返回特征解读

2016 年高考数学（全国卷 1理科）

exams_raw/普通高考/2016/2016全国1理(河南,河北,山西,江西,湖北,湖南,广东,安徽,福建).pdf · HS-MATH-1024-v2.1-solution-aware

24 题0 个小问/题组
1

一、选择题 · 组合数学

设集合 A={x∣x2−4x+3<0}\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}{x}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{x}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{<}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=24}{\}}，B={x∣2x−3>0}\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=10}{\mid }\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{>}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=18}{\}}，则 A∩B=\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=6}{\cap }\htmlData{tutor-start=6,tutor-end=7}{B}\htmlData{tutor-start=7,tutor-end=8}{=}（ ） (A) (−3,−32)\left(\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{-}\dfrac{\htmlData{tutor-start=17,tutor-end=18}{3}}{\htmlData{tutor-start=20,tutor-end=21}{2}}\right) (B) (−3,32)\left(\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{,}\dfrac{\htmlData{tutor-start=16,tutor-end=17}{3}}{\htmlData{tutor-start=19,tutor-end=20}{2}}\right) (C) (1,32)\left(\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,}\dfrac{\htmlData{tutor-start=15,tutor-end=16}{3}}{\htmlData{tutor-start=18,tutor-end=19}{2}}\right) (D) (32,3)\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{3}}{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{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 1 题

解题过程

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

2

一、选择题 · 数学竞赛/待细分

设 (1+i)x=1+yi\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{y}\mathrm{\htmlData{tutor-start=27,tutor-end=28}{i}}，其中 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y} 是实数，则 ∣x+yi∣=\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{y}\mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}\htmlData{tutor-start=14,tutor-end=15}{|}\htmlData{tutor-start=15,tutor-end=16}{=}（ ） (A) 1\htmlData{tutor-start=0,tutor-end=1}{1} (B) 2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}} (C) 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}} (D) 2\htmlData{tutor-start=0,tutor-end=1}{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。

3

一、选择题 · 代数

已知等差数列 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 前 9 项的和为 27，a10=8\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{8}，则 a100=\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0}}\htmlData{tutor-start=7,tutor-end=8}{=}（ ） (A) 100\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0} (B) 99\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{9} (C) 98\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{8} (D) 97\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{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 3 题

解题过程

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

4

一、选择题 · 数学竞赛/待细分

某公司的班车在 7:30，8:00，8:30 发车，小明在 7:50 至 8:30 之间到达发车站乘坐班车，且到达发车站的时刻是随机的，则他等车时间不超过 10 分钟的概率是（ ） (A) 13\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{3}} (B) 12\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}} (C) 23\dfrac{\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{3}} (D) 34\dfrac{\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{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 4 题

解题过程

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

5

一、选择题 · 数学竞赛/待细分

已知方程 x2m2+n−y23m2−n=1\dfrac{\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}}{\htmlData{tutor-start=14,tutor-end=15}{m}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{n}}\htmlData{tutor-start=22,tutor-end=23}{-}\dfrac{\htmlData{tutor-start=30,tutor-end=31}{y}^{\htmlData{tutor-start=33,tutor-end=34}{2}}}{\htmlData{tutor-start=37,tutor-end=38}{3}\htmlData{tutor-start=38,tutor-end=39}{m}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{n}}\htmlData{tutor-start=46,tutor-end=47}{=}\htmlData{tutor-start=47,tutor-end=48}{1} 表示双曲线，且该双曲线两焦点间的距离为 4，则 n\htmlData{tutor-start=0,tutor-end=1}{n} 的取值范围是（ ） (A) (−1,3)\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=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{)} (B) (−1,3)\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}{,}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{)} (C) (0,3)\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}{3}\htmlData{tutor-start=4,tutor-end=5}{)} (D) (0,3)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{3}}\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 5 题

