1、当a=0时,f(x)=x∈[0,1],满足|f(x)| ≤12、当a>0时,开口向上,且f(x)≥0所以只需f(x)≤1ax²+x ≤ 1a ≤ 1/x² - 1/x = (1/x - 1/2)² - 1/400矛盾3、当a0(1)当-1/2a ≥1 即 a≥ -1/2时f(x)在[0,1]单调递增f(x) ∈[0,a+1]-1/2 ≤ a
