Let N be the set of numbers and two functions f and g be defined as `f,g:N to N` such that `f(n)={((n+1)/(2), ,"if n is odd"),((n)/(2),,"if n is even"):}` and `g(n)=n-(-1)^(n)`. Then, fog is(A) one-one but not onto(B) onto but not one-one(C) both one-one and onto(D) neither one-one nor ontoA. one-one but not ontoB. onto but not one-oneC. both one-one and ontoD. neither one-one nor onto

Let N be the set of numbers and two functions f and g be defined as `f

1.	Let N be the set of numbers and two functions f and g be defined as `f,g:N to N` such that `f(n)={((n+1)/(2), ,"if n is odd"),((n)/(2),,"if n is even"):}` and `g(n)=n-(-1)^(n)`. Then, fog is(A) one-one but not onto(B) onto but not one-one(C) both one-one and onto(D) neither one-one nor ontoA. one-one but not ontoB. onto but not one-oneC. both one-one and ontoD. neither one-one nor onto
Answer» Correct Answer - B Given, `f(n)={((n+1)/(2)", if n is odd"),((n)/(2)", if n is even,"):}` and `g(n)=n-(-1)^(n)={(n+1", if n is odd"),(n-1", if n is even"):}` Now, `f(g(n))={(f(n+1)", if n is odd"),(f(n-1)", if n is even"):}` `={((n+1)/(2)", if n is odd"),((n-1+1)/(2)=(n)/(2)", if n is even"):}` `=f(x)` [ `because ` if in odd, then (n+1) is even and if n is even, then (n-1) is odd] Clearly, function is not one-one as f(2) = f(1) = 1 But it is onto function. [ `because " If " m in N ` (codomain) is odd, then `2m in N` (doamain) if `m in N` codomain is even, then `2m-1 in N`(domain) such that f(2m-1) = m] `therefore` Function is onto but not one-one.