解题过程

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

6

一、选择题 · 平面几何

如图，某几何体的三视图是三个半径相等的圆及每个圆中两条相互垂直的半径. 若该几何体的体积是 28π3\dfrac{\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{8}\htmlData{tutor-start=9,tutor-end=12}{\pi}}{\htmlData{tutor-start=14,tutor-end=15}{3}}，则它的表面积是（ ） (A) 17π\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{7}\htmlData{tutor-start=2,tutor-end=5}{\pi} (B) 18π\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=5}{\pi} (C) 20π\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=5}{\pi} (D) 28π\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=5}{\pi}

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 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。

7

一、选择题 · 代数

函数 y=2x2−e∣x∣\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{x}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{-}\mathrm{\htmlData{tutor-start=17,tutor-end=18}{e}}^{\htmlData{tutor-start=21,tutor-end=22}{|}\htmlData{tutor-start=22,tutor-end=23}{x}\htmlData{tutor-start=23,tutor-end=24}{|}} 在 [−2,2]\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=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{]} 的图象大致为（ ） (A) （图） (B) （图） (C) （图） (D) （图）

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

解题过程

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

8

一、选择题 · 数学竞赛/待细分

若 a>b>1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{>}\htmlData{tutor-start=4,tutor-end=5}{1}，0<c<1\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}，则（ ） (A) ac<bc\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{c}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{c}} (B) abc<bac\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}^{\htmlData{tutor-start=4,tutor-end=5}{c}}\htmlData{tutor-start=6,tutor-end=7}{<}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{a}^{\htmlData{tutor-start=11,tutor-end=12}{c}} (C) alog⁡bc<blog⁡ac\htmlData{tutor-start=0,tutor-end=1}{a}\log_{\htmlData{tutor-start=7,tutor-end=8}{b}}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{b}\log_{\htmlData{tutor-start=18,tutor-end=19}{a}}\htmlData{tutor-start=20,tutor-end=21}{c} (D) log⁡ac<log⁡bc\log_{\htmlData{tutor-start=6,tutor-end=7}{a}}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{<}\log_{\htmlData{tutor-start=16,tutor-end=17}{b}}\htmlData{tutor-start=18,tutor-end=19}{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 8 题

解题过程

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

9

一、选择题 · 数学竞赛/待细分

执行下面的程序图，如果输入的 x=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}，y=1\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，则输出 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y} 的值满足（ ） (A) y=2x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{x} (B) y=3x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{x} (C) y=4x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{x} (D) y=5x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{x}

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 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。

10

一、选择题 · 平面几何

以抛物线 C\htmlData{tutor-start=0,tutor-end=1}{C} 的顶点为圆心的圆交 C\htmlData{tutor-start=0,tutor-end=1}{C} 于 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{B} 两点，交 C\htmlData{tutor-start=0,tutor-end=1}{C} 的准线于 D,E\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E} 两点. 已知 ∣AB∣=42\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{4}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}，∣DE∣=25\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{5}}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的焦点到准线的距离为（ ） (A) 2\htmlData{tutor-start=0,tutor-end=1}{2} (B) 4\htmlData{tutor-start=0,tutor-end=1}{4} (C) 6\htmlData{tutor-start=0,tutor-end=1}{6} (D) 8\htmlData{tutor-start=0,tutor-end=1}{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 10 题

解题过程

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

11

一、选择题 · 组合数学

平面 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 过正方体 ABCD−A1B1C1D1\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}} 的顶点 A\htmlData{tutor-start=0,tutor-end=1}{A}，α∥\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=15}{\parallel} 平面 CB1D1\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{D}_{\htmlData{tutor-start=9,tutor-end=10}{1}}，α∩\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=10}{\cap} 平面 ABCD=m\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}{m}，α∩\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=10}{\cap} 平面 ABB1A1=n\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{B}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{1}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{n}，则 m,n\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{n} 所成角的正弦值为（ ） (A) 32\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{2}} (B) 22\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{2}}}{\htmlData{tutor-start=17,tutor-end=18}{2}} (C) 33\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{3}} (D) 13\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{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 11 题

