返回特征解读

2014 年高考数学（上海卷理科）

exams_raw/普通高考/2014/2014上海理.pdf · HS-MATH-1024-v2.1-solution-aware

23 题29 个小问/题组
1

一、填空题 · 二倍角公式与周期

函数 y=1−2cos⁡2(2x)\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{2}\cos^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{)} 的最小正周期是______.

答案：最小正周期为 π/2。

题目标签：三角函数的最小正周期

解题过程

化为单一余弦函数并求周期

求函数的最小正周期。

(1)
识别结构并建立关系

式中出现 1−2cos⁡2(2x)\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\cos^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{)}，它正好对应恒等式 2cos⁡2u=1+cos⁡2u\htmlData{tutor-start=0,tutor-end=1}{2}\cos^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{u}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{+}\cos\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{u}；先降幂可以直接看出角频率。

为什么从这里入手：统计量都来自明确的样本计数或加权。“式中出现 1−2cos⁡2(2x)\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\cos^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{)}，它正好对应恒等式 2cos⁡2u=1+cos⁡2u\htmlData{tutor-start=0,tutor-end=1}{2}\cos^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{u}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{+}\cos\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{u}；先降幂可以直接看出角频率。”给出了数据与权重，先识别结构并建立关系可以避免把频数、频率和概率混为一谈，并能用总数或概率和反查计算。

详细展开：令 u=2x\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{x}，则 2cos⁡2(2x)=1+cos⁡4x\htmlData{tutor-start=0,tutor-end=1}{2}\cos^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{x}\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}{+}\cos\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{x}，所以原函数化为 y=−cos⁡4x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\cos\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{x}。余弦函数自变量增加 2π\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\pi} 才完成一个最小周期，因此令 4T=2π\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{T}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=7}{\pi}，得到 T=π/2\htmlData{tutor-start=0,tutor-end=1}{T}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}。负号只改变图象关于横轴的翻折，不改变周期。

y=1−(1+cos⁡4x)=−cos⁡4x,4T=2π\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{+}\cos\htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{-}\cos\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{x}\htmlData{tutor-start=22,tutor-end=23}{,}\qquad \htmlData{tutor-start=30,tutor-end=31}{4}\htmlData{tutor-start=31,tutor-end=32}{T}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=37}{\pi}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：代入核验：f(x+π/2)=−cos⁡(4x+2π)=f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=7}{\pi}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{-}\cos\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{4}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=24}{\pi}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{f}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{x}\htmlData{tutor-start=29,tutor-end=30}{)}。若存在更小正周期，就会使 cos⁡u\cos \htmlData{tutor-start=5,tutor-end=6}{u} 具有小于 2π\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\pi} 的正周期，与余弦函数的最小正周期矛盾。

Tmin⁡=π2\boxed{\htmlData{tutor-start=7,tutor-end=8}{T}_{\min}\htmlData{tutor-start=15,tutor-end=16}{=}\frac{\htmlData{tutor-start=22,tutor-end=25}{\pi}}{\htmlData{tutor-start=27,tutor-end=28}{2}}}
2

一、填空题 · 复数的模与共轭

若复数 z=1+2i\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{2}\text{\htmlData{tutor-start=11,tutor-end=12}{i}}，其中 i\text{\htmlData{tutor-start=6,tutor-end=7}{i}} 是虚数单位，则 (z+1zˉ)⋅zˉ=\left(\htmlData{tutor-start=6,tutor-end=7}{z}\htmlData{tutor-start=7,tutor-end=8}{+}\frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\bar{\htmlData{tutor-start=22,tutor-end=23}{z}}}\right)\htmlData{tutor-start=32,tutor-end=37}{\cdot}\bar{\htmlData{tutor-start=42,tutor-end=43}{z}}\htmlData{tutor-start=44,tutor-end=45}{=}______.

答案：结果为 6。

题目标签：复数与共轭复数运算

解题过程

利用共轭乘积消去分式

计算给定复数式的值。

(1)
识别结构并建立关系

括号外恰好乘有 zˉ\bar{\htmlData{tutor-start=5,tutor-end=6}{z}}，分配后 zzˉ\htmlData{tutor-start=0,tutor-end=1}{z}\bar{\htmlData{tutor-start=6,tutor-end=7}{z}} 可化为实数 ∣z∣2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{z}\htmlData{tutor-start=2,tutor-end=3}{|}^{\htmlData{tutor-start=5,tutor-end=6}{2}}，而 1zˉzˉ=1\frac1{\bar{z}}\bar{z}=1，无需先求倒数。

为什么从这里入手：目标是“识别结构并建立关系”，而“括号外恰好乘有 zˉ\bar{\htmlData{tutor-start=5,tutor-end=6}{z}}，分配后 zzˉ\htmlData{tutor-start=0,tutor-end=1}{z}\bar{\htmlData{tutor-start=6,tutor-end=7}{z}} 可化为实数 ∣z∣2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{z}\htmlData{tutor-start=2,tutor-end=3}{|}^{\htmlData{tutor-start=5,tutor-end=6}{2}}，而 1zˉzˉ=1\frac1{\bar{z}}\bar{z}=1，无需先求倒数。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：由分配律，(z+1zˉ)zˉ=zzˉ+1\left(z+\frac1{\bar{z}}\right)\bar{z}=z\bar{z}+1。因为 z=1+2i\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{2}\mathrm{\htmlData{tutor-start=13,tutor-end=14}{i}}，故 zˉ=1−2i\bar{\htmlData{tutor-start=5,tutor-end=6}{z}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}\mathrm{\htmlData{tutor-start=19,tutor-end=20}{i}}，且 zzˉ=12+22=5\htmlData{tutor-start=0,tutor-end=1}{z}\bar{\htmlData{tutor-start=6,tutor-end=7}{z}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{2}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{5}。于是原式等于 5+1=6\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{6}。

(z+1zˉ)zˉ=zzˉ+1=∣z∣2+1=12+22+1=6\left(z+\frac1{\bar{z}}\right)\bar{z}=z\bar{z}+1=|z|^{2}+1=1^{2}+2^{2}+1=6
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：zˉ=1−2i≠0\bar{\htmlData{tutor-start=5,tutor-end=6}{z}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}\mathrm{\htmlData{tutor-start=19,tutor-end=20}{i}}\ne\htmlData{tutor-start=24,tutor-end=25}{0}，原分式有意义；直接展开 (1+2i)(1−2i)+1\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{2}\mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{2}\mathrm{\htmlData{tutor-start=27,tutor-end=28}{i}}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{1} 也得到 5+1=6\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{6}。

6\boxed{\htmlData{tutor-start=7,tutor-end=8}{6}}
3

一、填空题 · 圆锥曲线的标准方程

若抛物线 y2=2px\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x} 的焦点与椭圆 x29+y25=1\frac{\htmlData{tutor-start=6,tutor-end=7}{x}^{\htmlData{tutor-start=9,tutor-end=10}{2}}}{\htmlData{tutor-start=13,tutor-end=14}{9}}\htmlData{tutor-start=15,tutor-end=16}{+}\frac{\htmlData{tutor-start=22,tutor-end=23}{y}^{\htmlData{tutor-start=25,tutor-end=26}{2}}}{\htmlData{tutor-start=29,tutor-end=30}{5}}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{1} 的右焦点重合，则该抛物线的准线方程为______.

答案：准线方程为 x=−2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}。

题目标签：抛物线焦点与准线

解题过程

由椭圆焦点确定抛物线参数

求抛物线的准线方程。

(1)
识别结构并建立关系

两条曲线通过同一个焦点发生联系。应先由椭圆的 a2−b2\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}} 求右焦点，再与 y2=2px\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x} 的焦点 (p/2,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}\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}\htmlData{tutor-start=6,tutor-end=7}{)} 对照。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“两条曲线通过同一个焦点发生联系。应先由椭圆的 a2−b2\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}} 求右焦点，再与 y2=2px\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x} 的焦点 (p/2,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}\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}\htmlData{tutor-start=6,tutor-end=7}{)} 对照。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：椭圆中 a2=9,b2=5\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}{9}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{b}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{5}，所以 c=9−5=2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{9}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{5}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{2}，右焦点为 (2,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}。抛物线 y2=2px\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{p}\htmlData{tutor-start=8,tutor-end=9}{x} 的焦点为 (p/2,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}\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}\htmlData{tutor-start=6,tutor-end=7}{)}，故 p/2=2\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{2}，即 p=4\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}。其准线与焦点关于顶点对称，方程为 x=−p/2=−2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{p}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}。

c=9−5=2,p2=2,x=−p2\htmlData{tutor-start=0,tutor-end=1}{c}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{9}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{5}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{,}\qquad \frac{\htmlData{tutor-start=28,tutor-end=29}{p}}{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{,}\qquad \htmlData{tutor-start=43,tutor-end=44}{x}\htmlData{tutor-start=44,tutor-end=45}{=}\htmlData{tutor-start=45,tutor-end=46}{-}\frac{\htmlData{tutor-start=52,tutor-end=53}{p}}{\htmlData{tutor-start=55,tutor-end=56}{2}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：将 p=4\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4} 代回得抛物线 y2=8x\htmlData{tutor-start=0,tutor-end=1}{y}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{x}，焦点确为 (2,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}，准线 x=−2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2} 与焦点到顶点的距离均为 2\htmlData{tutor-start=0,tutor-end=1}{2}。

x=−2\boxed{\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}}
4

一、填空题 · 分段函数

设 f(x)={x,x∈(−∞,a),x2,x∈[a,+∞),\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}{=}\begin{cases} \htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{,} & \htmlData{tutor-start=24,tutor-end=25}{x}\htmlData{tutor-start=25,tutor-end=28}{\in}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=36}{\infty}\htmlData{tutor-start=36,tutor-end=37}{,}\htmlData{tutor-start=37,tutor-end=38}{a}\htmlData{tutor-start=38,tutor-end=39}{)}\htmlData{tutor-start=39,tutor-end=40}{,} \\ \htmlData{tutor-start=44,tutor-end=45}{x}^{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{,} & \htmlData{tutor-start=53,tutor-end=54}{x}\htmlData{tutor-start=54,tutor-end=57}{\in}\htmlData{tutor-start=57,tutor-end=58}{[}\htmlData{tutor-start=58,tutor-end=59}{a}\htmlData{tutor-start=59,tutor-end=60}{,}\htmlData{tutor-start=60,tutor-end=61}{+}\htmlData{tutor-start=61,tutor-end=67}{\infty}\htmlData{tutor-start=67,tutor-end=68}{)}\htmlData{tutor-start=68,tutor-end=69}{,} \end{cases} 若 f(2)=4\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{4}，则 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围为______.

答案：a≤2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{2}。

题目标签：分段函数值确定分界参数

解题过程

判断 x=2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 落在哪一段

由 f(2)=4\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{4} 求参数范围。

(1)
识别结构并建立关系

分段点由参数 a\htmlData{tutor-start=0,tutor-end=1}{a} 决定，代入 x=2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 前必须先判断 2<a\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a} 还是 2≥a\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{a}；两种分支给出的函数值不同。

为什么从这里入手：代数题要寻找能减少未知量或降低次数的关系。“分段点由参数 a\htmlData{tutor-start=0,tutor-end=1}{a} 决定，代入 x=2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 前必须先判断 2<a\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a} 还是 2≥a\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{a}；两种分支给出的函数值不同。”正好提供了因式、根或参数之间的等式，所以先识别结构并建立关系，再代入、消元或比较系数，计算链条会更短也更容易复核。

详细展开：若 2<a\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}，则 2∈(−∞,a)\htmlData{tutor-start=0,tutor-end=1}{2}\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=12}{\infty}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{)}，按第一段有 f(2)=2\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}，与题设 4\htmlData{tutor-start=0,tutor-end=1}{4} 矛盾。若 2≥a\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{a}，则 2∈[a,+∞)\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\in}\htmlData{tutor-start=4,tutor-end=5}{[}\htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=14}{\infty}\htmlData{tutor-start=14,tutor-end=15}{)}，按第二段有 f(2)=22=4\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{4}，恰好符合。因此所需且充分的条件是 a≤2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{2}。注意 a=2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 时第二段含左端点，仍取平方分支。

2<a⇒f(2)=2;a≤2⇒f(2)=4\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=15}{\Rightarrow }\htmlData{tutor-start=15,tutor-end=16}{f}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{;}\qquad \htmlData{tutor-start=29,tutor-end=30}{a}\htmlData{tutor-start=30,tutor-end=33}{\le}\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=46}{\Rightarrow }\htmlData{tutor-start=46,tutor-end=47}{f}\htmlData{tutor-start=47,tutor-end=48}{(}\htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=50}{)}\htmlData{tutor-start=50,tutor-end=51}{=}\htmlData{tutor-start=51,tutor-end=52}{4}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：当且仅当 a≤2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{2} 时，数字 2\htmlData{tutor-start=0,tutor-end=1}{2} 属于含等号的第二段，回代确得 4\htmlData{tutor-start=0,tutor-end=1}{4}；a>2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2} 全部被第一段的函数值 2\htmlData{tutor-start=0,tutor-end=1}{2} 排除。

a∈(−∞,2]\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=11}{\in}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=19}{\infty}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{]}}
5

一、填空题 · 基本不等式

若实数 x,y\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 xy=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}，则 x2+2y2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{y}^{\htmlData{tutor-start=10,tutor-end=11}{2}} 的最小值为______.

