Appeared in JEE ADVANCED 2018

The differential equation becomes $e^{-f(x)}f'(x) = e^{-g(x)}g'(x)$ , and hence $e^{-f(x)} - e^{-g(x)} = c$ . Thus we deduce that $\begin{aligned} e^{-f(2018)} - e^{-1} & = \; e^{-1} - e^{-g(0)} \\ e^{-f(2018)} & = \; 2e^{-1} - e^{-g(0)} \\ e^{f(2018)} & = \; \frac{e}{2 - e^{1-g(0)}} \end{aligned}$ Thus functions of this type are only possible for $0 < e^{1-g(0)} < 2$ , and hence $e^{f(2018)} > \tfrac12e$ , so that $f(2018) > \boxed{1 - \ln2}$ .