For some $A \in \left(-\frac{1}{2},\frac{1}{2} \right)$ and $c \in (0,1), \; f(1)=f(0)+f'\left(\frac{1}{2}\right)+Af''(c)$ 
Suppose $f(x)$ is continuous on $[0,1]$ and twice differentiable on $(0,1)$.
Show that there exists $A \in \left(-\frac{1}{2},\frac{1}{2} \right)$ and $c \in (0,1)$ such that
\begin{equation}
f(1)=f(0)+f'\left(\frac{1}{2}\right)+Af''(c)
\end{equation}

By MVT, there exists $c \in (0,1)$ such that
\begin{equation}
f'(c)=\frac{f(1)-f(0)}{1-0} = f(1)-f(0)
\end{equation}
Comparing with the required statement I see that it suffices to show, for some $A \in \left(-\frac{1}{2},\frac{1}{2} \right)$,
\begin{equation}
f'(c)=f'\left(\frac{1}{2}\right)+Af''(c)
\end{equation}
Rearranging,
\begin{equation}
\frac{f'(c)-f'\left(\frac{1}{2}\right)}{A}=f''(c)
\end{equation}
This looks like an application of MVT again, but I'm not very sure how to proceed, as $c$ appears on both LHS and RHS.
Any help would be appreciated!

$f'(c)-f'(\frac 1 2) =(c-\frac 1 2)f''(d)$ for some $d$ between $c$ and $\frac 1 2$ Note that $|c-\frac 1 2| \leq \frac 1 2$. Hence the required equation holds with $A=c-\frac 1 2$ and $c$ changed to $d$.
