Sol. The given inequality

|a 2x + a x +2 –1| ≥ 1 ...(1)

Using the notation t = a x , we can write inequality (1) in

the form

|t 2 + a 2 t –1| ≥ 1 … (2)

Where t > 0. Since t 2 + a 2 t –1 =

∀ a,

inequality (2) is equivalent to the collection of two

systems (since t > 0)

Let us solve the first system. The inequality –t 2 –a 2 t + 1 ≥ 1 is equivalent to the inequality t 2 + a 2 t ≤ 0 which has no solutions for t > 0, and consequently, the first system has no solutions. We solve the second system, the inequality t 2 + a 2 t –1 ≥ 1, is equivalent to the collection of two inequalities

t ≤ (–a 2 – ), t ≥ (–a 2

+ ), Since (–a 2 + ) > (–a 2 + ), the second system of the collection is

equivalent to the inequality t ≥ (–a 2 + ).

Consequently, for a > 0, a ≠ 1 and any

x inequality (1) is equivalent to the inequality

a x ≥ (–a 2 + ).

Hence we find that

for 0 < a < 1 inequality (1) has a solution x ≤ log a

,

for a > 1 inequality (1) has a solution x ≥ log a