返回特征解读

2023 年高考数学（新高考卷 1）

exams_raw/普通高考/2023/2023新高考1(山东,广东,湖南,湖北,河北,江苏,福建,浙江).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

已知集合 M={−2,−1,0,1,2}\htmlData{tutor-start=0,tutor-end=1}{M} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{2}\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}{,} \htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=23}{\}}, N={x∣x2−x−6⩾0}\htmlData{tutor-start=0,tutor-end=1}{N} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=13}{\mid }\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=24}{-} \htmlData{tutor-start=25,tutor-end=26}{6} \htmlData{tutor-start=27,tutor-end=37}{\geqslant }\htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=40}{\}}, 则 M∩N=()\htmlData{tutor-start=0,tutor-end=1}{M} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{N} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{(}\quad\htmlData{tutor-start=17,tutor-end=18}{)} (A) {−2,−1,0,1}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=16}{\}} (B) {0,1,2}\htmlData{tutor-start=0,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}{1}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=11}{\}} (C) {−2}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=6}{\}} (D) {2}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,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 1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

已知 z=1−i2+2i\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{-}\mathrm{\htmlData{tutor-start=20,tutor-end=21}{i}}}{\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{2}\mathrm{\htmlData{tutor-start=35,tutor-end=36}{i}}}, 则 z−zˉ=()\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{-} \bar{\htmlData{tutor-start=9,tutor-end=10}{z}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{(}\quad\htmlData{tutor-start=20,tutor-end=21}{)} (A) −i\htmlData{tutor-start=0,tutor-end=1}{-}\mathrm{\htmlData{tutor-start=9,tutor-end=10}{i}} (B) i\mathrm{\htmlData{tutor-start=8,tutor-end=9}{i}} (C) 0\htmlData{tutor-start=0,tutor-end=1}{0} (D) 1\htmlData{tutor-start=0,tutor-end=1}{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 1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 · 数学竞赛/待细分

已知向量 a=(1,1)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{)}, b=(1,−1)\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{)}. 若 (a+λb)⊥(a+μb)\htmlData{tutor-start=0,tutor-end=1}{(}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=26}{\lambda }\boldsymbol{\htmlData{tutor-start=38,tutor-end=39}{b}}\htmlData{tutor-start=40,tutor-end=41}{)} \htmlData{tutor-start=42,tutor-end=48}{\perp }\htmlData{tutor-start=48,tutor-end=49}{(}\boldsymbol{\htmlData{tutor-start=61,tutor-end=62}{a}} \htmlData{tutor-start=64,tutor-end=65}{+} \htmlData{tutor-start=66,tutor-end=70}{\mu }\boldsymbol{\htmlData{tutor-start=82,tutor-end=83}{b}}\htmlData{tutor-start=84,tutor-end=85}{)}, 则 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) λ+μ=1\htmlData{tutor-start=0,tutor-end=8}{\lambda }\htmlData{tutor-start=8,tutor-end=9}{+} \htmlData{tutor-start=10,tutor-end=14}{\mu }\htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{1} (B) λ+μ=−1\htmlData{tutor-start=0,tutor-end=8}{\lambda }\htmlData{tutor-start=8,tutor-end=9}{+} \htmlData{tutor-start=10,tutor-end=14}{\mu }\htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1} (C) λμ=1\htmlData{tutor-start=0,tutor-end=8}{\lambda }\htmlData{tutor-start=8,tutor-end=12}{\mu }\htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{1} (D) λμ=−1\htmlData{tutor-start=0,tutor-end=8}{\lambda }\htmlData{tutor-start=8,tutor-end=12}{\mu }\htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{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 1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 · 代数

设函数 f(x)=2x(x−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}{2}^{\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{a}\htmlData{tutor-start=15,tutor-end=16}{)}} 在区间 (0,1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} 单调递减, 则 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围是 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) (−∞,−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}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{]} (B) [−2,0)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)} (C) (0,2]\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{]} (D) [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}{)}

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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 · 平面几何

