Four answers.
The number of subsets in the set of Secret elements is 2 n, the number of proper subset is 2 n- 1, and the number of nonempty proper subset is 2 n-2.
2. Given the set A = {(x, y) 乸 0≤x≤ 1, y = 0} and b = {(x, y) 乸 y = ax+b}, then all real numbers of A∩B= empty sets A and B should satisfy the condition _.
Answer: b>0 and A+B >; 0 or not b
Process: Because A and B are both point sets, A∩B= empty set means that they have no common points, that is, the straight line y=ax+b has no common points in [0, 1] and y=0, so f(0)f( 1)>0, (that is, when X is [0 0 or not b
3. Let A={x|x=3n+ 1, n belongs to Z}, B={x|x=3n-2, n belongs to Z}, C={x|x=6n+3, n belongs to Z}.
(1) If C belongs to C, then the proof that A belongs to A and B belongs to B must exist, and C = A-B;
(2) Does a+b necessarily belong to C for any A belongs to A and B belongs to B? Why?
Answer: (1) A-B = 3n+1-3k+2 = 3 (n-k)+3 = 3m+3, where k and m are integers. So 6n+3=3*2n+3 can obviously represent the form of 3m+3, (2) a+b = 3n+1+3k-2 = 3 (n+k)-1= 3z-1,which is a number divided by 3.
4. let the complete set u = {x | x >- 10} and a = {x |-2 < x ≤ 4}, then cua = _ _ _ _ _ _ _ _ _
Let A={x|x? -ax+a？ - 19=0}，B={ x| x？ -5x+6=0}，C={x|x？ +2x-8=0}A∩B≠ empty set, A∩C= empty set, find the value of A.
(1) cua = {x |-10 ≤ x ≤-2 and x >；; 4}
(2) Because B = {2 2,3} C = [-4,2} and because A∩B≠ is an empty set and A∩C= is an empty set, there must be an element in A that is 3, but there cannot be an element that is 2, so bringing 3 into A is 9-3a+a? -19=0 gives a=5 or a=-2, but when a=5, a is {x| x? The two solutions of -5x+6=0} are 2 and 3, which does not conform to the empty set A∩C=, so A can't be 5, so the final answer is a=-2.
5.A={ diamond}, B={ rectangle} looking for A∩B?
Answer: Square.
6.A={x|x？ -4X+2m+6=0，x∈R}。 If A∩R≠ is empty, find the range of m.
Answer: solution: the intersection of a and r is not equal to an empty set.
The equation x*2-4x+2m+6=0 has a real root.
That is, (-4) * 2-4 (2m+6) >: =0
The solution is 16-8m-24 >: =0 for m.
7. There are 50 students in Class A, Grade One, Middle School A, 25 students in the math group and 32 students in the physics group. Find out the maximum and minimum number of students who join the math group and the physics group.
Answer: solution: let the students who participate in mathematics and physics be x people;
Then x is 25 at most, which is X.
The number of people in math and physics is 32+25=57.
Number of students 50 57-50=7
In other words, at least five people must participate in two activities.
So put these together: 7
8.A={x|x？ -3x+2=0}, B={x|ax-2=0} and A∪B=A, the set c consisting of the values of real number a.
Answer: Solution: The solution of the equation starts from the set A and x= 1 or x=2, that is, A={ 1, 2}
B={x/ax-2=0} is set because
Set a and set b = a.
Then when x= 1, substitute ax-2=0 to get a=2.
When x=2, a= 1.
So the set C={a/a= 1 or a=2}
9. know the complete set I, set a and b, and find CIA ∩ B.
( 1).I={ x，y)}，A = {(x，y) | y-4/x-2 = 3}，B = {(x，y) | y = 3x-2}
(2).I=R, A={a| quadratic equation ax? -x+ 1=0 has a real root} B={a| quadratic equation x? -ax+ 1=0 has real roots}
Answer: (1) solution: A = {(x, y)/y-2/x-2 = 3} = {(x, y)/y = 3x-2, and x is not equal to 2}
B={(x，y)/y=3x-2}
A cross B={(x, y)/x=2, y=4}
(2) solution: A={a/ quadratic equation ax*2-x+ 1=0 has real roots}
That is, (-1)-4a >: =0 squared.
