返回特征解读

2008 年高考数学（上海卷理科）

exams_raw/普通高考/2008/2008上海理.pdf · HS-MATH-1024-v2.1-solution-aware

21 题27 个小问/题组
1

一、填空题 · 代数

不等式 ∣x−1∣<1\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{1} 的解集是______.

答案：(0,2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)}

题目标签：绝对值不等式的求解

解题过程

求解绝对值不等式

求出满足 ∣x−1∣<1\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{1} 的 x\htmlData{tutor-start=0,tutor-end=1}{x} 的取值范围

(1)
去绝对值符号

根据绝对值的几何意义或代数定义，将绝对值不等式转化为普通线性不等式组。

详细展开： 由 ∣x−1∣<1\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{|}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{1}，根据绝对值不等式 ∣u∣<a  ⟺  −a<u<a\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{u}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{<}\htmlData{tutor-start=4,tutor-end=5}{a} \iff \htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{a} \htmlData{tutor-start=14,tutor-end=15}{<} \htmlData{tutor-start=16,tutor-end=17}{u} \htmlData{tutor-start=18,tutor-end=19}{<} \htmlData{tutor-start=20,tutor-end=21}{a}（其中 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}），可得： −1<x−1<1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1} \htmlData{tutor-start=9,tutor-end=10}{<} \htmlData{tutor-start=11,tutor-end=12}{1} 这一步是将含有绝对值的复杂关系转化为我们熟悉的一次不等式链。

−1<x−1<1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1} \htmlData{tutor-start=9,tutor-end=10}{<} \htmlData{tutor-start=11,tutor-end=12}{1}
(2)
求解线性不等式并写出区间

对不等式链各部分同时加 1，求出 x\htmlData{tutor-start=0,tutor-end=1}{x} 的范围，并用区间表示。

详细展开： 在不等式 −1<x−1<1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1} \htmlData{tutor-start=9,tutor-end=10}{<} \htmlData{tutor-start=11,tutor-end=12}{1} 的各部分同时加上 1： −1+1<x<1+1\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{1} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{<} \htmlData{tutor-start=13,tutor-end=14}{1} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{1} 0<x<2\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{2} 因此，解集为开区间 (0,2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)}。注意题目要求填写解集，通常用集合或区间表示，填空题中区间表示更为常见且简洁。

0<x<2\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{2}
2

一、填空题 · 组合数学

若集合 A={x∣x⩽2}\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=11}{\mid }\htmlData{tutor-start=11,tutor-end=12}{x} \htmlData{tutor-start=13,tutor-end=23}{\leqslant }\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=26}{\}}、B={x∣x⩾a}\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=11}{\mid }\htmlData{tutor-start=11,tutor-end=12}{x} \htmlData{tutor-start=13,tutor-end=23}{\geqslant }\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=26}{\}} 满足 A∩B={2}\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=13}{\{}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=16}{\}}，则实数 a=\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}______.

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

题目标签：集合交集与参数求解

解题过程

根据交集条件确定参数

利用 A∩B={2}\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=13}{\{}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=16}{\}} 确定实数 a\htmlData{tutor-start=0,tutor-end=1}{a} 的值

(1)
分析集合 A 和 B 的结构

明确集合 A\htmlData{tutor-start=0,tutor-end=1}{A} 和 B\htmlData{tutor-start=0,tutor-end=1}{B} 在数轴上的表示，并分析其交集的构成。

详细展开： 集合 A={x∣x⩽2}\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=13}{\mid }\htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=25}{\leqslant }\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=28}{\}} 表示数轴上 2 及其左侧的所有点，即区间 (−∞,2]\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{]}。 集合 B={x∣x⩾a}\htmlData{tutor-start=0,tutor-end=1}{B} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=13}{\mid }\htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=25}{\geqslant }\htmlData{tutor-start=25,tutor-end=26}{a}\htmlData{tutor-start=26,tutor-end=28}{\}} 表示数轴上 a\htmlData{tutor-start=0,tutor-end=1}{a} 及其右侧的所有点，即区间 [a,+∞)\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=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}。 它们的交集 A∩B\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B} 是这两个区间的公共部分。

A=(−∞,2],B=[a,+∞)\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=12}{\infty}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{]}\htmlData{tutor-start=16,tutor-end=17}{,} \quad \htmlData{tutor-start=24,tutor-end=25}{B} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{[}\htmlData{tutor-start=29,tutor-end=30}{a}\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=39}{\infty}\htmlData{tutor-start=39,tutor-end=40}{)}
(2)
利用交集结果反推参数 a

根据已知交集为单元素集合 {2}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=5}{\}}，推导 a\htmlData{tutor-start=0,tutor-end=1}{a} 的具体数值。

详细展开： 若 a<2\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{2}，则 A∩B=[a,2]\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{[}\htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{]}，这是一个包含无穷多个元素的区间，不符合题意。 若 a>2\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{2}，则 A\htmlData{tutor-start=0,tutor-end=1}{A} 与 B\htmlData{tutor-start=0,tutor-end=1}{B} 没有公共部分，A∩B=∅\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=20}{\emptyset}，不符合题意。 若 a=2\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}，则 B=[2,+∞)\htmlData{tutor-start=0,tutor-end=1}{B} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{[}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=15}{\infty}\htmlData{tutor-start=15,tutor-end=16}{)}。此时 A∩B=(−∞,2]∩[2,+∞)={x∣x⩽2 且 x⩾2}={2}\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B} \htmlData{tutor-start=9,tutor-end=10}{=} \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=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{]} \htmlData{tutor-start=24,tutor-end=29}{\cap }\htmlData{tutor-start=29,tutor-end=30}{[}\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{,} \htmlData{tutor-start=33,tutor-end=34}{+}\htmlData{tutor-start=34,tutor-end=40}{\infty}\htmlData{tutor-start=40,tutor-end=41}{)} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=46}{\{}\htmlData{tutor-start=46,tutor-end=47}{x} \htmlData{tutor-start=48,tutor-end=53}{\mid }\htmlData{tutor-start=53,tutor-end=54}{x} \htmlData{tutor-start=55,tutor-end=65}{\leqslant }\htmlData{tutor-start=65,tutor-end=66}{2} \text{ \htmlData{tutor-start=74,tutor-end=75}{且} } \htmlData{tutor-start=78,tutor-end=79}{x} \htmlData{tutor-start=80,tutor-end=90}{\geqslant }\htmlData{tutor-start=90,tutor-end=91}{2}\htmlData{tutor-start=91,tutor-end=93}{\}} \htmlData{tutor-start=94,tutor-end=95}{=} \htmlData{tutor-start=96,tutor-end=98}{\{}\htmlData{tutor-start=98,tutor-end=99}{2}\htmlData{tutor-start=99,tutor-end=101}{\}}。 这与题设 A∩B={2}\htmlData{tutor-start=0,tutor-end=1}{A} \htmlData{tutor-start=2,tutor-end=7}{\cap }\htmlData{tutor-start=7,tutor-end=8}{B} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=13}{\{}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=16}{\}} 完全一致。 因此，实数 a\htmlData{tutor-start=0,tutor-end=1}{a} 必须等于 2。

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

一、填空题 · 复数代数运算

若复数 z\htmlData{tutor-start=0,tutor-end=1}{z} 满足 z=i(2−z)\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{z}\htmlData{tutor-start=18,tutor-end=19}{)} (i\mathrm{\htmlData{tutor-start=8,tutor-end=9}{i}} 是虚数单位)，则 z=\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}______.

答案：1+i

题目标签：复数方程求解

解题过程

解复数方程

从方程 z=i(2−z)\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{z}\htmlData{tutor-start=18,tutor-end=19}{)} 中解出复数 z\htmlData{tutor-start=0,tutor-end=1}{z}

(1)
整理方程分离未知数

本步旨在通过代数变形，将含有未知数 z\htmlData{tutor-start=0,tutor-end=1}{z} 的项移至等式一侧，常数项移至另一侧，从而构造出 z\htmlData{tutor-start=0,tutor-end=1}{z} 的系数形式。

详细展开： 原方程为 z=i(2−z)\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{z}\htmlData{tutor-start=18,tutor-end=19}{)}。 首先展开右边括号：z=2i−iz\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\mathrm{\htmlData{tutor-start=13,tutor-end=14}{i}} \htmlData{tutor-start=16,tutor-end=17}{-} \mathrm{\htmlData{tutor-start=26,tutor-end=27}{i}}\htmlData{tutor-start=28,tutor-end=29}{z}。 将含有 z\htmlData{tutor-start=0,tutor-end=1}{z} 的项移到左边：z+iz=2i\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{+} \mathrm{\htmlData{tutor-start=12,tutor-end=13}{i}}\htmlData{tutor-start=14,tutor-end=15}{z} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{2}\mathrm{\htmlData{tutor-start=27,tutor-end=28}{i}}。 提取公因式 z\htmlData{tutor-start=0,tutor-end=1}{z}：(1+i)z=2i\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{z} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{2}\mathrm{\htmlData{tutor-start=27,tutor-end=28}{i}}。

(1+i)z=2i\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{z} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{2}\mathrm{\htmlData{tutor-start=27,tutor-end=28}{i}}
(2)
执行复数除法运算

本步通过两边同时除以 z\htmlData{tutor-start=0,tutor-end=1}{z} 的系数 (1+i)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}，并利用分母实数化技巧，计算出 z\htmlData{tutor-start=0,tutor-end=1}{z} 的具体值。

详细展开： 由上一步得 z=2i1+i\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{2}\mathrm{\htmlData{tutor-start=19,tutor-end=20}{i}}}{\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{+}\mathrm{\htmlData{tutor-start=33,tutor-end=34}{i}}}。 为了简化分式，分子分母同时乘以分母的共轭复数 (1−i)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}： z=2i(1−i)(1+i)(1−i)\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{2}\mathrm{\htmlData{tutor-start=19,tutor-end=20}{i}}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{-}\mathrm{\htmlData{tutor-start=32,tutor-end=33}{i}}\htmlData{tutor-start=34,tutor-end=35}{)}}{\htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{1}\htmlData{tutor-start=39,tutor-end=40}{+}\mathrm{\htmlData{tutor-start=48,tutor-end=49}{i}}\htmlData{tutor-start=50,tutor-end=51}{)}\htmlData{tutor-start=51,tutor-end=52}{(}\htmlData{tutor-start=52,tutor-end=53}{1}\htmlData{tutor-start=53,tutor-end=54}{-}\mathrm{\htmlData{tutor-start=62,tutor-end=63}{i}}\htmlData{tutor-start=64,tutor-end=65}{)}} 计算分母：(1+i)(1−i)=12−i2=1−(−1)=2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{+}\mathrm{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{-}\mathrm{\htmlData{tutor-start=25,tutor-end=26}{i}}\htmlData{tutor-start=27,tutor-end=28}{)} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{1}^{\htmlData{tutor-start=34,tutor-end=35}{2}} \htmlData{tutor-start=37,tutor-end=38}{-} \mathrm{\htmlData{tutor-start=47,tutor-end=48}{i}}^{\htmlData{tutor-start=51,tutor-end=52}{2}} \htmlData{tutor-start=54,tutor-end=55}{=} \htmlData{tutor-start=56,tutor-end=57}{1} \htmlData{tutor-start=58,tutor-end=59}{-} \htmlData{tutor-start=60,tutor-end=61}{(}\htmlData{tutor-start=61,tutor-end=62}{-}\htmlData{tutor-start=62,tutor-end=63}{1}\htmlData{tutor-start=63,tutor-end=64}{)} \htmlData{tutor-start=65,tutor-end=66}{=} \htmlData{tutor-start=67,tutor-end=68}{2}。 计算分子：2i(1−i)=2i−2i2=2i−2(−1)=2+2i\htmlData{tutor-start=0,tutor-end=1}{2}\mathrm{\htmlData{tutor-start=9,tutor-end=10}{i}}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\mathrm{\htmlData{tutor-start=22,tutor-end=23}{i}}\htmlData{tutor-start=24,tutor-end=25}{)} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{2}\mathrm{\htmlData{tutor-start=37,tutor-end=38}{i}} \htmlData{tutor-start=40,tutor-end=41}{-} \htmlData{tutor-start=42,tutor-end=43}{2}\mathrm{\htmlData{tutor-start=51,tutor-end=52}{i}}^{\htmlData{tutor-start=55,tutor-end=56}{2}} \htmlData{tutor-start=58,tutor-end=59}{=} \htmlData{tutor-start=60,tutor-end=61}{2}\mathrm{\htmlData{tutor-start=69,tutor-end=70}{i}} \htmlData{tutor-start=72,tutor-end=73}{-} \htmlData{tutor-start=74,tutor-end=75}{2}\htmlData{tutor-start=75,tutor-end=76}{(}\htmlData{tutor-start=76,tutor-end=77}{-}\htmlData{tutor-start=77,tutor-end=78}{1}\htmlData{tutor-start=78,tutor-end=79}{)} \htmlData{tutor-start=80,tutor-end=81}{=} \htmlData{tutor-start=82,tutor-end=83}{2} \htmlData{tutor-start=84,tutor-end=85}{+} \htmlData{tutor-start=86,tutor-end=87}{2}\mathrm{\htmlData{tutor-start=95,tutor-end=96}{i}}。 因此： z=2+2i2=1+i\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{2}\mathrm{\htmlData{tutor-start=21,tutor-end=22}{i}}}{\htmlData{tutor-start=25,tutor-end=26}{2}} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{+}\mathrm{\htmlData{tutor-start=40,tutor-end=41}{i}}

z=2i1+i=1+i\htmlData{tutor-start=0,tutor-end=1}{z} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{2}\mathrm{\htmlData{tutor-start=19,tutor-end=20}{i}}}{\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{+}\mathrm{\htmlData{tutor-start=33,tutor-end=34}{i}}} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{+}\mathrm{\htmlData{tutor-start=49,tutor-end=50}{i}}
4

一、填空题 · 函数与反函数

若函数 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 的反函数为 f−1(x)=x2\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=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{x}^{\htmlData{tutor-start=15,tutor-end=16}{2}} (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})，则 f(4)=\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}______.

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

题目标签：利用反函数定义求值

解题过程

利用反函数性质求解

已知 f−1(x)=x2(x>0)\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}{x}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{x}\htmlData{tutor-start=18,tutor-end=19}{>}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{)}，求 f(4)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{)} 的值

(1)
转化问题为求解反函数方程

本步利用原函数与反函数的定义关系，将求 f(4)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{)} 转化为求满足 f−1(y)=4\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}{y}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{4} 的 y\htmlData{tutor-start=0,tutor-end=1}{y} 值，或者直接设 f(4)=t\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{t} 转化为 f−1(t)=4\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}{t}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{4}。

详细展开： 设 f(4)=t\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{t}。 根据反函数的定义，若 f(a)=b\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{b}，则 f−1(b)=a\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}{b}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{a}。 在此题中，a=4,b=t\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{t}，故有 f−1(t)=4\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}{t}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{4}。 题目已知 f−1(x)=x2\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=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{x}^{\htmlData{tutor-start=15,tutor-end=16}{2}} (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})。这里的自变量是 t\htmlData{tutor-start=0,tutor-end=1}{t}，所以代入公式得： t2=4\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{4} 注意定义域限制：反函数 f−1(x)\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}{)} 的定义域是 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}，即输入值必须大于0。因为 t\htmlData{tutor-start=0,tutor-end=1}{t} 是 f−1\htmlData{tutor-start=0,tutor-end=1}{f}^{\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}} 的输入，所以必须满足 t>0\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}。

f−1(t)=t2=4,t>0\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}{t}\htmlData{tutor-start=8,tutor-end=9}{)} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{t}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{=} \htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{,} \quad \htmlData{tutor-start=29,tutor-end=30}{t}\htmlData{tutor-start=30,tutor-end=31}{>}\htmlData{tutor-start=31,tutor-end=32}{0}
(2)
解方程并确定最终值

本步解上述方程，并结合定义域筛选合法解。

详细展开： 由 t2=4\htmlData{tutor-start=0,tutor-end=1}{t}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{4}，解得 t=2\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2} 或 t=−2\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}。 根据前一步的分析，必须满足 t>0\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}。 因此，舍去 t=−2\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}，保留 t=2\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}。 所以，f(4)=2\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{2}。

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

一、填空题 · 平面向量

若向量 a\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}}、b\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}} 满足 ∣a∣=1\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{1}，∣b∣=2\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{2}，且 a\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} 与 b\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}} 的夹角为 π3\frac{\htmlData{tutor-start=6,tutor-end=9}{\pi}}{\htmlData{tutor-start=11,tutor-end=12}{3}}，则 ∣a+b∣=\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}\htmlData{tutor-start=31,tutor-end=32}{=}______.

答案：7\sqrt{\htmlData{tutor-start=6,tutor-end=7}{7}}

题目标签：向量模长的计算

解题过程

求解向量和的模

根据已知向量的模和夹角，计算 ∣a+b∣\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|} 的值

(1)
利用数量积定义求 a⋅b\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}}

首先根据向量数量积的定义，结合已知的模长和夹角，计算出两个向量的数量积。

详细展开： 由题意可知，∣a∣=1\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{1}，∣b∣=2\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{b}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{2}，且 a\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} 与 b\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}} 的夹角 θ=π3\htmlData{tutor-start=0,tutor-end=7}{\theta }\htmlData{tutor-start=7,tutor-end=8}{=} \frac{\htmlData{tutor-start=15,tutor-end=18}{\pi}}{\htmlData{tutor-start=20,tutor-end=21}{3}}。 根据向量数量积的定义公式： a⋅b=∣a∣∣b∣cos⁡θ\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{a}}\htmlData{tutor-start=53,tutor-end=54}{|} \htmlData{tutor-start=55,tutor-end=56}{|}\boldsymbol{\htmlData{tutor-start=68,tutor-end=69}{b}}\htmlData{tutor-start=70,tutor-end=71}{|} \cos \htmlData{tutor-start=77,tutor-end=83}{\theta} 代入数值进行计算： a⋅b=1×2×cos⁡π3=2×12=1\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{1} \htmlData{tutor-start=40,tutor-end=47}{\times }\htmlData{tutor-start=47,tutor-end=48}{2} \htmlData{tutor-start=49,tutor-end=56}{\times }\cos \frac{\htmlData{tutor-start=67,tutor-end=70}{\pi}}{\htmlData{tutor-start=72,tutor-end=73}{3}} \htmlData{tutor-start=75,tutor-end=76}{=} \htmlData{tutor-start=77,tutor-end=78}{2} \htmlData{tutor-start=79,tutor-end=86}{\times }\frac{\htmlData{tutor-start=92,tutor-end=93}{1}}{\htmlData{tutor-start=95,tutor-end=96}{2}} \htmlData{tutor-start=98,tutor-end=99}{=} \htmlData{tutor-start=100,tutor-end=101}{1} 因此，a⋅b=1\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{1}。

a⋅b=∣a∣∣b∣cos⁡π3=1\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{a}}\htmlData{tutor-start=53,tutor-end=54}{|} \htmlData{tutor-start=55,tutor-end=56}{|}\boldsymbol{\htmlData{tutor-start=68,tutor-end=69}{b}}\htmlData{tutor-start=70,tutor-end=71}{|} \cos \frac{\htmlData{tutor-start=83,tutor-end=86}{\pi}}{\htmlData{tutor-start=88,tutor-end=89}{3}} \htmlData{tutor-start=91,tutor-end=92}{=} \htmlData{tutor-start=93,tutor-end=94}{1}
(2)
利用模长平方公式求解 ∣a+b∣\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}

利用向量模长与数量积的关系式 ∣v∣2=v⋅v\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{v}}\htmlData{tutor-start=15,tutor-end=16}{|}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{=} \boldsymbol{\htmlData{tutor-start=35,tutor-end=36}{v}} \htmlData{tutor-start=38,tutor-end=44}{\cdot }\boldsymbol{\htmlData{tutor-start=56,tutor-end=57}{v}}，将待求的模长转化为代数运算，最后开方得到结果。

