Let N denote the set of all natural numbers,a | vedclass

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(A) For $f: N \rightarrow Z$,$f(1)=1, f(2)=1, f(3)=2, f(4)=0, f(5)=3, f(6)=-1, \dots$. Since $f(1)=f(2)=1$,$f$ is not one-one. Since the range of $f$

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$A, D$

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(A) For $f: N \rightarrow Z$,$f(1)=1, f(2)=1, f(3)=2, f(4)=0, f(5)=3, f(6)=-1, \dots$. Since $f(1)=f(2)=1$,$f$ is not one-one. Since the range of $f$ is $Z$,$f$ is onto. Thus,statement $(D)$ is true.For $g: Z \rightarrow N$,$g(0)=3, g(1)=5, g(-1)=2, g(-2)=4, g(-3)=6$. Since $g$ is strictly increasing for $n \geq 0$ and strictly decreasing for $n For $g \circ f: N \rightarrow N$,$(g \circ f)(1) = g(f(1)) = g(1) = 5$ and $(g \circ f)(2) = g(f(2)) = g(1) = 5$. Since $(g \circ f)(1) = (g \circ f)(2)$,$g \circ f$ is not one-one. The range of $g \circ f$ does not cover all of $N$,so it is not onto. Thus,statement $(A)$ is true.For $f \circ g: Z \rightarrow Z$,if $n \geq 0$,$f(g(n)) = f(3+2n) = (3+2n+1)/2 = n+2$. If $n Therefore,statements $(A)$ and $(D)$ are true.

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