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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 $f(0) = 0 + 0 - 0 + 1$,so $f(0) = 1$. To

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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 $f(0) = 0 + 0 - 0 + 1$,so $f(0) = 1$. To find $f'(x)$,we differentiate the given equation with respect to $x$ keeping $y$ constant: $f'(x + 2y) = 2yf'(x) + f(y) - 3y$. Setting $x = 0$ in this derivative equation: $f'(2y) = 2yf'(0) + f(y) - 3y$. Since $f'(0) = 1$,we have $f'(2y) = 2y(1) + f(y) - 3y = f(y) - y$. Let $2y = t$,then $y = t/2$. Thus,$f'(t) = f(t/2) - t/2$. Alternatively,differentiate the original equation with respect to $y$ keeping $x$ constant: $2f'(x + 2y) = 2f(x) + xf'(y) - 3x$. Setting $y = 0$: $2f'(x) = 2f(x) + xf'(0) - 3x$. 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 $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = -x$. The integrating factor is $IF = e^{\int -1 dx} = e^{-x}$. Multiplying by $IF$: $e^{-x} f'(x) - e^{-x} f(x) = -x e^{-x}$. Integrating both sides: $\int \frac{d}{dx} (f(x) e^{-x}) dx = \int -x e^{-x} dx$. $f(x) e^{-x} = -[x(-e^{-x}) - \int 1(-e^{-x}) dx] = x e^{-x} + e^{-x} + C$. $f(x) = x + 1 + Ce^x$. Using $f(0) = 1$: $1 = 0 + 1 + C(1) \Rightarrow C = 0$. Thus,$f(x) = x + 1$. Therefore,$f(2) = 2 + 1 = 3$.

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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