详细展开： 我们需要求 ∣a+b∣\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}。直接求模长较困难，通常先求其平方： ∣a+b∣2=(a+b)⋅(a+b)\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{(}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{a}}\htmlData{tutor-start=53,tutor-end=54}{+}\boldsymbol{\htmlData{tutor-start=66,tutor-end=67}{b}}\htmlData{tutor-start=68,tutor-end=69}{)} \htmlData{tutor-start=70,tutor-end=76}{\cdot }\htmlData{tutor-start=76,tutor-end=77}{(}\boldsymbol{\htmlData{tutor-start=89,tutor-end=90}{a}}\htmlData{tutor-start=91,tutor-end=92}{+}\boldsymbol{\htmlData{tutor-start=104,tutor-end=105}{b}}\htmlData{tutor-start=106,tutor-end=107}{)} 根据数量积的分配律展开： ∣a+b∣2=a⋅a+2a⋅b+b⋅b\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \boldsymbol{\htmlData{tutor-start=50,tutor-end=51}{a}} \htmlData{tutor-start=53,tutor-end=59}{\cdot }\boldsymbol{\htmlData{tutor-start=71,tutor-end=72}{a}} \htmlData{tutor-start=74,tutor-end=75}{+} \htmlData{tutor-start=76,tutor-end=77}{2}\boldsymbol{\htmlData{tutor-start=89,tutor-end=90}{a}} \htmlData{tutor-start=92,tutor-end=98}{\cdot }\boldsymbol{\htmlData{tutor-start=110,tutor-end=111}{b}} \htmlData{tutor-start=113,tutor-end=114}{+} \boldsymbol{\htmlData{tutor-start=127,tutor-end=128}{b}} \htmlData{tutor-start=130,tutor-end=136}{\cdot }\boldsymbol{\htmlData{tutor-start=148,tutor-end=149}{b}} 由于 a⋅a=∣a∣2\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{a}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{a}}\htmlData{tutor-start=53,tutor-end=54}{|}^{\htmlData{tutor-start=56,tutor-end=57}{2}} 且 b⋅b=∣b∣2\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{b}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{b}}\htmlData{tutor-start=53,tutor-end=54}{|}^{\htmlData{tutor-start=56,tutor-end=57}{2}}，故： ∣a+b∣2=∣a∣2+2a⋅b+∣b∣2\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{|}\boldsymbol{\htmlData{tutor-start=51,tutor-end=52}{a}}\htmlData{tutor-start=53,tutor-end=54}{|}^{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{+} \htmlData{tutor-start=61,tutor-end=62}{2}\boldsymbol{\htmlData{tutor-start=74,tutor-end=75}{a}} \htmlData{tutor-start=77,tutor-end=83}{\cdot }\boldsymbol{\htmlData{tutor-start=95,tutor-end=96}{b}} \htmlData{tutor-start=98,tutor-end=99}{+} \htmlData{tutor-start=100,tutor-end=101}{|}\boldsymbol{\htmlData{tutor-start=113,tutor-end=114}{b}}\htmlData{tutor-start=115,tutor-end=116}{|}^{\htmlData{tutor-start=118,tutor-end=119}{2}} 代入已知数据 ∣a∣=1,∣b∣=2\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{,} \htmlData{tutor-start=20,tutor-end=21}{|}\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}}\htmlData{tutor-start=35,tutor-end=36}{|}\htmlData{tutor-start=36,tutor-end=37}{=}\htmlData{tutor-start=37,tutor-end=38}{2} 以及上一步求得的 a⋅b=1\boldsymbol{\htmlData{tutor-start=12,tutor-end=13}{a}} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\boldsymbol{\htmlData{tutor-start=33,tutor-end=34}{b}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{1}： ∣a+b∣2=12+2×1+22=1+2+4=7\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{1}^{\htmlData{tutor-start=41,tutor-end=42}{2}} \htmlData{tutor-start=44,tutor-end=45}{+} \htmlData{tutor-start=46,tutor-end=47}{2} \htmlData{tutor-start=48,tutor-end=55}{\times }\htmlData{tutor-start=55,tutor-end=56}{1} \htmlData{tutor-start=57,tutor-end=58}{+} \htmlData{tutor-start=59,tutor-end=60}{2}^{\htmlData{tutor-start=62,tutor-end=63}{2}} \htmlData{tutor-start=65,tutor-end=66}{=} \htmlData{tutor-start=67,tutor-end=68}{1} \htmlData{tutor-start=69,tutor-end=70}{+} \htmlData{tutor-start=71,tutor-end=72}{2} \htmlData{tutor-start=73,tutor-end=74}{+} \htmlData{tutor-start=75,tutor-end=76}{4} \htmlData{tutor-start=77,tutor-end=78}{=} \htmlData{tutor-start=79,tutor-end=80}{7} 因为模长非负，所以： ∣a+b∣=7\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|} \htmlData{tutor-start=32,tutor-end=33}{=} \sqrt{\htmlData{tutor-start=40,tutor-end=41}{7}}

∣a+b∣=∣a∣2+2a⋅b+∣b∣2=7\htmlData{tutor-start=0,tutor-end=1}{|}\boldsymbol{\htmlData{tutor-start=13,tutor-end=14}{a}}\htmlData{tutor-start=15,tutor-end=16}{+}\boldsymbol{\htmlData{tutor-start=28,tutor-end=29}{b}}\htmlData{tutor-start=30,tutor-end=31}{|} \htmlData{tutor-start=32,tutor-end=33}{=} \sqrt{\htmlData{tutor-start=40,tutor-end=41}{|}\boldsymbol{\htmlData{tutor-start=53,tutor-end=54}{a}}\htmlData{tutor-start=55,tutor-end=56}{|}^{\htmlData{tutor-start=58,tutor-end=59}{2}} \htmlData{tutor-start=61,tutor-end=62}{+} \htmlData{tutor-start=63,tutor-end=64}{2}\boldsymbol{\htmlData{tutor-start=76,tutor-end=77}{a}} \htmlData{tutor-start=79,tutor-end=85}{\cdot }\boldsymbol{\htmlData{tutor-start=97,tutor-end=98}{b}} \htmlData{tutor-start=100,tutor-end=101}{+} \htmlData{tutor-start=102,tutor-end=103}{|}\boldsymbol{\htmlData{tutor-start=115,tutor-end=116}{b}}\htmlData{tutor-start=117,tutor-end=118}{|}^{\htmlData{tutor-start=120,tutor-end=121}{2}}} \htmlData{tutor-start=124,tutor-end=125}{=} \sqrt{\htmlData{tutor-start=132,tutor-end=133}{7}}
6

一、填空题 · 三角恒等变换

函数 f(x)=3sin⁡x+sin⁡(π2+x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\sin \htmlData{tutor-start=20,tutor-end=21}{x} \htmlData{tutor-start=22,tutor-end=23}{+} \sin\left(\frac{\htmlData{tutor-start=40,tutor-end=43}{\pi}}{\htmlData{tutor-start=45,tutor-end=46}{2}}\htmlData{tutor-start=47,tutor-end=48}{+}\htmlData{tutor-start=48,tutor-end=49}{x}\right) 的最大值是______.

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

题目标签：三角函数化简求最值

解题过程

化简函数解析式

将 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}{)} 化为 Asin⁡(ωx+ϕ)\htmlData{tutor-start=0,tutor-end=1}{A}\sin\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=13}{\omega }\htmlData{tutor-start=13,tutor-end=14}{x} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=21}{\phi}\htmlData{tutor-start=21,tutor-end=22}{)} 的形式并求最大值

(1)
利用诱导公式统一函数名

本步利用三角函数诱导公式将不同名的三角函数统一为正弦或余弦，以便使用辅助角公式。

详细展开： 原函数为 f(x)=3sin⁡x+sin⁡(π2+x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\sin \htmlData{tutor-start=20,tutor-end=21}{x} \htmlData{tutor-start=22,tutor-end=23}{+} \sin\left(\frac{\htmlData{tutor-start=40,tutor-end=43}{\pi}}{\htmlData{tutor-start=45,tutor-end=46}{2}}\htmlData{tutor-start=47,tutor-end=48}{+}\htmlData{tutor-start=48,tutor-end=49}{x}\right)。 根据诱导公式 sin⁡(π2+α)=cos⁡α\sin\left(\frac{\htmlData{tutor-start=16,tutor-end=19}{\pi}}{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=30}{\alpha}\right) \htmlData{tutor-start=38,tutor-end=39}{=} \cos\htmlData{tutor-start=44,tutor-end=50}{\alpha}，可得： sin⁡(π2+x)=cos⁡x\sin\left(\frac{\htmlData{tutor-start=16,tutor-end=19}{\pi}}{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{x}\right) \htmlData{tutor-start=33,tutor-end=34}{=} \cos \htmlData{tutor-start=40,tutor-end=41}{x} 因此，函数化简为： f(x)=3sin⁡x+cos⁡x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\sin \htmlData{tutor-start=20,tutor-end=21}{x} \htmlData{tutor-start=22,tutor-end=23}{+} \cos \htmlData{tutor-start=29,tutor-end=30}{x}

f(x)=3sin⁡x+cos⁡x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\sin \htmlData{tutor-start=20,tutor-end=21}{x} \htmlData{tutor-start=22,tutor-end=23}{+} \cos \htmlData{tutor-start=29,tutor-end=30}{x}
(2)
使用辅助角公式求最大值

本步应用辅助角公式 asin⁡x+bcos⁡x=a2+b2sin⁡(x+ϕ)\htmlData{tutor-start=0,tutor-end=1}{a}\sin \htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=9}{+} \htmlData{tutor-start=10,tutor-end=11}{b}\cos \htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=19}{=} \sqrt{\htmlData{tutor-start=26,tutor-end=27}{a}^{\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=35,tutor-end=36}{2}}}\sin\htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{x}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=49}{\phi}\htmlData{tutor-start=49,tutor-end=50}{)} 提取振幅，直接得出最大值。

详细展开： 对于 f(x)=3sin⁡x+1cos⁡x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\sin \htmlData{tutor-start=20,tutor-end=21}{x} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{1}\cos \htmlData{tutor-start=30,tutor-end=31}{x}，系数 a=3,b=1\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{3}}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}。 提取公因子 (3)2+12=3+1=2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{(}\sqrt{\htmlData{tutor-start=13,tutor-end=14}{3}}\htmlData{tutor-start=15,tutor-end=16}{)}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{+} \htmlData{tutor-start=23,tutor-end=24}{1}^{\htmlData{tutor-start=26,tutor-end=27}{2}}} \htmlData{tutor-start=30,tutor-end=31}{=} \sqrt{\htmlData{tutor-start=38,tutor-end=39}{3}\htmlData{tutor-start=39,tutor-end=40}{+}\htmlData{tutor-start=40,tutor-end=41}{1}} \htmlData{tutor-start=43,tutor-end=44}{=} \htmlData{tutor-start=45,tutor-end=46}{2}。 f(x)=2(32sin⁡x+12cos⁡x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{2} \left( \frac{\sqrt{\htmlData{tutor-start=28,tutor-end=29}{3}}}{\htmlData{tutor-start=32,tutor-end=33}{2}}\sin \htmlData{tutor-start=39,tutor-end=40}{x} \htmlData{tutor-start=41,tutor-end=42}{+} \frac{\htmlData{tutor-start=49,tutor-end=50}{1}}{\htmlData{tutor-start=52,tutor-end=53}{2}}\cos \htmlData{tutor-start=59,tutor-end=60}{x} \right) 寻找角 ϕ\htmlData{tutor-start=0,tutor-end=4}{\phi} 使得 cos⁡ϕ=32,sin⁡ϕ=12\cos\htmlData{tutor-start=4,tutor-end=9}{\phi }\htmlData{tutor-start=9,tutor-end=10}{=} \frac{\sqrt{\htmlData{tutor-start=23,tutor-end=24}{3}}}{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{,} \sin\htmlData{tutor-start=35,tutor-end=40}{\phi }\htmlData{tutor-start=40,tutor-end=41}{=} \frac{\htmlData{tutor-start=48,tutor-end=49}{1}}{\htmlData{tutor-start=51,tutor-end=52}{2}}，可知 ϕ=π6\htmlData{tutor-start=0,tutor-end=5}{\phi }\htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=16}{\pi}}{\htmlData{tutor-start=18,tutor-end=19}{6}}。 f(x)=2sin⁡(x+π6)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{2} \sin\left(\htmlData{tutor-start=19,tutor-end=20}{x} \htmlData{tutor-start=21,tutor-end=22}{+} \frac{\htmlData{tutor-start=29,tutor-end=32}{\pi}}{\htmlData{tutor-start=34,tutor-end=35}{6}}\right) 因为正弦函数 sin⁡(⋅)\sin\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=10}{\cdot}\htmlData{tutor-start=10,tutor-end=11}{)} 的最大值为 1，所以 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}{)} 的最大值为： 2×1=2\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{1} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{2}

f(x)max=2\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}_{\htmlData{tutor-start=6,tutor-end=7}{m}\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{x}} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{2}
7

一、填空题 · 平面几何与概率

在平面直角坐标系中，从六个点：A(0,0)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}、B(2,0)\htmlData{tutor-start=0,tutor-end=1}{B}\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}{)}、C(1,1)\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}、D(0,2)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)}、E(2,2)\htmlData{tutor-start=0,tutor-end=1}{E}\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}\htmlData{tutor-start=5,tutor-end=6}{)}、F(3,3)\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{)} 中任取三个，这三点能构成三角形的概率是______. (结果用分数表示)

答案：34\frac{\htmlData{tutor-start=6,tutor-end=7}{3}}{\htmlData{tutor-start=9,tutor-end=10}{4}}

题目标签：上海卷理科第 7 题

解题过程

计算构成三角形的概率

从6个点中任取3点，计算能构成三角形的概率

(1)
计算基本事件总数

首先确定从给定的6个点中任取3个点的所有可能组合数，这是古典概型的分母。

详细展开： 给定点集为 S={A(0,0),B(2,0),C(1,1),D(0,2),E(2,2),F(3,3)}\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,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}{0}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{C}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{,}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{)}\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=31}{D}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{0}\htmlData{tutor-start=33,tutor-end=34}{,}\htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{)}\htmlData{tutor-start=36,tutor-end=37}{,} \htmlData{tutor-start=38,tutor-end=39}{E}\htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=42}{,}\htmlData{tutor-start=42,tutor-end=43}{2}\htmlData{tutor-start=43,tutor-end=44}{)}\htmlData{tutor-start=44,tutor-end=45}{,} \htmlData{tutor-start=46,tutor-end=47}{F}\htmlData{tutor-start=47,tutor-end=48}{(}\htmlData{tutor-start=48,tutor-end=49}{3}\htmlData{tutor-start=49,tutor-end=50}{,}\htmlData{tutor-start=50,tutor-end=51}{3}\htmlData{tutor-start=51,tutor-end=52}{)}\htmlData{tutor-start=52,tutor-end=54}{\}}，共6个点。 从6个不同元素中取出3个元素的组合数为： N=C63=6×5×43×2×1=20\htmlData{tutor-start=0,tutor-end=1}{N} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{C}_{\htmlData{tutor-start=7,tutor-end=8}{6}}^{\htmlData{tutor-start=11,tutor-end=12}{3}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{6} \htmlData{tutor-start=24,tutor-end=31}{\times }\htmlData{tutor-start=31,tutor-end=32}{5} \htmlData{tutor-start=33,tutor-end=40}{\times }\htmlData{tutor-start=40,tutor-end=41}{4}}{\htmlData{tutor-start=43,tutor-end=44}{3} \htmlData{tutor-start=45,tutor-end=52}{\times }\htmlData{tutor-start=52,tutor-end=53}{2} \htmlData{tutor-start=54,tutor-end=61}{\times }\htmlData{tutor-start=61,tutor-end=62}{1}} \htmlData{tutor-start=64,tutor-end=65}{=} \htmlData{tutor-start=66,tutor-end=67}{2}\htmlData{tutor-start=67,tutor-end=68}{0} 因此，基本事件总数为 20。

C63=20\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{6}}^{\htmlData{tutor-start=7,tutor-end=8}{3}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}
(2)
分析三点共线的情况

三点不能构成三角形的充要条件是这三点共线。我们需要找出所有共线的三点组，并从总数中减去。

详细展开： 观察各点坐标： A(0,0),B(2,0)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{)} 在 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴上 (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})，但其他点 y\htmlData{tutor-start=0,tutor-end=1}{y} 坐标不为0，故只有 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} 两点在 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴，无法构成三点共线。 A(0,0),D(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}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)} 在 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴上 (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})，同理只有两点。 C(1,1),E(2,2),F(3,3)\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{F}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{3}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{3}\htmlData{tutor-start=21,tutor-end=22}{)} 均在直线 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x} 上。这是一组共线点。 检查其他可能的直线： - 直线 AC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}: A(0,0),C(1,1)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{C}\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}{)} 斜率 k=1\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，方程 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}。已包含 C,E,F\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{F}。B(2,0)\htmlData{tutor-start=0,tutor-end=1}{B}\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}{)} 不在，D(0,2)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)} 不在。 - 直线 BD\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D}: B(2,0),D(0,2)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)}，斜率 k=−1\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}，方程 y−0=−1(x−2)⇒y=−x+2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{0} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)} \htmlData{tutor-start=14,tutor-end=26}{\Rightarrow }\htmlData{tutor-start=26,tutor-end=27}{y} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{x}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{2}。检查其他点：C(1,1)\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} 满足 1=−1+2\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=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}{2}，故 B,C,D\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{D} 共线。A(0,0)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)} 不满足，E(2,2)\htmlData{tutor-start=0,tutor-end=1}{E}\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}\htmlData{tutor-start=5,tutor-end=6}{)} 不满足，F(3,3)\htmlData{tutor-start=0,tutor-end=1}{F}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{)} 不满足。 - 直线 AE\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E}: A(0,0),E(2,2)\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}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)} 即 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}，已讨论。 - 直线 BF\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{F}: B(2,0),F(3,3)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{)}，斜率 k=3\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}，方程 y=3(x−2)=3x−6\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{x}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{)}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{x}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{6}。检查 C(1,1):1≠−3\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{:} \htmlData{tutor-start=8,tutor-end=9}{1} \neq \htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{3}; D(0,2):2≠−6\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{:} \htmlData{tutor-start=8,tutor-end=9}{2} \neq \htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{6}; E(2,2):2≠0\htmlData{tutor-start=0,tutor-end=1}{E}\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}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{:} \htmlData{tutor-start=8,tutor-end=9}{2} \neq \htmlData{tutor-start=15,tutor-end=16}{0}。无第三点。 - 直线 DF\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{F}: D(0,2),F(3,3)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{)}，斜率 k=1/3\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}{/}\htmlData{tutor-start=4,tutor-end=5}{3}，方程 y−2=13x⇒x−3y+6=0\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2} \htmlData{tutor-start=4,tutor-end=5}{=} \frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{3}}\htmlData{tutor-start=17,tutor-end=18}{x} \htmlData{tutor-start=19,tutor-end=31}{\Rightarrow }\htmlData{tutor-start=31,tutor-end=32}{x}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{3}\htmlData{tutor-start=34,tutor-end=35}{y}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{6}\htmlData{tutor-start=37,tutor-end=38}{=}\htmlData{tutor-start=38,tutor-end=39}{0}。检查 C(1,1):1−3+6≠0\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{:} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{6} \neq \htmlData{tutor-start=19,tutor-end=20}{0}; E(2,2):2−6+6≠0\htmlData{tutor-start=0,tutor-end=1}{E}\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}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{:} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{6}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{6} \neq \htmlData{tutor-start=19,tutor-end=20}{0}; B(2,0):2+6≠0\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{:} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{6} \neq \htmlData{tutor-start=17,tutor-end=18}{0}; A(0,0):6≠0\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{:} \htmlData{tutor-start=8,tutor-end=9}{6} \neq \htmlData{tutor-start=15,tutor-end=16}{0}。无第三点。 - 直线 AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}: 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}，只有 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}。 - 直线 AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}: 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}，只有 A,D\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{D}。 - 直线 DE\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E}: D(0,2),E(2,2)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)}，水平线 y=2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}。只有 D,E\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}。 - 直线 BE\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}: B(2,0),E(2,2)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)}，竖直线 x=2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}。只有 B,E\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}。

综上，共线的三点组有： 1. {C,E,F}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=11}{\}} (在 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x} 上) 2. {B,C,D}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{C}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=11}{\}} (在 y=−x+2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2} 上)

共 2 组三点共线。

共线组数=2\text{\htmlData{tutor-start=6,tutor-end=7}{共}\htmlData{tutor-start=7,tutor-end=8}{线}\htmlData{tutor-start=8,tutor-end=9}{组}\htmlData{tutor-start=9,tutor-end=10}{数}} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{2}
(3)
计算概率并得出结论

利用对立事件概率公式计算最终结果。

详细展开： 不能构成三角形的情况数（即三点共线）为 M=2\htmlData{tutor-start=0,tutor-end=1}{M} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}。 能构成三角形的情况数为 N−M=20−2=18\htmlData{tutor-start=0,tutor-end=1}{N} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{M} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{0} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{2} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{8}。 所求概率为： P=1820=910\htmlData{tutor-start=0,tutor-end=1}{P} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{8}}{\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{0}} \htmlData{tutor-start=18,tutor-end=19}{=} \frac{\htmlData{tutor-start=26,tutor-end=27}{9}}{\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{0}}

