Let \( f:\mathrm{N}\rightarrow\mathrm{N} \) defined by \( f(x)=3x^2+2x+7 \), where N is set of natural numbers. Then \( f(x) \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → injective but not surjective.
One-one: let \( f(x_1)=f(x_2) \) for natural numbers \( x_1,x_2 \).
\( \displaystyle 3x_1^2+2x_1=3x_2^2+2x_2 \)
\( \displaystyle 3(x_1^2-x_2^2)+2(x_1-x_2)=0 \)
\( \displaystyle (x_1-x_2)\left(3(x_1+x_2)+2\right)=0 \)
Since \( x_1+x_2\gt 0 \), the second factor is positive, so \( x_1=x_2 \). Hence f is injective.
Onto: the smallest value is f(1) = 3 + 2 + 7 = 12, and f increases on N. So natural numbers such as 1, 2, …, 11 are not images of any natural number. Hence f is not surjective.
So f is injective but not surjective.