Given that point a (2-A, a + 1) is in the fourth quadrant, the value range of a is () A. a＜1B. a＜-1C. -1＜a＜2D. a＞2

∵ point is in the fourth quadrant, the abscissa of the point is a positive number, and the ordinate is also a negative number, that is, 2-A ＞ 0, a + 1 ＜ 0, the solution is a ＜ - 1

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