2015 年高考数学（全国卷 1理科）

exams_raw/普通高考/2015/2015全国1理(河南,河北,山西,江西).pdf · HS-MATH-1024-v2.1-solution-aware

24 题0 个小问/题组

一、选择题 · 数学竞赛/待细分

设复数 $\htmlData{tutor-start=0,tutor-end=1}{z}$ 满足 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{z}}{\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{z}}\htmlData{tutor-start=15,tutor-end=16}{=}\mathrm{\htmlData{tutor-start=24,tutor-end=25}{i}}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{z}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{=}$ ( ) (A) 1 (B) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}$ (C) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}$ (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 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 · 数学竞赛/待细分

$\sin \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{0}^{\htmlData{tutor-start=9,tutor-end=14}{\circ}} \cos \htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{0}^{\htmlData{tutor-start=25,tutor-end=30}{\circ}}\htmlData{tutor-start=31,tutor-end=32}{-}\cos \htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{6}\htmlData{tutor-start=39,tutor-end=40}{0}^{\htmlData{tutor-start=42,tutor-end=47}{\circ}} \sin \htmlData{tutor-start=54,tutor-end=55}{1}\htmlData{tutor-start=55,tutor-end=56}{0}^{\htmlData{tutor-start=58,tutor-end=63}{\circ}}\htmlData{tutor-start=64,tutor-end=65}{=}$ ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{2}}$ (B) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}}$ (D) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$

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 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

设命题 $\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=11}{\exists }\htmlData{tutor-start=11,tutor-end=12}{n} \htmlData{tutor-start=13,tutor-end=17}{\in }\mathbf{\htmlData{tutor-start=25,tutor-end=26}{N}}\htmlData{tutor-start=27,tutor-end=28}{,} \htmlData{tutor-start=29,tutor-end=30}{n}^{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{>}\htmlData{tutor-start=35,tutor-end=36}{2}^{\htmlData{tutor-start=38,tutor-end=39}{n}}$ ，则 $\htmlData{tutor-start=0,tutor-end=5}{\neg }\htmlData{tutor-start=5,tutor-end=6}{p}$ 为 ( ) (A) $\htmlData{tutor-start=0,tutor-end=8}{\forall }\htmlData{tutor-start=8,tutor-end=9}{n} \htmlData{tutor-start=10,tutor-end=14}{\in }\mathbf{\htmlData{tutor-start=22,tutor-end=23}{N}}\htmlData{tutor-start=24,tutor-end=25}{,} \htmlData{tutor-start=26,tutor-end=27}{n}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{>}\htmlData{tutor-start=32,tutor-end=33}{2}^{\htmlData{tutor-start=35,tutor-end=36}{n}}$ (B) $\htmlData{tutor-start=0,tutor-end=8}{\exists }\htmlData{tutor-start=8,tutor-end=9}{n} \htmlData{tutor-start=10,tutor-end=14}{\in }\mathbf{\htmlData{tutor-start=22,tutor-end=23}{N}}\htmlData{tutor-start=24,tutor-end=25}{,} \htmlData{tutor-start=26,tutor-end=27}{n}^{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=42}{\leqslant }\htmlData{tutor-start=42,tutor-end=43}{2}^{\htmlData{tutor-start=45,tutor-end=46}{n}}$ (C) $\htmlData{tutor-start=0,tutor-end=8}{\forall }\htmlData{tutor-start=8,tutor-end=9}{n} \htmlData{tutor-start=10,tutor-end=14}{\in }\mathbf{\htmlData{tutor-start=22,tutor-end=23}{N}}\htmlData{tutor-start=24,tutor-end=25}{,} \htmlData{tutor-start=26,tutor-end=27}{n}^{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=42}{\leqslant }\htmlData{tutor-start=42,tutor-end=43}{2}^{\htmlData{tutor-start=45,tutor-end=46}{n}}$ (D) $\htmlData{tutor-start=0,tutor-end=8}{\exists }\htmlData{tutor-start=8,tutor-end=9}{n} \htmlData{tutor-start=10,tutor-end=14}{\in }\mathbf{\htmlData{tutor-start=22,tutor-end=23}{N}}\htmlData{tutor-start=24,tutor-end=25}{,} \htmlData{tutor-start=26,tutor-end=27}{n}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{2}^{\htmlData{tutor-start=35,tutor-end=36}{n}}$

