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2016 年高考数学（全国卷 3理科）

exams_raw/普通高考/2016/2016全国3理(云南,广西,贵州).pdf · HS-MATH-1024-v2.1-solution-aware

24 题0 个小问/题组
1

一、选择题 · 组合数学

设集合 S={x∣(x−2)(x−3)⩾0}\htmlData{tutor-start=0,tutor-end=1}{S} \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}{(}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{3}\htmlData{tutor-start=22,tutor-end=23}{)} \htmlData{tutor-start=24,tutor-end=34}{\geqslant }\htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=37}{\}}, T={x∣x>0}\htmlData{tutor-start=0,tutor-end=1}{T} \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=15,tutor-end=16}{>} \htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=20}{\}}, 则 S∩T=\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{T} \htmlData{tutor-start=9,tutor-end=10}{=} （ ） (A) [2,3]\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}{3}\htmlData{tutor-start=5,tutor-end=6}{]} (B) (−∞,2]∪[3,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{]} \htmlData{tutor-start=13,tutor-end=18}{\cup }\htmlData{tutor-start=18,tutor-end=19}{[}\htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=29}{\infty}\htmlData{tutor-start=29,tutor-end=30}{)} (C) [3,+∞)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} (D) (0,2]∪[3,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{]} \htmlData{tutor-start=7,tutor-end=12}{\cup }\htmlData{tutor-start=12,tutor-end=13}{[}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=23}{\infty}\htmlData{tutor-start=23,tutor-end=24}{)}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 1 题

解题过程

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

2

一、选择题 · 数学竞赛/待细分

若 z=1+2i\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{2}\text{\htmlData{tutor-start=15,tutor-end=16}{i}}, 则 4izzˉ−1=\dfrac{\htmlData{tutor-start=7,tutor-end=8}{4}\text{\htmlData{tutor-start=14,tutor-end=15}{i}}}{\htmlData{tutor-start=18,tutor-end=19}{z}\bar{\htmlData{tutor-start=24,tutor-end=25}{z}} \htmlData{tutor-start=27,tutor-end=28}{-} \htmlData{tutor-start=29,tutor-end=30}{1}} \htmlData{tutor-start=32,tutor-end=33}{=} （ ） (A) 1\htmlData{tutor-start=0,tutor-end=1}{1} (B) −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} (C) i\text{\htmlData{tutor-start=6,tutor-end=7}{i}} (D) −i\htmlData{tutor-start=0,tutor-end=1}{-}\text{\htmlData{tutor-start=7,tutor-end=8}{i}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 2 题

解题过程

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

3

一、选择题 · 平面几何

已知向量 BA→=(12,32)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{A}} \htmlData{tutor-start=20,tutor-end=21}{=} \left(\dfrac{\htmlData{tutor-start=35,tutor-end=36}{1}}{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{,} \dfrac{\sqrt{\htmlData{tutor-start=55,tutor-end=56}{3}}}{\htmlData{tutor-start=59,tutor-end=60}{2}}\right), BC→=(32,12)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{C}} \htmlData{tutor-start=20,tutor-end=21}{=} \left(\dfrac{\sqrt{\htmlData{tutor-start=41,tutor-end=42}{3}}}{\htmlData{tutor-start=45,tutor-end=46}{2}}\htmlData{tutor-start=47,tutor-end=48}{,} \dfrac{\htmlData{tutor-start=56,tutor-end=57}{1}}{\htmlData{tutor-start=59,tutor-end=60}{2}}\right), 则 ∠ABC=\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{C} \htmlData{tutor-start=11,tutor-end=12}{=} （ ） (A) 30∘30^\circ (B) 45∘45^\circ (C) 60∘60^\circ (D) 120∘120^\circ

