Question

If $f\left(x\right)$ is a differentiable function in the interval $(0,\infty )$ such that

$f\left(1\right)=1$ and $\underset{t\to x}{lim}\frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1$, for each $x>0$, then $f\left(\frac{3}{2}\right)$ is equal to:

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

Answer 1

The correct option is D $\frac{31}{18}$

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

$L=\underset{t\to x}{lim}\frac{2tf\left(x\right)-x^{2}f\left(x\right)}{1}=1$

or $2xf\left(x\right)-x^2{f}^{\prime }\left(x\right)=1$

solving above diffrential equation we get

$f\left(x\right)=\frac{2}{3}{x}^2+\frac{1}{3x}$

Put $x=\frac{3}{2}$

$f\left(\frac{3}{2}\right)=\frac{2}{3}\times {\left(\frac{3}{2}\right)}^2+\frac{1}{3}\times 3\times 2$

$=\frac{3}{2}+\frac{2}{9}=\frac{27+4}{18}=\frac{31}{18}$