答案：最小值为 22\htmlData{tutor-start=0,tutor-end=1}{2}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{2}}。

题目标签：乘积约束下的最值

解题过程

消元后使用基本不等式

求 x2+2y2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{y}^{\htmlData{tutor-start=10,tutor-end=11}{2}} 的最小值。

(1)
识别结构并建立关系

xy=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1} 使 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y} 均非零，并把两个平方项的乘积固定为 2x2y2=2\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}，适合直接使用基本不等式并检查等号能否达到。

为什么从这里入手：最值与范围题要先找到变量受限方式以及等号何时成立。“xy=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1} 使 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{y} 均非零，并把两个平方项的乘积固定为 2x2y2=2\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}，适合直接使用基本不等式并检查等号能否达到。”给出了可行域或关键界，先识别结构并建立关系能把‘可能有多大’转成单调性、基本不等式或边界比较。

详细展开：因为 x2>0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{>}\htmlData{tutor-start=6,tutor-end=7}{0}、2y2>0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{y}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{>}\htmlData{tutor-start=7,tutor-end=8}{0}，由基本不等式得 x2+2y2≥22x2y2=22\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{y}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=15}{\ge}\htmlData{tutor-start=15,tutor-end=16}{2}\sqrt{\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{x}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{y}^{\htmlData{tutor-start=31,tutor-end=32}{2}}}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{2}\sqrt{\htmlData{tutor-start=42,tutor-end=43}{2}}。等号要求 x2=2y2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{y}^{\htmlData{tutor-start=10,tutor-end=11}{2}}。与 xy=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1} 联立可得 x4=2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}，例如取 x=21/4,y=2−1/4\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}^{\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{4}}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{y}\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}{1}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{4}}，确实满足全部条件。

x2+2y2≥22x2y2=22\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{y}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=15}{\ge}\htmlData{tutor-start=15,tutor-end=16}{2}\sqrt{\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{x}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{y}^{\htmlData{tutor-start=31,tutor-end=32}{2}}}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{2}\sqrt{\htmlData{tutor-start=42,tutor-end=43}{2}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：下界不仅由不等式得到，而且存在实数 x=21/4,y=2−1/4\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}^{\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{4}}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{y}\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}{1}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{4}} 使等号成立，因此它是最小值而非仅是下界。

min⁡(x2+2y2)=22\boxed{\min\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{x}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{y}^{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{2}\sqrt{\htmlData{tutor-start=33,tutor-end=34}{2}}}
6

一、填空题 · 圆锥侧面积

若圆锥的侧面积是底面积的3倍，则其母线与底面夹角的大小为______.（结果用反三角函数值表示）

答案：夹角为 arccos⁡(1/3)\arccos\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{)}。

题目标签：圆锥母线与底面的夹角

解题过程

由面积比求母线与半径之比

求母线与底面所成角。

(1)
识别结构并建立关系

侧面积和底面积之比先确定 l=3r\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{r}。母线在底面上的正投影是半径 r\htmlData{tutor-start=0,tutor-end=1}{r}，因此题问“与底面的角”时应使用邻边比 r/l\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{l}。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“侧面积和底面积之比先确定 l=3r\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{r}。母线在底面上的正投影是半径 r\htmlData{tutor-start=0,tutor-end=1}{r}，因此题问“与底面的角”时应使用邻边比 r/l\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{l}。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：设底面半径为 r\htmlData{tutor-start=0,tutor-end=1}{r}、母线为 l\htmlData{tutor-start=0,tutor-end=1}{l}，由 πrl=3πr2\htmlData{tutor-start=0,tutor-end=4}{\pi }\htmlData{tutor-start=4,tutor-end=5}{r}\htmlData{tutor-start=5,tutor-end=6}{l}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{3}\htmlData{tutor-start=8,tutor-end=12}{\pi }\htmlData{tutor-start=12,tutor-end=13}{r}^{\htmlData{tutor-start=15,tutor-end=16}{2}} 得 l=3r\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{r}。取含轴和母线的轴截面，母线、圆锥高和半径组成直角三角形。设母线与底面夹角为 φ\htmlData{tutor-start=0,tutor-end=7}{\varphi}，母线在底面的投影长为 r\htmlData{tutor-start=0,tutor-end=1}{r}，故 cos⁡φ=r/l=1/3\cos\htmlData{tutor-start=4,tutor-end=11}{\varphi}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{r}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{l}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{3}。夹角为锐角，所以 φ=arccos⁡(1/3)\htmlData{tutor-start=0,tutor-end=7}{\varphi}\htmlData{tutor-start=7,tutor-end=8}{=}\arccos\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{)}。

l=3r,cos⁡φ=rl=13\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{r}\htmlData{tutor-start=4,tutor-end=5}{,}\qquad \cos\htmlData{tutor-start=16,tutor-end=23}{\varphi}\htmlData{tutor-start=23,tutor-end=24}{=}\frac{\htmlData{tutor-start=30,tutor-end=31}{r}}{\htmlData{tutor-start=33,tutor-end=34}{l}}\htmlData{tutor-start=35,tutor-end=36}{=}\frac{\htmlData{tutor-start=42,tutor-end=43}{1}}{\htmlData{tutor-start=45,tutor-end=46}{3}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：这里不是求母线与轴的夹角；与底面的夹角使用母线及其正投影，故余弦值为 1/3\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}，答案唯一落在 (0,π/2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=6}{\pi}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{)}。

φ=arccos⁡13\boxed{\htmlData{tutor-start=7,tutor-end=14}{\varphi}\htmlData{tutor-start=14,tutor-end=15}{=}\arccos\frac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{3}}}
7

一、填空题 · 极坐标与直角坐标

已知曲线 C\htmlData{tutor-start=0,tutor-end=1}{C} 的极坐标方程为 ρ(3cos⁡θ−4sin⁡θ)=1\htmlData{tutor-start=0,tutor-end=4}{\rho}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{3}\cos\htmlData{tutor-start=10,tutor-end=16}{\theta}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{4}\sin\htmlData{tutor-start=22,tutor-end=28}{\theta}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{1}，则 C\htmlData{tutor-start=0,tutor-end=1}{C} 与极轴的交点到极点的距离是______.

答案：距离为 1/3\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}。

题目标签：极坐标曲线与极轴交点

解题过程

转成直角坐标后与横轴联立

求交点到极点的距离。

(1)
识别结构并建立关系

利用 x=ρcos⁡θ,y=ρsin⁡θ\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=6}{\rho}\cos\htmlData{tutor-start=10,tutor-end=16}{\theta}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{y}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=23}{\rho}\sin\htmlData{tutor-start=27,tutor-end=33}{\theta}，原方程左侧正好是 3x−4y\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{y}；极轴就是直角坐标的 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“利用 x=ρcos⁡θ,y=ρsin⁡θ\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=6}{\rho}\cos\htmlData{tutor-start=10,tutor-end=16}{\theta}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{y}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=23}{\rho}\sin\htmlData{tutor-start=27,tutor-end=33}{\theta}，原方程左侧正好是 3x−4y\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{y}；极轴就是直角坐标的 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：由极坐标与直角坐标关系，ρ(3cos⁡θ−4sin⁡θ)=3x−4y\htmlData{tutor-start=0,tutor-end=4}{\rho}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{3}\cos\htmlData{tutor-start=10,tutor-end=16}{\theta}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{4}\sin\htmlData{tutor-start=22,tutor-end=28}{\theta}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{=}\htmlData{tutor-start=30,tutor-end=31}{3}\htmlData{tutor-start=31,tutor-end=32}{x}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{4}\htmlData{tutor-start=34,tutor-end=35}{y}，所以曲线 C\htmlData{tutor-start=0,tutor-end=1}{C} 是直线 3x−4y=1\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}。极轴满足 y=0\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}，联立得 3x=1\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}，交点为 (1/3,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}。极点是原点，因此两点距离为 ∣1/3∣=1/3\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{/}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{3}。

3x−4y=1,y=0⇒(x,y)=(13,0)\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,}\quad \htmlData{tutor-start=14,tutor-end=15}{y}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=29}{\Rightarrow }\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{x}\htmlData{tutor-start=31,tutor-end=32}{,}\htmlData{tutor-start=32,tutor-end=33}{y}\htmlData{tutor-start=33,tutor-end=34}{)}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{(}\tfrac{\htmlData{tutor-start=43,tutor-end=44}{1}}{\htmlData{tutor-start=46,tutor-end=47}{3}}\htmlData{tutor-start=48,tutor-end=49}{,}\htmlData{tutor-start=49,tutor-end=50}{0}\htmlData{tutor-start=50,tutor-end=51}{)}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：把交点写回极坐标可取 ρ=1/3,θ=0\htmlData{tutor-start=0,tutor-end=4}{\rho}\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}{3}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=15}{\theta}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{0}，代入原方程左端等于 1\htmlData{tutor-start=0,tutor-end=1}{1}；距离按非负长度取 1/3\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}。

13\boxed{\frac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{3}}}
8

一、填空题 · 等比数列与极限

设无穷等比数列 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 的公比为 q\htmlData{tutor-start=0,tutor-end=1}{q}，若 a1=lim⁡n→∞(a3+a4+⋯+an)\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\lim_{\htmlData{tutor-start=12,tutor-end=13}{n}\htmlData{tutor-start=13,tutor-end=16}{\to}\htmlData{tutor-start=16,tutor-end=22}{\infty}}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{3}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{4}}\htmlData{tutor-start=35,tutor-end=36}{+}\cdots\htmlData{tutor-start=42,tutor-end=43}{+}\htmlData{tutor-start=43,tutor-end=44}{a}_{\htmlData{tutor-start=46,tutor-end=47}{n}}\htmlData{tutor-start=48,tutor-end=49}{)}，则 q=\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}______.

答案：q=(5−1)/2\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{5}}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{2}。

题目标签：无穷等比级数确定公比

解题过程

把尾项极限化为无穷级数

求等比数列的公比。

(1)
识别结构并建立关系

右侧是从 a3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}} 开始的无穷等比和。极限存在要求 ∣q∣<1\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{q}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}，再用首项 a3=a1q2\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}{1}}\htmlData{tutor-start=11,tutor-end=12}{q}^{\htmlData{tutor-start=14,tutor-end=15}{2}} 写出和。

为什么从这里入手：数列题要先识别相邻项之间保持不变的结构。“右侧是从 a3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}} 开始的无穷等比和。极限存在要求 ∣q∣<1\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{q}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}，再用首项 a3=a1q2\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}{1}}\htmlData{tutor-start=11,tutor-end=12}{q}^{\htmlData{tutor-start=14,tutor-end=15}{2}} 写出和。”提示了差、比或可累加关系，先识别结构并建立关系能把第 n\htmlData{tutor-start=0,tutor-end=1}{n} 项问题改写成已经会处理的等差、等比或裂项模型。

详细展开：在通常的非零等比数列中 a1≠0\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\ne\htmlData{tutor-start=8,tutor-end=9}{0}。由收敛性 ∣q∣<1\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{q}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}，且 a3+a4+⋯=a1q2/(1−q)\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}{4}}\htmlData{tutor-start=11,tutor-end=12}{+}\cdots\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{1}}\htmlData{tutor-start=24,tutor-end=25}{q}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{/}\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{q}\htmlData{tutor-start=34,tutor-end=35}{)}。题设给出 a1=a1q2/(1−q)\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{q}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{q}\htmlData{tutor-start=21,tutor-end=22}{)}，约去 a1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 得 q2+q−1=0\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{q}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{0}。两根为 (−1±5)/2\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=6}{\pm}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{5}}\htmlData{tutor-start=14,tutor-end=15}{)}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{2}，其中只有 (5−1)/2\htmlData{tutor-start=0,tutor-end=1}{(}\sqrt{\htmlData{tutor-start=7,tutor-end=8}{5}}\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}{2} 的绝对值小于 1\htmlData{tutor-start=0,tutor-end=1}{1}。

1=q21−q⇒q2+q−1=0\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{q}^{\htmlData{tutor-start=11,tutor-end=12}{2}}}{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{q}}\htmlData{tutor-start=19,tutor-end=31}{\Rightarrow }\htmlData{tutor-start=31,tutor-end=32}{q}^{\htmlData{tutor-start=34,tutor-end=35}{2}}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{q}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{=}\htmlData{tutor-start=41,tutor-end=42}{0}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：另一根 (−1−5)/2<−1\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{-}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{5}}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{<}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1} 会使无穷等比和发散，必须舍去；保留根约为 0.618\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{6}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{8}，满足收敛条件。

q=5−12\boxed{\htmlData{tutor-start=7,tutor-end=8}{q}\htmlData{tutor-start=8,tutor-end=9}{=}\frac{\sqrt{\htmlData{tutor-start=21,tutor-end=22}{5}}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{2}}}
9

一、填空题 · 幂函数与定义域

若 f(x)=x23−x−12\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}^{\frac{\htmlData{tutor-start=14,tutor-end=15}{2}}{\htmlData{tutor-start=17,tutor-end=18}{3}}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{x}^{\htmlData{tutor-start=24,tutor-end=25}{-}\frac{\htmlData{tutor-start=31,tutor-end=32}{1}}{\htmlData{tutor-start=34,tutor-end=35}{2}}}，则满足 f(x)<0\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{<}\htmlData{tutor-start=5,tutor-end=6}{0} 的 x\htmlData{tutor-start=0,tutor-end=1}{x} 的取值范围是______.

