Question

已知f(x)是定义在R上的函数,对于任意x,y属于R都有f(x+y)+(x-y)=2f(x)f(y),且f(o)不等于0

Answer 1

1.令x=y=0,所以由题意：

f(0)+f(0)=2(f(o))^2

---->2f(0)=2(f(o))^2

由于f(0)≠0

---->f(0)=1

2.2f(x)f(y)=f(x+y)+f(x-y)

2f(x)f(-y)=f(x-y)+f(x+y)

--->2f(x)f(y)=2f(x)f(-y)

--->f(0)f(y)=f(0)f(-y)

--->f(y)=f(-y)

由y的任意性,所以f(x)为偶函数