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Let \mathrm{f, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N}} be functions defined by \mathrm{ f(a)=\alpha,}  where \mathrm{\alpha } is the maximum of the powers of those primes \mathrm { p} such that \mathrm { p^{\alpha}} divides  \mathrm { a,}  and \mathrm { g(a)=a+1,}  for all \mathrm { a \in \mathbb{N}-\{1\}. }Then, the function \mathrm { f+g} is

Option: 1

 one-one but not onto 


Option: 2

 onto but not one-one 


Option: 3

 both one-one and onto 


Option: 4

 neither one-one nor onto 


Answers (1)

best_answer

\mathrm{f^{\prime}: N-\{1\} \rightarrow N \quad f(a)=\alpha}

\mathrm{where\; \alpha\; is\; max\; of\; powers\; of\; prime\; p\; such\; that \;p^{\alpha}}\mathrm{divides a. Also\; g(a)=a+1}

\mathrm{\therefore f(2)=1 \quad g(2)=3} \\

\mathrm{f(3)=1\quad g(3)=4 } \\

\mathrm{ f(4)=2 \quad g(4)=5} \\

\mathrm{f(5)=1 \quad g(5)=6} \\

\mathrm{\Rightarrow f(2)+g(2)=4 }\\

\mathrm{ (f(3)+g(3))=5} \\

\mathrm{f(4)+g(4)=7}

\mathrm{f(5)+g(5)=7}

\mathrm{\because\; many\; one f(x)+g(x) does\; not\; contain\; 1}

\mathrm{\Rightarrow into \;function}

Hence correct option is 4

Posted by

Ajit Kumar Dubey

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