JEE Advanced 2024 · Paper 1, Mathematics question 1

Turn the limit into a derivative. Numerator and denominator both vanish at $t = x$, so by L'Hospital's rule (differentiating in $t$):
$$\lim_{t\to x}\frac{t^{10}f(x)-x^{10}f(t)}{t^9-x^9} = \frac{10x^9f(x)-x^{10}f'(x)}{9x^8} = 1.$$
Hence $10xf - x^2f' = 9$, i.e.
$$f'(x) - \frac{10}{x}f(x) = -\frac{9}{x^2}.$$

Solve the linear ODE. Integrating factor $x^{-10}$:
$$\frac{d}{dx}\left(x^{-10}f\right) = -9x^{-12} \implies x^{-10}f = \frac{9}{11}x^{-11} + C \implies f(x) = \frac{9}{11x} + Cx^{10}.$$

$f(1) = 2$ gives $C = 2 - \frac{9}{11} = \frac{13}{11}$, so $f(x) = \dfrac{9}{11x} + \dfrac{13}{11}x^{10}$.