To determine the conditions under which the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \lambda x + \sin x \) is onto and one-to-one, we will analyze the function based on the given conditions for \( \lambda \).
### Step 1: Determine when \( f(x) \) is onto
A function is onto if its range is equal to its codomain. Here, the codomain is \( \mathbb{R} \).
To check if \( f(x) \) is onto, we need to analyze the behavior of \( f(x) \) as \( x \) varies over all real numbers.
1. **Case when \( \lambda = 0 \)**:
\[
f(x) = \sin x
\]
The range of \( \sin x \) is \( [-1, 1] \), which is not equal to \( \mathbb{R} \). Thus, \( f(x) \) is not onto.
2. **Case when \( \lambda \neq 0 \)**:
The term \( \lambda x \) dominates \( \sin x \) as \( x \) approaches \( \pm \infty \). Therefore, as \( x \to \infty \), \( f(x) \to \infty \) and as \( x \to -\infty \), \( f(x) \to -\infty \). This means that \( f(x) \) can take all real values when \( \lambda \neq 0 \).
Thus, for \( f(x) \) to be onto, \( \lambda \) must be in the set \( P = \mathbb{R} \setminus \{0\} \).
### Step 2: Determine when \( f(x) \) is one-to-one
A function is one-to-one if it is either strictly increasing or strictly decreasing. We will find the derivative of \( f(x) \):
\[
f'(x) = \lambda + \cos x
\]
1. **Condition for strictly increasing**:
\[
f'(x) > 0 \implies \lambda + \cos x > 0
\]
The maximum value of \( \cos x \) is 1, hence:
\[
\lambda + 1 > 0 \implies \lambda > -1
\]
2. **Condition for strictly decreasing**:
\[
f'(x) < 0 \implies \lambda + \cos x < 0
\]
The minimum value of \( \cos x \) is -1, hence:
\[
\lambda - 1 < 0 \implies \lambda < 1
\]
Combining these inequalities, we find:
\[
-1 < \lambda < 1
\]
Thus, for \( f(x) \) to be one-to-one, \( \lambda \) must be in the set \( Q = (-1, 1) \).
### Conclusion
- For \( f(x) \) to be onto, \( \lambda \in P = \mathbb{R} \setminus \{0\} \).
- For \( f(x) \) to be one-to-one, \( \lambda \in Q = (-1, 1) \).
### Final Result
The correct answer is that \( Q \) is a subset of \( P \).
