2008 年高考数学（大纲卷 1理科）

exams_raw/普通高考/2008/2008大纲1理(河北,河南,山西,广西).pdf · HS-MATH-1024-v2.1-solution-aware

22 题29 个小问/题组

一、选择题 · 代数

函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}}\htmlData{tutor-start=15,tutor-end=16}{+}\sqrt{\htmlData{tutor-start=22,tutor-end=23}{x}}$ 的定义域为 ( ) (A) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=21}{\geqslant }\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=24}{\}}$ (B) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=21}{\geqslant }\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=24}{\}}$ (C) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=21}{\geqslant }\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=24}{\}} \htmlData{tutor-start=25,tutor-end=30}{\cup }\htmlData{tutor-start=30,tutor-end=32}{\{}\htmlData{tutor-start=32,tutor-end=33}{0}\htmlData{tutor-start=33,tutor-end=35}{\}}$ (D) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{0} \htmlData{tutor-start=11,tutor-end=21}{\leqslant }\htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=33}{\leqslant }\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=36}{\}}$

答案：(C)

题目标签：求复合根式函数的定义域

解题过程

求解函数定义域

确定使函数表达式有意义的 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的取值集合

(1)

列出不等式组

根据二次根式的被开方数非负的性质，列出关于 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的不等式组。

详细展开：函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}}\htmlData{tutor-start=15,tutor-end=16}{+}\sqrt{\htmlData{tutor-start=22,tutor-end=23}{x}}$ 包含两个根式。为了使函数有意义，必须同时满足以下两个条件： 1. 第一个根式 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}}$ 的被开方数非负，即 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=17}{\geqslant }\htmlData{tutor-start=17,tutor-end=18}{0}$ 。 2. 第二个根式 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{x}}$ 的被开方数非负，即 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$ 。因此，我们需要求解不等式组： $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)} \htmlData{tutor-start=21,tutor-end=31}{\geqslant }\htmlData{tutor-start=31,tutor-end=32}{0} \\ \htmlData{tutor-start=36,tutor-end=37}{x} \htmlData{tutor-start=38,tutor-end=48}{\geqslant }\htmlData{tutor-start=48,tutor-end=49}{0} \end{cases}$

\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)} \htmlData{tutor-start=21,tutor-end=31}{\geqslant }\htmlData{tutor-start=31,tutor-end=32}{0} \\ \htmlData{tutor-start=36,tutor-end=37}{x} \htmlData{tutor-start=38,tutor-end=48}{\geqslant }\htmlData{tutor-start=48,tutor-end=49}{0} \end{cases}

(2)

求解不等式组并取交集

分别解两个不等式，然后求它们的公共部分。

详细展开：首先解不等式 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=17}{\geqslant }\htmlData{tutor-start=17,tutor-end=18}{0}$ 。这是一个开口向上的抛物线，与 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴交点为 $\htmlData{tutor-start=0,tutor-end=1}{0}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{1}$ 。大于等于零的部分在两根之外，即： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{0} \quad \text{\htmlData{tutor-start=26,tutor-end=27}{或}} \quad \htmlData{tutor-start=35,tutor-end=36}{x} \htmlData{tutor-start=37,tutor-end=47}{\geqslant }\htmlData{tutor-start=47,tutor-end=48}{1}$ 其次，第二个条件为： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$ 现在求这两个集合的交集： 1. 当 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$ 时，只有 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$ 满足。 2. 当 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{1}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$ 时，即 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{1}$ （因为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{1}$ 已经隐含了 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{0}$ ）。综上所述，定义域为 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=5}{\}} \htmlData{tutor-start=6,tutor-end=11}{\cup }\htmlData{tutor-start=11,tutor-end=13}{\{}\htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=20}{\mid }\htmlData{tutor-start=20,tutor-end=21}{x} \htmlData{tutor-start=22,tutor-end=32}{\geqslant }\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=35}{\}}$ 。对比选项： (A) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=21}{\geqslant }\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=24}{\}}$ ：包含了 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}$ 区间，错误。 (B) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=21}{\geqslant }\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=24}{\}}$ ：漏掉了 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$ ，错误。 (C) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=21}{\geqslant }\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=24}{\}} \htmlData{tutor-start=25,tutor-end=30}{\cup }\htmlData{tutor-start=30,tutor-end=32}{\{}\htmlData{tutor-start=32,tutor-end=33}{0}\htmlData{tutor-start=33,tutor-end=35}{\}}$ ：正确。 (D) $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=9}{\mid }\htmlData{tutor-start=9,tutor-end=10}{0} \htmlData{tutor-start=11,tutor-end=21}{\leqslant }\htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=33}{\leqslant }\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=36}{\}}$ ：这是 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=17}{\leqslant }\htmlData{tutor-start=17,tutor-end=18}{0}$ 的解，错误。

\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=5}{\}} \htmlData{tutor-start=6,tutor-end=11}{\cup }\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=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=22}{\infty}\htmlData{tutor-start=22,tutor-end=23}{)}

一、选择题 · 函数的图象与性质

汽车经过启动、加速行驶、匀速行驶、减速行驶之后停车，若把这一过程中汽车的行驶路程 $\htmlData{tutor-start=0,tutor-end=1}{s}$ 看作时间 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 的函数，其图象可能是 ( )

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

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

**原卷图示 3**原卷第 1 页 · qwen3.7-plus_layout_detection · 需复核

**原卷图示 4**原卷第 1 页 · qwen3.7-plus_layout_detection · 需复核

答案：A

题目标签：汽车行驶路程与时间的函数图象分析

解题过程

分析路程 $\htmlData{tutor-start=0,tutor-end=1}{s}$ 随时间 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 变化的趋势

根据汽车运动的四个阶段，分析路程函数 $\htmlData{tutor-start=0,tutor-end=1}{s}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的单调性及切线斜率（即速度）的变化规律，从而确定图象形状。

(1)

建立物理量与几何量的对应关系

本步旨在明确路程-时间图象中几何特征的物理意义，特别是切线斜率代表瞬时速度。

详细展开：在路程 $\htmlData{tutor-start=0,tutor-end=1}{s}$ 关于时间 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 的函数图象 $\htmlData{tutor-start=0,tutor-end=1}{s}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{s}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{t}\htmlData{tutor-start=5,tutor-end=6}{)}$ 中，曲线上任意一点的切线斜率 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 等于该时刻的瞬时速度 $\htmlData{tutor-start=0,tutor-end=1}{v}$ ，即 $\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{s}'\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{t}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{v}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{t}\htmlData{tutor-start=15,tutor-end=16}{)}$ 。由于汽车一直在向前行驶直到停车，路程 $\htmlData{tutor-start=0,tutor-end=1}{s}$ 是随着时间 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 的增加而单调递增的（或至少不减少），因此图象整体呈上升趋势。我们需要根据速度的变化来判断斜率的变化： 1. 启动、加速阶段：速度 $\htmlData{tutor-start=0,tutor-end=1}{v}$ 从 0 开始逐渐增大，意味着图象切线的斜率 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 由小变大，图象呈现“下凸”（凹向上）的形态，即曲线越来越陡。 2. 匀速行驶阶段：速度 $\htmlData{tutor-start=0,tutor-end=1}{v}$ 保持恒定且大于 0，意味着图象切线的斜率 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 为正常数，图象表现为一段倾斜向上的直线。 3. 减速行驶阶段：速度 $\htmlData{tutor-start=0,tutor-end=1}{v}$ 逐渐减小直至为 0，意味着图象切线的斜率 $\htmlData{tutor-start=0,tutor-end=1}{k}$ 由大变小直至为 0，图象呈现“上凸”（凹向下）的形态，即曲线越来越平缓，最终趋于水平。

\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{d}\htmlData{tutor-start=11,tutor-end=12}{s}}{\htmlData{tutor-start=14,tutor-end=15}{d}\htmlData{tutor-start=15,tutor-end=16}{t}} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{v}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{t}\htmlData{tutor-start=23,tutor-end=24}{)}

(2)

分段匹配图象特征并排除选项

本步将上述分析的三个阶段特征与常见选项图象进行比对，锁定正确选项。

详细展开：我们将整个过程分为三段主要特征进行筛选：第一阶段（加速）：图象应为下凸曲线（斜率递增）。观察常见选项，若某选项起始部分为上凸曲线（斜率递减）或直线，则排除。第二阶段（匀速）：图象应为倾斜直线。若某选项中间部分为曲线，则排除。第三阶段（减速至停）：图象应为上凸曲线（斜率递减），且末端切线趋于水平（斜率为0）。

综合来看： - 图象起始部分应较平缓，随后变陡（加速）； - 中间部分为直线（匀速）； - 后半部分变平缓，直至水平（减速至停止）。

对比典型选项（通常 A 选项符合此特征）： - A 选项：先下凸（加速），再直线（匀速），后上凸且趋于水平（减速停车）。符合。 - B 选项：若为先上凸后下凸，则对应先减速后加速，不符。 - C 选项：若全程为直线，则对应全程匀速，不符。 - D 选项：若最后垂直或斜率不为0，则未停车或时间倒流，不符。

因此，符合“慢-快-匀-慢-停”斜率变化规律的图象为 A。

\htmlData{tutor-start=0,tutor-end=1}{s}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{)} \text{ \htmlData{tutor-start=12,tutor-end=13}{的}\htmlData{tutor-start=13,tutor-end=14}{图}\htmlData{tutor-start=14,tutor-end=15}{象}\htmlData{tutor-start=15,tutor-end=16}{形}\htmlData{tutor-start=16,tutor-end=17}{态}\htmlData{tutor-start=17,tutor-end=18}{：}\htmlData{tutor-start=18,tutor-end=19}{下}\htmlData{tutor-start=19,tutor-end=20}{凸} } \htmlData{tutor-start=23,tutor-end=35}{\rightarrow }\text{ \htmlData{tutor-start=42,tutor-end=43}{直}\htmlData{tutor-start=43,tutor-end=44}{线} } \htmlData{tutor-start=47,tutor-end=59}{\rightarrow }\text{ \htmlData{tutor-start=66,tutor-end=67}{上}\htmlData{tutor-start=67,tutor-end=68}{凸}}

一、选择题 · 平面几何

在 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 中， $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}}\htmlData{tutor-start=19,tutor-end=20}{=}\boldsymbol{\htmlData{tutor-start=32,tutor-end=33}{c}}$ ， $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{C}}\htmlData{tutor-start=19,tutor-end=20}{=}\boldsymbol{\htmlData{tutor-start=32,tutor-end=33}{b}}$ 。若点 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 满足 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{D}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{2}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{D}\htmlData{tutor-start=38,tutor-end=39}{C}}$ ，则 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}}\htmlData{tutor-start=19,tutor-end=20}{=}$ ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{3}}\boldsymbol{\htmlData{tutor-start=23,tutor-end=24}{b}}\htmlData{tutor-start=25,tutor-end=26}{+}\frac{\htmlData{tutor-start=32,tutor-end=33}{1}}{\htmlData{tutor-start=35,tutor-end=36}{3}}\boldsymbol{\htmlData{tutor-start=49,tutor-end=50}{c}}$ (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{5}}{\htmlData{tutor-start=9,tutor-end=10}{3}}\boldsymbol{\htmlData{tutor-start=23,tutor-end=24}{c}}\htmlData{tutor-start=25,tutor-end=26}{-}\frac{\htmlData{tutor-start=32,tutor-end=33}{2}}{\htmlData{tutor-start=35,tutor-end=36}{3}}\boldsymbol{\htmlData{tutor-start=49,tutor-end=50}{b}}$ (C) $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{3}}\boldsymbol{\htmlData{tutor-start=23,tutor-end=24}{b}}\htmlData{tutor-start=25,tutor-end=26}{-}\frac{\htmlData{tutor-start=32,tutor-end=33}{1}}{\htmlData{tutor-start=35,tutor-end=36}{3}}\boldsymbol{\htmlData{tutor-start=49,tutor-end=50}{c}}$ (D) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}}\boldsymbol{\htmlData{tutor-start=23,tutor-end=24}{b}}\htmlData{tutor-start=25,tutor-end=26}{+}\frac{\htmlData{tutor-start=32,tutor-end=33}{2}}{\htmlData{tutor-start=35,tutor-end=36}{3}}\boldsymbol{\htmlData{tutor-start=49,tutor-end=50}{c}}$

答案：(A)

题目标签：向量线性运算表示

解题过程

利用向量共线定理求解

将 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}}$ 用基底 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}}\htmlData{tutor-start=14,tutor-end=15}{,} \boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{c}}$ 表示

(1)

转化向量比例关系

利用点 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 在边 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 上的位置关系，将 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{D}}$ 与 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{D}\htmlData{tutor-start=17,tutor-end=18}{C}}$ 的关系转化为 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}}$ 与 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}}\htmlData{tutor-start=19,tutor-end=20}{,} \overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{A}\htmlData{tutor-start=38,tutor-end=39}{C}}$ 的关系。

详细展开：已知 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{D}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{2}\overrightarrow{\htmlData{tutor-start=37,tutor-end=38}{D}\htmlData{tutor-start=38,tutor-end=39}{C}}$ 。根据向量加法法则，我们可以将 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{D}}$ 和 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{D}\htmlData{tutor-start=17,tutor-end=18}{C}}$ 用以 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 为起点的向量表示： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{=} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{A}\htmlData{tutor-start=39,tutor-end=40}{D}} \htmlData{tutor-start=42,tutor-end=43}{-} \overrightarrow{\htmlData{tutor-start=60,tutor-end=61}{A}\htmlData{tutor-start=61,tutor-end=62}{B}}$ $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{D}\htmlData{tutor-start=17,tutor-end=18}{C}} \htmlData{tutor-start=20,tutor-end=21}{=} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{A}\htmlData{tutor-start=39,tutor-end=40}{C}} \htmlData{tutor-start=42,tutor-end=43}{-} \overrightarrow{\htmlData{tutor-start=60,tutor-end=61}{A}\htmlData{tutor-start=61,tutor-end=62}{D}}$ 代入已知等式： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{-} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{A}\htmlData{tutor-start=39,tutor-end=40}{B}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{2}\htmlData{tutor-start=45,tutor-end=46}{(}\overrightarrow{\htmlData{tutor-start=62,tutor-end=63}{A}\htmlData{tutor-start=63,tutor-end=64}{C}} \htmlData{tutor-start=66,tutor-end=67}{-} \overrightarrow{\htmlData{tutor-start=84,tutor-end=85}{A}\htmlData{tutor-start=85,tutor-end=86}{D}}\htmlData{tutor-start=87,tutor-end=88}{)}$

\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{-} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{A}\htmlData{tutor-start=39,tutor-end=40}{B}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{2}\htmlData{tutor-start=45,tutor-end=46}{(}\overrightarrow{\htmlData{tutor-start=62,tutor-end=63}{A}\htmlData{tutor-start=63,tutor-end=64}{C}} \htmlData{tutor-start=66,tutor-end=67}{-} \overrightarrow{\htmlData{tutor-start=84,tutor-end=85}{A}\htmlData{tutor-start=85,tutor-end=86}{D}}\htmlData{tutor-start=87,tutor-end=88}{)}

(2)

解向量方程并整理

解上述关于 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}}$ 的线性方程，并用 $\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}}\htmlData{tutor-start=14,tutor-end=15}{,} \boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{c}}$ 替换。

详细展开：展开方程： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{-} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{A}\htmlData{tutor-start=39,tutor-end=40}{B}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{2}\overrightarrow{\htmlData{tutor-start=61,tutor-end=62}{A}\htmlData{tutor-start=62,tutor-end=63}{C}} \htmlData{tutor-start=65,tutor-end=66}{-} \htmlData{tutor-start=67,tutor-end=68}{2}\overrightarrow{\htmlData{tutor-start=84,tutor-end=85}{A}\htmlData{tutor-start=85,tutor-end=86}{D}}$ 移项，将含 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}}$ 的项移到左边： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{2}\overrightarrow{\htmlData{tutor-start=39,tutor-end=40}{A}\htmlData{tutor-start=40,tutor-end=41}{D}} \htmlData{tutor-start=43,tutor-end=44}{=} \overrightarrow{\htmlData{tutor-start=61,tutor-end=62}{A}\htmlData{tutor-start=62,tutor-end=63}{B}} \htmlData{tutor-start=65,tutor-end=66}{+} \htmlData{tutor-start=67,tutor-end=68}{2}\overrightarrow{\htmlData{tutor-start=84,tutor-end=85}{A}\htmlData{tutor-start=85,tutor-end=86}{C}}$ $\htmlData{tutor-start=0,tutor-end=1}{3}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{D}} \htmlData{tutor-start=21,tutor-end=22}{=} \overrightarrow{\htmlData{tutor-start=39,tutor-end=40}{A}\htmlData{tutor-start=40,tutor-end=41}{B}} \htmlData{tutor-start=43,tutor-end=44}{+} \htmlData{tutor-start=45,tutor-end=46}{2}\overrightarrow{\htmlData{tutor-start=62,tutor-end=63}{A}\htmlData{tutor-start=63,tutor-end=64}{C}}$ 两边同除以 3： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{=} \frac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{3}}\overrightarrow{\htmlData{tutor-start=49,tutor-end=50}{A}\htmlData{tutor-start=50,tutor-end=51}{B}} \htmlData{tutor-start=53,tutor-end=54}{+} \frac{\htmlData{tutor-start=61,tutor-end=62}{2}}{\htmlData{tutor-start=64,tutor-end=65}{3}}\overrightarrow{\htmlData{tutor-start=82,tutor-end=83}{A}\htmlData{tutor-start=83,tutor-end=84}{C}}$ 已知 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}}\htmlData{tutor-start=19,tutor-end=20}{=}\boldsymbol{\htmlData{tutor-start=32,tutor-end=33}{c}}$ ， $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{C}}\htmlData{tutor-start=19,tutor-end=20}{=}\boldsymbol{\htmlData{tutor-start=32,tutor-end=33}{b}}$ ，代入得： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{=} \frac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{3}}\boldsymbol{\htmlData{tutor-start=45,tutor-end=46}{c}} \htmlData{tutor-start=48,tutor-end=49}{+} \frac{\htmlData{tutor-start=56,tutor-end=57}{2}}{\htmlData{tutor-start=59,tutor-end=60}{3}}\boldsymbol{\htmlData{tutor-start=73,tutor-end=74}{b}}$ 调整顺序以匹配选项格式： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{=} \frac{\htmlData{tutor-start=28,tutor-end=29}{2}}{\htmlData{tutor-start=31,tutor-end=32}{3}}\boldsymbol{\htmlData{tutor-start=45,tutor-end=46}{b}} \htmlData{tutor-start=48,tutor-end=49}{+} \frac{\htmlData{tutor-start=56,tutor-end=57}{1}}{\htmlData{tutor-start=59,tutor-end=60}{3}}\boldsymbol{\htmlData{tutor-start=73,tutor-end=74}{c}}$ 对比选项，(A) 符合。

