Q58 / 90 · JEE Main 2024 (05 Apr Shift 1) · Mathematics
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 _______
Correct answer: 24
Solution
$\begin{aligned}
& \lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1 \\
& \lim _{t \rightarrow x} \frac{2 t . f(x)-x^2 f^{\prime}(x)}{1}=1 \\
& 2 x \cdot f(x)-x 2 f^{\prime}(x)=1 \\
& \frac{d y}{d x}-\frac{2}{x} \cdot y=\frac{-1}{x^2} \\
& \text { I.f. }=e^{\int-\frac{2}{x} d x}=\frac{1}{x^2} \\
& \therefore \frac{y}{x^2}=\int-\frac{1}{x^4} d x+C \\
& \frac{y}{x^2}=\frac{1}{3 x^3}+C
\end{aligned}$
Put $f(1)=1$
$\begin{aligned} & C=\frac{2}{3} \\ & y=\frac{1}{3 x}+\frac{2 x^2}{3} \\ & y=\frac{2 x^3+1}{3 x} \\ & f(2)=\frac{17}{6} \\ & f(3)=\frac{55}{9} \\ & 2 f(2)+3 f(3)=\frac{17}{3}+\frac{55}{3}=\frac{72}{3}=24\end{aligned}$
Hence, the correct answer is 24.