1.2集合
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集合
基本概念
定义 1.2.1(集合)
集合是不同对象的一个无序的聚集,对象也称为集合的元素.用\(\displaystyle a\in A\)来表示\(\displaystyle a\)是集合\(\displaystyle A\)中的一个元素,用\(\displaystyle a\notin A\)来表示\(\displaystyle a\)不是集合\(\displaystyle A\)中的一个元素.
定义 1.2.2(集合的基数)
对于集合\(\displaystyle S\),若 \(\displaystyle S\) 中恰有 \(\displaystyle n\) 个不同的元素,则称 \(\displaystyle S\) 是有限集, \(\displaystyle n\) 是 \(\displaystyle S\) 的基数,记为 \(\displaystyle |S|\)或\(\displaystyle \text{card}(S)\).如果一个集合不是有限集,那么它就是无限集.
为表示某些集合,我们可以列举出其所包含的全部元素,当一个集合的元素较多,或者它是无限集时,可以只写出几个元素,其他元素用省略号表示,要注意,写出的元素必须让人明白省略号表示了哪些元素,例如小于100的正整数集合可以记为\(\displaystyle \{1,2,\cdots,99\}\).
但是,有相当一部分集合是不可列的,其中一个是不等式\(\displaystyle x-7<3\)的解集.解该不等式得到\(\displaystyle x<10\),无法用列举法表示(无法以某一顺序列出所有小于10的实数),我们可以换种方法,抓住该集合中元素具有的共同特征:它们是实数且小于\(\displaystyle 10\),于是可以把这个集合写成\(\displaystyle \{x\in\mathbb{R}\mid x<10\}\).
定义 1.2.3(集合的描述法)
设 \(\displaystyle A\) 是一个集合,我们把集合 \(\displaystyle A\) 中所有具有共同特征 \(\displaystyle P(x)\) 的元素 \(\displaystyle x\) 所组成的集合表示为 \(\displaystyle \{x \mid P(x)\}\), 这种表示集合的方法称为描述法.
集合也可以用韦恩图形象地表示.在韦恩图中,全集\(\displaystyle U\)包含了所考虑的全部对象,用矩形框表示.在矩形框内部,一般用圆形表示集合,用点表示集合中特定的元素.韦恩图常用于直观表示集合之间的关系.
定义 1.2.4(子集)
若\(\displaystyle A,B\)满足$\(\displaystyle \forall x\in A\implies x\in B\)$
则称$\displaystyle A$是$\displaystyle B$的**子集**,记作$\displaystyle A\subseteq B$,否则称$\displaystyle A$不是$\displaystyle B$的子集,记作$\displaystyle A\nsubseteq B$.
若$\displaystyle A\subseteq B$且$\displaystyle A\neq B$,则称$\displaystyle A$是$\displaystyle B$的**真子集**,记作$\displaystyle A\subsetneq B$
集合运算
定义 1.2.5(集合的交、并、补、差运算)
对于集合\(\displaystyle A,B\)与全集\(\displaystyle U\),
由同时属于$\displaystyle A$与$\displaystyle B$的元素构成的集合称为$\displaystyle A$与$\displaystyle B$的**交集**,记作
$$\displaystyle A\cap B=\{x\mid x\in A\text{且}x\in B\}$$
由属于$\displaystyle A$或$\displaystyle B$的元素构成的集合称为$\displaystyle A$与$\displaystyle B$的**并集**,记作
\(\displaystyle A\cup B=\{x\mid x\in A\text{或}x\in B\}\).
多个集合的交,并运算可做如下简写:
\(\displaystyle A_1\cap A_2\cdots\cap A_n=\bigcap_{i=1}^{n} A_i \quad\quad A_1\cup A_2\cdots\cup A_n=\bigcup_{i=1}^{n} A_i\)
在全集$\displaystyle U$中,不属于$\displaystyle A$的所有元素构成的集合称为$\displaystyle A$在$\displaystyle U$中的**补集**,记作$\displaystyle \complement_U A$.
由属于$\displaystyle A$而不属于$\displaystyle B$的元素构成的集合称为$\displaystyle A$与$\displaystyle B$的**差集**,记作
$$\displaystyle A-B=\{x\mid x\in A\text{且}x\notin B\}\quad \text{或}\quad A\setminus B$$
问题
解答下述问题:
(1)证明集合的\(\displaystyle \cap,\cup\)运算满足分配律,即
\(\displaystyle A\cap(B\cup C)=(A\cap B)\cup (A\cap C)\)
$\displaystyle A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$
(2)证明德摩根律(De Morgan's laws):设\(\displaystyle U\)是全集,则集合\(\displaystyle A,B\)满足
\(\displaystyle \complement_U(A\cup B)=\complement_UA\cap \complement_UB\)
$\displaystyle \complement_U(A\cap B)=\complement_UA\cup \complement_UB$
提示:要证明集合\(\displaystyle A=B\),一般考虑证明“\(\displaystyle A\subseteq B\)且\(\displaystyle B\subseteq A\)”.
定义 1.2.6(笛卡尔积)
两个集合\(\displaystyle X,Y\)的笛卡尔积记作\(\displaystyle X\times Y\).
\(\displaystyle X\times Y=\{(x,y)\mid x\in X,y\in Y\}\)
可将\(\displaystyle \underbrace{X\times X\times\cdots\times X}_{n\text{个}X}\)简写为\(\displaystyle X^n\),例如二维实平面可表示为\(\displaystyle \mathbb{R}^2\);
A 组习题
习\(\displaystyle \quad\)题
A组
-
【2013 上海学考22】设全集\(\displaystyle U=\mathbb{R}\),下列集合运算结果为\(\displaystyle \mathbb{R}\)的是
- \(\displaystyle Z\cup\complement_UN\)
- \(\displaystyle N\cap\complement_UN\)
- \(\displaystyle \complement_U\left(\complement_U\varnothing\right)\)
- \(\displaystyle \complement_U\left\{0\right\}\)
答案
A. \(\displaystyle Z\cup\complement_UN=\mathbb{R}\),故\(\displaystyle A\)选项正确. \(\displaystyle N\cap\complement_UN=\varnothing\),故\(\displaystyle B\)选项错误. \(\displaystyle \complement_U\left(\complement_U\varnothing\right)=\complement_UU=\varnothing\),故\(\displaystyle C\)选项错误. \(\displaystyle \complement_U\left\{0\right\}=\left\{x\left|x\ne0\right.\right\}\),故\(\displaystyle D\)选项错误.
-
【1987全国卷理1】非空集合\(\displaystyle S,T\)满足\(\displaystyle S\nsubseteq T,T\nsubseteq S\),令\(\displaystyle X=S\cap T\),则\(\displaystyle S\cup X=\)
\(\displaystyle X\)
- \(\displaystyle T\)
- \(\displaystyle \varnothing\)
- \(\displaystyle S\)
答案
D. 由\(\displaystyle X=S\cap T\)可推出\(\displaystyle X\subseteq S\),因此 \(\displaystyle S\cup X=S\),所以D正确.
注:本节的2到8题基本均可使用韦恩图直观判断或用于辅助,答案主要从集合运算的角度给出.
-
【2005全国I卷理2】 设 \(\displaystyle I\) 为全集, \(\displaystyle S_1\)、\(\displaystyle S_2\)、\(\displaystyle S_3\) 是 \(\displaystyle I\) 的三个非空子集,满足\(\displaystyle S_1 \cup S_2 \cup S_3 = I\), 则下列命题正确的是
\(\displaystyle \complement_I S_1 \cap (S_2 \cup S_3) = \varnothing\)
- \(\displaystyle S_1 \subseteq (\complement_I S_2 \cap \complement_I S_3)\)
- \(\displaystyle \complement_I S_1 \cap \complement_I S_2 \cap \complement_I S_3 = \varnothing\)
- \(\displaystyle S_1 \subseteq (\complement_I S_2 \cup \complement_I S_3)\)
答案
C. 者对 \(\displaystyle S_1\cup S_2\cup S_3=I\)用德摩根律,得 \(\displaystyle \complement_I S_1\cap\complement_I S_2\cap\complement_I S_3=\varnothing\),直接得到C正确.
- 【2024深圳二模3】对于集合\(\displaystyle M,N\),下列命题正确的是
- \(\displaystyle M\cup\complement_{M\cup N}N=M\cup N\)
- \(\displaystyle \complement_{M\cup N}(M\cap N)=(\complement_{M\cup N}M)\cup(\complement_{M\cup N}N)\)
- \(\displaystyle M\cap\complement_{M\cup N}N=M\cap N\)
- \(\displaystyle \complement_{M\cup N}(M\cap N)=(\complement_{M\cup N}M)\cap(\complement_{M\cup N}N)\)
答案
B. 对B选项等式的左侧用德摩根律,其结果与右侧相同,直接得到B正确.
- 【2005山东理10】 设集合 \(\displaystyle A\)、\(\displaystyle B\) 是全集 \(\displaystyle U\) 的两个子集, 则 \(\displaystyle A \subsetneqq B\) 是 \(\displaystyle (\complement_U A) \cup B = U\) 的
- 充分不必要条件
- 必要不充分条件
- 充要条件
- 既不充分也不必要条件
答案
A. 充分性满足,当 \(\displaystyle A\subset B\) 时,可以推出 \(\displaystyle (\complement_U A)\cup B=U\).必要性不满足,\(\displaystyle (\complement_U A)\cup B=U\) 等价于 \(\displaystyle A\subseteq B\).这不等价于 \(\displaystyle A\subset B\),反例为\(\displaystyle A=B\),答案为充分不必要条件.
- 【2006江苏7】集合\(\displaystyle A\)、\(\displaystyle B\)、\(\displaystyle C\)满足\(\displaystyle A \cup B = B \cap C\),则下列命题正确的是
- \(\displaystyle A \subseteq C\)
- \(\displaystyle C \subseteq A\)
- \(\displaystyle A \neq C\)
- \(\displaystyle A \neq \varnothing\)
答案
A. 由 \(\displaystyle A\subseteq A\cup B=B\cap C\),得 \(\displaystyle A\subseteq C\),直接得到A 正确.
- 【2014 湖北3】 设 \(\displaystyle U\) 为全集,\(\displaystyle A, B\) 是集合,则“存在集合 \(\displaystyle C\) 使得 \(\displaystyle A \subseteq C, B \subseteq \complement_U C\)”是“\(\displaystyle A \cap B = \varnothing\)”的
- 充分不必要条件
- 必要不充分条件
- 充要条件
- 既不充分也不必要条件
答案
C. 充分性满足,采取反证法,假设\(\displaystyle A\cap B\neq \varnothing\),则存在\(\displaystyle x\in A\)且\(\displaystyle x\in B\),由\(\displaystyle A\subseteq C,B\subseteq \complement_U C\)可知\(\displaystyle x\in C\)且\(\displaystyle x\in \complement_U C\),矛盾,故\(\displaystyle A\cap B= \varnothing\).必要性满足,这是容易验证的.于是答案为充要条件.
- 【2007广东理8】 设集合\(\displaystyle S\)至少含有两个元素,定义\(\displaystyle S\) 上的二元运算“\(\displaystyle *\)”,即对任意\(\displaystyle a,b \in S\),有序元素对 \(\displaystyle (a,b)\)在 \(\displaystyle S\) 中有唯一确定的元素 \(\displaystyle a*b\) 与之对应.若对任意\(\displaystyle a,b \in S\),有 \(\displaystyle a*(b*a)=b\),则对任意 \(\displaystyle a,b \in S\), 下列等式中不恒成立的是
- \(\displaystyle (a*b)*a = a\)
- \(\displaystyle [a*(b*a)]*(a*b) = a\)
- \(\displaystyle b*(b*b) = b\)
- \(\displaystyle (a*b)*[b*(a*b)] = b\)
答案
由已知式交换 \(\displaystyle a,b\),得 \(\displaystyle b*(a*b)=a\);令 \(\displaystyle a=b\),得 \(\displaystyle b*(b*b)=b\),所以 C 正确.又 \(\displaystyle [a*(b*a)]*(a*b)=b*(a*b)=a\),所以 B 正确;而 \(\displaystyle (a*b)*[b*(a*b)]=(a*b)*a\),题设不能推出它等于 \(\displaystyle a\). 例如取任意指数为 \(\displaystyle 2\) 的非平凡阿贝尔群作运算,有 \(\displaystyle a*(b*a)=b\),但 \(\displaystyle (a*b)*a=b\),一般不等于 \(\displaystyle a\),故不恒成立的是 A.
- 设\(\displaystyle m,n\in\mathbb{N^*}\),集合\(\displaystyle A=\{1,2,\cdots ,m\},B=\{1,2,\cdots,n\}\),集合\(\displaystyle C\)是\(\displaystyle A\)的子集,且\(\displaystyle B\cap C\neq \varnothing\),讨论符合题意的\(\displaystyle C\)的个数.
答案
当\(\displaystyle m\leqslant n\)时有\(\displaystyle 2^m-1\)个,当 \(\displaystyle m>n\) 时有 \(\displaystyle 2^m-2^{m-n}\) 个. 若 \(\displaystyle m\leqslant n\),则 \(\displaystyle B\cap A=A\),所以 \(\displaystyle B\cap C\neq\varnothing\) 等价于 \(\displaystyle C\neq\varnothing\).\(\displaystyle A\) 的子集共有 \(\displaystyle 2^m\) 个,其中空集不符合条件,故符合条件的 \(\displaystyle C\) 有 \(\displaystyle 2^m-1\) 个.
若 $\displaystyle m>n$,则 $\displaystyle B\cap A=\{1,2,\ldots,n\}$.不符合条件的 $\displaystyle C$ 必须不含这 $\displaystyle n$ 个元素,只能从 $\displaystyle \{n+1,\ldots,m\}$ 中任取,故有 $\displaystyle 2^{m-n}$ 个.于是符合条件的 $\displaystyle C$ 有 $\displaystyle 2^m-2^{m-n}$ 个.- (多选)设\(\displaystyle X\)为全集,\(\displaystyle A\subseteq X\),定义\(\displaystyle f^S_A=\begin{cases} 1, & S\in A\\0,& S\notin A \end{cases}\),对于\(\displaystyle X\)的真子集\(\displaystyle A,B,C\),下列命题正确的是
- 若\(\displaystyle B\subseteq A\),则\(\displaystyle f^S_B\leqslant f^S_A\)
- 若\(\displaystyle B\cap A\neq \varnothing\),则\(\displaystyle f^S_{B\cap A}\leqslant f^S_A+f^S_B\)
- \(\displaystyle f^S_{B\cup A}=f^S_A+f^S_B\)
- \(\displaystyle f^S_{A\cup B\cup C}=1-(1-f^S_A)(1-f^S_B)(1-f^S_C)\)
答案
ABD. 若 \(\displaystyle B\subseteq A\),当\(\displaystyle S\notin B\)时,有\(\displaystyle f^S_B=0\leqslant f^S_A\),当\(\displaystyle S\in B\)时一定有\(\displaystyle S\in A\),此时\(\displaystyle f^S_B=f^S_A=1\),A 正确. 若\(\displaystyle S\notin B\cap A\)时,不等式一定满足.若 \(\displaystyle S\in B\cap A\),\(\displaystyle f^S_{B\cap A}=1<2=f_A^S+f_B^S\) , B正确.若\(\displaystyle A\cap B\neq \varnothing\),当\(\displaystyle x\in A\cap B\)时,\(\displaystyle f_{A\cup B}^S=1\neq 2=f^S_A+f^S_B\),C错误.对于D,\(\displaystyle f_{A\cup B\cup C}^S=1\)当且仅当\(\displaystyle f_A^S=1\)或\(\displaystyle f_B^S=1\)或\(\displaystyle f_C^S=1\),即\(\displaystyle 1-(1-f_A^S)(1-f_B^S)(1-f_C^S)=1\),\(\displaystyle f_{A\cup B\cup C}^S=0\)当且仅当\(\displaystyle f_A^S=f_B^S=f_C^S=0\),即\(\displaystyle 1-(1-f_A^S)(1-f_B^S)(1-f_C^S)=0\),故D 正确.
- 【2011福建文12改编】(多选)在整数集\(\displaystyle \mathbb{Z}\) 中,被5除所得余数为 \(\displaystyle k\) 的所有整数组成一个“类”,记为 \(\displaystyle [k]\),则下列命题正确的是
- \(\displaystyle 2011 \in [1],[7]=[92]\)
- \(\displaystyle \mathbb{Z} = [0] \cup [1] \cup [2] \cup [3] \cup [4]\)
- 整数 \(\displaystyle a, b\) 属于同一‘类’当且仅当 \(\displaystyle a - b \in [0]\)
- 对任意\(\displaystyle n\in\mathbb{Z}\),\(\displaystyle n^2\notin[2]\)且\(\displaystyle n^2\notin [3]\)
答案
ABCD. \(\displaystyle 2011=5\cdot402+1\),于是\(\displaystyle 2011\in [1]\), \(\displaystyle 7,92\) 除以 \(\displaystyle 5\) 的余数都为 \(\displaystyle 2\),于是\(\displaystyle [7]=[92]=[2]\),A 正确.每个整数的余数只能为 \(\displaystyle 0,1,2,3,4\),显然有\(\displaystyle \mathbb{Z}\subseteq [0]\cup [1]\cup [2]\cup [3]\cup [4]\),反向的包含关系是显然的,于是可得出 B 正确. \(\displaystyle a,b\) 属于同一类等价于它们模5同余,即\(\displaystyle 5\mid(a-b)\),可得 \(\displaystyle a-b\in[0]\),故 C 正确. 设\(\displaystyle x=5k+p(p=0,1,2,3,4)\),则\(\displaystyle x^2\equiv (5k+p)^2\equiv p^2\pmod 5\)而\(\displaystyle p^2\)模\(\displaystyle 5\) 只能为 \(\displaystyle 0,1,4\),所以 D 正确.
- 【2010四川16改编】(多选)设\(\displaystyle S\)是复数集\(\displaystyle \mathbb{C}\)的非空子集,若对任意\(\displaystyle x,y\in S\),都有\(\displaystyle x+y,x-y,xy\in S\),则称\(\displaystyle S\)为封闭集,下列叙述正确的是
- 集合\(\displaystyle S=\left \{a+b\text{i}\mid a,b\in\mathbb{Z} \right \}\)是封闭集
- 若\(\displaystyle S\)为封闭集,则一定有\(\displaystyle 0\in S\)
- 封闭集一定是无限集
- 若\(\displaystyle S\)为封闭集,则满足\(\displaystyle S\subseteq T\subseteq\mathbb{C}\)的任意集合\(\displaystyle T\)也是封闭集
答案
AB. 可以验证复数集关于加、减、乘均封闭,故A正确.任取封闭集中的 \(\displaystyle x\),封闭集关于减法封闭,于是\(\displaystyle x-x=0\in S\),故B正确.取集合\(\displaystyle \{0\}\) ,它是有限封闭集,于是C错误;设\(\displaystyle S=\mathbb{Q}\),可以证明\(\displaystyle S\)是封闭集,取\(\displaystyle T=S\cup \{\sqrt{2}\}\),而\(\displaystyle 1+\sqrt{2}\notin T\),故\(\displaystyle T\)不是封闭集,D错误.
- 【2006四川理16】若非空集合 \(\displaystyle G\) 关于运算 \(\displaystyle \oplus\) 满足:
(1)对任意 \(\displaystyle a,b \in G\),有\(\displaystyle a \oplus b \in G\);
(2)存在 \(\displaystyle \mathrm{e} \in G\), 使得对一切 \(\displaystyle a \in G\),都有 \(\displaystyle a \oplus \mathrm{e} = \mathrm{e} \oplus a = a\),
则称 \(\displaystyle G\) 关于运算 \(\displaystyle \oplus\) 为“融洽集”. 现给出下列集合和运算:
\settasks{ label=(\arabic*), label-width=2em, label-offset=0.2em, column-sep=2em, after-item-skip=0.5ex}
(2)
\task $\displaystyle G = \{\text{非负整数}\}$, $\displaystyle \oplus$ 为整数加法 \hfill \task $\displaystyle G = \{\text{偶数}\}$, $\displaystyle \oplus$ 为整数乘法 \hfill \task $\displaystyle G = \{\text{平面向量}\}$, $\displaystyle \oplus$ 为平面向量加法 \hfill \task $\displaystyle G = \{\text{二次三项式}\}$, $\displaystyle \oplus$ 为多项式加法 \hfill \task $\displaystyle G=\{\text{虚数}\}$, $\displaystyle \oplus$ 为复数乘法 \hfill 其中 $\displaystyle G$ 关于运算 $\displaystyle \oplus$ 为“融洽集”的是$\displaystyle (\triangle)$答案
选(1)(3). 该代数系统满足封闭性且存在单位元。
(1)封闭性满足,单位元为0,正确。
(2)封闭性满足但不存在单位元,错误。
(3)封闭性满足,单位元为零向量\(\displaystyle \vv{0}\),正确.
(4)封闭性不满足,\(\displaystyle x^2+x+1+(-x^2-x-1)=0\)不属于该集合,错误。
(5)封闭性不满足,\(\displaystyle \mathrm{i}\times \mathrm{i}=-1\)不属于该集合,错误。
- 【2025“漫游数海”回归课本联赛12】设\(\displaystyle a\in\mathbb{R}\),若\(\displaystyle \{1,\sqrt{a}\}\subseteq \{1,2,4,a^2\}\),求\(\displaystyle a\)所有可能的值.
答案
\(\displaystyle a=0\)或\(\displaystyle a=4\)或\(\displaystyle a=16\). 分类讨论 \(\displaystyle \sqrt a\)与\(\displaystyle 2,4,a^2\)相等的情况,可得\(\displaystyle a=0\)或\(\displaystyle a=4\)或\(\displaystyle a=16\).
- 解答下述问题:
- 【2007湖北3】 已知集合 \(\displaystyle P = \{x \mid \log_2 x < 1\},\quad Q = \{x \mid |x - 2| < 1\}\),求 \(\displaystyle P - Q\);
- 【2004湖北文1】 已知集合 \(\displaystyle A = \{x \mid x = \sqrt{5k + 1}, k \in \mathbb{N}\},\quad B = \{x \mid x \leqslant 6, x \in \mathbb{Q}\}\),求\(\displaystyle A \cap B\).
- 【2005浙江9】设 \(\displaystyle f(n) = 2n + 1 (n \in \mathbb{N}),P = \{1,2,3,4,5\},Q = \{3,4,5,6,7\}\).记 \(\displaystyle \hat{P} = \{n \in \mathbb{N} \mid f(n) \in P\},\quad \hat{Q} = \{n \in \mathbb{N} \mid f(n) \in Q\}\),求\(\displaystyle (\hat{P} \cap \complement_{\mathbb{N}}\hat{Q}) \cup (\hat{Q} \cap \complement_{\mathbb{N}}\hat{P})\);
- 【2021福建省质检2】已知集合\(\displaystyle A=\{1,2,4\},B=\{x\in\mathbb{Z}\mid x^2-4x+m<0\}\),\(\displaystyle A\cap B=\{1,2\}\),求\(\displaystyle A\cup B\);
- 【2026新高考I卷3】已知集合 \(\displaystyle A=\left\{\sin\frac{7\pi}{6},\cos\frac{5\pi}{3},\tan\frac{5\pi}{4}\right\}\),\(\displaystyle B=\left\{-\frac{\sqrt{3}}{2},-\frac{1}{2},1\right\}\),求 \(\displaystyle A\cap B\)。
- 【1990全国卷理9】已知集合\(\displaystyle I=\{(x,y)\mid x,y\in\mathbb{R}\},M=\{(x,y)\mid \frac{y-3}{x-2}=1\},N=\{(x,y)\mid y\neq x+1\}\),求\(\displaystyle \complement_IM\cap \complement_IN\);
- 【2004安徽理11】 已知集合 \(\displaystyle M = \{\vec{a} \mid \vec{a} = (1,2) + \lambda(3,4), \lambda \in \mathbb{R}\},N = \{\vec{a} \mid \vec{a} = (-2,-2) + \lambda(4,5), \lambda \in \mathbb{R}\}\), 求 \(\displaystyle M \cap N\);
答案
(1) \(\displaystyle (0,1]\).
