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2014 年高考数学（大纲卷文科）

exams_raw/普通高考/2014/2014大纲文(广西).pdf · HS-MATH-1024-v2.1-solution-aware

22 题27 个小问/题组
1

一、选择题 · 集合的基本运算

设集合 M={1,2,4,6,8}\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{6}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{8}\htmlData{tutor-start=17,tutor-end=19}{\}}，N={1,2,3,5,6,7}\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{5}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{6}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{7}\htmlData{tutor-start=20,tutor-end=22}{\}}，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} 中元素的个数为 ( )

答案：B，交集有3个元素。

题目标签：集合交集元素个数

解题过程

逐项求交集并计数

找出同时属于两个集合的元素，再求元素个数。

(1)
识别结构并建立关系

题目问的是交集中的元素个数。交集只保留在两个集合中都出现的元素，不能把并集中的元素也算进去。

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“题目问的是交集中的元素个数。交集只保留在两个集合中都出现的元素，不能把并集中的元素也算进去。”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：先看集合 M 的元素。1 在 N 中，保留；2 在 N 中，保留；4 不在 N 中，舍去；6 在 N 中，保留；8 不在 N 中，舍去。因此 M 与 N 共有的元素恰好是 1、2、6，即 M∩N={1,2,6}\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=6}{\cap }\htmlData{tutor-start=6,tutor-end=7}{N}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,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}{2}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{6}\htmlData{tutor-start=15,tutor-end=17}{\}}。集合中有三个互不相同的元素，所以交集的元素个数为 3。

M∩N={1,2,6},∣M∩N∣=3\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=6}{\cap }\htmlData{tutor-start=6,tutor-end=7}{N}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,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}{2}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{6}\htmlData{tutor-start=15,tutor-end=17}{\}}\htmlData{tutor-start=17,tutor-end=18}{,}\qquad \htmlData{tutor-start=25,tutor-end=26}{|}\htmlData{tutor-start=26,tutor-end=27}{M}\htmlData{tutor-start=27,tutor-end=32}{\cap }\htmlData{tutor-start=32,tutor-end=33}{N}\htmlData{tutor-start=33,tutor-end=34}{|}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{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：核验时反向检查：1、2、6 都分别出现在 M 和 N 中；M 中剩余的 4、8 不属于 N，N 中剩余的 3、5、7 不属于 M，没有漏项或多项。因此选择 B。

M∩N={1,2,6},3 (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=17}{\{}\htmlData{tutor-start=17,tutor-end=18}{1}\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}{6}\htmlData{tutor-start=22,tutor-end=24}{\}}}\htmlData{tutor-start=25,tutor-end=26}{,}\qquad \boxed{\htmlData{tutor-start=40,tutor-end=41}{3}\htmlData{tutor-start=41,tutor-end=43}{\ }\htmlData{tutor-start=43,tutor-end=44}{(}\mathrm{\htmlData{tutor-start=52,tutor-end=53}{B}}\htmlData{tutor-start=54,tutor-end=55}{)}}
2

一、选择题 · 任意角的三角函数

已知角 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 的终边经过点 (−4,3)\htmlData{tutor-start=0,tutor-end=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=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{)}，则 cos⁡α=\cos \htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{=} ( )

答案：D，-4/5。

题目标签：终边上一点求余弦

解题过程

由点的坐标计算余弦

利用终边上一点的横坐标与到原点距离求 cos⁡α\cos\htmlData{tutor-start=4,tutor-end=10}{\alpha}。

(1)
识别结构并建立关系

已知角的终边经过点 (x,y)\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}{y}\htmlData{tutor-start=4,tutor-end=5}{)} 时，应立即想到定义 cos⁡α=x/r\cos\htmlData{tutor-start=4,tutor-end=10}{\alpha}\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}{r}，其中 r\htmlData{tutor-start=0,tutor-end=1}{r} 是该点到原点的距离。横坐标为负还可以先判断余弦应为负。

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。“已知角的终边经过点 (x,y)\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}{y}\htmlData{tutor-start=4,tutor-end=5}{)} 时，应立即想到定义 cos⁡α=x/r\cos\htmlData{tutor-start=4,tutor-end=10}{\alpha}\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}{r}，其中 r\htmlData{tutor-start=0,tutor-end=1}{r} 是该点到原点的距离。横坐标为负还可以先判断余弦应为负。”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：终边上的点为 P(−4,3)\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}{4}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{)}。它到原点的距离为 r=(−4)2+32=5\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{(}\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=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{3}^{\htmlData{tutor-start=20,tutor-end=21}{2}}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{5}。按照任意角三角函数的坐标定义，cos⁡α\cos\htmlData{tutor-start=4,tutor-end=10}{\alpha} 等于横坐标除以 r\htmlData{tutor-start=0,tutor-end=1}{r}，因此 cos⁡α=−4/5\cos\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{4}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{5}。点 P 位于第二象限，第二象限的余弦为负，这也与计算结果的符号一致。

r=(−4)2+32=5,cos⁡α=xr=−45\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{(}\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=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{3}^{\htmlData{tutor-start=20,tutor-end=21}{2}}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{5}\htmlData{tutor-start=25,tutor-end=26}{,}\qquad \cos\htmlData{tutor-start=37,tutor-end=43}{\alpha}\htmlData{tutor-start=43,tutor-end=44}{=}\frac{\htmlData{tutor-start=50,tutor-end=51}{x}}{\htmlData{tutor-start=53,tutor-end=54}{r}}\htmlData{tutor-start=55,tutor-end=56}{=}\htmlData{tutor-start=56,tutor-end=57}{-}\frac{\htmlData{tutor-start=63,tutor-end=64}{4}}{\htmlData{tutor-start=66,tutor-end=67}{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：核验：由同一点还可得 sin⁡α=3/5\sin\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{5}，于是 sin⁡2α+cos⁡2α=9/25+16/25=1\sin^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=14}{\alpha}\htmlData{tutor-start=14,tutor-end=15}{+}\cos^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=29}{\alpha}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{9}\htmlData{tutor-start=31,tutor-end=32}{/}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{5}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{6}\htmlData{tutor-start=37,tutor-end=38}{/}\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{5}\htmlData{tutor-start=40,tutor-end=41}{=}\htmlData{tutor-start=41,tutor-end=42}{1}，满足平方关系。故选择 D。

cos⁡α=−45D\boxed{\cos\htmlData{tutor-start=11,tutor-end=17}{\alpha}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{-}\frac{\htmlData{tutor-start=25,tutor-end=26}{4}}{\htmlData{tutor-start=28,tutor-end=29}{5}}}\quad\boxed{\mathrm{\htmlData{tutor-start=51,tutor-end=52}{D}}}
3

一、选择题 · 不等式的解法

不等式组 {x(x+2)>0,∣x∣<1\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{)}\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=27,tutor-end=28}{|}\htmlData{tutor-start=28,tutor-end=29}{x}\htmlData{tutor-start=29,tutor-end=30}{|}\htmlData{tutor-start=30,tutor-end=31}{<}\htmlData{tutor-start=31,tutor-end=32}{1} \end{cases} 的解集为 ( )

答案：C，(0,1)。

题目标签：一元不等式组求交集

解题过程

分别求解后取公共部分

求两个不等式解集的交集。

(1)
识别结构并建立关系

大括号表示两个条件必须同时成立。先分别解因式乘积不等式和绝对值不等式，再在数轴上取公共部分最稳妥。

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。“大括号表示两个条件必须同时成立。先分别解因式乘积不等式和绝对值不等式，再在数轴上取公共部分最稳妥。”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：由 x(x+2)>0，两个零点是 -2 和 0。二次项系数为正，所以乘积在两根外侧为正，得到 x<-2 或 x>0。由 |x|<1 得 -1<x<1。前一解集的 x<-2 部分与 (-1,1) 没有交集，x>0 与 (-1,1) 的公共部分为 0<x<1。因此不等式组的解集是 (0,1)。

x(x+2)>0⟹x<−2 或 x>0;∣x∣<1⟹−1<x<1;x∈(0,1)\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=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}{>}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=24}{\Longrightarrow }\htmlData{tutor-start=24,tutor-end=25}{x}\htmlData{tutor-start=25,tutor-end=26}{<}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{2}\text{ \htmlData{tutor-start=35,tutor-end=36}{或} }\htmlData{tutor-start=38,tutor-end=39}{x}\htmlData{tutor-start=39,tutor-end=40}{>}\htmlData{tutor-start=40,tutor-end=41}{0}\htmlData{tutor-start=41,tutor-end=42}{;}\qquad \htmlData{tutor-start=49,tutor-end=50}{|}\htmlData{tutor-start=50,tutor-end=51}{x}\htmlData{tutor-start=51,tutor-end=52}{|}\htmlData{tutor-start=52,tutor-end=53}{<}\htmlData{tutor-start=53,tutor-end=54}{1}\htmlData{tutor-start=54,tutor-end=69}{\Longrightarrow}\htmlData{tutor-start=69,tutor-end=70}{-}\htmlData{tutor-start=70,tutor-end=71}{1}\htmlData{tutor-start=71,tutor-end=72}{<}\htmlData{tutor-start=72,tutor-end=73}{x}\htmlData{tutor-start=73,tutor-end=74}{<}\htmlData{tutor-start=74,tutor-end=75}{1}\htmlData{tutor-start=75,tutor-end=76}{;}\qquad \htmlData{tutor-start=83,tutor-end=84}{x}\htmlData{tutor-start=84,tutor-end=87}{\in}\htmlData{tutor-start=87,tutor-end=88}{(}\htmlData{tutor-start=88,tutor-end=89}{0}\htmlData{tutor-start=89,tutor-end=90}{,}\htmlData{tutor-start=90,tutor-end=91}{1}\htmlData{tutor-start=91,tutor-end=92}{)}
(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 x=1/2 代入，两个条件均成立；端点 x=0 使乘积等于 0，x=1 使 |x|=1，都不满足严格不等号，所以两个端点均不能取。故选择 C。

x∈(0,1)C\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}{0}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)}}\quad\boxed{\mathrm{\htmlData{tutor-start=37,tutor-end=38}{C}}}
4

一、选择题 · 空间向量

已知正四面体 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} 中，E\htmlData{tutor-start=0,tutor-end=1}{E} 是 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 的中点，则异面直线 CE\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{E} 与 BD\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D} 所成角的余弦值为 ( )

答案：B，sqrt(3)/6。

题目标签：正四面体异面直线所成角

解题过程

建立对称坐标计算方向向量夹角

把异面直线 CE、BD 平移到同一起点，用向量夹角求余弦。

(1)
识别结构并建立关系

正四面体具有高度对称性，用立方体四个交错顶点建立坐标可以让所有棱自动等长。异面直线所成角等于它们方向向量夹角中的锐角。

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。“正四面体具有高度对称性，用立方体四个交错顶点建立坐标可以让所有棱自动等长。异面直线所成角等于它们方向向量夹角中的锐角。”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：取 A(1,1,1)\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}{1}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}、B(1,−1,−1)\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}{-}\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}{1}\htmlData{tutor-start=9,tutor-end=10}{)}、C(−1,1,−1)\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=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}{1}\htmlData{tutor-start=9,tutor-end=10}{)}、D(−1,−1,1)\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}{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}{1}\htmlData{tutor-start=9,tutor-end=10}{)}，任意两点间距离都为 22\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}，所以构成正四面体。AB 的中点 E=(1,0,0)\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=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}{)}。可取方向向量 CE→=(2,−1,1)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{E}}\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}{,}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{,}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{)}，BD→=(−2,0,2)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{D}}\htmlData{tutor-start=19,tutor-end=20}{=}\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}{0}\htmlData{tutor-start=25,tutor-end=26}{,}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{)}。点积为 -2，两个模分别为 6\sqrt{\htmlData{tutor-start=6,tutor-end=7}{6}}、22\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}。异面直线所成角取锐角，故余弦取点积绝对值，得到 2/(6⋅22)=3/6\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{(}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{6}}\htmlData{tutor-start=11,tutor-end=16}{\cdot}\htmlData{tutor-start=16,tutor-end=17}{2}\sqrt{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{=}\sqrt{\htmlData{tutor-start=33,tutor-end=34}{3}}\htmlData{tutor-start=35,tutor-end=36}{/}\htmlData{tutor-start=36,tutor-end=37}{6}。

E=(1,0,0),CE→=(2,−1,1),BD→=(−2,0,2),cos⁡theta=∣−2∣6⋅22=36\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=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}{,}\quad \overrightarrow{\htmlData{tutor-start=32,tutor-end=33}{C}\htmlData{tutor-start=33,tutor-end=34}{E}}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{2}\htmlData{tutor-start=38,tutor-end=39}{,}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{1}\htmlData{tutor-start=41,tutor-end=42}{,}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{)}\htmlData{tutor-start=44,tutor-end=45}{,}\quad \overrightarrow{\htmlData{tutor-start=67,tutor-end=68}{B}\htmlData{tutor-start=68,tutor-end=69}{D}}\htmlData{tutor-start=70,tutor-end=71}{=}\htmlData{tutor-start=71,tutor-end=72}{(}\htmlData{tutor-start=72,tutor-end=73}{-}\htmlData{tutor-start=73,tutor-end=74}{2}\htmlData{tutor-start=74,tutor-end=75}{,}\htmlData{tutor-start=75,tutor-end=76}{0}\htmlData{tutor-start=76,tutor-end=77}{,}\htmlData{tutor-start=77,tutor-end=78}{2}\htmlData{tutor-start=78,tutor-end=79}{)}\htmlData{tutor-start=79,tutor-end=80}{,}\quad \cos\\\htmlData{tutor-start=92,tutor-end=93}{t}\htmlData{tutor-start=93,tutor-end=94}{h}\htmlData{tutor-start=94,tutor-end=95}{e}\htmlData{tutor-start=95,tutor-end=96}{t}\htmlData{tutor-start=96,tutor-end=97}{a}\htmlData{tutor-start=97,tutor-end=98}{=}\frac{\htmlData{tutor-start=104,tutor-end=105}{|}{\htmlData{tutor-start=106,tutor-end=107}{-}\htmlData{tutor-start=107,tutor-end=108}{2}}\htmlData{tutor-start=109,tutor-end=110}{|}}{\sqrt{\htmlData{tutor-start=118,tutor-end=119}{6}}\htmlData{tutor-start=120,tutor-end=125}{\cdot}\htmlData{tutor-start=125,tutor-end=126}{2}\sqrt{\htmlData{tutor-start=132,tutor-end=133}{2}}}\htmlData{tutor-start=135,tutor-end=136}{=}\frac{\sqrt{\htmlData{tutor-start=148,tutor-end=149}{3}}}{\htmlData{tutor-start=152,tutor-end=153}{6}}
(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：所得余弦在 0 与 1 之间，对应锐角；若改变任一方向向量的方向，点积只改变符号，绝对值不变，说明结果与直线定向无关。故选择 B。

cos⁡theta=36B\boxed{\cos\\\htmlData{tutor-start=13,tutor-end=14}{t}\htmlData{tutor-start=14,tutor-end=15}{h}\htmlData{tutor-start=15,tutor-end=16}{e}\htmlData{tutor-start=16,tutor-end=17}{t}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{=}\frac{\sqrt{\htmlData{tutor-start=31,tutor-end=32}{3}}}{\htmlData{tutor-start=35,tutor-end=36}{6}}}\quad\boxed{\mathrm{\htmlData{tutor-start=58,tutor-end=59}{B}}}
5

一、选择题 · 反函数

函数 y=ln⁡(x3+1) (x>−1)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=6}{(}\sqrt[\htmlData{tutor-start=12,tutor-end=13}{3}]{\htmlData{tutor-start=15,tutor-end=16}{x}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=22}{\ }\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{x}\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}{)} 的反函数是 ( )

答案：D，y=(e^x-1)^3，定义域为 R。

题目标签：含对数函数的反函数

解题过程

由 y 解出 x 并交换字母

求 y=ln(cuberoot(x)+1) 的反函数及其定义域。

(1)
识别结构并建立关系

反函数的核心不是套公式，而是从 y=f(x) 中把 x 单独解出来。反函数的定义域等于原函数的值域，因此还要判断对数表达式能取到哪些 y。

为什么从这里入手：含分母、根号或对数的式子必须先合法，后续变形才有意义。“反函数的核心不是套公式，而是从 y=f(x) 中把 x 单独解出来。反函数的定义域等于原函数的值域，因此还要判断对数表达式能取到哪些 y。”正好暴露了限制条件；先识别结构并建立关系，可以防止约分或平方时把禁值偷偷带回答案。

详细展开：令 y=ln⁡(x3+1)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=6}{(}\sqrt[\htmlData{tutor-start=12,tutor-end=13}{3}]{\htmlData{tutor-start=15,tutor-end=16}{x}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}。两边取指数得 ey=x3+1\htmlData{tutor-start=0,tutor-end=1}{e}^{\htmlData{tutor-start=3,tutor-end=4}{y}}\htmlData{tutor-start=5,tutor-end=6}{=}\sqrt[\htmlData{tutor-start=12,tutor-end=13}{3}]{\htmlData{tutor-start=15,tutor-end=16}{x}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}，所以 x3=ey−1\sqrt[\htmlData{tutor-start=6,tutor-end=7}{3}]{\htmlData{tutor-start=9,tutor-end=10}{x}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{e}^{\htmlData{tutor-start=15,tutor-end=16}{y}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}，进而 x=(ey−1)3\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}{e}^{\htmlData{tutor-start=6,tutor-end=7}{y}}\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=13,tutor-end=14}{3}}。交换自变量与因变量，得到 f−1(x)=(ex−1)3\htmlData{tutor-start=0,tutor-end=1}{f}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}}\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}{(}\htmlData{tutor-start=11,tutor-end=12}{e}^{\htmlData{tutor-start=14,tutor-end=15}{x}}\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=21,tutor-end=22}{3}}。原函数要求 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}；当 x\htmlData{tutor-start=0,tutor-end=1}{x} 从 -1 右侧增大到正无穷时，x3+1\sqrt[\htmlData{tutor-start=6,tutor-end=7}{3}]{\htmlData{tutor-start=9,tutor-end=10}{x}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1} 从 0 正向增大到正无穷，对数值覆盖整个实数集，所以反函数定义域为 R\mathbb{\htmlData{tutor-start=8,tutor-end=9}{R}}。

y=ln⁡(x3+1)⟺ey=x3+1⟺x=(ey−1)3\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=6}{(}\sqrt[\htmlData{tutor-start=12,tutor-end=13}{3}]{\htmlData{tutor-start=15,tutor-end=16}{x}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=40}{\Longleftrightarrow }\htmlData{tutor-start=40,tutor-end=41}{e}^{\htmlData{tutor-start=43,tutor-end=44}{y}}\htmlData{tutor-start=45,tutor-end=46}{=}\sqrt[\htmlData{tutor-start=52,tutor-end=53}{3}]{\htmlData{tutor-start=55,tutor-end=56}{x}}\htmlData{tutor-start=57,tutor-end=58}{+}\htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=79}{\Longleftrightarrow }\htmlData{tutor-start=79,tutor-end=80}{x}\htmlData{tutor-start=80,tutor-end=81}{=}\htmlData{tutor-start=81,tutor-end=82}{(}\htmlData{tutor-start=82,tutor-end=83}{e}^{\htmlData{tutor-start=85,tutor-end=86}{y}}\htmlData{tutor-start=87,tutor-end=88}{-}\htmlData{tutor-start=88,tutor-end=89}{1}\htmlData{tutor-start=89,tutor-end=90}{)}^{\htmlData{tutor-start=92,tutor-end=93}{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：把反函数代回原函数：ln⁡((ex−1)33+1)=ln⁡(ex)=x\ln\htmlData{tutor-start=3,tutor-end=4}{(}\sqrt[\htmlData{tutor-start=10,tutor-end=11}{3}]{\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{e}^{\htmlData{tutor-start=17,tutor-end=18}{x}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{3}}}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{=}\ln\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{e}^{\htmlData{tutor-start=38,tutor-end=39}{x}}\htmlData{tutor-start=40,tutor-end=41}{)}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{x}，恒对所有实数 x\htmlData{tutor-start=0,tutor-end=1}{x} 成立。定义域也与原函数值域一致，故选择 D。

f−1(x)=(ex−1)3, x∈RD\boxed{\htmlData{tutor-start=7,tutor-end=8}{f}^{\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}{x}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{e}^{\htmlData{tutor-start=21,tutor-end=22}{x}}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{1}\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=33}{\ }\htmlData{tutor-start=33,tutor-end=34}{x}\htmlData{tutor-start=34,tutor-end=37}{\in}\mathbb{\htmlData{tutor-start=45,tutor-end=46}{R}}}\quad\boxed{\mathrm{\htmlData{tutor-start=68,tutor-end=69}{D}}}
6

一、选择题 · 平面向量数量积

已知 a\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}、b\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}} 为单位向量，其夹角为 60∘60^\circ，则 (2a−b)⋅b=\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\boldsymbol{\htmlData{tutor-start=14,tutor-end=15}{a}}\htmlData{tutor-start=16,tutor-end=17}{-}\boldsymbol{\htmlData{tutor-start=29,tutor-end=30}{b}}\htmlData{tutor-start=31,tutor-end=32}{)}\htmlData{tutor-start=32,tutor-end=38}{\cdot }\boldsymbol{\htmlData{tutor-start=50,tutor-end=51}{b}} \htmlData{tutor-start=53,tutor-end=54}{=} ( )

答案：B，0。

题目标签：单位向量点积运算

解题过程

把夹角转化为点积

利用单位向量和夹角计算线性组合的点积。

(1)
识别结构并建立关系

看到单位向量及夹角，应先写 a 点 b=|a||b|cos60 度，再利用点积对加减法的分配律展开。

为什么从这里入手：空间关系只靠观察容易漏条件，而“看到单位向量及夹角，应先写 a 点 b=|a||b|cos60 度，再利用点积对加减法的分配律展开。”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：因为 a\boldsymbol a、b\boldsymbol b 都是单位向量，所以 ∣a∣=∣b∣=1|\boldsymbol a|=|\boldsymbol b|=1。它们夹角为 60 度，故 a⋅b=1⋅1⋅cos⁡60∘=1/2\boldsymbol a\cdot\boldsymbol b=1\cdot1\cdot\cos60^\circ=1/2。按分配律，(2a−b)⋅b=2(a⋅b)−(b⋅b)(2\boldsymbol a-\boldsymbol b)\cdot\boldsymbol b=2(\boldsymbol a\cdot\boldsymbol b)-(\boldsymbol b\cdot\boldsymbol b)。又 b⋅b=∣b∣2=1\boldsymbol b\cdot\boldsymbol b=|\boldsymbol b|^{2}=1，因此结果为 2×1/2−1=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=7}{\times}\htmlData{tutor-start=7,tutor-end=8}{1}\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}{1}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{0}。

a⋅b=12,b⋅b=1,(2a−b)⋅b=2⋅12−1=0\boldsymbol a\cdot\boldsymbol b=\frac{1}{2},\quad \boldsymbol b\cdot\boldsymbol b=1,\quad (2\boldsymbol a-\boldsymbol b)\cdot\boldsymbol b=2\cdot\frac{1}{2}-1=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：结果为 0 意味着向量 2a-b 与 b 垂直。直接计算其在 b 方向上的投影系数也是 2 cos60 度-1=0，二者一致，故选择 B。