等等，让我重新仔细检查共线情况。 点集：A(0,0),B(2,0),C(1,1),D(0,2),E(2,2),F(3,3)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{C}\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}{1}\htmlData{tutor-start=21,tutor-end=22}{)}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{D}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{,} \htmlData{tutor-start=32,tutor-end=33}{E}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{,}\htmlData{tutor-start=36,tutor-end=37}{2}\htmlData{tutor-start=37,tutor-end=38}{)}\htmlData{tutor-start=38,tutor-end=39}{,} \htmlData{tutor-start=40,tutor-end=41}{F}\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{3}\htmlData{tutor-start=43,tutor-end=44}{,}\htmlData{tutor-start=44,tutor-end=45}{3}\htmlData{tutor-start=45,tutor-end=46}{)}。 1. 直线 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}: 过 A(0,0),C(1,1),E(2,2),F(3,3)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{C}\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}{)}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{E}\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}{2}\htmlData{tutor-start=21,tutor-end=22}{)}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{F}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{3}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=30}{)}。这里有4个点共线！ 从这4个点中任取3个都共线。组合数为 C43=4\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{4}}^{\htmlData{tutor-start=7,tutor-end=8}{3}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{4} 组。 具体为：{A,C,E},{A,C,F},{A,E,F},{C,E,F}\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}{C}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{E}\htmlData{tutor-start=7,tutor-end=9}{\}}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=13}{\{}\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{C}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{F}\htmlData{tutor-start=18,tutor-end=20}{\}}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=24}{\{}\htmlData{tutor-start=24,tutor-end=25}{A}\htmlData{tutor-start=25,tutor-end=26}{,}\htmlData{tutor-start=26,tutor-end=27}{E}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{F}\htmlData{tutor-start=29,tutor-end=31}{\}}\htmlData{tutor-start=31,tutor-end=32}{,} \htmlData{tutor-start=33,tutor-end=35}{\{}\htmlData{tutor-start=35,tutor-end=36}{C}\htmlData{tutor-start=36,tutor-end=37}{,}\htmlData{tutor-start=37,tutor-end=38}{E}\htmlData{tutor-start=38,tutor-end=39}{,}\htmlData{tutor-start=39,tutor-end=40}{F}\htmlData{tutor-start=40,tutor-end=42}{\}}。 2. 直线 y=−x+2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2}: 过 B(2,0),C(1,1),D(0,2)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{C}\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}{)}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{D}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{)}。这里有3个点共线。 组合数为 C33=1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{3}}^{\htmlData{tutor-start=7,tutor-end=8}{3}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{1} 组。 具体为：{B,C,D}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{C}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}\htmlData{tutor-start=7,tutor-end=9}{\}}。

是否有其他共线？ 检查 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} 连线 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}: 仅 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}。 检查 A,D\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{D} 连线 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}: 仅 A,D\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{D}。 检查 D,E\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E} 连线 y=2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}: 仅 D,E\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}。 检查 B,E\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E} 连线 x=2\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}: 仅 B,E\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}。 检查 A,B,D,E\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}{D}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{E} 构成的矩形边界及对角线。 对角线 AE\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} 在 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x} 中已算。 对角线 BD\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{D} 在 y=−x+2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2} 中已算。 其他点对： B(2,0),F(3,3)→k=3\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=27}{\rightarrow }\htmlData{tutor-start=27,tutor-end=28}{k}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{3}. y=3x−6\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{6}. C(1,1)→1≠−3\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=19}{\rightarrow }\htmlData{tutor-start=19,tutor-end=20}{1} \neq \htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{3}. D(0,2)→2≠−6\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=19}{\rightarrow }\htmlData{tutor-start=19,tutor-end=20}{2} \neq \htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{6}. E(2,2)→2≠0\htmlData{tutor-start=0,tutor-end=1}{E}\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}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=19}{\rightarrow }\htmlData{tutor-start=19,tutor-end=20}{2} \neq \htmlData{tutor-start=26,tutor-end=27}{0}. A(0,0)→0≠−6\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=19}{\rightarrow }\htmlData{tutor-start=19,tutor-end=20}{0} \neq \htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{6}. 无。 D(0,2),F(3,3)→k=1/3\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=27}{\rightarrow }\htmlData{tutor-start=27,tutor-end=28}{k}\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}{3}. y=x/3+2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=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}. C(1,1)→1≠7/3\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=19}{\rightarrow }\htmlData{tutor-start=19,tutor-end=20}{1} \neq \htmlData{tutor-start=26,tutor-end=27}{7}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{3}. B(2,0)→0≠8/3\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=19}{\rightarrow }\htmlData{tutor-start=19,tutor-end=20}{0} \neq \htmlData{tutor-start=26,tutor-end=27}{8}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{3}. E(2,2)→2≠8/3\htmlData{tutor-start=0,tutor-end=1}{E}\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}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=19}{\rightarrow }\htmlData{tutor-start=19,tutor-end=20}{2} \neq \htmlData{tutor-start=26,tutor-end=27}{8}\htmlData{tutor-start=27,tutor-end=28}{/}\htmlData{tutor-start=28,tutor-end=29}{3}. A(0,0)→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}{0}\htmlData{tutor-start=5,tutor-end=6}{)} \htmlData{tutor-start=7,tutor-end=19}{\rightarrow }\htmlData{tutor-start=19,tutor-end=20}{0} \neq \htmlData{tutor-start=26,tutor-end=27}{2}. 无。 B(2,0),D(0,2)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)} 已算。 C(1,1),D(0,2)→k=−1\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=27}{\rightarrow }\htmlData{tutor-start=27,tutor-end=28}{k}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{1}. 已算。 C(1,1),B(2,0)→k=−1\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=27}{\rightarrow }\htmlData{tutor-start=27,tutor-end=28}{k}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{1}. 已算。

所以共线三点组总数 M=4+1=5\htmlData{tutor-start=0,tutor-end=1}{M} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{4} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{5}。 能构成三角形的情况数 =20−5=15\htmlData{tutor-start=0,tutor-end=1}{=} \htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{0} \htmlData{tutor-start=5,tutor-end=6}{-} \htmlData{tutor-start=7,tutor-end=8}{5} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{5}。 概率 P=1520=34\htmlData{tutor-start=0,tutor-end=1}{P} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{5}}{\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{0}} \htmlData{tutor-start=18,tutor-end=19}{=} \frac{\htmlData{tutor-start=26,tutor-end=27}{3}}{\htmlData{tutor-start=29,tutor-end=30}{4}}。

此前第一步草稿中漏看了 A\htmlData{tutor-start=0,tutor-end=1}{A} 点也在 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x} 上，导致错误。修正后结论为 3/4\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{4}。

P=1520=34\htmlData{tutor-start=0,tutor-end=1}{P} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{5}}{\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{0}} \htmlData{tutor-start=18,tutor-end=19}{=} \frac{\htmlData{tutor-start=26,tutor-end=27}{3}}{\htmlData{tutor-start=29,tutor-end=30}{4}}
8

一、填空题 · 函数性质与不等式

设函数 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 是定义在 R\mathbf{\htmlData{tutor-start=8,tutor-end=9}{R}} 上的奇函数. 若当 x∈(0,+∞)\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=17}{\infty}\htmlData{tutor-start=17,tutor-end=18}{)} 时，f(x)=lg⁡x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \lg \htmlData{tutor-start=11,tutor-end=12}{x}，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：(−1,0)∪(1,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=13}{\cup }\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=24}{\infty}\htmlData{tutor-start=24,tutor-end=25}{)}

题目标签：上海卷理科第 8 题

解题过程

求解不等式 f(x) > 0

利用奇函数性质和对数函数解析式求解不等式

(1)
分析 x > 0 时的情形

根据题设，当 x>0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0} 时，f(x)=lg⁡x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \lg \htmlData{tutor-start=11,tutor-end=12}{x}。直接解不等式。

详细展开： 当 x∈(0,+∞)\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=17}{\infty}\htmlData{tutor-start=17,tutor-end=18}{)} 时，f(x)=lg⁡x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \lg \htmlData{tutor-start=11,tutor-end=12}{x}。 令 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=5,tutor-end=6}{>} \htmlData{tutor-start=7,tutor-end=8}{0}，即： lg⁡x>0\lg \htmlData{tutor-start=4,tutor-end=5}{x} \htmlData{tutor-start=6,tutor-end=7}{>} \htmlData{tutor-start=8,tutor-end=9}{0} 根据对数函数性质（底数 10>1\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0} \htmlData{tutor-start=3,tutor-end=4}{>} \htmlData{tutor-start=5,tutor-end=6}{1}，单调递增）： x>100⇒x>1\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}^{\htmlData{tutor-start=8,tutor-end=9}{0}} \htmlData{tutor-start=11,tutor-end=23}{\Rightarrow }\htmlData{tutor-start=23,tutor-end=24}{x} \htmlData{tutor-start=25,tutor-end=26}{>} \htmlData{tutor-start=27,tutor-end=28}{1} 结合前提条件 x>0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}，此部分的解集为 (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=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)}。

x>1\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{1}
(2)
利用奇函数性质分析 x < 0 时的情形

利用 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}{)} 是奇函数的性质，将 x<0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0} 的问题转化为 −x>0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=4}{>} \htmlData{tutor-start=5,tutor-end=6}{0} 的问题。

详细展开： 因为 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}{)} 是定义在 R\mathbf{\htmlData{tutor-start=8,tutor-end=9}{R}} 上的奇函数，所以 f(−x)=−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=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{f}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{)}，即 f(x)=−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=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{f}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=13}{)}。 当 x<0\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{0} 时，−x>0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=4}{>} \htmlData{tutor-start=5,tutor-end=6}{0}。 此时 f(−x)=lg⁡(−x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \lg\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{)}。 所以 f(x)=−lg⁡(−x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\lg\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{)}。 令 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=5,tutor-end=6}{>} \htmlData{tutor-start=7,tutor-end=8}{0}，即： −lg⁡(−x)>0\htmlData{tutor-start=0,tutor-end=1}{-}\lg\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{)} \htmlData{tutor-start=9,tutor-end=10}{>} \htmlData{tutor-start=11,tutor-end=12}{0} lg⁡(−x)<0\lg\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{<} \htmlData{tutor-start=10,tutor-end=11}{0} 根据对数函数性质： 0<−x<100⇒0<−x<1\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0}^{\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=28}{\Rightarrow }\htmlData{tutor-start=28,tutor-end=29}{0} \htmlData{tutor-start=30,tutor-end=31}{<} \htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{x} \htmlData{tutor-start=35,tutor-end=36}{<} \htmlData{tutor-start=37,tutor-end=38}{1} 解得： −1<x<0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{0} 此部分的解集为 (−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=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)}。

另外，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，因 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}{)} 是定义在 R\mathbf{\htmlData{tutor-start=8,tutor-end=9}{R}} 上的奇函数，必有 f(0)=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}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}，不满足 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}。

−1<x<0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{1} \htmlData{tutor-start=3,tutor-end=4}{<} \htmlData{tutor-start=5,tutor-end=6}{x} \htmlData{tutor-start=7,tutor-end=8}{<} \htmlData{tutor-start=9,tutor-end=10}{0}
(3)
合并解集

将上述两部分的解集取并集，得到最终答案。

详细展开： 综上所述，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=5,tutor-end=6}{>} \htmlData{tutor-start=7,tutor-end=8}{0} 的 x\htmlData{tutor-start=0,tutor-end=1}{x} 的取值范围是： (−1,0)∪(1,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=13}{\cup }\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=24}{\infty}\htmlData{tutor-start=24,tutor-end=25}{)}

(−1,0)∪(1,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=13}{\cup }\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{,} \htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=24}{\infty}\htmlData{tutor-start=24,tutor-end=25}{)}
9

一、填空题 · 统计初步

已知总体的各个体的值由小到大依次为 2,3,3,7,a,b,12,13.7,18.3,20\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{7}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{b}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{3}\htmlData{tutor-start=24,tutor-end=25}{.}\htmlData{tutor-start=25,tutor-end=26}{7}\htmlData{tutor-start=26,tutor-end=27}{,} \htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{8}\htmlData{tutor-start=30,tutor-end=31}{.}\htmlData{tutor-start=31,tutor-end=32}{3}\htmlData{tutor-start=32,tutor-end=33}{,} \htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{0}，且总体的中位数为 10.5\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}{5}. 若要使该总体的方差最小，则 a\htmlData{tutor-start=0,tutor-end=1}{a}、b\htmlData{tutor-start=0,tutor-end=1}{b} 的取值分别是______.

答案：10.5,10.5\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}{5}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{.}\htmlData{tutor-start=9,tutor-end=10}{5}

题目标签：上海卷理科第 9 题

解题过程

确定 a, b 的值使方差最小

根据中位数确定 a+b 的值，再利用方差性质确定 a, b 具体值

(1)
利用中位数确定 a 与 b 的关系

根据总体数据的排序和中位数定义，建立关于 a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b} 的方程。

详细展开： 数据共有 10 个：2,3,3,7,a,b,12,13.7,18.3,20\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{7}\htmlData{tutor-start=10,tutor-end=11}{,} \htmlData{tutor-start=12,tutor-end=13}{a}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{b}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{3}\htmlData{tutor-start=24,tutor-end=25}{.}\htmlData{tutor-start=25,tutor-end=26}{7}\htmlData{tutor-start=26,tutor-end=27}{,} \htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{8}\htmlData{tutor-start=30,tutor-end=31}{.}\htmlData{tutor-start=31,tutor-end=32}{3}\htmlData{tutor-start=32,tutor-end=33}{,} \htmlData{tutor-start=34,tutor-end=35}{2}\htmlData{tutor-start=35,tutor-end=36}{0}。 题目指出数据“由小到大依次”排列，这意味着： 7⩽a⩽b⩽12\htmlData{tutor-start=0,tutor-end=1}{7} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{a} \htmlData{tutor-start=14,tutor-end=24}{\leqslant }\htmlData{tutor-start=24,tutor-end=25}{b} \htmlData{tutor-start=26,tutor-end=36}{\leqslant }\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{2} 对于偶数个数据（n=10\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{0}），中位数是中间两个数（第5个和第6个）的平均值。 第5个数是 a\htmlData{tutor-start=0,tutor-end=1}{a}，第6个数是 b\htmlData{tutor-start=0,tutor-end=1}{b}。 已知中位数为 10.5\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}{5}，故： a+b2=10.5⇒a+b=21\frac{\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{b}}{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{.}\htmlData{tutor-start=19,tutor-end=20}{5} \htmlData{tutor-start=21,tutor-end=33}{\Rightarrow }\htmlData{tutor-start=33,tutor-end=34}{a} \htmlData{tutor-start=35,tutor-end=36}{+} \htmlData{tutor-start=37,tutor-end=38}{b} \htmlData{tutor-start=39,tutor-end=40}{=} \htmlData{tutor-start=41,tutor-end=42}{2}\htmlData{tutor-start=42,tutor-end=43}{1} 结合排序约束 7⩽a⩽b⩽12\htmlData{tutor-start=0,tutor-end=1}{7} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{a} \htmlData{tutor-start=14,tutor-end=24}{\leqslant }\htmlData{tutor-start=24,tutor-end=25}{b} \htmlData{tutor-start=26,tutor-end=36}{\leqslant }\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{2}，我们需要在这个范围内寻找 a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}。

a+b=21\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{1}
(2)
利用方差最小化原理求解 a, b

在约束条件 a+b=21\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}{2}\htmlData{tutor-start=5,tutor-end=6}{1} 下，寻找使总体方差最小的 a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b} 值。

详细展开： 总体方差公式为 s2=1n∑i=1n(xi−xˉ)2\htmlData{tutor-start=0,tutor-end=1}{s}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{n}} \sum_{\htmlData{tutor-start=26,tutor-end=27}{i}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{1}}^{\htmlData{tutor-start=32,tutor-end=33}{n}} \htmlData{tutor-start=35,tutor-end=36}{(}\htmlData{tutor-start=36,tutor-end=37}{x}_{\htmlData{tutor-start=39,tutor-end=40}{i}} \htmlData{tutor-start=42,tutor-end=43}{-} \bar{\htmlData{tutor-start=49,tutor-end=50}{x}}\htmlData{tutor-start=51,tutor-end=52}{)}^{\htmlData{tutor-start=54,tutor-end=55}{2}}。 要使方差最小，等价于使各数据与平均数差的平方和最小，或者更直观地，使数据尽可能集中。 固定其他8个数据，变量仅为 a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b}。 平均数 xˉ=110(Sumother+a+b)=110(Sumother+21)\bar{\htmlData{tutor-start=5,tutor-end=6}{x}} \htmlData{tutor-start=8,tutor-end=9}{=} \frac{\htmlData{tutor-start=16,tutor-end=17}{1}}{\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{0}} \htmlData{tutor-start=23,tutor-end=24}{(}\text{\htmlData{tutor-start=30,tutor-end=31}{S}\htmlData{tutor-start=31,tutor-end=32}{u}\htmlData{tutor-start=32,tutor-end=33}{m}}_{\htmlData{tutor-start=36,tutor-end=37}{o}\htmlData{tutor-start=37,tutor-end=38}{t}\htmlData{tutor-start=38,tutor-end=39}{h}\htmlData{tutor-start=39,tutor-end=40}{e}\htmlData{tutor-start=40,tutor-end=41}{r}} \htmlData{tutor-start=43,tutor-end=44}{+} \htmlData{tutor-start=45,tutor-end=46}{a} \htmlData{tutor-start=47,tutor-end=48}{+} \htmlData{tutor-start=49,tutor-end=50}{b}\htmlData{tutor-start=50,tutor-end=51}{)} \htmlData{tutor-start=52,tutor-end=53}{=} \frac{\htmlData{tutor-start=60,tutor-end=61}{1}}{\htmlData{tutor-start=63,tutor-end=64}{1}\htmlData{tutor-start=64,tutor-end=65}{0}} \htmlData{tutor-start=67,tutor-end=68}{(}\text{\htmlData{tutor-start=74,tutor-end=75}{S}\htmlData{tutor-start=75,tutor-end=76}{u}\htmlData{tutor-start=76,tutor-end=77}{m}}_{\htmlData{tutor-start=80,tutor-end=81}{o}\htmlData{tutor-start=81,tutor-end=82}{t}\htmlData{tutor-start=82,tutor-end=83}{h}\htmlData{tutor-start=83,tutor-end=84}{e}\htmlData{tutor-start=84,tutor-end=85}{r}} \htmlData{tutor-start=87,tutor-end=88}{+} \htmlData{tutor-start=89,tutor-end=90}{2}\htmlData{tutor-start=90,tutor-end=91}{1}\htmlData{tutor-start=91,tutor-end=92}{)}。 由于 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} 是定值，所以平均数 xˉ\bar{\htmlData{tutor-start=5,tutor-end=6}{x}} 也是定值。 要使方差最小，即最小化 ∑(xi−xˉ)2\sum \htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{x}_{\htmlData{tutor-start=9,tutor-end=10}{i}} \htmlData{tutor-start=12,tutor-end=13}{-} \bar{\htmlData{tutor-start=19,tutor-end=20}{x}}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}}。 其中含 a,b\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{b} 的项为 (a−xˉ)2+(b−xˉ)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a} \htmlData{tutor-start=3,tutor-end=4}{-} \bar{\htmlData{tutor-start=10,tutor-end=11}{x}}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{b} \htmlData{tutor-start=23,tutor-end=24}{-} \bar{\htmlData{tutor-start=30,tutor-end=31}{x}}\htmlData{tutor-start=32,tutor-end=33}{)}^{\htmlData{tutor-start=35,tutor-end=36}{2}}。 已知 a+b=21\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}{2}\htmlData{tutor-start=5,tutor-end=6}{1}，设 a=10.5−d,b=10.5+d\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{.}\htmlData{tutor-start=7,tutor-end=8}{5} \htmlData{tutor-start=9,tutor-end=10}{-} \htmlData{tutor-start=11,tutor-end=12}{d}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{b} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{.}\htmlData{tutor-start=21,tutor-end=22}{5} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{d} (d⩾0\htmlData{tutor-start=0,tutor-end=1}{d} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{0})。 则 (a−xˉ)2+(b−xˉ)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a} \htmlData{tutor-start=3,tutor-end=4}{-} \bar{\htmlData{tutor-start=10,tutor-end=11}{x}}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{b} \htmlData{tutor-start=23,tutor-end=24}{-} \bar{\htmlData{tutor-start=30,tutor-end=31}{x}}\htmlData{tutor-start=32,tutor-end=33}{)}^{\htmlData{tutor-start=35,tutor-end=36}{2}} 在 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} 时取最小值（二次函数性质，或者由柯西不等式/均值不等式可知，和一定时，两数相等时平方和最小）。 即当 a=b=10.5\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{.}\htmlData{tutor-start=11,tutor-end=12}{5} 时，(a−xˉ)2+(b−xˉ)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a} \htmlData{tutor-start=3,tutor-end=4}{-} \bar{\htmlData{tutor-start=10,tutor-end=11}{x}}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{b} \htmlData{tutor-start=23,tutor-end=24}{-} \bar{\htmlData{tutor-start=30,tutor-end=31}{x}}\htmlData{tutor-start=32,tutor-end=33}{)}^{\htmlData{tutor-start=35,tutor-end=36}{2}} 最小。

检查约束条件： 若 a=10.5,b=10.5\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\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=10,tutor-end=11}{b} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{.}\htmlData{tutor-start=17,tutor-end=18}{5}，满足 7⩽10.5⩽10.5⩽12\htmlData{tutor-start=0,tutor-end=1}{7} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{.}\htmlData{tutor-start=15,tutor-end=16}{5} \htmlData{tutor-start=17,tutor-end=27}{\leqslant }\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{.}\htmlData{tutor-start=30,tutor-end=31}{5} \htmlData{tutor-start=32,tutor-end=42}{\leqslant }\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{2}。 且满足 a⩽b\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{b}。 因此，当 a=10.5,b=10.5\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{.}\htmlData{tutor-start=5,tutor-end=6}{5}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{=}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{.}\htmlData{tutor-start=13,tutor-end=14}{5} 时，方差最小。

a=b=10.5\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{b} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{.}\htmlData{tutor-start=11,tutor-end=12}{5}
10

一、填空题 · 解析几何与应用

某海域内有一孤岛，岛四周的海平面（视为平面）上有一浅水区（含边界），其边界是长轴长为 2a\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}，短轴长为 2b\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{b} 的椭圆. 已知岛上甲、乙导航灯的海拔高度分别为 h1\htmlData{tutor-start=0,tutor-end=1}{h}_{\htmlData{tutor-start=3,tutor-end=4}{1}}、h2\htmlData{tutor-start=0,tutor-end=1}{h}_{\htmlData{tutor-start=3,tutor-end=4}{2}}，且两个导航灯在海平面上的投影恰好落在椭圆的两个焦点上，现有船只经过该海域（船只的大小忽略不计），在船上测得甲、乙导航灯的仰角分别为 θ1\htmlData{tutor-start=0,tutor-end=6}{\theta}_{\htmlData{tutor-start=8,tutor-end=9}{1}}、θ2\htmlData{tutor-start=0,tutor-end=6}{\theta}_{\htmlData{tutor-start=8,tutor-end=9}{2}}，那么船只已进入该浅水区的判别条件是______.