(2)\(\displaystyle \{1,4,6\}\). 列出集合\(\displaystyle A\)中小于6的元素:\(\displaystyle 1,\sqrt{6},\sqrt{11},4,\sqrt{21},\sqrt{26},\sqrt{31},\sqrt{36}\),后略。
(3)\(\displaystyle \{0,3\}\).
(4)\(\displaystyle \{1,2,3,4\}\). 由于\(\displaystyle y=x^2-4x+m\)关于\(\displaystyle x=2\)对称,由题意知\(\displaystyle 1,2\in B\),根据对称性有\(\displaystyle 3\in B\),而\(\displaystyle 4\notin B\),所以\(\displaystyle 0\notin B\),集合\(\displaystyle B\)的解集可视为某区间与\(\displaystyle \mathbb{Z}\)的交集,综上\(\displaystyle B=\{1,2,3\}\),于是\(\displaystyle A\cup B=\{1,2,3,4\}\)
(5)\(\displaystyle \{(2,3)\}\). 化简集合\(\displaystyle M\)的描述:\(\displaystyle y=x+1\)且\(\displaystyle x\neq 2\)。由德摩根律,\(\displaystyle \complement_I M\cap\complement_I N=\complement_I(M\cup N)\),可知\(\displaystyle M\cup N=I\setminus \{(2,3)\}\),于是所求集合为\(\displaystyle \{(2,3)\}\).
(6)\(\displaystyle \{(-2,-2)\}\). 将此题涉及的所有向量的始点均记为原点,集合\(\displaystyle M,N\)中向量的终点分别在某直线上运动,这两条直线的交点即是属于\(\displaystyle M\cap N\)的向量的终点,设\(\displaystyle x\in M,x=(1+3\lambda_1,2+4\lambda_1),y\in N,y=(-2+4\lambda_2,-2+5\lambda_2)\),由\(\displaystyle x=y\)解得\(\displaystyle \lambda_1=-1,\lambda_2=0\),可得\(\displaystyle M\cap N=\{(-2,-2)\}\).
- 解答下述问题:
- 【2020全国III卷1】已知集合 \(\displaystyle A = \{(x,y) \mid x,y \in \mathbb{N}^*, y \geqslant x\},B = \{(x,y) \mid x + y = 8\}\),求 \(\displaystyle |A \cap B|\);
- 【2018全国II卷2】 已知集合 \(\displaystyle A = \{(x,y) \mid x^2 + y^2 \leqslant 3, x \in \mathbb{Z}, y \in \mathbb{Z}\}\),求 \(\displaystyle |A|\);
- 【2013 重庆 22(1)】对正整数 \(\displaystyle n\),记 $\(\displaystyle I_n = \{1, 2, 3, \cdots, n\},P_n = \left\{\frac{m}{\sqrt{k}} \;\middle|\; m \in I_n, k \in I_n \right\}\)$ 求\(\displaystyle |P_7|\);
- 【2015湖北文10,理9】 已知集合 $\(\displaystyle A = \{(x,y) \mid x^2 + y^2 \leqslant 1, x, y \in \mathbb{Z}\},B = \{(x,y) \mid |x| \leqslant 2, |y| \leqslant 2, x, y \in \mathbb{Z}\}\)$定义集合 \(\displaystyle A \oplus B = \{(x_1+x_2, y_1+y_2) \mid (x_1, y_1) \in A, (x_2, y_2) \in B\}\).求\(\displaystyle |A \oplus B|\).
答案
(1) 4. \(\displaystyle A\)中包含\(\displaystyle \{(1,7),(2,6),(3,5),(4,4)\}\).
(2)9.
(3) 46. 当 $\displaystyle k=1$ 时,得到 $\displaystyle 1,2,3,4,5,6,7$.当 $\displaystyle k=4$ 时,得到 $\displaystyle \frac12,1,\frac32,2,\frac52,3,\frac72$. 这两种情形合并后共有 $\displaystyle 11$ 个有理数. 当 $\displaystyle k=2,3,5,6,7$ 时,约分后得到的5组数,它们的分母分别含有 $\displaystyle \sqrt2,\sqrt3,\sqrt5,\sqrt6,\sqrt7$. 组内的数各不相同,组间的数也不可能出现相等的情况,而每种 $\displaystyle k$ 给出 $\displaystyle 7$ 个不同的数. 因此 $\displaystyle |P_7|=11+5\times 7=46$. (4)45. \(\displaystyle A=\{(0,0),(1,0),(-1,0),(0,1),(0,-1)\}\).集合 \(\displaystyle B\) 是正方形 \(\displaystyle [-2,2]\times[-2,2]\) 内的整数点集合.可以按下述方式构造集合\(\displaystyle A\oplus B\):将 \(\displaystyle B\) 中的每个元素分别向上、向下、向左、向右平移一个单位,所有平移结果的并集等于横向的 \(\displaystyle 7\times5\) 矩形与纵向的 \(\displaystyle 5\times7\) 矩形的并.计数得\(\displaystyle |A\oplus B|=45\).
- 解答下述问题:
- 【2022北京9】已知正三棱锥 \(\displaystyle P - ABC\) 的六条棱长均为 \(\displaystyle 6\), \(\displaystyle S\) 是 \(\displaystyle \triangle ABC\) 及其内部的点构成的集合.设集合 \(\displaystyle T = \{Q \in S \mid PQ \leqslant 5\}\),求\(\displaystyle T\) 表示的区域的面积;
- 【2012重庆10】 设平面点集 \(\displaystyle A = \{(x,y) \mid (y - x)(y - \frac{1}{x}) \geqslant 0\},B = \{(x,y) \mid (x - 1)^2 + (y - 1)^2 \leqslant 1\}\),求 \(\displaystyle A \cap B\) 所表示的平面图形的面积;
- 【2013浙江会考39】设集合 \(\displaystyle A = \{(x,y) \mid x^2 + y^2 \leqslant 1\},\quad B = \{(x,y) \mid [x]^2 + [y]^2 \leqslant 1\}\),求\(\displaystyle A \cup B\) 所表示的平面区域的面积.
答案
(1) \(\displaystyle \pi\). 设\(\displaystyle P\)在底面\(\displaystyle ABC\)的射影为\(\displaystyle O\),所求区域为底面内以 \(\displaystyle O\) 为圆心、半径为 \(\displaystyle 1\) 的圆盘.面积为 \(\displaystyle \pi\).
(2) $\displaystyle \pi /2.$ 做出示意图,阴影部分为$\displaystyle A\cap B$表示的区域,面积为$\displaystyle \frac{\pi}{2}$ \centering {0.46\textwidth} \centering  \bookcaption{第$\displaystyle \arabic{enumi}$题(2)图示} \hfill {0.46\textwidth} \centering  \bookcaption{第$\displaystyle \arabic{enumi}$题(3)图示} (3) $\displaystyle 5+\pi /4.$ 由 $\displaystyle [x]^2+[y]^2\leqslant1$,可知 $\displaystyle ([x],[y])$ 只能是 $\displaystyle (0,0),(1,0),(-1,0),(0,1),(0,-1)$.而集合$\displaystyle G=\{(x,y)\mid [x]=p,[y]=q\}$等价于$\displaystyle \{(x,y)\mid p\leqslant x<p+1,q\leqslant x<q+1\}$,其表示区域为一个边长为1的正方形.所以 $\displaystyle B$ 是五个互不重叠的单位方格,面积为 $\displaystyle 5$.作出$\displaystyle A\cup B$表示的区域如上: 可得 $\displaystyle A\cup B$ 的面积为 $\displaystyle 5+\frac{\pi}{4}$. 新答案(来源:1.25 曲线的方程和平面点集.md):\(\displaystyle 5+\frac{\pi}{4}\)
【解题思路】因为\(\displaystyle \left[x\right],\left[y\right]\)的取值范围是整数,所以在集合\(\displaystyle B\)中,\(\displaystyle \left(\left[x\right],\left[y\right]\right)\)表示的点只可能为\(\displaystyle \left(-1,0\right)\),\(\displaystyle \left(0,0\right)\),\(\displaystyle \left(1,0\right)\),\(\displaystyle \left(0,1\right)\),\(\displaystyle \left(0,-1\right)\).
而对任意整数\(\displaystyle m,n\),满足\(\displaystyle \left(\left[x\right],\left[y\right]\right)=\left(m,n\right)\)的点\(\displaystyle \left(x,y\right)\)是点\(\displaystyle \left(m,n\right)\)右上角的边长为\(\displaystyle 1\)的正方形.
如图所示,\(\displaystyle A\cup B\)所表示的区域的面积是\(\displaystyle 5\)个正方形的面积加上四分之一圆的面积,面积为\(\displaystyle 5+\frac{\pi}{4}\).
- 【2016上海学考29】对于函数 \(\displaystyle f(x)\) 与 \(\displaystyle g(x)\),记集合 \(\displaystyle D_{f>g} = \{x \mid f(x) > g(x)\}\).设 \(\displaystyle f_1(x) = x - 1, f_2(x) = \left(\frac{1}{3}\right)^x + a \cdot 3^x + 1,h(x) = 0\).如果 \(\displaystyle D_{f_1>h} \cup D_{f_2>h} = \mathbb{R}\),求实数 \(\displaystyle a\) 的取值范围.
答案
\(\displaystyle a>-4/9\). 可知\(\displaystyle D_{f_1}>h=(1,+\infty)\),由 \(\displaystyle D_{f_1>h}\cup D_{f_2>h}=\mathbb{R}\)可以得到\(\displaystyle (-\infty,1]\subseteq D_{f_2>h}\),所以 \(\displaystyle \left(\frac13\right)^x+a\cdot3^x+1>0\) 对于 \(\displaystyle x\leqslant1\) 恒成立\(\displaystyle (*)\)。设\(\displaystyle 3^x=t\),于是原恒等式变为\(\displaystyle at^2+t+1>0\),其解集必须包括\(\displaystyle (0,3]\).当\(\displaystyle a\geqslant 0\)时显然成立,当\(\displaystyle a<0\)时,将左式视为二次函数\(\displaystyle y=f(x)\),其对应图象的对称轴\(\displaystyle x=d\)满足\(\displaystyle d>0\),而\(\displaystyle f(0)=1>0\),\(\displaystyle (0,3]\)属于解集当且仅当\(\displaystyle f(3)\geqslant 0\),解得\(\displaystyle -\frac{4}{9}<a<0\),综合可得\(\displaystyle a>-\frac{4}{9}\).
思路二:从$\displaystyle (*)$处开始,即 $\displaystyle -a<\left(\frac19\right)^x+\left(\frac13\right)^x$ 对于 $\displaystyle x\leqslant1$ 恒成立。令\(\displaystyle T\left(x\right)=\left(\frac19\right)^x+\left(\frac13\right)^x.\)
由于 \(\displaystyle y=\left(\frac19\right)^x\) 与 \(\displaystyle y=\left(\frac13\right)^x\) 在 \(\displaystyle \left(-\infty,1\right]\) 上单调递减, 所以当 \(\displaystyle x\leqslant1\) 时, $\(\displaystyle T\left(x\right)=\left(\frac19\right)^x+\left(\frac13\right)^x\geqslant\frac19+\frac13=\frac49,\)$
当且仅当 \(\displaystyle x=1\) 时等号成立。 因此 \(\displaystyle T\left(x\right)_{\min}=\frac49\),从而 \(\displaystyle -a<\frac49\),即 \(\displaystyle a>-\frac49\)。
- 【2010湖北理2】已知集合 \(\displaystyle A = \left\{(x,y) \mid \frac{x^2}{4} + \frac{y^2}{16} = 1\right\},\quad B = \{(x,y) \mid y=3^x\}\), 求 \(\displaystyle A \cap B\) 的子集的个数.
答案
4. 画图可知\(\displaystyle A\cap B\) 有两个元素,其子集个数为 \(\displaystyle 2^2=4\).
-
【2023北京第二次学考28】已知数集 \(\displaystyle A\) 含有 \(\displaystyle n\) (\(\displaystyle n \in \mathbb{N}^*\))个元素,定义集合 \(\displaystyle A^* = \{x + y \mid x, y \in A\}\).
-
写出一个集合 \(\displaystyle A\),使得 \(\displaystyle A = A^*\);
- 当 \(\displaystyle n = 4\) 时,是否存在集合 \(\displaystyle A\),使得 \(\displaystyle A^* = \{2,3,4,6,7,8,10\}\)?若存在,写出一个符合条件的集合 \(\displaystyle A\);若不存在,说明理由.
答案
(1) 取 \(\displaystyle A=\{0\}\).此时任取 \(\displaystyle x,y\in A\),都有 \(\displaystyle x+y=0\),所以 \(\displaystyle A^*=\{0\}=A\).
(2) 设集合 $\displaystyle A$中的最大元为$\displaystyle p$,最小元为$\displaystyle q$,则$\displaystyle p+p$一定是$\displaystyle A^*$中的最大元,$\displaystyle q+q$一定是$\displaystyle A^*$中的最小元.由题意,$\displaystyle p=5$,$\displaystyle q=1$. 若$\displaystyle A=\{1,5\}$,$\displaystyle A^*=\{2,6,10\}$,不符题意,于是一定存在$\displaystyle x\in A,1<x<5$,此时$\displaystyle 6<5+x<10$且$\displaystyle 5+x\in A^*$,解得$\displaystyle x=2$或$\displaystyle x=3$,于是分别验证三个集合:$\displaystyle \{1,2,5\},\{1,3,5\},\{1,2,3,5\}$,均不满足题意.因此不存在这样的集合 $\displaystyle A$.- 设 \(\displaystyle f(x)\) 是定义在 \(\displaystyle \mathbb{R}\) 上的函数,定义$\(\displaystyle S_f = \{x \in D \mid f(x) = f(-x)\}\)$
- 证明: \(\displaystyle S_f \neq \varnothing\);
- 证明: \(\displaystyle f(x)\) 是偶函数当且仅当 \(\displaystyle S_f = \mathbb{R}\);
- 若 \(\displaystyle f(x)\) 是奇函数,证明: \(\displaystyle x_0 \in S_f\) 当且仅当 \(\displaystyle x_0\) 是 \(\displaystyle f(x)\) 的零点;
- 若对任意 \(\displaystyle x, y \in \mathbb{R}\),当 \(\displaystyle x < y\) 时, \(\displaystyle f(x) \leqslant f(y)\),判断是否成立:\(\displaystyle \forall x \in S_f\), \(\displaystyle f(x) = f(0)\);
- 设 \(\displaystyle f(x), g(x)\) 都是定义在 \(\displaystyle \mathbb{R}\) 上的函数, \(\displaystyle h(x) = f(x)g(x)\).判断下述命题是否成立:$\(\displaystyle (a).S_h \subseteq S_f \cap S_g\quad (b).S_f \cap S_g \subseteq S_h\quad (c).S_h \subseteq S_f \cup S_g\quad (d).S_f \cup S_g \subseteq S_h\)$
答案
(1)因为 \(\displaystyle 0=-0\),所以 \(\displaystyle f(0)=f(-0)\),从而 \(\displaystyle 0\in S_f\),故 \(\displaystyle S_f\neq\varnothing\).
(2)若 $\displaystyle f$ 是偶函数,则对任意 $\displaystyle x\in\mathbb R$,都有 $\displaystyle f(x)=f(-x)$,所以 $\displaystyle S_f=\mathbb R$. 反过来,若 $\displaystyle S_f=\mathbb R$,则对任意 $\displaystyle x\in\mathbb R$,都有 $\displaystyle f(x)=f(-x)$,这正是偶函数的定义. 因此 $\displaystyle f$ 为偶函数当且仅当 $\displaystyle S_f=\mathbb R$. (3)若 $\displaystyle x_0\in S_f$,因为 $\displaystyle f$ 为奇函数,所以$\displaystyle f(-x_0)=-f(x_0)$. 于是 $\displaystyle f(x_0)=f(-x_0)=-f(x_0)$,可得$\displaystyle f(x_0)=0$. 反过来,若 $\displaystyle f(x_0)=0$,则 $\displaystyle f(-x_0)=-f(x_0)=0=f(x_0)$,所以 $\displaystyle x_0\in S_f$. 因此 $\displaystyle x_0\in S_f$ 当且仅当 $\displaystyle x_0$ 是零点. (4)设 $\displaystyle f$ 单调不减,且 $\displaystyle x\in S_f$. 若 $\displaystyle x>0$,则 $\displaystyle -x<0<x$,所以 $\displaystyle f(-x)\leqslant f(0)\leqslant f(x)$. 又因为 $\displaystyle f(-x)=f(x)$,所以三者相等,特别有 $\displaystyle f(x)=f(0)$. 若 $\displaystyle x<0$,则 $\displaystyle x<0<-x$,所以 $\displaystyle f(x)\leqslant f(0)\leqslant f(-x)$. 同样利用 $\displaystyle f(x)=f(-x)$,得到 $\displaystyle f(x)=f(0)$. $\displaystyle x=0$ 时结论显然成立.因此命题成立. (5)若 $\displaystyle x\in S_f\cap S_g$,则 $\displaystyle f(x)=f(-x)$ 且 $\displaystyle g(x)=g(-x)$. 两式相乘得 $\displaystyle h(x)=f(x)g(x)=f(-x)g(-x)=h(-x)$, 所以 $\displaystyle x\in S_h$.因此 $\displaystyle S_f\cap S_g\subseteq S_h$,命题 b 正确. 命题 a 不成立.取 $\displaystyle f(x)=x$、$\displaystyle g(x)=0$,则 $\displaystyle h(x)=0$,所以 $\displaystyle S_h=\mathbb R$,但 $\displaystyle S_f=\{0\}$,故 $\displaystyle S_h\nsubseteq S_f\cap S_g$. 命题 c 不成立.取 $\displaystyle f(x)=g(x)=x$,则 $\displaystyle h(x)=x^2$,所以 $\displaystyle S_h=\mathbb R$,而 $\displaystyle S_f=S_g=\{0\}$,故 $\displaystyle S_h\nsubseteq S_f\cup S_g$. 命题 d 不成立.仍取 $\displaystyle f(x)=g(x)=x$,则 $\displaystyle S_f\cup S_g=\{0\}$,而 $\displaystyle S_h=\mathbb R$;为否定 d,应取 $\displaystyle f(x)=x$、$\displaystyle g(x)=1$,此时 $\displaystyle S_f\cup S_g=\mathbb R$,但 $\displaystyle S_h=S_f=\{0\}$. 因此四个命题中只有 b 正确.-
给定正整数\(\displaystyle n\),已知集合 $\(\displaystyle M_n = \{(x_1, x_2, \dots,x_{n}) \mid x_i \in \{0, 1\}, i = 1, 2, 3, \dots,n\}\)$ 对于\(\displaystyle X = (x_1, x_2, \dots, x_n), Y = (y_1, y_2, \dots, y_n) \in M_n\),定义 \(\displaystyle d(X, Y) = \sum_{i=1}^{n} |x_i - y_i|\).
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已知 \(\displaystyle X = (1, 0, 1, 0) \in M_4\),求所有的 \(\displaystyle Y \in M_4\),使得 \(\displaystyle d(X, Y) = 3\);
- 已知 \(\displaystyle X, Y, Z \in M_n\),求证:\(\displaystyle d(X, Y) + d(Y, Z) + d(Z, X)\) 为偶数.
答案
(1) 所有可能的 \(\displaystyle Y\) 为 \(\displaystyle (1,1,0,1),(0,0,0,1),(0,1,1,1),(0,1,0,0)\).
(2) 对每个坐标 $\displaystyle i$,因为 $\displaystyle x_i,y_i,z_i$ 都只能取 $\displaystyle 0$ 或 $\displaystyle 1$,通过枚举可以验证,三数 $\displaystyle |x_i-y_i|,\ |y_i-z_i|,\ |z_i-x_i|$ 中为 $\displaystyle 1$ 的个数只能是 $\displaystyle 0$ 或 $\displaystyle 2$.当三个数相同时,三个差均为 $\displaystyle 0$;当其中一个与另外两个不同,恰有两个差为 $\displaystyle 1$. 因此对每个坐标都有 $\displaystyle |x_i-y_i|+|y_i-z_i|+|z_i-x_i|\equiv0\pmod2$. 对 $\displaystyle i=1,2,\ldots,n$ 求和,得到 $\displaystyle d(X,Y)+d(Y,Z)+d(Z,X)\equiv0\pmod2$.所以该和为偶数. 或注意到$\displaystyle |x_i-y_i|$在此种情况下等价于$\displaystyle (x_i+y_i)$模2,于是$$\displaystyle |x_i-y_i|+|y_i-z_i|+|z_i-x_i|\equiv 2(x_i+y_i+z_i)\equiv 0\pmod 2$$-
【2023 北京第一次学考28】 给定正整数 \(\displaystyle k \geqslant 2\),设集合 $\(\displaystyle M = \{ (x_1, x_2, \dots, x_k) \mid x_i \in \{0, 1\}, i = 1, 2, \dots, k \}\)$ 对于集合 \(\displaystyle M\) 的子集 \(\displaystyle A\),若\(\displaystyle |A|\geqslant 2\)且\(\displaystyle A\) 中任意两个不同元素 \(\displaystyle (y_1, y_2, \dots, y_k), (z_1, z_2, \dots, z_k)\)满足\(\displaystyle y_1+y_2+\dots+y_k = z_1+z_2+\dots+z_k\)且 \(\displaystyle y_1+z_1, y_2+z_2, \dots, y_k+z_k\) 中有且只有一个为 \(\displaystyle 2\),则称 \(\displaystyle A\) 具有性质 \(\displaystyle \vv{P}\).
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当 \(\displaystyle k=2\) 时,判断 \(\displaystyle A = \{ (1, 0), (0, 1) \}\) 是否具有性质 \(\displaystyle \vv{P}\);(结论无需证明)
- 当 \(\displaystyle k=3\) 时,写出一个具有性质 \(\displaystyle \vv{P}\) 的集合 \(\displaystyle A\);
- 当 \(\displaystyle k=4\) 时,求证:若 \(\displaystyle A\) 中的元素个数为 \(\displaystyle 4\),则 \(\displaystyle A\) 不具有性质 \(\displaystyle \vv{P}\)
- (选做)对于正整数 \(\displaystyle k \geqslant 4\),若 \(\displaystyle A\) 具有性质 \(\displaystyle \vv{P}\),则 \(\displaystyle |A|\) 的最大值是多少?
答案
(1) 不具有性质 \(\displaystyle \mathbf P\).