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 · 数学竞赛/待细分

投篮测试中，每人投 3 次，至少投中 2 次才能通过测试. 已知某同学每次投篮投中的概率为 0.6，且各次投篮是否投中相互独立，则该同学通过测试的概率为 ( ) (A) 0.648 (B) 0.432 (C) 0.36 (D) 0.312

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 · 数学竞赛/待细分

已知 $\htmlData{tutor-start=0,tutor-end=1}{M}\left(\htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{0}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{y}_{\htmlData{tutor-start=17,tutor-end=18}{0}}\right)$ 是双曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \frac{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{y}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{1}$ 上的一点， $\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{F}_{\htmlData{tutor-start=10,tutor-end=11}{2}}$ 是 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的两个焦点. 若 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{M} \htmlData{tutor-start=18,tutor-end=19}{F}_{\htmlData{tutor-start=21,tutor-end=22}{1}}} \htmlData{tutor-start=25,tutor-end=31}{\cdot }\overrightarrow{\htmlData{tutor-start=47,tutor-end=48}{M} \htmlData{tutor-start=49,tutor-end=50}{F}_{\htmlData{tutor-start=52,tutor-end=53}{2}}}\htmlData{tutor-start=55,tutor-end=56}{<}\htmlData{tutor-start=56,tutor-end=57}{0}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{0}}$ 的取值范围是 ( ) (A) $\left(\htmlData{tutor-start=6,tutor-end=7}{-}\frac{\sqrt{\htmlData{tutor-start=19,tutor-end=20}{3}}}{\htmlData{tutor-start=23,tutor-end=24}{3}}\htmlData{tutor-start=25,tutor-end=26}{,} \frac{\sqrt{\htmlData{tutor-start=39,tutor-end=40}{3}}}{\htmlData{tutor-start=43,tutor-end=44}{3}}\right)$ (B) $\left(\htmlData{tutor-start=6,tutor-end=7}{-}\frac{\sqrt{\htmlData{tutor-start=19,tutor-end=20}{3}}}{\htmlData{tutor-start=23,tutor-end=24}{6}}\htmlData{tutor-start=25,tutor-end=26}{,} \frac{\sqrt{\htmlData{tutor-start=39,tutor-end=40}{3}}}{\htmlData{tutor-start=43,tutor-end=44}{6}}\right)$ (C) $\left(\htmlData{tutor-start=6,tutor-end=7}{-}\frac{\htmlData{tutor-start=13,tutor-end=14}{2} \sqrt{\htmlData{tutor-start=21,tutor-end=22}{2}}}{\htmlData{tutor-start=25,tutor-end=26}{3}}\htmlData{tutor-start=27,tutor-end=28}{,} \frac{\htmlData{tutor-start=35,tutor-end=36}{2} \sqrt{\htmlData{tutor-start=43,tutor-end=44}{2}}}{\htmlData{tutor-start=47,tutor-end=48}{3}}\right)$ (D) $\left(\htmlData{tutor-start=6,tutor-end=7}{-}\frac{\htmlData{tutor-start=13,tutor-end=14}{2} \sqrt{\htmlData{tutor-start=21,tutor-end=22}{3}}}{\htmlData{tutor-start=25,tutor-end=26}{3}}\htmlData{tutor-start=27,tutor-end=28}{,} \frac{\htmlData{tutor-start=35,tutor-end=36}{2} \sqrt{\htmlData{tutor-start=43,tutor-end=44}{3}}}{\htmlData{tutor-start=47,tutor-end=48}{3}}\right)$

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

《九章算术》是我国古代内容极为丰富的数学名著，书中有如下问题：“今有委米依垣内角，下周八尺，高五尺. 问：积及为米几何？”其意思为：“在屋内墙角处堆放米（如图，米堆为一个圆锥的四分之一），米堆底部的弧长为 8 尺，米堆的高为 5 尺，问米堆的体积和堆放的米各为多少？”已知 1 斛米的体积约为 1.62 立方尺，圆周率约为 3，估算出堆放的米约有 ( ) (A) 14 斛 (B) 22 斛 (C) 36 斛 (D) 66 斛

