Merge sort and binary search are both divide-and-conquer, yet one is O(n log n) and the other O(log n). Why does the same strategy give such different costs?

A. Because merge sort recurses into both halves and merges O(n) per level.

Both halve the problem, so both have about log n levels; what differs is the work per level. Binary search discards half and does constant work; merge sort recurses into both halves and touches every element merging, giving O(n) per level. That is what the Master Theorem formalises — how many subproblems, how much smaller, how much combining — and it lets you predict a design change.