2022 年高考数学（全国乙卷理科）

exams_raw/普通高考/2022/2022全国乙理(河南,山西,江西,安徽,陕西,甘肃,青海,内蒙古,黑龙江,吉林,.宁夏,新疆).pdf · HS-MATH-1024-v2.1-solution-aware

23 题0 个小问/题组

一、单选题 · 组合数学

设全集 $\htmlData{tutor-start=0,tutor-end=1}{U}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{5}\htmlData{tutor-start=13,tutor-end=15}{\}}$ ，集合 $\htmlData{tutor-start=0,tutor-end=1}{M}$ 满足 $\htmlData{tutor-start=0,tutor-end=11}{\complement}_{\htmlData{tutor-start=13,tutor-end=14}{U}} \htmlData{tutor-start=16,tutor-end=17}{M}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=20}{\{}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=25}{\}}$ ，则（） (A) $\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{M}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{M}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{4} \notin \htmlData{tutor-start=9,tutor-end=10}{M}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{5} \notin \htmlData{tutor-start=9,tutor-end=10}{M}$

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 1 题

解题过程

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

一、单选题 · 数学竞赛/待细分

已知 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{2}\text{\htmlData{tutor-start=11,tutor-end=12}{i}}$ ，且 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{a}\bar{\htmlData{tutor-start=8,tutor-end=9}{z}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{0}$ ，其中 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}$ 为实数，则（） (A) $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{2}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{2}$

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 2 题

解题过程

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

一、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 $\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=21}{|}\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}}\htmlData{tutor-start=35,tutor-end=36}{|}\htmlData{tutor-start=36,tutor-end=37}{=}\sqrt{\htmlData{tutor-start=43,tutor-end=44}{3}}\htmlData{tutor-start=45,tutor-end=46}{,} \htmlData{tutor-start=47,tutor-end=48}{|}\boldsymbol{\htmlData{tutor-start=60,tutor-end=61}{a}}\htmlData{tutor-start=62,tutor-end=63}{-}\htmlData{tutor-start=63,tutor-end=64}{2}\boldsymbol{\htmlData{tutor-start=76,tutor-end=77}{b}}\htmlData{tutor-start=78,tutor-end=79}{|}\htmlData{tutor-start=79,tutor-end=80}{=}\htmlData{tutor-start=80,tutor-end=81}{3}$ ，则 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=}$ （） (A) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{1}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{2}$

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 3 题

解题过程

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

一、单选题 · 代数

嫦娥二号卫星在完成探月任务后，继续进行深空探测，成为我国第一颗环绕太阳飞行的人造行星. 为研究嫦娥二号绕日周期与地球绕日周期的比值，用到数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{b}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ ： $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=23}{\alpha}_{\htmlData{tutor-start=25,tutor-end=26}{1}}}\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=31}{b}_{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{+}\frac{\htmlData{tutor-start=44,tutor-end=45}{1}}{\htmlData{tutor-start=47,tutor-end=53}{\alpha}_{\htmlData{tutor-start=55,tutor-end=56}{1}}\htmlData{tutor-start=57,tutor-end=58}{+}\frac{\htmlData{tutor-start=64,tutor-end=65}{1}}{\htmlData{tutor-start=67,tutor-end=73}{\alpha}_{\htmlData{tutor-start=75,tutor-end=76}{2}}}}\htmlData{tutor-start=79,tutor-end=80}{,} \htmlData{tutor-start=81,tutor-end=82}{b}_{\htmlData{tutor-start=84,tutor-end=85}{3}}\htmlData{tutor-start=86,tutor-end=87}{=}\htmlData{tutor-start=87,tutor-end=88}{1}\htmlData{tutor-start=88,tutor-end=89}{+}\frac{\htmlData{tutor-start=95,tutor-end=96}{1}}{\htmlData{tutor-start=98,tutor-end=104}{\alpha}_{\htmlData{tutor-start=106,tutor-end=107}{1}}\htmlData{tutor-start=108,tutor-end=109}{+}\frac{\htmlData{tutor-start=115,tutor-end=116}{1}}{\htmlData{tutor-start=118,tutor-end=124}{\alpha}_{\htmlData{tutor-start=126,tutor-end=127}{2}}\htmlData{tutor-start=128,tutor-end=129}{+}\frac{\htmlData{tutor-start=135,tutor-end=136}{1}}{\htmlData{tutor-start=138,tutor-end=144}{\alpha}_{\htmlData{tutor-start=146,tutor-end=147}{3}}}}}$ ， $\cdots$ ，依此类推，其中 $\alpha_{k} \in \mathbf{N}^* (k=1,2,\cdots)$ . 则（） (A) $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{b}_{\htmlData{tutor-start=11,tutor-end=12}{5}}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{b}_{\htmlData{tutor-start=11,tutor-end=12}{8}}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{6}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{b}_{\htmlData{tutor-start=11,tutor-end=12}{2}}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{b}_{\htmlData{tutor-start=11,tutor-end=12}{7}}$

