Final answer:
The minimum number of comparisons required to find if an element occurs more than n/2 times in a sorted array is O(log n), which is achieved using a modified binary search algorithm.
Step-by-step explanation:
The correct answer is option B. ⋅(log n). Since the array is sorted, a modified binary search algorithm can be used to find the first and last occurrence of the given element. If the element in question is to occur more than n/2 times in an array of size n, these occurrences are guaranteed to span the middle of the array.
Therefore, by performing a binary search to find the index of the first and the last occurrence of the element, we can determine the total number of occurrences of that element.
If the distance between the first and last occurrence indices is greater than n/2, the element occurs more than n/2 times. This approach requires only O(log n) comparisons, where n is the number of elements in the array.