If f(x+y)=f(xy)∀x,yεR, and f(2013) = 2013, then f(– 2013) equals

If $f (x + y) = f (x y) \forall x, y ε R,$ and $f (2013)$ = 2013, then f(– 2013) equals

A
2013
B
0
C
-2013
D
none of these

Solution:

$f (2013) = f (2013 + 0) = f [(2013 (0)] = f (0)$

$\Rightarrow f (0) = 2013$

Also, $f (- 2013) = f (- 2013 + 0) = f (0) = 2013 .$

It can also be seen that $f$ is a constant function.

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