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 4 题

解题过程

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

一、单选题 · 数学竞赛/待细分

设 $\htmlData{tutor-start=0,tutor-end=1}{F}$ 为抛物线 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{x}$ 的焦点，点 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 上，点 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$ ，若 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{B}\htmlData{tutor-start=7,tutor-end=8}{F}\htmlData{tutor-start=8,tutor-end=9}{|}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}$ （） (A) $\htmlData{tutor-start=0,tutor-end=1}{2}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{3}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{3}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}$

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 5 题

解题过程

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

一、单选题 · 数学竞赛/待细分

执行如图的程序框图，输出的 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}$ （） (A) $\htmlData{tutor-start=0,tutor-end=1}{3}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{4}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{5}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{6}$

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

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 6 题

解题过程

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

一、单选题 · 数学竞赛/待细分

在正方体 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{A}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{B}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{C}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{D}_{\htmlData{tutor-start=23,tutor-end=24}{1}}$ 中， $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{F}$ 分别为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{C}$ 的中点，则（） (A) 平面 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{F} \htmlData{tutor-start=8,tutor-end=13}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{D}_{\htmlData{tutor-start=5,tutor-end=6}{1}}$ (B) 平面 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{F} \htmlData{tutor-start=8,tutor-end=13}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{D}$ (C) 平面 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{F} \htmlData{tutor-start=8,tutor-end=17}{\parallel}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{C}$ (D) 平面 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{F} \htmlData{tutor-start=8,tutor-end=17}{\parallel}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{D}$

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 7 题

解题过程

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

一、单选题 · 代数

已知等比数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的前 3 项和为 168， $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{4}\htmlData{tutor-start=13,tutor-end=14}{2}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{6}}\htmlData{tutor-start=5,tutor-end=6}{=}$ （） (A) 14 (B) 12 (C) 6 (D) 3

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 8 题

解题过程

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

一、单选题 · 组合数学

已知球 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 的半径为 1，四棱锥的顶点为 $\htmlData{tutor-start=0,tutor-end=1}{O}$ ，底面的四个顶点均在球 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 的球面上，则当该四棱锥的体积最大时，其高为（） (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$ (C) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{3}}$ (D) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 9 题

解题过程

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

一、单选题 · 数学竞赛/待细分

某棋手与甲、乙、丙三位棋手各比赛一盘，各盘比赛结果相互独立. 已知该棋手与甲、乙、丙比赛获胜的概率分别为 $\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{p}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{p}_{\htmlData{tutor-start=17,tutor-end=18}{3}}$ ，且 $\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{p}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{>} \htmlData{tutor-start=16,tutor-end=17}{p}_{\htmlData{tutor-start=19,tutor-end=20}{1}} \htmlData{tutor-start=22,tutor-end=23}{>} \htmlData{tutor-start=24,tutor-end=25}{0}$ . 记该棋手连胜两盘的概率为 $\htmlData{tutor-start=0,tutor-end=1}{p}$ ，则（） (A) $\htmlData{tutor-start=0,tutor-end=1}{p}$ 与该棋手和甲、乙、丙的比赛次序无关 (B) 该棋手在第二盘与甲比赛， $\htmlData{tutor-start=0,tutor-end=1}{p}$ 最大 (C) 该棋手在第二盘与乙比赛， $\htmlData{tutor-start=0,tutor-end=1}{p}$ 最大 (D) 该棋手在第二盘与丙比赛， $\htmlData{tutor-start=0,tutor-end=1}{p}$ 最大

