Given that P: 1 / 2 ≤ x ≤ 1, Q: (x-a) (x-a-1) ≤ 0, if P is a sufficient and unnecessary condition of Q, then the value range of real number a is Methods ~ this kind of problem is embarrassing

Given that P: 1 / 2 ≤ x ≤ 1, Q: (x-a) (x-a-1) ≤ 0, if P is a sufficient and unnecessary condition of Q, then the value range of real number a is Methods ~ this kind of problem is embarrassing

If P is a sufficient and unnecessary condition of Q, then the solution of Q contains the solution of P
The inequality of solution q is x ≤ A and X ≥ a + 1, or a ≤ x ≤ a + 1
Then the value of a is: a ≤ 1 / 2 and a + 1 ≥ 1, that is, 0 ≤ a ≤ 1 / 2

The known propositions are p: | X-2 | 0), Q: | x ^ 2-4|

Since proposition p: | x – 2 | < a (a > 0), proposition q: | x ^ 2 – 4 | < 1, if P is a sufficient and unnecessary condition of Q,
So p can deduce Q, that is, the solution set of X in P is the proper subset of the solution set of X in Q
|The solution of X – 2 | < A is - a 2

It is known that P: x2-8x-20 ＞ 0, Q: x2-2x + 1-a2 ＞ 0. If P is a sufficient and unnecessary condition of Q, the value range of positive real number a is obtained

p: X ＜ - 2 or ＞ 10, Q: X ＜ 1-A or X ＞ 1 + a ∵ where p is a sufficient and unnecessary condition for Q, ∵ 1 − a ≥ − 21 + a ≤ 10, that is, 0 ＜ a ≤ 3