Let the funciton `f:R to R` be defined by `f(x)=2x+sin x`. Then, f is

Options

A. one-to-one and into
B. one-to-one but not onto
C. onto but not one-to-one
D. neither one-to-one nor onto

Explanation

To determine the properties of the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = 2x + \sin x \), we need to check if the function is one-to-one (injective) and onto (surjective).

### Step 1: Check if the function is one-to-one (injective)

To check if \( f \) is one-to-one, we can analyze its derivative.

1. **Find the derivative of \( f(x) \)**:
\[ f'(x) = \frac{d}{dx}(2x + \sin x) = 2 + \cos x \]

2. **Analyze the derivative**:
- The cosine function \( \cos x \) oscillates between -1 and 1.
- Therefore, \( f'(x) = 2 + \cos x \) will oscillate between \( 2 - 1 = 1 \) and \( 2 + 1 = 3 \).
- This means \( f'(x) \) is always greater than 0:
\[ f'(x) \geq 1 > 0 \]

3. **Conclusion about injectivity**:
Since \( f'(x) > 0 \) for all \( x \in \mathbb{R} \), the function \( f(x) \) is strictly increasing. A strictly increasing function is one-to-one.

### Step 2: Check if the function is onto (surjective)

To check if \( f \) is onto, we need to determine the range of \( f(x) \).

1. **Determine the range of \( f(x) \)**:
- The term \( 2x \) can take any value from \( -\infty \) to \( \infty \) as \( x \) varies over \( \mathbb{R} \).
- The term \( \sin x \) oscillates between -1 and 1.
- Therefore, the function \( f(x) = 2x + \sin x \) can be expressed as:
\[ f(x) \in (-\infty, \infty) + [-1, 1] \]
- This means the range of \( f(x) \) is still \( (-\infty, \infty) \).

2. **Conclusion about surjectivity**:
Since the range of \( f(x) \) is \( (-\infty, \infty) \), which covers all real numbers, the function \( f(x) \) is onto.

### Final Conclusion

Since \( f(x) \) is both one-to-one and onto, we conclude that:

- The function \( f(x) = 2x + \sin x \) is **one-to-one and onto**.