返回特征解读

2010 年高考数学（大纲卷 1文科）

exams_raw/普通高考/2010/2010大纲1文(河北,河南,山西,广西).pdf · HS-MATH-1024-v2.1-solution-aware

22 题26 个小问/题组
1

一、选择题 · 三角函数

cos⁡300∘=\cos \htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{0}^{\htmlData{tutor-start=10,tutor-end=15}{\circ}}\htmlData{tutor-start=16,tutor-end=17}{=} ( )

答案：1/2

题目标签：特殊角余弦

解题过程

求特殊角余弦值

化到第一象限参考角

(1)
写成周角之差

300∘=360∘−60∘300^\circ=360^\circ-60^\circ。

为什么从这里入手：“300∘=360∘−60∘300^\circ=360^\circ-60^\circ。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：余弦关于 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴对称，第四象限余弦为正，所以 cos⁡(360∘−θ)=cos⁡θ\cos(360^\circ-\theta)=\cos\theta。

cos⁡300∘=cos⁡60∘\cos300^\circ=\cos60^\circ
(2)
代入特殊角值

60∘60^\circ 是基本特殊角。

为什么从这里入手：“60∘60^\circ 是基本特殊角。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：cos⁡60∘=1/2\cos60^\circ=1/2。

12\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{2}}}
2

一、选择题 · 集合

设全集 U={1,2,3,4,5}\htmlData{tutor-start=0,tutor-end=1}{U}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{5}\htmlData{tutor-start=13,tutor-end=15}{\}}, 集合 M={1,4}\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=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=9}{\}}, N={1,3,5}\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=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=11}{\}}, 则 N∩(∁UM)=\htmlData{tutor-start=0,tutor-end=1}{N} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=19}{\complement}_{\htmlData{tutor-start=21,tutor-end=22}{U}} \htmlData{tutor-start=24,tutor-end=25}{M}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{=} ( )

答案：{3,5}

题目标签：集合补集与交集

解题过程

求补集与交集

先求全集中的补集，再和 N\htmlData{tutor-start=0,tutor-end=1}{N} 取公共元素

(1)
求 M\htmlData{tutor-start=0,tutor-end=1}{M} 的补集

全集是 U={1,2,3,4,5}\htmlData{tutor-start=0,tutor-end=1}{U}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{5}\htmlData{tutor-start=13,tutor-end=15}{\}}，M\htmlData{tutor-start=0,tutor-end=1}{M} 含 1 和 4。

为什么从这里入手：“全集是 U={1,2,3,4,5}\htmlData{tutor-start=0,tutor-end=1}{U}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{5}\htmlData{tutor-start=13,tutor-end=15}{\}}，M\htmlData{tutor-start=0,tutor-end=1}{M} 含 1 和 4。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：从全集删除 1、4，得到 ∁UM={2,3,5}\htmlData{tutor-start=0,tutor-end=11}{\complement}_{\htmlData{tutor-start=13,tutor-end=14}{U}}\htmlData{tutor-start=15,tutor-end=16}{M}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,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}{3}\htmlData{tutor-start=22,tutor-end=23}{,}\htmlData{tutor-start=23,tutor-end=24}{5}\htmlData{tutor-start=24,tutor-end=26}{\}}。

∁UM={2,3,5}\htmlData{tutor-start=0,tutor-end=11}{\complement}_{\htmlData{tutor-start=13,tutor-end=14}{U}}\htmlData{tutor-start=15,tutor-end=16}{M}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,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}{3}\htmlData{tutor-start=22,tutor-end=23}{,}\htmlData{tutor-start=23,tutor-end=24}{5}\htmlData{tutor-start=24,tutor-end=26}{\}}
(2)
与 N\htmlData{tutor-start=0,tutor-end=1}{N} 求交

N={1,3,5}\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=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=11}{\}}。

为什么从这里入手：“N={1,3,5}\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=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=11}{\}}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。目标是“与 N\htmlData{tutor-start=0,tutor-end=1}{N} 求交”，而“N={1,3,5}\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=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=11}{\}}。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：两个集合共有的元素是 3 和 5。

N∩(∁UM)={3,5}\boxed{\htmlData{tutor-start=7,tutor-end=8}{N}\htmlData{tutor-start=8,tutor-end=12}{\cap}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=24}{\complement}_{\htmlData{tutor-start=26,tutor-end=27}{U}}\htmlData{tutor-start=28,tutor-end=29}{M}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=33}{\{}\htmlData{tutor-start=33,tutor-end=34}{3}\htmlData{tutor-start=34,tutor-end=35}{,}\htmlData{tutor-start=35,tutor-end=36}{5}\htmlData{tutor-start=36,tutor-end=38}{\}}}
3

一、选择题 · 线性规划

若变量 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y} 满足约束条件 {y⩽1,x+y⩾0,x−y−2⩽0,\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{y} \htmlData{tutor-start=16,tutor-end=26}{\leqslant }\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{,} \\ \htmlData{tutor-start=32,tutor-end=33}{x}\htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=35}{y} \htmlData{tutor-start=36,tutor-end=46}{\geqslant }\htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{,} \\ \htmlData{tutor-start=52,tutor-end=53}{x}\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{y}\htmlData{tutor-start=55,tutor-end=56}{-}\htmlData{tutor-start=56,tutor-end=57}{2} \htmlData{tutor-start=58,tutor-end=68}{\leqslant }\htmlData{tutor-start=68,tutor-end=69}{0}\htmlData{tutor-start=69,tutor-end=70}{,} \end{cases} 则 z=x−2y\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}{2}\htmlData{tutor-start=5,tutor-end=6}{y} 的最大值为 ( )

答案：3\htmlData{tutor-start=0,tutor-end=1}{3}

题目标签：线性规划最大值

解题过程

求线性目标函数最大值

求可行域顶点并逐点代入

(1)
确定三角形可行域

三条边界分别是两条斜线和水平线。

为什么从这里入手：“三条边界分别是两条斜线和水平线。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：两两联立得到顶点 (−1,1),(3,1),(1,−1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{)}；不等式方向确定可行域是这三个点围成的三角形。

D=△((−1,1),(3,1),(1,−1))\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=11}{\triangle}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{3}\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}{1}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{)}
(2)
计算顶点目标值

线性函数在多边形上的最值出现在顶点。

为什么从这里入手：“线性函数在多边形上的最值出现在顶点。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：z=x−2y\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}{2}\htmlData{tutor-start=5,tutor-end=6}{y} 在三个顶点的值依次为 −3,1,3\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{3}，故最大值为 3。

zmax⁡=3\boxed{\htmlData{tutor-start=7,tutor-end=8}{z}_{\max}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{3}}
4

一、选择题 · 数列

已知各项均为正数的等比数列 {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}{\}}, a1a2a3=5\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{3}}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{5}, a7a8a9=10\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{7}} \htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{8}} \htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{9}}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{0}, 则 a4a5a6=\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{5}} \htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{6}}\htmlData{tutor-start=17,tutor-end=18}{=} ( )

答案：5√2

题目标签：等比数列连续项积

解题过程

求等比数列中间三项之积

把连续三项积化为中项立方

(1)
识别中项对称性

等比数列满足前后等距两项之积等于中项平方。

为什么从这里入手：“等比数列满足前后等距两项之积等于中项平方。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：a1a2a3=a23=5\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{a}_{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{2}}^{\htmlData{tutor-start=23,tutor-end=24}{3}}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{5}，a7a8a9=a83=10\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{7}}\htmlData{tutor-start=5,tutor-end=6}{a}_{\htmlData{tutor-start=8,tutor-end=9}{8}}\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{9}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{8}}^{\htmlData{tutor-start=23,tutor-end=24}{3}}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{0}；又 a52=a2a8\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{5}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{8}}。

a23=5,a83=10,a52=a2a8\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}^{\htmlData{tutor-start=7,tutor-end=8}{3}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{,}\quad \htmlData{tutor-start=18,tutor-end=19}{a}_{\htmlData{tutor-start=21,tutor-end=22}{8}}^{\htmlData{tutor-start=25,tutor-end=26}{3}}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=31}{,}\quad \htmlData{tutor-start=37,tutor-end=38}{a}_{\htmlData{tutor-start=40,tutor-end=41}{5}}^{\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{=}\htmlData{tutor-start=47,tutor-end=48}{a}_{\htmlData{tutor-start=50,tutor-end=51}{2}}\htmlData{tutor-start=52,tutor-end=53}{a}_{\htmlData{tutor-start=55,tutor-end=56}{8}}
(2)
求正的中间乘积

