Question

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

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

Answer 1

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

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

Differentiate w.r.t. $t$

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

$\Rightarrow x^2{f}^{\prime }\left(x\right)-2xf\left(x\right)+1=0$

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

$\Rightarrow \frac{dy}{dx}=\frac{2y}{x}-\frac{1}{x^2}$

I.F$=e^{-\int \frac{2}{x}dx}=e^{-2\ln x}=\frac{1}{x^2}$

$\frac{y}{x^2}=\frac{1}{3x^3}+c$ at $x=1,y=1$

$\Rightarrow 1=\frac{1}{3}+c$

$\Rightarrow c=1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}$

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

$\therefore f\left(\frac{3}{2}\right)=\frac{1}{3\times \frac{3}{2}}+\frac{2{\left(\frac{3}{2}\right)}^2}{3}=\frac{2}{9}+\frac{3}{2}=\frac{4+27}{18}=\frac{31}{18}$