Given that the real number x y satisfies the constraint condition x > = 1, Y > = 0, X-Y > = 0, then the value range of Z = (Y-1) / X is? A [- 1,0] B (negative infinity, 0] C [- 1, positive infinity) d [- 1,1)

D draw a plane region satisfying the constraint condition: {x ≥ 1 {y ≥ 0 {X-Y ≥ 0}. The geometric meaning of Z = (Y-1) / X indicates the slope of the line between the point in the region and (0,1) point. When x = 1, y = 0, z = - 1 straight line is parallel to X-Y = 0, z = 1 ∧ z = (Y-1) / X is [- 1,1]

Let real numbers x and y satisfy {x-y-2 ≤ 0, x + 2y-5 ≥ 0, Y-2 ≤ 0. (1) find the value range of Z = x + y; (2) find the value range of u = (x + y) / X

Using linear programming, we can get that the value range of (1) z = x + y is 3

If the real numbers x and y satisfy x ^ 2 + y ^ 2-4x + 1 = 0, then the value range of x ^ 2 + y ^ 2 + 2Y is 0

X ^ 2 + y ^ 2-4x + 1 = 0 (X-2) ^ 2 + y ^ 2 = 3 let x = 2 + √ 3 * cos θ, y = √ 3 * sin θ, then x ^ 2 + y ^ 2 + 2Y = (2 + √ 3 * cos θ) ^ 2 + (√ 3 * sin θ) ^ 2 + 2 √ 3 * sin θ = 4 + 4 √ 3 * cos θ + 3 (COS θ) ^ 2 + 2 √ 3 * sin θ = 7 + 2 √ 3 * sin θ + 4 √ 3 * cos θ = 7 + 2 √ 15 * sin (θ + α) (where

It is known that a = {x | x + A is less than or equal to 1}, B = {x | x is less than or equal to 3}, if a is less than or equal to B, the value range of real number a is given

From x + a ≤ 1, the solution is x ≤ 1-A,
Because a is a subset of B,
So 1-A ≤ 3, a ≥ - 2,
That is, the value range of a is [- 2, + ∞)

Given that the real number x y satisfies the constraint condition x > = 1, Y > = 0, X-Y > = 0, then the value range of Z = (Y-1) / X is? A [- 1,0] B (negative infinity, 0] C [- 1, positive infinity) d [- 1,1)