O}$ 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，且二面角 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{P}\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{C}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{°}$ ，求 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}$ 与平面 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{M}$ 所成角的正弦值.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 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 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \mathrm{\htmlData{tutor-start=15,tutor-end=16}{e}}^{\htmlData{tutor-start=19,tutor-end=20}{x}} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{a}\htmlData{tutor-start=25,tutor-end=26}{x}^{\htmlData{tutor-start=28,tutor-end=29}{2}}$ . (1) 若 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}$ ，证明：当 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$ 时， $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=15}{\geqslant }\htmlData{tutor-start=15,tutor-end=16}{1}$ ； (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}$ 只有一个零点，求 $\htmlData{tutor-start=0,tutor-end=1}{a}$ .

题解状态：标准答案与规范题解待补充

题目标签：全国卷 2理科第 21 题

解题过程

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

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 $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{2}\cos\htmlData{tutor-start=23,tutor-end=29}{\theta}\htmlData{tutor-start=29,tutor-end=30}{,} \\ \htmlData{tutor-start=34,tutor-end=35}{y} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{4}\sin\htmlData{tutor-start=43,tutor-end=49}{\theta}\htmlData{tutor-start=49,tutor-end=50}{,} \end{cases}$ （ $\htmlData{tutor-start=0,tutor-end=6}{\theta}$ 为参数），直线 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的参数方程为 $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{1} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{t}\cos\htmlData{tutor-start=27,tutor-end=33}{\alpha}\htmlData{tutor-start=33,tutor-end=34}{,} \\ \htmlData{tutor-start=38,tutor-end=39}{y} \htmlData{tutor-start=40,tutor-end=41}{=} \htmlData{tutor-start=42,tutor-end=43}{2} \htmlData{tutor-start=44,tutor-end=45}{+} \htmlData{tutor-start=46,tutor-end=47}{t}\sin\htmlData{tutor-start=51,tutor-end=58}{\alpha }\end{cases}$ （ $\htmlData{tutor-start=0,tutor-end=1}{t}$ 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理科第 22 题

解题过程

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

解答题 · 代数

设函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{5} \htmlData{tutor-start=9,tutor-end=10}{-} \htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{x} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{a}\htmlData{tutor-start=17,tutor-end=18}{|} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{|}\htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{|}$ . (1) 当 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}$ 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理科第 23 题

解题过程

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