`f:R to R"given by"f(x)=2x+|cos x|,` is

Options

A. one-one and into
B. one-one and onto
C. many-one and into
D. many-one and onto

Explanation

To analyze the function \( f(x) = 2x + |\cos x| \), we will determine its nature by finding its derivative and examining its behavior.

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

The function is given by:
\[ f(x) = 2x + |\cos x| \]

To differentiate \( f(x) \), we need to consider the derivative of \( |\cos x| \). The derivative of \( |\cos x| \) depends on the sign of \( \cos x \):

- If \( \cos x \geq 0 \), then \( |\cos x| = \cos x \) and \( \frac{d}{dx} |\cos x| = -\sin x \).
- If \( \cos x < 0 \), then \( |\cos x| = -\cos x \) and \( \frac{d}{dx} |\cos x| = \sin x \).

Thus, we can write the derivative \( f'(x) \) as:
\[ f'(x) = 2 + \frac{d}{dx} |\cos x| \]

This gives us two cases:

1. **Case 1**: When \( \cos x \geq 0 \):
\[ f'(x) = 2 - \sin x \]

2. **Case 2**: When \( \cos x < 0 \):
\[ f'(x) = 2 + \sin x \]

### Step 2: Analyze the derivative

Now, we will analyze the two cases to understand the behavior of the function.

- **For Case 1**: \( f'(x) = 2 - \sin x \)
- The sine function oscillates between -1 and 1. Therefore, \( f'(x) \) will oscillate between \( 2 - 1 = 1 \) and \( 2 + 1 = 3 \).
- Hence, \( f'(x) > 0 \) for all \( x \) in this case.

- **For Case 2**: \( f'(x) = 2 + \sin x \)
- Similarly, \( f'(x) \) will oscillate between \( 2 - 1 = 1 \) and \( 2 + 1 = 3 \).
- Therefore, \( f'(x) > 0 \) for all \( x \) in this case as well.

### Step 3: Conclusion about the function

Since \( f'(x) > 0 \) for all \( x \) in both cases, we can conclude that the function \( f(x) \) is **strictly increasing** over its entire domain \( \mathbb{R} \).

### Final Answer

The function \( f(x) = 2x + |\cos x| \) is strictly increasing.

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