返回特征解读

2020 年高考数学（新高考卷 2）

exams_raw/普通高考/2020/2020新高考2(海南).pdf · HS-MATH-1024-v2.1-solution-aware

22 题0 个小问/题组
1

一、单选题 · 组合数学

设集合 A={2,3,5,7}\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}{2}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{7}\htmlData{tutor-start=14,tutor-end=16}{\}}，B={1,2,3,5,8}\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}{1}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{5}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{8}\htmlData{tutor-start=17,tutor-end=19}{\}}，则 A∩B=\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{=} ( ) (A) {1,8}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{8}\htmlData{tutor-start=6,tutor-end=8}{\}} (B) {2,5}\htmlData{tutor-start=0,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}{5}\htmlData{tutor-start=6,tutor-end=8}{\}} (C) {2,3,5}\htmlData{tutor-start=0,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}{3}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=11}{\}} (D) {1,2,3,5,8}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{5}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{8}\htmlData{tutor-start=15,tutor-end=17}{\}}

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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、单选题 · 数学竞赛/待细分

(1+2i)(2+i)=\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=3,tutor-end=4}{2}\mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{+}\mathrm{\htmlData{tutor-start=26,tutor-end=27}{i}}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{=} ( ) (A) −5i\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{5}\mathrm{\htmlData{tutor-start=10,tutor-end=11}{i}} (B) 5i\htmlData{tutor-start=0,tutor-end=1}{5}\mathrm{\htmlData{tutor-start=9,tutor-end=10}{i}} (C) −5\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{5} (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：新高考卷 2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、单选题 · 平面几何

若 D\htmlData{tutor-start=0,tutor-end=1}{D} 为 △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} 的边 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 的中点，则 CB→=\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{B}}\htmlData{tutor-start=19,tutor-end=20}{=} ( ) (A) 2CD→−CA→\htmlData{tutor-start=0,tutor-end=1}{2}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{C}\htmlData{tutor-start=18,tutor-end=19}{D}}\htmlData{tutor-start=20,tutor-end=21}{-}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{C}\htmlData{tutor-start=38,tutor-end=39}{A}} (B) 2CA→−CD→\htmlData{tutor-start=0,tutor-end=1}{2}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{C}\htmlData{tutor-start=18,tutor-end=19}{A}}\htmlData{tutor-start=20,tutor-end=21}{-}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{C}\htmlData{tutor-start=38,tutor-end=39}{D}} (C) 2CD→+CA→\htmlData{tutor-start=0,tutor-end=1}{2}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{C}\htmlData{tutor-start=18,tutor-end=19}{D}}\htmlData{tutor-start=20,tutor-end=21}{+}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{C}\htmlData{tutor-start=38,tutor-end=39}{A}} (D) 2CA→+CD→\htmlData{tutor-start=0,tutor-end=1}{2}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{C}\htmlData{tutor-start=18,tutor-end=19}{A}}\htmlData{tutor-start=20,tutor-end=21}{+}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{C}\htmlData{tutor-start=38,tutor-end=39}{D}}

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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、单选题 · 数学竞赛/待细分

日晷是中国古代用来测定时间的仪器，利用与晷面垂直的晷针投射到晷面的影子来测定时间. 把地球看成一个球（球心记为 O\htmlData{tutor-start=0,tutor-end=1}{O}），地球上一点 A\htmlData{tutor-start=0,tutor-end=1}{A} 的纬度是指 OA\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{A} 与地球赤道所在平面所成角，点 A\htmlData{tutor-start=0,tutor-end=1}{A} 处的水平面是指过点 A\htmlData{tutor-start=0,tutor-end=1}{A} 且与 OA\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{A} 垂直的平面. 在点 A\htmlData{tutor-start=0,tutor-end=1}{A} 处放置一个日晷，若晷面与赤道所在平面平行，点 A\htmlData{tutor-start=0,tutor-end=1}{A} 处的纬度为北纬 40∘40^\circ，则晷针与点 A\htmlData{tutor-start=0,tutor-end=1}{A} 处的水平面所成角为 ( ) (A) 20∘20^\circ (B) 40∘40^\circ (C) 50∘50^\circ (D) 90∘90^\circ

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：新高考卷 2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、单选题 · 数学竞赛/待细分

某中学的学生积极参加体育锻炼，其中有 96%\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=4}{\%} 的学生喜欢足球或游泳，60%\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=4}{\%} 的学生喜欢足球，82%\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=4}{\%} 的学生喜欢游泳，则该中学既喜欢足球又喜欢游泳的学生数占该校学生总数的比例是 ( ) (A) 62%\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=4}{\%} (B) 56%\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=4}{\%} (C) 46%\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=4}{\%} (D) 42%\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=4}{\%}

