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2010 年高考数学（大纲卷 1理科）

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

22 题27 个小问/题组
1

一、选择题 · 复数

复数 3+2i2−3i=\frac{\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\mathrm{\htmlData{tutor-start=17,tutor-end=18}{i}}}{\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{3}\mathrm{\htmlData{tutor-start=32,tutor-end=33}{i}}}\htmlData{tutor-start=35,tutor-end=36}{=} ( ) (A) i\mathrm{\htmlData{tutor-start=8,tutor-end=9}{i}} (B) −i\htmlData{tutor-start=0,tutor-end=1}{-}\mathrm{\htmlData{tutor-start=9,tutor-end=10}{i}} (C) 12−13i\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{3}\mathrm{\htmlData{tutor-start=13,tutor-end=14}{i}} (D) 12+13i\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{3}\mathrm{\htmlData{tutor-start=13,tutor-end=14}{i}}

答案：A；i

题目标签：复数商

解题过程

计算复数商

乘以分母的共轭复数

(1)
实化分母

分母为 2−3i\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{i}，共轭复数是 2+3i\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{i}。

为什么从这里入手：“分母为 2−3i\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{i}，共轭复数是 2+3i\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{i}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：分子分母同乘 2+3i\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{i}，分母成为 22+32=13\htmlData{tutor-start=0,tutor-end=1}{2}^{\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=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{3}。

3+2i2−3i=(3+2i)(2+3i)13\frac{\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{i}}{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{i}}\htmlData{tutor-start=17,tutor-end=18}{=}\frac{\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{i}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{3}\htmlData{tutor-start=34,tutor-end=35}{i}\htmlData{tutor-start=35,tutor-end=36}{)}}{\htmlData{tutor-start=38,tutor-end=39}{1}\htmlData{tutor-start=39,tutor-end=40}{3}}
(2)
展开分子

复数乘法按分配律，并使用 i2=−1\htmlData{tutor-start=0,tutor-end=1}{i}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}。

为什么从这里入手：“复数乘法按分配律，并使用 i2=−1\htmlData{tutor-start=0,tutor-end=1}{i}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：分子为 6+9i+4i+6i2=13i\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{9}\htmlData{tutor-start=3,tutor-end=4}{i}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=10}{i}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{i}，所以商为 i\htmlData{tutor-start=0,tutor-end=1}{i}，选 A。

iA\boxed{\htmlData{tutor-start=7,tutor-end=8}{i}}\quad\text{\htmlData{tutor-start=20,tutor-end=21}{A}}
2

一、选择题 · 三角函数

记 cos⁡(−80∘)=k\cos(-80^\circ)=k，那么 tan⁡100∘=\tan 100^\circ= ( ) (A) 1−k2k\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{k}^{\htmlData{tutor-start=17,tutor-end=18}{2}}}}{\htmlData{tutor-start=22,tutor-end=23}{k}} (B) −1−k2k\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\sqrt{\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{k}^{\htmlData{tutor-start=18,tutor-end=19}{2}}}}{\htmlData{tutor-start=23,tutor-end=24}{k}} (C) k1−k2\frac{\htmlData{tutor-start=6,tutor-end=7}{k}}{\sqrt{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{k}^{\htmlData{tutor-start=20,tutor-end=21}{2}}}} (D) −k1−k2\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{k}}{\sqrt{\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{k}^{\htmlData{tutor-start=21,tutor-end=22}{2}}}}

答案：B；-√(1-k²)/k

题目标签：参考角正切

解题过程

用余弦值表示正切

把 100∘100^\circ 化到参考角 80∘80^\circ

(1)
确定正弦余弦符号

k=cos⁡(−80∘)=cos⁡80∘>0k=\cos(-80^\circ)=\cos80^\circ>0。

为什么从这里入手：“k=cos⁡(−80∘)=cos⁡80∘>0k=\cos(-80^\circ)=\cos80^\circ>0。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：100∘100^\circ 在第二象限，sin⁡100∘=sin⁡80∘=1−k2\sin100^\circ=\sin80^\circ=\sqrt{1-k^{2}}，cos⁡100∘=−cos⁡80∘=−k\cos100^\circ=-\cos80^\circ=-k。

sin⁡100∘=1−k2,cos⁡100∘=−k\sin100^\circ=\sqrt{1-k^{2}},\quad\cos100^\circ=-k
(2)
作商

