If `A={1,2,3,4}" and "B={1,2,3,4,5,6}` are two sets and function `f: A to B` is defined by `f(x)=x+2,AA x in A`, then the function f is

Options

A. bijective
B. onto
C. one-one
D. many-one

Explanation

To determine the nature of the function \( f: A \to B \) defined by \( f(x) = x + 2 \), where \( A = \{1, 2, 3, 4\} \) and \( B = \{1, 2, 3, 4, 5, 6\} \), we will analyze whether the function is one-to-one (injective), onto (surjective), or bijective.

### Step 1: Calculate the function values
We will compute \( f(x) \) for each element \( x \) in set \( A \):

- For \( x = 1 \):
\[ f(1) = 1 + 2 = 3 \]

- For \( x = 2 \):
\[ f(2) = 2 + 2 = 4 \]

- For \( x = 3 \):
\[ f(3) = 3 + 2 = 5 \]

- For \( x = 4 \):
\[ f(4) = 4 + 2 = 6 \]

### Step 2: Determine the range of the function
The range of the function \( f \) is the set of all output values we calculated:
\[ \text{Range} = \{f(1), f(2), f(3), f(4)\} = \{3, 4, 5, 6\} \]

### Step 3: Compare the range with the co-domain
The co-domain of the function is set \( B \):
\[ B = \{1, 2, 3, 4, 5, 6\} \]
The range we found is:
\[ \text{Range} = \{3, 4, 5, 6\} \]
Since the range does not cover all elements of the co-domain \( B \) (specifically, 1 and 2 are missing), the function is not onto.

### Step 4: Check if the function is one-to-one
To check if the function is one-to-one, we need to verify that different inputs produce different outputs.

- \( f(1) = 3 \)
- \( f(2) = 4 \)
- \( f(3) = 5 \)
- \( f(4) = 6 \)

Since all outputs are distinct, the function is one-to-one.

### Conclusion
The function \( f \) is one-to-one but not onto, hence it is not bijective.

### Final Answer
The function \( f \) is **one-to-one** (injective) but **not onto** (surjective).

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