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Question Number 33649 by rahul 19 last updated on 21/Apr/18

$$Considerf:R^{+}\to Rsuchthat$$$$f\left(3\right)=1fora\in R^{+}and$$$$f\left(x\right).f\left(y\right)+f\left(\frac{3}{x}\right).f\left(\frac{3}{y}\right)=2f\left(xy\right)$$$$\mathrm{\forall }x,y\in R^{+}.Thenfindf\left(x\right)?$$

Commented by rahul 19 last updated on 21/Apr/18

$$prof\mathit{A}bdoimad,MJSsirplshelp.$$

Commented by math khazana by abdo last updated on 21/Apr/18

$$\Rightarrow {\left(f\left(1\right)\right)}^2+{\left(f\left(3\right)\right)}^2=2f\left(1\right)\Rightarrow {\left(f\left(1\right)\right)}^2-2f\left(1\right)+1=0$$$$\Rightarrow {(f\left(1\right)-1)}^2=0\Rightarrow f\left(1\right)=1fory=\frac{1}{x}\Rightarrow$$$$f\left(x\right)f\left(\frac{1}{x}\right)+f\left(\frac{3}{x}\right).f\left(3x\right)=2letx\to +\mathrm{\infty }$$$$f(+\mathrm{\infty })f\left(0\right)+f(+\mathrm{\infty })f\left(0\right)=2\Rightarrow f\left(0\right)f(+\mathrm{\infty })=1$$$$thisconditionprovethatf\left(x\right)=c\left(constant\right)$$$$\mathrm{\forall }x\in R^{+}f\left(x\right)f\left(\frac{1}{x}\right)+f\left(\frac{3}{x}\right)f\left(3x\right)=2\Rightarrow$$$$c^2+c^2=2\Rightarrow 2c^2=2\Rightarrow c=\stackrel{-}{+}1butf\left(1\right)=f\left(3\right)=1$$$$\Rightarrow f\left(x\right)=1\mathrm{\forall }x\in R^{+}$$

Commented by rahul 19 last updated on 22/Apr/18

$$thankusir!$$

Answered by MJS last updated on 21/Apr/18

$$x=1;y=1$$$$f\left(1\right)f\left(1\right)+f\left(3\right)f\left(3\right)=2f\left(1\right)$$$$f{\left(1\right)}^2-2f\left(1\right)+1=0$$$${(f\left(1\right)-1)}^2=0$$$$f\left(1\right)=1$$$$$$$$x=3;y=3$$$$f\left(3\right)f\left(3\right)+f\left(1\right)f\left(1\right)=2f\left(9\right)$$$$1+1=2f\left(9\right)$$$$f\left(9\right)=1$$$$$$$$\mathrm{so}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{solution}\mathrm{is}$$$$f\left(x\right)=1$$$$$$$$\mathrm{we}\mathrm{could}\mathrm{also}\mathrm{start}\mathrm{with}$$$$x=1;y=3$$$$f\left(1\right)f\left(3\right)+f\left(\frac{3}{1}\right)f\left(\frac{3}{3}\right)=2f\left(1\times 3\right)$$$$f\left(1\right)+f\left(1\right)=2\Rightarrow f\left(1\right)=1$$

Commented by rahul 19 last updated on 21/Apr/18

$$sir,actuallyialsogotoneofthesol.$$$$asf\left(x\right)=1.$$$$Butiamfindinghardtoprovethat$$$$foranyvalueofx,y\in {\mathit{R}}^{+}.$$$$f\left(x\right)=1only!$$$$imeanonlyonesolution.$$$$$$

Commented by MJS last updated on 21/Apr/18

$$\mathrm{if}\mathrm{the}\mathrm{function}\mathrm{is}y=1\mathrm{then}\mathrm{the}\mathrm{given}\mathrm{equation}$$$$\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}x\in \mathbb{R}\mathrm{because}$$$$f\left(x\right)=f\left(y\right)=f\left(\frac{3}{x}\right)=f\left(\frac{3}{y}\right)=f\left(xy\right)=1\mathrm{and}$$$$1\times 1+1\times 1=2\times 1$$$$$$$$\mathrm{if}x\ne y\mathrm{was}\mathrm{demanded}{\mathrm{we}}^{\prime }\mathrm{d}\mathrm{need}\mathrm{a}\mathrm{different}$$$$\mathrm{function},\mathrm{but}\mathrm{I}\mathrm{could}\mathrm{not}\mathrm{find}\mathrm{one}$$

Commented by rahul 19 last updated on 22/Apr/18

$$thankusir.$$

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