(2) $\displaystyle A=\{(0,1,1),(1,0,1)\}$. (3) 任取集合$\displaystyle A$中的两个不同元素$\displaystyle (x_1,x_2,x_3,x_4),(y_1,y_2,y_3,y_4)$,假设它们都满足$\displaystyle x_1+x_2+x_3+x_4=y_1+y_2+y_3+y_4=r$,其中$\displaystyle r=0,1,2,3,4$. 当$\displaystyle r=0$时,$\displaystyle A$中有1个元素$\displaystyle (0,0,0,0)$,不具有性质 $\displaystyle \mathbf P$.当$\displaystyle r=4$时,$\displaystyle A$中有1个元素$\displaystyle (1,1,1,1)$,不具有性质 $\displaystyle \mathbf P$. 当$\displaystyle r=1$时,$\displaystyle A$中有4个元素$\displaystyle (1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)$,取元素$\displaystyle (1,0,0,0),(0,1,0,0)$发现不符题意,故集合$\displaystyle A$不具有性质 $\displaystyle \mathbf P$. 当$\displaystyle r=2$时,$\displaystyle A$中有6个元素$\displaystyle (1,1,0,0),(1,0,1,0),(1,0,0,1),(0,1,1,0),(0,1,0,1),(0,0,1,1)$,取元素$\displaystyle (1,0,1,0),(0,1,0,1)$发现不符题意,故集合$\displaystyle A$不具有性质 $\displaystyle \mathbf P$. 当$\displaystyle r=3$时,$\displaystyle A$中有4个元素$\displaystyle (1,1,1,0),(1,1,0,1),(1,0,1,1),(0,1,1,1)$,取$\displaystyle (1,1,1,0),(1,1,0,1)$发现不符题意,故集合$\displaystyle A$不具有性质 $\displaystyle \mathbf P$. 综上,$\displaystyle A$ 不具有性质 $\displaystyle \mathbf P$.
B 组习题
B组
-
设甲:\(\displaystyle A\subseteq B\);乙:对任意集合\(\displaystyle C\),有\(\displaystyle (B\cap C)\cup A=B\cap (C\cup A)\),则
- 甲是乙的充分不必要条件
- 甲是乙的必要不充分条件
- 甲是乙的充要条件
- 甲是乙既不充分也不必要条件
答案
C. 由分配律,有 \(\displaystyle B\cap(C\cup A)=(B\cap C)\cup(B\cap A)\).充分性满足,由\(\displaystyle A\subseteq B\)有 \(\displaystyle (B\cap C)\cup(B\cap A)=(B\cap C)\cup A\),所以命题乙成立.必要性满足,可知\(\displaystyle A\subseteq (B\cap C)\cup A=B\cap (C\cup A)\subseteq B\).
- 【2013广东理8】已知集合 $\(\displaystyle S=\{(x,y,z) \mid x,y,z \in \mathbb{N^*}, \text{且} x<y<z, y<z<x, z<x<y \text{ 三个条件中恰有一个成立}\}\)$若 \(\displaystyle (x,y,z),(z,w,x)\in S\),则下列命题正确的是
- \(\displaystyle (y,z,w) \in S, (x,y,w) \notin S\)
- \(\displaystyle (y,z,w) \in S, (x,y,w) \in S\)
- \(\displaystyle (y,z,w) \notin S, (x,y,w) \in S\)
- \(\displaystyle (y,z,w) \notin S, (x,y,w) \notin S\)
答案
B. 方法一:令 \(\displaystyle w=1\),\(\displaystyle x=2\),\(\displaystyle y=3\),\(\displaystyle z=4\),则 \(\displaystyle \left(2,3,4\right)\),\(\displaystyle \left(4,1,2\right)\in S\)。可以判断 \(\displaystyle \left(y,z,w\right)=\left(3,4,1\right)\in S\),\(\displaystyle \left(x,y,w\right)=\left(2,3,1\right)\in S\)。故选 B。
方法二:由条件,\ding{172} \(\displaystyle x<y<z\),\ding{173} \(\displaystyle y<z<x\),\ding{174} \(\displaystyle z<x<y\) 必有一个成立;\ding{175} \(\displaystyle z<w<x\),\ding{176} \(\displaystyle w<x<z\),\ding{177} \(\displaystyle x<z<w\) 中必有一个成立。
经过简单分析可知只能有下面四种情况:
\(\displaystyle \left(1\right)\) \ding{172}\ding{176}同时成立;\(\displaystyle \left(2\right)\) \ding{172}\ding{177}同时成立;\(\displaystyle \left(3\right)\) \ding{173}\ding{175}同时成立;\(\displaystyle \left(4\right)\) \ding{174}\ding{175}同时成立。这四种情况都能推出 B 选项的结果。
证明:因为 \(\displaystyle \left(x,y,z\right)\in S\),所以\ding{172} \(\displaystyle x<y<z\),\ding{173} \(\displaystyle y<z<x\),\ding{174} \(\displaystyle z<x<y\) 必有一个成立;因为 \(\displaystyle \left(z,w,x\right)\in S\),所以\ding{175} \(\displaystyle z<w<x\),\ding{176} \(\displaystyle w<x<z\),\ding{177} \(\displaystyle x<z<w\) 必有一个成立。
\(\displaystyle \left(1\right)\) 若\ding{172}成立,则\ding{176}或\ding{177}成立,从而 \(\displaystyle w<x<y<z\) 或 \(\displaystyle x<y<z<w\)。
当 \(\displaystyle w<x<y<z\) 时,因为 \(\displaystyle y<z<w\),\(\displaystyle z<w<y\),\(\displaystyle w<y<z\) 中恰好只有 \(\displaystyle w<y<z\) 成立,所以 \(\displaystyle \left(y,z,w\right)\in S\)。因为 \(\displaystyle x<y<w\),\(\displaystyle y<w<x\),\(\displaystyle w<x<y\) 中恰好只有 \(\displaystyle w<x<y\) 成立,所以 \(\displaystyle \left(x,y,w\right)\in S\)。
当 \(\displaystyle x<y<z<w\) 时,因为 \(\displaystyle y<z<w\),\(\displaystyle z<w<y\),\(\displaystyle w<y<z\) 中恰好只有 \(\displaystyle y<z<w\) 成立,所以 \(\displaystyle \left(y,z,w\right)\in S\)。因为 \(\displaystyle x<y<w\),\(\displaystyle y<w<x\),\(\displaystyle w<x<y\) 中恰好只有 \(\displaystyle x<y<w\) 成立,所以 \(\displaystyle \left(x,y,w\right)\in S\)。
\(\displaystyle \left(2\right)\) 若\ding{173}成立,则\ding{175}成立,从而 \(\displaystyle y<z<w<x\)。因为 \(\displaystyle y<z<w\),\(\displaystyle z<w<y\),\(\displaystyle w<y<z\) 中恰好只有 \(\displaystyle y<z<w\) 成立,所以 \(\displaystyle \left(y,z,w\right)\in S\)。因为 \(\displaystyle x<y<w\),\(\displaystyle y<w<x\),\(\displaystyle w<x<y\) 中恰好只有 \(\displaystyle y<w<x\) 成立,所以 \(\displaystyle \left(x,y,w\right)\in S\)。
\(\displaystyle \left(3\right)\) 若\ding{174}成立,则\ding{175}成立,从而 \(\displaystyle z<w<x<y\)。因为 \(\displaystyle y<z<w\),\(\displaystyle z<w<y\),\(\displaystyle w<y<z\) 中恰好只有 \(\displaystyle z<w<y\) 成立,所以 \(\displaystyle \left(y,z,w\right)\in S\)。因为 \(\displaystyle x<y<w\),\(\displaystyle y<w<x\),\(\displaystyle w<x<y\) 中恰好只有 \(\displaystyle w<x<y\) 成立,所以 \(\displaystyle \left(x,y,w\right)\in S\)。
综上所述,\(\displaystyle \left(y,z,w\right)\in S\) 且 \(\displaystyle \left(x,y,w\right)\in S\)。
- 【2020浙江10】 已知集合 \(\displaystyle S,T\subseteq\mathbb{N}^*\), \(\displaystyle S,T\) 中至少有两个元素, 且满足:(1)对任意 \(\displaystyle x,y\in S\), 若 \(\displaystyle x\neq y\), 则 \(\displaystyle xy\in T\);(2)对任意 \(\displaystyle x,y\in T\), 若 \(\displaystyle x<y\), 则 \(\displaystyle \frac{y}{x}\in S\),则下列命题正确的是
- 若 \(\displaystyle |S|=4\), 则 \(\displaystyle |S\cup T|=7\)
- 若\(\displaystyle |S|=4\), 则 \(\displaystyle |S\cup T|=6\)
- 若 \(\displaystyle |S|=3\), 则\(\displaystyle |S\cup T|=4\)
- 若 \(\displaystyle |S|=3\), 则 \(\displaystyle |S\cup T|=5\)
附加练习 1. 若\(\displaystyle S\)中恰有2个元素, 1. 若\(\displaystyle 1\in S\),证明:\(\displaystyle S\cap T\neq \varnothing\),并求\(\displaystyle S\cup T\)的元素个数的所有可能值. 2. 若\(\displaystyle 1\notin S\),求\(\displaystyle S\cup T\)的元素个数的所有可能值. 2. 若\(\displaystyle S\)中恰有3个元素\(\displaystyle a,b,c(a<b<c)\). 1. 若\(\displaystyle 1\in S\),证明\(\displaystyle a,b,c\)成公比为\(\displaystyle b\)的等比数列; 2. 若\(\displaystyle 1\notin S\),证明\(\displaystyle a,b,c\)成公比为\(\displaystyle a\)的等比数列; 3. 求\(\displaystyle S\cup T\)的元素个数的所有可能值. 3. 若\(\displaystyle S\)中恰有4个元素\(\displaystyle a,b,c,d(a<b<c<d)\),由条件(1)知\(\displaystyle ab,ac,ad,bc,bd,cd\in S\),由条件(2)知\(\displaystyle \frac{cd}{ab},\frac{d}{a},\frac{d}{b},\frac{d}{c},\frac{c}{b},\frac{c}{a},\frac{b}{a}\in S\). 1. 若\(\displaystyle 1\notin S\),证明\(\displaystyle a,b,c,d\)成公比为\(\displaystyle a\)的等比数列; 2. 是否可能有\(\displaystyle 1\in S\)? 3. 求\(\displaystyle S\cup T\)的元素个数.
答案
附加练习解答如下:
(1.1)设$\displaystyle S=\{1,a\}(a>1)$,则$\displaystyle a\in T,S\cap T\neq \varnothing$,若$\displaystyle |T|=3$,设$\displaystyle T=\{p,q,r\}(p<q<r)$,则$\displaystyle \frac{r}{p},\frac{r}{q}\in S$,这两个数不相同且不等于1,而$\displaystyle S$中只有$\displaystyle a\neq 1$,矛盾,当$\displaystyle |T|>3$时显然也不成立,故$\displaystyle |T|=2$,可得$\displaystyle T=\{a,a^2\}$或$\displaystyle T=\{1,a\}$,于是$\displaystyle |S\cup T|=2$或$\displaystyle 3$。 (1.2)设$\displaystyle S=\{a,b\}(1<a<b)$,则$\displaystyle ab\in T$。若$\displaystyle |T| = 2$,设 $\displaystyle T = \{ab, x\}$,可得$\displaystyle x=a$或$\displaystyle b$或$\displaystyle a^2b$或$\displaystyle ab^2$,于是$\displaystyle |S\cup T|=3$或$\displaystyle 4$。若 \(\displaystyle |T| = 3\),设 \(\displaystyle T = \{x, y, z\},ab\in T\) ,可知\(\displaystyle \frac{y}{x} , \frac{z}{x} ,\frac{z}{y} \in S\),而\(\displaystyle \frac{y}{x}<\frac{z}{x},\frac{z}{y}<\frac{z}{x}\),则一定有\(\displaystyle \frac{y}{x}=\frac{z}{y}\),故\(\displaystyle x,y,z\)成等比数列,公比是\(\displaystyle a\)。由于\(\displaystyle \frac{z}{x}=a^2\in S\),故\(\displaystyle b=a^2\),\(\displaystyle S=\{a,a^2\}\)。由上述讨论可知\(\displaystyle T=\{a,a^2,a^3\}\)或\(\displaystyle \{a^2,a^3,a^4\}\)或\(\displaystyle \{a^3,a^4,a^5\}\),故\(\displaystyle |S\cup T|=3\)或4或5.
综上,\(\displaystyle |S\cup T|=3\text{或}4\text{或}5\)。
(2.1)已知\(\displaystyle a = 1\),设\(\displaystyle S = \{1, b, c\}\),可知 \(\displaystyle b, c, bc \in T\),又因为 \(\displaystyle \frac{c}{b} \in S\),且 \(\displaystyle \frac{c}{b} \neq c, \frac{c}{b} \neq 1\),于是\(\displaystyle \frac{c}{b} = b\),可得\(\displaystyle c = b^2\),于是\(\displaystyle a,b,c\)成公比为\(\displaystyle b\)的等比数列。
(2.2)设\(\displaystyle S=\{a,b,c\}(1<a<b<c)\),可知\(\displaystyle ab<ac<bc\in T\),于是\(\displaystyle \frac{c}{b},\frac{c}{a},\frac{b}{a}\in S\), 由于\(\displaystyle \frac{b}{a}<b<c\),故\(\displaystyle \frac{b}{a}=a\),可得\(\displaystyle b=a^2\)。
已知\(\displaystyle \frac{c}{a}\neq c\),假设\(\displaystyle \frac{c}{a}=a\),则\(\displaystyle c=a^2=b\),矛盾,故\(\displaystyle \frac{c}{a}=b\),可得\(\displaystyle c=a^3\),于是\(\displaystyle a,b,c\)成公比为\(\displaystyle a\)的等比数列。
(2.3)若 \(\displaystyle 1 \in S\),由问题(2.2)的结论可设 \(\displaystyle S = \{1, b, b^2\}\),此时 \(\displaystyle b, b^2, b^3 \in T\)。
若存在\(\displaystyle k \in \mathbb{N^*}\setminus \{1, 2, 3\}\) 满足 \(\displaystyle b^k \in T\),则有 \(\displaystyle \frac{b^k}{b^3} , \frac{b^k}{b^2} , \frac{b^k}{b} \in S\),而\(\displaystyle \frac{b^k}{b}>b^2\),矛盾,故\(\displaystyle T=\{b,b^2,b^3\}\),该情况下\(\displaystyle |S\cup T|=4\)
若 \(\displaystyle 1 \notin S\),由问题(2.3)的结论可设 \(\displaystyle S = \{a, a^2, a^3\}(a>1)\),则\(\displaystyle a^3,a^4,a^5\in T\),讨论得\(\displaystyle T\)可为\(\displaystyle \{a^3,a^4,a^5\}\)或\(\displaystyle \{a^3,a^4,a^5,a^6\}\)或\(\displaystyle \{a^2,a^3,a^4,a^5\}\),在该情况下,\(\displaystyle |S\cup T|=5\)或\(\displaystyle 6\).
综上,\(\displaystyle |S\cup T|=4\)或\(\displaystyle 5\)或\(\displaystyle 6\)。
(3.1)设\(\displaystyle S = \{a, b, c, d\}(1<a<b<c<d)\)。由题设:\(\displaystyle \frac{cd}{ab}, \frac{d}{b}, \frac{d}{c}, \frac{c}{b} , \frac{c}{a} , \frac{b}{a} \in S\),由\(\displaystyle \frac{b}{a}<\frac{c}{a}<\frac{d}{a}\)与\(\displaystyle a>1\)可得\(\displaystyle \frac{b}{a}<b,c,d\),因此\(\displaystyle \frac{b}{a}=a\),于是\(\displaystyle b=a^2\).
因为\(\displaystyle \frac{c}{b}<\frac{c}{a}<\frac{d}{a}<\frac{cd}{ab}\)且这些元素都属于\(\displaystyle S\),所以\(\displaystyle \frac{c}{b}=a\),可得\(\displaystyle c=a^3\),此时有 \(\displaystyle d, \frac{d}{a}, \frac{d}{a^2}, \frac{d}{a^3}, a, a^2 \in S = \{a, a^2, a^3, d\}\),可得\(\displaystyle d=a^4\),于是\(\displaystyle a,b,c,d\)成公比为\(\displaystyle a\)的等比数列。
(3.2)不可能。假设 \(\displaystyle S = \{1, b, c, d\}\),则 \(\displaystyle \frac{cd}{b}, d, \frac{d}{b}, \frac{d}{c}, \frac{c}{b}, c, b \in S\),而 \(\displaystyle \frac{cd}{b} > d\) ,矛盾。
(3.3)由问题(3.1)(3.2)的结论,可知\(\displaystyle S = \{a, a^2, a^3, a^4\}\) ,此时\(\displaystyle a^3,a^4,a^5,a^6,a^7\in T\),\(\displaystyle T\)中也不可能存在其他元素(已有5个元素,彼此作商至少得到4个结果),故\(\displaystyle T=\{a^3,a^4,a^5,a^6,a^7\}\),于是\(\displaystyle |S\cap T|=7\)
- 【2010福建文15改编】(多选)在平面直角坐标系\(\displaystyle xOy\)中,定义点\(\displaystyle A(x_1,y_1),B(x_2,y_2)\)的运算:
(1)加法运算:\(\displaystyle A+B=(x_1+x_2,y_1+y_2)\);
(2)数乘运算:\(\displaystyle \forall \lambda\in\mathbb{R},\lambda A=(\lambda x_1,\lambda y_1)\).
对于集合\(\displaystyle S\subseteq\mathbb{R}^2\),若对任意\(\displaystyle P,Q\in S\)与满足\(\displaystyle 0\leqslant \theta\leqslant 1\)的\(\displaystyle \theta\)都有\(\displaystyle \theta P+(1-\theta)Q\in S\),那么称集合\(\displaystyle S\)为凸集.特别地,定义空集\(\displaystyle \varnothing\)是凸集.则下列集合中是凸集的有
- \(\displaystyle \{(x,y)\mid y=x+1\}\)
- \(\displaystyle \{(x,y)\mid \frac{x^2}{9}+y^2\leqslant1\}\)
- \(\displaystyle \{(x,y)\mid y\geqslant x-\frac{3}{x},x>0\}\)
- \(\displaystyle \{(x,y)\mid x^2+y^2-2|x|\leqslant0\}\)
答案
AB. 原题意指:若对\(\displaystyle S\)中的任意两个不同的点,其连线段上的所有点均属于集合\(\displaystyle S\),则\(\displaystyle S\)是凸集,依次画出四个选项表示的区域,据该定义判断即可.
- 【2006上海理16改编】(多选)设空间中存在两个相交平面\(\displaystyle \alpha,\beta\).对于平面上任意一点\(\displaystyle M\),定义\(\displaystyle d_M=(p,q)\),其中\(\displaystyle p,q\)分别是\(\displaystyle M\)到平面\(\displaystyle \alpha,\beta\)的距离.现给定常数\(\displaystyle x,y>0\),对于空间中的某个平面\(\displaystyle \gamma\),记集合\(\displaystyle A_{\gamma}=\{M\in\gamma\mid d_{M}=(x,y)\}\),\(\displaystyle A_{\gamma}\)中可能有\(\displaystyle (\triangle)\)个元素.
- 0
- 2
- 4
- 无穷多
答案
ACD. 到平面\(\displaystyle \alpha\)的距离为 \(\displaystyle |x|\)的点构成平行于\(\displaystyle \alpha\)的两个平面,它们位于平面\(\displaystyle \alpha\)的两侧,同理,到平面\(\displaystyle \beta\)的距离为 \(\displaystyle |y|\)的点构成平行于\(\displaystyle \beta\)的两个平面,它们位于平面\(\displaystyle \beta\)的两侧,同时满足上述两个条件的点构成这两组平面的四条平行交线,记为集合\(\displaystyle L\),可知\(\displaystyle A_{\gamma}=L\cap \gamma\),其元素个数可能为0(与这四条直线无交点),4(与这四条直线各有一个交点)或无穷多个(包含这四条直线中的至少一条).
- 【2009北京文8加强】【2024“fiddie”第二次模考11】(多选)在正方形\(\displaystyle A_1A_2A_3A_4\)中,设\(\displaystyle D\)是正方形\(\displaystyle A_1A_2A_3A_4\)内部的点构成的集合,\(\displaystyle P_0\in D\),则集合$\(\displaystyle S=\{P\mid P\in D,|PP_0|\leqslant |PA_i|.i=1,2,3,4\}\)$表示的平面区域可能是
- 四边形区域
- 五边形区域
- 六边形区域
- 八边形区域
答案
ABC. 对于条件 \(\displaystyle |PP_0|\leqslant|PA_i|\) ,当\(\displaystyle |PP_0|=|PA_i|\)时,满足条件的点构成线段\(\displaystyle P_0A_i\)的垂直平分线,它将平面划分为两部分,满足\(\displaystyle |PP_0|\leqslant |PA_i|\)的所有点构成靠近\(\displaystyle P_0\)侧的这一部分,记为\(\displaystyle S_i\),要判断\(\displaystyle S_1\cap S_2\cap S_3\cap S_4\)构成区域的形状,只需依次作出\(\displaystyle P_0A_1,P_0A_2,P_0A_3,P_0A_4\)的垂直平分线,它们与正方形边界围成的图形就是我们要求的区域,通过作图反复尝试,可知该区域可能为四边形区域,五边形区域,六边形区域.
-
【2004北京理8改编】(多选) 已知函数 \(\displaystyle f(x) = \begin{cases} x, & x \in P, \\ -x, & x \in M, \end{cases}\) 其中 \(\displaystyle P, M\) 为实数集 \(\displaystyle \mathbb{R}\) 的两个非空子集.设\(\displaystyle f(P) = \{y \mid y = f(x), x \in P\}\), \(\displaystyle f(M) = \{y \mid y = f(x), x \in M\}\), 则下列叙述正确的是
若 \(\displaystyle P \cap M = \varnothing\), 则 \(\displaystyle f(P) \cap f(M) =\varnothing\)
- 若 \(\displaystyle P \cap M \neq \varnothing\), 则 \(\displaystyle f(P) \cap f(M) \neq \varnothing\)
- 若 \(\displaystyle P \cup M = \mathbb{R}\), 则 \(\displaystyle f(P) \cup f(M) = \mathbb{R}\)
- 若 \(\displaystyle P \cup M \neq \mathbb{R}\), 则 \(\displaystyle f(P) \cup f(M) \neq \mathbb{R}\)
答案
A 错.例如 \(\displaystyle P=\{1\},M=\{-1\}\) 时,\(\displaystyle P\cap M=\varnothing\),但 \(\displaystyle f(P)=f(M)=\{1\}\). 由于 \(\displaystyle f\) 是函数,若 \(\displaystyle P\cap M\neq\varnothing\),交集中的元素只能是 \(\displaystyle 0\);于是 \(\displaystyle 0\in f(P)\cap f(M)\),B 正确. C 错,例如 \(\displaystyle P=(0,+\infty)\)、\(\displaystyle M=(-\infty,0]\) 时虽有 \(\displaystyle P\cup M=\mathbb R\),但 \(\displaystyle f(P)\cup f(M)=[0,+\infty)\). D 正确:若 \(\displaystyle P\cup M\neq\mathbb R\),取 \(\displaystyle x\notin P\cup M\).若 \(\displaystyle x\) 在像并集中,则必须有 \(\displaystyle -x\in M\);若 \(\displaystyle -x\) 不在像并集中,则矛盾,若 \(\displaystyle -x\in P\),再考察 \(\displaystyle -x\) 可知函数定义要求相应的另一侧也出现,最终仍与 \(\displaystyle x\notin P\cup M\) 矛盾.因此选 B、D.
