How to solve this problem of functional equation? $f:\mathbb{R}\to\mathbb{R}$ satisfying $f(x + y)=f(x)+\lambda xy+3x^2y^2$ for all $x,y\in\mathbb{R}$

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a function satisfying $$ f(x + y) = f(x) + \lambda x y + 3x^2y^2 \text{ for all } x, y \in \mathbb{R}. $$ If $f(3) = 4$ and $f(5) = 52$, then $f'(x)$ is equal to

$ \text{(a)}10x \qquad \text{(b)} -10x \qquad \text{(a)}20x \qquad \text{(a)}128x $

What I did:-

$f'(x) = \displaystyle\lim_{h \to 0}\dfrac{f(x+h) - f(x)}{h} = \displaystyle\lim_{h \to 0}(\lambda x + 3x^2h) = \lambda x$

Integrating both the sides we get $f(x) = \dfrac{\lambda x^2}{2} + c$

$ f(3) = \dfrac{9\lambda}{2} + c = 4 $ and $ f(5) = \dfrac{25\lambda}{2} + c = 52 $. Solving these equations we get $\lambda = 6$ and hence $f(x) = 6x$. But it's not even in the options. I have seen another approach where the answer obtained was correct but what is wrong in this approach? This is a question from the book R.D. Sharma's Objective mathematics

Answer 1

I think that $f(x)$ does not exists.

Let $f(x)=kx^2+c$ form, because all of the option $f'(x)$ are $\alpha x$ form.
Then, $f(x+y)-f(x)=\{k(x+y)^2+c\}-(kx^2+c)=2kxy+ky^2$.
It is contradiction with the question's condition.