(2a−b)⋅b=0B\boxed{(2\boldsymbol a-\boldsymbol b)\cdot\boldsymbol b=0}\quad\boxed{\mathrm{B}}
7

一、选择题 · 排列组合

有 6 名男医生、5 名女医生，从中选出 2 名男医生、1 名女医生组成一个医疗小组，则不同的选法共有 ( )

答案：C，75种。

题目标签：分步组合计数

解题过程

分别选男医生和女医生

计算不计顺序的医疗小组选法数。

(1)
识别结构并建立关系

题目只关心选出哪些人，不安排职务或次序，所以使用组合。男、女医生的选择相互独立，应用分步乘法计数原理。

为什么从这里入手：目标是“识别结构并建立关系”，而“题目只关心选出哪些人，不安排职务或次序，所以使用组合。男、女医生的选择相互独立，应用分步乘法计数原理。”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：从 6 名男医生中选 2 名，有 C(6,2)=6×5/2=15\htmlData{tutor-start=0,tutor-end=1}{C}\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}{2}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{6}\htmlData{tutor-start=8,tutor-end=14}{\times}\htmlData{tutor-start=14,tutor-end=15}{5}\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}{1}\htmlData{tutor-start=19,tutor-end=20}{5} 种；从 5 名女医生中选 1 名，有 C(5,1)=5\htmlData{tutor-start=0,tutor-end=1}{C}\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}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{5} 种。每一种男医生组合都可与任一种女医生选择配成一个小组，所以总数为 15×5=75\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{5}\htmlData{tutor-start=2,tutor-end=8}{\times}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{7}\htmlData{tutor-start=11,tutor-end=12}{5}。不能再乘 3!\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{!}，因为小组成员没有排列顺序。

N=(62)(51)=15×5=75N=\binom62\binom51=15\times5=75
(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：先选女医生有 5 种，对每位女医生都能搭配 15 对男医生，共 5 组各 15 种，仍为 75。故选择 C。

75C\boxed{\htmlData{tutor-start=7,tutor-end=8}{7}\htmlData{tutor-start=8,tutor-end=9}{5}}\quad\boxed{\mathrm{\htmlData{tutor-start=30,tutor-end=31}{C}}}
8

一、选择题 · 等比数列前 n 项和

设等比数列 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和为 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}，若 S2=3,S4=15\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{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=9,tutor-end=10}{S}_{\htmlData{tutor-start=12,tutor-end=13}{4}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{5}，则 S6=\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{6}} \htmlData{tutor-start=6,tutor-end=7}{=} ( )

答案：C，63。

题目标签：等比数列分段和

解题过程

按每两项一组利用公比平方

由 S2、S4 求 S6。

(1)
识别结构并建立关系

S2、S4、S6 的下标都相差 2。对等比数列，把相邻两项作为一组后，每组恰好是前一组的 q^2 倍，可以避开 q=1 等公式分类。

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。“S2、S4、S6 的下标都相差 2。对等比数列，把相邻两项作为一组后，每组恰好是前一组的 q^2 倍，可以避开 q=1 等公式分类。”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：前两项之和为 S2=3。第 3、4 项分别是第 1、2 项的 q^2 倍，所以 a3+a4=q^2S2。于是 S4=S2(1+q^2)=15，代入 S2=3 得 1+q^2=5，即 q^2=4。第 5、6 项之和又是前两项之和的 q^4 倍，因此 S6=S2(1+q^2+q^4)=3(1+4+16)=63。这个方法对 q=2 和 q=-2 都给出同一结果。

15=S4=S2(1+q2)=3(1+q2)⇒q2=4,S6=3(1+4+16)=63\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}{S}_{\htmlData{tutor-start=6,tutor-end=7}{4}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{S}_{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{q}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{)}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{3}\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}{q}^{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{)}\htmlData{tutor-start=34,tutor-end=46}{\Rightarrow }\htmlData{tutor-start=46,tutor-end=47}{q}^{\htmlData{tutor-start=49,tutor-end=50}{2}}\htmlData{tutor-start=51,tutor-end=52}{=}\htmlData{tutor-start=52,tutor-end=53}{4}\htmlData{tutor-start=53,tutor-end=54}{,}\qquad \htmlData{tutor-start=61,tutor-end=62}{S}_{\htmlData{tutor-start=64,tutor-end=65}{6}}\htmlData{tutor-start=66,tutor-end=67}{=}\htmlData{tutor-start=67,tutor-end=68}{3}\htmlData{tutor-start=68,tutor-end=69}{(}\htmlData{tutor-start=69,tutor-end=70}{1}\htmlData{tutor-start=70,tutor-end=71}{+}\htmlData{tutor-start=71,tutor-end=72}{4}\htmlData{tutor-start=72,tutor-end=73}{+}\htmlData{tutor-start=73,tutor-end=74}{1}\htmlData{tutor-start=74,tutor-end=75}{6}\htmlData{tutor-start=75,tutor-end=76}{)}\htmlData{tutor-start=76,tutor-end=77}{=}\htmlData{tutor-start=77,tutor-end=78}{6}\htmlData{tutor-start=78,tutor-end=79}{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：核验：若取 q=2，可令 a1+a2=3，三组两项和依次为 3、12、48，总和 63；若 q=-2，q^2 仍为 4，三组和不变。故选择 C。

S6=63C\boxed{\htmlData{tutor-start=7,tutor-end=8}{S}_{\htmlData{tutor-start=10,tutor-end=11}{6}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{6}\htmlData{tutor-start=14,tutor-end=15}{3}}\quad\boxed{\mathrm{\htmlData{tutor-start=36,tutor-end=37}{C}}}
9

一、选择题 · 椭圆的定义与标准方程

已知椭圆 C:x2a2+y2b2=1 (a>b>0)\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \frac{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}}\htmlData{tutor-start=22,tutor-end=23}{+}\frac{\htmlData{tutor-start=29,tutor-end=30}{y}^{\htmlData{tutor-start=32,tutor-end=33}{2}}}{\htmlData{tutor-start=36,tutor-end=37}{b}^{\htmlData{tutor-start=39,tutor-end=40}{2}}}\htmlData{tutor-start=42,tutor-end=43}{=}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=46}{\ }\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{a}\htmlData{tutor-start=48,tutor-end=49}{>}\htmlData{tutor-start=49,tutor-end=50}{b}\htmlData{tutor-start=50,tutor-end=51}{>}\htmlData{tutor-start=51,tutor-end=52}{0}\htmlData{tutor-start=52,tutor-end=53}{)} 的左、右焦点分别为 F1\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}、F2\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{2}}，离心率为 33\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{3}}，过 F2\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 交 C\htmlData{tutor-start=0,tutor-end=1}{C} 于 A\htmlData{tutor-start=0,tutor-end=1}{A}、B\htmlData{tutor-start=0,tutor-end=1}{B} 两点，若 △AF1B\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{F}_{\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{B} 的周长为 43\htmlData{tutor-start=0,tutor-end=1}{4}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{3}}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的方程为 ( )

答案：A，x^2/3+y^2/2=1。

题目标签：椭圆焦点弦与周长

解题过程

用椭圆定义化简三角形周长

由离心率与三角形周长确定 a、b。

(1)
识别结构并建立关系

直线经过焦点 F2 并与椭圆交于 A、B，焦点在弦 AB 内部。把 AB 拆成 AF2+BF2 后，三角形周长会出现两组椭圆定义中的焦距和。

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。“直线经过焦点 F2 并与椭圆交于 A、B，焦点在弦 AB 内部。把 AB 拆成 AF2+BF2 后，三角形周长会出现两组椭圆定义中的焦距和。”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：由于 F2 位于弦 AB 上，AB=AF2+BF2\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}{A}\htmlData{tutor-start=4,tutor-end=5}{F}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=10}{2}。于是三角形 AF1B 的周长等于 AF1+BF1+AF2+BF2=(AF1+AF2)+(BF1+BF2)=2a+2a=4a\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{F}\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}{F}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{F}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{F}\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}{A}\htmlData{tutor-start=18,tutor-end=19}{F}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{A}\htmlData{tutor-start=22,tutor-end=23}{F}\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}{(}\htmlData{tutor-start=27,tutor-end=28}{B}\htmlData{tutor-start=28,tutor-end=29}{F}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{B}\htmlData{tutor-start=32,tutor-end=33}{F}\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=35}{)}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{2}\htmlData{tutor-start=37,tutor-end=38}{a}\htmlData{tutor-start=38,tutor-end=39}{+}\htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=41}{a}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{4}\htmlData{tutor-start=43,tutor-end=44}{a}。题设周长为 43\htmlData{tutor-start=0,tutor-end=1}{4}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{3}}，所以 a=3\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}，a2=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}{3}。离心率 e=c/a=3/3\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}{=}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{3}，故 c=1\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，c2=1\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}。由 b2=a2−c2\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}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{c}^{\htmlData{tutor-start=15,tutor-end=16}{2}} 得 b2=2\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}{2}，因此椭圆方程为 x2/3+y2/2=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}{3}\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}{2}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{1}。

4a=43⇒a2=3,c=ae=1,b2=a2−c2=2\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{4}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=24}{\Rightarrow }\htmlData{tutor-start=24,tutor-end=25}{a}^{\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}{,}\qquad \htmlData{tutor-start=39,tutor-end=40}{c}\htmlData{tutor-start=40,tutor-end=41}{=}\htmlData{tutor-start=41,tutor-end=42}{a}\htmlData{tutor-start=42,tutor-end=43}{e}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{1}\htmlData{tutor-start=45,tutor-end=46}{,}\qquad \htmlData{tutor-start=53,tutor-end=54}{b}^{\htmlData{tutor-start=56,tutor-end=57}{2}}\htmlData{tutor-start=58,tutor-end=59}{=}\htmlData{tutor-start=59,tutor-end=60}{a}^{\htmlData{tutor-start=62,tutor-end=63}{2}}\htmlData{tutor-start=64,tutor-end=65}{-}\htmlData{tutor-start=65,tutor-end=66}{c}^{\htmlData{tutor-start=68,tutor-end=69}{2}}\htmlData{tutor-start=70,tutor-end=71}{=}\htmlData{tutor-start=71,tutor-end=72}{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：核验：方程 x2/3+y2/2=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}{3}\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}{2}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{1} 的焦距参数 c=3−2=1\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{1}，离心率 c/a=1/3=3/3\htmlData{tutor-start=0,tutor-end=1}{c}\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}{/}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{=}\sqrt{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{/}\htmlData{tutor-start=24,tutor-end=25}{3}；任一过焦点的弦所得上述三角形周长均为 4a=43\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{4}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}。故选择 A。

x23+y22=1A\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{x}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}{\htmlData{tutor-start=20,tutor-end=21}{3}}\htmlData{tutor-start=22,tutor-end=23}{+}\frac{\htmlData{tutor-start=29,tutor-end=30}{y}^{\htmlData{tutor-start=32,tutor-end=33}{2}}}{\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=39}{=}\htmlData{tutor-start=39,tutor-end=40}{1}}\quad\boxed{\mathrm{\htmlData{tutor-start=61,tutor-end=62}{A}}}
10

