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The function f : N->N given by f(n)=n-(-...

The function `f : N->N` given by `f(n)=n-(-1)^n` is

A

one-one and into

B

one-one and onto

C

many-one and into

D

many-one and onto

Text Solution

Verified by Experts

The correct Answer is:
A
We have,
`f(n)=n(-1)^(n)={{:(,n-1,"if n is even"),(,n+1,"if n is odd"):}`
Injectivity Let n, m be any two even natural number. Then,
`f(n)=f(m) Rightarrow n-1=m-1Rightarrow m`
If n, m are any two odd natural numbers. Then,
`f(n)=f(m) Rightarrow n+1=m+1 Rightarrow n=m`
Thus, in both the cases, we have
`f(n)=f(m)Rightarrow n=m`
If n is even and m is odd, then `n ne m`. Also, f(n) is odd f(m) is even. So, `f(n)nef(m)`
`"Thus", n ne m Rightarrow f(n) ne f(m)`
So, f is an injective map.
Surjectivity Let n be an arbitary natural number.
If n is odd natural number, then there exists an even natural number n+1 such that
`f(n+1)=n+1-1=n`
If n is an even natural number, then there exists an odd natural number (n-1) such that
`f(n-1)=n-1+1=n`
Thus, every `n in N` has its pre-image in N`.
So, `f: N to N is a surjection`.
Hence, `f:N to N` is a bijection.
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