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 10 题

解题过程

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

一、单选题 · 平面几何

双曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的两个焦点为 $\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{F}_{\htmlData{tutor-start=10,tutor-end=11}{2}}$ ，以 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的实轴为直径的圆记为 $\htmlData{tutor-start=0,tutor-end=1}{D}$ ，过 $\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 作 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 的切线与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 交于 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{N}$ 两点，且 $\cos \htmlData{tutor-start=5,tutor-end=12}{\angle }\htmlData{tutor-start=12,tutor-end=13}{F}_{\htmlData{tutor-start=15,tutor-end=16}{1}}\htmlData{tutor-start=17,tutor-end=18}{N}\htmlData{tutor-start=18,tutor-end=19}{F}_{\htmlData{tutor-start=21,tutor-end=22}{2}} \htmlData{tutor-start=24,tutor-end=25}{=} \frac{\htmlData{tutor-start=32,tutor-end=33}{3}}{\htmlData{tutor-start=35,tutor-end=36}{5}}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的离心率为（） (A) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{5}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$ (C) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{2}}$ (D) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{7}}}{\htmlData{tutor-start=17,tutor-end=18}{2}}$

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 11 题

解题过程

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

一、单选题 · 平面几何

已知函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{g}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}$ 的定义域均为 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{R}}$ ，且 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{g}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{5}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{g}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{f}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{x}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{4}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{7}$ . 若 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{g}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}$ 的图象关于直线 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$ 对称， $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{4}$ ，则 $\sum_{\htmlData{tutor-start=6,tutor-end=7}{k}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{f}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{k}\htmlData{tutor-start=19,tutor-end=20}{)} \htmlData{tutor-start=21,tutor-end=22}{=}$ （） (A) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{1}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{2}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{3}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{4}$

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 12 题

解题过程

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

二、填空题 · 数学竞赛/待细分

从甲、乙等 5 名同学中随机选 3 名参加社区服务工作，则甲、乙都入选的概率为______.

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 13 题

解题过程

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

二、填空题 · 平面几何

过四点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{4}\htmlData{tutor-start=24,tutor-end=25}{,}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{)}$ 中的三点的一个圆的方程为______.

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 14 题

解题过程

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

二、填空题 · 代数

记函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\cos\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=17}{\omega }\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=26}{\varphi}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=35}{\omega}\htmlData{tutor-start=35,tutor-end=36}{>}\htmlData{tutor-start=36,tutor-end=37}{0}\htmlData{tutor-start=37,tutor-end=38}{,} \htmlData{tutor-start=39,tutor-end=40}{0}\htmlData{tutor-start=40,tutor-end=41}{<}\htmlData{tutor-start=41,tutor-end=48}{\varphi}\htmlData{tutor-start=48,tutor-end=49}{<}\htmlData{tutor-start=49,tutor-end=52}{\pi}\htmlData{tutor-start=52,tutor-end=53}{)}$ 的最小正周期为 $\htmlData{tutor-start=0,tutor-end=1}{T}$ . 若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{T}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\frac{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}}{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{,} \htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{=}\frac{\htmlData{tutor-start=33,tutor-end=36}{\pi}}{\htmlData{tutor-start=38,tutor-end=39}{9}}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的零点，则 $\htmlData{tutor-start=0,tutor-end=6}{\omega}$ 的最小值为______.

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 15 题

解题过程

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

二、填空题 · 代数

已知 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}_{\htmlData{tutor-start=5,tutor-end=6}{1}}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}_{\htmlData{tutor-start=5,tutor-end=6}{2}}$ 分别是函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{x}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{e}\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{>}\htmlData{tutor-start=22,tutor-end=23}{0}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{a} \neq \htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}$ 的极小值点和极大值点. 若 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{2}}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的取值范围是______.

