The function `f : N->N` given by `f(n)=n-(-1)^n` isA. one-one and intoB. one-one and ontoC. many-one and intoD. many-one and onto

The function `f : N->N` given by `f(n)=n-(-1)^n` isA. one-one and into

1.	The function `f : N->N` given by `f(n)=n-(-1)^n` isA. one-one and intoB. one-one and ontoC. many-one and intoD. many-one and onto
Answer» Correct Answer - 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.