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 3 题

解题过程

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

4

一、选择题 · 数学竞赛/待细分

某旅游城市为向游客介绍本地的气温情况，绘制了一年中月平均最高气温和平均最低气温的雷达图。图中 A\htmlData{tutor-start=0,tutor-end=1}{A} 点表示十月的平均最高气温约为 15∘C15^\circ\text{C}, B\htmlData{tutor-start=0,tutor-end=1}{B} 点表示四月的平均最低气温约为 5∘C5^\circ\text{C}。下面叙述不正确的是 （ ） (A) 各月的平均最低气温都在 0∘C0^\circ\text{C} 以上 (B) 七月的平均温差比一月的平均温差大 (C) 三月和十一月的平均最高气温基本相同 (D) 平均气温高于 20∘C20^\circ\text{C} 的月份有 5\htmlData{tutor-start=0,tutor-end=1}{5} 个

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 4 题

解题过程

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

5

一、选择题 · 数学竞赛/待细分

若 tan⁡α=34\tan\htmlData{tutor-start=4,tutor-end=11}{\alpha }\htmlData{tutor-start=11,tutor-end=12}{=} \dfrac{\htmlData{tutor-start=20,tutor-end=21}{3}}{\htmlData{tutor-start=23,tutor-end=24}{4}}, 则 cos⁡2α+2sin⁡2α=\cos^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=15}{\alpha }\htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{2}\sin \htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=31}{\alpha }\htmlData{tutor-start=31,tutor-end=32}{=} （ ） (A) 6425\dfrac{\htmlData{tutor-start=7,tutor-end=8}{6}\htmlData{tutor-start=8,tutor-end=9}{4}}{\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{5}} (B) 4825\dfrac{\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{8}}{\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{5}} (C) 1\htmlData{tutor-start=0,tutor-end=1}{1} (D) 1625\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{6}}{\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{5}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 5 题

解题过程

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

6

一、选择题 · 数学竞赛/待细分

已知 a=243\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}^{\frac{\htmlData{tutor-start=13,tutor-end=14}{4}}{\htmlData{tutor-start=16,tutor-end=17}{3}}}, b=425\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{4}^{\frac{\htmlData{tutor-start=13,tutor-end=14}{2}}{\htmlData{tutor-start=16,tutor-end=17}{5}}}, c=2513\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{5}^{\frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{3}}}, 则 （ ） (A) b<a<c\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{c} (B) a<b<c\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{c} (C) b<c<a\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{c} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{a} (D) c<a<b\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{b}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 6 题

解题过程

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

7

一、选择题（续） · 数学竞赛/待细分

执行下图的程序框图，如果输入的 a=4\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{4}, b=6\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{6}, 那么输出的 n=\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=3}{=} （ ） (A) 3\htmlData{tutor-start=0,tutor-end=1}{3} (B) 4\htmlData{tutor-start=0,tutor-end=1}{4} (C) 5\htmlData{tutor-start=0,tutor-end=1}{5} (D) 6\htmlData{tutor-start=0,tutor-end=1}{6}

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 7 题

解题过程

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

8

一、选择题（续） · 平面几何

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，B=π4\htmlData{tutor-start=0,tutor-end=1}{B} \htmlData{tutor-start=2,tutor-end=3}{=} \dfrac{\htmlData{tutor-start=11,tutor-end=14}{\pi}}{\htmlData{tutor-start=16,tutor-end=17}{4}}, BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} 边上的高等于 13BC\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{C}, 则 cos⁡A=\cos \htmlData{tutor-start=5,tutor-end=6}{A} \htmlData{tutor-start=7,tutor-end=8}{=} （ ） (A) 31010\dfrac{\htmlData{tutor-start=7,tutor-end=8}{3}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{0}}}{\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{0}} (B) 1010\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{0}}}{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{0}} (C) −1010\htmlData{tutor-start=0,tutor-end=1}{-}\dfrac{\sqrt{\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{0}}}{\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{0}} (D) −31010\htmlData{tutor-start=0,tutor-end=1}{-}\dfrac{\htmlData{tutor-start=8,tutor-end=9}{3}\sqrt{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{0}}}{\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{0}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 8 题

