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(B) To determine if $f(x) = x^3 + 2$ is one-one and onto on the set of integers $\mathbb{Z}$:$1$. One-one check: Let $f(x_1) = f(x_2)$. Then $x_1^3 +

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one-one but not onto

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(B) To determine if $f(x) = x^3 + 2$ is one-one and onto on the set of integers $\mathbb{Z}$:$1$. One-one check: Let $f(x_1) = f(x_2)$. Then $x_1^3 + 2 = x_2^3 + 2$,which implies $x_1^3 = x_2^3$. Since the cube function is strictly increasing,$x_1 = x_2$. Thus,$f$ is one-one.$2$. Onto check: For $f$ to be onto,for every $y \in \mathbb{Z}$,there must exist $x \in \mathbb{Z}$ such that $y = x^3 + 2$. This means $x^3 = y - 2$,or $x = \sqrt[3]{y - 2}$. For $x$ to be an integer,$y - 2$ must be a perfect cube. For example,if $y = 3$,then $x^3 = 3 - 2 = 1$,so $x = 1 \in \mathbb{Z}$. However,if $y = 0$,then $x^3 = 0 - 2 = -2$. Since $-2$ is not a perfect cube of any integer,there is no $x \in \mathbb{Z}$ such that $f(x) = 0$. Therefore,$f$ is not onto.Conclusion: $f$ is one-one but not onto.

Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be defined by $f(x) = x^3 + 2$. Then,$f$ is . . . . . . .

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