Analytic function on an open disc. Let $\mathbb{D}=\{ z\in\mathbb{C}:|z|<1\}$. Which of the following are correct?


*

*There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(0)=0$ and $f'(0)=2$

*There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(3/4)=3/4$ and $f'(2/3)=3/4$

*There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(3/4)=-3/4$ and $f'(3/4)=-3/4$

*There exist a holomorphic function $f:\mathbb{D}\rightarrow\mathbb{D}$ with $f(1/2)=-1/2$ and $f'(1/4)=1.$
Option one is not true by Schwartz-lemma. But i don't know how to think about other options. I  don't know which theorem or result gives condition of existence of this type of holomorphic function. Please help me to solve this problem. If possible please solve this problem. Thanks in advance.   
 A: For 1. Incorrect.  Since Using  Schwarz Lemma we must have $|f'(0)|\leq 1$.
For 2. Correct. consider $f(z) = \frac{3}{4}z + \frac{3}{16}$
For 3. Correct. consider $f(z) = -\frac{3}{4}z - \frac{3}{16}$
For 4. Incorrect. Using Schwarz Lemma  (derivative version) i.e 
$|f'(z)| \leq \frac{1-|f(z)|^2}{1-|z|^2}$. Here given that $f'(\frac{1}{4})=1$. So we have $|f(\frac{1}{4})| \leq \frac{1}{4}$.
Also using Schwarz Pick Lemma we have $$\frac{|f(z) -f(w)|}{|1-f(z)\overline{f(w)}|} \leq \frac{|z-w|}{|1-z\bar{w}|}$$ 
Using above inequality and the fact that $f(\frac{1}{2}) =-\frac{1}{2}$, one will get 
\begin{align}
\frac{|f(\frac{1}{4}) + \frac{1}{2}|}{|1+ \frac{1}{2}f(\frac{1}{4})| } &\leq \frac{2}{7} \\
\implies |2 - \frac{3}{f(\frac{1}{4})+2}|&\leq \frac{2}{7} \\
\implies \frac{12}{7} \leq \frac{3}{|f(\frac{1}{4})+2|} &\leq \frac{16}{7}\\
\implies \frac{21}{16}\leq |f(\frac{1}{4})+2| &\leq \frac{21}{12} 
\end{align}
So $f(\frac{1}{4})$ belong to Annulus with centre at $-2$ and with inner radius $\frac{21}{16}$ and outer radius $\frac{21}{12}$.
we also have $|f(\frac{1}{4})| \leq \frac{1}{4}$ which gives us $f(\frac{1}{4})$ lies in disk with centre $0$ and radius $\frac{1}{4}$. This two domain intersect at $-\frac{1}{4}$. So, one must have $f(\frac{1}{4}) = -\frac{1}{4}$. 
Then we have $|f(\frac{1}{4})|= \frac{1}{4}$ and $f'(\frac{1}{4}) =1$. we have equality in the inequality $|f'(z)| \leq \frac{1-|f(z)|^2}{1-|z|^2}$ for $z = \frac{1}{4}$. Hence $f$ has to be scalar times identity function i.e $f(z) = cz$ for all $z$ in unit disk with $c$ is an unimodulur constant. But $f'(\frac{1}{2}) =1 $ gives us $c=1$. But that contradict $f(\frac{1}{2}) = - \frac{1}{2}$.
