Consider the following problems:
Let be the number of points of discontinuity of in the interval .
Let be the number of solutions to .
Find .
Notation : denotes the floor function .
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Plotting the graph (or a little bit of investigation of the function f(x) ) shows that G = 1 because the graph is discontinuous at all points where | x | = n where n is a non-negative integer, except for n = 1 . So in the given interval it is discontinuous at 0 < 2 < 1 . 6 only.
Now G − 1 = 0 implies ( G − 1 ) ( H − 1 ) = 0
so that G + H − G H = 1 .
Also if we want, we can find the value of H by considering that x = 0 is a solution to the given equation. Also the number of positive solutions is equal to the number of negative solutions which can be identified from the shape of the graph only. Also from the fact that y = x and y = 9 9 s i n ( π x ) both are odd, if x 0 is a solution then − x 0 is a solution.
The number of positive solutions is 99, which can be found from sketching the graph of the two functions itself.