2024 年高考数学（新高考卷 2）

exams_raw/普通高考/2024/2024新高考2(辽宁,重庆,海南,吉林,黑龙江,山西,云南,广西,甘肃,贵州,新疆).pdf · HS-MATH-1024-v2.1-solution-aware

19 题0 个小问/题组

一、单选题 · 数学竞赛/待细分

已知 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1} \htmlData{tutor-start=7,tutor-end=8}{-} \mathrm{\htmlData{tutor-start=17,tutor-end=18}{i}}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{z}\htmlData{tutor-start=2,tutor-end=3}{|} \htmlData{tutor-start=4,tutor-end=5}{=}$ ( ) (A) 0 (B) 1 (C) $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}$ (D) 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 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、单选题 · 数学竞赛/待细分

已知命题 $\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=11}{\forall }\htmlData{tutor-start=11,tutor-end=12}{x} \htmlData{tutor-start=13,tutor-end=17}{\in }\mathbf{\htmlData{tutor-start=25,tutor-end=26}{R}}\htmlData{tutor-start=27,tutor-end=28}{,} \htmlData{tutor-start=29,tutor-end=30}{|}\htmlData{tutor-start=30,tutor-end=31}{x}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{|} \htmlData{tutor-start=35,tutor-end=36}{>} \htmlData{tutor-start=37,tutor-end=38}{1}$ ; 命题 $\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=11}{\exists }\htmlData{tutor-start=11,tutor-end=12}{x} \htmlData{tutor-start=13,tutor-end=14}{>} \htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{x}^{\htmlData{tutor-start=21,tutor-end=22}{3}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{x}$ . 则 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{p}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{q}$ 都是真命题 (B) $\htmlData{tutor-start=0,tutor-end=5}{\neg }\htmlData{tutor-start=5,tutor-end=6}{p}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{q}$ 都是真命题 (C) $\htmlData{tutor-start=0,tutor-end=1}{p}$ 和 $\htmlData{tutor-start=0,tutor-end=5}{\neg }\htmlData{tutor-start=5,tutor-end=6}{q}$ 都是真命题 (D) $\htmlData{tutor-start=0,tutor-end=5}{\neg }\htmlData{tutor-start=5,tutor-end=6}{p}$ 和 $\htmlData{tutor-start=0,tutor-end=5}{\neg }\htmlData{tutor-start=5,tutor-end=6}{q}$ 都是真命题

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。

一、单选题 · 数学竞赛/待细分

已知向量 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}\htmlData{tutor-start=14,tutor-end=15}{,} \boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}$ 满足 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{|}\boldsymbol{\htmlData{tutor-start=35,tutor-end=36}{a}} \htmlData{tutor-start=38,tutor-end=39}{+} \htmlData{tutor-start=40,tutor-end=41}{2}\boldsymbol{\htmlData{tutor-start=53,tutor-end=54}{b}}\htmlData{tutor-start=55,tutor-end=56}{|} \htmlData{tutor-start=57,tutor-end=58}{=} \htmlData{tutor-start=59,tutor-end=60}{2}$ , 且 $\htmlData{tutor-start=0,tutor-end=1}{(}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{2}\boldsymbol{\htmlData{tutor-start=31,tutor-end=32}{a}}\htmlData{tutor-start=33,tutor-end=34}{)} \htmlData{tutor-start=35,tutor-end=41}{\perp }\boldsymbol{\htmlData{tutor-start=53,tutor-end=54}{b}}$ , 则 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|} \htmlData{tutor-start=17,tutor-end=18}{=}$ ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$ (B) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$ (C) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$ (D) 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 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、单选题 · 数学竞赛/待细分

某农业研究部门在面积相等的 100 块稻田上种植一种新型水稻，得到各块稻田的亩产量（单位：kg）并整理得下表：根据表中数据，下列结论中正确的是 ( ) (A) 100 块稻田亩产量的中位数小于 1050 kg (B) 100 块稻田中亩产量低于 1100 kg 的稻田所占比例超过 80% (C) 100 块稻田亩产量的极差介于 200 kg 至 300 kg 之间 (D) 100 块稻田亩产量的平均值介于 900 kg 至 1000 kg 之间

