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2014 年高考数学（新课标卷 1文科）

exams_raw/普通高考/2014/2014新课标1文(河南,河北,山西).pdf · HS-MATH-1024-v2.1-solution-aware

24 题33 个小问/题组
1

一、选择题 · 集合的基本运算

已知集合 M={x∣−1<x<3}\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}{x} \htmlData{tutor-start=8,tutor-end=13}{\mid }\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{x} \htmlData{tutor-start=20,tutor-end=21}{<} \htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=25}{\}}, N={x∣−2<x<1}\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}{-}\htmlData{tutor-start=14,tutor-end=15}{2} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{x} \htmlData{tutor-start=20,tutor-end=21}{<} \htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=25}{\}}, 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)\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}{1}\htmlData{tutor-start=6,tutor-end=7}{)} (B) (−1,1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)} (C) (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}{3}\htmlData{tutor-start=5,tutor-end=6}{)} (D) (−2,3)\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}{3}\htmlData{tutor-start=6,tutor-end=7}{)}

答案：B，(-1,1)。

题目标签：区间表示集合的交集

解题过程

求两个开区间的交集

找出同时满足两组不等式的实数范围。

(1)
识别结构并建立关系

交集符号要求两个条件同时成立。把集合先看成数轴上的开区间，再取它们重叠的部分，比逐个比较选项更直接。

为什么从这里入手：集合题不能只看式子的外形，必须回到“元素是否满足条件”。这里的“交集符号要求两个条件同时成立。把集合先看成数轴上的开区间，再取它们重叠的部分，比逐个比较选项更直接。”给出了元素筛选规则，因此先识别结构并建立关系，再逐项保留或删除元素，思路最直接。

详细展开：由题意，属于交集的实数必须同时满足−1<x<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=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{<}\htmlData{tutor-start=5,tutor-end=6}{3}与−2<x<1\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=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{<}\htmlData{tutor-start=5,tutor-end=6}{1}。下界要取两个下界中较大的−1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}，上界要取两个上界中较小的1\htmlData{tutor-start=0,tutor-end=1}{1}。原不等式都是严格不等式，所以两个端点都不取。

x∈M∩N⟺−1<x<3 且 −2<x<1⟺−1<x<1.\begin{aligned}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=20}{\in }\htmlData{tutor-start=20,tutor-end=21}{M}\htmlData{tutor-start=21,tutor-end=26}{\cap }\htmlData{tutor-start=26,tutor-end=27}{N}&\htmlData{tutor-start=28,tutor-end=48}{\Longleftrightarrow }\htmlData{tutor-start=48,tutor-end=49}{-}\htmlData{tutor-start=49,tutor-end=50}{1}\htmlData{tutor-start=50,tutor-end=51}{<}\htmlData{tutor-start=51,tutor-end=52}{x}\htmlData{tutor-start=52,tutor-end=53}{<}\htmlData{tutor-start=53,tutor-end=54}{3}\text{ \htmlData{tutor-start=61,tutor-end=62}{且} }\htmlData{tutor-start=64,tutor-end=65}{-}\htmlData{tutor-start=65,tutor-end=66}{2}\htmlData{tutor-start=66,tutor-end=67}{<}\htmlData{tutor-start=67,tutor-end=68}{x}\htmlData{tutor-start=68,tutor-end=69}{<}\htmlData{tutor-start=69,tutor-end=70}{1}\\&\htmlData{tutor-start=73,tutor-end=93}{\Longleftrightarrow }\htmlData{tutor-start=93,tutor-end=94}{-}\htmlData{tutor-start=94,tutor-end=95}{1}\htmlData{tutor-start=95,tutor-end=96}{<}\htmlData{tutor-start=96,tutor-end=97}{x}\htmlData{tutor-start=97,tutor-end=98}{<}\htmlData{tutor-start=98,tutor-end=99}{1}\htmlData{tutor-start=99,tutor-end=100}{.}\end{aligned}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：核验端点与内部点：0\htmlData{tutor-start=0,tutor-end=1}{0}同时属于两个集合；−1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}和1\htmlData{tutor-start=0,tutor-end=1}{1}均因严格不等号而不属于交集；区间(−1,1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}外的点至少违反一个条件。因此选择B。

M∩N=(−1,1)B\boxed{\htmlData{tutor-start=7,tutor-end=8}{M}\htmlData{tutor-start=8,tutor-end=13}{\cap }\htmlData{tutor-start=13,tutor-end=14}{N}\htmlData{tutor-start=14,tutor-end=15}{=}\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=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{)}}\quad\boxed{\mathrm{\htmlData{tutor-start=42,tutor-end=43}{B}}}
2

一、选择题 · 三角函数符号与二倍角

若 tan⁡α>0\tan \htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{>} \htmlData{tutor-start=14,tutor-end=15}{0}, 则 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) sin⁡2α>0\sin \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}{0} (B) cos⁡α>0\cos \htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{>} \htmlData{tutor-start=14,tutor-end=15}{0} (C) sin⁡α>0\sin \htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{>} \htmlData{tutor-start=14,tutor-end=15}{0} (D) cos⁡2α>0\cos \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}{0}

答案：A，sin 2α>0。

题目标签：由正切符号判断二倍角正弦

解题过程

确定必然为正的三角函数

从tan⁡α>0\tan\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{>}\htmlData{tutor-start=11,tutor-end=12}{0}推出四个选项中恒成立的结论。

(1)
识别结构并建立关系

tan⁡α>0\tan\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{>}\htmlData{tutor-start=11,tutor-end=12}{0}只说明正弦与余弦同号，不能确定它们各自是正还是负；应使用只含tan⁡α\tan\htmlData{tutor-start=4,tutor-end=10}{\alpha}的二倍角公式消除象限不确定性。

为什么从这里入手：三角题先把未知角转成特殊角、同一角或已知边角关系。“tan⁡α>0\tan\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{>}\htmlData{tutor-start=11,tutor-end=12}{0}只说明正弦与余弦同号，不能确定它们各自是正还是负；应使用只含tan⁡α\tan\htmlData{tutor-start=4,tutor-end=10}{\alpha}的二倍角公式消除象限不确定性。”指出了可用的象限、恒等式或几何关系，先识别结构并建立关系可以同时确定数值和正负号。

详细展开：因为tan⁡α\tan\htmlData{tutor-start=4,tutor-end=10}{\alpha}有定义，1+tan⁡2α>0\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\tan^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=16}{\alpha}\htmlData{tutor-start=16,tutor-end=17}{>}\htmlData{tutor-start=17,tutor-end=18}{0}。由sin⁡2α=2tan⁡α1+tan⁡2α\sin\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\frac{\htmlData{tutor-start=18,tutor-end=19}{2}\tan\htmlData{tutor-start=23,tutor-end=29}{\alpha}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{+}\tan^{\htmlData{tutor-start=39,tutor-end=40}{2}}\htmlData{tutor-start=41,tutor-end=47}{\alpha}}，分子为正、分母为正，所以sin⁡2α>0\sin\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{>}\htmlData{tutor-start=12,tutor-end=13}{0}。例如第三象限也有tan⁡α>0\tan\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{>}\htmlData{tutor-start=11,tutor-end=12}{0}，却有sin⁡α<0,cos⁡α<0\sin\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{,}\cos\htmlData{tutor-start=17,tutor-end=23}{\alpha}\htmlData{tutor-start=23,tutor-end=24}{<}\htmlData{tutor-start=24,tutor-end=25}{0}，可排除B、C；cos⁡2α\cos\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}还随∣tan⁡α∣\htmlData{tutor-start=0,tutor-end=1}{|}\tan\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{|}变化，符号不定。

sin⁡2α=2tan⁡α1+tan⁡2α>0\sin\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\frac{\htmlData{tutor-start=18,tutor-end=19}{2}\tan\htmlData{tutor-start=23,tutor-end=29}{\alpha}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{+}\tan^{\htmlData{tutor-start=39,tutor-end=40}{2}}\htmlData{tutor-start=41,tutor-end=47}{\alpha}}\htmlData{tutor-start=48,tutor-end=49}{>}\htmlData{tutor-start=49,tutor-end=50}{0}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：取α=π/4\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=10}{\pi}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{4}时题设成立，A成立而cos⁡2α=0\cos\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0}，说明D不恒成立；取α=5π/4\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=11}{\pi}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{4}时正弦、余弦均为负，说明B、C不恒成立。唯一正确选项为A。

sin⁡2α>0A\boxed{\sin\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=18}{\alpha}\htmlData{tutor-start=18,tutor-end=19}{>}\htmlData{tutor-start=19,tutor-end=20}{0}}\quad\boxed{\mathrm{\htmlData{tutor-start=41,tutor-end=42}{A}}}
3

一、选择题 · 复数的运算

设 z=11+i+i\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \dfrac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{+}\mathrm{\htmlData{tutor-start=24,tutor-end=25}{i}}} \htmlData{tutor-start=28,tutor-end=29}{+} \mathrm{\htmlData{tutor-start=38,tutor-end=39}{i}}, 则 ∣z∣=()\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{z}\htmlData{tutor-start=2,tutor-end=3}{|} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{(}\quad\htmlData{tutor-start=12,tutor-end=13}{)} (A) 12\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}} (B) 22\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{2}}}{\htmlData{tutor-start=17,tutor-end=18}{2}} (C) 32\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{2}} (D) 2\htmlData{tutor-start=0,tutor-end=1}{2}

答案：B，√2/2。

题目标签：复数除法与模

解题过程

化为代数形式后求模

把复数写成a+bi\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}{i}并计算模长。

(1)
识别结构并建立关系

分母含i\htmlData{tutor-start=0,tutor-end=1}{i}时先乘共轭复数有理化。模不能对实部和虚部分别取绝对值，而应使用∣a+bi∣=a2+b2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{i}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{=}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{b}^{\htmlData{tutor-start=22,tutor-end=23}{2}}}。

为什么从这里入手：复数题先判断目标需要代数形式还是模与辐角。“分母含i\htmlData{tutor-start=0,tutor-end=1}{i}时先乘共轭复数有理化。模不能对实部和虚部分别取绝对值，而应使用∣a+bi∣=a2+b2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{i}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{=}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{b}^{\htmlData{tutor-start=22,tutor-end=23}{2}}}。”与“识别结构并建立关系”直接相连，先利用共轭、模或实虚部关系，通常能避免把复数完全展开成冗长乘积。

详细展开：给11+i\frac1{1+i}的分子分母同乘1−i\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{i}，分母变为(1+i)(1−i)=2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{i}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{i}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{2}，故该分式等于1−i2\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{i}}{\htmlData{tutor-start=11,tutor-end=12}{2}}。再加i\htmlData{tutor-start=0,tutor-end=1}{i}得到z=12+12i\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{+}\frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{i}，于是模为两个坐标平方和的平方根。

z=1−i2+i=1+i2,∣z∣=(12)2+(12)2=22\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{i}}{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{i}\htmlData{tutor-start=17,tutor-end=18}{=}\frac{\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{i}}{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{,}\qquad \htmlData{tutor-start=39,tutor-end=40}{|}\htmlData{tutor-start=40,tutor-end=41}{z}\htmlData{tutor-start=41,tutor-end=42}{|}\htmlData{tutor-start=42,tutor-end=43}{=}\sqrt{\left(\frac{\htmlData{tutor-start=61,tutor-end=62}{1}}{\htmlData{tutor-start=64,tutor-end=65}{2}}\right)^{\htmlData{tutor-start=75,tutor-end=76}{2}}\htmlData{tutor-start=77,tutor-end=78}{+}\left(\frac{\htmlData{tutor-start=90,tutor-end=91}{1}}{\htmlData{tutor-start=93,tutor-end=94}{2}}\right)^{\htmlData{tutor-start=104,tutor-end=105}{2}}}\htmlData{tutor-start=107,tutor-end=108}{=}\frac{\sqrt{\htmlData{tutor-start=120,tutor-end=121}{2}}}{\htmlData{tutor-start=124,tutor-end=125}{2}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：反算z(1+i)=i\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}{i}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{i}，与原式乘以1+i\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{i}所得结果一致；且∣z∣2=1/4+1/4=1/2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{z}\htmlData{tutor-start=2,tutor-end=3}{|}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\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}{4}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{4}\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}{2}，所以正值模长确为2/2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}，选择B。

∣z∣=22B\boxed{\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{z}\htmlData{tutor-start=9,tutor-end=10}{|}\htmlData{tutor-start=10,tutor-end=11}{=}\frac{\sqrt{\htmlData{tutor-start=23,tutor-end=24}{2}}}{\htmlData{tutor-start=27,tutor-end=28}{2}}}\quad\boxed{\mathrm{\htmlData{tutor-start=50,tutor-end=51}{B}}}
4

一、选择题 · 双曲线的标准方程

已知双曲线 x2a2−y23=1\dfrac{\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}}{\htmlData{tutor-start=14,tutor-end=15}{a}^{\htmlData{tutor-start=17,tutor-end=18}{2}}} \htmlData{tutor-start=21,tutor-end=22}{-} \dfrac{\htmlData{tutor-start=30,tutor-end=31}{y}^{\htmlData{tutor-start=33,tutor-end=34}{2}}}{\htmlData{tutor-start=37,tutor-end=38}{3}} \htmlData{tutor-start=40,tutor-end=41}{=} \htmlData{tutor-start=42,tutor-end=43}{1} (a>0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a} \htmlData{tutor-start=3,tutor-end=4}{>} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)} 的离心率为 2\htmlData{tutor-start=0,tutor-end=1}{2}, 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) 2\htmlData{tutor-start=0,tutor-end=1}{2} (B) 62\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{6}}}{\htmlData{tutor-start=17,tutor-end=18}{2}} (C) 52\dfrac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{5}}}{\htmlData{tutor-start=17,tutor-end=18}{2}} (D) 1\htmlData{tutor-start=0,tutor-end=1}{1}

答案：D，a=1。

题目标签：由双曲线离心率求参数

解题过程

建立离心率与半轴的关系

利用e=c/a\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{a}和c2=a2+b2\htmlData{tutor-start=0,tutor-end=1}{c}^{\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}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{2}}求a\htmlData{tutor-start=0,tutor-end=1}{a}。

(1)
识别结构并建立关系

方程已经是焦点在x\htmlData{tutor-start=0,tutor-end=1}{x}轴上的双曲线标准式，其中b2=3\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}。题目给离心率，应立刻联立e2=1+b2/a2\htmlData{tutor-start=0,tutor-end=1}{e}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{a}^{\htmlData{tutor-start=17,tutor-end=18}{2}}。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“方程已经是焦点在x\htmlData{tutor-start=0,tutor-end=1}{x}轴上的双曲线标准式，其中b2=3\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}。题目给离心率，应立刻联立e2=1+b2/a2\htmlData{tutor-start=0,tutor-end=1}{e}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{a}^{\htmlData{tutor-start=17,tutor-end=18}{2}}。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：双曲线的实半轴平方为a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}，虚半轴平方为3\htmlData{tutor-start=0,tutor-end=1}{3}，故c2=a2+3\htmlData{tutor-start=0,tutor-end=1}{c}^{\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}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{3}。由离心率e=2\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}可得c=2a\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{a}。平方后代入焦距关系，得到4a2=a2+3\htmlData{tutor-start=0,tutor-end=1}{4}\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}{a}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{3}，即3a2=3\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}。题设a>0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}，所以只取正根。

4=e2=c2a2=a2+3a2=1+3a2⟹a2=1⟹a=14=e^{2}=\frac{c^{2}}{a^{2}}=\frac{a^{2}+3}{a^{2}}=1+\frac3{a^{2}}\Longrightarrow a^{2}=1\Longrightarrow a=1
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：把a=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}代回，得到c2=1+3=4\htmlData{tutor-start=0,tutor-end=1}{c}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{4}，故c=2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}且e=c/a=2\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}，完全满足题设。负根不符合a>0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}，选择D。

a=1D\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}}\quad\boxed{\mathrm{\htmlData{tutor-start=31,tutor-end=32}{D}}}
5

一、选择题 · 函数的奇偶性

设函数 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}{)}, g(x)\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 的定义域都为 R\mathbf{\htmlData{tutor-start=8,tutor-end=9}{R}}, 且 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}{)} 是奇函数, g(x)\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}{)} 是偶函数, 则下列结论正确的是 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) f(x)∣g(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=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}{x}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{|} 是奇函数 (B) ∣f(x)∣g(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}{|}\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}{)} 是奇函数 (C) f(x)g(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=4,tutor-end=5}{g}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)} 是偶函数 (D) ∣f(x)g(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}{g}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{|} 是奇函数