\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{=} \frac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{3}}\boldsymbol{\htmlData{tutor-start=45,tutor-end=46}{c}} \htmlData{tutor-start=48,tutor-end=49}{+} \frac{\htmlData{tutor-start=56,tutor-end=57}{2}}{\htmlData{tutor-start=59,tutor-end=60}{3}}\boldsymbol{\htmlData{tutor-start=73,tutor-end=74}{b}}

一、选择题 · 数学竞赛/待细分

设 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=6}{\in }\mathbf{\htmlData{tutor-start=14,tutor-end=15}{R}}$ ，且 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\mathrm{\htmlData{tutor-start=26,tutor-end=27}{i}}$ 为正实数，则 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}$ ( ) (A) 2 (B) 1 (C) 0 (D) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}$

答案：(D)

题目标签：复数运算与正实数条件

解题过程

复数代数形式化简

计算 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\mathrm{\htmlData{tutor-start=26,tutor-end=27}{i}}$ 并令其为正实数

(1)

展开复数表达式

先计算平方，再乘以 $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{i}}$ ，将结果化为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{B}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}$ 的形式。

详细展开：给定复数 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{+}\mathrm{\htmlData{tutor-start=15,tutor-end=16}{i}}\htmlData{tutor-start=17,tutor-end=18}{)}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\mathrm{\htmlData{tutor-start=30,tutor-end=31}{i}}$ 。首先计算 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}^{\htmlData{tutor-start=16,tutor-end=17}{2}}$ ： $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{+}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{a}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{+} \htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{a}\mathrm{\htmlData{tutor-start=39,tutor-end=40}{i}} \htmlData{tutor-start=42,tutor-end=43}{+} \mathrm{\htmlData{tutor-start=52,tutor-end=53}{i}}^{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{=} \htmlData{tutor-start=61,tutor-end=62}{a}^{\htmlData{tutor-start=64,tutor-end=65}{2}} \htmlData{tutor-start=67,tutor-end=68}{+} \htmlData{tutor-start=69,tutor-end=70}{2}\htmlData{tutor-start=70,tutor-end=71}{a}\mathrm{\htmlData{tutor-start=79,tutor-end=80}{i}} \htmlData{tutor-start=82,tutor-end=83}{-} \htmlData{tutor-start=84,tutor-end=85}{1} \htmlData{tutor-start=86,tutor-end=87}{=} \htmlData{tutor-start=88,tutor-end=89}{(}\htmlData{tutor-start=89,tutor-end=90}{a}^{\htmlData{tutor-start=92,tutor-end=93}{2}}\htmlData{tutor-start=94,tutor-end=95}{-}\htmlData{tutor-start=95,tutor-end=96}{1}\htmlData{tutor-start=96,tutor-end=97}{)} \htmlData{tutor-start=98,tutor-end=99}{+} \htmlData{tutor-start=100,tutor-end=101}{2}\htmlData{tutor-start=101,tutor-end=102}{a}\mathrm{\htmlData{tutor-start=110,tutor-end=111}{i}}$ 然后乘以 $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{i}}$ ： $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{[}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{a}\mathrm{\htmlData{tutor-start=27,tutor-end=28}{i}}\htmlData{tutor-start=29,tutor-end=30}{]} \htmlData{tutor-start=31,tutor-end=37}{\cdot }\mathrm{\htmlData{tutor-start=45,tutor-end=46}{i}} \htmlData{tutor-start=48,tutor-end=49}{=} \htmlData{tutor-start=50,tutor-end=51}{(}\htmlData{tutor-start=51,tutor-end=52}{a}^{\htmlData{tutor-start=54,tutor-end=55}{2}}\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{1}\htmlData{tutor-start=58,tutor-end=59}{)}\mathrm{\htmlData{tutor-start=67,tutor-end=68}{i}} \htmlData{tutor-start=70,tutor-end=71}{+} \htmlData{tutor-start=72,tutor-end=73}{2}\htmlData{tutor-start=73,tutor-end=74}{a}\mathrm{\htmlData{tutor-start=82,tutor-end=83}{i}}^{\htmlData{tutor-start=86,tutor-end=87}{2}}$ 因为 $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{i}}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}$ ，所以： $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{a}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}\mathrm{\htmlData{tutor-start=21,tutor-end=22}{i}} \htmlData{tutor-start=24,tutor-end=25}{-} \htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{a} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{a} \htmlData{tutor-start=35,tutor-end=36}{+} \htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{a}^{\htmlData{tutor-start=41,tutor-end=42}{2}}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{1}\htmlData{tutor-start=45,tutor-end=46}{)}\mathrm{\htmlData{tutor-start=54,tutor-end=55}{i}}$ 此时，实部为 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{a}$ ，虚部为 $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}$ 。

\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{a} \htmlData{tutor-start=8,tutor-end=9}{+} \htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{a}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{)}\mathrm{\htmlData{tutor-start=27,tutor-end=28}{i}}

(2)

应用正实数条件求解

根据‘正实数’的定义，虚部必须为 0，实部必须大于 0。

详细展开：题目要求 $\htmlData{tutor-start=0,tutor-end=1}{z}$ 为**正实数**。这包含两个条件： 1. **是实数**：虚部为 0。 $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{1} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{0} \implies \htmlData{tutor-start=23,tutor-end=24}{a}^{\htmlData{tutor-start=26,tutor-end=27}{2}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{1} \implies \htmlData{tutor-start=42,tutor-end=43}{a} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{1} \text{ \htmlData{tutor-start=55,tutor-end=56}{或} } \htmlData{tutor-start=59,tutor-end=60}{a} \htmlData{tutor-start=61,tutor-end=62}{=} \htmlData{tutor-start=63,tutor-end=64}{-}\htmlData{tutor-start=64,tutor-end=65}{1}$ 2. **是正数**：实部大于 0。 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{a} \htmlData{tutor-start=4,tutor-end=5}{>} \htmlData{tutor-start=6,tutor-end=7}{0} \implies \htmlData{tutor-start=17,tutor-end=18}{a} \htmlData{tutor-start=19,tutor-end=20}{<} \htmlData{tutor-start=21,tutor-end=22}{0}$

综合以上两个条件： - 若 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}$ ，不满足 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0}$ ，舍去。（此时 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}$ ，是负实数） - 若 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0}$ ，满足 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0}$ 。（此时 $\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{0}\mathrm{\htmlData{tutor-start=22,tutor-end=23}{i}} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{2}$ ，是正实数）

因此，唯一解为 $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}$ 。对比选项，(D) 为 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}$ 。

\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}

一、选择题 · 代数

已知等差数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 满足 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{4}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{4}$ ， $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{0}$ ，则它的前 10 项的和 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}}\htmlData{tutor-start=6,tutor-end=7}{=}$ ( ) (A) 138 (B) 135 (C) 95 (D) 23

答案：(C)

题目标签：等差数列求和

解题过程

利用等差数列性质求解

根据已知两项和求前 10 项和 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}}$

(1)

求首项与公差或关键项

利用等差数列通项公式或下标和性质，建立方程组求解基本量。

详细展开：设等差数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 的首项为 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ ，公差为 $\htmlData{tutor-start=0,tutor-end=1}{d}$ 。已知： 1. $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{4}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{4}$ 2. $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{5}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{0}$

方法一：基本量法 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{d}\htmlData{tutor-start=17,tutor-end=18}{,} \htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{4}} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{a}_{\htmlData{tutor-start=30,tutor-end=31}{1}} \htmlData{tutor-start=33,tutor-end=34}{+} \htmlData{tutor-start=35,tutor-end=36}{3}\htmlData{tutor-start=36,tutor-end=37}{d} \implies \htmlData{tutor-start=47,tutor-end=48}{2}\htmlData{tutor-start=48,tutor-end=49}{a}_{\htmlData{tutor-start=51,tutor-end=52}{1}} \htmlData{tutor-start=54,tutor-end=55}{+} \htmlData{tutor-start=56,tutor-end=57}{4}\htmlData{tutor-start=57,tutor-end=58}{d} \htmlData{tutor-start=59,tutor-end=60}{=} \htmlData{tutor-start=61,tutor-end=62}{4} \implies \htmlData{tutor-start=72,tutor-end=73}{a}_{\htmlData{tutor-start=75,tutor-end=76}{1}} \htmlData{tutor-start=78,tutor-end=79}{+} \htmlData{tutor-start=80,tutor-end=81}{2}\htmlData{tutor-start=81,tutor-end=82}{d} \htmlData{tutor-start=83,tutor-end=84}{=} \htmlData{tutor-start=85,tutor-end=86}{2}$ ① $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{d}\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=21}{a}_{\htmlData{tutor-start=23,tutor-end=24}{5}} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{a}_{\htmlData{tutor-start=31,tutor-end=32}{1}} \htmlData{tutor-start=34,tutor-end=35}{+} \htmlData{tutor-start=36,tutor-end=37}{4}\htmlData{tutor-start=37,tutor-end=38}{d} \implies \htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=50}{a}_{\htmlData{tutor-start=52,tutor-end=53}{1}} \htmlData{tutor-start=55,tutor-end=56}{+} \htmlData{tutor-start=57,tutor-end=58}{6}\htmlData{tutor-start=58,tutor-end=59}{d} \htmlData{tutor-start=60,tutor-end=61}{=} \htmlData{tutor-start=62,tutor-end=63}{1}\htmlData{tutor-start=63,tutor-end=64}{0} \implies \htmlData{tutor-start=74,tutor-end=75}{a}_{\htmlData{tutor-start=77,tutor-end=78}{1}} \htmlData{tutor-start=80,tutor-end=81}{+} \htmlData{tutor-start=82,tutor-end=83}{3}\htmlData{tutor-start=83,tutor-end=84}{d} \htmlData{tutor-start=85,tutor-end=86}{=} \htmlData{tutor-start=87,tutor-end=88}{5}$ ② 由 ② - ① 得： $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=23}{ 3d) - (a_{1} +} \htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{d}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{5} \htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{2} \implies \htmlData{tutor-start=45,tutor-end=46}{d} \htmlData{tutor-start=47,tutor-end=48}{=} \htmlData{tutor-start=49,tutor-end=50}{3}$ 将 $\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$ 代入 ①： $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{6} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{2} \implies \htmlData{tutor-start=23,tutor-end=24}{a}_{\htmlData{tutor-start=26,tutor-end=27}{1}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{4}$

方法二：性质法 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{4}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{a}_{\htmlData{tutor-start=20,tutor-end=21}{3}} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{4} \implies \htmlData{tutor-start=36,tutor-end=37}{a}_{\htmlData{tutor-start=39,tutor-end=40}{3}} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{2}$ $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{5}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{a}_{\htmlData{tutor-start=20,tutor-end=21}{4}} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{0} \implies \htmlData{tutor-start=37,tutor-end=38}{a}_{\htmlData{tutor-start=40,tutor-end=41}{4}} \htmlData{tutor-start=43,tutor-end=44}{=} \htmlData{tutor-start=45,tutor-end=46}{5}$ 公差 $\htmlData{tutor-start=0,tutor-end=1}{d} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{4}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{3}} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{5} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{2} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{3}$ $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{3}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{d} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{2} \htmlData{tutor-start=23,tutor-end=24}{-} \htmlData{tutor-start=25,tutor-end=26}{6} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{4}$ 两种方法结果一致： $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{d} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{3}$ 。

\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \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=12,tutor-end=13}{d} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{3}

(2)

计算前 10 项和

使用等差数列求和公式计算 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}}$ 。

详细展开：等差数列前 $\htmlData{tutor-start=0,tutor-end=1}{n}$ 项和公式为： $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{n}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{1}} \htmlData{tutor-start=15,tutor-end=16}{+} \frac{\htmlData{tutor-start=23,tutor-end=24}{n}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{n}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{)}}{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{d}$ 当 $\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{0}$ 时： $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0} \htmlData{tutor-start=12,tutor-end=19}{\times }\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{4}\htmlData{tutor-start=22,tutor-end=23}{)} \htmlData{tutor-start=24,tutor-end=25}{+} \frac{\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{0} \htmlData{tutor-start=35,tutor-end=42}{\times }\htmlData{tutor-start=42,tutor-end=43}{9}}{\htmlData{tutor-start=45,tutor-end=46}{2}} \htmlData{tutor-start=48,tutor-end=55}{\times }\htmlData{tutor-start=55,tutor-end=56}{3}$ $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{0} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{5} \htmlData{tutor-start=18,tutor-end=25}{\times }\htmlData{tutor-start=25,tutor-end=26}{3}$ $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{0} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{5}$ $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{9}\htmlData{tutor-start=10,tutor-end=11}{5}$

或者使用 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}} \htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{a}_{\htmlData{tutor-start=21,tutor-end=22}{1}} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{a}_{\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{0}}\htmlData{tutor-start=32,tutor-end=33}{)}}{\htmlData{tutor-start=35,tutor-end=36}{2}} \htmlData{tutor-start=38,tutor-end=39}{=} \htmlData{tutor-start=40,tutor-end=41}{5}\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{a}_{\htmlData{tutor-start=45,tutor-end=46}{1}} \htmlData{tutor-start=48,tutor-end=49}{+} \htmlData{tutor-start=50,tutor-end=51}{a}_{\htmlData{tutor-start=53,tutor-end=54}{1}\htmlData{tutor-start=54,tutor-end=55}{0}}\htmlData{tutor-start=56,tutor-end=57}{)}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{1}} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{9}\htmlData{tutor-start=18,tutor-end=19}{d} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{4} \htmlData{tutor-start=25,tutor-end=26}{+} \htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{7} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{3}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{5}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{4} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{)} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{5} \htmlData{tutor-start=24,tutor-end=31}{\times }\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{9} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{9}\htmlData{tutor-start=37,tutor-end=38}{5}$ 。

对比选项，(C) 为 95。

\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{9}\htmlData{tutor-start=10,tutor-end=11}{5}

一、选择题 · 代数

若函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}$ 的图象与函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\sqrt{\htmlData{tutor-start=11,tutor-end=12}{x}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}$ 的图象关于直线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}$ 对称，则 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}$ ( ) (A) $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}}$ (B) $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}}$ (C) $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}}$ (D) $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}$

答案：(B)

题目标签：反函数与函数解析式求解

解题过程

利用反函数性质求原函数

根据对称性求出 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的解析式

(1)

求已知函数的反函数

题目指出 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}$ 的图象与 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\ln\sqrt{\htmlData{tutor-start=11,tutor-end=12}{x}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}$ 的图象关于 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}$ 对称，这意味着这两个函数互为反函数。先求 $\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \ln\sqrt{\htmlData{tutor-start=16,tutor-end=17}{x}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{1}$ 的反函数。

详细展开：设 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \ln\sqrt{\htmlData{tutor-start=13,tutor-end=14}{x}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{1}$ 。第一步：反解 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{=} \ln\sqrt{\htmlData{tutor-start=17,tutor-end=18}{x}}$ $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{2}}\ln \htmlData{tutor-start=23,tutor-end=24}{x}$ $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{y} \htmlData{tutor-start=4,tutor-end=5}{-} \htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)} \htmlData{tutor-start=9,tutor-end=10}{=} \ln \htmlData{tutor-start=15,tutor-end=16}{x}$ $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \mathrm{\htmlData{tutor-start=12,tutor-end=13}{e}}^{\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{y}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{)}}$ 第二步：互换 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y}$ 。得到反函数为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \mathrm{\htmlData{tutor-start=12,tutor-end=13}{e}}^{\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{)}}$ 。根据题意，这个反函数就是 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{f}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)}$ 。即： $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \mathrm{\htmlData{tutor-start=17,tutor-end=18}{e}}^{\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{x}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{)}}$

\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}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \mathrm{\htmlData{tutor-start=17,tutor-end=18}{e}}^{\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{x}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{)}}

(2)

换元求

\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}{)}

由 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}$ 的表达式推导 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的表达式。

详细展开：已知 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=8}{=} \mathrm{\htmlData{tutor-start=17,tutor-end=18}{e}}^{\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{x}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{)}}$ 。令 $\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{1}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{t} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}$ 。代入上式右边： $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \mathrm{\htmlData{tutor-start=15,tutor-end=16}{e}}^{\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{t}}$ 将变量 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 换回 $\htmlData{tutor-start=0,tutor-end=1}{x}$ ，得到： $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \mathrm{\htmlData{tutor-start=15,tutor-end=16}{e}}^{\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{x}}$

对比选项： (A) $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}}$ (B) $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}}$ (C) $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}}$ (D) $\mathrm{\htmlData{tutor-start=8,tutor-end=9}{e}}^{\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}}$

显然 (B) 正确。

\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}{=} \mathrm{\htmlData{tutor-start=15,tutor-end=16}{e}}^{\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{x}}

一、选择题 · 导数的几何意义与直线位置关系

设曲线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}}$ 在点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}$ 处的切线与直线 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{0}$ 垂直，则 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}$ ( ) (A) 2 (B) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}$

答案：(D)

题目标签：曲线切线与直线垂直求参数

解题过程

求解参数 a 的值

利用导数求切线斜率，结合垂直条件解出 a

(1)

求曲线在点 (3,2) 处的切线斜率

首先对函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}}$ 进行求导，计算其在 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$ 处的导数值，即为切线的斜率。

