High Score Two questions for high school collections

x∈m，m=

So x=0 or x=1

So p= so p=m

a= 4≤2x≤2a

1≤2x+3≤2a+3

i.e. b = = 1 when a 0.

2≤x≤aa²≤x²≤4

i.e. c== because c is a true subset of b, it is related.

a²≥-1，4≤2a+3

Solution, A 1 2, contradict A 0, discard.

2. When 02.

2≤x≤a0≤x²≤a²

i.e. c== because c is a true subset of b, it is related.

0≥-1，a²≤2a+3

a²-2a-3≤0

a-3)(a+1)≤0

Solution, -1 a 3

Therefore, 2 synthesizes the three cases and takes their union to obtain, 1 2 a 3 supplements: a cannot be less than -2, otherwise at this time a = b = c = empty set.

1.It is known that m= p=.Find the relationship between m and p I think it's m=p but I don't know how to prove it, please help prove p=

m 2.Gather.

a= b=c= and c is a true subset of b. Find the range of values of the real number a.

c is the true subset of b b=

c= when a>2.

When 0< a<=2.

When -2 = empty set a<-2.

A is less than -2 or a is greater than or equal to, less than or equal to 3 pairs.

p is a subset of m, and p has three cases, 2

The minimum in b is -1, the downline contains c, only look at the upline, the upline in b is 2a+3, and (-2) in c 2=4 and a 2

There are 2a+3>=4 and 2a+3>=a 2, and a is greater than or equal to 3, less than or equal to 3, a cannot be less than -2, and the range a in a is greater than -2

1.Because element x in the p set belongs to m

So p=, p=, p }, p=

So p is contained in m, i.e., p is a subset of m.

Just let the equation have a unique solution, which is =0

b^2 - 4ac = 2^2 - 4 * a * 1 = 0

A = 1 element is x = 1

The meaning of the function is the same x, only the only y corresponds to it, where x is equivalent to a and a has two elements, so the range c is divided into two cases, c is a single element set, and there are three kinds.

c is a two-element set, and there are three kinds.

The answer is c

Take a look at the map and be patient whenever you want to figure it out.

The number of ranges is actually the number of non-empty subsets of b.

b has 3 elements.

So the non-empty subset has 2 -1 = 7 elective d

Solution: This problem is very simple, you might as well think about it from another angle, it is a function from a to b, then the value of the function must be in the set of b, and the case that requires the range c is to find the non-empty subset of b, because the value of the function cannot be empty.

So, there are 7 non-empty subsets of b, and the answer is D, if there is anything you don't understand, ask me on hi.

There are eight types of value ranges, such as:

And so on, i.e. for every element in a, there is a correspondence in element b. Since the title doesn't say how to correspond to it, the range can be a true subset of any set of elements in b = >.

If there are n elements, the number of true subsets is 2 n-1.

1. A 2-A≠2A is fine, that is: a≠0, a≠3

2. a={-3,0,1,2,4,5,6,9}, by enumeration.

It is obtained by f(x-y)=f(x)-y(2x-y+1) and f(0)=1 so that x=0.

f(0-y)=f(0)-y(2*0-y+1)=1-y(-y+1)=1+y(y-1)=y^2-y+1

So f(-y)=y 2-y+1, so u=-y then f(u)=u 2+u+1

So f(x)=x2+x+1

a, young people do not have a definite boundary, (how old they are, they must be young), there is no certainty, b, there is no certainty, how small positive numbers are.

dWhich numbers do these numbers refer to, 1,0,5,1|2

4|6, root number 1|Although there are five different numbers, the set they form may only have four of them, and it doesn't have to be five.

And these numbers don't make a set, because 2|3 = 4l6, 1l2 = root number 1|4. There can be no two identical numbers in the set, which contradicts each other.

The answer to c reflects the disordered nature of the set.