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) = cases 3 + 2n & if n 0, \ -2n & if n < 0. cases 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?
- g f is NOT one-one and g f is NOT onto
- f g is NOT one-one but f g is onto
- g is one-one and g is onto
- f is NOT one-one but f is onto
Answer
(1) g f is NOT one-one and g f is NOT onto, (4) f is NOT one-one but f is onto
Solution
f: odd n = 1, 3, 5, 1, 2, 3, ; even n = 2, 4, 6, 1, 0, -1,
So f is onto Z but f(1) = f(2) = 1: not one-one. (D) true.
g: n 0 3, 5, 7, ; n < 0 2, 4, 6, — one-one, but 1 is never attained: not onto. (C) false.
g f: f not one-one g f not one-one; range range of g, which misses 1 not onto. (A) true.
f g: range of g is N - \1\. On it, odd n 3 2, 3, 4, and even n 2 1, 0, -1, — no overlap, every integer once. So f g is one-one and onto. (B) false.
Answer: (A), (D)