The fucntion f(x)`=(sin x)/(x)` is decreasing in the interval

Options

A. `(-pi/2, 0 )`
B. `(0, pi//2)`
C. `(0, pi)`
D. none of these

Explanation

To determine the interval in which the function \( f(x) = \frac{\sin x}{x} \) is decreasing, we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( \frac{\sin x}{x} \right) \]
Using the quotient rule, we have:
\[ f'(x) = \frac{x \cdot \cos x - \sin x \cdot 1}{x^2} = \frac{x \cos x - \sin x}{x^2} \]

### Step 2: Set the derivative less than zero
For the function to be decreasing, we need:
\[ f'(x) < 0 \]
This implies:
\[ \frac{x \cos x - \sin x}{x^2} < 0 \]
Since \( x^2 \) is always positive for \( x \neq 0 \), we can focus on the numerator:
\[ x \cos x - \sin x < 0 \]
This simplifies to:
\[ x \cos x < \sin x \]

### Step 3: Rearranging the inequality
Rearranging gives us:
\[ x < \frac{\sin x}{\cos x} = \tan x \]
Thus, we need to find the intervals where \( x < \tan x \).

### Step 4: Analyze the function \( g(x) = \tan x - x \)
To find where \( x < \tan x \), we can analyze the function:
\[ g(x) = \tan x - x \]
We need to find where \( g(x) > 0 \).

### Step 5: Determine intervals of \( g(x) \)
1. **At \( x = 0 \)**:
\[ g(0) = \tan(0) - 0 = 0 \]
2. **For \( x \) in \( (0, \frac{\pi}{2}) \)**:
The function \( \tan x \) increases faster than \( x \), so \( g(x) > 0 \).
3. **At \( x = \frac{\pi}{2} \)**:
\( g(x) \) approaches infinity.
4. **For \( x \) in \( (\frac{\pi}{2}, \pi) \)**:
The function \( \tan x \) becomes negative, and \( g(x) < 0 \).

### Conclusion
From our analysis, we conclude that \( f(x) \) is decreasing in the interval:
\[ (0, \frac{\pi}{2}) \]