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Question Number 33649 by rahul 19 last updated on 21/Apr/18
Consider f:R^+ →R such that f(3)=1 for a∈R^+ and f(x).f(y) + f((3/x)).f((3/y)) = 2f(xy) ∀ x,y ∈ R^+ . Then find f(x) ?
$${Consider}\:{f}:{R}^{+} \rightarrow{R}\:{such}\:{that} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{1}\:{for}\:{a}\in{R}^{+} \:{and}\: \\ $$$${f}\left({x}\right).{f}\left({y}\right)\:+\:{f}\left(\frac{\mathrm{3}}{{x}}\right).{f}\left(\frac{\mathrm{3}}{{y}}\right)\:=\:\mathrm{2}{f}\left({xy}\right) \\ $$$$\forall\:{x},{y}\:\in\:{R}^{+} .\:{Then}\:{find}\:{f}\left({x}\right)\:? \\ $$
Commented by rahul 19 last updated on 21/Apr/18
prof Abdo imad , MJS sir pls help .
$${prof}\:\boldsymbol{{A}}{bdo}\:{imad}\:,\:{MJS}\:{sir}\:\:{pls}\:{help}\:. \\ $$
Commented by math khazana by abdo last updated on 21/Apr/18
⇒(f(1))^2 +(f(3))^2 =2f(1)⇒(f(1))^2 −2f(1) +1 =0 ⇒ (f(1)−1)^2 =0 ⇒ f(1)=1 for y=(1/x) ⇒ f(x)f((1/x)) +f((3/x)).f(3x) =2 let x→+∞ f(+∞)f(0) +f(+∞)f(0) =2 ⇒ f(0) f(+∞)=1 this condition prove that f(x)=c(constant) ∀ x ∈R^+ f(x)f((1/x)) +f((3/x))f(3x) =2 ⇒ c^2 +c^2 =2 ⇒ 2c^2 =2 ⇒ c =+^− 1 but f(1)=f(3) =1 ⇒ f(x)=1 ∀x ∈R^+
$$\Rightarrow\left({f}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+\left({f}\left(\mathrm{3}\right)\right)^{\mathrm{2}} \:=\mathrm{2}{f}\left(\mathrm{1}\right)\Rightarrow\left({f}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:−\mathrm{2}{f}\left(\mathrm{1}\right)\:+\mathrm{1}\:=\mathrm{0} \\ $$$$\Rightarrow\:\left({f}\left(\mathrm{1}\right)−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{f}\left(\mathrm{1}\right)=\mathrm{1}\:{for}\:{y}=\frac{\mathrm{1}}{{x}}\:\Rightarrow \\ $$$${f}\left({x}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)\:\:+{f}\left(\frac{\mathrm{3}}{{x}}\right).{f}\left(\mathrm{3}{x}\right)\:=\mathrm{2}\:\:{let}\:{x}\rightarrow+\infty \\ $$$${f}\left(+\infty\right){f}\left(\mathrm{0}\right)\:+{f}\left(+\infty\right){f}\left(\mathrm{0}\right)\:=\mathrm{2}\:\Rightarrow\:{f}\left(\mathrm{0}\right)\:{f}\left(+\infty\right)=\mathrm{1} \\ $$$${this}\:{condition}\:{prove}\:{that}\:{f}\left({x}\right)={c}\left({constant}\right) \\ $$$$\forall\:{x}\:\in{R}^{+} \:\:{f}\left({x}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)\:+{f}\left(\frac{\mathrm{3}}{{x}}\right){f}\left(\mathrm{3}{x}\right)\:=\mathrm{2}\:\Rightarrow \\ $$$${c}^{\mathrm{2}} \:+{c}^{\mathrm{2}} \:=\mathrm{2}\:\Rightarrow\:\mathrm{2}{c}^{\mathrm{2}} \:=\mathrm{2}\:\Rightarrow\:{c}\:=\overset{−} {+}\mathrm{1}\:\:{but}\:{f}\left(\mathrm{1}\right)={f}\left(\mathrm{3}\right)\:=\mathrm{1} \\ $$$$\Rightarrow\:{f}\left({x}\right)=\mathrm{1}\:\forall{x}\:\in{R}^{+} \\ $$
Commented by rahul 19 last updated on 22/Apr/18
thank u sir !
