IIT-JEE 2007 · Paper 1, Mathematics question 46

Evaluate the limit. Write $t^2f(x)-x^2f(t)=\left(t^2-x^2\right)f(x)-x^2\left(f(t)-f(x)\right)$. Dividing by $t-x$ and letting $t\to x$:
$$2xf(x)-x^2f'(x)=1.$$

Solve the ODE. Dividing by $-x^4$: $\left(\dfrac f{x^2}\right)'=\dfrac{x^2f'-2xf}{x^4}=-\dfrac1{x^4}$, so
$$\frac{f}{x^2}=\frac1{3x^3}+C,\qquad f(x)=\frac1{3x}+Cx^2.$$

Condition. $f(1)=\frac13+C=1$ gives $C=\frac23$: $f(x)=\dfrac1{3x}+\dfrac{2x^2}3$.

(A)