Let g:N→ N be defined as g(3n+1)=3n+2 g(3n+2)=3n+3 g(3n+3)=3n+1, for all n≥ 0 Then which of the following statements is true ?

$$g:N\rightarrow N$$g:N→N

$$g(3n+1)=3n+2$$g(3n+1)=3n+2

$$g(3n+2)=3n+3$$g(3n+2)=3n+3

$$g(3n+3)=3n+1$$g(3n+3)=3n+1

g(x)={\begin{matrix}\begin{matrix}x+1; & x=3k+1 \\ x+1; & x=3k+2 \\ x-2; & x=3k+3\end{matrix}\end{matrix}\begin{matrix} \\ \\ \end{matrix}

g(g(x))={\begin{matrix}\begin{matrix}x+2; & x=3k+1 \\ x-1; & x=3k+2 \\ x-1; & x=3k+3\end{matrix}\end{matrix}\begin{matrix} \\ \\ \end{matrix}

g(g(g(x)))={\begin{matrix}x; & x=3k+1 \\ x; & x=3k+2 \\ x; & x=3k+3\end{matrix}\begin{matrix} \\ \\ \end{matrix}\begin{matrix} \\ \\ \end{matrix}

If $f:N\rightarrow N,f$f:N→N,f is a one-one function such that $f(g(x))=f(x)\Rightarrow g(x)=x$f(g(x))=f(x)⇒g(x)=x, which is not the case

If $f:N\rightarrow N,f$f:N→N,f is an onto function such that $f(g(x))=f(x)$f(g(x))=f(x) one possibility is

f(x)={\begin{matrix}n; & x=3n+1 \\ n; & x=3n+2 \\ n; & x=3n+3\end{matrix};n\in N

Here $f(x)$f(x) is onto, also $f(g(x))=f(x)\forall x\in N$f(g(x))=f(x)∀x∈N