(I)∵f (x)=x3-3ax+1, ∴f′(x)=3x2-3a, 当a≤0时,f′(x)≥0恒成立,f (x)的单调增区间为R; 当a>0时,由f′(x)>0得x<-或x> 故f (x)的单调增区间为(-∞,-)和(,+∞),f (x)的单调减区间为(-,) (II)当a≤0时,由(I)可知f (x)在[0,]递增,且f(0)=1,此时无解; 当0<a<3时,由(I)可知f (x)在∈[0,-)上递减,在(,]递增, ∴f (x)在[0,]的最小值为f()=1-2a ∴,即 解得:a=1 当a≥3时,由(I)可知f (x)在[0,]上递减,且f(0)=1, ∴f()=3-3a+1≥-1 解得:a≤1+ 此时无解 综上a=1 |