If the quadratic function f (x) = - x ^ 2 + BX + C is a decreasing function in the interval (2, + ∞), and an increasing function in the interval ((∞), 2), its function and X axis intersect at two points a and B, and | ab | = 6, let g (x) be defined as an even function in R, when x > = 0, G (x) = f (x), and write the monotone interval of G (x)

If the quadratic function f (x) = - x ^ 2 + BX + C is a decreasing function in the interval (2, + ∞), and an increasing function in the interval ((∞), 2), its function and X axis intersect at two points a and B, and | ab | = 6, let g (x) be defined as an even function in R, when x > = 0, G (x) = f (x), and write the monotone interval of G (x)

A:
F (x) = - x ^ 2 + BX + C is a decreasing function when x > 2. When x = 0, G (x) = f (x) = - x ^ 2 + 4x + 5
x=0,g(-x)=-x^2-4x+5=g(x)
So: when x = 0, G (x) = - x ^ 2 + 4x + 5, the opening is downward, and the axis of symmetry x = 2
x

Let the quadratic function f (x) = ax ^ 2 + BX (a ≠ 0) satisfy the condition; 1. F (x) = f (- 2-x) 2. The image of F (x) is tangent to the line y = x, and the analytic expression of F (x) is obtained
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From F (x) = f (- 2-x), we know that the axis of symmetry of F (x) is x = - 1 (the sum of two in brackets divided by 2)
That is - B / 2A = - 1, so B = 2A
f(x)=ax²+2ax
Because the f (x) image is tangent to the line y = X
Substituting y = x into the analytic formula, we can get x = ax & sup2; + 2aX, that is, ax & sup2; + (2a-1) x = 0
Δ=(2a-1)²-4a×0=0
A = 1 / 2
f(x)=(1/2)x²+x

Let f (x) = ax ^ 2 + BX (a ≠ 0) satisfy the condition 1. F (- 1 + x) = f (- 1-x); there is only one common point between the image of 2 function f (x) and the straight line y = x, (1) the analytic expression of F (x) (2) if the 2-tx power of F (x) > (1 / π) of inequality π is constant in t ∈ [- 2,2], find the value range of real number X

(1) F (x) = 0.5x ^ 2 + X (2) if the power of F (x) > (1 / π) of inequality π is constant at t ∈ [- 2,2], that is, the power of F (x) > π is constant at t ∈ [- 2,2], that is, the power of F (x) > TX-2 is constant at t ∈ [- 2,2], that is, the problem above is changed to 0.5x ^ 2 + x > TX-2 at t ∈ [- 2,2]