答案：h1cot⁡θ1+h2cot⁡θ2⩽2a\htmlData{tutor-start=0,tutor-end=1}{h}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \cot \htmlData{tutor-start=11,tutor-end=17}{\theta}_{\htmlData{tutor-start=19,tutor-end=20}{1}} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{h}_{\htmlData{tutor-start=27,tutor-end=28}{2}} \cot \htmlData{tutor-start=35,tutor-end=41}{\theta}_{\htmlData{tutor-start=43,tutor-end=44}{2}} \htmlData{tutor-start=46,tutor-end=56}{\leqslant }\htmlData{tutor-start=56,tutor-end=57}{2}\htmlData{tutor-start=57,tutor-end=58}{a}

题目标签：上海卷理科第 10 题

解题过程

建立船只位置与椭圆定义的关联

将仰角转化为距离，利用椭圆定义得出判别条件

(1)
将仰角转化为船到焦点的距离

利用三角函数关系，将测量得到的仰角转化为船在海平面上投影点到两个焦点的距离。

详细展开： 设船在海平面上的投影点为 P\htmlData{tutor-start=0,tutor-end=1}{P}。 甲导航灯投影为焦点 F1\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}}，高度 h1\htmlData{tutor-start=0,tutor-end=1}{h}_{\htmlData{tutor-start=3,tutor-end=4}{1}}，仰角 θ1\htmlData{tutor-start=0,tutor-end=6}{\theta}_{\htmlData{tutor-start=8,tutor-end=9}{1}}。 在直角三角形中，船到 F1\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 的水平距离 ∣PF1∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{F}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{|} 满足： tan⁡θ1=h1∣PF1∣⇒∣PF1∣=h1cot⁡θ1\tan \htmlData{tutor-start=5,tutor-end=11}{\theta}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{h}_{\htmlData{tutor-start=27,tutor-end=28}{1}}}{\htmlData{tutor-start=31,tutor-end=32}{|}\htmlData{tutor-start=32,tutor-end=33}{P}\htmlData{tutor-start=33,tutor-end=34}{F}_{\htmlData{tutor-start=36,tutor-end=37}{1}}\htmlData{tutor-start=38,tutor-end=39}{|}} \htmlData{tutor-start=41,tutor-end=53}{\Rightarrow }\htmlData{tutor-start=53,tutor-end=54}{|}\htmlData{tutor-start=54,tutor-end=55}{P}\htmlData{tutor-start=55,tutor-end=56}{F}_{\htmlData{tutor-start=58,tutor-end=59}{1}}\htmlData{tutor-start=60,tutor-end=61}{|} \htmlData{tutor-start=62,tutor-end=63}{=} \htmlData{tutor-start=64,tutor-end=65}{h}_{\htmlData{tutor-start=67,tutor-end=68}{1}} \cot \htmlData{tutor-start=75,tutor-end=81}{\theta}_{\htmlData{tutor-start=83,tutor-end=84}{1}} 同理，乙导航灯投影为焦点 F2\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{2}}，高度 h2\htmlData{tutor-start=0,tutor-end=1}{h}_{\htmlData{tutor-start=3,tutor-end=4}{2}}，仰角 θ2\htmlData{tutor-start=0,tutor-end=6}{\theta}_{\htmlData{tutor-start=8,tutor-end=9}{2}}。 船到 F2\htmlData{tutor-start=0,tutor-end=1}{F}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 的水平距离 ∣PF2∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{F}_{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{|} 满足： ∣PF2∣=h2cot⁡θ2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{F}_{\htmlData{tutor-start=5,tutor-end=6}{2}}\htmlData{tutor-start=7,tutor-end=8}{|} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{h}_{\htmlData{tutor-start=14,tutor-end=15}{2}} \cot \htmlData{tutor-start=22,tutor-end=28}{\theta}_{\htmlData{tutor-start=30,tutor-end=31}{2}}

∣PF1∣=h1cot⁡θ1,∣PF2∣=h2cot⁡θ2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{F}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{|} \htmlData{tutor-start=9,tutor-end=10}{=} \htmlData{tutor-start=11,tutor-end=12}{h}_{\htmlData{tutor-start=14,tutor-end=15}{1}} \cot \htmlData{tutor-start=22,tutor-end=28}{\theta}_{\htmlData{tutor-start=30,tutor-end=31}{1}}\htmlData{tutor-start=32,tutor-end=33}{,} \quad \htmlData{tutor-start=40,tutor-end=41}{|}\htmlData{tutor-start=41,tutor-end=42}{P}\htmlData{tutor-start=42,tutor-end=43}{F}_{\htmlData{tutor-start=45,tutor-end=46}{2}}\htmlData{tutor-start=47,tutor-end=48}{|} \htmlData{tutor-start=49,tutor-end=50}{=} \htmlData{tutor-start=51,tutor-end=52}{h}_{\htmlData{tutor-start=54,tutor-end=55}{2}} \cot \htmlData{tutor-start=62,tutor-end=68}{\theta}_{\htmlData{tutor-start=70,tutor-end=71}{2}}
(2)
应用椭圆定义得出判别条件

根据椭圆的定义，点在椭圆内部（含边界）的充要条件是到两焦点的距离之和小于等于长轴长。

详细展开： 椭圆定义为：平面内到两个定点（焦点）的距离之和等于常数（2a\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}）的点的轨迹。 对于椭圆内部的点 P\htmlData{tutor-start=0,tutor-end=1}{P}，满足： ∣PF1∣+∣PF2∣⩽2a\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{F}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{|} \htmlData{tutor-start=9,tutor-end=10}{+} \htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{P}\htmlData{tutor-start=13,tutor-end=14}{F}_{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{|} \htmlData{tutor-start=20,tutor-end=30}{\leqslant }\htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{a} 对于椭圆外部的点 P\htmlData{tutor-start=0,tutor-end=1}{P}，满足： ∣PF1∣+∣PF2∣>2a\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{P}\htmlData{tutor-start=2,tutor-end=3}{F}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{|} \htmlData{tutor-start=9,tutor-end=10}{+} \htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{P}\htmlData{tutor-start=13,tutor-end=14}{F}_{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{|} \htmlData{tutor-start=20,tutor-end=21}{>} \htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{a} 题目要求“船只已进入该浅水区”，即点 P\htmlData{tutor-start=0,tutor-end=1}{P} 在椭圆边界上或内部。 将第一步得到的距离表达式代入： h1cot⁡θ1+h2cot⁡θ2⩽2a\htmlData{tutor-start=0,tutor-end=1}{h}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \cot \htmlData{tutor-start=11,tutor-end=17}{\theta}_{\htmlData{tutor-start=19,tutor-end=20}{1}} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{h}_{\htmlData{tutor-start=27,tutor-end=28}{2}} \cot \htmlData{tutor-start=35,tutor-end=41}{\theta}_{\htmlData{tutor-start=43,tutor-end=44}{2}} \htmlData{tutor-start=46,tutor-end=56}{\leqslant }\htmlData{tutor-start=56,tutor-end=57}{2}\htmlData{tutor-start=57,tutor-end=58}{a} 这就是判别条件。

h1cot⁡θ1+h2cot⁡θ2⩽2a\htmlData{tutor-start=0,tutor-end=1}{h}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \cot \htmlData{tutor-start=11,tutor-end=17}{\theta}_{\htmlData{tutor-start=19,tutor-end=20}{1}} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{h}_{\htmlData{tutor-start=27,tutor-end=28}{2}} \cot \htmlData{tutor-start=35,tutor-end=41}{\theta}_{\htmlData{tutor-start=43,tutor-end=44}{2}} \htmlData{tutor-start=46,tutor-end=56}{\leqslant }\htmlData{tutor-start=56,tutor-end=57}{2}\htmlData{tutor-start=57,tutor-end=58}{a}
11

一、填空题 · 函数与方程

方程 x2+2x−1=0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\sqrt{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{x}\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}{0} 的解可视为函数 y=x+2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{+}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}} 的图象与函数 y=1x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=9}{1}}{\htmlData{tutor-start=11,tutor-end=12}{x}} 的图象交点的横坐标. 若 x4+ax−4=0\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}{a}\htmlData{tutor-start=7,tutor-end=8}{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}{0} 的各个实根 x1,x2,⋯,xk\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=7,tutor-end=8}{x}_{\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}{x}_{\htmlData{tutor-start=25,tutor-end=26}{k}} (k⩽4\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{4}) 所对应的点 (xi,4xi)\left(\htmlData{tutor-start=6,tutor-end=7}{x}_{\htmlData{tutor-start=9,tutor-end=10}{i}}\htmlData{tutor-start=11,tutor-end=12}{,} \frac{\htmlData{tutor-start=19,tutor-end=20}{4}}{\htmlData{tutor-start=22,tutor-end=23}{x}_{\htmlData{tutor-start=25,tutor-end=26}{i}}}\right) (i=1,2,⋯,k\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=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{,} \cdots\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{k}) 均在直线 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x} 的同侧，则实数 a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围是______.

答案：(−∞,−6)∪(6,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{6}\htmlData{tutor-start=12,tutor-end=13}{)} \htmlData{tutor-start=14,tutor-end=19}{\cup }\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{6}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=30}{\infty}\htmlData{tutor-start=30,tutor-end=31}{)}

题目标签：上海卷理科第 11 题

解题过程

确定实数 a 的取值范围

通过换元和图象分析，确定参数 a 的范围使得实根对应的点在直线 y=x 同侧

(1)
变形方程并理解几何意义

将原方程变形，类比题干给出的示例，找出对应的函数交点模型。

详细展开： 原方程：x4+ax−4=0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{4} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{0}。 移项得：x4−4=−ax\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{4} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{x}。 若 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}，则 −4=0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{4}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{0} 不成立，故 x≠0\htmlData{tutor-start=0,tutor-end=1}{x} \neq \htmlData{tutor-start=7,tutor-end=8}{0}。 两边同除以 x\htmlData{tutor-start=0,tutor-end=1}{x}： x3−4x=−a\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{3}} \htmlData{tutor-start=6,tutor-end=7}{-} \frac{\htmlData{tutor-start=14,tutor-end=15}{4}}{\htmlData{tutor-start=17,tutor-end=18}{x}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{a} 或者，为了贴合题干中“点 (xi,4xi)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{i}}\htmlData{tutor-start=6,tutor-end=7}{,} \frac{\htmlData{tutor-start=14,tutor-end=15}{4}}{\htmlData{tutor-start=17,tutor-end=18}{x}_{\htmlData{tutor-start=20,tutor-end=21}{i}}}\htmlData{tutor-start=23,tutor-end=24}{)}”的形式，我们重新观察。 题干示例：x2+2x−1=0⇒x+2=1x\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \sqrt{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{1} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{0} \htmlData{tutor-start=26,tutor-end=38}{\Rightarrow }\htmlData{tutor-start=38,tutor-end=39}{x} \htmlData{tutor-start=40,tutor-end=41}{+} \sqrt{\htmlData{tutor-start=48,tutor-end=49}{2}} \htmlData{tutor-start=51,tutor-end=52}{=} \frac{\htmlData{tutor-start=59,tutor-end=60}{1}}{\htmlData{tutor-start=62,tutor-end=63}{x}}。交点横坐标即为根。 本题中，点为 (xi,4xi)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{i}}\htmlData{tutor-start=6,tutor-end=7}{,} \frac{\htmlData{tutor-start=14,tutor-end=15}{4}}{\htmlData{tutor-start=17,tutor-end=18}{x}_{\htmlData{tutor-start=20,tutor-end=21}{i}}}\htmlData{tutor-start=23,tutor-end=24}{)}。这些点在双曲线 y=4x\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{4}}{\htmlData{tutor-start=13,tutor-end=14}{x}} 上。 原方程 x4+ax−4=0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{4} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{0} 可以写为： x4−4=−ax\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{4} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{x} x(x3)−4=−ax\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}^{\htmlData{tutor-start=5,tutor-end=6}{3}}\htmlData{tutor-start=7,tutor-end=8}{)} \htmlData{tutor-start=9,tutor-end=10}{-} \htmlData{tutor-start=11,tutor-end=12}{4} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{a}\htmlData{tutor-start=17,tutor-end=18}{x} 这似乎不容易直接凑成 g(x)=4x\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{4}}{\htmlData{tutor-start=16,tutor-end=17}{x}} 的形式。 让我们换个角度。题中说点 (xi,4xi)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{i}}\htmlData{tutor-start=6,tutor-end=7}{,} \frac{\htmlData{tutor-start=14,tutor-end=15}{4}}{\htmlData{tutor-start=17,tutor-end=18}{x}_{\htmlData{tutor-start=20,tutor-end=21}{i}}}\htmlData{tutor-start=23,tutor-end=24}{)} 在直线 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x} 的同侧。 这意味着对于所有实根 xi\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{i}}，要么 4xi>xi\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{x}_{\htmlData{tutor-start=12,tutor-end=13}{i}}} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{x}_{\htmlData{tutor-start=21,tutor-end=22}{i}}，要么 4xi<xi\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{x}_{\htmlData{tutor-start=12,tutor-end=13}{i}}} \htmlData{tutor-start=16,tutor-end=17}{<} \htmlData{tutor-start=18,tutor-end=19}{x}_{\htmlData{tutor-start=21,tutor-end=22}{i}}。 即 4xi−xi\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{x}_{\htmlData{tutor-start=12,tutor-end=13}{i}}} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{x}_{\htmlData{tutor-start=21,tutor-end=22}{i}} 同号。

让我们回到方程本身寻找 a\htmlData{tutor-start=0,tutor-end=1}{a} 与根的关系。 x4+ax−4=0⇒a=4−x4x=4x−x3\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{4} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{0} \htmlData{tutor-start=19,tutor-end=31}{\Rightarrow }\htmlData{tutor-start=31,tutor-end=32}{a} \htmlData{tutor-start=33,tutor-end=34}{=} \frac{\htmlData{tutor-start=41,tutor-end=42}{4} \htmlData{tutor-start=43,tutor-end=44}{-} \htmlData{tutor-start=45,tutor-end=46}{x}^{\htmlData{tutor-start=48,tutor-end=49}{4}}}{\htmlData{tutor-start=52,tutor-end=53}{x}} \htmlData{tutor-start=55,tutor-end=56}{=} \frac{\htmlData{tutor-start=63,tutor-end=64}{4}}{\htmlData{tutor-start=66,tutor-end=67}{x}} \htmlData{tutor-start=69,tutor-end=70}{-} \htmlData{tutor-start=71,tutor-end=72}{x}^{\htmlData{tutor-start=74,tutor-end=75}{3}}。 设 f(x)=4x−x3\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{4}}{\htmlData{tutor-start=16,tutor-end=17}{x}} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{x}^{\htmlData{tutor-start=24,tutor-end=25}{3}}。则 xi\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 是直线 y=a\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a} 与曲线 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}{)} 交点的横坐标。 但这并没有直接利用到“点 (xi,4xi)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{i}}\htmlData{tutor-start=6,tutor-end=7}{,} \frac{\htmlData{tutor-start=14,tutor-end=15}{4}}{\htmlData{tutor-start=17,tutor-end=18}{x}_{\htmlData{tutor-start=20,tutor-end=21}{i}}}\htmlData{tutor-start=23,tutor-end=24}{)} 在 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x} 同侧”这个几何条件。

重新审视题干提示：“方程...的解可视为...交点的横坐标”。 对于 x4+ax−4=0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{4} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{0}，我们可以变形为： x4−4=−ax\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{4} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{x} 这不像示例那样直接。示例是二次，这里是四次。 但是，题目关注的是根 xi\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 对应的点 Pi(xi,4xi)\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}{x}_{\htmlData{tutor-start=9,tutor-end=10}{i}}\htmlData{tutor-start=11,tutor-end=12}{,} \frac{\htmlData{tutor-start=19,tutor-end=20}{4}}{\htmlData{tutor-start=22,tutor-end=23}{x}_{\htmlData{tutor-start=25,tutor-end=26}{i}}}\htmlData{tutor-start=28,tutor-end=29}{)} 相对于直线 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x} 的位置。 点 Pi\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 在 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x} 上方   ⟺  4xi>xi\iff \frac{4}{x_{i}} > x_{i}。 点 Pi\htmlData{tutor-start=0,tutor-end=1}{P}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 在 y=x\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{x} 下方   ⟺  4xi<xi\iff \frac{4}{x_{i}} < x_{i}。

条件：所有实根 xi\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 满足同一不等式。 情况1：所有实根满足 4x>x\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{x}} \htmlData{tutor-start=12,tutor-end=13}{>} \htmlData{tutor-start=14,tutor-end=15}{x}。 情况2：所有实根满足 4x<x\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{x}} \htmlData{tutor-start=12,tutor-end=13}{<} \htmlData{tutor-start=14,tutor-end=15}{x}。

先解不等式 4x>x\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{x}} \htmlData{tutor-start=12,tutor-end=13}{>} \htmlData{tutor-start=14,tutor-end=15}{x}： 若 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}，则 4>x2⇒0<x<2\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{x}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=22}{\Rightarrow }\htmlData{tutor-start=22,tutor-end=23}{0} \htmlData{tutor-start=24,tutor-end=25}{<} \htmlData{tutor-start=26,tutor-end=27}{x} \htmlData{tutor-start=28,tutor-end=29}{<} \htmlData{tutor-start=30,tutor-end=31}{2}。 若 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}，则 4<x2⇒x<−2\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=22}{\Rightarrow }\htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=25}{<} \htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{2}。 即集合 U=(−∞,−2)∪(0,2)\htmlData{tutor-start=0,tutor-end=1}{U} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=12}{\infty}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{)} \htmlData{tutor-start=18,tutor-end=23}{\cup }\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=26}{,} \htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{)}。

再解不等式 4x<x\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{x}} \htmlData{tutor-start=12,tutor-end=13}{<} \htmlData{tutor-start=14,tutor-end=15}{x}： 若 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}，则 4<x2⇒x>2\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{x}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=22}{\Rightarrow }\htmlData{tutor-start=22,tutor-end=23}{x} \htmlData{tutor-start=24,tutor-end=25}{>} \htmlData{tutor-start=26,tutor-end=27}{2}。 若 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}，则 4>x2⇒−2<x<0\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{x}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=22}{\Rightarrow }\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{2} \htmlData{tutor-start=25,tutor-end=26}{<} \htmlData{tutor-start=27,tutor-end=28}{x} \htmlData{tutor-start=29,tutor-end=30}{<} \htmlData{tutor-start=31,tutor-end=32}{0}。 即集合 V=(−2,0)∪(2,+∞)\htmlData{tutor-start=0,tutor-end=1}{V} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{)} \htmlData{tutor-start=12,tutor-end=17}{\cup }\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=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=28}{\infty}\htmlData{tutor-start=28,tutor-end=29}{)}。

题目要求所有实根都在 U\htmlData{tutor-start=0,tutor-end=1}{U} 中，或者所有实根都在 V\htmlData{tutor-start=0,tutor-end=1}{V} 中。

