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. one one into
C. many oneonto
D. None of these

Answer

To determine the nature of the function \( f: A \to B \) defined by \( f(x) = x + 2 \) for \( x \in A \), where \( A = \{1, 2, 3, 4\} \) and \( B = \{1, 2, 3, 4, 5, 6\} \), we will analyze the function step by step.

### Step 1: Define the function
The function is defined as:
\[ f(x) = x + 2 \]
for all \( x \) in set \( A \).

### Step 2: Calculate the function values
We will calculate the values of \( f(x) \) for each element 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 3: Determine the range of the function
The outputs of the function \( f \) for the inputs from set \( A \) are:
\[ f(A) = \{f(1), f(2), f(3), f(4)\} = \{3, 4, 5, 6\} \]
Thus, the range of the function \( f \) is \( \{3, 4, 5, 6\} \).

### Step 4: Check if the function is one-to-one (injective)
A function is one-to-one if different inputs map to different outputs. We check the outputs:
- \( f(1) = 3 \)
- \( f(2) = 4 \)
- \( f(3) = 5 \)
- \( f(4) = 6 \)

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

### Step 5: Check if the function is onto (surjective)
A function is onto if every element in the codomain \( B \) is mapped by some element in the domain \( A \). The codomain \( B \) is:
\[ B = \{1, 2, 3, 4, 5, 6\} \]
The range \( f(A) = \{3, 4, 5, 6\} \) does not cover all elements of \( B \) (specifically, 1 and 2 are not in the range). Therefore, the function is not onto.

### Conclusion
The function \( f \) is one-to-one but not onto. Thus, it is classified as an "into" function.

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

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