First slide
Functions (XII)
Question

 Let f:NZ and f(x)==x12    ; when x is odd x2 ;     when x is even , then : 

Moderate
Solution

f(x)=x12x= odd x2x= even  f(x):NZ

 Let x=odd=(2n+1);n>0

f(x)=2n+112=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

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