Given that the even function f (x) = loga ∣ ax + B ∣ increases monotonically on (0, + ∞), then the size relation between F (b-2) and f (a + 1) is obtained

Given that the even function f (x) = loga ∣ ax + B ∣ increases monotonically on (0, + ∞), then the size relation between F (b-2) and f (a + 1) is obtained

If f (x) = loga ∣ ax + B ∣ is an even function, then:
f(-x)= f(x) loga∣-ax+b∣=loga∣ax+b∣
∣ - ax + B ∣ = ∣ ax + B ∣ so B = 0
Then f (x) = loga ∣ ax|
A is a time-increasing function with base greater than 0, ∣ ax | on (0, + ∞),
According to the principle of compound function "same increase different decrease", the base a must be greater than 1
Because f (b-2) = f (- 2) = f (2) and a + 1 > 2
And f (x) increases monotonically on (0, + ∞), so f (a + 1) > F (2)
That is, f (a + 1) > F (b-2)