答案：x∈(0,1)\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}{)}。

题目标签：分数指数不等式

解题过程

先定定义域再比较幂

解不等式 x2/3−x−1/2<0\htmlData{tutor-start=0,tutor-end=1}{x}^{\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=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{<}\htmlData{tutor-start=17,tutor-end=18}{0}。

(1)
识别结构并建立关系

x−1/2=1/x\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{/}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{x}} 同时要求根式有意义且分母非零，所以必须先写 x>0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}；在正数域上可以乘正幂而不改变不等号方向。

为什么从这里入手：含分母、根号或对数的式子必须先合法，后续变形才有意义。“x−1/2=1/x\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{/}\sqrt{\htmlData{tutor-start=17,tutor-end=18}{x}} 同时要求根式有意义且分母非零，所以必须先写 x>0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}；在正数域上可以乘正幂而不改变不等号方向。”正好暴露了限制条件；先识别结构并建立关系，可以防止约分或平方时把禁值偷偷带回答案。

详细展开：定义域为 x>0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}。原不等式等价于 x2/3<x−1/2\htmlData{tutor-start=0,tutor-end=1}{x}^{\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=7,tutor-end=8}{<}\htmlData{tutor-start=8,tutor-end=9}{x}^{\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{2}}。两边同乘正数 x1/2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{2}}，得到 x7/6<1\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{7}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{6}}\htmlData{tutor-start=7,tutor-end=8}{<}\htmlData{tutor-start=8,tutor-end=9}{1}。幂函数 x7/6\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{7}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{6}} 在 (0,+∞)\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}{)} 上严格递增，故恰有 0<x<1\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{1}。

x>0,x2/3<x−1/2⟺x7/6<1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\qquad \htmlData{tutor-start=11,tutor-end=12}{x}^{\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{3}}\htmlData{tutor-start=18,tutor-end=19}{<}\htmlData{tutor-start=19,tutor-end=20}{x}^{\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=27,tutor-end=47}{\Longleftrightarrow }\htmlData{tutor-start=47,tutor-end=48}{x}^{\htmlData{tutor-start=50,tutor-end=51}{7}\htmlData{tutor-start=51,tutor-end=52}{/}\htmlData{tutor-start=52,tutor-end=53}{6}}\htmlData{tutor-start=54,tutor-end=55}{<}\htmlData{tutor-start=55,tutor-end=56}{1}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：x=1\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 时两项相等，不满足严格小于；x=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 不在定义域，负数也因 x−1/2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}} 无实数意义而排除。

x∈(0,1)\boxed{\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=11}{\in}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)}}
10

一、填空题 · 组合计数与概率

为强化安全意识，某商场拟在未来的连续10天中随机选择3天进行紧急疏散演练，则选择的3天恰好为连续3天的概率是______.（结果用最简分数表示）

答案：概率为 1/15\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{5}。

题目标签：连续日期的古典概型

解题过程

分别计数全部组合与连续三天

求随机选择的三天连续的概率。

(1)
识别结构并建立关系

选择三天不考虑先后顺序，总样本数用组合数；连续三天则只需确定这段连续区间的第一天。

为什么从这里入手：统计量都来自明确的样本计数或加权。“选择三天不考虑先后顺序，总样本数用组合数；连续三天则只需确定这段连续区间的第一天。”给出了数据与权重，先识别结构并建立关系可以避免把频数、频率和概率混为一谈，并能用总数或概率和反查计算。

详细展开：从 10\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0} 天任取 3\htmlData{tutor-start=0,tutor-end=1}{3} 天共有 (103)=120\binom{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{0}}{\htmlData{tutor-start=11,tutor-end=12}{3}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{0} 种等可能结果。连续三天可以从第 1\htmlData{tutor-start=0,tutor-end=1}{1} 天开始到第 8\htmlData{tutor-start=0,tutor-end=1}{8} 天开始，共有 10−3+1=8\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{3}\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}{8} 种。因此所求概率为 8/120=1/15\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{5}。

P=8(103)=8120=115P=\frac{8}{\binom{10}{3}}=\frac8{120}=\frac1{15}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：连续段的起点不能是第 9\htmlData{tutor-start=0,tutor-end=1}{9} 或第 10\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0} 天，否则会超出十天范围；八个有利组合互不重复，分数已约为最简。

115\boxed{\frac1{15}}
11

一、填空题 · 复数与集合

已知互异的复数 a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b} 满足 ab≠0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\neq \htmlData{tutor-start=7,tutor-end=8}{0}，集合 {a,b}={a2,b2}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=7}{\}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{b}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=23}{\}}，则 a+b=\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}______.

答案：a+b=−1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}。

题目标签：集合相等约束下的复数

解题过程

利用二元集合相等分类配对

求两个复数之和。

(1)
识别结构并建立关系

两个只含两个互异元素的集合相等，元素配对只有“各自对应”或“交叉对应”两种；互异与非零条件会排除第一种。

为什么从这里入手：集合题不能只看式子的外形，必须回到“元素是否满足条件”。这里的“两个只含两个互异元素的集合相等，元素配对只有“各自对应”或“交叉对应”两种；互异与非零条件会排除第一种。”给出了元素筛选规则，因此先识别结构并建立关系，再逐项保留或删除元素，思路最直接。

详细展开：若 a=a2,b=b2\htmlData{tutor-start=0,tutor-end=1}{a}\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}{b}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{b}^{\htmlData{tutor-start=13,tutor-end=14}{2}}，由非零得 a=b=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}，与互异矛盾。因此只能有 a=b2,b=a2\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=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{a}^{\htmlData{tutor-start=13,tutor-end=14}{2}}。代入得 a=a4\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}^{\htmlData{tutor-start=5,tutor-end=6}{4}}，由 a≠0\htmlData{tutor-start=0,tutor-end=1}{a}\ne\htmlData{tutor-start=4,tutor-end=5}{0} 得 a3=1\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}{1}。若 a=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 仍会有 b=1\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，故 a\htmlData{tutor-start=0,tutor-end=1}{a} 是非实三次单位根，b=a2\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}{2}}，而 1+a+a2=0\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{a}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{0}。所以 a+b=−1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}。

a=b2, b=a2⇒a3=1, a≠1,a+a2=−1a=b^{2},\ b=a^{2}\Rightarrow a^{3}=1,\ a\ne1,\quad a+a^{2}=-1
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：两枚非实三次单位根互异且乘积 ab=a3=1≠0\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=6,tutor-end=7}{3}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}\ne\htmlData{tutor-start=13,tutor-end=14}{0}，平方后恰好互换，确实满足原集合等式，因此填入 −1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}。

a+b=−1\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{b}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}}
12

一、填空题 · 辅助角公式与区间根

设常数 a\htmlData{tutor-start=0,tutor-end=1}{a} 使方程 sin⁡x+3cos⁡x=a\sin \htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{+}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\cos \htmlData{tutor-start=20,tutor-end=21}{x}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{a} 在闭区间 [0,2π]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=7}{\pi}\htmlData{tutor-start=7,tutor-end=8}{]} 上恰有三个解 x1,x2,x3\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}{x}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{x}_{\htmlData{tutor-start=15,tutor-end=16}{3}}，则 x1+x2+x3=\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}{x}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{x}_{\htmlData{tutor-start=15,tutor-end=16}{3}}\htmlData{tutor-start=17,tutor-end=18}{=}______.

答案：x1+x2+x3=7π/3\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}{x}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{x}_{\htmlData{tutor-start=15,tutor-end=16}{3}}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{7}\htmlData{tutor-start=19,tutor-end=22}{\pi}\htmlData{tutor-start=22,tutor-end=23}{/}\htmlData{tutor-start=23,tutor-end=24}{3}。

题目标签：三角方程恰有三个解时的根和

解题过程

利用端点重复函数值识别特殊参数

在恰有三个解的条件下求三根之和。

(1)
识别结构并建立关系

函数 2sin⁡(x+π/3)\htmlData{tutor-start=0,tutor-end=1}{2}\sin\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=11}{\pi}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{)} 在长度为 2π\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\pi} 的闭区间通常给同一水平线两个交点；恰有三个不同解只能来自两个端点 0,2π\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=6}{\pi} 同时被计入，因此先算端点函数值。

为什么从这里入手：目标是“识别结构并建立关系”，而“函数 2sin⁡(x+π/3)\htmlData{tutor-start=0,tutor-end=1}{2}\sin\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=11}{\pi}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{)} 在长度为 2π\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=4}{\pi} 的闭区间通常给同一水平线两个交点；恰有三个不同解只能来自两个端点 0,2π\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=6}{\pi} 同时被计入，因此先算端点函数值。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：原式为 2sin⁡(x+π/3)\htmlData{tutor-start=0,tutor-end=1}{2}\sin\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=11}{\pi}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{)}，且在 x=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 与 x=2π\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=6}{\pi} 时都等于 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}。若水平线不经过端点，完整周期内通常有两个交点；若取极值则只有一个交点。因此恰有三个解时必须 a=3\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}。解 2sin⁡(x+π/3)=3\htmlData{tutor-start=0,tutor-end=1}{2}\sin\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=11}{\pi}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{=}\sqrt{\htmlData{tutor-start=21,tutor-end=22}{3}}，得到 x=0,π/3,2π\htmlData{tutor-start=0,tutor-end=1}{x}\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=7}{\pi}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=14}{\pi}，三者都在闭区间内。

a=3,{x1,x2,x3}={0,π3,2π}a=\sqrt{3},\qquad \{x_{1},x_{2},x_{3}\}=\{0,\tfrac\pi3,2\pi\}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：三个值互不相同并逐一满足原方程，其和为 0+π/3+2π=7π/3\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=12}{\pi}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{7}\htmlData{tutor-start=14,tutor-end=17}{\pi}\htmlData{tutor-start=17,tutor-end=18}{/}\htmlData{tutor-start=18,tutor-end=19}{3}；其他参数不能在一个完整周期的闭区间中产生恰好三个不同根。

x1+x2+x3=7π3\boxed{\htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{1}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{x}_{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{x}_{\htmlData{tutor-start=22,tutor-end=23}{3}}\htmlData{tutor-start=24,tutor-end=25}{=}\frac{\htmlData{tutor-start=31,tutor-end=32}{7}\htmlData{tutor-start=32,tutor-end=35}{\pi}}{\htmlData{tutor-start=37,tutor-end=38}{3}}}
13

一、填空题 · 离散型随机变量

某游戏的得分为1, 2, 3, 4, 5，随机变量 ξ\htmlData{tutor-start=0,tutor-end=3}{\xi} 表示小白玩该游戏的得分. 若 E(ξ)=4.2\htmlData{tutor-start=0,tutor-end=1}{E}\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}{=}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{.}\htmlData{tutor-start=9,tutor-end=10}{2}，则小白得5分的概率至少为______.

答案：至少为 1/5\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}。

题目标签：由数学期望估计最高分概率

解题过程

用非最高分的最大值建立上界

求得 5 分概率的最小可能值。

(1)
识别结构并建立关系

要让 5 分概率尽量小，同时维持较高期望 4.2\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{2}，其余概率应尽可能集中在非最高分中的最大值 4\htmlData{tutor-start=0,tutor-end=1}{4}；这给出期望关于 p5\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{5}} 的最强上界。

为什么从这里入手：概率题的关键不是立刻代数，而是先说清随机试验与事件。“要让 5 分概率尽量小，同时维持较高期望 4.2\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{2}，其余概率应尽可能集中在非最高分中的最大值 4\htmlData{tutor-start=0,tutor-end=1}{4}；这给出期望关于 p5\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{5}} 的最强上界。”确定了基本事件或随机变量，所以先识别结构并建立关系，再依据独立、互斥、对立或期望线性性计算。

详细展开：记 p=P(ξ=5)\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{P}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=7}{\xi}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{5}\htmlData{tutor-start=9,tutor-end=10}{)}。当没有得到 5 分时，得分至多为 4\htmlData{tutor-start=0,tutor-end=1}{4}，所以 E(ξ)≤5p+4(1−p)=4+p\htmlData{tutor-start=0,tutor-end=1}{E}\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=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{5}\htmlData{tutor-start=10,tutor-end=11}{p}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{4}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{p}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{4}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{p}。由 E(ξ)=4.2\htmlData{tutor-start=0,tutor-end=1}{E}\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}{=}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{.}\htmlData{tutor-start=9,tutor-end=10}{2} 得 4.2≤4+p\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=6}{\le}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{p}，故 p≥0.2=1/5\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{.}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{/}\htmlData{tutor-start=10,tutor-end=11}{5}。取 P(ξ=5)=1/5\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}{5}\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}{5}、P(ξ=4)=4/5\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}{4}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{5}、其余概率为零时，期望正好为 5/5+16/5=21/5=4.2\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{6}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{5}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{.}\htmlData{tutor-start=16,tutor-end=17}{2}，下界可以达到。

4.2=E(ξ)≤5p+4(1−p)=4+p⇒p≥15\htmlData{tutor-start=0,tutor-end=1}{4}\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}{E}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=9}{\xi}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=13}{\le}\htmlData{tutor-start=13,tutor-end=14}{5}\htmlData{tutor-start=14,tutor-end=15}{p}\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}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{p}\htmlData{tutor-start=21,tutor-end=22}{)}\htmlData{tutor-start=22,tutor-end=23}{=}\htmlData{tutor-start=23,tutor-end=24}{4}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{p}\htmlData{tutor-start=26,tutor-end=38}{\Rightarrow }\htmlData{tutor-start=38,tutor-end=39}{p}\htmlData{tutor-start=39,tutor-end=42}{\ge}\frac{\htmlData{tutor-start=48,tutor-end=49}{1}}{\htmlData{tutor-start=51,tutor-end=52}{5}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：构造的概率分布各概率非负且总和为 1\htmlData{tutor-start=0,tutor-end=1}{1}，并达到期望 4.2\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{2}，所以 1/5\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{5} 是“至少”值而不是仅有的粗略下界。

P(ξ=5)min⁡=15\boxed{\htmlData{tutor-start=7,tutor-end=8}{P}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=12}{\xi}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{5}\htmlData{tutor-start=14,tutor-end=15}{)}_{\min}\htmlData{tutor-start=22,tutor-end=23}{=}\frac{\htmlData{tutor-start=29,tutor-end=30}{1}}{\htmlData{tutor-start=32,tutor-end=33}{5}}}
14

一、填空题 · 解析几何与向量

已知曲线 C:x=−4−y2\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{-}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{4}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{y}^{\htmlData{tutor-start=17,tutor-end=18}{2}}}，直线 l:x=6\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{6}. 若对于点 A(m,0)\htmlData{tutor-start=0,tutor-end=1}{A}\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}{0}\htmlData{tutor-start=5,tutor-end=6}{)}，存在 C\htmlData{tutor-start=0,tutor-end=1}{C} 上的点 P\htmlData{tutor-start=0,tutor-end=1}{P} 和 l\htmlData{tutor-start=0,tutor-end=1}{l} 上的 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 使得 AP→+AQ→=0→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{P}}\htmlData{tutor-start=19,tutor-end=20}{+}\overrightarrow{\htmlData{tutor-start=36,tutor-end=37}{A}\htmlData{tutor-start=37,tutor-end=38}{Q}}\htmlData{tutor-start=39,tutor-end=40}{=}\overrightarrow{\htmlData{tutor-start=56,tutor-end=57}{0}}，则 m\htmlData{tutor-start=0,tutor-end=1}{m} 的取值范围为______.