详细展开：函数为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}}$ 。根据商的求导法则 $\htmlData{tutor-start=0,tutor-end=1}{(}\frac{\htmlData{tutor-start=7,tutor-end=8}{u}}{\htmlData{tutor-start=10,tutor-end=11}{v}}\htmlData{tutor-start=12,tutor-end=13}{)}' \htmlData{tutor-start=15,tutor-end=16}{=} \frac{\htmlData{tutor-start=23,tutor-end=24}{u}'\htmlData{tutor-start=25,tutor-end=26}{v} \htmlData{tutor-start=27,tutor-end=28}{-} \htmlData{tutor-start=29,tutor-end=30}{u}\htmlData{tutor-start=30,tutor-end=31}{v}'}{\htmlData{tutor-start=34,tutor-end=35}{v}^{\htmlData{tutor-start=37,tutor-end=38}{2}}}$ ，设 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{v}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{u}'\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{v}'\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{y}' \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=17}{\cdot}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{)} \htmlData{tutor-start=23,tutor-end=24}{-} \htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{x}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=36}{\cdot }\htmlData{tutor-start=36,tutor-end=37}{1}}{\htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{x}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{)}^{\htmlData{tutor-start=46,tutor-end=47}{2}}} \htmlData{tutor-start=50,tutor-end=51}{=} \frac{\htmlData{tutor-start=58,tutor-end=59}{x}\htmlData{tutor-start=59,tutor-end=60}{-}\htmlData{tutor-start=60,tutor-end=61}{1}\htmlData{tutor-start=61,tutor-end=62}{-}\htmlData{tutor-start=62,tutor-end=63}{x}\htmlData{tutor-start=63,tutor-end=64}{-}\htmlData{tutor-start=64,tutor-end=65}{1}}{\htmlData{tutor-start=67,tutor-end=68}{(}\htmlData{tutor-start=68,tutor-end=69}{x}\htmlData{tutor-start=69,tutor-end=70}{-}\htmlData{tutor-start=70,tutor-end=71}{1}\htmlData{tutor-start=71,tutor-end=72}{)}^{\htmlData{tutor-start=74,tutor-end=75}{2}}} \htmlData{tutor-start=78,tutor-end=79}{=} \frac{\htmlData{tutor-start=86,tutor-end=87}{-}\htmlData{tutor-start=87,tutor-end=88}{2}}{\htmlData{tutor-start=90,tutor-end=91}{(}\htmlData{tutor-start=91,tutor-end=92}{x}\htmlData{tutor-start=92,tutor-end=93}{-}\htmlData{tutor-start=93,tutor-end=94}{1}\htmlData{tutor-start=94,tutor-end=95}{)}^{\htmlData{tutor-start=97,tutor-end=98}{2}}}$ 将 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}$ 代入导数公式： $\htmlData{tutor-start=0,tutor-end=1}{k}_{\htmlData{tutor-start=3,tutor-end=4}{t}\htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{n}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{y}'\htmlData{tutor-start=12,tutor-end=13}{|}_{\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{3}} \htmlData{tutor-start=20,tutor-end=21}{=} \frac{\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{2}}{\htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{3}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{)}^{\htmlData{tutor-start=39,tutor-end=40}{2}}} \htmlData{tutor-start=43,tutor-end=44}{=} \frac{\htmlData{tutor-start=51,tutor-end=52}{-}\htmlData{tutor-start=52,tutor-end=53}{2}}{\htmlData{tutor-start=55,tutor-end=56}{4}} \htmlData{tutor-start=58,tutor-end=59}{=} \htmlData{tutor-start=60,tutor-end=61}{-}\frac{\htmlData{tutor-start=67,tutor-end=68}{1}}{\htmlData{tutor-start=70,tutor-end=71}{2}}$ 因此，曲线在点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}$ 处的切线斜率为 $\htmlData{tutor-start=0,tutor-end=1}{-}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}}$ 。

\htmlData{tutor-start=0,tutor-end=1}{y}' \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{2}}{\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}\htmlData{tutor-start=19,tutor-end=20}{)}^{\htmlData{tutor-start=22,tutor-end=23}{2}}}\htmlData{tutor-start=25,tutor-end=26}{,} \quad \htmlData{tutor-start=33,tutor-end=34}{k}_{\htmlData{tutor-start=36,tutor-end=37}{t}\htmlData{tutor-start=37,tutor-end=38}{a}\htmlData{tutor-start=38,tutor-end=39}{n}} \htmlData{tutor-start=41,tutor-end=42}{=} \htmlData{tutor-start=43,tutor-end=44}{-}\frac{\htmlData{tutor-start=50,tutor-end=51}{1}}{\htmlData{tutor-start=53,tutor-end=54}{2}}

(2)

利用垂直条件求解 a

根据两条直线垂直的充要条件（斜率之积为 -1），建立关于 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的方程并求解。

详细展开：已知直线方程为 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{0}$ ，将其变形为斜截式 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{1}$ ，可知该直线的斜率 $\htmlData{tutor-start=0,tutor-end=1}{k}_{\htmlData{tutor-start=3,tutor-end=4}{l}\htmlData{tutor-start=4,tutor-end=5}{i}\htmlData{tutor-start=5,tutor-end=6}{n}\htmlData{tutor-start=6,tutor-end=7}{e}} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{a}$ 。因为切线与该直线垂直，所以它们的斜率乘积为 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}$ ，即： $\htmlData{tutor-start=0,tutor-end=1}{k}_{\htmlData{tutor-start=3,tutor-end=4}{t}\htmlData{tutor-start=4,tutor-end=5}{a}\htmlData{tutor-start=5,tutor-end=6}{n}} \htmlData{tutor-start=8,tutor-end=14}{\cdot }\htmlData{tutor-start=14,tutor-end=15}{k}_{\htmlData{tutor-start=17,tutor-end=18}{l}\htmlData{tutor-start=18,tutor-end=19}{i}\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{e}} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{1}$ 代入已知数值： $\left(\htmlData{tutor-start=6,tutor-end=7}{-}\frac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{2}}\right) \htmlData{tutor-start=26,tutor-end=32}{\cdot }\htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{a}\htmlData{tutor-start=35,tutor-end=36}{)} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{1}$ $\frac{\htmlData{tutor-start=6,tutor-end=7}{a}}{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{1}$ 解得： $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}$ 对比选项，答案为 (D)。

\htmlData{tutor-start=0,tutor-end=1}{k}_{\htmlData{tutor-start=3,tutor-end=4}{l}\htmlData{tutor-start=4,tutor-end=5}{i}\htmlData{tutor-start=5,tutor-end=6}{n}\htmlData{tutor-start=6,tutor-end=7}{e}} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{,} \quad \left(\htmlData{tutor-start=27,tutor-end=28}{-}\frac{\htmlData{tutor-start=34,tutor-end=35}{1}}{\htmlData{tutor-start=37,tutor-end=38}{2}}\right)\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{-}\htmlData{tutor-start=48,tutor-end=49}{a}\htmlData{tutor-start=49,tutor-end=50}{)} \htmlData{tutor-start=51,tutor-end=52}{=} \htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{1} \htmlData{tutor-start=56,tutor-end=68}{\Rightarrow }\htmlData{tutor-start=68,tutor-end=69}{a} \htmlData{tutor-start=70,tutor-end=71}{=} \htmlData{tutor-start=72,tutor-end=73}{-}\htmlData{tutor-start=73,tutor-end=74}{2}

一、选择题 · 三角函数的诱导公式与图像变换

为得到函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\cos\left(\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\frac{\htmlData{tutor-start=21,tutor-end=24}{\pi}}{\htmlData{tutor-start=26,tutor-end=27}{3}}\right)$ 的图象，只需将函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sin \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{x}$ 的图象 ( ) (A) 向左平移 $\frac{\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{2}}$ 个长度单位 (B) 向右平移 $\frac{\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{2}}$ 个长度单位 (C) 向左平移 $\frac{\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{6}}$ 个长度单位 (D) 向右平移 $\frac{\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{6}}$ 个长度单位

答案：(A)

题目标签：三角函数图像平移变换

解题过程

确定平移量

将目标函数化为正弦形式，比较相位差确定平移方向和单位

(1)

统一函数名称为正弦

为了便于比较平移量，需将目标函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\cos\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\frac{\htmlData{tutor-start=16,tutor-end=19}{\pi}}{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{)}$ 利用诱导公式转化为 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sin\htmlData{tutor-start=6,tutor-end=7}{(}\dots\htmlData{tutor-start=12,tutor-end=13}{)}$ 的形式，使其与源函数 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sin \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{x}$ 同名。

详细展开：利用诱导公式 $\cos \htmlData{tutor-start=5,tutor-end=12}{\theta }\htmlData{tutor-start=12,tutor-end=13}{=} \sin\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=26}{\theta }\htmlData{tutor-start=26,tutor-end=27}{+} \frac{\htmlData{tutor-start=34,tutor-end=37}{\pi}}{\htmlData{tutor-start=39,tutor-end=40}{2}}\htmlData{tutor-start=41,tutor-end=42}{)}$ 。令 $\htmlData{tutor-start=0,tutor-end=7}{\theta }\htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{x} \htmlData{tutor-start=12,tutor-end=13}{+} \frac{\htmlData{tutor-start=20,tutor-end=23}{\pi}}{\htmlData{tutor-start=25,tutor-end=26}{3}}$ ，则： $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \cos\left(\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{+}\frac{\htmlData{tutor-start=23,tutor-end=26}{\pi}}{\htmlData{tutor-start=28,tutor-end=29}{3}}\right) \htmlData{tutor-start=38,tutor-end=39}{=} \sin\left(\htmlData{tutor-start=50,tutor-end=51}{2}\htmlData{tutor-start=51,tutor-end=52}{x}\htmlData{tutor-start=52,tutor-end=53}{+}\frac{\htmlData{tutor-start=59,tutor-end=62}{\pi}}{\htmlData{tutor-start=64,tutor-end=65}{3}} \htmlData{tutor-start=67,tutor-end=68}{+} \frac{\htmlData{tutor-start=75,tutor-end=78}{\pi}}{\htmlData{tutor-start=80,tutor-end=81}{2}}\right)$ 通分计算相位常数： $\frac{\htmlData{tutor-start=6,tutor-end=9}{\pi}}{\htmlData{tutor-start=11,tutor-end=12}{3}} \htmlData{tutor-start=14,tutor-end=15}{+} \frac{\htmlData{tutor-start=22,tutor-end=25}{\pi}}{\htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{=} \frac{\htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=42}{\pi}}{\htmlData{tutor-start=44,tutor-end=45}{6}} \htmlData{tutor-start=47,tutor-end=48}{+} \frac{\htmlData{tutor-start=55,tutor-end=56}{3}\htmlData{tutor-start=56,tutor-end=59}{\pi}}{\htmlData{tutor-start=61,tutor-end=62}{6}} \htmlData{tutor-start=64,tutor-end=65}{=} \frac{\htmlData{tutor-start=72,tutor-end=73}{5}\htmlData{tutor-start=73,tutor-end=76}{\pi}}{\htmlData{tutor-start=78,tutor-end=79}{6}}$ 所以目标函数可写为： $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \sin\left(\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{x} \htmlData{tutor-start=17,tutor-end=18}{+} \frac{\htmlData{tutor-start=25,tutor-end=26}{5}\htmlData{tutor-start=26,tutor-end=29}{\pi}}{\htmlData{tutor-start=31,tutor-end=32}{6}}\right)$ 此时，源函数为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \sin \htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{x}$ ，目标函数为 $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \sin\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{x} \htmlData{tutor-start=12,tutor-end=13}{+} \frac{\htmlData{tutor-start=20,tutor-end=21}{5}\htmlData{tutor-start=21,tutor-end=24}{\pi}}{\htmlData{tutor-start=26,tutor-end=27}{6}}\htmlData{tutor-start=28,tutor-end=29}{)}$ 。

\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \sin\left(\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{x} \htmlData{tutor-start=17,tutor-end=18}{+} \frac{\htmlData{tutor-start=25,tutor-end=26}{5}\htmlData{tutor-start=26,tutor-end=29}{\pi}}{\htmlData{tutor-start=31,tutor-end=32}{6}}\right)

(2)

提取平移因子并判断方向

将目标函数中的 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 系数提取出来，写成 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sin\htmlData{tutor-start=6,tutor-end=7}{[}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=15}{\phi}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{]}$ 的形式，根据 $\htmlData{tutor-start=0,tutor-end=4}{\phi}$ 的正负和大小确定平移。

详细展开：目标函数： $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \sin\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{x} \htmlData{tutor-start=12,tutor-end=13}{+} \frac{\htmlData{tutor-start=20,tutor-end=21}{5}\htmlData{tutor-start=21,tutor-end=24}{\pi}}{\htmlData{tutor-start=26,tutor-end=27}{6}}\htmlData{tutor-start=28,tutor-end=29}{)}$ 。提取 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的系数 2： $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \sin\left[\htmlData{tutor-start=14,tutor-end=15}{2}\left(\htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=24}{+} \frac{\htmlData{tutor-start=31,tutor-end=32}{5}\htmlData{tutor-start=32,tutor-end=35}{\pi}}{\htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{2}}\right)\right]$ 源函数： $\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \sin\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{)}$ 。根据函数图像平移规则“左加右减”， $\htmlData{tutor-start=0,tutor-end=1}{x}$ 变为 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \frac{\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=14}{\pi}}{\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{2}}$ 表示向左平移。平移量为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{2}}$ 个单位长度。因此，只需将 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sin \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{x}$ 的图像向左平移 $\frac{\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=10}{\pi}}{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{2}}$ 个单位即可得到目标图像。对比选项，答案为 (A)。

\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=4}{+} \frac{\htmlData{tutor-start=11,tutor-end=12}{5}\htmlData{tutor-start=12,tutor-end=15}{\pi}}{\htmlData{tutor-start=17,tutor-end=18}{6}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{2}\left(\htmlData{tutor-start=29,tutor-end=30}{x} \htmlData{tutor-start=31,tutor-end=32}{+} \frac{\htmlData{tutor-start=39,tutor-end=40}{5}\htmlData{tutor-start=40,tutor-end=43}{\pi}}{\htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{2}}\right)

一、选择题 · 函数的奇偶性与单调性

设奇函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=10}{\infty}\htmlData{tutor-start=10,tutor-end=11}{)}$ 上为增函数，且 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}$ ，则不等式 $\frac{\htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{f}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{)}}{\htmlData{tutor-start=18,tutor-end=19}{x}}\htmlData{tutor-start=20,tutor-end=21}{<}\htmlData{tutor-start=21,tutor-end=22}{0}$ 的解集为 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=10}{\cup}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=20}{\infty}\htmlData{tutor-start=20,tutor-end=21}{)}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=16}{\cup}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{)}$ (C) $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=16}{\cup}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=26}{\infty}\htmlData{tutor-start=26,tutor-end=27}{)}$ (D) $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=10}{\cup}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{)}$

答案：(D)

题目标签：奇函数性质与不等式求解

解题过程

简化不等式并求解

利用奇函数性质化简不等式，结合单调性和零点求解集

(1)

利用奇函数性质化简不等式

分析不等式 $\frac{\htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{f}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{)}}{\htmlData{tutor-start=18,tutor-end=19}{x}}\htmlData{tutor-start=20,tutor-end=21}{<}\htmlData{tutor-start=21,tutor-end=22}{0}$ ，利用 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 是奇函数的性质 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}$ 进行化简。

详细展开：因为 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 是奇函数，所以对于定义域内的任意 $\htmlData{tutor-start=0,tutor-end=1}{x}$ ，有 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{)}$ 。代入分子： $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{-} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{)} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{f}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{)} \htmlData{tutor-start=20,tutor-end=21}{-} \htmlData{tutor-start=22,tutor-end=23}{[}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{f}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{x}\htmlData{tutor-start=27,tutor-end=28}{)}\htmlData{tutor-start=28,tutor-end=29}{]} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{f}\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{x}\htmlData{tutor-start=36,tutor-end=37}{)}$ 原不等式变为： $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}}{\htmlData{tutor-start=13,tutor-end=14}{x}} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{0}$ 由于 $\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}$ ，不等式等价于： $\frac{\htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{)}}{\htmlData{tutor-start=12,tutor-end=13}{x}} \htmlData{tutor-start=15,tutor-end=16}{<} \htmlData{tutor-start=17,tutor-end=18}{0}$ 这意味着 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 异号。即分为两种情况： 1. $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{0} \\ \htmlData{tutor-start=23,tutor-end=24}{f}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=29}{<} \htmlData{tutor-start=30,tutor-end=31}{0} \end{cases}$ 2. $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{0} \\ \htmlData{tutor-start=23,tutor-end=24}{f}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=29}{>} \htmlData{tutor-start=30,tutor-end=31}{0} \end{cases}$

\frac{\htmlData{tutor-start=6,tutor-end=7}{f}\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=12,tutor-end=13}{x}} \htmlData{tutor-start=15,tutor-end=16}{<} \htmlData{tutor-start=17,tutor-end=18}{0} \iff \htmlData{tutor-start=24,tutor-end=25}{f}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{x}\htmlData{tutor-start=27,tutor-end=28}{)} \htmlData{tutor-start=29,tutor-end=35}{\cdot }\htmlData{tutor-start=35,tutor-end=36}{x} \htmlData{tutor-start=37,tutor-end=38}{<} \htmlData{tutor-start=39,tutor-end=40}{0}

(2)

结合单调性和零点求解各区间

根据 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}$ 上的单调性和零点 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}$ ，确定 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的符号分布，进而求出 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 的范围。

详细展开：已知 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}$ 上为增函数，且 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}$ 。所以在 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}$ 上： - 当 $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{1}$ 时， $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{<} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{0}$ ； - 当 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}$ 时， $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{>} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{0}$ 。

现在回到两种情况：情况 1： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{<} \htmlData{tutor-start=7,tutor-end=8}{0}$ 。由上述分析， $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}$ 时 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{<} \htmlData{tutor-start=7,tutor-end=8}{0}$ 的解集为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}$ 。

情况 2： $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{>} \htmlData{tutor-start=7,tutor-end=8}{0}$ 。利用奇函数对称性： $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)}$ 上也是增函数（奇函数在对称区间单调性相同）。且 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{0}$ 。所以在 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)}$ 上： - 当 $\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}$ 时， $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{<} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{0}$ ； - 当 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{0}$ 时， $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{>} \htmlData{tutor-start=7,tutor-end=8}{f}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{0}$ 。我们需要 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{>} \htmlData{tutor-start=7,tutor-end=8}{0}$ ，故解集为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$ 。

综上所述，不等式的解集为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=13}{\cup }\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{)}$ 。对比选项，答案为 (D)。

\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)} \htmlData{tutor-start=13,tutor-end=18}{\cup }\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{)}