**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 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

设 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 为 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A} \htmlData{tutor-start=12,tutor-end=13}{B} \htmlData{tutor-start=14,tutor-end=15}{C}$ 所在平面内一点， $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B} \htmlData{tutor-start=18,tutor-end=19}{C}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{3} \overrightarrow{\htmlData{tutor-start=39,tutor-end=40}{C} \htmlData{tutor-start=41,tutor-end=42}{D}}$ ，则 ( ) (A) $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A} \htmlData{tutor-start=18,tutor-end=19}{D}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{-}\frac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{3}} \overrightarrow{\htmlData{tutor-start=50,tutor-end=51}{A} \htmlData{tutor-start=52,tutor-end=53}{B}}\htmlData{tutor-start=54,tutor-end=55}{+}\frac{\htmlData{tutor-start=61,tutor-end=62}{4}}{\htmlData{tutor-start=64,tutor-end=65}{3}} \overrightarrow{\htmlData{tutor-start=83,tutor-end=84}{A} \htmlData{tutor-start=85,tutor-end=86}{C}}$ (B) $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A} \htmlData{tutor-start=18,tutor-end=19}{D}}\htmlData{tutor-start=20,tutor-end=21}{=}\frac{\htmlData{tutor-start=27,tutor-end=28}{1}}{\htmlData{tutor-start=30,tutor-end=31}{3}} \overrightarrow{\htmlData{tutor-start=49,tutor-end=50}{A} \htmlData{tutor-start=51,tutor-end=52}{B}}\htmlData{tutor-start=53,tutor-end=54}{-}\frac{\htmlData{tutor-start=60,tutor-end=61}{4}}{\htmlData{tutor-start=63,tutor-end=64}{3}} \overrightarrow{\htmlData{tutor-start=82,tutor-end=83}{A} \htmlData{tutor-start=84,tutor-end=85}{C}}$ (C) $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A} \htmlData{tutor-start=18,tutor-end=19}{D}}\htmlData{tutor-start=20,tutor-end=21}{=}\frac{\htmlData{tutor-start=27,tutor-end=28}{4}}{\htmlData{tutor-start=30,tutor-end=31}{3}} \overrightarrow{\htmlData{tutor-start=49,tutor-end=50}{A} \htmlData{tutor-start=51,tutor-end=52}{B}}\htmlData{tutor-start=53,tutor-end=54}{+}\frac{\htmlData{tutor-start=60,tutor-end=61}{1}}{\htmlData{tutor-start=63,tutor-end=64}{3}} \overrightarrow{\htmlData{tutor-start=82,tutor-end=83}{A} \htmlData{tutor-start=84,tutor-end=85}{C}}$ (D) $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A} \htmlData{tutor-start=18,tutor-end=19}{D}}\htmlData{tutor-start=20,tutor-end=21}{=}\frac{\htmlData{tutor-start=27,tutor-end=28}{4}}{\htmlData{tutor-start=30,tutor-end=31}{3}} \overrightarrow{\htmlData{tutor-start=49,tutor-end=50}{A} \htmlData{tutor-start=51,tutor-end=52}{B}}\htmlData{tutor-start=53,tutor-end=54}{-}\frac{\htmlData{tutor-start=60,tutor-end=61}{1}}{\htmlData{tutor-start=63,tutor-end=64}{3}} \overrightarrow{\htmlData{tutor-start=82,tutor-end=83}{A} \htmlData{tutor-start=84,tutor-end=85}{C}}$

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 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 · 代数