所求 a4a5a6=a53\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{a}_{\htmlData{tutor-start=8,tutor-end=9}{5}}\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{6}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{5}}^{\htmlData{tutor-start=23,tutor-end=24}{3}}，且各项为正。

为什么从这里入手：“所求 a4a5a6=a53\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{a}_{\htmlData{tutor-start=8,tutor-end=9}{5}}\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{6}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{5}}^{\htmlData{tutor-start=23,tutor-end=24}{3}}，且各项为正。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：(a53)2=a23a83=50\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{5}}^{\htmlData{tutor-start=8,tutor-end=9}{3}}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{2}}^{\htmlData{tutor-start=23,tutor-end=24}{3}}\htmlData{tutor-start=25,tutor-end=26}{a}_{\htmlData{tutor-start=28,tutor-end=29}{8}}^{\htmlData{tutor-start=32,tutor-end=33}{3}}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{5}\htmlData{tutor-start=36,tutor-end=37}{0}，所以 a53=50=52\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{5}}^{\htmlData{tutor-start=7,tutor-end=8}{3}}\htmlData{tutor-start=9,tutor-end=10}{=}\sqrt{\htmlData{tutor-start=16,tutor-end=17}{5}\htmlData{tutor-start=17,tutor-end=18}{0}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{5}\sqrt{\htmlData{tutor-start=27,tutor-end=28}{2}}。

a4a5a6=52\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{4}}\htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{5}}\htmlData{tutor-start=17,tutor-end=18}{a}_{\htmlData{tutor-start=20,tutor-end=21}{6}}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{5}\sqrt{\htmlData{tutor-start=30,tutor-end=31}{2}}}
5

一、选择题 · 二项式定理

(1−x)4(1−x)3\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{4}}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{-}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{x}}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{3}} 的展开式中 x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 的系数是 ( )

答案：−6\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{6}

题目标签：二项展开系数

解题过程

求含根式二项展开的系数

令 t=x\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{x}} 把分数指数变成整数指数

(1)
改写多项式

要求 x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 系数等价于要求 t4\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{4}} 系数。

为什么从这里入手：“要求 x2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 系数等价于要求 t4\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{4}} 系数。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：1−x=1−t2=(1−t)(1+t)\htmlData{tutor-start=0,tutor-end=1}{1}\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}{1}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{t}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{t}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{t}\htmlData{tutor-start=21,tutor-end=22}{)}，原式变为 (1−t)7(1+t)4\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{t}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{7}}\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}{t}\htmlData{tutor-start=13,tutor-end=14}{)}^{\htmlData{tutor-start=16,tutor-end=17}{4}}。

(1−x)4(1−x)3=(1−t)7(1+t)4\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{4}}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{-}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{x}}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{3}}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{t}\htmlData{tutor-start=30,tutor-end=31}{)}^{\htmlData{tutor-start=33,tutor-end=34}{7}}\htmlData{tutor-start=35,tutor-end=36}{(}\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{t}\htmlData{tutor-start=39,tutor-end=40}{)}^{\htmlData{tutor-start=42,tutor-end=43}{4}}
(2)
卷积求四次项

从两因子所取次数之和必须为 4。

为什么从这里入手：“从两因子所取次数之和必须为 4。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：系数为 ∑j=04(−1)j(7j)(44−j)=1−28+126−140+35=−6\sum_{j=0}^{4}(-1)^{j}\binom7j\binom4{4-j}=1-28+126-140+35=-6。

−6\boxed{\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{6}}
6

一、选择题 · 立体几何

直三棱柱 ABC−A1B1C1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\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=10,tutor-end=11}{B}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{C}_{\htmlData{tutor-start=19,tutor-end=20}{1}} 中, 若 ∠BAC=90∘\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{9}\htmlData{tutor-start=12,tutor-end=13}{0}^{\htmlData{tutor-start=15,tutor-end=20}{\circ}}, AB=AC=AA1\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}{C}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{A}_{\htmlData{tutor-start=10,tutor-end=11}{1}}, 则异面直线 BA1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 与 AC1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 所成的角等于 ( )

答案：60°

题目标签：异面直线所成角

解题过程

求异面直线所成角

平移方向向量后用数量积

(1)
建立直三棱柱坐标

三条互相垂直且等长的棱适合作为坐标轴。

为什么从这里入手：“三条互相垂直且等长的棱适合作为坐标轴。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：令公共棱长为 1，取 A(0,0,0),B(1,0,0),C(0,1,0),A1(0,0,1),C1(0,1,1)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{B}\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}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{C}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=23}{1}\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}{,}\htmlData{tutor-start=27,tutor-end=28}{A}_{\htmlData{tutor-start=30,tutor-end=31}{1}}\htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{0}\htmlData{tutor-start=34,tutor-end=35}{,}\htmlData{tutor-start=35,tutor-end=36}{0}\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}{,}\htmlData{tutor-start=40,tutor-end=41}{C}_{\htmlData{tutor-start=43,tutor-end=44}{1}}\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}{1}\htmlData{tutor-start=49,tutor-end=50}{,}\htmlData{tutor-start=50,tutor-end=51}{1}\htmlData{tutor-start=51,tutor-end=52}{)}。

BA1→=(−1,0,1),AC1→=(0,1,1)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{A}_{\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}{-}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{,}\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{)}\htmlData{tutor-start=32,tutor-end=33}{,}\quad\overrightarrow{\htmlData{tutor-start=54,tutor-end=55}{A}\htmlData{tutor-start=55,tutor-end=56}{C}_{\htmlData{tutor-start=58,tutor-end=59}{1}}}\htmlData{tutor-start=61,tutor-end=62}{=}\htmlData{tutor-start=62,tutor-end=63}{(}\htmlData{tutor-start=63,tutor-end=64}{0}\htmlData{tutor-start=64,tutor-end=65}{,}\htmlData{tutor-start=65,tutor-end=66}{1}\htmlData{tutor-start=66,tutor-end=67}{,}\htmlData{tutor-start=67,tutor-end=68}{1}\htmlData{tutor-start=68,tutor-end=69}{)}
(2)
计算方向夹角

异面直线所成角定义为方向向量的锐角。

为什么从这里入手：“异面直线所成角定义为方向向量的锐角。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：两向量点积为 1，模都为 2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}，故余弦为 1/2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2}，所成角为 60∘60^\circ。

60∘\boxed{60^\circ}
7

一、选择题 · 函数

已知函数 f(x)=∣lg⁡x∣\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{|}\lg \htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{|}. 若 a≠b\htmlData{tutor-start=0,tutor-end=1}{a} \neq \htmlData{tutor-start=7,tutor-end=8}{b} 且 f(a)=f(b)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{)}, 则 a+b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b} 的取值范围是 ( )

答案：(2,+∞)

题目标签：绝对值对数等值

解题过程

求绝对值对数等值时的和

利用绝对值相等的两个分支

(1)
排除同号分支

f(a)=f(b)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{)} 给出 ∣lg⁡a∣=∣lg⁡b∣\htmlData{tutor-start=0,tutor-end=1}{|}\lg \htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{|}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{|}\lg \htmlData{tutor-start=13,tutor-end=14}{b}\htmlData{tutor-start=14,tutor-end=15}{|}。

为什么从这里入手：“f(a)=f(b)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{)} 给出 ∣lg⁡a∣=∣lg⁡b∣\htmlData{tutor-start=0,tutor-end=1}{|}\lg \htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{|}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{|}\lg \htmlData{tutor-start=13,tutor-end=14}{b}\htmlData{tutor-start=14,tutor-end=15}{|}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：若 lg⁡a=lg⁡b\lg \htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{=}\lg \htmlData{tutor-start=10,tutor-end=11}{b}，则 a=b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{b}，与题设矛盾；所以只能有 lg⁡a=−lg⁡b\lg \htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\lg \htmlData{tutor-start=11,tutor-end=12}{b}，即 ab=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}。

a>0,b>0,ab=1,a≠ba>0,b>0,\quad ab=1,\quad a\ne b
(2)
用均值不等式求范围

两个正数积固定为 1。

为什么从这里入手：“两个正数积固定为 1。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。最值与范围题要先找到变量受限方式以及等号何时成立。“两个正数积固定为 1。”给出了可行域或关键界，先用均值不等式求范围能把‘可能有多大’转成单调性、基本不等式或边界比较。

