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Question Number 195954 by sonukgindia last updated on 13/Aug/23
Answered by mahdipoor last updated on 13/Aug/23
(1/4)×(((4−a)−x)/(x+(a−2)))×(((4−a)+x)/(−x+(a−2)))=c c=(1/4) ⇒ { ((4−a=2−a⇒∄)),((4−a=a−2⇒a=3)) :}
$$\frac{\mathrm{1}}{\mathrm{4}}×\frac{\left(\mathrm{4}−{a}\right)−{x}}{{x}+\left({a}−\mathrm{2}\right)}×\frac{\left(\mathrm{4}−{a}\right)+{x}}{−{x}+\left({a}−\mathrm{2}\right)}={c} \\ $$$${c}=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:\begin{cases}{\mathrm{4}−{a}=\mathrm{2}−{a}\Rightarrow\nexists}\\{\mathrm{4}−{a}={a}−\mathrm{2}\Rightarrow{a}=\mathrm{3}}\end{cases} \\ $$
Answered by Rasheed.Sindhi last updated on 14/Aug/23
f(x)=((4−x)/(2x−4)), f(a+x)f(a−x)=c x=a: f(a+x)f(a−x)=c ⇒f(2a)f(0)=c ⇒((4−2a)/(2(2a)−4))×((4−0)/(2(0)−4))=c c=((2a−4)/(4a−4))=((a−2)/(2a−2))...........(i) x=c: f(a+x)f(a−x)=c⇒ f(a+c)f(a−c)=c ((4−(a+c))/(2(a+c)−4))×((4−(a−c))/(2(a−c)−4))=c ((4−a−c)/(2a+2c−4))×((4−a+c)/(2a−2c−4))=c (((4−a)−c)/((2a−4)+2c))×(((4−a)+c)/((2a−4)−2c))=c (((4−a)^2 −(((a−2)/(2a−2)))^2 )/((2a−4)^2 −4(((a−2)/(2a−2)))^2 ))=((a−2)/(2a−2)) (((4−a)^2 (2a−2)^2 −(a−2)^2 )/((2a−4)^2 (2a−2)^2 −4(a−2)^2 ))=((a−2)/(2a−2)) yeilds fifth order equation with a=3 ✓ (With the help of sir mahdipoor′s answer) c=((a−2)/(2a−2))=((3−2)/(2(3)−2))=(1/4)✓
$$ \\ $$$${f}\left({x}\right)=\frac{\mathrm{4}−{x}}{\mathrm{2}{x}−\mathrm{4}},\:{f}\left({a}+{x}\right){f}\left({a}−{x}\right)={c} \\ $$$${x}={a}:\:\: \\ $$$${f}\left({a}+{x}\right){f}\left({a}−{x}\right)={c} \\ $$$$\Rightarrow{f}\left(\mathrm{2}{a}\right){f}\left(\mathrm{0}\right)={c} \\ $$$$\Rightarrow\frac{\mathrm{4}−\mathrm{2}{a}}{\mathrm{2}\left(\mathrm{2}{a}\right)−\mathrm{4}}×\frac{\mathrm{4}−\mathrm{0}}{\mathrm{2}\left(\mathrm{0}\right)−\mathrm{4}}={c} \\ $$$$\:\:\:\:\:\:{c}=\frac{\mathrm{2}{a}−\mathrm{4}}{\mathrm{4}{a}−\mathrm{4}}=\frac{{a}−\mathrm{2}}{\mathrm{2}{a}−\mathrm{2}}………..