Find the number of values of `f(x)=[x/15][-15 /x]` can take where `x in (0,90)` where [.] =GIF

Options

A. 5
B. 6
C. 7
D. Infinite

Answer

To solve the problem of finding the number of values that the function \( f(x) = \left[\frac{x}{15}\right] \left[-\frac{15}{x}\right] \) can take for \( x \in (0, 90) \), we will analyze the function step by step.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) \) is defined as the product of two greatest integer functions (GIFs):
- \( \left[\frac{x}{15}\right] \) is the greatest integer less than or equal to \( \frac{x}{15} \).
- \( \left[-\frac{15}{x}\right] \) is the greatest integer less than or equal to \( -\frac{15}{x} \).

2. **Finding the Range of \( x \)**:
We will consider intervals of \( x \) from \( 0 \) to \( 90 \) and analyze the function in segments.

3. **Interval \( (0, 15) \)**:
- For \( x \in (0, 15) \), \( \frac{x}{15} \) ranges from \( 0 \) to \( 1 \).
- Thus, \( \left[\frac{x}{15}\right] = 0 \).
- \( -\frac{15}{x} \) ranges from \( -\infty \) to \( -1 \), so \( \left[-\frac{15}{x}\right] \) will be \( -1, -2, -3, \ldots \).
- Therefore, \( f(x) = 0 \cdot \text{(any integer)} = 0 \).

4. **Interval \( [15, 30) \)**:
- For \( x \in [15, 30) \), \( \frac{x}{15} \) ranges from \( 1 \) to \( 2 \).
- Thus, \( \left[\frac{x}{15}\right] = 1 \).
- \( -\frac{15}{x} \) ranges from \( -1 \) to \( -0.5 \), so \( \left[-\frac{15}{x}\right] = -1 \).
- Therefore, \( f(x) = 1 \cdot (-1) = -1 \).

5. **Interval \( [30, 45) \)**:
- For \( x \in [30, 45) \), \( \frac{x}{15} \) ranges from \( 2 \) to \( 3 \).
- Thus, \( \left[\frac{x}{15}\right] = 2 \).
- \( -\frac{15}{x} \) ranges from \( -0.5 \) to \( -0.333 \), so \( \left[-\frac{15}{x}\right] = -1 \).
- Therefore, \( f(x) = 2 \cdot (-1) = -2 \).

6. **Interval \( [45, 60) \)**:
- For \( x \in [45, 60) \), \( \frac{x}{15} \) ranges from \( 3 \) to \( 4 \).
- Thus, \( \left[\frac{x}{15}\right] = 3 \).
- \( -\frac{15}{x} \) ranges from \( -0.333 \) to \( -0.25 \), so \( \left[-\frac{15}{x}\right] = -1 \).
- Therefore, \( f(x) = 3 \cdot (-1) = -3 \).

7. **Interval \( [60, 75) \)**:
- For \( x \in [60, 75) \), \( \frac{x}{15} \) ranges from \( 4 \) to \( 5 \).
- Thus, \( \left[\frac{x}{15}\right] = 4 \).
- \( -\frac{15}{x} \) ranges from \( -0.25 \) to \( -0.2 \), so \( \left[-\frac{15}{x}\right] = -1 \).
- Therefore, \( f(x) = 4 \cdot (-1) = -4 \).

8. **Interval \( [75, 90) \)**:
- For \( x \in [75, 90) \), \( \frac{x}{15} \) ranges from \( 5 \) to \( 6 \).
- Thus, \( \left[\frac{x}{15}\right] = 5 \).
- \( -\frac{15}{x} \) ranges from \( -0.2 \) to \( -0.1667 \), so \( \left[-\frac{15}{x}\right] = -1 \).
- Therefore, \( f(x) = 5 \cdot (-1) = -5 \).

9. **Summary of Values**:
The possible values of \( f(x) \) in the intervals are:
- From \( (0, 15) \): \( 0 \)
- From \( [15, 30) \): \( -1 \)
- From \( [30, 45) \): \( -2 \)
- From \( [45, 60) \): \( -3 \)
- From \( [60, 75) \): \( -4 \)
- From \( [75, 90) \): \( -5 \)

10. **Counting Unique Values**:
The unique values that \( f(x) \) can take are \( 0, -1, -2, -3, -4, -5 \). Thus, there are a total of **6 unique values**.

### Final Answer:
The number of values that \( f(x) \) can take is **6**.