If the solution set of inequality | x + 1 | - | X-1 | ≤ m is r, then the value range of real number m is r

If the solution set of inequality | x + 1 | - | X-1 | ≤ m is r, then the value range of real number m is r

The range of X has three segments: (negative infinity, - 1), (- 1,1), (1, positive infinity)
In (negative infinity, - 1), the original inequality is reduced to - X-1 + X-1 ≤ m, and - 2 ≤ M
In (- 1,1), the original inequality is reduced to + X + 1 + X-1 ≤ m, and 2x ≤ M
In (1, positive infinity), the original inequality is reduced to + X + 1-x + 1 ≤ m, and 2x + 2 ≤ M

The range of function y = (log14x) 2-log14x2 + 5 & nbsp; when & nbsp; 2 ≤ x ≤ 4 is______ ．

Let t = log14x, & nbsp; because 2 ≤ x ≤ 4, so - 1 ≤ t ≤ - 12, then y = (log14x) 2 − 2log14x + 5 = (t-1) 2 + 4, and because the function decreases monotonically in [- 1, - 12], when t = - 12 is the minimum value of 254, when t = - 1, the function has the maximum value of 8; so the answer is: {y | 254 ≤ y ≤ 8}

Given the set a = {x y = Log & # 8322; (1 + x) - Log & # 8322; (1-x)} (I) finding the set a
(II) if B = {y y = 2 Λ x, X ∈ a}, find the set B

(I) 1 + x > 0, so x > - 1
1-x > 0 so x