\left({i}\right) \\ $$$${x}={c}: \\ $$$${f}\left({a}+{x}\right){f}\left({a}−{x}\right)={c}\Rightarrow \\ $$$$\:\:{f}\left({a}+{c}\right){f}\left({a}−{c}\right)={c} \\ $$$$\frac{\mathrm{4}−\left({a}+{c}\right)}{\mathrm{2}\left({a}+{c}\right)−\mathrm{4}}×\frac{\mathrm{4}−\left({a}−{c}\right)}{\mathrm{2}\left({a}−{c}\right)−\mathrm{4}}={c} \\ $$$$\frac{\mathrm{4}−{a}−{c}}{\mathrm{2}{a}+\mathrm{2}{c}−\mathrm{4}}×\frac{\mathrm{4}−{a}+{c}}{\mathrm{2}{a}−\mathrm{2}{c}−\mathrm{4}}={c} \\ $$$$\frac{\left(\mathrm{4}−{a}\right)−{c}}{\left(\mathrm{2}{a}−\mathrm{4}\right)+\mathrm{2}{c}}×\frac{\left(\mathrm{4}−{a}\right)+{c}}{\left(\mathrm{2}{a}−\mathrm{4}\right)−\mathrm{2}{c}}={c} \\ $$$$\frac{\left(\mathrm{4}−{a}\right)^{\mathrm{2}} −\left(\frac{{a}−\mathrm{2}}{\mathrm{2}{a}−\mathrm{2}}\right)^{\mathrm{2}} }{\left(\mathrm{2}{a}−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{{a}−\mathrm{2}}{\mathrm{2}{a}−\mathrm{2}}\right)^{\mathrm{2}} }=\frac{{a}−\mathrm{2}}{\mathrm{2}{a}−\mathrm{2}} \\ $$$$\frac{\left(\mathrm{4}−{a}\right)^{\mathrm{2}} \left(\mathrm{2}{a}−\mathrm{2}\right)^{\mathrm{2}} −\left({a}−\mathrm{2}\right)^{\mathrm{2}} }{\left(\mathrm{2}{a}−\mathrm{4}\right)^{\mathrm{2}} \left(\mathrm{2}{a}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left({a}−\mathrm{2}\right)^{\mathrm{2}} }=\frac{{a}−\mathrm{2}}{\mathrm{2}{a}−\mathrm{2}} \\ $$$${yeilds}\:{fifth}\:{order}\:{equation}\:{with} \\ $$$${a}=\mathrm{3}\:\checkmark\:\left({With}\:{the}\:{help}\:{of}\:{sir}\:{mahdipoor}'{s}\:{answer}\right) \\ $$$${c}=\frac{{a}−\mathrm{2}}{\mathrm{2}{a}−\mathrm{2}}=\frac{\mathrm{3}−\mathrm{2}}{\mathrm{2}\left(\mathrm{3}\right)−\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 14/Aug/23
f(x)=((4−x)/(2x−4)), f(a+x)f(a−x)=c x=0: f(a+0)f(a−0)=c f^( 2) (a)=c (((4−a)/(2a−4)))^2 =c..................(i) x=1: f(a+1)f(a−1)=c ((4−(a+1))/(2(a+1)−4)) ∙ ((4−(a−1))/(2(a−1)−4)) ((3−a)/(2a−2)) ∙ ((5−a)/(2a−6))=c..........(ii) from (i) & (ii): (((4−a)/(2a−4)))^2 =((3−a)/(2a−2)) ∙ ((5−a)/(2a−6)) ((a^2 −8a+16)/(4a^2 −16a+16))=((a^2 −8a+15)/(4a^2 −16a+12)) (a^2 −8a+16)(4a^2 −16a+12) =(a^2 −8a+15)(4a^2 −16a+16) 4a^4 −48a^3 +204a^2 −352a+192 =4a^4 −48a^3 +204a^2 −368a+240 −352a+192=−368a+240 −352a+368a=240−192 16a=48⇒a=3✓ c=(((4−a)/(2a−4)))^2 =(((4−3)/(2(3)−4)))^2 =(1/4)✓