一、选择题 · 球与空间几何体

正四棱锥的顶点都在同一球面上，若该棱锥的高为 4，底面边长为 2，则该球的表面积为 ( )

答案：A，81pi/4。

题目标签：正四棱锥外接球

解题过程

在棱锥轴线上确定球心

由球心到底面顶点和棱锥顶点等距求半径。

(1)
识别结构并建立关系

正四棱锥关于高所在直线旋转对称，所以外接球球心必在这条高上。设球心到底面中心的有向距离，再列两种半径平方相等。

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。“正四棱锥关于高所在直线旋转对称，所以外接球球心必在这条高上。设球心到底面中心的有向距离，再列两种半径平方相等。”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：设底面中心为 O，棱锥顶点为 V，球心在 VO 上，且球心到底面平面的距离为 x\htmlData{tutor-start=0,tutor-end=1}{x}。底面边长为 2，所以 O 到任一底面顶点的距离是底面对角线的一半，即 2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}。球半径平方一方面为 R2=x2+2\htmlData{tutor-start=0,tutor-end=1}{R}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}，另一方面为顶点到球心距离平方 R2=(4−x)2\htmlData{tutor-start=0,tutor-end=1}{R}^{\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}{4}\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=13,tutor-end=14}{2}}。令二者相等，得 x2+2=16−8x+x2\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}{2}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{6}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{8}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{x}^{\htmlData{tutor-start=17,tutor-end=18}{2}}，所以 x=7/4\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{7}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{4}，R2=49/16+2=81/16\htmlData{tutor-start=0,tutor-end=1}{R}^{\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}{9}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{6}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{8}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{6}。球表面积为 4πR2=81π/4\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=5}{\pi }\htmlData{tutor-start=5,tutor-end=6}{R}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{8}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=16}{\pi}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{4}。

x2+2=(4−x)2⇒x=74,R2=8116,4πR2=81π4\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}{2}\htmlData{tutor-start=7,tutor-end=8}{=}\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=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=29}{\Rightarrow }\htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{=}\frac{\htmlData{tutor-start=37,tutor-end=38}{7}}{\htmlData{tutor-start=40,tutor-end=41}{4}}\htmlData{tutor-start=42,tutor-end=43}{,}\qquad \htmlData{tutor-start=50,tutor-end=51}{R}^{\htmlData{tutor-start=53,tutor-end=54}{2}}\htmlData{tutor-start=55,tutor-end=56}{=}\frac{\htmlData{tutor-start=62,tutor-end=63}{8}\htmlData{tutor-start=63,tutor-end=64}{1}}{\htmlData{tutor-start=66,tutor-end=67}{1}\htmlData{tutor-start=67,tutor-end=68}{6}}\htmlData{tutor-start=69,tutor-end=70}{,}\qquad \htmlData{tutor-start=77,tutor-end=78}{4}\htmlData{tutor-start=78,tutor-end=82}{\pi }\htmlData{tutor-start=82,tutor-end=83}{R}^{\htmlData{tutor-start=85,tutor-end=86}{2}}\htmlData{tutor-start=87,tutor-end=88}{=}\frac{\htmlData{tutor-start=94,tutor-end=95}{8}\htmlData{tutor-start=95,tutor-end=96}{1}\htmlData{tutor-start=96,tutor-end=99}{\pi}}{\htmlData{tutor-start=101,tutor-end=102}{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：核验：球心到底面顶点的距离平方为 (7/4)^2+2=81/16，到棱锥顶点的距离为 4-7/4=9/4，平方也为 81/16，五个顶点确实在同一球面上。故选择 A。

S球=81π4A\boxed{\htmlData{tutor-start=7,tutor-end=8}{S}_{\text{\htmlData{tutor-start=16,tutor-end=17}{球}}}\htmlData{tutor-start=19,tutor-end=20}{=}\frac{\htmlData{tutor-start=26,tutor-end=27}{8}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=31}{\pi}}{\htmlData{tutor-start=33,tutor-end=34}{4}}}\quad\boxed{\mathrm{\htmlData{tutor-start=56,tutor-end=57}{A}}}
11

一、选择题 · 双曲线的几何性质

双曲线 C:x2a2−y2b2=1 (a>0,b>0)\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \frac{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}}\htmlData{tutor-start=22,tutor-end=23}{-}\frac{\htmlData{tutor-start=29,tutor-end=30}{y}^{\htmlData{tutor-start=32,tutor-end=33}{2}}}{\htmlData{tutor-start=36,tutor-end=37}{b}^{\htmlData{tutor-start=39,tutor-end=40}{2}}}\htmlData{tutor-start=42,tutor-end=43}{=}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=46}{\ }\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{a}\htmlData{tutor-start=48,tutor-end=49}{>}\htmlData{tutor-start=49,tutor-end=50}{0}\htmlData{tutor-start=50,tutor-end=51}{,} \htmlData{tutor-start=52,tutor-end=53}{b}\htmlData{tutor-start=53,tutor-end=54}{>}\htmlData{tutor-start=54,tutor-end=55}{0}\htmlData{tutor-start=55,tutor-end=56}{)} 的离心率为 2，焦点到渐近线的距离为 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的焦距等于 ( )

答案：C，4。

题目标签：双曲线焦距

解题过程

用焦点到渐近线距离确定参数

由离心率和点到直线距离求 c。

(1)
识别结构并建立关系

双曲线的渐近线可写成 bx-ay=0。把焦点 (c,0) 代入点到直线距离公式后，利用 c^2=a^2+b^2，距离会直接化为 b。

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。“双曲线的渐近线可写成 bx-ay=0。把焦点 (c,0) 代入点到直线距离公式后，利用 c^2=a^2+b^2，距离会直接化为 b。”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：离心率 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}，所以 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}。又 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}}，代入得 4a2=a2+b2\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}{b}^{\htmlData{tutor-start=16,tutor-end=17}{2}}，从而 b2=3a2\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}\htmlData{tutor-start=7,tutor-end=8}{a}^{\htmlData{tutor-start=10,tutor-end=11}{2}}。渐近线 bx−ay=0\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，焦点 (c,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}{0}\htmlData{tutor-start=4,tutor-end=5}{)} 到它的距离为 bc/a2+b2=bc/c=b\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{/}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{b}^{\htmlData{tutor-start=18,tutor-end=19}{2}}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{b}\htmlData{tutor-start=23,tutor-end=24}{c}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{c}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{b}。题设该距离为 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}，因此 b=3\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}。由 b2=3a2\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}\htmlData{tutor-start=7,tutor-end=8}{a}^{\htmlData{tutor-start=10,tutor-end=11}{2}} 得 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}，进而 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}，焦距为两焦点间距离 2c=4\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{4}。

c=2a,b2=c2−a2=3a2,d((c,0),bx−ay=0)=b=3⇒a=1,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}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{,}\quad \htmlData{tutor-start=11,tutor-end=12}{b}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{c}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{a}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{3}\htmlData{tutor-start=30,tutor-end=31}{a}^{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=36}{,}\quad \htmlData{tutor-start=42,tutor-end=43}{d}\htmlData{tutor-start=43,tutor-end=44}{(}\htmlData{tutor-start=44,tutor-end=45}{(}\htmlData{tutor-start=45,tutor-end=46}{c}\htmlData{tutor-start=46,tutor-end=47}{,}\htmlData{tutor-start=47,tutor-end=48}{0}\htmlData{tutor-start=48,tutor-end=49}{)}\htmlData{tutor-start=49,tutor-end=50}{,}\htmlData{tutor-start=50,tutor-end=51}{b}\htmlData{tutor-start=51,tutor-end=52}{x}\htmlData{tutor-start=52,tutor-end=53}{-}\htmlData{tutor-start=53,tutor-end=54}{a}\htmlData{tutor-start=54,tutor-end=55}{y}\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}{=}\htmlData{tutor-start=59,tutor-end=60}{b}\htmlData{tutor-start=60,tutor-end=61}{=}\sqrt{\htmlData{tutor-start=67,tutor-end=68}{3}}\htmlData{tutor-start=69,tutor-end=81}{\Rightarrow }\htmlData{tutor-start=81,tutor-end=82}{a}\htmlData{tutor-start=82,tutor-end=83}{=}\htmlData{tutor-start=83,tutor-end=84}{1}\htmlData{tutor-start=84,tutor-end=85}{,}\htmlData{tutor-start=85,tutor-end=86}{c}\htmlData{tutor-start=86,tutor-end=87}{=}\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：核验：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}、b=3\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}} 时 c=1+3=2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{2}，离心率为 2；焦点到渐近线的距离确为 b=3\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}。题目问焦距而不是半焦距，所以答案是 2c=4\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{c}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{4}。

2c=4C\boxed{\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{4}}\quad\boxed{\mathrm{\htmlData{tutor-start=32,tutor-end=33}{C}}}
12

一、选择题 · 函数的奇偶性和周期性

奇函数 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 的定义域为 R\mathbf{\htmlData{tutor-start=8,tutor-end=9}{R}}，若 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=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)} 为偶函数，f(1)=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}{1}，则 f(8)+f(9)=\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}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{9}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=11}{=} ( )

答案：D，1。

题目标签：奇偶性与对称性求函数值

解题过程

由两种对称性推出周期

把 f(8)、f(9) 化为已知的 f(0)、f(1)。

(1)
识别结构并建立关系

f(x+2) 为偶函数不是说 f 本身关于 y 轴对称，而是说 f 的图象关于直线 x=2 对称。再与奇函数关于原点对称结合，通常能推出平移周期。

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。“f(x+2) 为偶函数不是说 f 本身关于 y 轴对称，而是说 f 的图象关于直线 x=2 对称。再与奇函数关于原点对称结合，通常能推出平移周期。”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：令 g(x)=f(x+2)。g 为偶函数，所以 g(x)=g(-x)，即 f(x+2)=f(2-x)。把自变量改写可得 f(t)=f(4-t)。因此 f(x+4)=f(-x)。又 f 是奇函数，f(-x)=-f(x)，故 f(x+4)=-f(x)，再平移 4 得 f(x+8)=f(x)，即 8 是一个周期。奇函数满足 f(0)=0，所以 f(8)=f(0)=0；又 f(9)=f(1)=1。

f(x+2)=f(2−x)⇒f(x+4)=f(−x)=−f(x)⇒f(x+8)=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}{+}\htmlData{tutor-start=4,tutor-end=5}{2}\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}{2}\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=25}{\Rightarrow }\htmlData{tutor-start=25,tutor-end=26}{f}\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}{4}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{f}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{x}\htmlData{tutor-start=36,tutor-end=37}{)}\htmlData{tutor-start=37,tutor-end=38}{=}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{f}\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{x}\htmlData{tutor-start=42,tutor-end=43}{)}\htmlData{tutor-start=43,tutor-end=55}{\Rightarrow }\htmlData{tutor-start=55,tutor-end=56}{f}\htmlData{tutor-start=56,tutor-end=57}{(}\htmlData{tutor-start=57,tutor-end=58}{x}\htmlData{tutor-start=58,tutor-end=59}{+}\htmlData{tutor-start=59,tutor-end=60}{8}\htmlData{tutor-start=60,tutor-end=61}{)}\htmlData{tutor-start=61,tutor-end=62}{=}\htmlData{tutor-start=62,tutor-end=63}{f}\htmlData{tutor-start=63,tutor-end=64}{(}\htmlData{tutor-start=64,tutor-end=65}{x}\htmlData{tutor-start=65,tutor-end=66}{)}
(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：由 f(x+4)=-f(x) 直接得到 f(8)=-f(4)=f(0)=0，f(9)=-f(5)=f(1)=1，与周期结论一致。因此和为 1，选择 D。

f(8)+f(9)=0+1=1D\boxed{\htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{8}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{f}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{9}\htmlData{tutor-start=15,tutor-end=16}{)}\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}{1}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{1}}\quad\boxed{\mathrm{\htmlData{tutor-start=43,tutor-end=44}{D}}}
13

