集合

(2020年上海市高考数学练习)设$a\in \mathbb{R}$,若存在定义域为$\mathbb{R}$的函数$f(x)$既满足“对于任意$x_0\in \mathbb{R}$, $f(x_0)$的值为$x_0^2$或$x_0$”,又满足“关于$x$的方程$f(x)=a$无实数解”,则$a$的取值范围为

$(-\infty,0)\cup (0,1)\cup (1,+\infty)$

唐代诗人李顾的诗《古从军行》开头两句说: “白日登山望烽火,黄昏饮马傍交河.”诗中隐含着一个有趣的数学问题——“将军饮马”问题,即将军在观望烽火之后从山脚下某处出发,先到河边饮马后再回到军营,怎样走才能使总路程最短?在平面直角坐标系中,设军营所在区域为$x^2+y^2\leqslant 1$,若将军从点$A(2,0)$处出发,河岸线所在直线方程为$x+y=3$,并假定将军只要到达军营所在区域即回到军营,则“将军饮马”的最短总路程为

(东城区2019—2020学年第二学期期末高一)对于任意实数$a,b,c,d$,表达式$ad-bc$称为二阶行列式(determinant),记作
$\left| \begin{matrix}
a& b\\
c& d\\
\end{matrix} \right|$.

(I)求下列行列式的值:

\ding{172} $\left| \begin{matrix}
1& 0\\
0& 1\\
\end{matrix} \right|$;\quad
\ding{173} $\left| \begin{matrix}
1& 3\\
2& 6\\
\end{matrix} \right|$;\quad
\ding{174} $\left| \begin{matrix}
-2& 5\\
10& -25\\
\end{matrix} \right|$;

(II)求证:向量$p=(a,b)$与向量$q=(c,d)$共线的充要条件是
$\left| \begin{matrix}
a& b\\
c& d\\
\end{matrix} \right|=0$;

(III)讨论关于$x,y$的二元一次方程组
$$
\begin{cases}
a_1x+b_1y=c_1,\\
a_2x+b_2y=c_2,
\end{cases}(a_1a_2b_1b_2\neq 0)
$$
有唯一解的条件,并求出解. (结果用二阶行列式的记号表示)

20.已知函数$f(x)=ax\ln x+b$ ($a,b$为常数),在点$(1,0)$处切线方程为$y=x-1$.

(1)试求$a,b$的值;

(2)若方程$f(x)=m$有两不等实数根,求$m$的范围.

(3)设$g(x)=f'(x)$, $A(x_1,y_1),B(x_2,y_2)$为曲线$y=g(x)$上不同两点,记直线$AB$的斜率为$k$,证明:
$k>g'\left(\frac{x_1+x_2}{2}\right)$.

21.已知$S_n=\left\{A|A=(a_1,a_2,a_3,\cdots,a_n),a_i=0\,\text{或}\,
1,1=1,2,\cdots,n\right\}\,(n\geqslant 2)$,对于$U,V\in S_n$, $d(U,V)$表示$U$和$V$中相对应的元素不同的个数.

(1)令$U=(0,0,0,0,0)$,存在$m$个$V\in S_5$,使得$d(U,V)=2$,写出$m$的值;

(2)令$W=\underbrace{(0,0,0,\cdots,0)}_{n\,\text{个}\,0}$,
若$U,V=S_n$,求证: $d(U,W)+d(V,W)\geqslant d(U,V)$;

(3)令$U=(a_1,a_2,a_3,\cdots,a_n)$,若$V\in S_n$,求所有$d(U,V)$之和.

(东城区2019—2020学年第二学期期末高二数学试题)设集合$S_n=\{n,n+1,\cdots,2n-1\}$,若$X$是$S_n$的子集,把$X$中所有数的和称为$X$的“容量” (规定空集的容量为$0$),若$X$的容量为奇(偶)数,则称$X$为$S_n$的奇(偶)子集.

(I)当$n=3$时,写出$S_n$的所有奇子集;

(II)求证:当$n\geqslant 3$时, $S_n$的所有奇子集的容量之和等于所有偶子集的容量之和;

(III)当$n\geqslant 3$时,求$S_n$的所有奇子集的容量之和.

