Let `f:N to Z and f (x)=[{:((x-1)/(2),"when x is odd"),(-(x)/(2),"when x is even"):},` then:
(a). f (x) is bijective (b).f (x) is injective but not surjective (c).f (x) is not injective but surjective (d).f (x) is neither injective nor subjective

Options

A. f (x) is bijective
B. f (x) is injective but not surjective
C. f (x) is not injective but surjective
D. f (x) is neither injective nor subjective

Answer

To determine the nature of the function \( f: \mathbb{N} \to \mathbb{Z} \) defined as:

\[ f(x) = \begin{cases} \frac{x - 1}{2} & \text{when } x \text{ is odd} \\ -\frac{x}{2} & \text{when } x \text{ is even} \end{cases} \]

we will analyze whether the function is injective (one-to-one) and surjective (onto).

### Step 1: Analyze the function for odd \( x \)

Let \( x \) be an odd natural number. We can express any odd number as \( x = 2n + 1 \) for some \( n \in \mathbb{N} \) (where \( n \geq 0 \)).

Substituting this into the function:

\[ f(x) = f(2n + 1) = \frac{(2n + 1) - 1}{2} = \frac{2n}{2} = n \]

Thus, when \( x \) is odd, \( f(x) \) maps to the set of non-negative integers \( \{0, 1, 2, \ldots\} \).

### Step 2: Analyze the function for even \( x \)

Now let \( x \) be an even natural number. We can express any even number as \( x = 2m \) for some \( m \in \mathbb{N} \) (where \( m \geq 1 \)).

Substituting this into the function:

\[ f(x) = f(2m) = -\frac{2m}{2} = -m \]

Thus, when \( x \) is even, \( f(x) \) maps to the set of negative integers \( \{-1, -2, -3, \ldots\} \).

### Step 3: Determine the range of \( f \)

From the analysis above, we can see that:
- The range of \( f \) when \( x \) is odd is \( \{0, 1, 2, \ldots\} \).
- The range of \( f \) when \( x \) is even is \( \{-1, -2, -3, \ldots\} \).

Thus, the overall range of \( f \) is \( \mathbb{Z} \), which includes all integers (both positive and negative).

### Step 4: Check if \( f \) is injective

To check if \( f \) is injective, we need to see if different inputs produce different outputs.

- For odd \( x \): If \( f(2n_1 + 1) = f(2n_2 + 1) \), then \( n_1 = n_2 \) implies \( x_1 = x_2 \).
- For even \( x \): If \( f(2m_1) = f(2m_2) \), then \( -m_1 = -m_2 \) implies \( m_1 = m_2 \) implies \( x_1 = x_2 \).

Thus, \( f \) is injective.

### Step 5: Check if \( f \) is surjective

To check if \( f \) is surjective, we need to see if every integer \( z \in \mathbb{Z} \) has a pre-image in \( \mathbb{N} \).

- For any non-negative integer \( n \), there exists an odd \( x = 2n + 1 \) such that \( f(x) = n \).
- For any negative integer \( -m \), there exists an even \( x = 2m \) such that \( f(x) = -m \).

Since every integer can be achieved by some natural number, \( f \) is surjective.

### Conclusion

Since \( f \) is both injective and surjective, we conclude that \( f \) is bijective.

### Final Answer

The correct option is:
**(a) \( f(x) \) is bijective.**