解题过程

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

9

一、选择题（续） · 数学竞赛/待细分

如图，网格纸上小正方形的边长为 1\htmlData{tutor-start=0,tutor-end=1}{1}，粗实线画出的是某多面体的三视图，则该多面体的表面积为 （ ） (A) 18+365\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{8} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{6}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{5}} (B) 54+185\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{4} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{8}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{5}} (C) 90\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{0} (D) 81\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{1}

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 9 题

解题过程

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

10

一、选择题（续） · 数学竞赛/待细分

在封闭的直三棱柱 ABC-A1B1C1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\text{\htmlData{tutor-start=9,tutor-end=10}{-}}\htmlData{tutor-start=11,tutor-end=12}{A}_{\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{B}_{\htmlData{tutor-start=19,tutor-end=20}{1}}\htmlData{tutor-start=21,tutor-end=22}{C}_{\htmlData{tutor-start=24,tutor-end=25}{1}} 内有一个体积为 V\htmlData{tutor-start=0,tutor-end=1}{V} 的球，若 AB⊥BC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{C}, AB=6\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{6}, BC=8\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{8}, AA1=3\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=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{3}, 则 V\htmlData{tutor-start=0,tutor-end=1}{V} 的最大值是 （ ） (A) 4π\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=4}{\pi} (B) 9π2\dfrac{\htmlData{tutor-start=7,tutor-end=8}{9}\htmlData{tutor-start=8,tutor-end=11}{\pi}}{\htmlData{tutor-start=13,tutor-end=14}{2}} (C) 6π\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=4}{\pi} (D) 32π3\dfrac{\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=12}{\pi}}{\htmlData{tutor-start=14,tutor-end=15}{3}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 10 题

解题过程

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

11

一、选择题（续） · 平面几何

已知 O\htmlData{tutor-start=0,tutor-end=1}{O} 为坐标原点，F\htmlData{tutor-start=0,tutor-end=1}{F} 是椭圆 C: x2a2+y2b2=1\htmlData{tutor-start=0,tutor-end=1}{C}\text{\htmlData{tutor-start=7,tutor-end=8}{:} } \dfrac{\htmlData{tutor-start=18,tutor-end=19}{x}^{\htmlData{tutor-start=21,tutor-end=22}{2}}}{\htmlData{tutor-start=25,tutor-end=26}{a}^{\htmlData{tutor-start=28,tutor-end=29}{2}}} \htmlData{tutor-start=32,tutor-end=33}{+} \dfrac{\htmlData{tutor-start=41,tutor-end=42}{y}^{\htmlData{tutor-start=44,tutor-end=45}{2}}}{\htmlData{tutor-start=48,tutor-end=49}{b}^{\htmlData{tutor-start=51,tutor-end=52}{2}}} \htmlData{tutor-start=55,tutor-end=56}{=} \htmlData{tutor-start=57,tutor-end=58}{1} (a>b>0\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{0}) 的左焦点，A\htmlData{tutor-start=0,tutor-end=1}{A}, B\htmlData{tutor-start=0,tutor-end=1}{B} 分别为 C\htmlData{tutor-start=0,tutor-end=1}{C} 的左、右顶点，P\htmlData{tutor-start=0,tutor-end=1}{P} 为 C\htmlData{tutor-start=0,tutor-end=1}{C} 上一点，且 PF⊥x\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{F} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{x} 轴。过点 A\htmlData{tutor-start=0,tutor-end=1}{A} 的直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与线段 PF\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{F} 交于点 M\htmlData{tutor-start=0,tutor-end=1}{M}, 与 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴交于点 E\htmlData{tutor-start=0,tutor-end=1}{E}。若直线 BM\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{M} 经过 OE\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{E} 的中点，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的离心率为 （ ） (A) 13\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{3}} (B) 12\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}} (C) 23\dfrac{\htmlData{tutor-start=7,tutor-end=8}{2}}{\htmlData{tutor-start=10,tutor-end=11}{3}} (D) 34\dfrac{\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{4}}

