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Question

Let g:NN be defined as
g(3n+1)=3n+2,
g(3n+2)=3n+3,
g(3n+3)=3n+1, for all n0.
Then which of the following statements is true?

A
gogog=g
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B
There exists an onto function f:NN such that fog=f
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C
There exists a one-one function f:NN such that fog=f
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D
There exists a function f:NN such that gof=f
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Solution

The correct option is B There exists an onto function f:NN such that fog=f
g(3n+1)=3n+2,g(3n+2)=3n+3 and g(3n+3)=3n+1
gogog(3n+1)=g(g(g(3n+1)))=g(g(3n+2))=g(3n+3)=3n+1
So we can see that gogog(x)=x (identity)
For fog=f to hold f must be an onto function.

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