Exam CTFL v4.0 topic 1 question 27 discussion

Can somebody explain this answer? Given that "In 2-value BVA (Craig 2002, Myers 2011), for each boundary value there are two coverage items: this boundary value and its closest neighbor belonging to the adjacent partition. To achieve 100% coverage with 2-value BVA, test cases must exercise all coverage items, i.e., all identified boundary values. Coverage is measured as the number of boundary values that were exercised, divided by the total number of identified boundary values, and is expressed as a percentage." it seems like it should be 0, 1, 99, 100 by the options given..

Boundary is 1 to 99 because 0 < X < 100 so below the boundary is 0 and beyond the boundary is 100 so it should be C) 0,1,99,100

Boundary value analysis focuses on testing values at the edges of the input domain. The two-point boundary value analysis technique suggests testing values just below, at, and just above the boundaries. For the given requirement (x > 0 and x < 100), the boundaries are 0 and 100. Here’s the breakdown: -1: Just below the lower boundary (0). 1: Just above the lower boundary (0). 99: Just below the upper boundary (100). 101: Just above the upper boundary (100).

Boundaries are 0, 1 - 99, 100 .. Using a 2 point BVA, the coverage items are the boundary values, hence 0,1,99, 100. Answer should be C

Invalid range- 0 -> test values: 0 Valid range 1 - 99 -> Test values 0,1, 99, 100 Invalid range 100 to inf -> test values: 99, 100 to sum up: 0,1,99,100

I thin the answer is C Invalid range- 0 -> test values: 0 Valid range 1 - 99 -> Test values 0,1, 99, 100 Invalid range 100 to inf -> test values: 99, 100 to sum up: 0,1,99,100 C has all there values, nothing more nothing less