The function `f:NrarrN` (N is the set of natural numbers defined by `f(n)=2n+3` is :

Options

A. a. surjective
B. b. injective
C. c. bijective
D. d. none of these

Answer

To solve the problem, we need to analyze the function \( f: \mathbb{N} \rightarrow \mathbb{N} \) defined by \( f(n) = 2n + 3 \). We will determine the nature of this function, specifically whether it is one-to-one (injective) and onto (surjective).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function is defined as \( f(n) = 2n + 3 \). Here, \( n \) is a natural number (i.e., \( n \in \mathbb{N} \)).

2. **Calculating Function Values**:
Let's calculate the values of the function for the first few natural numbers:
- For \( n = 1 \):
\[ f(1) = 2 \times 1 + 3 = 2 + 3 = 5 \]
- For \( n = 2 \):
\[ f(2) = 2 \times 2 + 3 = 4 + 3 = 7 \]
- For \( n = 3 \):
\[ f(3) = 2 \times 3 + 3 = 6 + 3 = 9 \]
- For \( n = 4 \):
\[ f(4) = 2 \times 4 + 3 = 8 + 3 = 11 \]

The outputs for \( n = 1, 2, 3, 4 \) are \( 5, 7, 9, 11 \) respectively.

3. **Identifying the Pattern**:
The outputs \( 5, 7, 9, 11 \) are all odd numbers. This indicates that the function only produces odd numbers for natural number inputs.

4. **Checking for Injectivity (One-to-One)**:
A function is one-to-one if different inputs produce different outputs. Let's assume \( f(a) = f(b) \):
\[ 2a + 3 = 2b + 3 \]
Subtracting 3 from both sides:
\[ 2a = 2b \]
Dividing by 2:
\[ a = b \]
Since \( a = b \), the function is injective.

5. **Checking for Surjectivity (Onto)**:
A function is onto if every element in the codomain (natural numbers) is mapped by some element in the domain. The outputs of the function are all odd numbers starting from 5. However, not all natural numbers are odd (e.g., 1, 2, 3, 4, etc., are missing). Therefore, the function is not surjective.

### Conclusion:
The function \( f(n) = 2n + 3 \) is injective (one-to-one) but not surjective (onto).

### Final Answer:
The function \( f(n) = 2n + 3 \) is injective.

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