which of the following functions from `ZtoZ` is a bijection ?

Options

A. `f(x)=x^(3)`
B. `f(x)=x+2`
C. `f(x)=2x+1`
D. `f(x)=x^(2)+1`

Answer

To determine which of the given functions from \( \mathbb{Z} \) to \( \mathbb{Z} \) is a bijection, we need to check each function for both injectivity (one-to-one) and surjectivity (onto).

### Step-by-Step Solution:

1. **Understand the Definitions**:
- A function \( f: A \to B \) is **injective** (one-to-one) if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
- A function \( f: A \to B \) is **surjective** (onto) if for every \( b \in B \), there exists at least one \( a \in A \) such that \( f(a) = b \).
- A function is **bijective** if it is both injective and surjective.

2. **Evaluate Each Function**:
- **Option A**: \( f(x) = x^2 + 1 \)
- **Injectivity**: Check if \( f(x_1) = f(x_2) \) leads to \( x_1 = x_2 \).
- \( x^2 + 1 = y^2 + 1 \) implies \( x^2 = y^2 \), which gives \( x = y \) or \( x = -y \). Thus, it is not injective (e.g., \( f(1) = f(-1) = 2 \)).
- **Conclusion**: Not a bijection.

- **Option B**: \( f(x) = x^3 \)
- **Injectivity**: Assume \( f(x_1) = f(x_2) \).
- \( x_1^3 = x_2^3 \) implies \( x_1 = x_2 \). Thus, it is injective.
- **Surjectivity**: Check if every integer \( y \) can be expressed as \( x^3 \).
- The range of \( f(x) = x^3 \) is \( \mathbb{Z} \), as every integer can be represented as the cube of some integer.
- **Conclusion**: This function is a bijection.

- **Option C**: \( f(x) = 2x + 1 \)
- **Injectivity**: Assume \( f(x_1) = f(x_2) \).
- \( 2x_1 + 1 = 2x_2 + 1 \) implies \( 2x_1 = 2x_2 \) or \( x_1 = x_2 \). Thus, it is injective.
- **Surjectivity**: Check if every integer \( y \) can be expressed as \( 2x + 1 \).
- The function only takes odd integers (since \( 2x + 1 \) is always odd). Thus, it is not surjective as it cannot produce even integers.
- **Conclusion**: Not a bijection.

- **Option D**: \( f(x) = x + 2 \)
- **Injectivity**: Assume \( f(x_1) = f(x_2) \).
- \( x_1 + 2 = x_2 + 2 \) implies \( x_1 = x_2 \). Thus, it is injective.
- **Surjectivity**: Check if every integer \( y \) can be expressed as \( x + 2 \).
- For any integer \( y \), we can find \( x = y - 2 \), which is also an integer. Thus, it is surjective.
- **Conclusion**: This function is a bijection.

3. **Final Answer**: The functions that are bijections from \( \mathbb{Z} \) to \( \mathbb{Z} \) are:
- Option B: \( f(x) = x^3 \)
- Option D: \( f(x) = x + 2 \)