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# If $f(x+y)=f(x).f(y) \forall x,y$ and $f(5)=2,f'(0)=3$ then $f'(5)$ is

$(a)\;0\qquad(b)\;1\qquad(c)\;6\qquad(d)\;2$

$f(x+y)=f(x)\times f(y)$
Differentiate with respect to $x$,treating y as constant $f'(x+y)=f'(x)((y)$
Putting $x=0$ and $y=x$
We get $f'(x)=f'(0)f(x)$
$\Rightarrow f'(5)=3f(5)=3\times 2=6$
Hence (c) is the correct answer.