A is true. Finding the maximum requires examining the elements, so it can be done in $\Theta(n)$
B is true. One simple method is:
- Sort the array in $O(n\log n)$.
- Return the element at position $i$.
So $O(n\log n)$ is certainly possible.
C is false.
Even if one call to selection takes $O(n)$ that does not mean the whole array can be sorted in $O(n)$.
To obtain all order statistics independently would require many selection operations.
Also, comparison-based sorting has a lower bound of $\Omega(n\log n)$ so a general linear-time comparison sort would contradict that lower bound.
D is true. The three largest elements can be maintained during one scan $O(n)$
E is also true. Minimum and maximum can be tracked simultaneously during one traversal.
Answer: C