Q32 / 75 · JEE Main 2025 (24 Jan Shift 2) · Mathematics

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 :

  • Correct answer
    A
    39
  • B
    19
  • C
    29
  • D
    23

Solution

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

Hence, the correct answer is option: 'A'.

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