- 设集合\(\displaystyle A=\{a_1,a_2,a_3,a_4,a_5\},B=\{a_1^2,a_2^2,a_3^2,a_4^2,a_5^2\}\),其中\(\displaystyle a_1,a_2,\cdots,a_5\)是从小到大排序的正整数,已知\(\displaystyle A\cap B=\{a_1,a_4\}\)且元素和是\(\displaystyle 10\),\(\displaystyle A\cup B\)的元素和是\(\displaystyle 224\),求集合\(\displaystyle A\).
答案
A={1,3,4,9,10}. 由\(\displaystyle a_1\leqslant a_1^2<a_2^2 ,a_1<a_2\)且\(\displaystyle a_1\in A\cap B\)可知,\(\displaystyle a_1=a_1^2\),于是\(\displaystyle a_1=1,a_4=9\).\(\displaystyle A\cup B\)的元素和为224,\(\displaystyle A\cap B\)的元素和为10,故\(\displaystyle a_1,a_2,\cdots,a_5,a_1^2,a_2^2,\cdots,a_5\)这10个元素的和为234,于是\(\displaystyle a_2+a_3+a_5+a_2^2+a_3^2+a_5^2=142\).下确定\(\displaystyle a_5\)的值,可知\(\displaystyle 10\leqslant a_5\leqslant 11\),当\(\displaystyle a_5=11\)时,\(\displaystyle a_2+a_3+a_2^2+a_3^2=10\),不存在满足条件的\(\displaystyle a_2,a_3\),于是\(\displaystyle a_5=10\),有\(\displaystyle a_2+a_3+a_2^2+a_3^2=32\),同上进行讨论,可得\(\displaystyle a_2=3,a_3=4\).故\(\displaystyle A=\{1,3,4,9,10\}\).
- 【2015浙江6】 设 \(\displaystyle A, B\) 是有限集,定义 \(\displaystyle d(A, B) = |A \cup B| - |A \cap B|\),证明或反驳:对任意有限集 \(\displaystyle A, B, C\),\(\displaystyle d(A, C) \leqslant d(A, B) + d(B, C)\).
答案
对任意有限集 \(\displaystyle E\),记 \(\displaystyle \mathbf1_E(x)\) 为其示性函数.对任意元素 \(\displaystyle x\),有 \(\displaystyle \mathbf1_{A\triangle B}(x)=|\mathbf1_A(x)-\mathbf1_B(x)|\).可知 $\(\displaystyle d(A,B)=\sum_x|\mathbf1_A(x)-\mathbf1_B(x)|\)$ 由三角不等式: \(\displaystyle |\mathbf1_A(x)-\mathbf1_C(x)| \leqslant|\mathbf1_A(x)-\mathbf1_B(x)| +|\mathbf1_B(x)-\mathbf1_C(x)|\).对所有可能的 \(\displaystyle x\) 求和,得到 \(\displaystyle d(A,C)\leqslant d(A,B)+d(B,C)\). 所以题述命题成立.
- 【2019上海春考12】已知集合\(\displaystyle A=[t,t+1]\cup [t+4,t+9],0\notin A\),且存在正数\(\displaystyle \lambda\),使得对任意\(\displaystyle a\in A\)都有\(\displaystyle \frac{\lambda}{a}\in A\),求\(\displaystyle t\)的值.
答案
\(\displaystyle 1\)或\(\displaystyle -3\). 对\(\displaystyle A\)中的每个元素\(\displaystyle a\)变换为\(\displaystyle \lambda /a\)后,新元素仍形成两段区间,并且每个新区间内的元素由同一区间元素变换得到,记区间\(\displaystyle A,B\)变换得到区间\(\displaystyle A^*,B^*\)。若\(\displaystyle A=A^*,B=B^*\),则\(\displaystyle \lambda /(t+1)=t,\lambda (t+4)=t+9\),解得\(\displaystyle t=-3\),此时\(\displaystyle A=[-3,-2],B=[1,6]\).若\(\displaystyle A=B^*,B=A^*\),则\(\displaystyle t=\lambda/(t+9),t+1=\lambda/(t+4)\),解得\(\displaystyle t=1\),此时\(\displaystyle A=[1,2],B=[5,10]\).
- 【2016浙江学考25】 已知函数 \(\displaystyle f(x) = \frac{1}{x - a} - \frac{1}{x - b}\) (\(\displaystyle a, b\) 为实常数且 \(\displaystyle a < b\)).设集合 \(\displaystyle M = \{(x,y) \mid y = f(x)\},N = \{(x,y) \mid y = \lambda(x - \frac{a+b}{2})^2, \lambda \in \mathbb{R}\}\),若 \(\displaystyle M \cap N = \varnothing\),求 \(\displaystyle \lambda\) 的取值范围.
答案
\(\displaystyle 0\leqslant\lambda<64/(b-a)^3\). 因为 \(\displaystyle M\cap N=\varnothing\),所以函数 \(\displaystyle y=f\left(x\right)\) 与 \(\displaystyle y=\lambda\left(x-\frac{a+b}{2}\right)^2\) 的图象无公共点,即方程 $\(\displaystyle \frac1{x-a}-\frac1{x-b}=\lambda\left(x-\frac{a+b}{2}\right)^2\)$
无实数解,两边同乘\(\displaystyle (x-a)(x-b)\),即方程 $\(\displaystyle a-b=\lambda\left(x-a\right)\left(x-b\right)\left(x-\frac{a+b}{2}\right)^2\quad\left(x\ne a\text{ 且 }x\ne b\right)\eqno{\left(*\right)}\)$
无实数解。
(i)当 \(\displaystyle \lambda=0\) 时 \(\displaystyle \left(*\right)\) 无解,显然符合题意。
(ii)当 \(\displaystyle \lambda\ne0\) 时,令 \(\displaystyle y=\left(x-a\right)\left(x-b\right)\left(x-\frac{a+b}{2}\right)^2,\)变形得 $\(\displaystyle y=\left[\left(x-\frac{a+b}{2}\right)^2-\frac{\left(a-b\right)^2}{4}\right]\left(x-\frac{a+b}{2}\right)^2.\)$
设\(\displaystyle t=\left(x-\frac{a+b}{2}\right)^2\) 得 \(\displaystyle y=t\left[t-\frac{\left(a-b\right)^2}{4}\right]=\left[t-\frac{\left(a-b\right)^2}{8}\right]^2-\frac{\left(a-b\right)^4}{64}.\)
于是当 \(\displaystyle t=\frac{\left(a-b\right)^2}{8}\),即 \(\displaystyle x=\frac{a+b}{2}\pm\frac{\sqrt2}{4}\left(a-b\right)\) 时,有 \(\displaystyle y_{\min}=-\frac{\left(a-b\right)^4}{64}\)。
要使 \(\displaystyle \left(*\right)\) 无实数解,只要 \(\displaystyle \frac{a-b}{\lambda}<-\frac{\left(a-b\right)^4}{64},\)解得 \(\displaystyle 0<\lambda<\frac{64}{\left(b-a\right)^3}\)。
综上可得 \(\displaystyle 0\leqslant\lambda<\frac{64}{\left(b-a\right)^3}\)。
- 【2007湖南文10,理10】 设集合 \(\displaystyle M = \{1, 2, 3, 4, 5, 6\}\).\(\displaystyle S_1, S_2, \cdots, S_k\) 都是 \(\displaystyle M\) 的含两个元素的子集,且满足对任意的 \(\displaystyle S_i = \{a_i, b_i\}, S_j = \{a_j, b_j\}\)(\(\displaystyle i \neq j, i, j \in \{1, 2, \cdots, k\}\)),都有 \(\displaystyle \min\left\{\frac{a_i}{b_i}, \frac{b_i}{a_i}\right\} \neq \min\left\{\frac{a_j}{b_j}, \frac{b_j}{a_j}\right\}\),求\(\displaystyle k\) 的最大值.
答案
11. 不妨设\(\displaystyle a_i < b_i\), 题目条件转化为: 对任意\(\displaystyle S_i,S_j(i,j\in\{1,\cdots,k\},i\ne j)\)都有\(\displaystyle \dfrac{a_i}{b_i} \neq \dfrac{a_j}{b_j}\). 相当于求由1,2,3,4,5,6之间可以构成的互不相等且小于1的分数的个数,考虑在取数时只保留既约分数,有 \(\displaystyle 1/2,1/3,1/4,1/5,1/6,2/3,2/5,3/4,3/5,4/5,5/6\)共11个.
- 【2024上海春考12改编】已知集合\(\displaystyle A=\{a_1,a_2,a_3,a_4\}=\{2,4,8,16\},B=\{b_1,b_2,b_3,b_4\}\)满足\(\displaystyle \{a_i+a_j\mid 1\leqslant i<j\leqslant 4\}=\{b_i+b_j\mid 1\leqslant i<j\leqslant 4 \}\),求集合\(\displaystyle B\).
答案
\(\displaystyle \{2,4,6,8\}\)或 \(\displaystyle \{-1,7,11,13\}\). 计算 \(\displaystyle A\)中两两元素之和,其构成集合\(\displaystyle \{6,10,12,18,20,24\}\).
对于集合\(\displaystyle B\),不妨设\(\displaystyle b_1<b_2<b_3<b_4\),可知$\(\displaystyle b_1+b_2<b_1+b_3<b_2+b_3<b_2+b_4<b_3+b_4\)$
其中\(\displaystyle b_1+b_3<b_1+b_4<b_2+b_4\),现讨论\(\displaystyle b_1+b_4\)与\(\displaystyle b_2+b_3\)的大小关系.
若 \(\displaystyle b_1+b_4<b_2+b_3\),则唯一确定了所有两两元素之和的值,在此条件下求得\(\displaystyle B=\{-1,7,11,13\}\)
若\(\displaystyle b_1+b_4>b_2+b_3\),同理可得\(\displaystyle B=\{2,4,6,8\}\)
- 【2026深圳一模14】已知 \(\displaystyle a_1,a_2,\cdots,a_8\in\mathbb{N^*}\).记 $\(\displaystyle S = \{a_{i_1}+a_{i_2}+\cdots+a_{i_7} \mid 1 \leqslant i_1 < i_2 < \cdots < i_7 \leqslant 8\}\quad \text{其中} i_1,i_2,\cdots,i_7 \in \mathbb{N}^*\)$ 若 \(\displaystyle S = \{82,83,84,85,86,87,89\}\),求这8个正整数中的最大数与最小数的和.
答案
23. 满足 \(\displaystyle 1\leqslant i_1 < i_2 < \cdots < i_7\leqslant 8\) 的正整数组 \(\displaystyle (i_1,\cdots,i_7)\) 最多只有 8 组,即从 \(\displaystyle 1,2,\cdots,8\) 中去掉某一个数后从小到大排列得到的数组.而 \(\displaystyle S\) 是 7 元集,故在所有 \(\displaystyle a_{i_1}+a_{i_2}+\cdots+a_{i_7}\) ( \(\displaystyle 1\leqslant i_1 < i_2 < \cdots < i_7\leqslant 8\) ) 中,恰有两个数相同.
设 $\displaystyle n=a_1+a_2+\cdots+a_8$.改写 $\displaystyle S$ 为 $$\displaystyle S=\{n-a_i\mid 1\leqslant i\leqslant 8\}.$$ 不妨设 $\displaystyle a_1 < a_2 < \cdots < a_7$,且 $\displaystyle a_8\in\{a_1,\cdots,a_7\}$,则 $\displaystyle S$ 中元素从小到大排列后是 $$\displaystyle n-a_7 < n-a_6 < \cdots < n-a_1,$$ 因此,$\displaystyle n-a_7=82$,$\displaystyle n-a_6=83$,$\displaystyle \cdots$,$\displaystyle n-a_2=87$,$\displaystyle n-a_1=89$. 设 $\displaystyle a_1=a$,则 $\displaystyle a_2=a+2$,$\displaystyle a_3=a+3$,$\displaystyle \cdots$,$\displaystyle a_7=a+7$.于是 $\displaystyle n = (a_1+a_2+\cdots+a_7)+a_8 = 7a+27+a_8$.所以 $\displaystyle n-a_7 = 6a+20+a_8=82$,得 $\displaystyle 6a+a_8 = 62$. 由 $\displaystyle a_1\leqslant a_8 \leqslant a_7$ ,即 $\displaystyle a \leqslant a_8 \leqslant a+7$,得 $\displaystyle 7a\leqslant 62$ 且 $\displaystyle 7a+7\geqslant 62$.又因为 $\displaystyle a\in\mathbb{N}^*$,所以 $\displaystyle a=8$,所以 $\displaystyle a_8=14$. 因此 $\displaystyle a_1,\cdots,a_8$ 是 8, 10, 11, 12, 13, 14, 14, 15,其中的最小数是 8,最大数是 $\displaystyle a_7=a+7=15$,最小数与最大数之和是 23.- 【2010湖南文15】 规定 \(\displaystyle E=\{a_1,a_2,\dots,a_{10}\}\) 的子集 \(\displaystyle \{a_{i_1},a_{i_2},\dots,a_{i_n}\}\) 为 \(\displaystyle E\) 的第 \(\displaystyle k\) 个子集,其中 \(\displaystyle k=2^{i_1-1}+2^{i_2-1}+2^{i_3-1}+\dots+2^{i_n-1}\).\(\displaystyle \{a_1,a_3\}\) 是 \(\displaystyle E\) 的第几个子集?\(\displaystyle E\) 的第 \(\displaystyle 211\) 个子集是什么?
答案
5,\(\displaystyle \quad\{a_1,a_2,a_5,a_7,a_8\}\). \(\displaystyle \{a_1,a_3\}\) 的编号为 \(\displaystyle 2^{1-1}+2^{3-1}=1+4=5\).又\(\displaystyle 211 =2^7+2^6+2^4+2^1+2^0\).可知第 \(\displaystyle 211\) 个子集为 \(\displaystyle \{a_1,a_2,a_5,a_7,a_8\}\).
- 设集合\(\displaystyle X=\{1,2,\cdots,2n+1\},n\in\mathbb{N^*}\),\(\displaystyle A\subseteq X\)且满足:\(\displaystyle \forall x,y\in A,x+y\notin A\),求\(\displaystyle |A|\)的最大值.(用含\(\displaystyle n\)的式子表示)
答案
\(\displaystyle n+1\). 方法一:构造集合\(\displaystyle A=\{n+1,n+2,\ldots,2n+1\}\),这是符合题意的,因为其中任意两个元素之和都大于 \(\displaystyle 2n+1\).此时\(\displaystyle |A|=n+1.\)
下面证明任何满足$\displaystyle |A|\geqslant n+2$的集合 $\displaystyle A$ 都不符合题意.可以证明,若任意满足$\displaystyle |A|=k$的集合$\displaystyle A$都不符合题意,则任意满足$\displaystyle |A|>k$的集合$\displaystyle A$也不符合题意.下证明,任何满足$\displaystyle |A|=n+2$的集合 $\displaystyle A$ 都不符合题意. 使用数学归纳法,当 $\displaystyle n=1$ 时,全集为 $\displaystyle \{1,2,3\}$.此时$\displaystyle |A|=3$,只能有$\displaystyle A=\{1,2,3\}.$由于 $\displaystyle 1+2=3$,所以 $\displaystyle A$ 不符合题意. 假设当 $\displaystyle n=k$ 时,任意满足$\displaystyle |A|=k+2$的集合$\displaystyle A\subseteq\{1,2,\ldots,2k+1\}$都不符合题意. 当 $\displaystyle n=k+1$ 时,设集合$\displaystyle A\subseteq\{1,2,\ldots,2k+3\}$满足$\displaystyle |A|=k+3$且符合题意,此时集合$\displaystyle A$中最多有$\displaystyle k-1$个元素属于$\displaystyle \{1,2,\ldots,2k+1\}$,故$\displaystyle 2k+2,2k+3\in A$.将集合$\displaystyle \{1,2,\ldots,2k+2\}$分成$\displaystyle k+1$对:\[\displaystyle \{1,2k+2\},\{2,2k+1\},\ldots, \{k+1,k+2\}.\]每一对中的两个元素之和均为\(\displaystyle 2k+3\),它们不能同时出现在集合\(\displaystyle A\)中,故集合\(\displaystyle A\)中的元素最多有\(\displaystyle (k+1)+1=k+2\)个,与\(\displaystyle |A|=k+3\)矛盾.故对于任意满足\(\displaystyle |A|=k+3\)的集合\(\displaystyle A\subseteq\{1,2,\ldots,2k+3\}\)都不符合题意.
根据数学归纳法,原命题得证.
法二:构造集合\(\displaystyle A=\{1,3,5,\ldots,2n+1\}.\)这是符合题意的,因为其中任意两个元素之和为偶数,一定不属于\(\displaystyle A\).此时\(\displaystyle |A|=n+1\).
下面证明任意符合题意的集合都满足\(\displaystyle |A|\leqslant n+1.\)
设 \(\displaystyle A\) 中共有 \(\displaystyle k(k\leqslant n+1)\) 个奇数\(\displaystyle a_1>a_2>\cdots>a_k\),则偶数\(\displaystyle a_1-a_2<a_1-a_3<\cdots<a_1-a_k\),这\(\displaystyle (k-1)\)个偶数互不相同且一定属于\(\displaystyle A\),假设其中一个偶数\(\displaystyle a_1-a_i(2\leqslant i\leqslant k)\in A\),则\(\displaystyle a_i+(a_1-a_i)=a_1\in A\),这与题设矛盾,故这\(\displaystyle k\)个偶数均不属于\(\displaystyle A\),于是\(\displaystyle |A|\leqslant k+[n-(k-1)]=n+1\).
- 【2011广东8改编】设\(\displaystyle S,T\)是整数集\(\displaystyle \mathbb{Z}\)的非空子集,如果\(\displaystyle \forall a,b\in S\),有\(\displaystyle ab\in S\),那么称\(\displaystyle S\)关于乘法封闭.设\(\displaystyle T,V\)是\(\displaystyle \mathbb{Z}\)的两个不相交的非空子集,满足\(\displaystyle T\cup V=\mathbb{Z}\),且\(\displaystyle \forall a,b,c\in T\)有\(\displaystyle abc\in T\),\(\displaystyle \forall x,y,z\in V\)有\(\displaystyle xyz\in V\).证明:\(\displaystyle T,V\)中至少有一个关于乘法是封闭的.
- 【2016清华大学自主招生30改编】有\(\displaystyle n\)项的数列\(\displaystyle \{a_n\}\)满足
(1)对任意的\(\displaystyle i,j(1\leqslant i<j\leqslant n)\),\(\displaystyle a_i<a_j\);
(2)对任意的\(\displaystyle i,j,k(1\leqslant i<j<k\leqslant n)\),\(\displaystyle a_i+a_j,a_j+a_k,a_k+a_i\)三者中至少有一个是\(\displaystyle \{a_n\}\)中的项.
求\(\displaystyle n\)的最大值.
答案
讨论 \(\displaystyle \{a_n\}\) 中的正数和负数的个数.
由条件(1),$\displaystyle \{a_n\}$ 各项互不相同.如果 $\displaystyle \{a_n\}$ 中至少有 8 项,那么 $\displaystyle \{a_n\}$ 中要么至少有 4 个正数,要么至少有 4 个负数. 不妨设 $\displaystyle \{a_n\}$ 中至少有 4 个正数.记 $\displaystyle \{a_n\}$ 中的 4 个最大的正数从小到大排列为$\displaystyle 0 < x < y < z < w$. 由条件(2),$\displaystyle y+z$, $\displaystyle z+w$, $\displaystyle w+y$ 中至少有一个是 $\displaystyle \{a_n\}$ 中的项.但是 $\displaystyle z+w$, $\displaystyle w+y$ 都大于 $\displaystyle w$,所以只能有 $\displaystyle y+z$ 是 $\displaystyle \{a_n\}$ 中的项,而且$\displaystyle y+z=w.$ 再由条件(2),$\displaystyle x+z$, $\displaystyle z+w$, $\displaystyle w+x$ 中至少有一个是 $\displaystyle \{a_n\}$ 中的项.但是 $\displaystyle z+w$, $\displaystyle w+x$ 都大于 $\displaystyle w$,所以只能有 $\displaystyle x+z$ 是 $\displaystyle \{a_n\}$ 中的项.但是$\displaystyle z < x+z < y+z = w$.根据前面的假设,$\displaystyle x,y,z,w$ 是 $\displaystyle \{a_n\}$ 中最大的四项,从而 $\displaystyle \{a_n\}$ 中没有数位于区间 $\displaystyle (z,w)$ 之间.所以 $\displaystyle x+z$ 不可能是 $\displaystyle \{a_n\}$ 中的项,矛盾. 综上,$\displaystyle \{a_n\}$ 不可能有 8 项. 另外可以举出 含有7项的满足两个条件的数列:$\displaystyle -3,-2,-1,0,1,2,3$.故$\displaystyle n$的最大值为 $\displaystyle 7$.-
【2026海淀期末21节选】已知有穷数列\(\displaystyle T:t_1,t_2,\cdots,t_m(m\geqslant 2)\),用\(\displaystyle \varphi(T)\)表示数列\(\displaystyle T\)中所有的项构成的集合,定义变换\(\displaystyle F\),\(\displaystyle F\)将数列\(\displaystyle T\)变换成数列\(\displaystyle F(T):|t_1-t_2|,|t_2-t_3|,\cdots ,|t_{n-1}-t_n|\). 对有穷数列\(\displaystyle A_1:a_1,a_2,\cdots,a_n(n\geqslant 2)\),令数列\(\displaystyle A_{k+1}=F(A_k)(k=1,2,\cdots ,n-1)\),若$\(\displaystyle \varphi(A_1)\cup \varphi(A_2)\cup\cdots\cup \varphi(A_n)=\left \{ 1,2,\cdots,\frac{n(n+1)}{2} \right \}\)$ 则称\(\displaystyle A_1\)为\(\displaystyle n\)阶完美数列.
-
写出所有的2阶完美数列;
- 若数列\(\displaystyle A_1\)为3阶完美数列,求集合\(\displaystyle \varphi(A_1)\cap \left \{ 1,2,3 \right \}\)的元素个数.
- 【2024海淀期末21(2)】给定奇数\(\displaystyle m(m\geqslant 3)\),设\(\displaystyle A\)是由\(\displaystyle m\times m\)个实数组成的\(\displaystyle m\)行\(\displaystyle m\)列的数表,且\(\displaystyle A\)中所有数不全相同,记\(\displaystyle A\)中第\(\displaystyle i\)行第\(\displaystyle j\)列的数为\(\displaystyle a_{ij}\in\left \{ -1,1 \right \}\),\(\displaystyle r(i)\)为\(\displaystyle A\)的第\(\displaystyle i\)行各数之和,\(\displaystyle c(j)\)为\(\displaystyle A\)的第\(\displaystyle j\)列各数之和,设集合$\(\displaystyle H=\left \{ (i,j)|a_{ij}\cdot r(i)<0\text{或}a_{ij}\cdot c(j)<0,i,j\in\left \{ 1,2,\cdots,m \right \} \right \}\)\(记\)\displaystyle H(A)\(为集合\)\displaystyle H\(所含元素的个数.若\)\displaystyle r(1),r(2),\cdots,r(m)\(中恰有\)\displaystyle s\(个正数,\)\displaystyle c(1),c(2),\cdots ,c(m)\(中恰有\)\displaystyle t\(个正数,求证:\)\displaystyle H(A)\geqslant mt+ms-2ts$.
