To answer this question, let's analyze the given expression |sin(2x)| as x increases from π/4 to 3π/4.
The function sin(2x) is an oscillating function with a period of π. That means it completes one full cycle from 0 to 2π.
When x increases from π/4 to 3π/4, we can see that the values of 2x also increase. Let's consider the values of 2x within the given range:
When x = π/4, 2x = 2(π/4) = π/2
When x = 3π/4, 2x = 2(3π/4) = 3π/2
Now, let's evaluate the absolute value of sin(2x) at these two points:
|sin(π/2)| = 1
|sin(3π/2)| = |sin(π + π/2)| = |sin(π/2)| = 1
We can observe that the absolute value of sin(2x) is equal to 1 at both x = π/4 and x = 3π/4.
Therefore, as x increases from π/4 to 3π/4, |sin(2x)| does not always increase or always decrease. Instead, it remains constant at 1 throughout this interval.
Hence, the correct answer is D) decreases then increases.