二、填空题 · 二项式定理

(x−2)6\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=7,tutor-end=8}{6}} 的展开式中 x3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} 的系数为______.（用数字作答）

答案：-160。

题目标签：二项展开式指定项系数

解题过程

确定产生 x 三次项的通项

求展开式中 x^3 的系数。

(1)
识别结构并建立关系

二项式展开中每选一次常数 -2，x 的次数就降低 1。先令通项中的 x 次数等于 3，再计算组合数和符号。

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。“二项式展开中每选一次常数 -2，x 的次数就降低 1。先令通项中的 x 次数等于 3，再计算组合数和符号。”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：展开式通项可写为 Tk+1=(6k)x6−k(−2)kT_{k+1}=\binom6k x^{6-k}(-2)^{k}。要得到 x3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}}，必须有 6−k=3\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3}，所以 k=3\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}。此时系数为 (63)(−2)3=20×(−8)=−160\binom63(-2)^{3}=20\times(-8)=-160。负号来自 -2 的奇数次幂，不能只计算绝对值。

Tk+1=(6k)x6−k(−2)k,6−k=3⇒k=3,(63)(−2)3=−160T_{k+1}=\binom6k x^{6-k}(-2)^{k},\quad 6-k=3\Rightarrow k=3,\quad \binom63(-2)^{3}=-160
(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：对应项确为 -160x^3；相邻的 k=2、4 分别产生 x^4、x^2，不会再对 x^3 项作贡献。因此系数唯一且为 -160。

−160\boxed{\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{6}\htmlData{tutor-start=10,tutor-end=11}{0}}
14

二、填空题 · 三角恒等变换与二次函数

函数 y=cos⁡2x+2sin⁡x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\cos \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{2}\sin \htmlData{tutor-start=16,tutor-end=17}{x} 的最大值为______.

答案：3/2。

题目标签：三角函数式的最大值

解题过程

化成关于 sin x 的完全平方

求 cos2x+2sinx 的全局最大值。

(1)
识别结构并建立关系

式中同时出现 cos⁡2x\cos\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{x} 和 sin⁡x\sin \htmlData{tutor-start=5,tutor-end=6}{x}，可用 cos⁡2x=1−2sin⁡2x\cos\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{2}\sin^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{x} 把它们统一成同一个变量，再配方判断最大值。

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。“式中同时出现 cos⁡2x\cos\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{x} 和 sin⁡x\sin \htmlData{tutor-start=5,tutor-end=6}{x}，可用 cos⁡2x=1−2sin⁡2x\cos\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{2}\sin^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{x} 把它们统一成同一个变量，再配方判断最大值。”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：令 t=sin⁡x\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\sin \htmlData{tutor-start=7,tutor-end=8}{x}，则 −1≤t≤1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{t}\htmlData{tutor-start=7,tutor-end=10}{\le}\htmlData{tutor-start=10,tutor-end=11}{1}。原式 y=1−2t2+2t\htmlData{tutor-start=0,tutor-end=1}{y}\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}{t}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{t}。配方得 y=3/2−2(t−1/2)2\htmlData{tutor-start=0,tutor-end=1}{y}\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}{2}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{t}\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}{2}\htmlData{tutor-start=13,tutor-end=14}{)}^{\htmlData{tutor-start=16,tutor-end=17}{2}}。平方项总不小于 0，因此 y≤3/2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}；当 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} 时等号成立。因为确实存在 x\htmlData{tutor-start=0,tutor-end=1}{x} 使 sin⁡x=1/2\sin \htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}，例如 x=π/6\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{6}，所以这个上界能够取得。

y=1−2sin⁡2x+2sin⁡x=32−2(sin⁡x−12)2≤32\htmlData{tutor-start=0,tutor-end=1}{y}\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}\sin^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}\sin \htmlData{tutor-start=21,tutor-end=22}{x}\htmlData{tutor-start=22,tutor-end=23}{=}\frac{\htmlData{tutor-start=29,tutor-end=30}{3}}{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{2}\left(\sin \htmlData{tutor-start=47,tutor-end=48}{x}\htmlData{tutor-start=48,tutor-end=49}{-}\frac{\htmlData{tutor-start=55,tutor-end=56}{1}}{\htmlData{tutor-start=58,tutor-end=59}{2}}\right)^{\htmlData{tutor-start=69,tutor-end=70}{2}}\htmlData{tutor-start=71,tutor-end=74}{\le}\frac{\htmlData{tutor-start=80,tutor-end=81}{3}}{\htmlData{tutor-start=83,tutor-end=84}{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：核验：取 x=π/6\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{6}，cos⁡2x=cos⁡(π/3)=1/2\cos\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{=}\cos\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=15}{\pi}\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}{=}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{/}\htmlData{tutor-start=21,tutor-end=22}{2}，2sin⁡x=1\htmlData{tutor-start=0,tutor-end=1}{2}\sin \htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}，两项之和为 3/2，说明不是仅有上界而是实际最大值。

ymax⁡=32\boxed{\htmlData{tutor-start=7,tutor-end=8}{y}_{\max}\htmlData{tutor-start=15,tutor-end=16}{=}\frac{\htmlData{tutor-start=22,tutor-end=23}{3}}{\htmlData{tutor-start=25,tutor-end=26}{2}}}
15

二、填空题 · 简单线性规划

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⩾0,x+2y⩽3,x−2y⩽1,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{y} \htmlData{tutor-start=18,tutor-end=28}{\geqslant }\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{,} \\ \htmlData{tutor-start=34,tutor-end=35}{x}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{2}\htmlData{tutor-start=37,tutor-end=38}{y} \htmlData{tutor-start=39,tutor-end=49}{\leqslant }\htmlData{tutor-start=49,tutor-end=50}{3}\htmlData{tutor-start=50,tutor-end=51}{,} \\ \htmlData{tutor-start=55,tutor-end=56}{x}\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{2}\htmlData{tutor-start=58,tutor-end=59}{y} \htmlData{tutor-start=60,tutor-end=70}{\leqslant }\htmlData{tutor-start=70,tutor-end=71}{1}\htmlData{tutor-start=71,tutor-end=72}{,} \end{cases} 则 z=x+4y\htmlData{tutor-start=0,tutor-end=1}{z}\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}{4}\htmlData{tutor-start=5,tutor-end=6}{y} 的最大值为______.

答案：5。

题目标签：线性规划求最大值

解题过程

求可行域顶点并逐点代入

在约束条件下求 z=x+4y 的最大值。

(1)
识别结构并建立关系

目标函数和约束都是一次式，最大值若存在必在可行多边形的顶点取得。先把三条边界直线两两联立，再判断顶点。

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。“目标函数和约束都是一次式，最大值若存在必在可行多边形的顶点取得。先把三条边界直线两两联立，再判断顶点。”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：三条边界为 y=x、x+2y=3、x-2y=1。前两条相交于 (1,1)；第一、三条相交于 (-1,-1)；后两条相交于 (2,1/2)。这三个点都满足全部不等式，构成可行三角形。分别计算 z：在 (1,1) 处 z=5，在 (-1,-1) 处 z=-5，在 (2,1/2) 处 z=4。比较可得最大值为 5。

z(1,1)=5,z(−1,−1)=−5,z(2,12)=4\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}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{,}\qquad \htmlData{tutor-start=16,tutor-end=17}{z}\htmlData{tutor-start=17,tutor-end=18}{(}\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}{1}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{5}\htmlData{tutor-start=27,tutor-end=28}{,}\qquad \htmlData{tutor-start=35,tutor-end=36}{z}\left(\htmlData{tutor-start=42,tutor-end=43}{2}\htmlData{tutor-start=43,tutor-end=44}{,}\frac{\htmlData{tutor-start=50,tutor-end=51}{1}}{\htmlData{tutor-start=53,tutor-end=54}{2}}\right)\htmlData{tutor-start=62,tutor-end=63}{=}\htmlData{tutor-start=63,tutor-end=64}{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：核验：把直线 x+4y=z 平行移动，z 增大时最后接触可行域的点是 (1,1)，与顶点代入结果一致。因此最大值 5 可以取得。

zmax⁡=5\boxed{\htmlData{tutor-start=7,tutor-end=8}{z}_{\max}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{5}}
16

二、填空题 · 直线与圆

直线 l1\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 和 l2\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 是圆 x2+y2=2\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}{2} 的两条切线，若 l1\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 与 l2\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的交点为 (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=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{)}，则 l1\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 与 l2\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的夹角的正切值等于______.

答案：4/3。

题目标签：圆外一点两切线夹角

解题过程

利用切点直角三角形求半角

求从点 (1,3) 引圆的两条切线的夹角正切。

(1)
识别结构并建立关系

两条切线关于圆心与外点的连线对称，因此连接圆心和一个切点后，可先求夹角的一半，再用正切二倍角公式。

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。“两条切线关于圆心与外点的连线对称，因此连接圆心和一个切点后，可先求夹角的一半，再用正切二倍角公式。”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：圆心 O=(0,0)\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}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}，半径 r=2\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}，外点 P=(1,3)\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}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{)}，所以 OP=10\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{=}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0}}。设切点为 T，两切线的较小夹角为 θ\htmlData{tutor-start=0,tutor-end=6}{\theta}，则 OP 平分该角，且 OT⊥PT\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{T}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{T}。在直角三角形 OPT 中，PT=OP2−OT2=10−2=22\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{T}\htmlData{tutor-start=2,tutor-end=3}{=}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{O}\htmlData{tutor-start=10,tutor-end=11}{P}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{O}\htmlData{tutor-start=17,tutor-end=18}{T}^{\htmlData{tutor-start=20,tutor-end=21}{2}}}\htmlData{tutor-start=23,tutor-end=24}{=}\sqrt{\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{0}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{2}\sqrt{\htmlData{tutor-start=43,tutor-end=44}{2}}，所以 tan⁡(θ/2)=OT/PT=1/2\tan\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=11}{\theta}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{O}\htmlData{tutor-start=16,tutor-end=17}{T}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{P}\htmlData{tutor-start=19,tutor-end=20}{T}\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}。由二倍角公式，tan⁡θ=2(1/2)/(1−(1/2)2)=4/3\tan\htmlData{tutor-start=4,tutor-end=10}{\theta}\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}{1}\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}{(}\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}{1}\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=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{4}\htmlData{tutor-start=33,tutor-end=34}{/}\htmlData{tutor-start=34,tutor-end=35}{3}。

OP=10,PT=22,tan⁡θ2=12,tan⁡theta=2tan⁡(theta/2)1−tan⁡2(theta/2)=43OP=\sqrt{10},\quad PT=2\sqrt{2},\quad \tan\frac\theta2=\frac{1}{2},\quad \tan\\theta=\frac{2\tan(\\theta/2)}{1-\tan^{2}(\\theta/2)}=\frac{4}{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：核验：sin⁡(θ/2)=r/OP=1/5\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=11}{\theta}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{r}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{O}\htmlData{tutor-start=18,tutor-end=19}{P}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{/}\sqrt{\htmlData{tutor-start=28,tutor-end=29}{5}}，故 cos⁡(θ/2)=2/5\cos\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=11}{\theta}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{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}{/}\sqrt{\htmlData{tutor-start=23,tutor-end=24}{5}}，仍得 tan⁡(θ/2)=1/2\tan\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=11}{\theta}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{2}\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}{/}\htmlData{tutor-start=17,tutor-end=18}{2}。由于半角小于 45 度，较小夹角为锐角，正切取正值 4/3。

tan⁡theta=43\boxed{\tan\\\htmlData{tutor-start=13,tutor-end=14}{t}\htmlData{tutor-start=14,tutor-end=15}{h}\htmlData{tutor-start=15,tutor-end=16}{e}\htmlData{tutor-start=16,tutor-end=17}{t}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{=}\frac{\htmlData{tutor-start=25,tutor-end=26}{4}}{\htmlData{tutor-start=28,tutor-end=29}{3}}}
17

三、解答题 · 数列递推与等差数列

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}{\}} 满足 a1=1,a2=2,an+2=2an+1−an+2\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}{1}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{2}}\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=18,tutor-end=19}{a}_{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{a}_{\htmlData{tutor-start=30,tutor-end=31}{n}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{1}}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{a}_{\htmlData{tutor-start=38,tutor-end=39}{n}}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{2}. (1) 设 bn=an+1−an\htmlData{tutor-start=0,tutor-end=1}{b}_{\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}{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}{a}_{\htmlData{tutor-start=17,tutor-end=18}{n}}，证明 {bn}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{b}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 是等差数列； (2) 求数列 {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}{\}} 的通项公式.

