Given the function f (x) = - x2 + ax + b2-b + 1, (a, B ∈ R), f (1-x) = f (1 + x) holds for any real number X. if f (x) > 0 holds for X ∈ [- 1, 1], then the value range of B is () A. - 1 ＜ B ＜ 0b. B ＞ 2C. B ＞ 2 or B ＜ - 1D. B ＜ - 1

Given the function f (x) = - x2 + ax + b2-b + 1, (a, B ∈ R), f (1-x) = f (1 + x) holds for any real number X. if f (x) > 0 holds for X ∈ [- 1, 1], then the value range of B is () A. - 1 ＜ B ＜ 0b. B ＞ 2C. B ＞ 2 or B ＜ - 1D. B ＜ - 1

∵ for any real number x, f (1-x) = f (1 + x) holds, the axis of symmetry of function f (x) is x = 1 = A2, the solution is a = 2, the axis of symmetry of function f (x) is x = 1, the opening is downward, the function f (x) is monotone increasing function on [- 1, 1], and f (x) > 0 holds, f (x) min = f (- 1) = b2-b-2 > 0, the solution is B < - 1 or b > 2, so choose C

Mathematical solution to quadratic inequality of one variable
The following univariate quadratic inequalities are solved and the solution set is represented by interval
(1) The square of 5 + 4x-x is more than 0;
(2) The square of 2x + 3x + 10

（1)5+4x-x^2>0；
x^2-4x-5

Find the solution of the inequality M2X + 2 ＞ 2mx + m about X

The original inequality can be changed into: m (m-2) x ＞ m-2 (1) when m-2 ＞ 0 is m ＞ 2, MX ＞ 1, the solution of the inequality is x ＞ 1m (2) when m-2 ＜ 0 is m ＜ 2, MX ＜ 1. ① when m ＜ 0 is m ＜ 2, the solution of the inequality is x ＜ 1m; ② when m ＜ 0, the solution of the inequality is x ＞ 1m; ③ when m = 0, the solution of the inequality is all real numbers. (3) when m-2 = 0 is m = 2, the inequality has no solution When m ＞ 2, the solution of the inequality is x ＞ 1m; when 0 ＜ m ＜ 2, the solution of the inequality is x ＜ 1m; when m = 0, the solution of the inequality is all real numbers; when m = 2, the inequality has no solution

M is a real number, solve the inequality about X: m ^ 2x ^ 2 + 2mx-3 > 0

M^2X^2+2MX-3>0
M = 0, no solution
M^2>0
4M^2+12m^2=16m^2>0
m> 0, the solution of the inequality is
x1/m
m