Let f(x)=`sin((pi)/x)` and `D={x:f(x)gt0)`, then D contains

Options

A. `(1/3,1/2)`
B. `(1/5,1/4)`
C. `(-1,-1/2)`
D. `(-pi,-1/2)`

Answer

To solve the problem, we need to determine the domain \( D \) where the function \( f(x) = \sin\left(\frac{\pi}{x}\right) \) is greater than 0.

### Step-by-Step Solution:

1. **Understand the Function**:
We have \( f(x) = \sin\left(\frac{\pi}{x}\right) \). The sine function is positive in certain intervals. Specifically, \( \sin(t) > 0 \) when \( t \) is in the intervals \( (2k\pi, (2k+1)\pi) \) for integers \( k \).

2. **Set the Condition**:
We need to find when \( \sin\left(\frac{\pi}{x}\right) > 0 \). This implies:
\[ \frac{\pi}{x} \in (2k\pi, (2k+1)\pi) \]
for integers \( k \).

3. **Solve for \( x \)**:
From the inequality \( 2k\pi < \frac{\pi}{x} < (2k+1)\pi \), we can multiply through by \( x \) (noting that \( x > 0 \)):
\[ 2kx < \pi < (2k+1)x \]

This gives us two inequalities:
- From \( 2kx < \pi \):
\[ x < \frac{\pi}{2k} \]
- From \( \pi < (2k+1)x \):
\[ x > \frac{\pi}{2k+1} \]

4. **Determine Valid Intervals**:
Thus, we have:
\[ \frac{\pi}{2k+1} < x < \frac{\pi}{2k} \]
for each integer \( k \).

5. **Find Specific Intervals**:
Let's calculate the intervals for different values of \( k \):
- For \( k = 1 \):
\[ \frac{\pi}{3} < x < \frac{\pi}{2} \]
- For \( k = 2 \):
\[ \frac{\pi}{5} < x < \frac{\pi}{4} \]
- For \( k = 3 \):
\[ \frac{\pi}{7} < x < \frac{\pi}{6} \]

6. **Combine Intervals**:
The domain \( D \) where \( f(x) > 0 \) is the union of all these intervals:
\[ D = \left( \frac{\pi}{3}, \frac{\pi}{2} \right) \cup \left( \frac{\pi}{5}, \frac{\pi}{4} \right) \cup \left( \frac{\pi}{7}, \frac{\pi}{6} \right) \cup \ldots \]

### Conclusion:
The intervals where \( f(x) > 0 \) are:
- \( \left( \frac{\pi}{3}, \frac{\pi}{2} \right) \)
- \( \left( \frac{\pi}{5}, \frac{\pi}{4} \right) \)
- \( \left( \frac{\pi}{7}, \frac{\pi}{6} \right) \)
- and so on.

Thus, \( D \) contains all \( x \) in these intervals.