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 11 题

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

12

一、选择题（续）/ 二、填空题 / 三、解答题 · 代数

定义"规范 01\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{1} 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}{\}} 如下：{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}{\}} 共有 2m\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{m} 项，其中 m\htmlData{tutor-start=0,tutor-end=1}{m} 项为 0\htmlData{tutor-start=0,tutor-end=1}{0}, m\htmlData{tutor-start=0,tutor-end=1}{m} 项为 1\htmlData{tutor-start=0,tutor-end=1}{1}，且对任意 k⩽2m\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{m}, a1,a2,⋯,ak\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{a}_{\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}{a}_{\htmlData{tutor-start=25,tutor-end=26}{k}} 中 0\htmlData{tutor-start=0,tutor-end=1}{0} 的个数不少于 1\htmlData{tutor-start=0,tutor-end=1}{1} 的个数。若 m=4\htmlData{tutor-start=0,tutor-end=1}{m} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{4}, 则不同的"规范 01\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{1} 数列"共有 （ ） (A) 18\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{8} 个 (B) 16\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{6} 个 (C) 14\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{4} 个 (D) 12\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{2} 个

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 12 题

解题过程

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

13

一、选择题（续）/ 二、填空题 / 三、解答题 · 数学竞赛/待细分

若 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y} 满足约束条件 {x−y+1⩾0,x−2y⩽0,x+2y−2⩽0,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{y} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{1} \htmlData{tutor-start=24,tutor-end=34}{\geqslant }\htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=36}{,} \\ \htmlData{tutor-start=40,tutor-end=41}{x} \htmlData{tutor-start=42,tutor-end=43}{-} \htmlData{tutor-start=44,tutor-end=45}{2}\htmlData{tutor-start=45,tutor-end=46}{y} \htmlData{tutor-start=47,tutor-end=57}{\leqslant }\htmlData{tutor-start=57,tutor-end=58}{0}\htmlData{tutor-start=58,tutor-end=59}{,} \\ \htmlData{tutor-start=63,tutor-end=64}{x} \htmlData{tutor-start=65,tutor-end=66}{+} \htmlData{tutor-start=67,tutor-end=68}{2}\htmlData{tutor-start=68,tutor-end=69}{y} \htmlData{tutor-start=70,tutor-end=71}{-} \htmlData{tutor-start=72,tutor-end=73}{2} \htmlData{tutor-start=74,tutor-end=84}{\leqslant }\htmlData{tutor-start=84,tutor-end=85}{0}\htmlData{tutor-start=85,tutor-end=86}{,} \end{cases} 则 z=x+y\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{y} 的最大值为______。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 13 题

解题过程

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

14

一、选择题（续）/ 二、填空题 / 三、解答题 · 代数

函数 y=sin⁡x−3cos⁡x\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \sin \htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{-} \sqrt{\htmlData{tutor-start=19,tutor-end=20}{3}}\cos \htmlData{tutor-start=26,tutor-end=27}{x} 的图象可由函数 y=sin⁡x+3cos⁡x\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \sin \htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{+} \sqrt{\htmlData{tutor-start=19,tutor-end=20}{3}}\cos \htmlData{tutor-start=26,tutor-end=27}{x} 的图象至少向右平移______个单位长度得到。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 14 题

解题过程

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

15

一、选择题（续）/ 二、填空题 / 三、解答题 · 代数

已知 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}{)} 为偶函数，当 x<0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0} 时，f(x)=ln⁡(−x)+3x\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}{-}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{x}, 则曲线 y=f(x)\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)} 在 (1,−3)\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}{-}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{)} 处的切线方程是______。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 15 题

