Existence of $\xi$ and $\eta$ such that $f'(\xi)+f'(\eta)=\xi+\eta$ Let $f$ be continuous on $[0,1]$, differentiable in $(0,1)$. Assume futher that $f(0)=0$, $f(1)=1/2$. Show that there exist $\xi,\eta\in (0,1)$ such that  $f(\xi)+f'(\eta)=\xi+\eta$.
I saw this problem in a draft. I do not know whether it is true. Up to now, I have not find a counterexample.
My idea is as follows:
$f(\xi)-\xi=\eta-f'(\eta)$.
Let $F(x)=f(x)-x$, then $F(0)=0, F(1)=-1/2$, so $f(\xi)-\xi$ can be chosen to be arbitrary $a\in (-1/2,0)$. The strategy is then to find $\eta$ such $a=\eta-f'(\eta)$. Roll's theorem may be applied. However, letting $G(x)=ax-x^2/2+f(x)$, then $G(0)=0$, $G(1)=a$. This fact does not verify that of Roll.
So how can I get across the difficulties?

If we change $f(\xi)+f'(\eta)=\xi+\eta$ to be $f'(\xi)+f'(\eta)=\xi+\eta$ for some $\xi,\eta\in (0,1), \xi\neq \eta$. Can we prove it?


For $f(\xi)+f'(\eta)=\xi+\eta$ there is a counterexample given in 
Prove $\exists \xi, \eta \in (0,1)$, such that $f(\xi)+f'(\eta)=\xi+\eta$.?
On the other hand the statement with $f'(\xi)+f'(\eta)=\xi+\eta$ is true.
Let $F(x)=f(x)-x^2/2$.
Now by the MVT there is $\xi\in (0,1/2)$ such that
$$\frac{F(1/2)-F(0)}{1/2-0}=F'(\xi)=f'(\xi)-\xi.$$
Similarly there is $\eta\in (1/2,1)$ such that
$$\frac{F(1)-F(1/2)}{1-1/2}=F'(\eta)=f'(\eta)-\eta.$$
Hence $0<\xi<1/2<\eta<1$ and
$$\begin{align*}
f'(\xi)+f'(\eta)-(\xi+\eta)&=F'(\xi)+F'(\eta)\\
&=\frac{F(1/2)-F(0)}{1/2-0}+\frac{F(1)-F(1/2)}{1-1/2}\\&=\frac{F(1/2)}{1/2}-\frac{F(1/2)}{1/2}=0.
\end{align*}$$
where we used $F(0)=0-0=0$ and $F(1)=1/2-1/2=0$.
