Question

Let \( f(x) \) be a differentiable function such that \( f(1) = 2 \), \( f(2) = 6 \), and \[ f(x+y) = f(x) + kxy + \frac{4}{3} y^2, \quad \forall x, y \in \mathbb{R}. \] Then \( f(x) = \) ?

• A. \( 4x - 2 \)
• B. \( y - 4x^2 + 2x - 4 \)
• C. \( \frac{8}{3}x^2 + 4 \)
• D. \( \frac{4}{3}x^2 + \frac{2}{3} \)

Hint:

Use functional equations and derivatives to determine the general form of a differentiable function.

Answer 1

The Correct Option is D

Given: \[ f(x+y) = f(x) + kxy + \frac{4}{3}y^2 \quad \forall x, y \in \mathbb{R} \] Differentiate both sides with respect to \( y \): \[ \frac{d}{dy}f(x+y) = \frac{d}{dy}[f(x) + kxy + \frac{4}{3}y^2] \Rightarrow f'(x+y) = kx + \frac{8}{3}y \] Now put \( y = 0 \): \[ f'(x) = kx \Rightarrow f'(x) = kx \Rightarrow f(x) = \frac{k}{2}x^2 + C \] Let us use \( f(1) = 2 \) and \( f(2) = 6 \): \[ f(1) = \frac{k}{2} + C = 2 \quad \text{(i)}
f(2) = 2k + C = 6 \quad \text{(ii)} \] Subtracting (i) from (ii): \[ (2k + C) - \left( \frac{k}{2} + C \right) = 4
\Rightarrow \frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3} \] Substitute \( k = \frac{8}{3} \) in (i): \[ \frac{4}{3} + C = 2 \Rightarrow C = \frac{2}{3} \] Therefore, \[ f(x) = \frac{4}{3}x^2 + \frac{2}{3} \]