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Is a function on the natural numbers one-one or onto

This type of question, asked 4 times: 2021 Paper 3, 28 Jun 2022, 27 Jul 2022, 27 Jan 2024. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 3

    70
    Let f,g:N→Nf,g:N \rightarrow N such that f(n+1)=f(n)+f(1) ∀n∈Nf(n + 1) = f(n) + f(1)\ \forall n \in N and gg be any arbitrary function. Which of the following statements is NOT true?

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  2. Question 2

    28 Jun 2022

    62
    Let a function f:N→Nf:\mathbb{N}\rightarrow\mathbb{N} be defined by
    f(n)=[2n,n=2,4,6,8,…..n−1,n=3,7,11,15,…..n+12,n=1,5,9,13,…..f(n) =\left\lbrack \begin{matrix} 2n, & n= 2,4,6,8,\ldots.. \\ n- 1, & n= 3,7,11,15,\ldots.. \\ \frac{n+ 1}{2}, & n= 1,5,9,13,\ldots.. \end{matrix} \right.
    then, ff is

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  3. Question 3

    27 Jul 2022

    62
    Let f,g:N−{1}→N\mathbf{f,g:N - \{ 1\} \rightarrow N} be functions defined by f(a)=αf(a) = \alpha, where α\alpha is the maximum of the powers of those primes pp such that pαp^{\alpha} divides aa, and g(a)=a+1g(a) = a + 1, for all a∈N−{1}\mathbf{a \in N - \{ 1\}}. Then, the function f+gf + g is

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  4. Question 4

    27 Jan 2024

    80
    The function f:N−{1}→N\mathbf{f:N - \{ 1\} \rightarrow N}; defined by f(n)=f(n) = the highest prime factor of nn, is :

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