解题过程

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

12

一、选择题 · 代数

已知函数 f(x)=sin⁡(ωx+φ)\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}{=}\sin\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=17}{\omega }\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=26}{\varphi}\htmlData{tutor-start=26,tutor-end=27}{)} (ω>0,∣φ∣⩽π2)\left(\htmlData{tutor-start=6,tutor-end=12}{\omega}\htmlData{tutor-start=12,tutor-end=13}{>}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=23}{\varphi}\htmlData{tutor-start=23,tutor-end=24}{|}\htmlData{tutor-start=24,tutor-end=33}{\leqslant}\dfrac{\htmlData{tutor-start=40,tutor-end=43}{\pi}}{\htmlData{tutor-start=45,tutor-end=46}{2}}\right)，x=−π4\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\dfrac{\htmlData{tutor-start=10,tutor-end=13}{\pi}}{\htmlData{tutor-start=15,tutor-end=16}{4}} 为 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 的零点，x=π4\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\dfrac{\htmlData{tutor-start=9,tutor-end=12}{\pi}}{\htmlData{tutor-start=14,tutor-end=15}{4}} 为 y=f(x)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)} 图象的对称轴，且 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 在 (π18,5π36)\left(\dfrac{\htmlData{tutor-start=13,tutor-end=16}{\pi}}{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{8}}\htmlData{tutor-start=21,tutor-end=22}{,}\dfrac{\htmlData{tutor-start=29,tutor-end=30}{5}\htmlData{tutor-start=30,tutor-end=33}{\pi}}{\htmlData{tutor-start=35,tutor-end=36}{3}\htmlData{tutor-start=36,tutor-end=37}{6}}\right) 单调，则 ω\htmlData{tutor-start=0,tutor-end=6}{\omega} 的最大值为（ ） (A) 11\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{1} (B) 9\htmlData{tutor-start=0,tutor-end=1}{9} (C) 7\htmlData{tutor-start=0,tutor-end=1}{7} (D) 5\htmlData{tutor-start=0,tutor-end=1}{5}

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。

13

二、填空题 · 数学竞赛/待细分

设向量 a=(m,1)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{m}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}，b=(1,2)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{)}，且 ∣a+b∣2=∣a∣2+∣b∣2\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{|}\boldsymbol{\htmlData{tutor-start=49,tutor-end=50}{a}}\htmlData{tutor-start=51,tutor-end=52}{|}^{\htmlData{tutor-start=54,tutor-end=55}{2}}\htmlData{tutor-start=56,tutor-end=57}{+}\htmlData{tutor-start=57,tutor-end=58}{|}\boldsymbol{\htmlData{tutor-start=70,tutor-end=71}{b}}\htmlData{tutor-start=72,tutor-end=73}{|}^{\htmlData{tutor-start=75,tutor-end=76}{2}}，则 m=\htmlData{tutor-start=0,tutor-end=1}{m}\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。

14

二、填空题 · 数学竞赛/待细分

(2x+x)5\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{+}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{x}}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{5}} 的展开式中，x3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} 的系数是 ______.（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 14 题

解题过程

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

15

二、填空题 · 代数

设等比数列满足 a1+a3=10\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}{a}_{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{0}，a2+a4=5\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{4}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{5}，则 a1a2⋯an\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{a}_{\htmlData{tutor-start=8,tutor-end=9}{2}}\cdots \htmlData{tutor-start=17,tutor-end=18}{a}_{\htmlData{tutor-start=20,tutor-end=21}{n}} 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。

16

二、填空题 · 数学竞赛/待细分

某高科技企业生产产品 A\htmlData{tutor-start=0,tutor-end=1}{A} 和产品 B\htmlData{tutor-start=0,tutor-end=1}{B}，需要甲、乙两种新型材料. 生产一件产品 A\htmlData{tutor-start=0,tutor-end=1}{A} 需要甲材料 1.5 kg，乙材料 1 kg，用 5 个工时；生产一件产品 B\htmlData{tutor-start=0,tutor-end=1}{B} 需要甲材料 0.5 kg，乙材料 0.3 kg，用 3 个工时，生产一件产品 A\htmlData{tutor-start=0,tutor-end=1}{A} 的利润为 2100 元，生产一件产品 B\htmlData{tutor-start=0,tutor-end=1}{B} 的利润为 900 元. 该企业现有甲材料 150 kg，乙材料 90 kg，则在不超过 600 个工时的条件下，生产产品 A\htmlData{tutor-start=0,tutor-end=1}{A}、产品 B\htmlData{tutor-start=0,tutor-end=1}{B} 的利润之和的最大值为 ______ 元.