正切等于正弦除以余弦。

为什么从这里入手：“正切等于正弦除以余弦。”能直接确定主变量之间的关系，先使用它可以避免无效展开。解析几何的目的，是把图形关系变成坐标等式。“正切等于正弦除以余弦。”提供了点、斜率、距离或焦点条件，先作商后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：tan⁡100∘=−1−k2/k\tan100^\circ=-\sqrt{1-k^{2}}/k，选 B。

−1−k2kB\boxed{-\frac{\sqrt{1-k^{2}}}k}\quad\text{B}
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 ( ) (A) 4 (B) 3 (C) 2 (D) 1

答案：B；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=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{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=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{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=18,tutor-end=19}{=} ( ) (A) 52\htmlData{tutor-start=0,tutor-end=1}{5}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}} (B) 7 (C) 6 (D) 42\htmlData{tutor-start=0,tutor-end=1}{4}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}

答案：A；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+2x)3(1−x3)5\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{2}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{x}}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{3}} \htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{-}\sqrt[\htmlData{tutor-start=27,tutor-end=28}{3}]{\htmlData{tutor-start=30,tutor-end=31}{x}}\htmlData{tutor-start=32,tutor-end=33}{)}^{\htmlData{tutor-start=35,tutor-end=36}{5}} 的展开式中 x\htmlData{tutor-start=0,tutor-end=1}{x} 的系数是 ( ) (A) -4 (B) -2 (C) 2 (D) 4

答案：C；2

题目标签：分数指数二项展开

解题过程

求分数指数展开中的 x\htmlData{tutor-start=0,tutor-end=1}{x} 系数

统一令 t=x1/6\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}^{\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{6}}

(1)
转成整数幂

此时 x=t3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{x}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{t}^{\htmlData{tutor-start=12,tutor-end=13}{3}}，x3=t2\sqrt[3]x=t^{2}，而 x=t6\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{t}^{\htmlData{tutor-start=5,tutor-end=6}{6}}。

为什么从这里入手：“此时 x=t3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{x}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{t}^{\htmlData{tutor-start=12,tutor-end=13}{3}}，x3=t2\sqrt[3]x=t^{2}，而 x=t6\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{t}^{\htmlData{tutor-start=5,tutor-end=6}{6}}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：原式变为 (1+2t3)3(1−t2)5\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{t}^{\htmlData{tutor-start=7,tutor-end=8}{3}}\htmlData{tutor-start=9,tutor-end=10}{)}^{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{t}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{)}^{\htmlData{tutor-start=25,tutor-end=26}{5}}，只需找 t6\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{6}} 项。

(1+2t3)3(1−t2)5\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{t}^{\htmlData{tutor-start=7,tutor-end=8}{3}}\htmlData{tutor-start=9,tutor-end=10}{)}^{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{t}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{)}^{\htmlData{tutor-start=25,tutor-end=26}{5}}
(2)
列出所有次数组合

第一因子只出现 3 的倍数次幂，第二因子只出现偶次幂。

为什么从这里入手：“第一因子只出现 3 的倍数次幂，第二因子只出现偶次幂。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：得到 t6\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{6}} 只有两种：第一因子取常数、第二取 t6\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{6}}，系数 −(53)=−10-\binom53=-10；第一取 t6\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{6}}、第二取常数，系数 (32)22=12\binom32 2^{2}=12。合为 2。

2C\boxed{\htmlData{tutor-start=7,tutor-end=8}{2}}\quad\text{\htmlData{tutor-start=20,tutor-end=21}{C}}
6

一、选择题 · 排列组合

某校开设 A 类选修课 3 门，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 ( ) (A) 30 种 (B) 35 种 (C) 42 种 (D) 48 种

答案：A；30

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}}
7

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 ( ) (A) 23\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{3}} (B) 33\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{3}} (C) 23\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{3}} (D) 63\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{6}}}{\htmlData{tutor-start=16,tutor-end=17}{3}}

答案：D；√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}}}
8

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 ( ) (A) a<b<c\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{c} (B) b<c<a\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{c} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{a} (C) c<a<b\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{b} (D) c<b<a\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{a}

