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A-function-f-is-such-that-f-R-R-where-f-xy-1-f-x-f-y-f-y-x-2-x-y-R-Find-value-of-10-f-2017-f-0-




Question Number 162116 by naka3546 last updated on 26/Dec/21
A function  f   is  such  that  f : R → R  where     f(xy+1) = f(x)∙f(y)−f(y)−x+2  ,  ∀x,y ∈ R .  Find  value  of  10∙f(2017)+f(0) .
Afunctionfissuchthatf:RRwheref(xy+1)=f(x)f(y)f(y)x+2,x,yR.Findvalueof10f(2017)+f(0).
Answered by Rasheed.Sindhi last updated on 27/Dec/21
f(xy+1) = f(x)∙f(y)−f(y)−x+2  ,  ∀x,y ∈ R .  10∙f(2017)+f(0)=?  ⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢  •x=0 : f(1)=f(0).f(y)−f(y)+2               f(1)−2=f(y)( f(0)−1 )  f(y)=((f(1)−2)/(f(0)−1))  ⇒f(x)=((f(1)−2)/(f(0)−1))................(i)  ⇒f(x) is constant       f(1)=((f(1)−2)/(f(0)−1)) & f(0)=((f(1)−2)/(f(0)−1))       f(1)=f(0)  ▶f(0)=((f(1)−2)/(f(0)−1))   ⇒f(0)=((f(0)−2)/(f(0)))⇒f^2 (0)−f(0)+2=0     f(0)=((1±(√(1−8)))/2)=((1±i(√7))/2)  ▶f(1)=((f(1)−2)/(f(0)−1))⇒f(0)=((f(0)−2)/(f(0)−1))        f^2 (0)−f(0)=f(0)−2         f^2 (0)−2f(0)+2=0          f(0)=((2±(√(4−8)))/2)=1±i          f(0) is not unique.Also f(0)∉R  •y=0:f(1)=f(x).f(0)−f(0)−x+2    f(x)=((f(1)+f(0)+x−2)/(f(0)))........(ii)  (i) & (ii):((f(1)−2)/(f(0)−1))=((f(1)+f(0)+x−2)/(f(0)))  f(1).f(0)+f^(  2) (0+xf(0)−2f(0)−f(1)−f(0)−x+2                                        =f(1).f(0)−2f(0)  f^(  2) (0+xf(0)−f(1)−f(0)−x+2=0  x( f(0)−1)=f^(  2) (0)−f(1)−f(0)+2       x=((f^(  2) (0)−f(1)−f(0)+2)/(f(0)−1))  x is constant  x & f(x) both are constant!!!  There′s a problem in question.
f(xy+1)=f(x)f(y)f(y)x+2,x,yR.10f(2017)+f(0)=?⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢x=0:f(1)=f(0).f(y)f(y)+2f(1)2=f(y)(f(0)1)f(y)=f(1)2f(0)1f(x)=f(1)2f(0)1.(i)f(x)isconstantf(1)=f(1)2f(0)1&f(0)=f(1)2f(0)1f(1)=f(0)f(0)=f(1)2f(0)1f(0)=f(0)2f(0)f2(0)f(0)+2=0f(0)=1±182=1±i72f(1)=f(1)2f(0)1f(0)=f(0)2f(0)1f2(0)f(0)=f(0)2f2(0)2f(0)+2=0f(0)=2±482=1±if(0)isnotunique.Alsof(0)Ry=0:f(1)=f(x).f(0)f(0)x+2f(x)=f(1)+f(0)+x2f(0)..(ii)(i)&(ii):f(1)2f(0)1=f(1)+f(0)+x2f(0)f(1).f(0)+f2(0+xf(0)2f(0)f(1)f(0)x+2=f(1).f(0)2f(0)f2(0+xf(0)f(1)f(0)x+2=0x(f(0)1)=f2(0)f(1)f(0)+2x=f2(0)f(1)f(0)+2f(0)1xisconstantx&f(x)bothareconstant!!!Theresaprobleminquestion.
Commented by naka3546 last updated on 27/Dec/21
Thank  you,  sir . Might  be  the  question is  not  completed.
Thankyou,sir.Mightbethequestionisnotcompleted.
Answered by Rasheed.Sindhi last updated on 27/Dec/21
 f(xy+1) = f(x)∙f(y)−f(y)−x+2  ,  ∀x,y ∈ R .  10∙f(2017)+f(0)=?  •x=0:   f(0.y+1) = f(0)∙f(y)−f(y)−0+2   f(1) = f(0)∙f(y)−f(y)+2  f(y)( f(0)−1)=f(1)−2  f(y)=((f(1)−2)/(f(0)−1))  f(x)=((f(1)−2)/(f(0)−1)) ⇒f(x) is constant...(i)  f(0)=f(1)=f(3)=...f(2017)  •y=0:   f(x.0+1) = f(x)∙f(0)−f(0)−x+2   f(1) = f(x)∙f(0)−f(0)−x+2  f(x)=((f(1)+f(0)+x−2)/(f(0)))⇒f(x) is not constant...(ii)  (i) & (ii) are contradictory.If they  occur symultaneously,then x is  constant.That means f(x) is defined  only for a single constant which   even does not exist itself.Which exist  only when f(0) & f(1) exist.But  f(0) & f(1) are not defined because f(x)  is defined for a single constant which  is again dependant on f(0) & f(1)...
f(xy+1)=f(x)f(y)f(y)x+2,x,yR.10f(2017)+f(0)=?x=0:f(0.y+1)=f(0)f(y)f(y)0+2f(1)=f(0)f(y)f(y)+2f(y)(f(0)1)=f(1)2f(y)=f(1)2f(0)1f(x)=f(1)2f(0)1f(x)isconstant(i)f(0)=f(1)=f(3)=f(2017)y=0:f(x.0+1)=f(x)f(0)f(0)x+2f(1)=f(x)f(0)f(0)x+2f(x)=f(1)+f(0)+x2f(0)f(x)isnotconstant(ii)(i)&(ii)arecontradictory.Iftheyoccursymultaneously,thenxisconstant.Thatmeansf(x)isdefinedonlyforasingleconstantwhichevendoesnotexistitself.Whichexistonlywhenf(0)&f(1)exist.Butf(0)&f(1)arenotdefinedbecausef(x)isdefinedforasingleconstantwhichisagaindependantonf(0)&f(1)
Commented by mr W last updated on 28/Dec/21
i also think the question is wrong or  not completely expressed.
ialsothinkthequestioniswrongornotcompletelyexpressed.
Commented by Rasheed.Sindhi last updated on 28/Dec/21
Thanks a lot sir!
Thanksalotsir!

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