**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。

一、单选题 · 数学竞赛/待细分

已知曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{x}^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{+} \htmlData{tutor-start=11,tutor-end=12}{y}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{6} \htmlData{tutor-start=22,tutor-end=24}{\ }\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{y} \htmlData{tutor-start=27,tutor-end=28}{>} \htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=31}{)}$ ，从 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 上任意一点 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 向 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴作垂线段 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{P}'$ ， $\htmlData{tutor-start=0,tutor-end=1}{P}'$ 为垂足，则线段 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{P}'$ 的中点 $\htmlData{tutor-start=0,tutor-end=1}{M}$ 的轨迹方程为 ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{6}} \htmlData{tutor-start=17,tutor-end=18}{+} \frac{\htmlData{tutor-start=25,tutor-end=26}{y}^{\htmlData{tutor-start=28,tutor-end=29}{2}}}{\htmlData{tutor-start=32,tutor-end=33}{4}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{1} \htmlData{tutor-start=39,tutor-end=41}{\ }\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{y} \htmlData{tutor-start=44,tutor-end=45}{>} \htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{)}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{6}} \htmlData{tutor-start=17,tutor-end=18}{+} \frac{\htmlData{tutor-start=25,tutor-end=26}{y}^{\htmlData{tutor-start=28,tutor-end=29}{2}}}{\htmlData{tutor-start=32,tutor-end=33}{8}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{1} \htmlData{tutor-start=39,tutor-end=41}{\ }\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{y} \htmlData{tutor-start=44,tutor-end=45}{>} \htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{)}$ (C) $\frac{\htmlData{tutor-start=6,tutor-end=7}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{6}} \htmlData{tutor-start=17,tutor-end=18}{+} \frac{\htmlData{tutor-start=25,tutor-end=26}{x}^{\htmlData{tutor-start=28,tutor-end=29}{2}}}{\htmlData{tutor-start=32,tutor-end=33}{4}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{1} \htmlData{tutor-start=39,tutor-end=41}{\ }\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{y} \htmlData{tutor-start=44,tutor-end=45}{>} \htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{)}$ (D) $\frac{\htmlData{tutor-start=6,tutor-end=7}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{6}} \htmlData{tutor-start=17,tutor-end=18}{+} \frac{\htmlData{tutor-start=25,tutor-end=26}{x}^{\htmlData{tutor-start=28,tutor-end=29}{2}}}{\htmlData{tutor-start=32,tutor-end=33}{8}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{1} \htmlData{tutor-start=39,tutor-end=41}{\ }\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{y} \htmlData{tutor-start=44,tutor-end=45}{>} \htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{)}$

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。

一、单选题 · 代数

设函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{g}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=29}{=} \cos \htmlData{tutor-start=35,tutor-end=36}{x} \htmlData{tutor-start=37,tutor-end=38}{+} \htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=41}{a}\htmlData{tutor-start=41,tutor-end=42}{x}$ . 当 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}$ 时，曲线 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{g}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)}$ 恰有一个交点. 则 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=}$ ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$ (C) 1 (D) 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 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、单选题 · 数学竞赛/待细分

已知正三棱台 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{A}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{B}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{C}_{\htmlData{tutor-start=17,tutor-end=18}{1}}$ 的体积为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{3}}$ ， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{6}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{A}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{B}_{\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{2}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{A}$ 与平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 所成角的正切值为 ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$ (B) 1 (C) 2 (D) 3

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。

一、单选题 · 代数

设函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{)}\ln\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{b}\htmlData{tutor-start=19,tutor-end=20}{)}$ . 若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=15}{\geqslant }\htmlData{tutor-start=15,tutor-end=16}{0}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}}$ 的最小值为 ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{8}}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{4}}$ (C) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$ (D) 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 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。

二、多选题 · 代数

对于函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sin \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sin\left(\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{x} \htmlData{tutor-start=20,tutor-end=21}{-} \frac{\htmlData{tutor-start=28,tutor-end=31}{\pi}}{\htmlData{tutor-start=33,tutor-end=34}{4}}\right)$ ，下列说法中正确的有 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 有相同的零点 (B) $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 有相同的最大值 (C) $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 有相同的最小正周期 (D) $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的图象有相同的对称轴