答案：(1) b_n 是首项1、公差2的等差数列；(2) a_n=n^2-2n+2。

题目标签：递推数列作差降阶

解题过程

（1）证明一阶差数列为等差数列

由原递推式推出 b_{n+1}-b_n 为常数。

(1)
识别结构并建立关系

原递推式同时含 a_{n+2}、a_{n+1}、a_n，而 b_n 正是相邻两项之差。把递推式移项成两个相邻差的关系即可降阶。

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。“原递推式同时含 a_{n+2}、a_{n+1}、a_n，而 b_n 正是相邻两项之差。把递推式移项成两个相邻差的关系即可降阶。”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：由定义 b_{n+1}=a_{n+2}-a_{n+1}。将递推式 a_{n+2}=2a_{n+1}-a_n+2 代入，得到 b_{n+1}=a_{n+1}-a_n+2=b_n+2，所以 b_{n+1}-b_n=2 对所有正整数 n 成立。又 b_1=a_2-a_1=1，因此 {b_n} 是首项为 1、公差为 2 的等差数列，并有 b_n=1+2(n-1)=2n-1。

bn+1=an+2−an+1=an+1−an+2=bn+2,b1=1\htmlData{tutor-start=0,tutor-end=1}{b}_{\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=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{1}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{n}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{1}}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{a}_{\htmlData{tutor-start=35,tutor-end=36}{n}}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{b}_{\htmlData{tutor-start=43,tutor-end=44}{n}}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{,}\qquad \htmlData{tutor-start=55,tutor-end=56}{b}_{\htmlData{tutor-start=58,tutor-end=59}{1}}\htmlData{tutor-start=60,tutor-end=61}{=}\htmlData{tutor-start=61,tutor-end=62}{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：核验：由递推式先算 a3=2a2-a1+2=5，因此 b2=a3-a2=3，而 b2-b1=3-1=2，与所证公差一致。

bn=2n−1，故 {bn}是等差数列\boxed{\htmlData{tutor-start=7,tutor-end=8}{b}_{\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}{n}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}\text{\htmlData{tutor-start=23,tutor-end=24}{，}\htmlData{tutor-start=24,tutor-end=25}{故} }\htmlData{tutor-start=27,tutor-end=29}{\{}\htmlData{tutor-start=29,tutor-end=30}{b}_{\htmlData{tutor-start=32,tutor-end=33}{n}}\htmlData{tutor-start=34,tutor-end=36}{\}}\text{\htmlData{tutor-start=42,tutor-end=43}{是}\htmlData{tutor-start=43,tutor-end=44}{等}\htmlData{tutor-start=44,tutor-end=45}{差}\htmlData{tutor-start=45,tutor-end=46}{数}\htmlData{tutor-start=46,tutor-end=47}{列}}}

（2）累加相邻差求通项

利用 a_n-a_1 的裂项和求 a_n。

(1)
识别结构并建立关系

已经知道 a_{k+1}-a_k=b_k，要求 a_n 时应把 k=1 到 n-1 的相邻差相加，中间项会全部消去。

为什么从这里入手：目标是“识别结构并建立关系”，而“已经知道 a_{k+1}-a_k=b_k，要求 a_n 时应把 k=1 到 n-1 的相邻差相加，中间项会全部消去。”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 b_k=a_{k+1}-a_k=2k-1，将 k=1,2,...,n-1 的等式相加，左边得到 a_n-a_1，右边是前 n-1 个正奇数之和。前 m 个正奇数之和为 m^2，因此 a_n-a_1=(n-1)^2。代入 a_1=1，得 a_n=1+(n-1)^2=n^2-2n+2。

an−a1=∑k=1n−1(ak+1−ak)=∑k=1n−1(2k−1)=(n−1)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}{a}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{=}\sum_{\htmlData{tutor-start=18,tutor-end=19}{k}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{1}}^{\htmlData{tutor-start=24,tutor-end=25}{n}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{1}}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{a}_{\htmlData{tutor-start=32,tutor-end=33}{k}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{1}}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{a}_{\htmlData{tutor-start=40,tutor-end=41}{k}}\htmlData{tutor-start=42,tutor-end=43}{)}\htmlData{tutor-start=43,tutor-end=44}{=}\sum_{\htmlData{tutor-start=50,tutor-end=51}{k}\htmlData{tutor-start=51,tutor-end=52}{=}\htmlData{tutor-start=52,tutor-end=53}{1}}^{\htmlData{tutor-start=56,tutor-end=57}{n}\htmlData{tutor-start=57,tutor-end=58}{-}\htmlData{tutor-start=58,tutor-end=59}{1}}\htmlData{tutor-start=60,tutor-end=61}{(}\htmlData{tutor-start=61,tutor-end=62}{2}\htmlData{tutor-start=62,tutor-end=63}{k}\htmlData{tutor-start=63,tutor-end=64}{-}\htmlData{tutor-start=64,tutor-end=65}{1}\htmlData{tutor-start=65,tutor-end=66}{)}\htmlData{tutor-start=66,tutor-end=67}{=}\htmlData{tutor-start=67,tutor-end=68}{(}\htmlData{tutor-start=68,tutor-end=69}{n}\htmlData{tutor-start=69,tutor-end=70}{-}\htmlData{tutor-start=70,tutor-end=71}{1}\htmlData{tutor-start=71,tutor-end=72}{)}^{\htmlData{tutor-start=74,tutor-end=75}{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=1 得 a1=1，代入 n=2 得 a2=2；并且把通项代入原递推式，两边都化为 n^2+2，说明初值和递推关系同时满足，通项正确。

an=n2−2n+2\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{n}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{n}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{2}}
18

三、解答题 · 正弦定理与和角公式

△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} 的内角 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} 的对边分别为 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}，已知 3acos⁡C=2ccos⁡A\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{a}\cos \htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{c}\cos \htmlData{tutor-start=16,tutor-end=17}{A}，tan⁡A=13\tan \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{=}\frac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{3}}，求 B\htmlData{tutor-start=0,tutor-end=1}{B}.

答案：B=3pi/4。

题目标签：三角恒等关系求角

解题过程

把边角关系化为正切关系

由已知等式和 tanA 求角 B。

(1)
识别结构并建立关系

等式中边 a、c 分别与角 A、C 对应，可用正弦定理把边之比换成 sinA、sinC，从而只剩角的关系。

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。“等式中边 a、c 分别与角 A、C 对应，可用正弦定理把边之比换成 sinA、sinC，从而只剩角的关系。”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：由正弦定理 a/c=sin⁡A/sin⁡C\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{=}\sin \htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{/}\sin \htmlData{tutor-start=16,tutor-end=17}{C}，把 3acos⁡C=2ccos⁡A\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{a}\cos \htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{c}\cos \htmlData{tutor-start=16,tutor-end=17}{A} 同除以 ccos⁡Acos⁡C\htmlData{tutor-start=0,tutor-end=1}{c}\cos \htmlData{tutor-start=6,tutor-end=7}{A}\cos \htmlData{tutor-start=12,tutor-end=13}{C}，可得 3tan⁡A=2tan⁡C\htmlData{tutor-start=0,tutor-end=1}{3}\tan \htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{2}\tan \htmlData{tutor-start=14,tutor-end=15}{C}。已知 tan⁡A=1/3\tan \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{3}，所以 tan⁡C=1/2\tan \htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}。三角形内 A、C 均在 (0,π)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=6}{\pi}\htmlData{tutor-start=6,tutor-end=7}{)}，正切为正说明二者都是锐角。于是 tan⁡(A+C)=(1/3+1/2)/(1−1/6)=1\tan\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}{C}\htmlData{tutor-start=8,tutor-end=9}{)}\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}{3}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}\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}{(}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{6}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{1}。又 0<A+C<π\htmlData{tutor-start=0,tutor-end=1}{0}\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}{C}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=9}{\pi}，且 A+C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{C} 为锐角，所以 A+C=π/4\htmlData{tutor-start=0,tutor-end=1}{A}\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=7}{\pi}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{4}，故 B=π−(A+C)=3π/4\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pi}\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}{C}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=16}{\pi}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{4}。

3sin⁡Acos⁡C=2sin⁡Ccos⁡A⇒3tan⁡A=2tan⁡C⇒tan⁡C=12,tan⁡(A+C)=1\htmlData{tutor-start=0,tutor-end=1}{3}\sin \htmlData{tutor-start=6,tutor-end=7}{A}\cos \htmlData{tutor-start=12,tutor-end=13}{C}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{2}\sin \htmlData{tutor-start=20,tutor-end=21}{C}\cos \htmlData{tutor-start=26,tutor-end=27}{A}\htmlData{tutor-start=27,tutor-end=39}{\Rightarrow }\htmlData{tutor-start=39,tutor-end=40}{3}\tan \htmlData{tutor-start=45,tutor-end=46}{A}\htmlData{tutor-start=46,tutor-end=47}{=}\htmlData{tutor-start=47,tutor-end=48}{2}\tan \htmlData{tutor-start=53,tutor-end=54}{C}\htmlData{tutor-start=54,tutor-end=66}{\Rightarrow }\tan \htmlData{tutor-start=71,tutor-end=72}{C}\htmlData{tutor-start=72,tutor-end=73}{=}\frac{\htmlData{tutor-start=79,tutor-end=80}{1}}{\htmlData{tutor-start=82,tutor-end=83}{2}}\htmlData{tutor-start=84,tutor-end=85}{,}\quad \tan\htmlData{tutor-start=95,tutor-end=96}{(}\htmlData{tutor-start=96,tutor-end=97}{A}\htmlData{tutor-start=97,tutor-end=98}{+}\htmlData{tutor-start=98,tutor-end=99}{C}\htmlData{tutor-start=99,tutor-end=100}{)}\htmlData{tutor-start=100,tutor-end=101}{=}\htmlData{tutor-start=101,tutor-end=102}{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：核验：取 tanA=1/3、tanC=1/2 时，tan(A+C)=1 且 A、C 均为锐角，所以 A+C=45 度；三个内角之和为 180 度，B=135 度，满足三角形角度范围。

B=3π4\boxed{\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{=}\frac{\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=19}{\pi}}{\htmlData{tutor-start=21,tutor-end=22}{4}}}
19

三、解答题 · 空间向量与立体几何

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，点 A1\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 在平面 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} 内的射影 D\htmlData{tutor-start=0,tutor-end=1}{D} 在 AC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} 上，∠ACB=90∘,BC=1,AC=CC1=2\angle ACB=90^\circ, BC=1, AC=CC_{1}=2. (1) 证明：AC1⊥A1B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=13}{\perp }\htmlData{tutor-start=13,tutor-end=14}{A}_{\htmlData{tutor-start=16,tutor-end=17}{1}}\htmlData{tutor-start=18,tutor-end=19}{B}； (2) 设直线 AA1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 与平面 BCC1B1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{C}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{B}_{\htmlData{tutor-start=10,tutor-end=11}{1}} 的距离为 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}，求二面角 A1−AB−C\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}{A}\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{C} 的大小.

