2018 年高考数学（全国卷 1理科）

exams_raw/普通高考/2018/2018全国1理(河南,河北,山西,江西,湖北,湖南,广东,安徽,福建,山东).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

设 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{-}\text{\htmlData{tutor-start=18,tutor-end=19}{i}}}{\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{+}\text{\htmlData{tutor-start=30,tutor-end=31}{i}}} \htmlData{tutor-start=34,tutor-end=35}{+} \htmlData{tutor-start=36,tutor-end=37}{2}\text{\htmlData{tutor-start=43,tutor-end=44}{i}}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{z}\htmlData{tutor-start=2,tutor-end=3}{|} \htmlData{tutor-start=4,tutor-end=5}{=}$ ( ) (A) 0 (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$ (C) 1 (D) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=13}{\mid }\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=24}{-} \htmlData{tutor-start=25,tutor-end=26}{2} \htmlData{tutor-start=27,tutor-end=28}{>} \htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=32}{\}}$ ，则 $\htmlData{tutor-start=0,tutor-end=11}{\complement}_{\mathbf{\htmlData{tutor-start=21,tutor-end=22}{R}}}\htmlData{tutor-start=24,tutor-end=25}{A} \htmlData{tutor-start=26,tutor-end=27}{=}$ (A) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1} \htmlData{tutor-start=12,tutor-end=13}{<} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=21}{\}}$ (B) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1} \htmlData{tutor-start=12,tutor-end=22}{\leqslant }\htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=34}{\leqslant }\htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=37}{\}}$ (C) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{<} \htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=17}{\}} \htmlData{tutor-start=18,tutor-end=23}{\cup }\htmlData{tutor-start=23,tutor-end=25}{\{}\htmlData{tutor-start=25,tutor-end=26}{x} \htmlData{tutor-start=27,tutor-end=32}{\mid }\htmlData{tutor-start=32,tutor-end=33}{x} \htmlData{tutor-start=34,tutor-end=35}{>} \htmlData{tutor-start=36,tutor-end=37}{2}\htmlData{tutor-start=37,tutor-end=39}{\}}$ (D) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=21}{\leqslant }\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=25}{\}} \htmlData{tutor-start=26,tutor-end=31}{\cup }\htmlData{tutor-start=31,tutor-end=33}{\{}\htmlData{tutor-start=33,tutor-end=34}{x} \htmlData{tutor-start=35,tutor-end=40}{\mid }\htmlData{tutor-start=40,tutor-end=41}{x} \htmlData{tutor-start=42,tutor-end=52}{\geqslant }\htmlData{tutor-start=52,tutor-end=53}{2}\htmlData{tutor-start=53,tutor-end=55}{\}}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 数学竞赛/待细分

某地区经过一年的新农村建设，农村的经济收入增加了一倍，实现翻番. 为更好地了解该地区农村的经济收入变化情况，统计了该地区新农村建设前后农村的经济收入构成比例，得到如下饼图：则下面结论中不正确的是 ( ) (A) 新农村建设后，种植收入减少 (B) 新农村建设后，其他收入增加了一倍以上 (C) 新农村建设后，养殖收入增加了一倍 (D) 新农村建设后，养殖收入与第三产业收入的总和超过了经济收入的一半

**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 1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 · 代数

设 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ 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，若 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{S}_{\htmlData{tutor-start=4,tutor-end=5}{3}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{S}_{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{S}_{\htmlData{tutor-start=20,tutor-end=21}{4}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{1}} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{2}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{5}} \htmlData{tutor-start=6,tutor-end=7}{=}$ ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{2}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{0}$ (C) 10 (D) 12