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。

二、多选题 · 数学竞赛/待细分

抛物线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y}^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{x}$ 的准线为 $\htmlData{tutor-start=0,tutor-end=1}{l}$ ， $\htmlData{tutor-start=0,tutor-end=1}{P}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 上动点. 过 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 作 $\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{:} \htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{y}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{1}$ 的一条切线， $\htmlData{tutor-start=0,tutor-end=1}{Q}$ 为切点. 过 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 作 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的垂线，垂足为 $\htmlData{tutor-start=0,tutor-end=1}{B}$ . 则 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{l}$ 与 $\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{A}$ 相切 (B) 当 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{B}$ 三点共线时， $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{Q}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{=} \sqrt{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{5}}$ (C) 当 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{2}$ 时， $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{B}$ (D) 满足 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{|}$ 的点 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 有且仅有 2 个

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。

二、多选题 · 代数

设函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{3}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{x}^{\htmlData{tutor-start=21,tutor-end=22}{2}} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{1}$ ，则 ( ) (A) 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， $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 有三个零点 (B) 当 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0}$ 时， $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$ 是 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的极大值点 (C) 存在 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}$ ，使得 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{b}$ 为曲线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}$ 的对称轴 (D) 存在 $\htmlData{tutor-start=0,tutor-end=1}{a}$ ，使得点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{)}$ 为曲线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}$ 的对称中心

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。

三、填空题 · 代数

记 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ 为等差数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的前 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项和. 若 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{4}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{7}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{a}_{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{a}_{\htmlData{tutor-start=31,tutor-end=32}{5}} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{5}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}} \htmlData{tutor-start=7,tutor-end=8}{=}$ ______.

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。

三、填空题 · 数学竞赛/待细分

已知 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 为第一象限角， $\htmlData{tutor-start=0,tutor-end=5}{\beta}$ 为第三象限角， $\tan \htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{+} \tan \htmlData{tutor-start=19,tutor-end=25}{\beta }\htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{4}\htmlData{tutor-start=28,tutor-end=29}{,} \tan \htmlData{tutor-start=35,tutor-end=42}{\alpha }\tan \htmlData{tutor-start=47,tutor-end=53}{\beta }\htmlData{tutor-start=53,tutor-end=54}{=} \sqrt{\htmlData{tutor-start=61,tutor-end=62}{2}} \htmlData{tutor-start=64,tutor-end=65}{+} \htmlData{tutor-start=66,tutor-end=67}{1}$ ，则 $\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=19}{\beta}\htmlData{tutor-start=19,tutor-end=20}{)} \htmlData{tutor-start=21,tutor-end=22}{=}$ ______.

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。

三、填空题 · 数学竞赛/待细分

在如图的 $\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{4}$ 的方格表中选 4 个方格，要求每行和每列均恰有一个方格被选中，则共有种选法. 在所有符合上述要求的选法中，选中方格中的 4 个数之和的最大值是.

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。

四、解答题 · 平面几何

记 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 的内角 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}$ 的对边分别为 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{c}$ ，已知 $\sin \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{+} \sqrt{\htmlData{tutor-start=15,tutor-end=16}{3}} \cos \htmlData{tutor-start=23,tutor-end=24}{A} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{2}$ . (1) 求 $\htmlData{tutor-start=0,tutor-end=1}{A}$ ; (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,} \sqrt{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{b} \sin \htmlData{tutor-start=20,tutor-end=21}{C} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{c} \sin \htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{B}$ ，求 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 的周长.

题解状态：标准答案与规范题解待补充

题目标签：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。

四、解答题 · 代数

已知函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{e}^{\htmlData{tutor-start=10,tutor-end=11}{x}} \htmlData{tutor-start=13,tutor-end=14}{-} \htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{a}^{\htmlData{tutor-start=23,tutor-end=24}{3}}$ . (1) 当 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$ 时，求曲线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}$ 在点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{)}$ 处的切线方程; (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 有极小值，且极小值小于 0，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：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。