答案：C；c<a<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}}
9

一、选择题 · 圆锥曲线

已知 F1,F2\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{F}_{\htmlData{tutor-start=10,tutor-end=11}{2}} 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=9,tutor-end=10}{-} \htmlData{tutor-start=11,tutor-end=12}{y}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{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∘\angle F_{1} P F_{2} = 60^\circ，则 P\htmlData{tutor-start=0,tutor-end=1}{P} 到 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的距离为 ( ) (A) 32\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{2}} (B) 62\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{6}}}{\htmlData{tutor-start=16,tutor-end=17}{2}} (C) 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}} (D) 6\sqrt{\htmlData{tutor-start=6,tutor-end=7}{6}}

答案：B；√6/2

题目标签：双曲线点到轴距离

解题过程

求双曲线上点到 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的距离

先由焦点三角形求焦距乘积

(1)
求两焦半径乘积

双曲线定义与余弦定理联合使用。

为什么从这里入手：“双曲线定义与余弦定理联合使用。”能直接确定主变量之间的关系，先使用它可以避免无效展开。解析几何的目的，是把图形关系变成坐标等式。“双曲线定义与余弦定理联合使用。”提供了点、斜率、距离或焦点条件，先求两焦半径乘积后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：仿照焦点三角形计算，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}{P}\htmlData{tutor-start=3,tutor-end=4}{F}_{\htmlData{tutor-start=6,tutor-end=7}{1}}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{v}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{P}\htmlData{tutor-start=12,tutor-end=13}{F}_{\htmlData{tutor-start=15,tutor-end=16}{2}}，有 (u−v)2=4\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}，且 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}，所以 uv=4\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{4}。

PF1⋅PF2=4\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{F}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=12}{\cdot }\htmlData{tutor-start=12,tutor-end=13}{P}\htmlData{tutor-start=13,tutor-end=14}{F}_{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{4}
(2)
把乘积化为纵坐标

设 P=(x,y)\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{y}\htmlData{tutor-start=6,tutor-end=7}{)}，由 x2−y2=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}。

为什么从这里入手：“设 P=(x,y)\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{y}\htmlData{tutor-start=6,tutor-end=7}{)}，由 x2−y2=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}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：直接相乘两焦距平方得 (PF1PF2)2=(2y2+1)2\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}{P}\htmlData{tutor-start=8,tutor-end=9}{F}_{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{)}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{y}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\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}}。因乘积为 4，故 2y2+1=4\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{y}^{\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}{4}，从而 ∣y∣=6/2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{=}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{6}}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{2}。

d(P,x轴)=62B\boxed{\htmlData{tutor-start=7,tutor-end=8}{d}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{P}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{x}\text{\htmlData{tutor-start=18,tutor-end=19}{轴}}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{=}\frac{\sqrt{\htmlData{tutor-start=34,tutor-end=35}{6}}}{\htmlData{tutor-start=38,tutor-end=39}{2}}}\quad\text{\htmlData{tutor-start=52,tutor-end=53}{B}}
10

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=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{|}\lg \htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{|}，若 0<a<b\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{b}，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=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{)}，则 a+2b\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}{b} 的取值范围是 ( ) (A) (22,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=19}{\infty}\htmlData{tutor-start=19,tutor-end=20}{)} (B) [22,+∞)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{2}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=19}{\infty}\htmlData{tutor-start=19,tutor-end=20}{)} (C) (3,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} (D) [3,+∞)\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}

答案：C；(3,+∞)

题目标签：绝对值对数范围

解题过程

求绝对值对数等值下的范围

先得到倒数关系，再做单变量最值

(1)
利用 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} 确定位置

∣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}{|} 且 a≠b\htmlData{tutor-start=0,tutor-end=1}{a}\ne \htmlData{tutor-start=5,tutor-end=6}{b}。

为什么从这里入手：“∣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}{|} 且 a≠b\htmlData{tutor-start=0,tutor-end=1}{a}\ne \htmlData{tutor-start=5,tutor-end=6}{b}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：因此 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}。又 0<a<b\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{b}，所以 0<a<1<b\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{b}，并有 b=1/a\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{a}。

0<a<1,b=1a\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,}\quad \htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{=}\frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{a}}
(2)
求 a+2b\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}{b} 的范围

目标变为 g(a)=a+2/a\htmlData{tutor-start=0,tutor-end=1}{g}\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}{a}\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}{a}，其中 0<a<1\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}。

