In binary search, average number of comparison required for searching an element in a list if nnumbers is# (A) n log2 .(B) 2/n .(C) n.(D) n – 1.(E)n/2

The correct answer is (A) n log2.

Binary search is an efficient algorithm for finding an item from a sorted list of items. It works by repeatedly dividing in half the portion of the list that could contain the item, until you've narrowed down the possible locations to just one.

Here's a step-by-step explanation:

1. Start with the middle element: Compare the middle element of the list with the target value. If it matches, return the index of the middle element.

2. If the target value is less than the middle element, repeat the process with the left half of the list. Otherwise, repeat the process with the right half of the list.

3. Repeat steps 1 and 2 until you find the target value or until the list cannot be halved any further.

The number of comparisons in binary search is given by log2(n) in the worst case scenario, where n is the number of elements in the list. This is because with each comparison, you're effectively halving the search space.

However, the question asks for the average number of comparisons, which is a bit trickier. In the average case, you might find the target value before you've halved the list all the way down to one element. But even in the average case, the number of comparisons is still proportional to log2(n), so (A) is the best answer.