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(x)={x+1,x=3k+1x+1,x=3k+2x−2,x=3k+3g(g(x))={x+2,x=3k+1x−1,x=3k+2x−1,x=3k+3g(g(g(x)))={x,x=3k+1x,x=3k+2x,x=3k+3If F:N→N, f is one−one function such thatf(g(x))=f(x)⇒ g(x)=x, which is not the caseIf F:N→N, f is an onto functionsuch that f(g(x))=f(x), one possibility isf(x)={n,x=3n+1n,x=3n+2,n∈Nn,x=3n+3∴ f(x) is onto, also f(g(x))=f(x) ∀ x∈N

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(x)={x+1,x=3k+1x+1,x=3k+2x−2,x=3k+3g(g(x))={x+2,x=3k+1x−1,x=3k+2x−1,x=3k+3g(g(g(x)))={x,x=3k+1x,x=3k+2x,x=3k+3If F:N→N, f is one−one function such thatf(g(x))=f(x)⇒ g(x)=x, which is not the caseIf F:N→N, f is an onto functionsuch that f(g(x))=f(x), one possibility isf(x)={n,x=3n+1n,x=3n+2,n∈Nn,x=3n+3∴ f(x) is onto, also f(g(x))=f(x) ∀ x∈N

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$\begin{array}{l} L e t g : N \to N b e d e f i n e d a s g (3 n + 1) = 3 n + 2, g (3 n + 2) = 3 n + 3, g (3 n + 3) = 3 n + 1, f o r a l l n \geq 0. T h e n w h i c h o f t h e f o l l o w i n g s t a t e m e n t s i s t r u e ? \end{array}$

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$T h e r e e x i s t a a o n e - o n e f u n c t i o n f : N \to N s u c h t h a t f o g = f$

$g o g o g = g$ $T h e r e e x i s t s a n o n t o f u n c t i o n s u c h t h a t f o g = f$

$T h e r e e x i s t s a f u n c t i o n s u c h t h a t f o g = f$

$T h e r e e x i s t s a n o n t o f u n c t i o n s u c h t h a t f o g = f$

answer is C.

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Detailed Solution

$\begin{array}{l} g (x) = {\begin{cases} x + 1, x = 3 k + 1 \\ x + 1, x = 3 k + 2 \\ x - 2, x = 3 k + 3 \end{cases} \\ g (g (x)) = {\begin{cases} x + 2, x = 3 k + 1 \\ x - 1, x = 3 k + 2 \\ x - 1, x = 3 k + 3 \end{cases} \\ g (g (g (x))) = {\begin{cases} x, x = 3 k + 1 \\ x, x = 3 k + 2 \\ x, x = 3 k + 3 \end{cases} \\ I f F : N \to N, f i s o n e - o n e f u n c t i o n s u c h t h a t \\ f (g (x)) = f (x) \Rightarrow g (x) = x, w h i c h i s n o t t h e c a s e \\ I f F : N \to N, f i s a n o n t o f u n c t i o n \\ s u c h t h a t f (g (x)) = f (x), o n e p o s s i b i l i t y i s \\ f (x) = {\begin{cases} n, x = 3 n + 1 \\ n, x = 3 n + 2, n \in N \\ n, x = 3 n + 3 \end{cases} \\ ∴ f (x) i s o n t o, a l s o f (g (x)) = f (x) \forall x \in N \end{array}$