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 16 题

解题过程

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

三、解答题 · 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 $\sin \htmlData{tutor-start=5,tutor-end=6}{C} \sin\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{)} \htmlData{tutor-start=17,tutor-end=18}{=} \sin \htmlData{tutor-start=24,tutor-end=25}{B} \sin\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{C}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{A}\htmlData{tutor-start=34,tutor-end=35}{)}$ . (1) 证明： $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{b}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{c}^{\htmlData{tutor-start=16,tutor-end=17}{2}}$ ； (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,} \cos \htmlData{tutor-start=10,tutor-end=11}{A} \htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{5}}{\htmlData{tutor-start=24,tutor-end=25}{3}\htmlData{tutor-start=25,tutor-end=26}{1}}$ ，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：全国乙卷理科第 17 题

解题过程

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

三、解答题 · 平面几何

如图，四面体 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}$ 中， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{D}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{D}\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=27}{\angle }\htmlData{tutor-start=27,tutor-end=28}{A}\htmlData{tutor-start=28,tutor-end=29}{D}\htmlData{tutor-start=29,tutor-end=30}{B} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=40}{\angle }\htmlData{tutor-start=40,tutor-end=41}{B}\htmlData{tutor-start=41,tutor-end=42}{D}\htmlData{tutor-start=42,tutor-end=43}{C}\htmlData{tutor-start=43,tutor-end=44}{,} \htmlData{tutor-start=45,tutor-end=46}{E}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}$ 的中点. (1) 证明：平面 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{D} \htmlData{tutor-start=4,tutor-end=9}{\perp}$ 平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}$ ； (2) 设 $AB=BD=2, \angle ACB=60^\circ$ ，点 $\htmlData{tutor-start=0,tutor-end=1}{F}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D}$ 上，当 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{F}\htmlData{tutor-start=12,tutor-end=13}{C}$ 的面积最小时，求 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{F}$ 与平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{D}$ 所成的角的正弦值.

