Let f : R → R be a function satisfying $f(x+y)=f(x)+2 y^2+k x y$ for all $x, y \in R$. If f(1) = 2 and f(2) = 8, then f(x) is equal to

$2 x^2$

$f(x+y)=f(x)+2 y^2+k x y$ for all $x, y \in R$

$\Rightarrow \frac{f(x+y)-f(x)}{y}=2 y+k x$ for all $x \in R$

$\Rightarrow \lim\limits_{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}=\lim\limits_{y \rightarrow 0}(2 y+k x)$

$\Rightarrow f'(x)=k x$ for all $x \in R$

$\Rightarrow f(x)=\frac{k x^2}{2}+C$ for all $x \in R$

But, f(1) = 2 and f(2) = 8. Therefore,

2 = $\frac{k}{2}$ + C and 8 = 2k + C

⇒ k = 4  and C = 0

Hence, f(x) = 2x² for all $x \in R$.