详细展开：a+b≥2ab=2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=6}{\ge}\htmlData{tutor-start=6,tutor-end=7}{2}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{b}}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{2}，等号仅在 a=b=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1} 时成立，而该情形被排除。反过来任意大于 2 的和都对应方程 t2−st+1=0\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{s}\htmlData{tutor-start=7,tutor-end=8}{t}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0} 的两个正根。

a+b∈(2,+∞)\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=13}{\in}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=23}{\infty}\htmlData{tutor-start=23,tutor-end=24}{)}}
8

一、选择题 · 圆锥曲线

已知 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}} 为双曲线 C:x2−y2=1\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{x}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=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}{1} 的左、右焦点, 点 P\htmlData{tutor-start=0,tutor-end=1}{P} 在 C\htmlData{tutor-start=0,tutor-end=1}{C} 上, ∠F1PF2=60∘\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{F}_{\htmlData{tutor-start=10,tutor-end=11}{1}} \htmlData{tutor-start=13,tutor-end=14}{P} \htmlData{tutor-start=15,tutor-end=16}{F}_{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{6}\htmlData{tutor-start=22,tutor-end=23}{0}^{\htmlData{tutor-start=25,tutor-end=30}{\circ}}, 则 ∣PF1∣⋅∣PF2∣=\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{F}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{|} \htmlData{tutor-start=9,tutor-end=15}{\cdot }\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{P}\htmlData{tutor-start=17,tutor-end=18}{F}_{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{|}\htmlData{tutor-start=23,tutor-end=24}{=} ( )

答案：4\htmlData{tutor-start=0,tutor-end=1}{4}

题目标签：双曲线焦点三角形

解题过程

求双曲线焦点距离乘积

联立距离差与焦点三角形余弦定理

(1)
写出双曲线定义

双曲线 x2−y2=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}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1} 中 a=1,c=2\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}{c}\htmlData{tutor-start=5,tutor-end=6}{=}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}。

为什么从这里入手：“双曲线 x2−y2=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}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1} 中 a=1,c=2\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}{c}\htmlData{tutor-start=5,tutor-end=6}{=}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：设 u=∣PF1∣,v=∣PF2∣\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{P}\htmlData{tutor-start=4,tutor-end=5}{F}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{|}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{v}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{|}\htmlData{tutor-start=14,tutor-end=15}{P}\htmlData{tutor-start=15,tutor-end=16}{F}_{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{|}，则 ∣u−v∣=2a=2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{2}，而 ∣F1F2∣=2c=22\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{F}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{c}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{2}\sqrt{\htmlData{tutor-start=23,tutor-end=24}{2}}。

(u−v)2=4,∣F1F2∣2=8\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{v}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{,}\quad \htmlData{tutor-start=18,tutor-end=19}{|}\htmlData{tutor-start=19,tutor-end=20}{F}_{\htmlData{tutor-start=22,tutor-end=23}{1}}\htmlData{tutor-start=24,tutor-end=25}{F}_{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{|}^{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{8}
(2)
应用余弦定理

焦点三角形在 P\htmlData{tutor-start=0,tutor-end=1}{P} 处夹角为 60∘60^\circ。

为什么从这里入手：“焦点三角形在 P\htmlData{tutor-start=0,tutor-end=1}{P} 处夹角为 60∘60^\circ。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：8=u2+v2−uv\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{u}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{v}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{u}\htmlData{tutor-start=15,tutor-end=16}{v}；又 u2+v2=(u−v)2+2uv=4+2uv\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{v}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{u}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{v}\htmlData{tutor-start=16,tutor-end=17}{)}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{u}\htmlData{tutor-start=24,tutor-end=25}{v}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{4}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{u}\htmlData{tutor-start=30,tutor-end=31}{v}。代入得 8=4+uv\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=6}{v}。

∣PF1∣ ∣PF2∣=4\boxed{\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{F}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{|}\,\htmlData{tutor-start=17,tutor-end=18}{|}\htmlData{tutor-start=18,tutor-end=19}{P}\htmlData{tutor-start=19,tutor-end=20}{F}_{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{|}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{4}}
9

一、选择题 · 立体几何

正方体 ABCD−A1B1C1D1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{A}_{\htmlData{tutor-start=8,tutor-end=9}{1}} \htmlData{tutor-start=11,tutor-end=12}{B}_{\htmlData{tutor-start=14,tutor-end=15}{1}} \htmlData{tutor-start=17,tutor-end=18}{C}_{\htmlData{tutor-start=20,tutor-end=21}{1}} \htmlData{tutor-start=23,tutor-end=24}{D}_{\htmlData{tutor-start=26,tutor-end=27}{1}} 中, BB1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 与平面 ACD1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}_{\htmlData{tutor-start=5,tutor-end=6}{1}} 所成角的余弦值为 ( )

答案：√6/3

题目标签：棱与平面所成角

解题过程

求正方体棱与平面所成角

用方向向量与平面法向量

(1)
建立单位正方体坐标

坐标轴沿正方体三条棱设置。

为什么从这里入手：“坐标轴沿正方体三条棱设置。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。解析几何的目的，是把图形关系变成坐标等式。“坐标轴沿正方体三条棱设置。”提供了点、斜率、距离或焦点条件，先建立单位正方体坐标后，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(0,0,0),C(1,1,0),D1(0,1,1)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{C}\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}{1}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{D}_{\htmlData{tutor-start=21,tutor-end=22}{1}}\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}{1}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{)}，则平面 ACD1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}_{\htmlData{tutor-start=5,tutor-end=6}{1}} 的法向量可取 n=(1,−1,1)\boldsymbol n=(1,-1,1)；BB1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 的方向向量为 v=(0,0,1)\boldsymbol v=(0,0,1)。

n=(1,−1,1),v=(0,0,1)\boldsymbol n=(1,-1,1),\quad\boldsymbol v=(0,0,1)
(2)
从正弦转成余弦

线面角的正弦等于方向向量与法向量夹角余弦的绝对值。

为什么从这里入手：“线面角的正弦等于方向向量与法向量夹角余弦的绝对值。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：sin⁡θ=∣v⋅n∣/(∣v∣∣n∣)=1/3\sin\theta=|\boldsymbol v\cdot\boldsymbol n|/(|\boldsymbol v||\boldsymbol n|)=1/\sqrt{3}，所以 cos⁡θ=1−1/3=6/3\cos\htmlData{tutor-start=4,tutor-end=10}{\theta}\htmlData{tutor-start=10,tutor-end=11}{=}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{/}\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{=}\sqrt{\htmlData{tutor-start=30,tutor-end=31}{6}}\htmlData{tutor-start=32,tutor-end=33}{/}\htmlData{tutor-start=33,tutor-end=34}{3}。

cos⁡theta=63\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}{6}}}{\htmlData{tutor-start=35,tutor-end=36}{3}}}
10

一、选择题 · 函数

设 a=log⁡32,b=ln⁡2,c=5−12\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\log_{\htmlData{tutor-start=8,tutor-end=9}{3}} \htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{b}\htmlData{tutor-start=15,tutor-end=16}{=}\ln \htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{c}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{5}^{\htmlData{tutor-start=28,tutor-end=29}{-}\frac{\htmlData{tutor-start=35,tutor-end=36}{1}}{\htmlData{tutor-start=38,tutor-end=39}{2}}}, 则 ( )

答案：c<a<b\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}{b}

题目标签：对数与幂比较

解题过程

比较对数与幂的大小

转成可估计的正数

(1)
确定三个数的近似位置

a=ln⁡2/ln⁡3\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{/}\ln\htmlData{tutor-start=10,tutor-end=11}{3}，b=ln⁡2\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=6}{2}，c=1/5\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{5}}。

为什么从这里入手：“a=ln⁡2/ln⁡3\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{/}\ln\htmlData{tutor-start=10,tutor-end=11}{3}，b=ln⁡2\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\htmlData{tutor-start=5,tutor-end=6}{2}，c=1/5\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{5}}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：利用 1<ln⁡3<2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{<}\ln\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{<}\htmlData{tutor-start=7,tutor-end=8}{2}、0<ln⁡2<1\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\ln\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{<}\htmlData{tutor-start=7,tutor-end=8}{1} 及常用估计，可得 c≈0.447\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=8}{\approx}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{7}，a≈0.631\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=8}{\approx}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{6}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{1}，b≈0.693\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=8}{\approx}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{6}\htmlData{tutor-start=11,tutor-end=12}{9}\htmlData{tutor-start=12,tutor-end=13}{3}。

c<a<b\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}{b}
(2)
写出排序

三个数均为正，估计区间彼此分离。

为什么从这里入手：“三个数均为正，估计区间彼此分离。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：因此 5−1/2<log⁡32<ln⁡2\htmlData{tutor-start=0,tutor-end=1}{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}{2}}\htmlData{tutor-start=8,tutor-end=9}{<}\log_{\htmlData{tutor-start=15,tutor-end=16}{3}}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{<}\ln\htmlData{tutor-start=22,tutor-end=23}{2}。

c<a<b\boxed{\htmlData{tutor-start=7,tutor-end=8}{c}\htmlData{tutor-start=8,tutor-end=9}{<}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{<}\htmlData{tutor-start=11,tutor-end=12}{b}}
11

