f(x)=(a*2^x-1)/(2^x+1)是奇函數(shù),則由f(0)=0得
(a-1)/2=0,得a=1
則f(x)=(2^x-1)/(2^x+1)=(2^x+1-2)/(2^x+1)=1-2/(2^x+1)
設(shè)x12/(2^x1+1)-2/(2^x1+1)=0
也即f(x1)>f(x2),故f(x)嚴(yán)格單調(diào)遞減.
f(t^2-2t)＋f(2t^2-k)=f(2t^2-k)-f(2t-t^2)＜0
故f(2t^2-k)2t-t^2,得
3t^2-2t-k>0
也即,對任意的t∈［-2,2］,均有3t^2-2t-k>0成立.
也即對任意的t∈［-2,2］,均有k