The math problem in senior one requires an explanation of the third problem.

CuB This is the complement of set B. Assuming that b is x≥ 1, its complement is x < 1.

Because the intersection of complement sets of set A and set B is-1 ≤ x < 1, and set A- 1 ≤ x < 2,

Just bring these directly into the competition one by one.

AC is all qualified, no problem.

The difference of BD lies in whether it is equal to-1.

Assuming that set B contains-1, then the complement of set B does not have-1, then the intersection of set A and the complement of set B does not have-1, which contradicts the topic, so set B does not have-1.

The answer should be D.

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