Let f: N to N be defined by f(x) = x^2 + x + | vedclass

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(A) To check if $f$ is one-one,let $x, y \in N$ such that $f(x) = f(y)$.$x^2 + x + 1 = y^2 + y + 1$$x^2 - y^2 + x - y = 0$$(x - y)(x + y) + (x - y) =

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One-one but not onto

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(A) To check if $f$ is one-one,let $x, y \in N$ such that $f(x) = f(y)$.$x^2 + x + 1 = y^2 + y + 1$$x^2 - y^2 + x - y = 0$$(x - y)(x + y) + (x - y) = 0$$(x - y)(x + y + 1) = 0$Since $x, y \in N$,$x + y + 1 > 0$,so we must have $x - y = 0$,which implies $x = y$. Thus,$f$ is one-one.To check if $f$ is onto,we look for the range of $f$. For $x = 1$,$f(1) = 1^2 + 1 + 1 = 3$. For $x = 2$,$f(2) = 2^2 + 2 + 1 = 7$. Since $f(x) = x^2 + x + 1$,the minimum value for $x \in N$ is $3$. Values like $1, 2, 4, 5, 6$ are not in the range. Therefore,$f$ is not onto.Hence,$f$ is one-one but not onto.

Let $f: N \to N$ be defined by $f(x) = x^2 + x + 1$,where $x \in N$. Then $f$ is:

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