答案：A，f(x)|g(x)|是奇函数。

题目标签：奇偶函数的乘积判定

解题过程

逐项计算函数在-x处的值

判断含绝对值与乘积的函数奇偶性。

(1)
识别结构并建立关系

奇偶性必须按定义检查h(−x)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}。偶函数取绝对值后仍为偶函数，奇函数取绝对值后则变为偶函数。

为什么从这里入手：目标是“识别结构并建立关系”，而“奇偶性必须按定义检查h(−x)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}。偶函数取绝对值后仍为偶函数，奇函数取绝对值后则变为偶函数。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：由f(−x)=−f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\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}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}、g(−x)=g(x)\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\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}{=}\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}{)}，可得∣g(−x)∣=∣g(x)∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{g}\htmlData{tutor-start=2,tutor-end=3}{(}\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=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{|}\htmlData{tutor-start=9,tutor-end=10}{g}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{|}。于是h(x)=f(x)∣g(x)∣\htmlData{tutor-start=0,tutor-end=1}{h}\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}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{|}\htmlData{tutor-start=10,tutor-end=11}{g}\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}{|}满足h(−x)=f(−x)∣g(−x)∣=−f(x)∣g(x)∣=−h(x)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\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}{=}\htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\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}{g}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{|}\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}{|}\htmlData{tutor-start=25,tutor-end=26}{g}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{x}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{|}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{h}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{x}\htmlData{tutor-start=35,tutor-end=36}{)}，所以A中的函数是奇函数。其余各项中，∣f∣g\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}{g}和∣fg∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{f}\htmlData{tutor-start=2,tutor-end=3}{g}\htmlData{tutor-start=3,tutor-end=4}{|}都是偶函数，而fg\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{g}是奇函数。

h(−x)=f(−x)∣g(−x)∣=−f(x)∣g(x)∣=−h(x)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\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}{=}\htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\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}{g}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{|}\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}{|}\htmlData{tutor-start=25,tutor-end=26}{g}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{x}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{|}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{h}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{x}\htmlData{tutor-start=35,tutor-end=36}{)}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：定义域均为R\mathbb{\htmlData{tutor-start=8,tutor-end=9}{R}}，关于原点对称，奇偶性判定前提满足。A恰好满足h(−x)=−h(x)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\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}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{h}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}；B、D误把偶函数判为奇函数，C误把奇函数判为偶函数。

f(x)∣g(x)∣ 是奇函数A\boxed{\htmlData{tutor-start=7,tutor-end=8}{f}\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}{g}\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}{|}\text{ \htmlData{tutor-start=24,tutor-end=25}{是}\htmlData{tutor-start=25,tutor-end=26}{奇}\htmlData{tutor-start=26,tutor-end=27}{函}\htmlData{tutor-start=27,tutor-end=28}{数}}}\quad\boxed{\mathrm{\htmlData{tutor-start=50,tutor-end=51}{A}}}
6

一、选择题 · 平面向量的线性运算

设 D\htmlData{tutor-start=0,tutor-end=1}{D}, E\htmlData{tutor-start=0,tutor-end=1}{E}, F\htmlData{tutor-start=0,tutor-end=1}{F} 分别为 △ABC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C} 的三边 BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}, CA\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{A}, AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 的中点, 则 EB→+FC→=()\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{E}\htmlData{tutor-start=17,tutor-end=18}{B}} \htmlData{tutor-start=20,tutor-end=21}{+} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{F}\htmlData{tutor-start=39,tutor-end=40}{C}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{(}\quad\htmlData{tutor-start=50,tutor-end=51}{)} (A) BC→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{C}} (B) 12AD→\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}}\overrightarrow{\htmlData{tutor-start=28,tutor-end=29}{A}\htmlData{tutor-start=29,tutor-end=30}{D}} (C) AD→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}} (D) 12BC→\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}}\overrightarrow{\htmlData{tutor-start=28,tutor-end=29}{B}\htmlData{tutor-start=29,tutor-end=30}{C}}

答案：C，向量AD。

题目标签：中点条件下的向量运算

解题过程

用位置向量表示各中点

把两个向量和化为一条几何向量。

(1)
识别结构并建立关系

出现多个中点时，选同一原点写位置向量最稳妥。中点的位置向量等于两个端点位置向量的平均。

为什么从这里入手：统计量都来自明确的样本计数或加权。“出现多个中点时，选同一原点写位置向量最稳妥。中点的位置向量等于两个端点位置向量的平均。”给出了数据与权重，先识别结构并建立关系可以避免把频数、频率和概率混为一谈，并能用总数或概率和反查计算。

详细展开：记A,B,C\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}{C}的位置向量分别为a,b,c\boldsymbol a,\boldsymbol b,\boldsymbol c。由中点定义，E\htmlData{tutor-start=0,tutor-end=1}{E}的位置向量为(c+a)/2(\boldsymbol c+\boldsymbol a)/2，F\htmlData{tutor-start=0,tutor-end=1}{F}为(a+b)/2(\boldsymbol a+\boldsymbol b)/2，D\htmlData{tutor-start=0,tutor-end=1}{D}为(b+c)/2(\boldsymbol b+\boldsymbol c)/2。代入EB→=b−e\overrightarrow{EB}=\boldsymbol b-\boldsymbol e与FC→=c−f\overrightarrow{FC}=\boldsymbol c-\boldsymbol f并合并同类项，结果正是d−a\boldsymbol d-\boldsymbol a。

EB→+FC→=b−c+a2+c−a+b2=b+c2−a=AD→.\begin{aligned}\overrightarrow{EB}+\overrightarrow{FC}&=\boldsymbol b-\frac{\boldsymbol c+\boldsymbol a}{2}+\boldsymbol c-\frac{\boldsymbol a+\boldsymbol b}{2}\\&=\frac{\boldsymbol b+\boldsymbol c}{2}-\boldsymbol a=\overrightarrow{AD}.\end{aligned}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：结果的起点为A\htmlData{tutor-start=0,tutor-end=1}{A}、终点为BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}中点D\htmlData{tutor-start=0,tutor-end=1}{D}，方向和长度均与中线向量一致；没有额外的1/2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2}系数。因此选择C。

EB→+FC→=AD→C\boxed{\overrightarrow{\htmlData{tutor-start=23,tutor-end=24}{E}\htmlData{tutor-start=24,tutor-end=25}{B}}\htmlData{tutor-start=26,tutor-end=27}{+}\overrightarrow{\htmlData{tutor-start=43,tutor-end=44}{F}\htmlData{tutor-start=44,tutor-end=45}{C}}\htmlData{tutor-start=46,tutor-end=47}{=}\overrightarrow{\htmlData{tutor-start=63,tutor-end=64}{A}\htmlData{tutor-start=64,tutor-end=65}{D}}}\quad\boxed{\mathrm{\htmlData{tutor-start=87,tutor-end=88}{C}}}
7

一、选择题 · 三角函数的周期性

在函数① y=cos⁡∣2x∣\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \cos\htmlData{tutor-start=8,tutor-end=9}{|}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{|}, ② y=∣cos⁡x∣\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{|}\cos \htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{|}, ③ y=cos⁡ ⁣(2x+π6)\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \cos\!\left(\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{x} \htmlData{tutor-start=19,tutor-end=20}{+} \dfrac{\htmlData{tutor-start=28,tutor-end=31}{\pi}}{\htmlData{tutor-start=33,tutor-end=34}{6}}\right), ④ y=tan⁡ ⁣(2x−π4)\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \tan\!\left(\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{x} \htmlData{tutor-start=19,tutor-end=20}{-} \dfrac{\htmlData{tutor-start=28,tutor-end=31}{\pi}}{\htmlData{tutor-start=33,tutor-end=34}{4}}\right) 中, 最小正周期为 π\htmlData{tutor-start=0,tutor-end=3}{\pi} 的所有函数为 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) ②④ (B) ①③④ (C) ①②③ (D) ①③

答案：C，函数①②③。

题目标签：复合三角函数的最小正周期

解题过程

分别化简并求最小正周期

从四个函数中筛出最小正周期为π\htmlData{tutor-start=0,tutor-end=3}{\pi}者。

(1)
识别结构并建立关系

绝对值可能改变周期，不能只套2π/∣ω∣\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\pi}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=12}{\omega}\htmlData{tutor-start=12,tutor-end=13}{|}。先利用余弦偶性化简①，再观察②取绝对值后正负半周重合。

为什么从这里入手：三角题先把未知角转成特殊角、同一角或已知边角关系。“绝对值可能改变周期，不能只套2π/∣ω∣\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\pi}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=12}{\omega}\htmlData{tutor-start=12,tutor-end=13}{|}。先利用余弦偶性化简①，再观察②取绝对值后正负半周重合。”指出了可用的象限、恒等式或几何关系，先识别结构并建立关系可以同时确定数值和正负号。

详细展开：①中cos⁡∣2x∣=cos⁡2x\cos\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{=}\cos\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{x}，最小正周期为π\htmlData{tutor-start=0,tutor-end=3}{\pi}。②中∣cos⁡(x+π)∣=∣−cos⁡x∣=∣cos⁡x∣\htmlData{tutor-start=0,tutor-end=1}{|}\cos\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=11}{\pi}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{|}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{|}\htmlData{tutor-start=15,tutor-end=16}{-}\cos \htmlData{tutor-start=21,tutor-end=22}{x}\htmlData{tutor-start=22,tutor-end=23}{|}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{|}\cos \htmlData{tutor-start=30,tutor-end=31}{x}\htmlData{tutor-start=31,tutor-end=32}{|}，且没有更小的公共正周期，故为π\htmlData{tutor-start=0,tutor-end=3}{\pi}。③的角频率为2，平移不改变周期，故周期为2π/2=π\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\pi}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=10}{\pi}。④中正切本身周期为π\htmlData{tutor-start=0,tutor-end=3}{\pi}，角频率为2，所以最小正周期为π/2\htmlData{tutor-start=0,tutor-end=3}{\pi}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}。

T1=π,T2=π,T3=π,T4=π2\htmlData{tutor-start=0,tutor-end=1}{T}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=9}{\pi}\htmlData{tutor-start=9,tutor-end=10}{,}\qquad \htmlData{tutor-start=17,tutor-end=18}{T}_{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=26}{\pi}\htmlData{tutor-start=26,tutor-end=27}{,}\qquad \htmlData{tutor-start=34,tutor-end=35}{T}_{\htmlData{tutor-start=37,tutor-end=38}{3}}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=43}{\pi}\htmlData{tutor-start=43,tutor-end=44}{,}\qquad \htmlData{tutor-start=51,tutor-end=52}{T}_{\htmlData{tutor-start=54,tutor-end=55}{4}}\htmlData{tutor-start=56,tutor-end=57}{=}\frac{\htmlData{tutor-start=63,tutor-end=66}{\pi}}{\htmlData{tutor-start=68,tutor-end=69}{2}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：代入x\htmlData{tutor-start=0,tutor-end=1}{x}与x+π\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=5}{\pi}可验证①②③函数值相同；④在平移π/2\htmlData{tutor-start=0,tutor-end=3}{\pi}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}后已经重复，故其最小正周期不是π\htmlData{tutor-start=0,tutor-end=3}{\pi}。所以应选①②③，即C。

最小正周期为 π 的是①②③C\boxed{\text{\htmlData{tutor-start=13,tutor-end=14}{最}\htmlData{tutor-start=14,tutor-end=15}{小}\htmlData{tutor-start=15,tutor-end=16}{正}\htmlData{tutor-start=16,tutor-end=17}{周}\htmlData{tutor-start=17,tutor-end=18}{期}\htmlData{tutor-start=18,tutor-end=19}{为} }\htmlData{tutor-start=21,tutor-end=24}{\pi}\text{ \htmlData{tutor-start=31,tutor-end=32}{的}\htmlData{tutor-start=32,tutor-end=33}{是}\htmlData{tutor-start=33,tutor-end=34}{①}\htmlData{tutor-start=34,tutor-end=35}{②}\htmlData{tutor-start=35,tutor-end=36}{③}}}\quad\boxed{\mathrm{\htmlData{tutor-start=58,tutor-end=59}{C}}}
8

一、选择题 · 空间几何体的三视图

如图, 网格纸的各小格都是正方形, 粗实线画出的是一个几何体的三视图, 则这个几何体是 ()\htmlData{tutor-start=0,tutor-end=1}{(}\quad\htmlData{tutor-start=6,tutor-end=7}{)} (A) 三棱锥 (B) 三棱柱 (C) 四棱锥 (D) 四棱柱

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

答案：B，三棱柱。

题目标签：由三视图识别几何体

解题过程

匹配三个投影的轮廓

根据正视图、侧视图和俯视图判断几何体类型。

(1)
识别结构并建立关系

原卷整页图中，一个投影为三角形，另两个投影为矩形。棱锥通常会在多个方向出现汇聚到顶点的斜棱，而棱柱有两个平行全等底面。

为什么从这里入手：空间关系只靠观察容易漏条件，而“原卷整页图中，一个投影为三角形，另两个投影为矩形。棱锥通常会在多个方向出现汇聚到顶点的斜棱，而棱柱有两个平行全等底面。”可以直接翻译成垂直、平行、数量积或体积公式。先识别结构并建立关系，是在建立可验证的几何链条，而不是凭图形猜结论。

详细展开：三棱柱以三角形为底面。沿平行于侧棱方向观察时可见三角形轮廓；沿另外两个正交方向观察，三角形底面被拉伸成矩形轮廓。因此三棱柱可以产生一组三角形投影和两组矩形投影。三棱锥没有平行全等的两个底面，不会同时给出这样的两个矩形轮廓；四棱柱三个主投影通常均为四边形。

三角形投影+矩形投影+矩形投影⟹三棱柱\text{\htmlData{tutor-start=6,tutor-end=7}{三}\htmlData{tutor-start=7,tutor-end=8}{角}\htmlData{tutor-start=8,tutor-end=9}{形}\htmlData{tutor-start=9,tutor-end=10}{投}\htmlData{tutor-start=10,tutor-end=11}{影}}\htmlData{tutor-start=12,tutor-end=13}{+}\text{\htmlData{tutor-start=19,tutor-end=20}{矩}\htmlData{tutor-start=20,tutor-end=21}{形}\htmlData{tutor-start=21,tutor-end=22}{投}\htmlData{tutor-start=22,tutor-end=23}{影}}\htmlData{tutor-start=24,tutor-end=25}{+}\text{\htmlData{tutor-start=31,tutor-end=32}{矩}\htmlData{tutor-start=32,tutor-end=33}{形}\htmlData{tutor-start=33,tutor-end=34}{投}\htmlData{tutor-start=34,tutor-end=35}{影}}\htmlData{tutor-start=36,tutor-end=51}{\Longrightarrow}\text{\htmlData{tutor-start=57,tutor-end=58}{三}\htmlData{tutor-start=58,tutor-end=59}{棱}\htmlData{tutor-start=59,tutor-end=60}{柱}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：逐一与选项核对：A、C为棱锥，斜棱应向一个顶点汇聚；D为四棱柱，底面不是三角形。只有B与原卷三视图的三角形底面和等长侧棱结构一致。

三棱柱B\boxed{\text{\htmlData{tutor-start=13,tutor-end=14}{三}\htmlData{tutor-start=14,tutor-end=15}{棱}\htmlData{tutor-start=15,tutor-end=16}{柱}}}\quad\boxed{\mathrm{\htmlData{tutor-start=38,tutor-end=39}{B}}}
9

一、选择题 · 算法与程序框图

执行如图的程序框图, 若输入的 a\htmlData{tutor-start=0,tutor-end=1}{a}, b\htmlData{tutor-start=0,tutor-end=1}{b}, k\htmlData{tutor-start=0,tutor-end=1}{k} 分别为 1\htmlData{tutor-start=0,tutor-end=1}{1}, 2\htmlData{tutor-start=0,tutor-end=1}{2}, 3\htmlData{tutor-start=0,tutor-end=1}{3}, 则输出的 M=()\htmlData{tutor-start=0,tutor-end=1}{M} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\quad\htmlData{tutor-start=10,tutor-end=11}{)} (A) 203\dfrac{\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{0}}{\htmlData{tutor-start=11,tutor-end=12}{3}} (B) 72\dfrac{\htmlData{tutor-start=7,tutor-end=8}{7}}{\htmlData{tutor-start=10,tutor-end=11}{2}} (C) 165\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{6}}{\htmlData{tutor-start=11,tutor-end=12}{5}} (D) 158\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{5}}{\htmlData{tutor-start=11,tutor-end=12}{8}}

