If f(x+y)=f(xy)∀x,yεR, and f(2013) = 2013, then f(– 2013) equals
2013
0
-2013
none of these
f(2013)=f(2013+0)=f[(2013(0)]=f(0)
⇒ f(0) = 2013
Also, f(– 2013)=f(– 2013+0)=f(0)=2013.
It can also be seen that f is a constant function.