- 【2026济南二模19】已知数集\(\displaystyle A=\{a_1,a_2,\cdots,a_n\}\)满足:对任意\(\displaystyle a_i,a_j\in A\),存在\(\displaystyle a_m\in A\)使得\(\displaystyle |a_i-a_j|=|a_m-a_1|\).
- 若\(\displaystyle A=\{1,2,x\},x\in\mathbb{N}^*\),求\(\displaystyle x\);
- 证明:对任意\(\displaystyle a_i,a_j\in A\),都有\(\displaystyle (a_i-a_1)(a_j-a_1)\geqslant0\);
- 若\(\displaystyle a_1=a(a\neq 0),a_2=2a\),且\(\displaystyle a_2<a_3<\cdots<a_n\),求数集\(\displaystyle A\)中所有元素的和.(用含\(\displaystyle a,n\)的式子表示)
答案
(1)由题设,\(\displaystyle A=\{1,2,x\}\) 的任意两元素之差的绝对值都必须等于某个\(\displaystyle |a_m-a_1|\),设前者构成的集合为\(\displaystyle B=\{1,|x-1|,|x-2|\}\),后者构成的集合为\(\displaystyle C\).必须有\(\displaystyle B\subseteq C\).
如果$\displaystyle x_1=1$,那么$\displaystyle C=\{0,1,|x-1|\}$,可得$\displaystyle x=3$; 如果$\displaystyle x_1=2$,那么$\displaystyle C=\{0,1,|x-2|\}$,没有满足条件的$\displaystyle x$; 如果$\displaystyle x_1=x$,那么$\displaystyle C=\{0,|x-1|,|x-2|\}$,可得$\displaystyle x=3$. 综上,$\displaystyle x=3$. (2)用反证法,假设存在 $\displaystyle a_i,a_j\in A$使得 $\displaystyle (a_i-a_1)(a_j-a_1)<0$. 不妨设 $\displaystyle a_i<a_1<a_j$,记$\displaystyle A$中的最小元为$\displaystyle u$,最大元为$\displaystyle v$,一定有$\displaystyle u\leqslant a_i<a_j\leqslant v$. 对任意$\displaystyle a_m\in A$,$$\displaystyle |a_m-a_1|\leqslant \max\{|a_u-a_1|,|a_v-a_1|\}<|(u-a_1)-(v-a_1)|=|u-v|$$ 于是对任意$\displaystyle a_m\in A$,$\displaystyle |a_u-a_v|\neq |a_m-a_1|$,矛盾,因此$\displaystyle (a_i-a_1)(a_j-a_1)\geqslant0$. (3)当$\displaystyle a>0$时,有$\displaystyle a_1<a_2<a_3<\cdots<a_n$. 首先确定$\displaystyle a_3-a_2$的值,可知 $$\displaystyle a_3-a_2<a_3-a_1<a_4-a_1<\cdots<a_n-a_1$$ 这里的$\displaystyle n-1$个元素互不相同,并且它们均属于集合$\displaystyle \{|a_2-a_1|,|a_3-a_1|,\ldots,|a_n-a_1|\}$,于是它们按从小到大的顺序对应相等,即有$\displaystyle a_3-a_2=a_2-a_1$。再确定$\displaystyle a_4-a_3$的值,可知 $$\displaystyle a_4-a_3<a_4-a_2<a_4-a_1<a_5-a_1<\cdots<a_n-a_1$$ 这里的$\displaystyle n-1$个元素互不相同,并且它们均属于集合$\displaystyle \{|a_2-a_1|,|a_3-a_1|,\ldots,|a_n-a_1|\}$,同上可得$\displaystyle a_4-a_3=a_2-a_1$,依次类推,可以证明对任意$\displaystyle 2\leqslant i\leqslant n$,$\displaystyle a_{i}-a_{i-1}=a_2-a_1$,于是$\displaystyle a_1,a_2,\cdots,a_n$成等差数列,其通项公式为$\displaystyle a_n=na$,元素和为$\displaystyle \frac{an(n+1)}{2}$ 当$\displaystyle a<0$时,有$\displaystyle a_2<a_3<\cdots a_n<a_1$,同理可证$\displaystyle a_2,a_3,\cdots,a_n,a_1$成等差数列,其首项为$\displaystyle a_2=2a$,末项为$\displaystyle a_1=a$,元素和为$\displaystyle \frac{3an}{2}$.注:如果考虑数学归纳法,则只能用于证明\(\displaystyle a>0\)的情况。
用数学归纳法证明命题:当\(\displaystyle a>0\)时,\(\displaystyle a_n\)的通项公式为\(\displaystyle a_n = na\)。当\(\displaystyle n=2\)时,\(\displaystyle a_1=a,a_2=2a\),命题成立。假设当\(\displaystyle n=k\)时,\(\displaystyle a_k=ka\),当\(\displaystyle n=k+1\)时,集合\(\displaystyle A=\{a,2a,\cdots,ka,a_{k+1}\}\),下证明\(\displaystyle a_{k+1}=(k+1)a\).\(\displaystyle a_{k+1}= (k+1)a\)当且仅当\(\displaystyle a_{k+1}-a_k=a=a_2-a_1\)。用反证法,假设\(\displaystyle a_{k+1}-a_k\neq a_2-a_1\),那么 $\(\displaystyle a_{k+1}-a_k\in\{a_3-a_1,a_4-a_1,\cdots,a_k-a_1\}\)\(即\)\displaystyle a_{k+1}-a_k\geqslant 2a\(,于是\)\displaystyle a_{k+1}\geqslant (k+2)a,a_{k+1}-a_2\geqslant ka$,而 $\(\displaystyle a_1-a_1<a_2-a_1<a_3-a_1<\cdots<a_{k}-a_1<a_{k+1}-a_2<a_{k+1}-a_1\)\(于是不存在\)\displaystyle a_m\in A\(使得\)\displaystyle |a_{k+1}-a_2|=|a_m-a_1|\(,矛盾.因此\)\displaystyle a_{k+1}-a_k=a_2-a_1\(,即\)\displaystyle a_{k+1}=(k+1)a$.由数学归纳法,命题得证。
观察此方法,从\(\displaystyle a_1,a_2,\cdots,a_k\)为等差数列可推出\(\displaystyle a_1,a_2,\cdots,a_{k+1}\)为等差数列,据此改进反证法。
用反证法,假设\(\displaystyle a_1,a_2,\cdots,a_{n}\)不是以\(\displaystyle a_2-a_1\)为公差的等差数列,则存在\(\displaystyle 2\leqslant i\leqslant n-1\)使得\(\displaystyle a_{i+1}-a_i\neq a_2-a_1\),取所有满足该条件的\(\displaystyle i\)中的最小值记为\(\displaystyle x\),于是$\(\displaystyle a_3-a_2=a_4-a_3=\cdots =a_{x-1}-a_{x-2}=a_2-a_1,\quad a_x-a_{x-1}\neq a_2-a_1\)$
由于\(\displaystyle a_2-a_1<a_3-a_1<\cdots\),于是\(\displaystyle a_x-a_{x-1}> a_2-a_1\),可推出$\(\displaystyle a_x-a_2=(a_x-a_{x-1})+(a_{x-1}-a_2)> a_{x-1}-a_1,\quad a_x-a_2<a_x-a_1\)$ 故不存在\(\displaystyle a_m\in A\)使得\(\displaystyle |a_{k+1}-a_2|=|a_m-a_1|\),矛盾.因此\(\displaystyle a_1,a_2,\cdots,a_n\)成等差数列.
教评:上课时询问了学生下述问题:
问题1:本题第(2)问为什么考虑反证法?(使用反证法的动机是什么?)
题目要求证明对任意 \(\displaystyle a_i, a_j \in A\)都有 \(\displaystyle (a_i - a_1)(a_j - a_1) \geqslant 0\)。直接证明所需要分析的范围过大,反证法通过假设存在一个反例来推导出矛盾,将难以利用的待证命题转化为可以直接上手使用的具体的假设。
学生不清楚使用反证法的动机,一方面是学生在此前的学习中对此接触较少,遇到需使用反证法(正面难以解答)的习题时产生了畏惧。实际上反证法就是一种寻常的证明方式,并非什么高级的技巧。
数学思维训练的最关键一步在于如何正确利用和吸收外部帮助,对高中学生而言,具体体现在理解并学习参考答案这一步,这里也往往伴随着诸多严重的问题,学生首先不能对参考答案的每一步都给出一个(或多个)足以说服自己的动机,他们要么认为这些步骤是凭空出现的,从而泄气,要么是草草了事,自认为完全搞懂了,实则一问三不知。
教师的职责是引导学生去思考这些步骤的动机,对于不理解的,学生要在教师的鼓励下积极提出自己的疑惑,而非给学生灌输现成的答案。
问题2:对于哪些问题可以考虑使用反证法?确定反证法后的下一步是什么?如何最大可能地暴露矛盾?
一个经典的使用反证法的习题是:证明\(\displaystyle \sqrt{2}\)是无理数。一个数是有理数,那么它可以写作\(\displaystyle \frac{p}{q}\)的形式,其中\(\displaystyle p\)和\(\displaystyle q\)是互质的整数。但我们很难用如此简洁的方式去描述无理数的特征,从无理数入手难而从有理数入手易,于是反设\(\displaystyle \sqrt{2}\)是有理数,推导出矛盾。这一个例子说明,当待证命题较为宽泛而其否命题较为具体时,反证法是一个不错的选择。
确定反证法的下一步就是尝试最大限度地暴露矛盾。对于本题第(2)问,虽然存在\(\displaystyle a_i<a_1<a_j\),但\(\displaystyle |a_i-a_j|\)仍可能等于\(\displaystyle |a_m-a_1|\),如何找到一组\(\displaystyle a_p,a_q\),使得\(\displaystyle |a_p-a_q|\)不等于任何\(\displaystyle |a_m-a_1|\)?我们尝试将它们的差推到最大值,即取集合中的最大元与最小元。再以本节B组习题第8题【2026“fiddie”模拟考19(3)】为例,要证明\(\displaystyle S\)不存在含5个元素的等差子集。
- 【2025 北京第一次学考25】 已知正整数\(\displaystyle n \geqslant 4\),集合 \(\displaystyle M = \{a_1, a_2, \cdots, a_n\}\subseteq \mathbb{Z}\) .若存在 \(\displaystyle M\) 的非空子集 \(\displaystyle A\),满足 \(\displaystyle 2 \leqslant |A| \leqslant n - 2\)且 \(\displaystyle A\) 中所有元素之积与 \(\displaystyle \complement_M A\) 中所有元素之和相等,则称 \(\displaystyle M\) 为“积和集合”.
- 若\(\displaystyle M = \{-1, 2, 3, 7\}\),判断 \(\displaystyle M\) 是否为“积和集合”;(结论无需证明)
- 若 \(\displaystyle M = \{1, 2, 4, a\}\) 是“积和集合”,写出 \(\displaystyle a\) 的所有可能取值;
- 若 \(\displaystyle M = \{-2, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}\),判断 \(\displaystyle M\) 是否为“积和集合”,并说明理由.
答案
(1)设 \(\displaystyle A=\{2,3\}\),\(\displaystyle \complement_MA=\{-1,7\}\),则 \(\displaystyle A\) 中元素之积等于 \(\displaystyle \complement_MA\) 中元素之和,故 \(\displaystyle M\) 为“积和集合”.
所以本题答案为:是(答题时无需写过程) (2) 分情况讨论. 若 $\displaystyle A=\{1,2\}$,$\displaystyle \complement_MA=\{4,a\}$,则 $\displaystyle a=-2$.若 \(\displaystyle A=\{1,4\}\),\(\displaystyle \complement_MA=\{2,a\}\),则 \(\displaystyle a=2\),但与互异性矛盾.
若 \(\displaystyle A=\{1,a\}\),\(\displaystyle \complement_MA=\{2,4\}\),则 \(\displaystyle a=6\).
若 \(\displaystyle A=\{2,4\}\),\(\displaystyle \complement_MA=\{1,a\}\),则 \(\displaystyle a=7\).
若 \(\displaystyle A=\{2,a\}\),\(\displaystyle \complement_MA=\{1,4\}\),则 \(\displaystyle 2a=5\),与 \(\displaystyle a\) 是整数矛盾.
若 \(\displaystyle A=\{4,a\}\),\(\displaystyle \complement_MA=\{1,2\}\),则 \(\displaystyle 4a=3\),与 \(\displaystyle a\) 是整数矛盾.
因此 \(\displaystyle a\) 的所有可能取值是 \(\displaystyle -2\)、\(\displaystyle 6\)、\(\displaystyle 7\).(答题时无需写上述分析过程)
(3)若 $\displaystyle M$ 是积和集合,则存在 $\displaystyle M$ 的非空真子集 $\displaystyle A$,使得 $\displaystyle A$ 中所有元素的乘积等于 $\displaystyle \complement_MA$ 中所有元素的和.设 $\displaystyle A$ 中所有元素的乘积为 $\displaystyle P$,$\displaystyle \complement_MA$ 中所有元素的和为 $\displaystyle S$.由题意,$\displaystyle P=S$.若 \(\displaystyle -2\in A\),则 \(\displaystyle P < 0\),\(\displaystyle S > 0\),矛盾.故 \(\displaystyle -2\in\complement_MA\).
若\(\displaystyle |A|\geqslant 4\),则\(\displaystyle P\geqslant 1\times 3\times 5\times 7 = 105\).而\(\displaystyle M\)中所有元素和为\(\displaystyle -2+1+3+\cdots +19=98\),所以\(\displaystyle S<98<P\),矛盾.下讨论\(\displaystyle |A|=2\),\(\displaystyle |A|=3\)两种情况.
若\(\displaystyle |A|=2\),则\(\displaystyle P\)一定是奇数,而\(\displaystyle S\)一定是偶数,矛盾.
若 \(\displaystyle |A|=3\),当\(\displaystyle 1\notin A\)时,\(\displaystyle P\geqslant 3\times 5\times 7=105\),同上可推出矛盾,故\(\displaystyle 1\in A\).设 \(\displaystyle A=\{1,x,y\}\),则 \(\displaystyle P=xy\),\(\displaystyle S=98-1-x-y=97-x-y\).
由 \(\displaystyle P=S\) 得 \(\displaystyle xy=97-x-y\),即 \(\displaystyle xy+x+y=97\).这一方程可以改写为 \(\displaystyle (x+1)(y+1)=98\),又\(\displaystyle x+1,y+1\)均为偶数,所以 \(\displaystyle (y+1)(z+1)\) 是 4 的倍数,而 98 不是 4 的倍数,矛盾.
综上,\(\displaystyle M\) 不是 “积和集合”.
-
【2012上海理23】 已知数集 \(\displaystyle X = \{-1, x_1, x_2, \cdots, x_n\}(0 < x_1 < x_2 < \cdots < x_n,n \geqslant 2)\),向量集 \(\displaystyle Y = \{\vv{a} \mid \vv{a} = (s, t), s, t \in X\}\),若对任意 \(\displaystyle \vv{a}_1 \in Y\),存在 \(\displaystyle \vv{a}_2 \in Y\),使得 \(\displaystyle \vv{a}_1 \cdot \vv{a}_2 = 0\),则称 \(\displaystyle X\) 具有性质 \(\displaystyle P\).
-
若 \(\displaystyle x > 2\),且 \(\displaystyle \{-1, 1, 2, x\}\) 具有性质 \(\displaystyle P\),求 \(\displaystyle x\) 的值;
- 若 \(\displaystyle X\) 具有性质 \(\displaystyle P\),求证:\(\displaystyle 1 \in X\),且当 \(\displaystyle x_n > 1\) 时,\(\displaystyle x_1 = 1\);
- 若 \(\displaystyle X\) 具有性质 \(\displaystyle P\),且 \(\displaystyle x_1 = 1, x_2 = q\)(\(\displaystyle q\) 为常数),求有穷数列 \(\displaystyle x_1, x_2, \cdots, x_n\) 的通项公式.
答案
(1)取向量 \(\displaystyle (2,x)\in Y\).设它的一个垂直向量为 \(\displaystyle (u,v)\),其满足\(\displaystyle 2u+xv=0\).\(\displaystyle u,v\)不可能均为正数或均为负数
若 $\displaystyle u=-1$,则 $\displaystyle xv=2$,而$\displaystyle xv\geqslant x>2$,不可能使等式成立.若 $\displaystyle v=-1$,则 $\displaystyle 2u=x$.只可能有$\displaystyle u=2$,此时$\displaystyle x=4$,检验 $\displaystyle X=\{-1,1,2,4\}$是符合题意的,所以 $\displaystyle x=4$. (2)取向量 $\displaystyle (-1,-1)\in Y$.由题意,设它的一个垂直向量为 $\displaystyle (u,v)$,其满足$\displaystyle (-1)u+(-1)v=0$,即 $\displaystyle u+v=0$.$\displaystyle u,v$不可能均为正数,不妨设$\displaystyle u=-1$,则$\displaystyle v=1$,于是$\displaystyle 1\in X$. 假设当$\displaystyle x_n>1$时,$\displaystyle x_1<1$,取向量$\displaystyle (x_1,x_n)\in A$,由题意,设它的一个垂直向量为 $\displaystyle (u,v)$,其满足 $\displaystyle ux_1+vx_n=0$.$\displaystyle u,v$不可能均为正数或均为负数 若$\displaystyle u=-1$,则$\displaystyle vx_n=x_1$,而$\displaystyle vx_n\geqslant x_1x_n>x_1$,矛盾;若$\displaystyle v=-1$,则$\displaystyle ux_1=x_n$,而$\displaystyle ux_1\leqslant x_1x_n<x_n$,矛盾. 于是当$\displaystyle x_n>1$时,必有$\displaystyle x_1\geqslant 1$. (3)猜测有穷数列$\displaystyle \{x_n\}$的通项公式只可能为$\displaystyle x_n=q^{n-1}(q>1)$,采用数学归纳法证明: 当$\displaystyle n=2$时,命题显然成立.当$\displaystyle n=3$时,假设$\displaystyle X=\{-1,1,q,x\}$,取向量$\displaystyle (q,x)\in X$,设它的一个垂直向量为$\displaystyle (u,v)$,其满足$\displaystyle uq+vx=0$.$\displaystyle u,v$不可能均为正数或均为负数 当 $\displaystyle u=-1$ 时,$\displaystyle vx=q$,而$\displaystyle vx\geqslant x>q$,不符题意.当$\displaystyle v=-1$ 时,$\displaystyle uq=x$,$\displaystyle u$只可能为$\displaystyle q$,于是$\displaystyle x=q^2$,命题成立. 假设当$\displaystyle n=k$时命题成立.当$\displaystyle n=k+1$时,考虑向量$\displaystyle (q,x_{k+1})$,设它的一个垂直向量为$\displaystyle (u,v)$,其满足$\displaystyle uq+vx_{k+1}=0$.$\displaystyle u,v$不可能均为正数或均为负数 当 $\displaystyle u=-1$ 时,$\displaystyle vx_{k+1}=q$,而$\displaystyle vx_{k+1}\geqslant x_{k+1}>q$,不符题意.当$\displaystyle v=-1$ 时,$\displaystyle uq=x_{k+1}$,当$\displaystyle u$取$\displaystyle x_1=1,x_2=q,\cdots,x_{k-1}=q^{k-2}$时均不满足该等式,于是$\displaystyle u$只可能为$\displaystyle x_k=q^{k-1}$.故$\displaystyle x_{k+1}=q^k$.由此完成了证明.评:本题与【2009北京 20】非常相似,只是改变了问题呈现的背景,条件复杂了、隐蔽了,如果将向量的数量积为 \(\displaystyle 0\) 关系:对任意向量 \(\displaystyle \vv{a}_{i, j}=(x_i, x_j)\) 必存在向量 \(\displaystyle (-1, x_k)\) 或 \(\displaystyle (x_m, -1)\) 使 \(\displaystyle x_i=x_k x_j\) 或 \(\displaystyle x_j=x_m x_i\) 成立,也就是对任意的 \(\displaystyle i, j\) (\(\displaystyle 1\leqslant i, j\leqslant n\)),都有 \(\displaystyle \frac{x_i}{x_j}\) 与 \(\displaystyle \frac{x_j}{x_i}\) 两数中至少有一个属于 \(\displaystyle X\).这个中间结论与北京卷的条件“对任意的 \(\displaystyle i, j\) (\(\displaystyle 1\leqslant i\leqslant j\leqslant n\)),\(\displaystyle a_i a_j\) 与 \(\displaystyle \frac{a_j}{a_i}\) 两数中至少有一个属于 \(\displaystyle A\)”何其相像!
解答高考难题或抽象度高的新颖题时,一定要学会“退”,退到最简单又不失本质的特殊问题上.我们寻求一个问题的解答没有成功的原因最可能是,有一个比手头问题更简单、更容易的问题没有完全解决或完全没有解决.
新答案(来源:4.3 多个量词的问题(2)数列.md):(1)选取\(\displaystyle \vv{a_1}=\left(x,2\right)\),\(\displaystyle Y\)中与\(\displaystyle \vv{a_1}\)垂直的元素必有形式\(\displaystyle \left(-1,b\right)\).
……2 分
所以\(\displaystyle x=2b\),从而\(\displaystyle x=4\).
……4 分
(2)证明:取\(\displaystyle \vv{a_1}=\left(x_1,x_1\right)\in Y\).设\(\displaystyle \vv{a_2}=\left(s,t\right)\in Y\)满足\(\displaystyle \vv{a_1}\cdot\vv{a_2}=0\).
由\(\displaystyle \left(s+t\right)x_1=0\)得\(\displaystyle s+t=0\),所以\(\displaystyle s,t\)异号.
因为\(\displaystyle -1\)是\(\displaystyle X\)中唯一的负数,则\(\displaystyle s,t\)之中一为\(\displaystyle -1\),另一为\(\displaystyle 1\),故\(\displaystyle 1\in X\).
……7 分
假设\(\displaystyle x_k=1\),其中\(\displaystyle 1<k<n\),则\(\displaystyle 0<x_1<1<x_n\).
选取\(\displaystyle \vv{a_1}=\left(x_1,x_n\right)\in Y\),并设\(\displaystyle \vv{a_2}=\left(s,t\right)\in Y\)满足\(\displaystyle \vv{a_1}\cdot\vv{a_2}=0\),即\(\displaystyle sx_1+tx_n=0\),
则\(\displaystyle s,t\)异号,从而\(\displaystyle s,t\)之中恰有一个为\(\displaystyle -1\).
若\(\displaystyle s=-1\),则\(\displaystyle x_1=tx_n>t\geqslant x_1\),矛盾.若\(\displaystyle t=-1\),则\(\displaystyle x_n=sx_1<s\leqslant x_n\),矛盾.所以\(\displaystyle x_1=1\).
……10 分
(3)解法一:设\(\displaystyle \vv{a_1}=\left(s_1,t_1\right)\),\(\displaystyle \vv{a_2}=\left(s_2,t_2\right)\),则\(\displaystyle \vv{a_1}\cdot\vv{a_2}=0\)等价于\(\displaystyle \frac{s_1}{t_1}=-\frac{t_2}{s_2}\).
