f(x+y)+f(x-y)=2f(x)f(y)
put y= -x
f(0)+f(2x)=2f(x)f(-x) -----1
putx= -y
f(0)+f(-2y)=2f(-y)f(y) -----2
since function is variable immaterial
subtracting 1 from 2 we get
f(2x)-f(-2x)=0
which means f is even
now in 1 put x=0
which gives f(0)=either 0 or 1.
case1 (f(0)=0)
put y=0 in main equation
f(x)+f(x)=2f(x)f(0)
2f(x)[1-f(0)]=0
since f(0)=0
f(x)=0 thus it is a constant function
case2 (f(0)=1)
there r some possibilities where each value is different
but one immediate conclusion is f assuming a constant value
of 1
Another way:
I think (correct me if i am wrong)
that the fn given is such that it can assume some constant value k
then according to question
k+k=2k*k
which gives value of k as either 1 or 0
f(x) can be a constant function taking wither 0 or 1