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

答案：D，15/8。

题目标签：程序框图的循环计算

解题过程

按变量更新顺序执行三轮循环

求循环结束时输出的M\htmlData{tutor-start=0,tutor-end=1}{M}。

(1)
识别结构并建立关系

程序框图先计算M=a+1/b\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,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}{b}，再依次更新a=b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{b}、b=M\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{M}。同一轮中后面的赋值使用更新前或刚更新的值，必须按箭头顺序列表，不能把三行同时替换。

为什么从这里入手：目标是“识别结构并建立关系”，而“程序框图先计算M=a+1/b\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,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}{b}，再依次更新a=b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{b}、b=M\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{M}。同一轮中后面的赋值使用更新前或刚更新的值，必须按箭头顺序列表，不能把三行同时替换。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：初始a=1,b=2,n=1\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=4,tutor-end=5}{b}\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}{n}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}。第一轮得M=3/2\htmlData{tutor-start=0,tutor-end=1}{M}\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}{2}，随后a=2,b=3/2,n=2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{n}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}。第二轮得M=2+2/3=8/3\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{8}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{3}，随后a=3/2,b=8/3,n=3\htmlData{tutor-start=0,tutor-end=1}{a}\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}{2}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{8}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{n}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{3}。第三轮得M=3/2+3/8=15/8\htmlData{tutor-start=0,tutor-end=1}{M}\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}{2}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{8}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{5}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{8}，随后n=4\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}。此时4≤3\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{3}为假，沿“否”分支输出当前M\htmlData{tutor-start=0,tutor-end=1}{M}。

(a,b,M): (1,2,32)→(2,32,83)→(32,83,158)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{M}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{:}\htmlData{tutor-start=8,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}{2}\htmlData{tutor-start=14,tutor-end=15}{,}\tfrac{\htmlData{tutor-start=22,tutor-end=23}{3}}{\htmlData{tutor-start=25,tutor-end=26}{2}}\htmlData{tutor-start=27,tutor-end=28}{)}\htmlData{tutor-start=28,tutor-end=31}{\to}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{,}\tfrac{\htmlData{tutor-start=41,tutor-end=42}{3}}{\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{,}\tfrac{\htmlData{tutor-start=54,tutor-end=55}{8}}{\htmlData{tutor-start=57,tutor-end=58}{3}}\htmlData{tutor-start=59,tutor-end=60}{)}\htmlData{tutor-start=60,tutor-end=63}{\to}\htmlData{tutor-start=63,tutor-end=64}{(}\tfrac{\htmlData{tutor-start=71,tutor-end=72}{3}}{\htmlData{tutor-start=74,tutor-end=75}{2}}\htmlData{tutor-start=76,tutor-end=77}{,}\tfrac{\htmlData{tutor-start=84,tutor-end=85}{8}}{\htmlData{tutor-start=87,tutor-end=88}{3}}\htmlData{tutor-start=89,tutor-end=90}{,}\tfrac{\htmlData{tutor-start=97,tutor-end=98}{1}\htmlData{tutor-start=98,tutor-end=99}{5}}{\htmlData{tutor-start=101,tutor-end=102}{8}}\htmlData{tutor-start=103,tutor-end=104}{)}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：循环条件在n=1,2,3\htmlData{tutor-start=0,tutor-end=1}{n}\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}{,}\htmlData{tutor-start=6,tutor-end=7}{3}时成立，恰执行三轮；第三轮后n=4\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}才退出，所以不能提前输出8/3\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}。最终值15/8\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{5}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{8}与选项D一致。

M=158D\boxed{\htmlData{tutor-start=7,tutor-end=8}{M}\htmlData{tutor-start=8,tutor-end=9}{=}\frac{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{5}}{\htmlData{tutor-start=19,tutor-end=20}{8}}}\quad\boxed{\mathrm{\htmlData{tutor-start=42,tutor-end=43}{D}}}
10

一、选择题 · 抛物线的定义与焦半径

已知抛物线 C:y2=x\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}{x} 的焦点为 F\htmlData{tutor-start=0,tutor-end=1}{F}, A(x0,y0)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}_{\htmlData{tutor-start=5,tutor-end=6}{0}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{y}_{\htmlData{tutor-start=12,tutor-end=13}{0}}\htmlData{tutor-start=14,tutor-end=15}{)} 是 C\htmlData{tutor-start=0,tutor-end=1}{C} 上一点, ∣AF∣=54x0\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=5,tutor-end=6}{=} \dfrac{\htmlData{tutor-start=14,tutor-end=15}{5}}{\htmlData{tutor-start=17,tutor-end=18}{4}}\htmlData{tutor-start=19,tutor-end=20}{x}_{\htmlData{tutor-start=22,tutor-end=23}{0}}, 则 x0=()\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\quad\htmlData{tutor-start=14,tutor-end=15}{)} (A) 4\htmlData{tutor-start=0,tutor-end=1}{4} (B) 2\htmlData{tutor-start=0,tutor-end=1}{2} (C) 1\htmlData{tutor-start=0,tutor-end=1}{1} (D) 8\htmlData{tutor-start=0,tutor-end=1}{8}

答案：C，x₀=1。

题目标签：抛物线定义求点的横坐标

解题过程

使用焦半径公式

由∣AF∣\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}{|}与x0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}的比例求x0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}。

(1)
识别结构并建立关系

y2=x\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{x}可写为y2=4px\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x}，其中p=1/4\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=4,tutor-end=5}{4}。抛物线上一点到焦点的距离等于它到准线x=−p\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{p}的距离，即x0+p\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{p}。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“y2=x\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{x}可写为y2=4px\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x}，其中p=1/4\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=4,tutor-end=5}{4}。抛物线上一点到焦点的距离等于它到准线x=−p\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{p}的距离，即x0+p\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{p}。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：比较y2=4px\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x}与y2=x\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{x}得4p=1\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{p}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}，所以p=1/4\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=4,tutor-end=5}{4}。点A\htmlData{tutor-start=0,tutor-end=1}{A}在抛物线上，故焦半径∣AF∣=x0+p=x0+1/4\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}{x}_{\htmlData{tutor-start=8,tutor-end=9}{0}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{p}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{x}_{\htmlData{tutor-start=16,tutor-end=17}{0}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{/}\htmlData{tutor-start=21,tutor-end=22}{4}。根据题设，它又等于5x0/4\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{0}}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{4}。移项后1/4=x0/4\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{x}_{\htmlData{tutor-start=7,tutor-end=8}{0}}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{4}，所以x0=1\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}。

x0+14=54x0⟹x0=1\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{+}\frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{4}}\htmlData{tutor-start=17,tutor-end=18}{=}\frac{\htmlData{tutor-start=24,tutor-end=25}{5}}{\htmlData{tutor-start=27,tutor-end=28}{4}}\htmlData{tutor-start=29,tutor-end=30}{x}_{\htmlData{tutor-start=32,tutor-end=33}{0}}\htmlData{tutor-start=34,tutor-end=50}{\Longrightarrow }\htmlData{tutor-start=50,tutor-end=51}{x}_{\htmlData{tutor-start=53,tutor-end=54}{0}}\htmlData{tutor-start=55,tutor-end=56}{=}\htmlData{tutor-start=56,tutor-end=57}{1}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：当x0=1\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}时，抛物线给出y02=1\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{0}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}，点确实存在；其焦距为1+1/4=5/4\htmlData{tutor-start=0,tutor-end=1}{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=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{4}，等于(5/4)x0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{5}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{x}_{\htmlData{tutor-start=8,tutor-end=9}{0}}。故选择C。

x0=1C\boxed{\htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{0}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{1}}\quad\boxed{\mathrm{\htmlData{tutor-start=35,tutor-end=36}{C}}}
11

一、选择题 · 线性规划与参数

设 x\htmlData{tutor-start=0,tutor-end=1}{x}, y\htmlData{tutor-start=0,tutor-end=1}{y} 满足约束条件 {x+y⩾a,x−y⩽−1,\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=30}{\geqslant }\htmlData{tutor-start=30,tutor-end=31}{a}\htmlData{tutor-start=31,tutor-end=32}{,} \\ \htmlData{tutor-start=36,tutor-end=37}{x} \htmlData{tutor-start=38,tutor-end=39}{-} \htmlData{tutor-start=40,tutor-end=41}{y} \htmlData{tutor-start=42,tutor-end=52}{\leqslant }\htmlData{tutor-start=52,tutor-end=53}{-}\htmlData{tutor-start=53,tutor-end=54}{1}\htmlData{tutor-start=54,tutor-end=55}{,} \end{cases} 且 z=x+ay\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}{a}\htmlData{tutor-start=9,tutor-end=10}{y} 的最小值为 7\htmlData{tutor-start=0,tutor-end=1}{7}, 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) −5\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{5} (B) 3\htmlData{tutor-start=0,tutor-end=1}{3} (C) −5\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{5} 或 3\htmlData{tutor-start=0,tutor-end=1}{3} (D) 5\htmlData{tutor-start=0,tutor-end=1}{5} 或 −3\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3}

答案：B，a=3。

题目标签：含参数线性规划的最小值

解题过程

先判定最小值存在条件再取顶点

由目标函数最小值为7反求参数a\htmlData{tutor-start=0,tutor-end=1}{a}。

(1)
识别结构并建立关系

约束区域向上无界，直接把交点代入会产生增根。应先检查目标函数沿两条无界边界的变化，确定何时最小值有限。

为什么从这里入手：最值与范围题要先找到变量受限方式以及等号何时成立。“约束区域向上无界，直接把交点代入会产生增根。应先检查目标函数沿两条无界边界的变化，确定何时最小值有限。”给出了可行域或关键界，先识别结构并建立关系能把‘可能有多大’转成单调性、基本不等式或边界比较。

详细展开：约束等价于y≥a−x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{x}且y≥x+1\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}，两条边界交于V((a−1)/2,(a+1)/2)\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{)}。若a<0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0}，令y\htmlData{tutor-start=0,tutor-end=1}{y}无限增大可使x+ay\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{y}趋于负无穷；若0≤a≤1\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{1}，沿边界y=a−x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{x}令x\htmlData{tutor-start=0,tutor-end=1}{x}趋于负无穷也不能取得有限最小值。只有a>1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}时，目标函数沿交点两侧均向上增大，最小值在V\htmlData{tutor-start=0,tutor-end=1}{V}处。代入得到zmin⁡=(a2+2a−1)/2=7\htmlData{tutor-start=0,tutor-end=1}{z}_{\min}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{/}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{7}。

a2+2a−12=7⟹a2+2a−15=0⟹a=3 或 −5\frac{\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{7}\htmlData{tutor-start=22,tutor-end=38}{\Longrightarrow }\htmlData{tutor-start=38,tutor-end=39}{a}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{2}\htmlData{tutor-start=45,tutor-end=46}{a}\htmlData{tutor-start=46,tutor-end=47}{-}\htmlData{tutor-start=47,tutor-end=48}{1}\htmlData{tutor-start=48,tutor-end=49}{5}\htmlData{tutor-start=49,tutor-end=50}{=}\htmlData{tutor-start=50,tutor-end=51}{0}\htmlData{tutor-start=51,tutor-end=67}{\Longrightarrow }\htmlData{tutor-start=67,tutor-end=68}{a}\htmlData{tutor-start=68,tutor-end=69}{=}\htmlData{tutor-start=69,tutor-end=70}{3}\text{ \htmlData{tutor-start=77,tutor-end=78}{或} }\htmlData{tutor-start=80,tutor-end=81}{-}\htmlData{tutor-start=81,tutor-end=82}{5}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：有限最小值的必要条件是a>1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}，因此舍去代数方程产生的a=−5\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}{5}。当a=3\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}时交点为(1,2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}，目标值1+3×2=7\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=9}{\times}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{7}，并且沿两条边界离开该点都增大，故答案为3。

a=3B\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{3}}\quad\boxed{\mathrm{\htmlData{tutor-start=31,tutor-end=32}{B}}}
12

一、选择题 · 导数与函数零点

已知函数 f(x)=ax3−3x2+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=8}{a}\htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{3}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{x}^{\htmlData{tutor-start=20,tutor-end=21}{2}} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{1}, 若 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}{)} 存在唯一的零点 x0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}, 且 x0>0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{0}, 则 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围为 ()\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) (1,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} (C) (2,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} (D) (−∞,−1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}

答案：A，a<-2。

题目标签：三次函数唯一零点的参数范围

解题过程

用极值控制零点个数

找出三次函数只有一个且为正零点时a\htmlData{tutor-start=0,tutor-end=1}{a}的范围。

(1)
识别结构并建立关系

三次函数的唯一零点不能只看f(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}{)}。先用导数找极值，再要求负半轴的局部最低值仍在x\htmlData{tutor-start=0,tutor-end=1}{x}轴上方，才能排除额外负根。

为什么从这里入手：函数的局部变化由导数控制。“三次函数的唯一零点不能只看f(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}{)}。先用导数找极值，再要求负半轴的局部最低值仍在x\htmlData{tutor-start=0,tutor-end=1}{x}轴上方，才能排除额外负根。”给出了函数值、斜率或导数符号的入口，因此先识别结构并建立关系，就能把图象语言转换成方程或符号表。

详细展开：f′(x)=3x(ax−2)\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}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)}。若a>0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}，函数在负无穷处趋于负无穷而f(0)=1\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=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{1}，负半轴必有零点，不可能只有正零点。令a<0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0}，则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}时f′(x)<0\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}{)}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{0}，函数从f(0)=1\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=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{1}严格下降到负无穷，恰有一个正根。负半轴的极小点为x=2/a\htmlData{tutor-start=0,tutor-end=1}{x}\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}{a}，其函数值是1−4/a2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{a}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。要使负半轴没有零点，必须令该极小值严格大于0。

f ⁣(2a)=1−4a2>0⟺a2>4;a<0⟹a<−2f\!\left(\frac{2}{a}\right)=1-\frac4{a^{2}}>0\Longleftrightarrow a^{2}>4;\quad a<0\Longrightarrow a<-2
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：边界a=−2\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}{2}时x=−1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}是二重零点，同时正半轴仍有零点，因此不能取等号。任取a<−2\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}{2}，负半轴最低值为正、正半轴严格下降穿过一次x\htmlData{tutor-start=0,tutor-end=1}{x}轴，故零点唯一且为正。选择A。

a∈(−∞,−2)A\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=11}{\in}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=19}{\infty}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{)}}\quad\boxed{\mathrm{\htmlData{tutor-start=44,tutor-end=45}{A}}}
13

二、填空题 · 排列组合与概率

将 2\htmlData{tutor-start=0,tutor-end=1}{2} 本不同的数学书和 1\htmlData{tutor-start=0,tutor-end=1}{1} 本语文书在书架上随机排成一行, 则 2\htmlData{tutor-start=0,tutor-end=1}{2} 本数学书相邻的概率为 ‾\underline{\hspace{2cm}}.