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 18 题

解题过程

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

解答题 · 数学竞赛/待细分

某地经过多年的环境治理，已将荒山改造成了绿水青山. 为估计一林区某种树木的总材积量，随机选取了 10 棵这种树木，测量每棵树的根部横截面积（单位： $\text{\htmlData{tutor-start=6,tutor-end=7}{m}}^{\htmlData{tutor-start=10,tutor-end=11}{2}}$ ）和材积量（单位： $\text{\htmlData{tutor-start=6,tutor-end=7}{m}}^{\htmlData{tutor-start=10,tutor-end=11}{3}}$ ），得到如下数据： | 样本号 $\htmlData{tutor-start=0,tutor-end=1}{i}$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 总和 | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | 根部横截面积 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$ | 0.04 | 0.06 | 0.04 | 0.08 | 0.08 | 0.05 | 0.05 | 0.07 | 0.07 | 0.06 | 0.6 | | 材积量 $\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{i}}$ | 0.25 | 0.40 | 0.22 | 0.54 | 0.51 | 0.34 | 0.36 | 0.46 | 0.42 | 0.40 | 3.9 | 并计算得 $\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{x}_{\htmlData{tutor-start=19,tutor-end=20}{i}}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{.}\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{3}\htmlData{tutor-start=32,tutor-end=33}{8}$ ， $\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{y}_{\htmlData{tutor-start=19,tutor-end=20}{i}}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{.}\htmlData{tutor-start=30,tutor-end=31}{6}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{5}\htmlData{tutor-start=33,tutor-end=34}{8}$ ， $\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{x}_{\htmlData{tutor-start=19,tutor-end=20}{i}} \htmlData{tutor-start=22,tutor-end=23}{y}_{\htmlData{tutor-start=25,tutor-end=26}{i}} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{.}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{4}\htmlData{tutor-start=34,tutor-end=35}{7}\htmlData{tutor-start=35,tutor-end=36}{4}$ . (1) 估计该林区这种树木平均一棵的根部横截面积与平均一棵的材积量； (2) 求该林区这种树木的根部横截面积与材积量的样本相关系数（精确到 0.01）； (3) 现测量了该林区所有这种树木的根部横截面积，并得到所有这种树木的根部横截面积总和为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{6}\text{ \htmlData{tutor-start=10,tutor-end=11}{m}}^{\htmlData{tutor-start=14,tutor-end=15}{2}}$ . 已知树木的材积量与其根部横截面积近似成正比. 利用以上数据给出该林区这种树木的总材积量的估计值. 附：相关系数 $\htmlData{tutor-start=0,tutor-end=1}{r} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\sum_{\htmlData{tutor-start=16,tutor-end=17}{i}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{1}}^{\htmlData{tutor-start=22,tutor-end=23}{n}} \htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{x}_{\htmlData{tutor-start=29,tutor-end=30}{i}} \htmlData{tutor-start=32,tutor-end=33}{-} \overline{\htmlData{tutor-start=44,tutor-end=45}{x}}\htmlData{tutor-start=46,tutor-end=47}{)}\htmlData{tutor-start=47,tutor-end=48}{(}\htmlData{tutor-start=48,tutor-end=49}{y}_{\htmlData{tutor-start=51,tutor-end=52}{i}} \htmlData{tutor-start=54,tutor-end=55}{-} \overline{\htmlData{tutor-start=66,tutor-end=67}{y}}\htmlData{tutor-start=68,tutor-end=69}{)}}{\sqrt{\sum_{\htmlData{tutor-start=83,tutor-end=84}{i}\htmlData{tutor-start=84,tutor-end=85}{=}\htmlData{tutor-start=85,tutor-end=86}{1}}^{\htmlData{tutor-start=89,tutor-end=90}{n}} \htmlData{tutor-start=92,tutor-end=93}{(}\htmlData{tutor-start=93,tutor-end=94}{x}_{\htmlData{tutor-start=96,tutor-end=97}{i}} \htmlData{tutor-start=99,tutor-end=100}{-} \overline{\htmlData{tutor-start=111,tutor-end=112}{x}}\htmlData{tutor-start=113,tutor-end=114}{)}^{\htmlData{tutor-start=116,tutor-end=117}{2}} \sum_{\htmlData{tutor-start=125,tutor-end=126}{i}\htmlData{tutor-start=126,tutor-end=127}{=}\htmlData{tutor-start=127,tutor-end=128}{1}}^{\htmlData{tutor-start=131,tutor-end=132}{n}} \htmlData{tutor-start=134,tutor-end=135}{(}\htmlData{tutor-start=135,tutor-end=136}{y}_{\htmlData{tutor-start=138,tutor-end=139}{i}} \htmlData{tutor-start=141,tutor-end=142}{-} \overline{\htmlData{tutor-start=153,tutor-end=154}{y}}\htmlData{tutor-start=155,tutor-end=156}{)}^{\htmlData{tutor-start=158,tutor-end=159}{2}}}}$ ， $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{.}\htmlData{tutor-start=8,tutor-end=9}{8}\htmlData{tutor-start=9,tutor-end=10}{9}\htmlData{tutor-start=10,tutor-end=11}{6}} \htmlData{tutor-start=13,tutor-end=21}{\approx }\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{.}\htmlData{tutor-start=23,tutor-end=24}{3}\htmlData{tutor-start=24,tutor-end=25}{7}\htmlData{tutor-start=25,tutor-end=26}{7}$ .

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 19 题

解题过程

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

解答题 · 平面几何

已知椭圆 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 的中心为坐标原点，对称轴为 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴、 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 轴，且过 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{)}$ ， $\htmlData{tutor-start=0,tutor-end=1}{B}\left(\frac{\htmlData{tutor-start=13,tutor-end=14}{3}}{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{1}\right)$ 两点. (1) 求 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 的方程； (2) 设过点 $\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{)}$ 的直线交 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 于 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{N}$ 两点，过 $\htmlData{tutor-start=0,tutor-end=1}{M}$ 且平行于 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴的直线与线段 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 交于点 $\htmlData{tutor-start=0,tutor-end=1}{T}$ ，点 $\htmlData{tutor-start=0,tutor-end=1}{H}$ 满足 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{M}\htmlData{tutor-start=17,tutor-end=18}{T}} \htmlData{tutor-start=20,tutor-end=21}{=} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{T}\htmlData{tutor-start=39,tutor-end=40}{H}}$ . 证明：直线 $\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{N}$ 过定点.