解题过程

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

16

一、选择题（续）/ 二、填空题 / 三、解答题 · 平面几何

已知直线 l: mx+y+3m−3=0\htmlData{tutor-start=0,tutor-end=1}{l}\text{\htmlData{tutor-start=7,tutor-end=8}{:} } \htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{x} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{y} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{3}\htmlData{tutor-start=21,tutor-end=22}{m} \htmlData{tutor-start=23,tutor-end=24}{-} \sqrt{\htmlData{tutor-start=31,tutor-end=32}{3}} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{0} 与圆 x2+y2=12\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}{1}\htmlData{tutor-start=17,tutor-end=18}{2} 交于 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B} 两点，过 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B} 分别做 l\htmlData{tutor-start=0,tutor-end=1}{l} 的垂线与 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴交于 C,D\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{D} 两点，若 AB=23\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{2}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}, 则 ∣CD∣=\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{=}______。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 16 题

解题过程

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

17

一、选择题（续）/ 二、填空题 / 三、解答题 · 代数

已知数列 {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 Sn=1+λan\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}{1} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=20}{\lambda }\htmlData{tutor-start=20,tutor-end=21}{a}_{\htmlData{tutor-start=23,tutor-end=24}{n}}, 其中 λ≠0\htmlData{tutor-start=0,tutor-end=8}{\lambda }\neq \htmlData{tutor-start=13,tutor-end=14}{0}; (1) 证明 {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}{\}} 是等比数列，并求其通项公式; (2) 若 S5=3132\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{5}} \htmlData{tutor-start=6,tutor-end=7}{=} \dfrac{\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{1}}{\htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{2}}, 求 λ\htmlData{tutor-start=0,tutor-end=7}{\lambda}。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 17 题

