Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 145534 by wassim last updated on 05/Jul/21

	Answered by Rasheed.Sindhi last updated on 06/Jul/21
	$C-1:Finite sets containing single member: Let x∈M ∴ x^2 −3∣x∣+4=x x^2 −3∣x∣+4=±∣x∣ ⇒ { (((∣x∣)^2 −4∣x∣+4=0....(i))),(((∣x∣)^2 −2∣x∣+4=0....(ii))) :} (i) ⇒∣x∣=2⇒x=±2 M={2},{−2} (ii)⇒∣x∣=((2±(√(4−16)))/2)∉R C−2:Finite sets containing more than one member. f(x)=x^2 −3∣x∣+4 x∈M⇒f(x)∈M (Given) Let f(x)≠f(x) ∵ f(x)∈M ∴ f( f(x) )∈M & f( f(x) )≠f(x) ∵ f( f(x) )∈M ∴ f( f( f(x) ) )∈M & f( f( f(x) ) )≠f( f(x) ) In this way x,f(x),f( f(x) ),f( f( f(x) ) ),... ∈M & they all are different ∴ M have infinite different members There exist infinite members of M • •^(•) M has only two finite values { (),() :}2 {: (),() } , { (),() :}−2 {: (),() }$
	$C - 1 : F i n i t e s e t s c o n t a i n i n g$ $s i n g l e m e m b e r :$ $L e t x \in M$ $∴ x^{2} - 3 ∣ x ∣ + 4 = x$ $x^{2} - 3 ∣ x ∣ + 4 = \pm ∣ x ∣$ $\Rightarrow {\begin{cases} {(∣ x ∣)}^{2} - 4 ∣ x ∣ + 4 = 0 . . . . (i) \\ {(∣ x ∣)}^{2} - 2 ∣ x ∣ + 4 = 0 . . . . (i i) \end{cases}$ $(i) \Rightarrow∣ x ∣= 2 \Rightarrow x = \pm 2$ $M = {2}, {- 2}$ $(i i) \Rightarrow∣ x ∣= \frac{2 \pm \sqrt{4 - 16}}{2} \notin R$ $C - 2 : F i n i t e s e t s c o n t a i n i n g m o r e$ $t h a n o n e m e m b e r .$ $f (x) = x^{2} - 3 ∣ x ∣ + 4$ $x \in M \Rightarrow f (x) \in M (Given)$ $Let f (x) \neq f (x)$ $∵ f (x) \in M$ $∴ f (f (x)) \in M & f (f (x)) \neq f (x)$ $∵ f (f (x)) \in M$ $∴ f (f (f (x))) \in M & f (f (f (x))) \neq f (f (x))$ $In this way$ $x, f (x), f (f (x)), f (f (f (x))), . . . \in M$ $& they all are different$ $∴ M have infinite different members$ $T h e r e e x i s t i n f i n i t e m e m b e r s o f M$ $\overset{∙}{∙ ∙} M has only two finite values$ ${\begin{cases} \end{cases} 2 \begin{matrix} \end{matrix}}, {\begin{cases} \end{cases} - 2 \begin{matrix} \end{matrix}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum