Given that X and y satisfy 2x − y ≥ 2x + y ≤ 2Y ≥ a (x − 1), and z = x + y can get the minimum value, then the value range of real number a is () A. a＜-1B. a≥2C. -1≤a≤0D. -1≤a＜2

Given that X and y satisfy 2x − y ≥ 2x + y ≤ 2Y ≥ a (x − 1), and z = x + y can get the minimum value, then the value range of real number a is () A. a＜-1B. a≥2C. -1≤a≤0D. -1≤a＜2

Let L: z = x + y, the slope of line l be - 1, and observe the intercept change of line L on y-axis. The more the line L moves down, the smaller the intercept of line L on y-axis, and the corresponding objective function Z becomes smaller. The plane region represented by the system of inequalities 2x − y ≥ 2x + y ≤ 2 is below the line 2x-y = 2, and below the line x + y = 2. ∵ the line y = a (x-1) represents the passing point a (1, 0) The plane region represented by the inequality system 2x − y ≥ 2x + y ≤ 2Y ≥ a (x − 1) is above the line y = a (x-1). It can be concluded that when the slope of the line y = a (x-1) is greater than - 1 and less than the slope of the line 2x-y = 2, the plane region represented by the inequality system 2x − y ≥ 2x + y ≤ 2Y ≥ a (x − 1) is the △ ABC and its interior as shown in the figure, and the objective function z = x + y achieves the best result at point a (1, 0) If the small value is 1, the value range of a is - 1 ＜ a ＜ 2; if a = - 1, the plane area represented by the inequality group in the question is the part between two parallel lines with slope equal to - 1 and below the straight line 2x-y = 2, z = x + y can also obtain the minimum value 1 at point a (1,0), and the value range of real number a is - 1 ≤ a ＜ 2

If y = - x ^ 2 + 2x + 2 has a maximum value of 3 on [- 1, M], and a minimum value of - 1, then the value range of real number m?

Draw an image based on the function,
The function has a downward opening and vertex coordinates (1,3)
When y = - 1, x = - 1 or 3,
Therefore, the range of M is [1,3]

Given that real numbers x and y satisfy x 2 + y 2 = 1, the value range of Y + 2x + 1 is obtained

Let k = y − (− 2) x − (− 1), then K can be regarded as the slope of the line from the moving point P on the circle x2 + y2 = 1 to the fixed point a (- 1, - 2) and the slope when it is tangent. Since the line is tangent to the circle at this time, let the equation of the line be y + 2 = K (x + 1), and change it into a straight line, the general formula is kx-y + K-2 = 0