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Let N denote the set of all natural numbers, and Z denote the set of all integers. Consider the functions f: N Z and g: Z N defined by f(n)= cases (n+1) / 2 & if n is odd (4-n) / 2 & if n is even cases And g(n)= array cc 3+2 n & if n 0 -2 n & if n <0 array . Define (g f)(n)=g(f(n)) for all n N , and (f g)(n)=f(g(n)) for all n Z . Then which of the following statements is (are) TRUE?

Options

  1. Ag f is NOT one-one and g f is NOT onto
  2. Bf ~g is NOT one-one but f g is onto
  3. Cg is one-one and g is onto
  4. Df is NOT one-one but f is onto

Correct answer

A. g f is NOT one-one and g f is NOT onto

Step-by-step solution

aligned & f(n)= cases (n+1) / 2 & if n is odd (4-n) / 2 & if n is even cases & f(n)= (1,1),(2,1),(3,2),(4,0),(5,3),(6,-1), aligned f ( n ) is many one and onto function aligned & g ( n )= array cc 3+2 n & if n 0 -2 n & if n <0 array . & g ( n )= (-3,6),(-2,4),(-1,2),(0,3),(1,5),(2,7),(3,9),(4,15), aligned g ( n ) is one-one and into function f ( ~g ( n ))=2+ n , n ~N fog is one-one and into g(f(n))= cases 4+n & if n is odd natural number 7-n & if n=2,4 n-4 & if n is even natural number and n 6 cases g(f(2))=g(f(1))

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