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Question Number 33375 by rahul 19 last updated on 15/Apr/18

$$Iff:R\to Risan\mathit{o}\mathit{d}\mathit{d}functionsuch$$$$that:$$$$a)f(1+x)=1+f\left(x\right).$$$$b)x^{2}f\left(\frac{1}{x}\right)=f\left(x\right),x\ne 0.$$$$Thenfind\mathit{f}\left(\mathit{x}\right)?$$

Commented by prof Abdo imad last updated on 15/Apr/18

$$fisodd\Rightarrow f\left(0\right)=0\Rightarrow f(1+0)=1+f\left(0\right)\Rightarrow f\left(1\right)=1$$$$f\left(2\right)=f(1+1)=1+f\left(1\right)=2$$$$letsupposef\left(n\right)=n\Rightarrow f(n+1)=1+f\left(n\right)=1+n\Rightarrow$$$$\mathrm{\forall }n\in Nf\left(n\right)=nifn\in Z^{-}n=-m$$$$f\left(n\right)=f(-m)=-f\left(m\right)=-m=n\Rightarrow$$$$\mathrm{\forall }n\in Zf\left(n\right)=n$$$$f\left(1\right)=f\left(\frac{n}{n}\right)=nf\left(\frac{1}{n}\right)=1\Rightarrow f\left(\frac{1}{n}\right)=\frac{1}{n}$$$$f\left(\frac{p}{n}\right)=f(\frac{1}{n}+..+\frac{1}{n})\left(p\times \right)=pf\left(\frac{1}{n}\right)=\frac{p}{n}$$$$\Rightarrow \mathrm{\forall }x\in Qf\left(x\right)=xbutQisdenseinsideR\Rightarrow$$$$\mathrm{\forall }x\in Rf\left(x\right)=xwehaveforx\ne o$$$$x^{2}f\left(\frac{1}{x}\right)=x^2.\frac{1}{x}=x=f\left(x\right)cond.bisverified.$$$$$$

Commented by rahul 19 last updated on 15/Apr/18

$$thankyousir!$$

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