Search any question & find its solution
Question:
Answered & Verified by Expert
Let $f(x)$ be differentiable on 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, $f(x)$ is
Options:
Solution:
2306 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{3 x}+\frac{2 x^2}{3}$
$\frac{1}{3 x}+\frac{2 x^2}{3}$
$$
\text { } \lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1
$$
$$
\begin{aligned}
& \Rightarrow & x^2 f^{\prime}(x)-2 x f(x)+1 & =0 \\
& \Rightarrow & \frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}+\frac{1}{x^4} & =0 \\
& \Rightarrow & \frac{d}{d x}\left(\frac{f(x)}{x^2}\right) & =-\frac{1}{x^4} \\
\Rightarrow & & f(x) & =c x^2+\frac{1}{3 x} \text { also } f(1)=1 \Rightarrow c=\frac{2}{3} . \\
& \text { Hence, } & f(x) & =\frac{2}{3} x^2+\frac{1}{3 x}
\end{aligned}
$$
\text { } \lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1
$$
$$
\begin{aligned}
& \Rightarrow & x^2 f^{\prime}(x)-2 x f(x)+1 & =0 \\
& \Rightarrow & \frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}+\frac{1}{x^4} & =0 \\
& \Rightarrow & \frac{d}{d x}\left(\frac{f(x)}{x^2}\right) & =-\frac{1}{x^4} \\
\Rightarrow & & f(x) & =c x^2+\frac{1}{3 x} \text { also } f(1)=1 \Rightarrow c=\frac{2}{3} . \\
& \text { Hence, } & f(x) & =\frac{2}{3} x^2+\frac{1}{3 x}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.