If function `f:NtoN` is defined by `f(x)=2x+3`, for all `x inN` then f is

Options

A. surjective
B. injective
C. bijective
D. none of these

Answer

To determine whether the function \( f: \mathbb{N} \to \mathbb{N} \) defined by \( f(x) = 2x + 3 \) is injective, surjective, or bijective, we will analyze the function step by step.

### Step 1: Check if \( f \) is Injective
A function is injective (one-to-one) if different inputs lead to different outputs. In other words, if \( f(a) = f(b) \) implies \( a = b \).

Let's assume:
\[ f(a) = f(b) \]
This means:
\[ 2a + 3 = 2b + 3 \]
Subtracting 3 from both sides gives:
\[ 2a = 2b \]
Dividing both sides by 2 gives:
\[ a = b \]
Since \( a = b \), we conclude that \( f \) is injective.

### Step 2: Check if \( f \) is Surjective
A function is surjective (onto) if every element in the codomain (in this case \( \mathbb{N} \)) has a pre-image in the domain.

To check if \( f \) is surjective, we need to see if for every \( y \in \mathbb{N} \), there exists an \( x \in \mathbb{N} \) such that:
\[ f(x) = y \]
This means:
\[ 2x + 3 = y \]
Rearranging gives:
\[ 2x = y - 3 \]
\[ x = \frac{y - 3}{2} \]
For \( x \) to be a natural number, \( y - 3 \) must be even and \( y \) must be greater than or equal to 3 (since \( x \) must be a natural number).

Thus, \( f \) is not surjective because not every natural number \( y \) can be expressed in this form (for example, \( y = 1 \) or \( y = 2 \) cannot be achieved).

### Step 3: Conclusion
Since \( f \) is injective but not surjective, we conclude that \( f \) is an injective function.

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

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