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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、单选题 · 数学竞赛/待细分

3 名大学生利用假期到 2 个山村参加扶贫工作，每名大学生只去 1 个村，每个村至少 1 人，则不同的分配方案共有 ( ) (A) 4 种 (B) 5 种 (C) 6 种 (D) 8 种

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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、单选题 · 代数

已知函数 f(x)=lg⁡(x2−4x−5)\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}{=}\lg\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{5}\htmlData{tutor-start=19,tutor-end=20}{)} 在 (a,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\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}{)} 单调递增，则 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围是 ( ) (A) (−∞,−1]\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{]} (B) (−∞,2]\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{]} (C) [2,+∞)\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}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} (D) [5,+∞)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{5}\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：新高考卷 2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、单选题 · 代数

若定义在 R\mathbf{\htmlData{tutor-start=8,tutor-end=9}{R}} 的奇函数 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 (−∞,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)} 单调递减，且 f(2)=0\htmlData{tutor-start=0,tutor-end=1}{f}\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}{=}\htmlData{tutor-start=5,tutor-end=6}{0}，则满足 xf(x−1)⩾0\htmlData{tutor-start=0,tutor-end=1}{x}\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}{1}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=18}{\geqslant }\htmlData{tutor-start=18,tutor-end=19}{0} 的 x\htmlData{tutor-start=0,tutor-end=1}{x} 的取值范围是 ( ) (A) [−1,1]∪[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=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{]} \htmlData{tutor-start=8,tutor-end=13}{\cup }\htmlData{tutor-start=13,tutor-end=14}{[}\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=24}{\infty}\htmlData{tutor-start=24,tutor-end=25}{)} (B) [−3,−1]∪[0,1]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{]} \htmlData{tutor-start=9,tutor-end=14}{\cup }\htmlData{tutor-start=14,tutor-end=15}{[}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{]} (C) [−1,0]∪[1,+∞)\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}{]} \htmlData{tutor-start=8,tutor-end=13}{\cup }\htmlData{tutor-start=13,tutor-end=14}{[}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=24}{\infty}\htmlData{tutor-start=24,tutor-end=25}{)} (D) [−1,0]∪[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=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{]} \htmlData{tutor-start=8,tutor-end=13}{\cup }\htmlData{tutor-start=13,tutor-end=14}{[}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{]}

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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

二、多选题 · 数学竞赛/待细分

我国新冠肺炎疫情进入常态化，各地有序推进复工复产，下面是某地连续 11 天复工复产指数折线图，下列说法正确的是 ( ) (A) 这 11 天复工指数和复产指数均逐日增加 (B) 这 11 天期间，复产指数增量大于复工指数的增量 (C) 第 3 天至第 11 天复工复产指数均超过 80%\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=4}{\%} (D) 第 9 天至第 11 天复产指数增量大于复工指数的增量

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：新高考卷 2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

二、多选题 · 平面几何

已知曲线 C:mx2+ny2=1\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{m}\htmlData{tutor-start=4,tutor-end=5}{x}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{n}\htmlData{tutor-start=11,tutor-end=12}{y}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{1} ( ) (A) 若 m>n>0\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{>}\htmlData{tutor-start=4,tutor-end=5}{0}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 是椭圆，其焦点在 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴上 (B) 若 m=n>0\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{>}\htmlData{tutor-start=4,tutor-end=5}{0}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 是圆，其半径为 n\sqrt{\htmlData{tutor-start=6,tutor-end=7}{n}} (C) 若 mn<0\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{<}\htmlData{tutor-start=3,tutor-end=4}{0}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 是双曲线，其渐近线方程为 y=±−mnx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pm}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{-}\frac{\htmlData{tutor-start=18,tutor-end=19}{m}}{\htmlData{tutor-start=21,tutor-end=22}{n}}}\htmlData{tutor-start=24,tutor-end=25}{x} (D) 若 m=0,n>0\htmlData{tutor-start=0,tutor-end=1}{m}\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=5,tutor-end=6}{n}\htmlData{tutor-start=6,tutor-end=7}{>}\htmlData{tutor-start=7,tutor-end=8}{0}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 是两条直线

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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