一、选择题 · 平面几何

已知圆 O\htmlData{tutor-start=0,tutor-end=1}{O} 的半径为 1, PA\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}、PB\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{B} 为该圆的两条切线, A\htmlData{tutor-start=0,tutor-end=1}{A}、B\htmlData{tutor-start=0,tutor-end=1}{B} 为两切点, 那么 PA→⋅PB→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{P}\htmlData{tutor-start=17,tutor-end=18}{A}} \htmlData{tutor-start=20,tutor-end=26}{\cdot }\overrightarrow{\htmlData{tutor-start=42,tutor-end=43}{P}\htmlData{tutor-start=43,tutor-end=44}{B}} 的最小值为 ( )

答案：-3+2√2

题目标签：圆的两切向量

解题过程

求两切向量数量积最小值

用圆心距表示切线长和夹角

(1)
把数量积化成单变量函数

设 OP=d>1\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{d}\htmlData{tutor-start=4,tutor-end=5}{>}\htmlData{tutor-start=5,tutor-end=6}{1}，则切线长平方为 d2−1\htmlData{tutor-start=0,tutor-end=1}{d}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}。

为什么从这里入手：“设 OP=d>1\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{d}\htmlData{tutor-start=4,tutor-end=5}{>}\htmlData{tutor-start=5,tutor-end=6}{1}，则切线长平方为 d2−1\htmlData{tutor-start=0,tutor-end=1}{d}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：在直角三角形 OAP\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{P} 中，cos⁡∠APO=PA/PO=d2−1/d\cos\htmlData{tutor-start=4,tutor-end=11}{\angle }\htmlData{tutor-start=11,tutor-end=12}{A}\htmlData{tutor-start=12,tutor-end=13}{P}\htmlData{tutor-start=13,tutor-end=14}{O}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{P}\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{P}\htmlData{tutor-start=19,tutor-end=20}{O}\htmlData{tutor-start=20,tutor-end=21}{=}\sqrt{\htmlData{tutor-start=27,tutor-end=28}{d}^{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{1}}\htmlData{tutor-start=35,tutor-end=36}{/}\htmlData{tutor-start=36,tutor-end=37}{d}，而 ∠APB=2∠APO\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{P}\htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=19}{\angle }\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{P}\htmlData{tutor-start=21,tutor-end=22}{O}。故点积为 (d2−1)(d2−2)/d2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{d}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{d}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{d}^{\htmlData{tutor-start=22,tutor-end=23}{2}}。

PA→⋅PB→=d2−3+2d2\overrightarrow{PA}\cdot\overrightarrow{PB}=d^{2}-3+\frac2{d^{2}}
(2)
求函数最小值

令 u=d2>1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{d}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{1}，目标为 u−3+2/u\htmlData{tutor-start=0,tutor-end=1}{u}\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}{u}。

为什么从这里入手：“令 u=d2>1\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{d}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{1}，目标为 u−3+2/u\htmlData{tutor-start=0,tutor-end=1}{u}\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}{u}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：由 u+2/u≥22\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{u}\htmlData{tutor-start=5,tutor-end=8}{\ge}\htmlData{tutor-start=8,tutor-end=9}{2}\sqrt{\htmlData{tutor-start=15,tutor-end=16}{2}}，当 u=2\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}} 时取等号，所以最小值为 22−3\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{3}。

−3+22\boxed{\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{2}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{2}}}
12

二、填空题 · 立体几何

已知在半径为 2 的球面上有 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}、D\htmlData{tutor-start=0,tutor-end=1}{D} 四点, 若 AB=CD=2\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{D}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}, 则四面体 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} 的体积的最大值为 ( )

答案：4√3/3

题目标签：球内四面体最大体积

解题过程

求球内四面体体积最大值

用一对对棱及其中点连线表示体积

(1)
求两弦中点的活动范围

半径为 2 的球中，长为 2 的弦到球心的距离固定。

为什么从这里入手：“半径为 2 的球中，长为 2 的弦到球心的距离固定。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：设 M,N\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{N} 为 AB,CD\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}{C}\htmlData{tutor-start=4,tutor-end=5}{D} 中点，则 OM=ON=22−12=3\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{O}\htmlData{tutor-start=4,tutor-end=5}{N}\htmlData{tutor-start=5,tutor-end=6}{=}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}^{\htmlData{tutor-start=21,tutor-end=22}{2}}}\htmlData{tutor-start=24,tutor-end=25}{=}\sqrt{\htmlData{tutor-start=31,tutor-end=32}{3}}，所以 MN≤23\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}\htmlData{tutor-start=2,tutor-end=5}{\le}\htmlData{tutor-start=5,tutor-end=6}{2}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}。

MN≤23\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{N}\htmlData{tutor-start=2,tutor-end=5}{\le}\htmlData{tutor-start=5,tutor-end=6}{2}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}
(2)
放大混合积并构造等号

四面体体积可写为 V=∣AB→×CD→⋅MN→∣/6\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{|}\overrightarrow{\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{B}}\htmlData{tutor-start=22,tutor-end=28}{\times}\overrightarrow{\htmlData{tutor-start=44,tutor-end=45}{C}\htmlData{tutor-start=45,tutor-end=46}{D}}\htmlData{tutor-start=47,tutor-end=52}{\cdot}\overrightarrow{\htmlData{tutor-start=68,tutor-end=69}{M}\htmlData{tutor-start=69,tutor-end=70}{N}}\htmlData{tutor-start=71,tutor-end=72}{|}\htmlData{tutor-start=72,tutor-end=73}{/}\htmlData{tutor-start=73,tutor-end=74}{6}。

为什么从这里入手：“四面体体积可写为 V=∣AB→×CD→⋅MN→∣/6\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{|}\overrightarrow{\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{B}}\htmlData{tutor-start=22,tutor-end=28}{\times}\overrightarrow{\htmlData{tutor-start=44,tutor-end=45}{C}\htmlData{tutor-start=45,tutor-end=46}{D}}\htmlData{tutor-start=47,tutor-end=52}{\cdot}\overrightarrow{\htmlData{tutor-start=68,tutor-end=69}{M}\htmlData{tutor-start=69,tutor-end=70}{N}}\htmlData{tutor-start=71,tutor-end=72}{|}\htmlData{tutor-start=72,tutor-end=73}{/}\htmlData{tutor-start=73,tutor-end=74}{6}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：故 V≤(2⋅2⋅23)/6=43/3\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=11}{\cdot}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=17}{\cdot}\htmlData{tutor-start=17,tutor-end=18}{2}\sqrt{\htmlData{tutor-start=24,tutor-end=25}{3}}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{6}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{4}\sqrt{\htmlData{tutor-start=37,tutor-end=38}{3}}\htmlData{tutor-start=39,tutor-end=40}{/}\htmlData{tutor-start=40,tutor-end=41}{3}。当 M,N\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{N} 关于球心对称且 AB,CD,MN\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}{C}\htmlData{tutor-start=4,tutor-end=5}{D}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{M}\htmlData{tutor-start=7,tutor-end=8}{N} 两两垂直时取等号。

Vmax⁡=433\boxed{\htmlData{tutor-start=7,tutor-end=8}{V}_{\max}\htmlData{tutor-start=15,tutor-end=16}{=}\frac{\htmlData{tutor-start=22,tutor-end=23}{4}\sqrt{\htmlData{tutor-start=29,tutor-end=30}{3}}}{\htmlData{tutor-start=33,tutor-end=34}{3}}}
13

二、填空题 · 不等式

不等式 x−2x2+3x+2>0\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}}{\htmlData{tutor-start=11,tutor-end=12}{x}^{\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=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{>}\htmlData{tutor-start=23,tutor-end=24}{0} 的解集是______.

