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),ninN`
C. `((1)/(2n+2),(1)/(2n+1)),ninN`
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 its sign.

### Step 1: Find the derivative \( f'(x) \)

Using the chain rule, we differentiate \( f(x) \):

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

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

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

Thus, the derivative becomes:

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

### Step 2: Determine when \( f'(x) < 0 \)

For the function to be decreasing, we need:

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

Since \( \frac{\pi}{x^2} > 0 \) for \( x \neq 0 \), we need:

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

### Step 3: Analyze when \( \sin\left(\frac{\pi}{x}\right) < 0 \)

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

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

This simplifies to:

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

Dividing through by \( \pi \):

\[ n < \frac{1}{x} < n+1 \]

Taking the reciprocal (and reversing the inequalities):

\[ \frac{1}{n+1} < x < \frac{1}{n} \]

### Step 4: Generalize the intervals

For odd \( n \), we can express the intervals as:

- For \( n = 1 \): \( \frac{1}{2} < x < 1 \)
- For \( n = 3 \): \( \frac{1}{4} < x < \frac{1}{3} \)
- For \( n = 5 \): \( \frac{1}{6} < x < \frac{1}{5} \)

Thus, the general intervals where \( f(x) \) is decreasing are:

\[ \left(\frac{1}{n+1}, \frac{1}{n}\right) \quad \text{for odd } n \]

### Final Result

The function \( f(x) = \cos\left(\frac{\pi}{x}\right) \) is decreasing in the intervals:

\[ \left(\frac{1}{2n+1}, \frac{1}{2n}\right) \quad \text{for } n \in \mathbb{Z}^+ \]