A) M=N P B) M N=P C) M N P D) N P M
Analysis 1: Start with judging the uniqueness and difference of elements.
Answer 1: For the set m: {x | x =, m? z }； For the set n: {x | x =, n? Z}
For the set p: {x | x =, p? Z}, because 3(n- 1)+ 1 and 3p+ 1 both represent numbers divided by 3, and 6m+ 1 represents numbers divided by 6, so choose Mn = p.
Analysis 2: Simply enumerate the elements in the collection.
Answer 2: M={? , ,? }，N={？ , , , ,? }，P={？ , , ,? }, then don't rush to judge the relationship between the three sets, you should analyze the different elements in each set.
= ? n，？ n，？ M N, again = M,? M N，
= P，？ N P and n,? P N, so P=N, so choose B.
Comments: Because the second way of thinking only stays in the initial inductive hypothesis and does not solve the problem theoretically, I advocate the first way of thinking, but the second way of thinking is easy to handle.
Variant: Set, and then (b)
A.M=N B.M N C.N M D。
Solution:
When 2k+ 1 is an odd number and k+2 is an integer. Choose B.
Example 2 defines the set A*B={x|x? A and x B}, if a = {1, 3,5,7} and b = {2 2,3,5}, then the number of subsets of A*B is
1 B)2 C)3 D)4
Analysis: To determine the number of subsets of set A*B, first determine the number of elements, and then use the formula: set A={a 1, a2,? , an} has a subset 2n that requires a solution.
Answer: ∫a* B = {x | x? A and x B},? A*B={ 1, 7} has two elements, so A*B * * has 22 subsets. Choose D.
Variant 1: known nonempty set M {1, 2,3,4,5}, if a? M, then 6? Answer? M, then the set number m is
A) Five b) Six c) Seven d) Eight.
Variant 2: Given {a, b} A {a, b, c, d, e}, find the set a. 。
Solution: A known set must contain elements A and B.
Set a can be {a, b}, {a, b, c}, {a, b, d}, {a, b, e}, {a, b, c, d}, {a, b, c, e}, {a, b, d, e}.
Comment on the number of set A in this question is actually the number of proper subset of set {c, d, e}, so * * * has one.
Example 3 sets a = {x | x2+px+q = 0} and b = {x | x2? 4x+r=0}, and a? B={ 1}，A？ B={？ 2, 1, 3}, the values of real numbers p, q, r q, r.
Answer: a? B={ 1}？ 1? b？ 12? 4? 1+r=0，r=3。
? B={x|x2？ 4x+r=0}={ 1，3}，∫A？ B={？ 2, 1,3},? 2 B，2？ A
∵A？ B={ 1}？ 1? Answer? The two roots of the equation x2+px+q=0 are -2 and 1.
Variant: known set a = {x | x2+bx+c = 0}, b = {x | x2+MX+6 = 0}, a? B={2}，A？ B=B, the values of real numbers b, c and m.
Solution: a? B={2}？ 1? b？ 22+m？ 2+6=0，m=-5
? B={x|x2-5x+6=0}={2，3 }∫A？ B=B？
A again? B={2}？ A={2}？ b=-(2+2)=4，c=2？ 2=4
? b=-4，c=4，m=-5
Example 4 the known set a = {x | (x-1) (x+1) (x+2) >; 0}, set b satisfies: a? B = {x | x & gt-2} and a? B={x| 1
Analysis: simplify the set a first, and then change it from a? B and a? B determines which elements on the number axis belong to B and which elements do not belong to B.
Answer: A={x|-2 1}. By a? B={x| 1-2} indicates [- 1, 1] B, while (-? ,-2)? B=ф.
Combining the above categories, there is B={x|- 1? x？ 5}
variant 1:If A = { x | x3+2 x2-8x & gt； 0}，B={x|x2+ax+b？ 0}, called a? B = { x | x & gt-4}，A？ B=？ , A, B. (Answer: a=-2, b=0)
Comments: When solving a kind of set problem about inequality solution set, we should pay attention to using the method of combining numbers and shapes to do the number axis to solve it.
Variant 2: Let m = {x | x2-2x-3 = 0} and n = {x | ax- 1 = 0}. If m? N=N, find the set of all A's that satisfy the conditions.
Answer: M={- 1 3}, ∫M? N=N，？ n·M
① When ax- 1=0 has no solution. a=0 ②
Synthesis ① ②: The required set is {- 1, 0,}
Example 5 Given a set, the domain of the function y=log2(ax2-2x+2) is Q. If p? Q, the range of the real number a.
Analysis: Firstly, the original problem is transformed into the inequality AX2-2x+2 >; 0, and then solve it by parameter separation.
Answer: (1) If yes, there is a solution.
When the time is right,
So a & gt-4, so the range of a is
Variant: if the equation about x has real roots, find the range of the number a.