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 代数

设函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{3}} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{a}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{x}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{a}\htmlData{tutor-start=29,tutor-end=30}{x}$ . 若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 为奇函数，则曲线 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)}$ 在点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}$ 处的切线方程为 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{x}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{x}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 边上的中线， $\htmlData{tutor-start=0,tutor-end=1}{E}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}$ 的中点，则 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{E}\htmlData{tutor-start=17,tutor-end=18}{B}} \htmlData{tutor-start=20,tutor-end=21}{=}$ ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{4}}\overrightarrow{\htmlData{tutor-start=27,tutor-end=28}{A}\htmlData{tutor-start=28,tutor-end=29}{B}} \htmlData{tutor-start=31,tutor-end=32}{-} \frac{\htmlData{tutor-start=39,tutor-end=40}{1}}{\htmlData{tutor-start=42,tutor-end=43}{4}}\overrightarrow{\htmlData{tutor-start=60,tutor-end=61}{A}\htmlData{tutor-start=61,tutor-end=62}{C}}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{4}}\overrightarrow{\htmlData{tutor-start=27,tutor-end=28}{A}\htmlData{tutor-start=28,tutor-end=29}{B}} \htmlData{tutor-start=31,tutor-end=32}{-} \frac{\htmlData{tutor-start=39,tutor-end=40}{3}}{\htmlData{tutor-start=42,tutor-end=43}{4}}\overrightarrow{\htmlData{tutor-start=60,tutor-end=61}{A}\htmlData{tutor-start=61,tutor-end=62}{C}}$ (C) $\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{4}}\overrightarrow{\htmlData{tutor-start=27,tutor-end=28}{A}\htmlData{tutor-start=28,tutor-end=29}{B}} \htmlData{tutor-start=31,tutor-end=32}{+} \frac{\htmlData{tutor-start=39,tutor-end=40}{1}}{\htmlData{tutor-start=42,tutor-end=43}{4}}\overrightarrow{\htmlData{tutor-start=60,tutor-end=61}{A}\htmlData{tutor-start=61,tutor-end=62}{C}}$ (D) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{4}}\overrightarrow{\htmlData{tutor-start=27,tutor-end=28}{A}\htmlData{tutor-start=28,tutor-end=29}{B}} \htmlData{tutor-start=31,tutor-end=32}{+} \frac{\htmlData{tutor-start=39,tutor-end=40}{3}}{\htmlData{tutor-start=42,tutor-end=43}{4}}\overrightarrow{\htmlData{tutor-start=60,tutor-end=61}{A}\htmlData{tutor-start=61,tutor-end=62}{C}}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 平面几何

某圆柱的高为 2，底面周长为 16，其三视图如图. 圆柱表面上的点 $\htmlData{tutor-start=0,tutor-end=1}{M}$ 在正视图上的对应点为 $\htmlData{tutor-start=0,tutor-end=1}{A}$ ，圆柱表面上的点 $\htmlData{tutor-start=0,tutor-end=1}{N}$ 在左视图上的对应点为 $\htmlData{tutor-start=0,tutor-end=1}{B}$ ，则在此圆柱侧面上，从 $\htmlData{tutor-start=0,tutor-end=1}{M}$ 到 $\htmlData{tutor-start=0,tutor-end=1}{N}$ 的路径中，最短路径的长度为 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{7}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{5}}$ (C) 3 (D) 2

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 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 $\htmlData{tutor-start=0,tutor-end=1}{F}$ ，过点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$ 且斜率为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{3}}$ 的直线与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 交于 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{N}$ 两点，则 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{F}\htmlData{tutor-start=17,tutor-end=18}{M}} \htmlData{tutor-start=20,tutor-end=26}{\cdot }\overrightarrow{\htmlData{tutor-start=42,tutor-end=43}{F}\htmlData{tutor-start=43,tutor-end=44}{N}} \htmlData{tutor-start=46,tutor-end=47}{=}$ ( ) (A) 5 (B) 6 (C) 7 (D) 8

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 代数

已知函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \begin{cases} \text{\htmlData{tutor-start=27,tutor-end=28}{e}}^{\htmlData{tutor-start=31,tutor-end=32}{x}}\htmlData{tutor-start=33,tutor-end=34}{,} & \htmlData{tutor-start=37,tutor-end=38}{x} \htmlData{tutor-start=39,tutor-end=49}{\leqslant }\htmlData{tutor-start=49,tutor-end=50}{0}\htmlData{tutor-start=50,tutor-end=51}{,} \\ \ln \htmlData{tutor-start=59,tutor-end=60}{x}\htmlData{tutor-start=60,tutor-end=61}{,} & \htmlData{tutor-start=64,tutor-end=65}{x} \htmlData{tutor-start=66,tutor-end=67}{>} \htmlData{tutor-start=68,tutor-end=69}{0}\htmlData{tutor-start=69,tutor-end=70}{,} \end{cases}$ $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{a}$ . 若 $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 存在 2 个零点，则 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的取值范围是 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=12}{\infty}\htmlData{tutor-start=12,tutor-end=13}{)}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 平面几何