$${f}\left({x}\right)=\frac{\mathrm{4}−{x}}{\mathrm{2}{x}−\mathrm{4}},\:{f}\left({a}+{x}\right){f}\left({a}−{x}\right)={c} \\ $$$${x}=\mathrm{0}: \\ $$$${f}\left({a}+\mathrm{0}\right){f}\left({a}−\mathrm{0}\right)={c} \\ $$$${f}^{\:\mathrm{2}} \left({a}\right)={c} \\ $$$$\left(\frac{\mathrm{4}−{a}}{\mathrm{2}{a}−\mathrm{4}}\right)^{\mathrm{2}} ={c}………………\left({i}\right) \\ $$$${x}=\mathrm{1}: \\ $$$${f}\left({a}+\mathrm{1}\right){f}\left({a}−\mathrm{1}\right)={c} \\ $$$$\frac{\mathrm{4}−\left({a}+\mathrm{1}\right)}{\mathrm{2}\left({a}+\mathrm{1}\right)−\mathrm{4}}\:\centerdot\:\frac{\mathrm{4}−\left({a}−\mathrm{1}\right)}{\mathrm{2}\left({a}−\mathrm{1}\right)−\mathrm{4}} \\ $$$$\frac{\mathrm{3}−{a}}{\mathrm{2}{a}−\mathrm{2}}\:\centerdot\:\frac{\mathrm{5}−{a}}{\mathrm{2}{a}−\mathrm{6}}={c}……….\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:\&\:\left({ii}\right): \\ $$$$\:\:\:\left(\frac{\mathrm{4}−{a}}{\mathrm{2}{a}−\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{3}−{a}}{\mathrm{2}{a}−\mathrm{2}}\:\centerdot\:\frac{\mathrm{5}−{a}}{\mathrm{2}{a}−\mathrm{6}} \\ $$$$\:\:\frac{{a}^{\mathrm{2}} −\mathrm{8}{a}+\mathrm{16}}{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{16}{a}+\mathrm{16}}=\frac{{a}^{\mathrm{2}} −\mathrm{8}{a}+\mathrm{15}}{\mathrm{4}{a}^{\mathrm{2}} −\mathrm{16}{a}+\mathrm{12}} \\ $$$$ \\ $$$$\:\:\:\:\:\left({a}^{\mathrm{2}} −\mathrm{8}{a}+\mathrm{16}\right)\left(\mathrm{4}{a}^{\mathrm{2}} −\mathrm{16}{a}+\mathrm{12}\right) \\ $$$$\:\:\:\:\:\:\:\:=\left({a}^{\mathrm{2}} −\mathrm{8}{a}+\mathrm{15}\right)\left(\mathrm{4}{a}^{\mathrm{2}} −\mathrm{16}{a}+\mathrm{16}\right) \\ $$$$\: \\ $$$$\:\:\mathrm{4}{a}^{\mathrm{4}} −\mathrm{48}{a}^{\mathrm{3}} +\mathrm{204}{a}^{\mathrm{2}} −\mathrm{352}{a}+\mathrm{192} \\ $$$$\:\:\:\:=\mathrm{4}{a}^{\mathrm{4}} −\mathrm{48}{a}^{\mathrm{3}} +\mathrm{204}{a}^{\mathrm{2}} −\mathrm{368}{a}+\mathrm{240} \\ $$$$ \\ $$$$\:\:−\mathrm{352}{a}+\mathrm{192}=−\mathrm{368}{a}+\mathrm{240} \\ $$$$\:\:−\mathrm{352}{a}+\mathrm{368}{a}=\mathrm{240}−\mathrm{192} \\ $$$$\:\:\:\:\:\mathrm{16}{a}=\mathrm{48}\Rightarrow{a}=\mathrm{3}\checkmark \\ $$$$\:\:\:\:\:\:{c}=\left(\frac{\mathrm{4}−{a}}{\mathrm{2}{a}−\mathrm{4}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{4}−\mathrm{3}}{\mathrm{2}\left(\mathrm{3}\right)−\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\checkmark \\ $$

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