The function `f:R rarr R` be defined by `f(x)=2x+cosx` then `f`

Options

A. has minimum at x=`pi`
B. has a maximum at x=0
C. is a decreasing function
D. is in increasing function

Explanation

To solve the problem, we need to analyze the function \( f(x) = 2x + \cos x \) and determine its behavior, specifically whether it is increasing or decreasing.

### Step-by-Step Solution:

1. **Identify the function**:
We have the function defined as:
\[ f(x) = 2x + \cos x \]

2. **Differentiate the function**:
We need to find the derivative \( f'(x) \) to analyze the function's behavior. The derivative of \( f(x) \) is calculated as follows:
\[ f'(x) = \frac{d}{dx}(2x) + \frac{d}{dx}(\cos x) \]
Using the rules of differentiation:
- The derivative of \( 2x \) is \( 2 \).
- The derivative of \( \cos x \) is \( -\sin x \).

Therefore, we have:
\[ f'(x) = 2 - \sin x \]

3. **Analyze the derivative**:
We need to determine the sign of \( f'(x) \):
\[ f'(x) = 2 - \sin x \]
The sine function, \( \sin x \), oscillates between -1 and 1 for all real \( x \). Therefore, the maximum value of \( \sin x \) is 1.

Substituting the maximum value of \( \sin x \) into the derivative:
\[ f'(x) = 2 - 1 = 1 \]
Since \( f'(x) \) is always greater than 0 (because \( 2 - \sin x \geq 2 - 1 = 1 > 0 \)), we conclude that:
\[ f'(x) > 0 \quad \text{for all } x \in \mathbb{R} \]

4. **Conclusion**:
Since the derivative \( f'(x) \) is positive for all real values of \( x \), the function \( f(x) \) is an increasing function for all \( x \in \mathbb{R} \).

### Final Answer:
The function \( f(x) = 2x + \cos x \) is an increasing function for all \( x \in \mathbb{R} \).