`f:N to N,` where `f(x)=x-(-1)^(x)`, Then `f` is

Options

A. one-one and into
B. many-one and into
C. one-one and onto
D. many-one and onto

Explanation

To analyze the function \( f: \mathbb{N} \to \mathbb{N} \) defined by \( f(x) = x - (-1)^x \), we will determine whether the function is one-one (injective) and onto (surjective).

### Step 1: Rewrite the function based on the parity of \( x \)

The function can be expressed differently depending on whether \( x \) is even or odd:

- If \( x \) is even, then \( (-1)^x = 1 \), so:
\[ f(x) = x - 1 \]

- If \( x \) is odd, then \( (-1)^x = -1 \), so:
\[ f(x) = x + 1 \]

### Step 2: Determine if the function is one-one (injective)

To check if \( f \) is one-one, we need to see if \( f(a) = f(b) \) implies \( a = b \).

1. **Case 1: Both \( a \) and \( b \) are even**
\[ f(a) = a - 1 \quad \text{and} \quad f(b) = b - 1 \]
If \( f(a) = f(b) \), then:
\[ a - 1 = b - 1 \implies a = b \]

2. **Case 2: Both \( a \) and \( b \) are odd**
\[ f(a) = a + 1 \quad \text{and} \quad f(b) = b + 1 \]
If \( f(a) = f(b) \), then:
\[ a + 1 = b + 1 \implies a = b \]

3. **Case 3: One is even, the other is odd**
If \( a \) is even and \( b \) is odd, then:
\[ f(a) = a - 1 \quad \text{and} \quad f(b) = b + 1 \]
Here, \( f(a) \) will always be odd (since \( a - 1 \) is odd) and \( f(b) \) will always be even (since \( b + 1 \) is even). Thus, \( f(a) \neq f(b) \).

Since in all cases \( f(a) = f(b) \) implies \( a = b \), the function \( f \) is one-one.

### Step 3: Determine if the function is onto (surjective)

To check if \( f \) is onto, we need to see if every element in the codomain \( \mathbb{N} \) can be expressed as \( f(x) \) for some \( x \in \mathbb{N} \).

1. **For even \( y \) in \( \mathbb{N} \)**:
Let \( y \) be even, say \( y = 2k \). Then:
\[ f(2k + 1) = (2k + 1) + 1 = 2k + 2 = y \]
So every even number can be achieved.

2. **For odd \( y \) in \( \mathbb{N} \)**:
Let \( y \) be odd, say \( y = 2k + 1 \). Then:
\[ f(2k) = 2k - 1 = y \]
So every odd number can also be achieved.

Since every natural number can be expressed as \( f(x) \) for some \( x \in \mathbb{N} \), the function \( f \) is onto.

### Conclusion

The function \( f(x) = x - (-1)^x \) is both one-one and onto. Therefore, the correct answer is that \( f \) is a bijection.