为什么从这里入手：“目标变为 g(a)=a+2/a\htmlData{tutor-start=0,tutor-end=1}{g}\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}{a}\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}{a}，其中 0<a<1\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：g′(a)=1−2/a2<0\htmlData{tutor-start=0,tutor-end=1}{g}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{<}\htmlData{tutor-start=16,tutor-end=17}{0}，故随 a\htmlData{tutor-start=0,tutor-end=1}{a} 增大而减小；当 a→1−a\to1^- 时趋近 3 但不能取到，a→0+a\to0^+ 时趋于无穷。

a+2b∈(3,+∞)C\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{b}\htmlData{tutor-start=11,tutor-end=14}{\in}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=24}{\infty}\htmlData{tutor-start=24,tutor-end=25}{)}}\quad\text{\htmlData{tutor-start=37,tutor-end=38}{C}}
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,PB\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{P}\htmlData{tutor-start=5,tutor-end=6}{B} 为该圆的两条切线，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} 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 ( ) (A) −4+2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{+}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{2}} (B) −3+2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{+}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{2}} (C) −4+22\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}{2}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}} (D) −3+22\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}{2}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}

答案：D；-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

一、选择题 · 立体几何

已知在半径为 2 的球面上有 A,B,C,D\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{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 ( ) (A) 233\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{3}} (B) 433\frac{\htmlData{tutor-start=6,tutor-end=7}{4}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{3}} (C) 23\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{3}} (D) 833\frac{\htmlData{tutor-start=6,tutor-end=7}{8}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}}{\htmlData{tutor-start=17,tutor-end=18}{3}}

答案：B；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 2x2+1−x⩽1\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}\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}{1}} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{x} \htmlData{tutor-start=20,tutor-end=30}{\leqslant }\htmlData{tutor-start=30,tutor-end=31}{1} 的解集是______.

答案：[0,2]\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}{2}\htmlData{tutor-start=4,tutor-end=5}{]}

题目标签：根式不等式

解题过程

解根式不等式

先保证平方合法，再平方化简

(1)
检查右端非负

2x2+1≤x+1\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}\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}{1}}\htmlData{tutor-start=15,tutor-end=19}{\le }\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{1} 必须有 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=6}{\ge}\htmlData{tutor-start=6,tutor-end=7}{0}。

为什么从这里入手：“2x2+1≤x+1\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}\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}{1}}\htmlData{tutor-start=15,tutor-end=19}{\le }\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{1} 必须有 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=6}{\ge}\htmlData{tutor-start=6,tutor-end=7}{0}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：原不等式等价于 x≥−1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1} 且平方后的不等式成立。

x≥−1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}
(2)
平方并求交集

两边均非负后平方不改变等价性。

为什么从这里入手：“两边均非负后平方不改变等价性。”能直接确定主变量之间的关系，先使用它可以避免无效展开。集合题不能只看式子的外形，必须回到“元素是否满足条件”。这里的“两边均非负后平方不改变等价性。”给出了元素筛选规则，因此先平方并求交集，再逐项保留或删除元素，思路最直接。

详细展开：2x2+1≤(x+1)2\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}^{\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=11}{\le}\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=18,tutor-end=19}{2}} 化为 x(x−2)≤0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{0}，得 0≤x≤2\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{2}，已满足前置条件。

[0,2]\boxed{\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}{]}}
14

二、填空题 · 三角函数

已知 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 为第三象限的角，cos⁡2α=−35\cos \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=13}{\alpha }\htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{-}\frac{\htmlData{tutor-start=22,tutor-end=23}{3}}{\htmlData{tutor-start=25,tutor-end=26}{5}}，则 tan⁡(π4+2α)=\tan\left(\frac{\htmlData{tutor-start=16,tutor-end=19}{\pi}}{\htmlData{tutor-start=21,tutor-end=22}{4}} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=33}{\alpha}\right) \htmlData{tutor-start=41,tutor-end=42}{=} ______.

