题目内容
设奇函数y=f(x)(x∈R),满足对任意t∈R都有f(t)=f(1-t),且
【答案】分析:由题设知f(3)=f(1-3)=f(-2)=-f(2)=-[f(1-2)]=-f(-1)=f(1)=f(0)=0.
=
=
=
=
=-
.所以
=-
.
解答:解:∵奇函数y=f(x)(x∈R),满足对任意t∈R都有f(t)=f(1-t),
且
时,f(x)=-x2,
∴f(3)=f(1-3)=f(-2)=-f(2)=-[f(1-2)]=-f(-1)=f(1)=f(0)=0.
=
=
=
=
=-
.
∴
=-
.
故答案为:-
.
点评:本题考查函数值的求法,解题时要注意孙数的奇偶性和对任意t∈R都有f(t)=f(1-t),且
时,f(x)=-x2的灵活运用,合理地进行等价转化.
解答:解:∵奇函数y=f(x)(x∈R),满足对任意t∈R都有f(t)=f(1-t),
且
∴f(3)=f(1-3)=f(-2)=-f(2)=-[f(1-2)]=-f(-1)=f(1)=f(0)=0.
∴
故答案为:-
点评:本题考查函数值的求法,解题时要注意孙数的奇偶性和对任意t∈R都有f(t)=f(1-t),且
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