If the function f satisfies the relation fx+y+fx-y=2fxfy∀x,y∈R and f0≠0, then
f(x) is an even function
f(x) is an odd function
If f2=a, then f-2=a
If f4=b, then f-4=-b
fx+y+fx-y=2fx.fyPut x=0. Then fy+f-y=2f0fyPut x=y=0. Then f0 +f0=2f0f0∴ f0=1as f0≠0∴f-y=fy