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。

17

三、解答题 · 平面几何

△ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C} 的内角 A,B,C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{C} 的对边分别为 a,b,c\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{c}，已知 2cos⁡C(acos⁡B+bcos⁡A)=c\htmlData{tutor-start=0,tutor-end=1}{2}\cos \htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{a}\cos \htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{b}\cos \htmlData{tutor-start=22,tutor-end=23}{A}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{c}. (1) 求 C\htmlData{tutor-start=0,tutor-end=1}{C}； (2) 若 c=7\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{7}}，△ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C} 的面积为 332\dfrac{\htmlData{tutor-start=7,tutor-end=8}{3}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{3}}}{\htmlData{tutor-start=18,tutor-end=19}{2}}，求 △ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C} 的周长.

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。

18

三、解答题 · 平面几何

如图，在以 A,B,C,D,E,F\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{F} 为顶点的五面体中，面 ABEF\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{F} 为正方形，AF=2FD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{F}\htmlData{tutor-start=5,tutor-end=6}{D}，∠AFD=90∘\angle AFD=90^\circ，且二面角 D−AF−E\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{F}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{E} 与二面角 C−BE−F\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{E}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{F} 都是 60∘60^\circ. (1) 证明：平面 ABEF⊥\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{F}\htmlData{tutor-start=4,tutor-end=9}{\perp} 平面 EFDC\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{C}； (2) 求二面角 E−BC−A\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{A} 的余弦值.

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 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。

19

解答题 · 数学竞赛/待细分

某公司计划购买 2 台机器，该种机器使用三年后即被淘汰，机器有一易损零件，在购进机器时，可以额外购买这种零件作为备件，每个 200 元。在机器使用期间，如果备件不足再购买，则每个 500 元。现需决策在购买机器时应同时购买几个易损零件，为此搜集并整理了 100 台这种机器在三年使用期内更换的易损零件数，得下面柱状图： 以这 100 台机器更换的易损零件数的频率代替 1 台机器更换的易损零件数发生的概率，记 X\htmlData{tutor-start=0,tutor-end=1}{X} 表示 2 台机器三年内共需更换的易损零件数，n\htmlData{tutor-start=0,tutor-end=1}{n} 表示购买 2 台机器的同时购买的易损零件数。 (1) 求 X\htmlData{tutor-start=0,tutor-end=1}{X} 的分布列； (2) 若要求 P(X⩽n)⩾0.5\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{X} \htmlData{tutor-start=4,tutor-end=14}{\leqslant }\htmlData{tutor-start=14,tutor-end=15}{n}\htmlData{tutor-start=15,tutor-end=16}{)} \htmlData{tutor-start=17,tutor-end=27}{\geqslant }\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{.}\htmlData{tutor-start=29,tutor-end=30}{5}，确定 n\htmlData{tutor-start=0,tutor-end=1}{n} 的最小值； (3) 以购买易损零件所需费用的期望值为决策依据，在 n=19\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{9} 与 n=20\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{0} 之中选其一，应选用哪个？

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 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。

20

解答题 · 平面几何

设圆 x2+y2+2x−15=0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{x} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{5} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{0} 的圆心为 A\htmlData{tutor-start=0,tutor-end=1}{A}，直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 过点 B(1,0)\htmlData{tutor-start=0,tutor-end=1}{B}\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}{)} 且与 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴不重合，l\htmlData{tutor-start=0,tutor-end=1}{l} 交圆 A\htmlData{tutor-start=0,tutor-end=1}{A} 于 C,D\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{D} 两点，过 B\htmlData{tutor-start=0,tutor-end=1}{B} 作 AC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} 的平行线交 AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} 于点 E\htmlData{tutor-start=0,tutor-end=1}{E}。 (1) 证明 ∣EA∣+∣EB∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{A}\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}{E}\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{|} 为定值，并写出点 E\htmlData{tutor-start=0,tutor-end=1}{E} 的轨迹方程； (2) 设点 E\htmlData{tutor-start=0,tutor-end=1}{E} 的轨迹为曲线 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}，直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 交 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 于 M,N\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{N} 两点，过 B\htmlData{tutor-start=0,tutor-end=1}{B} 且与 l\htmlData{tutor-start=0,tutor-end=1}{l} 垂直的直线与圆 A\htmlData{tutor-start=0,tutor-end=1}{A} 交于 P,Q\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{Q} 两点，求四边形 MPNQ\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{N}\htmlData{tutor-start=3,tutor-end=4}{Q} 面积的取值范围。