答案：-1/7

题目标签：和角正切

解题过程

计算和角正切

先由象限确定 sin⁡2α\sin\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha} 的符号

(1)
补出二倍角正弦

α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 在第三象限，所以 2α∈(2π,3π)\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=7}{\alpha}\htmlData{tutor-start=7,tutor-end=10}{\in}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=15}{\pi}\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=20}{\pi}\htmlData{tutor-start=20,tutor-end=21}{)}。

为什么从这里入手：“α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 在第三象限，所以 2α∈(2π,3π)\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=7}{\alpha}\htmlData{tutor-start=7,tutor-end=10}{\in}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=15}{\pi}\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=20}{\pi}\htmlData{tutor-start=20,tutor-end=21}{)}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：结合 cos⁡2α=−3/5\cos\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{5}，可知 2α\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=7}{\alpha} 对应第二象限方向，故 sin⁡2α=4/5\sin\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{4}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{5}、tan⁡2α=−4/3\tan\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{3}。

tan⁡2α=−43\tan\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{-}\frac{\htmlData{tutor-start=19,tutor-end=20}{4}}{\htmlData{tutor-start=22,tutor-end=23}{3}}
(2)
使用正切和角公式

tan⁡(π/4)=1\tan\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=8}{\pi}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}。

为什么从这里入手：“tan⁡(π/4)=1\tan\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=8}{\pi}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：tan⁡(π/4+2α)=1−4/31+4/3=−1/7\tan\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=8}{\pi}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=18}{\alpha}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{=}\frac{\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{4}\htmlData{tutor-start=29,tutor-end=30}{/}\htmlData{tutor-start=30,tutor-end=31}{3}}{\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{4}\htmlData{tutor-start=36,tutor-end=37}{/}\htmlData{tutor-start=37,tutor-end=38}{3}}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{/}\htmlData{tutor-start=43,tutor-end=44}{7}。

−17\boxed{\htmlData{tutor-start=7,tutor-end=8}{-}\frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{7}}}
15

二、填空题 · 函数

直线 y=1\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 与曲线 y=x2−∣x∣+a\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{|}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{|}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{a} 有四个交点，则 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围是______.

答案：(1,5/4)

题目标签：绝对值曲线四交点

解题过程

由四个交点求参数范围

令 u=∣x∣\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}{x}\htmlData{tutor-start=4,tutor-end=5}{|} 把偶函数方程降为二次方程

(1)
转化交点数量

交点满足 x2−∣x∣+a=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}{|}\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}{a}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}。

为什么从这里入手：“交点满足 x2−∣x∣+a=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}{|}\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}{a}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：令 u=∣x∣≥0\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}{x}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=8}{\ge}\htmlData{tutor-start=8,tutor-end=9}{0}，得到 u2−u+a−1=0\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}{u}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0}。要产生四个不同的 x\htmlData{tutor-start=0,tutor-end=1}{x}，该方程必须有两个不同的正根。

u2−u+a−1=0\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}{u}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0}
(2)
使用根的判别条件

两根和为 1，故只需积为正且判别式为正。

为什么从这里入手：“两根和为 1，故只需积为正且判别式为正。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：a−1>0\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}{0} 且 1−4(a−1)>0\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{>}\htmlData{tutor-start=9,tutor-end=10}{0}，所以 1<a<5/4\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{5}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{4}。

1<a<54\boxed{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{<}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{<}\frac{\htmlData{tutor-start=17,tutor-end=18}{5}}{\htmlData{tutor-start=20,tutor-end=21}{4}}}
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=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{2}\overrightarrow{\htmlData{tutor-start=39,tutor-end=40}{F}\htmlData{tutor-start=40,tutor-end=41}{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 · 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=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{a} \cot \htmlData{tutor-start=13,tutor-end=14}{A} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{b} \cot \htmlData{tutor-start=24,tutor-end=25}{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}
18

三、解答题 · 概率统计

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) 记 X\htmlData{tutor-start=0,tutor-end=1}{X} 表示投到该杂志的 4 篇稿件中被录用的篇数，求 X\htmlData{tutor-start=0,tutor-end=1}{X} 的分布列及期望。

答案：p=2/5；分布列见解析；EX=8/5

题目标签：稿件录用数分布

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）求四篇稿件录用数的分布列与期望

使用二项分布公式

(1)
列出各取值概率

每篇独立录用概率为 2/5\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}，故 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}{)}。