答案：m∈[2,3]\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=4}{\in}\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}{]}。

题目标签：向量中点条件与曲线横坐标范围

解题过程

把向量和为零翻译为中点

求点 A\htmlData{tutor-start=0,tutor-end=1}{A} 的横坐标范围。

(1)
识别结构并建立关系

AP→+AQ→=0\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{P}}\htmlData{tutor-start=19,tutor-end=20}{+}\overrightarrow{\htmlData{tutor-start=36,tutor-end=37}{A}\htmlData{tutor-start=37,tutor-end=38}{Q}}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{0} 表示两个向量互为相反向量，因此 A\htmlData{tutor-start=0,tutor-end=1}{A} 是线段 PQ\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{Q} 的中点；只比较横坐标即可连接半圆和直线。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“AP→+AQ→=0\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{P}}\htmlData{tutor-start=19,tutor-end=20}{+}\overrightarrow{\htmlData{tutor-start=36,tutor-end=37}{A}\htmlData{tutor-start=37,tutor-end=38}{Q}}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{0} 表示两个向量互为相反向量，因此 A\htmlData{tutor-start=0,tutor-end=1}{A} 是线段 PQ\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{Q} 的中点；只比较横坐标即可连接半圆和直线。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：设 P=(xP,yP)\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}_{\htmlData{tutor-start=6,tutor-end=7}{P}}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{y}_{\htmlData{tutor-start=12,tutor-end=13}{P}}\htmlData{tutor-start=14,tutor-end=15}{)}。曲线 C\htmlData{tutor-start=0,tutor-end=1}{C} 是圆 x2+y2=4\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{y}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{4} 的左半圆，所以 −2≤xP≤0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{x}_{\htmlData{tutor-start=9,tutor-end=10}{P}}\htmlData{tutor-start=11,tutor-end=14}{\le}\htmlData{tutor-start=14,tutor-end=15}{0}。由中点关系 Q=2A−P\htmlData{tutor-start=0,tutor-end=1}{Q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{P}，其横坐标为 2m−xP\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}_{\htmlData{tutor-start=6,tutor-end=7}{P}}。因 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 在 x=6\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{6} 上，故 2m−xP=6\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}_{\htmlData{tutor-start=6,tutor-end=7}{P}}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{6}，即 xP=2m−6\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{P}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{m}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{6}。要求该横坐标落在 [−2,0]\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}{]}，得到 −2≤2m−6≤0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=5}{\le}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{m}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=12}{\le}\htmlData{tutor-start=12,tutor-end=13}{0}，即 2≤m≤3\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{m}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{3}。

−2≤2m−6≤0⟺2≤m≤3\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=5}{\le}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{m}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=12}{\le}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=33}{\Longleftrightarrow }\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=38}{\le }\htmlData{tutor-start=38,tutor-end=39}{m}\htmlData{tutor-start=39,tutor-end=42}{\le}\htmlData{tutor-start=42,tutor-end=43}{3}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：对任意 m∈[2,3]\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=4}{\in}\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}{]}，可在左半圆取横坐标 2m−6\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{6} 的点 P\htmlData{tutor-start=0,tutor-end=1}{P}，再令 Q=2A−P\htmlData{tutor-start=0,tutor-end=1}{Q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{A}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{P}，其横坐标就是 6\htmlData{tutor-start=0,tutor-end=1}{6}；端点对应半圆上的合法点，故闭区间全部可达。

m∈[2,3]\boxed{\htmlData{tutor-start=7,tutor-end=8}{m}\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=15}{3}\htmlData{tutor-start=15,tutor-end=16}{]}}
15

二、选择题 · 简易逻辑

设 a,b∈R\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=6}{\in}\mathbf{\htmlData{tutor-start=14,tutor-end=15}{R}}，则“a+b>4\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}{4}”是“a>2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2} 且 b>2\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2}”的 ( )

答案：B，必要非充分条件。

题目标签：充分条件与必要条件

解题过程

双向检验两个命题

判断 a+b>4\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}{4} 与 a>2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2} 且 b>2\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2} 的条件关系。

(1)
识别结构并建立关系

判断条件关系必须检查两个方向：后者能否推出前者，以及前者能否推出后者。一个方向可直接相加，另一个方向适合用反例。

为什么从这里入手：目标是“识别结构并建立关系”，而“判断条件关系必须检查两个方向：后者能否推出前者，以及前者能否推出后者。一个方向可直接相加，另一个方向适合用反例。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：若 a>2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2} 且 b>2\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2}，相加得 a+b>4\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}{4}，所以 a+b>4\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}{4} 是后者成立的必要条件。反过来不成立，例如 a=5,b=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0} 时 a+b=5>4\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}{5}\htmlData{tutor-start=5,tutor-end=6}{>}\htmlData{tutor-start=6,tutor-end=7}{4}，但 b>2\htmlData{tutor-start=0,tutor-end=1}{b}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{2} 不成立。因此它不是充分条件。

(a>2∧b>2)⇒a+b>4,(a,b)=(5,0)\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}{2}\htmlData{tutor-start=4,tutor-end=10}{\land }\htmlData{tutor-start=10,tutor-end=11}{b}\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=26}{\Rightarrow }\htmlData{tutor-start=26,tutor-end=27}{a}\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}{4}\htmlData{tutor-start=31,tutor-end=32}{,}\qquad \htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{a}\htmlData{tutor-start=41,tutor-end=42}{,}\htmlData{tutor-start=42,tutor-end=43}{b}\htmlData{tutor-start=43,tutor-end=44}{)}\htmlData{tutor-start=44,tutor-end=45}{=}\htmlData{tutor-start=45,tutor-end=46}{(}\htmlData{tutor-start=46,tutor-end=47}{5}\htmlData{tutor-start=47,tutor-end=48}{,}\htmlData{tutor-start=48,tutor-end=49}{0}\htmlData{tutor-start=49,tutor-end=50}{)}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：直接推导证明了必要性，合法反例否定了充分性，所以只能选择“必要非充分条件”，即 B。

B\boxed{\mathrm{\htmlData{tutor-start=15,tutor-end=16}{B}}}
16

二、选择题 · 空间向量数量积

如图，四个棱长为1的正方体排成一个正四棱柱，AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 是一条侧棱，Pi\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{i}} (i=1,2,⋯,8\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,}\cdots\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{8}) 是上底面上其余的八个点，则 AB→⋅APi→\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}}\htmlData{tutor-start=19,tutor-end=24}{\cdot}\overrightarrow{\htmlData{tutor-start=40,tutor-end=41}{A}\htmlData{tutor-start=41,tutor-end=42}{P}_{\htmlData{tutor-start=44,tutor-end=45}{i}}} (i=1,2,⋯,8\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,}\cdots\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{8}) 的不同值的个数为 ( )

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

答案：不同值共有 1 个，选 D。

题目标签：空间网格中的向量点积

解题过程

利用上下底面平行确定投影

统计八个点积的不同取值。

(1)
识别结构并建立关系

AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 是竖直侧棱，八个 Pi\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 全在上底面；点积只取 APi\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{P}_{\htmlData{tutor-start=4,tutor-end=5}{i}} 在 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 方向上的投影，而所有上底面点的这一高度都相同。

为什么从这里入手：目标是“识别结构并建立关系”，而“AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 是竖直侧棱，八个 Pi\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 全在上底面；点积只取 APi\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{P}_{\htmlData{tutor-start=4,tutor-end=5}{i}} 在 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 方向上的投影，而所有上底面点的这一高度都相同。”恰好给出了它所缺的前置量或等价关系。先把这条信息写成可计算的等式，后续只需沿等价变形逐步推进；若跳过这一步，就无法说明答案为何由题设唯一确定。

详细展开：建立坐标系，令 A=(0,0,0)\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}{)}，B=(0,0,1)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{)}，则 AB→=(0,0,1)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=23}{,}\htmlData{tutor-start=23,tutor-end=24}{0}\htmlData{tutor-start=24,tutor-end=25}{,}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{)}。每个 Pi\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 都在上底面 z=1\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 上，可写为 Pi=(ui,vi,1)\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{u}_{\htmlData{tutor-start=10,tutor-end=11}{i}}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{v}_{\htmlData{tutor-start=16,tutor-end=17}{i}}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{)}，于是 APi→=(ui,vi,1)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{P}_{\htmlData{tutor-start=20,tutor-end=21}{i}}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{u}_{\htmlData{tutor-start=28,tutor-end=29}{i}}\htmlData{tutor-start=30,tutor-end=31}{,}\htmlData{tutor-start=31,tutor-end=32}{v}_{\htmlData{tutor-start=34,tutor-end=35}{i}}\htmlData{tutor-start=36,tutor-end=37}{,}\htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{)}。因此 AB→⋅APi→=1\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}}\htmlData{tutor-start=19,tutor-end=24}{\cdot}\overrightarrow{\htmlData{tutor-start=40,tutor-end=41}{A}\htmlData{tutor-start=41,tutor-end=42}{P}_{\htmlData{tutor-start=44,tutor-end=45}{i}}}\htmlData{tutor-start=47,tutor-end=48}{=}\htmlData{tutor-start=48,tutor-end=49}{1}，与 ui,vi\htmlData{tutor-start=0,tutor-end=1}{u}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{v}_{\htmlData{tutor-start=9,tutor-end=10}{i}} 无关，八个点积完全相同。

(0,0,1)⋅(ui,vi,1)=1(i=1,…,8)\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}{1}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=12}{\cdot}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{u}_{\htmlData{tutor-start=16,tutor-end=17}{i}}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=20}{v}_{\htmlData{tutor-start=22,tutor-end=23}{i}}\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=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{1}\quad\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{i}\htmlData{tutor-start=36,tutor-end=37}{=}\htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{,}\htmlData{tutor-start=39,tutor-end=45}{\ldots}\htmlData{tutor-start=45,tutor-end=46}{,}\htmlData{tutor-start=46,tutor-end=47}{8}\htmlData{tutor-start=47,tutor-end=48}{)}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：各点在上底面内的水平位置只改变与 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 垂直的分量，不影响点积，所以不同值恰有一个，选择 D。

1D\boxed{\htmlData{tutor-start=7,tutor-end=8}{1}}\quad\boxed{\mathrm{\htmlData{tutor-start=29,tutor-end=30}{D}}}
17

二、选择题 · 线性方程组与行列式

已知 P1(a1,b1)\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{b}_{\htmlData{tutor-start=15,tutor-end=16}{1}}\htmlData{tutor-start=17,tutor-end=18}{)} 与 P2(a2,b2)\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{b}_{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{)} 是直线 y=kx+1\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1} (k\htmlData{tutor-start=0,tutor-end=1}{k} 为常数) 上两个不同的点，则关于 x\htmlData{tutor-start=0,tutor-end=1}{x} 和 y\htmlData{tutor-start=0,tutor-end=1}{y} 的方程组 {a1x+b1y=1,a2x+b2y=1\begin{cases} \htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{1}}\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{b}_{\htmlData{tutor-start=24,tutor-end=25}{1}}\htmlData{tutor-start=26,tutor-end=27}{y}\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=34,tutor-end=35}{a}_{\htmlData{tutor-start=37,tutor-end=38}{2}}\htmlData{tutor-start=39,tutor-end=40}{x}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{b}_{\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{y}\htmlData{tutor-start=47,tutor-end=48}{=}\htmlData{tutor-start=48,tutor-end=49}{1} \end{cases} 的解的情况是 ( )

答案：无论如何总有唯一解，选 B。

题目标签：点在线上与二元方程组

解题过程

用直线关系化简系数行列式

判断方程组解的情况。

(1)
识别结构并建立关系

二元一次方程组是否唯一可由系数行列式判断；两点都在 y=kx+1\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1} 上，可把 bi\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 写成 kai+1\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{i}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}，从而消去 k\htmlData{tutor-start=0,tutor-end=1}{k}。

为什么从这里入手：代数题要寻找能减少未知量或降低次数的关系。“二元一次方程组是否唯一可由系数行列式判断；两点都在 y=kx+1\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1} 上，可把 bi\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 写成 kai+1\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{i}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}，从而消去 k\htmlData{tutor-start=0,tutor-end=1}{k}。”正好提供了因式、根或参数之间的等式，所以先识别结构并建立关系，再代入、消元或比较系数，计算链条会更短也更容易复核。

详细展开：由 Pi(ai,bi)\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{i}}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{b}_{\htmlData{tutor-start=15,tutor-end=16}{i}}\htmlData{tutor-start=17,tutor-end=18}{)} 在直线上，bi=kai+1\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{i}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{k}\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{i}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}。系数行列式 Δ=a1b2−a2b1=a1(ka2+1)−a2(ka1+1)=a1−a2\htmlData{tutor-start=0,tutor-end=6}{\Delta}\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}{b}_{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{a}_{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{b}_{\htmlData{tutor-start=26,tutor-end=27}{1}}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{a}_{\htmlData{tutor-start=32,tutor-end=33}{1}}\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{k}\htmlData{tutor-start=36,tutor-end=37}{a}_{\htmlData{tutor-start=39,tutor-end=40}{2}}\htmlData{tutor-start=41,tutor-end=42}{+}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{)}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{a}_{\htmlData{tutor-start=48,tutor-end=49}{2}}\htmlData{tutor-start=50,tutor-end=51}{(}\htmlData{tutor-start=51,tutor-end=52}{k}\htmlData{tutor-start=52,tutor-end=53}{a}_{\htmlData{tutor-start=55,tutor-end=56}{1}}\htmlData{tutor-start=57,tutor-end=58}{+}\htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=60}{)}\htmlData{tutor-start=60,tutor-end=61}{=}\htmlData{tutor-start=61,tutor-end=62}{a}_{\htmlData{tutor-start=64,tutor-end=65}{1}}\htmlData{tutor-start=66,tutor-end=67}{-}\htmlData{tutor-start=67,tutor-end=68}{a}_{\htmlData{tutor-start=70,tutor-end=71}{2}}。两点互异，而同一直线是 x\htmlData{tutor-start=0,tutor-end=1}{x} 的函数，若 a1=a2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{2}} 则 b1=b2\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{b}_{\htmlData{tutor-start=9,tutor-end=10}{2}}，会使两点相同；故 a1≠a2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\ne \htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{2}}，于是 Δ≠0\htmlData{tutor-start=0,tutor-end=6}{\Delta}\ne\htmlData{tutor-start=9,tutor-end=10}{0}。