设椭圆 C1:x2a2+y2=1 (a>1)\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{:} \frac{\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}{\htmlData{tutor-start=20,tutor-end=21}{a}^{\htmlData{tutor-start=23,tutor-end=24}{2}}} \htmlData{tutor-start=27,tutor-end=28}{+} \htmlData{tutor-start=29,tutor-end=30}{y}^{\htmlData{tutor-start=32,tutor-end=33}{2}} \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}{a} \htmlData{tutor-start=44,tutor-end=45}{>} \htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{)}, C2:x24+y2=1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{:} \frac{\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}{\htmlData{tutor-start=20,tutor-end=21}{4}} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{y}^{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=34}{1} 的离心率分别为 e1,e2\htmlData{tutor-start=0,tutor-end=1}{e}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{e}_{\htmlData{tutor-start=10,tutor-end=11}{2}}. 若 e2=3e1\htmlData{tutor-start=0,tutor-end=1}{e}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \sqrt{\htmlData{tutor-start=14,tutor-end=15}{3}}\htmlData{tutor-start=16,tutor-end=17}{e}_{\htmlData{tutor-start=19,tutor-end=20}{1}}, 则 a=()\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\quad\htmlData{tutor-start=10,tutor-end=11}{)} (A) 233\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{3}} (B) 2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}} (C) 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}} (D) 6\sqrt{\htmlData{tutor-start=6,tutor-end=7}{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 1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 · 平面几何

过点 (0,−2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{)} 与圆 x2+y2−4x−1=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}{4}\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=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{0} 相切的两条直线的夹角为 α\htmlData{tutor-start=0,tutor-end=6}{\alpha}, 则 sin⁡α=()\sin \htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{(}\quad\htmlData{tutor-start=20,tutor-end=21}{)} (A) 1\htmlData{tutor-start=0,tutor-end=1}{1} (B) 154\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{5}}}{\htmlData{tutor-start=17,tutor-end=18}{4}} (C) 104\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{0}}}{\htmlData{tutor-start=17,tutor-end=18}{4}} (D) 64\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{6}}}{\htmlData{tutor-start=16,tutor-end=17}{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 1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

记 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}} 为数列 {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}{\}} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和, 设甲: {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}{\}} 为等差数列; 乙: {Snn}\left\{\frac{\htmlData{tutor-start=13,tutor-end=14}{S}_{\htmlData{tutor-start=16,tutor-end=17}{n}}}{\htmlData{tutor-start=20,tutor-end=21}{n}}\right\} 为等差数列, 则 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) 甲是乙的充分条件但不是必要条件 (B) 甲是乙的必要条件但不是充分条件 (C) 甲是乙的充要条件 (D) 甲既不是乙的充分条件也不是乙的必要条件

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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 · 数学竞赛/待细分

已知 sin⁡(α−β)=13\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}{=} \frac{\htmlData{tutor-start=29,tutor-end=30}{1}}{\htmlData{tutor-start=32,tutor-end=33}{3}}, cos⁡αsin⁡β=16\cos \htmlData{tutor-start=5,tutor-end=12}{\alpha }\sin \htmlData{tutor-start=17,tutor-end=23}{\beta }\htmlData{tutor-start=23,tutor-end=24}{=} \frac{\htmlData{tutor-start=31,tutor-end=32}{1}}{\htmlData{tutor-start=34,tutor-end=35}{6}}, 则 cos⁡(2α+2β)=()\cos\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=13}{\alpha }\htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=21}{\beta}\htmlData{tutor-start=21,tutor-end=22}{)} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{(}\quad\htmlData{tutor-start=31,tutor-end=32}{)} (A) 79\frac{\htmlData{tutor-start=6,tutor-end=7}{7}}{\htmlData{tutor-start=9,tutor-end=10}{9}} (B) 19\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{9}} (C) −19\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{9}} (D) −79\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{7}}{\htmlData{tutor-start=10,tutor-end=11}{9}}

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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