如图来自古希腊数学家希波克拉底所研究的几何图形. 此图由三个半圆构成，三个半圆的直径分别为直角三角形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 的斜边 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ ，直角边 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{A}\htmlData{tutor-start=5,tutor-end=6}{C}$ . $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 的三边所围成的区域记为 I，黑色部分记为 II，其余部分记为 III. 在整个图形中随机取一点，此点取自 I, II, III 的概率分别记为 $\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{p}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{p}_{\htmlData{tutor-start=17,tutor-end=18}{3}}$ ，则 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{p}_{\htmlData{tutor-start=11,tutor-end=12}{2}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{p}_{\htmlData{tutor-start=11,tutor-end=12}{3}}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{p}_{\htmlData{tutor-start=11,tutor-end=12}{3}}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{p}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{p}_{\htmlData{tutor-start=19,tutor-end=20}{3}}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 平面几何

已知双曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \frac{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{3}} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{y}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{1}$ ， $\htmlData{tutor-start=0,tutor-end=1}{O}$ 为坐标原点， $\htmlData{tutor-start=0,tutor-end=1}{F}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的右焦点，过 $\htmlData{tutor-start=0,tutor-end=1}{F}$ 的直线与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的两条渐近线的交点分别为 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{N}$ . 若 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{O}\htmlData{tutor-start=11,tutor-end=12}{M}\htmlData{tutor-start=12,tutor-end=13}{N}$ 为直角三角形，则 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{N}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{=}$ ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$ (B) 3 (C) $\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{3}}$ (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 1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

已知正方体的棱长为 1，每条棱所在直线与平面 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 所成的角相等，则 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 截此正方体所得截面面积的最大值为 ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{3}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{4}}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{3}}$ (C) $\frac{\htmlData{tutor-start=6,tutor-end=7}{3}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{2}}}{\htmlData{tutor-start=17,tutor-end=18}{4}}$ (D) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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

若 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y}$ 满足约束条件 $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{y} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{2} \htmlData{tutor-start=25,tutor-end=35}{\leqslant }\htmlData{tutor-start=35,tutor-end=36}{0}\htmlData{tutor-start=36,tutor-end=37}{,} \\ \htmlData{tutor-start=41,tutor-end=42}{x} \htmlData{tutor-start=43,tutor-end=44}{-} \htmlData{tutor-start=45,tutor-end=46}{y} \htmlData{tutor-start=47,tutor-end=48}{+} \htmlData{tutor-start=49,tutor-end=50}{1} \htmlData{tutor-start=51,tutor-end=61}{\geqslant }\htmlData{tutor-start=61,tutor-end=62}{0}\htmlData{tutor-start=62,tutor-end=63}{,} \\ \htmlData{tutor-start=67,tutor-end=68}{y} \htmlData{tutor-start=69,tutor-end=79}{\leqslant }\htmlData{tutor-start=79,tutor-end=80}{0}\htmlData{tutor-start=80,tutor-end=81}{,} \end{cases}$ 则 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{y}$ 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 1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 · 代数

记 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ 为数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的前 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项和，若 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{1}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{6}} \htmlData{tutor-start=6,tutor-end=7}{=}$ \_\_\_\_\_\_.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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

从 2 位女生，4 位男生中选 3 人参加科技比赛，且至少有 1 位女生入选，则不同的选法共有 \_\_\_\_\_\_ 种. (用数字填写答案)

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 代数

已知函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{2}\sin \htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=16}{+} \sin \htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{x}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的最小值是 \_\_\_\_\_\_.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 平面几何

在平面四边形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}$ 中， $\angle ADC = 90^\circ, \angle A = 45^\circ, AB = 2, BD = 5$ . (1) 求 $\cos \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{D}\htmlData{tutor-start=14,tutor-end=15}{B}$ ; (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{2}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}$ ，求 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ .

