Prove the following statements 
Let $f:[0,1] \rightarrow \mathbb{R}$ be continuous with $f(0)=0$ and $f(1)=1$. For the following you may apply standard results without proof provided you state them carefully;
  $(1)$ If $f(x)>0$ for $x>0$ show that there exists $\delta \in (0,1)$ with $f(x) \neq \delta$ for all $x \in [1/2,1]$
  $(2)$ If also $f$ is differentiable on $(0,1)$ prove that there exists $\theta \in (0,1)$ such that $$\dfrac{f'(\theta)}{\cos(\dfrac{\pi}{2}\theta)}=\dfrac{2}{\pi}$$

My Attempt
(1)
By the way of Intermediate value theorem, if  there exists $\delta<f(1/2)$ then there doesn't necessarily exist $c \in (0.5,1)$ such that $f(c)=\delta$ .
I am not very happy with this answer, as $f$ might have a minimum between $(0.5,1)$ but I don't know how to incorporate this in my proof.  
(2)  Im quite lost on how to approach this problem;  
Any helps/tips will be much appreciated
 A: *

*Replace your $f(1/2)$ with $m=\min_{x \in [1/2,1]}f(x)$. Then pick $\delta = \frac{m}{2}$ and conclude that $f(x) > \delta$ for every $x \in [1/2,1]$.

*Apply Cauchy's theorem to the functions $x \mapsto f(x)$ and $x \mapsto \sin \left( \frac{\pi}{2}x \right)$ on the interval $[0,1]$: for some $\theta \in (0,1)$, $$1=\frac{f(1)-f(0)}{\sin \left( \frac{\pi}{2} \right) - \sin \left( \frac{\pi}{2}\cdot 0 \right)} = \frac{f'(\theta)}{\frac{\pi}{2} \cos \left( \frac{\pi}{2} \theta \right)}.$$

A: *

*For (1):
since $f$ is continuous on the compact $[1/2, 1]$, its image $f([1/2, 1])$ is also a compact, i.e. there exists $m\leq M$ such that for all $x\in[1/2, 1]$ $m\leq f(x) \leq M$; and, additionally, $m$ and $M$ are reached by some $x_m$ and $x_M$. But since $f>0$ on $[1/2, 1]$, $m=f(x_m) > 0$> Take $\delta\stackrel{\rm def}{=} \frac{m}{2}$.

*For (2): consider 
$$
g\colon x\in[0,1]\mapsto \sin\left(\frac{\pi}{2}x\right)
$$
so that $g(0)=f(0)=0$, $g(1)=f(1)=1$ and $g$ is differentiable on $[0,1]$ with $g^\prime(x) = \frac{\pi}{2}\cos\left(\frac{\pi}{2}x\right)$. What you now have to prove is that there exists $\delta \in(0,1)$ such that $(f-g)^\prime(\delta)=f^\prime(\delta) - g^\prime(\delta)=0$ -- does Rolle's Theorem ring a bell?
