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 $x0$. Then, for all $x0, f(x)$ is equal to

Question

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

Aicharya · Accepted Answer

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

Let $f(x)$ be a continuously differentiable function on the interval $(0, \infty)$ such that $f(1)=2$ and…

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