Given the function f (x) = {X & sup2; + 4x, X ≥ 0.4x-x & sup2; x < 0. If f (2-A & sup2;) > F (a), then the value range of real number a x²4x,x≥0 Given function f (x)={ 4x-x² ,x＜0. If f (2-A & # 178;) > F (a), then the value range of real number a

Let x1 ＞ x2 ≥ 0f (x1) - f (x2) = = X1 & # 178; + 4x1 - (x2 & # 178; + 4x2) = X1 & # 178; - x2 & # 178; + 4 (x1 - x2) > 0, so when x ≥ 0, the function is an increasing function 2-A & # 178; > a ≥ 0, and the solution is a > 1. When x < 0, f (x) = 4x-x & # 178;, the opening downward parabolic

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