函数满足f(X)+f(-x)=0 且f(x)在【-2,2】上是减函数 f(2)=1若函数f(x)小于等于t2+2at+1

f(x)+f(-x)=0, ∴f(-x)=-f(x),f(x)是奇函数， ∴f(-2)=-f(2)=1, f(x)在[-2,2]上是减函数, ∴f(x)<=f(-2)=1, f(x)<=t^2+2at+1对所有x∈[-2,2],a∈[-1,1]成立， <==>1<=t^2+2at+1对所有a∈[-1,1]成立, <==>g(a)=2at+t^2>=0对所有a∈[-1,1]成立, <==>g(1)=2t+t^2>=0,且g(-1)=-2t+t^2>=0, <==>"t<=-2或t>=0",且“t<=0,或t>=2", <==>t<=-2,或t=0,或t>=2,为所求.