记\(\displaystyle B=\left\{\frac{s}{t}\left|s\in X,t\in X,\left|s\right|>\left|t\right|\right.\right\}\),则数集\(\displaystyle X\)具有性质\(\displaystyle P\)当且仅当数集\(\displaystyle B\)关于原点对称.
……14 分
注意到\(\displaystyle -1\)是\(\displaystyle X\)中的唯一负数,\(\displaystyle B\cap\left(-\infty,0\right)=\left\{-x_2,-x_3,\cdots,-x_n\right\}\)具有\(\displaystyle n-1\)个数,所以\(\displaystyle B\cap\left(0,+\infty\right)\)也只有\(\displaystyle n-1\)个数.
由于\(\displaystyle \frac{x_n}{x_{n-1}}<\frac{x_n}{x_{n-2}}<\cdots<\frac{x_n}{x_2}<\frac{x_n}{x_1}\),已有\(\displaystyle n-1\)个数,对以下三角数阵 $\(\displaystyle \frac{x_n}{x_{n-1}}<\frac{x_n}{x_{n-2}}<\cdots<\frac{x_n}{x_2}<\frac{x_n}{x_1}\)$ $\(\displaystyle \frac{x_{n-1}}{x_{n-2}}<\frac{x_{n-1}}{x_{n-3}}<\cdots<\frac{x_{n-1}}{x_1}\)$ $\(\displaystyle \cdots\cdots\)$ $\(\displaystyle \frac{x_2}{x_1}\)$ 注意到\(\displaystyle \frac{x_n}{x_1}>\frac{x_{n-1}}{x_1}>\cdots>\frac{x_2}{x_1}\),则\(\displaystyle \frac{x_n}{x_{n-1}}=\frac{x_{n-1}}{x_{n-2}}=\cdots=\frac{x_2}{x_1}\),从而数列的通项为\(\displaystyle x_k=x_1\left(\frac{x_2}{x_1}\right)^{k-1}=q^{k-1}\),\(\displaystyle k=1,2,\cdots,n\).
……18 分
解法二:猜测\(\displaystyle x_i=q^{i-1}\),\(\displaystyle i=1,2,\cdots,n\).
……12 分
记\(\displaystyle A_k=\left\{-1,1,x_2,\cdots,x_k\right\}\),\(\displaystyle k=2,3,\cdots,n\).
先证明:若\(\displaystyle A_{k+1}\)具有性质\(\displaystyle P\),则\(\displaystyle A_k\)也具有性质\(\displaystyle P\).
任取\(\displaystyle \vv{a_1}=\left(s,t\right)\),\(\displaystyle s,t\in A_k\),当\(\displaystyle s,t\)中出现\(\displaystyle -1\)时,显然有\(\displaystyle \vv{a_2}\)满足\(\displaystyle \vv{a_1}\cdot\vv{a_2}=0\).
当\(\displaystyle s\ne-1\)且\(\displaystyle t\ne-1\)时,则\(\displaystyle s,t\geqslant1\).
因为\(\displaystyle A_{k+1}\)具有性质\(\displaystyle P\),所以有\(\displaystyle \vv{a_2}=\left(s_1,t_1\right)\),\(\displaystyle s_1,t_1\in A_{k+1}\),使得\(\displaystyle \vv{a_1}\cdot\vv{a_2}=0\),从而\(\displaystyle s_1\)和\(\displaystyle t_1\)中有一个是\(\displaystyle -1\),不妨设\(\displaystyle s_1=-1\).
假设\(\displaystyle t_1\in A_{k+1}\)且\(\displaystyle t_1\notin A_k\),则\(\displaystyle t_1=x_{k+1}\).由\(\displaystyle \left(s,t\right)\cdot\left(-1,x_{k+1}\right)=0\),得\(\displaystyle s=tx_{k+1}\geqslant x_{k+1}\),与\(\displaystyle s\in A_k\)矛盾.
所以\(\displaystyle t_1\in A_k\).从而\(\displaystyle A_k\)也具有性质\(\displaystyle P\).
……15 分
现用数学归纳法证明:\(\displaystyle x_i=q^{i-1}\),\(\displaystyle i=1,2,\cdots,n\).
当\(\displaystyle n=2\)时,结论显然成立;
假设\(\displaystyle n=k\)时,\(\displaystyle A_k=\left\{-1,1,x_2,\cdots,x_k\right\}\)有性质\(\displaystyle P\),则\(\displaystyle x_i=q^{i-1}\),\(\displaystyle i=1,2,\cdots,k\);
当\(\displaystyle n=k+1\)时,若\(\displaystyle A_{k+1}=\left\{-1,1,x_2,\cdots,x_k,x_{k+1}\right\}\)有性质\(\displaystyle P\),则\(\displaystyle A_k=\left\{-1,1,x_2,\cdots,x_k\right\}\)也有性质\(\displaystyle P\),
所以\(\displaystyle A_{k+1}=\left\{-1,1,q,\cdots,q^{k-1},x_{k+1}\right\}\).
取\(\displaystyle \vv{a_1}=\left(x_{k+1},q\right)\),并设\(\displaystyle \vv{a_2}=\left(s,t\right)\)满足\(\displaystyle \vv{a_1}\cdot\vv{a_2}=0\),由此可得\(\displaystyle s=-1\)或\(\displaystyle t=-1\).
若\(\displaystyle t=-1\),则\(\displaystyle x_{k+1}=\frac{q}{s}\leqslant q\),不可能;
所以\(\displaystyle s=-1\),\(\displaystyle x_{k+1}=qt=q^j\leqslant q^k\)且\(\displaystyle x_{k+1}>q^{k-1}\),所以\(\displaystyle x_{k+1}=q^k\).
综上所述,\(\displaystyle x_i=q^{i-1}\),\(\displaystyle i=1,2,\cdots,n\).
……18 分
- 【2009北京理20】已知数集 \(\displaystyle A=\{a_1,a_2,\dots,a_n\}\ (1\leqslant a_1<a_2<\cdots<a_n,n\geqslant 2)\) 具有性质 \(\displaystyle P\): 对任意 \(\displaystyle i,j\ (1\leqslant i\leqslant j\leqslant n)\), \(\displaystyle a_i a_j\) 与 \(\displaystyle \frac{a_j}{a_i}\) 两数中至少有一个属于 \(\displaystyle A\).
- 分别判断数集 \(\displaystyle \{1,3,4\}\) 与 \(\displaystyle \{1,2,3,6\}\) 是否具有性质 \(\displaystyle P\);
- 证明: \(\displaystyle a_1=1\), 且 \(\displaystyle \frac{a_1+a_2+\cdots+a_n}{a_1^{-1}+a_2^{-1}+\cdots+a_n^{-1}}=a_n\);
- 证明: 当 \(\displaystyle n=5\) 时, \(\displaystyle a_1,a_2,a_3,a_4,a_5\) 成等比数列.
答案
(1) 集合\(\displaystyle \{1,3,4\}\)不具有性质 \(\displaystyle \mathbf P\).集合 \(\displaystyle \{1,2,3,6\}\)具有性质 \(\displaystyle \mathbf P\).
(2) 取 $\displaystyle i=j=n$.因为 $\displaystyle a_n^2>a_n$,所以 $\displaystyle a_n^2\notin A$;题设的两个数中至少一个属于 $\displaystyle A$,因此只能有 $\displaystyle \frac{a_n}{a_n}=1\in A$,即 $\displaystyle a_1=1$. 当 $\displaystyle j=n$时,对任意$\displaystyle i>1$,因为 $\displaystyle a_i a_n>a_n$,于是$\displaystyle a_ia_n\notin A$,所以$\displaystyle \frac{a_n}{a_i}\in A$.于是可以找到这$\displaystyle n$个互不相等的属于$\displaystyle A$的数,按递增排列 $$\displaystyle \frac{a_n}{a_n}<\frac{a_n}{a_{n-1}}<\cdots<\frac{a_n}{a_2}<\frac{a_n}{a_1}$$ 于是有$\displaystyle \frac{a_n}{a_n}=a_1,\frac{a_n}{a_{n-1}}=a_2,\ldots,\frac{a_n}{a_2}=a_{n-1},\frac{a_n}{a_1}=a_n$(若要严谨证明,可参考本节习题第37题(3)的证明思路) 因此$$\displaystyle \frac{a_n}{a_1}+\frac{a_n}{a_2}+\cdots+\frac{a_n}{a_n}=a_1+a_2+\cdots+a_n\iff \frac{a_1+a_2+\cdots+a_n}{a_1^{-1}+a_2^{-1}+\cdots+a_n^{-1}}=a_n$$ (3) 当 $\displaystyle n=5$ 时,由上一问有 $\displaystyle a_1a_5=a_5$,所以 $\displaystyle a_1=1$; 还有 $\displaystyle a_2a_4=a_3^2=a_5$,$\displaystyle a_2,a_3,a_4$成等比数列,设$\displaystyle a_2=x,a_3=xq(x>1,q>1)$,则$\displaystyle a_4=xq^2,a_5=x^2q^2$. 考察 $\displaystyle a_2,a_3$.若 $\displaystyle a_2a_3\in A$,由于$\displaystyle a_2a_3=x^2q$,而$\displaystyle a_3<x^2q<x_5$,只可能有$\displaystyle x^2q=a_4=xq^2$,即$\displaystyle x=q$.于是$\displaystyle (a_1,a_2,a_3,a_4,a_5)=(1,q,q^2,q^3,q^4)$,$\displaystyle a_1,a_2,a_3,a_4,a_5$成等比数列. 若 $\displaystyle a_2a_3\notin A$,则 $\displaystyle \frac{a_3}{a_2}\in A$.而$\displaystyle \frac{a_3}{a_2}=q$,$\displaystyle a_1<q<a_3$,只可能有$\displaystyle a_2=q$,同上可证$\displaystyle a_1,a_2,a_3,a_4,a_5$成等比数列.- 【2013 江苏23】 设数列 \(\displaystyle \{a_n\}\): $\(\displaystyle 1,-2,-2,3,3,3,-4,-4,-4,-4,\cdots,\overbrace{(-1)^{k-1}k, \ \dots, \ (-1)^{k-1}k}^{k\text{个}},\cdots\)$即当 \(\displaystyle \frac{(k-1)k}{2} < n \leqslant \frac{k(k+1)}{2} (k \in \mathbb{N}^*)\) 时, \(\displaystyle a_n = (-1)^{k-1}k\).记 \(\displaystyle S_n = a_1 + a_2 + \cdots + a_n (n \in \mathbb{N}^*)\).对于 \(\displaystyle l \in \mathbb{N}^*\),定义集合 $\(\displaystyle P_l = \{n \mid S_n \text{是} a_n \text{的整数倍},1 \leqslant n \leqslant l\}\)$ 求\(\displaystyle |P_{11}|,|P_{2000}|\).
答案
通过逐个讨论可以求出\(\displaystyle |P_{11}|\),此处直接介绍一般方法。考虑连续\(\displaystyle k\)个\(\displaystyle (-1)^{k-1}k\),记它们对应原数列中的\(\displaystyle a_{t+1},a_{t+2},\cdots,a_{t+k}\)。可知,\(\displaystyle a_{t+1}\mid S_{t+1}\)当且仅当对任意\(\displaystyle 1\leqslant p\leqslant k\),有\(\displaystyle a_{t+k}\mid S_{t+k}\),于是这\(\displaystyle k\)个数要么同时属于\(\displaystyle P_l\)或同时不属于\(\displaystyle P_l\)。
记$\displaystyle T_k$表示数列中$\displaystyle k$个$\displaystyle (-1)^{k-1}k$的和。可以证明,当$\displaystyle k$为偶数时,$\displaystyle \sum_{i=1}^{k}T_i=(-1)^kk^2=-\frac{k(k+1)}{2}$,此时$\displaystyle k+1\mid \sum_{i=1}^{k}T_i$,当$\displaystyle k$为奇数时,$\displaystyle \sum_{i=1}^{k}T_i=(-1)^kk^2=\frac{k(k+1)}{2}$,此时$\displaystyle k+1\nmid \sum_{i=1}^{k}T_k$,由此推出所有的奇数都属于$\displaystyle P_l$,所有的偶数都不属于$\displaystyle P_l$ 故$\displaystyle |P_{11}|=5$。再计算得$\displaystyle |a_{2000}|=63$,且前2000项中末尾的47项均为63,于是$\displaystyle |P_{2000}|=(1+3+5+\cdots +61)+47=1008$- 【2003全国卷22】设\(\displaystyle \{a_n\}\)是集合\(\displaystyle \{2^t+2^s\mid s,t\in\mathbb{Z},0\leqslant s<t\}\)中所有的数从小到大排列成的数列, 将数列\(\displaystyle \{a_n\}\)各项按照上小下大、左小右大的原则写成如下的三角形数表: $\(\displaystyle \begin{matrix} & & 3 & & \\ & 5 & & 6 & \\ 9 & & 10 & & 12 \\ \cdots & & \cdots & & \cdots \\ \end{matrix}\)$
- 写出这个三角形数表的第四行、第五行各数;
- 求\(\displaystyle a_{100}\);
- 设\(\displaystyle \{b_n\}\)是集合\(\displaystyle \{2^r+2^s+2^t\mid 0\leqslant r<s<t, \text{且} r,s,t\in\mathbb{Z}\}\)中所有的数从小到大排列成的数列,已知\(\displaystyle b_k=1160\),求\(\displaystyle k\).
答案
(1)可知,第 \(\displaystyle r\) 行的数从左到右分别为\(\displaystyle 2^r+2^s\),其中 \(\displaystyle s=0,1,\ldots,r-1\).于是第四行的数从左到右分别为\(\displaystyle 17,18,20,24\).第五行的数从左到右分别为 \(\displaystyle 33,34,36,40,48\).
(2)\(\displaystyle a_{100}=2^{14}+2^8=16384+256=16640\).
(3)受前述问题启发,将 \(\displaystyle 1160\) 写成三个不同的二次幂之和:\(\displaystyle 1160=1024+128+8=2^{10}+2^7+2^3\).它位于最大指数为 \(\displaystyle 10\) 的那一组.
最大指数小于 $\displaystyle 10$ 的项数为 $\displaystyle \sum_{t=2}^{9}\mathrm C_t^2=\mathrm C_{10}^{3}=120$.在最大指数为 $\displaystyle 10$ 的组中,按中间指数从小到大排列. 中间指数小于 $\displaystyle 7$ 时,项数为 $\displaystyle \mathrm C_7^2=21$.中间指数等于 $\displaystyle 7$ 时,较小指数依次为 $\displaystyle 0,1,2,3,\ldots,6$.目标项的较小指数为 $\displaystyle 3$,所以在它之前有 $\displaystyle 3$ 项.因此 $\displaystyle k=120+21+3+1=145$.- 【2026北京第一次学考25】 已知集合\(\displaystyle A=\{a_1,a_2,\cdots,a_n\}\),\(\displaystyle A\)中恰有\(\displaystyle s\)个元素是奇数,若\(\displaystyle \frac{s}{2}\)是偶数,则称\(\displaystyle A\)为“分偶集合”.
(1)判断\(\displaystyle \{-1,2,5,7\},\{2,3,5,6,7,9,16\}\)是否为“分偶集合”;(结论无需证明);
(2)若对任意整数\(\displaystyle a\),\(\displaystyle \{2,3,5,6,7,a^2,(a+b)^2\}\)均为分偶集合,写出\(\displaystyle b\)的三个取值;
(3)设\(\displaystyle n=30,A\subseteq\{x\in \mathbb{Z}\mid -29\leqslant x\leqslant 30\}\),若\(\displaystyle A\)中任意两个元素之和不等于1,且\(\displaystyle \frac{1}{2}\sum_{i=1}^{30}a_i\)是偶数,判断\(\displaystyle A\)是否可能为“分偶集合”,并说明理由.
答案
(1)集合 \(\displaystyle \{-1,2,5,7\}\)不是“分偶集合”.集合 \(\displaystyle \{2,3,5,6,7,9,16\}\)是“分偶集合”.
(2)因为 $\displaystyle 3,5,7$ 是奇数且不是平方数,要使得 $\displaystyle \{2, 3, 5, 6, 7, a^2, (a+b)^2\}$为“分偶集合”,必须 $\displaystyle a^2$ 与 $\displaystyle (a+b)^2$ 中有且只有一个奇数,即 $\displaystyle a$ 与 $\displaystyle a+b$ 有且只有一个奇数.而当 \(\displaystyle b\) 为奇数时,若 \(\displaystyle a\) 为偶数,则 \(\displaystyle a+b\) 为奇数;若 \(\displaystyle a\) 为奇数,则 \(\displaystyle a+b\) 为偶数,因此 \(\displaystyle a\) 与 \(\displaystyle a+b\) 有且只有一个奇数.
因此本题的答案中,\(\displaystyle b\) 可以是任意的奇数.
(3)我们要证明 \(\displaystyle A\) 不可能是“分偶集合”.采用反证法,假设 \(\displaystyle A\) 是“分偶集合”,则 \(\displaystyle A\) 恰有 \(\displaystyle s\) 个元素是奇数,且 \(\displaystyle \dfrac{s}{2}\) 是偶数,所以 \(\displaystyle s=4m(m\in\mathbb{N}^*)\).
把集合 \(\displaystyle \{x\in\mathbb{Z}|-29\leqslant x\leqslant 30\}\) 写成 30 个二元集的并:\(\displaystyle \{-29,30\}\),\(\displaystyle \{-28,29\}\),\(\displaystyle \{-27,28\}\),\(\displaystyle \cdots\),\(\displaystyle \{-1,2\}\),\(\displaystyle \{0,1\}\).因为 \(\displaystyle A\) 中任意两个元素之和均不等于 1,且 \(\displaystyle A\) 是 30 元集,所以在这 30 个二元集中,每个二元集有且只有一个元素属于 \(\displaystyle A\).
然后再把这 30 个二元集分成两组:
甲组:\(\displaystyle \{-29,30\}\),\(\displaystyle \{-26,27\}\),\(\displaystyle \{-25,26\}\),\(\displaystyle \{-22,23\}\),\(\displaystyle \{-21,22\}\),\(\displaystyle \cdots\),\(\displaystyle \{-2,3\}\),\(\displaystyle \{-1,2\}\).这里共 15 个集合,每个元素要么是形如 \(\displaystyle 4k-1\) 的奇数,要么是形如 \(\displaystyle 4k-2\) 的偶数.设 \(\displaystyle A\) 的元素中有 \(\displaystyle p\) 个偶数和 \(\displaystyle 15-p\) 个奇数位于甲组.
乙组:\(\displaystyle \{-28,29\}\),\(\displaystyle \{-27,28\}\),\(\displaystyle \{-24,25\}\),\(\displaystyle \{-23,24\}\),\(\displaystyle \cdots\),\(\displaystyle \{-3,4\}\),\(\displaystyle \{0,1\}\).这里共 15 个集合,每个元素要么是形如 \(\displaystyle 4k+1\) 的奇数,要么是形如 \(\displaystyle 4k\) 的偶数.设 \(\displaystyle A\) 的元素中有 \(\displaystyle q\) 个偶数和 \(\displaystyle 15-q\) 个奇数位于乙组.
于是,\(\displaystyle s=(15-p)+(15-q)\),即 \(\displaystyle p+q = 30-4m\).现在, $\(\displaystyle \begin{aligned} \sum\limits_{i=1}^{30}a_i &=\sum\limits_{a_i\text{形如}4k-1}a_i+\sum\limits_{a_i\text{形如}4k-2}a_i+\sum\limits_{a_i\text{形如}4k}a_i+\sum\limits_{a_i\text{形如}4k+1}a_i \\ &\equiv \Big[-2p + (-1)(15-p) + q\cdot 0 + (15-q)\cdot 1\Big] (\mathrm{mod}\,4) \\ &\equiv (-p-q) (\mathrm{mod}\,4) \\ &\equiv (30-4m) (\mathrm{mod}\,4) \\ &\equiv 30 (\mathrm{mod}\,4). \end{aligned}\)$
所以 \(\displaystyle \dfrac{1}{2}\sum\limits_{i=1}^{30}a_i\equiv 15(\mathrm{mod}\,2)\),即 \(\displaystyle \dfrac{1}{2}\sum\limits_{i=1}^{30}a_i\) 与 15 同奇偶.
然而由题干,\(\displaystyle \dfrac{1}{2}\sum\limits_{i=1}^{30}a_i\) 是偶数,但 15 是奇数,矛盾.
综上,\(\displaystyle A\) 不可能为“分偶集合”.
新答案(来源:4.3 多个量词的问题(2)数列.md):首先证明一个"基本事实":
一个等差数列中,若有连续三项成等比数列,则这个数列的公差\(\displaystyle d_0=0\).
事实上,设这个数列中的连续三项\(\displaystyle a-d_0\),\(\displaystyle a\),\(\displaystyle a+d_0\)成等比数列,则 $\(\displaystyle a^2=\left(a-d_0\right)\left(a+d_0\right),\)$ 由此得\(\displaystyle d_0=0\).
(1)(ⅰ)当\(\displaystyle n=4\)时,由于数列的公差\(\displaystyle d\ne0\),故由"基本事实"推知,删去的项只可能为\(\displaystyle a_2\)或\(\displaystyle a_3\).
①若删去\(\displaystyle a_2\),则由\(\displaystyle a_1,a_3,a_4\)成等比数列,得\(\displaystyle \left(a_1+2d\right)^2=a_1\left(a_1+3d\right)\).
因\(\displaystyle d\ne0\),故由上式得\(\displaystyle a_1=-4d\),即\(\displaystyle \frac{a_1}{d}=-4\),此时数列为\(\displaystyle -4d,-3d,-2d,-d\),满足题设.
②若删去\(\displaystyle a_3\),则由\(\displaystyle a_1,a_2,a_4\)成等比数列,得\(\displaystyle \left(a_1+d\right)^2=a_1\left(a_1+3d\right)\).
因\(\displaystyle d\ne0\),故由上式得\(\displaystyle a_1=d\),即\(\displaystyle \frac{a_1}{d}=1\).此时数列为\(\displaystyle d,2d,3d,4d\),满足题设.
综上可知,\(\displaystyle \frac{a_1}{d}\)的值为\(\displaystyle -4\)或\(\displaystyle 1\).
(ⅱ)若\(\displaystyle n\geqslant6\),则从满足题设的数列\(\displaystyle a_1,a_2,\cdots,a_n\)中删去一项后得到的数列,必有原数列中的连续三项,从而这三项既成等差数列又成等比数列,故由"基本事实"知,数列\(\displaystyle a_1,a_2,\cdots,a_n\)的公差必为\(\displaystyle 0\),这与题设矛盾.所以满足题设的数列的项数\(\displaystyle n\leqslant5\).又因题设\(\displaystyle n\geqslant4\),故\(\displaystyle n=4\)或\(\displaystyle 5\).
当\(\displaystyle n=4\)时,由(ⅰ)中的讨论知存在满足题设的数列.
当\(\displaystyle n=5\)时,若存在满足题设的数列\(\displaystyle a_1,a_2,a_3,a_4,a_5\),则由"基本事实"知,删去的项只能是\(\displaystyle a_3\),
从而\(\displaystyle a_1,a_2,a_4,a_5\)成等比数列,故 $\(\displaystyle \left(a_1+d\right)^2=a_1\left(a_1+3d\right),\)$ 及 $\(\displaystyle \left(a_1+3d\right)^2=\left(a_1+d\right)\left(a_1+4d\right).\)$ 分别化简上述两个等式,得\(\displaystyle a_1d=d^2\)及\(\displaystyle a_1d=-5d^2\),故\(\displaystyle d=0\),矛盾.因此,不存在满足题设的项数为\(\displaystyle 5\)的等差数列.
