Let f be a differentiable function in the interval (0, ∞) such that f(1)=1 and lim t rightarrow x frac t 2 f(x)-x 2 f(t) t-x =1 for each x 0 . Then 2 f(2)+3 f(3) is equal to

Mathematics · Differential Equations

JEE Main 2024 — 5 April, Shift 1 — Question 26

Let f be a differentiable function in the interval $(0, \infty)$ such that $f(1)=1$ and $\lim _{t \rightarrow x} \frac{t^{2} f(x)-x^{2} f(t)}{t-x}=1$

for each $x>0$ . Then $2 f(2)+3 f(3)$ is equal to

Numerical answer — enter this value.

$\lim _{t \rightarrow x} \frac{t^{2} f(x)-x^{2} f(t)}{t-x}=1$

$\lim _{t \rightarrow x} \frac{2 t . f(x)-x^{2} f^{\prime}(x)}{1}=1$

$2 x . f(x)-x 2 f^{\prime}(x)=1$

$\frac{d y}{d x}-\frac{2}{x} \cdot y=\frac{-1}{x^{2}}$

I.f. $=\mathrm{e}^{\int-\frac{2}{x} d x}=\frac{1}{x^{2}}$

$\therefore \frac{\mathrm{y}}{\mathrm{x}^{2}}=\int-\frac{1}{\mathrm{x}^{4}} \mathrm{dx}+\mathrm{C}$

$\frac{\mathrm{y}}{\mathrm{x}^{2}}=\frac{1}{3 \mathrm{x}^{3}}+C$

Put $f(1)=1$

$\mathrm{C}=\frac{2}{3}$

$y=\frac{1}{3 x}+\frac{2 x^{2}}{3}$

$y=\frac{2 x^{3}+1}{3 x}$

$f(2)=\frac{17}{6}$

$\mathrm{f}(3)=\frac{55}{9}$

$2 \mathrm{f}(2)+3 \mathrm{f}(3)=\frac{17}{3}+\frac{55}{3}=\frac{72}{3}=24$

Answer key and solution verified before publishing.

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