(1): 当a = 0时, f (x)=x3-4x2+5x , >0, 所以f (x)的单调递增区间为, . (2)解: 一方面由题意, 得 即; 另一方面当时, f (x) = (-2x3+9x2-12x+4)a+x3-4x2+5x , 令g(a) = (-2x3+9x2-12x+4)a+x3-4x2+5x, 则 g(a)≤ max{ g(0), g() } = max{x3-4x2+5x , (-2x3+9x2-12x+4)+x3-4x2+5x } = max{x3-4x2+5x , x2-x+2 }, f (x) = g(a) ≤ max{x3-4x2+5x , x2-x+2 }, 又{x3-4x2+5x}="2," {x2-x+2}="2," 且f (2)=2, 所以当时, f (x)在区间[0,2]上的最大值是2. 综上, 所求a的取值范围是. |