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Question Number 86160 by jagoll last updated on 27/Mar/20
If (x−1) f(x) + f((1/x)) = (1/(x−1)) find f(x)
$$\mathrm{If}\:\left(\mathrm{x}−\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{x}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}} \\ $$$$\mathrm{find}\:\mathrm{f}\left(\mathrm{x}\right)\: \\ $$
Commented by john santu last updated on 27/Mar/20
replace x by (1/x) ⇒((1/x)−1)f((1/x))+f(x)= (1/((1/x)−1)) (((1−x)/x))f((1/x)) +f(x) = (x/(1−x)) [ × x ] (1−x) f((1/x)) + xf(x) = (x^2 /(1−x)) →(ii) multiply eq (i) by (1−x) (1−x)f((1/x)) −(1−x)^2 f(x)= −1←(i) (ii)−(i) (x+(1−x)^2 ) f(x) =(x^2 /(1−x))+1 (x^2 −x+1)f(x) = ((x^2 −x+1)/(1−x)) f(x) = (1/(1−x))
$${replace}\:{x}\:{by}\:\frac{\mathrm{1}}{{x}} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)+{f}\left({x}\right)=\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{{x}}−\mathrm{1}} \\ $$$$\left(\frac{\mathrm{1}−{x}}{{x}}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)\:+{f}\left({x}\right)\:=\:\frac{{x}}{\mathrm{1}−{x}}\:\left[\:×\:{x}\:\right] \\ $$$$\left(\mathrm{1}−{x}\right)\:{f}\left(\frac{\mathrm{1}}{{x}}\right)\:+\:{xf}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}}\:\rightarrow\left({ii}\right) \\ $$$${multiply}\:{eq}\:\left({i}\right)\:{by}\:\left(\mathrm{1}−{x}\right) \\ $$$$\left(\mathrm{1}−{x}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)\:−\left(\mathrm{1}−{x}\right)^{\mathrm{2}} {f}\left({x}\right)=\:−\mathrm{1}\leftarrow\left({i}\right) \\ $$$$\left({ii}\right)−\left({i}\right) \\ $$$$\left({x}+\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \right)\:{f}\left({x}\right)\:=\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}}+\mathrm{1} \\ $$$$\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right){f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$ \\ $$
Commented by jagoll last updated on 27/Mar/20
thank you mr john & tanmay
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mr}\:\mathrm{john}\:\&\:\mathrm{tanmay} \\ $$
Commented by mathmax by abdo last updated on 27/Mar/20
(x−1)f(x)+f((1/x))=(1/(x−1)) ⇒((1/x)−1)f((1/x))+f(x) =(1/((1/x)−1)) =(x/(1−x)) we get the system { (((x−1)f(x)+f((1/x))=(1/(x−1)))),((f(x)+((1−x)/x)f((1/x)) =(x/(1−x)))) :} Δ_s = determinant (((x−1 1)),((1 ((1−x)/x))))=(((x−1)(1−x))/x)−1 =((x−x^2 −1+x−x)/x) =((−x^2 +x−1)/x) ⇒f(x) =((Δf(x))/Δ) =( determinant ((((1/(x−1)) 1)),(((x/(1−x)) ((1−x)/x))))/((−x^2 +x−1)/x)) =((−(1/x)−(x/(1−x)))/((−x^2 +x−1)/x)) =((−1+x−x^2 )/(x(1−x)))×(x/(−x^2 +x−1)) =(1/(1−x)) ⇒f(x) =(1/(1−x))
$$\left({x}−\mathrm{1}\right){f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}−\mathrm{1}}\:\Rightarrow\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)+{f}\left({x}\right)\:=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{x}}−\mathrm{1}}\:=\frac{{x}}{\mathrm{1}−{x}} \\ $$$${we}\:{get}\:{the}\:{system}\:\begin{cases}{\left({x}−\mathrm{1}\right){f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}−\mathrm{1}}}\\{{f}\left({x}\right)+\frac{\mathrm{1}−{x}}{{x}}{f}\left(\frac{\mathrm{1}}{{x}}\right)\:=\frac{{x}}{\mathrm{1}−{x}}}\end{cases} \\ $$$$\Delta_{{s}} =\begin{vmatrix}{{x}−\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}−{x}}{{x}}}\end{vmatrix}=\frac{\left({x}−\mathrm{1}\right)\left(\mathrm{1}−{x}\right)}{{x}}−\mathrm{1}\:=\frac{{x}−{x}^{\mathrm{2}} −\mathrm{1}+{x}−{x}}{{x}} \\ $$$$=\frac{−{x}^{\mathrm{2}} +{x}−\mathrm{1}}{{x}}\:\Rightarrow{f}\left({x}\right)\:=\frac{\Delta{f}\left({x}\right)}{\Delta} \\ $$$$=\frac{\begin{vmatrix}{\frac{\mathrm{1}}{{x}−\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\frac{{x}}{\mathrm{1}−{x}}\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}−{x}}{{x}}}\end{vmatrix}}{\frac{−{x}^{\mathrm{2}} +{x}−\mathrm{1}}{{x}}}\:=\frac{−\frac{\mathrm{1}}{{x}}−\frac{{x}}{\mathrm{1}−{x}}}{\frac{−{x}^{\mathrm{2}} +{x}−\mathrm{1}}{{x}}}\:=\frac{−\mathrm{1}+{x}−{x}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{x}\right)}×\frac{{x}}{−{x}^{\mathrm{2}} +{x}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow{f}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$ \\ $$$$ \\ $$
