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Q.
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 :
JEE MainJEE Main 2016Continuity and Differentiability
Solution:
let $L=\displaystyle \lim_{t \to x}$ $\frac{t^{2}f \left(x\right)-x^{2} f\left(t\right)}{t-x}=1$
Applying L.H. rule
$L=\displaystyle \lim_{t \to x}$ $\frac{2t f\left(x\right)-x^{2}f \left(x\right)}{1}$
or $2x f\left(x\right)-x^{2} f '\left(x\right)=1$
solving above differential equation we get
$f \left(x\right)=\frac{2}{3}x^{2}+\frac{1}{3x}$
Put $x=\frac{3}{2}$
$f \left(\frac{3}{2}\right)=\frac{2}{3}\times\left(\frac{3}{2}\right)^{2}+\frac{1}{3}\times3\times2$
$=\frac{3}{2}+\frac{2}{9}=\frac{27+4}{18}=\frac{31}{18}$