Let `f:RrarrR` be defined by `f(x)="cos"(5x+2)`. If f invertible ?

Answer

To determine whether the function \( f(x) = \cos(5x + 2) \) is invertible, we need to check if the function is both one-to-one (injective) and onto (surjective).

### Step-by-step Solution:

1. **Understanding the Function**:
The function \( f(x) = \cos(5x + 2) \) is a cosine function, which is periodic. The cosine function has a range of \([-1, 1]\) and repeats every \(2\pi\).

2. **Checking for One-to-One**:
To check if \( f \) is one-to-one, we need to see if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
- Assume \( f(x_1) = f(x_2) \):
\[ \cos(5x_1 + 2) = \cos(5x_2 + 2) \]
- Using the property of cosine, we can say:
\[ 5x_1 + 2 = 2n\pi + (5x_2 + 2) \quad \text{or} \quad 5x_1 + 2 = 2n\pi - (5x_2 + 2) \]
- Simplifying the first equation:
\[ 5x_1 = 2n\pi + 5x_2 \implies 5x_1 - 5x_2 = 2n\pi \implies 5(x_1 - x_2) = 2n\pi \]
This implies:
\[ x_1 - x_2 = \frac{2n\pi}{5} \]
- Since \( n \) can be any integer, \( x_1 \) can take multiple values for the same \( f(x) \). Thus, \( f \) is not one-to-one.

3. **Conclusion on One-to-One**:
Since we found that \( x_1 \neq x_2 \) can yield the same function value, \( f \) is not one-to-one.

4. **Checking for Onto**:
The function \( f(x) = \cos(5x + 2) \) has a range of \([-1, 1]\). For \( f \) to be onto, it must cover all real numbers \( \mathbb{R} \).
- Since the range of \( f \) is limited to \([-1, 1]\), it does not cover all real numbers.

5. **Conclusion on Onto**:
Since \( f \) does not cover all real numbers, \( f \) is not onto.

### Final Conclusion:
Since \( f \) is neither one-to-one nor onto, we conclude that the function \( f(x) = \cos(5x + 2) \) is not invertible.

### Summary:
- **Not One-to-One**: \( f(x_1) = f(x_2) \) does not imply \( x_1 = x_2 \).
- **Not Onto**: The range of \( f \) is limited to \([-1, 1]\).

Thus, **the function is not invertible**.

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