四、解答题 · 平面几何

如图，平面四边形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}$ 中， $AB=8, CD=3, AD=5\sqrt{3}, \angle ADC=90^\circ, \angle BAD=30^\circ$ ，点 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{F}$ 满足 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{E}}\htmlData{tutor-start=19,tutor-end=20}{=}\frac{\htmlData{tutor-start=26,tutor-end=27}{2}}{\htmlData{tutor-start=29,tutor-end=30}{5}}\overrightarrow{\htmlData{tutor-start=47,tutor-end=48}{A}\htmlData{tutor-start=48,tutor-end=49}{D}}\htmlData{tutor-start=50,tutor-end=51}{,} \overrightarrow{\htmlData{tutor-start=68,tutor-end=69}{A}\htmlData{tutor-start=69,tutor-end=70}{F}}\htmlData{tutor-start=71,tutor-end=72}{=}\frac{\htmlData{tutor-start=78,tutor-end=79}{1}}{\htmlData{tutor-start=81,tutor-end=82}{2}}\overrightarrow{\htmlData{tutor-start=99,tutor-end=100}{A}\htmlData{tutor-start=100,tutor-end=101}{B}}$ ，将 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 沿 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}$ 翻折至 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ ，使得 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{4}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}$ 。 (1) 证明： $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{P}\htmlData{tutor-start=10,tutor-end=11}{D}$ ； (2) 求面 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}$ 与面 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{F}$ 所成的二面角的正弦值。

**原卷题面及图示 1**原卷第 2 页 · question_region_fallback · 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 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。

四、解答题 · 数学竞赛/待细分

某投篮比赛分为两个阶段，每个参赛队由两名队员组成。比赛具体规则如下：第一阶段由参赛队中一名队员投篮 3 次，若 3 次都未投中，则该队被淘汰，比赛成绩为 0 分；若至少投中 1 次，则该队进入第二阶段。第二阶段由该队的另一名队员投篮 3 次，每次投篮投中得 5 分，未投中得 0 分。该队的比赛成绩为第二阶段的得分总和。某参赛队由甲、乙两名队员组成，设甲每次投中的概率为 $\htmlData{tutor-start=0,tutor-end=1}{p}$ ，乙每次投中的概率为 $\htmlData{tutor-start=0,tutor-end=1}{q}$ ，各次投中与否相互独立。 (1) 若 $\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{q}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{.}\htmlData{tutor-start=11,tutor-end=12}{5}$ ，甲参加第一阶段比赛，求甲、乙所在队的比赛成绩不少于 5 分的概率； (2) 假设 $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{p} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{q}$ 。 ① 为使得甲、乙所在队的比赛成绩为 15 分的概率最大，应该由谁参加第一阶段比赛？ ② 为使得甲、乙所在队的比赛成绩的数学期望最大，应该由谁参加第一阶段比赛？

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。

四、解答题 · 平面几何

已知双曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{x}^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{-} \htmlData{tutor-start=11,tutor-end=12}{y}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{m} \htmlData{tutor-start=21,tutor-end=23}{\ }\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{m} \htmlData{tutor-start=26,tutor-end=27}{>} \htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{)}$ ，点 $\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{)}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 上， $\htmlData{tutor-start=0,tutor-end=1}{k}$ 为常数， $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{k} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{1}$ 。按照如下方式依次构造点 $\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=8}{\ }\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{,} \cdots\htmlData{tutor-start=23,tutor-end=24}{)}$ ：过点 $\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}}$ 作斜率为 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 的直线与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的左支点交于点 $\htmlData{tutor-start=0,tutor-end=1}{Q}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}}$ ，令 $\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{Q}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}}$ 关于 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 轴的对称点。记 $\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ 的坐标为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{n}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{y}_{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=14}{)}$ 。 (1) 若 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{2}}$ ，求 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{y}_{\htmlData{tutor-start=10,tutor-end=11}{2}}$ ； (2) 证明：数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x}_{\htmlData{tutor-start=5,tutor-end=6}{n}} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{y}_{\htmlData{tutor-start=13,tutor-end=14}{n}}\htmlData{tutor-start=15,tutor-end=17}{\}}$ 是公比为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{k}}{\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{k}}$ 的等比数列； (3) 设 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}$ 为 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}_{\htmlData{tutor-start=13,tutor-end=14}{n}} \htmlData{tutor-start=16,tutor-end=17}{P}_{\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{1}} \htmlData{tutor-start=24,tutor-end=25}{P}_{\htmlData{tutor-start=27,tutor-end=28}{n}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{2}}$ 的面积。证明：对于任意正整数 $\htmlData{tutor-start=0,tutor-end=1}{n}$ ， $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{S}_{\htmlData{tutor-start=11,tutor-end=12}{n}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}}$ 。

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 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。