If $f(x)$ is differentiable function in 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 $f\left(\frac{3}{2}\right)$ is equal to |
$\frac{13}{6}$ $\frac{23}{18}$ $\frac{25}{9}$ $\frac{31}{18}$ |
$\frac{31}{18}$ |
We have, $\lim\limits _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ $\Rightarrow \lim\limits _{t \rightarrow x} \frac{t^2\{f(x)-f(t)\}-\left(x^2-t^2\right) f(t)}{t-x}=1$ $\Rightarrow \lim\limits _{t \rightarrow x}\left\{\frac{f(x)-f(t)}{t-x}\right\} t^2+\lim\limits _{t \rightarrow x}(x+t) f(t)=1$ $\Rightarrow -x^2 f'(x)+2 x f(x)=1$ [∵ f is differentiable and so continuous also] $\Rightarrow f'(x)-\frac{2}{x} f(x)=-\frac{1}{x^2}$ .....(i) This is a linear differential equation with integrating factor $e^{-\int \frac{2}{x} d x}=\frac{1}{x^2}$ Multiplying (i) by integrating factor $=\frac{1}{x^2}$ integrating, we get $\frac{f(x)}{x^2}=\frac{1}{3 x^3}+C$ .....(ii) It is given that $f(1)=1$. Putting $x=1$ in (ii), we get $1=\frac{1}{3}+C \Rightarrow C=\frac{2}{3}$ ∴ $\frac{f(x)}{x^2}=\frac{1}{3 x^3}+\frac{2}{3}$ $\Rightarrow f(x)=\frac{1}{3 x}+\frac{2 x^2}{3} \Rightarrow f\left(\frac{3}{2}\right)=\frac{2}{9}+\frac{3}{2}=\frac{31}{18}$ |