If the equation | x | = ax + 1 about X has and has only one negative root, the value range of a is obtained ||It's absolute

If the equation | x | = ax + 1 about X has and has only one negative root, the value range of a is obtained ||It's absolute

With negative roots, X-1

Given that the equation | x | = ax + 1 has a negative root but no positive root, the value range of a is ()
A. A ≥ 1b. A ＜ 1C. - 1 ＜ a ＜ 1D. A ＞ - 1 and a ≠ 0

∵ the equation | x | = ax + 1 has a negative root but no positive root, ∵ x ＜ 0, the equation is changed to: - x = ax + 1, X (a + 1) = - 1, x = − 1A + 1 ＜ 0, ∵ a + 1 ＞ 0, ∵ a ＞ - 1, and a ≠ 0. If x ＞ 0, | x | = x, x = ax + 1, x = 11 − a ＞ 0, then 1-A ＞ 0, the solution is a ＜ 1. ∵ a ＜ 1 is not established without positive root, ∵ a ＞ 1. Therefore, select a

If LXL ≤ 1, the value of y = ax + 2A + 1 has positive and negative, then the value range of a is

The maximum and minimum of a function are at the end points
-1

It is known that the equation | x | = ax + 1 about X has a negative root and no positive root
The answer to this question is that a is greater than or equal to 1. But I calculated that a is greater than - 1

|x|=ax+1
x=ax + 1 (x>=0) -x = ax+1 (x=0 a+1>0
a>=1 a>-1
If we take the intersection of the two, we get a > = 1