(1)y=f(x)=2x+1+-8,设t=2x+1,1≤t≤3 则y=t+-8,t∈[1,3]. 任取t1、t2∈[1,3],且t1<t2,f(t1)-f(t2)=, 当1≤t≤2,即0≤x≤时,f(x)单调递减; 当2<t≤3,即<x≤1时,f(x)单调递增. 由f(0)=-3,f()=-4,f(1)=-,得f(x)的值域为[-4,-3]. (2)设x1、x2∈[0,1],且x1<x2, 则g(x1)-g(x2)=(x1-x2)(x12+x1x2+x22-3a2)>0, 所以g(x)单调递减. (3)由g(x)的值域为:1-3a2-2a=g(1)≤g(x)≤g(0)=-2a, 所以满足题设仅需:1-3a2-2a≤-4≤-3≤-2a, 解得,1≤a≤. |