答案：2/3。

题目标签：相邻排列的古典概型

解题过程

用捆绑法计算有利排列

求两本不同数学书相邻的概率。

(1)
识别结构并建立关系

三本书互不相同且等可能排列。要求两本指定书相邻时，把它们捆成一个整体，同时保留整体内部的两种次序。

为什么从这里入手：目标是“识别结构并建立关系”，而“三本书互不相同且等可能排列。要求两本指定书相邻时，把它们捆成一个整体，同时保留整体内部的两种次序。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：全部排列有3!=6\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{!}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{6}种。把两本数学书记作M1,M2\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{M}_{\htmlData{tutor-start=9,tutor-end=10}{2}}并视为一个书块，该书块与语文书共有2!\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{!}种先后位置；书块内部M1,M2\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{M}_{\htmlData{tutor-start=9,tutor-end=10}{2}}又有2!\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{!}种次序，所以相邻排列共2!×2!=4\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{!}\htmlData{tutor-start=2,tutor-end=8}{\times}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{!}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{4}种。概率为有利数除以总数。

P=2!⋅2!3!=46=23\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{!}\htmlData{tutor-start=10,tutor-end=15}{\cdot}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{!}}{\htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{!}}\htmlData{tutor-start=22,tutor-end=23}{=}\frac{\htmlData{tutor-start=29,tutor-end=30}{4}}{\htmlData{tutor-start=32,tutor-end=33}{6}}\htmlData{tutor-start=34,tutor-end=35}{=}\frac{\htmlData{tutor-start=41,tutor-end=42}{2}}{\htmlData{tutor-start=44,tutor-end=45}{3}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：也可列出数学书不相邻的两种排列M1YM2,M2YM1\htmlData{tutor-start=0,tutor-end=1}{M}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{Y}\htmlData{tutor-start=6,tutor-end=7}{M}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{M}_{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{Y}\htmlData{tutor-start=18,tutor-end=19}{M}_{\htmlData{tutor-start=21,tutor-end=22}{1}}，其概率为2/6=1/3\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{6}\htmlData{tutor-start=3,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}{3}，取补集仍得2/3\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}，结果核验一致。

23\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{2}}{\htmlData{tutor-start=16,tutor-end=17}{3}}}
14

二、填空题 · 逻辑推理与集合

甲、乙、丙三位同学被问到是否去过 A\htmlData{tutor-start=0,tutor-end=1}{A}, B\htmlData{tutor-start=0,tutor-end=1}{B}, C\htmlData{tutor-start=0,tutor-end=1}{C} 三个城市时, 甲说: 我去过的城市比乙多, 但没去过 B\htmlData{tutor-start=0,tutor-end=1}{B} 城市; 乙说: 我没去过 C\htmlData{tutor-start=0,tutor-end=1}{C} 城市; 丙说: 我们三人去过同一个城市. 由此可判断乙去过的城市为 ‾\underline{\hspace{2cm}}.

答案：乙只去过A城市。

题目标签：集合信息的逻辑推断

解题过程

由共同城市和人数关系逐步排除

确定乙去过哪些城市。

(1)
识别结构并建立关系

“三人去过同一个城市”给出三个城市集合的公共元素。先与甲没去过B、乙没去过C结合，就能锁定公共城市。

为什么从这里入手：集合题不能只看式子的外形，必须回到“元素是否满足条件”。这里的““三人去过同一个城市”给出三个城市集合的公共元素。先与甲没去过B、乙没去过C结合，就能锁定公共城市。”给出了元素筛选规则，因此先识别结构并建立关系，再逐项保留或删除元素，思路最直接。

详细展开：共同城市必须既是甲去过的、又是乙去过的。甲没去过B，乙没去过C，所以共同城市既不能是B也不能是C，只能是A，因此乙一定去过A。乙若还去过B，则乙去过两个城市；甲没去B，最多只能去A、C两个城市，不可能比乙去得更多。故乙不能去B。题设又已说明乙没去C，所以乙只去过A。

A∈S乙,B,C∉S乙⟹S乙={A}A\in S_{\text{乙}},\quad B,C\notin S_{\text{乙}}\Longrightarrow S_{\text{乙}}=\{A\}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：构造核验：乙只去A时，甲可去A、C，确实比乙多且没去B；丙只要也去A即可满足三人有共同城市。所有陈述能够同时成立，结论无矛盾。

乙只去过 A 城市\boxed{\text{\htmlData{tutor-start=13,tutor-end=14}{乙}\htmlData{tutor-start=14,tutor-end=15}{只}\htmlData{tutor-start=15,tutor-end=16}{去}\htmlData{tutor-start=16,tutor-end=17}{过} }\htmlData{tutor-start=19,tutor-end=20}{A}\text{ \htmlData{tutor-start=27,tutor-end=28}{城}\htmlData{tutor-start=28,tutor-end=29}{市}}}
15

二、填空题 · 指数函数与幂函数

设函数 f(x)={ex−1,x<1,x13,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}{=} \begin{cases} \mathrm{\htmlData{tutor-start=29,tutor-end=30}{e}}^{\htmlData{tutor-start=33,tutor-end=34}{x}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}}\htmlData{tutor-start=37,tutor-end=38}{,} & \htmlData{tutor-start=41,tutor-end=42}{x} \htmlData{tutor-start=43,tutor-end=44}{<} \htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{,} \\ \htmlData{tutor-start=51,tutor-end=52}{x}^{\frac{\htmlData{tutor-start=60,tutor-end=61}{1}}{\htmlData{tutor-start=63,tutor-end=64}{3}}}\htmlData{tutor-start=66,tutor-end=67}{,} & \htmlData{tutor-start=70,tutor-end=71}{x} \htmlData{tutor-start=72,tutor-end=82}{\geqslant }\htmlData{tutor-start=82,tutor-end=83}{1}\htmlData{tutor-start=83,tutor-end=84}{,} \end{cases} 则使得 f(x)⩽2\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}{2} 成立的 x\htmlData{tutor-start=0,tutor-end=1}{x} 的取值范围是 ‾\underline{\hspace{2cm}}.

答案：(-∞,8]。

题目标签：分段函数不等式

解题过程

在两个定义区间分别求解

求f(x)≤2\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=7}{\le}\htmlData{tutor-start=7,tutor-end=8}{2}的完整解集。

(1)
识别结构并建立关系

函数表达式在x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}处改变，必须按x<1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}和x≥1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}分别求解，最后取并集；不能把两个公式混在同一不等式中。

为什么从这里入手：集合题不能只看式子的外形，必须回到“元素是否满足条件”。这里的“函数表达式在x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}处改变，必须按x<1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}和x≥1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}分别求解，最后取并集；不能把两个公式混在同一不等式中。”给出了元素筛选规则，因此先识别结构并建立关系，再逐项保留或删除元素，思路最直接。

详细展开：当x<1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}时，x−1<0\htmlData{tutor-start=0,tutor-end=1}{x}\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}{0}，所以ex−1<1<2\htmlData{tutor-start=0,tutor-end=1}{e}^{\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=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}{2}，这一整段都满足。当x≥1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}时，f(x)=x1/3\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=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{x}^{\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{3}}，立方根函数严格递增，x1/3≤2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{3}}\htmlData{tutor-start=7,tutor-end=10}{\le}\htmlData{tutor-start=10,tutor-end=11}{2}等价于x≤8\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{8}，与x≥1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}合并得到1≤x≤8\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{8}。两段解集合并即(−∞,8]\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=9,tutor-end=10}{8}\htmlData{tutor-start=10,tutor-end=11}{]}。

{x<1}∪{1≤x≤8}=(−∞,8]\htmlData{tutor-start=0,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}{1}\htmlData{tutor-start=5,tutor-end=7}{\}}\htmlData{tutor-start=7,tutor-end=11}{\cup}\htmlData{tutor-start=11,tutor-end=13}{\{}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=18}{\le }\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=22}{\le}\htmlData{tutor-start=22,tutor-end=23}{8}\htmlData{tutor-start=23,tutor-end=25}{\}}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=34}{\infty}\htmlData{tutor-start=34,tutor-end=35}{,}\htmlData{tutor-start=35,tutor-end=36}{8}\htmlData{tutor-start=36,tutor-end=37}{]}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：分界点x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}使用第二段，f(1)=1≤2\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=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{2}，已包含；右端点x=8\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{8}满足f(8)=2\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{8}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}，应保留；任意x>8\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{8}时立方根大于2。故端点开闭正确。

x∈(−∞,8]\boxed{\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=11}{\in}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=19}{\infty}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{8}\htmlData{tutor-start=21,tutor-end=22}{]}}
16

二、填空题 · 解三角形的实际应用

如图, 为测量山高 MN\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}, 选择 A\htmlData{tutor-start=0,tutor-end=1}{A} 和另一座山的山顶 C\htmlData{tutor-start=0,tutor-end=1}{C} 为测量观测点. 从 A\htmlData{tutor-start=0,tutor-end=1}{A} 点测得 M\htmlData{tutor-start=0,tutor-end=1}{M} 点的仰角 ∠MAN=60°\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{M}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{N} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{6}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{°}, C\htmlData{tutor-start=0,tutor-end=1}{C} 点的仰角 ∠CAB=45°\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{B} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=16}{°} 以及 ∠MAC=75°\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{M}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{C} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{7}\htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=16}{°}; 从 C\htmlData{tutor-start=0,tutor-end=1}{C} 点测得 ∠MCA=60°\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{M}\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{A} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{6}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{°}. 已知山高 BC=100\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}{1}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{0} m, 则山高 MN=‾\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N} \htmlData{tutor-start=3,tutor-end=4}{=} \underline{\hspace{2cm}} m.

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

答案：150 m。

题目标签：仰角测高中的空间三角形

解题过程

先解观测点三角形再取竖直分量

由角度和已知山高BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}求MN\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}。

(1)
识别结构并建立关系

BC=100\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0}与∠CAB=45∘\angle CAB=45^\circ先确定斜距AC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}；随后在已知两角的△AMC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{M}\htmlData{tutor-start=12,tutor-end=13}{C}中用正弦定理求AM\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M}，最后在直角三角形AMN\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{N}中取竖直分量。

为什么从这里入手：三角题先把未知角转成特殊角、同一角或已知边角关系。“BC=100\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0}与∠CAB=45∘\angle CAB=45^\circ先确定斜距AC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}；随后在已知两角的△AMC\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{M}\htmlData{tutor-start=12,tutor-end=13}{C}中用正弦定理求AM\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M}，最后在直角三角形AMN\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{N}中取竖直分量。”指出了可用的象限、恒等式或几何关系，先识别结构并建立关系可以同时确定数值和正负号。

详细展开：在直角三角形ABC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}中，∠CAB=45∘\angle CAB=45^\circ且BC=100\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0}，所以AB=100,AC=1002\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}{1}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{C}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{0}\sqrt{\htmlData{tutor-start=19,tutor-end=20}{2}}。在三角形AMC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{C}中，∠MAC=75∘,∠MCA=60∘\angle MAC=75^\circ,\angle MCA=60^\circ，故∠AMC=45∘\angle AMC=45^\circ。由正弦定理，AM/sin⁡60∘=AC/sin⁡45∘AM/\sin60^\circ=AC/\sin45^\circ，得到AM=1003\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}。山高MN\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}垂直于水平线AN\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{N}，而∠MAN=60∘\angle MAN=60^\circ，所以MN=AMsin⁡60∘MN=AM\sin60^\circ。

AC=1002,AM=1002sin⁡60∘sin⁡45∘=1003,MN=1003⋅32=150AC=100\sqrt{2},\quad AM=\frac{100\sqrt{2}\sin60^\circ}{\sin45^\circ}=100\sqrt{3},\quad MN=100\sqrt{3}\cdot\frac{\sqrt{3}}{2}=150
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：由AM=1003\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}还可得水平距离AN=AMcos⁡60∘=503AN=AM\cos60^\circ=50\sqrt{3}，于是MN/AN=150/(503)=3=tan⁡60∘MN/AN=150/(50\sqrt{3})=\sqrt{3}=\tan60^\circ，与A点仰角完全一致。

MN=150 m\boxed{\htmlData{tutor-start=7,tutor-end=8}{M}\htmlData{tutor-start=8,tutor-end=9}{N}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{5}\htmlData{tutor-start=12,tutor-end=13}{0}\text{ \htmlData{tutor-start=20,tutor-end=21}{m}}}
17

三、解答题 · 数列通项与前n项和

已知 {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}{\}} 是递增的等差数列, a2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}, a4\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}} 是方程 x2−5x+6=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}{5}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{6} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{0} 的根. (1) 求 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 的通项公式; (2) 求数列 {an2n}\left\{\dfrac{\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{n}}}{\htmlData{tutor-start=21,tutor-end=22}{2}^{\htmlData{tutor-start=24,tutor-end=25}{n}}}\right\} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和.

答案：(1) a_n=(n+2)/2；(2) S_n=2-(n+4)/2^{n+1}。

题目标签：等差数列通项与错位相减求和

解题过程

（1）由方程两根确定等差数列

求递增等差数列{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}{\}}的通项。

(1)
识别结构并建立关系

先解方程得到两根2、3。因为数列递增且a2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}在a4\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}}之前，所以必须对应a2=2,a4=3\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}{2}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{4}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{3}，顺序不能交换。

为什么从这里入手：数列题要先识别相邻项之间保持不变的结构。“先解方程得到两根2、3。因为数列递增且a2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}在a4\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}}之前，所以必须对应a2=2,a4=3\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}{2}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{4}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{3}，顺序不能交换。”提示了差、比或可累加关系，先识别结构并建立关系能把第 n\htmlData{tutor-start=0,tutor-end=1}{n} 项问题改写成已经会处理的等差、等比或裂项模型。

详细展开：方程x2−5x+6=0\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=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{6}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0}分解为(x−2)(x−3)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\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=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\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}{=}\htmlData{tutor-start=11,tutor-end=12}{0}。递增性给出a2<a4\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{4}}，故a2=2,a4=3\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}{2}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{4}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{3}。设公差为d\htmlData{tutor-start=0,tutor-end=1}{d}，则a4−a2=2d=1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{d}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{1}，所以d=1/2\htmlData{tutor-start=0,tutor-end=1}{d}\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}。由an=a2+(n−2)d\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{d}得到an=2+(n−2)/2=(n+2)/2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\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}{(}\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{n}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{/}\htmlData{tutor-start=22,tutor-end=23}{2}。

a2=2,a4=3,d=a4−a22=12,an=n+22\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}{2}\htmlData{tutor-start=7,tutor-end=8}{,}\quad \htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{4}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{3}\htmlData{tutor-start=21,tutor-end=22}{,}\quad \htmlData{tutor-start=28,tutor-end=29}{d}\htmlData{tutor-start=29,tutor-end=30}{=}\frac{\htmlData{tutor-start=36,tutor-end=37}{a}_{\htmlData{tutor-start=39,tutor-end=40}{4}}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{a}_{\htmlData{tutor-start=45,tutor-end=46}{2}}}{\htmlData{tutor-start=49,tutor-end=50}{2}}\htmlData{tutor-start=51,tutor-end=52}{=}\frac{\htmlData{tutor-start=58,tutor-end=59}{1}}{\htmlData{tutor-start=61,tutor-end=62}{2}}\htmlData{tutor-start=63,tutor-end=64}{,}\quad \htmlData{tutor-start=70,tutor-end=71}{a}_{\htmlData{tutor-start=73,tutor-end=74}{n}}\htmlData{tutor-start=75,tutor-end=76}{=}\frac{\htmlData{tutor-start=82,tutor-end=83}{n}\htmlData{tutor-start=83,tutor-end=84}{+}\htmlData{tutor-start=84,tutor-end=85}{2}}{\htmlData{tutor-start=87,tutor-end=88}{2}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：代入n=2\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}得a2=2\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}{2}，代入n=4\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}得a4=3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}；公差an+1−an=1/2>0\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{>}\htmlData{tutor-start=18,tutor-end=19}{0}，满足递增条件，因此根的对应与通项均正确。

an=n+22\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\htmlData{tutor-start=12,tutor-end=13}{=}\frac{\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{2}}{\htmlData{tutor-start=24,tutor-end=25}{2}}}

（2）求数列a_n/2^n的前n项和

计算Sn=∑k=1nak/2k\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\sum_{\htmlData{tutor-start=12,tutor-end=13}{k}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}}^{\htmlData{tutor-start=18,tutor-end=19}{n}} \htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{k}}\htmlData{tutor-start=26,tutor-end=27}{/}\htmlData{tutor-start=27,tutor-end=28}{2}^{\htmlData{tutor-start=30,tutor-end=31}{k}}。

(1)
识别结构并建立关系

通项是一次式乘等比因子，既可错位相减，也可拆成∑k/2k\sum \htmlData{tutor-start=5,tutor-end=6}{k}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{2}^{\htmlData{tutor-start=10,tutor-end=11}{k}}与等比和。拆分后公式更清楚，但要统一分母避免漏掉常数项。

为什么从这里入手：数列题要先识别相邻项之间保持不变的结构。“通项是一次式乘等比因子，既可错位相减，也可拆成∑k/2k\sum \htmlData{tutor-start=5,tutor-end=6}{k}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{2}^{\htmlData{tutor-start=10,tutor-end=11}{k}}与等比和。拆分后公式更清楚，但要统一分母避免漏掉常数项。”提示了差、比或可累加关系，先识别结构并建立关系能把第 n\htmlData{tutor-start=0,tutor-end=1}{n} 项问题改写成已经会处理的等差、等比或裂项模型。