有一组样本数据 x1,x2,⋯,x6\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}}\htmlData{tutor-start=12,tutor-end=13}{,} \cdots\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{x}_{\htmlData{tutor-start=25,tutor-end=26}{6}}, 其中 x1\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 是最小值, x6\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{6}} 是最大值, 则 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) x2,x3,x4,x5\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}{x}_{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{x}_{\htmlData{tutor-start=17,tutor-end=18}{4}}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{x}_{\htmlData{tutor-start=24,tutor-end=25}{5}} 的平均数等于 x1,x2,⋯,x6\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}}\htmlData{tutor-start=12,tutor-end=13}{,} \cdots\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{x}_{\htmlData{tutor-start=25,tutor-end=26}{6}} 的平均数 (B) x2,x3,x4,x5\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}{x}_{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{x}_{\htmlData{tutor-start=17,tutor-end=18}{4}}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{x}_{\htmlData{tutor-start=24,tutor-end=25}{5}} 的中位数等于 x1,x2,⋯,x6\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}}\htmlData{tutor-start=12,tutor-end=13}{,} \cdots\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{x}_{\htmlData{tutor-start=25,tutor-end=26}{6}} 的中位数 (C) x2,x3,x4,x5\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}{x}_{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{x}_{\htmlData{tutor-start=17,tutor-end=18}{4}}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{x}_{\htmlData{tutor-start=24,tutor-end=25}{5}} 的标准差不小于 x1,x2,⋯,x6\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}}\htmlData{tutor-start=12,tutor-end=13}{,} \cdots\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{x}_{\htmlData{tutor-start=25,tutor-end=26}{6}} 的标准差 (D) x2,x3,x4,x5\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}{x}_{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{x}_{\htmlData{tutor-start=17,tutor-end=18}{4}}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{x}_{\htmlData{tutor-start=24,tutor-end=25}{5}} 的极差不大于 x1,x2,⋯,x6\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}}\htmlData{tutor-start=12,tutor-end=13}{,} \cdots\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{x}_{\htmlData{tutor-start=25,tutor-end=26}{6}} 的极差

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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 · 数学竞赛/待细分

噪声污染问题越来越受到重视. 用声压级来度量声音的强弱, 定义声压级 Lp=20×lg⁡pp0\htmlData{tutor-start=0,tutor-end=1}{L}_{\htmlData{tutor-start=3,tutor-end=4}{p}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{0} \htmlData{tutor-start=11,tutor-end=18}{\times }\lg \frac{\htmlData{tutor-start=28,tutor-end=29}{p}}{\htmlData{tutor-start=31,tutor-end=32}{p}_{\htmlData{tutor-start=34,tutor-end=35}{0}}}, 其中常数 p0 (p0>0)\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=8}{\ }\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{p}_{\htmlData{tutor-start=12,tutor-end=13}{0}} \htmlData{tutor-start=15,tutor-end=16}{>} \htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{)} 是听觉下限阈值, p\htmlData{tutor-start=0,tutor-end=1}{p} 是实际声压. 下表为不同声源的声压级: | 声源 | 与声源的距离 /m | 声压级 /dB | |---|---|---| | 燃油汽车 | 10 | 60∼90\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{0} \htmlData{tutor-start=3,tutor-end=8}{\sim }\htmlData{tutor-start=8,tutor-end=9}{9}\htmlData{tutor-start=9,tutor-end=10}{0} | | 混合动力汽车 | 10 | 50∼60\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{0} \htmlData{tutor-start=3,tutor-end=8}{\sim }\htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=10}{0} | | 电动汽车 | 10 | 40 | 已知在距离燃油汽车, 混合动力汽车, 电动汽车 10 m 处测得实际声压分别为 p1,p2,p3\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{p}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{p}_{\htmlData{tutor-start=17,tutor-end=18}{3}}, 则 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) p1⩾p2\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=16}{\geqslant }\htmlData{tutor-start=16,tutor-end=17}{p}_{\htmlData{tutor-start=19,tutor-end=20}{2}} (B) p2>10p3\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{p}_{\htmlData{tutor-start=13,tutor-end=14}{3}} (C) p3=100p0\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{p}_{\htmlData{tutor-start=14,tutor-end=15}{0}} (D) p1⩽100p2\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=16}{\leqslant }\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{p}_{\htmlData{tutor-start=22,tutor-end=23}{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 1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