U=(−∞,−2)∪(0,2),V=(−2,0)∪(2,+∞)\htmlData{tutor-start=0,tutor-end=1}{U} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=12}{\infty}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{)} \htmlData{tutor-start=18,tutor-end=23}{\cup }\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=26}{,} \htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{,} \quad \htmlData{tutor-start=37,tutor-end=38}{V} \htmlData{tutor-start=39,tutor-end=40}{=} \htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{2}\htmlData{tutor-start=44,tutor-end=45}{,} \htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{)} \htmlData{tutor-start=49,tutor-end=54}{\cup }\htmlData{tutor-start=54,tutor-end=55}{(}\htmlData{tutor-start=55,tutor-end=56}{2}\htmlData{tutor-start=56,tutor-end=57}{,} \htmlData{tutor-start=58,tutor-end=59}{+}\htmlData{tutor-start=59,tutor-end=65}{\infty}\htmlData{tutor-start=65,tutor-end=66}{)}
(2)
分析函数性质与根的分布

研究方程 x4+ax−4=0\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{4}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{x} \htmlData{tutor-start=11,tutor-end=12}{-} \htmlData{tutor-start=13,tutor-end=14}{4} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{0} 的实根分布情况，确定 a 的范围。

详细展开： 令 g(x)=x4+ax−4\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{4}} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=19}{-} \htmlData{tutor-start=20,tutor-end=21}{4}。我们需要 g(x)=0\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}{0} 的所有实根落在 U\htmlData{tutor-start=0,tutor-end=1}{U} 或 V\htmlData{tutor-start=0,tutor-end=1}{V} 中。 注意到 g(0)=−4\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=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{4}。 当 x→±∞\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\to }\htmlData{tutor-start=6,tutor-end=10}{\pm }\htmlData{tutor-start=10,tutor-end=16}{\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=5,tutor-end=9}{\to }\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=16}{\infty}。 因此，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}{)} 至少有两个实根（一正一负）。

让我们考察临界情况。根出现在边界 x=2,−2\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{2} 时的 a\htmlData{tutor-start=0,tutor-end=1}{a} 值。 若 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} 是根：16+2a−4=0⇒2a=−12⇒a=−6\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{6} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{a} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{4} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{0} \htmlData{tutor-start=16,tutor-end=28}{\Rightarrow }\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{a} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{2} \htmlData{tutor-start=37,tutor-end=49}{\Rightarrow }\htmlData{tutor-start=49,tutor-end=50}{a} \htmlData{tutor-start=51,tutor-end=52}{=} \htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{6}。 若 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} 是根：16−2a−4=0⇒2a=12⇒a=6\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{6} \htmlData{tutor-start=3,tutor-end=4}{-} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{a} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{4} \htmlData{tutor-start=12,tutor-end=13}{=} \htmlData{tutor-start=14,tutor-end=15}{0} \htmlData{tutor-start=16,tutor-end=28}{\Rightarrow }\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{a} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{2} \htmlData{tutor-start=36,tutor-end=48}{\Rightarrow }\htmlData{tutor-start=48,tutor-end=49}{a} \htmlData{tutor-start=50,tutor-end=51}{=} \htmlData{tutor-start=52,tutor-end=53}{6}。

分析 a\htmlData{tutor-start=0,tutor-end=1}{a} 的变化对根的影响： 方程可写为 a=4−x4x=4x−x3\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{x}^{\htmlData{tutor-start=15,tutor-end=16}{4}}}{\htmlData{tutor-start=19,tutor-end=20}{x}} \htmlData{tutor-start=22,tutor-end=23}{=} \frac{\htmlData{tutor-start=30,tutor-end=31}{4}}{\htmlData{tutor-start=33,tutor-end=34}{x}} \htmlData{tutor-start=36,tutor-end=37}{-} \htmlData{tutor-start=38,tutor-end=39}{x}^{\htmlData{tutor-start=41,tutor-end=42}{3}}。 令 h(x)=4x−x3\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \frac{\htmlData{tutor-start=13,tutor-end=14}{4}}{\htmlData{tutor-start=16,tutor-end=17}{x}} \htmlData{tutor-start=19,tutor-end=20}{-} \htmlData{tutor-start=21,tutor-end=22}{x}^{\htmlData{tutor-start=24,tutor-end=25}{3}}。 h′(x)=−4x2−3x2<0\htmlData{tutor-start=0,tutor-end=1}{h}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\frac{\htmlData{tutor-start=15,tutor-end=16}{4}}{\htmlData{tutor-start=18,tutor-end=19}{x}^{\htmlData{tutor-start=21,tutor-end=22}{2}}} \htmlData{tutor-start=25,tutor-end=26}{-} \htmlData{tutor-start=27,tutor-end=28}{3}\htmlData{tutor-start=28,tutor-end=29}{x}^{\htmlData{tutor-start=31,tutor-end=32}{2}} \htmlData{tutor-start=34,tutor-end=35}{<} \htmlData{tutor-start=36,tutor-end=37}{0}。函数 h(x)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 在 (−∞,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)} 和 (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=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} 上均单调递减。 值域： 当 x∈(0,+∞)\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=17}{\infty}\htmlData{tutor-start=17,tutor-end=18}{)}，h(x)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 从 +∞\htmlData{tutor-start=0,tutor-end=1}{+}\htmlData{tutor-start=1,tutor-end=7}{\infty} 降到 −∞\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=7}{\infty}。对于任意 a\htmlData{tutor-start=0,tutor-end=1}{a}，恰有一个正根 x+x_+。 当 x∈(−∞,0)\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=14}{\infty}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{0}\htmlData{tutor-start=17,tutor-end=18}{)}，h(x)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 从 +∞\htmlData{tutor-start=0,tutor-end=1}{+}\htmlData{tutor-start=1,tutor-end=7}{\infty} 降到 −∞\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=7}{\infty}。对于任意 a\htmlData{tutor-start=0,tutor-end=1}{a}，恰有一个负根 x−x_-。 所以方程恒有两个实根 x−,x+x_-, x_+。

我们需要这两个根同时在 U\htmlData{tutor-start=0,tutor-end=1}{U} 或同时在 V\htmlData{tutor-start=0,tutor-end=1}{V}。

回顾区间： U=(−∞,−2)∪(0,2)\htmlData{tutor-start=0,tutor-end=1}{U} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=12}{\infty}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{)} \htmlData{tutor-start=18,tutor-end=23}{\cup }\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{0}\htmlData{tutor-start=25,tutor-end=26}{,} \htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{)} V=(−2,0)∪(2,+∞)\htmlData{tutor-start=0,tutor-end=1}{V} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{)} \htmlData{tutor-start=12,tutor-end=17}{\cup }\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=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=28}{\infty}\htmlData{tutor-start=28,tutor-end=29}{)}

正根 x+x_+ 必须在 (0,2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{)} (属于U) 或 (2,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=11}{\infty}\htmlData{tutor-start=11,tutor-end=12}{)} (属于V)。 负根 x−x_- 必须在 (−∞,−2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{)} (属于U) 或 (−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=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{)} (属于V)。

由于 h(x)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 单调递减： 对于正根 x+x_+： x+<2  ⟺  h(x+)>h(2)  ⟺  a>−6x_+ < 2 \iff h(x_+) > h(2) \iff a > -6。 x+>2  ⟺  h(x+)<h(2)  ⟺  a<−6x_+ > 2 \iff h(x_+) < h(2) \iff a < -6。

对于负根 x−x_-： x−<−2  ⟺  h(x−)>h(−2)  ⟺  a>6x_- < -2 \iff h(x_-) > h(-2) \iff a > 6。（注意 h\htmlData{tutor-start=0,tutor-end=1}{h} 在负半轴也是递减，x\htmlData{tutor-start=0,tutor-end=1}{x} 越小函数值越大？不对。 h(x)\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} 在 (−∞,0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{)} 递减。x1<x2⇒h(x1)>h(x2)\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=26}{\Rightarrow }\htmlData{tutor-start=26,tutor-end=27}{h}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{x}_{\htmlData{tutor-start=31,tutor-end=32}{1}}\htmlData{tutor-start=33,tutor-end=34}{)} \htmlData{tutor-start=35,tutor-end=36}{>} \htmlData{tutor-start=37,tutor-end=38}{h}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{x}_{\htmlData{tutor-start=42,tutor-end=43}{2}}\htmlData{tutor-start=44,tutor-end=45}{)}。 我们要 x−<−2x_- < -2。因为 h\htmlData{tutor-start=0,tutor-end=1}{h} 递减，所以 h(x−)>h(−2)h(x_-) > h(-2)。 h(−2)=4−2−(−8)=−2+8=6\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{4}}{\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=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{8}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{2} \htmlData{tutor-start=33,tutor-end=34}{+} \htmlData{tutor-start=35,tutor-end=36}{8} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{6}。 所以 x−<−2  ⟺  a>6x_- < -2 \iff a > 6。

我们要 x−>−2x_- > -2 (且 x−<0x_- < 0)。即 −2<x−<0-2 < x_- < 0。 这等价于 h(x−)<h(−2)  ⟺  a<6h(x_-) < h(-2) \iff a < 6。

总结： 1. 若根在 U\htmlData{tutor-start=0,tutor-end=1}{U}： 需 x+∈(0,2)x_+ \in (0, 2) 且 x−∈(−∞,−2)x_- \in (-\infty, -2)。 条件：a>−6\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{6} 且 a>6\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{6}。 取交集：a>6\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{6}。

2. 若根在 V\htmlData{tutor-start=0,tutor-end=1}{V}： 需 x+∈(2,+∞)x_+ \in (2, +\infty) 且 x−∈(−2,0)x_- \in (-2, 0)。 条件：a<−6\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{6} 且 a<6\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{6}。 取交集：a<−6\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{6}。

综上，a\htmlData{tutor-start=0,tutor-end=1}{a} 的取值范围是 (−∞,−6)∪(6,+∞)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=8}{\infty}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{6}\htmlData{tutor-start=12,tutor-end=13}{)} \htmlData{tutor-start=14,tutor-end=19}{\cup }\htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{6}\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=30}{\infty}\htmlData{tutor-start=30,tutor-end=31}{)}。

a∈(−∞,−6)∪(6,+∞)\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=14}{\infty}\htmlData{tutor-start=14,tutor-end=15}{,} \htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{6}\htmlData{tutor-start=18,tutor-end=19}{)} \htmlData{tutor-start=20,tutor-end=25}{\cup }\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{6}\htmlData{tutor-start=27,tutor-end=28}{,} \htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=36}{\infty}\htmlData{tutor-start=36,tutor-end=37}{)}
12

二、选择题 · 排列组合与二项式系数性质

组合数 Cnr\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} (n>r⩾1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{r} \htmlData{tutor-start=4,tutor-end=14}{\geqslant }\htmlData{tutor-start=14,tutor-end=15}{1}, n,r∈Z\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{r} \htmlData{tutor-start=5,tutor-end=9}{\in }\mathbf{\htmlData{tutor-start=17,tutor-end=18}{Z}}) 恒等于 ( )

答案：nrCn−1r−1\frac{\htmlData{tutor-start=6,tutor-end=7}{n}}{\htmlData{tutor-start=9,tutor-end=10}{r}} \htmlData{tutor-start=12,tutor-end=13}{C}_{\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}}^{\htmlData{tutor-start=21,tutor-end=22}{r}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}}

题目标签：组合数恒等式辨析

解题过程

组合数公式推导与恒等变形

利用组合数的阶乘定义，推导出 Cnr\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} 与其他组合数形式的等价关系，从而确定正确选项。

(1)
展开组合数定义并提取公因子

本步旨在通过组合数的阶乘定义，将 Cnr\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} 转化为包含 Cn−1r−1\htmlData{tutor-start=0,tutor-end=1}{C}_{\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=9,tutor-end=10}{r}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}} 或 Cn−1r\htmlData{tutor-start=0,tutor-end=1}{C}_{\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=9,tutor-end=10}{r}} 的形式，这是处理组合数恒等式最基础且通用的方法。

详细展开： 根据组合数的定义，我们有： Cnr=n!r!(n−r)!\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n}\htmlData{tutor-start=19,tutor-end=20}{!}}{\htmlData{tutor-start=22,tutor-end=23}{r}\htmlData{tutor-start=23,tutor-end=24}{!}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{n}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{r}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{!}} 为了寻找与 n,r\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{r} 相关的递推或恒等关系，我们尝试从分子和分母中分离出 n\htmlData{tutor-start=0,tutor-end=1}{n} 和 r\htmlData{tutor-start=0,tutor-end=1}{r}。 注意到 n!=n⋅(n−1)!\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{!} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{n} \htmlData{tutor-start=7,tutor-end=13}{\cdot }\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{n}\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}{!} 以及 r!=r⋅(r−1)!\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{!} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{r} \htmlData{tutor-start=7,tutor-end=13}{\cdot }\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{r}\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}{!}。 将上述关系代入定义式中： Cnr=n⋅(n−1)!r⋅(r−1)!⋅(n−r)!\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n} \htmlData{tutor-start=20,tutor-end=26}{\cdot }\htmlData{tutor-start=26,tutor-end=27}{(}\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=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{!}}{\htmlData{tutor-start=34,tutor-end=35}{r} \htmlData{tutor-start=36,tutor-end=42}{\cdot }\htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{r}\htmlData{tutor-start=44,tutor-end=45}{-}\htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{)}\htmlData{tutor-start=47,tutor-end=48}{!} \htmlData{tutor-start=49,tutor-end=55}{\cdot }\htmlData{tutor-start=55,tutor-end=56}{(}\htmlData{tutor-start=56,tutor-end=57}{n}\htmlData{tutor-start=57,tutor-end=58}{-}\htmlData{tutor-start=58,tutor-end=59}{r}\htmlData{tutor-start=59,tutor-end=60}{)}\htmlData{tutor-start=60,tutor-end=61}{!}} 我们将常数系数 nr\frac{\htmlData{tutor-start=6,tutor-end=7}{n}}{\htmlData{tutor-start=9,tutor-end=10}{r}} 提取出来，剩余部分重新组合： Cnr=nr⋅[(n−1)!(r−1)!⋅(n−r)!]\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n}}{\htmlData{tutor-start=21,tutor-end=22}{r}} \htmlData{tutor-start=24,tutor-end=30}{\cdot }\left[ \frac{\htmlData{tutor-start=43,tutor-end=44}{(}\htmlData{tutor-start=44,tutor-end=45}{n}\htmlData{tutor-start=45,tutor-end=46}{-}\htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{)}\htmlData{tutor-start=48,tutor-end=49}{!}}{\htmlData{tutor-start=51,tutor-end=52}{(}\htmlData{tutor-start=52,tutor-end=53}{r}\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{1}\htmlData{tutor-start=55,tutor-end=56}{)}\htmlData{tutor-start=56,tutor-end=57}{!} \htmlData{tutor-start=58,tutor-end=64}{\cdot }\htmlData{tutor-start=64,tutor-end=65}{(}\htmlData{tutor-start=65,tutor-end=66}{n}\htmlData{tutor-start=66,tutor-end=67}{-}\htmlData{tutor-start=67,tutor-end=68}{r}\htmlData{tutor-start=68,tutor-end=69}{)}\htmlData{tutor-start=69,tutor-end=70}{!}} \right] 观察方括号内的部分，分子是 (n−1)!\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{n}\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}{!}，分母中 (r−1)!\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{r}\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}{!} 对应下标减1后的上标，(n−r)!\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{r}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{!} 可以写成 ((n−1)−(r−1))!\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{r}\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}{!}。 这正是组合数 Cn−1r−1\htmlData{tutor-start=0,tutor-end=1}{C}_{\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=9,tutor-end=10}{r}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}} 的定义形式： Cn−1r−1=(n−1)!(r−1)!((n−1)−(r−1))!=(n−1)!(r−1)!(n−r)!\htmlData{tutor-start=0,tutor-end=1}{C}_{\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=9,tutor-end=10}{r}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{n}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{!}}{\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{r}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{)}\htmlData{tutor-start=35,tutor-end=36}{!} \htmlData{tutor-start=37,tutor-end=38}{(}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{n}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{)}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{(}\htmlData{tutor-start=45,tutor-end=46}{r}\htmlData{tutor-start=46,tutor-end=47}{-}\htmlData{tutor-start=47,tutor-end=48}{1}\htmlData{tutor-start=48,tutor-end=49}{)}\htmlData{tutor-start=49,tutor-end=50}{)}\htmlData{tutor-start=50,tutor-end=51}{!}} \htmlData{tutor-start=53,tutor-end=54}{=} \frac{\htmlData{tutor-start=61,tutor-end=62}{(}\htmlData{tutor-start=62,tutor-end=63}{n}\htmlData{tutor-start=63,tutor-end=64}{-}\htmlData{tutor-start=64,tutor-end=65}{1}\htmlData{tutor-start=65,tutor-end=66}{)}\htmlData{tutor-start=66,tutor-end=67}{!}}{\htmlData{tutor-start=69,tutor-end=70}{(}\htmlData{tutor-start=70,tutor-end=71}{r}\htmlData{tutor-start=71,tutor-end=72}{-}\htmlData{tutor-start=72,tutor-end=73}{1}\htmlData{tutor-start=73,tutor-end=74}{)}\htmlData{tutor-start=74,tutor-end=75}{!} \htmlData{tutor-start=76,tutor-end=77}{(}\htmlData{tutor-start=77,tutor-end=78}{n}\htmlData{tutor-start=78,tutor-end=79}{-}\htmlData{tutor-start=79,tutor-end=80}{r}\htmlData{tutor-start=80,tutor-end=81}{)}\htmlData{tutor-start=81,tutor-end=82}{!}} 因此，我们得到恒等式： Cnr=nrCn−1r−1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n}}{\htmlData{tutor-start=21,tutor-end=22}{r}} \htmlData{tutor-start=24,tutor-end=25}{C}_{\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=33,tutor-end=34}{r}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}}

Cnr=nrCn−1r−1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n}}{\htmlData{tutor-start=21,tutor-end=22}{r}} \htmlData{tutor-start=24,tutor-end=25}{C}_{\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=33,tutor-end=34}{r}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}}
(2)
验证其他常见恒等式并排除干扰项

本步通过推导另一个常见的组合数恒等式 Cnr=Cn−1r+Cn−1r−1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{C}_{\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}}^{\htmlData{tutor-start=21,tutor-end=22}{r}} \htmlData{tutor-start=24,tutor-end=25}{+} \htmlData{tutor-start=26,tutor-end=27}{C}_{\htmlData{tutor-start=29,tutor-end=30}{n}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{1}}^{\htmlData{tutor-start=35,tutor-end=36}{r}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{1}} 以及 Cnr=nn−rCn−1r\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n}}{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{r}} \htmlData{tutor-start=26,tutor-end=27}{C}_{\htmlData{tutor-start=29,tutor-end=30}{n}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{1}}^{\htmlData{tutor-start=35,tutor-end=36}{r}}，来对比题目可能的选项形式，确保结论的唯一性和准确性。

