Question-145534

Question Number 145534 by wassim last updated on 05/Jul/21

Answered by Rasheed.Sindhi last updated on 06/Jul/21

C-1:Finite sets containing single member: Let x∈M ∴ x^2 −3∣x∣+4=x x^2 −3∣x∣+4=±∣x∣ ⇒ { (((∣x∣)^2 −4∣x∣+4=0....(i))),(((∣x∣)^2 −2∣x∣+4=0....(ii))) :} (i) ⇒∣x∣=2⇒x=±2 M={2},{−2} (ii)⇒∣x∣=((2±(√(4−16)))/2)∉R C−2:Finite sets containing more than one member. f(x)=x^2 −3∣x∣+4 x∈M⇒f(x)∈M (Given) Let f(x)≠f(x) ∵ f(x)∈M ∴ f( f(x) )∈M & f( f(x) )≠f(x) ∵ f( f(x) )∈M ∴ f( f( f(x) ) )∈M & f( f( f(x) ) )≠f( f(x) ) In this way x,f(x),f( f(x) ),f( f( f(x) ) ),... ∈M & they all are different ∴ M have infinite different members There exist infinite members of M • •^(•) M has only two finite values { (),() :}2 {: (),() } , { (),() :}−2 {: (),() }

$${C}-\mathrm{1}:{Finite}\:{sets}\:{containing}\: \\ $$$$\boldsymbol{{single}}\:\boldsymbol{{member}}: \\ $$$${Let}\:{x}\in\mathrm{M} \\ $$$$\therefore\:{x}^{\mathrm{2}} −\mathrm{3}\mid{x}\mid+\mathrm{4}={x} \\ $$$$\:\:\:\:{x}^{\mathrm{2}} −\mathrm{3}\mid{x}\mid+\mathrm{4}=\pm\mid{x}\mid \\ $$$$\Rightarrow\begin{cases}{\left(\mid{x}\mid\right)^{\mathrm{2}} −\mathrm{4}\mid{x}\mid+\mathrm{4}=\mathrm{0}….\left({i}\right)}\\{\left(\mid{x}\mid\right)^{\mathrm{2}} −\mathrm{2}\mid{x}\mid+\mathrm{4}=\mathrm{0}….\left({ii}\right)}\end{cases} \\ $$$$\left({i}\right)\:\Rightarrow\mid{x}\mid=\mathrm{2}\Rightarrow{x}=\pm\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{M}=\left\{\mathrm{2}\right\},\left\{−\mathrm{2}\right\} \\ $$$$\left({ii}\right)\Rightarrow\mid{x}\mid=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{16}}}{\mathrm{2}}\notin\mathbb{R} \\ $$$${C}−\mathrm{2}:{Finite}\:{sets}\:{containing}\:{more} \\ $$$${than}\:{one}\:{member}. \\ $$$$\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)={x}^{\mathrm{2}} −\mathrm{3}\mid{x}\mid+\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}\in\mathrm{M}\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\in\mathrm{M}\:\left(\mathrm{Given}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Let}\:\mathrm{f}\left(\mathrm{x}\right)\neq\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\:\:\:\because\:\mathrm{f}\left(\mathrm{x}\right)\in\mathrm{M} \\ $$$$\:\:\:\therefore\:\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{x}\right)\:\right)\in\mathrm{M}\:\&\:\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{x}\right)\:\right)\neq\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\:\: \\ $$$$\:\:\:\because\:\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{x}\right)\:\right)\in\mathrm{M} \\ $$$$\:\:\:\therefore\:\mathrm{f}\left(\:\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{x}\right)\:\right)\:\right)\in\mathrm{M}\:\&\:\mathrm{f}\left(\:\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{x}\right)\:\right)\:\right)\neq\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{x}\right)\:\right) \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{way}\: \\ $$$$\:\:\:\:\mathrm{x},\mathrm{f}\left(\mathrm{x}\right),\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{x}\right)\:\right),\mathrm{f}\left(\:\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{x}\right)\:\right)\:\right),…\:\in\mathrm{M} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\&\:\mathrm{they}\:\mathrm{all}\:\mathrm{are}\:\mathrm{different} \\ $$$$\therefore\:\mathrm{M}\:\mathrm{have}\:\mathrm{infinite}\:\:\mathrm{different}\:\mathrm{members} \\ $$$${There}\:{exist}\:{infinite}\:{members}\:{of}\:\mathrm{M} \\ $$$$ \\ $$$$\overset{\bullet} {\bullet\:\:\bullet}\:\:\mathrm{M}\:\mathrm{has}\:\mathrm{only}\:\mathrm{two}\:\mathrm{finite}\:\mathrm{values} \\ $$$$\:\:\:\:\:\:\:\:\:\begin{cases}{}\\{}\end{cases}\mathrm{2\left.\begin{matrix}{}\\{}\end{matrix}\right\}}\:,\:\:\:\begin{cases}{}\\{}\end{cases}−\mathrm{2\left.\begin{matrix}{}\\{}\end{matrix}\right\}} \\ $$

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