Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 24 January, Evening Shift — Question 4

Let $\mathrm{f}:(0, \infty) \rightarrow \mathbf{R}$ 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:

Answer: C

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

$\left(\frac{x^{2} f^{\prime}(x)-2 x f(x)}{\left(x^{2}\right)^{2}}\right)=\frac{3}{\left(x^{2}\right)^{2}}$

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

Integrating both sides

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

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

put $\mathrm{x}=1$

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

$\mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+5 \mathrm{x}^{2}$

Now $2 \times f(2)=2 \times\left[-\frac{1}{2}+5 \times 2^{2}\right]$

$=39$

Answer key and solution verified before publishing.

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