If the real-valued function f(x) = px + sinx is a bijective function, then the set of all possible values of `p in R` is

Options

A. R – {0}
B. R
C. (0, `prop`)
D. None of these

Answer

To determine the set of all possible values of \( p \) in the function \( f(x) = px + \sin x \) such that it is a bijective function, we need to ensure that the function is both one-to-one (injective) and onto (surjective).

### Step-by-step Solution:

1. **Understanding Bijective Functions**:
A function is bijective if it is both one-to-one (injective) and onto (surjective). For our function \( f(x) \), we will focus primarily on the one-to-one property, as it is crucial for establishing the function's behavior.

2. **Finding the Derivative**:
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(px + \sin x) = p + \cos x \]

3. **Condition for Monotonicity**:
For \( f(x) \) to be one-to-one, \( f'(x) \) must not change sign. This means \( f'(x) \) should either be always positive or always negative.

4. **Analyzing the Derivative**:
- For \( f'(x) \) to be always positive:
\[ p + \cos x > 0 \quad \forall x \]
The maximum value of \( \cos x \) is 1, and the minimum value is -1. Therefore, for the minimum value of \( \cos x \) (which is -1), we have:
\[ p - 1 > 0 \implies p > -1 \]

- For \( f'(x) \) to be always negative:
\[ p + \cos x < 0 \quad \forall x \]
The maximum value of \( \cos x \) is 1. Therefore, for the maximum value of \( \cos x \):
\[ p + 1 < 0 \implies p < -1 \]

5. **Combining Conditions**:
The function can be one-to-one if \( p > -1 \) (always increasing) or \( p < -1 \) (always decreasing). Therefore, the values of \( p \) that make \( f(x) \) bijective are:
\[ p \in (-\infty, -1) \cup (-1, \infty) \]

6. **Final Result**:
Thus, the set of all possible values of \( p \) in \( \mathbb{R} \) such that \( f(x) = px + \sin x \) is a bijective function is:
\[ p \in \mathbb{R} \setminus \{-1\} \]