Let f(x) be differentiable on the interval (0, ∞) such that f(1) = 1 and $\underset{t→x}{\lim}\frac{t^2f(x)-x^2t(t)}{t-x}=1$ for x > 0. Then f(x) is

$\frac{1}{3x}+\frac{2x^2}{3}$

Using L’Hospital rule, we get

$1 = 2 x f(x) - x^2 f'(x)$

or, $f'(x) -\frac{2}{x}f(x)=-\frac{1}{x^2}$

The integrating factor is $\frac{1}{x^2}$ and hence $\frac{f(x)}{x^2}=∫-\frac{1}{x^4}dx+c=\frac{1}{3x^3}+c$.

Also f(1) = 1 

$⇒c =\frac{2}{3}$

Hence (A) is the correct answer.