答案：(-2,-1)∪(2,+∞)

题目标签：分式不等式

解题过程

解分式不等式

因式分解并作符号表

(1)
找临界点

分母 x2+3x+2=(x+1)(x+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}{3}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{)}。

为什么从这里入手：“分母 x2+3x+2=(x+1)(x+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}{3}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{)}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：零点或无定义点为 −2,−1,2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=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}{2}，其中 −2,−1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1} 必须排除，2 因严格不等式也排除。

x−2(x+1)(x+2)>0\frac{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}}{\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{)}}\htmlData{tutor-start=22,tutor-end=23}{>}\htmlData{tutor-start=23,tutor-end=24}{0}
(2)
读取正号区间

按临界点把数轴分成四段。

为什么从这里入手：“按临界点把数轴分成四段。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。目标是“读取正号区间”，而“按临界点把数轴分成四段。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：逐段判断符号，正号出现在 (−2,−1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)} 与 (2,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=10}{\infty}\htmlData{tutor-start=10,tutor-end=11}{)}。

(−2,−1)∪(2,+∞)\boxed{\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=18}{\cup}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=28}{\infty}\htmlData{tutor-start=28,tutor-end=29}{)}}
14

二、填空题 · 三角函数

已知 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 为第二象限的角, sin⁡α=35\sin \htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\frac{\htmlData{tutor-start=18,tutor-end=19}{3}}{\htmlData{tutor-start=21,tutor-end=22}{5}}, 则 tan⁡2α=\tan \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=13}{=}______.

答案：-24/7

题目标签：二倍角正切

解题过程

由象限求二倍角正切

先补出余弦和正切

(1)
确定一倍角正切

第二象限中正弦为正、余弦为负。

为什么从这里入手：“第二象限中正弦为正、余弦为负。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：由 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} 得 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}，所以 tan⁡α=−3/4\tan\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}{3}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{4}。

tan⁡α=−34\tan\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{-}\frac{\htmlData{tutor-start=18,tutor-end=19}{3}}{\htmlData{tutor-start=21,tutor-end=22}{4}}
(2)
使用二倍角公式

正切二倍角只需一倍角正切。

为什么从这里入手：“正切二倍角只需一倍角正切。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。三角题先把未知角转成特殊角、同一角或已知边角关系。“正切二倍角只需一倍角正切。”指出了可用的象限、恒等式或几何关系，先使用二倍角公式可以同时确定数值和正负号。

详细展开：tan⁡2α=2(−3/4)1−9/16=−24/7\tan\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\frac{\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{3}\htmlData{tutor-start=22,tutor-end=23}{/}\htmlData{tutor-start=23,tutor-end=24}{4}\htmlData{tutor-start=24,tutor-end=25}{)}}{\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{9}\htmlData{tutor-start=30,tutor-end=31}{/}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{6}}\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}{4}\htmlData{tutor-start=38,tutor-end=39}{/}\htmlData{tutor-start=39,tutor-end=40}{7}。

−247\boxed{\htmlData{tutor-start=7,tutor-end=8}{-}\frac{\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{4}}{\htmlData{tutor-start=18,tutor-end=19}{7}}}
15

二、填空题 · 排列组合

某学校开设 A\htmlData{tutor-start=0,tutor-end=1}{A} 类选修课 3 门, B\htmlData{tutor-start=0,tutor-end=1}{B} 类选修课 4 门, 一位同学从中共选 3 门, 若要求两类课程中各至少选一门, 则不同的选法共有______种. (用数字作答)

答案：30\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{0}

题目标签：两类选修课组合

解题过程

计算两类课程均非空的选法

用总数减去只选一类

(1)
计算不加限制的总数

7 门不同课程中选 3 门。

为什么从这里入手：“7 门不同课程中选 3 门。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。目标是“计算不加限制的总数”，而“7 门不同课程中选 3 门。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：共有 (73)=35\binom73=35 种。

Nall=35\htmlData{tutor-start=0,tutor-end=1}{N}_{\rm \htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{l}\htmlData{tutor-start=9,tutor-end=10}{l}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{5}
(2)
排除缺少一类的情况

只选 A 类必须三门全选，只有 1 种；只选 B 类有 (43)=4\binom43=4 种。

为什么从这里入手：“只选 A 类必须三门全选，只有 1 种；只选 B 类有 (43)=4\binom43=4 种。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：所求为 35−1−4=30\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{5}\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}{4}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{0}。

30\boxed{\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{0}}
16

二、填空题 · 圆锥曲线

已知 F\htmlData{tutor-start=0,tutor-end=1}{F} 是椭圆 C\htmlData{tutor-start=0,tutor-end=1}{C} 的一个焦点, B\htmlData{tutor-start=0,tutor-end=1}{B} 是短轴的一个端点, 线段 BF\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{F} 的延长线交 C\htmlData{tutor-start=0,tutor-end=1}{C} 于点 D\htmlData{tutor-start=0,tutor-end=1}{D}, 且 BF→=2FD→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{F}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{2}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{F}\htmlData{tutor-start=38,tutor-end=39}{D}}, 则 C\htmlData{tutor-start=0,tutor-end=1}{C} 的离心率为______.

答案：√3/3

题目标签：椭圆离心率

解题过程

求椭圆离心率

给焦点和短轴端点设标准坐标

(1)
表示延长线上的点

不妨设椭圆焦点 F=(c,0)\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}、短轴端点 B=(0,b)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{)}。

为什么从这里入手：“不妨设椭圆焦点 F=(c,0)\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}、短轴端点 B=(0,b)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{)}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：由 BF→=2FD→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{F}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{2}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{F}\htmlData{tutor-start=38,tutor-end=39}{D}} 得 D=F+12(F−B)=(3c/2,−b/2)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{+}\frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{F}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{B}\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}{3}\htmlData{tutor-start=23,tutor-end=24}{c}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{b}\htmlData{tutor-start=29,tutor-end=30}{/}\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{)}。

D=(3c2,−b2)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{=}\left(\frac{\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{c}}{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=22}{-}\frac{\htmlData{tutor-start=28,tutor-end=29}{b}}{\htmlData{tutor-start=31,tutor-end=32}{2}}\right)
(2)
代入椭圆方程

点 D\htmlData{tutor-start=0,tutor-end=1}{D} 在椭圆上，且 e=c/a\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{a}。

为什么从这里入手：“点 D\htmlData{tutor-start=0,tutor-end=1}{D} 在椭圆上，且 e=c/a\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{a}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：9c24a2+14=1\frac{\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{c}^{\htmlData{tutor-start=10,tutor-end=11}{2}}}{\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{a}^{\htmlData{tutor-start=18,tutor-end=19}{2}}}\htmlData{tutor-start=21,tutor-end=22}{+}\frac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{4}}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{1}，所以 9e2=3\htmlData{tutor-start=0,tutor-end=1}{9}\htmlData{tutor-start=1,tutor-end=2}{e}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{3}，取正值得 e=3/3\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{3}。

e=33\boxed{\htmlData{tutor-start=7,tutor-end=8}{e}\htmlData{tutor-start=8,tutor-end=9}{=}\frac{\sqrt{\htmlData{tutor-start=21,tutor-end=22}{3}}}{\htmlData{tutor-start=25,tutor-end=26}{3}}}
17

三、解答题 · 数列

记等差数列 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 的前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项的和为 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}, 设 S3=12\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{2}, 且 2a1,a2,a3+1\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{1} 成等比数列, 求 Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}.

答案：n(3n-1)/2 或 2n(5-n)

题目标签：等差数列前n项和

解题过程

求等差数列前 n\htmlData{tutor-start=0,tutor-end=1}{n} 项和

用 S3\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{3}} 锁定中项，再解公差方程

(1)
设公差并使用等比条件

S3=3a2=12\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\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}{1}\htmlData{tutor-start=14,tutor-end=15}{2}，所以 a2=4\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}{4}。

为什么从这里入手：“S3=3a2=12\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\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}{1}\htmlData{tutor-start=14,tutor-end=15}{2}，所以 a2=4\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}{4}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：设公差为 d\htmlData{tutor-start=0,tutor-end=1}{d}，则 a1=4−d,a3=4+d\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}{4}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{d}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{d}。2a1,a2,a3+1\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=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}{a}_{\htmlData{tutor-start=16,tutor-end=17}{3}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{1} 成等比数列给出 a22=2a1(a3+1)\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{1}}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{a}_{\htmlData{tutor-start=20,tutor-end=21}{3}}\htmlData{tutor-start=22,tutor-end=23}{+}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}。

16=2(4−d)(5+d)\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{d}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{d}\htmlData{tutor-start=13,tutor-end=14}{)}
(2)
解出两种数列

题设没有规定公差符号，所以两个实根都要保留。

为什么从这里入手：“题设没有规定公差符号，所以两个实根都要保留。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：d2+d−12=0\htmlData{tutor-start=0,tutor-end=1}{d}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{d}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0}，得 d=3\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3} 或 d=−4\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{4}。相应地 Sn=n2[2a1+(n−1)d]\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{n}}{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{[}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{1}}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{n}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{d}\htmlData{tutor-start=31,tutor-end=32}{]}。

Sn=n(3n−1)2或Sn=2n(5−n)\boxed{S_{n}=\frac{n(3n-1)}2\quad\text{或}\quad S_{n}=2n(5-n)}
18

三、解答题 · 解三角形

已知 △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,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B} 及其对边 a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b} 满足 a+b=acot⁡A+bcot⁡B\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{a} \cot \htmlData{tutor-start=11,tutor-end=12}{A}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{b} \cot \htmlData{tutor-start=20,tutor-end=21}{B}, 求内角 C\htmlData{tutor-start=0,tutor-end=1}{C}.

