Let  $$f:(0,\infty)\to R$$ be a function which is differentiable at all points of its domain and satisfies the condition $$x^2 f'(x) = 2x f(x) + 3,$$  with  $$f(1)=4.$$ Then  $$2f(2)$$  is equal to:

$$x^2 \frac{dy}{dx} - 2xy = 3 \implies \frac{x^2 \frac{dy}{dx} - y(2x)}{x^4} = \frac{3}{x^4}$$

This is the derivative of a quotient: $\frac{d}{dx}(\frac{y}{x^2}) = \frac{3}{x^4}$

Integrate both sides:

$$\frac{f(x)}{x^2} = \int 3x^{-4} dx = \frac{3x^{-3}}{-3} + C = -\frac{1}{x^3} + C$$

Use $f(1) = 4$:

$$\frac{4}{1} = -1 + C \implies C = 5$$

Function: $$f(x) = x^2 (5 - \frac{1}{x^3}) = 5x^2 - \frac{1}{x}$$

Calculate $$2f(2)$$:

$$2 \left( 5(4) - \frac{1}{2} \right) = 2 \left( 20 - 0.5 \right) = 2(19.5) = \mathbf{39}$$

(Option A).