To apply one-dimensional equivalence partitioning to the ID requirement, we need to consider each condition individually:
* Length of the ID: Valid partitions (5-10 characters), Invalid partitions (less than 5, more than 10) = 3 partitions.
* Type of characters: Valid partitions (alphanumeric), Invalid partitions (non-alphanumeric) = 2 partitions.
* First character: Valid partitions (letter), Invalid partitions (non-letter) = 2 partitions. Adding these partitions, we get a total of 3 (length) + 2 (character type) + 2 (first character) = 7 partitions. Thus, the correct answer is A.