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

# If $f\left(x+y\right)=f\left(xy\right)\mathrm{\forall }x,y\epsilon \mathbf{R},$ and $f\left(2013\right)$ = 2013, then f(– 2013) equals

1. A

2013

2. B

0

3. C

-2013

4. D

none of these

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### Solution:

$f\left(2013\right)=f\left(2013+0\right)=f\left[\left(2013\left(0\right)\right]=f\left(0\right)$

Also,

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

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