Question

If $f(x)$ is a differentiable function in the interval $(0,\infty )$ such that $f(1)=1$ and ${\lim }_{t\to x}\frac{t^{2}f(x)-x^{2}f(t)}{t-x}=1$, for each $x>0$, then $f(3/2)$ is equal to :

• A. $\frac{13}{6}$
• B. $\frac{23}{18}$
• C. $\frac{25}{9}$
• D. $\frac{31}{18}$

Answer 1

Answer: (D)

let $L=\underset{t\to x}{\lim }$ $\frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1$
Applying L.H. rule
$L=\underset{t\to x}{\lim }$ $\frac{2tf\left(x\right)-x^{2}f\left(x\right)}{1}$
or $2xf\left(x\right)-x^2{f}^{\prime }\left(x\right)=1$
solving above differential equation we get
$f\left(x\right)=\frac{2}{3}{x}^2+\frac{1}{3x}$
Put $x=\frac{3}{2}$
$f\left(\frac{3}{2}\right)=\frac{2}{3}\times {\left(\frac{3}{2}\right)}^2+\frac{1}{3}\times 3\times 2$
$=\frac{3}{2}+\frac{2}{9}=\frac{27+4}{18}=\frac{31}{18}$