综上可知,\(\displaystyle n\)只能为\(\displaystyle 4\).
(2)假设对于某个正整数\(\displaystyle n\),存在一个公差为\(\displaystyle d'\)的\(\displaystyle n\)项等差数列\(\displaystyle b_1,b_1+d',\cdots,b_1+\left(n-1\right)d'\left(b_1d'\ne0\right)\),其中三项\(\displaystyle b_1+m_1d', b_1+m_2d', b_1+m_3d'\)成等比数列,这里\(\displaystyle 0\leqslant m_1<m_2<m_3\leqslant n-1\).则有 $\(\displaystyle \left(b_1+m_2d'\right)^2=\left(b_1+m_1d'\right)\left(b_1+m_3d'\right),\)$ 化简得 $\(\displaystyle \left(m_1+m_3-2m_2\right)b_1d'=\left(m_2^2-m_1m_3\right)\left(d'\right)^2.\)$ (*)
由\(\displaystyle b_1d'\ne0\)知,\(\displaystyle m_1+m_3-2m_2\)与\(\displaystyle m_2^2-m_1m_3\)或同时为零,或均不为零.
若\(\displaystyle m_1+m_3-2m_2=0\),且\(\displaystyle m_2^2-m_1m_3=0\),则有\(\displaystyle \left(\frac{m_1+m_3}{2}\right)^2-m_1m_3=0\),
即\(\displaystyle \left(m_1-m_3\right)^2=0\),得\(\displaystyle m_1=m_3\),从而\(\displaystyle m_1=m_2=m_3\),矛盾.
因此,\(\displaystyle m_1+m_3-2m_2\)与\(\displaystyle m_2^2-m_1m_3\)都不为零,故由(*)得 $\(\displaystyle \frac{b_1}{d'}=\frac{m_2^2-m_1m_3}{m_1+m_3-2m_2}.\)$ 因为\(\displaystyle m_1,m_2,m_3\)均为非负整数,所以上式右边为有理数,从而\(\displaystyle \frac{b_1}{d'}\)是一个有理数.
于是,对于任意的正整数\(\displaystyle n\geqslant4\),只要取\(\displaystyle \frac{b_1}{d'}\)为无理数,则相应的数列\(\displaystyle b_1,b_2,\cdots,b_n\)就是满足要求的数列.例如,取\(\displaystyle b_1=1\),\(\displaystyle d'=\sqrt{2}\),那么,\(\displaystyle n\)项数列\(\displaystyle 1,1+\sqrt{2},1+2\sqrt{2},\cdots,1+\left(n-1\right)\sqrt{2}\)满足要求.
【解题思路】第(2)问在解答时也可以先给出数列\(\displaystyle 1,1+\sqrt{2},1+2\sqrt{2},\cdots,1+\left(n-1\right)\sqrt{2}\)(只要公差为无理数均可),然后再证明其中任意三项都不构成等比数列(最终会得出\(\displaystyle \sqrt{2}\)等于一个有理数的形式,就导出矛盾).
新答案(来源:1.29 数列综合.md):①③④
【解题思路】①设\(\displaystyle a_n=a+nd\),\(\displaystyle b_n=b+nc\).则\(\displaystyle k\in M\)当且仅当\(\displaystyle a+kd=b+kc\),即\(\displaystyle a-b=k\left(c-d\right)\).这是个一次方程,这个关于\(\displaystyle k\)的方程只可能有\(\displaystyle 0\)个或\(\displaystyle 1\)个正整数解.特别地,对\(\displaystyle a_n=2n-1\),\(\displaystyle b_n=3n-2\),此时\(\displaystyle M=\left\{1\right\}\).故①正确.
②考虑\(\displaystyle a_n=2^n\),\(\displaystyle b_n=\left(-2\right)^n\),则\(\displaystyle \left\{a_n\right\}\)与\(\displaystyle \left\{b_n\right\}\)都是等比数列,但是\(\displaystyle M=\left\{2,4,6,8,\cdots\right\}=\left\{\text{全体偶数}\right\}\),故②错误.
③设\(\displaystyle a_n=a+nd\),\(\displaystyle b_n=q^n\)(不妨设\(\displaystyle b_1=q\)).则\(\displaystyle k\in M\)当且仅当\(\displaystyle a+kd=q^k\).
设\(\displaystyle f\left(x\right)=\left|q\right|^x-dx-a\),\(\displaystyle g\left(x\right)=\left|q\right|^x+dx+a\).
(i)若\(\displaystyle q>0\),则\(\displaystyle a+kd=q^k\)当且仅当\(\displaystyle f\left(k\right)=0\).求导可知\(\displaystyle f\left(x\right)\)最多只有两段单调区间,从而最多只有\(\displaystyle 2\)个零点,即\(\displaystyle M\)的元素个数不超过\(\displaystyle 2\).
(ii)若\(\displaystyle q<0\),则当\(\displaystyle k\)为奇数时,\(\displaystyle a+kd=q^k\)等价于\(\displaystyle a+kd=-\left|q\right|^k\),等价于\(\displaystyle g\left(k\right)=0\).
当\(\displaystyle k\)为偶数时,\(\displaystyle a+kd=q^k\)等价于\(\displaystyle f\left(k\right)=0\).
注意到,通过讨论\(\displaystyle q,d\)的取值可发现,\(\displaystyle f\left(x\right)\)和\(\displaystyle g\left(x\right)\)必有一个在\(\displaystyle \mathbb{R}\)上为单调函数,另一个在\(\displaystyle \mathbb{R}\)上有两段单调区间.于是,\(\displaystyle f\left(x\right)\),\(\displaystyle g\left(x\right)\)的零点个数之和不超过\(\displaystyle 3\).
特别地,取\(\displaystyle b_n=\left(-\frac{1}{2}\right)^n\),则\(\displaystyle b_1=-\frac{1}{2}\),\(\displaystyle b_3=-\frac{1}{8}\),\(\displaystyle b_4=\frac{1}{16}\).再取\(\displaystyle a_n=-\frac{1}{2}+\frac{3}{16}\left(n-1\right)\),则\(\displaystyle a_1=-\frac{1}{2}\),\(\displaystyle a_3=-\frac{1}{8}\),\(\displaystyle a_4=\frac{1}{16}\),所以\(\displaystyle M=\left\{1,3,4\right\}\).所以③正确.
④设\(\displaystyle c_n=a_n-b_n\),则\(\displaystyle \left\{c_n\right\}\)是递增数列,\(\displaystyle M=\left\{n\left|c_n=0\right.\right\}\).取一个连续函数\(\displaystyle f\left(x\right)\)满足\(\displaystyle f\left(n\right)=c_n\),使得\(\displaystyle f\left(x\right)\)是单调递增函数(构造方法:连接\(\displaystyle \left(n,c_n\right)\)和\(\displaystyle \left(n+1,c_{n+1}\right)\)的折线).由零点存在定理,\(\displaystyle f\left(x\right)\)最多只有一个零点,\(\displaystyle M\)最多只有\(\displaystyle 1\)个元素.故④正确.
新答案(来源:061-080极值零点数列与切割函数图像.md):- 显然构成公比为 \(\displaystyle 2^d\) 的等比数列,过程略.
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假设存在 \(\displaystyle a_1,d\) 满足题意.
\(\displaystyle a_1,a_2^2,a_3^3,a_4^4\) 依次构成等比数列且均为正数因此
\(\displaystyle \ln a_1,2\ln\left(a_1+d\right),3\ln\left(a_1+2d\right),4\ln\left(a_1+3d\right)\) 依次构成等差数列,设其公差为 \(\displaystyle d_1\).
令 \(\displaystyle b=\ln a_1\),则 \(\displaystyle n\ln\left[a_1+\left(n-1\right)d\right]=b+\left(n-1\right)d_1\left(n=1,2,3,4\right)\)
即方程 \(\displaystyle \ln\left[a_1+\left(x-1\right)d\right]=\frac{b-d_1}{x}+d_1\) 至少存在四个解 \(\displaystyle x=1,x=2,x=3,x=4\)
令 \(\displaystyle \varphi\left(x\right)=\ln\left[a_1+\left(x-1\right)d\right]-\frac{b-d_1}{x}-d_1\left(1\leqslant x\leqslant 4\right)\),则 \(\displaystyle \varphi\left(x\right)\) 至少有 \(\displaystyle 4\) 个零点.
\(\displaystyle \varphi'\left(x\right)=\frac{d}{a_1+\left(x-1\right)d}+\frac{b-d_1}{x^2}=\frac{dx^2+\left(b-d_1\right)\left[a_1+\left(x-1\right)d\right]}{x^2\left[a_1+\left(x-1\right)d\right]}\)
显然 \(\displaystyle x\geqslant 1\) 时 \(\displaystyle x^2\left[a_1+\left(x-1\right)d\right]>0\).
令 \(\displaystyle g\left(x\right)=dx^2+\left(b-d_1\right)\left[a_1+\left(x-1\right)d\right]\),\(\displaystyle g\left(x\right)\) 是二次函数至多有 \(\displaystyle 2\) 个变号零点,进而 \(\displaystyle \varphi\left(x\right)\) 至多有 \(\displaystyle 3\) 个单调区间,即 \(\displaystyle \varphi\left(x\right)\) 至多有 \(\displaystyle 3\) 个零点,矛盾.
综上不存在 \(\displaystyle a_1,d\) 满足题意.
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问与(2)问做法如出一辙:
假设存在 \(\displaystyle a_1,d\) 以及正整数 \(\displaystyle n,k\) 满足题意.
\(\displaystyle a_1^n,a_2^{n+k},a_3^{n+2k},a_4^{n+3k}\) 依次构成等比数列且均为正数因此
\(\displaystyle n\ln a_1,\left(n+k\right)\ln\left(a_1+d\right),\left(n+2k\right)\ln\left(a_1+2d\right),\left(n+3k\right)\ln\left(a_1+3d\right)\) 依次构成等差数列,设其公差为 \(\displaystyle d_2\).
令 \(\displaystyle b=n\ln a_1\),则 \(\displaystyle \left[n+\left(i-1\right)k\right]\ln\left[a_1+\left(i-1\right)d\right]=b+\left(i-1\right)d_2\left(i=1,2,3,4\right)\)
即方程 \(\displaystyle \ln\left(a_1+dx\right)=\frac{b+d_2x}{n+kx}\) 至少存在四个解 \(\displaystyle x=0,x=1,x=2,x=3\)
令 \(\displaystyle \phi\left(x\right)=\ln\left(a_1+dx\right)-\frac{b+d_2x}{n+kx}\left(0\leqslant x\leqslant 3\right)\),则 \(\displaystyle \phi\left(x\right)\) 至少有 \(\displaystyle 4\) 个零点.
\(\displaystyle \phi'\left(x\right)=\frac{d}{a_1+dx}+\frac{bk-d_2n}{\left(n+kx\right)^2}=\frac{d\left(n+kx\right)^2+\left(bk-d_2n\right)\left(a_1+dx\right)}{\left(n+kx\right)^2\left(a_1+dx\right)}\)
显然 \(\displaystyle x\geqslant 0\) 时 \(\displaystyle \left(n+kx\right)^2\left(a_1+dx\right)>0\)
令 \(\displaystyle h\left(x\right)=d\left(n+kx\right)^2+\left(bk-d_2n\right)\left(a_1+dx\right)\),\(\displaystyle h\left(x\right)\) 是二次函数至多有 \(\displaystyle 2\) 个变号零点,进而 \(\displaystyle \phi\left(x\right)\) 至多有 \(\displaystyle 3\) 个单调区间,即 \(\displaystyle \phi\left(x\right)\) 至多有 \(\displaystyle 3\) 个零点,矛盾.
综上,不存在 \(\displaystyle a_1,d\) 以及正整数 \(\displaystyle n,k\) 满足题意.
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【2016江苏20】已知数列 \(\displaystyle \{a_n\}\)是公比为3的等比数列,有限集\(\displaystyle T\in\mathbb{N^*}\).若 \(\displaystyle T=\varnothing\),定义 \(\displaystyle S_T=0\);若 \(\displaystyle T=\{t_1,t_2,\cdots,t_k\}\),定义 \(\displaystyle S_T=a_{t_1}+a_{t_2}+\cdots+a_{t_k}\).当 \(\displaystyle T=\{2,4\}\) 时, \(\displaystyle S_T=30\).
- 求 \(\displaystyle \{a_n\}\) 的通项公式;
- 求证:对任意正整数 \(\displaystyle k\),若 \(\displaystyle T\subseteq\{1,2,\cdots,k\}\),则\(\displaystyle S_T<a_{k+1}\);
- 设\(\displaystyle C\subseteq \mathbb{N^*}\), \(\displaystyle D\subseteq \mathbb{N^*}\), \(\displaystyle S_C\geqslant S_D\), 求证: \(\displaystyle S_C+S_{C\cap D}\geqslant 2S_D\).
- 对集合 \(\displaystyle A = \{1, 2, \dots, n\}\) 的每一个子集,定义其“交替和”如下:按照递减的次序重新排列该子集,然后从最大数开始交替地减、加后续的数. 例如集合 \(\displaystyle \{1, 2, 4, 6, 9\}\) 的“交替和”是 \(\displaystyle 9-6+4-2+1=6\),集合 \(\displaystyle \{5\}\) 的“交替和”是 \(\displaystyle 5\),规定空集的“交替和”为 \(\displaystyle 0\).记 \(\displaystyle A\) 所有子集“交替和”的和为 \(\displaystyle S_n\),求 $\(\displaystyle (a).S_3\quad (b). S_n \quad (c).\sum_{k=1}^n S_k\)$
\item 【2026“fiddie”模拟考19】设集合\(\displaystyle S=\{2^t+2^s\mid 0\leqslant s<t\text{且}s,t\in\mathbb{Z}\}\).例如\(\displaystyle 3,6,9,12,18,24\)属于\(\displaystyle S\),若\(\displaystyle T\subseteq S\),且\(\displaystyle T\)的所有元素从小到大排列后成等差数列,则称\(\displaystyle T\)是\(\displaystyle S\)的一个等差子集.
- 是否存在奇数\(\displaystyle x,y,z\in S(x<y<z)\),使得\(\displaystyle \{x,y,z\}\)是\(\displaystyle S\)的等差子集?并说明理由;
- 证明:\(\displaystyle S\)的含4个元素的等差子集有无穷多个;
- 证明:\(\displaystyle S\)不存在含5个元素的等差子集.
- 求 \(\displaystyle \{a_n\}\) 的通项公式;
答案
(1)不存在奇数 \(\displaystyle x,y,z \in S\) 使得 \(\displaystyle \{x,y,z\}\) 是 \(\displaystyle S\) 的等差子集. 证明如下:
当 \(\displaystyle t > s \geqslant 1\) 时,\(\displaystyle 2^t + 2^s\) 是偶数,所以 \(\displaystyle S\) 中的三个奇数形如 \(\displaystyle x=2^a+1\),\(\displaystyle y=2^b+1,z=2^c+1\)(\(\displaystyle 1 \leqslant a < b < c\),\(\displaystyle a,b,c \in \mathbb{Z}\)).
因为 \(\displaystyle c > b\),所以 \(\displaystyle c \geqslant b+1\),从而 \(\displaystyle 2^c - 2^{b+1} \geqslant 0\). 因此 \(\displaystyle (2^a+1)+(2^c+1)-2(2^b+1) = 2^a+2^c-2^{b+1} \geqslant 2^a > 0\),即 \(\displaystyle x+z > 2y\), 故 \(\displaystyle x,y,z\) 不构成等差数列,\(\displaystyle \{x,y,z\}\) 不可能是 \(\displaystyle S\) 的等差子集.
(2)设集合 \(\displaystyle T_m = \{3 \cdot 2^m, 6 \cdot 2^m, 9 \cdot 2^m, 12 \cdot 2^m\}\),\(\displaystyle m \in \mathbb{N}^*\). \(\displaystyle 3 \cdot 2^m = 2^{m+1} + 2^m \in S\),\(\displaystyle 6 \cdot 2^m = 2^{m+2} + 2^{m+1} \in S\), \(\displaystyle 9 \cdot 2^m = 2^{m+3} + 2^m \in S\),\(\displaystyle 12 \cdot 2^m = 2^{m+3} + 2^{m+2} \in S\), 故 \(\displaystyle T_m \subseteq S\),且 \(\displaystyle T_m\) 中元素从小到大排列后成公差为 \(\displaystyle 3 \cdot 2^m\) 的等差数列.
故 \(\displaystyle S\) 的含 4 个元素的等差子集有 \(\displaystyle T_1,T_2,\cdots,T_m,\cdots\),这里有无穷多个.
(3)假设 \(\displaystyle T = \{b_1,b_2,b_3,b_4,b_5\}(b_1 < b_2 < \cdots < b_5)\) 是 \(\displaystyle S\) 的等差子集,设其公差 \(\displaystyle d \in \mathbb{N}^*\).
若 \(\displaystyle T\) 中元素均为偶数,则 \(\displaystyle b_i = 2^{t_i} + 2^{s_i}\),\(\displaystyle t_i > s_i \geqslant 1\),\(\displaystyle t_i,s_i \in \mathbb{N}^*\),\(\displaystyle i = 1,2,3,4,5\). 于是 \(\displaystyle \frac{b_i}{2} = 2^{t_i-1} + 2^{s_i-1} \in S\),且 \(\displaystyle \frac{b_1}{2},\cdots,\frac{b_5}{2}\) 成公差为 \(\displaystyle \frac{d}{2}\) 的等差数列. 不断将 \(\displaystyle T\) 中所有元素除以 2,直至得到的子集中至少有一个奇数. 因此可不妨设原子集中至少有一个奇数.
若 \(\displaystyle d\) 为偶数,则 \(\displaystyle b_1,b_2,b_3,b_4,b_5\) 同奇偶. 若全为奇数,则奇数 \(\displaystyle b_1,b_2,b_3\) 成等差数列,与(1)矛盾;若全为偶数,与子集中至少有一个奇数矛盾. 故 \(\displaystyle d\) 为奇数.
若 \(\displaystyle b_1\) 为奇数,则 \(\displaystyle b_3 = b_1+2d\),\(\displaystyle b_5 = b_1+4d\) 均为奇数,且 \(\displaystyle b_1,b_3,b_5\) 成等差数列,与(1)矛盾. 故 \(\displaystyle b_1\) 为偶数,则 \(\displaystyle b_3,b_5\) 为偶数,\(\displaystyle b_2 = b_1+d\),\(\displaystyle b_4 = b_1+3d\) 为奇数.
因为 \(\displaystyle b_2,b_4 \in S\),所以 \(\displaystyle b_2 = 2^p+1\),\(\displaystyle b_4 = 2^q+1\),其中 \(\displaystyle 1 \leqslant p<q\),\(\displaystyle p,q \in \mathbb{Z}\).
于是 \(\displaystyle 2d = (2^q+1) - (2^p+1) = 2^p(2^{q-p}-1)\). 若 \(\displaystyle p \geqslant 2\),则 \(\displaystyle 2^p\) 是 4 的倍数,从而 \(\displaystyle 2d\) 是 4 的倍数,\(\displaystyle d\) 为偶数,矛盾. 故 \(\displaystyle p=1\),所以 \(\displaystyle b_2 = 3\). 但 3 是 \(\displaystyle S\) 中的最小元素,而 \(\displaystyle b_1 = b_2-d < 3\),这与 \(\displaystyle b_1 \in S\) 矛盾. 综上,\(\displaystyle S\) 不存在含有 5 个元素的等差子集.
评:本题以正整数的二进制表示为背景,并以考生熟悉的等差数列为知识素材,考查集合、等差数列的概念与性质、数列的递推关系;考查学生的逻辑思维、运算求解、自主探究和创新应用等能力,同时重点考查学生应用所学知识分析问题、解决问题的能力;考查数列递推的基本思想与方法,同时重点考查从特殊到一般、转化与化归、分类讨论等数学思想与方法. 本题是一道综合性很强、能检验考生数学思维的高质量的题目.
本题的命题灵感来源是 【2003 全国旧课程卷22】(收录在本书的1.1节习题),该题的数列是 OEIS 数列 A018900. 该题研究的是满足下面条件的正整数 \(\displaystyle N\) 构成的集合:将 \(\displaystyle N\) 表示成二进制后,只有两位数是 1,其余位均为 0. 与 2003 年高考题不一样的是,本题在命时仅采用 2003 年高考题中出现的集合,并没有考虑研究由这个集合里面的正整数构成的数表,而是转为研究这个集合里面最多有多少个数字能构成等差数列. “等差子列”、“等比子列”的长度问题同样也是非常著名的第一类“加性组合”的问题. 这种问题在往年高考也经常考查,例如【2008江苏19】,【2024北京15】,【2015江苏20】等.
命制此题的灵感来自Erdös-Turan 的等差数列猜想:对于正整数数列 \(\displaystyle \{n\}\) 的任意子列 \(\displaystyle \{A_n\}\),若其所有元素的倒数和发散,即 \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{A_n} = \infty\),则 \(\displaystyle \{A_n\}\) 含有任意长度的等差子列.
在 2004 年,Ben Green 和陶哲轩宣布证明了上述猜想的弱化版本——“存在任意长度的素数等差数列”,并于同年 4 月 9 日将论文提交至《美国数学年鉴》,该论文于 2005 年 9 月 12 日被接受. 2004 年已知最长的等差素数列包含 23 个素数.本题研究集合 \(\displaystyle S = \{2^s + 2^t \mid 0 \leqslant s < t,\text{且} s,t \in \mathbb{Z}\}\) 的等差子列,并设计了一个中学生能用中学数学知识解决的问题.
第(1)问将新定义的包装去除之后,就会发现 \(\displaystyle S\) 中元素 \(\displaystyle 2^t + 2^s(0 \leqslant s < t)\) 在 \(\displaystyle s \geqslant 1\) 时一定是偶数,故奇数必须满足 \(\displaystyle s=0\),需要学生对集合元素结构的初步分析能力. 因此,本题转化为:当 \(\displaystyle a < b < c\) 时,\(\displaystyle 2^a + 1, 2^b + 1, 2^c + 1\) 是否成等差数列,即 \(\displaystyle 2^a, 2^b, 2^c\) 是否成等差数列. 这是中学生比较熟悉的问题,
第(2)问需要学生证明等差子集有无穷多个,首先可以用数列递推的思想方法,注意到 \(\displaystyle S\) 中元素具有乘以 2 的封闭性,也就是说,将一个 4 元等差子集的所有元素都乘以 2 之后,可以得到新的 4 元等差子集. 从等差(比)数列构造另一个等差(比)数列的结论在教材中也有提及. 题干已经提示了 \(\displaystyle 3,6,9,12,18,24 \in S\),于是可以从初始的等差子集 \(\displaystyle \{3,6,9,12\}\) 出发,乘以 2 后得到 \(\displaystyle \{6,12,18,24\}\),一直下去按上述递推的方式即可得到无穷多个等差子集.
这个构造转化思想和数列递推的思想与 【2024新高考I卷19(2)】不谋而合,这样的设问能有效考查学生面对新问题时的逻辑思维能力,能有效区分学生分析问题、解决问题的能力水平. 这一问并不复杂,体现了由特殊到一般的数学思维,同时也为后续证明不存在元素个数为 5 的等差子集提供了方法上的对比.