Δ=a1(ka2+1)−a2(ka1+1)=a1−a2≠0\htmlData{tutor-start=0,tutor-end=6}{\Delta}\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=13,tutor-end=14}{k}\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{2}}\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}{-}\htmlData{tutor-start=23,tutor-end=24}{a}_{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{k}\htmlData{tutor-start=30,tutor-end=31}{a}\htmlData{tutor-start=31,tutor-end=36}{_{1}+}\htmlData{tutor-start=36,tutor-end=37}{1}\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}}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{a}_{\htmlData{tutor-start=48,tutor-end=49}{2}}\ne\htmlData{tutor-start=53,tutor-end=54}{0}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：非零行列式保证方程组对任意允许的 k,P1,P2\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{P}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{P}_{\htmlData{tutor-start=11,tutor-end=12}{2}} 都有且只有一组解，不存在无解或无穷多解的例外。

总有唯一解B\boxed{\text{\htmlData{tutor-start=13,tutor-end=14}{总}\htmlData{tutor-start=14,tutor-end=15}{有}\htmlData{tutor-start=15,tutor-end=16}{唯}\htmlData{tutor-start=16,tutor-end=17}{一}\htmlData{tutor-start=17,tutor-end=18}{解}}}\quad\boxed{\mathrm{\htmlData{tutor-start=40,tutor-end=41}{B}}}
18

二、选择题 · 含参数函数最值

设 f(x)={(x−a)2,x⩽0,x+1x+a,x>0,\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\begin{cases} \htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{x}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{a}\htmlData{tutor-start=23,tutor-end=24}{)}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{,} & \htmlData{tutor-start=32,tutor-end=33}{x}\htmlData{tutor-start=33,tutor-end=43}{\leqslant }\htmlData{tutor-start=43,tutor-end=44}{0}\htmlData{tutor-start=44,tutor-end=45}{,} \\ \htmlData{tutor-start=49,tutor-end=50}{x}\htmlData{tutor-start=50,tutor-end=51}{+}\frac{\htmlData{tutor-start=57,tutor-end=58}{1}}{\htmlData{tutor-start=60,tutor-end=61}{x}}\htmlData{tutor-start=62,tutor-end=63}{+}\htmlData{tutor-start=63,tutor-end=64}{a}\htmlData{tutor-start=64,tutor-end=65}{,} & \htmlData{tutor-start=68,tutor-end=69}{x}\htmlData{tutor-start=69,tutor-end=70}{>}\htmlData{tutor-start=70,tutor-end=71}{0}\htmlData{tutor-start=71,tutor-end=72}{,} \end{cases} 若 f(0)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{)} 是 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 的最小值，则 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围为 ( )

答案：a∈[0,2]\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\in}\htmlData{tutor-start=4,tutor-end=5}{[}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{]}。

题目标签：分段函数全局最小值

解题过程

分别控制零点两侧的函数值

求使 f(0)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{)} 成为全局最小值的参数范围。

(1)
识别结构并建立关系

分段点 x=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 的函数值为 a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}。要成为全局最小值，必须同时不大于左段和右段的最小值，因此要分两段建立条件。

为什么从这里入手：最值与范围题要先找到变量受限方式以及等号何时成立。“分段点 x=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 的函数值为 a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}。要成为全局最小值，必须同时不大于左段和右段的最小值，因此要分两段建立条件。”给出了可行域或关键界，先识别结构并建立关系能把‘可能有多大’转成单调性、基本不等式或边界比较。

详细展开：左段为 (x−a)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。当 a>0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0} 时，其顶点在定义域 x≤0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{0} 的右侧，左段在 x=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 取得最小值 a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}；当 a<0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0} 时可取 x=a\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a} 得 0<a2\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}{2}}，不合题意；a=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 可行，所以先得 a≥0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{0}。右段中 x+1/x≥2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=8}{\ge}\htmlData{tutor-start=8,tutor-end=9}{2}，故其下确界且最小值为 a+2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}，要求 a2≤a+2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=9}{\le }\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{2}，即 (a−2)(a+1)≤0\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}{2}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=13}{\le}\htmlData{tutor-start=13,tutor-end=14}{0}。与 a≥0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{0} 取交集得 0≤a≤2\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{a}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{2}。

a≥0,a2≤a+2⟺−1≤a≤2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{,}\qquad \htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=22}{\le }\htmlData{tutor-start=22,tutor-end=23}{a}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=45}{\Longleftrightarrow }\htmlData{tutor-start=45,tutor-end=46}{-}\htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=51}{\le }\htmlData{tutor-start=51,tutor-end=52}{a}\htmlData{tutor-start=52,tutor-end=55}{\le}\htmlData{tutor-start=55,tutor-end=56}{2}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：当 a∈[0,2]\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\in}\htmlData{tutor-start=4,tutor-end=5}{[}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{]} 时，左段均不小于 a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}，右段均不小于 a+2≥a2\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=7}{\ge }\htmlData{tutor-start=7,tutor-end=8}{a}^{\htmlData{tutor-start=10,tutor-end=11}{2}}；端点 a=0,2\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}{2} 也满足非严格不等式。因此范围含两个端点。

a∈[0,2]\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}{0}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{]}}
19

三、解答题 · 空间几何体与展开图

底面边长为2的正三棱锥 P−ABC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{A}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}，其表面展开图是三角形 P1P2P3\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{P}_{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{P}_{\htmlData{tutor-start=13,tutor-end=14}{3}}，如图，求 △P1P2P3\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{P}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{P}_{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{P}_{\htmlData{tutor-start=23,tutor-end=24}{3}} 的各边长及此三棱锥的体积 V\htmlData{tutor-start=0,tutor-end=1}{V}.

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

答案：P1P2=P2P3=P3P1=4\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{P}_{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{P}_{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{P}_{\htmlData{tutor-start=19,tutor-end=20}{3}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{P}_{\htmlData{tutor-start=25,tutor-end=26}{3}}\htmlData{tutor-start=27,tutor-end=28}{P}_{\htmlData{tutor-start=30,tutor-end=31}{1}}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{4}，V=22/3\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{3}。

题目标签：由展开图确定正三棱锥

解题过程

由共线角确定侧棱并计算体积

求展开图外三角形边长和三棱锥体积。

(1)
识别结构并建立关系

折叠后 P1,P2,P3\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{P}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{P}_{\htmlData{tutor-start=15,tutor-end=16}{3}} 重合为顶点 P\htmlData{tutor-start=0,tutor-end=1}{P}，正三棱锥的三个侧面全等。展开图中 P1,A,P3\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{P}_{\htmlData{tutor-start=11,tutor-end=12}{3}} 共线，能把两个侧面底角与底面 60∘60^\circ 角相加。

为什么从这里入手：空间关系只靠观察容易漏条件，而“折叠后 P1,P2,P3\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{P}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{P}_{\htmlData{tutor-start=15,tutor-end=16}{3}} 重合为顶点 P\htmlData{tutor-start=0,tutor-end=1}{P}，正三棱锥的三个侧面全等。展开图中 P1,A,P3\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{P}_{\htmlData{tutor-start=11,tutor-end=12}{3}} 共线，能把两个侧面底角与底面 60∘60^\circ 角相加。”可以直接翻译成垂直、平行、数量积或体积公式。先识别结构并建立关系，是在建立可验证的几何链条，而不是凭图形猜结论。

详细展开：设侧棱长为 s\htmlData{tutor-start=0,tutor-end=1}{s}，侧面等腰三角形的底角为 θ\htmlData{tutor-start=0,tutor-end=6}{\theta}。沿直线 P1AP3\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{P}_{\htmlData{tutor-start=9,tutor-end=10}{3}} 有 θ+60∘+θ=180∘\theta+60^\circ+\theta=180^\circ，故 θ=60∘\theta=60^\circ，侧面成为边长 2\htmlData{tutor-start=0,tutor-end=1}{2} 的等边三角形，所以 s=2\htmlData{tutor-start=0,tutor-end=1}{s}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}。外三角形每边由两条侧棱首尾组成，均为 2s=4\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{s}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{4}。三棱锥实际为棱长 2\htmlData{tutor-start=0,tutor-end=1}{2} 的正四面体；底面中心到顶点距离为 2/3\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{/}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}，高 h=22−(2/3)2=26/3\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{2}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{/}\sqrt{\htmlData{tutor-start=23,tutor-end=24}{3}}\htmlData{tutor-start=25,tutor-end=26}{)}^{\htmlData{tutor-start=28,tutor-end=29}{2}}}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{2}\sqrt{\htmlData{tutor-start=39,tutor-end=40}{6}}\htmlData{tutor-start=41,tutor-end=42}{/}\htmlData{tutor-start=42,tutor-end=43}{3}。底面积为 3\sqrt{\htmlData{tutor-start=6,tutor-end=7}{3}}，故 V=133⋅26/3=22/3\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{3}}\sqrt{\htmlData{tutor-start=19,tutor-end=20}{3}}\htmlData{tutor-start=21,tutor-end=26}{\cdot}\htmlData{tutor-start=26,tutor-end=27}{2}\sqrt{\htmlData{tutor-start=33,tutor-end=34}{6}}\htmlData{tutor-start=35,tutor-end=36}{/}\htmlData{tutor-start=36,tutor-end=37}{3}\htmlData{tutor-start=37,tutor-end=38}{=}\htmlData{tutor-start=38,tutor-end=39}{2}\sqrt{\htmlData{tutor-start=45,tutor-end=46}{2}}\htmlData{tutor-start=47,tutor-end=48}{/}\htmlData{tutor-start=48,tutor-end=49}{3}。

2theta+60∘=180∘,s=2,h=2632\\theta+60^\circ=180^\circ,\quad s=2,\quad h=\frac{2\sqrt{6}}{3}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：棱长 2\htmlData{tutor-start=0,tutor-end=1}{2} 的正四面体展开时每条外边确由两条长 2\htmlData{tutor-start=0,tutor-end=1}{2} 的侧棱组成，外边为 4\htmlData{tutor-start=0,tutor-end=1}{4}；体积也与公式 a3/(62)\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}{(}\htmlData{tutor-start=7,tutor-end=8}{6}\sqrt{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{)} 在 a=2\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 时的结果一致。

P1P2=P2P3=P3P1=4,V=223\boxed{\htmlData{tutor-start=7,tutor-end=8}{P}_{\htmlData{tutor-start=10,tutor-end=11}{1}}\htmlData{tutor-start=12,tutor-end=13}{P}_{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{P}_{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{P}_{\htmlData{tutor-start=26,tutor-end=27}{3}}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{P}_{\htmlData{tutor-start=32,tutor-end=33}{3}}\htmlData{tutor-start=34,tutor-end=35}{P}_{\htmlData{tutor-start=37,tutor-end=38}{1}}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{4}}\htmlData{tutor-start=42,tutor-end=43}{,}\qquad \boxed{\htmlData{tutor-start=57,tutor-end=58}{V}\htmlData{tutor-start=58,tutor-end=59}{=}\frac{\htmlData{tutor-start=65,tutor-end=66}{2}\sqrt{\htmlData{tutor-start=72,tutor-end=73}{2}}}{\htmlData{tutor-start=76,tutor-end=77}{3}}}
20

三、解答题 · 反函数与函数奇偶性

设常数 a⩾0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=11}{\geqslant }\htmlData{tutor-start=11,tutor-end=12}{0}，函数 f(x)=2x+a2x−a\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\frac{\htmlData{tutor-start=11,tutor-end=12}{2}^{\htmlData{tutor-start=14,tutor-end=15}{x}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{a}}{\htmlData{tutor-start=20,tutor-end=21}{2}^{\htmlData{tutor-start=23,tutor-end=24}{x}}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{a}}. (1) 若 a=4\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}，求函数 y=f(x)\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}{)} 的反函数 y=f−1(x)\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=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{)}； (2) 根据 a\htmlData{tutor-start=0,tutor-end=1}{a} 的不同取值，讨论函数 y=f(x)\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}{)} 的奇偶性，并说明理由.

答案：f−1(x)=2+log⁡2x+1x−1\htmlData{tutor-start=0,tutor-end=1}{f}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{+}\log_{\htmlData{tutor-start=18,tutor-end=19}{2}}\frac{\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=31,tutor-end=32}{x}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{1}}，定义域为 (−∞,−1)∪(1,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=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}{)}；a=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 时偶，a=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 时奇，其余时非奇非偶。

题目标签：分式型指数函数的反函数与奇偶性

解题过程

（1）求 a=4\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4} 时的反函数

解出反函数表达式及其定义域。

(1)
识别结构并建立关系

令 t=2x>0\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}^{\htmlData{tutor-start=5,tutor-end=6}{x}}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{0} 后原函数是关于 t\htmlData{tutor-start=0,tutor-end=1}{t} 的分式一次式，可以先由 y\htmlData{tutor-start=0,tutor-end=1}{y} 反解 t\htmlData{tutor-start=0,tutor-end=1}{t}，再取以 2\htmlData{tutor-start=0,tutor-end=1}{2} 为底的对数；反函数定义域必须等于原函数值域。

为什么从这里入手：含分母、根号或对数的式子必须先合法，后续变形才有意义。“令 t=2x>0\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}^{\htmlData{tutor-start=5,tutor-end=6}{x}}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{0} 后原函数是关于 t\htmlData{tutor-start=0,tutor-end=1}{t} 的分式一次式，可以先由 y\htmlData{tutor-start=0,tutor-end=1}{y} 反解 t\htmlData{tutor-start=0,tutor-end=1}{t}，再取以 2\htmlData{tutor-start=0,tutor-end=1}{2} 为底的对数；反函数定义域必须等于原函数值域。”正好暴露了限制条件；先识别结构并建立关系，可以防止约分或平方时把禁值偷偷带回答案。

