Let $f(x)$ be differentiable on the interval $(0, \infty)$ such that $f(1)=1$, and $\lim\limits_{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ for each $x>0$. Then $f(x)$ is
Answer & explanation
Correct answer: option 1
We have,
$\lim\limits_{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$
$\Rightarrow \lim\limits_{t \rightarrow x} \frac{t^2 f(x)-x^2 f(x)+x^2 f(x)-x^2 f(t)}{t-x}=1$
$\Rightarrow \lim\limits_{t \rightarrow x}\left(\frac{t^2-x^2}{t-x}\right) f(x)-x^2 \lim\limits_{t \rightarrow x} \frac{f(t)-f(x)}{t-x}=1$
$\Rightarrow f(x) \lim\limits_{t \rightarrow x}(t+x)-x^2 \lim\limits_{t \rightarrow x} \frac{f(t)-f(x)}{t-x}=1$
$\Rightarrow 2 x f(x)-x^2 f'(x)=1$
$\Rightarrow x^2 f'(x)-2 x f(x)=-1$
$\Rightarrow f'(x)-\frac{2}{x} f(x)=-\frac{1}{x^2}$ ......(i)
This is a linear differential equation with I.F. $=e^{-\int \frac{2}{x} d x}=\frac{1}{x^2}$
Multiplying both sides of (i) by I.F. $=\frac{1}{x^2}$ and integrating with respect to $x$, we get
$\frac{f(x)}{x^2}=\frac{1}{3 x^3}+C$ .....(ii)
It is given that $f(1)=1$. So, putting $x=1$ and $f(1)=1$ in (ii), we get $C=2 / 3$.
Putting $C=2 / 3$ in (i), we get
$\frac{f(x)}{x^2}=\frac{1}{3 x^3}+\frac{2}{3} \Rightarrow f(x)=\frac{1}{3 x}+\frac{2}{3} x^2$