已知定义在R上的函数f(x)满足:①对任意实数x,y都有f(x)+f(y)=1+f(x+y)②对任意非零实数x都有
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由条件1可知:
令 x=y=0 f(0) + f(0) =1 + f(0) 所以f(0) =1;
令y=-x ; f(x) + f(-x) = 1 + f(0) 得出f(x) + f(-x) =2;
由条件2可知:
f(-1) = -f(-1) 得出f(-1) = 0; 所以 f(1) = 2 - f(-1) =2;
所以 f(2012) = f(1) + f(2011) -1
= 2f(1) + f(2000)-2
= ........
=2012f(1) - 2011
=2012 *2 -2011
=2013
令 x=y=0 f(0) + f(0) =1 + f(0) 所以f(0) =1;
令y=-x ; f(x) + f(-x) = 1 + f(0) 得出f(x) + f(-x) =2;
由条件2可知:
f(-1) = -f(-1) 得出f(-1) = 0; 所以 f(1) = 2 - f(-1) =2;
所以 f(2012) = f(1) + f(2011) -1
= 2f(1) + f(2000)-2
= ........
=2012f(1) - 2011
=2012 *2 -2011
=2013
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