Q. Let $f:N \rightarrow N$ defined by $f(x) = x^{2} + x + 1,\ x \in N$, then $f$ is
A) One-one onto
B) Many-one onto
C) One -one but not onto
D) None of these
Explanation
(c)
Let $x,\ y \in N$ such that $f(x) = f(y)$
$\Rightarrow \ \ \ \ x^{2} + x + 1 = y^{2} + y + 1$
$\Rightarrow \ \ \ (x - y)(x + y + 1) = 0$
$\Rightarrow \ \ \ \ \ x = y$ or $x = ( - y - 1) \notin N$
$\therefore f$ one-one.
Also, $f$ is not onto.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.