If the equation is given as: $f(x + y) = f(x)f(y),{\text{ }}f(3) = 3,{\text{ }}f'(0) = 11$. Then $f'(3)$ is equal to

${\text{(A) }}11.{e^{33}}$

${\text{ (B) }}33$

${\text{(C) }}11$

${\text{ (D) }}g(33)$

Hint: First use the value of x=3 and y=0 in the given function to find the value of f(0). Later differentiate the given function w.r.t. x and use x=0 and y=3 and use the value of f(0) in this equation to find the required solution.

Complete step by step answer:

As given in the question,

$\Rightarrow f(x + y) = f(x)f(y) $............................(1)

$ \Rightarrow f(3) = 3$ and $f'(0) = 11$

Now for finding the value of f'(3) first we have to find the value of f(0).

So, for finding the value of f(0).

Put x = 3 and y = 0 in equation (1) we get,

$\Rightarrow f(3 + 0) = f(3)f(0)$

$\Rightarrow f(0) = 1$……………………..(2)

Now for finding the value of f'(3) we have to differentiate equation (1) w.r.t x

On differentiating equation (1) w.r.t x we get,

$\Rightarrow f'(x + y)(1 + y') = f'(x)f(y) + f(x)f'(y)y'$...................................(3)

Putting value of x = 0 and y = 3 in equation (3) we get,

$\Rightarrow f'(3)(1 + y') = f'(0)f(3) + f(0)f'(3)y'$.......................................(4)

Solving equation (4) by putting value of $f(3) = 3$, $f'(0) = 11$ and $f(0) = 1$ we get,

$\Rightarrow f'(3)(1 + y') = 33 + f'(3)y' $

Taking $f'(3)y'$ to LHS of the above equation we get,

$\Rightarrow f'(3) = 33 $

Hence the correct answer will be option B.

NOTE: - Whenever we come up with this type of problem then put values of x and y according to the given values and then differentiate the given equation and put values of x and y as required.