Given the set a = {(x, y) y = f (x), X belongs to [a, b]}, B = {(x, y) x = 1}, then the number of elements contained in a intersection B is?

Given the set a = {(x, y) y = f (x), X belongs to [a, b]}, B = {(x, y) x = 1}, then the number of elements contained in a intersection B is?

If a > 1, or B

What is the number of elements of the set a = {x | y = 12 / (x + 3), X ∈ Z, y ∈ Z}
The answer is 12, but when x = - 3, 12 / (x + 3) is meaningless

Not counting x = - 3, there are 12
x1=-2
x2=-1
x3=0
x4=1
x5=3
x6=9
x7=-4
x8=-5
x9=-6
x10=-7
x11=-9
x12=-15

The number of elements of set a = {x belongs to Z | y = 12 / x + 3, y belongs to Z}
Detailed analysis

X = 12 / (Y-3) x, y is an integer. The divisor of 12 has 1,2,3,4,6,12. If | Y-3 | = 1,2,3,4,6,12, then x = - 12, - 2, - 3, - 6, - 4, - 2, - 1,1,2,3,4,6,12. Then the number of elements of a is 12

The number of elements in the set a = {x ︱ y = 12 / x + 3, X ∈ Z, y ∈ Z} is 12, but why are there 12 troublesome writing processes,
That's the problem
Write down what you want to know,

y=12/x+3
x1=-2
x2=-1
x3=0
x4=1
x5=3
x6=9
x7=-4
x8=-5
x9=-6
x10=-7
x11=-9
x12=-15

A = {x y = 12 / (x + 3), X ∈ Z, y ∈ Z}, then the number of elements of set a is? How to solve,

Z is an integer, one by one, 0,1,3,9
I don't know whether Z also includes negative numbers

Given the set a = {1,2,3,4}, B = {(x, y)} | x ∈ a, y ∈ a, XY ∈ a}, then the number of all proper subsets of B is ()
A.512 B.256 C.255 D.254
How can there be so many?
(x, y) there are only 16 in all

When x = 1, y can be 1,2,3,4.4
When x = 2, y can be 1,2.2
When x = 3, y can take 1
When x = 4, y can take 1
That is to say, there are 4 + 2 + 1 + 1 = 8 elements in B
The number of all proper subsets of B is
C(8,1)+C(8,2)+.+C(8,8)
=2^8-C(8,0)
=255

Given that the solution set of inequality ax & # 178; + BX-1 ＞ 0 is {x ｜ 3 ＜ x ＜ 4}, find the value of real numbers a and B

The solution set of ax & # 178; + BX-1 > 0 is {x 3 < x 4}
a

Given the set a = {x ^ 2 + BX + C = 0}, B = {x ^ 2 + MX + 6 = 0}, and a ∪ B = B, a ∩ B = {2}, find the value of real number B, C, M

If a intersects B = {2}, then 2 is the common root of equation a and B. thus, M = - 5, and the root of equation B is 3,2
Because a is not b = B, then a is either a subset of B or equal to B. If a intersects B with only one element, then a is not equal to B. then there is only one equation in A. then B ^ 2-4c = 0,4 + 2B + C = 0; the simultaneous solution is b = - 4, C = 4

It is known that for any real number x, KX & # 178; - 2x + k > 0, the value range of real number k is obtained?

Because the original formula is always greater than 0
therefore
There is no real solution
therefore
b²-4ac1
k> 1 or K0
So K

If the solution of equation x2-5x + 6 = 0 and equation x2-x-2 = 0 is m, then the number of elements in M is ()
A. 1B. 2C. 3D. 4

∵ the solution of equation x2-5x + 6 = 0 is X1 = 2, X2 = 3 ∪ the set of solution of equation x2-5x + 6 = 0 is {2, 3}. Similarly, the set of solution of equation x2-x-2 = 0 is {- 1, 2}. Therefore, the set M = {2, 3} ∪ {- 1, 2} = {- 1, 2, 3} has three elements in total