If `f(x)`=int_o^x sint/t dt` which of the following is true?

Options

A. `f (0) gt f (1.1)`
B. `f (0) lt f (1.1) gt f (2.1)`
C. `f (0) lt f (1.1)lt f (2.1) gt f (3.1)`
D. `f (0) lt f (1.1) lt f (2.1) lt f (3.1) gt f (4.1)`

Answer

To solve the problem where \( f(x) = \int_0^x \frac{\sin t}{t} \, dt \), we need to analyze the function and determine which of the given statements is true.

### Step-by-Step Solution:

1. **Define the Function**:
We start with the definition of the function:
\[ f(x) = \int_0^x \frac{\sin t}{t} \, dt \]

2. **Differentiate the Function**:
To find \( f'(x) \), we can apply the Fundamental Theorem of Calculus:
\[ f'(x) = \frac{\sin x}{x} \]

3. **Evaluate \( f'(0) \)**:
We need to find the limit of \( f'(x) \) as \( x \) approaches 0:
\[ f'(0) = \lim_{x \to 0} \frac{\sin x}{x} \]
Using the standard limit, we know:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \implies f'(0) = 1 \]

4. **Analyze the Behavior of \( f'(x) \)**:
The function \( f'(x) = \frac{\sin x}{x} \) is positive for \( x > 0 \) and decreases as \( x \) increases because \( \sin x < x \) for \( x > 0 \). This means \( f'(x) < 1 \) for \( x > 0 \).

5. **Conclude on the Monotonicity of \( f(x) \)**:
Since \( f'(x) \) is positive and decreasing, \( f(x) \) is increasing but at a decreasing rate. Thus, for \( x_1 < x_2 \), we have:
\[ f(x_1) > f(x_2) \]

6. **Evaluate Specific Values**:
Specifically, we can evaluate:
\[ f(0) > f(1.1) > f(2.1) > f(3.1) \]
This indicates that as \( x \) increases, \( f(x) \) decreases.

### Conclusion:
From the analysis, we conclude that:
\[ f(0) > f(1.1) > f(2.1) > f(3.1) \]
Thus, the correct statement is that \( f(0) \) is greater than \( f(1.1) \).