The function` f(x) = cos (pi / x)` is decreasing in the interval

Options

A. `[2n + 1,2n),n in N`
B. ` ](1)/(2n+1) , 2n [, n in N`
C. `](1)/(2n +2),(1)/(2n +1) [ n in N`
D. none of these

Answer

To determine the interval in which the function \( f(x) = \cos\left(\frac{\pi}{x}\right) \) is decreasing, we need to find the derivative of the function and analyze where it is less than zero.

### Step 1: Differentiate the function
The derivative of \( f(x) \) can be found using the chain rule.

\[ f'(x) = \frac{d}{dx} \left( \cos\left(\frac{\pi}{x}\right) \right) = -\sin\left(\frac{\pi}{x}\right) \cdot \frac{d}{dx}\left(\frac{\pi}{x}\right) \]

Now, we differentiate \( \frac{\pi}{x} \):

\[ \frac{d}{dx}\left(\frac{\pi}{x}\right) = -\frac{\pi}{x^2} \]

Thus, we have:

\[ f'(x) = -\sin\left(\frac{\pi}{x}\right) \cdot \left(-\frac{\pi}{x^2}\right) = \frac{\pi \sin\left(\frac{\pi}{x}\right)}{x^2} \]

### Step 2: Set the derivative less than zero
To find the intervals where \( f(x) \) is decreasing, we need to solve the inequality:

\[ f'(x) < 0 \implies \frac{\pi \sin\left(\frac{\pi}{x}\right)}{x^2} < 0 \]

Since \( \frac{\pi}{x^2} > 0 \) for \( x > 0 \), we can simplify the inequality to:

\[ \sin\left(\frac{\pi}{x}\right) < 0 \]

### Step 3: Determine where the sine function is negative
The sine function is negative in the intervals where its argument is in the third and fourth quadrants. This occurs when:

\[ \frac{\pi}{x} \in (n\pi, (n+1)\pi) \quad \text{for } n \in \mathbb{Z} \]

This translates to:

\[ n\pi < \frac{\pi}{x} < (n+1)\pi \]

### Step 4: Solve for \( x \)
Rearranging the inequalities gives:

1. From \( n\pi < \frac{\pi}{x} \):
\[ x < \frac{1}{n} \]

2. From \( \frac{\pi}{x} < (n+1)\pi \):
\[ x > \frac{1}{n+1} \]

### Step 5: Combine the intervals
Thus, for each integer \( n \), the function \( f(x) \) is decreasing in the intervals:

\[ \left(\frac{1}{n+1}, \frac{1}{n}\right) \]

### Conclusion
The function \( f(x) = \cos\left(\frac{\pi}{x}\right) \) is decreasing in the intervals of the form \( \left(\frac{1}{n+1}, \frac{1}{n}\right) \) for \( n \in \mathbb{N} \).