|f(x)|>1的解集为空集,则|f(x)|<=1,只需|f(x)max|<=1且|f(x)min|<=1即可满足。
f(x)=4x^2-4ax,x∈[0,1]
对称轴x=a/2
当a>信核轿2时,f(x)为减函数,f(x)min=f(1)=4-4a,f(x)max=f(0)=0,则|4-4a|<=1,解得3/4当a<0时,f(x)为增函数,f(x)min=f(0)=0,f(x)max=f(1)=4-4a,则|4-4a|<=1,解得3/4当1>=a>=0时,f(x)min=f(a/2)=-a^2,f(x)max=f(1)=4-4a,则|4-4a|<=1,解得3/4当2>a>1时,f(x)min=f(a/2)=-a^2,f(x)max=f(0)=0,|-a^2|<=1,解得-1<=a<=1,上述的交集为空集;
所以氏纤3/4请采纳,O(∩_∩)O谢谢