二、多选题 · 代数

下图是函数 y=sin⁡(ωx+φ)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sin\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=14}{\omega }\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=23}{\varphi}\htmlData{tutor-start=23,tutor-end=24}{)} 的部分图象，则 sin⁡(ωx+φ)=\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=12}{\omega }\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=21}{\varphi}\htmlData{tutor-start=21,tutor-end=22}{)}\htmlData{tutor-start=22,tutor-end=23}{=} ( ) (A) sin⁡(x+π3)\sin\left(\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{+}\frac{\htmlData{tutor-start=18,tutor-end=21}{\pi}}{\htmlData{tutor-start=23,tutor-end=24}{3}}\right) (B) sin⁡(π3−2x)\sin\left(\frac{\htmlData{tutor-start=16,tutor-end=19}{\pi}}{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{x}\right) (C) cos⁡(2x+π6)\cos\left(\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\frac{\htmlData{tutor-start=19,tutor-end=22}{\pi}}{\htmlData{tutor-start=24,tutor-end=25}{6}}\right) (D) cos⁡(5π6−2x)\cos\left(\frac{\htmlData{tutor-start=16,tutor-end=17}{5}\htmlData{tutor-start=17,tutor-end=20}{\pi}}{\htmlData{tutor-start=22,tutor-end=23}{6}}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{x}\right)

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：新高考卷 2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

二、多选题 · 数学竞赛/待细分

已知 a>0,b>0\htmlData{tutor-start=0,tutor-end=1}{a}\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=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{>}\htmlData{tutor-start=7,tutor-end=8}{0}，且 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}，则 ( ) (A) a2+b2⩾12\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}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=22}{\geqslant }\frac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{2}} (B) 2a−b>12\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{b}} \htmlData{tutor-start=8,tutor-end=9}{>} \frac{\htmlData{tutor-start=16,tutor-end=17}{1}}{\htmlData{tutor-start=19,tutor-end=20}{2}} (C) log⁡2a+log⁡2b⩾−2\log_{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{a} \htmlData{tutor-start=11,tutor-end=12}{+} \log_{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{b} \htmlData{tutor-start=24,tutor-end=34}{\geqslant }\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{2} (D) a+b⩽2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{a}}\htmlData{tutor-start=8,tutor-end=9}{+}\sqrt{\htmlData{tutor-start=15,tutor-end=16}{b}} \htmlData{tutor-start=18,tutor-end=28}{\leqslant }\sqrt{\htmlData{tutor-start=34,tutor-end=35}{2}}

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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

三、填空题 · 数学竞赛/待细分

棱长为 2 的正方体 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}} 中，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} 分别为棱 BB1,AB\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{B} 的中点，则三棱锥 A−NMD1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{N}\htmlData{tutor-start=3,tutor-end=4}{M}\htmlData{tutor-start=4,tutor-end=5}{D}_{\htmlData{tutor-start=7,tutor-end=8}{1}} 的体积为______.

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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

三、填空题 · 平面几何

斜率为 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}} 的直线过抛物线 C:y2=4x\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{x} 的焦点，且与 C\htmlData{tutor-start=0,tutor-end=1}{C} 交于 A,B\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，则 ∣AB∣=\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}{=}______.

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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

三、填空题 · 代数

将数列 {2n−1}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=8}{\}} 与 {3n−2}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=8}{\}} 的公共项从小到大排列得到数列 {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}{\}}，则 {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}{\}} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和为______.

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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

三、填空题 · 平面几何

某中学开展劳动实习，学生加工制作零件，零件的截面如图所示. O\htmlData{tutor-start=0,tutor-end=1}{O} 为圆孔及轮廓圆弧 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 所在圆的圆心，A\htmlData{tutor-start=0,tutor-end=1}{A} 是圆弧 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 与直线 AG\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{G} 的切点，B\htmlData{tutor-start=0,tutor-end=1}{B} 是圆弧 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 与直线 BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} 的切点，四边形 DEFG\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{G} 为矩形，BC⊥DG\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{D}\htmlData{tutor-start=10,tutor-end=11}{G}，垂足为 C\htmlData{tutor-start=0,tutor-end=1}{C}，tan⁡∠ODC=35\tan \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{O}\htmlData{tutor-start=13,tutor-end=14}{D}\htmlData{tutor-start=14,tutor-end=15}{C} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{3}}{\htmlData{tutor-start=27,tutor-end=28}{5}}，BH∥DG\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{H} \htmlData{tutor-start=3,tutor-end=13}{\parallel }\htmlData{tutor-start=13,tutor-end=14}{D}\htmlData{tutor-start=14,tutor-end=15}{G}，EF=12 cm\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{2}\text{ \htmlData{tutor-start=12,tutor-end=13}{c}\htmlData{tutor-start=13,tutor-end=14}{m}}，DE=2 cm\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\text{ \htmlData{tutor-start=11,tutor-end=12}{c}\htmlData{tutor-start=12,tutor-end=13}{m}}，A\htmlData{tutor-start=0,tutor-end=1}{A} 到直线 DE\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E} 和 EF\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F} 的距离均为 7 cm\htmlData{tutor-start=0,tutor-end=1}{7}\text{ \htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{m}}，圆孔半径为 1 cm\htmlData{tutor-start=0,tutor-end=1}{1}\text{ \htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{m}}，则图中阴影部分的面积为______cm2\text{\htmlData{tutor-start=6,tutor-end=7}{c}\htmlData{tutor-start=7,tutor-end=8}{m}}^{\htmlData{tutor-start=11,tutor-end=12}{2}}.