(I)解:当$n=3$时, $S_n=\{3,4,5\}$.

$S_n$的所有奇子集为$\{3\},\{5\},\{3,4\},\{4,5\}$.

(II)证明:首先证明$S_n$,的奇子集与偶子集个数相等.

设奇数$k\in S_n$,对于$S_n$的每个奇子集$A$,

当$k\in A$时,取$B=\{x|x\in A\text{且}x\neq k\}$.

当$k\notin A$时,取$B=A\cup\{k\}$,则$B$为$S_n$的偶子集.

反之,亦然.

所以, $S_n$的奇子集与偶子集是一一对应的.

所以, $S_n$的奇子集与偶子集个数相等.

对于$\forall i\in S_n,i>1$,含$i$的$S_n$的子集共有$2^{n-1}$个.

其中必有一半是奇子集,一半是偶子集,从而对于每个数$i$,在奇子集的和与偶子集的和中, $i$所占的个数是一样的.

所以$S_n$的所有奇子集的容量的和与所有偶子集的容量的和相等.

(III)解:由于每个元素在奇子集中都出现$2^{n-2}$次,故奇子集的容量和为$(n+n+1+\cdots +2n-1)\times 2^{n-2}=n(3n-1)\times 2^{n-3}$.

(北京市西城区 2019—2020学年度第二学期期末试卷)设函数$f(x)$定义域为$D$,若函数$f(x)$满足:对任意$c\in D$,存在$a,b\in D$,使得$\frac{f(a)-f(b)}{a-b}=f'(c)$成立,则称函数$f(x)$满足性质$\Gamma$.下列函数不满足性质$\Gamma$的是

(A) $f(x)=x^2$ (B) $f(x)=x^3$

(C) $f(x)=e^x$ (D) $f(x)=\ln x$

选B.可取$c=0$.

(北京市西城区 2019—2020学年度第二学期期末试卷)已知函数$f(x)=lnx+ax-a$.

(I)求函数$f(x)$的单调区间;

(II)求证:当$a>1$时,函数$g(x)=e^{x-1}-f(x)$存在最小值,且最小值小于$1$.

(2019-2020房山区第二学期期末高二数学试题)设$[x]$表示不大于$x$的最大整数,则对任意实数$x$,给出以下四个命题:
\ding{172} $[-x]=-[x]$; \ding{173} $[x+1]=[x]$; \ding{174} $[2x]=2[x]$; \ding{175} $[x]+\left[x+\frac{1}{2}\right]=[2x]$.
则假命题是(填上所有假命题的序号).

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1.集合的概念

(1)集合是数学中的基本概念,很多后续数学知识或理论(函数,集合论等)都是建立在集合的基础之上,但集合是不可定义(第三次数学危机)的概念.具有某种性质的对象全体称为一个集合,对象称为集合的元素.

集合中的元素具有确定性、互异性和无序性.

(2)集合的分类:元素个数为有限个的集合称为有限集;元素个数为无限个的集合称为无限集.

特别地,不含任何元素的集合称为空集,用$\emptyset$表示.

(3)集合的表示方法：列举法、描述法和韦恩(Venn)图法.

2.元素与集合、集合之间的关系

(1)元素与集合之间的关系有“属于$\in$”或“不属于$\notin$”.

(2)集合之间的关系

(i)包含关系:包含关系与子集的概念等价,即集合$A$包含于集合$B$等价于$A$是$B$的子集,记作$A\subseteq B$;集合$A$真包含于集合$B$等价于$A$是$B$的真子集,记作$A\varsubsetneqq B$;特别地,空集$\emptyset$是任何集合的子集,是任何非空集合的真子集.

注:包含关系是用来刻画集合元素个数的多少,与不等号$<$与$\leqslant$类似.

(ii)不包含关系:若集合$A$中至少有一个元素不属于集合$B$,就称集合$A$不包含于集合$B$,记作$A\nsubseteq B$.

(iii)集合的相等:若$A\subseteq B$且$B\subseteq A$,则集合$A,B$相等,记作$A=B$.