函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\cos \htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=18}{\omega }\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=27}{\varphi}\htmlData{tutor-start=27,tutor-end=28}{)}$ 的部分图象如图所示，则 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的单调递减区间为 ( ) (A) $\left(\htmlData{tutor-start=6,tutor-end=7}{k} \htmlData{tutor-start=8,tutor-end=11}{\pi}\htmlData{tutor-start=11,tutor-end=12}{-}\frac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{4}}\htmlData{tutor-start=23,tutor-end=24}{,} \htmlData{tutor-start=25,tutor-end=26}{k} \htmlData{tutor-start=27,tutor-end=30}{\pi}\htmlData{tutor-start=30,tutor-end=31}{+}\frac{\htmlData{tutor-start=37,tutor-end=38}{3}}{\htmlData{tutor-start=40,tutor-end=41}{4}}\right)\htmlData{tutor-start=49,tutor-end=50}{,} \htmlData{tutor-start=51,tutor-end=52}{k} \htmlData{tutor-start=53,tutor-end=57}{\in }\mathbf{\htmlData{tutor-start=65,tutor-end=66}{Z}}$ (B) $\left(\htmlData{tutor-start=6,tutor-end=7}{2} \htmlData{tutor-start=8,tutor-end=9}{k} \htmlData{tutor-start=10,tutor-end=13}{\pi}\htmlData{tutor-start=13,tutor-end=14}{-}\frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{4}}\htmlData{tutor-start=25,tutor-end=26}{,} \htmlData{tutor-start=27,tutor-end=28}{2} \htmlData{tutor-start=29,tutor-end=30}{k} \htmlData{tutor-start=31,tutor-end=34}{\pi}\htmlData{tutor-start=34,tutor-end=35}{+}\frac{\htmlData{tutor-start=41,tutor-end=42}{3}}{\htmlData{tutor-start=44,tutor-end=45}{4}}\right)\htmlData{tutor-start=53,tutor-end=54}{,} \htmlData{tutor-start=55,tutor-end=56}{k} \htmlData{tutor-start=57,tutor-end=61}{\in }\mathbf{\htmlData{tutor-start=69,tutor-end=70}{Z}}$ (C) $\left(\htmlData{tutor-start=6,tutor-end=7}{k}\htmlData{tutor-start=7,tutor-end=8}{-}\frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{4}}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{k}\htmlData{tutor-start=22,tutor-end=23}{+}\frac{\htmlData{tutor-start=29,tutor-end=30}{3}}{\htmlData{tutor-start=32,tutor-end=33}{4}}\right)\htmlData{tutor-start=41,tutor-end=42}{,} \htmlData{tutor-start=43,tutor-end=44}{k} \htmlData{tutor-start=45,tutor-end=49}{\in }\mathbf{\htmlData{tutor-start=57,tutor-end=58}{Z}}$ (D) $\left(\htmlData{tutor-start=6,tutor-end=7}{2} \htmlData{tutor-start=8,tutor-end=9}{k}\htmlData{tutor-start=9,tutor-end=10}{-}\frac{\htmlData{tutor-start=16,tutor-end=17}{1}}{\htmlData{tutor-start=19,tutor-end=20}{4}}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{2} \htmlData{tutor-start=25,tutor-end=26}{k}\htmlData{tutor-start=26,tutor-end=27}{+}\frac{\htmlData{tutor-start=33,tutor-end=34}{3}}{\htmlData{tutor-start=36,tutor-end=37}{4}}\right)\htmlData{tutor-start=45,tutor-end=46}{,} \htmlData{tutor-start=47,tutor-end=48}{k} \htmlData{tutor-start=49,tutor-end=53}{\in }\mathbf{\htmlData{tutor-start=61,tutor-end=62}{Z}}$

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 · 数学竞赛/待细分

执行如图的程序框图，如果输入的 $\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{1}$ ，则输出的 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}$ ( ) (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 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 · 数学竞赛/待细分

$\left(\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{y}\right)^{\htmlData{tutor-start=24,tutor-end=25}{5}}$ 的展开式中， $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{5}} \htmlData{tutor-start=6,tutor-end=7}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}$ 的系数为 ( ) (A) 10 (B) 20 (C) 30 (D) 60

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

圆柱被一个平面截去一部分后与半球（半径为 $\htmlData{tutor-start=0,tutor-end=1}{r}$ ）组成一个几何体，该几何体三视图中的正视图和俯视图如图所示. 若该几何体的表面积为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{0} \htmlData{tutor-start=6,tutor-end=9}{\pi}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{=}$ ( ) (A) 1 (B) 2 (C) 4 (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 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 · 数论

