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Let N be the set of numbers and two func...

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 onto

A

one-one but not onto

B

onto but not one-one

C

both one-one and onto

D

neither one-one nor onto

Text Solution

Verified by Experts

The correct Answer is:
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.
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