Let f:(0, ) R be a function which is differentiable at all points of its domain and satisf…

Question

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 :

Accepted Answer

$$\begin{aligned}
& x^2 f^{\prime}(x)-2 x f(x)=3 \
& \left(\frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}\right)=\frac{3}{\left(x^2\right)^2} \
& \Rightarrow \frac{d}{d x}\left(\frac{f(x)}{x^2}\right)=\frac{3}{x^4}
\end{aligned}$$

Integrating both sides

$$\begin{aligned}
& \frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^2}=-\frac{1}{\mathrm{x}^3}+\mathrm{C} \
& \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+\mathrm{Cx}^2 \
& \text { put } \mathrm{x}=1 \
& 4=-1+\mathrm{C} \Rightarrow \mathrm{C}=5 \
& \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+5 \mathrm{x}^2
\end{aligned}$$

Now $2 \times f(2)=2 \times\left[-\frac{1}{2}+5 \times 2^2\right]$

$=39$

Solutions · 1