1, given the complete set U = {1, 2, 3, 4, 5, 6, 7, 8}, A= {3, 4, 5}, B= {1, 3, 6}, then the set {2, 7,
2. If there is only one element in the set A={x|ax2+2x+ 1=0}, the value of a is ().
A.0 B.0 or1C.1D. Not sure.
3. Let the set A={x| 1.
A.{a|a？ 2} B.{a|a？ 1} C.{a|a？ 1}.D.{a|a？ 2}.
5. The number of sets m satisfying {1, 2,3} m {1,2,3,4,5,6} is ().
A.8 B.7 C.6 D.5
6. Let A={a2, a+ 1,-1}, B={2a- 1, | a-2 |, 3a2+4}, a? B={- 1}, then the value of a is ().
A.- 1 B.0 or1c.2d.0.
7. Complete set I=N and set A={x|x=2n, n? N}，B={x|x=4n，n？ N}, then ()
A.I=A？ B B.I=()？ b C I = A？ ()D.I=()？ ( )
8. Let the set M=, then ()
A.M =N B. M N C.M N D. N
9. Let A={x|x=2n+ 1, n? Z}，B={y|y=4k？ 1，k？ Z}, then the relationship between A and B is ()
A.A B B A B C A = B D A？ B
10. Let U={ 1, 2,3,4,5}, if a? B={2}，(UA)？ B={4}，(UA)？ (UB) = {1, 5}, then the following conclusion is correct ().
A.3 A and 3 B B.3 B and 3? A C.3 A and 3? B D.3？ A and 3? B
2. Fill in the blanks (5 points? 5=25 points)
1 1. There are 55 students in a class, including 34 music lovers and 43 sports lovers. There are 4 people who like neither sports nor music, so there are people who like both sports and music in the class.
12. Let the set U={(x, y)|y=3x- 1} and A={(x, y)| =3}, then A=.
13. Set M={y∣y= x2+1, x? R},N={y∣ y=5- x2，x？ R}, then m? N=_ __。
14. Set M={a|? N, and a? Z}, the set M=_ is expressed by enumeration.
15, known set a = {- 1, 1}, b = {x | MX = 1}, a? B=A, then the value of m is
Three. Solve the problem. 10+ 10+ 10=30
16. let the set a = {x, x2 x, y2- 1}, b = {0, | x |, y} and A=B, and find the values of x and y.
17. let the set A={x|x2+4x=0}, b = {x | x2+2 (a+1) x+a2-1= 0}, a? B=B, the value of the real number a.
18. let A = {x | x2-ax+A2- 19 = 0}, B = {x | x2-5x+6 = 0}, and C = {x | x2+2x-8 = 0}.
(1) if a? B=A？ B, find the value of a;
(2) if a? b，A？ C=, find the value of a.
19. (The full mark of this small question is 10) The known set A = {x | x2-3x+2 = 0}, and B = {x | x2-ax+3a-5 = 0}. If a? B=B, the range of the real number A. 。
20, known as A={x|x2+3x+2? 0}，B = { x | mx2-4x+m- 1 & gt； 0，m？ R}, if a? B=？ What about a? B=A, find the range of m.
2 1, known set, B={x|2.
Reference answer
developing country
26 {( 1, 2)} r {4,3,2,-1} 1 or-1 or 0.
16、x=- 1 y=- 1
17, solution: A={0, -4} again.
(1) If B=, then,
(2) If B={0}, substitute x=0 into the equation to get a= When a= 1, B=
(3) If B={-4}, substitute x=-4 to get a= 1 or a=7.
When a= 1, B={0, -4}? {-4},? Answer? 1.
When a=7, B={-4,-12}? {-4}, ? Answer? 7.
(4) If B={0, -4}, then a= 1, and when a= 1, B={0, -4},? a= 1
To sum up: A.
18,. solution: from the known, we get b = {2 2,3} and c = {2 2,4}.
( 1)∵A？ B=A？ b，？ A=B
Therefore, 2,3 are two roots of the unary quadratic equation x2-ax+a2- 19=0, which can be known from Vieta's theorem:
The solution is a=5.
2 by a? b？ , another one? C=，3？ A, 2 A, -4 A, from 3? One,
Get 32-3a+a2- 19=0, get a=5 or a=-2?
When a=5, a = {x | x2-5x+6 = 0} = {2,3}, which contradicts 2 A;
When a=-2, a = {x | x2+2x- 15 = 0} = {3, -5}, which is consistent with the meaning of the question.
? a=-2。
19, solution: A={x|x2-3x+2=0}={ 1, 2},
From x2-ax+3a-5=0, you know? = a2-4(3a-5)= a2- 12a+20 =(a-2)(a- 10)。
(1) When 2
(2) when a? 2 or a? 10,0, what about b? .
If x= 1, then 1-a+3a-5=0 and a=2,
At this time, b = {x | x2-2x+1= 0} = {1} a;
If x=2, then 4-2a+3a-5=0, a= 1,
At this time, b = {2,-1} a.