解题过程

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

18

一、选择题（续）/ 二、填空题 / 三、解答题 · 数学竞赛/待细分

下图是我国 2008\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{8} 年至 2014\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{4} 年生活垃圾无害化处理量（单位：亿吨）的折线图。 (1) 由折线图看出，可用线性回归模型拟合 y\htmlData{tutor-start=0,tutor-end=1}{y} 与 t\htmlData{tutor-start=0,tutor-end=1}{t} 的关系，请用相关系数加以说明； (2) 建立 y\htmlData{tutor-start=0,tutor-end=1}{y} 关于 t\htmlData{tutor-start=0,tutor-end=1}{t} 的回归方程（系数精确到 0.01），预测 2016 年我国生活垃圾无害化处理量。 参考数据：∑i=17yi=9.32\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}{7}} \htmlData{tutor-start=15,tutor-end=16}{y}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{9}\htmlData{tutor-start=24,tutor-end=25}{.}\htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{2}，∑i=17tiyi=40.17\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}{7}} \htmlData{tutor-start=15,tutor-end=16}{t}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{y}_{\htmlData{tutor-start=24,tutor-end=25}{i}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{4}\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{.}\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{7}，∑i=17(yi−yˉ)2=0.55\sqrt{\sum_{\htmlData{tutor-start=12,tutor-end=13}{i}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}}^{\htmlData{tutor-start=18,tutor-end=19}{7}} \htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{y}_{\htmlData{tutor-start=25,tutor-end=26}{i}} \htmlData{tutor-start=28,tutor-end=29}{-} \bar{\htmlData{tutor-start=35,tutor-end=36}{y}}\htmlData{tutor-start=37,tutor-end=38}{)}^{\htmlData{tutor-start=40,tutor-end=41}{2}}} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{.}\htmlData{tutor-start=48,tutor-end=49}{5}\htmlData{tutor-start=49,tutor-end=50}{5}，7≈2.646\sqrt{\htmlData{tutor-start=6,tutor-end=7}{7}} \htmlData{tutor-start=9,tutor-end=17}{\approx }\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{.}\htmlData{tutor-start=19,tutor-end=20}{6}\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{6}。 参考公式：相关系数 r=∑i=1n(ti−tˉ)(yi−yˉ)∑i=1n(ti−tˉ)2∑i=1n(yi−yˉ)2\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}{t}_{\htmlData{tutor-start=29,tutor-end=30}{i}} \htmlData{tutor-start=32,tutor-end=33}{-} \bar{\htmlData{tutor-start=39,tutor-end=40}{t}}\htmlData{tutor-start=41,tutor-end=42}{)}\htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{y}_{\htmlData{tutor-start=46,tutor-end=47}{i}} \htmlData{tutor-start=49,tutor-end=50}{-} \bar{\htmlData{tutor-start=56,tutor-end=57}{y}}\htmlData{tutor-start=58,tutor-end=59}{)}}{\sqrt{\sum_{\htmlData{tutor-start=73,tutor-end=74}{i}\htmlData{tutor-start=74,tutor-end=75}{=}\htmlData{tutor-start=75,tutor-end=76}{1}}^{\htmlData{tutor-start=79,tutor-end=80}{n}} \htmlData{tutor-start=82,tutor-end=83}{(}\htmlData{tutor-start=83,tutor-end=84}{t}_{\htmlData{tutor-start=86,tutor-end=87}{i}} \htmlData{tutor-start=89,tutor-end=90}{-} \bar{\htmlData{tutor-start=96,tutor-end=97}{t}}\htmlData{tutor-start=98,tutor-end=99}{)}^{\htmlData{tutor-start=101,tutor-end=102}{2}} \sum_{\htmlData{tutor-start=110,tutor-end=111}{i}\htmlData{tutor-start=111,tutor-end=112}{=}\htmlData{tutor-start=112,tutor-end=113}{1}}^{\htmlData{tutor-start=116,tutor-end=117}{n}} \htmlData{tutor-start=119,tutor-end=120}{(}\htmlData{tutor-start=120,tutor-end=121}{y}_{\htmlData{tutor-start=123,tutor-end=124}{i}} \htmlData{tutor-start=126,tutor-end=127}{-} \bar{\htmlData{tutor-start=133,tutor-end=134}{y}}\htmlData{tutor-start=135,tutor-end=136}{)}^{\htmlData{tutor-start=138,tutor-end=139}{2}}}}。 回归方程 y^=a^+b^t\hat{\htmlData{tutor-start=5,tutor-end=6}{y}} \htmlData{tutor-start=8,tutor-end=9}{=} \hat{\htmlData{tutor-start=15,tutor-end=16}{a}} \htmlData{tutor-start=18,tutor-end=19}{+} \hat{\htmlData{tutor-start=25,tutor-end=26}{b}}\htmlData{tutor-start=27,tutor-end=28}{t} 中斜率和截距的最小二乘估计公式分别为： b^=∑i=1n(ti−tˉ)(yi−yˉ)∑i=1n(ti−tˉ)2\hat{\htmlData{tutor-start=5,tutor-end=6}{b}} \htmlData{tutor-start=8,tutor-end=9}{=} \frac{\sum_{\htmlData{tutor-start=22,tutor-end=23}{i}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{1}}^{\htmlData{tutor-start=28,tutor-end=29}{n}} \htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{t}_{\htmlData{tutor-start=35,tutor-end=36}{i}} \htmlData{tutor-start=38,tutor-end=39}{-} \bar{\htmlData{tutor-start=45,tutor-end=46}{t}}\htmlData{tutor-start=47,tutor-end=48}{)}\htmlData{tutor-start=48,tutor-end=49}{(}\htmlData{tutor-start=49,tutor-end=50}{y}_{\htmlData{tutor-start=52,tutor-end=53}{i}} \htmlData{tutor-start=55,tutor-end=56}{-} \bar{\htmlData{tutor-start=62,tutor-end=63}{y}}\htmlData{tutor-start=64,tutor-end=65}{)}}{\sum_{\htmlData{tutor-start=73,tutor-end=74}{i}\htmlData{tutor-start=74,tutor-end=75}{=}\htmlData{tutor-start=75,tutor-end=76}{1}}^{\htmlData{tutor-start=79,tutor-end=80}{n}} \htmlData{tutor-start=82,tutor-end=83}{(}\htmlData{tutor-start=83,tutor-end=84}{t}_{\htmlData{tutor-start=86,tutor-end=87}{i}} \htmlData{tutor-start=89,tutor-end=90}{-} \bar{\htmlData{tutor-start=96,tutor-end=97}{t}}\htmlData{tutor-start=98,tutor-end=99}{)}^{\htmlData{tutor-start=101,tutor-end=102}{2}}}，a^=yˉ−b^tˉ\hat{\htmlData{tutor-start=5,tutor-end=6}{a}} \htmlData{tutor-start=8,tutor-end=9}{=} \bar{\htmlData{tutor-start=15,tutor-end=16}{y}} \htmlData{tutor-start=18,tutor-end=19}{-} \hat{\htmlData{tutor-start=25,tutor-end=26}{b}}\bar{\htmlData{tutor-start=32,tutor-end=33}{t}}。

