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(B) Given $\lim _{t \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1$.Applying $L$'$H$ôpital's rule with respect to $t$:$\lim _{t \rightarrow

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$\frac{9}{11 x}+\frac{13}{11} x^{10}$

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(B) Given $\lim _{t \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1$.Applying $L$'$H$ôpital's rule with respect to $t$:$\lim _{t \rightarrow x} \frac{10 t^9 f(x)-x^{10} f^{\prime}(t)}{9 t^8}=1$$\Rightarrow \frac{10 x^9 f(x)-x^{10} f^{\prime}(x)}{9 x^8}=1$$\Rightarrow 10 x f(x)-x^2 f^{\prime}(x)=9 x^8 \cdot \frac{9}{x^8} = 9$$\Rightarrow f^{\prime}(x)-\frac{10}{x} f(x)=-\frac{9}{x^2}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{10}{x}$ and $Q(x) = -\frac{9}{x^2}$.The integrating factor $IF = e^{\int P(x) dx} = e^{-10 \ln x} = x^{-10} = \frac{1}{x^{10}}$.The solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.$\frac{f(x)}{x^{10}} = \int -\frac{9}{x^2} \cdot \frac{1}{x^{10}} dx = -9 \int x^{-12} dx = -9 \left( \frac{x^{-11}}{-11} \right) + C = \frac{9}{11 x^{11}} + C$.Given $f(1)=2$,we have $\frac{2}{1} = \frac{9}{11} + C \Rightarrow C = 2 - \frac{9}{11} = \frac{13}{11}$.Thus,$f(x) = x^{10} \left( \frac{9}{11 x^{11}} + \frac{13}{11} \right) = \frac{9}{11 x} + \frac{13}{11} x^{10}$.Therefore,option $(B)$ is correct.

Let $f(x)$ be a continuously differentiable function on the interval $(0, \infty)$ such that $f(1)=2$ and $\lim _{t \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1$ for each $x>0$. Then,for all $x>0, f(x)$ is equal to

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