To sum up, when 2? A< is in 10, and there is an a? B=B。
20. solution: obtained from the known A={x|x2+3x+2}. (1)∵A is not empty,? b =； (2)∵A={x|x }？ On the other hand, the above (2) is not valid, otherwise, it contradicts the topic. From the above analysis, we know that B=. By combining the known B= and B=, it holds for all x constants, so the range of values is
2 1、∫A = { x |(x- 1)(x+2)？ 0}={x|-2？ x？ 1},
B={x| 1
∵，(A？ b)？ C=R，
? Complete works u = R.
? The solution of is X.
That is, the two roots of the equation are x=-2 and x=3 respectively.
From the relationship between the roots and coefficients of a quadratic equation in one variable, we get
b=-(-2+3)=- 1，c=(-2)？ 3=-6
The knowledge point about set in senior high school mathematics (1) Set is a basic concept in mathematics, so-called? Basic concepts? It can't be defined by other concepts, but can only be understood by describing its characteristics and properties.
(2) The set must be regarded as a whole. For example, by? Students in our class? A set A is a whole, that is, a class group;
(3) The objects that make up a set must be? Are you sure? And then what? Different? Yes
(4) Pay attention to the composition of the scenery? Object? On the one hand, any definite object can form a set, such as people, animals, numbers, equations, inequalities and so on. A that can be used as an object of the collection; On the other hand, the set itself can also be used as the object of the set, such as the above-mentioned set A, which can be used as the object of the set. What classes do we have in Grade One? The elements of set B.
1, certainty:
That is, given a set, whether each object is an element in the set must have a clear criterion, and there can be no ambiguity.
For example, students who are taller, people who run faster, and people with very high quality, do all the objects described above constitute a collection?
Because these expressions cannot find a clear criterion, the objects they describe cannot form a set.
2. Heterogeneity:
The elements in the collection are different from each other. If two or more identical elements appear, they can only be counted as one, and the elements in the collection are not repeated.
3. Disorder:
In other words, the elements in the collection have no order. As long as the elements of two sets have the same Quan Wang, then the two sets are the same set.
Knowledge interpretation:
The elements in a set must be deterministic, different and unordered. On the other hand, if a group of objects does not have these three properties, then this group of objects cannot form a set. These three characteristics of elements in a set are the basis for us to judge whether a group of objects can form a set.
When solving problems related to set, we should make full use of it? Three natures? To analyze and solve, that is, on the one hand, to use the collection of elements? Three natures? Find a solution? Breakthrough? ; On the other hand, when solving the problem, we should pay attention to checking whether the elements are satisfied. Three natures? .
The following are several groups of numbers commonly used in high school mathematics and their corresponding letters, which are easy to be confused in the learning process:
Set of rational numbers (n), set of integers (z), set of rational numbers (q), set of real numbers (r)
In fact, we just need to list them in turn according to the scope they represent, and then memorize four English letters, which is very concise and efficient.
note:
(1) natural number set is the same as the non-negative integer set, that is, natural number set contains the number 0.
(2) The set excluding 0 in the non-negative integer set is recorded as N* or N+, and Q+ represents a non-negative rational number.
1, the concept of set
Set is an undefined original concept in set theory, and the concept of set is described in the textbook: Generally speaking, if some identifiable different objects are regarded as a whole, it is said that the whole is a set (or set) composed of all these objects? . To understand this sentence, we should grasp four key words: object, certainty, difference and wholeness.
Objects, that is, elements in the collection. A collection is uniquely identified by its elements.
The whole set does not study a single object, but focuses on the whole of these objects.
Certainty ―― Certainty of set elements ―― Elements and sets? Subordinate? Relationship.
Difference-the mutual difference between elements of a set.
2. The meaning of finite set, infinite set and empty set.
Finite sets and infinite sets are for non-empty sets. It is not difficult for us to understand.
We call a set without any elements an empty set, remember? . Think about it before you think about it? 0 and and and {？ }? The relationship.
Several commonly used number sets N, N*, N+, Z, Q and R should be remembered.
3. Representation method of set
The expression of (1) enumeration method is easy to master, and not all sets can be expressed by enumeration method. Students need to know three sets that can be expressed by enumeration:
① finite sets with few elements, such as {0, 1, 8}
(2) Finite sets with many elements but certain regularity, such as {1, 2, 3,? , 100}
(3) Show some laws of infinite sets, such as {1, 2, 3,? ，n，？ }
Pay attention to the difference between a and {a}
Pay attention to the enumeration method to represent sets, set elements? Disorder? .
(2) The key of the characterization method is to classify the studied sets. Characteristic nature? Just find the right one and express it appropriately. But the point is also difficult. Just practice more when you study. Besides, find out? Representing elements? It's also important. For example, {x|y=x2}, {y|y=x2} and {(x, y)|y=x2} are three different sets.
4, the relationship between sets