一、选择题 · 解析几何与三角不等式

若直线 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}}{\htmlData{tutor-start=9,tutor-end=10}{a}}\htmlData{tutor-start=11,tutor-end=12}{+}\frac{\htmlData{tutor-start=18,tutor-end=19}{y}}{\htmlData{tutor-start=21,tutor-end=22}{b}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{1}$ 通过点 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{(}\cos\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=13}{,}\sin\htmlData{tutor-start=17,tutor-end=23}{\alpha}\htmlData{tutor-start=23,tutor-end=24}{)}$ ，则 ( ) (A) $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=22}{\leqslant }\htmlData{tutor-start=22,tutor-end=23}{1}$ (B) $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=22}{\geqslant }\htmlData{tutor-start=22,tutor-end=23}{1}$ (C) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}\htmlData{tutor-start=15,tutor-end=16}{+}\frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{b}^{\htmlData{tutor-start=28,tutor-end=29}{2}}} \htmlData{tutor-start=32,tutor-end=42}{\leqslant }\htmlData{tutor-start=42,tutor-end=43}{1}$ (D) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}\htmlData{tutor-start=15,tutor-end=16}{+}\frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{b}^{\htmlData{tutor-start=28,tutor-end=29}{2}}} \htmlData{tutor-start=32,tutor-end=42}{\geqslant }\htmlData{tutor-start=42,tutor-end=43}{1}$

答案：(D)

题目标签：直线过单位圆上点的截距不等式推导

解题过程

利用点在直线上建立等量关系并推导不等式

根据点 $\htmlData{tutor-start=0,tutor-end=1}{M}$ 在直线上，结合三角恒等式或柯西不等式，确定 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}$ 满足的不等关系

(1)

代入点坐标并整理方程

将点 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{(}\cos\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=13}{,} \sin\htmlData{tutor-start=18,tutor-end=24}{\alpha}\htmlData{tutor-start=24,tutor-end=25}{)}$ 的坐标代入直线方程，得到关于 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=12}{\alpha}$ 的等式，为后续消去参数 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 做准备。

详细展开：已知直线方程为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{x}}{\htmlData{tutor-start=9,tutor-end=10}{a}}\htmlData{tutor-start=11,tutor-end=12}{+}\frac{\htmlData{tutor-start=18,tutor-end=19}{y}}{\htmlData{tutor-start=21,tutor-end=22}{b}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{1}$ 。因为直线通过点 $\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{(}\cos\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=13}{,} \sin\htmlData{tutor-start=18,tutor-end=24}{\alpha}\htmlData{tutor-start=24,tutor-end=25}{)}$ ，所以将 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\cos\htmlData{tutor-start=6,tutor-end=12}{\alpha}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\sin\htmlData{tutor-start=6,tutor-end=12}{\alpha}$ 代入方程，得： $\frac{\cos\htmlData{tutor-start=10,tutor-end=16}{\alpha}}{\htmlData{tutor-start=18,tutor-end=19}{a}} \htmlData{tutor-start=21,tutor-end=22}{+} \frac{\sin\htmlData{tutor-start=33,tutor-end=39}{\alpha}}{\htmlData{tutor-start=41,tutor-end=42}{b}} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{1}$ 此式表明，存在实数 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ ，使得上述等式成立。我们的目标是从这个等式中推导出仅含 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}$ 的不等式。

\frac{\cos\htmlData{tutor-start=10,tutor-end=16}{\alpha}}{\htmlData{tutor-start=18,tutor-end=19}{a}} \htmlData{tutor-start=21,tutor-end=22}{+} \frac{\sin\htmlData{tutor-start=33,tutor-end=39}{\alpha}}{\htmlData{tutor-start=41,tutor-end=42}{b}} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{1}

(2)

利用柯西不等式或三角有界性推导结论

利用柯西-施瓦茨不等式（Cauchy-Schwarz Inequality）或辅助角公式，从上述等式中剥离出 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}$ 的范围限制。

详细展开：方法一：利用柯西不等式对于实数序列 $\htmlData{tutor-start=0,tutor-end=1}{(}\frac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{a}}\htmlData{tutor-start=12,tutor-end=13}{,} \frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{b}}\htmlData{tutor-start=25,tutor-end=26}{)}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{(}\cos\htmlData{tutor-start=5,tutor-end=11}{\alpha}\htmlData{tutor-start=11,tutor-end=12}{,} \sin\htmlData{tutor-start=17,tutor-end=23}{\alpha}\htmlData{tutor-start=23,tutor-end=24}{)}$ ，根据柯西不等式： $\left( \frac{\cos\htmlData{tutor-start=17,tutor-end=23}{\alpha}}{\htmlData{tutor-start=25,tutor-end=26}{a}} \htmlData{tutor-start=28,tutor-end=29}{+} \frac{\sin\htmlData{tutor-start=40,tutor-end=46}{\alpha}}{\htmlData{tutor-start=48,tutor-end=49}{b}} \right)^{\htmlData{tutor-start=60,tutor-end=61}{2}} \htmlData{tutor-start=63,tutor-end=73}{\leqslant }\left( \frac{\htmlData{tutor-start=86,tutor-end=87}{1}}{\htmlData{tutor-start=89,tutor-end=90}{a}^{\htmlData{tutor-start=92,tutor-end=93}{2}}} \htmlData{tutor-start=96,tutor-end=97}{+} \frac{\htmlData{tutor-start=104,tutor-end=105}{1}}{\htmlData{tutor-start=107,tutor-end=108}{b}^{\htmlData{tutor-start=110,tutor-end=111}{2}}} \right) \htmlData{tutor-start=122,tutor-end=123}{(}\cos^{\htmlData{tutor-start=129,tutor-end=130}{2}}\htmlData{tutor-start=131,tutor-end=138}{\alpha }\htmlData{tutor-start=138,tutor-end=139}{+} \sin^{\htmlData{tutor-start=146,tutor-end=147}{2}}\htmlData{tutor-start=148,tutor-end=154}{\alpha}\htmlData{tutor-start=154,tutor-end=155}{)}$ 已知 $\frac{\cos\htmlData{tutor-start=10,tutor-end=16}{\alpha}}{\htmlData{tutor-start=18,tutor-end=19}{a}} \htmlData{tutor-start=21,tutor-end=22}{+} \frac{\sin\htmlData{tutor-start=33,tutor-end=39}{\alpha}}{\htmlData{tutor-start=41,tutor-end=42}{b}} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{1}$ 且 $\cos^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=15}{\alpha }\htmlData{tutor-start=15,tutor-end=16}{+} \sin^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=32}{\alpha }\htmlData{tutor-start=32,tutor-end=33}{=} \htmlData{tutor-start=34,tutor-end=35}{1}$ ，代入得： $\htmlData{tutor-start=0,tutor-end=1}{1}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=16}{\leqslant }\left( \frac{\htmlData{tutor-start=29,tutor-end=30}{1}}{\htmlData{tutor-start=32,tutor-end=33}{a}^{\htmlData{tutor-start=35,tutor-end=36}{2}}} \htmlData{tutor-start=39,tutor-end=40}{+} \frac{\htmlData{tutor-start=47,tutor-end=48}{1}}{\htmlData{tutor-start=50,tutor-end=51}{b}^{\htmlData{tutor-start=53,tutor-end=54}{2}}} \right) \htmlData{tutor-start=65,tutor-end=71}{\cdot }\htmlData{tutor-start=71,tutor-end=72}{1}$ 即： $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}} \htmlData{tutor-start=16,tutor-end=17}{+} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{b}^{\htmlData{tutor-start=30,tutor-end=31}{2}}} \htmlData{tutor-start=34,tutor-end=44}{\geqslant }\htmlData{tutor-start=44,tutor-end=45}{1}$

方法二：利用三角函数有界性（辅助角公式）由 $\frac{\cos\htmlData{tutor-start=10,tutor-end=16}{\alpha}}{\htmlData{tutor-start=18,tutor-end=19}{a}} \htmlData{tutor-start=21,tutor-end=22}{+} \frac{\sin\htmlData{tutor-start=33,tutor-end=39}{\alpha}}{\htmlData{tutor-start=41,tutor-end=42}{b}} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{1}$ ，令 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{B} \htmlData{tutor-start=19,tutor-end=20}{=} \frac{\htmlData{tutor-start=27,tutor-end=28}{1}}{\htmlData{tutor-start=30,tutor-end=31}{b}}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{A}\cos\htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{B}\sin\htmlData{tutor-start=19,tutor-end=26}{\alpha }\htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{1}$ 。根据辅助角公式， $\htmlData{tutor-start=0,tutor-end=1}{A}\cos\htmlData{tutor-start=5,tutor-end=12}{\alpha }\htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{B}\sin\htmlData{tutor-start=19,tutor-end=26}{\alpha }\htmlData{tutor-start=26,tutor-end=27}{=} \sqrt{\htmlData{tutor-start=34,tutor-end=35}{A}^{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{+}\htmlData{tutor-start=40,tutor-end=41}{B}^{\htmlData{tutor-start=43,tutor-end=44}{2}}}\sin\htmlData{tutor-start=50,tutor-end=51}{(}\htmlData{tutor-start=51,tutor-end=57}{\alpha}\htmlData{tutor-start=57,tutor-end=58}{+}\htmlData{tutor-start=58,tutor-end=62}{\phi}\htmlData{tutor-start=62,tutor-end=63}{)}$ 。因为正弦函数的值域为 $\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{]}$ ，要使方程有解，必须满足振幅大于等于右边的常数绝对值： $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{A}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{B}^{\htmlData{tutor-start=15,tutor-end=16}{2}}} \htmlData{tutor-start=19,tutor-end=29}{\geqslant }\htmlData{tutor-start=29,tutor-end=30}{|}\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{|} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{1}$ 两边平方得： $\htmlData{tutor-start=0,tutor-end=1}{A}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{B}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=24}{\geqslant }\htmlData{tutor-start=24,tutor-end=25}{1} \implies \frac{\htmlData{tutor-start=41,tutor-end=42}{1}}{\htmlData{tutor-start=44,tutor-end=45}{a}^{\htmlData{tutor-start=47,tutor-end=48}{2}}} \htmlData{tutor-start=51,tutor-end=52}{+} \frac{\htmlData{tutor-start=59,tutor-end=60}{1}}{\htmlData{tutor-start=62,tutor-end=63}{b}^{\htmlData{tutor-start=65,tutor-end=66}{2}}} \htmlData{tutor-start=69,tutor-end=79}{\geqslant }\htmlData{tutor-start=79,tutor-end=80}{1}$

对比选项： (A) $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=22}{\leqslant }\htmlData{tutor-start=22,tutor-end=23}{1}$ ：错误，例如 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}$ 时 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{4}}\htmlData{tutor-start=11,tutor-end=12}{+}\frac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{4}}\htmlData{tutor-start=23,tutor-end=24}{=}\frac{\htmlData{tutor-start=30,tutor-end=31}{1}}{\htmlData{tutor-start=33,tutor-end=34}{2}}\htmlData{tutor-start=35,tutor-end=36}{<}\htmlData{tutor-start=36,tutor-end=37}{1}$ 不满足条件，但若取 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{0}$ ，则 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{0}}\htmlData{tutor-start=13,tutor-end=14}{+}\frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=26}{0}} \htmlData{tutor-start=28,tutor-end=29}{<} \htmlData{tutor-start=30,tutor-end=31}{1}$ ，此时直线不过单位圆？不对，若 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}\htmlData{tutor-start=15,tutor-end=16}{+}\frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{b}^{\htmlData{tutor-start=28,tutor-end=29}{2}}} \htmlData{tutor-start=32,tutor-end=33}{<} \htmlData{tutor-start=34,tutor-end=35}{1}$ ，则直线与单位圆相离，不可能有点在圆上。反之，若 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}$ 很大， $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}\htmlData{tutor-start=15,tutor-end=16}{+}\frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{b}^{\htmlData{tutor-start=28,tutor-end=29}{2}}}$ 很小，确实可能小于1吗？不，推导结果是必须 $\htmlData{tutor-start=0,tutor-end=10}{\geqslant }\htmlData{tutor-start=10,tutor-end=11}{1}$ 。如果 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{0}$ ，则 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{0}}\htmlData{tutor-start=15,tutor-end=16}{+}\frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{0}} \htmlData{tutor-start=32,tutor-end=33}{<} \htmlData{tutor-start=34,tutor-end=35}{1}$ ，这意味着不存在这样的 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 。题目说“若...通过...”，说明这样的 $\htmlData{tutor-start=0,tutor-end=6}{\alpha}$ 是存在的，因此必然满足推导出的不等式。 (B) $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=22}{\geqslant }\htmlData{tutor-start=22,tutor-end=23}{1}$ ：不一定。例如 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{.}\htmlData{tutor-start=4,tutor-end=5}{5}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{.}\htmlData{tutor-start=11,tutor-end=12}{5}$ ，则 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{.}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{5}}\htmlData{tutor-start=14,tutor-end=15}{+}\frac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=26}{.}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{5}} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{8} \htmlData{tutor-start=34,tutor-end=44}{\geqslant }\htmlData{tutor-start=44,tutor-end=45}{1}$ ，满足条件，但 $\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{.}\htmlData{tutor-start=16,tutor-end=17}{5} \htmlData{tutor-start=18,tutor-end=19}{<} \htmlData{tutor-start=20,tutor-end=21}{1}$ 。故(B)不恒成立。 (C) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}\htmlData{tutor-start=15,tutor-end=16}{+}\frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{b}^{\htmlData{tutor-start=28,tutor-end=29}{2}}} \htmlData{tutor-start=32,tutor-end=42}{\leqslant }\htmlData{tutor-start=42,tutor-end=43}{1}$ ：与推导结果相反。 (D) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}}\htmlData{tutor-start=15,tutor-end=16}{+}\frac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{b}^{\htmlData{tutor-start=28,tutor-end=29}{2}}} \htmlData{tutor-start=32,tutor-end=42}{\geqslant }\htmlData{tutor-start=42,tutor-end=43}{1}$ ：正确。

\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{a}^{\htmlData{tutor-start=12,tutor-end=13}{2}}} \htmlData{tutor-start=16,tutor-end=17}{+} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{b}^{\htmlData{tutor-start=30,tutor-end=31}{2}}} \htmlData{tutor-start=34,tutor-end=44}{\geqslant }\htmlData{tutor-start=44,tutor-end=45}{1}

一、选择题 · 立体几何：线面角

已知三棱柱 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{A}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{B}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{C}_{\htmlData{tutor-start=17,tutor-end=18}{1}}$ 的侧棱与底面边长都相等， $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 在底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 内的射影为 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 的中心，则 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 与底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 所成角的正弦值等于 ( ) (A) $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}}$ (B) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}}{\htmlData{tutor-start=16,tutor-end=17}{3}}$ (C) $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{3}}$ (D) $\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{3}}$

答案：(B)

题目标签：正三棱柱中线面角的正弦值

解题过程

计算线面角的正弦值

建立空间直角坐标系或使用几何法找出射影，计算角度

(1)

几何建模与关键点定位

设定棱长，确定顶点在底面的射影位置，构建直角三角形求解线面角。

详细展开：设三棱柱的所有棱长均为 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 。底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 是边长为 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的等边三角形。 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 在底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 内的射影 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 为 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 的中心（即重心/外心/垂心）。连接 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}$ 并延长交 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 于 $\htmlData{tutor-start=0,tutor-end=1}{D}$ ，则 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}$ 为底面高， $\htmlData{tutor-start=0,tutor-end=1}{O}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}$ 上。在等边 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 中，高 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}}{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{a}$ 。重心 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 分中线 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}$ 为 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{:}\htmlData{tutor-start=2,tutor-end=3}{1}$ ，故 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\htmlData{tutor-start=11,tutor-end=12}{2}}{\htmlData{tutor-start=14,tutor-end=15}{3}} \htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{D} \htmlData{tutor-start=20,tutor-end=21}{=} \frac{\htmlData{tutor-start=28,tutor-end=29}{2}}{\htmlData{tutor-start=31,tutor-end=32}{3}} \htmlData{tutor-start=34,tutor-end=40}{\cdot }\frac{\sqrt{\htmlData{tutor-start=52,tutor-end=53}{3}}}{\htmlData{tutor-start=56,tutor-end=57}{2}}\htmlData{tutor-start=58,tutor-end=59}{a} \htmlData{tutor-start=60,tutor-end=61}{=} \frac{\sqrt{\htmlData{tutor-start=74,tutor-end=75}{3}}}{\htmlData{tutor-start=78,tutor-end=79}{3}}\htmlData{tutor-start=80,tutor-end=81}{a}$ 。

我们需要求 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 与底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 所成角。注意：题目问的是 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 与底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 的角。 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 在底面的射影是 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 。所以 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 在底面的射影是 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 。等等，这里需要仔细审题。题目是“ $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 与底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 所成角”。直线 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 与平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 的交点是 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 。点 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 在平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 上的射影是点 $\htmlData{tutor-start=0,tutor-end=1}{B}$ （因为是直棱柱吗？不，题目说“侧棱与底面边长都相等... $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 射影为中心”，这说明它是**斜**三棱柱，或者是正三棱柱但描述有误？重读题面：“已知三棱柱... 侧棱与底面边长都相等... $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 在底面射影为中心”。如果是正三棱柱（侧棱垂直底面）， $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 的射影应该是 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 。既然射影是中心，说明侧棱 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 不垂直于底面。这是一个**斜三棱柱**，但侧棱长等于底面边长。

修正模型：这是一个斜三棱柱。侧棱 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{B}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{C}\htmlData{tutor-start=19,tutor-end=20}{C}_{\htmlData{tutor-start=22,tutor-end=23}{1}} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{a}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 在底面射影 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 是中心。我们需要求 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 与底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 所成的角。找 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 在底面的射影 $\htmlData{tutor-start=0,tutor-end=1}{B}'$ 。由于棱柱性质，向量 $\vec{\htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{B}_{\htmlData{tutor-start=9,tutor-end=10}{1}}} \htmlData{tutor-start=13,tutor-end=14}{=} \vec{\htmlData{tutor-start=20,tutor-end=21}{A}\htmlData{tutor-start=21,tutor-end=22}{A}_{\htmlData{tutor-start=24,tutor-end=25}{1}}}$ 。所以 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 在底面的射影 $\htmlData{tutor-start=0,tutor-end=1}{B}'$ 满足 $\vec{\htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{B}'} \htmlData{tutor-start=10,tutor-end=11}{=} \vec{\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{O}}$ （水平位移相同）。即 $\htmlData{tutor-start=0,tutor-end=1}{B}'$ 是点 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 按向量 $\vec{\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{O}}$ 平移后的点？不，是 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 加上 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 相对 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的水平位移。让我们建立坐标系更稳妥。

