Choosing MLE(Maximum likelihood estimator) from three different estimators(Uniform distribution) Let $X_1,X_2....X_{2n}$ be random sample from $U(\theta,\theta+1)$. Consider the following three estimators for $\theta,\theta\in R$
$$T_1=x_{(n)}$$
$$T_2=\dfrac{x_{(1)}+x_{(n)}}{2}$$
$$T_3=\dfrac{x_{(1)}+x_{(n)}}{2}-\dfrac{1}{2}$$
Then 
$(A)$ $T_1$ and $T_3$ are both MLE of $\theta$ while $T_2$ is not 
$(B)$ $T_3$ is unique  MLE of $\theta$ 
$(C)$ $T_3$ is an MLE of $\theta$ while $T_1$ and $T_2$ are not 
$(D)$$T_1$ and $T_2$ are both MLE of $\theta$ while $T_3$ is not 
My input:
Every statistic satisfying following condition are MLE of $\theta $
$ \theta \leq x_{(1)}\leq x_{(2)}....x_{(n)}\leq\theta +1$
$ \theta \leq x_{(1)}\leq x_{(2)}....(x_{(n)}-1)\leq\theta$
That means $x_{(n)}$ can't be MLE of $\theta$ where as $(x_{(n)}-1)$ can. 
$T_2 \ \ $and $\ \  T_3$  follow my condition $T_1$ doesn't.(Please correct me if there is something wrong in my input). And any of my option doesn't matches up with my thought process. Did i do something wrong or it's just typo in the options. And what could be an alternative way to think about these types of problems. 
Original image:http://prntscr.com/i8a5py
 A: You are right to care for the following:
\begin{align}
\theta \leq x_{(1)} \leq ... \leq x_{(n)} \leq \theta +1 
\end{align}
That is exactly what you would obtain if you would look at the Likelihood (that is what you did right?). 
So we want that $$\tag{1}\theta\leq x_{(1)} \ \ \ \text{ and } \ \ \  x_{(n)}\leq \theta+1$$ and that implies the inequalities you had in the middle. The likelihood function is constant (which constant?) and it is nonzero if the inequality in $(1)$ is statisfied. That means that an MLE $T$ satisfies:
\begin{align}
x_{(n)}-1\leq T \leq x_{(1)}
\end{align} 
Now you immediately see that it cannot be unique, so we can forget $(B)$.
Let's look at the estimators you have. First $T_1$ does not satisfy $T_1\leq x_{(1)}$ in general so we drop it. Secondly $T_2$ does not satisfy $T_2\leq x_{(1)}$ either so we throw it away. There is no need to check if $T_3$ is an MLE, there is only one option left, namely $(C)$. I'll advice you to check it yourself that it indeed satisfies $(1)$.
A: First, note that $X_{(n)}-X_{(1)} \le 1$. Second, if $T_i$ is an MLE of $\theta$ then $$X_{(n)}-1 \le T_i \le X_{(1)},$$
as you found.
Amongst the given $T_i$'s, only $T_3$ satisfies the above condition. Therefore, the correct option is (C).
