Given the function y = (A-2) x-3a-1, when the value range of the independent variable x is 3 ≤ x ≤ 5, y can reach a value greater than 5 and a value less than 3, then the value range of the real number a is () A. A ＜ 3B. A ＞ 5C. A ＞ 8D. Any real number

Given the function y = (A-2) x-3a-1, when the value range of the independent variable x is 3 ≤ x ≤ 5, y can reach a value greater than 5 and a value less than 3, then the value range of the real number a is () A. A ＜ 3B. A ＞ 5C. A ＞ 8D. Any real number

If A-2 ＞ 0, that is, a ＞ 2, the function is an increasing function. According to the meaning of the problem, when x = 5, y ＞ 5, that is, (A-2) × 5-3a-1 ＞ 5, the solution is a ＞ 8; when x = 3, y ＜ 3, that is, (A-2) × 3-3a-1 ＜ 3, no matter what the real number inequality of a is, a ＞ 8; if a = 2, y = - 7, it is not the meaning of the problem; if A-2 ＜ 0, that is a ＜ 2, the function is a decreasing function; when x = 3, y ＞ 5, that is (A-2) × 3-3a-1 ＞ 5, it is a ＞ 2 The inequality does not hold. Therefore, there is no such case