答案：π/2

题目标签：边角关系求角

解题过程

由边角混合等式求第三角

用正弦定理消去边长

(1)
统一成角的三角函数

边 a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b} 分别与 sin⁡A,sin⁡B\sin \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{,}\sin \htmlData{tutor-start=12,tutor-end=13}{B} 成比例。

为什么从这里入手：“边 a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b} 分别与 sin⁡A,sin⁡B\sin \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{,}\sin \htmlData{tutor-start=12,tutor-end=13}{B} 成比例。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：设 a=2Rsin⁡A,b=2Rsin⁡B\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{R}\sin \htmlData{tutor-start=9,tutor-end=10}{A}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{R}\sin \htmlData{tutor-start=20,tutor-end=21}{B}。原式除以 2R\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{R} 后变为 sin⁡A+sin⁡B=cos⁡A+cos⁡B\sin \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{+}\sin \htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{=}\cos \htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{+}\cos \htmlData{tutor-start=26,tutor-end=27}{B}。

sin⁡A+sin⁡B=cos⁡A+cos⁡B\sin \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{+}\sin \htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{=}\cos \htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{+}\cos \htmlData{tutor-start=26,tutor-end=27}{B}
(2)
使用和差化积

两边都有公共因子 cos⁡((A−B)/2)>0\cos\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{>}\htmlData{tutor-start=14,tutor-end=15}{0}。

为什么从这里入手：“两边都有公共因子 cos⁡((A−B)/2)>0\cos\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{>}\htmlData{tutor-start=14,tutor-end=15}{0}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：化简得 sin⁡((A+B)/2)=cos⁡((A+B)/2)\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{)}\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}{=}\cos\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{A}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{B}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{)}。因 (A+B)/2∈(0,π/2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=10}{\in}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=16}{\pi}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}，故 A+B=π/2\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=7}{\pi}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}，所以 C=π/2\htmlData{tutor-start=0,tutor-end=1}{C}\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}{2}。

C=π2\boxed{C=\frac\pi2}
19

三、解答题 · 概率

投到某杂志的稿件, 先由两位初审专家进行评审. 若能通过两位初审专家的评审, 则予以录用; 若两位初审专家都未予通过, 则不予录用; 若恰能通过一位初审专家的评审, 则再由第三位专家进行复审, 若能通过复审专家的评审, 则予以录用, 否则不予录用. 设稿件能通过各初审专家评审的概率均为 0.5, 复审的稿件能通过评审的概率为 0.3. 各专家独立评审. (1) 求投到该杂志的 1 篇稿件被录用的概率; (2) 求投到该杂志的 4 篇稿件中, 至少有 2 篇被录用的概率.

答案：2/5；328/625

题目标签：稿件录用概率

解题过程

（1）求一篇稿件被录用的概率

按初审结果互斥分类

(1)
计算直接录用与复审录用

录用有两条互斥路径：两位初审都通过，或恰一位通过且复审通过。

为什么从这里入手：“录用有两条互斥路径：两位初审都通过，或恰一位通过且复审通过。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：前者概率为 0.52=0.25\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{5}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{.}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{5}；恰一位初审通过概率为 2⋅0.5⋅0.5=0.5\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=6}{\cdot}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{.}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=14}{\cdot}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{.}\htmlData{tutor-start=16,tutor-end=17}{5}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{.}\htmlData{tutor-start=20,tutor-end=21}{5}，再乘复审概率 0.3。

p=0.25+0.5×0.3\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{5}\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}{5}\htmlData{tutor-start=10,tutor-end=16}{\times}\htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{.}\htmlData{tutor-start=18,tutor-end=19}{3}
(2)
合并概率

互斥事件概率相加。

为什么从这里入手：“互斥事件概率相加。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。概率题的关键不是立刻代数，而是先说清随机试验与事件。“互斥事件概率相加。”确定了基本事件或随机变量，所以先合并概率，再依据独立、互斥、对立或期望线性性计算。

详细展开：p=0.25+0.15=0.4=2/5\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{5}\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}{1}\htmlData{tutor-start=10,tutor-end=11}{5}\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}{4}\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}{5}。

p=25\boxed{\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{=}\frac{\htmlData{tutor-start=15,tutor-end=16}{2}}{\htmlData{tutor-start=18,tutor-end=19}{5}}}

（2）求四篇中至少两篇录用的概率

识别二项分布并用对立事件

(1)
建立二项模型

各篇稿件按同样规则独立评审。

为什么从这里入手：“各篇稿件按同样规则独立评审。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。目标是“建立二项模型”，而“各篇稿件按同样规则独立评审。”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：录用篇数 X∼B(4,2/5)\htmlData{tutor-start=0,tutor-end=1}{X}\htmlData{tutor-start=1,tutor-end=6}{\sim }\htmlData{tutor-start=6,tutor-end=7}{B}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{5}\htmlData{tutor-start=13,tutor-end=14}{)}，所以“至少 2 篇”的对立情形是 0 篇或 1 篇。

P(X≥2)=1−P(X=0)−P(X=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=6}{\ge}\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}{1}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{P}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{X}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{P}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{X}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{)}
(2)
代入二项概率

不录用概率为 3/5\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}。

为什么从这里入手：“不录用概率为 3/5\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。概率题的关键不是立刻代数，而是先说清随机试验与事件。“不录用概率为 3/5\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}。”确定了基本事件或随机变量，所以先代入二项概率，再依据独立、互斥、对立或期望线性性计算。

详细展开：P(X≥2)=1−(3/5)4−4(2/5)(3/5)3=328/625\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=6}{\ge}\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}{1}\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=14}{/}\htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=16}{)}^{\htmlData{tutor-start=18,tutor-end=19}{4}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{4}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{5}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=30}{/}\htmlData{tutor-start=30,tutor-end=31}{5}\htmlData{tutor-start=31,tutor-end=32}{)}^{\htmlData{tutor-start=34,tutor-end=35}{3}}\htmlData{tutor-start=36,tutor-end=37}{=}\htmlData{tutor-start=37,tutor-end=38}{3}\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{8}\htmlData{tutor-start=40,tutor-end=41}{/}\htmlData{tutor-start=41,tutor-end=42}{6}\htmlData{tutor-start=42,tutor-end=43}{2}\htmlData{tutor-start=43,tutor-end=44}{5}。

328625\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{8}}{\htmlData{tutor-start=18,tutor-end=19}{6}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{5}}}
20

解答题 · 立体几何

如图，四棱锥 S−ABCD\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{D} 中，SD⊥\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=8}{\perp} 底面 ABCD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}，AB//DC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{/} \htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=8}{C}，AD⊥DC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{D}\htmlData{tutor-start=10,tutor-end=11}{C}，AB=AD=1\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}{D}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}，DC=SD=2\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{S}\htmlData{tutor-start=4,tutor-end=5}{D}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}，E\htmlData{tutor-start=0,tutor-end=1}{E} 为棱 SB\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{B} 上的一点，平面 EDC⊥\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{C} \htmlData{tutor-start=4,tutor-end=9}{\perp} 平面 SBC\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}. (1) 证明：SE=2EB\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{E}\htmlData{tutor-start=5,tutor-end=6}{B}; (2) 求二面角 A−DE−C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{E}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{C} 的大小.