已知函数 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}{)} 的定义域为 R\mathbf{\htmlData{tutor-start=8,tutor-end=9}{R}}, f(xy)=y2f(x)+x2f(y)\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}{y}\htmlData{tutor-start=4,tutor-end=5}{)} \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}{f}\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=19,tutor-end=20}{+} \htmlData{tutor-start=21,tutor-end=22}{x}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{f}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{y}\htmlData{tutor-start=30,tutor-end=31}{)}, 则 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) f(0)=0\htmlData{tutor-start=0,tutor-end=1}{f}\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}{=} \htmlData{tutor-start=7,tutor-end=8}{0} (B) f(1)=0\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{0} (C) 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}{)} 是偶函数 (D) 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} 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}{)} 的极小值点

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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 · 平面几何

下列物体中, 能被整体放入棱长为 1 (单位: m) 的正方体容器 (容器壁厚度忽略不计) 内的有 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) 直径为 0.99 m 的球体 (B) 所有棱长均为 1.4 m 的四面体 (C) 底面直径为 0.01 m, 高为 1.8 m 的圆柱体 (D) 底面直径为 1.2 m, 高为 0.01 m 的圆柱体

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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

某学校开设了 4 门体育类选修课和 4 门艺术类选修课, 学生需从这 8 门课中选修 2 门或 3 门课, 并且每类选修课至少选修 1 门, 则不同的选课方案共有______种. (用数字作答)

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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 · 数学竞赛/待细分

在正四棱台 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, AB=2,A1B1=1,AA1=2\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}{2}\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}{1}\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=21}{A}\htmlData{tutor-start=21,tutor-end=22}{A}_{\htmlData{tutor-start=24,tutor-end=25}{1}}\htmlData{tutor-start=26,tutor-end=27}{=}\sqrt{\htmlData{tutor-start=33,tutor-end=34}{2}}, 则该棱台的体积为______.

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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

已知函数 f(x)=cos⁡ωx−1 (ω>0)\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}{=} \cos \htmlData{tutor-start=12,tutor-end=19}{\omega }\htmlData{tutor-start=19,tutor-end=20}{x} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{1} \htmlData{tutor-start=25,tutor-end=27}{\ }\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=35}{\omega }\htmlData{tutor-start=35,tutor-end=36}{>} \htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{)} 在区间 [0,2π]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=8}{\pi}\htmlData{tutor-start=8,tutor-end=9}{]}, 有且仅有 3 个零点, 则 ω\htmlData{tutor-start=0,tutor-end=6}{\omega} 的取值范围是______.

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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 · 数学竞赛/待细分

已知双曲线 C:x2a2−y2b2=1 (a>0,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=54,tutor-end=55}{>} \htmlData{tutor-start=56,tutor-end=57}{0}\htmlData{tutor-start=57,tutor-end=58}{,} \htmlData{tutor-start=59,tutor-end=60}{b} \htmlData{tutor-start=61,tutor-end=62}{>} \htmlData{tutor-start=63,tutor-end=64}{0}\htmlData{tutor-start=64,tutor-end=65}{)} 的左、右焦点分别为 F1,F2\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{F}_{\htmlData{tutor-start=10,tutor-end=11}{2}}. 点 A\htmlData{tutor-start=0,tutor-end=1}{A} 在 C\htmlData{tutor-start=0,tutor-end=1}{C} 上, 点 B\htmlData{tutor-start=0,tutor-end=1}{B} 在 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴上, F1A→⊥F1B→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{F}_{\htmlData{tutor-start=19,tutor-end=20}{1}}\htmlData{tutor-start=21,tutor-end=22}{A}} \htmlData{tutor-start=24,tutor-end=30}{\perp }\overrightarrow{\htmlData{tutor-start=46,tutor-end=47}{F}_{\htmlData{tutor-start=49,tutor-end=50}{1}}\htmlData{tutor-start=51,tutor-end=52}{B}}, F2A→=−23F2B→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{F}_{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{A}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{-}\frac{\htmlData{tutor-start=33,tutor-end=34}{2}}{\htmlData{tutor-start=36,tutor-end=37}{3}} \overrightarrow{\htmlData{tutor-start=55,tutor-end=56}{F}_{\htmlData{tutor-start=58,tutor-end=59}{2}}\htmlData{tutor-start=60,tutor-end=61}{B}}, 则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的离心率为______.

