Analysis of the problem

Solution 1:

This is a nice puzzle at the calculus level. Since nobody seems interested in helping here is my advice.

First of all your answer is perfect for half of the problem, but missing a key half. You show that this maximum is at least as big as 50. You don't prove that it cannot be any larger.

Here is a sketch that you can flesh out with appropriate computation.

Preliminaries. If $f\in M$ first note that $f(0)=f(1)$ implies there is a number $c$ so that $f'(c)=0$. Moreover, this also implies that $$0 = f(1)-f(0) = \int_0^1 f'(t)\,dt = \int_0^c f'(t)\,dt + \int_c^1 f'(t)\,dt $$ and hence $$\int_0^c f'(t)\,dt = - \int_c^1 f'(t)\,dt \tag{1}.$$

Step 1. Consider the case $c=\frac12$. The condition $|f''(x)|\leq 100$ and the mean-value theorem applied on $[0,\frac12]$ and on $[\frac12,1]$ shows that $|f'(x)|\leq 50$ for all values of $x$ in $[0,1]$.

But your most excellent example $p(x)=50x(1-x)$ has $p'(x) = 100x -50$, $p''(x)=100$, $p'(0)=-50$, and $p'(1)=50$.

Thus the value $50$ is attained by at least one function.

Step 2. Consider the case $c> \frac12$. By the mean-value theorem as before we must have $|f'(x)|<50$ for all $x$ in $[c,1]$.

But if there is a value $x_0$ in $[0,c]$ with $f'(x_0)> 50$ it must be the case, since $|f''(x)|\leq 100$, that $$ \int_0^c f'(x) > \left|\int_c^1 f'(x)\right|.$$ This contradicts (1) above.

Step 3. The case $c< \frac12$ is simiilar.

Thus no value of $|f'(x)|$ larger than $50$ can occur if $c\not=\frac12$. We have already checked that the value $50$ is the largest possible for the case $c=\frac12$ and is actually attained for the function $p(x)$ given.