Let $f:[0,1]\to R$ be a real-valued function which is continuous on $[0, 1]$, differentiable on $(0,1)$ and $f(0)=f(1)=0$, $f(1/2)=1$. prove the following statements
$a$. There exists $\epsilon \in (1/2, 1)$ such that $f(\epsilon)=\epsilon$
$b$. For any $\lambda \in R$, there exists $\tau \in (0, \epsilon)$ such that $f'(\tau)-\lambda(f(\tau)-\tau)=1$
For $(b)$, Hint: use Rolle's theorem
Update: I finished part a, for part b, I have done the following.
Given $\epsilon\in(1/2,1)$ such that $f(\epsilon)=\epsilon$, then by M.V.T, there exist $\tau \in (0, \epsilon)$ such that $f'(\tau)=[f(\epsilon)-f(0)]/\epsilon=1$. Now I suffice to prove $f(\tau)=\tau$ so that for any $\lambda \in R$, $f'(\tau)-\lambda(f(\tau)-\tau)=1$.
Now I fix $\tau$, define $g(x):=f(x)-\tau$, so $g(0)=\tau<0$ and $g(\epsilon)=\epsilon-\tau>0$. By I.V.T, there exist $\tau' \in (0,\epsilon)$ such that $g(\tau')=f(\tau')-\tau'=0$, i.e. $f(\tau')=\tau'$. Now I claim $\tau'=\tau$
However, I have no idea how to prove this last claim.
For $a)$ you can use the function $g(x) = f(x) - x$, and apply mean value theorem to get the answer. For $b)$, you define the function $h(\tau) = e^{-\lambda \tau}f(\tau) - \displaystyle \int_{0}^{\tau} (1-\lambda t)e^{-\lambda t} dt$, and use Rolle's theorem on $(0,\epsilon)$.