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 20 题

解题过程

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

解答题 · 代数

已知函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \ln\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{1} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{)} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{x}\htmlData{tutor-start=22,tutor-end=23}{e}^{\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{x}}$ . (1) 当 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}$ 时，求曲线 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)}$ 在点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{)}$ 处的切线方程； (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 在区间 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=20}{\infty}\htmlData{tutor-start=20,tutor-end=21}{)}$ 各恰有一个零点，求 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的取值范围.

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 21 题

解题过程

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

解答题 · 平面几何

在直角坐标系 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{y}$ 中，曲线 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的参数方程为 $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{=} \sqrt{\htmlData{tutor-start=24,tutor-end=25}{3}}\cos \htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{t}\htmlData{tutor-start=33,tutor-end=34}{,} \\ \htmlData{tutor-start=38,tutor-end=39}{y} \htmlData{tutor-start=40,tutor-end=41}{=} \htmlData{tutor-start=42,tutor-end=43}{2}\sin \htmlData{tutor-start=48,tutor-end=49}{t} \end{cases}$ ( $\htmlData{tutor-start=0,tutor-end=1}{t}$ 为参数). 以坐标原点为极点， $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴正半轴为极轴建立极坐标系，已知直线 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的极坐标方程为 $\htmlData{tutor-start=0,tutor-end=4}{\rho}\sin\left(\htmlData{tutor-start=14,tutor-end=21}{\theta }\htmlData{tutor-start=21,tutor-end=22}{+} \frac{\htmlData{tutor-start=29,tutor-end=32}{\pi}}{\htmlData{tutor-start=34,tutor-end=35}{3}}\right) \htmlData{tutor-start=44,tutor-end=45}{+} \htmlData{tutor-start=46,tutor-end=47}{m} \htmlData{tutor-start=48,tutor-end=49}{=} \htmlData{tutor-start=50,tutor-end=51}{0}$ . (1) 写出 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 的直角坐标方程； (2) 若 $\htmlData{tutor-start=0,tutor-end=1}{l}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 有公共点，求 $\htmlData{tutor-start=0,tutor-end=1}{m}$ 的取值范围.

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 22 题

解题过程

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

解答题 · 数学竞赛/待细分

已知 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{c}$ 都是正数，且 $\htmlData{tutor-start=0,tutor-end=1}{a}^{\frac{\htmlData{tutor-start=9,tutor-end=10}{3}}{\htmlData{tutor-start=12,tutor-end=13}{2}}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{b}^{\frac{\htmlData{tutor-start=27,tutor-end=28}{3}}{\htmlData{tutor-start=30,tutor-end=31}{2}}} \htmlData{tutor-start=34,tutor-end=35}{+} \htmlData{tutor-start=36,tutor-end=37}{c}^{\frac{\htmlData{tutor-start=45,tutor-end=46}{3}}{\htmlData{tutor-start=48,tutor-end=49}{2}}} \htmlData{tutor-start=52,tutor-end=53}{=} \htmlData{tutor-start=54,tutor-end=55}{1}$ ，证明： (1) $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{c} \htmlData{tutor-start=4,tutor-end=14}{\leqslant }\frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{9}}$ ； (2) $\frac{\htmlData{tutor-start=6,tutor-end=7}{a}}{\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{c}} \htmlData{tutor-start=14,tutor-end=15}{+} \frac{\htmlData{tutor-start=22,tutor-end=23}{b}}{\htmlData{tutor-start=25,tutor-end=26}{a}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{c}} \htmlData{tutor-start=30,tutor-end=31}{+} \frac{\htmlData{tutor-start=38,tutor-end=39}{c}}{\htmlData{tutor-start=41,tutor-end=42}{a}\htmlData{tutor-start=42,tutor-end=43}{+}\htmlData{tutor-start=43,tutor-end=44}{b}} \htmlData{tutor-start=46,tutor-end=56}{\leqslant }\frac{\htmlData{tutor-start=62,tutor-end=63}{1}}{\htmlData{tutor-start=65,tutor-end=66}{2}\sqrt{\htmlData{tutor-start=72,tutor-end=73}{a}\htmlData{tutor-start=73,tutor-end=74}{b}\htmlData{tutor-start=74,tutor-end=75}{c}}}$ .

题解状态：标准答案与规范题解待补充

题目标签：全国乙卷理科第 23 题

解题过程

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