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f(x+1)=f(x) + f(1) , f(x), g(x) : N to N...

`f(x+1)=f(x) + f(1) , f(x), g(x) : N to N`
`g(x)` = any arbitrary function and `fog(x)` is one-one

A

`f(x)` is many-one

B

`g(x)` is many-one

C

`g(x)` is one-one

D

`f(x)` and `g(x)` both are many-one

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