If a function f: Z rightarrow Z is defined by | vedclass

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(C) Given $f(x) = x - (-1)^x$ where $x \in Z$.Case $1$: If $x$ is even,let $x = 2k$ for some $k \in Z$.Then $f(2k) = 2k - (-1)^{2k} = 2k - 1$.Case $2$

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both one-one and onto

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(C) Given $f(x) = x - (-1)^x$ where $x \in Z$.Case $1$: If $x$ is even,let $x = 2k$ for some $k \in Z$.Then $f(2k) = 2k - (-1)^{2k} = 2k - 1$.Case $2$: If $x$ is odd,let $x = 2k+1$ for some $k \in Z$.Then $f(2k+1) = (2k+1) - (-1)^{2k+1} = 2k+1 - (-1) = 2k+2$.For one-one: Let $f(x_1) = f(x_2)$. If $x_1$ is even and $x_2$ is odd,$2k_1 - 1 = 2k_2 + 2 \implies 2(k_1 - k_2) = 3$,which has no integer solution. If both are even,$2k_1 - 1 = 2k_2 - 1 \implies k_1 = k_2 \implies x_1 = x_2$. If both are odd,$2k_1 + 2 = 2k_2 + 2 \implies k_1 = k_2 \implies x_1 = x_2$. Thus,$f$ is one-one.For onto: The range consists of all odd numbers (from even inputs) and all even numbers (from odd inputs). Since every integer $y \in Z$ is either even or odd,$f$ is onto.Therefore,$f$ is both one-one and onto.

If a function $f: Z \rightarrow Z$ is defined by $f(x) = x - (-1)^x$,then $f(x)$ is

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