Question

f(X)是定义在R上的函数,对任意X,y∈R都有f(X+y)+f(X-y)=2f(X)f(y),且f(0)≠0,(1)求

f(X)是定义在R上的函数,对任意X,y∈R都有f(X+y)+f(X-y)=2f(X)f(y),且f(0)≠0,(1)求证:f(0)=1(2)判断函数的奇偶性.

Answer 1

解：(1)令x=y=0, f(0)+f(0)=2f(0)f(0), 2f(0)=2f^2(0),
因f(0)≠0, 所以f(0)=1

(2)定义域为R，

令x=0, f(y)+f(-y)=2f(0)f(y)=2f(y),
[因f(0)=1]

所以f(-y)=f(y), 为偶函数