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Question Number 162116 by naka3546 last updated on 26/Dec/21

$$Afunctionfissuchthatf:\mathbb{R}\to \mathbb{R}where$$$$f(xy+1)=f\left(x\right)\cdot f\left(y\right)-f\left(y\right)-x+2,\mathrm{\forall }x,y\in \mathbb{R}.$$$$Findvalueof10\cdot f\left(2017\right)+f\left(0\right).$$

Answered by Rasheed.Sindhi last updated on 27/Dec/21

$$f(xy+1)=f\left(x\right)\cdot f\left(y\right)-f\left(y\right)-x+2,\mathrm{\forall }x,y\in \mathbb{R}.$$$$10\cdot f\left(2017\right)+f\left(0\right)=?$$$$⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢$$$$\bullet x=0:f\left(1\right)=f\left(0\right).f\left(y\right)-f\left(y\right)+2$$$$f\left(1\right)-2=f\left(y\right)(f\left(0\right)-1)$$$$f\left(y\right)=\frac{f\left(1\right)-2}{f\left(0\right)-1}$$$$\Rightarrow f\left(x\right)=\frac{f\left(1\right)-2}{f\left(0\right)-1}................\left(i\right)$$$$\Rightarrow f\left(x\right)isconstant$$$$f\left(1\right)=\frac{f\left(1\right)-2}{f\left(0\right)-1}\mathrm{\&}f\left(0\right)=\frac{f\left(1\right)-2}{f\left(0\right)-1}$$$$f\left(1\right)=f\left(0\right)$$$$▸f\left(0\right)=\frac{f\left(1\right)-2}{f\left(0\right)-1}$$$$\Rightarrow f\left(0\right)=\frac{f\left(0\right)-2}{f\left(0\right)}\Rightarrow f^2\left(0\right)-f\left(0\right)+2=0$$$$f\left(0\right)=\frac{1\pm \sqrt{1-8}}{2}=\frac{1\pm i\sqrt{7}}{2}$$$$▸f\left(1\right)=\frac{f\left(1\right)-2}{f\left(0\right)-1}\Rightarrow f\left(0\right)=\frac{f\left(0\right)-2}{f\left(0\right)-1}$$$$f^2\left(0\right)-f\left(0\right)=f\left(0\right)-2$$$$f^2\left(0\right)-2f\left(0\right)+2=0$$$$f\left(0\right)=\frac{2\pm \sqrt{4-8}}{2}=1\pm i$$$$f\left(0\right)isnotunique.Alsof\left(0\right)\notin \mathbb{R}$$$$\bullet y=0:f\left(1\right)=f\left(x\right).f\left(0\right)-f\left(0\right)-x+2$$$$f\left(x\right)=\frac{f\left(1\right)+f\left(0\right)+x-2}{f\left(0\right)}........\left(ii\right)$$$$\left(i\right)\mathrm{\&}\left(ii\right):\frac{f\left(1\right)-2}{f\left(0\right)-1}=\frac{f\left(1\right)+f\left(0\right)+x-2}{f\left(0\right)}$$$$f\left(1\right).f\left(0\right)+f^2(0+xf\left(0\right)-2f\left(0\right)-f\left(1\right)-f\left(0\right)-x+2$$$$=f\left(1\right).f\left(0\right)-2f\left(0\right)$$$$f^2(0+xf\left(0\right)-f\left(1\right)-f\left(0\right)-x+2=0$$$$x(f\left(0\right)-1)=f^2\left(0\right)-f\left(1\right)-f\left(0\right)+2$$$$x=\frac{f^2\left(0\right)-f\left(1\right)-f\left(0\right)+2}{f\left(0\right)-1}$$$$xisconstant$$$$x\mathrm{\&}f\left(x\right)bothareconstant!!!$$$$\mathcal{T}{here}^{\prime }saprobleminquestion.$$

Commented by naka3546 last updated on 27/Dec/21

$$Thankyou,sir.Mightbethequestionisnotcompleted.$$

Answered by Rasheed.Sindhi last updated on 27/Dec/21

$$f(xy+1)=f\left(x\right)\cdot f\left(y\right)-f\left(y\right)-x+2,\mathrm{\forall }x,y\in \mathbb{R}.$$$$10\cdot f\left(2017\right)+f\left(0\right)=?$$$$\bullet x=0:$$$$f(0.y+1)=f\left(0\right)\cdot f\left(y\right)-f\left(y\right)-0+2$$$$f\left(1\right)=f\left(0\right)\cdot f\left(y\right)-f\left(y\right)+2$$$$f\left(y\right)(f\left(0\right)-1)=f\left(1\right)-2$$$$f\left(y\right)=\frac{f\left(1\right)-2}{f\left(0\right)-1}$$$$f\left(x\right)=\frac{f\left(1\right)-2}{f\left(0\right)-1}\Rightarrow f\left(x\right)isconstant...\left(i\right)$$$$f\left(0\right)=f\left(1\right)=f\left(3\right)=...f\left(2017\right)$$$$\bullet y=0:$$$$f(x.0+1)=f\left(x\right)\cdot f\left(0\right)-f\left(0\right)-x+2$$$$f\left(1\right)=f\left(x\right)\cdot f\left(0\right)-f\left(0\right)-x+2$$$$f\left(x\right)=\frac{f\left(1\right)+f\left(0\right)+x-2}{f\left(0\right)}\Rightarrow f\left(x\right)isnotconstant...\left(ii\right)$$$$\left(i\right)\mathrm{\&}\left(ii\right)arecontradictory.Ifthey$$$$occursymultaneously,thenxis$$$$constant.\mathcal{T}hatmeansf\left(x\right)isdefined$$$$onlyforasingleconstantwhich$$$$evendoesnotexistitself.Whichexist$$$$onlywhenf\left(0\right)\mathrm{\&}f\left(1\right)exist.But$$$$f\left(0\right)\mathrm{\&}f\left(1\right)arenotdefinedbecausef\left(x\right)$$$$isdefinedforasingleconstantwhich$$$$isagaindependantonf\left(0\right)\mathrm{\&}f\left(1\right)...$$

Commented by mr W last updated on 28/Dec/21

$$ialsothinkthequestioniswrongor$$$$notcompletelyexpressed.$$

Commented by Rasheed.Sindhi last updated on 28/Dec/21

$$\mathbb{T}\mathrm{han}\mathbb{k}\mathrm{s}\mathrm{a}\mathrm{lot}\mathbf{sir}!$$

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