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) AC1 垂直 A1B；(2) 二面角为 arccos(1/4)。

题目标签：斜三棱柱中的垂直与二面角

解题过程

（1）建立与投影相适配的坐标系证明垂直

用方向向量点积证明 AC1 垂直 A1B。

(1)
识别结构并建立关系

图中 A1 到底面的垂足 D 在 AC 上，且角 ACB 为直角。以 C 为原点、CA 和 CB 为坐标轴，可把 D、A1 都写成只含两个参数的简单坐标。

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。“图中 A1 到底面的垂足 D 在 AC 上，且角 ACB 为直角。以 C 为原点、CA 和 CB 为坐标轴，可把 D、A1 都写成只含两个参数的简单坐标。”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：令底面 ABC 为 z=0，取 C(0,0,0)、A(2,0,0)、B(0,1,0)。设 D(d,0,0)，A1(d,0,h)，h>0。棱柱的侧棱向量 v=AA1=(d-2,0,h)，且 |v|=CC1=2，所以 (d-2)^2+h^2=4，即 d^2-4d+h^2=0。又 C1=C+v=(d-2,0,h)。于是 AC1=(d-4,0,h)，A1B=(-d,1,-h)，二者点积为 -d(d-4)-h^2=-(d^2-4d+h^2)=0，故两直线垂直。

AC1→=(d−4,0,h),A1B→=(−d,1,−h),AC1→⋅A1B→=0\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{C}_{\htmlData{tutor-start=20,tutor-end=21}{1}}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{d}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{4}\htmlData{tutor-start=28,tutor-end=29}{,}\htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=31}{,}\htmlData{tutor-start=31,tutor-end=32}{h}\htmlData{tutor-start=32,tutor-end=33}{)}\htmlData{tutor-start=33,tutor-end=34}{,}\quad \overrightarrow{\htmlData{tutor-start=56,tutor-end=57}{A}_{\htmlData{tutor-start=59,tutor-end=60}{1}}\htmlData{tutor-start=61,tutor-end=62}{B}}\htmlData{tutor-start=63,tutor-end=64}{=}\htmlData{tutor-start=64,tutor-end=65}{(}\htmlData{tutor-start=65,tutor-end=66}{-}\htmlData{tutor-start=66,tutor-end=67}{d}\htmlData{tutor-start=67,tutor-end=68}{,}\htmlData{tutor-start=68,tutor-end=69}{1}\htmlData{tutor-start=69,tutor-end=70}{,}\htmlData{tutor-start=70,tutor-end=71}{-}\htmlData{tutor-start=71,tutor-end=72}{h}\htmlData{tutor-start=72,tutor-end=73}{)}\htmlData{tutor-start=73,tutor-end=74}{,}\quad \overrightarrow{\htmlData{tutor-start=96,tutor-end=97}{A}\htmlData{tutor-start=97,tutor-end=98}{C}_{\htmlData{tutor-start=100,tutor-end=101}{1}}}\htmlData{tutor-start=103,tutor-end=108}{\cdot}\overrightarrow{\htmlData{tutor-start=124,tutor-end=125}{A}_{\htmlData{tutor-start=127,tutor-end=128}{1}}\htmlData{tutor-start=129,tutor-end=130}{B}}\htmlData{tutor-start=131,tutor-end=132}{=}\htmlData{tutor-start=132,tutor-end=133}{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：核验：证明只使用了侧棱长度 CC1=2、投影位置和底面边长关系，未依赖第 (2) 问的距离条件；点积严格为 0，因此结论对题设允许的整个棱柱族都成立。

AC1⊥A1B\boxed{\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{C}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=19}{\perp }\htmlData{tutor-start=19,tutor-end=20}{A}_{\htmlData{tutor-start=22,tutor-end=23}{1}}\htmlData{tutor-start=24,tutor-end=25}{B}}

（2）由线面距离定参数并求二面角

求平面 A1AB 与底面 ABC 的夹角。

(1)
识别结构并建立关系

AA1 平行于侧面 BCC1B1，因此线面距离可转成点 A 到该平面的距离。先由距离求出投影高度 h 和 d，再用两个平面的法向量求夹角。

为什么从这里入手：空间关系只靠观察容易漏条件，而“AA1 平行于侧面 BCC1B1，因此线面距离可转成点 A 到该平面的距离。先由距离求出投影高度 h 和 d，再用两个平面的法向量求夹角。”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：侧面 BCC1B1 由 CB=(0,1,0) 与 CC1=(d-2,0,h) 张成，其法向量可取 n=(h,0,2-d)，模为 2，平面过 C。点 A 到该平面的距离为 |n 点 CA|/|n|=2h/2=h。题设距离为 sqrt3，故 h=sqrt3。由 (d-2)^2+h^2=4 得 (d-2)^2=1；D 在线段 AC 上，所以 0<=d<=2，只能 d=1。底面法向量 n0=(0,0,1)。平面 A1AB 内有 AB=(-2,1,0)、AA1=(-1,0,sqrt3)，其法向量 n1=AB 叉 AA1=(sqrt3,2sqrt3,1)，模为 4。因此二面角 theta 的余弦为 |n0 点 n1|/(|n0||n1|)=1/4。

h=3,d=1,n0=(0,0,1),n1=(3,23,1),cos⁡theta=14h=\sqrt{3},\quad d=1,\quad \boldsymbol n_{0}=(0,0,1),\quad \boldsymbol n_{1}=(\sqrt{3},2\sqrt{3},1),\quad \cos\\theta=\frac{1}{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：为确认取到题目所指的两个半平面，可在 AB 两侧分别取指向 A1 和 C 的垂直于 AB 的向量，它们的点积同样给出正的 1/4，所以二面角是锐角而非其补角。

theta=arccos⁡14\boxed{\\\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{h}\htmlData{tutor-start=11,tutor-end=12}{e}\htmlData{tutor-start=12,tutor-end=13}{t}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{=}\arccos\frac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{4}}}
20

解答题 · 相互独立事件的概率

设每个工作日甲、乙、丙、丁4人需使用某种设备的概率分别为0.6、0.5、0.5、0.4，各人是否需使用设备相互独立. (1) 求同一工作日至少3人需使用设备的概率; (2) 实验室计划购买 k\htmlData{tutor-start=0,tutor-end=1}{k} 台设备供甲、乙、丙、丁使用，若要求“同一工作日需使用设备的人数大于 k\htmlData{tutor-start=0,tutor-end=1}{k}”的概率小于0.1，求 k\htmlData{tutor-start=0,tutor-end=1}{k} 的最小值.

答案：(1) 0.31；(2) k的最小值为3。

题目标签：独立事件计数与设备配置

解题过程

（1）枚举恰有三人和四人需要设备

求同一工作日至少三人需要设备的概率。

(1)
识别结构并建立关系

至少三人在总人数只有四人时只包含两种互斥情况：恰有三人和四人全都需要。恰有三人可按唯一不需要设备的人分类。

为什么从这里入手：目标是“识别结构并建立关系”，而“至少三人在总人数只有四人时只包含两种互斥情况：恰有三人和四人全都需要。恰有三人可按唯一不需要设备的人分类。”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：记四人的需要概率依次为 0.6、0.5、0.5、0.4。四人都需要的概率为 0.6乘0.5乘0.5乘0.4=0.06。恰有三人需要时，分别令甲、乙、丙、丁为唯一不需要者，概率为 0.04、0.06、0.06、0.09，合计 0.25。这五种情形互斥，所以至少三人需要的概率为 0.25+0.06=0.31。

P(X=3)=0.04+0.06+0.06+0.09=0.25,P(X=4)=0.06\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{X}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3}\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}{0}\htmlData{tutor-start=10,tutor-end=11}{4}\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}{6}\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}{0}\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}{9}\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}{2}\htmlData{tutor-start=30,tutor-end=31}{5}\htmlData{tutor-start=31,tutor-end=32}{,}\qquad \htmlData{tutor-start=39,tutor-end=40}{P}\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{X}\htmlData{tutor-start=42,tutor-end=43}{=}\htmlData{tutor-start=43,tutor-end=44}{4}\htmlData{tutor-start=44,tutor-end=45}{)}\htmlData{tutor-start=45,tutor-end=46}{=}\htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{.}\htmlData{tutor-start=48,tutor-end=49}{0}\htmlData{tutor-start=49,tutor-end=50}{6}
(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：每一项都包含三个需要概率和一个不需要概率，且四种唯一缺席者无重叠；再加四人全需的情形，完整覆盖 X>=3。因此结果为 0.31。

P(X⩾3)=0.31\boxed{\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{X}\htmlData{tutor-start=10,tutor-end=19}{\geqslant}\htmlData{tutor-start=19,tutor-end=20}{3}\htmlData{tutor-start=20,tutor-end=21}{)}\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}{3}\htmlData{tutor-start=25,tutor-end=26}{1}}

（2）比较相邻设备数的超额概率

求满足 P(X>k)<0.1 的最小整数 k。

(1)
识别结构并建立关系

k 是设备台数，只需从相邻整数开始比较。由第 (1) 问已经知道 P(X>2)=P(X>=3)，再计算 k=3 时只剩四人全需。

为什么从这里入手：目标是“识别结构并建立关系”，而“k 是设备台数，只需从相邻整数开始比较。由第 (1) 问已经知道 P(X>2)=P(X>=3)，再计算 k=3 时只剩四人全需。”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 k=2 时，设备不足等价于 X>2，即至少三人需要，其概率为 0.31，不小于 0.1，所以两台不够。当 k=3 时，只有 X=4 才会设备不足，概率为 0.06，小于 0.1，所以三台满足要求。由于 k=2 不满足而 k=3 满足，最小值就是 3。

P(X>2)=0.31>0.1,P(X>3)=P(X=4)=0.06<0.1\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{X}\htmlData{tutor-start=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}{=}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{.}\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{1}\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}{1}\htmlData{tutor-start=15,tutor-end=16}{,}\qquad \htmlData{tutor-start=23,tutor-end=24}{P}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{X}\htmlData{tutor-start=26,tutor-end=27}{>}\htmlData{tutor-start=27,tutor-end=28}{3}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{P}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{X}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{4}\htmlData{tutor-start=35,tutor-end=36}{)}\htmlData{tutor-start=36,tutor-end=37}{=}\htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{.}\htmlData{tutor-start=39,tutor-end=40}{0}\htmlData{tutor-start=40,tutor-end=41}{6}\htmlData{tutor-start=41,tutor-end=42}{<}\htmlData{tutor-start=42,tutor-end=43}{0}\htmlData{tutor-start=43,tutor-end=44}{.}\htmlData{tutor-start=44,tutor-end=45}{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：核验：概率 P(X>k) 随 k 增大不会上升，因此一旦 k=3 满足，所有更大 k 也满足；而前一个整数 2 已被排除，最小性得到证明。

kmin⁡=3\boxed{\htmlData{tutor-start=7,tutor-end=8}{k}_{\min}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{3}}
21

解答题 · 导数与函数单调性

函数 f(x)=ax3+3x2+3x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \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}{3}\htmlData{tutor-start=26,tutor-end=27}{x} (a≠0\htmlData{tutor-start=0,tutor-end=1}{a} \neq \htmlData{tutor-start=7,tutor-end=8}{0}). (1) 讨论函数 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 的单调性; (2) 若函数 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}{)} 在区间 (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=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)} 是增函数，求 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围.

