Check the injectivity and surjectivity of the | vedclass

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(C) The function $f: Z \rightarrow Z$ is defined by $f(x) = x^{2}$.For injectivity:Consider $f(-1) = (-1)^{2} = 1$ and $f(1) = (1)^{2} = 1$.Since $f(-

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Neither injective nor surjective

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(C) The function $f: Z ightarrow Z$ is defined by $f(x) = x^{2}$.For injectivity:Consider $f(-1) = (-1)^{2} = 1$ and $f(1) = (1)^{2} = 1$.Since $f(-1) = f(1)$ but $-1 
eq 1$,the function is not injective.For surjectivity:Consider the codomain $Z$. For any negative integer,such as $-2 \in Z$,there exists no $x \in Z$ such that $f(x) = x^{2} = -2$,because the square of any integer is always non-negative.Thus,the range of $f$ is ${0, 1, 4, 9, ...}$,which is not equal to the codomain $Z$.Therefore,$f$ is not surjective.Conclusion: The function $f$ is neither injective nor surjective.

Check the injectivity and surjectivity of the function $f: Z \rightarrow Z$ defined by $f(x) = x^{2}$.

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