It is known that the function f (x) on R satisfies f (x) = f (4-x), and the function f (x + 2) monotonically decreases in [0, + ∞). (1) find the solution set of the inequality f (3x) > F (2x-1); (2) let the solution set in (1) be a, and for any t ∈ a, the inequality x2 + (T-2) x + 1-T > 0 holds, and find the value range of real number X

It is known that the function f (x) on R satisfies f (x) = f (4-x), and the function f (x + 2) monotonically decreases in [0, + ∞). (1) find the solution set of the inequality f (3x) > F (2x-1); (2) let the solution set in (1) be a, and for any t ∈ a, the inequality x2 + (T-2) x + 1-T > 0 holds, and find the value range of real number X

(1) ∵ f (x) = f (4-x) ∵ f (x) image is symmetric with respect to line x = 2, and ∵ f (x + 2) monotonically decreases on [0, + ∞) ∵ f (x) monotonically decreases on [2, + ∞) ∵ inequality f (3x) > f (2x-1) is equivalent to: | 3x-2 | 2x-1-2 | 3x-2 ⇔ (2x-2) 2 ＜ (2x-3) 2 ⇔ (5x-5) (x + 1) ∵