为什么从这里入手：“每篇独立录用概率为 2/5\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}，故 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}{)}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：P(X=k)=(4k)(2/5)k(3/5)4−kP(X=k)=\binom4k(2/5)^{k}(3/5)^{4-k}，依次为 81,216,216,96,16\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{6}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{6}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{9}\htmlData{tutor-start=12,tutor-end=13}{6}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{6} 除以 625。

k01234P81/625216/625216/62596/62516/625\begin{array}{c|ccccc}\htmlData{tutor-start=22,tutor-end=23}{k}&\htmlData{tutor-start=24,tutor-end=25}{0}&\htmlData{tutor-start=26,tutor-end=27}{1}&\htmlData{tutor-start=28,tutor-end=29}{2}&\htmlData{tutor-start=30,tutor-end=31}{3}&\htmlData{tutor-start=32,tutor-end=33}{4}\\\hline \htmlData{tutor-start=42,tutor-end=43}{P}&\htmlData{tutor-start=44,tutor-end=45}{8}\htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{/}\htmlData{tutor-start=47,tutor-end=48}{6}\htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=50}{5}&\htmlData{tutor-start=51,tutor-end=52}{2}\htmlData{tutor-start=52,tutor-end=53}{1}\htmlData{tutor-start=53,tutor-end=54}{6}\htmlData{tutor-start=54,tutor-end=55}{/}\htmlData{tutor-start=55,tutor-end=56}{6}\htmlData{tutor-start=56,tutor-end=57}{2}\htmlData{tutor-start=57,tutor-end=58}{5}&\htmlData{tutor-start=59,tutor-end=60}{2}\htmlData{tutor-start=60,tutor-end=61}{1}\htmlData{tutor-start=61,tutor-end=62}{6}\htmlData{tutor-start=62,tutor-end=63}{/}\htmlData{tutor-start=63,tutor-end=64}{6}\htmlData{tutor-start=64,tutor-end=65}{2}\htmlData{tutor-start=65,tutor-end=66}{5}&\htmlData{tutor-start=67,tutor-end=68}{9}\htmlData{tutor-start=68,tutor-end=69}{6}\htmlData{tutor-start=69,tutor-end=70}{/}\htmlData{tutor-start=70,tutor-end=71}{6}\htmlData{tutor-start=71,tutor-end=72}{2}\htmlData{tutor-start=72,tutor-end=73}{5}&\htmlData{tutor-start=74,tutor-end=75}{1}\htmlData{tutor-start=75,tutor-end=76}{6}\htmlData{tutor-start=76,tutor-end=77}{/}\htmlData{tutor-start=77,tutor-end=78}{6}\htmlData{tutor-start=78,tutor-end=79}{2}\htmlData{tutor-start=79,tutor-end=80}{5}\end{array}
(2)
计算期望

二项分布期望等于试验次数乘成功概率。

为什么从这里入手：“二项分布期望等于试验次数乘成功概率。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：EX=4⋅2/5=8/5\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{X}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=9}{\cdot}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{5}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{8}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{5}；也可用分布列逐项加权复核。

EX=85\boxed{\htmlData{tutor-start=7,tutor-end=8}{E}\htmlData{tutor-start=8,tutor-end=9}{X}\htmlData{tutor-start=9,tutor-end=10}{=}\frac{\htmlData{tutor-start=16,tutor-end=17}{8}}{\htmlData{tutor-start=19,tutor-end=20}{5}}}
19

三、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=13}{\parallel }\htmlData{tutor-start=13,tutor-end=14}{D}\htmlData{tutor-start=14,tutor-end=15}{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,DC=SD=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}{A}\htmlData{tutor-start=4,tutor-end=5}{D}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{D}\htmlData{tutor-start=10,tutor-end=11}{C}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{S}\htmlData{tutor-start=13,tutor-end=14}{D}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{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 1 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}}}
20

三、解答题 · 导数

已知函数 f(x)=(x+1)ln⁡x−x+1\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)}\ln \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}. (1) 若 xf′(x)⩽x2+ax+1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{f}'\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=16}{\leqslant }\htmlData{tutor-start=16,tutor-end=17}{x}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{a}\htmlData{tutor-start=23,tutor-end=24}{x}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{1}, 求 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围; (2) 证明: (x−1)f(x)⩾0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=19}{\geqslant }\htmlData{tutor-start=19,tutor-end=20}{0}.