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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

已知在 △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} 中, A+B=3C,2sin⁡(A−C)=sin⁡B\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}{3}\htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{2}\sin\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{)} \htmlData{tutor-start=19,tutor-end=20}{=} \sin \htmlData{tutor-start=26,tutor-end=27}{B}. (1) 求 sin⁡A\sin \htmlData{tutor-start=5,tutor-end=6}{A}; (2) 设 AB=5\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}{5}, 求 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 边上的高.

题解状态：标准答案与规范题解待补充

题目标签：新高考卷 1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 · 数学竞赛/待细分

如图, 在正四棱柱 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, AB=2,AA1=4\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}{2}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{A}\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}{4}. 点 A2,B2,C2,D2\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=7,tutor-end=8}{B}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{C}_{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{D}_{\htmlData{tutor-start=24,tutor-end=25}{2}} 分别在棱 AA1,BB1,CC1,DD1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{B}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{C}_{\htmlData{tutor-start=20,tutor-end=21}{1}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{D}\htmlData{tutor-start=25,tutor-end=26}{D}_{\htmlData{tutor-start=28,tutor-end=29}{1}} 上, AA2=1,BB2=DD2=2,CC2=3\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{B}_{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{D}\htmlData{tutor-start=18,tutor-end=19}{D}_{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{,} \htmlData{tutor-start=27,tutor-end=28}{C}\htmlData{tutor-start=28,tutor-end=29}{C}_{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{3}. (1) 证明: B2C2∥A2D2\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=21}{\parallel }\htmlData{tutor-start=21,tutor-end=22}{A}_{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{D}_{\htmlData{tutor-start=29,tutor-end=30}{2}}; (2) 点 P\htmlData{tutor-start=0,tutor-end=1}{P} 在棱 BB1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 上, 当二面角 P−A2C2−D2\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=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{C}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{D}_{\htmlData{tutor-start=16,tutor-end=17}{2}} 为 150∘150^\circ 时, 求 B2P\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{P}.

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 1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 · 代数

已知函数 f(x)=a(ex+a)−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}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{(}\mathrm{\htmlData{tutor-start=17,tutor-end=18}{e}}^{\htmlData{tutor-start=21,tutor-end=22}{x}} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{a}\htmlData{tutor-start=27,tutor-end=28}{)} \htmlData{tutor-start=29,tutor-end=30}{-} \htmlData{tutor-start=31,tutor-end=32}{x}. (1) 讨论 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}{)} 的单调性; (2) 证明: 当 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} 时, 求证: f(x)>2ln⁡a+32\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}\ln \htmlData{tutor-start=12,tutor-end=13}{a} \htmlData{tutor-start=14,tutor-end=15}{+} \frac{\htmlData{tutor-start=22,tutor-end=23}{3}}{\htmlData{tutor-start=25,tutor-end=26}{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 1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 · 代数

