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As we'll see, this problem requires an understanding of how graphs work.
Let 2 0 2 1 a + sin 4 ( 2 0 2 1 2 a 6 1 x ) − ∣ x 2 − 2 0 2 1 2 0 2 1 a 1 2 ∣ = 0 . We have:
2 0 2 1 a + sin 4 ( 2 0 2 1 2 a 6 1 x ) = ∣ x 2 − 2 0 2 1 2 0 2 1 a 1 2 ∣
There is no point in trying to manipulate this equation any further, so let's think about how this equation looks visually. It's an intersection of a trigonometric function (which looks like similar to sine, but it's restricted to positive values) and an "absolute parabola". This absolute parabola is x 2 , but translated 1 unit down 2 0 2 1 2 0 2 1 a 1 2 times, and everything below the x-axis is mirrored to be above the x-axis. We need to find how many times these two graphs intersect. Notice that since 2 0 2 1 2 0 2 1 a 1 2 (which lifts the remainder of the parabola up) is much bigger than 2 0 2 1 a (which lifts the trigonometric function up), these 2 graphs will always intersect at least 4 times, which should be clear intuitively. Another important characteristic about this trigonometric function is that regardless of a , its frequency is so small that there is no chance that the parabola will ever intersect one of its waves more than 4 times, just because of how fast it grows - i.e. how high its derivative is, compared to the trigonometric function. So from this we conclude that the number of intersections is 4 .
Here is how it should look graphically. Note that I put much smaller values than the ones in this exercise, for presentation purposes.