The function ` f : N to N ` defined f(x) = a+bx, where a and b are natural numbers is

Options

A. not a bijection
B. is may- one and onto
C. is many - one
D. None of these

Answer

To determine the nature of the function \( f: \mathbb{N} \to \mathbb{N} \) defined by \( f(x) = a + bx \), where \( a \) and \( b \) are natural numbers, we will analyze whether the function is one-to-one (injective), onto (surjective), or both (bijective).

### Step 1: Understand the function
The function is given as \( f(x) = a + bx \). Here, \( a \) and \( b \) are natural numbers, and \( x \) is a natural number.

### Step 2: Determine if the function is one-to-one (injective)
To check if the function is one-to-one, we need to see if different inputs produce different outputs.

Assume \( f(x_1) = f(x_2) \) for \( x_1, x_2 \in \mathbb{N} \):
\[ a + bx_1 = a + bx_2 \]
Subtracting \( a \) from both sides gives:
\[ bx_1 = bx_2 \]
Since \( b \) is a natural number and cannot be zero, we can divide both sides by \( b \):
\[ x_1 = x_2 \]
This shows that if \( f(x_1) = f(x_2) \), then \( x_1 \) must equal \( x_2 \). Therefore, the function is one-to-one.

### Step 3: Determine if the function is onto (surjective)
Next, we check if the function can produce every natural number in its range.

The output of the function is:
\[ f(x) = a + bx \]
As \( x \) varies over all natural numbers, the smallest value of \( f(x) \) occurs when \( x = 1 \):
\[ f(1) = a + b \]
As \( x \) increases, \( f(x) \) will take values \( a + b, a + 2b, a + 3b, \ldots \).

The outputs are in the form of an arithmetic sequence starting from \( a + b \) with a common difference of \( b \). This means that not all natural numbers can be reached unless \( a = 1 \) and \( b = 1 \).

Thus, the function is not onto for general values of \( a \) and \( b \).

### Step 4: Conclusion
Since the function is one-to-one but not onto, we conclude that the function \( f(x) = a + bx \) is not a bijection.

### Final Answer
The function is **one-to-one but not onto**.

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