详细展开：由ak=(k+2)/2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{k}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{2}，有ak/2k=(k+2)/2k+1=k/2k+1+1/2k\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}^{\htmlData{tutor-start=9,tutor-end=10}{k}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{k}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{2}^{\htmlData{tutor-start=21,tutor-end=22}{k}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{1}}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{k}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{2}^{\htmlData{tutor-start=31,tutor-end=32}{k}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{1}}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{/}\htmlData{tutor-start=38,tutor-end=39}{2}^{\htmlData{tutor-start=41,tutor-end=42}{k}}。利用有限和∑k=1nk/2k=2−(n+2)/2n\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}{n}} \htmlData{tutor-start=15,tutor-end=16}{k}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{2}^{\htmlData{tutor-start=20,tutor-end=21}{k}}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{n}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{/}\htmlData{tutor-start=31,tutor-end=32}{2}^{\htmlData{tutor-start=34,tutor-end=35}{n}}，第一部分等于1−(n+2)/2n+1\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}^{\htmlData{tutor-start=11,tutor-end=12}{n}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}}；第二部分∑k=1n2−k=1−2−n\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}{n}}\htmlData{tutor-start=14,tutor-end=15}{2}^{\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{k}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{2}^{\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{n}}。相加并化为同分母得到2−(n+4)/2n+1\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}^{\htmlData{tutor-start=11,tutor-end=12}{n}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}}。

Sn=12∑k=1nk2k+∑k=1n12k=1−n+22n+1+1−12n=2−n+42n+1.\begin{aligned}S_{n}&=\frac{1}{2}\sum_{k=1}^{n}\frac{k}{2^{k}}+\sum_{k=1}^{n}\frac1{2^{k}}\\&=1-\frac{n+2}{2^{n+1}}+1-\frac1{2^{n}}=2-\frac{n+4}{2^{n+1}}.\end{aligned}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：当n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}时公式给出2−5/4=3/4\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{4}，而a1/2=(3/2)/2=3/4\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=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{4}。并且Sn−Sn−1=(n+2)/2n+1=an/2n\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}{S}_{\htmlData{tutor-start=9,tutor-end=10}{n}\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=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{2}^{\htmlData{tutor-start=23,tutor-end=24}{n}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{1}}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{a}_{\htmlData{tutor-start=31,tutor-end=32}{n}}\htmlData{tutor-start=33,tutor-end=34}{/}\htmlData{tutor-start=34,tutor-end=35}{2}^{\htmlData{tutor-start=37,tutor-end=38}{n}}，说明每次增加的正是第n\htmlData{tutor-start=0,tutor-end=1}{n}项。

Sn=2−n+42n+1\boxed{\htmlData{tutor-start=7,tutor-end=8}{S}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{-}\frac{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{4}}{\htmlData{tutor-start=26,tutor-end=27}{2}^{\htmlData{tutor-start=29,tutor-end=30}{n}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{1}}}}
18

三、解答题 · 统计与频率分布

从某企业生产的某种产品中抽取 100\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0} 件, 测量这些产品的一项质量指标值, 由测量结果得如下频数分布表: | 质量指标值分组 | [75,85)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{7}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{8}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{)} | [85,95)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{9}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{)} | [95,105)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{9}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{)} | [105,115)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=10}{)} | [115,125)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=10}{)} | | 频数 | 6\htmlData{tutor-start=0,tutor-end=1}{6} | 26\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{6} | 38\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{8} | 22\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{2} | 8\htmlData{tutor-start=0,tutor-end=1}{8} | (1) 作出这些数据的频率分布直方图; (2) 估计这种产品质量指标值的平均数及方差（同一组中的数据用该组区间的中点值作代表）; (3) 根据以上抽样调查数据, 能否认为该企业生产的这种产品符合"质量指标值不低于 95\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{5} 的产品至少要占全部产品的 80%\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=4}{\%}"的规定?

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

答案：(1) 各柱高依次为0.006、0.026、0.038、0.022、0.008；(2) 平均数100，方差104；(3) 不能。

题目标签：频数表的直方图、数字特征与比例判断

解题过程

（1）作频率分布直方图

确定五个等组距区间的柱高。

(1)
识别结构并建立关系

纵轴是“频率/组距”，不是频率。每个柱子的面积才等于该组频率，因此柱高应为频数除以样本量再除以组距10。

为什么从这里入手：统计量都来自明确的样本计数或加权。“纵轴是“频率/组距”，不是频率。每个柱子的面积才等于该组频率，因此柱高应为频数除以样本量再除以组距10。”给出了数据与权重，先识别结构并建立关系可以避免把频数、频率和概率混为一谈，并能用总数或概率和反查计算。

详细展开：样本量为100，各组频率依次为0.06,0.26,0.38,0.22,0.08\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{6}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{6}\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}{3}\htmlData{tutor-start=13,tutor-end=14}{8}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{.}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{.}\htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=24}{8}。所有组距均为10，所以柱高依次为这些频率除以10，即0.006,0.026,0.038,0.022,0.008\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{6}\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}{0}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{6}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{.}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{8}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{.}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{,}\htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=26}{.}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{8}。在对应五个区间上画等宽矩形即可。

hi=fi/10010,(h1,…,h5)=(0.006,0.026,0.038,0.022,0.008)\htmlData{tutor-start=0,tutor-end=1}{h}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{f}_{\htmlData{tutor-start=15,tutor-end=16}{i}}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{0}}{\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{0}}\htmlData{tutor-start=26,tutor-end=27}{,}\qquad \htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{h}_{\htmlData{tutor-start=38,tutor-end=39}{1}}\htmlData{tutor-start=40,tutor-end=41}{,}\htmlData{tutor-start=41,tutor-end=47}{\ldots}\htmlData{tutor-start=47,tutor-end=48}{,}\htmlData{tutor-start=48,tutor-end=49}{h}_{\htmlData{tutor-start=51,tutor-end=52}{5}}\htmlData{tutor-start=53,tutor-end=54}{)}\htmlData{tutor-start=54,tutor-end=55}{=}\htmlData{tutor-start=55,tutor-end=56}{(}\htmlData{tutor-start=56,tutor-end=57}{0}\htmlData{tutor-start=57,tutor-end=58}{.}\htmlData{tutor-start=58,tutor-end=59}{0}\htmlData{tutor-start=59,tutor-end=60}{0}\htmlData{tutor-start=60,tutor-end=61}{6}\htmlData{tutor-start=61,tutor-end=62}{,}\htmlData{tutor-start=62,tutor-end=63}{0}\htmlData{tutor-start=63,tutor-end=64}{.}\htmlData{tutor-start=64,tutor-end=65}{0}\htmlData{tutor-start=65,tutor-end=66}{2}\htmlData{tutor-start=66,tutor-end=67}{6}\htmlData{tutor-start=67,tutor-end=68}{,}\htmlData{tutor-start=68,tutor-end=69}{0}\htmlData{tutor-start=69,tutor-end=70}{.}\htmlData{tutor-start=70,tutor-end=71}{0}\htmlData{tutor-start=71,tutor-end=72}{3}\htmlData{tutor-start=72,tutor-end=73}{8}\htmlData{tutor-start=73,tutor-end=74}{,}\htmlData{tutor-start=74,tutor-end=75}{0}\htmlData{tutor-start=75,tutor-end=76}{.}\htmlData{tutor-start=76,tutor-end=77}{0}\htmlData{tutor-start=77,tutor-end=78}{2}\htmlData{tutor-start=78,tutor-end=79}{2}\htmlData{tutor-start=79,tutor-end=80}{,}\htmlData{tutor-start=80,tutor-end=81}{0}\htmlData{tutor-start=81,tutor-end=82}{.}\htmlData{tutor-start=82,tutor-end=83}{0}\htmlData{tutor-start=83,tutor-end=84}{0}\htmlData{tutor-start=84,tutor-end=85}{8}\htmlData{tutor-start=85,tutor-end=86}{)}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：五个矩形面积之和为10(0.006+0.026+0.038+0.022+0.008)=1\htmlData{tutor-start=0,tutor-end=1}{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}{0}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{6}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{.}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{6}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{.}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{8}\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}{0}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{.}\htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{8}\htmlData{tutor-start=32,tutor-end=33}{)}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{1}，说明总频率正确；最高柱位于[95,105)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{9}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{)}，与该组频数38最大一致。

h=(0.006,0.026,0.038,0.022,0.008)\boxed{\htmlData{tutor-start=7,tutor-end=8}{h}\htmlData{tutor-start=8,tutor-end=9}{=}\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}{0}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{6}\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{.}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{6}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=24}{.}\htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{8}\htmlData{tutor-start=27,tutor-end=28}{,}\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}{2}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{,}\htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=36}{.}\htmlData{tutor-start=36,tutor-end=37}{0}\htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{8}\htmlData{tutor-start=39,tutor-end=40}{)}}

（2）估计平均数与方差

以组中值代表组内数据计算xˉ\bar{\htmlData{tutor-start=5,tutor-end=6}{x}}与s2\htmlData{tutor-start=0,tutor-end=1}{s}^{\htmlData{tutor-start=3,tutor-end=4}{2}}。

(1)
识别结构并建立关系

组中值分别为80、90、100、110、120。平均数是组中值的频率加权和，方差是各组中值与平均数差的平方的频率加权和。

为什么从这里入手：统计量都来自明确的样本计数或加权。“组中值分别为80、90、100、110、120。平均数是组中值的频率加权和，方差是各组中值与平均数差的平方的频率加权和。”给出了数据与权重，先识别结构并建立关系可以避免把频数、频率和概率混为一谈，并能用总数或概率和反查计算。

详细展开：用频数计算，平均数为(80×6+90×26+100×38+110×22+120×8)/100=100\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=9}{\times}\htmlData{tutor-start=9,tutor-end=10}{6}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{9}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=19}{\times}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{6}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{0}\htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=31}{\times}\htmlData{tutor-start=31,tutor-end=32}{3}\htmlData{tutor-start=32,tutor-end=33}{8}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{0}\htmlData{tutor-start=37,tutor-end=43}{\times}\htmlData{tutor-start=43,tutor-end=44}{2}\htmlData{tutor-start=44,tutor-end=45}{2}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{2}\htmlData{tutor-start=48,tutor-end=49}{0}\htmlData{tutor-start=49,tutor-end=55}{\times}\htmlData{tutor-start=55,tutor-end=56}{8}\htmlData{tutor-start=56,tutor-end=57}{)}\htmlData{tutor-start=57,tutor-end=58}{/}\htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=60}{0}\htmlData{tutor-start=60,tutor-end=61}{0}\htmlData{tutor-start=61,tutor-end=62}{=}\htmlData{tutor-start=62,tutor-end=63}{1}\htmlData{tutor-start=63,tutor-end=64}{0}\htmlData{tutor-start=64,tutor-end=65}{0}。相对100的偏差为−20,−10,0,10,20\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{0}，所以方差为[6×400+26×100+38×0+22×100+8×400]/100=104\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=8}{\times}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{6}\htmlData{tutor-start=14,tutor-end=20}{\times}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{3}\htmlData{tutor-start=25,tutor-end=26}{8}\htmlData{tutor-start=26,tutor-end=32}{\times}\htmlData{tutor-start=32,tutor-end=33}{0}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=42}{\times}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{0}\htmlData{tutor-start=44,tutor-end=45}{0}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{8}\htmlData{tutor-start=47,tutor-end=53}{\times}\htmlData{tutor-start=53,tutor-end=54}{4}\htmlData{tutor-start=54,tutor-end=55}{0}\htmlData{tutor-start=55,tutor-end=56}{0}\htmlData{tutor-start=56,tutor-end=57}{]}\htmlData{tutor-start=57,tutor-end=58}{/}\htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=60}{0}\htmlData{tutor-start=60,tutor-end=61}{0}\htmlData{tutor-start=61,tutor-end=62}{=}\htmlData{tutor-start=62,tutor-end=63}{1}\htmlData{tutor-start=63,tutor-end=64}{0}\htmlData{tutor-start=64,tutor-end=65}{4}。

xˉ=100,s2=2400+2600+0+2200+3200100=104\bar{\htmlData{tutor-start=5,tutor-end=6}{x}}\htmlData{tutor-start=7,tutor-end=8}{=}\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}{,}\qquad \htmlData{tutor-start=19,tutor-end=20}{s}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{=}\frac{\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{4}\htmlData{tutor-start=33,tutor-end=34}{0}\htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{2}\htmlData{tutor-start=37,tutor-end=38}{6}\htmlData{tutor-start=38,tutor-end=39}{0}\htmlData{tutor-start=39,tutor-end=40}{0}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{0}\htmlData{tutor-start=42,tutor-end=43}{+}\htmlData{tutor-start=43,tutor-end=44}{2}\htmlData{tutor-start=44,tutor-end=45}{2}\htmlData{tutor-start=45,tutor-end=46}{0}\htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{+}\htmlData{tutor-start=48,tutor-end=49}{3}\htmlData{tutor-start=49,tutor-end=50}{2}\htmlData{tutor-start=50,tutor-end=51}{0}\htmlData{tutor-start=51,tutor-end=52}{0}}{\htmlData{tutor-start=54,tutor-end=55}{1}\htmlData{tutor-start=55,tutor-end=56}{0}\htmlData{tutor-start=56,tutor-end=57}{0}}\htmlData{tutor-start=58,tutor-end=59}{=}\htmlData{tutor-start=59,tutor-end=60}{1}\htmlData{tutor-start=60,tutor-end=61}{0}\htmlData{tutor-start=61,tutor-end=62}{4}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：加权偏差和6(−20)+26(−10)+38(0)+22(10)+8(20)=0\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,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}{0}\htmlData{tutor-start=5,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}{6}\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}{0}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{8}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{0}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{8}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{0}\htmlData{tutor-start=32,tutor-end=33}{)}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{0}，与平均数100相符；方差单位平方且非负，计算结构正确。

xˉ=100,s2=104\boxed{\bar{\htmlData{tutor-start=12,tutor-end=13}{x}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{,}\quad \htmlData{tutor-start=25,tutor-end=26}{s}^{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{0}\htmlData{tutor-start=33,tutor-end=34}{4}}

（3）用样本频率判断是否符合规定

估计质量指标不低于95的产品比例并与80%比较。

(1)
识别结构并建立关系

“不低于95”包括[95,105)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{9}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{)}、[105,115)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{)}和[115,125)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{)}三组。题目要求依据抽样数据作判断，应先合并这些频数。

为什么从这里入手：目标是“识别结构并建立关系”，而““不低于95”包括[95,105)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{9}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{)}、[105,115)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{)}和[115,125)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{)}三组。题目要求依据抽样数据作判断，应先合并这些频数。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：样本中质量指标值不低于95的产品有38+22+8=68\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=10}{8}件，占68/100=68%\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{6}\htmlData{tutor-start=8,tutor-end=9}{8}\htmlData{tutor-start=9,tutor-end=11}{\%}。规定要求至少达到80%\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=4}{\%}，而68%<80%\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=4}{\%}\htmlData{tutor-start=4,tutor-end=5}{<}\htmlData{tutor-start=5,tutor-end=6}{8}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=9}{\%}，所以这份抽样数据不支持企业产品符合该规定。

p^=38+22+8100=0.68<0.80\widehat p=\frac{38+22+8}{100}=0.68<0.80
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：即使把95作为闭端点，它已归入[95,105)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{9}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{)}组，计数没有遗漏。样本估计比例低于规定下限12个百分点，因此回答“不能认为符合”。

不能认为该产品符合规定\boxed{\text{\htmlData{tutor-start=13,tutor-end=14}{不}\htmlData{tutor-start=14,tutor-end=15}{能}\htmlData{tutor-start=15,tutor-end=16}{认}\htmlData{tutor-start=16,tutor-end=17}{为}\htmlData{tutor-start=17,tutor-end=18}{该}\htmlData{tutor-start=18,tutor-end=19}{产}\htmlData{tutor-start=19,tutor-end=20}{品}\htmlData{tutor-start=20,tutor-end=21}{符}\htmlData{tutor-start=21,tutor-end=22}{合}\htmlData{tutor-start=22,tutor-end=23}{规}\htmlData{tutor-start=23,tutor-end=24}{定}}}
19

三、解答题 · 空间几何中的垂直与距离

如图，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}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{A}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{B}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{C}_{\htmlData{tutor-start=17,tutor-end=18}{1}} 中，侧面 BB1C1C\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{C}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{C} 为菱形，B1C\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C} 的中点为 O\htmlData{tutor-start=0,tutor-end=1}{O}，且 AO⊥\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} \htmlData{tutor-start=3,tutor-end=8}{\perp} 平面 BB1C1C\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{C}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{C}. (1) 证明：B1C⊥AB\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C} \htmlData{tutor-start=7,tutor-end=13}{\perp }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}; (2) 若 AC⊥AB1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{B}_{\htmlData{tutor-start=13,tutor-end=14}{1}}, ∠CBB1=60∘\angle CBB_{1} = 60^\circ, BC=1\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}{1}, 求三棱柱 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}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{A}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{B}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{C}_{\htmlData{tutor-start=17,tutor-end=18}{1}} 的高.