3.集合的运算(可用韦恩图直观呈现)

(1)集合的交、并、补的运算

$A\cap B=\{x|x\in A\text{且}x\in B\},A\cup B=\{x|x\in A\text{或}x\in B\}$,

$\complement_UA=\{x|x\in U,x\notin A, A\subseteq U\}$.也可记为$A'$或$A^c$.

(2)集合运算中一些常用的结论

(i)交换律: $A\cap B=B\cap A,A\cup B=B\cup A$;

(ii)结合律: $A\cap (B\cap C)=(A\cap B)\cap C, A\cup (B\cup C)=(A\cup B)\cup C$;

(iii)分配律: $A\cap (B\cup C)=(A\cap B)\cup (A\cap C), A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$;

(iv)吸收律: $A\cup (A\cap B)=A,A\cap (A\cup B)=A$;

(v)反演律(摩根律): $\complement_U(A\cap B)=\complement_U A\cup \complement_UB,\complement_U (A\cup B)=\complement_U A\cap \complement_UB$.

4.有限集子集的个数
若有限集$A$含有$n$个元素,则$A$的子集有$2^n$个,真子集有$2^n-1$个.

5.有限集的元素个数$\mathrm{card}(A)$或$|A|$

对任意两个有限集合$A,B$,有$|A\cup B|=|A|+|B|-|A\cap B|$,

$|\complement_U(A\cup B)|=|U|-|A|-|B|+|\complement_U(A\cap B)|$.

此结论可以推广到任意$n$个有限集合$A_1,A_2,\cdots,A_n$ (称为容斥原理),并在组合数学、数论等领域有非常重要的应用.

(2019 AMC 10A) Problem
For how many integers $n$ between $1$ and $50$, inclusive, is\[\frac{(n^2-1)!}{(n!)^n}\]an integer? (Recall that $0! = 1$.)

$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

Solution
Solution 1
The main insight is that

\[\frac{(n^2)!}{(n!)^{n+1}}\]
is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$. Thus,

\[\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}\]
is an integer if $n^2 \mid n!$, or in other words, if $n \mid (n-1)!$. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson's Theorem. There are $15$ primes between $1$ and $50$, inclusive, so there are $15 + 1 = 16$ terms for which

\[\frac{(n^2-1)!}{(n!)^{n}}\]
is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of $2$ in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$, as there are more factors of p in the denominator than the numerator. Thus all $16$ values of n make the expression not an integer and the answer is $50-16=\boxed{\mathbf{(D)}\ 34}$.

Solution 2
We can use the P-Adic Valuation (more info could be found here: Mathematicial notation) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by $v_p (n)$ and is defined as the greatest power of some prime 'p' that divides n. For example, $v_2 (6)=1$ or $v_7 (245)=2$ .) Using Legendre's formula, we know that :

\[v_p (n!)= \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor\]
Seeing factorials involved in the problem, this prompts us to use Legendre's formula where n is a power of a prime.

We also know that , $v_p (m^n) = n \cdot v_p (m)$ . Knowing that $a\mid b$ if $v_p (a) \le v_p (b)$ , we have that :

\[n \cdot v_p (n!) \le v_p ((n^2 -1 )!)\]and we must find all n for which this is true.

If we plug in $n=p$, by Legendre's we get two equations:

\[v_p ((n^2 -1)!) = \sum_{i=1}^\infty \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1\]
And we also get :

\[v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p\]
But we are asked to prove that $n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1$ which is false for all 'n' where n is prime.

Now we try the same for $n=p^2$ , where p is a prime. By Legendre we arrive at:

\[v_p ((p^4 -1)!) = p^3 + p^2 + p -3\]and\[p^2 \cdot v_p (p^2 !) = p^3 + p^2\]
Then we get:

\[p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3\]Which is true for all primes except for 2, so $2^2 = 4$ doesn't work. It can easily be verified that for all $n=p^i$ where $i$ is an integer greater than 2, satisfies the inequality :\[n \cdot v_p (n!) \le v_p ((n^2 -1 )!).\]
Therefore, there are 16 values that don't work and $50-16 = \boxed{\mathbf{(D)}\ 34}$ values that work.

Solution 3 (Guessing)
First, we see that $n=1, 6, 8, 9, 10, 12, 14$ work. This leads us to the conclusion of $50-16 = \boxed{\textbf{(D)}\ 34}$