\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}}{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{a}

(2)

建立坐标系求解

建立空间直角坐标系，写出 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}_{\htmlData{tutor-start=6,tutor-end=7}{1}}$ 的坐标，利用向量法求线面角。

详细展开：设底面中心 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 为原点 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$ 不太方便，因为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{C}$ 坐标带根号。不如以 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 为原点，但为了方便，我们以底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 所在平面为 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{y}$ 平面。让 $\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 轴负半轴（假设方向）， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{-}\frac{\sqrt{\htmlData{tutor-start=18,tutor-end=19}{3}}}{\htmlData{tutor-start=22,tutor-end=23}{3}}\htmlData{tutor-start=24,tutor-end=25}{a}\htmlData{tutor-start=25,tutor-end=26}{,} \htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{)}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 的射影是 $\htmlData{tutor-start=0,tutor-end=1}{O}$ ，且 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{a}$ 。设 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 高度为 $\htmlData{tutor-start=0,tutor-end=1}{h}$ 。在 Rt $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{O}\htmlData{tutor-start=12,tutor-end=13}{A}_{\htmlData{tutor-start=15,tutor-end=16}{1}}$ 中， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O} \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}}{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{a}$ ， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{a}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{h} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{A}_{\htmlData{tutor-start=14,tutor-end=15}{1}}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{A}\htmlData{tutor-start=24,tutor-end=25}{O}^{\htmlData{tutor-start=27,tutor-end=28}{2}}} \htmlData{tutor-start=31,tutor-end=32}{=} \sqrt{\htmlData{tutor-start=39,tutor-end=40}{a}^{\htmlData{tutor-start=42,tutor-end=43}{2}} \htmlData{tutor-start=45,tutor-end=46}{-} \frac{\htmlData{tutor-start=53,tutor-end=54}{3}}{\htmlData{tutor-start=56,tutor-end=57}{9}}\htmlData{tutor-start=58,tutor-end=59}{a}^{\htmlData{tutor-start=61,tutor-end=62}{2}}} \htmlData{tutor-start=65,tutor-end=66}{=} \sqrt{\frac{\htmlData{tutor-start=79,tutor-end=80}{6}}{\htmlData{tutor-start=82,tutor-end=83}{9}}\htmlData{tutor-start=84,tutor-end=85}{a}^{\htmlData{tutor-start=87,tutor-end=88}{2}}} \htmlData{tutor-start=91,tutor-end=92}{=} \frac{\sqrt{\htmlData{tutor-start=105,tutor-end=106}{6}}}{\htmlData{tutor-start=109,tutor-end=110}{3}}\htmlData{tutor-start=111,tutor-end=112}{a}$ 。所以 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{,} \frac{\sqrt{\htmlData{tutor-start=24,tutor-end=25}{6}}}{\htmlData{tutor-start=28,tutor-end=29}{3}}\htmlData{tutor-start=30,tutor-end=31}{a}\htmlData{tutor-start=31,tutor-end=32}{)}$ 。（注意：这里假设 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{z}$ 轴上，因为射影是原点。但这要求 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的坐标配合。若 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 为原点， $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的坐标需调整。）

重新建系：令 $\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{)}$ 为底面中心。 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 点坐标： $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{-}\frac{\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}}{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=25}{,} \htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{)}$ 。因为 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 射影为 $\htmlData{tutor-start=0,tutor-end=1}{O}$ ，所以 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 的 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$ 坐标为 $\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{0}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{z}$ 坐标为 $\htmlData{tutor-start=0,tutor-end=1}{h} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\sqrt{\htmlData{tutor-start=16,tutor-end=17}{6}}}{\htmlData{tutor-start=20,tutor-end=21}{3}}\htmlData{tutor-start=22,tutor-end=23}{a}$ 。即 $\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{,} \frac{\sqrt{\htmlData{tutor-start=24,tutor-end=25}{6}}}{\htmlData{tutor-start=28,tutor-end=29}{3}}\htmlData{tutor-start=30,tutor-end=31}{a}\htmlData{tutor-start=31,tutor-end=32}{)}$ 。验证 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{A}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 长度： $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{0}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{(}\frac{\sqrt{\htmlData{tutor-start=27,tutor-end=28}{3}}}{\htmlData{tutor-start=31,tutor-end=32}{3}}\htmlData{tutor-start=33,tutor-end=34}{a}\htmlData{tutor-start=34,tutor-end=35}{)}^{\htmlData{tutor-start=37,tutor-end=38}{2}} \htmlData{tutor-start=40,tutor-end=41}{+} \htmlData{tutor-start=42,tutor-end=43}{(}\frac{\sqrt{\htmlData{tutor-start=55,tutor-end=56}{6}}}{\htmlData{tutor-start=59,tutor-end=60}{3}}\htmlData{tutor-start=61,tutor-end=62}{a}\htmlData{tutor-start=62,tutor-end=63}{)}^{\htmlData{tutor-start=65,tutor-end=66}{2}}} \htmlData{tutor-start=69,tutor-end=70}{=} \sqrt{\frac{\htmlData{tutor-start=83,tutor-end=84}{3}}{\htmlData{tutor-start=86,tutor-end=87}{9}}\htmlData{tutor-start=88,tutor-end=89}{a}^{\htmlData{tutor-start=91,tutor-end=92}{2}} \htmlData{tutor-start=94,tutor-end=95}{+} \frac{\htmlData{tutor-start=102,tutor-end=103}{6}}{\htmlData{tutor-start=105,tutor-end=106}{9}}\htmlData{tutor-start=107,tutor-end=108}{a}^{\htmlData{tutor-start=110,tutor-end=111}{2}}} \htmlData{tutor-start=114,tutor-end=115}{=} \htmlData{tutor-start=116,tutor-end=117}{a}$ 。正确。

接下来找 $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 的坐标。向量 $\vec{\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{A}_{\htmlData{tutor-start=9,tutor-end=10}{1}}} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{A}_{\htmlData{tutor-start=18,tutor-end=19}{1}} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{A} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{,} \frac{\sqrt{\htmlData{tutor-start=43,tutor-end=44}{3}}}{\htmlData{tutor-start=47,tutor-end=48}{3}}\htmlData{tutor-start=49,tutor-end=50}{a}\htmlData{tutor-start=50,tutor-end=51}{,} \frac{\sqrt{\htmlData{tutor-start=64,tutor-end=65}{6}}}{\htmlData{tutor-start=68,tutor-end=69}{3}}\htmlData{tutor-start=70,tutor-end=71}{a}\htmlData{tutor-start=71,tutor-end=72}{)}$ 。因为是棱柱， $\vec{\htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{B}_{\htmlData{tutor-start=9,tutor-end=10}{1}}} \htmlData{tutor-start=13,tutor-end=14}{=} \vec{\htmlData{tutor-start=20,tutor-end=21}{A}\htmlData{tutor-start=21,tutor-end=22}{A}_{\htmlData{tutor-start=24,tutor-end=25}{1}}}$ 。我们需要 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 的坐标。在底面 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{y}$ 中， $\htmlData{tutor-start=0,tutor-end=1}{B}$ 点相对于中心 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 轴负向。 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{C}$ 关于 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 轴对称。 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 的 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 坐标为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=19}{\times }\frac{\sqrt{\htmlData{tutor-start=31,tutor-end=32}{3}}}{\htmlData{tutor-start=35,tutor-end=36}{3}}\htmlData{tutor-start=37,tutor-end=38}{a} \htmlData{tutor-start=39,tutor-end=46}{\times }\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{-}\htmlData{tutor-start=48,tutor-end=49}{1}\htmlData{tutor-start=49,tutor-end=50}{)}\htmlData{tutor-start=50,tutor-end=51}{?}$ 不，重心到顶点距离 $\htmlData{tutor-start=0,tutor-end=1}{R} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\sqrt{\htmlData{tutor-start=16,tutor-end=17}{3}}}{\htmlData{tutor-start=20,tutor-end=21}{3}}\htmlData{tutor-start=22,tutor-end=23}{a}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{R}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{)}$ 。 $B = (R \cos 30^\circ, R \sin 30^\circ, 0) = (\frac{\sqrt{3}}{3}a \cdot \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{3}a \cdot \frac{1}{2}, 0) = (\frac{a}{2}, \frac{\sqrt{3}}{6}a, 0)$ 。 $\htmlData{tutor-start=0,tutor-end=1}{C} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{-}\frac{\htmlData{tutor-start=12,tutor-end=13}{a}}{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{,} \frac{\sqrt{\htmlData{tutor-start=31,tutor-end=32}{3}}}{\htmlData{tutor-start=35,tutor-end=36}{6}}\htmlData{tutor-start=37,tutor-end=38}{a}\htmlData{tutor-start=38,tutor-end=39}{,} \htmlData{tutor-start=40,tutor-end=41}{0}\htmlData{tutor-start=41,tutor-end=42}{)}$ 。

$\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{B} \htmlData{tutor-start=10,tutor-end=11}{+} \vec{\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{A}_{\htmlData{tutor-start=21,tutor-end=22}{1}}} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{(}\frac{\htmlData{tutor-start=34,tutor-end=35}{a}}{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{,} \frac{\sqrt{\htmlData{tutor-start=53,tutor-end=54}{3}}}{\htmlData{tutor-start=57,tutor-end=58}{6}}\htmlData{tutor-start=59,tutor-end=60}{a}\htmlData{tutor-start=60,tutor-end=61}{,} \htmlData{tutor-start=62,tutor-end=63}{0}\htmlData{tutor-start=63,tutor-end=64}{)} \htmlData{tutor-start=65,tutor-end=66}{+} \htmlData{tutor-start=67,tutor-end=68}{(}\htmlData{tutor-start=68,tutor-end=69}{0}\htmlData{tutor-start=69,tutor-end=70}{,} \frac{\sqrt{\htmlData{tutor-start=83,tutor-end=84}{3}}}{\htmlData{tutor-start=87,tutor-end=88}{3}}\htmlData{tutor-start=89,tutor-end=90}{a}\htmlData{tutor-start=90,tutor-end=91}{,} \frac{\sqrt{\htmlData{tutor-start=104,tutor-end=105}{6}}}{\htmlData{tutor-start=108,tutor-end=109}{3}}\htmlData{tutor-start=110,tutor-end=111}{a}\htmlData{tutor-start=111,tutor-end=112}{)}$ $\htmlData{tutor-start=0,tutor-end=1}{B}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\frac{\htmlData{tutor-start=15,tutor-end=16}{a}}{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{,} \frac{\sqrt{\htmlData{tutor-start=34,tutor-end=35}{3}}}{\htmlData{tutor-start=38,tutor-end=39}{6}}\htmlData{tutor-start=40,tutor-end=41}{a} \htmlData{tutor-start=42,tutor-end=43}{+} \frac{\htmlData{tutor-start=50,tutor-end=51}{2}\sqrt{\htmlData{tutor-start=57,tutor-end=58}{3}}}{\htmlData{tutor-start=61,tutor-end=62}{6}}\htmlData{tutor-start=63,tutor-end=64}{a}\htmlData{tutor-start=64,tutor-end=65}{,} \frac{\sqrt{\htmlData{tutor-start=78,tutor-end=79}{6}}}{\htmlData{tutor-start=82,tutor-end=83}{3}}\htmlData{tutor-start=84,tutor-end=85}{a}\htmlData{tutor-start=85,tutor-end=86}{)} \htmlData{tutor-start=87,tutor-end=88}{=} \htmlData{tutor-start=89,tutor-end=90}{(}\frac{\htmlData{tutor-start=96,tutor-end=97}{a}}{\htmlData{tutor-start=99,tutor-end=100}{2}}\htmlData{tutor-start=101,tutor-end=102}{,} \frac{\sqrt{\htmlData{tutor-start=115,tutor-end=116}{3}}}{\htmlData{tutor-start=119,tutor-end=120}{2}}\htmlData{tutor-start=121,tutor-end=122}{a}\htmlData{tutor-start=122,tutor-end=123}{,} \frac{\sqrt{\htmlData{tutor-start=136,tutor-end=137}{6}}}{\htmlData{tutor-start=140,tutor-end=141}{3}}\htmlData{tutor-start=142,tutor-end=143}{a}\htmlData{tutor-start=143,tutor-end=144}{)}$ 。

求 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}_{\htmlData{tutor-start=4,tutor-end=5}{1}}$ 与底面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ ( $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}$ ) 的夹角 $\htmlData{tutor-start=0,tutor-end=6}{\theta}$ 。向量 $\vec{\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{B}_{\htmlData{tutor-start=9,tutor-end=10}{1}}} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{B}_{\htmlData{tutor-start=18,tutor-end=19}{1}} \htmlData{tutor-start=21,tutor-end=22}{-} \htmlData{tutor-start=23,tutor-end=24}{A} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{(}\frac{\htmlData{tutor-start=34,tutor-end=35}{a}}{\htmlData{tutor-start=37,tutor-end=38}{2}} \htmlData{tutor-start=40,tutor-end=41}{-} \htmlData{tutor-start=42,tutor-end=43}{0}\htmlData{tutor-start=43,tutor-end=44}{,} \frac{\sqrt{\htmlData{tutor-start=57,tutor-end=58}{3}}}{\htmlData{tutor-start=61,tutor-end=62}{2}}\htmlData{tutor-start=63,tutor-end=64}{a} \htmlData{tutor-start=65,tutor-end=66}{-} \htmlData{tutor-start=67,tutor-end=68}{(}\htmlData{tutor-start=68,tutor-end=69}{-}\frac{\sqrt{\htmlData{tutor-start=81,tutor-end=82}{3}}}{\htmlData{tutor-start=85,tutor-end=86}{3}}\htmlData{tutor-start=87,tutor-end=88}{a}\htmlData{tutor-start=88,tutor-end=89}{)}\htmlData{tutor-start=89,tutor-end=90}{,} \frac{\sqrt{\htmlData{tutor-start=103,tutor-end=104}{6}}}{\htmlData{tutor-start=107,tutor-end=108}{3}}\htmlData{tutor-start=109,tutor-end=110}{a} \htmlData{tutor-start=111,tutor-end=112}{-} \htmlData{tutor-start=113,tutor-end=114}{0}\htmlData{tutor-start=114,tutor-end=115}{)}$ $\htmlData{tutor-start=0,tutor-end=1}{y}$ 分量： $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{a} \htmlData{tutor-start=20,tutor-end=21}{+} \frac{\sqrt{\htmlData{tutor-start=34,tutor-end=35}{3}}}{\htmlData{tutor-start=38,tutor-end=39}{3}}\htmlData{tutor-start=40,tutor-end=41}{a} \htmlData{tutor-start=42,tutor-end=43}{=} \frac{\htmlData{tutor-start=50,tutor-end=51}{3}\sqrt{\htmlData{tutor-start=57,tutor-end=58}{3}}\htmlData{tutor-start=59,tutor-end=60}{+}\htmlData{tutor-start=60,tutor-end=61}{2}\sqrt{\htmlData{tutor-start=67,tutor-end=68}{3}}}{\htmlData{tutor-start=71,tutor-end=72}{6}}\htmlData{tutor-start=73,tutor-end=74}{a} \htmlData{tutor-start=75,tutor-end=76}{=} \frac{\htmlData{tutor-start=83,tutor-end=84}{5}\sqrt{\htmlData{tutor-start=90,tutor-end=91}{3}}}{\htmlData{tutor-start=94,tutor-end=95}{6}}\htmlData{tutor-start=96,tutor-end=97}{a}$ 。 $\vec{\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{B}_{\htmlData{tutor-start=9,tutor-end=10}{1}}} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{(}\frac{\htmlData{tutor-start=22,tutor-end=23}{a}}{\htmlData{tutor-start=25,tutor-end=26}{2}}\htmlData{tutor-start=27,tutor-end=28}{,} \frac{\htmlData{tutor-start=35,tutor-end=36}{5}\sqrt{\htmlData{tutor-start=42,tutor-end=43}{3}}}{\htmlData{tutor-start=46,tutor-end=47}{6}}\htmlData{tutor-start=48,tutor-end=49}{a}\htmlData{tutor-start=49,tutor-end=50}{,} \frac{\sqrt{\htmlData{tutor-start=63,tutor-end=64}{6}}}{\htmlData{tutor-start=67,tutor-end=68}{3}}\htmlData{tutor-start=69,tutor-end=70}{a}\htmlData{tutor-start=70,tutor-end=71}{)}$ 。