A<= one quarter
B={a/ Quadratic equation x*2-ax+ 1=0 has real roots}
Get a & gt=2 or a.
So the complement set of A in I should satisfy a>= 1/4.
The intersection of it and B. The intersection of CIA and B should satisfy a>= 1/4 and a & gt=2 or A.
Their cross is a > = 2, then the corresponding set about A is
CIA to B = { a/a & gt； =2}
10. The known set A = {(x, y) | 2x-y = 0}. B = {(x，y) | 3x+y = 0}。 C = {(x，y) | 2x-y = 3}，
Find A∩B. A∩C .(A∩B)∩( B∩C).
Answer: intersection b is the intersection of two straight lines a and B.
When solving binary linear equations, X=0 and y=0.
So a crosses B={(x, y)|x=0, y=0}
From a to c
The two straight lines A and C are parallel and have no intersection.
So a goes through C= empty set.
(a to b) and (b to c)
The intersection of b and c is (3/5, -9/5).
So b crosses C={(x, y)|x=3/5, y=-9/5}
A cross B={(x, y)|x=0, y=0}
So (a crosses b) and (b crosses C)={(x, y) | (0,0), (3/5, -9/5)}
1 1. The known function f (x) =1-x/1+x]. Find:
( 1)f(a)+ 1(a≦- 1)
(2)f(a+ 1)(a≦-2)
Answer: (1) f (a)+1= [1-a/1+a]+1= 2/1+a.
②f(a+ 1)=[ 1-(a+ 1)/ 1+(a+ 1)]+ 1 = 2/2+a
12. It is known that m = {x | x = m+ 1/6, m ∈ z}, n = {x | x = n/2- 1/3, n ∈ z} M, n, n.
Answer: m takes all integers.
K/2 can be divided into decimals or integers.
So m is really included in p.
n/2- 1/3 =(n- 1)/2+ 1/2- 1/3 =(n- 1)/2+ 1/6
(n- 1)/2 is the same as k/2.
So N = P
Attachment: test questions
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Class (in English)
First, multiple choice questions
1. let a = {x | 1 < x < 2 =, and B = {x | x < a =. If so, the range of a is ().
(A)[2，+∞] (B)(-∞， 1)
(C)[ 1，+∞] (C)(-∞，2]
2. It means that the set of shaded parts in the right picture is ().
A∪B(B)A∪B
(C) (D)
3. if the set A={ 1, 3, x}, A∪B={ 1, 3, x}, then the number of real numbers x satisfying the conditions is
1 (B)2
(C)3 (D)4
4. If it is set, it must be an empty set ().
(A) (B)
A∩B
5. Let A={x|x≠0, x∈R }∩{ y | y≠2, y∈R}, and B=(-∞, 0) ∩ (0,2) ∩ (2).
(A)B∈A (B)
A=B
Second, fill in the blanks
1. The number of satisfaction sets A is _ _ _ _ _ _ _ _.
2. Let the complete set I = {2 2,3,5} and A={|a-5| 2}, then the value of A is _ _ _ _ _ _ _ _.
3. Suppose that given a ∪ b = {-2, 1, 5} and A∪B = {-2}, then p = _ _ _ _ _ _ _ _ _ _ _ _ _ _.
4. Let the complete set be I={(x, y)|x, y∈R}, set,
N={(x, y)|y≠x+ 1}, which is equal to _ _ _ _ _ _.
5. Given that set A has 10 elements, set B has 8 elements and set A∩B has 4 elements, then the set
A∪B has _ _ _ _ _ _ elements.
Third, answer questions.
1. Let A={x|ax+ 1=0}. If so, find the value of the number A..
2. Given that,,, and, a ∩ c = φ, find the value of a..
3. Collectively and known, find the range of the real number p..
Answer or prompt
collection
One,
Chronology of a.d.
Second,
1.7;
2.2 or 8;
3.- 1,-3,- 10;
4.{(2,3)};
5. 14.
Third,
1.a = 0 or-1 or;
2 . a =-2；
3.(-4,+∞)。
Can you solve your problem?