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：新高考卷 2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

四、解答题 · 平面几何

在① ac=3\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{=}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{3}}，② csin⁡A=3\htmlData{tutor-start=0,tutor-end=1}{c}\sin \htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{3}，③ c=3b\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}\htmlData{tutor-start=10,tutor-end=11}{b} 这三个条件中任选一个，补充在下面问题中，若问题中的三角形存在，求 c\htmlData{tutor-start=0,tutor-end=1}{c} 的值；若问题中的三角形不存在，说明理由. 问题：是否存在 △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}，它的内角 A,B,C\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} 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=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{c}，且 sin⁡A=3sin⁡B,C=π6\sin \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{=}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\sin \htmlData{tutor-start=20,tutor-end=21}{B}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{C}\htmlData{tutor-start=24,tutor-end=25}{=}\frac{\htmlData{tutor-start=31,tutor-end=34}{\pi}}{\htmlData{tutor-start=36,tutor-end=37}{6}}，______？

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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

四、解答题 · 代数

已知公比大于 1 的等比数列 {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}{\}} 满足 a2+a4=20,a3=8\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}{2}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{3}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{8}. (1) 求 {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}{\}} 的通项公式； (2) 求 a1a2−a2a3+⋯+(−1)n−1anan+1\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}}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{3}}\htmlData{tutor-start=21,tutor-end=22}{+}\cdots\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{)}^{\htmlData{tutor-start=35,tutor-end=36}{n}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{1}}\htmlData{tutor-start=39,tutor-end=40}{a}_{\htmlData{tutor-start=42,tutor-end=43}{n}}\htmlData{tutor-start=44,tutor-end=45}{a}_{\htmlData{tutor-start=47,tutor-end=48}{n}\htmlData{tutor-start=48,tutor-end=49}{+}\htmlData{tutor-start=49,tutor-end=50}{1}}.

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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

四、解答题 · 数学竞赛/待细分

为加强环境保护，治理空气污染，环境监测部门对某市空气质量进行调研，随机抽查了 100 天空气中的 PM2.5 和 SO2_{\htmlData{tutor-start=2,tutor-end=3}{2}} 浓度（单位：μg/m3\htmlData{tutor-start=0,tutor-end=3}{\mu}\text{\htmlData{tutor-start=9,tutor-end=10}{g}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{m}}^{\htmlData{tutor-start=15,tutor-end=16}{3}}），得下表： | | SO2_{\htmlData{tutor-start=2,tutor-end=3}{2}} [0, 50] | (50, 150] | (150, 475] | |---|---|---|---| | PM2.5 [0, 35] | 32 | 18 | 4 | | (35, 75] | 6 | 8 | 12 | | (75, 115] | 3 | 7 | 10 | (1) 估计事件“该市一天空气中 PM2.5 浓度不超过 75，且 SO2_{\htmlData{tutor-start=2,tutor-end=3}{2}} 浓度不超过 150”的概率； (2) 根据所给数据，完成下面的 2×2\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{2} 列联表： | | SO2_{\htmlData{tutor-start=2,tutor-end=3}{2}} [0, 150] | (150, 475] | |---|---|---| | PM2.5 [0, 75] | | | | (75, 115] | | | (3) 根据 (2) 中的列联表，判断是否有 99% 的把握认为该市一天空气中 PM2.5 浓度与 SO2_{\htmlData{tutor-start=2,tutor-end=3}{2}} 浓度有关？ 附：K2=n(ad−bc)2(a+b)(c+d)(a+c)(b+d)\htmlData{tutor-start=0,tutor-end=1}{K}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{n}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{a}\htmlData{tutor-start=17,tutor-end=18}{d}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{b}\htmlData{tutor-start=20,tutor-end=21}{c}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}}}{\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{a}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{b}\htmlData{tutor-start=32,tutor-end=33}{)}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{c}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{d}\htmlData{tutor-start=37,tutor-end=38}{)}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{a}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{c}\htmlData{tutor-start=42,tutor-end=43}{)}\htmlData{tutor-start=43,tutor-end=44}{(}\htmlData{tutor-start=44,tutor-end=45}{b}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{d}\htmlData{tutor-start=47,tutor-end=48}{)}}， | P(K2⩾k)\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{K}^{\htmlData{tutor-start=5,tutor-end=6}{2}} \htmlData{tutor-start=8,tutor-end=18}{\geqslant }\htmlData{tutor-start=18,tutor-end=19}{k}\htmlData{tutor-start=19,tutor-end=20}{)} | 0.050 | 0.010 | 0.001 | |---|---|---|---| | k\htmlData{tutor-start=0,tutor-end=1}{k} | 3.841 | 6.635 | 10.828 |

