Let f : (0, $\infty$ ) → R be a function which is differentiable at all points of its domain and satisfies the condition x²f '(x) = 2xf(x) + 3, with f(1) = 4. Then 2f(2) is equal to:

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answer is C.

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Detailed Solution

$\begin{array}{l} x^{2} f^{'} (x) - 2 xf (x) = 3 \\ (\frac{x^{2} f^{'} (x) - 2 xf (x)}{{(x^{2})}^{2}}) = \frac{3}{{(x^{2})}^{2}} \\ \Rightarrow \frac{d}{dx} (\frac{f (x)}{x^{2}}) = \frac{3}{x^{4}} \end{array}$
Integrating both sides
$\begin{aligned} \frac{f (x)}{x^{2}} = - \frac{1}{x^{3}} + C \\ f (x) = - \frac{1}{x} + {Cx}^{2} \end{aligned}$
put x = 1
$\begin{array}{l} 4 = - 1 + C \Rightarrow C = 5 \\ f (x) = - \frac{1}{x} + 5 x^{2} \end{array}$
Now $2 \times f (2) = 2 \times [- \frac{1}{2} + 5 \times 2^{2}] = 39$