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 18 题

解题过程

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

19

解答题 · 平面几何

如图，四棱锥 P−ABCD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{D} 中，PA⊥\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=8}{\perp} 平面 ABCD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}，AD//BC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{/} \htmlData{tutor-start=6,tutor-end=7}{B}\htmlData{tutor-start=7,tutor-end=8}{C}，AB=AD=AC=3\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{D} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{C} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{3}，PA=BC=4\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{C} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{4}，M\htmlData{tutor-start=0,tutor-end=1}{M} 为线段 AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} 上一点，AM=2MD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{M}\htmlData{tutor-start=7,tutor-end=8}{D}，N\htmlData{tutor-start=0,tutor-end=1}{N} 为 PC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C} 的中点。 (1) 证明：MN//\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N} \htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{/} 平面 PAB\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}； (2) 求直线 AN\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{N} 与平面 PMN\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{N} 所成角的正弦值。

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 19 题

解题过程

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

20

解答题 · 平面几何

已知抛物线 C:y2=2x\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}{2}\htmlData{tutor-start=12,tutor-end=13}{x} 的焦点为 F\htmlData{tutor-start=0,tutor-end=1}{F}，平行于 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的两条直线 l1,l2\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{l}_{\htmlData{tutor-start=10,tutor-end=11}{2}} 分别交 C\htmlData{tutor-start=0,tutor-end=1}{C} 于 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B} 两点，交 C\htmlData{tutor-start=0,tutor-end=1}{C} 的准线于 P,Q\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{Q} 两点。 (1) 若 F\htmlData{tutor-start=0,tutor-end=1}{F} 在线段 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 上，R\htmlData{tutor-start=0,tutor-end=1}{R} 是 PQ\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{Q} 的中点，证明 AR//FQ\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{R} \htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{/} \htmlData{tutor-start=6,tutor-end=7}{F}\htmlData{tutor-start=7,tutor-end=8}{Q}； (2) 若 △PQF\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{Q}\htmlData{tutor-start=12,tutor-end=13}{F} 的面积是 △ABF\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}{F} 的面积的两倍，求 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 中点的轨迹方程。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 20 题

解题过程

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

21

解答题 · 代数

设函数 f(x)=αcos⁡2x+(α−1)(cos⁡x+1)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=14}{\alpha }\cos \htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{x} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=32}{\alpha }\htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{)}\htmlData{tutor-start=36,tutor-end=37}{(}\cos \htmlData{tutor-start=42,tutor-end=43}{x} \htmlData{tutor-start=44,tutor-end=45}{+} \htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{)}，其中 α>0\htmlData{tutor-start=0,tutor-end=7}{\alpha }\htmlData{tutor-start=7,tutor-end=8}{>} \htmlData{tutor-start=9,tutor-end=10}{0}，∣f(x)∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{f}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{|} 的最大值为 A\htmlData{tutor-start=0,tutor-end=1}{A}。 (1) 求 f′(x)\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}； (2) 求 A\htmlData{tutor-start=0,tutor-end=1}{A}； (3) 证明 ∣f′(x)∣⩽2A\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{f}'\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{|} \htmlData{tutor-start=8,tutor-end=18}{\leqslant }\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{A}。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 21 题

