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Let N be the set of natural numbers and two functions f and g be defined as f, g : N → N such that f(n) = { (n+1)/2 : if n is odd, n/2 : if n is even } and g(n) = n - (-1)^n. Then fog is

1. neither one-one nor onto
2. onto but not one-one
3. both one-one and onto
4. one-one but not onto

Correct answer: onto but not one-one

Solution

The function fog is onto because for every natural number m, there exists a natural number n such that fog(n) = m, covering all possible outputs. However, it is not one-one since multiple inputs can produce the same output, particularly when n is even, leading to the same result after applying both functions.

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