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

答案：SE=2EB；2π/3

题目标签：四棱锥分点与二面角

解题过程

（1）证明棱上分点比例

坐标化两个垂直平面的法向量

(1)
给分点设参数

底面是直角梯形，且 SD\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{D} 垂直底面。

为什么从这里入手：“底面是直角梯形，且 SD\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{D} 垂直底面。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：取 D(0,0,0),C(2,0,0),A(0,1,0),B(1,1,0),S(0,0,2)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{C}\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}{0}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{A}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=23}{1}\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}{,}\htmlData{tutor-start=27,tutor-end=28}{B}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{,}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{,}\htmlData{tutor-start=33,tutor-end=34}{0}\htmlData{tutor-start=34,tutor-end=35}{)}\htmlData{tutor-start=35,tutor-end=36}{,}\htmlData{tutor-start=36,tutor-end=37}{S}\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{0}\htmlData{tutor-start=39,tutor-end=40}{,}\htmlData{tutor-start=40,tutor-end=41}{0}\htmlData{tutor-start=41,tutor-end=42}{,}\htmlData{tutor-start=42,tutor-end=43}{2}\htmlData{tutor-start=43,tutor-end=44}{)}。设 E=S+t(B−S)=(t,t,2−2t)\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{S}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{B}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{S}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{t}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{t}\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}{2}\htmlData{tutor-start=19,tutor-end=20}{t}\htmlData{tutor-start=20,tutor-end=21}{)}，则 t=SE/SB\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{S}\htmlData{tutor-start=3,tutor-end=4}{E}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{S}\htmlData{tutor-start=6,tutor-end=7}{B}。

E=(t,t,2−2t)\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}{t}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{t}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{t}\htmlData{tutor-start=11,tutor-end=12}{)}
(2)
使用面面垂直

两平面垂直等价于其法向量垂直。

为什么从这里入手：“两平面垂直等价于其法向量垂直。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：平面 EDC\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{C} 法向量可取 (0,−2+2t,t)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{-}\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}{t}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{)}，平面 SBC\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C} 法向量可取 (1,1,1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{)}。点积为 −2+3t=0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，故 t=2/3\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{3}。

SE=2EB\boxed{\htmlData{tutor-start=7,tutor-end=8}{S}\htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{B}}

（2）求二面角 A−DE−C\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{E}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{C}

在垂直于棱 DE\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E} 的截面中比较两侧方向

(1)
构造两个垂棱向量

代入 t=2/3\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{3} 后，DE→∥(1,1,1)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{D}\htmlData{tutor-start=17,tutor-end=18}{E}}\htmlData{tutor-start=19,tutor-end=28}{\parallel}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{,}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{,}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{)}。

为什么从这里入手：“代入 t=2/3\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{3} 后，DE→∥(1,1,1)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{D}\htmlData{tutor-start=17,tutor-end=18}{E}}\htmlData{tutor-start=19,tutor-end=28}{\parallel}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{,}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{,}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{)}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：在平面 ADE\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{E} 内指向 A\htmlData{tutor-start=0,tutor-end=1}{A} 且垂直 DE\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E} 的向量可取 (−1,2,−1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}；在平面 CDE\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{E} 内指向 C\htmlData{tutor-start=0,tutor-end=1}{C} 的对应向量可取 (2,−1,−1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{-}\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}{1}\htmlData{tutor-start=8,tutor-end=9}{)}。

u=(−1,2,−1),v=(2,−1,−1)\boldsymbol u=(-1,2,-1),\quad\boldsymbol v=(2,-1,-1)
(2)
计算二面角

二面角由两向量的夹角给出，方向必须分别指向 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}。

为什么从这里入手：“二面角由两向量的夹角给出，方向必须分别指向 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}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：cos⁡θ=−366=−1/2\cos\htmlData{tutor-start=4,tutor-end=10}{\theta}\htmlData{tutor-start=10,tutor-end=11}{=}\frac{\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{3}}{\sqrt{\htmlData{tutor-start=27,tutor-end=28}{6}}\sqrt{\htmlData{tutor-start=35,tutor-end=36}{6}}}\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}{2}，故 θ=2π/3\htmlData{tutor-start=0,tutor-end=6}{\theta}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=11}{\pi}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{3}。

∠(A−DE−C)=2π3\boxed{\htmlData{tutor-start=7,tutor-end=13}{\angle}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{D}\htmlData{tutor-start=17,tutor-end=18}{E}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{C}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{=}\frac{\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=32}{\pi}}{\htmlData{tutor-start=34,tutor-end=35}{3}}}
21

解答题 · 导数

已知函数 f(x)=3ax4−2(3a+1)x2+4x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{4}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{a}\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}{x}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{4}\htmlData{tutor-start=27,tutor-end=28}{x}. (1) 当 a=16\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{6}} 时，求 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 的极值; (2) 若 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 在 (−1,1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} 上是增函数，求 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围.

答案：极小值-12；-4/3≤a≤1/6

题目标签：四次函数极值与单调

解题过程

（1）求指定参数下的极值

因式分解导函数并观察变号

(1)
求导并分解

当 a=1/6\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}{6} 时，f(x)=12x4−3x2+4x\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}{=}\frac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{x}^{\htmlData{tutor-start=19,tutor-end=20}{4}}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{x}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{4}\htmlData{tutor-start=30,tutor-end=31}{x}。

为什么从这里入手：“当 a=1/6\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}{6} 时，f(x)=12x4−3x2+4x\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}{=}\frac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{x}^{\htmlData{tutor-start=19,tutor-end=20}{4}}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{x}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{4}\htmlData{tutor-start=30,tutor-end=31}{x}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：f′(x)=2x3−6x+4=2(x+2)(x−1)2\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{6}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{x}\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}{(}\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{)}^{\htmlData{tutor-start=31,tutor-end=32}{2}}。

f′(x)=2(x+2)(x−1)2\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{x}\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=19,tutor-end=20}{2}}
(2)
用导数变号判断极值

二重根 x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 两侧符号不变，不产生极值。

为什么从这里入手：“二重根 x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 两侧符号不变，不产生极值。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：f′\htmlData{tutor-start=0,tutor-end=1}{f}' 在 x=−2\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}{2} 处由负变正，所以只有极小值 f(−2)=8−12−8=−12\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{8}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{2}。

x=−2 时极小值为 −12\boxed{\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}{2}\text{ \htmlData{tutor-start=18,tutor-end=19}{时}\htmlData{tutor-start=19,tutor-end=20}{极}\htmlData{tutor-start=20,tutor-end=21}{小}\htmlData{tutor-start=21,tutor-end=22}{值}\htmlData{tutor-start=22,tutor-end=23}{为} }\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{2}}

（2）求在 (−1,1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} 上递增的参数

把导函数分解为固定符号因子与含参二次式

(1)
分解导函数

区间内 x−1<0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{0}。

为什么从这里入手：“区间内 x−1<0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{0}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。目标是“分解导函数”，而“区间内 x−1<0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{0}。”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：f′(x)=4[3ax3−(3a+1)x+1]=4(x−1)(3ax2+3ax−1)\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}{4}\htmlData{tutor-start=7,tutor-end=8}{[}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{x}^{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{a}\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=22,tutor-end=23}{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=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}{x}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{)}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{3}\htmlData{tutor-start=35,tutor-end=36}{a}\htmlData{tutor-start=36,tutor-end=37}{x}^{\htmlData{tutor-start=39,tutor-end=40}{2}}\htmlData{tutor-start=41,tutor-end=42}{+}\htmlData{tutor-start=42,tutor-end=43}{3}\htmlData{tutor-start=43,tutor-end=44}{a}\htmlData{tutor-start=44,tutor-end=45}{x}\htmlData{tutor-start=45,tutor-end=46}{-}\htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{)}。因此递增要求 h(x)=3ax2+3ax−1≤0\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{a}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=21}{\le}\htmlData{tutor-start=21,tutor-end=22}{0} 对 −1<x<1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{<}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{<}\htmlData{tutor-start=5,tutor-end=6}{1} 成立。

h(x)=3ax2+3ax−1≤0\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{a}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=21}{\le}\htmlData{tutor-start=21,tutor-end=22}{0}
(2)
按抛物线开口求最大值

只需控制 h\htmlData{tutor-start=0,tutor-end=1}{h} 在区间上的最大值。

为什么从这里入手：“只需控制 h\htmlData{tutor-start=0,tutor-end=1}{h} 在区间上的最大值。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：当 a≥0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{0}，最大值由 x→1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\to}\htmlData{tutor-start=4,tutor-end=5}{1} 给出，要求 6a−1≤0\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=7}{\le}\htmlData{tutor-start=7,tutor-end=8}{0}；当 a<0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0}，顶点 x=−1/2\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}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2} 处最大，要求 −3a/4−1≤0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=10}{\le}\htmlData{tutor-start=10,tutor-end=11}{0}。合并得 −4/3≤a≤1/6\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=8}{\le }\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=12}{\le}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{6}。