解题过程

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

22

解答题 · 平面几何

如图，⊙O\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O} 中 AB⌢\overset{\htmlData{tutor-start=9,tutor-end=15}{\frown}}{\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{B}} 的中点为 P\htmlData{tutor-start=0,tutor-end=1}{P}，弦 PC,PD\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=4,tutor-end=5}{P}\htmlData{tutor-start=5,tutor-end=6}{D} 分别交 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 于 E,F\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{F} 两点。 (1) 若 ∠PFB=2∠PCD\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=10}{B} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=21}{\angle }\htmlData{tutor-start=21,tutor-end=22}{P}\htmlData{tutor-start=22,tutor-end=23}{C}\htmlData{tutor-start=23,tutor-end=24}{D}，求 ∠PCD\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{D} 的大小； (2) 若 EC\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{C} 的垂直平分线与 FD\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{D} 的垂直平分线交于点 G\htmlData{tutor-start=0,tutor-end=1}{G}，证明 OG⊥CD\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{G} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{D}。

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

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 22 题

解题过程

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

23

解答题 · 数学竞赛/待细分

在直角坐标系 xOy\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{y} 中，曲线 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的参数方程为 {x=3cos⁡α,y=sin⁡α,\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=32,tutor-end=38}{\alpha}\htmlData{tutor-start=38,tutor-end=39}{,} \\ \htmlData{tutor-start=43,tutor-end=44}{y} \htmlData{tutor-start=45,tutor-end=46}{=} \sin \htmlData{tutor-start=52,tutor-end=58}{\alpha}\htmlData{tutor-start=58,tutor-end=59}{,} \end{cases} (α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 为参数)，以坐标原点为极点，以 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的正半轴为极轴，建立极坐标系，曲线 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的极坐标方程为 ρsin⁡(θ+π4)=22\htmlData{tutor-start=0,tutor-end=5}{\rho }\sin\left(\htmlData{tutor-start=15,tutor-end=22}{\theta }\htmlData{tutor-start=22,tutor-end=23}{+} \frac{\htmlData{tutor-start=30,tutor-end=33}{\pi}}{\htmlData{tutor-start=35,tutor-end=36}{4}}\right) \htmlData{tutor-start=45,tutor-end=46}{=} \htmlData{tutor-start=47,tutor-end=48}{2}\sqrt{\htmlData{tutor-start=54,tutor-end=55}{2}}。 (1) 写出 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的普通方程和 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的直角坐标方程； (2) 设点 P\htmlData{tutor-start=0,tutor-end=1}{P} 在 C1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 上，点 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 在 C2\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 上，求 ∣PQ∣\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}{|} 的最小值及此时 P\htmlData{tutor-start=0,tutor-end=1}{P} 的直角坐标。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 23 题

解题过程

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

24

解答题 · 代数

已知函数 f(x)=∣2x−a∣+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}{|}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{|} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{a}。 (1) 当 a=2\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2} 时，求不等式 f(x)⩽6\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}{\leqslant }\htmlData{tutor-start=15,tutor-end=16}{6} 的解集； (2) 设函数 g(x)=∣2x−1∣\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}{=} \htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{|}，当 x∈R\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\mathbf{\htmlData{tutor-start=14,tutor-end=15}{R}} 时，f(x)+g(x)⩾3\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{+} \htmlData{tutor-start=7,tutor-end=8}{g}\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=12,tutor-end=22}{\geqslant }\htmlData{tutor-start=22,tutor-end=23}{3}，求 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围。

题解状态：标准答案与规范题解待补充

题目标签：全国卷 3理科第 24 题

解题过程

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