原卷图示 1
原卷图示 1原卷第 2 页 · qwen3.7-plus_layout_detection · 需复核
原卷图示 2
原卷图示 2原卷第 2 页 · qwen3.7-plus_layout_detection · 需复核
原卷图示 3
原卷图示 3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：新高考卷 2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

四、解答题 · 数学竞赛/待细分

如图，四棱锥 P−ABCD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{D} 的底面为正方形，PD⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=8}{\perp} 底面 ABCD\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}。设平面 PAD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{D} 与平面 PBC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C} 的交线为 l\htmlData{tutor-start=0,tutor-end=1}{l}。 (1) 证明：l⊥\htmlData{tutor-start=0,tutor-end=1}{l} \htmlData{tutor-start=2,tutor-end=7}{\perp} 平面 PDC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{C}； (2) 已知 PD=AD=1\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{D}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}，Q\htmlData{tutor-start=0,tutor-end=1}{Q} 为 l\htmlData{tutor-start=0,tutor-end=1}{l} 上的点，QB=2\htmlData{tutor-start=0,tutor-end=1}{Q}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{2}}，求 PB\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B} 与平面 QCD\htmlData{tutor-start=0,tutor-end=1}{Q}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D} 所成角的正弦值。

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：新高考卷 2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

四、解答题 · 平面几何

已知椭圆 C:x2a2+y2b2=1 (a>b>0)\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \frac{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}} \htmlData{tutor-start=23,tutor-end=24}{+} \frac{\htmlData{tutor-start=31,tutor-end=32}{y}^{\htmlData{tutor-start=34,tutor-end=35}{2}}}{\htmlData{tutor-start=38,tutor-end=39}{b}^{\htmlData{tutor-start=41,tutor-end=42}{2}}} \htmlData{tutor-start=45,tutor-end=46}{=} \htmlData{tutor-start=47,tutor-end=48}{1} \htmlData{tutor-start=49,tutor-end=51}{\ }\htmlData{tutor-start=51,tutor-end=52}{(}\htmlData{tutor-start=52,tutor-end=53}{a}\htmlData{tutor-start=53,tutor-end=54}{>}\htmlData{tutor-start=54,tutor-end=55}{b}\htmlData{tutor-start=55,tutor-end=56}{>}\htmlData{tutor-start=56,tutor-end=57}{0}\htmlData{tutor-start=57,tutor-end=58}{)} 过点 M(2,3)\htmlData{tutor-start=0,tutor-end=1}{M}\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}{3}\htmlData{tutor-start=5,tutor-end=6}{)}，点 A\htmlData{tutor-start=0,tutor-end=1}{A} 为其左顶点，且 AM\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M} 的斜率为 12\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}。 (1) 求 C\htmlData{tutor-start=0,tutor-end=1}{C} 方程； (2) 点 N\htmlData{tutor-start=0,tutor-end=1}{N} 为椭圆上任意一点，求 △AMN\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{M}\htmlData{tutor-start=12,tutor-end=13}{N} 的面积的最大值。

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 2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

四、解答题 · 平面几何

已知函数 f(x)=aex−1−ln⁡x+ln⁡a\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}{a}\htmlData{tutor-start=8,tutor-end=9}{e}^{\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{-} \ln \htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=25}{+} \ln \htmlData{tutor-start=30,tutor-end=31}{a}。 (1) 当 a=e\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{e} 时，求曲线 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}{)} 在点 (1,f(1))\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}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{)} 处的切线与两坐标轴围成的三角形的面积； (2) 若 f(x)⩾1\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=15}{\geqslant }\htmlData{tutor-start=15,tutor-end=16}{1}，求 a\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：新高考卷 2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。