Let $A$ be a priority queue for maintaining a set of elements. Suppose $A$ is implemented using a max-heap data structure. The operation $\text{EXTRACT-MAX}(A)$ extracts and deletes the maximum element from $A$. The operation $\text{INSERT}(A, key)$ inserts a new element $key$ in $A$. The properties of a max-heap are preserved at the end of each of these operations. When $A$ contains $n$ elements, which one of the following statements about the worst case running time of these two operations is TRUE?

In a binary max-heap containing $n$ elements:

1. $\text{EXTRACT-MAX}(A)$:

   • The maximum element is always at the root ($A[1]$).

   • To delete it and maintain the max-heap property, we replace the root with the last element of the heap and decrease the heap size by 1.

   • Then, we call MAX-HEAPIFY on the root, which traverses down the height of the tree in the worst case.

   • Since the height of a binary heap with $n$ elements is $\Theta(\log n)$, the worst-case running time of $\text{EXTRACT-MAX}$ is $O(\log n)$.

2. $\text{INSERT}(A, key)$:

   • The new element is placed at the end of the heap (the next available leaf node).

   • Then, it bubbles up along the path to the root until the max-heap property is satisfied.

   • The maximum distance it can travel is the height of the heap, which is $\Theta(\log n)$.

   • Thus, the worst-case running time of $\text{INSERT}$ is $O(\log n)$.

Therefore, both operations run in $O(\log(n))$ in the worst case. Option (B) is correct.