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

\(\begin{aligned} & \lim _{t \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1 \\ & \Rightarrow \lim _{t \rightarrow x} \frac{10 t^9 f(x)-x^{10} f^{\prime}(t)}{9 t^8}=1\end{aligned}\)
\(\begin{aligned} & \Rightarrow 10 x f(x)-x^2 f^{\prime}(x)=9 \\ & \Rightarrow x^2 f^{\prime}(x)=10 x f(x)-9 \\ & \Rightarrow f^{\prime}(x)=\frac{10 f(x)}{x}-\frac{9}{x^2} \\ & \Rightarrow \frac{d y}{d x}-\frac{10}{x} y=-\frac{9}{x^2} \\ & \Rightarrow y \cdot \frac{1}{x^{10}}=\int-\frac{9}{x^2} \cdot \frac{1}{x^{10}} d x \\ & \Rightarrow \frac{y}{x^{10}}=\frac{9}{11 x^{11}}+c...(1)\end{aligned}\)
\(\begin{aligned} & \because \mathrm{f}(1)=2 \Rightarrow \frac{2}{1}=\frac{9}{11}+\mathrm{c} \Rightarrow \mathrm{c}=\frac{13}{11} \\ & \therefore \mathrm{f}(\mathrm{x})=\frac{9}{11 \mathrm{x}}+\frac{13}{11} \mathrm{x}^{10}\end{aligned}\)
\(\Rightarrow\) Option (2) is correct.