设函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\mathrm{\htmlData{tutor-start=13,tutor-end=14}{e}}^{\htmlData{tutor-start=17,tutor-end=18}{x}}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{2} \htmlData{tutor-start=22,tutor-end=23}{x}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{a} \htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{a}$ ，其中 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}$ ，若存在唯一的整数 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}$ 使得 $\htmlData{tutor-start=0,tutor-end=1}{f}\left(\htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{0}}\right)\htmlData{tutor-start=19,tutor-end=20}{<}\htmlData{tutor-start=20,tutor-end=21}{0}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的取值范围是 ( ) (A) $\left[\htmlData{tutor-start=6,tutor-end=7}{-}\frac{\htmlData{tutor-start=13,tutor-end=14}{3}}{\htmlData{tutor-start=16,tutor-end=17}{2} \mathrm{\htmlData{tutor-start=26,tutor-end=27}{e}}}\htmlData{tutor-start=29,tutor-end=30}{,} \htmlData{tutor-start=31,tutor-end=32}{1}\right)$ (B) $\left[\htmlData{tutor-start=6,tutor-end=7}{-}\frac{\htmlData{tutor-start=13,tutor-end=14}{3}}{\htmlData{tutor-start=16,tutor-end=17}{2} \mathrm{\htmlData{tutor-start=26,tutor-end=27}{e}}}\htmlData{tutor-start=29,tutor-end=30}{,} \frac{\htmlData{tutor-start=37,tutor-end=38}{3}}{\htmlData{tutor-start=40,tutor-end=41}{4}}\right)$ (C) $\left[\frac{\htmlData{tutor-start=12,tutor-end=13}{3}}{\htmlData{tutor-start=15,tutor-end=16}{2} \mathrm{\htmlData{tutor-start=25,tutor-end=26}{e}}}\htmlData{tutor-start=28,tutor-end=29}{,} \frac{\htmlData{tutor-start=36,tutor-end=37}{3}}{\htmlData{tutor-start=39,tutor-end=40}{4}}\right)$ (D) $\left[\frac{\htmlData{tutor-start=12,tutor-end=13}{3}}{\htmlData{tutor-start=15,tutor-end=16}{2} \mathrm{\htmlData{tutor-start=25,tutor-end=26}{e}}}\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=31}{1}\right)$

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 · 代数

若函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{x} \ln \left(\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\sqrt{\htmlData{tutor-start=25,tutor-end=26}{a}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{x}^{\htmlData{tutor-start=30,tutor-end=31}{2}}}\right)$ 为偶函数，则 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}$ ______.

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 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 · 平面几何

一个圆经过椭圆 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{6}}\htmlData{tutor-start=16,tutor-end=17}{+}\frac{\htmlData{tutor-start=23,tutor-end=24}{y}^{\htmlData{tutor-start=26,tutor-end=27}{2}}}{\htmlData{tutor-start=30,tutor-end=31}{4}}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{1}$ 的三个顶点，且圆心在 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴的正半轴上，则该圆的标准方程为______.

题解状态：标准答案与规范题解待补充

题目标签：全国卷 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

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 $\left\{\begin{array}{l}\htmlData{tutor-start=23,tutor-end=24}{x}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1} \htmlData{tutor-start=27,tutor-end=37}{\geqslant }\htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{,} \\ \htmlData{tutor-start=43,tutor-end=44}{x}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{y} \htmlData{tutor-start=47,tutor-end=57}{\leqslant }\htmlData{tutor-start=57,tutor-end=58}{0}\htmlData{tutor-start=58,tutor-end=59}{,} \\ \htmlData{tutor-start=63,tutor-end=64}{x}\htmlData{tutor-start=64,tutor-end=65}{+}\htmlData{tutor-start=65,tutor-end=66}{y}\htmlData{tutor-start=66,tutor-end=67}{-}\htmlData{tutor-start=67,tutor-end=68}{4} \htmlData{tutor-start=69,tutor-end=79}{\leqslant }\htmlData{tutor-start=79,tutor-end=80}{0}\htmlData{tutor-start=80,tutor-end=81}{,}\end{array}\right.$ 则 $\frac{\htmlData{tutor-start=6,tutor-end=7}{y}}{\htmlData{tutor-start=9,tutor-end=10}{x}}$ 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 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

