Is there an NP-complete problem that can be solved in $O(n^{\log n})$ time? I'm following an online course which has the following (multiple-choice) quiz question:


  Which of the following statements cannot be true, given the current
  state of knowledge?
  
  
  Some NP-complete problems are polynomial-time solvable, and some NP-complete problems are not polynomial-time solvable.
  There is an NP-complete problem that is polynomial-time solvable.
  There is an NP-complete problem that can be solved in $ O(n^{\log n}) $ time, where $ n $ is the size of the input.
  There is no NP-complete problem that can be solved in $ O(n^{\log n}) $ time, where $ n $ is the size of the input.
  


It seems to me that (1) and (2) are false since it has not been proven that any NP-complete problem is polynomial-time solvable. So I'm vacillating between (3) and (4).

In particular, I'm a bit nonplussed where the $O(n^{\log n})$ comes from; I haven't seen this expression in any of the lecture notes. However, I do know about the traveling salesman problem which can be solved in $O(n^2 2^n)$ time using the Held-Karp algorithm. It would seem to me like this function grows less faster asymptotically than $O(n^{\log n})$, since it is usually the expression in the exponent that matters the most, in which case the answer would be (3).

Is this correct? Also, how would I show this formally?
The accepted answer is incorrect. I've taken the same course, and option 1 is correct. They didn't say why, but I have the following argument.
Quoting Wikipedia:

polynomial time: An algorithm is said to be of polynomial time if its running time is upper bounded by a polynomial expression in the size of the input for the algorithm, i.e., $ T(n) = O(n^k) $ for some positive constant $ k $.
exponential time: An algorithm is said to be exponential time, if $ T(n) $ is upper bounded by $ 2^{poly(n)} $, where $ poly(n) $ is some polynomial in $ n $. More formally, an algorithm is exponential time if $ T(n) $ is bounded by $ O(2^{n^k}) $ for some constant $ k $.

We don't know $ P \neq NP $, so option 2 can't be ruled out.
For all we know, the running time required to solve NP-complete problems could be anywhere between polynomial and exponential (note that $ n^{\log n} $ is more than polynomial but less than exponential). Thus, options 3 and 4 can't be ruled out.
That only leaves option 1. The claim in option 1 is incorrect because if there's a polynomial-time algorithm for any NP-complete problem X, then every other NP problem has a polynomial-time algorithm by reduction to X.