If the quadratic function y = x2 + MX + (M + 3) has two different zeros, then the value range of M is () A. (-2,6)B. [-2,6]C. {-2,6}D. (-∞,-2)∪(6,+∞)
∵ the quadratic function y = x2 + MX + (M + 3) has two different zeros, namely M2-4 (M + 3) > 0. The solution is m ∈ (- ∞, - 2) ∪ (6, + ∞), so D is chosen
RELATED INFORMATIONS
- 1. If the quadratic function y = - x ^ 2 + MX + M-3 has two different zeros, then the value range of M is The first floor is not △ 0, is it Please be more specific on the second floor
- 2. Given that the zeros of the function f (x) = x ^ 2 + 2mx + 2m + 3 are x1, X2, find the minimum value of X1 ^ 2 + x2 ^ 2 It's a process!
- 3. The quadratic function y = - x2 + 6x 1 is known. The function is reduced to the form of y = a (X-H) 2 + K (where a, h, K are all constants and a is not equal to 0) by using the collocation method
- 4. If the independent variable x takes two different values X1 and X2, and the function values are equal, then X1 + X2 is equal
- 5. Let f (x) = e ^ x + SiNx. G (x) = 1 / 3x. If x 1 and x 2 belong to 0 to positive infinity, if f (x 1) = g (x 2), then x 1-x 2 is the minimum
- 6. It is known that the function f (x) = x (x ~ 2 + 2aX + 2A + 8) has three zeros x1, X2 and X3
- 7. Two real roots x1, X2 of 2 + 2x + T (t is all real numbers), if lx1l + lx2l = f (T), find the analytic expression of function f (T) (x ~ 2 is the root of x)
- 8. Mathematics: the following statements are correct: 1. X = 4 is the solution set of inequality X-1 > 2. 2. The solution of inequality - 2x > 4 is X
- 9. Given the set a = {x | x is less than or equal to 2, X belongs to R}, B = {x | x is greater than a} if AUB = R, then the value range of A
- 10. Let a = [(x, y) | Y > = | X-2 | x > = 0], B = [(x, y) | y
- 11. If the image of function f (x) = x ^ 2 + MX + 1 is symmetric with respect to x = 1, what is the value set of real number m?
- 12. U = {x | x = 2N-1, n ∈ n +, n less than or equal to 7}, a ∩ (cub) = {3,7}, CUA ∩ B = {9,13}, CUA ∩ cub = {1,11}, find a, B
- 13. Cosx = root sign 3 / 3, X belongs to the set of [0,2 π] finding X Such as the title
- 14. According to Pythagorean theorem, what is the equivalent of B & # 178; = 8 & # 178; - 4 & # 178;. B?
- 15. Can the value of y = x & # 178; - 2x-3 be equal to - 5? Why? Help!
- 16. If f (x) = 2x + 3 and G (x + 2) = f (x), then G (x) equals () A. 2x+1B. 2x-1C. 2x-3D. 2x+7
- 17. If f (x) = 1-2x, G [f (x)] = 1-x & # 178; (x is not equal to 0), then G (1 / 2) = g (1 / 2) brings 1 / 2 directly into G [f (x)] = 1-x & # 178/ If the function f (x) = 1-2x, G [f (x)] = 1-x & # 178 / / X & # 178; (x is not equal to 0), then G (1 / 2)= G (1 / 2) directly takes 1 / 2 into G [f (x)] = 1-x & # 178 / / X & # 178; why not take it into f (x) and then calculate x into another Isn't this the wrong way to get x
- 18. Given the function f (x) = a ^ x, G (x) = (a ^ 2x) + m, where m > 0, a > 0 and a is not equal to 1, when x belongs to [- 1,1], the sum of the maximum and minimum value of y = f (x) is 5 / 2 (1) Find the value of A; (2) If a > 1, note the function H (x) = g (x) - 2mf (x), and find the minimum value H (m) of H (x) when a belongs to [0,1]; (3) If a > 1 and the inequality | [f (x) - Mg (x)] / F (x) | is less than or equal to 1, then M is obtained when x belongs to [0,1]
- 19. If the equation a2x + (1 + LGM) ax + 1 = 0 (a > 0 and a ≠ 1) of X has a solution, then the value range of M is______ .
- 20. Given that XY ≠ 1, if 5x2 + 2011x + 9 = 0, 9y2 + 2011y + 5 = 0, then the value of XY is equal to () A. 59B. 95C. −20115D. −20119