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# Let $f(x+y)=f(x)f(y)$ $\forall x,y\in R$. Suppose that $f(3)=3$ and $f'(0)=11$, then $f'(3)$ is given by

$(a)\;22\qquad(b)\;44\qquad(c)\;28\qquad(d)\;33$

$f'(3)=\lim\limits_{h\to 0}\large\frac{f(3+h)-f(3)}{h}$
$\qquad=\lim\limits_{h\to 0}\large\frac{f(3)f(h)-f(3)}{h}$
$\Rightarrow f'(3)=3\lim\limits_{\large h\to 0}\large\frac{f(h)-1}{h}$-----(1)
$f(3)=3$
Now $f(x+0)=f(x)f(0)$
$\Rightarrow f(x)(f(0)-1)=0$
$\Rightarrow$ either $f(x)=0$ or $f(0)=1$
But $f(3)=3\neq 0$
$\therefore f(0)=1$
From (1),$f'(3)=3f'(0)$
$\qquad\qquad\quad\;\;=3\times 11$
$\qquad\qquad\quad\;\;=33$
Hence (d) is the correct option.