设等差数列 {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}{\}} 的公差为 d\htmlData{tutor-start=0,tutor-end=1}{d}，且 d>1\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}。令 bn=n2+nan\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{n}}{\htmlData{tutor-start=22,tutor-end=23}{a}_{\htmlData{tutor-start=25,tutor-end=26}{n}}}，记 Sn,Tn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{T}_{\htmlData{tutor-start=9,tutor-end=10}{n}} 分别为数列 {an},{bn}\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}{\}}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=12}{\{}\htmlData{tutor-start=12,tutor-end=13}{b}_{\htmlData{tutor-start=15,tutor-end=16}{n}}\htmlData{tutor-start=17,tutor-end=19}{\}} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和。 (1) 若 3a2=3a1+a3\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{3}}，S3+T3=21\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{T}_{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{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) 若 {bn}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{b}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 为等差数列，且 S99−T99=99\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{9}\htmlData{tutor-start=4,tutor-end=5}{9}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{T}_{\htmlData{tutor-start=10,tutor-end=11}{9}\htmlData{tutor-start=11,tutor-end=12}{9}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{9}\htmlData{tutor-start=15,tutor-end=16}{9}，求 d\htmlData{tutor-start=0,tutor-end=1}{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 1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 · 数学竞赛/待细分

甲、乙两人投篮，每次由其中一人投篮，规则如下：若命中则此人继续投篮，若未命中则换为对方投篮。无论之前投篮情况如何，甲每次投篮的命中率均为 0.6\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{6}，乙每次投篮的命中率均为 0.8\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{8}。由抽签决定第 1\htmlData{tutor-start=0,tutor-end=1}{1} 次投篮的人选，第 1\htmlData{tutor-start=0,tutor-end=1}{1} 次投篮的人是甲、乙的概率各为 0.5\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{5}。 (1) 求第 2\htmlData{tutor-start=0,tutor-end=1}{2} 次投篮的人是乙的概率； (2) 求第 i\htmlData{tutor-start=0,tutor-end=1}{i} 次投篮的人是甲的概率； (3) 已知：若随机变量 Xi\htmlData{tutor-start=0,tutor-end=1}{X}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 服从两点分布，且 P(Xi=1)=1−P(Xi=0)=qi\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=5,tutor-end=6}{i}}\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=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{P}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{X}_{\htmlData{tutor-start=18,tutor-end=19}{i}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=23}{)}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{q}_{\htmlData{tutor-start=27,tutor-end=28}{i}}，i=1,2,⋯,n\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,}\cdots\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{n}，则 E(∑i=1nXi)=∑i=1nqi\htmlData{tutor-start=0,tutor-end=1}{E}\left(\displaystyle\sum_{\htmlData{tutor-start=26,tutor-end=27}{i}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{1}}^{\htmlData{tutor-start=32,tutor-end=33}{n}}\htmlData{tutor-start=34,tutor-end=35}{X}_{\htmlData{tutor-start=37,tutor-end=38}{i}}\right)\htmlData{tutor-start=46,tutor-end=47}{=}\displaystyle\sum_{\htmlData{tutor-start=66,tutor-end=67}{i}\htmlData{tutor-start=67,tutor-end=68}{=}\htmlData{tutor-start=68,tutor-end=69}{1}}^{\htmlData{tutor-start=72,tutor-end=73}{n}}\htmlData{tutor-start=74,tutor-end=75}{q}_{\htmlData{tutor-start=77,tutor-end=78}{i}}。记前 n\htmlData{tutor-start=0,tutor-end=1}{n} 次（即从第 1\htmlData{tutor-start=0,tutor-end=1}{1} 次到第 n\htmlData{tutor-start=0,tutor-end=1}{n} 次投篮）中甲投篮的次数为 Y\htmlData{tutor-start=0,tutor-end=1}{Y}，求 E(Y)\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{Y}\htmlData{tutor-start=3,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 1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 · 组合数学

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，点 P\htmlData{tutor-start=0,tutor-end=1}{P} 到 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的距离等于点 P\htmlData{tutor-start=0,tutor-end=1}{P} 到点 (0,12)\left(\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,}\dfrac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}}\right) 的距离，记动点 P\htmlData{tutor-start=0,tutor-end=1}{P} 的轨迹为 W\htmlData{tutor-start=0,tutor-end=1}{W}。 (1) 求 W\htmlData{tutor-start=0,tutor-end=1}{W} 的方程； (2) 已知矩形 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} 有三个顶点在 W\htmlData{tutor-start=0,tutor-end=1}{W} 上，证明：矩形 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} 的周长大于 33\htmlData{tutor-start=0,tutor-end=1}{3}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{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 1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。