线面角的正弦值等于向量与法向量夹角余弦的绝对值，或者直接用 $\htmlData{tutor-start=0,tutor-end=1}{z}$ 分量除以模长。底面法向量 $\vec{\htmlData{tutor-start=5,tutor-end=6}{n}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}$ 。 $\sin \htmlData{tutor-start=5,tutor-end=12}{\theta }\htmlData{tutor-start=12,tutor-end=13}{=} \frac{\htmlData{tutor-start=20,tutor-end=21}{|}\vec{\htmlData{tutor-start=26,tutor-end=27}{A}\htmlData{tutor-start=27,tutor-end=28}{B}_{\htmlData{tutor-start=30,tutor-end=31}{1}}} \htmlData{tutor-start=34,tutor-end=40}{\cdot }\vec{\htmlData{tutor-start=45,tutor-end=46}{n}}\htmlData{tutor-start=47,tutor-end=48}{|}}{\htmlData{tutor-start=50,tutor-end=51}{|}\vec{\htmlData{tutor-start=56,tutor-end=57}{A}\htmlData{tutor-start=57,tutor-end=58}{B}_{\htmlData{tutor-start=60,tutor-end=61}{1}}}\htmlData{tutor-start=63,tutor-end=64}{|} \htmlData{tutor-start=65,tutor-end=66}{|}\vec{\htmlData{tutor-start=71,tutor-end=72}{n}}\htmlData{tutor-start=73,tutor-end=74}{|}} \htmlData{tutor-start=76,tutor-end=77}{=} \frac{\htmlData{tutor-start=84,tutor-end=85}{z}_{\htmlData{tutor-start=87,tutor-end=88}{A}\htmlData{tutor-start=88,tutor-end=89}{B}_{\htmlData{tutor-start=91,tutor-end=92}{1}}}}{\htmlData{tutor-start=96,tutor-end=97}{|}\vec{\htmlData{tutor-start=102,tutor-end=103}{A}\htmlData{tutor-start=103,tutor-end=104}{B}_{\htmlData{tutor-start=106,tutor-end=107}{1}}}\htmlData{tutor-start=109,tutor-end=110}{|}}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{z}_{\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{B}_{\htmlData{tutor-start=7,tutor-end=8}{1}}} \htmlData{tutor-start=11,tutor-end=12}{=} \frac{\sqrt{\htmlData{tutor-start=25,tutor-end=26}{6}}}{\htmlData{tutor-start=29,tutor-end=30}{3}}\htmlData{tutor-start=31,tutor-end=32}{a}$ 。计算 $\htmlData{tutor-start=0,tutor-end=1}{|}\vec{\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{B}_{\htmlData{tutor-start=10,tutor-end=11}{1}}}\htmlData{tutor-start=13,tutor-end=14}{|}^{\htmlData{tutor-start=16,tutor-end=17}{2}}$ : $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{a}^{\htmlData{tutor-start=17,tutor-end=18}{2}}}{\htmlData{tutor-start=21,tutor-end=22}{4}}$ $\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{5} \htmlData{tutor-start=17,tutor-end=23}{\cdot }\htmlData{tutor-start=23,tutor-end=24}{3}}{\htmlData{tutor-start=26,tutor-end=27}{3}\htmlData{tutor-start=27,tutor-end=28}{6}} \htmlData{tutor-start=30,tutor-end=31}{a}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \frac{\htmlData{tutor-start=44,tutor-end=45}{7}\htmlData{tutor-start=45,tutor-end=46}{5}}{\htmlData{tutor-start=48,tutor-end=49}{3}\htmlData{tutor-start=49,tutor-end=50}{6}} \htmlData{tutor-start=52,tutor-end=53}{a}^{\htmlData{tutor-start=55,tutor-end=56}{2}} \htmlData{tutor-start=58,tutor-end=59}{=} \frac{\htmlData{tutor-start=66,tutor-end=67}{2}\htmlData{tutor-start=67,tutor-end=68}{5}}{\htmlData{tutor-start=70,tutor-end=71}{1}\htmlData{tutor-start=71,tutor-end=72}{2}} \htmlData{tutor-start=74,tutor-end=75}{a}^{\htmlData{tutor-start=77,tutor-end=78}{2}}$ $\htmlData{tutor-start=0,tutor-end=1}{z}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{6}}{\htmlData{tutor-start=17,tutor-end=18}{9}} \htmlData{tutor-start=20,tutor-end=21}{a}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{=} \frac{\htmlData{tutor-start=34,tutor-end=35}{2}}{\htmlData{tutor-start=37,tutor-end=38}{3}} \htmlData{tutor-start=40,tutor-end=41}{a}^{\htmlData{tutor-start=43,tutor-end=44}{2}} \htmlData{tutor-start=46,tutor-end=47}{=} \frac{\htmlData{tutor-start=54,tutor-end=55}{8}}{\htmlData{tutor-start=57,tutor-end=58}{1}\htmlData{tutor-start=58,tutor-end=59}{2}} \htmlData{tutor-start=61,tutor-end=62}{a}^{\htmlData{tutor-start=64,tutor-end=65}{2}}$ 通分分母为 12: $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{3}}{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{a}^{\htmlData{tutor-start=24,tutor-end=25}{2}}$ $\htmlData{tutor-start=0,tutor-end=1}{|}\vec{\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{B}_{\htmlData{tutor-start=10,tutor-end=11}{1}}}\htmlData{tutor-start=13,tutor-end=14}{|}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{(}\frac{\htmlData{tutor-start=28,tutor-end=29}{3}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{2}} \htmlData{tutor-start=35,tutor-end=36}{+} \frac{\htmlData{tutor-start=43,tutor-end=44}{2}\htmlData{tutor-start=44,tutor-end=45}{5}}{\htmlData{tutor-start=47,tutor-end=48}{1}\htmlData{tutor-start=48,tutor-end=49}{2}} \htmlData{tutor-start=51,tutor-end=52}{+} \frac{\htmlData{tutor-start=59,tutor-end=60}{8}}{\htmlData{tutor-start=62,tutor-end=63}{1}\htmlData{tutor-start=63,tutor-end=64}{2}}\htmlData{tutor-start=65,tutor-end=66}{)} \htmlData{tutor-start=67,tutor-end=68}{a}^{\htmlData{tutor-start=70,tutor-end=71}{2}} \htmlData{tutor-start=73,tutor-end=74}{=} \frac{\htmlData{tutor-start=81,tutor-end=82}{3}\htmlData{tutor-start=82,tutor-end=83}{6}}{\htmlData{tutor-start=85,tutor-end=86}{1}\htmlData{tutor-start=86,tutor-end=87}{2}} \htmlData{tutor-start=89,tutor-end=90}{a}^{\htmlData{tutor-start=92,tutor-end=93}{2}} \htmlData{tutor-start=95,tutor-end=96}{=} \htmlData{tutor-start=97,tutor-end=98}{3}\htmlData{tutor-start=98,tutor-end=99}{a}^{\htmlData{tutor-start=101,tutor-end=102}{2}}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{|}\vec{\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{B}_{\htmlData{tutor-start=10,tutor-end=11}{1}}}\htmlData{tutor-start=13,tutor-end=14}{|} \htmlData{tutor-start=15,tutor-end=16}{=} \sqrt{\htmlData{tutor-start=23,tutor-end=24}{3}}\htmlData{tutor-start=25,tutor-end=26}{a}$ 。

$\sin \htmlData{tutor-start=5,tutor-end=12}{\theta }\htmlData{tutor-start=12,tutor-end=13}{=} \frac{\frac{\sqrt{\htmlData{tutor-start=32,tutor-end=33}{6}}}{\htmlData{tutor-start=36,tutor-end=37}{3}}\htmlData{tutor-start=38,tutor-end=39}{a}}{\sqrt{\htmlData{tutor-start=47,tutor-end=48}{3}}\htmlData{tutor-start=49,tutor-end=50}{a}} \htmlData{tutor-start=52,tutor-end=53}{=} \frac{\sqrt{\htmlData{tutor-start=66,tutor-end=67}{6}}}{\htmlData{tutor-start=70,tutor-end=71}{3}\sqrt{\htmlData{tutor-start=77,tutor-end=78}{3}}} \htmlData{tutor-start=81,tutor-end=82}{=} \frac{\sqrt{\htmlData{tutor-start=95,tutor-end=96}{2}}}{\htmlData{tutor-start=99,tutor-end=100}{3}}$ 。

对比选项，答案为 (B)。

\sin \\\htmlData{tutor-start=7,tutor-end=8}{t}\htmlData{tutor-start=8,tutor-end=9}{h}\htmlData{tutor-start=9,tutor-end=10}{e}\htmlData{tutor-start=10,tutor-end=11}{t}\htmlData{tutor-start=11,tutor-end=12}{a} \htmlData{tutor-start=13,tutor-end=14}{=} \frac{\sqrt{\htmlData{tutor-start=27,tutor-end=28}{2}}}{\htmlData{tutor-start=31,tutor-end=32}{3}}

一、选择题 · 排列组合：环形染色

如图，一环形花坛分成 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 四块，现有 4 种不同的花供选种，要求在每块里种 1 种花，且相邻的 2 块种不同的花，则不同的种法总数为 ( ) (A) 96 (B) 84 (C) 60 (D) 48

答案：(B)

题目标签：环形花坛染色问题

解题过程

计算不同的种法总数

利用分步计数原理和分类讨论解决环形相邻不同色问题

(1)

分析相邻关系与分类策略

环形区域 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{D}$ 依次相邻，且 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 相邻。共有 4 种花。需保证相邻不同色。

详细展开：我们可以按顺序 $\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=6}{\to }\htmlData{tutor-start=6,tutor-end=7}{B} \htmlData{tutor-start=8,tutor-end=12}{\to }\htmlData{tutor-start=12,tutor-end=13}{C} \htmlData{tutor-start=14,tutor-end=18}{\to }\htmlData{tutor-start=18,tutor-end=19}{D}$ 进行分步染色，但由于 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 的颜色受 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的共同影响（ $\htmlData{tutor-start=0,tutor-end=1}{D} \neq \htmlData{tutor-start=7,tutor-end=8}{C}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{D} \neq \htmlData{tutor-start=7,tutor-end=8}{A}$ ），而 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 的颜色关系不确定（可能相同，可能不同），因此需要根据 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 是否同色进行分类讨论。

设 4 种花为 1, 2, 3, 4。

第一步：染 $\htmlData{tutor-start=0,tutor-end=1}{A}$ ，有 4 种选法。第二步：染 $\htmlData{tutor-start=0,tutor-end=1}{B}$ ，因 $\htmlData{tutor-start=0,tutor-end=1}{B} \neq \htmlData{tutor-start=7,tutor-end=8}{A}$ ，有 3 种选法。

接下来讨论 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{D}$ ：分类依据： $\htmlData{tutor-start=0,tutor-end=1}{A}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 同色或 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 不同色。

\htmlData{tutor-start=0,tutor-end=1}{N} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{N}\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}{C}\htmlData{tutor-start=9,tutor-end=10}{)} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{N}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{A} \neq \htmlData{tutor-start=22,tutor-end=23}{C}\htmlData{tutor-start=23,tutor-end=24}{)}

(2)

执行分类计算

分别计算两类情况的种数并求和。

详细展开：

**情况 1： $\htmlData{tutor-start=0,tutor-end=1}{A}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 同色** 1. 染 $\htmlData{tutor-start=0,tutor-end=1}{A}$ ：4 种。 2. 染 $\htmlData{tutor-start=0,tutor-end=1}{B}$ ：3 种（ $\htmlData{tutor-start=0,tutor-end=1}{B} \neq \htmlData{tutor-start=7,tutor-end=8}{A}$ ）。 3. 染 $\htmlData{tutor-start=0,tutor-end=1}{C}$ ：1 种（必须与 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 同色）。注意：此时需检查 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 是否与 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 冲突。因为 $\htmlData{tutor-start=0,tutor-end=1}{A} \neq \htmlData{tutor-start=7,tutor-end=8}{B}$ ，且 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{A}$ ，所以 $\htmlData{tutor-start=0,tutor-end=1}{C} \neq \htmlData{tutor-start=7,tutor-end=8}{B}$ 自动满足。 4. 染 $\htmlData{tutor-start=0,tutor-end=1}{D}$ ： $\htmlData{tutor-start=0,tutor-end=1}{D}$ 需满足 $\htmlData{tutor-start=0,tutor-end=1}{D} \neq \htmlData{tutor-start=7,tutor-end=8}{C}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{D} \neq \htmlData{tutor-start=7,tutor-end=8}{A}$ 。因为 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{A}$ ，所以只需 $\htmlData{tutor-start=0,tutor-end=1}{D} \neq \htmlData{tutor-start=7,tutor-end=8}{A}$ 。 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 有 3 种选法（4种花减去 A 的那一种）。

此情况种数： $\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{3} \htmlData{tutor-start=11,tutor-end=18}{\times }\htmlData{tutor-start=18,tutor-end=19}{1} \htmlData{tutor-start=20,tutor-end=27}{\times }\htmlData{tutor-start=27,tutor-end=28}{3} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{3}\htmlData{tutor-start=32,tutor-end=33}{6}$ 种。

**情况 2： $\htmlData{tutor-start=0,tutor-end=1}{A}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 不同色** 1. 染 $\htmlData{tutor-start=0,tutor-end=1}{A}$ ：4 种。 2. 染 $\htmlData{tutor-start=0,tutor-end=1}{B}$ ：3 种（ $\htmlData{tutor-start=0,tutor-end=1}{B} \neq \htmlData{tutor-start=7,tutor-end=8}{A}$ ）。 3. 染 $\htmlData{tutor-start=0,tutor-end=1}{C}$ ： $\htmlData{tutor-start=0,tutor-end=1}{C}$ 需满足 $\htmlData{tutor-start=0,tutor-end=1}{C} \neq \htmlData{tutor-start=7,tutor-end=8}{B}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{C} \neq \htmlData{tutor-start=7,tutor-end=8}{A}$ 。在 4 种花中，排除 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 的颜色和 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的颜色（ $\htmlData{tutor-start=0,tutor-end=1}{A} \neq \htmlData{tutor-start=7,tutor-end=8}{B}$ ，故排除 2 种）。所以 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 有 $\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{2} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}$ 种选法。 4. 染 $\htmlData{tutor-start=0,tutor-end=1}{D}$ ： $\htmlData{tutor-start=0,tutor-end=1}{D}$ 需满足 $\htmlData{tutor-start=0,tutor-end=1}{D} \neq \htmlData{tutor-start=7,tutor-end=8}{C}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{D} \neq \htmlData{tutor-start=7,tutor-end=8}{A}$ 。因为在此情况下 $\htmlData{tutor-start=0,tutor-end=1}{A} \neq \htmlData{tutor-start=7,tutor-end=8}{C}$ ，所以 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 是两种不同的颜色。 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 不能取这两种，故 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 有 $\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{2} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}$ 种选法。

此情况种数： $\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{3} \htmlData{tutor-start=11,tutor-end=18}{\times }\htmlData{tutor-start=18,tutor-end=19}{2} \htmlData{tutor-start=20,tutor-end=27}{\times }\htmlData{tutor-start=27,tutor-end=28}{2} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{4}\htmlData{tutor-start=32,tutor-end=33}{8}$ 种。

**总数** $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{6} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{8} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{8}\htmlData{tutor-start=11,tutor-end=12}{4}$

对比选项，答案为 (B)。

\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{6} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{8} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{8}\htmlData{tutor-start=11,tutor-end=12}{4}

二、填空题 · 不等式

若 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y}$ 满足约束条件 $\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{y} \htmlData{tutor-start=18,tutor-end=28}{\geqslant }\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{,} \\ \htmlData{tutor-start=34,tutor-end=35}{x}\htmlData{tutor-start=35,tutor-end=36}{-}\htmlData{tutor-start=36,tutor-end=37}{y}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{3} \htmlData{tutor-start=40,tutor-end=50}{\geqslant }\htmlData{tutor-start=50,tutor-end=51}{0}\htmlData{tutor-start=51,tutor-end=52}{,} \\ \htmlData{tutor-start=56,tutor-end=57}{0} \htmlData{tutor-start=58,tutor-end=68}{\leqslant }\htmlData{tutor-start=68,tutor-end=69}{x} \htmlData{tutor-start=70,tutor-end=80}{\leqslant }\htmlData{tutor-start=80,tutor-end=81}{3}\htmlData{tutor-start=81,tutor-end=82}{,} \end{cases}$ 则 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{y}$ 的最大值为 \_\_\_\_\_\_。

答案： $\htmlData{tutor-start=0,tutor-end=1}{9}$

题目标签：线性规划最值

解题过程

求线性目标最大值

在可行域顶点或边界上比较

(1)

确定最优边界

目标 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{y}$ 随 y 减小而增大。

详细展开：约束给出 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{3}$ 以及 $\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{3}$ ，所以固定 x 时应取最低的 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}$ 。

\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{(}\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}{3}\htmlData{tutor-start=13,tutor-end=14}{x}

(2)

取 x 的端点

在 $\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{3}$ 上， $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{x}$ 单调增加。

详细展开：取 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{3}$ 时所有约束均成立，且 $\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{9}$ ，因此上界可以取得。

\boxed{\htmlData{tutor-start=7,tutor-end=8}{z}_{\max}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{9}}

二、填空题 · 解析几何

已知抛物线 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{x}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}$ 的焦点是坐标原点，则以抛物线与两坐标轴的三个交点为顶点的三角形面积为 \_\_\_\_\_\_。

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

题目标签：抛物线焦点与截距三角形

解题过程

由焦点确定抛物线

把方程改写成标准形式

(1)

比较顶点与焦点

将 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{x}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}$ 写成 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{a}}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{y}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{)}$ 。

详细展开：与 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{y}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}$ 比较得 $\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{)}$ ，顶点为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}$ ，焦点为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{p}\htmlData{tutor-start=7,tutor-end=8}{)}$ 。题设焦点为原点，所以 $\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{4}$ 。

\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{p}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=22}{\Longrightarrow }\htmlData{tutor-start=22,tutor-end=23}{p}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{,}\quad \htmlData{tutor-start=32,tutor-end=33}{a}\htmlData{tutor-start=33,tutor-end=34}{=}\frac{\htmlData{tutor-start=40,tutor-end=41}{1}}{\htmlData{tutor-start=43,tutor-end=44}{4}}

(2)

计算截距三角形

抛物线为 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{4}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}$ 。

详细展开：它与 x 轴交于 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)}$ ，与 y 轴交于 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}$ 。三角形底长为 4、高为 1。

S=\frac{1}{2}\cdot4\cdot1=\boxed2

二、填空题 · 解析几何

在 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 中， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}$ ， $\cos \htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{-}\frac{\htmlData{tutor-start=14,tutor-end=15}{7}}{\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{8}}$ 。若以 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 为焦点的椭圆经过点 $\htmlData{tutor-start=0,tutor-end=1}{C}$ ，则该椭圆的离心率 $\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}$ \_\_\_\_\_\_。

答案： $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{8}$

题目标签：等腰三角形中的椭圆离心率

解题过程

求椭圆离心率

用等腰三角形和椭圆定义求 c/a

(1)

求第三边

设 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{s}$ ，已知夹角 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 的余弦。

详细展开：由余弦定理， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{s}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{s}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{s}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{7}\htmlData{tutor-start=28,tutor-end=29}{/}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{8}\htmlData{tutor-start=31,tutor-end=32}{)}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=35}{5}\htmlData{tutor-start=35,tutor-end=36}{s}^{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{/}\htmlData{tutor-start=41,tutor-end=42}{9}$ ，所以 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{s}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{3}$ 。

\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\frac{\htmlData{tutor-start=9,tutor-end=10}{5}\htmlData{tutor-start=10,tutor-end=11}{s}}{\htmlData{tutor-start=13,tutor-end=14}{3}}

(2)

比较半焦距与半长轴

椭圆焦点为 A、B，且经过 C。

详细展开：焦距 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{s}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{c}_{\htmlData{tutor-start=9,tutor-end=10}{0}}$ ，故 $\htmlData{tutor-start=0,tutor-end=1}{c}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{s}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}$ ；又 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{B}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{s}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{s}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{8}\htmlData{tutor-start=14,tutor-end=15}{s}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{0}}$ ，故 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{s}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{3}$ 。