答案：a≥-1；见证明

题目标签：对数函数不等式

解题过程

（1）求参数 a\htmlData{tutor-start=0,tutor-end=1}{a} 的范围

把原不等式化为关于 a\htmlData{tutor-start=0,tutor-end=1}{a} 的恒成立条件

(1)
计算导数并约去正变量

函数定义域为 x>0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}。

为什么从这里入手：“函数定义域为 x>0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。含分母、根号或对数的式子必须先合法，后续变形才有意义。“函数定义域为 x>0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}。”正好暴露了限制条件；先计算导数并约去正变量，可以防止约分或平方时把禁值偷偷带回答案。

详细展开：f′(x)=ln⁡x+1/x\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\ln \htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{x}，所以 xf′(x)=xln⁡x+1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{f}'\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{x}\ln \htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}。题设不等式等价于 ln⁡x−x≤a\ln \htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=11}{\le }\htmlData{tutor-start=11,tutor-end=12}{a} 对所有 x>0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0} 成立。

a≥ln⁡x−x(x>0)\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\ge}\ln \htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{x}\quad\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}{0}\htmlData{tutor-start=20,tutor-end=21}{)}
(2)
求右侧最大值

令 g(x)=ln⁡x−x\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\ln \htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{x}。

为什么从这里入手：“令 g(x)=ln⁡x−x\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\ln \htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{x}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：g′(x)=1/x−1\htmlData{tutor-start=0,tutor-end=1}{g}'\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}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}，故在 x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 取得最大值 −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}，所以 a≥−1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}。

a≥−1\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=11}{\ge}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}}

（2）证明乘积非负

证明 f\htmlData{tutor-start=0,tutor-end=1}{f} 在 x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 两侧与 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} 同号

(1)
证明 f\htmlData{tutor-start=0,tutor-end=1}{f} 单调递增

f′(x)=ln⁡x+1/x\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\ln \htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{x}。

为什么从这里入手：“f′(x)=ln⁡x+1/x\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\ln \htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{x}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：设 h(x)=ln⁡x+1/x\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\ln \htmlData{tutor-start=9,tutor-end=10}{x}\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}{x}，则 h′(x)=(x−1)/x2\htmlData{tutor-start=0,tutor-end=1}{h}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\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}{x}^{\htmlData{tutor-start=15,tutor-end=16}{2}}，所以 h\htmlData{tutor-start=0,tutor-end=1}{h} 在 x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 取得最小值 h(1)=1>0\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}{=}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{>}\htmlData{tutor-start=7,tutor-end=8}{0}。故 f′(x)>0\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{>}\htmlData{tutor-start=6,tutor-end=7}{0}。

f′(x)>0(x>0)\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{>}\htmlData{tutor-start=6,tutor-end=7}{0}\quad\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}{)}
(2)
利用零点比较符号

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}{1}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}。

为什么从这里入手：“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}{1}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。目标是“利用零点比较符号”，而“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}{1}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{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：因 f\htmlData{tutor-start=0,tutor-end=1}{f} 严格递增，0<x<1\htmlData{tutor-start=0,tutor-end=1}{0}\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} 时 f(x)<0\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}{0}，x>1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1} 时 f(x)>0\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}{0}；这与 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} 的符号完全一致。

(x−1)f(x)≥0\boxed{\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{f}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=19}{\ge}\htmlData{tutor-start=19,tutor-end=20}{0}}
21

三、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,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B} 两点, 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=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}{=}\frac{\htmlData{tutor-start=50,tutor-end=51}{8}}{\htmlData{tutor-start=53,tutor-end=54}{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}}}
22

三、解答题 · 数列

已知数列 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 中, a1=1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}, an+1=c−1an\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{c}\htmlData{tutor-start=9,tutor-end=10}{-}\frac{\htmlData{tutor-start=16,tutor-end=17}{1}}{\htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{n}}}. (1) 设 c=52\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{5}}{\htmlData{tutor-start=11,tutor-end=12}{2}}, bn=1an−2\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{n}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{2}}, 求数列 {bn}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{b}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 的通项公式; (2) 求使不等式 an<an+1<3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{n}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{3} 成立的 c\htmlData{tutor-start=0,tutor-end=1}{c} 的取值范围.

答案：bₙ=-(4ⁿ⁻¹+2)/3；2<c≤10/3

题目标签：分式递推数列

解题过程

（1）求辅助数列通项

把递推式代入 bn+1\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}

(1)
推出一阶线性递推

当 c=5/2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2} 时，an+1−2=(an−2)/(2an)\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{n}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{a}_{\htmlData{tutor-start=25,tutor-end=26}{n}}\htmlData{tutor-start=27,tutor-end=28}{)}。