答案：(1) 按 a<0、0<a<1、a>=1 分类；(2) a>=-5/4 且 a不等于0。

题目标签：含参数三次函数的单调性

解题过程

（1）按导函数二次式的根分类

讨论 f 在实数域上的增减区间。

(1)
识别结构并建立关系

三次函数的单调性由导函数 f'(x)=3(ax^2+2x+1) 的符号决定。参数 a 同时影响开口方向和判别式，分类点来自 a=0 与判别式为零的 a=1。

为什么从这里入手：函数的局部变化由导数控制。“三次函数的单调性由导函数 f'(x)=3(ax^2+2x+1) 的符号决定。参数 a 同时影响开口方向和判别式，分类点来自 a=0 与判别式为零的 a=1。”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：有 f'(x)=3(ax^2+2x+1)。当 a<0 时，令 x1=(-1+sqrt(1-a))/a、x2=(-1-sqrt(1-a))/a，则 x1<x2，导函数在 (x1,x2) 为正、两侧为负，故 f 先减、后增、再减。当 0<a<1 时，令 x1=(-1-sqrt(1-a))/a、x2=(-1+sqrt(1-a))/a，则 x1<x2，导函数在两根外为正、两根间为负，故 f 先增、后减、再增。当 a=1 时 f'=3(x+1)^2>=0；当 a>1 时判别式小于 0 且开口向上，f'>0。因此 a>=1 时 f 在 R 上递增。

f′(x)=3(ax2+2x+1),Δ=4(1−a)\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}{(}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{,}\qquad \htmlData{tutor-start=28,tutor-end=34}{\Delta}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{4}\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}{)}
(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：核验分类覆盖了题设允许的全部 a不等于0：a<0、0<a<1、a=1、a>1。各区间的增减次序都与二次函数开口方向一致，重根情形也单独处理。

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（2）把区间递增转为导数非负

求 f 在 (1,2) 上递增时 a 的范围。

(1)
识别结构并建立关系

在固定区间内，不必重复完整分类，只需保证 q(x)=ax^2+2x+1 在每个 x属于(1,2) 上非负。由于 q 对 a 是一次式，可直接解成 a 的统一下界。

为什么从这里入手：目标是“识别结构并建立关系”，而“在固定区间内，不必重复完整分类，只需保证 q(x)=ax^2+2x+1 在每个 x属于(1,2) 上非负。由于 q 对 a 是一次式，可直接解成 a 的统一下界。”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 x>0，条件 q(x)>=0 等价于 a>=-(2x+1)/x^2。令 h(x)=-(2x+1)/x^2=-2/x-1/x^2，则 h'(x)=2/x^2+2/x^3>0，所以 h 在 (1,2) 上递增，其上确界为 h(2)=-5/4。故必须且只需 a>=-5/4。当 a=-5/4 时，q(2)=0 而对 1<x<2 有 q(x)>0，仍严格递增。结合原题 a不等于0，最终为 a>=-5/4 且 a不等于0。

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(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：核验边界：a=-5/4 时 f'(x)>0 对所有 1<x<2 成立；若 a<-5/4，则 q(2)<0，并由连续性知在足够靠近 2 的左侧也有 f'<0，不能递增。必要性与充分性均已验证。

a⩾−54,a≠0\boxed{a\geqslant-\frac{5}{4},\quad a\ne0}
22

解答题 · 解析几何

已知抛物线 C:y2=2px\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y}^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{p}\htmlData{tutor-start=13,tutor-end=14}{x} (p>0\htmlData{tutor-start=0,tutor-end=1}{p} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}) 的焦点为 F\htmlData{tutor-start=0,tutor-end=1}{F}，直线 y=4\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{4} 与 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴的交点为 P\htmlData{tutor-start=0,tutor-end=1}{P}，与 C\htmlData{tutor-start=0,tutor-end=1}{C} 的交点为 Q\htmlData{tutor-start=0,tutor-end=1}{Q}，且 ∣QF∣=54∣PQ∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{Q}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{|} \htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{5}}{\htmlData{tutor-start=16,tutor-end=17}{4}}\htmlData{tutor-start=18,tutor-end=19}{|}\htmlData{tutor-start=19,tutor-end=20}{P}\htmlData{tutor-start=20,tutor-end=21}{Q}\htmlData{tutor-start=21,tutor-end=22}{|}. (1) 求 C\htmlData{tutor-start=0,tutor-end=1}{C} 的方程; (2) 过 F\htmlData{tutor-start=0,tutor-end=1}{F} 的直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与 C\htmlData{tutor-start=0,tutor-end=1}{C} 相交于 A\htmlData{tutor-start=0,tutor-end=1}{A}、B\htmlData{tutor-start=0,tutor-end=1}{B} 两点，若 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 的垂直平分线 l′\htmlData{tutor-start=0,tutor-end=1}{l}' 与 C\htmlData{tutor-start=0,tutor-end=1}{C} 相交于 M\htmlData{tutor-start=0,tutor-end=1}{M}、N\htmlData{tutor-start=0,tutor-end=1}{N} 两点，且 A\htmlData{tutor-start=0,tutor-end=1}{A}、M\htmlData{tutor-start=0,tutor-end=1}{M}、B\htmlData{tutor-start=0,tutor-end=1}{B}、N\htmlData{tutor-start=0,tutor-end=1}{N} 四点在同一圆上，求 l\htmlData{tutor-start=0,tutor-end=1}{l} 的方程.

答案：(1) y^2=4x；(2) y=x-1或y=-x+1。

题目标签：抛物线焦点弦与四点共圆

解题过程

（1）利用抛物线定义确定参数

求抛物线 C 的方程。

(1)
识别结构并建立关系

Q 的纵坐标已知为 4，先代入抛物线求横坐标。Q 到焦点的距离可用抛物线定义改写成 Q 到准线的距离，计算最短。

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。“Q 的纵坐标已知为 4，先代入抛物线求横坐标。Q 到焦点的距离可用抛物线定义改写成 Q 到准线的距离，计算最短。”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：直线 y=4 与 y 轴交于 P(0,4)。设 Q=(x_Q,4)，由 16=2px_Q 得 x_Q=8/p，所以 |PQ|=8/p。抛物线焦点为 F(p/2,0)，准线为 x=-p/2。由抛物线定义，|QF| 等于 Q 到准线的距离，即 x_Q+p/2=8/p+p/2。题设给出 8/p+p/2=(5/4)(8/p)=10/p，化简得 p^2=4。因 p>0，所以 p=2，方程为 y^2=4x。

Q=(8p,4),8p+p2=54⋅8p⇒p=2\htmlData{tutor-start=0,tutor-end=1}{Q}\htmlData{tutor-start=1,tutor-end=2}{=}\left(\frac{\htmlData{tutor-start=14,tutor-end=15}{8}}{\htmlData{tutor-start=17,tutor-end=18}{p}}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{4}\right)\htmlData{tutor-start=28,tutor-end=29}{,}\quad \frac{\htmlData{tutor-start=41,tutor-end=42}{8}}{\htmlData{tutor-start=44,tutor-end=45}{p}}\htmlData{tutor-start=46,tutor-end=47}{+}\frac{\htmlData{tutor-start=53,tutor-end=54}{p}}{\htmlData{tutor-start=56,tutor-end=57}{2}}\htmlData{tutor-start=58,tutor-end=59}{=}\frac{\htmlData{tutor-start=65,tutor-end=66}{5}}{\htmlData{tutor-start=68,tutor-end=69}{4}}\htmlData{tutor-start=70,tutor-end=75}{\cdot}\frac{\htmlData{tutor-start=81,tutor-end=82}{8}}{\htmlData{tutor-start=84,tutor-end=85}{p}}\htmlData{tutor-start=86,tutor-end=98}{\Rightarrow }\htmlData{tutor-start=98,tutor-end=99}{p}\htmlData{tutor-start=99,tutor-end=100}{=}\htmlData{tutor-start=100,tutor-end=101}{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：核验：当 p=2 时 Q=(4,4)、F=(1,0)，|QF|=5，|PQ|=4，确有 |QF|=(5/4)|PQ|。因此参数和方程均满足原条件。

C:y2=4x\boxed{\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{:}\htmlData{tutor-start=9,tutor-end=10}{y}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{x}}

（2）用相交弦定理化简共圆条件

求所有满足条件的焦点弦 l。

(1)
识别结构并建立关系

焦点 F=(1,0)。一般非竖直焦点弦可设为 y=k(x-1)。垂直平分线与 AB 交于中点 H；若 A、M、B、N 共圆，则两弦 AB、MN 在 H 相交，可用有向线段乘积相等代替直接求圆方程。

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。“焦点 F=(1,0)。一般非竖直焦点弦可设为 y=k(x-1)。垂直平分线与 AB 交于中点 H；若 A、M、B、N 共圆，则两弦 AB、MN 在 H 相交，可用有向线段乘积相等代替直接求圆方程。”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 l:y=k(x-1)，k不等于0。以交点的纵坐标为未知数，代入 y^2=4x 得 y^2-4y/k-4=0，所以 A、B 的纵坐标和为 4/k、积为 -4，中点 H=(1+2/k^2,2/k)。AB 的垂直平分线为 l':x+ky=3+2/k^2。设 M、N 的纵坐标为 u、v，代入抛物线得 u+v=-4k，uv=-4(3+2/k^2)。计算得 HA^2=4(1+k^2)^2/k^4，而有向积 HM乘HN=(1+k^2)(u-2/k)(v-2/k)=-4(1+k^2)^2/k^2。共圆的相交弦定理给出 -HA^2=HM乘HN，从而 k^2=1。故 k=正负1。另查竖直线 x=1 时垂直平分线 y=0 只与抛物线相切，k=0 时 l 本身也相切，都不能产生题设两对交点。

−4(1+k2)2k4=−4(1+k2)2k2⇒k2=1\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{4}\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}{k}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{)}^{\htmlData{tutor-start=19,tutor-end=20}{2}}}{\htmlData{tutor-start=23,tutor-end=24}{k}^{\htmlData{tutor-start=26,tutor-end=27}{4}}}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{-}\frac{\htmlData{tutor-start=37,tutor-end=38}{4}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{k}^{\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{)}^{\htmlData{tutor-start=49,tutor-end=50}{2}}}{\htmlData{tutor-start=53,tutor-end=54}{k}^{\htmlData{tutor-start=56,tutor-end=57}{2}}}\htmlData{tutor-start=59,tutor-end=71}{\Rightarrow }\htmlData{tutor-start=71,tutor-end=72}{k}^{\htmlData{tutor-start=74,tutor-end=75}{2}}\htmlData{tutor-start=76,tutor-end=77}{=}\htmlData{tutor-start=77,tutor-end=78}{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 k=1 或 -1 代回，l 都与抛物线有两个不同交点，l' 也有两个不同交点，且相交弦乘积相等，四点确实共圆。排除的水平线和竖直线均缺少两个不同交点，因此答案完整。

y=x−1或y=−x+1\boxed{\htmlData{tutor-start=7,tutor-end=8}{y}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\quad\text{\htmlData{tutor-start=23,tutor-end=24}{或}}\quad \htmlData{tutor-start=31,tutor-end=32}{y}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{x}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{1}}