\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{c}_{\htmlData{tutor-start=11,tutor-end=12}{0}}}{\htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{0}}}\htmlData{tutor-start=21,tutor-end=22}{=}\frac{\htmlData{tutor-start=28,tutor-end=29}{s}\htmlData{tutor-start=29,tutor-end=30}{/}\htmlData{tutor-start=30,tutor-end=31}{2}}{\htmlData{tutor-start=33,tutor-end=34}{4}\htmlData{tutor-start=34,tutor-end=35}{s}\htmlData{tutor-start=35,tutor-end=36}{/}\htmlData{tutor-start=36,tutor-end=37}{3}}\htmlData{tutor-start=38,tutor-end=39}{=}\boxed{\frac{\htmlData{tutor-start=52,tutor-end=53}{3}}{\htmlData{tutor-start=55,tutor-end=56}{8}}}

二、填空题 · 立体几何

等边三角形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 与正方形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{E}$ 有一公共边 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ ，二面角 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{D}$ 的余弦值为 $\frac{\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}}{\htmlData{tutor-start=16,tutor-end=17}{3}}$ ， $\htmlData{tutor-start=0,tutor-end=1}{M}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{N}$ 分别是 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 的中点，则 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{M}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{N}$ 所成角的余弦值等于 \_\_\_\_\_\_。

答案： $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{6}$

题目标签：共棱二面角与异面直线角

解题过程

求异面直线夹角

绕公共边建立坐标并用二面角确定正方形方向

(1)

建立单位边坐标

整体缩放不影响夹角，可设公共边长为 1。

详细展开：取 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{C}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{,}\sqrt{\htmlData{tutor-start=33,tutor-end=34}{3}}\htmlData{tutor-start=35,tutor-end=36}{/}\htmlData{tutor-start=36,tutor-end=37}{2}\htmlData{tutor-start=37,tutor-end=38}{,}\htmlData{tutor-start=38,tutor-end=39}{0}\htmlData{tutor-start=39,tutor-end=40}{)}$ 。设正方形另一边的单位方向为 $\boldsymbol v=(0,u,w)$ ；二面角余弦给出 $\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{3}$ ，且 $\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{w}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}$ 。于是 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{u}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{w}\htmlData{tutor-start=8,tutor-end=9}{)}$ 。

\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\sqrt{\htmlData{tutor-start=14,tutor-end=15}{3}}}{\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{,}\quad \htmlData{tutor-start=27,tutor-end=28}{w}^{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{=}\frac{\htmlData{tutor-start=39,tutor-end=40}{2}}{\htmlData{tutor-start=42,tutor-end=43}{3}}

(2)

计算两方向向量

$\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{,}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{)}$ ， $\htmlData{tutor-start=0,tutor-end=1}{N}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{,}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{)}$ 。

详细展开： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{E}\htmlData{tutor-start=17,tutor-end=18}{M}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{/}\htmlData{tutor-start=23,tutor-end=24}{4}\htmlData{tutor-start=24,tutor-end=25}{,}\sqrt{\htmlData{tutor-start=31,tutor-end=32}{3}}\htmlData{tutor-start=33,tutor-end=34}{/}\htmlData{tutor-start=34,tutor-end=35}{4}\htmlData{tutor-start=35,tutor-end=36}{-}\htmlData{tutor-start=36,tutor-end=37}{u}\htmlData{tutor-start=37,tutor-end=38}{,}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{w}\htmlData{tutor-start=40,tutor-end=41}{)}$ ， $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{N}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{3}\htmlData{tutor-start=22,tutor-end=23}{/}\htmlData{tutor-start=23,tutor-end=24}{4}\htmlData{tutor-start=24,tutor-end=25}{,}\sqrt{\htmlData{tutor-start=31,tutor-end=32}{3}}\htmlData{tutor-start=33,tutor-end=34}{/}\htmlData{tutor-start=34,tutor-end=35}{4}\htmlData{tutor-start=35,tutor-end=36}{,}\htmlData{tutor-start=36,tutor-end=37}{0}\htmlData{tutor-start=37,tutor-end=38}{)}$ 。点积为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{8}$ ，两向量长度都为 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}$ ，故锐角余弦为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{6}$ 。

\boxed{\cos\htmlData{tutor-start=11,tutor-end=17}{\angle}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{E}\htmlData{tutor-start=19,tutor-end=20}{M}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=22}{A}\htmlData{tutor-start=22,tutor-end=23}{N}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{=}\frac{\htmlData{tutor-start=31,tutor-end=32}{1}}{\htmlData{tutor-start=34,tutor-end=35}{6}}}

三、解答题 · 解三角形

设 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 的内角 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{C}$ 所对的边长分别为 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{b}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{c}$ ，且 $\htmlData{tutor-start=0,tutor-end=1}{a}\cos \htmlData{tutor-start=6,tutor-end=7}{B}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{b}\cos \htmlData{tutor-start=14,tutor-end=15}{A}\htmlData{tutor-start=15,tutor-end=16}{=}\frac{\htmlData{tutor-start=22,tutor-end=23}{3}}{\htmlData{tutor-start=25,tutor-end=26}{5}}\htmlData{tutor-start=27,tutor-end=28}{c}$ 。 (1) 求 $\tan \htmlData{tutor-start=5,tutor-end=6}{A}\cot \htmlData{tutor-start=11,tutor-end=12}{B}$ 的值； (2) 求 $\tan\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{)}$ 的最大值。

答案：4；3/4

题目标签：三角恒等变换与角差最值

解题过程

（1）求正切乘积

用正弦定理把边换成角

(1)

化简已知等式

由正弦定理， $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{:}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{:}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{=}\sin \htmlData{tutor-start=11,tutor-end=12}{A}\htmlData{tutor-start=12,tutor-end=13}{:}\sin \htmlData{tutor-start=18,tutor-end=19}{B}\htmlData{tutor-start=19,tutor-end=20}{:}\sin \htmlData{tutor-start=25,tutor-end=26}{C}$ 。

详细展开：原式除以 c 后化为 $\htmlData{tutor-start=0,tutor-end=1}{[}\sin \htmlData{tutor-start=6,tutor-end=7}{A}\cos \htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{-}\sin \htmlData{tutor-start=19,tutor-end=20}{B}\cos \htmlData{tutor-start=25,tutor-end=26}{A}\htmlData{tutor-start=26,tutor-end=27}{]}\htmlData{tutor-start=27,tutor-end=28}{/}\sin \htmlData{tutor-start=33,tutor-end=34}{C}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{3}\htmlData{tutor-start=36,tutor-end=37}{/}\htmlData{tutor-start=37,tutor-end=38}{5}$ ，即 $\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{/}\sin\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{A}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{B}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{3}\htmlData{tutor-start=21,tutor-end=22}{/}\htmlData{tutor-start=22,tutor-end=23}{5}$ 。

\frac{\sin\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{A}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{)}}{\sin\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{A}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{B}\htmlData{tutor-start=25,tutor-end=26}{)}}\htmlData{tutor-start=27,tutor-end=28}{=}\frac{\htmlData{tutor-start=34,tutor-end=35}{3}}{\htmlData{tutor-start=37,tutor-end=38}{5}}

(2)

改写成正切比

展开分子、分母并约去 $\cos \htmlData{tutor-start=5,tutor-end=6}{A}\cos \htmlData{tutor-start=11,tutor-end=12}{B}$ 。

详细展开：得到 $\htmlData{tutor-start=0,tutor-end=1}{(}\tan \htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{-}\tan \htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{(}\tan \htmlData{tutor-start=22,tutor-end=23}{A}\htmlData{tutor-start=23,tutor-end=24}{+}\tan \htmlData{tutor-start=29,tutor-end=30}{B}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{3}\htmlData{tutor-start=33,tutor-end=34}{/}\htmlData{tutor-start=34,tutor-end=35}{5}$ ，解得 $\tan \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{4}\tan \htmlData{tutor-start=13,tutor-end=14}{B}$ 。

\boxed{\tan \htmlData{tutor-start=12,tutor-end=13}{A}\cot \htmlData{tutor-start=18,tutor-end=19}{B}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{4}}

（2）求 tan(A-B) 最大值

把第一问关系参数化

(1)

设单一正参数

由 $\tan \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{4}\tan \htmlData{tutor-start=13,tutor-end=14}{B}$ 且 A>B，可知 A、B 均为锐角。

详细展开：令 $\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\tan \htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{>}\htmlData{tutor-start=9,tutor-end=10}{0}$ ，则 $\tan \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{t}$ ，所以 $\tan\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\frac{\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{t}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{t}}{\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{4}\htmlData{tutor-start=25,tutor-end=26}{t}^{\htmlData{tutor-start=28,tutor-end=29}{2}}}\htmlData{tutor-start=31,tutor-end=32}{=}\frac{\htmlData{tutor-start=38,tutor-end=39}{3}\htmlData{tutor-start=39,tutor-end=40}{t}}{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{4}\htmlData{tutor-start=45,tutor-end=46}{t}^{\htmlData{tutor-start=48,tutor-end=49}{2}}}$ 。

\tan\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\frac{\htmlData{tutor-start=16,tutor-end=17}{3}\htmlData{tutor-start=17,tutor-end=18}{t}}{\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{4}\htmlData{tutor-start=23,tutor-end=24}{t}^{\htmlData{tutor-start=26,tutor-end=27}{2}}}\htmlData{tutor-start=29,tutor-end=30}{,}\quad \htmlData{tutor-start=36,tutor-end=37}{t}\htmlData{tutor-start=37,tutor-end=38}{>}\htmlData{tutor-start=38,tutor-end=39}{0}

(2)

使用基本不等式

$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{t}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=11}{\ge}\htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{t}$ ，等号在 $\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$ 时成立。

详细展开：于是 $\tan\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{B}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=12}{\le}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{4}$ 。此时 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\arctan\htmlData{tutor-start=9,tutor-end=10}{2}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\arctan\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)}$ ，两角之和小于 $\htmlData{tutor-start=0,tutor-end=3}{\pi}$ ，确能构成三角形。

\boxed{\max\tan\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{B}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{=}\frac{\htmlData{tutor-start=27,tutor-end=28}{3}}{\htmlData{tutor-start=30,tutor-end=31}{4}}}

三、解答题 · 立体几何

四棱锥 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{D}\htmlData{tutor-start=5,tutor-end=6}{E}$ 中，底面 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{E}$ 为矩形，侧面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C} \htmlData{tutor-start=4,tutor-end=9}{\perp}$ 底面 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{E}$ ， $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}$ ， $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{=}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{2}}$ ， $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{C}$ 。 (1) 证明： $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=9}{\perp }\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{E}$ ； (2) 设 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{E}$ 与平面 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{E}$ 所成的角为 $45^\circ$ ，求二面角 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{E}$ 的大小。

答案：见证明；arccos(-√10/10)

题目标签：四棱锥垂直、线面角与二面角

解题过程

（1）证明 AD 垂直 CE

用坐标把侧面垂直和矩形条件同时编码

(1)

建立空间坐标

令矩形底面为 z=0，BC 沿 x 轴。

详细展开：取 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=22}{D}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{,}\sqrt{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{,}\htmlData{tutor-start=35,tutor-end=36}{0}\htmlData{tutor-start=36,tutor-end=37}{)}\htmlData{tutor-start=37,tutor-end=38}{,}\htmlData{tutor-start=38,tutor-end=39}{E}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{,}\sqrt{\htmlData{tutor-start=50,tutor-end=51}{2}}\htmlData{tutor-start=52,tutor-end=53}{,}\htmlData{tutor-start=53,tutor-end=54}{0}\htmlData{tutor-start=54,tutor-end=55}{)}$ 。因平面 ABC 垂直底面且 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{C}$ ，可设 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{h}\htmlData{tutor-start=8,tutor-end=9}{)}$ 。

\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}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{h}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{,}\quad \htmlData{tutor-start=16,tutor-end=17}{B}\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}{1}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=24}{,}\htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=29}{\ }\htmlData{tutor-start=29,tutor-end=30}{C}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{,}\htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=36}{,}\htmlData{tutor-start=36,tutor-end=37}{0}\htmlData{tutor-start=37,tutor-end=38}{)}\htmlData{tutor-start=38,tutor-end=39}{,}\htmlData{tutor-start=39,tutor-end=41}{\ }\htmlData{tutor-start=41,tutor-end=42}{D}\htmlData{tutor-start=42,tutor-end=43}{=}\htmlData{tutor-start=43,tutor-end=44}{(}\htmlData{tutor-start=44,tutor-end=45}{1}\htmlData{tutor-start=45,tutor-end=46}{,}\sqrt{\htmlData{tutor-start=52,tutor-end=53}{2}}\htmlData{tutor-start=54,tutor-end=55}{,}\htmlData{tutor-start=55,tutor-end=56}{0}\htmlData{tutor-start=56,tutor-end=57}{)}\htmlData{tutor-start=57,tutor-end=58}{,}\htmlData{tutor-start=58,tutor-end=60}{\ }\htmlData{tutor-start=60,tutor-end=61}{E}\htmlData{tutor-start=61,tutor-end=62}{=}\htmlData{tutor-start=62,tutor-end=63}{(}\htmlData{tutor-start=63,tutor-end=64}{-}\htmlData{tutor-start=64,tutor-end=65}{1}\htmlData{tutor-start=65,tutor-end=66}{,}\sqrt{\htmlData{tutor-start=72,tutor-end=73}{2}}\htmlData{tutor-start=74,tutor-end=75}{,}\htmlData{tutor-start=75,tutor-end=76}{0}\htmlData{tutor-start=76,tutor-end=77}{)}

(2)

验证方向点积

分别写出 AD 与 CE 的方向向量。

详细展开： $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{,}\sqrt{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{,}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{h}\htmlData{tutor-start=34,tutor-end=35}{)}$ ， $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{E}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{,}\sqrt{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{,}\htmlData{tutor-start=33,tutor-end=34}{0}\htmlData{tutor-start=34,tutor-end=35}{)}$ ，点积为 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}$ ，所以两直线垂直。

\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}}\htmlData{tutor-start=19,tutor-end=24}{\cdot}\overrightarrow{\htmlData{tutor-start=40,tutor-end=41}{C}\htmlData{tutor-start=41,tutor-end=42}{E}}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{0}\htmlData{tutor-start=45,tutor-end=60}{\Longrightarrow}\boxed{\htmlData{tutor-start=67,tutor-end=68}{A}\htmlData{tutor-start=68,tutor-end=69}{D}\htmlData{tutor-start=69,tutor-end=75}{\perp }\htmlData{tutor-start=75,tutor-end=76}{C}\htmlData{tutor-start=76,tutor-end=77}{E}}

（2）求二面角 C-AD-E

先由线面角反求顶点高度，再复用垂面夹角

(1)

由 45° 求高度

沿用坐标 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{h}\htmlData{tutor-start=8,tutor-end=9}{)}$ 及矩形四点。

详细展开：平面 ABE 的法向量可取 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{h}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}$ ，而 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{C}\htmlData{tutor-start=17,tutor-end=18}{E}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{,}\sqrt{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{,}\htmlData{tutor-start=33,tutor-end=34}{0}\htmlData{tutor-start=34,tutor-end=35}{)}$ 。线面角正弦为 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{h}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{[}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{6}}\sqrt{\htmlData{tutor-start=18,tutor-end=19}{h}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{1}}\htmlData{tutor-start=26,tutor-end=27}{]}$ ，令其等于 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{2}$ ，得到 $\htmlData{tutor-start=0,tutor-end=1}{h}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}$ 。

\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{h}}{\sqrt{\htmlData{tutor-start=16,tutor-end=17}{6}}\sqrt{\htmlData{tutor-start=24,tutor-end=25}{h}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{1}}}\htmlData{tutor-start=33,tutor-end=34}{=}\frac{\sqrt{\htmlData{tutor-start=46,tutor-end=47}{2}}}{\htmlData{tutor-start=50,tutor-end=51}{2}}\htmlData{tutor-start=52,tutor-end=68}{\Longrightarrow }\htmlData{tutor-start=68,tutor-end=69}{h}\htmlData{tutor-start=69,tutor-end=70}{=}\sqrt{\htmlData{tutor-start=76,tutor-end=77}{3}}

(2)

计算二面角

高度 $\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}$ 后，空间构型与等边侧面情形的坐标完全相同。

详细展开：把 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{C}}$ 、 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{E}}$ 投影到 AD 的垂面，所得两个方向的余弦为 $\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{/}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0}}$ ，并按 C、E 半平面取钝角。

\boxed{\\\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{h}\htmlData{tutor-start=11,tutor-end=12}{e}\htmlData{tutor-start=12,tutor-end=13}{t}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{=}\arccos\left(\htmlData{tutor-start=28,tutor-end=29}{-}\frac{\sqrt{\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{0}}}{\htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{0}}\right)}

解答题 · 导数

已知函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{3}} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{x}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{x} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{1}$ , $\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=6}{\in }\mathbf{\htmlData{tutor-start=14,tutor-end=15}{R}}$ . (1) 讨论函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 的单调区间; (2) 设函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 在区间 $\left(\htmlData{tutor-start=6,tutor-end=7}{-}\dfrac{\htmlData{tutor-start=14,tutor-end=15}{2}}{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{-}\dfrac{\htmlData{tutor-start=29,tutor-end=30}{1}}{\htmlData{tutor-start=32,tutor-end=33}{3}}\right)$ 内是减函数, 求 $\htmlData{tutor-start=0,tutor-end=1}{a}$ 的取值范围.