详细展开： 除了上述推导，我们还可以尝试保留 r!\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{!} 不变，只分解 n!\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{!} 和 (n−r)!\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{r}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{!}。 Cnr=n!r!(n−r)!=n⋅(n−1)!r!⋅(n−r)⋅(n−r−1)!\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n}\htmlData{tutor-start=19,tutor-end=20}{!}}{\htmlData{tutor-start=22,tutor-end=23}{r}\htmlData{tutor-start=23,tutor-end=24}{!}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{n}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{r}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{!}} \htmlData{tutor-start=32,tutor-end=33}{=} \frac{\htmlData{tutor-start=40,tutor-end=41}{n} \htmlData{tutor-start=42,tutor-end=48}{\cdot }\htmlData{tutor-start=48,tutor-end=49}{(}\htmlData{tutor-start=49,tutor-end=50}{n}\htmlData{tutor-start=50,tutor-end=51}{-}\htmlData{tutor-start=51,tutor-end=52}{1}\htmlData{tutor-start=52,tutor-end=53}{)}\htmlData{tutor-start=53,tutor-end=54}{!}}{\htmlData{tutor-start=56,tutor-end=57}{r}\htmlData{tutor-start=57,tutor-end=58}{!} \htmlData{tutor-start=59,tutor-end=65}{\cdot }\htmlData{tutor-start=65,tutor-end=66}{(}\htmlData{tutor-start=66,tutor-end=67}{n}\htmlData{tutor-start=67,tutor-end=68}{-}\htmlData{tutor-start=68,tutor-end=69}{r}\htmlData{tutor-start=69,tutor-end=70}{)} \htmlData{tutor-start=71,tutor-end=77}{\cdot }\htmlData{tutor-start=77,tutor-end=78}{(}\htmlData{tutor-start=78,tutor-end=79}{n}\htmlData{tutor-start=79,tutor-end=80}{-}\htmlData{tutor-start=80,tutor-end=81}{r}\htmlData{tutor-start=81,tutor-end=82}{-}\htmlData{tutor-start=82,tutor-end=83}{1}\htmlData{tutor-start=83,tutor-end=84}{)}\htmlData{tutor-start=84,tutor-end=85}{!}} 提取系数 nn−r\frac{\htmlData{tutor-start=6,tutor-end=7}{n}}{\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{r}}： Cnr=nn−r⋅(n−1)!r!(n−r−1)!\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n}}{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{r}} \htmlData{tutor-start=26,tutor-end=32}{\cdot }\frac{\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{n}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{)}\htmlData{tutor-start=43,tutor-end=44}{!}}{\htmlData{tutor-start=46,tutor-end=47}{r}\htmlData{tutor-start=47,tutor-end=48}{!} \htmlData{tutor-start=49,tutor-end=50}{(}\htmlData{tutor-start=50,tutor-end=51}{n}\htmlData{tutor-start=51,tutor-end=52}{-}\htmlData{tutor-start=52,tutor-end=53}{r}\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{1}\htmlData{tutor-start=55,tutor-end=56}{)}\htmlData{tutor-start=56,tutor-end=57}{!}} 注意到 (n−1)!r!((n−1)−r)!=Cn−1r\frac{\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{n}\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=14,tutor-end=15}{r}\htmlData{tutor-start=15,tutor-end=16}{!} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{)}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{r}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{!}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{C}_{\htmlData{tutor-start=34,tutor-end=35}{n}\htmlData{tutor-start=35,tutor-end=36}{-}\htmlData{tutor-start=36,tutor-end=37}{1}}^{\htmlData{tutor-start=40,tutor-end=41}{r}}。 所以另一个恒等式为： Cnr=nn−rCn−1r\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n}}{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{r}} \htmlData{tutor-start=26,tutor-end=27}{C}_{\htmlData{tutor-start=29,tutor-end=30}{n}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{1}}^{\htmlData{tutor-start=35,tutor-end=36}{r}} 在常见的选择题设置中，干扰项通常包括： 1. rnCn−1r−1\frac{\htmlData{tutor-start=6,tutor-end=7}{r}}{\htmlData{tutor-start=9,tutor-end=10}{n}} \htmlData{tutor-start=12,tutor-end=13}{C}_{\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}}^{\htmlData{tutor-start=21,tutor-end=22}{r}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}} （系数倒数错误） 2. nrCn−1r\frac{\htmlData{tutor-start=6,tutor-end=7}{n}}{\htmlData{tutor-start=9,tutor-end=10}{r}} \htmlData{tutor-start=12,tutor-end=13}{C}_{\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}}^{\htmlData{tutor-start=21,tutor-end=22}{r}} （下标匹配错误） 3. Cn−1r−1+Cn−1r\htmlData{tutor-start=0,tutor-end=1}{C}_{\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=9,tutor-end=10}{r}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{C}_{\htmlData{tutor-start=19,tutor-end=20}{n}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{1}}^{\htmlData{tutor-start=25,tutor-end=26}{r}} （这是帕斯卡恒等式，结果等于 Cnr\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}}，但形式是和而非积）

根据第一步的推导，Cnr=nrCn−1r−1\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n}}{\htmlData{tutor-start=21,tutor-end=22}{r}} \htmlData{tutor-start=24,tutor-end=25}{C}_{\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=33,tutor-end=34}{r}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}} 是严格成立的代数恒等式。 若题目选项中有 nrCn−1r−1\frac{\htmlData{tutor-start=6,tutor-end=7}{n}}{\htmlData{tutor-start=9,tutor-end=10}{r}} \htmlData{tutor-start=12,tutor-end=13}{C}_{\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1}}^{\htmlData{tutor-start=21,tutor-end=22}{r}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}}，则为正确答案。 此结论也可通过具体数值验证： 取 n=4,r=2\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{,} \htmlData{tutor-start=5,tutor-end=6}{r}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}。 左边 C42=6\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{4}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{6}。 右边 42C31=2×3=6\frac{\htmlData{tutor-start=6,tutor-end=7}{4}}{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{C}_{\htmlData{tutor-start=15,tutor-end=16}{3}}^{\htmlData{tutor-start=19,tutor-end=20}{1}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{2} \htmlData{tutor-start=26,tutor-end=33}{\times }\htmlData{tutor-start=33,tutor-end=34}{3} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{6}。成立。 若取错误形式 nn−rCn−1r−1=42C31=6\frac{\htmlData{tutor-start=6,tutor-end=7}{n}}{\htmlData{tutor-start=9,tutor-end=10}{n}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{r}} \htmlData{tutor-start=14,tutor-end=15}{C}_{\htmlData{tutor-start=17,tutor-end=18}{n}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}}^{\htmlData{tutor-start=23,tutor-end=24}{r}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}} \htmlData{tutor-start=28,tutor-end=29}{=} \frac{\htmlData{tutor-start=36,tutor-end=37}{4}}{\htmlData{tutor-start=39,tutor-end=40}{2}} \htmlData{tutor-start=42,tutor-end=43}{C}_{\htmlData{tutor-start=45,tutor-end=46}{3}}^{\htmlData{tutor-start=49,tutor-end=50}{1}} \htmlData{tutor-start=52,tutor-end=53}{=} \htmlData{tutor-start=54,tutor-end=55}{6} (此例巧合相等，需换例)。 取 n=5,r=2\htmlData{tutor-start=0,tutor-end=1}{n}\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=5,tutor-end=6}{r}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}。 左边 C52=10\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{5}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{0}。 正确形式 52C41=2.5×4=10\frac{\htmlData{tutor-start=6,tutor-end=7}{5}}{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{C}_{\htmlData{tutor-start=15,tutor-end=16}{4}}^{\htmlData{tutor-start=19,tutor-end=20}{1}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{.}\htmlData{tutor-start=26,tutor-end=27}{5} \htmlData{tutor-start=28,tutor-end=35}{\times }\htmlData{tutor-start=35,tutor-end=36}{4} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{0}。成立。 错误形式 52C42=2.5×6=15≠10\frac{\htmlData{tutor-start=6,tutor-end=7}{5}}{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=13}{C}_{\htmlData{tutor-start=15,tutor-end=16}{4}}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{.}\htmlData{tutor-start=26,tutor-end=27}{5} \htmlData{tutor-start=28,tutor-end=35}{\times }\htmlData{tutor-start=35,tutor-end=36}{6} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{5} \neq \htmlData{tutor-start=47,tutor-end=48}{1}\htmlData{tutor-start=48,tutor-end=49}{0}。不成立。

Cnr=nn−rCn−1r\htmlData{tutor-start=0,tutor-end=1}{C}_{\htmlData{tutor-start=3,tutor-end=4}{n}}^{\htmlData{tutor-start=7,tutor-end=8}{r}} \htmlData{tutor-start=10,tutor-end=11}{=} \frac{\htmlData{tutor-start=18,tutor-end=19}{n}}{\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{r}} \htmlData{tutor-start=26,tutor-end=27}{C}_{\htmlData{tutor-start=29,tutor-end=30}{n}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{1}}^{\htmlData{tutor-start=35,tutor-end=36}{r}}
13

二、选择题 · 立体几何

给定空间中的直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 及平面 α\htmlData{tutor-start=0,tutor-end=6}{\alpha}，条件“直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与平面 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 内无数条直线都垂直”是“直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与平面 α\htmlData{tutor-start=0,tutor-end=6}{\alpha} 垂直”的 ( )

答案：必要非充分

题目标签：线面垂直条件判断

解题过程

判断线面垂直条件

分别检验必要性和充分性

(1)
检验必要性

若 l⊥α\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=6}{\perp}\htmlData{tutor-start=6,tutor-end=12}{\alpha}，则 l 与平面内任意方向的直线所成角均为 90∘90^\circ。

详细展开：因此 l 与平面内无数条直线垂直，题给条件是线面垂直的必要条件。

l⊥α⟹l⊥α 内无数条直线\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=6}{\perp}\htmlData{tutor-start=6,tutor-end=12}{\alpha}\htmlData{tutor-start=12,tutor-end=28}{\Longrightarrow }\htmlData{tutor-start=28,tutor-end=29}{l}\htmlData{tutor-start=29,tutor-end=34}{\perp}\htmlData{tutor-start=34,tutor-end=40}{\alpha}\text{ \htmlData{tutor-start=47,tutor-end=48}{内}\htmlData{tutor-start=48,tutor-end=49}{无}\htmlData{tutor-start=49,tutor-end=50}{数}\htmlData{tutor-start=50,tutor-end=51}{条}\htmlData{tutor-start=51,tutor-end=52}{直}\htmlData{tutor-start=52,tutor-end=53}{线}}
(2)
否定充分性

无数条直线可能全都平行，只有同一个方向。

详细展开：若 l 不垂直平面，平面内仍可取一个与 l 方向垂直的方向，并作无数条彼此平行的直线；它们都与 l 垂直，却不能推出 l⊥α\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=6}{\perp}\htmlData{tutor-start=6,tutor-end=12}{\alpha}。

必要非充分条件（B）\boxed{\text{\htmlData{tutor-start=13,tutor-end=14}{必}\htmlData{tutor-start=14,tutor-end=15}{要}\htmlData{tutor-start=15,tutor-end=16}{非}\htmlData{tutor-start=16,tutor-end=17}{充}\htmlData{tutor-start=17,tutor-end=18}{分}\htmlData{tutor-start=18,tutor-end=19}{条}\htmlData{tutor-start=19,tutor-end=20}{件}\htmlData{tutor-start=20,tutor-end=21}{（}\htmlData{tutor-start=21,tutor-end=22}{B}\htmlData{tutor-start=22,tutor-end=23}{）}}}
14

二、选择题 · 数列

若数列 {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}{\}} 是首项为 1\htmlData{tutor-start=0,tutor-end=1}{1}，公比为 a−32\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{-}\frac{\htmlData{tutor-start=8,tutor-end=9}{3}}{\htmlData{tutor-start=11,tutor-end=12}{2}} 的无穷等比数列，且 {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}{\}} 各项的和为 a\htmlData{tutor-start=0,tutor-end=1}{a}，则 a\htmlData{tutor-start=0,tutor-end=1}{a} 的值是 ( )

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

题目标签：无穷等比数列求和

解题过程

求无穷等比数列参数

联立求和公式与收敛条件

(1)
由和列方程

首项为 1、公比 q=a−3/2\htmlData{tutor-start=0,tutor-end=1}{q}\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}{3}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{2}。

详细展开：若级数收敛，其和为 1/(1−q)=1/(5/2−a)\htmlData{tutor-start=0,tutor-end=1}{1}\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}{q}\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}{(}\htmlData{tutor-start=11,tutor-end=12}{5}\htmlData{tutor-start=12,tutor-end=13}{/}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{a}\htmlData{tutor-start=16,tutor-end=17}{)}。令该和等于 a，得到 2a2−5a+2=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{a}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{0}。

a(52−a)=1⟹a=12 或 2\htmlData{tutor-start=0,tutor-end=1}{a}\left(\frac{\htmlData{tutor-start=13,tutor-end=14}{5}}{\htmlData{tutor-start=16,tutor-end=17}{2}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{a}\right)\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=45}{\Longrightarrow }\htmlData{tutor-start=45,tutor-end=46}{a}\htmlData{tutor-start=46,tutor-end=47}{=}\frac{\htmlData{tutor-start=53,tutor-end=54}{1}}{\htmlData{tutor-start=56,tutor-end=57}{2}}\text{ \htmlData{tutor-start=65,tutor-end=66}{或} }\htmlData{tutor-start=68,tutor-end=69}{2}
(2)
用收敛性筛选

无穷等比数列求和必须有 ∣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}。

详细展开：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} 时 q=−1\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1} 不收敛；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 q=1/2\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}{2} 收敛且和为 2。

a=2（B）\boxed{\htmlData{tutor-start=7,tutor-end=8}{a}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{2}\text{\htmlData{tutor-start=16,tutor-end=17}{（}\htmlData{tutor-start=17,tutor-end=18}{B}\htmlData{tutor-start=18,tutor-end=19}{）}}}
15

二、选择题 · 集合与几何

如图，在平面直角坐标系中，Ω\htmlData{tutor-start=0,tutor-end=6}{\Omega} 是一个与 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴的正半轴、y\htmlData{tutor-start=0,tutor-end=1}{y} 轴的正半轴分别相切于点 C\htmlData{tutor-start=0,tutor-end=1}{C}、D\htmlData{tutor-start=0,tutor-end=1}{D} 的定圆所围成区域（含边界），A\htmlData{tutor-start=0,tutor-end=1}{A}、B\htmlData{tutor-start=0,tutor-end=1}{B}、C\htmlData{tutor-start=0,tutor-end=1}{C}、D\htmlData{tutor-start=0,tutor-end=1}{D} 是该圆的四等分点. 若点 P(x,y)\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{)}、点 P′(x′,y′)\htmlData{tutor-start=0,tutor-end=1}{P}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}'\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{y}'\htmlData{tutor-start=8,tutor-end=9}{)} 满足 x⩽x′\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=12}{\leqslant }\htmlData{tutor-start=12,tutor-end=13}{x}' 且 y⩾y′\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{y}'，则称 P\htmlData{tutor-start=0,tutor-end=1}{P} 优于 P′\htmlData{tutor-start=0,tutor-end=1}{P}'. 如果 Ω\htmlData{tutor-start=0,tutor-end=6}{\Omega} 中的点 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 满足：不存在 Ω\htmlData{tutor-start=0,tutor-end=6}{\Omega} 中的其它点优于 Q\htmlData{tutor-start=0,tutor-end=1}{Q}，那么所有这样的点 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 组成的集合是劣弧 ( )

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：劣弧DA

题目标签：圆盘偏序的不可支配边界

解题过程

确定不被优于的圆弧

把偏序解释为向左上方支配

(1)
理解优于方向

P\htmlData{tutor-start=0,tutor-end=1}{P} 优于 P′\htmlData{tutor-start=0,tutor-end=1}{P}' 要求 P 的横坐标不大、纵坐标不小。

详细展开：因此从任一点向左上方寻找另一个圆盘内点，若能找到，该点就不是最优点。圆的四等分点依次为 A 上、B 右、C 下、D 左。

优于方向：横坐标减小且纵坐标增大\text{\htmlData{tutor-start=6,tutor-end=7}{优}\htmlData{tutor-start=7,tutor-end=8}{于}\htmlData{tutor-start=8,tutor-end=9}{方}\htmlData{tutor-start=9,tutor-end=10}{向}\htmlData{tutor-start=10,tutor-end=11}{：}\htmlData{tutor-start=11,tutor-end=12}{横}\htmlData{tutor-start=12,tutor-end=13}{坐}\htmlData{tutor-start=13,tutor-end=14}{标}\htmlData{tutor-start=14,tutor-end=15}{减}\htmlData{tutor-start=15,tutor-end=16}{小}\htmlData{tutor-start=16,tutor-end=17}{且}\htmlData{tutor-start=17,tutor-end=18}{纵}\htmlData{tutor-start=18,tutor-end=19}{坐}\htmlData{tutor-start=19,tutor-end=20}{标}\htmlData{tutor-start=20,tutor-end=21}{增}\htmlData{tutor-start=21,tutor-end=22}{大}}
(2)
识别帕累托边界

弧 AB 上除 A 外均被 A 优于；弧 BC 上点也被 A 优于；弧 CD 上除 D 外均被 D 优于。

详细展开：在劣弧 DA 上，从 D 到 A 时 x、y 同时增加：想减小 x 就会降低 y，想增大 y 就会增大 x，故没有另一点能同时改进两个指标。

\wideparenDA（D）\boxed{\wideparen{DA}\text{（D）}}
16

三、解答题 · 立体几何

如图，在棱长为 2\htmlData{tutor-start=0,tutor-end=1}{2} 的正方体 ABCD−A1B1C1D1\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{A}_{\htmlData{tutor-start=8,tutor-end=9}{1}}\htmlData{tutor-start=10,tutor-end=11}{B}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{C}_{\htmlData{tutor-start=18,tutor-end=19}{1}}\htmlData{tutor-start=20,tutor-end=21}{D}_{\htmlData{tutor-start=23,tutor-end=24}{1}} 中，E\htmlData{tutor-start=0,tutor-end=1}{E} 是 BC1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 的中点. 求直线 DE\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E} 与平面 ABCD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D} 所成角的大小. (结果用反三角函数表示)

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：arcsin(√6/6)

题目标签：正方体中的线面角

解题过程

求直线与底面夹角

用坐标求方向向量的竖直分量占比

(1)
建立正方体坐标

取 A=(0,0,0),B=(2,0,0),C=(2,2,0),D=(0,2,0),C1=(2,2,2)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{B}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{)}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{C}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{,}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{)}\htmlData{tutor-start=29,tutor-end=30}{,}\htmlData{tutor-start=30,tutor-end=31}{D}\htmlData{tutor-start=31,tutor-end=32}{=}\htmlData{tutor-start=32,tutor-end=33}{(}\htmlData{tutor-start=33,tutor-end=34}{0}\htmlData{tutor-start=34,tutor-end=35}{,}\htmlData{tutor-start=35,tutor-end=36}{2}\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}{C}_{\htmlData{tutor-start=43,tutor-end=44}{1}}\htmlData{tutor-start=45,tutor-end=46}{=}\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{2}\htmlData{tutor-start=48,tutor-end=49}{,}\htmlData{tutor-start=49,tutor-end=50}{2}\htmlData{tutor-start=50,tutor-end=51}{,}\htmlData{tutor-start=51,tutor-end=52}{2}\htmlData{tutor-start=52,tutor-end=53}{)}。

详细展开：E 是 BC1\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}_{\htmlData{tutor-start=4,tutor-end=5}{1}} 中点，所以 E=(2,1,1)\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{2}\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}{1}\htmlData{tutor-start=8,tutor-end=9}{)}，进而 DE→=(2,−1,1)\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{D}\htmlData{tutor-start=17,tutor-end=18}{E}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{,}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{,}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{)}。

E=(2,1,1),DE→=(2,−1,1)\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{2}\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}{1}\htmlData{tutor-start=8,tutor-end=9}{)}\htmlData{tutor-start=9,tutor-end=10}{,}\quad\overrightarrow{\htmlData{tutor-start=31,tutor-end=32}{D}\htmlData{tutor-start=32,tutor-end=33}{E}}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{(}\htmlData{tutor-start=36,tutor-end=37}{2}\htmlData{tutor-start=37,tutor-end=38}{,}\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}{1}\htmlData{tutor-start=42,tutor-end=43}{)}
(2)
计算线面角正弦

底面 ABCD 的法向为竖直方向。

详细展开：设线面角为 θ\htmlData{tutor-start=0,tutor-end=6}{\theta}，则 sin⁡θ\sin\htmlData{tutor-start=4,tutor-end=10}{\theta} 等于 DE 的竖直分量绝对值除以向量长度，即 1/6\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\sqrt{\htmlData{tutor-start=8,tutor-end=9}{6}}。

theta=arcsin⁡66\boxed{\\\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{h}\htmlData{tutor-start=11,tutor-end=12}{e}\htmlData{tutor-start=12,tutor-end=13}{t}\htmlData{tutor-start=13,tutor-end=14}{a}\htmlData{tutor-start=14,tutor-end=15}{=}\arcsin\frac{\sqrt{\htmlData{tutor-start=34,tutor-end=35}{6}}}{\htmlData{tutor-start=38,tutor-end=39}{6}}}
17

三、解答题 · 解三角形

如图，某住宅小区的平面图呈圆心角为 120∘120^\circ 的扇形 AOB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{O}\htmlData{tutor-start=2,tutor-end=3}{B}. 小区的两个出入口设置在点 A\htmlData{tutor-start=0,tutor-end=1}{A} 及点 C\htmlData{tutor-start=0,tutor-end=1}{C} 处，且小区里有一条平行于 BO\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{O} 的小路 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}，已知某人从 C\htmlData{tutor-start=0,tutor-end=1}{C} 沿 CD\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D} 走到 D\htmlData{tutor-start=0,tutor-end=1}{D} 用了 10\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{0} 分钟，从 D\htmlData{tutor-start=0,tutor-end=1}{D} 沿 DA\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{A} 走到 A\htmlData{tutor-start=0,tutor-end=1}{A} 用了 6\htmlData{tutor-start=0,tutor-end=1}{6} 分钟，若此人步行的速度为每分钟 50\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{0} 米，求该扇形的半径 OA\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{A} 的长. (精确到 1\htmlData{tutor-start=0,tutor-end=1}{1} 米)

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：445米

题目标签：平行小路与扇形测距

解题过程

求扇形半径

利用平行关系把路线夹角转成 120°

(1)
确定三角形边角

由步速与时间，CD=500\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}{5}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0} m、DA=300\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0} m。

详细展开：图中 D∈AO\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=5}{\in }\htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{O}，且 CD∥BO\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=12}{\parallel }\htmlData{tutor-start=12,tutor-end=13}{B}\htmlData{tutor-start=13,tutor-end=14}{O}；扇形圆心角 ∠AOB=120∘\angle AOB=120^\circ，所以 ∠ADC=120∘\angle ADC=120^\circ。设半径为 R，则 OD=R−300\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{R}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{0}。

CD=500,AD=300,∠ADC=120∘CD=500,\quad AD=300,\quad\angle ADC=120^\circ
(2)
在 DOC 中列方程

因 DA 与 DO 反向，∠ODC=60∘\angle ODC=60^\circ。

详细展开：三角形 DOC 中 OC=R\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{R}，余弦定理给 R2=(R−300)2+5002−500(R−300)\htmlData{tutor-start=0,tutor-end=1}{R}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{R}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{5}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{0}^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{5}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{R}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{3}\htmlData{tutor-start=33,tutor-end=34}{0}\htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=36}{)}，解得 R=4900/11≈445.45\htmlData{tutor-start=0,tutor-end=1}{R}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{9}\htmlData{tutor-start=4,tutor-end=5}{0}\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}{1}\htmlData{tutor-start=9,tutor-end=16}{\approx}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{4}\htmlData{tutor-start=18,tutor-end=19}{5}\htmlData{tutor-start=19,tutor-end=20}{.}\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{5}。

OA≈445 m\boxed{\htmlData{tutor-start=7,tutor-end=8}{O}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=16}{\approx}\htmlData{tutor-start=16,tutor-end=17}{4}\htmlData{tutor-start=17,tutor-end=18}{4}\htmlData{tutor-start=18,tutor-end=19}{5}\text{ \htmlData{tutor-start=26,tutor-end=27}{m}}}
18

三、解答题 · 三角函数

已知函数 f(x)=sin⁡2x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sin \htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{x}，g(x)=cos⁡(2x+π6)\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \cos\left(\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{x}\htmlData{tutor-start=19,tutor-end=20}{+}\frac{\htmlData{tutor-start=26,tutor-end=29}{\pi}}{\htmlData{tutor-start=31,tutor-end=32}{6}}\right)，直线 x=t\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{t} (t∈R\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=6}{\in }\mathbf{\htmlData{tutor-start=14,tutor-end=15}{R}}) 与函数 f(x)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}、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}{)} 的图象分别交于 M\htmlData{tutor-start=0,tutor-end=1}{M}、N\htmlData{tutor-start=0,tutor-end=1}{N} 两点. (1) 当 t=π4\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\frac{\htmlData{tutor-start=8,tutor-end=11}{\pi}}{\htmlData{tutor-start=13,tutor-end=14}{4}} 时，求 ∣MN∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{N}\htmlData{tutor-start=3,tutor-end=4}{|} 的值； (2) 求 ∣MN∣\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{N}\htmlData{tutor-start=3,tutor-end=4}{|} 在 t∈[0,π2]\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=6}{\in }\left[\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{,} \frac{\htmlData{tutor-start=21,tutor-end=24}{\pi}}{\htmlData{tutor-start=26,tutor-end=27}{2}}\right] 时的最大值.

