Answer:
Option D
Explanation:
We have,
f(x+y)=f(x)f(y)
differentiate with respect to x, y as constant , we get
$f'(x+y)=f'(x) f(y) $
Put x=0 , y=3 we get
f'(0+3)=f'(0).f(3)
$\Rightarrow$ f'(3) =f'(0).f(3)
$\Rightarrow$ f'(3)=11 x3 =33 $[\because f'(0)=11, f(3)=3]$