Let f:N⟶Z and f(x)==x−12 ; when x is odd −x2 ; when x is even , then :
f(x) is bijective
f(x) is injective but not surjective
f(x) is not injective but surjective
f(x) is neither injective nor surjective
f(x)=x−12x= odd −x2x= even f(x):N→Z
Let x=odd=(2n+1);n>0
f(x)=2n+1−12=n⇒+ ve integer
Let x= even =2m;m>0
f(x)=−2m2=−m⇒− ve integer
⇒ Range = codomains ⇒ onto and clearly f(x) is one-one function.
Hence, bijective