在平面四边形 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{B} \htmlData{tutor-start=4,tutor-end=5}{C} \htmlData{tutor-start=6,tutor-end=7}{D}$ 中， $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=16}{\angle }\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=25}{\angle }\htmlData{tutor-start=25,tutor-end=26}{C}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{7}\htmlData{tutor-start=28,tutor-end=29}{5}^{\htmlData{tutor-start=31,tutor-end=36}{\circ}}\htmlData{tutor-start=37,tutor-end=38}{,} \htmlData{tutor-start=39,tutor-end=40}{B} \htmlData{tutor-start=41,tutor-end=42}{C}\htmlData{tutor-start=42,tutor-end=43}{=}\htmlData{tutor-start=43,tutor-end=44}{2}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{B}$ 的取值范围是______.

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 · 代数

$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ 为数列 $\left\{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\right\}$ 的前 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项和，已知 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{>}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{n}}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2} \htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{n}}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{4} \htmlData{tutor-start=29,tutor-end=30}{S}_{\htmlData{tutor-start=32,tutor-end=33}{n}}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{3}$ . (1) 求 $\left\{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\right\}$ 的通项公式； (2) 设 $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{n}} \htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{n}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{1}}}$ ，求数列 $\left\{\htmlData{tutor-start=7,tutor-end=8}{b}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\right\}$ 的前 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项和.

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 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{B} \htmlData{tutor-start=4,tutor-end=5}{C} \htmlData{tutor-start=6,tutor-end=7}{D}$ 为菱形， $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A} \htmlData{tutor-start=9,tutor-end=10}{B} \htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{0}^{\htmlData{tutor-start=18,tutor-end=23}{\circ}}\htmlData{tutor-start=24,tutor-end=25}{,} \htmlData{tutor-start=26,tutor-end=27}{E}\htmlData{tutor-start=27,tutor-end=28}{,} \htmlData{tutor-start=29,tutor-end=30}{F}$ 是平面 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{B} \htmlData{tutor-start=4,tutor-end=5}{C} \htmlData{tutor-start=6,tutor-end=7}{D}$ 同一侧的两点， $\htmlData{tutor-start=0,tutor-end=1}{B} \htmlData{tutor-start=2,tutor-end=3}{E} \htmlData{tutor-start=4,tutor-end=9}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{B} \htmlData{tutor-start=4,tutor-end=5}{C} \htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{D} \htmlData{tutor-start=11,tutor-end=12}{F} \htmlData{tutor-start=13,tutor-end=18}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{B} \htmlData{tutor-start=4,tutor-end=5}{C} \htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{B} \htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{2} \htmlData{tutor-start=15,tutor-end=16}{D} \htmlData{tutor-start=17,tutor-end=18}{F}\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=21}{A} \htmlData{tutor-start=22,tutor-end=23}{E} \htmlData{tutor-start=24,tutor-end=30}{\perp }\htmlData{tutor-start=30,tutor-end=31}{E} \htmlData{tutor-start=32,tutor-end=33}{C}$ . (1) 证明：平面 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{E} \htmlData{tutor-start=4,tutor-end=5}{C} \htmlData{tutor-start=6,tutor-end=11}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{F} \htmlData{tutor-start=4,tutor-end=5}{C}$ ； (2) 求直线 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{E}$ 与直线 $\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{F}$ 所成角的余弦值.

