Let be a function which is differentiable at all points of its domain and satisfies the condition , with f(1) = 4. Then 2f(2) is equal to : [2025]

Question

Let $f : (0, \infty) \to R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^{2} f' (x) = 2 x f (x) + 3$ , with f(1) = 4. Then 2f(2) is equal to : [2025]

Smart Achievers · Accepted Answer

(4)

Given, $x^{2} f' (x) = 2 x f (x) + 3$ and f(1) = 4

$\Rightarrow x^{2} f' (x) - 2 x f (x) = 3$

$\Rightarrow \frac{x^{2} f' (x) - 2 x f (x)}{x^{4}} = \frac{3}{x^{4}}$ (Divide both sides by $x^{4}$ )

$\Rightarrow \frac{d}{d x} (\frac{f (x)}{x^{2}}) = \frac{3}{x^{4}}$

Using integration on both sides,

$\Rightarrow \frac{f (x)}{x^{2}} = \int \frac{3}{x^{4}} d x$

$\Rightarrow \frac{f (x)}{x^{2}} = 3 \times \frac{- 1}{3 x^{3}} + C$

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

$\Rightarrow f (x) = \frac{- 1}{x} + C x^{2}$

Since f(1) = 4

$\Rightarrow 4 = - 1 + C \Rightarrow C = 4 + 1 = 5$

Now, we get f(x) $= \frac{- 1}{x} + 5 x^{2}$

$\Rightarrow 2 f (2) = 2 \cdot (\frac{- 1}{2} + 5 \times 4)$

$= 2 \times \frac{39}{2} = 39$ .

Solution

Q.
Let $f : (0, \infty) \to R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^{2} f' (x) = 2 x f (x) + 3$ , with f(1) = 4. Then 2f(2) is equal to : [2025]

1 23

2 19

3 29

4 39

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