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

<p>$$\begin{aligned} & x^2 f^{\prime}(x)-2 x f(x)=3 \\ & \left(\frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}\right)=\frac{3}{\left(x^2\right)^2} \\ & \Rightarrow \frac{d}{d x}\left(\frac{f(x)}{x^2}\right)=\frac{3}{x^4} \end{aligned}$$</p> <p>Integrating both sides</p> <p>$$\begin{aligned} & \frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^2}=-\frac{1}{\mathrm{x}^3}+\mathrm{C} \\ & \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+\mathrm{Cx}^2 \\ & \text { put } \mathrm{x}=1 \\ & 4=-1+\mathrm{C} \Rightarrow \mathrm{C}=5 \\ & \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+5 \mathrm{x}^2 \end{aligned}$$</p> <p>Now $2 \times f(2)=2 \times\left[-\frac{1}{2}+5 \times 2^2\right]$</p> <p>$=39$</p>