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(B) Given $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$. Applying $L$'$H$ôpital's rule with respect to $t$:$\lim _{t \rightarrow x} \frac{

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$24$

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(B) Given $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$. Applying $L$'$H$ôpital's rule with respect to $t$:$\lim _{t \rightarrow x} \frac{2t f(x)-x^2 f'(t)}{1}=1$$2x f(x)-x^2 f'(x)=1$$f'(x) - \frac{2}{x} f(x) = -\frac{1}{x^2}$This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{1}{x^2}$.Integrating factor $I.F. = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}$.The solution is $f(x) \cdot \frac{1}{x^2} = \int -\frac{1}{x^2} \cdot \frac{1}{x^2} dx + C = \int -x^{-4} dx + C = \frac{1}{3x^3} + C$.$f(x) = \frac{1}{3x} + Cx^2$.Given $f(1) = 1$,we have $1 = \frac{1}{3} + C$,so $C = \frac{2}{3}$.Thus,$f(x) = \frac{1}{3x} + \frac{2x^2}{3} = \frac{1+2x^3}{3x}$.$f(2) = \frac{1+2(8)}{3(2)} = \frac{17}{6}$.$f(3) = \frac{1+2(27)}{3(3)} = \frac{55}{9}$.$2f(2) + 3f(3) = 2(\frac{17}{6}) + 3(\frac{55}{9}) = \frac{17}{3} + \frac{55}{3} = \frac{72}{3} = 24$.

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 ....................

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