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 20 题

解题过程

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

21

解答题 · 代数

已知函数 f(x)=(x−2)ex+a(x−1)2\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}{2}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{e}^{\htmlData{tutor-start=15,tutor-end=16}{x}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{(}\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=28,tutor-end=29}{2}} 有两个零点。 (1) 求 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围； (2) 设 x1,x2\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}} 是 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 的两个零点，证明：x1+x2<2\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{<} \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 21 题

解题过程

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

22

解答题 · 平面几何

如图，△OAB\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{O}\htmlData{tutor-start=11,tutor-end=12}{A}\htmlData{tutor-start=12,tutor-end=13}{B} 是等腰三角形，∠AOB=120∘\angle AOB = 120^\circ，以 O\htmlData{tutor-start=0,tutor-end=1}{O} 为圆心，12OA\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{O}\htmlData{tutor-start=12,tutor-end=13}{A} 为半径作圆。 (1) 证明：直线 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 与 ⊙O\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O} 相切； (2) 点 C,D\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{D} 在 ⊙O\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O} 上，且 A,B,C,D\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{D} 四点共圆，证明：AB//CD\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=4,tutor-end=5}{/} \htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{D}。

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 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。

23

解答题 · 平面几何

在直角坐标系 xOy\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 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的参数方程为 {x=acos⁡t,y=1+asin⁡t,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{a} \cos \htmlData{tutor-start=25,tutor-end=26}{t}\htmlData{tutor-start=26,tutor-end=27}{,} \\ \htmlData{tutor-start=31,tutor-end=32}{y} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{1} \htmlData{tutor-start=37,tutor-end=38}{+} \htmlData{tutor-start=39,tutor-end=40}{a} \sin \htmlData{tutor-start=46,tutor-end=47}{t}\htmlData{tutor-start=47,tutor-end=48}{,} \end{cases} (t\htmlData{tutor-start=0,tutor-end=1}{t} 为参数，a>0\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0})。在以坐标原点为极点，x\htmlData{tutor-start=0,tutor-end=1}{x} 轴正半轴为极轴的极坐标系中，曲线 C2:ρ=4cos⁡θ\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=12}{\rho }\htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{4} \cos \htmlData{tutor-start=21,tutor-end=27}{\theta}。 (1) 说明 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 是哪种曲线，并将 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的方程化为极坐标方程； (2) 直线 C3\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{3}} 的极坐标方程为 θ=α0\htmlData{tutor-start=0,tutor-end=7}{\theta }\htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=15}{\alpha}_{\htmlData{tutor-start=17,tutor-end=18}{0}}，其中 α0\htmlData{tutor-start=0,tutor-end=6}{\alpha}_{\htmlData{tutor-start=8,tutor-end=9}{0}} 满足 tan⁡α0=2\tan \htmlData{tutor-start=5,tutor-end=11}{\alpha}_{\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{2}，若曲线 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 与 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的公共点都在 C3\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{3}} 上，求 a\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 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。

24

解答题 · 代数

已知函数 f(x)=∣x+1∣−∣2x−3∣\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}{|}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{|}。 (1) 在图中画出 y=f(x)\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}{)} 的图象； (2) 求不等式 ∣f(x)∣>1\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{f}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{|} \htmlData{tutor-start=7,tutor-end=8}{>} \htmlData{tutor-start=9,tutor-end=10}{1} 的解集。

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 1理科第 24 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。