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

答案：(1) B₁C⊥AB；(2) 棱柱高为√21/7。

题目标签：菱形侧面的垂直证明与棱柱高

解题过程

（1）证明B₁C垂直AB

由菱形对角线与面面垂直条件推出线线垂直。

(1)
识别结构并建立关系

要证一条直线垂直AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}，可先证它垂直包含AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}的平面AOB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{B}。菱形对角线提供一条垂线，AO⊥\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=7}{\perp}侧面提供另一条垂线。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“要证一条直线垂直AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}，可先证它垂直包含AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}的平面AOB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{B}。菱形对角线提供一条垂线，AO⊥\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=7}{\perp}侧面提供另一条垂线。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：在菱形BB1C1C\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{C}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{C}中，两条对角线B1C\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}与BC1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}}互相垂直且在中点O\htmlData{tutor-start=0,tutor-end=1}{O}相交，所以B1C⊥BO\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=12}{\perp }\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{O}。又因AO⊥\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=7}{\perp}平面BB1C1C\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{C}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{C}，而B1C\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}在该平面内，故AO⊥B1C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{B}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{C}。直线AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}与BO\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{O}相交并同在平面AOB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{B}内，因此B1C⊥\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=11}{\perp}平面AOB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{B}。AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}位于该平面内，遂有B1C⊥AB\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=12}{\perp }\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{B}。

B1C⊥AO,B1C⊥BO⟹B1C⊥平面 AOB⟹B1C⊥AB\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=12}{\perp }\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{O}\htmlData{tutor-start=14,tutor-end=15}{,}\quad \htmlData{tutor-start=21,tutor-end=22}{B}_{\htmlData{tutor-start=24,tutor-end=25}{1}}\htmlData{tutor-start=26,tutor-end=27}{C}\htmlData{tutor-start=27,tutor-end=33}{\perp }\htmlData{tutor-start=33,tutor-end=34}{B}\htmlData{tutor-start=34,tutor-end=35}{O}\htmlData{tutor-start=35,tutor-end=51}{\Longrightarrow }\htmlData{tutor-start=51,tutor-end=52}{B}_{\htmlData{tutor-start=54,tutor-end=55}{1}}\htmlData{tutor-start=56,tutor-end=57}{C}\htmlData{tutor-start=57,tutor-end=62}{\perp}\text{\htmlData{tutor-start=68,tutor-end=69}{平}\htmlData{tutor-start=69,tutor-end=70}{面} }\htmlData{tutor-start=72,tutor-end=73}{A}\htmlData{tutor-start=73,tutor-end=74}{O}\htmlData{tutor-start=74,tutor-end=75}{B}\htmlData{tutor-start=75,tutor-end=91}{\Longrightarrow }\htmlData{tutor-start=91,tutor-end=92}{B}_{\htmlData{tutor-start=94,tutor-end=95}{1}}\htmlData{tutor-start=96,tutor-end=97}{C}\htmlData{tutor-start=97,tutor-end=103}{\perp }\htmlData{tutor-start=103,tutor-end=104}{A}\htmlData{tutor-start=104,tutor-end=105}{B}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：两条用于判定线面垂直的直线AO,BO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{O}确实相交于O\htmlData{tutor-start=0,tutor-end=1}{O}；AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}确实在平面AOB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{B}内，故线面垂直定理的条件完整，结论成立。

B1C⊥AB\boxed{\htmlData{tutor-start=7,tutor-end=8}{B}_{\htmlData{tutor-start=10,tutor-end=11}{1}}\htmlData{tutor-start=12,tutor-end=13}{C}\htmlData{tutor-start=13,tutor-end=19}{\perp }\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{B}}

（2）用坐标求三棱柱的高

在给定边长和垂直条件下求两个底面间距离。

(1)
识别结构并建立关系

侧面菱形的边长和夹角已知，适合先把菱形放在坐标平面内。注意AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}不是棱柱高；棱柱高是平移向量BB1→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{B}_{\htmlData{tutor-start=20,tutor-end=21}{1}}}在底面法向量上的投影。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“侧面菱形的边长和夹角已知，适合先把菱形放在坐标平面内。注意AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}不是棱柱高；棱柱高是平移向量BB1→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{B}_{\htmlData{tutor-start=20,tutor-end=21}{1}}}在底面法向量上的投影。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：令B=(0,0,0),C=(1,0,0),B1=(1/2,3/2,0)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\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}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{C}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{B}_{\htmlData{tutor-start=23,tutor-end=24}{1}}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{/}\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{,}\sqrt{\htmlData{tutor-start=37,tutor-end=38}{3}}\htmlData{tutor-start=39,tutor-end=40}{/}\htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=42}{,}\htmlData{tutor-start=42,tutor-end=43}{0}\htmlData{tutor-start=43,tutor-end=44}{)}，则O=(3/4,3/4,0)\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{,}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{)}。因AO\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}垂直侧面，设A=(3/4,3/4,t)\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}{3}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{,}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{t}\htmlData{tutor-start=19,tutor-end=20}{)}。条件AC⊥AB1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{B}_{\htmlData{tutor-start=12,tutor-end=13}{1}}给出(C−A)⋅(B1−A)=t2−1/4=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=10}{\cdot}\htmlData{tutor-start=10,tutor-end=11}{(}\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}{A}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{t}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{4}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{0}，故t=1/2\htmlData{tutor-start=0,tutor-end=1}{t}\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}。底面ABC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}的一个法向量可取n=(0,−1/2,3/4)\boldsymbol n=(0,-1/2,\sqrt{3}/4)，平移向量为v=BB1→=(1/2,3/2,0)\boldsymbol v=\overrightarrow{BB_{1}}=(1/2,\sqrt{3}/2,0)。棱柱高等于∣v⋅n∣/∣n∣|\boldsymbol v\cdot\boldsymbol n|/|\boldsymbol n|。

h=∣v⋅n∣∣n∣=3/47/4=37=217h=\frac{|\boldsymbol v\cdot\boldsymbol n|}{|\boldsymbol n|}=\frac{\sqrt{3}/4}{\sqrt{7}/4}=\sqrt{\frac{3}{7}}=\frac{\sqrt{21}}7
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：由t=1/2\htmlData{tutor-start=0,tutor-end=1}{t}\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}回代可得(C−A)⋅(B1−A)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=10}{\cdot}\htmlData{tutor-start=10,tutor-end=11}{(}\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}{A}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{0}。所得高度3/7\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{7}}小于侧棱BB1=1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}，符合斜棱柱中“高不超过侧棱”的几何边界。

h=217\boxed{h=\frac{\sqrt{21}}7}
20

三、解答题 · 圆的方程与轨迹

已知点 P(2,2)\htmlData{tutor-start=0,tutor-end=1}{P}\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}{2}\htmlData{tutor-start=6,tutor-end=7}{)}, 圆 C:x2+y2−8y=0\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{x}^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{+} \htmlData{tutor-start=11,tutor-end=12}{y}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{-} \htmlData{tutor-start=19,tutor-end=20}{8}\htmlData{tutor-start=20,tutor-end=21}{y} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{0}, 过点 P\htmlData{tutor-start=0,tutor-end=1}{P} 的动直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与圆 C\htmlData{tutor-start=0,tutor-end=1}{C} 交于 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B} 两点，线段 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 的中点为 M\htmlData{tutor-start=0,tutor-end=1}{M}, O\htmlData{tutor-start=0,tutor-end=1}{O} 为坐标原点. (1) 求 M\htmlData{tutor-start=0,tutor-end=1}{M} 的轨迹方程; (2) 当 ∣OP∣=∣OM∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{P}\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}{O}\htmlData{tutor-start=9,tutor-end=10}{M}\htmlData{tutor-start=10,tutor-end=11}{|} 时，求 l\htmlData{tutor-start=0,tutor-end=1}{l} 的方程及 △POM\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{O}\htmlData{tutor-start=12,tutor-end=13}{M} 的面积.

答案：(1) (x-1)^2+(y-3)^2=2；(2) l为x-y=0或x+3y-8=0，面积分别为0或16/5；若三角形限非退化，则取后一组。

题目标签：动弦中点轨迹与面积

解题过程

（1）求弦中点M的轨迹

消去动直线，得到M\htmlData{tutor-start=0,tutor-end=1}{M}满足的定点几何关系。

(1)
识别结构并建立关系

圆心到弦的连线垂直于弦。由于P,M\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{M}都在弦所在直线上，所以圆心C0\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{0}}到M\htmlData{tutor-start=0,tutor-end=1}{M}的线段垂直PM\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{M}，从而出现以PC0\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{0}}为直径的圆。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“圆心到弦的连线垂直于弦。由于P,M\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{M}都在弦所在直线上，所以圆心C0\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{0}}到M\htmlData{tutor-start=0,tutor-end=1}{M}的线段垂直PM\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{M}，从而出现以PC0\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{0}}为直径的圆。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：原圆化为x2+(y−4)2=16\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=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{y}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{6}，圆心C0=(0,4)\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{)}。M\htmlData{tutor-start=0,tutor-end=1}{M}是弦AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}中点，故C0M⊥AB\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{M}\htmlData{tutor-start=6,tutor-end=12}{\perp }\htmlData{tutor-start=12,tutor-end=13}{A}\htmlData{tutor-start=13,tutor-end=14}{B}；又P,M\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{M}在直线AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}上，所以C0M⊥PM\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{M}\htmlData{tutor-start=6,tutor-end=12}{\perp }\htmlData{tutor-start=12,tutor-end=13}{P}\htmlData{tutor-start=13,tutor-end=14}{M}，即∠PMC0=90∘\angle PMC_{0}=90^\circ。因此M\htmlData{tutor-start=0,tutor-end=1}{M}在以PC0\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{0}}为直径的圆上。该圆圆心为(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=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{)}，半径为∣PC0∣/2=2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{C}_{\htmlData{tutor-start=5,tutor-end=6}{0}}\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{=}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{2}}。

(x−1)2+(y−3)2=2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\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=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{y}\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=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{2}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：点P\htmlData{tutor-start=0,tutor-end=1}{P}在原圆内部，因为PC0=22<4\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{0}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{<}\htmlData{tutor-start=17,tutor-end=18}{4}，过P\htmlData{tutor-start=0,tutor-end=1}{P}任意方向均与圆有两交点；反过来轨迹圆上的每个M\htmlData{tutor-start=0,tutor-end=1}{M}都满足C0M⊥PM\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{M}\htmlData{tutor-start=6,tutor-end=12}{\perp }\htmlData{tutor-start=12,tutor-end=13}{P}\htmlData{tutor-start=13,tutor-end=14}{M}，可作为相应弦中点，故轨迹没有缺支。

(x−1)2+(y−3)2=2\boxed{\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{y}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{3}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{2}}

（2）由等距条件求直线与面积

联立轨迹与∣OM∣=∣OP∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{M}\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}{O}\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{|}，恢复对应直线l\htmlData{tutor-start=0,tutor-end=1}{l}并求面积。

(1)
识别结构并建立关系

∣OP∣=22\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{P}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}固定，所以M\htmlData{tutor-start=0,tutor-end=1}{M}还在以O\htmlData{tutor-start=0,tutor-end=1}{O}为圆心、半径22\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}的圆上。先求两个交点，再分别根据“M\htmlData{tutor-start=0,tutor-end=1}{M}是弦中点”恢复直线，尤其要单独处理M=P\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{P}。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“∣OP∣=22\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{P}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}固定，所以M\htmlData{tutor-start=0,tutor-end=1}{M}还在以O\htmlData{tutor-start=0,tutor-end=1}{O}为圆心、半径22\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}的圆上。先求两个交点，再分别根据“M\htmlData{tutor-start=0,tutor-end=1}{M}是弦中点”恢复直线，尤其要单独处理M=P\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{P}。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：设M=(x,y)\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\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}{y}\htmlData{tutor-start=6,tutor-end=7}{)}。由x2+y2=8\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=6,tutor-end=7}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{8}和轨迹方程相减得x+3y=8\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{8}，联立可得M=P=(2,2)\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{P}\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}{,}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{)}或M=(−2/5,14/5)\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{5}\htmlData{tutor-start=12,tutor-end=13}{)}。第二种情形中P,M\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{M}确定l:x+3y−8=0\htmlData{tutor-start=0,tutor-end=1}{l}\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}{3}\htmlData{tutor-start=5,tutor-end=6}{y}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{8}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{0}，面积为∣det⁡(OP→,OM→)∣/2=16/5\htmlData{tutor-start=0,tutor-end=1}{|}\det\htmlData{tutor-start=5,tutor-end=6}{(}\overrightarrow{\htmlData{tutor-start=22,tutor-end=23}{O}\htmlData{tutor-start=23,tutor-end=24}{P}}\htmlData{tutor-start=25,tutor-end=26}{,}\overrightarrow{\htmlData{tutor-start=42,tutor-end=43}{O}\htmlData{tutor-start=43,tutor-end=44}{M}}\htmlData{tutor-start=45,tutor-end=46}{)}\htmlData{tutor-start=46,tutor-end=47}{|}\htmlData{tutor-start=47,tutor-end=48}{/}\htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=50}{=}\htmlData{tutor-start=50,tutor-end=51}{1}\htmlData{tutor-start=51,tutor-end=52}{6}\htmlData{tutor-start=52,tutor-end=53}{/}\htmlData{tutor-start=53,tutor-end=54}{5}。第一种情形中弦中点就是P\htmlData{tutor-start=0,tutor-end=1}{P}，故l\htmlData{tutor-start=0,tutor-end=1}{l}应过P\htmlData{tutor-start=0,tutor-end=1}{P}且垂直C0P\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{P}，得到l:y=x\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{:}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{x}；此时P=M\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{M}，面积为0。若按“△POM\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{O}\htmlData{tutor-start=12,tutor-end=13}{M}”只允许非退化三角形，则排除第一种情形。

M=(2,2) 或 (−25,145);l:x−y=0 或 x+3y−8=0;S=0 或 165\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,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=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{)}\text{ \htmlData{tutor-start=14,tutor-end=15}{或} }\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{-}\tfrac{\htmlData{tutor-start=26,tutor-end=27}{2}}{\htmlData{tutor-start=29,tutor-end=30}{5}}\htmlData{tutor-start=31,tutor-end=32}{,}\tfrac{\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{4}}{\htmlData{tutor-start=43,tutor-end=44}{5}}\htmlData{tutor-start=45,tutor-end=46}{)}\htmlData{tutor-start=46,tutor-end=47}{;}\quad \htmlData{tutor-start=53,tutor-end=54}{l}\htmlData{tutor-start=54,tutor-end=55}{:}\htmlData{tutor-start=55,tutor-end=56}{x}\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{y}\htmlData{tutor-start=58,tutor-end=59}{=}\htmlData{tutor-start=59,tutor-end=60}{0}\text{ \htmlData{tutor-start=67,tutor-end=68}{或} }\htmlData{tutor-start=70,tutor-end=71}{x}\htmlData{tutor-start=71,tutor-end=72}{+}\htmlData{tutor-start=72,tutor-end=73}{3}\htmlData{tutor-start=73,tutor-end=74}{y}\htmlData{tutor-start=74,tutor-end=75}{-}\htmlData{tutor-start=75,tutor-end=76}{8}\htmlData{tutor-start=76,tutor-end=77}{=}\htmlData{tutor-start=77,tutor-end=78}{0}\htmlData{tutor-start=78,tutor-end=79}{;}\quad \htmlData{tutor-start=85,tutor-end=86}{S}\htmlData{tutor-start=86,tutor-end=87}{=}\htmlData{tutor-start=87,tutor-end=88}{0}\text{ \htmlData{tutor-start=95,tutor-end=96}{或} }\frac{\htmlData{tutor-start=104,tutor-end=105}{1}\htmlData{tutor-start=105,tutor-end=106}{6}}{\htmlData{tutor-start=108,tutor-end=109}{5}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：对l:y=x\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{:}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{x}，圆心到直线的垂足为P\htmlData{tutor-start=0,tutor-end=1}{P}，故P\htmlData{tutor-start=0,tutor-end=1}{P}确为弦中点；对l:x+3y−8=0\htmlData{tutor-start=0,tutor-end=1}{l}\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}{3}\htmlData{tutor-start=5,tutor-end=6}{y}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{8}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{0}，圆心到该线的垂足为(−2/5,14/5)\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=4,tutor-end=5}{5}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{5}\htmlData{tutor-start=10,tutor-end=11}{)}。两条线均过P\htmlData{tutor-start=0,tutor-end=1}{P}并与原圆相交，回代核验完整。

l:x−y=0, S=0或l:x+3y−8=0, S=165\boxed{\htmlData{tutor-start=7,tutor-end=8}{l}\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}{y}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=17}{\ }\htmlData{tutor-start=17,tutor-end=18}{S}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{0}}\quad\text{\htmlData{tutor-start=32,tutor-end=33}{或}}\quad\boxed{\htmlData{tutor-start=46,tutor-end=47}{l}\htmlData{tutor-start=47,tutor-end=48}{:}\htmlData{tutor-start=48,tutor-end=49}{x}\htmlData{tutor-start=49,tutor-end=50}{+}\htmlData{tutor-start=50,tutor-end=51}{3}\htmlData{tutor-start=51,tutor-end=52}{y}\htmlData{tutor-start=52,tutor-end=53}{-}\htmlData{tutor-start=53,tutor-end=54}{8}\htmlData{tutor-start=54,tutor-end=55}{=}\htmlData{tutor-start=55,tutor-end=56}{0}\htmlData{tutor-start=56,tutor-end=57}{,}\htmlData{tutor-start=57,tutor-end=59}{\ }\htmlData{tutor-start=59,tutor-end=60}{S}\htmlData{tutor-start=60,tutor-end=61}{=}\frac{\htmlData{tutor-start=67,tutor-end=68}{1}\htmlData{tutor-start=68,tutor-end=69}{6}}{\htmlData{tutor-start=71,tutor-end=72}{5}}}
21

