A function \( f:N\to N \), (\( N \): set of natural numbers) defined by \( f(x)=x^2+x+1 \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → one-one but not onto
One-one: let \( f(x_1)=f(x_2) \) for \( x_1,x_2\in N \).
\( \displaystyle x_1^2+x_1+1=x_2^2+x_2+1 \)
\( \displaystyle (x_1-x_2)(x_1+x_2+1)=0 \)
Since \( x_1+x_2+1\gt 0 \), \( x_1=x_2 \). So f is one-one.
Onto: the smallest value is f(1) = 3, and the values are 3, 7, 13, 21, …
The natural numbers 1, 2, 4, 5, … have no pre-image, so f is not onto.
Hence f is one-one but not onto.