第(3)问是本题的核心,要求证明不存在 5 元等差子集. 证明过程需要综合运用多个层面的推理:首先,解决第(2)问的思路是对一个等差子集不断乘以 2 就可以得到新的等差子集,受这个启发“反向操作”,先通过除以 2 将问题化归为至少含有一个奇数项的情形,再分析公差的奇偶性并结合第(1)问排除全奇数和首项为奇数的情形,最后利用奇数项在 \(\displaystyle S\) 中的唯一表示导出矛盾. 这一过程不仅考查了对集合性质的深入理解,还要求考生具备较强的逻辑分类与构造性推理能力.
整体来看,本题的三问层层递进,逐步深入,使得每个学生都能有所收获,都能获得相应的分数. 前两问为最后一问提供了必要的铺垫和方法示范,体现了数学问题设计中“引导—深化—综合”的常见结构,对培养学生对数学的兴趣,激发学生的探究精神,引导学生开展科学研究有很好的引领作用. 试题以离散数学中的简单表示为载体,考查学生从基本定义出发进行推理和构造的能力,具有较强的灵活性和选拔功能. 通过解答试题展现数学探究的过程,实现对分析、推理、判断等关键能力的考查,引导学生用规范的数学语言、新定义、新符号表达推理与论证过程.
本题设计新颖,极具探索性、创新性,作为试卷的最后一题,难度合理,考查学生利用新情境、新问题的创新能力,也考查学生的情感、态度和价值观. 试题引导学生面对问题要“多思考,少运算”,体现了对数学基础知识与数学核心素养的有机整合,具有很好的区分和选拔功能,能够有效地助力选拔创新人才. 试题情境、形式的创新,着力于“反套路、反刷题”,既使试题规避重题模式,又引导中学教学破除题海,消除套路,重视培养学生思维能力,真正注重能力和素养的培养.
数列题,特别是以等差等比数列为背景的数列题作为高考中的难题,通常没有“秒杀大招”来解决. 对数列知识的研究和复习,要关注以下几点:
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熟练掌握等差、等比数列中的一些基本问题,如通项公式与求和公式的各种形式、结构特点;等差、等比数列的常见性质;判断数列是等差、等比数列的方法,常见数列求和的方法等.
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理解和掌握等差、等比数列通项与求和公式推导中蕴含的基本方法,如累加(乘)等.
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体会“从特殊到一般”的思想在解决数列问题中的作用.
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用函数的观点认识和理解等差、等比等一些特殊的数列.
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把一些特殊的数列,如 \(\displaystyle a_n = 2n-1\) 作为平时学习中研究的对象,挖掘其背后所蕴含的特征,例如
- 数列 \(\displaystyle \{2n-1\}\) 的前 \(\displaystyle n\) 项和为完全平方数 \(\displaystyle n^2\);
- 正整数均有二进制表示,即表示成数列 \(\displaystyle \{2^{n-1}\}\) 的某些项之和;
- 正整数也可以表示成 Fibonacci 数列的某些非相邻项之和,这个结论叫 Zeckendorf 定理.
6.在数列的学习过程中,一定要意识到数感在学习过程中所起到的作用. 证明数列的存在性问题时,有时会用到构造性证明的方法(例如【2008江苏19(2)】),如何将符合要求的数列构造出来,是需要学生在平时学习中不断积累各种例子的.
- 【2025北京21】已知集合\(\displaystyle A=\left \{ 1,2,3,4,5,6,7,8\right \},M=\{ (x,y)|x,y\in A \}\),从\(\displaystyle M\)中选出\(\displaystyle n\)个不同的元素组成序列:\(\displaystyle (x_1,y_1),(x_2,y_2),\cdots,(x_n,y_n)\),若该序列的相邻项\(\displaystyle (x_i,y_i),(x_{i+1},y_{i+1})(1\leqslant i\leqslant n-1)\)满足$\(\displaystyle \begin{cases} |x_{i+1}-x_i|=3 \\ |y_{i+1}-y_i|=4 \end{cases}\text{或}\quad\begin{cases} |x_{i+1}-x_i|=4 \\ |y_{i+1}-y_i|=3 \end{cases}\)$
则将该列称为K列.
- (A)对于第1项为\(\displaystyle (3,3)\)的K列,写出它的第2项;
- (B)设\(\displaystyle \Gamma\)为K列,且\(\displaystyle \Gamma\)中的项\(\displaystyle (x_i,y_i)(i=1,2,\cdots,n)\)满足:当\(\displaystyle i\)为奇数时,\(\displaystyle x_i\in \left \{ 1,2,7,8 \right \}\),\(\displaystyle i\)为偶数时,\(\displaystyle x_i\in \left \{ 3,4,5,6 \right \}\),判断\(\displaystyle (3,2),(4,4)\)能否同时为\(\displaystyle \Gamma\)中的项,并说明理由.
- (C)证明:由\(\displaystyle M\)中全部元素组成的序列不是K列.
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【2010北京理20】 集合 \(\displaystyle S_n = \{X \mid X=(x_1,x_2,\dots,x_n), x_i \in \{0,1\}, i=1,2,\dots,n\}\) (\(\displaystyle n \geqslant 2\)), 对于 \(\displaystyle A=(a_1,a_2,\dots,a_n), B=(b_1,b_2,\dots,\) \(\displaystyle b_n) \in S_n\), 定义 \(\displaystyle A\) 与 \(\displaystyle B\) 的差为 \(\displaystyle A-B = (|a_1-b_1|,|a_2-b_2|,\dots,|a_n-b_n|)\); \(\displaystyle A\) 与 \(\displaystyle B\) 之间的距离为 \(\displaystyle d(A,B) = \sum_{i=1}^n |a_i-b_i|\).
- 证明: \(\displaystyle \forall A,B,C \in S_n\), 有 \(\displaystyle A-B \in S_n\), 且 \(\displaystyle d(A-C,B-C) = d(A,B)\);
- 证明: \(\displaystyle \forall A,B,C \in S_n\), \(\displaystyle d(A,B), d(A,C), d(B,C)\) 三个数中至少有一个是偶数;
- 设 \(\displaystyle P \subseteq S_n\), \(\displaystyle P\) 中有 \(\displaystyle m\) (\(\displaystyle m \geqslant 2\)) 个元素, 记 \(\displaystyle P\) 中所有两元素间距离的平均值为 \(\displaystyle \overline{d}(P)\). 证明: \(\displaystyle \overline{d}(P) \leqslant \frac{mn}{2(m-1)}\).
- 【2018北京理20】设 \(\displaystyle n\) 为正整数, 集合 $\(\displaystyle A=\{\alpha\mid\alpha=(t_1,t_2,\cdots,t_n),\ t_k\in\{0,1\},\ k=1,2,\cdots,n\}\)$ 对于集合 \(\displaystyle A\) 中的任意元素 \(\displaystyle \alpha=(x_1,x_2,\cdots,x_n)\) 和 \(\displaystyle \beta=(y_1,y_2,\cdots,y_n)\), 记 $\(\displaystyle M(\alpha,\beta)=\frac{1}{2}[(x_1+y_1-|x_1-y_1|)+(x_2+y_2-|x_2-y_2|)+\cdots+(x_n+y_n-|x_n-y_n|)]\)$
- 证明: \(\displaystyle \forall A,B,C \in S_n\), 有 \(\displaystyle A-B \in S_n\), 且 \(\displaystyle d(A-C,B-C) = d(A,B)\);
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当 \(\displaystyle n=3\) 时, 若 \(\displaystyle \alpha=(1,1,0)\), \(\displaystyle \beta=(0,1,1)\), 求 \(\displaystyle M(\alpha,\alpha)\) 和 \(\displaystyle M(\alpha,\beta)\) 的值;
- 当 \(\displaystyle n=4\) 时, 设 \(\displaystyle B\) 是 \(\displaystyle A\) 的子集, 且满足: 对于 \(\displaystyle B\) 中的任意元素 \(\displaystyle \alpha,\beta\), 当 \(\displaystyle \alpha,\beta\) 相同时, \(\displaystyle M(\alpha,\beta)\) 是奇数; 当 \(\displaystyle \alpha,\beta\) 不同时, \(\displaystyle M(\alpha,\beta)\) 是偶数. 求集合 \(\displaystyle B\) 中元素个数的最大值;
- 给定不小于 \(\displaystyle 2\) 的 \(\displaystyle n\), 设 \(\displaystyle B\) 是 \(\displaystyle A\) 的子集, 且满足: 对于 \(\displaystyle B\) 中的任意两个不同的元素 \(\displaystyle \alpha,\beta\), \(\displaystyle M(\alpha,\beta)=0\). 写出一个集合 \(\displaystyle B\), 使其元素个数最多, 并说明理由.
C 组习题
C组
1. 证明集合的笛卡尔积对集合的交、并运算满足分配律,即对任意集合$\displaystyle A,B,C$有 $$\displaystyle A\times(B\cap C) &= (A\times B)\cap(A\times C), A\times(B\cup C) &= (A\times B)\cup(A\times C). (A\times B)\cap(A\times C) &= A\times(B\cap C), (A\times B)\cup(A\times C) &= A\times(B\cup C).$$- 是否存在正整数 \(\displaystyle n\)使得集合 \(\displaystyle \{n,n+1,n+2,n+3,n+4,n+5\}\) 可以划分成两个无公共元素的非空子集,使得一个子集的所有元素的积等于另一个子集的所有元素的积.
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【2025STEP2第8题】设有限集合 \(\displaystyle S \subseteq \mathbb{Z}\) . 对于 \(\displaystyle S\) 的两个子集 \(\displaystyle A, B\) , 定义如下性质:
性质1: \(\displaystyle A \cap B = \varnothing\) , \(\displaystyle A \cup B = S\) .
性质2: \(\displaystyle A\) 中所有元素的和等于 \(\displaystyle B\) 中所有元素的和.
性质3: \(\displaystyle A\) 中所有元素的平方和等于 \(\displaystyle B\) 中所有元素的平方和.
若 \(\displaystyle A, B\) 同时满足性质1,2,3, 则称 \(\displaystyle A, B\) 是 \(\displaystyle S\) 的一个平衡划分.
- 写出 \(\displaystyle \{1, 2, 3, 4, 5, 6, 7, 8\}\) 的一个平衡划分 \(\displaystyle A, B\) , 其中 \(\displaystyle A, B\) 各有 \(\displaystyle 4\) 个元素.
- 设实数 \(\displaystyle a_1, a_2, \cdots, a_m, b_1, b_2, \cdots, b_m\) 满足 $\(\displaystyle \sum_{k=1}^m a_k = \sum_{k=1}^m b_k, \quad \sum_{k=1}^m a_k^2 = \sum_{k=1}^m b_k^2.\)$ 证明: 对任意实数 \(\displaystyle c\) , 都有 $\(\displaystyle \sum_{k=1}^m a_k^3 + \sum_{k=1}^m (c + b_k)^3 = \sum_{k=1}^m b_k^3 + \sum_{k=1}^m (c + a_k)^3.\)$
- 求 \(\displaystyle \{1, 2, 3, \cdots, 16\}\) 的一个平衡划分 \(\displaystyle A, B\) , 使得 \(\displaystyle A, B\) 各有 \(\displaystyle 8\) 个元素, 且满足\(\displaystyle A\) 中所有元素的立方和等于 \(\displaystyle B\) 中所有元素的立方和.
- 已知 \(\displaystyle \{1, 2, 3, \cdots, 11\}\) 的一个平衡划分分为 \(\displaystyle \{1, 3, 4, 5, 9, 11\}\) , \(\displaystyle \{2, 6, 7, 8, 10\}\) .设 \(\displaystyle S = \{n^2, (n + 1)^2, (n + 2)^2, \cdots, (n + 11)^2\}\) , 其中 \(\displaystyle n\) 是正整数.求 \(\displaystyle S\) 的一对子集 \(\displaystyle A, B\) 使得性质1,2成立.
答案
(1)\(\displaystyle A=\{2,3,5,8\}\) 和 \(\displaystyle B=\{1,4,6,7\}\) 是 \(\displaystyle S=\{1,2,3,4,5,6,7,8\}\) 的一个平衡划分.因为它们中元素的和与平方和均分别为18与102.
分析过程如下(不用写出):因为 \(\displaystyle 1+2+\cdots+8=36\),\(\displaystyle 1^2+2^2+\cdots+8^2 = 204\),由平衡划分的定义, \(\displaystyle A,B\) 的所有元素之和都为 18,所有元素的平方和都为 102.
进行穷举.假设\(\displaystyle 8\in A\),分析\(\displaystyle A\)中另外三个元素,它们的平方和是 38,和是 10,那么 \(\displaystyle 6,7\notin A\).而\(\displaystyle 5^2+2^2+3^2 = 38\),所以可以写 \(\displaystyle A=\{2,3,5,8\}\),\(\displaystyle B=\{1,4,6,7\}\) .
(2)展开待证式中的立方式 $\(\displaystyle \begin{aligned} a_k^3+(c+b_k)^3&=a_k^3+b_k^3+c^3+3c^2b_k+3cb_k^2, \\ b_k^3+(c+a_k)^3&=a_k^3+b_k^3+c^3+3c^2a_k+3ca_k^2, \\ \end{aligned}\)$ 两边对 \(\displaystyle k=1,2,\cdots,m\) 求和,并由条件 $\(\displaystyle \sum\limits_{k=1}^ma_k = \sum\limits_{k=1}^mb_k, \quad \sum\limits_{k=1}^ma_k^2=\sum\limits_{k=1}^mb_k^2.\)$
即可完成证明.
(3)取 \(\displaystyle (a_1,a_2,a_3,a_4) = (8,5,3,2)\),\(\displaystyle (b_1,b_2,b_3,b_4) = (1,4,6,7)\),并取 $\(\displaystyle \begin{aligned} A&=\{a_1,a_2,a_3,a_4,c+b_1,c+b_2,c+b_3,c+b_4\}, \\ B&=\{b_1,b_2,b_3,b_4,c+a_1,c+a_2,c+a_3,c+a_4\}, \end{aligned}\)$
即 $\(\displaystyle \begin{aligned} A&=\{2,3,5,8,9,12,14,15\}, \\ B&=\{1,4,6,7,10,11,13,16\}. \end{aligned}\)$
易知性质1成立.
由(1)得
\[\displaystyle \sum\limits_{k=1}^ma_k = \sum\limits_{k=1}^mb_k, \quad \sum\limits_{k=1}^ma_k^2=\sum\limits_{k=1}^mb_k^2.\]由(2),取 \(\displaystyle c=8\) 得
\[\displaystyle \sum\limits_{k=1}^ma_k^3 + \sum\limits_{k=1}^m(c+b_k)^3 = \sum\limits_{k=1}^mb_k^3+\sum\limits_{k=1}^m(c+a_k)^3.\]所以性质\ding{175}成立.
因为 \(\displaystyle A\) 中元素的和以及 \(\displaystyle B\) 中元素的和都是 \(\displaystyle \sum\limits_{k=1}^4(a_k+b_k+c)\),所以性质\ding{173}成立.
因为当 \(\displaystyle \sum\limits_{k=1}^ma_k=\sum\limits_{k=1}^mb_k\) 时,
\[\displaystyle \begin{aligned} &\quad\,\,\sum\limits_{k=1}^ma_k^2+\sum\limits_{k=1}^m(b_k+c)^2 \\ &=\sum\limits_{k=1}^m(a_k^2+b_k^2+c^2+2cb_k) \\ &=\sum\limits_{k=1}^m(a_k^2+b_k^2+c^2+2ca_k) \\ &=\sum\limits_{k=1}^mb_k^2+\sum\limits_{k=1}^m(a_k+c)^2, \end{aligned}\]所以 \(\displaystyle A\) 中元素的平方和以及 \(\displaystyle B\) 中元素的平方和相等.所以性质\ding{174}成立.
综上,如上构造的 \(\displaystyle A\),\(\displaystyle B\) 是 \(\displaystyle S\) 的一个平衡划分,并且满足性质\ding{175}.
分析:这里还要求立方和相同,考虑用(2)的结论.注意(2)中的 \(\displaystyle c\) 是任取的.要构造集合 \(\displaystyle A\) , \(\displaystyle B\) 满足第三问新给出的条件,观察到(2)中结论的等号两边都是 \(\displaystyle 2m\) 个数字相加,所以可以尝试取 \(\displaystyle m=4\).这样就需要取 \(\displaystyle c\) 以及合适的 \(\displaystyle a_1,\cdots,a_4, b_1,\cdots,b_4\) ,然后取 $\(\displaystyle \begin{aligned} A&=\{a_1,a_2,a_3,a_4,c+b_1,c+b_2,c+b_3,c+b_4\}, \\ B&=\{b_1,b_2,b_3,b_4,c+a_1,c+a_2,c+a_3,c+a_4\}, \end{aligned}\)$ 使得 \(\displaystyle A\), \(\displaystyle B\) 满足性质1. 在(1)中已经构造了 \(\displaystyle \{1,2,3,4,5,6,7,8\}\) 的平衡划分,自然可以取 \(\displaystyle (a_1,a_2,a_3,a_4) = (8,5,3,2)\),\(\displaystyle (b_1,b_2,b_3,b_4) = (1,4,6,7)\), 然后取 \(\displaystyle c=8\),这样就能构造满足性质1的两个集合: $\(\displaystyle \begin{aligned} A&=\{2,3,5,8,9,12,14,15\}, \\ B&=\{1,4,6,7,10,11,13,16\}. \end{aligned}\)$ 接下来尝试验证结论成立即可。
(4)补 0 元素,则 \(\displaystyle \{1,3,4,5,9,11\}\), \(\displaystyle \{0,2,6,7,8,10\}\) 是集合 \(\displaystyle \{0,1,2,\cdots,11\}\) 的一个平衡划分.
由(3)中的证明过程,取 \(\displaystyle m=6\), \(\displaystyle (a_1,\cdots,a_6)=(1,3,4,5,9,11)\),\(\displaystyle (b_1,\cdots,b_6) = (0,2,6,7,8,10)\),\(\displaystyle c=n\),
则 $\(\displaystyle \sum\limits_{k=1}^6a_k^2+\sum\limits_{k=1}^6(n+b_k)^2 = \sum\limits_{k=1}^6b_k^2+\sum\limits_{k=1}^6(n+a_k)^2.\)$
取 \(\displaystyle A = \{(n+1)^2,(n+3)^2,(n+4)^2,(n+5)^2,(n+9)^2,(n+11)^2\}\),
\(\displaystyle B=\{n^2,(n+2)^2,(n+6)^2,(n+7)^2,(n+8)^2,(n+10)^2\}\).
则 \(\displaystyle A\),\(\displaystyle B\) 满足性质\ding{172}.
因为 \(\displaystyle \sum\limits_{k=1}^6a_k^2 = \sum\limits_{k=1}^6b_k^2\),所以
\[\displaystyle \sum\limits_{k=1}^6(n+b_k)^2 = \sum\limits_{k=1}^6(n+a_k)^2.\]即 \(\displaystyle A\),\(\displaystyle B\) 满足性质\ding{173}.\(\displaystyle \square\)
思考题
考虑集合 \(\displaystyle I_n = \{1,2,\cdots,n\}\),则 \(\displaystyle I_n\) 的元素之和为 \(\displaystyle \dfrac{n(n+1)}{2}\).如果 \(\displaystyle I_n\) 存在平衡划分,则 \(\displaystyle \dfrac{n(n+1)}{2}\) 是偶数,从而 \(\displaystyle n(n+1)\) 是 4 的倍数,即 \(\displaystyle n\) 形如 \(\displaystyle 4k\) 或 \(\displaystyle 4k-1\),\(\displaystyle k\in\vv{N}^*\).
显然,\(\displaystyle \{1,2,3\}\) 和 \(\displaystyle \{1,2,3,4\}\) 都不存在平衡划分.
\(\displaystyle \{1,2,3,4,5,6,7\}\) 的一个平衡划分是 \(\displaystyle \{1,2,4,7\}\) , \(\displaystyle \{3,5,6\}\).
对于哪些正整数 \(\displaystyle n\), 集合 \(\displaystyle I_n\) 存在平衡划分?若存在,如何给出一个平衡划分?
根据前面的观察,当 \(\displaystyle n=7,8,11,16\) 时集合 \(\displaystyle I_n\) 存在平衡划分.更一般的,可以自行思考.思考这个问题应该有助于提升构造集合(数列)的经验.
- 【2026“神算杯一模”(B卷)(网络联考)19】设正整数\(\displaystyle n\geqslant 3\),集合\(\displaystyle A=\left \{ x\in\mathbb{N}\mid 3\leqslant x\leqslant n \right \}\),集合\(\displaystyle S\in\mathbb{N}\)满足:对任意\(\displaystyle a\in A\),总存在两两不相等的\(\displaystyle s_1,s_2,\cdots ,s_k\in S(k\geqslant 3)\),使得\(\displaystyle \sum_{i=1}^{k}s_i=a\),记\(\displaystyle |S|=m\).
- 证明:\(\displaystyle \left \{ 0,1,2 \right \}\subseteq S\);
- 若\(\displaystyle m=5\),求\(\displaystyle n\)的最大值;
- 证明:\(\displaystyle m\)的最小值为\(\displaystyle \lceil \log_2(\frac{n}{3})\rceil+3\).
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已知正整数\(\displaystyle n\geqslant 4\),有序数组\(\displaystyle \alpha=(x_1,x_2,\cdots,x_n)\)满足\(\displaystyle x_i\in\left \{ 0,1 \right \}(i=1,2,\cdots,n)\),且\(\displaystyle \sum_{i=1}^{n}x_i=m\),并称\(\displaystyle \alpha\)是\(\displaystyle n\)元\(\displaystyle m\)阶\(\displaystyle 0-1\)数组,设\(\displaystyle \beta(y_1,y_2,\cdots,y_n)\)是另一个\(\displaystyle n\)元\(\displaystyle m\)阶\(\displaystyle 0-1\)数组,若\(\displaystyle \alpha\)和\(\displaystyle \beta\)满足\(\displaystyle \alpha\neq \beta\),且\(\displaystyle \sum_{i=1}^{n}|x_i-y_i|=m\),则称\(\displaystyle \alpha\)与\(\displaystyle \beta\)孪生.
- 若\(\displaystyle n=4,\alpha=(1,0,0,1)\),写出2个不同的\(\displaystyle \beta\),使得\(\displaystyle \alpha\)与\(\displaystyle \beta\)孪生;
- 证明:若\(\displaystyle \alpha\)与\(\displaystyle \beta\)孪生,则\(\displaystyle m\)是偶数;
- 设\(\displaystyle \gamma(z_1,z_2,\cdots,z_n)\)是\(\displaystyle n\)元\(\displaystyle m\)阶\(\displaystyle 0-1\)数组,证明:若\(\displaystyle \alpha\)与\(\displaystyle \beta\)孪生,则存在\(\displaystyle \gamma\)与\(\displaystyle \alpha,\beta\)都孪生,且对任意与\(\displaystyle \alpha ,\beta\)都孪生的\(\displaystyle \gamma\),\(\displaystyle \sum_{i=1}^{n}(x_i+y_i)(y_i+z_i)(z_i+x_i)\leqslant 2n\)
- 若\(\displaystyle n=4,\alpha=(1,0,0,1)\),写出2个不同的\(\displaystyle \beta\),使得\(\displaystyle \alpha\)与\(\displaystyle \beta\)孪生;
D 组习题