三、解答题 · 导数与参数范围

设函数 f(x)=aln⁡x+1−a2x2−bx(a≠1)f(x) = a \ln x + \frac{1-a}{2}x^{2} - bx (a \neq 1), 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}{)} 在点 (1,f(1))\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{)} 处的切线斜率为 0. (1) 求 b\htmlData{tutor-start=0,tutor-end=1}{b}; (2) 若存在 x0⩾1\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=16}{\geqslant }\htmlData{tutor-start=16,tutor-end=17}{1} 使得 f(x0)<aa−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=5,tutor-end=6}{0}}\htmlData{tutor-start=7,tutor-end=8}{)} \htmlData{tutor-start=9,tutor-end=10}{<} \frac{\htmlData{tutor-start=17,tutor-end=18}{a}}{\htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{1}}, 求 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围.

答案：(1) b=1；(2) a∈(-1-√2,-1+√2)∪(1,+∞)。

题目标签：含参数函数的切线与存在性不等式

解题过程

（1）由切线斜率求b

利用f′(1)=0\htmlData{tutor-start=0,tutor-end=1}{f}'\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=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}确定b\htmlData{tutor-start=0,tutor-end=1}{b}。

(1)
识别结构并建立关系

曲线在x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}处切线斜率为0，直接翻译为导数值f′(1)=0\htmlData{tutor-start=0,tutor-end=1}{f}'\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=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，无需先求切线方程。

为什么从这里入手：函数的局部变化由导数控制。“曲线在x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}处切线斜率为0，直接翻译为导数值f′(1)=0\htmlData{tutor-start=0,tutor-end=1}{f}'\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=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，无需先求切线方程。”给出了函数值、斜率或导数符号的入口，因此先识别结构并建立关系，就能把图象语言转换成方程或符号表。

详细展开：函数定义域为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}。求导得f′(x)=a/x+(1−a)x−b\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}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{b}。代入x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，有f′(1)=a+1−a−b=1−b\htmlData{tutor-start=0,tutor-end=1}{f}'\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=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}\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}{a}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{b}。由切线斜率为0可得1−b=0\htmlData{tutor-start=0,tutor-end=1}{1}\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}{0}，所以b=1\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}。

f′(1)=1−b=0⟹b=1\htmlData{tutor-start=0,tutor-end=1}{f}'\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=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=27}{\Longrightarrow }\htmlData{tutor-start=27,tutor-end=28}{b}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{1}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：结果与参数a\htmlData{tutor-start=0,tutor-end=1}{a}无关；把b=1\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}代回，任意允许的a≠1\htmlData{tutor-start=0,tutor-end=1}{a}\ne\htmlData{tutor-start=4,tutor-end=5}{1}都有f′(1)=0\htmlData{tutor-start=0,tutor-end=1}{f}'\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=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，所以没有漏掉附加条件。

b=1\boxed{\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}}

（2）判断何时存在x₀使函数低于给定值

在x≥1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}上比较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}{)}与a/(a−1)\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}{a}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}并求a\htmlData{tutor-start=0,tutor-end=1}{a}范围。

(1)
识别结构并建立关系

这是“存在”问题，应研究x≥1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}上f\htmlData{tutor-start=0,tutor-end=1}{f}的最小行为。a>1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}时二次项系数为负，函数趋于负无穷；a<1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}时用导数判定最小点。

为什么从这里入手：函数的局部变化由导数控制。“这是“存在”问题，应研究x≥1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}上f\htmlData{tutor-start=0,tutor-end=1}{f}的最小行为。a>1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}时二次项系数为负，函数趋于负无穷；a<1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}时用导数判定最小点。”给出了函数值、斜率或导数符号的入口，因此先识别结构并建立关系，就能把图象语言转换成方程或符号表。

详细展开：代入b=1\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}。若a>1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}，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}{)}中的(1−a)x2/2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{x}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{2}主导且趋于负无穷，因此必存在所需x0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}。若a<1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}，f′(x)=[(x−1)((1−a)x−a)]/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}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{[}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{a}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{x}\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}{]}\htmlData{tutor-start=23,tutor-end=24}{/}\htmlData{tutor-start=24,tutor-end=25}{x}。当a≤1/2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}时，f\htmlData{tutor-start=0,tutor-end=1}{f}在[1,+∞)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=10}{\infty}\htmlData{tutor-start=10,tutor-end=11}{)}递增，最小值为f(1)=−(a+1)/2\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=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{2}；不等式f(1)<a/(a−1)\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=4,tutor-end=5}{<}\htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}等价于a2+2a−1<0\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}{2}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{0}，即−1−2<a<−1+2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{2}}\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}{-}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{+}\sqrt{\htmlData{tutor-start=23,tutor-end=24}{2}}。当1/2<a<1\htmlData{tutor-start=0,tutor-end=1}{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=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{1}时，最小点x=a/(1−a)>1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{>}\htmlData{tutor-start=10,tutor-end=11}{1}，代入后fmin⁡−a/(a−1)=a[ln⁡(a/(1−a))+a/(2(1−a))]>0\htmlData{tutor-start=0,tutor-end=1}{f}_{\min}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{/}\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}{1}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{[}\ln\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{a}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{a}\htmlData{tutor-start=33,tutor-end=34}{/}\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{a}\htmlData{tutor-start=40,tutor-end=41}{)}\htmlData{tutor-start=41,tutor-end=42}{)}\htmlData{tutor-start=42,tutor-end=43}{]}\htmlData{tutor-start=43,tutor-end=44}{>}\htmlData{tutor-start=44,tutor-end=45}{0}，不存在。

a∈(−1−2,−1+2)∪(1,+∞)\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\in}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{+}\sqrt{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=33}{\cup}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{,}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=43}{\infty}\htmlData{tutor-start=43,tutor-end=44}{)}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：两个边界a=−1±2\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=7}{\pm}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{2}}处只能取得等号，不满足严格小于；a=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}被题设排除。a>1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}由极限保证存在，a<1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{1}的各单调区间均已覆盖，故范围完整。

a∈(−1−2,−1+2)∪(1,+∞)\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=11}{\in}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{-}\sqrt{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{,}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{+}\sqrt{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=36}{)}\htmlData{tutor-start=36,tutor-end=40}{\cup}\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{,}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=50}{\infty}\htmlData{tutor-start=50,tutor-end=51}{)}}
22

三、解答题 · 圆的角度关系

如图，四边形 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} 是 ⊙O\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O} 的内接四边形，AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 的延长线与 DC\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{C} 的延长线交于点 E\htmlData{tutor-start=0,tutor-end=1}{E}, 且 CB=CE\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{E}. (1) 证明：∠D=∠E\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{E}; (2) 设 AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} 不是 ⊙O\htmlData{tutor-start=0,tutor-end=6}{\odot }\htmlData{tutor-start=6,tutor-end=7}{O} 的直径，AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} 的中点为 M\htmlData{tutor-start=0,tutor-end=1}{M}, 且 MB=MC\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{M}\htmlData{tutor-start=6,tutor-end=7}{C}, 证明：△ADE\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{D}\htmlData{tutor-start=12,tutor-end=13}{E} 为等边三角形.

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

答案：(1) ∠D=∠E；(2) △ADE为等边三角形。

题目标签：圆内接四边形与等边三角形

解题过程

（1）证明外角相等

利用CB=CE\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{E}和圆内接四边形证明∠D=∠E\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=16}{\angle }\htmlData{tutor-start=16,tutor-end=17}{E}。

(1)
识别结构并建立关系

CB=CE\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{E}先给出等腰三角形CBE\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{E}的两个底角相等；圆内接四边形的对角互补，再利用A,B,E\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}{E}共线把补角连接起来。

为什么从这里入手：目标是“识别结构并建立关系”，而“CB=CE\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{E}先给出等腰三角形CBE\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{E}的两个底角相等；圆内接四边形的对角互补，再利用A,B,E\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}{E}共线把补角连接起来。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：在△CBE\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{C}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{E}中，CB=CE\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{E}，所以∠CBE=∠BEC=∠E\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{E}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{B}\htmlData{tutor-start=19,tutor-end=20}{E}\htmlData{tutor-start=20,tutor-end=21}{C}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=29}{\angle }\htmlData{tutor-start=29,tutor-end=30}{E}。因A,B,E\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}{E}共线，∠CBE\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{E}与∠CBA\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{A}互补。又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}为圆内接四边形，∠D=∠ADC\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=16}{\angle }\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}\htmlData{tutor-start=18,tutor-end=19}{C}与∠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}互补。两个角都等于180∘−∠ABC180^\circ-\angle ABC，故∠D=∠CBE=∠E\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=16}{\angle }\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{B}\htmlData{tutor-start=18,tutor-end=19}{E}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=27}{\angle }\htmlData{tutor-start=27,tutor-end=28}{E}。

∠D=180∘−∠ABC=∠CBE=∠E\angle D=180^\circ-\angle ABC=\angle CBE=\angle E
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：共线关系决定了这里使用的是B\htmlData{tutor-start=0,tutor-end=1}{B}点外角而非内角；等腰三角形底角和圆内接四边形对角互补两条关系方向一致，结论成立。

∠D=∠E\boxed{\htmlData{tutor-start=7,tutor-end=14}{\angle }\htmlData{tutor-start=14,tutor-end=15}{D}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=23}{\angle }\htmlData{tutor-start=23,tutor-end=24}{E}}

（2）证明三角形ADE为等边三角形

由中点和等距条件推出三个角均为60度。

(1)
识别结构并建立关系

MB=MC\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{M}\htmlData{tutor-start=4,tutor-end=5}{C}说明M\htmlData{tutor-start=0,tutor-end=1}{M}在弦BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}的垂直平分线上，圆心O\htmlData{tutor-start=0,tutor-end=1}{O}也在该线上；M\htmlData{tutor-start=0,tutor-end=1}{M}又是弦AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}中点，所以OM\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}同时垂直BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}和AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}，从而得到关键平行关系AD∥BC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=12}{\parallel }\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{C}。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“MB=MC\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{M}\htmlData{tutor-start=4,tutor-end=5}{C}说明M\htmlData{tutor-start=0,tutor-end=1}{M}在弦BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}的垂直平分线上，圆心O\htmlData{tutor-start=0,tutor-end=1}{O}也在该线上；M\htmlData{tutor-start=0,tutor-end=1}{M}又是弦AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}中点，所以OM\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}同时垂直BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}和AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}，从而得到关键平行关系AD∥BC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=12}{\parallel }\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{C}。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：由MB=MC\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{M}\htmlData{tutor-start=4,tutor-end=5}{C}及OB=OC\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{O}\htmlData{tutor-start=4,tutor-end=5}{C}，直线OM\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}是BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}的垂直平分线，故OM⊥BC\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{C}。又M\htmlData{tutor-start=0,tutor-end=1}{M}为弦AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}中点，圆心到弦中点的连线垂直弦，故OM⊥AD\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{D}。因AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}不是直径，M≠O\htmlData{tutor-start=0,tutor-end=1}{M}\ne \htmlData{tutor-start=5,tutor-end=6}{O}，所以OM\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}是确定直线，进而AD∥BC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=12}{\parallel }\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{C}。于是∠DAE=∠CBA\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{E}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{C}\htmlData{tutor-start=19,tutor-end=20}{B}\htmlData{tutor-start=20,tutor-end=21}{A}；同弧CA\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{A}所对圆周角相等，∠CBA=∠CDA=∠ADE\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{C}\htmlData{tutor-start=19,tutor-end=20}{D}\htmlData{tutor-start=20,tutor-end=21}{A}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=29}{\angle }\htmlData{tutor-start=29,tutor-end=30}{A}\htmlData{tutor-start=30,tutor-end=31}{D}\htmlData{tutor-start=31,tutor-end=32}{E}。结合第(1)问∠ADE=∠AED\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{E}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{A}\htmlData{tutor-start=19,tutor-end=20}{E}\htmlData{tutor-start=20,tutor-end=21}{D}，三角形三个内角相等。

AD∥BC⟹∠DAE=∠CBA=∠ADE=∠AED=60∘AD\parallel BC\Longrightarrow \angle DAE=\angle CBA=\angle ADE=\angle AED=60^\circ
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：三角形ADE\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{E}内角和为180∘180^\circ，三个角相等就各为60∘60^\circ，因此三边也相等。题设排除AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}为直径，保证OM\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}不是退化为一点的论证漏洞。

△ADE 为等边三角形\boxed{\htmlData{tutor-start=7,tutor-end=17}{\triangle }\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{D}\htmlData{tutor-start=19,tutor-end=20}{E}\text{ \htmlData{tutor-start=27,tutor-end=28}{为}\htmlData{tutor-start=28,tutor-end=29}{等}\htmlData{tutor-start=29,tutor-end=30}{边}\htmlData{tutor-start=30,tutor-end=31}{三}\htmlData{tutor-start=31,tutor-end=32}{角}\htmlData{tutor-start=32,tutor-end=33}{形}}}
23

三、解答题 · 参数方程与直线距离

已知曲线 C:x24+y29=1\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}{4}} \htmlData{tutor-start=19,tutor-end=20}{+} \frac{\htmlData{tutor-start=27,tutor-end=28}{y}^{\htmlData{tutor-start=30,tutor-end=31}{2}}}{\htmlData{tutor-start=34,tutor-end=35}{9}} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{1}, 直线 l:{x=2+t, y=2−2t,\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{:} \begin{cases} \htmlData{tutor-start=17,tutor-end=18}{x} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{2} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{t}\htmlData{tutor-start=26,tutor-end=27}{,} \htmlData{tutor-start=28,tutor-end=30}{\ }\htmlData{tutor-start=30,tutor-end=31}{y} \htmlData{tutor-start=32,tutor-end=33}{=} \htmlData{tutor-start=34,tutor-end=35}{2} \htmlData{tutor-start=36,tutor-end=37}{-} \htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{t}\htmlData{tutor-start=40,tutor-end=41}{,} \end{cases} (t\htmlData{tutor-start=0,tutor-end=1}{t} 为参数). (1) 写出曲线 C\htmlData{tutor-start=0,tutor-end=1}{C} 的参数方程，直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 的普通方程; (2) 过曲线 C\htmlData{tutor-start=0,tutor-end=1}{C} 上任一点 P\htmlData{tutor-start=0,tutor-end=1}{P} 作与 l\htmlData{tutor-start=0,tutor-end=1}{l} 夹角为 30∘30^\circ 的直线，交 l\htmlData{tutor-start=0,tutor-end=1}{l} 于点 A\htmlData{tutor-start=0,tutor-end=1}{A}, 求 ∣PA∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{|} 的最大值与最小值.

