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
Answer & explanation
Correct answer: option 1
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.