《2025年高考帮数学》

1.下列说法正确的是 ( )
A.$\{x|y = x^{2}\}=\{y|y = x^{2}\}=\{(x,y)|y = x^{2}\}$
B.方程$\sqrt{x - 2024}+(y + 2025)^{2}=0$的解集为$\{2024,-2025\}$
C.若$\{x^{2},1\}=\{0,1\}$,则$x = 0$或1
D.对任意两个集合A,B,$(A\cap B)\subseteq(A\cup B)$恒成立

答案： 1.D

2.若集合$P = \{x\in\mathbf{N}|x\leqslant\sqrt{2025}\},a = 2\sqrt{2}$,则 ( )
A.$a\in P$
B.$\{a\}\in P$
C.$\{a\}\subseteq P$
D.$a\notin P$

答案： 2.D

3.集合$\{a,b\}$的真子集的个数为______.

答案： 3.3 解法一 集合$\{a,b\}$的真子集为$\varnothing,\{a\},\{b\}$，有3个.
解法二 集合$\{a,b\}$有2个元素，则集合$\{a,b\}$的真子集的个数为$2^2 - 1 = 3$.

4.设$a,b\in\mathbf{R},P = \{2,a\},Q = \{-1,-b\}$,若$P = Q$,则$a - b =$______.

答案： 4.1 $\because P = Q$，$\therefore\begin{cases}a = -1\\-b = 2\end{cases}$，$\therefore a - b = -1 - (-2) = 1$.

5.已知集合$U = \{1,2,3,4,5,6,7\},A = \{2,4,5\},B = \{1,3,5,7\}$,则$A\cap(\complement_{U}B)=$______,$(\complement_{U}A)\cap(\complement_{U}B)=$______.

答案： 5.$\{2,4\}$ $\{6\}$ $\because\complement_U A = \{1,3,6,7\}$，$\complement_U B = \{2,4,6\}$，$\therefore A\cap(\complement_U B)=\{2,4,5\}\cap\{2,4,6\}=\{2,4\}$，$(\complement_U A)\cap(\complement_U B)=\{1,3,6,7\}\cap\{2,4,6\}=\{6\}$.

例1 (1)[2022全国卷乙]设全集U = {1,2,3,4,5}，集合M满足∁UM = {1,3}，则 ( )
A. 2∈M
B. 3∈M
C. 4 M
D. 5 M

答案： 例1
(1)A　由题意知M = {2,4,5},故选A.

例1 (2)[全国卷Ⅲ]已知集合A = {(x,y)|x,y∈N*,y≥x}，B = {(x,y)|x + y = 8}，则A∩B中元素的个数为 ( )
A. 2
B. 3
C. 4
D. 6

答案：
(2)C　由题意得,A∩B = {(1,7),(2,6),(3,5),(4,4)},所以A∩B中元素的个数为4,故选C.

训练1 (1)[多选/2024黑龙江模拟]已知集合A = {x|4ax² - 4(a + 2)x + 9 = 0}中只有一个元素，则实数a的可能取值为 ( )
A. 0
B. 1
C. 2
D. 4

答案： 高考帮训练1
(1)ABD　当a = 0时, - 8x + 9 = 0,解得x = $\frac{9}{8}$,所以A = {$\frac{9}{8}$},符合题意;当a≠0时,由题意,得Δ = [4(a + 2)]² - 4×4a×9 = 0,解得a = 1或a = 4.故选ABD.

训练1 (2)[多选/2023江苏省镇江中学模拟]已知集合A = {y|y = x² + 2}，集合B = {(x,y)|y = x² + 2}，下列关系正确的是 ( )
A. (1,3)∈B
B. (0,0) B
C. 0∈A
D. A = B

答案： 高考帮
(2)AB
∵集合A = {y|y≥2} = [2,+∞),集合B = {(x,y)|y = x² + 2}是由抛物线y = x² + 2上的点组成的集合,
∴ AB正确,CD错误,故选AB.

训练1 (3)已知集合A = {0,m,m² - 5m + 6}，且2∈A，则实数m的值为________.

答案： 高考帮
(3)1或4　因为A = {0,m,m² - 5m + 6},2∈A,所以m = 2或m² - 5m + 6 = 2.当m = 2时,m² - 5m + 6 = 0,不满足集合中元素互异性,所以m = 2不符合题意.当m² - 5m + 6 = 2时,m = 1或m = 4,若m = 1,A = {0,1,2}符合题意;若m = 4,A = {0,4,2}符合题意.所以实数m的值为1或4.

例2 (1)[2023新高考卷Ⅱ]设集合A = {0, - a}，B = {1,a - 2,2a - 2}，若A⊆B，则a = ( )
A. 2
B. 1
C. $\frac{2}{3}$
D. - 1

答案： 高考帮例2
(1)B　依题意,有a - 2 = 0或2a - 2 = 0.当a - 2 = 0时,解得a = 2,此时A = {0, - 2},B = {1,0,2},不满足A⊆B;当2a - 2 = 0时,解得a = 1,此时A = {0, - 1},B = { - 1,0,1},满足A⊆B.所以a = 1,故选B.

例2 (2)[2024山西太原模拟]满足条件{1,2}⊆A⫋{1,2,3,4,5}的集合A的个数是 ( )
A. 5
B. 6
C. 7
D. 8

答案： 高考帮
(2)C　解法一　因为集合{1,2}⊆A⫋{1,2,3,4,5},所以集合A可以是{1,2},{1,2,3},{1,2,4},{1,2,5},{1,2,3,4},{1,2,3,5},{1,2,4,5},共7个.故选C.
解法二　问题等价于求集合{3,4,5}的真子集的个数,则共有2³ - 1 = 7个.故选C.