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

某公司为确定下一年度投入某产品的宣传费，需了解年宣传费 $\htmlData{tutor-start=0,tutor-end=1}{x}$ （单位：千元）对年销售量 $\htmlData{tutor-start=0,tutor-end=1}{y}$ （单位：t）和年利润 $\htmlData{tutor-start=0,tutor-end=1}{z}$ （单位：千元）的影响. 对近 8 年的年宣传费 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$ 和年销售量 $\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$ ( $\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \cdots\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{8}$ ) 数据作了初步处理，得到下面的散点图及一些统计量的值. 表中 $\htmlData{tutor-start=0,tutor-end=1}{w}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \sqrt{\htmlData{tutor-start=14,tutor-end=15}{x}_{\htmlData{tutor-start=17,tutor-end=18}{i}}}$ , $\bar{\htmlData{tutor-start=5,tutor-end=6}{w}} \htmlData{tutor-start=8,tutor-end=9}{=} \frac{\htmlData{tutor-start=16,tutor-end=17}{1}}{\htmlData{tutor-start=19,tutor-end=20}{8}} \sum_{\htmlData{tutor-start=28,tutor-end=29}{i}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{1}}^{\htmlData{tutor-start=34,tutor-end=35}{8}} \htmlData{tutor-start=37,tutor-end=38}{w}_{\htmlData{tutor-start=40,tutor-end=41}{i}}$ . (1) 根据散点图判断， $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{a} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{x}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{c} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{d}\sqrt{\htmlData{tutor-start=15,tutor-end=16}{x}}$ 哪一个适宜作为年销售量 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 关于年宣传费 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的回归方程类型？（给出判断即可，不必说明理由） (2) 根据 (1) 的判断结果及表中数据，建立 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 关于 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的回归方程； (3) 已知这种产品的年利润 $\htmlData{tutor-start=0,tutor-end=1}{z}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y}$ 的关系为 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{.}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{y} \htmlData{tutor-start=9,tutor-end=10}{-} \htmlData{tutor-start=11,tutor-end=12}{x}$ . 根据 (2) 的结果回答下列问题： ①年宣传费 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{9}$ 时，年销售量及年利润的预报值是多少？ ②年宣传费 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 为何值时，年利润的预报值最大？附：对于一组数据 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{v}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{u}_{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{v}_{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{,} \cdots\htmlData{tutor-start=38,tutor-end=39}{,} \htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{u}_{\htmlData{tutor-start=44,tutor-end=45}{n}}\htmlData{tutor-start=46,tutor-end=47}{,} \htmlData{tutor-start=48,tutor-end=49}{v}_{\htmlData{tutor-start=51,tutor-end=52}{n}}\htmlData{tutor-start=53,tutor-end=54}{)}$ ，其回归直线 $\htmlData{tutor-start=0,tutor-end=1}{v} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=11}{\alpha }\htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=19}{\beta }\htmlData{tutor-start=19,tutor-end=20}{u}$ 的斜率和截距的最小二乘估计分别为 $\hat{\htmlData{tutor-start=5,tutor-end=10}{\beta}} \htmlData{tutor-start=12,tutor-end=13}{=} \frac{\sum_{\htmlData{tutor-start=26,tutor-end=27}{i}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{1}}^{\htmlData{tutor-start=32,tutor-end=33}{n}} \htmlData{tutor-start=35,tutor-end=36}{(}\htmlData{tutor-start=36,tutor-end=37}{u}_{\htmlData{tutor-start=39,tutor-end=40}{i}} \htmlData{tutor-start=42,tutor-end=43}{-} \bar{\htmlData{tutor-start=49,tutor-end=50}{u}}\htmlData{tutor-start=51,tutor-end=52}{)}\htmlData{tutor-start=52,tutor-end=53}{(}\htmlData{tutor-start=53,tutor-end=54}{v}_{\htmlData{tutor-start=56,tutor-end=57}{i}} \htmlData{tutor-start=59,tutor-end=60}{-} \bar{\htmlData{tutor-start=66,tutor-end=67}{v}}\htmlData{tutor-start=68,tutor-end=69}{)}}{\sum_{\htmlData{tutor-start=77,tutor-end=78}{i}\htmlData{tutor-start=78,tutor-end=79}{=}\htmlData{tutor-start=79,tutor-end=80}{1}}^{\htmlData{tutor-start=83,tutor-end=84}{n}} \htmlData{tutor-start=86,tutor-end=87}{(}\htmlData{tutor-start=87,tutor-end=88}{u}_{\htmlData{tutor-start=90,tutor-end=91}{i}} \htmlData{tutor-start=93,tutor-end=94}{-} \bar{\htmlData{tutor-start=100,tutor-end=101}{u}}\htmlData{tutor-start=102,tutor-end=103}{)}^{\htmlData{tutor-start=105,tutor-end=106}{2}}}$ , $\hat{\htmlData{tutor-start=5,tutor-end=11}{\alpha}} \htmlData{tutor-start=13,tutor-end=14}{=} \bar{\htmlData{tutor-start=20,tutor-end=21}{v}} \htmlData{tutor-start=23,tutor-end=24}{-} \hat{\htmlData{tutor-start=30,tutor-end=35}{\beta}}\bar{\htmlData{tutor-start=41,tutor-end=42}{u}}$ .

