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(D) Given the functional equation: $f(x + 2y) = 2yf(x) + xf(y) - 3xy + 1$. Setting $x = 0$ and $y = 0$,we get $f(0) = 0 \cdot f(0) + 0 \cdot f(0) - 0

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$3$

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(D) Given the functional equation: $f(x + 2y) = 2yf(x) + xf(y) - 3xy + 1$. Setting $x = 0$ and $y = 0$,we get $f(0) = 0 \cdot f(0) + 0 \cdot f(0) - 0 + 1$,which implies $f(0) = 1$. Now,differentiate the given equation with respect to $y$ using the chain rule: $\frac{\partial}{\partial y} [f(x + 2y)] = \frac{\partial}{\partial y} [2yf(x) + xf(y) - 3xy + 1]$ $2f'(x + 2y) = 2f(x) + xf'(y) - 3x$. Setting $y = 0$ in this derivative equation: $2f'(x) = 2f(x) + xf'(0) - 3x$. Since $f'(0) = 1$,we have $2f'(x) = 2f(x) + x - 3x$,which simplifies to $2f'(x) = 2f(x) - 2x$,or $f'(x) - f(x) = -x$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = -1$ and $Q(x) = -x$. The integrating factor is $I.F. = e^{\int -1 dx} = e^{-x}$. Multiplying by $I.F.$,we get $e^{-x} f'(x) - e^{-x} f(x) = -x e^{-x}$. Integrating both sides: $\int \frac{d}{dx} [f(x) e^{-x}] dx = \int -x e^{-x} dx$. $f(x) e^{-x} = -[x(-e^{-x}) - \int 1(-e^{-x}) dx] = x e^{-x} + \int e^{-x} dx = x e^{-x} - e^{-x} + C$. $f(x) = x - 1 + Ce^x$. Using $f(0) = 1$: $1 = 0 - 1 + C \Rightarrow C = 2$. Thus,$f(x) = x - 1 + 2e^x$. Calculating $f(2)$: $f(2) = 2 - 1 + 2e^2 = 1 + 2e^2$. Wait,re-evaluating the derivative step: If $f(x) = x+1$,then $f'(x)=1$. Checking $f(x)=x+1$: $x+2y+1 = 2y(x+1) + x(y+1) - 3xy + 1 = 2xy + 2y + xy + x - 3xy + 1 = x + 2y + 1$. This satisfies the equation. Thus $f(2) = 2 + 1 = 3$.

Let $f$ be a differentiable function satisfying $f(x + 2y) = 2yf(x) + xf(y) - 3xy + 1$ for all $x, y \in R$ such that $f'(0) = 1$. Then $f(2)$ is equal to:

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