Answered by TANMAY PANACEA. last updated on 27/Mar/20
(x−1)f(x)+f((1/x))=(1/(x−1)) replacing x by (1/x) ((1/x)−1)f((1/x))+f(x)=(1/((1/x)−1)) f(x)=(x/(1−x))−((1−x)/x)f((1/x)) putting the value of f(x) in given eqn [(x−1)f(x)+f((1/x))=(1/(1−x))] so we get (x−1)[(x/(1−x))−((1−x)/x)f((1/x))]+f((1/x))=(1/(x−1)) −x+(((x−1)^2 )/x)f((1/x))+f((1/x))=(1/(x−1)) f((1/x))[1+((x^2 −2x+1)/x)]=(1/(x−1))+x f((1/x))[((x+x^2 −2x+1)/x)]=((1+x^2 −x)/(x−1)) f((1/x))=(x/(x−1))=(1/(1−(1/x))) f(x)=(1/(1−x))
$$\left({x}−\mathrm{1}\right){f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$$${replacing}\:{x}\:{by}\:\frac{\mathrm{1}}{{x}} \\ $$$$\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)+{f}\left({x}\right)=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{x}}−\mathrm{1}} \\ $$$${f}\left({x}\right)=\frac{{x}}{\mathrm{1}−{x}}−\frac{\mathrm{1}−{x}}{{x}}{f}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$${putting}\:{the}\:{value}\:{of}\:{f}\left({x}\right)\:\:{in}\:{given}\:{eqn} \\ $$$$\left[\left({x}−\mathrm{1}\right){f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}\right] \\ $$$${so}\:{we}\:{get} \\ $$$$\left({x}−\mathrm{1}\right)\left[\frac{{x}}{\mathrm{1}−{x}}−\frac{\mathrm{1}−{x}}{{x}}{f}\left(\frac{\mathrm{1}}{{x}}\right)\right]+{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$$$−{x}+\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}}{f}\left(\frac{\mathrm{1}}{{x}}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}−\mathrm{1}} \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)\left[\mathrm{1}+\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{{x}}\right]=\frac{\mathrm{1}}{{x}−\mathrm{1}}+{x} \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)\left[\frac{{x}+{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{{x}}\right]=\frac{\mathrm{1}+{x}^{\mathrm{2}} −{x}}{{x}−\mathrm{1}} \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{{x}}{{x}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{x}}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$
Commented by jagoll last updated on 27/Mar/20
thank you mister
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$
Commented by Serlea last updated on 27/Mar/20
I think Something is wrong Sir Line number 4: How did u get ur Second f((1/x)) 2) Why didn′t you multiply the both side by (x−1) only the left hand side of the equation.
$$ \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{Something}\:\mathrm{is}\:\mathrm{wrong}\:\mathrm{Sir} \\ $$$$\mathrm{Line}\:\mathrm{number}\:\mathrm{4}:\:\mathrm{How}\:\mathrm{did}\:\mathrm{u}\:\mathrm{get}\:\mathrm{ur}\:\mathrm{Second}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Why}\:\mathrm{didn}'\mathrm{t}\:\mathrm{you}\:\mathrm{multiply}\:\mathrm{the}\:\mathrm{both}\:\mathrm{side}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{1}\right)\:\mathrm{only}\:\mathrm{the} \\ $$$$\mathrm{left}\:\mathrm{hand}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$ \\ $$
Commented by john santu last updated on 27/Mar/20
yes sir.
$${yes}\:{sir}.\: \\ $$

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