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 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 · 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 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y} \htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}{\htmlData{tutor-start=20,tutor-end=21}{4}}$ 与直线 $\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{k}\htmlData{tutor-start=8,tutor-end=9}{x} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{a}$ ( $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}$ ) 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. (1) 当 $\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{0}$ 时，分别求 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 在点 $\htmlData{tutor-start=0,tutor-end=1}{M}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{N}$ 处的切线方程； (2) $\htmlData{tutor-start=0,tutor-end=1}{y}$ 轴上是否存在点 $\htmlData{tutor-start=0,tutor-end=1}{P}$ ，使得当 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 变动时，总有 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{M} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=20}{\angle }\htmlData{tutor-start=20,tutor-end=21}{O}\htmlData{tutor-start=21,tutor-end=22}{P}\htmlData{tutor-start=22,tutor-end=23}{N}$ ？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 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 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{3}} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=19}{+} \frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{4}}$ , $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\ln \htmlData{tutor-start=12,tutor-end=13}{x}$ . (1) 当 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 为何值时， $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴为曲线 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)}$ 的切线； (2) 用 $\min\htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{m}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=12}{\}}$ 表示 $\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{n}$ 中的最小值，设函数 $\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \min\htmlData{tutor-start=11,tutor-end=13}{\{}\htmlData{tutor-start=13,tutor-end=14}{f}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{g}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{x}\htmlData{tutor-start=22,tutor-end=23}{)}\htmlData{tutor-start=23,tutor-end=25}{\}}$ ( $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}$ )，讨论 $\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 零点的个数.

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 · 平面几何

如图， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 是 $\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O}$ 的直径， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}$ 是 $\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O}$ 的切线， $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 交 $\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O}$ 于点 $\htmlData{tutor-start=0,tutor-end=1}{E}$ . (1) 若 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}$ 的中点，证明： $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E}$ 是 $\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O}$ 的切线； (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=4}{=} \sqrt{\htmlData{tutor-start=11,tutor-end=12}{3}}\htmlData{tutor-start=13,tutor-end=14}{C}\htmlData{tutor-start=14,tutor-end=15}{E}$ ，求 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{B}$ 的大小.

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 · 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，直线 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{:} \htmlData{tutor-start=7,tutor-end=8}{x} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}$ ，圆 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{:} \htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{y}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{)}^{\htmlData{tutor-start=26,tutor-end=27}{2}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{1}$ ，以坐标原点为极点， $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴的正半轴为极轴建立极坐标系. (1) 求 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{C}_{\htmlData{tutor-start=10,tutor-end=11}{2}}$ 的极坐标方程； (2) 若直线 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{3}}$ 的极坐标方程为 $\htmlData{tutor-start=0,tutor-end=7}{\theta }\htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=18}{\pi}}{\htmlData{tutor-start=20,tutor-end=21}{4}}$ ( $\htmlData{tutor-start=0,tutor-end=5}{\rho }\htmlData{tutor-start=5,tutor-end=9}{\in }\mathbf{\htmlData{tutor-start=17,tutor-end=18}{R}}$ )，设 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{3}}$ 的交点为 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{N}$ ，求 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{C}_{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{M}\htmlData{tutor-start=16,tutor-end=17}{N}$ 的面积.

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。

解答题 · 代数

已知函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{|} \htmlData{tutor-start=13,tutor-end=14}{-} \htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{|}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{a}\htmlData{tutor-start=20,tutor-end=21}{|}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{a} \htmlData{tutor-start=25,tutor-end=26}{>} \htmlData{tutor-start=27,tutor-end=28}{0}$ . (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 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的图象与 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴围成的三角形面积大于 6，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 24 题

解题过程

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