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Question Number 120325 by Ar Brandon last updated on 30/Oct/20
Let f:R→R be a function satisfying the functional relation (f(x))^y +(f(y))^x =2f(xy) for all x, y ∈R and it is given that f(1)=1/2. Answer the following questions. (i) f(x+y)= (A) f(x)+f(y) (B) f(x)f(y) (C) f(x^y y^x ) (D) ((f(x))/(f(y))) (ii) f(xy)= (A) f(x)f(y) (B) f(x)+f(y) (C) (f(x))^y (D) (f(xy))^(xy) (iii) Σ_(k=0) ^∞ f(k)= (A) 5/2 (B) 3/2 (C) 3 (D) 2
$$\mathrm{Let}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the} \\ $$$$\mathrm{functional}\:\mathrm{relation} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({f}\left(\mathrm{x}\right)\right)^{\mathrm{y}} +\left({f}\left(\mathrm{y}\right)\right)^{\mathrm{x}} =\mathrm{2}{f}\left(\mathrm{xy}\right) \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{x},\:\mathrm{y}\:\in\mathbb{R}\:\mathrm{and}\:\mathrm{it}\:\mathrm{is}\:\mathrm{given}\:\mathrm{that}\:{f}\left(\mathrm{1}\right)=\mathrm{1}/\mathrm{2}.\:\mathrm{Answer} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{questions}. \\ $$$$\left(\boldsymbol{\mathrm{i}}\right)\:\:\:\:{f}\left(\mathrm{x}+\mathrm{y}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{A}\right)\:{f}\left(\mathrm{x}\right)+{f}\left(\mathrm{y}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\:{f}\left(\mathrm{x}\right){f}\left(\mathrm{y}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:{f}\left(\mathrm{x}^{\mathrm{y}} \mathrm{y}^{\mathrm{x}} \right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\:\:\frac{{f}\left(\mathrm{x}\right)}{{f}\left(\mathrm{y}\right)} \\ $$$$\left(\boldsymbol{\mathrm{ii}}\right)\:\:\:\:{f}\left(\mathrm{xy}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{A}\right)\:{f}\left(\mathrm{x}\right){f}\left(\mathrm{y}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:{f}\left(\mathrm{x}\right)+{f}\left(\mathrm{y}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\left({f}\left(\mathrm{x}\right)\right)^{\mathrm{y}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\:\left({f}\left(\mathrm{xy}\right)\right)^{\mathrm{xy}} \\ $$$$\left(\boldsymbol{\mathrm{iii}}\right)\:\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}{f}\left({k}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{A}\right)\:\mathrm{5}/\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{3}/\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{2} \\ $$
Commented by Dwaipayan Shikari last updated on 30/Oct/20
f(x)=(Ce)^x f(x+y)=(Ce)^(x+y) =C^x e^x .C^y e^y =f(x)f(y) f(xy)=C^(xy) e^(xy) =(f(x))^y f(1)=Ce=(1/2)⇒C=(1/(2e)) Σ_(k=0) ^∞ f(k)=1+(1/2^1 )+(1/2^2 )+...=(1/(1−(1/2)))=2
$${f}\left({x}\right)=\left({Ce}\right)^{{x}} \\ $$$${f}\left({x}+{y}\right)=\left({Ce}\right)^{{x}+{y}} ={C}^{{x}} {e}^{{x}} .{C}^{{y}} {e}^{{y}} ={f}\left({x}\right){f}\left({y}\right) \\ $$$${f}\left({xy}\right)={C}^{{xy}} {e}^{{xy}} =\left({f}\left({x}\right)\right)^{{y}} \\ $$$${f}\left(\mathrm{1}\right)={Ce}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{C}=\frac{\mathrm{1}}{\mathrm{2}{e}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{f}\left({k}\right)=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+…=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{2} \\ $$
Commented by Ar Brandon last updated on 30/Oct/20
Wow ! Thanks bro. Are there any calculations to get f(x)=(Ce)^x ?
$$\mathrm{Wow}\:!\:\mathrm{Thanks}\:\mathrm{bro}.\:\mathrm{Are}\:\mathrm{there}\:\mathrm{any}\: \\ $$$$\:\mathrm{calculations}\:\mathrm{to}\:\mathrm{get}\:{f}\left(\mathrm{x}\right)=\left(\mathrm{Ce}\right)^{\mathrm{x}} \:? \\ $$
Commented by Dwaipayan Shikari last updated on 30/Oct/20
I just assumed it
$${I}\:{just}\:{assumed}\:{it} \\ $$
Commented by mindispower last updated on 30/Oct/20
f(x)^y +f(y)^x =2f(xy),x=1 ⇒f(1)^y +f(y)=2f(y) f(y)=f(1)^y =(1/2^y )
$${f}\left({x}\right)^{{y}} +{f}\left({y}\right)^{{x}} =\mathrm{2}{f}\left({xy}\right),{x}=\mathrm{1} \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)^{{y}} +{f}\left({y}\right)=\mathrm{2}{f}\left({y}\right) \\ $$$${f}\left({y}\right)={f}\left(\mathrm{1}\right)^{{y}} =\frac{\mathrm{1}}{\mathrm{2}^{{y}} } \\ $$
Commented by Dwaipayan Shikari last updated on 30/Oct/20
f(1)^y +f(y)^1 =2(f(y)) ((1/2))^y =f(y) So((1/2))^(xy) =f(xy)=(((1/2))^x )^y =f(x)^y ((1/2))^x =f(x) So Σ_(k=0) ^∞ f(k)=((1/2))^0 +((1/2))^1 +..=2 ((1/2))^(x+y) =f(x+y) f(x)f(y)=f(x+y)
$${f}\left(\mathrm{1}\right)^{{y}} +{f}\left({y}\right)^{\mathrm{1}} =\mathrm{2}\left({f}\left({y}\right)\right) \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{y}} ={f}\left({y}\right)\:\:{So}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{xy}} ={f}\left({xy}\right)=\left(\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} \right)^{{y}} ={f}\left({x}\right)^{{y}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} ={f}\left({x}\right)\:\:{So}\:\:\:\:\:\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{f}\left({k}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{0}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}} +..=\mathrm{2} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}+{y}} ={f}\left({x}+{y}\right) \\ $$$${f}\left({x}\right){f}\left({y}\right)={f}\left({x}+{y}\right) \\ $$
Commented by Ar Brandon last updated on 30/Oct/20
Thanks��

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