Question

Let $f:N\to N$ defined by $f\left(x\right)=x^2+x+1$, $x\in N$ then $f$ is

• A. One-one onto
• B. Many-one onto
• C. One-one but not onto
• D. None of the above

Answer 1

Answer: (C)

One-one but not onto

Let $x,y\in N$ such that $f\left(x\right)=f\left(y\right)$

$x^2+x+1=y^2+y+1$

$\Rightarrow x^2+x=y^2+y$

$\Rightarrow x^2+x+xy=y^2+y+xy$

$\Rightarrow x(x+y+1)=y(y+x+1)$

$\Rightarrow (x-y)(x+y+1)=0$

$\Rightarrow x=y$ or $x=-y-1$

But $x=(-y-1)=-(y+1)$ does not belong to $N$

$\therefore x=y$

$\therefore f$ is one -one

Also, $f$ is not onto