$f''(x)\ge 0$, $f'(0)=1$ and $f(x)\le 100$. Does such a function exist? $f''(x)\ge 0$, $f'(0)=1$ and $f(x)\le 100$. Does such a function exist?
I can show that $100-e^{-x}$ does not satisfy the given condition. But I have to show that no function can satisfy the initial condition. 
I can also say that since $f'$ is increasing, using Lagrange's MVT,
$$f'(x)-f(0)\ge1$$

Let $g(x)=f(x)-x-f(0)$.  Then $g'(x)=f'(x)-1$ and $g''(x)=f''(x)$.  The assumption $f''(x)\ge0$ implies $g''(x)\ge0$ which implies $g'(x)\ge g'(0)=f'(0)-1$ for $x\ge0$.  The assumption $f'(0)=1$ now tells us that $g'(x)\ge0$ for $x\ge0$, from which is follows that $g(x)\ge g(0)=f(0)-0-f(0)=0$ for all $x\ge0$.  In particular, 
$$g(101+|f(0)|)=f(101+|f(0)|)-(101+|f(0)|)-f(0)\ge0$$
hence
$$f(101+|f(0)|)\ge101+|f(0)|-f(0)\ge101\gt100$$
