Functionally impossible.

I don't have the most powerful understanding of differential equations, but my reasoning was this. If I could find the value of all the derivatives of $f$ evaluated at $0$ , then I could get a Maclaurin series expansion for the solution(s). So I substituted $0$ into the original equation in order to find $f''(0)$ . Then the equation could be differentiated, and then evaluated at $x=0$ to find $f'''(0)$ . This process could be repeated indefinitely. It was easily seen that the values of the derivatives at $0$ were always unique because the differential equation obtained after $n$ differentiations was always linear in the highest derivative which always had a coefficient of $2$ (and the lower derivatives evaluated at $0$ would all be known beforehand if you followed the procedure correctly). I did not prove that the Maclaurin series solution for $f$ converged (perhaps there is a theorem someone could show me regarding this). Since there is one unique solution (the Maclaurin series), $f(1)$ takes a unique value. This eliminates any answer that is just a single number because it implies the other numbers are possible values of $f(1)$ . If you choose the answer "None of the above options are correct" (which I interpret to mean "none of the other answers are correct"), then that implies all of the numbers are possible values of $f(1)$ , but we know $f(1)$ can only take on one value. So, the only answer that makes sense is "All of the above numbers.