Let A = {x, y, z), B = {u, v, w} and f : A `rarr` B be defined by f(x) = u,
f(y) = v, f(z) = w. Then, f is

Options

A. surjective but not injective
B. injective but not surjective
C. bijective
D. None of the above

Explanation

To determine the nature of the function \( f: A \rightarrow B \) defined by \( f(x) = u \), \( f(y) = v \), and \( f(z) = w \), we will analyze whether the function is one-to-one (injective) and onto (surjective).

### Step 1: Identify the sets and the function
Let:
- \( A = \{x, y, z\} \)
- \( B = \{u, v, w\} \)
- The function \( f \) is defined as:
- \( f(x) = u \)
- \( f(y) = v \)
- \( f(z) = w \)

### Step 2: Check if \( f \) is one-to-one (injective)
A function is one-to-one if different elements in the domain map to different elements in the codomain.

- Here, we have:
- \( f(x) = u \)
- \( f(y) = v \)
- \( f(z) = w \)

Since \( u \), \( v \), and \( w \) are distinct, it follows that:
- \( f(x) \neq f(y) \)
- \( f(x) \neq f(z) \)
- \( f(y) \neq f(z) \)

Thus, \( f \) is one-to-one.

### Step 3: Check if \( f \) is onto (surjective)
A function is onto if every element in the codomain has a pre-image in the domain.

- The codomain of \( f \) is \( B = \{u, v, w\} \).
- The range of \( f \) is also \( \{u, v, w\} \) since:
- \( f(x) = u \)
- \( f(y) = v \)
- \( f(z) = w \)

Since the range of \( f \) is equal to the codomain \( B \), we conclude that \( f \) is onto.

### Step 4: Conclusion
Since \( f \) is both one-to-one and onto, we can conclude that \( f \) is a bijective function.

### Final Answer
Thus, the function \( f \) is bijective.

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