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# If $f\;(xy)=f\;(x+y)\;\forall\;x,y\;\in\;R$ and $\;f\;(2009)=2009$, then $f\;(-2009)$ equals

$(a)\;-2009\qquad(b)\;2009\qquad(c)\;0\qquad(d)\;None$

Can you answer this question?

Answer : (b) 2009
Explanation : $f\;(2009)=f\;(2009+0)=f\;(2009*0)$
$=f\;(0)=2009$
$f\;(-2009)=f\;(-2009+0)=f(-2009*0)=f(0)=2009.$
answered Jan 20, 2014 by