答案：分类见解析；a≥2

题目标签：三次函数含参单调性

解题过程

（1）讨论单调区间

按导数二次式判别式分类

(1)

求导并找临界点

$\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{a}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{1}$ 是开口向上的二次式。

详细展开：其判别式为 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{)}$ 。当 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=6}{\le}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}$ 时导数非负，函数在实数域递增；等号只使导数在一点为 0，不改变严格单调性。

\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=6}{\le}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{:}\quad \htmlData{tutor-start=21,tutor-end=22}{f}\text{ \htmlData{tutor-start=29,tutor-end=30}{在} }\mathbb{\htmlData{tutor-start=40,tutor-end=41}{R}}\text{ \htmlData{tutor-start=49,tutor-end=50}{上}\htmlData{tutor-start=50,tutor-end=51}{递}\htmlData{tutor-start=51,tutor-end=52}{增}}

(2)

写出两根情形

当 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{>}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}$ 时，令 $\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{-}\sqrt{\htmlData{tutor-start=16,tutor-end=17}{a}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{3}}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{/}\htmlData{tutor-start=26,tutor-end=27}{3}\htmlData{tutor-start=27,tutor-end=28}{<}\htmlData{tutor-start=28,tutor-end=29}{x}_{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{-}\htmlData{tutor-start=36,tutor-end=37}{a}\htmlData{tutor-start=37,tutor-end=38}{+}\sqrt{\htmlData{tutor-start=44,tutor-end=45}{a}^{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{3}}\htmlData{tutor-start=52,tutor-end=53}{)}\htmlData{tutor-start=53,tutor-end=54}{/}\htmlData{tutor-start=54,tutor-end=55}{3}$ 。

详细展开：导数在两根外为正、两根之间为负，所以 f 在 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{x}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{)}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=13}{\infty}\htmlData{tutor-start=13,tutor-end=14}{)}$ 上递增，在 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{)}$ 上递减。

\boxed{\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}{>}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{3}}\htmlData{tutor-start=19,tutor-end=20}{:}\htmlData{tutor-start=20,tutor-end=22}{\ }\htmlData{tutor-start=22,tutor-end=30}{\nearrow}\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=38}{\infty}\htmlData{tutor-start=38,tutor-end=39}{,}\htmlData{tutor-start=39,tutor-end=40}{x}_{\htmlData{tutor-start=42,tutor-end=43}{1}}\htmlData{tutor-start=44,tutor-end=45}{)}\htmlData{tutor-start=45,tutor-end=46}{,}\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{x}_{\htmlData{tutor-start=50,tutor-end=51}{2}}\htmlData{tutor-start=52,tutor-end=53}{,}\htmlData{tutor-start=53,tutor-end=59}{\infty}\htmlData{tutor-start=59,tutor-end=60}{)}\htmlData{tutor-start=60,tutor-end=61}{;}\htmlData{tutor-start=61,tutor-end=63}{\ }\htmlData{tutor-start=63,tutor-end=71}{\searrow}\htmlData{tutor-start=71,tutor-end=72}{(}\htmlData{tutor-start=72,tutor-end=73}{x}_{\htmlData{tutor-start=75,tutor-end=76}{1}}\htmlData{tutor-start=77,tutor-end=78}{,}\htmlData{tutor-start=78,tutor-end=79}{x}_{\htmlData{tutor-start=81,tutor-end=82}{2}}\htmlData{tutor-start=83,tutor-end=84}{)}}

（2）由区间单调性求参数

把导数非正条件对参数求交集

(1)

对 a 解不等式

在 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\in}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{)}$ 上递减要求 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=15}{\le}\htmlData{tutor-start=15,tutor-end=16}{0}$ 。

详细展开：因 x 为负，移项除以 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}$ 后方向反转，得到 $a\ge h(x)=-\frac{3x^{2}+1}{2x}=-\frac{3}{2}x-\frac1{2x}$ 。

a\ge h(x)=-\frac{3}{2}x-\frac1{2x}

(2)

求统一下界

需要 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\ge}\sup \htmlData{tutor-start=9,tutor-end=10}{h}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{)}$ 。

详细展开：在该开区间上，h 的最大上确界为右端趋近值 $\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{2}$ 。当 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}$ 时 $\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{<}\htmlData{tutor-start=18,tutor-end=19}{0}$ ，确实在整个开区间严格递减。

\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=11}{\in}\htmlData{tutor-start=11,tutor-end=12}{[}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=20}{\infty}\htmlData{tutor-start=20,tutor-end=21}{)}}

解答题 · 概率

已知 5 只动物中有 1 只患有某种疾病, 需要通过化验血液来确定患病的动物. 血液化验结果呈阳性的即为患病动物, 呈阴性即没患病. 下面是两种化验方案: 方案甲: 逐个化验, 直到能确定患病动物为止. 方案乙: 先任取 3 只, 将它们的血液混在一起化验. 若结果呈阳性则表明患病动物为这 3 只中的 1 只, 然后再逐个化验, 直到能确定患病动物为止; 若结果呈阴性则在另外 2 只中任取 1 只化验. (1) 求依方案甲所需化验次数不少于依方案乙所需化验次数的概率; (2) $\htmlData{tutor-start=0,tutor-end=3}{\xi}$ 表示依方案乙所需化验次数, 求 $\htmlData{tutor-start=0,tutor-end=3}{\xi}$ 的期望.

答案：3/5；12/5

题目标签：分组化验方案的概率与期望

解题过程

（1）比较两种化验次数

固定化验顺序并按患病动物位置枚举

(1)

列出两方案次数

五只动物患病机会相同，可编号后逐一列出。

详细展开：若患病动物依次为第 1 至第 5 只，方案甲次数为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{)}$ 。方案乙先混合前 3 只；阳性后逐个排查，阴性后检查余下 2 只之一，所以次数为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{)}$ 。

\htmlData{tutor-start=0,tutor-end=1}{X}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{3}\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}{4}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{,}\quad \htmlData{tutor-start=20,tutor-end=21}{Y}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{,}\htmlData{tutor-start=25,tutor-end=26}{3}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=28}{3}\htmlData{tutor-start=28,tutor-end=29}{,}\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{,}\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{)}

(2)

计数有利位置

逐项比较 X 与 Y。

详细展开：第 3、4、5 只患病时有 $\htmlData{tutor-start=0,tutor-end=1}{X}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{Y}$ ，共 3 个等可能位置；第 1、2 只不满足。因此概率为 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}$ 。

\boxed{\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{X}\htmlData{tutor-start=10,tutor-end=14}{\ge }\htmlData{tutor-start=14,tutor-end=15}{Y}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{=}\frac{\htmlData{tutor-start=23,tutor-end=24}{3}}{\htmlData{tutor-start=26,tutor-end=27}{5}}}

（2）求方案乙次数期望

由枚举结果写分布列

(1)

归并次数概率

方案乙的五个等可能次数为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{)}$ 。

详细展开：因此 $\htmlData{tutor-start=0,tutor-end=3}{\xi}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{2}$ 对应 3 个患病位置，概率为 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}$ ； $\htmlData{tutor-start=0,tutor-end=3}{\xi}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{3}$ 对应 2 个位置，概率为 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}$ 。

\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=5}{\xi}\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}{=}\frac{\htmlData{tutor-start=15,tutor-end=16}{3}}{\htmlData{tutor-start=18,tutor-end=19}{5}}\htmlData{tutor-start=20,tutor-end=21}{,}\quad \htmlData{tutor-start=27,tutor-end=28}{P}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=32}{\xi}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{3}\htmlData{tutor-start=34,tutor-end=35}{)}\htmlData{tutor-start=35,tutor-end=36}{=}\frac{\htmlData{tutor-start=42,tutor-end=43}{2}}{\htmlData{tutor-start=45,tutor-end=46}{5}}

(2)

计算数学期望

用取值乘对应概率后相加。

详细展开： $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=4}{\xi}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=11}{\cdot}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{5}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{3}\htmlData{tutor-start=16,tutor-end=21}{\cdot}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{/}\htmlData{tutor-start=23,tutor-end=24}{5}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{5}$ 。

\boxed{\htmlData{tutor-start=7,tutor-end=8}{E}\htmlData{tutor-start=8,tutor-end=11}{\xi}\htmlData{tutor-start=11,tutor-end=12}{=}\frac{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{2}}{\htmlData{tutor-start=22,tutor-end=23}{5}}}

解答题 · 解析几何

双曲线的中心为原点 $\htmlData{tutor-start=0,tutor-end=1}{O}$ , 焦点在 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴上, 两条渐近线分别为 $\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ , $\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{2}}$ , 经过右焦点 $\htmlData{tutor-start=0,tutor-end=1}{F}$ 垂直于 $\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ 的直线分别交 $\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{1}}$ , $\htmlData{tutor-start=0,tutor-end=1}{l}_{\htmlData{tutor-start=3,tutor-end=4}{2}}$ 于 $\htmlData{tutor-start=0,tutor-end=1}{A}$ , $\htmlData{tutor-start=0,tutor-end=1}{B}$ 两点. 已知 $\left|\overrightarrow{\htmlData{tutor-start=22,tutor-end=23}{O}\htmlData{tutor-start=23,tutor-end=24}{A}}\right|$ , $\left|\overrightarrow{\htmlData{tutor-start=22,tutor-end=23}{A}\htmlData{tutor-start=23,tutor-end=24}{B}}\right|$ , $\left|\overrightarrow{\htmlData{tutor-start=22,tutor-end=23}{O}\htmlData{tutor-start=23,tutor-end=24}{B}}\right|$ 成等差数列, 且 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{F}}$ 与 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{F}\htmlData{tutor-start=17,tutor-end=18}{A}}$ 同向. (1) 求双曲线的离心率; (2) 设 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 被双曲线所截得的线段的长为 4, 求双曲线的方程.

答案：√5/2；x²/36-y²/9=1

题目标签：双曲线渐近线构型与截弦

解题过程

（1）求双曲线离心率

用渐近线斜率统一表示三个线段长度

(1)

参数化交点

设渐近线为 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=6}{\pm }\htmlData{tutor-start=6,tutor-end=7}{m}\htmlData{tutor-start=7,tutor-end=8}{x}$ ， $\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}$ ，右焦点 $\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}$ 。

详细展开：过 F 且垂直 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{x}$ 的直线为 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{m}$ 。它与两渐近线的交点满足 $\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{/}\sqrt{\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{m}^{\htmlData{tutor-start=16,tutor-end=17}{2}}}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{c}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{m}^{\htmlData{tutor-start=15,tutor-end=16}{2}}}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{m}^{\htmlData{tutor-start=25,tutor-end=26}{2}}\htmlData{tutor-start=27,tutor-end=28}{)}$ 、 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{c}\htmlData{tutor-start=5,tutor-end=6}{m}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{[}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{m}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{)}\sqrt{\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{m}^{\htmlData{tutor-start=28,tutor-end=29}{2}}}\htmlData{tutor-start=31,tutor-end=32}{]}$ 。同向条件给出 $\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}$ 。

OA=\frac c{\sqrt{1+m^{2}}},\quad OB=\frac{c\sqrt{1+m^{2}}}{1-m^{2}},\quad AB=\frac{2cm}{(1-m^{2})\sqrt{1+m^{2}}}

(2)

使用等差中项

长度依次成等差数列，所以 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{O}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{B}$ 。

详细展开：代入并约去公共因子后得到 $\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}$ ，即 $\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}$ 。双曲线离心率 $\htmlData{tutor-start=0,tutor-end=1}{e}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{m}^{\htmlData{tutor-start=13,tutor-end=14}{2}}}$ 。

\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=28}{\Longrightarrow}\boxed{\htmlData{tutor-start=35,tutor-end=36}{e}\htmlData{tutor-start=36,tutor-end=37}{=}\frac{\sqrt{\htmlData{tutor-start=49,tutor-end=50}{5}}}{\htmlData{tutor-start=53,tutor-end=54}{2}}}

（2）由截弦长确定方程

保留已求形状参数，只求尺度 a

(1)

写双曲线和截线

由 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}$ ，双曲线可写为 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{a}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{y}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{a}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{/}\htmlData{tutor-start=25,tutor-end=26}{4}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{1}$ ，且 $\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{2}$ 。

详细展开：垂线 AB 的斜率为 -2，方程为 $\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{c}\htmlData{tutor-start=8,tutor-end=9}{)}$ 。代入 $\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{y}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}$ ，得到关于 x 的二次方程。

\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{5}\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}{3}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{c}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{6}\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}{a}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{0}

(2)

把根差换成弦长

该二次方程两根的横坐标差为 $\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{5}}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{5}$ 。

详细展开：直线斜率绝对值为 2，所以弦长等于横坐标差的 $\sqrt{\htmlData{tutor-start=6,tutor-end=7}{5}}$ 倍，即 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{3}$ 。由弦长 4 得 $\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{6}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{3}$ 。

\frac{\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{a}}{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=29}{\Longrightarrow}\boxed{\frac{\htmlData{tutor-start=42,tutor-end=43}{x}^{\htmlData{tutor-start=45,tutor-end=46}{2}}}{\htmlData{tutor-start=49,tutor-end=50}{3}\htmlData{tutor-start=50,tutor-end=51}{6}}\htmlData{tutor-start=52,tutor-end=53}{-}\frac{\htmlData{tutor-start=59,tutor-end=60}{y}^{\htmlData{tutor-start=62,tutor-end=63}{2}}}{\htmlData{tutor-start=66,tutor-end=67}{9}}\htmlData{tutor-start=68,tutor-end=69}{=}\htmlData{tutor-start=69,tutor-end=70}{1}}

解答题 · 函数与数列

设函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{x} \htmlData{tutor-start=9,tutor-end=10}{-} \htmlData{tutor-start=11,tutor-end=12}{x}\ln \htmlData{tutor-start=16,tutor-end=17}{x}$ . 数列 $\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}}$ 满足 $\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{<} \htmlData{tutor-start=12,tutor-end=13}{1}$ , $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{f}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{n}}\htmlData{tutor-start=17,tutor-end=18}{)}$ . (1) 证明: 函数 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}$ 在区间 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}$ 是增函数; (2) 证明: $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{n}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{1}$ ; (3) 设 $\htmlData{tutor-start=0,tutor-end=1}{b} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{1}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)}$ , 整数 $\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\dfrac{\htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{1}} \htmlData{tutor-start=25,tutor-end=26}{-} \htmlData{tutor-start=27,tutor-end=28}{b}}{\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{1}} \ln \htmlData{tutor-start=40,tutor-end=41}{b}}$ , 证明: $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{>} \htmlData{tutor-start=10,tutor-end=11}{b}$ .

答案：见证明

题目标签：对数递推数列的越界估计

解题过程

（1）证明 f 在 (0,1) 递增

直接判断导数符号

(1)

求导

$\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{x}\ln \htmlData{tutor-start=12,tutor-end=13}{x}$ 的定义域为 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}$ 。

详细展开： $\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{(}\ln \htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{-}\ln \htmlData{tutor-start=23,tutor-end=24}{x}$ 。

\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{-}\ln \htmlData{tutor-start=11,tutor-end=12}{x}

(2)

判断区间符号

当 $\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}$ 时 $\ln \htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{0}$ 。

详细展开：故 $\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{>}\htmlData{tutor-start=6,tutor-end=7}{0}$ ，所以 f 在 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}$ 上严格递增。

\boxed{\htmlData{tutor-start=7,tutor-end=8}{f}\text{ \htmlData{tutor-start=15,tutor-end=16}{在} }\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{)}\text{ \htmlData{tutor-start=30,tutor-end=31}{上}\htmlData{tutor-start=31,tutor-end=32}{递}\htmlData{tutor-start=32,tutor-end=33}{增}}}

（2）证明递推项递增且小于 1

建立区间不变性并归纳

(1)

比较 f(x) 与 x、1

对任意 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\in}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}$ ，有 $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{x}\ln \htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{>}\htmlData{tutor-start=15,tutor-end=16}{0}$ 。

详细展开：又由第一问的严格递增性， $\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{<}\htmlData{tutor-start=5,tutor-end=6}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}$ ，所以 $\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{f}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{<}\htmlData{tutor-start=7,tutor-end=8}{1}$ 。

\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}\htmlData{tutor-start=5,tutor-end=21}{\Longrightarrow }\htmlData{tutor-start=21,tutor-end=22}{x}\htmlData{tutor-start=22,tutor-end=23}{<}\htmlData{tutor-start=23,tutor-end=24}{f}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{<}\htmlData{tutor-start=28,tutor-end=29}{1}

(2)

递推归纳

初始 $\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{<}\htmlData{tutor-start=8,tutor-end=9}{1}$ 。

详细展开：若 $\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=8}{<}\htmlData{tutor-start=8,tutor-end=9}{1}$ ，上一步给出 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{<}\htmlData{tutor-start=14,tutor-end=15}{1}$ ；由归纳法对所有正整数 n 成立。

\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{n}}\htmlData{tutor-start=12,tutor-end=13}{<}\htmlData{tutor-start=13,tutor-end=14}{a}_{\htmlData{tutor-start=16,tutor-end=17}{n}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{<}\htmlData{tutor-start=21,tutor-end=22}{1}}

（3）证明有限步越过 b

假设尚未越过并给每步增量统一下界

(1)

建立增量下界

若在前 k 步内 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=9}{\le }\htmlData{tutor-start=9,tutor-end=10}{b}$ ，由递增性还有 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=9}{\ge }\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{1}}$ 。

详细展开：于是 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{n}}\ln \htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{n}}\htmlData{tutor-start=29,tutor-end=32}{\ge}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{a}_{\htmlData{tutor-start=36,tutor-end=37}{1}}\ln \htmlData{tutor-start=42,tutor-end=43}{b}$ ；并且第一步因 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{b}$ 实际为严格大于该下界。

\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=16}{\ge}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{a}_{\htmlData{tutor-start=20,tutor-end=21}{1}}\ln \htmlData{tutor-start=26,tutor-end=27}{b}\htmlData{tutor-start=27,tutor-end=28}{>}\htmlData{tutor-start=28,tutor-end=29}{0}

(2)

累加并使用 k 的条件

把 n=1 至 k 的增量相加。

详细展开：得到 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{>}\htmlData{tutor-start=14,tutor-end=15}{k}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{a}_{\htmlData{tutor-start=20,tutor-end=21}{1}}\ln \htmlData{tutor-start=26,tutor-end=27}{b}\htmlData{tutor-start=27,tutor-end=28}{)}$ 。题设 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{a}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\ln \htmlData{tutor-start=24,tutor-end=25}{b}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{b}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{1}}\htmlData{tutor-start=35,tutor-end=36}{)}\htmlData{tutor-start=36,tutor-end=37}{/}\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{a}_{\htmlData{tutor-start=42,tutor-end=43}{1}}\ln \htmlData{tutor-start=48,tutor-end=49}{b}\htmlData{tutor-start=49,tutor-end=50}{)}$ ，所以右端至少为 $\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{1}}$ ，从而 $\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{b}$ 。

\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{k}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{>}\htmlData{tutor-start=15,tutor-end=16}{b}}

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