Question

A function \(f: N \to N\), (\(N\): set of natural numbers) defined by \(f(x) = x^2 + x + 1\) is

• A. one-one and onto
• B. one-one but not onto
• C. many-one and onto
• D. many-one but not onto

Hint:

Solve \(f(a)=f(b)\) to test one-one. Check whether 1 or 2 is ever an output to test onto.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A function is one-one if different inputs always give different outputs. It is onto if every element of the codomain is an output. Here the domain and codomain are both \(N=\{1,2,3,\ldots\}\).

Step 2: Test one-one.
Suppose \(f(a)=f(b)\). Then \(a^2+a+1=b^2+b+1\), so \(a^2-b^2+a-b=0\).
\[ (a-b)(a+b+1)=0 \]
Since \(a,b\) are natural numbers, \(a+b+1 \geq 3\), so it is never zero. Hence \(a=b\). The function is one-one.

Step 3: Test onto.
Compute a few values: \(f(1)=3\), \(f(2)=7\), \(f(3)=13\). The smallest output is 3. So the numbers 1 and 2 in the codomain are never reached. The function is not onto.

Step 4: Check the options.
Option 1 says onto, which is false. Option 3 and option 4 say many-one, which is false. Option 2, one-one but not onto, is correct.

Final Answer:
The function is one-one but not onto. \[ \boxed{\text{Option 2: one-one but not onto}} \]