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(D) Given $f(1)=2$,$f(2)=6$ and $f(x+y)=f(x)+kxy+\frac{4}{3}y^2$. Substitute $x=1$ and $y=1$ into the functional equation: $f(1+1) = f(1) + k(1)(1) +

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$\frac{4}{3}x^2+\frac{2}{3}$

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(D) Given $f(1)=2$,$f(2)=6$ and $f(x+y)=f(x)+kxy+\frac{4}{3}y^2$. Substitute $x=1$ and $y=1$ into the functional equation: $f(1+1) = f(1) + k(1)(1) + \frac{4}{3}(1)^2$ $f(2) = f(1) + k + \frac{4}{3}$ $6 = 2 + k + \frac{4}{3}$ $4 = k + \frac{4}{3}$ $k = 4 - \frac{4}{3} = \frac{8}{3}$. Now,substitute $k = \frac{8}{3}$ into the original equation: $f(x+y) = f(x) + \frac{8}{3}xy + \frac{4}{3}y^2$. To find $f(x)$,we use the definition of the derivative: $f'(x) = \lim_{h 	o 0} \frac{f(x+h)-f(x)}{h} = \lim_{h 	o 0} \frac{\frac{8}{3}xh + \frac{4}{3}h^2}{h} = \lim_{h 	o 0} (\frac{8}{3}x + \frac{4}{3}h) = \frac{8}{3}x$. Integrating $f'(x) = \frac{8}{3}x$ with respect to $x$: $f(x) = \int \frac{8}{3}x \, dx = \frac{8}{3} \cdot \frac{x^2}{2} + C = \frac{4}{3}x^2 + C$. Using $f(1)=2$: $2 = \frac{4}{3}(1)^2 + C \Rightarrow C = 2 - \frac{4}{3} = \frac{2}{3}$. Thus,$f(x) = \frac{4}{3}x^2 + \frac{2}{3}$.

Let $f(x)$ be a differentiable function such that $f(1)=2$,$f(2)=6$ and $f(x+y)=f(x)+kxy+\frac{4}{3}y^2$ for all $x, y \in R$. Then $f(x)$ is:

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