Verified Solution • 2025–26 Session moderate

QJEEM25Mabc2c: JEE Main 2025 Mathematics - Class 12 JEEM25Mabc2c Solution: Let be a function which is differentiable at all points of its…

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 :

Multiple Choice Options

  • A 19
  • B 23
  • C 29
  • D 39

Step-by-Step Verified Solution

amp;x2f′(x)−2xf(x)=3amp;(x2f′(x)−2xf(x)(x2)2)=3(x2)2amp;⇒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

amp;f(x)x2=−1x3+Camp;f(x)=−1x+Cx2amp; put x=1amp;4=−1+C⇒C=5amp;f(x)=−1x+5x2\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×f(2)=2×[−12+5×22]2 \times f(2)=2 \times\left[-\frac{1}{2}+5 \times 2^2\right]

=39=39

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