答案：(1) x=2cosα,y=3sinα；l:2x+y-6=0；(2) 最小值2√5/5，最大值22√5/5。

题目标签：椭圆参数方程与斜线段极值

解题过程

（1）写出曲线与直线的目标形式

把椭圆参数化并消去直线参数t\htmlData{tutor-start=0,tutor-end=1}{t}。

(1)
识别结构并建立关系

椭圆x2/a2+y2/b2=1\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=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{y}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{b}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{1}的标准参数是x=acos⁡α,y=bsin⁡α\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\cos\htmlData{tutor-start=7,tutor-end=13}{\alpha}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{y}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{b}\sin\htmlData{tutor-start=21,tutor-end=27}{\alpha}。直线参数式则从x=2+t\htmlData{tutor-start=0,tutor-end=1}{x}\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}{t}解出t\htmlData{tutor-start=0,tutor-end=1}{t}并代入y\htmlData{tutor-start=0,tutor-end=1}{y}。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“椭圆x2/a2+y2/b2=1\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=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{y}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{b}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{1}的标准参数是x=acos⁡α,y=bsin⁡α\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\cos\htmlData{tutor-start=7,tutor-end=13}{\alpha}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{y}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{b}\sin\htmlData{tutor-start=21,tutor-end=27}{\alpha}。直线参数式则从x=2+t\htmlData{tutor-start=0,tutor-end=1}{x}\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}{t}解出t\htmlData{tutor-start=0,tutor-end=1}{t}并代入y\htmlData{tutor-start=0,tutor-end=1}{y}。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：椭圆半轴分别为2和3，所以可令x=2cos⁡α,y=3sin⁡α\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\cos\htmlData{tutor-start=7,tutor-end=13}{\alpha}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{y}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{3}\sin\htmlData{tutor-start=21,tutor-end=27}{\alpha}。直线中t=x−2\htmlData{tutor-start=0,tutor-end=1}{t}\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}{2}，代入y=2−2t\htmlData{tutor-start=0,tutor-end=1}{y}\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}{2}\htmlData{tutor-start=5,tutor-end=6}{t}得y=2−2(x−2)=6−2x\htmlData{tutor-start=0,tutor-end=1}{y}\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}{2}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{6}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{x}，整理为2x+y−6=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{6}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{0}。

C:{x=2cos⁡α,\y=3sin⁡α,l:2x+y−6=0C:\begin{cases}x=2\cos\alpha,\\\y=3\sin\alpha,\end{cases}\quad l:2x+y-6=0
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：代入参数方程有x2/4+y2/9=cos⁡2α+sin⁡2α=1\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=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{9}\htmlData{tutor-start=15,tutor-end=16}{=}\cos^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=30}{\alpha}\htmlData{tutor-start=30,tutor-end=31}{+}\sin^{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=45}{\alpha}\htmlData{tutor-start=45,tutor-end=46}{=}\htmlData{tutor-start=46,tutor-end=47}{1}；把x=2+t\htmlData{tutor-start=0,tutor-end=1}{x}\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}{t}代入普通方程也得到y=2−2t\htmlData{tutor-start=0,tutor-end=1}{y}\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}{2}\htmlData{tutor-start=5,tutor-end=6}{t}，两种形式互相验证。

x=2cos⁡α, y=3sin⁡α2x+y−6=0\boxed{\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{2}\cos\htmlData{tutor-start=14,tutor-end=20}{\alpha}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=23}{\ }\htmlData{tutor-start=23,tutor-end=24}{y}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{3}\sin\htmlData{tutor-start=30,tutor-end=36}{\alpha}}\quad\boxed{\htmlData{tutor-start=49,tutor-end=50}{2}\htmlData{tutor-start=50,tutor-end=51}{x}\htmlData{tutor-start=51,tutor-end=52}{+}\htmlData{tutor-start=52,tutor-end=53}{y}\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{6}\htmlData{tutor-start=55,tutor-end=56}{=}\htmlData{tutor-start=56,tutor-end=57}{0}}

（2）求与直线成30度的线段长度极值

将∣PA∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{|}转化为点P\htmlData{tutor-start=0,tutor-end=1}{P}到l\htmlData{tutor-start=0,tutor-end=1}{l}的距离并求三角函数极值。

(1)
识别结构并建立关系

PA\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}与l\htmlData{tutor-start=0,tutor-end=1}{l}夹角为30∘30^\circ，所以点P\htmlData{tutor-start=0,tutor-end=1}{P}到l\htmlData{tutor-start=0,tutor-end=1}{l}的垂距等于∣PA∣sin⁡30∘|PA|\sin30^\circ。这样不必求交点A\htmlData{tutor-start=0,tutor-end=1}{A}，只需优化点到定直线距离。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“PA\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}与l\htmlData{tutor-start=0,tutor-end=1}{l}夹角为30∘30^\circ，所以点P\htmlData{tutor-start=0,tutor-end=1}{P}到l\htmlData{tutor-start=0,tutor-end=1}{l}的垂距等于∣PA∣sin⁡30∘|PA|\sin30^\circ。这样不必求交点A\htmlData{tutor-start=0,tutor-end=1}{A}，只需优化点到定直线距离。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：设P=(2cos⁡α,3sin⁡α)\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{2}\cos\htmlData{tutor-start=8,tutor-end=14}{\alpha}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{3}\sin\htmlData{tutor-start=20,tutor-end=26}{\alpha}\htmlData{tutor-start=26,tutor-end=27}{)}。点到直线距离d=∣4cos⁡α+3sin⁡α−6∣/5\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{4}\cos\htmlData{tutor-start=8,tutor-end=14}{\alpha}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{3}\sin\htmlData{tutor-start=20,tutor-end=26}{\alpha}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{6}\htmlData{tutor-start=28,tutor-end=29}{|}\htmlData{tutor-start=29,tutor-end=30}{/}\sqrt{\htmlData{tutor-start=36,tutor-end=37}{5}}，而∣PA∣=d/sin⁡30∘=2d|PA|=d/\sin30^\circ=2d。由于4cos⁡α+3sin⁡α\htmlData{tutor-start=0,tutor-end=1}{4}\cos\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{3}\sin\htmlData{tutor-start=17,tutor-end=23}{\alpha}的范围是[−5,5]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{5}\htmlData{tutor-start=5,tutor-end=6}{]}，表达式4cos⁡α+3sin⁡α−6\htmlData{tutor-start=0,tutor-end=1}{4}\cos\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{3}\sin\htmlData{tutor-start=17,tutor-end=23}{\alpha}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{6}恒为负，绝对值范围为[1,11]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{]}。因此∣PA∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{|}范围为[2/5,22/5]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{/}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{/}\sqrt{\htmlData{tutor-start=21,tutor-end=22}{5}}\htmlData{tutor-start=23,tutor-end=24}{]}。

∣PA∣=2∣4cos⁡α+3sin⁡α−6∣5∈[255,2255]\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\frac{\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{|}\htmlData{tutor-start=13,tutor-end=14}{4}\cos\htmlData{tutor-start=18,tutor-end=24}{\alpha}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{3}\sin\htmlData{tutor-start=30,tutor-end=36}{\alpha}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{6}\htmlData{tutor-start=38,tutor-end=39}{|}}{\sqrt{\htmlData{tutor-start=47,tutor-end=48}{5}}}\htmlData{tutor-start=50,tutor-end=53}{\in}\left[\frac{\htmlData{tutor-start=65,tutor-end=66}{2}\sqrt{\htmlData{tutor-start=72,tutor-end=73}{5}}}{\htmlData{tutor-start=76,tutor-end=77}{5}}\htmlData{tutor-start=78,tutor-end=79}{,}\frac{\htmlData{tutor-start=85,tutor-end=86}{2}\htmlData{tutor-start=86,tutor-end=87}{2}\sqrt{\htmlData{tutor-start=93,tutor-end=94}{5}}}{\htmlData{tutor-start=97,tutor-end=98}{5}}\right]
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：4cos⁡α+3sin⁡α=5\htmlData{tutor-start=0,tutor-end=1}{4}\cos\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{3}\sin\htmlData{tutor-start=17,tutor-end=23}{\alpha}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{5}和−5\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{5}都能由适当α\htmlData{tutor-start=0,tutor-end=6}{\alpha}取得，所以两个边界都可达，不只是上、下界估计。故最小值和最大值分别如上。

∣PA∣min⁡=255,∣PA∣max⁡=2255\boxed{\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{|}_{\min}\htmlData{tutor-start=18,tutor-end=19}{=}\frac{\htmlData{tutor-start=25,tutor-end=26}{2}\sqrt{\htmlData{tutor-start=32,tutor-end=33}{5}}}{\htmlData{tutor-start=36,tutor-end=37}{5}}\htmlData{tutor-start=38,tutor-end=39}{,}\quad \htmlData{tutor-start=45,tutor-end=46}{|}\htmlData{tutor-start=46,tutor-end=47}{P}\htmlData{tutor-start=47,tutor-end=48}{A}\htmlData{tutor-start=48,tutor-end=49}{|}_{\max}\htmlData{tutor-start=56,tutor-end=57}{=}\frac{\htmlData{tutor-start=63,tutor-end=64}{2}\htmlData{tutor-start=64,tutor-end=65}{2}\sqrt{\htmlData{tutor-start=71,tutor-end=72}{5}}}{\htmlData{tutor-start=75,tutor-end=76}{5}}}
24

三、解答题 · 基本不等式

若 a>0,b>0\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{b} \htmlData{tutor-start=9,tutor-end=10}{>} \htmlData{tutor-start=11,tutor-end=12}{0}, 且 1a+1b=ab\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}} \htmlData{tutor-start=12,tutor-end=13}{+} \frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{b}} \htmlData{tutor-start=26,tutor-end=27}{=} \sqrt{\htmlData{tutor-start=34,tutor-end=35}{a}\htmlData{tutor-start=35,tutor-end=36}{b}}. (1) 求 a3+b3\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{3}} 的最小值; (2) 是否存在 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}, 使得 2a+3b=6\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{b} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{6}? 并说明理由.

答案：(1) 最小值4√2；(2) 不存在。

题目标签：正数条件下的代数最值与存在性

解题过程

（1）求a³+b³的最小值

把分式条件改写为和与积的关系，再求对称式最值。

(1)
识别结构并建立关系

条件和目标都关于a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}对称，适合设p=a+b,q=ab\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}{+}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{q}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{b}。原条件清除分母后会直接给出p=q3/2\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{q}^{\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{2}}。

为什么从这里入手：目标是“识别结构并建立关系”，而“条件和目标都关于a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}对称，适合设p=a+b,q=ab\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}{+}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{q}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{b}。原条件清除分母后会直接给出p=q3/2\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{q}^{\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{2}}。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：由1/a+1/b=ab\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,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}{b}\htmlData{tutor-start=7,tutor-end=8}{=}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{a}\htmlData{tutor-start=15,tutor-end=16}{b}}得a+b=(ab)3/2\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}{(}\htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{)}^{\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{2}}，即p=q3/2\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{q}^{\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{2}}。又p≥2q\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{2}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{q}}，所以q3/2≥2q\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=10}{\ge}\htmlData{tutor-start=10,tutor-end=11}{2}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{q}}，从而q≥2\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{2}。目标a3+b3=p3−3pq=q9/2−3q5/2=q5/2(q2−3)\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{p}^{\htmlData{tutor-start=15,tutor-end=16}{3}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{p}\htmlData{tutor-start=20,tutor-end=21}{q}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{q}^{\htmlData{tutor-start=25,tutor-end=26}{9}\htmlData{tutor-start=26,tutor-end=27}{/}\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{3}\htmlData{tutor-start=31,tutor-end=32}{q}^{\htmlData{tutor-start=34,tutor-end=35}{5}\htmlData{tutor-start=35,tutor-end=36}{/}\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=39}{=}\htmlData{tutor-start=39,tutor-end=40}{q}^{\htmlData{tutor-start=42,tutor-end=43}{5}\htmlData{tutor-start=43,tutor-end=44}{/}\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{q}^{\htmlData{tutor-start=50,tutor-end=51}{2}}\htmlData{tutor-start=52,tutor-end=53}{-}\htmlData{tutor-start=53,tutor-end=54}{3}\htmlData{tutor-start=54,tutor-end=55}{)}。该函数在q≥2\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{2}上严格递增，故最小值在q=2\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}处取得，此时基本不等式取等要求a=b=2\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}{=}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}。

a3+b3≥2(2)3=42\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=15}{\ge }\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{(}\sqrt{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{)}^{\htmlData{tutor-start=28,tutor-end=29}{3}}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{4}\sqrt{\htmlData{tutor-start=38,tutor-end=39}{2}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：取a=b=2\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}{=}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}时，1/a+1/b=2=ab\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,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}{b}\htmlData{tutor-start=7,tutor-end=8}{=}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{=}\sqrt{\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=25}{b}}，满足原约束，且a3+b3=42\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{4}\sqrt{\htmlData{tutor-start=19,tutor-end=20}{2}}。等号条件可达，所以这是最小值而非仅有下界。

(a3+b3)min⁡=42\boxed{\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{a}^{\htmlData{tutor-start=11,tutor-end=12}{3}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{b}^{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{)}_{\min}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{4}\sqrt{\htmlData{tutor-start=35,tutor-end=36}{2}}}

（2）判断线性条件能否同时成立

检验是否存在正数a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}同时满足原条件与2a+3b=6\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{6}。

(1)
识别结构并建立关系

第(1)问已经推出必要条件ab≥2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=5}{\ge}\htmlData{tutor-start=5,tutor-end=6}{2}。只要证明线性约束下ab\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}的最大值小于2，就能直接否定存在性。

为什么从这里入手：最值与范围题要先找到变量受限方式以及等号何时成立。“第(1)问已经推出必要条件ab≥2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=5}{\ge}\htmlData{tutor-start=5,tutor-end=6}{2}。只要证明线性约束下ab\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}的最大值小于2，就能直接否定存在性。”给出了可行域或关键界，先识别结构并建立关系能把‘可能有多大’转成单调性、基本不等式或边界比较。

详细展开：若2a+3b=6\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{6}，则a=(6−3b)/2\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}{6}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{3}\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}，其中0<b<2\htmlData{tutor-start=0,tutor-end=1}{0}\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}{2}。于是ab=(6b−3b2)/2=3/2−(3/2)(b−1)2≤3/2\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}{(}\htmlData{tutor-start=4,tutor-end=5}{6}\htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{3}\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=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{b}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)}^{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=38}{\le}\htmlData{tutor-start=38,tutor-end=39}{3}\htmlData{tutor-start=39,tutor-end=40}{/}\htmlData{tutor-start=40,tutor-end=41}{2}。这与原条件必然推出的ab≥2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=5}{\ge}\htmlData{tutor-start=5,tutor-end=6}{2}矛盾，因此不存在这样的正数a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}。

2a+3b=6⟹ab≤32<2≤ab\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{6}\htmlData{tutor-start=7,tutor-end=23}{\Longrightarrow }\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=25}{b}\htmlData{tutor-start=25,tutor-end=28}{\le}\frac{\htmlData{tutor-start=34,tutor-end=35}{3}}{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{<}\htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=45}{\le }\htmlData{tutor-start=45,tutor-end=46}{a}\htmlData{tutor-start=46,tutor-end=47}{b}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

为什么从这里入手：目标是“完成计算并核验结论”，而“上一阶段已经得到可直接求值、判号或证明的核心关系。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：线性约束下的最大乘积在b=1,a=3/2\htmlData{tutor-start=0,tutor-end=1}{b}\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}{a}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}处取得，仍只有3/2\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2}，所以不是等号边界遗漏造成的矛盾，结论为不存在。

不存在满足条件的正数 a,b\boxed{\text{\htmlData{tutor-start=13,tutor-end=14}{不}\htmlData{tutor-start=14,tutor-end=15}{存}\htmlData{tutor-start=15,tutor-end=16}{在}\htmlData{tutor-start=16,tutor-end=17}{满}\htmlData{tutor-start=17,tutor-end=18}{足}\htmlData{tutor-start=18,tutor-end=19}{条}\htmlData{tutor-start=19,tutor-end=20}{件}\htmlData{tutor-start=20,tutor-end=21}{的}\htmlData{tutor-start=21,tutor-end=22}{正}\htmlData{tutor-start=22,tutor-end=23}{数} }\htmlData{tutor-start=25,tutor-end=26}{a}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=28}{b}}