Q103 / 160 · AP EAMCET 2023 (17 May Shift 1) · Mathematics

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 $f(x)=$

  • A
    $4 x-2$
  • B
    $y-4 x^2+2 x-4$
  • C
    $\frac{8}{3} x^2+\frac{4}{3}$
  • Correct answer
    D
    $\frac{4}{3} x^2+\frac{2}{3}$

Solution

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

Hence, the correct answer is option: 'D'.

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