Let `f:R to R` be a function defined b f(x)=cos(5x+2). Then,f is

Options

A. injective
B. surjective
C. bijective
D. none of these

Answer

To determine the nature of the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \cos(5x + 2) \), we need to analyze whether the function is injective, surjective, or bijective.

### Step-by-step Solution:

1. **Understanding the Function**:
The function is given as \( f(x) = \cos(5x + 2) \). The cosine function oscillates between -1 and 1 for all real numbers.

**Hint**: Recall the basic properties of the cosine function and its range.

2. **Finding the Range**:
Since \( f(x) = \cos(5x + 2) \), the output values (range) of this function will also oscillate between -1 and 1. Therefore, the range of \( f \) is \( [-1, 1] \).

**Hint**: Consider the maximum and minimum values of the cosine function.

3. **Checking for Injectivity**:
A function is injective (one-to-one) if different inputs produce different outputs. However, since the cosine function is periodic, there will be multiple values of \( x \) that yield the same value of \( f(x) \). For example, \( f(0) = \cos(2) \) and \( f(\frac{2\pi}{5}) = \cos(2) \).

**Hint**: Think about the periodic nature of the cosine function and how it affects the uniqueness of outputs.

4. **Checking for Surjectivity**:
A function is surjective (onto) if every element in the codomain has a pre-image in the domain. The codomain of \( f \) is \( \mathbb{R} \), but the range of \( f \) is only \( [-1, 1] \). Therefore, there are many values in \( \mathbb{R} \) that are not covered by \( f(x) \).

**Hint**: Compare the codomain and the range to see if they match.

5. **Conclusion**:
Since \( f(x) \) is neither injective nor surjective, it cannot be bijective either. Thus, the function \( f \) is classified as "none of these".

**Final Answer**: The function \( f(x) = \cos(5x + 2) \) is neither injective nor surjective, hence it is "none of these".