详细展开：当 a=4\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}，y=(2x+4)/(2x−4)\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=6,tutor-end=7}{x}}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{2}^{\htmlData{tutor-start=16,tutor-end=17}{x}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{4}\htmlData{tutor-start=20,tutor-end=21}{)}，定义域排除 x=2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}。整理得 (y−1)2x=4(y+1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{2}^{\htmlData{tutor-start=8,tutor-end=9}{x}}\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}{y}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{)}，所以 2x=4(y+1)/(y−1)\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{x}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{y}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{y}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{)}，进而 x=2+log⁡2((y+1)/(y−1))\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{+}\log_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{y}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{)}\htmlData{tutor-start=18,tutor-end=19}{/}\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{y}\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{)}。因右侧指数必须为正，(y+1)/(y−1)>0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{1}\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}{y}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{>}\htmlData{tutor-start=12,tutor-end=13}{0}，即 y<−1\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}{1} 或 y>1\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1}。交换自变量、函数值即得反函数。

f−1(x)=2+log⁡2x+1x−1,x∈(−∞,−1)∪(1,+∞)\htmlData{tutor-start=0,tutor-end=1}{f}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{+}\log_{\htmlData{tutor-start=18,tutor-end=19}{2}}\frac{\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=31,tutor-end=32}{x}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{1}}\htmlData{tutor-start=35,tutor-end=36}{,}\quad \htmlData{tutor-start=42,tutor-end=43}{x}\htmlData{tutor-start=43,tutor-end=46}{\in}\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{-}\htmlData{tutor-start=48,tutor-end=54}{\infty}\htmlData{tutor-start=54,tutor-end=55}{,}\htmlData{tutor-start=55,tutor-end=56}{-}\htmlData{tutor-start=56,tutor-end=57}{1}\htmlData{tutor-start=57,tutor-end=58}{)}\htmlData{tutor-start=58,tutor-end=62}{\cup}\htmlData{tutor-start=62,tutor-end=63}{(}\htmlData{tutor-start=63,tutor-end=64}{1}\htmlData{tutor-start=64,tutor-end=65}{,}\htmlData{tutor-start=65,tutor-end=66}{+}\htmlData{tutor-start=66,tutor-end=72}{\infty}\htmlData{tutor-start=72,tutor-end=73}{)}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：把 2f−1(x)=4(x+1)/(x−1)\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{f}^{\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{)}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{x}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{/}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{x}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{)} 代回原分式，分子分母约化后得到 x\htmlData{tutor-start=0,tutor-end=1}{x}；定义域又避开了对数真数非正和分母为零的情况。

f−1(x)=2+log⁡2x+1x−1\boxed{\htmlData{tutor-start=7,tutor-end=8}{f}^{\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{+}\log_{\htmlData{tutor-start=25,tutor-end=26}{2}}\frac{\htmlData{tutor-start=33,tutor-end=34}{x}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{1}}{\htmlData{tutor-start=38,tutor-end=39}{x}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{1}}}

（2）按参数讨论奇偶性

确定各参数范围内函数的奇偶性。

(1)
识别结构并建立关系

奇偶函数的定义域必须关于原点对称。这里唯一被排除的点是 log⁡2a\log_{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{a}（当 a>0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}），先用定义域对称性可迅速筛出 a=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}；a=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 要单独化简。

为什么从这里入手：含分母、根号或对数的式子必须先合法，后续变形才有意义。“奇偶函数的定义域必须关于原点对称。这里唯一被排除的点是 log⁡2a\log_{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{a}（当 a>0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0}），先用定义域对称性可迅速筛出 a=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}；a=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 要单独化简。”正好暴露了限制条件；先识别结构并建立关系，可以防止约分或平方时把禁值偷偷带回答案。

详细展开：当 a=0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 时 f(x)=1\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{1}，定义域为 R\mathbb{\htmlData{tutor-start=8,tutor-end=9}{R}}，故为偶函数。当 a>0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0} 时定义域为 R∖{log⁡2a}\mathbb{\htmlData{tutor-start=8,tutor-end=9}{R}}\htmlData{tutor-start=10,tutor-end=19}{\setminus}\htmlData{tutor-start=19,tutor-end=21}{\{}\log_{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{a}\htmlData{tutor-start=30,tutor-end=32}{\}}，它关于原点对称只可能有 log⁡2a=0\log_{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{0}，即 a=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}。此时 f(−x)=(2−x+1)/(2−x−1)=(1+2x)/(1−2x)=−f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{2}^{\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{x}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{2}^{\htmlData{tutor-start=21,tutor-end=22}{-}\htmlData{tutor-start=22,tutor-end=23}{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=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{2}^{\htmlData{tutor-start=34,tutor-end=35}{x}}\htmlData{tutor-start=36,tutor-end=37}{)}\htmlData{tutor-start=37,tutor-end=38}{/}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{2}^{\htmlData{tutor-start=44,tutor-end=45}{x}}\htmlData{tutor-start=46,tutor-end=47}{)}\htmlData{tutor-start=47,tutor-end=48}{=}\htmlData{tutor-start=48,tutor-end=49}{-}\htmlData{tutor-start=49,tutor-end=50}{f}\htmlData{tutor-start=50,tutor-end=51}{(}\htmlData{tutor-start=51,tutor-end=52}{x}\htmlData{tutor-start=52,tutor-end=53}{)}，故为奇函数。若 a>0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0} 且 a≠1\htmlData{tutor-start=0,tutor-end=1}{a}\ne\htmlData{tutor-start=4,tutor-end=5}{1}，定义域不关于原点对称，函数非奇非偶。

a=0: f(x)=1;a=1: f(−x)=−f(x)\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=6}{\ }\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}{1}\htmlData{tutor-start=12,tutor-end=13}{;}\qquad \htmlData{tutor-start=20,tutor-end=21}{a}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{:}\htmlData{tutor-start=24,tutor-end=26}{\ }\htmlData{tutor-start=26,tutor-end=27}{f}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{-}\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}{)}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：三个参数情形覆盖 a≥0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{0} 的全部取值；常函数 1\htmlData{tutor-start=0,tutor-end=1}{1} 不是奇函数，而 a=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 时定义域 R∖{0}\mathbb{\htmlData{tutor-start=8,tutor-end=9}{R}}\htmlData{tutor-start=10,tutor-end=19}{\setminus}\htmlData{tutor-start=19,tutor-end=21}{\{}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=24}{\}} 确实关于原点对称。

a=0 时为偶函数; a=1 时为奇函数; a>0,a≠1 时非奇非偶\boxed{a=0\text{ 时为偶函数};\ a=1\text{ 时为奇函数};\ a>0,a\ne1\text{ 时非奇非偶}}
21

解答题 · 解三角形与实际应用

如图，某公司要在 A\htmlData{tutor-start=0,tutor-end=1}{A}、B\htmlData{tutor-start=0,tutor-end=1}{B} 两地连线上的定点 C\htmlData{tutor-start=0,tutor-end=1}{C} 处建造广告牌 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}，其中 D\htmlData{tutor-start=0,tutor-end=1}{D} 为顶端，AC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} 长 35 米，CB\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{B} 长 80 米. 设点 A\htmlData{tutor-start=0,tutor-end=1}{A}、B\htmlData{tutor-start=0,tutor-end=1}{B} 在同一水平面上，从 A\htmlData{tutor-start=0,tutor-end=1}{A} 和 B\htmlData{tutor-start=0,tutor-end=1}{B} 看 D\htmlData{tutor-start=0,tutor-end=1}{D} 的仰角分别为 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 和 β\htmlData{tutor-start=0,tutor-end=5}{\beta}. (1) 设计中 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 是铅垂方向. 若要求 α⩾2β\htmlData{tutor-start=0,tutor-end=7}{\alpha }\htmlData{tutor-start=7,tutor-end=17}{\geqslant }\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=23}{\beta}，问 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 的长至多为多少（结果精确到 0.01 米）？ (2) 施工完成后，CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 与铅垂方向有偏差. 现在实测得 α=38.12∘\alpha = 38.12^\circ，β=18.45∘\beta = 18.45^\circ，求 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 的长（结果精确到 0.01 米）.

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

答案：（1）至多 28.28\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{.}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{8} 米；（2）CD=26.93\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{6}\htmlData{tutor-start=5,tutor-end=6}{.}\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{3} 米。

题目标签：仰角模型中的长度计算

解题过程

（1）铅垂广告牌的最大高度

在 α≥2β\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=9}{\ge}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=15}{\beta} 下求 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 的最大值。

(1)
识别结构并建立关系

设计状态中 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 垂直地面，两个直角三角形立即给出 tan⁡α=h/35\tan\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{h}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{5}、tan⁡β=h/80\tan\htmlData{tutor-start=4,tutor-end=9}{\beta}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{h}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{8}\htmlData{tutor-start=13,tutor-end=14}{0}。最大值出现在角度约束取等号的边界。

为什么从这里入手：最值与范围题要先找到变量受限方式以及等号何时成立。“设计状态中 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 垂直地面，两个直角三角形立即给出 tan⁡α=h/35\tan\htmlData{tutor-start=4,tutor-end=10}{\alpha}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{h}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{5}、tan⁡β=h/80\tan\htmlData{tutor-start=4,tutor-end=9}{\beta}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{h}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{8}\htmlData{tutor-start=13,tutor-end=14}{0}。最大值出现在角度约束取等号的边界。”给出了可行域或关键界，先识别结构并建立关系能把‘可能有多大’转成单调性、基本不等式或边界比较。

详细展开：设 h=CD>0\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{>}\htmlData{tutor-start=5,tutor-end=6}{0}。由 α≥2β\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=9}{\ge}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=15}{\beta} 且 α<90∘\alpha<90^\circ 可知 2β<90∘2\beta<90^\circ，从而 h<80\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{8}\htmlData{tutor-start=3,tutor-end=4}{0}，二倍角公式分母为正。正切在该范围递增，故 h/35=tan⁡α≥tan⁡2β=2(h/80)/(1−h2/802)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{5}\htmlData{tutor-start=4,tutor-end=5}{=}\tan\htmlData{tutor-start=9,tutor-end=15}{\alpha}\htmlData{tutor-start=15,tutor-end=18}{\ge}\tan\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=28}{\beta}\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}{h}\htmlData{tutor-start=32,tutor-end=33}{/}\htmlData{tutor-start=33,tutor-end=34}{8}\htmlData{tutor-start=34,tutor-end=35}{0}\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}{1}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{h}^{\htmlData{tutor-start=43,tutor-end=44}{2}}\htmlData{tutor-start=45,tutor-end=46}{/}\htmlData{tutor-start=46,tutor-end=47}{8}\htmlData{tutor-start=47,tutor-end=48}{0}^{\htmlData{tutor-start=50,tutor-end=51}{2}}\htmlData{tutor-start=52,tutor-end=53}{)}。约去 h\htmlData{tutor-start=0,tutor-end=1}{h} 并整理得 1−h2/6400≥7/8\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{h}^{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=15}{\ge}\htmlData{tutor-start=15,tutor-end=16}{7}\htmlData{tutor-start=16,tutor-end=17}{/}\htmlData{tutor-start=17,tutor-end=18}{8}，即 h2≤800\htmlData{tutor-start=0,tutor-end=1}{h}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=8}{\le}\htmlData{tutor-start=8,tutor-end=9}{8}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{0}。所以最大 h=202\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{0}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}。

h35≥2(h/80)1−(h/80)2⟹h≤202\frac h{35}\ge\frac{2(h/80)}{1-(h/80)^{2}}\Longrightarrow h\le20\sqrt{2}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：取 h=202\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{0}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}} 时上式各步均取等号，因而 α=2β\htmlData{tutor-start=0,tutor-end=6}{\alpha}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=13}{\beta}，几何状态可达。按题意保留两位小数为 28.28\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{8}\htmlData{tutor-start=2,tutor-end=3}{.}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{8} 米。

CDmax⁡=202 m≈28.28 m\boxed{\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{D}_{\max}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{0}\sqrt{\htmlData{tutor-start=25,tutor-end=26}{2}}\text{ \htmlData{tutor-start=34,tutor-end=35}{m}}\htmlData{tutor-start=36,tutor-end=43}{\approx}\htmlData{tutor-start=43,tutor-end=44}{2}\htmlData{tutor-start=44,tutor-end=45}{8}\htmlData{tutor-start=45,tutor-end=46}{.}\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{8}\text{ \htmlData{tutor-start=55,tutor-end=56}{m}}}

（2）广告牌倾斜后的实际长度

由两个实测仰角求 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}。

(1)
识别结构并建立关系

倾斜后不能再把 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 当作高，但三角形 ABD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{D} 的底边和两个底角已知，可先用正弦定理求 AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}，再在三角形 ACD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D} 中用余弦定理。

为什么从这里入手：三角题先把未知角转成特殊角、同一角或已知边角关系。“倾斜后不能再把 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 当作高，但三角形 ABD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{D} 的底边和两个底角已知，可先用正弦定理求 AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}，再在三角形 ACD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{D} 中用余弦定理。”指出了可用的象限、恒等式或几何关系，先识别结构并建立关系可以同时确定数值和正负号。

详细展开：AB=35+80=115\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}{3}\htmlData{tutor-start=4,tutor-end=5}{5}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{8}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{5}。在 △ABD\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}{D} 中，∠D=180∘−α−β\angle D=180^\circ-\alpha-\beta，由正弦定理 AD=115sin⁡β/sin⁡(α+β)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{5}\sin\htmlData{tutor-start=10,tutor-end=15}{\beta}\htmlData{tutor-start=15,tutor-end=16}{/}\sin\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=27}{\alpha}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=33}{\beta}\htmlData{tutor-start=33,tutor-end=34}{)}。代入 α=38.12∘,β=18.45∘\alpha=38.12^\circ,\beta=18.45^\circ 得 AD≈43.60965\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=9}{\approx}\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{.}\htmlData{tutor-start=12,tutor-end=13}{6}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{9}\htmlData{tutor-start=15,tutor-end=16}{6}\htmlData{tutor-start=16,tutor-end=17}{5}。在 △ACD\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=13}{D} 中，AC=35\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}{3}\htmlData{tutor-start=4,tutor-end=5}{5} 且夹角 ∠CAD=α\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{D}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=17}{\alpha}，故 CD2=AD2+352−70ADcos⁡α\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{D}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{5}^{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{7}\htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=24}{A}\htmlData{tutor-start=24,tutor-end=25}{D}\cos\htmlData{tutor-start=29,tutor-end=35}{\alpha}，计算得 CD≈26.92958\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=9}{\approx}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{6}\htmlData{tutor-start=11,tutor-end=12}{.}\htmlData{tutor-start=12,tutor-end=13}{9}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{9}\htmlData{tutor-start=15,tutor-end=16}{5}\htmlData{tutor-start=16,tutor-end=17}{8}。