−43≤a≤16\boxed{\htmlData{tutor-start=7,tutor-end=8}{-}\frac{\htmlData{tutor-start=14,tutor-end=15}{4}}{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=23}{\le }\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=27}{\le}\frac{\htmlData{tutor-start=33,tutor-end=34}{1}}{\htmlData{tutor-start=36,tutor-end=37}{6}}}
22

解答题 · 解析几何

已知抛物线 C:y2=4x\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{y}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{x} 的焦点为 F\htmlData{tutor-start=0,tutor-end=1}{F}，过点 K(−1,0)\htmlData{tutor-start=0,tutor-end=1}{K}\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}{)} 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，点 A\htmlData{tutor-start=0,tutor-end=1}{A} 关于 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的对称点为 D\htmlData{tutor-start=0,tutor-end=1}{D}. (1) 证明：点 F\htmlData{tutor-start=0,tutor-end=1}{F} 在直线 BD\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D} 上; (2) 设 FA→⋅FB→=89\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{F}\htmlData{tutor-start=17,tutor-end=18}{A}} \htmlData{tutor-start=20,tutor-end=26}{\cdot }\overrightarrow{\htmlData{tutor-start=42,tutor-end=43}{F}\htmlData{tutor-start=43,tutor-end=44}{B}} \htmlData{tutor-start=46,tutor-end=47}{=} \frac{\htmlData{tutor-start=54,tutor-end=55}{8}}{\htmlData{tutor-start=57,tutor-end=58}{9}}，求 △BDK\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{D}\htmlData{tutor-start=12,tutor-end=13}{K} 的内切圆 M\htmlData{tutor-start=0,tutor-end=1}{M} 的方程.

答案：见证明；(x-1/9)²+y²=4/9

题目标签：抛物线弦与内切圆

解题过程

（1）证明焦点在指定直线上

用抛物线参数表示两个交点

(1)
写出过定点弦的参数关系

设 A=(t2,2t),B=(s2,2s)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{t}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{t}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{s}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{s}\htmlData{tutor-start=24,tutor-end=25}{)}。

为什么从这里入手：“设 A=(t2,2t),B=(s2,2s)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{t}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{t}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{s}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{s}\htmlData{tutor-start=24,tutor-end=25}{)}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：抛物线两参数点的弦方程为 (t+s)y=2x+2ts\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{s}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{y}\htmlData{tutor-start=6,tutor-end=7}{=}\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}\htmlData{tutor-start=11,tutor-end=12}{t}\htmlData{tutor-start=12,tutor-end=13}{s}。代入 K(−1,0)\htmlData{tutor-start=0,tutor-end=1}{K}\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}{)} 得 ts=1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{s}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}，故可写 s=1/t\htmlData{tutor-start=0,tutor-end=1}{s}\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}{t}；D=(t2,−2t)\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}{t}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{t}\htmlData{tutor-start=12,tutor-end=13}{)}。

A=(t2,2t),B=(t−2,2t−1),ts=1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{t}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{t}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{,}\quad \htmlData{tutor-start=19,tutor-end=20}{B}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{t}^{\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{,}\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{t}^{\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}{,}\quad \htmlData{tutor-start=44,tutor-end=45}{t}\htmlData{tutor-start=45,tutor-end=46}{s}\htmlData{tutor-start=46,tutor-end=47}{=}\htmlData{tutor-start=47,tutor-end=48}{1}
(2)
比较两条直线斜率

焦点为 F=(1,0)\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\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}{)}。

为什么从这里入手：“焦点为 F=(1,0)\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\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}{)}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。解析几何的目的，是把图形关系变成坐标等式。“焦点为 F=(1,0)\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\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}{)}。”提供了点、斜率、距离或焦点条件，先比较两条直线斜率后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：kDF=2t/(1−t2)\htmlData{tutor-start=0,tutor-end=1}{k}_{\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{F}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{t}\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}{t}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{)}；直接用 B,D\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{D} 坐标计算也得 kDB=2t/(1−t2)\htmlData{tutor-start=0,tutor-end=1}{k}_{\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{B}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{t}\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}{t}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{)}，所以 D,B,F\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{F} 共线。

F∈BD\boxed{\htmlData{tutor-start=7,tutor-end=8}{F}\htmlData{tutor-start=8,tutor-end=12}{\in }\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{D}}

（2）求三角形内切圆

由向量条件确定参数对称量，再算内切圆半径和圆心

(1)
化简向量点积条件

已有 s=1/t\htmlData{tutor-start=0,tutor-end=1}{s}\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}{t}。

为什么从这里入手：“已有 s=1/t\htmlData{tutor-start=0,tutor-end=1}{s}\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}{t}。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。空间关系只靠观察容易漏条件，而“已有 s=1/t\htmlData{tutor-start=0,tutor-end=1}{s}\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}{t}。”可以直接翻译成垂直、平行、数量积或体积公式。先化简向量点积条件，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：FA→⋅FB→=6−(t2+t−2)=8/9\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{F}\htmlData{tutor-start=17,tutor-end=18}{A}}\htmlData{tutor-start=19,tutor-end=24}{\cdot}\overrightarrow{\htmlData{tutor-start=40,tutor-end=41}{F}\htmlData{tutor-start=41,tutor-end=42}{B}}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{6}\htmlData{tutor-start=45,tutor-end=46}{-}\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{t}^{\htmlData{tutor-start=50,tutor-end=51}{2}}\htmlData{tutor-start=52,tutor-end=53}{+}\htmlData{tutor-start=53,tutor-end=54}{t}^{\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{2}}\htmlData{tutor-start=59,tutor-end=60}{)}\htmlData{tutor-start=60,tutor-end=61}{=}\htmlData{tutor-start=61,tutor-end=62}{8}\htmlData{tutor-start=62,tutor-end=63}{/}\htmlData{tutor-start=63,tutor-end=64}{9}，故 (t+t−1)2=64/9\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{t}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{t}^{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{)}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{6}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{9}。由抛物线距离公式可得 BD=64/9\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{6}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{9}，BK+DK=80/9\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{K}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{K}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{9}。

BD=649,BK+DK=809\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{=}\frac{\htmlData{tutor-start=9,tutor-end=10}{6}\htmlData{tutor-start=10,tutor-end=11}{4}}{\htmlData{tutor-start=13,tutor-end=14}{9}}\htmlData{tutor-start=15,tutor-end=16}{,}\quad \htmlData{tutor-start=22,tutor-end=23}{B}\htmlData{tutor-start=23,tutor-end=24}{K}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{D}\htmlData{tutor-start=26,tutor-end=27}{K}\htmlData{tutor-start=27,tutor-end=28}{=}\frac{\htmlData{tutor-start=34,tutor-end=35}{8}\htmlData{tutor-start=35,tutor-end=36}{0}}{\htmlData{tutor-start=38,tutor-end=39}{9}}
(2)
求内心与半径

直线 KF\htmlData{tutor-start=0,tutor-end=1}{K}\htmlData{tutor-start=1,tutor-end=2}{F} 是顶角 K\htmlData{tutor-start=0,tutor-end=1}{K} 的角平分线，因此内心在 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴上。

为什么从这里入手：“直线 KF\htmlData{tutor-start=0,tutor-end=1}{K}\htmlData{tutor-start=1,tutor-end=2}{F} 是顶角 K\htmlData{tutor-start=0,tutor-end=1}{K} 的角平分线，因此内心在 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴上。”是题面中最能直接转成公式的条件，先处理它可以缩短推理链。

详细展开：三角形半周长为 8；K\htmlData{tutor-start=0,tutor-end=1}{K} 到 BD\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D} 的距离为 3/2\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2}，面积为 16/3\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{3}，故内切圆半径 r=2/3\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{3}。设内心 (h,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{h}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}，到 BD\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D} 距离为 2/3\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3} 得 h=1/9\htmlData{tutor-start=0,tutor-end=1}{h}\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}{9}。

(x−19)2+y2=49\boxed{\left(\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{-}\frac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{9}}\right)^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{y}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{=}\frac{\htmlData{tutor-start=50,tutor-end=51}{4}}{\htmlData{tutor-start=53,tutor-end=54}{9}}}