为什么从这里入手：“当 c=5/2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2} 时，an+1−2=(an−2)/(2an)\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{n}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{a}_{\htmlData{tutor-start=25,tutor-end=26}{n}}\htmlData{tutor-start=27,tutor-end=28}{)}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：取倒数得 bn+1=2an/(an−2)=2+4bn\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{n}}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{n}}\htmlData{tutor-start=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}{2}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{4}\htmlData{tutor-start=28,tutor-end=29}{b}_{\htmlData{tutor-start=31,tutor-end=32}{n}}，且 b1=−1\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}。

bn+1=4bn+2,b1=−1\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{b}_{\htmlData{tutor-start=12,tutor-end=13}{n}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{,}\quad \htmlData{tutor-start=23,tutor-end=24}{b}_{\htmlData{tutor-start=26,tutor-end=27}{1}}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{1}
(2)
平移成等比数列

常数递推 bn+1=4bn+2\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{b}_{\htmlData{tutor-start=12,tutor-end=13}{n}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2} 的不动点为 −2/3\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{3}。

为什么从这里入手：“常数递推 bn+1=4bn+2\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{b}_{\htmlData{tutor-start=12,tutor-end=13}{n}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2} 的不动点为 −2/3\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{3}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：bn+1+2/3=4(bn+2/3)\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{4}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{b}_{\htmlData{tutor-start=17,tutor-end=18}{n}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{/}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{)}，故 bn+2/3=(−1/3)4n−1\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{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}{(}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{4}^{\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{1}}。

bn=−4n−1+23\boxed{\htmlData{tutor-start=7,tutor-end=8}{b}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{-}\frac{\htmlData{tutor-start=20,tutor-end=21}{4}^{\htmlData{tutor-start=23,tutor-end=24}{n}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{2}}{\htmlData{tutor-start=31,tutor-end=32}{3}}}

（2）求递增且小于 3 的参数范围

研究递推函数的不动点

(1)
求固定点并定位初值

要对所有 n\htmlData{tutor-start=0,tutor-end=1}{n} 有 an<an+1<3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{<}\htmlData{tutor-start=14,tutor-end=15}{3}，先由 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 得 c>2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2}。

为什么从这里入手：“要对所有 n\htmlData{tutor-start=0,tutor-end=1}{n} 有 an<an+1<3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{<}\htmlData{tutor-start=14,tutor-end=15}{3}，先由 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 得 c>2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：当 c>2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2} 时，方程 x=c−1/x\htmlData{tutor-start=0,tutor-end=1}{x}\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}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{x} 有两个正根 α<1<β\htmlData{tutor-start=0,tutor-end=6}{\alpha}\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=14}{\beta}，且 β=(c+c2−4)/2\htmlData{tutor-start=0,tutor-end=5}{\beta}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{c}\htmlData{tutor-start=8,tutor-end=9}{+}\sqrt{\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}{4}}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{2}。从 a1=1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1} 出发，只要 an<β\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=11}{\beta}，就有 an+1−an>0\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=14}{>}\htmlData{tutor-start=14,tutor-end=15}{0}。

α<1=a1<β\htmlData{tutor-start=0,tutor-end=6}{\alpha}\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}{a}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{<}\htmlData{tutor-start=15,tutor-end=20}{\beta}
(2)
用上界限制大固定点

递增序列趋向较大固定点 β\htmlData{tutor-start=0,tutor-end=5}{\beta}，所以严格小于 3 等价于 β≤3\htmlData{tutor-start=0,tutor-end=5}{\beta}\htmlData{tutor-start=5,tutor-end=8}{\le}\htmlData{tutor-start=8,tutor-end=9}{3}。

为什么从这里入手：“递增序列趋向较大固定点 β\htmlData{tutor-start=0,tutor-end=5}{\beta}，所以严格小于 3 等价于 β≤3\htmlData{tutor-start=0,tutor-end=5}{\beta}\htmlData{tutor-start=5,tutor-end=8}{\le}\htmlData{tutor-start=8,tutor-end=9}{3}。”能直接确定主变量之间的关系，先使用它可以避免无效展开。

详细展开：由 c=β+1/β\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=7}{\beta}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=15}{\beta} 且 1<β≤3\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=7}{\beta}\htmlData{tutor-start=7,tutor-end=10}{\le}\htmlData{tutor-start=10,tutor-end=11}{3}，得到 2<c≤3+1/3=10/3\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=6}{\le}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{3}；端点时各项仍严格小于极限 3。

2<c≤103\boxed{\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{<}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=13}{\le}\frac{\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{0}}{\htmlData{tutor-start=23,tutor-end=24}{3}}}