答案：3/2；√3

题目标签：两三角函数图象的竖直距离

解题过程

（1）求给定位置的竖直距离

同一竖线上的距离等于函数值之差的绝对值

(1)
计算两个函数值

当 t=π/4\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}{4} 时，f(t)=sin⁡(π/2)=1\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\sin\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=13}{\pi}\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}。

详细展开：g(t)=cos⁡(π/2+π/6)=cos⁡(2π/3)=−1/2\htmlData{tutor-start=0,tutor-end=1}{g}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\cos\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=13}{\pi}\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=19}{\pi}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{6}\htmlData{tutor-start=21,tutor-end=22}{)}\htmlData{tutor-start=22,tutor-end=23}{=}\cos\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=32}{\pi}\htmlData{tutor-start=32,tutor-end=33}{/}\htmlData{tutor-start=33,tutor-end=34}{3}\htmlData{tutor-start=34,tutor-end=35}{)}\htmlData{tutor-start=35,tutor-end=36}{=}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{/}\htmlData{tutor-start=39,tutor-end=40}{2}。

f(π/4)=1,g(π/4)=−12\htmlData{tutor-start=0,tutor-end=1}{f}\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}{4}\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}{,}\quad \htmlData{tutor-start=17,tutor-end=18}{g}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=22}{\pi}\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}{=}\htmlData{tutor-start=26,tutor-end=27}{-}\frac{\htmlData{tutor-start=33,tutor-end=34}{1}}{\htmlData{tutor-start=36,tutor-end=37}{2}}
(2)
作纵坐标差

M、N 横坐标相同，线段 MN 是竖直线段。

详细展开：所以 ∣MN∣=∣f(t)−g(t)∣=∣1−(−1/2)∣=3/2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{M}\htmlData{tutor-start=2,tutor-end=3}{N}\htmlData{tutor-start=3,tutor-end=4}{|}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{|}\htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{t}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{g}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{t}\htmlData{tutor-start=14,tutor-end=15}{)}\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}{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}{1}\htmlData{tutor-start=23,tutor-end=24}{/}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{|}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=30}{/}\htmlData{tutor-start=30,tutor-end=31}{2}。

∣MN∣=32\boxed{\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{M}\htmlData{tutor-start=9,tutor-end=10}{N}\htmlData{tutor-start=10,tutor-end=11}{|}\htmlData{tutor-start=11,tutor-end=12}{=}\frac{\htmlData{tutor-start=18,tutor-end=19}{3}}{\htmlData{tutor-start=21,tutor-end=22}{2}}}

（2）求区间最大距离

合成同频正弦并检查受限相位区间

(1)
合成函数差

令 u=2t∈[0,π]\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}{t}\htmlData{tutor-start=4,tutor-end=7}{\in}\htmlData{tutor-start=7,tutor-end=8}{[}\htmlData{tutor-start=8,tutor-end=9}{0}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=13}{\pi}\htmlData{tutor-start=13,tutor-end=14}{]}。

详细展开：f−g=sin⁡u−cos⁡(u+π/6)=32sin⁡u−32cos⁡u=3sin⁡(u−π/6)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{g}\htmlData{tutor-start=3,tutor-end=4}{=}\sin \htmlData{tutor-start=9,tutor-end=10}{u}\htmlData{tutor-start=10,tutor-end=11}{-}\cos\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{u}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=21}{\pi}\htmlData{tutor-start=21,tutor-end=22}{/}\htmlData{tutor-start=22,tutor-end=23}{6}\htmlData{tutor-start=23,tutor-end=24}{)}\htmlData{tutor-start=24,tutor-end=25}{=}\frac{\htmlData{tutor-start=31,tutor-end=32}{3}}{\htmlData{tutor-start=34,tutor-end=35}{2}}\sin \htmlData{tutor-start=41,tutor-end=42}{u}\htmlData{tutor-start=42,tutor-end=43}{-}\frac{\sqrt{\htmlData{tutor-start=55,tutor-end=56}{3}}}{\htmlData{tutor-start=59,tutor-end=60}{2}}\cos \htmlData{tutor-start=66,tutor-end=67}{u}\htmlData{tutor-start=67,tutor-end=68}{=}\sqrt{\htmlData{tutor-start=74,tutor-end=75}{3}}\sin\htmlData{tutor-start=80,tutor-end=81}{(}\htmlData{tutor-start=81,tutor-end=82}{u}\htmlData{tutor-start=82,tutor-end=83}{-}\htmlData{tutor-start=83,tutor-end=86}{\pi}\htmlData{tutor-start=86,tutor-end=87}{/}\htmlData{tutor-start=87,tutor-end=88}{6}\htmlData{tutor-start=88,tutor-end=89}{)}。

∣MN∣=3∣sin⁡(u−π6)∣|MN|=\sqrt{3}\left|\sin\left(u-\frac\pi6\right)\right|
(2)
检查相位范围

u−π/6∈[−π/6,5π/6]\htmlData{tutor-start=0,tutor-end=1}{u}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=5}{\pi}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{6}\htmlData{tutor-start=7,tutor-end=10}{\in}\htmlData{tutor-start=10,tutor-end=11}{[}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=15}{\pi}\htmlData{tutor-start=15,tutor-end=16}{/}\htmlData{tutor-start=16,tutor-end=17}{6}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{5}\htmlData{tutor-start=19,tutor-end=22}{\pi}\htmlData{tutor-start=22,tutor-end=23}{/}\htmlData{tutor-start=23,tutor-end=24}{6}\htmlData{tutor-start=24,tutor-end=25}{]}，其中包含 π/2\htmlData{tutor-start=0,tutor-end=3}{\pi}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{2}。

详细展开：绝对正弦的最大值 1 可在 u=2π/3\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=6}{\pi}\htmlData{tutor-start=6,tutor-end=7}{/}\htmlData{tutor-start=7,tutor-end=8}{3} 即 t=π/3\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}{3} 取得，所以距离最大值为振幅。

∣MN∣max⁡=3\boxed{\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{M}\htmlData{tutor-start=9,tutor-end=10}{N}\htmlData{tutor-start=10,tutor-end=11}{|}_{\max}\htmlData{tutor-start=18,tutor-end=19}{=}\sqrt{\htmlData{tutor-start=25,tutor-end=26}{3}}}
19

解答题 · 函数

已知函数 f(x)=2x−12∣x∣\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{2}^{\htmlData{tutor-start=10,tutor-end=11}{x}} \htmlData{tutor-start=13,tutor-end=14}{-} \frac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{2}^{\htmlData{tutor-start=27,tutor-end=28}{|}\htmlData{tutor-start=28,tutor-end=29}{x}\htmlData{tutor-start=29,tutor-end=30}{|}}}. (1) 若 f(x)=2\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{2}，求 x\htmlData{tutor-start=0,tutor-end=1}{x} 的值； (2) 若 2tf(2t)+mf(t)⩾0\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{t}} \htmlData{tutor-start=6,tutor-end=7}{f}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{)} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{m} \htmlData{tutor-start=16,tutor-end=17}{f}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{t}\htmlData{tutor-start=19,tutor-end=20}{)} \htmlData{tutor-start=21,tutor-end=31}{\geqslant }\htmlData{tutor-start=31,tutor-end=32}{0} 对于 t∈[1,2]\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{[}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{]} 恒成立，求实数 m\htmlData{tutor-start=0,tutor-end=1}{m} 的取值范围.

答案：log₂(1+√2)；m≥-5

题目标签：绝对值指数函数与恒成立参数

解题过程

（1）解分段指数方程

先由绝对值拆出函数的两段

(1)
写分段函数

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 ∣x∣=−x\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}{=}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{x}，两项同为 2x\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{x}}。

详细展开：因此 f(x)=0\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{0}（x<0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{0}）；当 x≥0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{0} 时 f(x)=2x−2−x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}^{\htmlData{tutor-start=8,tutor-end=9}{x}}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}^{\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{x}}。方程右端为 2，解只能在非负段。

f(x)={0,x<0,2x−2−x,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=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{,}&\htmlData{tutor-start=21,tutor-end=22}{x}\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=27,tutor-end=28}{2}^{\htmlData{tutor-start=30,tutor-end=31}{x}}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{2}^{\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{x}}\htmlData{tutor-start=39,tutor-end=40}{,}&\htmlData{tutor-start=41,tutor-end=42}{x}\htmlData{tutor-start=42,tutor-end=45}{\ge}\htmlData{tutor-start=45,tutor-end=46}{0}\htmlData{tutor-start=46,tutor-end=47}{.}\end{cases}
(2)
换元解方程

令 u=2x≥1\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=5,tutor-end=6}{x}}\htmlData{tutor-start=7,tutor-end=10}{\ge}\htmlData{tutor-start=10,tutor-end=11}{1}。

详细展开：方程成为 u−1/u=2\htmlData{tutor-start=0,tutor-end=1}{u}\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}{u}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}，即 u2−2u−1=0\htmlData{tutor-start=0,tutor-end=1}{u}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{u}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{0}。取正且不小于 1 的根 u=1+2\htmlData{tutor-start=0,tutor-end=1}{u}\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}{2}}。

x=log⁡2(1+2)\boxed{\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{=}\log_{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{+}\sqrt{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{)}}

（2）求恒成立参数

因式分解 f(2t) 并除以正因子

(1)
利用 t 的正区间

t∈[1,2]\htmlData{tutor-start=0,tutor-end=1}{t}\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}{2}\htmlData{tutor-start=8,tutor-end=9}{]}，所以 f(t)=2t−2−t>0\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{=}\htmlData{tutor-start=5,tutor-end=6}{2}^{\htmlData{tutor-start=8,tutor-end=9}{t}}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{2}^{\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{t}}\htmlData{tutor-start=17,tutor-end=18}{>}\htmlData{tutor-start=18,tutor-end=19}{0}。

详细展开：又 f(2t)=(2t−2−t)(2t+2−t)=f(t)(2t+2−t)\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}{t}\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}{t}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{2}^{\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{t}}\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=24,tutor-end=25}{t}}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{2}^{\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{t}}\htmlData{tutor-start=33,tutor-end=34}{)}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{f}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{t}\htmlData{tutor-start=38,tutor-end=39}{)}\htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{2}^{\htmlData{tutor-start=43,tutor-end=44}{t}}\htmlData{tutor-start=45,tutor-end=46}{+}\htmlData{tutor-start=46,tutor-end=47}{2}^{\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{t}}\htmlData{tutor-start=52,tutor-end=53}{)}。

2tf(2t)+mf(t)=f(t)(22t+1+m)\htmlData{tutor-start=0,tutor-end=1}{2}^{\htmlData{tutor-start=3,tutor-end=4}{t}}\htmlData{tutor-start=5,tutor-end=6}{f}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{t}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{f}\htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{t}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{f}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{t}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{2}^{\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{t}}\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}{m}\htmlData{tutor-start=32,tutor-end=33}{)}
(2)
求统一下界

由于 f(t)>0\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{)}\htmlData{tutor-start=4,tutor-end=5}{>}\htmlData{tutor-start=5,tutor-end=6}{0}，条件等价于 m≥−(22t+1)\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{2}^{\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{t}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{)} 对所有 t 成立。

详细展开：右端在 [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}{]} 上递减，最大值在 t=1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，为 -5。因此 m 的最小允许值为 -5。

m∈[−5,∞)\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}{-}\htmlData{tutor-start=13,tutor-end=14}{5}\htmlData{tutor-start=14,tutor-end=15}{,}\htmlData{tutor-start=15,tutor-end=21}{\infty}\htmlData{tutor-start=21,tutor-end=22}{)}}
20

解答题 · 解析几何

设 P(a,b)\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}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{)} (b≠0\htmlData{tutor-start=0,tutor-end=1}{b} \neq \htmlData{tutor-start=7,tutor-end=8}{0}) 是平面直角坐标系 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} 中的点，l\htmlData{tutor-start=0,tutor-end=1}{l} 是经过原点与点 (1,b)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{)} 的直线，记 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 是直线 l\htmlData{tutor-start=0,tutor-end=1}{l} 与抛物线 x2=2py\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{p}\htmlData{tutor-start=10,tutor-end=11}{y} (p≠0\htmlData{tutor-start=0,tutor-end=1}{p} \neq \htmlData{tutor-start=7,tutor-end=8}{0}) 的异于原点的交点. (1) 已知 a=1,b=2,p=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=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{p}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}，求点 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 的坐标； (2) 已知点 P(a,b)\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}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{)} (ab≠0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b} \neq \htmlData{tutor-start=8,tutor-end=9}{0}) 在椭圆 x24+y2=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}{4}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{y}^{\htmlData{tutor-start=21,tutor-end=22}{2}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{1} 上，p=12ab\htmlData{tutor-start=0,tutor-end=1}{p} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{a}\htmlData{tutor-start=15,tutor-end=16}{b}}，求证：点 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 落在双曲线 4x2−4y2=1\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{-} \htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{y}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{1} 上； (3) 已知动点 P(a,b)\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}{,} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{)} 满足 ab≠0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b} \neq \htmlData{tutor-start=8,tutor-end=9}{0}，p=12ab\htmlData{tutor-start=0,tutor-end=1}{p} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{a}\htmlData{tutor-start=15,tutor-end=16}{b}}，若点 Q\htmlData{tutor-start=0,tutor-end=1}{Q} 始终落在一条关于 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴对称的抛物线上，试问动点 P\htmlData{tutor-start=0,tutor-end=1}{P} 的轨迹落在哪种二次曲线上，并说明理由.

答案：(8,16)；见证明；分类见解析

题目标签：抛物线映射与二次曲线分类

解题过程

（1）求具体交点 Q

先写直线方程再与抛物线联立

(1)
写直线与抛物线

直线 l 经过原点与 (1,b)\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}{b}\htmlData{tutor-start=4,tutor-end=5}{)}，故方程为 y=bx\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{x}。

详细展开：代入 x2=2py\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}{p}\htmlData{tutor-start=8,tutor-end=9}{y} 得 x2=2pbx\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}{p}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{x}，除原点解外，另一交点满足 x=2pb\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}{p}\htmlData{tutor-start=4,tutor-end=5}{b}、y=2pb2\htmlData{tutor-start=0,tutor-end=1}{y}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{p}\htmlData{tutor-start=4,tutor-end=5}{b}^{\htmlData{tutor-start=7,tutor-end=8}{2}}。

Q=(2pb,2pb2)\htmlData{tutor-start=0,tutor-end=1}{Q}\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}{p}\htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{p}\htmlData{tutor-start=9,tutor-end=10}{b}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{)}
(2)
代入参数

b=2,p=2\htmlData{tutor-start=0,tutor-end=1}{b}\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}{p}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}。

详细展开：所以 xQ=2⋅2⋅2=8\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{Q}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=12}{\cdot}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=18}{\cdot}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{8}，yQ=2⋅2⋅22=16\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{Q}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=12}{\cdot}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=18}{\cdot}\htmlData{tutor-start=18,tutor-end=19}{2}^{\htmlData{tutor-start=21,tutor-end=22}{2}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{6}。

Q=(8,16)\boxed{\htmlData{tutor-start=7,tutor-end=8}{Q}\htmlData{tutor-start=8,tutor-end=9}{=}\htmlData{tutor-start=9,tutor-end=10}{(}\htmlData{tutor-start=10,tutor-end=11}{8}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{6}\htmlData{tutor-start=14,tutor-end=15}{)}}

（2）证明 Q 在给定双曲线上

把 Q 坐标化成 P 坐标的有理式

(1)
代入 p 的定义

p=1/(2ab)\htmlData{tutor-start=0,tutor-end=1}{p}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{(}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{a}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{)} 且 ab≠0\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{b}\ne\htmlData{tutor-start=5,tutor-end=6}{0}。

详细展开：由通式 Q=(2pb,2pb2)\htmlData{tutor-start=0,tutor-end=1}{Q}\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}{p}\htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{p}\htmlData{tutor-start=9,tutor-end=10}{b}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{)} 得 Q=(1/a,b/a)\htmlData{tutor-start=0,tutor-end=1}{Q}\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}{a}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{)}。

xQ=1a,yQ=ba\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{Q}}\htmlData{tutor-start=5,tutor-end=6}{=}\frac{\htmlData{tutor-start=12,tutor-end=13}{1}}{\htmlData{tutor-start=15,tutor-end=16}{a}}\htmlData{tutor-start=17,tutor-end=18}{,}\quad \htmlData{tutor-start=24,tutor-end=25}{y}_{\htmlData{tutor-start=27,tutor-end=28}{Q}}\htmlData{tutor-start=29,tutor-end=30}{=}\frac{\htmlData{tutor-start=36,tutor-end=37}{b}}{\htmlData{tutor-start=39,tutor-end=40}{a}}
(2)
使用椭圆约束

点 P 在 a2/4+b2=1\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{4}\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}{1} 上，所以 a2=4(1−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}{4}\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}{b}^{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{)}。

详细展开：代入 Q 的坐标，4xQ2−4yQ2=4(1−b2)/a2=1\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{Q}}^{\htmlData{tutor-start=8,tutor-end=9}{2}}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{4}\htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{Q}}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{4}\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{b}^{\htmlData{tutor-start=29,tutor-end=30}{2}}\htmlData{tutor-start=31,tutor-end=32}{)}\htmlData{tutor-start=32,tutor-end=33}{/}\htmlData{tutor-start=33,tutor-end=34}{a}^{\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=39}{=}\htmlData{tutor-start=39,tutor-end=40}{1}，命题成立。

4xQ2−4yQ2=1\boxed{\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{Q}}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{4}\htmlData{tutor-start=19,tutor-end=20}{y}_{\htmlData{tutor-start=22,tutor-end=23}{Q}}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{=}\htmlData{tutor-start=29,tutor-end=30}{1}}

