Let $f(x)$ be a differentiable function such that $f(1)=2$, $f(2)=6$ and $f(x+y)=f(x)+k x y+\frac{4}{3} y^2 \forall x, y \in \mathbb{R}$ then $f(x)=$

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$f(1)=2, f(2)=6$ $f(x+y)=f(x)+k x y+\frac{4}{3} y^2... (I)$ Put $x=1 \& y=1$ $\begin{aligned} & 	herefore f(1+1)=f(1)+k+\frac{4}{3} \Rightarrow 6-\frac{4}{3}=2+k \ & \Rightarrow k=\frac{8}{3} \end{aligned}$ Put $x=1$ and $y=m-1$ in equation (i), we get $\begin{aligned} & 	herefore f(1+m-1)=f(1)+k 	imes 1(m-1)+\frac{4}{3}(m-1)^2 \ & \Rightarrow f(m)=\frac{4}{3} m^2+\frac{2}{3} \ & \Rightarrow f(x)=\frac{4}{3} x^2+\frac{2}{3} \end{aligned}$

Let $f(x)$ be a differentiable function such that $f(1)=2$, $f(2)=6$ and $f(x+y)=f(x)+k x y+\frac{4}{3} y^2…

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