Given that the function f (x) = x2 + (a + 2) x + B satisfies f (- 1) = - 2 (1) if the equation f (x) = 2x has a unique solution, find the value of real numbers a and B; (2) if the function f (x) is not a monotone function in the interval [- 2,2], find the value range of real numbers a

Given that the function f (x) = x2 + (a + 2) x + B satisfies f (- 1) = - 2 (1) if the equation f (x) = 2x has a unique solution, find the value of real numbers a and B; (2) if the function f (x) is not a monotone function in the interval [- 2,2], find the value range of real numbers a

(1) ∵ f (- 1) = - 2 ∵ 1 - (a + 2) + B = - 2, that is, B-A = - 1 & nbsp; & nbsp; ①∵ the equation f (x) = 2x has a unique solution, that is, X2 + ax + B = 0, and the unique solution ∵ = a2-4b = 0 & nbsp; & nbsp; & nbsp; ② A = 2 and B = 1 can be obtained from (1) and (2). From (1), it can be seen that B = A-1  f (x) = x2 + (a + 2) x + B = x2 + (a + 2) x + A-1, and its axis of symmetry is x = - A + 22 ∵ function f (x) is not monotone function in the interval [- 2, 2]. The solution of - 6 ≤ a ≤ 2 ∵ real number a is - 6 ≤ a ≤ 2