Let a = [(x, y) | Y > = | X-2 | x > = 0], B = [(x, y) | y

Let a = [(x, y) | Y > = | X-2 | x > = 0], B = [(x, y) | y

Know from the title
Set a = {(x, y) | y ≥ | X-2 |, X ≥ 0}, B = {(x, y) | y ≤ - x + B},
If a ∩ B ≠ empty set
That is, y = - x + B intersects with the region {y ≥ | X-2 |, X ≥ 0}
We can see by drawing
b≥2

Set a = {(x, y) x ^ 2 + mx-y + 2 = 0}, set B = {(x, y) X-Y + 1 = 0 and 0 ≤ x ≤ 2}, and the intersection of a and B is not equal to the empty set, so the value range of M is obtained

This question means:
When x + 1 = x ^ 2 + MX + 2 is 0 = 0, we get m > = 3 or m

Let a = {x | - 1 ≤ X

If a ≤ - 1
Then the interval of B is on the left side of A
Intersection is an empty set
So the intersection of a and B is not an empty set
Then a ＞ - 1

Let a = {x | - 1 less than or equal to x less than 2}, B = {x | - x less than a}, if a intersects B not equal to an empty set, find the value range of A
Why the answer is a ＞ - 1 a ＜ 2 why not

Graphically, when a