AD=115sin⁡18.45∘sin⁡56.57∘,CD=AD2+352−70ADcos⁡38.12∘AD=\frac{115\sin18.45^\circ}{\sin56.57^\circ},\quad CD=\sqrt{AD^{2}+35^{2}-70AD\cos38.12^\circ}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：未使用 CD⊥AB\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{B}，因此适用于施工后的倾斜状态；按四舍五入精确到 0.01\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{.}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{1} 米得到 26.93\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=3}{.}\htmlData{tutor-start=3,tutor-end=4}{9}\htmlData{tutor-start=4,tutor-end=5}{3} 米。

CD≈26.93 m\boxed{\htmlData{tutor-start=7,tutor-end=8}{C}\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=16}{\approx}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{6}\htmlData{tutor-start=18,tutor-end=19}{.}\htmlData{tutor-start=19,tutor-end=20}{9}\htmlData{tutor-start=20,tutor-end=21}{3}\text{ \htmlData{tutor-start=28,tutor-end=29}{m}}}
22

解答题 · 解析几何与存在性证明

在平面直角坐标系 xOy\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，对于直线 l:ax+by+c=0\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{:} \htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{b}\htmlData{tutor-start=7,tutor-end=8}{y}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0} 和点 P1(x1,y1)\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}_{\htmlData{tutor-start=9,tutor-end=10}{1}}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{y}_{\htmlData{tutor-start=16,tutor-end=17}{1}}\htmlData{tutor-start=18,tutor-end=19}{)}，P2(x2,y2)\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{y}_{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{)}，记 η=(ax1+by1+c)(ax2+by2+c)\htmlData{tutor-start=0,tutor-end=5}{\eta }\htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{x}_{\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{b}\htmlData{tutor-start=16,tutor-end=17}{y}_{\htmlData{tutor-start=19,tutor-end=20}{1}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{c}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{a}\htmlData{tutor-start=26,tutor-end=27}{x}_{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{b}\htmlData{tutor-start=33,tutor-end=34}{y}_{\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=39}{+}\htmlData{tutor-start=39,tutor-end=40}{c}\htmlData{tutor-start=40,tutor-end=41}{)}. 若 η<0\htmlData{tutor-start=0,tutor-end=5}{\eta }\htmlData{tutor-start=5,tutor-end=6}{<} \htmlData{tutor-start=7,tutor-end=8}{0}，则称点 P1,P2\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{P}_{\htmlData{tutor-start=10,tutor-end=11}{2}} 被直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 分隔. 若曲线 C\htmlData{tutor-start=0,tutor-end=1}{C} 与直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 没有公共点，且曲线 C\htmlData{tutor-start=0,tutor-end=1}{C} 上存在点 P1,P2\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{P}_{\htmlData{tutor-start=10,tutor-end=11}{2}} 被直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 分隔，则称直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 为曲线 C\htmlData{tutor-start=0,tutor-end=1}{C} 的一条分隔线. (1) 求证：点 A(1,2),B(−1,0)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{B}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{)} 被直线 x+y−1=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0} 分隔； (2) 若直线 y=kx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x} 是曲线 x2−4y2=1\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{y}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{1} 的分隔线，求实数 k\htmlData{tutor-start=0,tutor-end=1}{k} 的取值范围； (3) 动点 M\htmlData{tutor-start=0,tutor-end=1}{M} 到点 Q(0,2)\htmlData{tutor-start=0,tutor-end=1}{Q}\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}{2}\htmlData{tutor-start=6,tutor-end=7}{)} 的距离与到 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴的距离之积为 1，设点 M\htmlData{tutor-start=0,tutor-end=1}{M} 的轨迹为 E\htmlData{tutor-start=0,tutor-end=1}{E}，求证：通过原点的直线中，有且仅有一条直线是 E\htmlData{tutor-start=0,tutor-end=1}{E} 的分割线.

答案：（1）乘积为 −4\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{4}；（2）∣k∣≥1/2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{k}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=6}{\ge}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}；（3）通过原点唯一的分隔线为 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴。

题目标签：曲线分隔线的存在与唯一性

解题过程

（1）按定义验证两点被分隔

证明点 A,B\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{B} 被给定直线分隔。

(1)
识别结构并建立关系

题目已经给出判定量 η\htmlData{tutor-start=0,tutor-end=4}{\eta}，证明只需把两个点代入直线左端并检验乘积是否为负。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“题目已经给出判定量 η\htmlData{tutor-start=0,tutor-end=4}{\eta}，证明只需把两个点代入直线左端并检验乘积是否为负。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：对直线 x+y−1=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{y}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{0}，点 A(1,2)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)} 代入得到 1+2−1=2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}\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}；点 B(−1,0)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)} 代入得到 −1+0−1=−2\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2}。二者乘积为 2×(−2)=−4<0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=7}{\times}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=15}{<}\htmlData{tutor-start=15,tutor-end=16}{0}，恰好满足分隔定义。

η=(1+2−1)(−1+0−1)=2⋅(−2)=−4<0\htmlData{tutor-start=0,tutor-end=4}{\eta}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=27}{\cdot}\htmlData{tutor-start=27,tutor-end=28}{(}\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}{=}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{4}\htmlData{tutor-start=34,tutor-end=35}{<}\htmlData{tutor-start=35,tutor-end=36}{0}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：两个代入值均非零且异号，所以两点分别位于直线两侧，而不是有点落在直线上，定义条件完全满足。

η=−4<0\boxed{\htmlData{tutor-start=7,tutor-end=11}{\eta}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=15}{<}\htmlData{tutor-start=15,tutor-end=16}{0}}

（2）求双曲线的分隔线斜率

求使 y=kx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x} 成为分隔线的 k\htmlData{tutor-start=0,tutor-end=1}{k}。

(1)
识别结构并建立关系

分隔线首先必须与曲线无公共点。把 y=kx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x} 代入双曲线即可由实数解条件确定范围；随后还要另找曲线上异侧两点验证定义的第二部分。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“分隔线首先必须与曲线无公共点。把 y=kx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x} 代入双曲线即可由实数解条件确定范围；随后还要另找曲线上异侧两点验证定义的第二部分。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：代入得 (1−4k2)x2=1\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{4}\htmlData{tutor-start=4,tutor-end=5}{k}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{x}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{1}。当 1−4k2>0\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{k}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{>}\htmlData{tutor-start=9,tutor-end=10}{0} 时有实交点，不能成为分隔线；无交点等价于 1−4k2≤0\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{k}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=11}{\le}\htmlData{tutor-start=11,tutor-end=12}{0}，即 ∣k∣≥1/2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{k}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=6}{\ge}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{2}。对这些 k\htmlData{tutor-start=0,tutor-end=1}{k}，曲线上的 (1,0)\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}{0}\htmlData{tutor-start=4,tutor-end=5}{)} 与 (−1,0)\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}{)} 代入直线函数 y−kx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x} 分别得 −k\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{k} 与 k\htmlData{tutor-start=0,tutor-end=1}{k}，乘积 −k2<0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{k}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{<}\htmlData{tutor-start=7,tutor-end=8}{0}，所以两点被该直线分隔。

(1−4k2)x2=1,∣k∣≥12\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}{4}\htmlData{tutor-start=4,tutor-end=5}{k}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{x}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{,}\qquad \htmlData{tutor-start=25,tutor-end=26}{|}\htmlData{tutor-start=26,tutor-end=27}{k}\htmlData{tutor-start=27,tutor-end=28}{|}\htmlData{tutor-start=28,tutor-end=31}{\ge}\frac{\htmlData{tutor-start=37,tutor-end=38}{1}}{\htmlData{tutor-start=40,tutor-end=41}{2}}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：边界 ∣k∣=1/2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{k}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2} 时方程化为 0=1\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，仍无交点；且此时 −k2<0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{k}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{<}\htmlData{tutor-start=7,tutor-end=8}{0}，故两个端点都应包含。

k∈(−∞,−12]∪[12,+∞)\boxed{\htmlData{tutor-start=7,tutor-end=8}{k}\htmlData{tutor-start=8,tutor-end=11}{\in}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=19}{\infty}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{-}\frac{\htmlData{tutor-start=27,tutor-end=28}{1}}{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{]}\htmlData{tutor-start=33,tutor-end=37}{\cup}\htmlData{tutor-start=37,tutor-end=38}{[}\frac{\htmlData{tutor-start=44,tutor-end=45}{1}}{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{,}\htmlData{tutor-start=50,tutor-end=51}{+}\htmlData{tutor-start=51,tutor-end=57}{\infty}\htmlData{tutor-start=57,tutor-end=58}{)}}

（3）证明过原点分隔线的唯一性

证明通过原点有且仅有一条直线是轨迹 E\htmlData{tutor-start=0,tutor-end=1}{E} 的分隔线。

(1)
识别结构并建立关系

先把距离乘积写成轨迹方程。过原点直线分为 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴和 y=kx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x} 两类：前者可直接验证分隔，后者只要证明必与 E\htmlData{tutor-start=0,tutor-end=1}{E} 相交就能全部排除。

为什么从这里入手：解析几何的目的，是把图形关系变成坐标等式。“先把距离乘积写成轨迹方程。过原点直线分为 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴和 y=kx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x} 两类：前者可直接验证分隔，后者只要证明必与 E\htmlData{tutor-start=0,tutor-end=1}{E} 相交就能全部排除。”提供了点、斜率、距离或焦点条件，先识别结构并建立关系后，交点、弦长和轨迹都能交给方程与韦达定理处理。

详细展开：设 M(x,y)\htmlData{tutor-start=0,tutor-end=1}{M}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{)}，由 ∣x∣x2+(y−2)2=1\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{|}\sqrt{\htmlData{tutor-start=9,tutor-end=10}{x}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{y}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{)}^{\htmlData{tutor-start=22,tutor-end=23}{2}}}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{1} 得 E:x2[x2+(y−2)2]=1\htmlData{tutor-start=0,tutor-end=1}{E}\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}{x}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{y}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{)}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{]}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{1}。当 x=0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0} 时左端为 0\htmlData{tutor-start=0,tutor-end=1}{0}，故 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴与 E\htmlData{tutor-start=0,tutor-end=1}{E} 无交点；而 (1,2),(−1,2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)}\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}{1}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)} 都在 E\htmlData{tutor-start=0,tutor-end=1}{E} 上且横坐标异号，所以 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴是分隔线。任取另一条过原点直线 y=kx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{x}，代入后令 g(x)=x2[x2+(kx−2)2]\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{x}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{[}\htmlData{tutor-start=11,tutor-end=12}{x}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{k}\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{)}^{\htmlData{tutor-start=25,tutor-end=26}{2}}\htmlData{tutor-start=27,tutor-end=28}{]}。函数 g\htmlData{tutor-start=0,tutor-end=1}{g} 连续，g(0)=0\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}，且 x→+∞\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\to}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty} 时 g(x)→+∞\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=7}{\to}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=14}{\infty}，由介值定理存在 x0>0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{>}\htmlData{tutor-start=6,tutor-end=7}{0} 使 g(x0)=1\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=5,tutor-end=6}{0}}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}，故该直线必与 E\htmlData{tutor-start=0,tutor-end=1}{E} 相交，不能成为分隔线。

g(x)=x2[x2+(kx−2)2],g(0)=0,lim⁡x→+∞g(x)=+∞\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{x}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{[}\htmlData{tutor-start=11,tutor-end=12}{x}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{+}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{k}\htmlData{tutor-start=19,tutor-end=20}{x}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{)}^{\htmlData{tutor-start=25,tutor-end=26}{2}}\htmlData{tutor-start=27,tutor-end=28}{]}\htmlData{tutor-start=28,tutor-end=29}{,}\quad \htmlData{tutor-start=35,tutor-end=36}{g}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{0}\htmlData{tutor-start=38,tutor-end=39}{)}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{0}\htmlData{tutor-start=41,tutor-end=42}{,}\quad \lim_{\htmlData{tutor-start=54,tutor-end=55}{x}\htmlData{tutor-start=55,tutor-end=58}{\to}\htmlData{tutor-start=58,tutor-end=59}{+}\htmlData{tutor-start=59,tutor-end=65}{\infty}}\htmlData{tutor-start=66,tutor-end=67}{g}\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}{=}\htmlData{tutor-start=71,tutor-end=72}{+}\htmlData{tutor-start=72,tutor-end=78}{\infty}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：y\htmlData{tutor-start=0,tutor-end=1}{y} 轴满足“无交点且存在异侧点”两项定义；所有非纵直的过原点直线都因至少一个公共点被排除，故唯一性成立。

过原点唯一的分隔线是 x=0\boxed{\text{\htmlData{tutor-start=13,tutor-end=14}{过}\htmlData{tutor-start=14,tutor-end=15}{原}\htmlData{tutor-start=15,tutor-end=16}{点}\htmlData{tutor-start=16,tutor-end=17}{唯}\htmlData{tutor-start=17,tutor-end=18}{一}\htmlData{tutor-start=18,tutor-end=19}{的}\htmlData{tutor-start=19,tutor-end=20}{分}\htmlData{tutor-start=20,tutor-end=21}{隔}\htmlData{tutor-start=21,tutor-end=22}{线}\htmlData{tutor-start=22,tutor-end=23}{是} }\htmlData{tutor-start=25,tutor-end=26}{x}\htmlData{tutor-start=26,tutor-end=27}{=}\htmlData{tutor-start=27,tutor-end=28}{0}}
23

解答题 · 数列不等式

已知数列 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 满足 13an⩽an+1⩽3an,n∈N∗,a1=1\frac{1}{3}a_{n} \leqslant a_{n+1} \leqslant 3a_{n}, n \in \mathbf{N}^*, a_{1}=1. (1) 若 a2=2,a3=x,a4=9\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{3}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{a}_{\htmlData{tutor-start=21,tutor-end=22}{4}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{9}，求 x\htmlData{tutor-start=0,tutor-end=1}{x} 的取值范围； (2) 设 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 是公比为 q\htmlData{tutor-start=0,tutor-end=1}{q} 的等比数列，Sn=a1+a2+⋯+an\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}{a}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{+}\cdots\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{a}_{\htmlData{tutor-start=30,tutor-end=31}{n}}. 若 13Sn⩽Sn+1⩽3Sn,n∈N∗\frac{1}{3}S_{n} \leqslant S_{n+1} \leqslant 3S_{n}, n \in \mathbf{N}^*，求 q\htmlData{tutor-start=0,tutor-end=1}{q} 的取值范围； (3) 若 a1,a2,⋯,ak\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=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \cdots\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{a}_{\htmlData{tutor-start=25,tutor-end=26}{k}} 成等差数列，且 a1+a2+⋯+ak=1000\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\cdots\htmlData{tutor-start=18,tutor-end=19}{+}\htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{k}}\htmlData{tutor-start=24,tutor-end=25}{=}\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}，求正整数 k\htmlData{tutor-start=0,tutor-end=1}{k} 的最大值，以及 k\htmlData{tutor-start=0,tutor-end=1}{k} 取最大值时相应数列 a1,a2,⋯,ak\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=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \cdots\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{a}_{\htmlData{tutor-start=25,tutor-end=26}{k}} 的公差.