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 平面几何

如图，四边形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}$ 为正方形， $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{F}$ 分别为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{C}$ 的中点，以 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{F}$ 为折痕，把 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{F}\htmlData{tutor-start=12,tutor-end=13}{C}$ 折起，使点 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 到达点 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 的位置，且 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{F} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{F}$ . (1) 证明：平面 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{F} \htmlData{tutor-start=4,tutor-end=9}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{D}$ ; (2) 求 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{P}$ 与平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{D}$ 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 1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 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \frac{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{y}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{1}$ 的右焦点为 $\htmlData{tutor-start=0,tutor-end=1}{F}$ ，过 $\htmlData{tutor-start=0,tutor-end=1}{F}$ 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，点 $\htmlData{tutor-start=0,tutor-end=1}{M}$ 的坐标为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$ . (1) 当 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴垂直时，求直线 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M}$ 的方程； (2) 设 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 为坐标原点，证明： $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{M}\htmlData{tutor-start=9,tutor-end=10}{A} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=20}{\angle }\htmlData{tutor-start=20,tutor-end=21}{O}\htmlData{tutor-start=21,tutor-end=22}{M}\htmlData{tutor-start=22,tutor-end=23}{B}$ .

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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

某工厂的某种产品成箱包装，每箱 200 件，每一箱产品在交付用户之前要对产品作检验，如检验出不合格品，则更换为合格品. 检验时，先从这箱产品中任取 20 件作检验，再根据检验结果决定是否对余下的所有产品作检验. 设每件产品为不合格品的概率都为 $\htmlData{tutor-start=0,tutor-end=1}{p}$ $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{p} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)}$ ，且各件产品是否为不合格品相互独立. (1) 记 20 件产品中恰有 2 件不合格品的概率为 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=4}{)}$ ，求 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{p}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的最大值点 $\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{0}}$ ； (2) 现对一箱产品检验了 20 件，结果恰有 2 件不合格品，以 (1) 中确定的 $\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{0}}$ 作为 $\htmlData{tutor-start=0,tutor-end=1}{p}$ 的值. 已知每件产品的检验费用为 2 元，若有不合格品进入用户手中，则工厂要对每件不合格品支付 25 元的赔偿费用. ① 若不对该箱余下的产品作检验，这一箱产品的检验费用与赔偿费用的和记为 $\htmlData{tutor-start=0,tutor-end=1}{X}$ ，求 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{X}$ ； ② 以检验费用与赔偿费用和的期望值为决策依据，是否该对这箱余下的所有产品作检验？

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{x}} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{a} \ln \htmlData{tutor-start=31,tutor-end=32}{x}$ . (1) 讨论 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的单调性； (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 存在两个极值点 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{2}}$ ，证明： $\frac{\htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{-} \htmlData{tutor-start=17,tutor-end=18}{f}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{x}_{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{)}}{\htmlData{tutor-start=27,tutor-end=28}{x}_{\htmlData{tutor-start=30,tutor-end=31}{1}} \htmlData{tutor-start=33,tutor-end=34}{-} \htmlData{tutor-start=35,tutor-end=36}{x}_{\htmlData{tutor-start=38,tutor-end=39}{2}}} \htmlData{tutor-start=42,tutor-end=43}{<} \htmlData{tutor-start=44,tutor-end=45}{a} \htmlData{tutor-start=46,tutor-end=47}{-} \htmlData{tutor-start=48,tutor-end=49}{2}$ .

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1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 · 数学竞赛/待细分

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 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 的方程为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{k}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{|} \htmlData{tutor-start=9,tutor-end=10}{+} \htmlData{tutor-start=11,tutor-end=12}{2}$ . 以坐标原点为极点， $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴正半轴为极轴建立极坐标系，曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}}$ 的极坐标方程为 $\htmlData{tutor-start=0,tutor-end=4}{\rho}^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{+} \htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=17}{\rho }\cos \htmlData{tutor-start=22,tutor-end=29}{\theta }\htmlData{tutor-start=29,tutor-end=30}{-} \htmlData{tutor-start=31,tutor-end=32}{3} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{0}$ . (1) 求 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}}$ 的直角坐标方程； (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}}$ 有且仅有三个公共点，求 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 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 1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

已知 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{x} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{|} \htmlData{tutor-start=15,tutor-end=16}{-} \htmlData{tutor-start=17,tutor-end=18}{|}\htmlData{tutor-start=18,tutor-end=19}{a}\htmlData{tutor-start=19,tutor-end=20}{x} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{|}$ . (1) 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 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{>} \htmlData{tutor-start=7,tutor-end=8}{1}$ 的解集； (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}$ 时不等式 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{>} \htmlData{tutor-start=7,tutor-end=8}{x}$ 成立，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 1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。