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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Differential Equations

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Differential Equations

Let f:(0, ∞) arrow R be a function which is differentiable at all...

Let f:(0,∞)→Rf:(0, \infty) \rightarrow \mathbf{R} be a function which is differentiable at all points of its domain and satisfies the condition x2f′(x)=2xf(x)+3x^2 f^{\prime}(x)=2 x f(x)+3, with f(1)=4f(1)=4. Then 2f(2)2 f(2) is equal to :

Options

  1. A39
  2. B19
  3. C29
  4. D23

Correct answer

A. 39

Step-by-step solution

x2f′(x)−2xf(x)=3(x2f′(x)−2xf(x)(x2)2)=3(x2)2⇒ddx(f(x)x2)=3x4\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 f(x)x2=−1x3+Cf(x)=−1x+Cx2 put x=14=−1+C⇒C=5f(x)=−1x+5x2 Now 2×f(2)=2×[−12+5×22]=39\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}
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