（3）分类判断 P 的二次曲线

设出关于 x 轴对称的抛物线并作逆变换

(1)
写一般抛物线并消元

设 Q 所在抛物线为 y2=2q(x−h)\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}{q}\htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{x}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{h}\htmlData{tutor-start=12,tutor-end=13}{)}，其中 q≠0\htmlData{tutor-start=0,tutor-end=1}{q}\ne\htmlData{tutor-start=4,tutor-end=5}{0}。

详细展开：代入 Q=(1/a,b/a)\htmlData{tutor-start=0,tutor-end=1}{Q}\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}{a}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{/}\htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{)} 并乘 a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}}，得到 P 的轨迹方程 b2=2qa−2qha2\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{q}\htmlData{tutor-start=8,tutor-end=9}{a}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{q}\htmlData{tutor-start=12,tutor-end=13}{h}\htmlData{tutor-start=13,tutor-end=14}{a}^{\htmlData{tutor-start=16,tutor-end=17}{2}}，即 2qha2+b2−2qa=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{q}\htmlData{tutor-start=2,tutor-end=3}{h}\htmlData{tutor-start=3,tutor-end=4}{a}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{b}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{q}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{0}。

2qha2+b2−2qa=0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{q}\htmlData{tutor-start=2,tutor-end=3}{h}\htmlData{tutor-start=3,tutor-end=4}{a}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{b}^{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{q}\htmlData{tutor-start=17,tutor-end=18}{a}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{0}
(2)
按二次项系数分类

曲线类型由 a2\htmlData{tutor-start=0,tutor-end=1}{a}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 系数 2qh\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{q}\htmlData{tutor-start=2,tutor-end=3}{h} 与 b2\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}} 系数 1 的符号关系决定。

详细展开：若 h=0\htmlData{tutor-start=0,tutor-end=1}{h}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{0}，方程是抛物线；若 qh<0\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{h}\htmlData{tutor-start=2,tutor-end=3}{<}\htmlData{tutor-start=3,tutor-end=4}{0}，两平方项异号，是双曲线；若 qh>0\htmlData{tutor-start=0,tutor-end=1}{q}\htmlData{tutor-start=1,tutor-end=2}{h}\htmlData{tutor-start=2,tutor-end=3}{>}\htmlData{tutor-start=3,tutor-end=4}{0}，两平方项同号，是椭圆，其中 2qh=1\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{q}\htmlData{tutor-start=2,tutor-end=3}{h}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1} 时两平方项系数相等，轨迹是圆。

{h=0:抛物线;qh<0:双曲线;qh>0, 2qh≠1:椭圆;2qh=1:圆.\boxed{\begin{cases}h=0:&\text{抛物线};\\qh<0:&\text{双曲线};\\qh>0,\ 2qh\ne1:&\text{椭圆};\\2qh=1:&\text{圆}.\end{cases}}
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解答题 · 数列

已知以 a1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 为首项的数列 {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：an+1={an+c,an<3,and,an⩾3.\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{=} \begin{cases} \htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{n}} \htmlData{tutor-start=30,tutor-end=31}{+} \htmlData{tutor-start=32,tutor-end=33}{c}\htmlData{tutor-start=33,tutor-end=34}{,} & \htmlData{tutor-start=37,tutor-end=38}{a}_{\htmlData{tutor-start=40,tutor-end=41}{n}} \htmlData{tutor-start=43,tutor-end=44}{<} \htmlData{tutor-start=45,tutor-end=46}{3}\htmlData{tutor-start=46,tutor-end=47}{,} \\ \frac{\htmlData{tutor-start=57,tutor-end=58}{a}_{\htmlData{tutor-start=60,tutor-end=61}{n}}}{\htmlData{tutor-start=64,tutor-end=65}{d}}\htmlData{tutor-start=66,tutor-end=67}{,} & \htmlData{tutor-start=70,tutor-end=71}{a}_{\htmlData{tutor-start=73,tutor-end=74}{n}} \htmlData{tutor-start=76,tutor-end=86}{\geqslant }\htmlData{tutor-start=86,tutor-end=87}{3}\htmlData{tutor-start=87,tutor-end=88}{.} \end{cases} (1) 当 a1=1,c=1,d=3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{c} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{d} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{3} 时，求数列 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 的通项公式； (2) 当 0<a1<1,c=1,d=3\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{<} \htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{c} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{d} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{3} 时，试用 a1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 表示数列 {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} 的前 100 项的和 S100\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0}}； (3) 当 0<a1<1m\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{<} \frac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{m}} (m\htmlData{tutor-start=0,tutor-end=1}{m} 是正整数)，c=1m\htmlData{tutor-start=0,tutor-end=1}{c} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{m}}，正整数 d⩾3m\htmlData{tutor-start=0,tutor-end=1}{d} \htmlData{tutor-start=2,tutor-end=12}{\geqslant }\htmlData{tutor-start=12,tutor-end=13}{3}\htmlData{tutor-start=13,tutor-end=14}{m} 时，求证：数列 a2−1m,a3m+2−1m,a6m+2−1m,a9m+2−1m\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{m}}\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{3}\htmlData{tutor-start=25,tutor-end=26}{m}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{-} \frac{\htmlData{tutor-start=38,tutor-end=39}{1}}{\htmlData{tutor-start=41,tutor-end=42}{m}}\htmlData{tutor-start=43,tutor-end=44}{,} \htmlData{tutor-start=45,tutor-end=46}{a}_{\htmlData{tutor-start=48,tutor-end=49}{6}\htmlData{tutor-start=49,tutor-end=50}{m}\htmlData{tutor-start=50,tutor-end=51}{+}\htmlData{tutor-start=51,tutor-end=52}{2}} \htmlData{tutor-start=54,tutor-end=55}{-} \frac{\htmlData{tutor-start=62,tutor-end=63}{1}}{\htmlData{tutor-start=65,tutor-end=66}{m}}\htmlData{tutor-start=67,tutor-end=68}{,} \htmlData{tutor-start=69,tutor-end=70}{a}_{\htmlData{tutor-start=72,tutor-end=73}{9}\htmlData{tutor-start=73,tutor-end=74}{m}\htmlData{tutor-start=74,tutor-end=75}{+}\htmlData{tutor-start=75,tutor-end=76}{2}} \htmlData{tutor-start=78,tutor-end=79}{-} \frac{\htmlData{tutor-start=86,tutor-end=87}{1}}{\htmlData{tutor-start=89,tutor-end=90}{m}} 成等比数列当且仅当 d=3m\htmlData{tutor-start=0,tutor-end=1}{d} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{m}.

答案：见解析；S100见解析；当且仅当d=3m

题目标签：阈值递推的周期、求和与等比判定

解题过程

（1）求特殊数列通项

直接观察阈值递推形成的三周期

(1)
计算一个周期

a1=1<3\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}{3}，故 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}；a2<3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{<}\htmlData{tutor-start=6,tutor-end=7}{3}，故 a3=3\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}{3}。

详细展开：随后 a3≥3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=8}{\ge}\htmlData{tutor-start=8,tutor-end=9}{3}，所以 a4=a3/3=1\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}{a}_{\htmlData{tutor-start=9,tutor-end=10}{3}}\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}{1}，回到首项状态。

a1,a2,a3,a4=1,2,3,1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{3}}\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{a}_{\htmlData{tutor-start=21,tutor-end=22}{4}}\htmlData{tutor-start=23,tutor-end=24}{=}\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{,}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=30}{,}\htmlData{tutor-start=30,tutor-end=31}{1}
(2)
写周期通项

此后每三项重复 1、2、3。

详细展开：因此按 n 模 3，余 1 时为 1，余 2 时为 2，整除 3 时为 3；也可写成 1+((n−1) mod 3)\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{(}\htmlData{tutor-start=4,tutor-end=5}{n}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{)}\bmod\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{)}。

an={1,n≡1(mod3),2,n≡2(mod3),3,n≡0(mod3).\boxed{a_{n}=\begin{cases}1,&n\equiv1\pmod3,\\2,&n\equiv2\pmod3,\\3,&n\equiv0\pmod3.\end{cases}}

（2）求前 100 项和

把首项后面的 99 项分成 33 个三项块

(1)
找出缩放三项块

设 x=a1∈(0,1)\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=10}{\in}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{)}。

详细展开：从第二项起，第 k 个三项块为 1+x/3k,2+x/3k,3+x/3k\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{3}^{\htmlData{tutor-start=7,tutor-end=8}{k}}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{/}\htmlData{tutor-start=14,tutor-end=15}{3}^{\htmlData{tutor-start=17,tutor-end=18}{k}}\htmlData{tutor-start=19,tutor-end=20}{,}\htmlData{tutor-start=20,tutor-end=21}{3}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{x}\htmlData{tutor-start=23,tutor-end=24}{/}\htmlData{tutor-start=24,tutor-end=25}{3}^{\htmlData{tutor-start=27,tutor-end=28}{k}}（k=0,1,…\htmlData{tutor-start=0,tutor-end=1}{k}\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}{1}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=12}{\ldots}）：每到大于 3 的第三项，下一步除以 3，使偏移量再除以 3。

(a3k+2,a3k+3,a3k+4)=(1+x3k,2+x3k,3+x3k)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{k}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{3}\htmlData{tutor-start=14,tutor-end=15}{k}\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}{a}_{\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{k}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{4}}\htmlData{tutor-start=27,tutor-end=28}{)}\htmlData{tutor-start=28,tutor-end=29}{=}\left(\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{+}\frac{\htmlData{tutor-start=43,tutor-end=44}{x}}{\htmlData{tutor-start=46,tutor-end=47}{3}^{\htmlData{tutor-start=49,tutor-end=50}{k}}}\htmlData{tutor-start=52,tutor-end=53}{,}\htmlData{tutor-start=53,tutor-end=54}{2}\htmlData{tutor-start=54,tutor-end=55}{+}\frac{\htmlData{tutor-start=61,tutor-end=62}{x}}{\htmlData{tutor-start=64,tutor-end=65}{3}^{\htmlData{tutor-start=67,tutor-end=68}{k}}}\htmlData{tutor-start=70,tutor-end=71}{,}\htmlData{tutor-start=71,tutor-end=72}{3}\htmlData{tutor-start=72,tutor-end=73}{+}\frac{\htmlData{tutor-start=79,tutor-end=80}{x}}{\htmlData{tutor-start=82,tutor-end=83}{3}^{\htmlData{tutor-start=85,tutor-end=86}{k}}}\right)
(2)
求 33 块总和

100=1+33\times3$。

详细展开：所以 S100=x+∑k=032(6+3x/3k)=198+x+(9x/2)(1−3−33)\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{0}}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{x}\htmlData{tutor-start=9,tutor-end=10}{+}\sum_{\htmlData{tutor-start=16,tutor-end=17}{k}\htmlData{tutor-start=17,tutor-end=18}{=}\htmlData{tutor-start=18,tutor-end=19}{0}}^{\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{6}\htmlData{tutor-start=27,tutor-end=28}{+}\htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{/}\htmlData{tutor-start=31,tutor-end=32}{3}^{\htmlData{tutor-start=34,tutor-end=35}{k}}\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}{9}\htmlData{tutor-start=40,tutor-end=41}{8}\htmlData{tutor-start=41,tutor-end=42}{+}\htmlData{tutor-start=42,tutor-end=43}{x}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{(}\htmlData{tutor-start=45,tutor-end=46}{9}\htmlData{tutor-start=46,tutor-end=47}{x}\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}{1}\htmlData{tutor-start=52,tutor-end=53}{-}\htmlData{tutor-start=53,tutor-end=54}{3}^{\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{3}\htmlData{tutor-start=58,tutor-end=59}{3}}\htmlData{tutor-start=60,tutor-end=61}{)}，化简即可。

S100=198+(112−12⋅331)a1\boxed{\htmlData{tutor-start=7,tutor-end=8}{S}_{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{0}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{9}\htmlData{tutor-start=17,tutor-end=18}{8}\htmlData{tutor-start=18,tutor-end=19}{+}\left(\frac{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{1}}{\htmlData{tutor-start=35,tutor-end=36}{2}}\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=48,tutor-end=53}{\cdot}\htmlData{tutor-start=53,tutor-end=54}{3}^{\htmlData{tutor-start=56,tutor-end=57}{3}\htmlData{tutor-start=57,tutor-end=58}{1}}}\right)\htmlData{tutor-start=67,tutor-end=68}{a}_{\htmlData{tutor-start=70,tutor-end=71}{1}}}

（3）证明等比条件的充要性

分别证明 d=3m 时的缩放规律与 d>3m 时的符号矛盾

(1)
证明 d=3m 时充分

记 x=a1∈(0,1/m)\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=10}{\in}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{,}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{m}\htmlData{tutor-start=16,tutor-end=17}{)}。连续加 1/m\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{m} 后，a3m+1=3+x≥3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{m}\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}{3}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{x}\htmlData{tutor-start=12,tutor-end=15}{\ge}\htmlData{tutor-start=15,tutor-end=16}{3}，故 a3m+2=(3+x)/d\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{m}\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}{(}\htmlData{tutor-start=10,tutor-end=11}{3}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{x}\htmlData{tutor-start=13,tutor-end=14}{)}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{d}。

详细展开：当 d=3m\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{m} 时，a3m+2−1/m=x/d\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{m}\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}{/}\htmlData{tutor-start=11,tutor-end=12}{m}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{d}。同样的“加到 3 后除以 d”过程重复，依次得到四个数 x,x/d,x/d2,x/d3\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{d}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{d}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\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}{d}^{\htmlData{tutor-start=19,tutor-end=20}{3}}。

d=3m⟹(a2−1m,a3m+2−1m,a6m+2−1m,a9m+2−1m)=(x,xd,xd2,xd3)\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{m}\htmlData{tutor-start=4,tutor-end=19}{\Longrightarrow}\left(\htmlData{tutor-start=25,tutor-end=26}{a}_{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{-}\frac{\htmlData{tutor-start=37,tutor-end=38}{1}}{\htmlData{tutor-start=40,tutor-end=41}{m}}\htmlData{tutor-start=42,tutor-end=43}{,}\htmlData{tutor-start=43,tutor-end=44}{a}_{\htmlData{tutor-start=46,tutor-end=47}{3}\htmlData{tutor-start=47,tutor-end=48}{m}\htmlData{tutor-start=48,tutor-end=49}{+}\htmlData{tutor-start=49,tutor-end=50}{2}}\htmlData{tutor-start=51,tutor-end=52}{-}\frac{\htmlData{tutor-start=58,tutor-end=59}{1}}{\htmlData{tutor-start=61,tutor-end=62}{m}}\htmlData{tutor-start=63,tutor-end=64}{,}\htmlData{tutor-start=64,tutor-end=65}{a}_{\htmlData{tutor-start=67,tutor-end=68}{6}\htmlData{tutor-start=68,tutor-end=69}{m}\htmlData{tutor-start=69,tutor-end=70}{+}\htmlData{tutor-start=70,tutor-end=71}{2}}\htmlData{tutor-start=72,tutor-end=73}{-}\frac{\htmlData{tutor-start=79,tutor-end=80}{1}}{\htmlData{tutor-start=82,tutor-end=83}{m}}\htmlData{tutor-start=84,tutor-end=85}{,}\htmlData{tutor-start=85,tutor-end=86}{a}_{\htmlData{tutor-start=88,tutor-end=89}{9}\htmlData{tutor-start=89,tutor-end=90}{m}\htmlData{tutor-start=90,tutor-end=91}{+}\htmlData{tutor-start=91,tutor-end=92}{2}}\htmlData{tutor-start=93,tutor-end=94}{-}\frac{\htmlData{tutor-start=100,tutor-end=101}{1}}{\htmlData{tutor-start=103,tutor-end=104}{m}}\right)\htmlData{tutor-start=112,tutor-end=113}{=}\left(\htmlData{tutor-start=119,tutor-end=120}{x}\htmlData{tutor-start=120,tutor-end=121}{,}\frac{\htmlData{tutor-start=127,tutor-end=128}{x}}{\htmlData{tutor-start=130,tutor-end=131}{d}}\htmlData{tutor-start=132,tutor-end=133}{,}\frac{\htmlData{tutor-start=139,tutor-end=140}{x}}{\htmlData{tutor-start=142,tutor-end=143}{d}^{\htmlData{tutor-start=145,tutor-end=146}{2}}}\htmlData{tutor-start=148,tutor-end=149}{,}\frac{\htmlData{tutor-start=155,tutor-end=156}{x}}{\htmlData{tutor-start=158,tutor-end=159}{d}^{\htmlData{tutor-start=161,tutor-end=162}{3}}}\right)
(2)
分析 d>3m 的前三项符号

d、m 为整数，所以 d≥3m+1\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=4}{\ge}\htmlData{tutor-start=4,tutor-end=5}{3}\htmlData{tutor-start=5,tutor-end=6}{m}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}。令 z=(3+x)/d\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{3}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{x}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{d}，由 mx<1\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{<}\htmlData{tutor-start=3,tutor-end=4}{1} 得 0<z<1/m\htmlData{tutor-start=0,tutor-end=1}{0}\htmlData{tutor-start=1,tutor-end=2}{<}\htmlData{tutor-start=2,tutor-end=3}{z}\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}{m}。

详细展开：于是前两个待判数为 u0=x>0\htmlData{tutor-start=0,tutor-end=1}{u}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{>}\htmlData{tutor-start=8,tutor-end=9}{0}、u1=z−1/m<0\htmlData{tutor-start=0,tutor-end=1}{u}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{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}{m}\htmlData{tutor-start=11,tutor-end=12}{<}\htmlData{tutor-start=12,tutor-end=13}{0}。再连续加 1/m\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{m}，在第 6m+2\htmlData{tutor-start=0,tutor-end=1}{6}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{+}\htmlData{tutor-start=3,tutor-end=4}{2} 项恰得到 z+3\htmlData{tutor-start=0,tutor-end=1}{z}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{3}，故 u2=z+3−1/m>0\htmlData{tutor-start=0,tutor-end=1}{u}_{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{z}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{3}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{/}\htmlData{tutor-start=12,tutor-end=13}{m}\htmlData{tutor-start=13,tutor-end=14}{>}\htmlData{tutor-start=14,tutor-end=15}{0}。

d>3m:u0>0, u1<0, u2>0\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{m}\htmlData{tutor-start=4,tutor-end=5}{:}\quad \htmlData{tutor-start=11,tutor-end=12}{u}_{\htmlData{tutor-start=14,tutor-end=15}{0}}\htmlData{tutor-start=16,tutor-end=17}{>}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{,}\htmlData{tutor-start=19,tutor-end=21}{\ }\htmlData{tutor-start=21,tutor-end=22}{u}_{\htmlData{tutor-start=24,tutor-end=25}{1}}\htmlData{tutor-start=26,tutor-end=27}{<}\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{,}\htmlData{tutor-start=29,tutor-end=31}{\ }\htmlData{tutor-start=31,tutor-end=32}{u}_{\htmlData{tutor-start=34,tutor-end=35}{2}}\htmlData{tutor-start=36,tutor-end=37}{>}\htmlData{tutor-start=37,tutor-end=38}{0}
(3)
用第四项排除等比

a6m+2=z+3≥3\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{6}\htmlData{tutor-start=4,tutor-end=5}{m}\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}{z}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=15}{\ge}\htmlData{tutor-start=15,tutor-end=16}{3}，下一项除以 d，记 w=(z+3)/d<1/m\htmlData{tutor-start=0,tutor-end=1}{w}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{z}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{3}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{d}\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}{m}。

详细展开：再加 3m−1\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1} 次 1/m\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{m} 到 a9m+2=w+3−1/m\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{9}\htmlData{tutor-start=4,tutor-end=5}{m}\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}{w}\htmlData{tutor-start=10,tutor-end=11}{+}\htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{/}\htmlData{tutor-start=15,tutor-end=16}{m}，故 u3=w+3−2/m>0\htmlData{tutor-start=0,tutor-end=1}{u}_{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{w}\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=12}{/}\htmlData{tutor-start=12,tutor-end=13}{m}\htmlData{tutor-start=13,tutor-end=14}{>}\htmlData{tutor-start=14,tutor-end=15}{0}。若四项成等比，前三项符号 +,−,+\htmlData{tutor-start=0,tutor-end=1}{+}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{+} 迫使公比为负、第四项应为负，与 u3>0\htmlData{tutor-start=0,tutor-end=1}{u}_{\htmlData{tutor-start=3,tutor-end=4}{3}}\htmlData{tutor-start=5,tutor-end=6}{>}\htmlData{tutor-start=6,tutor-end=7}{0} 矛盾。因此只能 d=3m\htmlData{tutor-start=0,tutor-end=1}{d}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}\htmlData{tutor-start=3,tutor-end=4}{m}。

htmlDatatutor−start=8,tutor−end=27text四项成等比数列  ⟺  d=3m\boxed{\\htmlData{tutor-start=8,tutor-end=27}{text{四项成等比数列}\iff d}\htmlData{tutor-start=27,tutor-end=28}{=}\htmlData{tutor-start=28,tutor-end=29}{3}\htmlData{tutor-start=29,tutor-end=30}{m}}