$${thank}\:{u}\:{sir}\:! \\ $$
Answered by MJS last updated on 21/Apr/18
x=1; y=1 f(1)f(1)+f(3)f(3)=2f(1) f(1)^2 −2f(1)+1=0 (f(1)−1)^2 =0 f(1)=1 x=3; y=3 f(3)f(3)+f(1)f(1)=2f(9) 1+1=2f(9) f(9)=1 so at least one solution is f(x)=1 we could also start with x=1; y=3 f(1)f(3)+f((3/1))f((3/3))=2f(1×3) f(1)+f(1)=2 ⇒ f(1)=1
$${x}=\mathrm{1};\:{y}=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right){f}\left(\mathrm{1}\right)+{f}\left(\mathrm{3}\right){f}\left(\mathrm{3}\right)=\mathrm{2}{f}\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{f}\left(\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\left({f}\left(\mathrm{1}\right)−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$ \\ $$$${x}=\mathrm{3};\:{y}=\mathrm{3} \\ $$$${f}\left(\mathrm{3}\right){f}\left(\mathrm{3}\right)+{f}\left(\mathrm{1}\right){f}\left(\mathrm{1}\right)=\mathrm{2}{f}\left(\mathrm{9}\right) \\ $$$$\mathrm{1}+\mathrm{1}=\mathrm{2}{f}\left(\mathrm{9}\right) \\ $$$${f}\left(\mathrm{9}\right)=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{is} \\ $$$${f}\left({x}\right)=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{could}\:\mathrm{also}\:\mathrm{start}\:\mathrm{with} \\ $$$${x}=\mathrm{1};\:{y}=\mathrm{3} \\ $$$${f}\left(\mathrm{1}\right){f}\left(\mathrm{3}\right)+{f}\left(\frac{\mathrm{3}}{\mathrm{1}}\right){f}\left(\frac{\mathrm{3}}{\mathrm{3}}\right)=\mathrm{2}{f}\left(\mathrm{1}×\mathrm{3}\right) \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(\mathrm{1}\right)=\mathrm{2}\:\Rightarrow\:{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$
Commented by rahul 19 last updated on 21/Apr/18
sir , actually i also got one of the sol. as f(x)=1. But i am finding hard to prove that for any value of x,y ∈ R^+ . f(x)=1 only ! i mean only one solution.
$${sir}\:,\:{actually}\:{i}\:{also}\:{got}\:{one}\:{of}\:{the}\:{sol}. \\ $$$${as}\:{f}\left({x}\right)=\mathrm{1}.\: \\ $$$${But}\:{i}\:{am}\:{finding}\:{hard}\:{to}\:{prove}\:{that} \\ $$$${for}\:{any}\:{value}\:{of}\:{x},{y}\:\in\:\boldsymbol{{R}}^{+} . \\ $$$${f}\left({x}\right)=\mathrm{1}\:{only}\:!\: \\ $$$${i}\:{mean}\:{only}\:{one}\:{solution}. \\ $$$$ \\ $$
Commented by MJS last updated on 21/Apr/18
if the function is y=1 then the given equation is true for all x∈R because f(x)=f(y)=f((3/x))=f((3/y))=f(xy)=1 and 1×1+1×1=2×1 if x≠y was demanded we′d need a different function, but I could not find one
$$\mathrm{if}\:\mathrm{the}\:\mathrm{function}\:\mathrm{is}\:{y}=\mathrm{1}\:\mathrm{then}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$$$\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{all}\:{x}\in\mathbb{R}\:\mathrm{because} \\ $$$${f}\left({x}\right)={f}\left({y}\right)={f}\left(\frac{\mathrm{3}}{{x}}\right)={f}\left(\frac{\mathrm{3}}{{y}}\right)={f}\left({xy}\right)=\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{1}×\mathrm{1}+\mathrm{1}×\mathrm{1}=\mathrm{2}×\mathrm{1} \\ $$$$ \\ $$$$\mathrm{if}\:{x}\neq{y}\:\mathrm{was}\:\mathrm{demanded}\:\mathrm{we}'\mathrm{d}\:\mathrm{need}\:\mathrm{a}\:\mathrm{different} \\ $$$$\mathrm{function},\:\mathrm{but}\:\mathrm{I}\:\mathrm{could}\:\mathrm{not}\:\mathrm{find}\:\mathrm{one} \\ $$
Commented by rahul 19 last updated on 22/Apr/18
thank u sir.
$${thank}\:{u}\:{sir}. \\ $$

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