We need to find the number of elements in the range set of the function \(f(x) = \left[ \frac{x}{15} \right] \left[ -\frac{15}{x} \right]\) for \(x \in (0, 90)\).
Let's break down the function \(f(x)\):
Let's analyze each component:
The function \(f(x) = \left[\frac{x}{15}\right]\left[-\frac{15}{x}\right]\) is thus a product of possible integers for each component.
Let's consider values which are possible for \(f(x)\):
| \(\left[\frac{x}{15}\right]\) | \(\left[-\frac{15}{x}\right]\) | \(\left[\frac{x}{15}\right] \left[-\frac{15}{x}\right]\;\) |
|---|---|---|
| 0 | -1 | 0 |
| 1 | -1, -2 | -1, -2 |
| 2 | -1, -2, -3, -4 | -2, -3, -4, -6 |
| 3 | -1, -2, -3, -4, -5 | -3, -4, -6, -9, -12 |
| 4 | -1, -2, -3, -4 | -4, -6, -8, -12 |
| 5 | -1, -2, -3 | -5, -10, -15 |
From this, possible values for \(f(x)\) are: \(0, -1, -2, -3, -4, -5, -6, -9\).
Counting distinct values gives us \(8\) possible values.
Therefore, the number of elements in the range set of \(f(x)\) is \(8\).
For a real number $y$, consider $[y]$ denotes the greatest integer less than or equal to $y$.
If $f(x) = \frac{\tan(\pi[x-\pi])}{1+[x]^2}$, then