Let f: R → R be a function satisfying f(x + y) = f(x) + 2y^ 2 + kxy...

We have, $f\left(x + y\right) = f\left(x\right) + 2{y}^{2} + kxy$ for all $x, y \in R$

$⇒ \dfrac{f\left(x + y\right) - f\left(x\right)}{y} = 2y + kx$ for all $x \in R$

$⇒ \lim_{y \to 0}\dfrac{f\left(x + y\right) - f\left(x\right)}{y} = \lim_{y \to 0}\left(2y + kx\right)$

$⇒ {f}^{'}\left(x\right) = kx$ for all $x \in R$

$⇒ f\left(x\right) = \dfrac{k{x}^{2}}{2} + C$ for all $x \in R$ [by integration]

But $f\left(1\right) = 2$ and $f\left(2\right) = 8$

$∴ 2 = \dfrac{k}{2} + C$ and $8 = 2k + C$

$k = 4$ and $C = 0$

Hence, $f\left(x\right) = 2{x}^{2}$ for all $x \in R$

So, option (a) is correct.