答案：（1）x∈[3,6]\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}{3}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{6}\htmlData{tutor-start=8,tutor-end=9}{]}；（2）q∈[1/3,2]\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=4}{\in}\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}{3}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{]}；（3）kmax⁡=1999\htmlData{tutor-start=0,tutor-end=1}{k}_{\max}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{9}\htmlData{tutor-start=11,tutor-end=12}{9}\htmlData{tutor-start=12,tutor-end=13}{9}，d=−1/1999\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{9}\htmlData{tutor-start=8,tutor-end=9}{9}。

题目标签：倍数约束下的等比与等差数列

解题过程

（1）由相邻两组约束夹住 x\htmlData{tutor-start=0,tutor-end=1}{x}

求 a3=x\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}{x} 的范围。

(1)
识别结构并建立关系

x\htmlData{tutor-start=0,tutor-end=1}{x} 同时与前一项 a2=2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2} 和后一项 a4=9\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{9} 相邻，必须分别在 n=2,3\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{3} 写出不等式后取交集。

为什么从这里入手：集合题不能只看式子的外形，必须回到“元素是否满足条件”。这里的“x\htmlData{tutor-start=0,tutor-end=1}{x} 同时与前一项 a2=2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2} 和后一项 a4=9\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{4}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{9} 相邻，必须分别在 n=2,3\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{3} 写出不等式后取交集。”给出了元素筛选规则，因此先识别结构并建立关系，再逐项保留或删除元素，思路最直接。

详细展开：由 n=2\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2} 得 2/3≤x≤6\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=7}{\le }\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=11}{\le}\htmlData{tutor-start=11,tutor-end=12}{6}。由 n=3\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3} 得 x/3≤9≤3x\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=6}{\le}\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=10}{\le}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{x}，即 3≤x≤27\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{7}。两组条件同时成立，所以 x∈[2/3,6]∩[3,27]=[3,6]\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}{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}{6}\htmlData{tutor-start=10,tutor-end=11}{]}\htmlData{tutor-start=11,tutor-end=15}{\cap}\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}{7}\htmlData{tutor-start=20,tutor-end=21}{]}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{[}\htmlData{tutor-start=23,tutor-end=24}{3}\htmlData{tutor-start=24,tutor-end=25}{,}\htmlData{tutor-start=25,tutor-end=26}{6}\htmlData{tutor-start=26,tutor-end=27}{]}。

23≤x≤6,3≤x≤27\frac{\htmlData{tutor-start=6,tutor-end=7}{2}}{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=15}{\le }\htmlData{tutor-start=15,tutor-end=16}{x}\htmlData{tutor-start=16,tutor-end=19}{\le}\htmlData{tutor-start=19,tutor-end=20}{6}\htmlData{tutor-start=20,tutor-end=21}{,}\qquad \htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=33}{\le }\htmlData{tutor-start=33,tutor-end=34}{x}\htmlData{tutor-start=34,tutor-end=37}{\le}\htmlData{tutor-start=37,tutor-end=38}{2}\htmlData{tutor-start=38,tutor-end=39}{7}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：端点 x=3\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3} 与 x=6\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{6} 逐项代入均满足四个非严格不等式，因此闭区间两端都保留。

x∈[3,6]\boxed{\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=11}{\in}\htmlData{tutor-start=11,tutor-end=12}{[}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{6}\htmlData{tutor-start=15,tutor-end=16}{]}}

（2）求同时约束项与部分和的公比

求等比数列公比 q\htmlData{tutor-start=0,tutor-end=1}{q} 的范围。

(1)
识别结构并建立关系

原相邻项条件先直接限制公比；再利用 Sn+1=Sn+qn\htmlData{tutor-start=0,tutor-end=1}{S}_{\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}{S}_{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{q}^{\htmlData{tutor-start=17,tutor-end=18}{n}}，其中部分和下界自动成立，最关键的上界可先看 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 并证明其充分性。

为什么从这里入手：逻辑题必须先拆成可独立判断的原子命题。“原相邻项条件先直接限制公比；再利用 Sn+1=Sn+qn\htmlData{tutor-start=0,tutor-end=1}{S}_{\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}{S}_{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{q}^{\htmlData{tutor-start=17,tutor-end=18}{n}}，其中部分和下界自动成立，最关键的上界可先看 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 并证明其充分性。”给出了真值判断依据，先识别结构并建立关系，再代入且、或、非或充分必要关系，能避免被复合句式干扰。

详细展开：由 an+1=qan\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}{q}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{n}} 及 a1=1>0\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{0}，原条件给 1/3≤q≤3\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=7}{\le }\htmlData{tutor-start=7,tutor-end=8}{q}\htmlData{tutor-start=8,tutor-end=11}{\le}\htmlData{tutor-start=11,tutor-end=12}{3}，故各项和 Sn>0\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{>}\htmlData{tutor-start=6,tutor-end=7}{0}。于是 Sn+1>Sn>13Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\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}{S}_{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=14}{>}\frac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{3}}\htmlData{tutor-start=25,tutor-end=26}{S}_{\htmlData{tutor-start=28,tutor-end=29}{n}}，下界自动满足。上界在 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 时给 S2=1+q≤3S1=3\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{q}\htmlData{tutor-start=9,tutor-end=12}{\le}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{S}_{\htmlData{tutor-start=16,tutor-end=17}{1}}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{3}，故必要条件为 q≤2\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{2}。反之若 1/3≤q≤1\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=7}{\le }\htmlData{tutor-start=7,tutor-end=8}{q}\htmlData{tutor-start=8,tutor-end=11}{\le}\htmlData{tutor-start=11,tutor-end=12}{1}，有 qn≤1≤2Sn\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=8}{\le}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=12}{\le}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{S}_{\htmlData{tutor-start=16,tutor-end=17}{n}}；若 1<q≤2\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{q}\htmlData{tutor-start=3,tutor-end=6}{\le}\htmlData{tutor-start=6,tutor-end=7}{2}，有 qn=q⋅qn−1≤2qn−1≤2Sn\htmlData{tutor-start=0,tutor-end=1}{q}^{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{q}\htmlData{tutor-start=7,tutor-end=13}{\cdot }\htmlData{tutor-start=13,tutor-end=14}{q}^{\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=23}{\le}\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{q}^{\htmlData{tutor-start=27,tutor-end=28}{n}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}}\htmlData{tutor-start=31,tutor-end=34}{\le}\htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{S}_{\htmlData{tutor-start=38,tutor-end=39}{n}}。所以总有 Sn+1=Sn+qn≤3Sn\htmlData{tutor-start=0,tutor-end=1}{S}_{\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}{S}_{\htmlData{tutor-start=11,tutor-end=12}{n}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{q}^{\htmlData{tutor-start=17,tutor-end=18}{n}}\htmlData{tutor-start=19,tutor-end=22}{\le}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{S}_{\htmlData{tutor-start=26,tutor-end=27}{n}}。

13≤q≤2\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}}\htmlData{tutor-start=11,tutor-end=15}{\le }\htmlData{tutor-start=15,tutor-end=16}{q}\htmlData{tutor-start=16,tutor-end=19}{\le}\htmlData{tutor-start=19,tutor-end=20}{2}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：必要条件由 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 得到，充分性已覆盖 q≤1\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{1} 与 q>1\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{1} 两段；端点 q=1/3,2\htmlData{tutor-start=0,tutor-end=1}{q}\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}{3}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{2} 均满足非严格不等式。

q∈[13,2]\boxed{\htmlData{tutor-start=7,tutor-end=8}{q}\htmlData{tutor-start=8,tutor-end=11}{\in}\left[\frac{\htmlData{tutor-start=23,tutor-end=24}{1}}{\htmlData{tutor-start=26,tutor-end=27}{3}}\htmlData{tutor-start=28,tutor-end=29}{,}\htmlData{tutor-start=29,tutor-end=30}{2}\right]}

（3）在固定总和下最大化等差项数

求最大正整数 k\htmlData{tutor-start=0,tutor-end=1}{k} 及对应公差。

(1)
识别结构并建立关系

等差通项和求和公式把公差表示为 k\htmlData{tutor-start=0,tutor-end=1}{k} 的函数。最大 k\htmlData{tutor-start=0,tutor-end=1}{k} 必超过 1000\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{0}，此时公差为负，原相邻项下界在靠后的项最严格，可据此给 k\htmlData{tutor-start=0,tutor-end=1}{k} 设置上限。

为什么从这里入手：数列题要先识别相邻项之间保持不变的结构。“等差通项和求和公式把公差表示为 k\htmlData{tutor-start=0,tutor-end=1}{k} 的函数。最大 k\htmlData{tutor-start=0,tutor-end=1}{k} 必超过 1000\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{0}，此时公差为负，原相邻项下界在靠后的项最严格，可据此给 k\htmlData{tutor-start=0,tutor-end=1}{k} 设置上限。”提示了差、比或可累加关系，先识别结构并建立关系能把第 n\htmlData{tutor-start=0,tutor-end=1}{n} 项问题改写成已经会处理的等差、等比或裂项模型。

详细展开：设 an=1+(n−1)d\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{d}。由 Sk=k2[2+(k−1)d]=1000\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{k}}\htmlData{tutor-start=5,tutor-end=6}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{k}}{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{[}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{k}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{d}\htmlData{tutor-start=26,tutor-end=27}{]}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{0} 得 d=2(1000−k)/[k(k−1)]\htmlData{tutor-start=0,tutor-end=1}{d}\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}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{k}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{[}\htmlData{tutor-start=13,tutor-end=14}{k}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{k}\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}{]}。为求最大值考虑 k>1000\htmlData{tutor-start=0,tutor-end=1}{k}\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}{0}，此时 d<0\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0}。对 n=1,…,k−1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=10}{\ldots}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{k}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{1}，原下界等价于 2+(2n+1)d≥0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{n}\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}{d}\htmlData{tutor-start=9,tutor-end=12}{\ge}\htmlData{tutor-start=12,tutor-end=13}{0}，最严格的是 n=k−1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}，即 d≥−2/(2k−1)\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{k}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{)}。代入 d\htmlData{tutor-start=0,tutor-end=1}{d} 并整理得 k2−2000k+1000≤0\htmlData{tutor-start=0,tutor-end=1}{k}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{0}\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=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=19}{\le}\htmlData{tutor-start=19,tutor-end=20}{0}。该二次式在 k≥1000\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{0} 递增，且 k=1999\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{9}\htmlData{tutor-start=4,tutor-end=5}{9}\htmlData{tutor-start=5,tutor-end=6}{9} 时为 −999\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{9}\htmlData{tutor-start=2,tutor-end=3}{9}\htmlData{tutor-start=3,tutor-end=4}{9}，k=2000\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0} 时为 1000\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{0}，故整数 k≤1999\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=4}{\le}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{9}\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{9}。取 k=1999\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{9}\htmlData{tutor-start=4,tutor-end=5}{9}\htmlData{tutor-start=5,tutor-end=6}{9} 时 d=−1/1999\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{/}\htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{9}\htmlData{tutor-start=7,tutor-end=8}{9}\htmlData{tutor-start=8,tutor-end=9}{9}，求和正好为 1000\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{0}。

d=2(1000−k)k(k−1),kmax⁡=1999,d=−11999d=\frac{2(1000-k)}{k(k-1)},\qquad k_{\max}=1999,\quad d=-\frac1{1999}
(2)
完成计算并核验结论

上一阶段已经得到可直接求值、判号或证明的核心关系。

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

详细展开：当 k=1999,d=−1/1999\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{9}\htmlData{tutor-start=4,tutor-end=5}{9}\htmlData{tutor-start=5,tutor-end=6}{9}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{d}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{9}\htmlData{tutor-start=14,tutor-end=15}{9}\htmlData{tutor-start=15,tutor-end=16}{9} 时，对所有 1≤n≤1998\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=5}{\le }\htmlData{tutor-start=5,tutor-end=6}{n}\htmlData{tutor-start=6,tutor-end=9}{\le}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{9}\htmlData{tutor-start=11,tutor-end=12}{9}\htmlData{tutor-start=12,tutor-end=13}{8}，两族条件 2+(2n+1)d≥0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{n}\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}{d}\htmlData{tutor-start=9,tutor-end=12}{\ge}\htmlData{tutor-start=12,tutor-end=13}{0}、2+(2n−3)d≥0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{n}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{d}\htmlData{tutor-start=9,tutor-end=12}{\ge}\htmlData{tutor-start=12,tutor-end=13}{0} 在最大 n\htmlData{tutor-start=0,tutor-end=1}{n} 处仍分别为 1/1999\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{9}\htmlData{tutor-start=4,tutor-end=5}{9}\htmlData{tutor-start=5,tutor-end=6}{9}、5/1999\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{9}\htmlData{tutor-start=4,tutor-end=5}{9}\htmlData{tutor-start=5,tutor-end=6}{9}，故全程可行；k=2000\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0} 已违反必要条件。

kmax⁡=1999,d=−11999\boxed{k_{\max}=1999,\qquad d=-\frac1{1999}}