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Question Number 198276 by essaad last updated on 16/Oct/23

Answered by witcher3 last updated on 16/Oct/23

$$(\mathrm{x}=\mathrm{y})\Rightarrow \mathrm{f}\left(2\mathrm{x}\right)+2\mathrm{f}\left(0\right)=1$$$$\iff \mathrm{f}\left(2\mathrm{x}\right)=1-2\mathrm{f}\left(0\right)$$$$\mathrm{x}\to 2\mathrm{x}\mathrm{surjective}\iff \mathrm{\forall }\mathrm{t}\in \mathbb{R}\mathrm{f}\left(\mathrm{t}\right)=1-2\mathrm{f}\left(0\right)\mathrm{constant}$$$$$$

Answered by Rasheed.Sindhi last updated on 16/Oct/23

$$f(x+y)+2f(x-y)=\frac{f\left(x\right)}{f\left(y\right)}.....\left(i\right)$$$$x=y=0:$$$$\left(i\right)\Rightarrow 3f\left(0\right)=\frac{f\left(0\right)}{f\left(0\right)}=1$$$$\Rightarrow f\left(0\right)=\frac{1}{3}$$$$y=x:$$$$\left(i\right)\Rightarrow f\left(2x\right)+2f\left(0\right)=1$$$$f\left(2x\right)=1-2\left(\frac{1}{3}\right)=\frac{1}{3}$$$$f\left(x\right)=\frac{1}{3}$$

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