The mapping f : `N to N` given by `f(n)=1+n^(2), n in N`, where N is the set of natural numbers, is

The mapping f:NrarrN given by f(n)=1+n^(2), n in N where N is the set of natural numbers, is

A mapping from N to N is defined as follows F:N rarrN given by f(n) = (n+5)^(2),n in N (N is the set of natural numbers) then

The functon f: N rarr N , defined by f(n)=2n+3 , for all n in N , is (here N is the set of natural numbers)

The funciton f: N to N given by f(n)=n-(-1)^(n) for all n in N is

If X = {4^(n) - 3n - 1: n in N} " and " Y = {9(n-1): n in N} , where N is the set of natural numbers, then X cup Y is equal to

If X={4^n -3n-1 , n in N} and Y={9(n-1), n in N} , where N is the set of natural numbers, then X uu Y is equal to :

If f:NrarrZ defined as f(n)={{:((n-1)/(2),":"," if n is odd"),((-n)/(2),":", " if n is even"):} and g:NrarrN defined as g(n)=n-(-1)^(n) , then fog is (where, N is the set of natural numbers and Z is the set of integers)

If f:NrarrZ defined as f(n)={{:((n-1)/(2),":"," if n is odd"),((-n)/(2),":", " if n is even"):} and g:NrarrN defined as g(n)=n-(-1)^(n) , then fog is (where, N is the set of natural numbers and Z is the set of integers)