Question Number 118513 by I want to learn more last updated on 18/Oct/20
$$\mathrm{If}\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:\:+\:\:\mathrm{f}\left(\frac{\mathrm{x}\:\:−\:\:\mathrm{1}}{\mathrm{x}}\right)\:\:\:=\:\:\:\mathrm{1}\:\:+\:\:\mathrm{x},\:\:\:\:\:\:\:\mathrm{find}\:\:\:\mathrm{f}\left(\mathrm{2}\right). \\ $$
Answered by Dwaipayan Shikari last updated on 18/Oct/20
$${f}\left(\mathrm{2}\right)+{f}\left(\frac{\mathrm{2}−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{1}+\mathrm{2}\rightarrow\left({a}\right)\:\:\:\left({Takng}\:{x}=\mathrm{2}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{f}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\right)=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\rightarrow\left({b}\right)\:\:\left({Taking}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${f}\left(\mathrm{2}\right)−{f}\left(−\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{2}}\rightarrow\left({a}\right)−\left({b}\right) \\ $$$${f}\left(−\mathrm{1}\right)+{f}\left(\frac{−\mathrm{1}−\mathrm{1}}{−\mathrm{1}}\right)=\mathrm{1}−\mathrm{1}\:\:\rightarrow\left({f}\right)\:\left({Taking}\:\:{x}=−\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{2}{f}\left(\mathrm{2}\right)=\frac{\mathrm{3}}{\mathrm{2}}\:\:\rightarrow\left({a}\right)−\left({b}\right)+\left({f}\right) \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by I want to learn more last updated on 18/Oct/20
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{but}\:\mathrm{please}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{of}\:\mathrm{nunbers}. \\ $$$$\mathrm{like}\:\mathrm{the}\:\mathrm{second}\:\mathrm{line}\:\mathrm{and}\:\mathrm{the}\:\mathrm{like}. \\ $$
Commented by I want to learn more last updated on 18/Oct/20
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{understand}\:\mathrm{now}. \\ $$
Answered by benjo_mathlover last updated on 18/Oct/20
![Given (1) f(x) + f (((x−1)/x)) = 1+x replacing x by ((x−1)/x) (2) f(((x−1)/x)) +f (((((x−1)/x)−1)/((x−1)/x))) = 1+((x−1)/x) ⇒ f(((x−1)/x)) + f ((1/(1−x))) = ((2x−1)/x) replacing ((x−1)/x) by x (3) f ((1/(1−x))) + f(x) = 1+ (1/(1−x)) ⇒f ((1/(1−x))) + f(x) = ((2−x)/x) adding (1),(2) and (3) ; gives 2 [ f(x) + f (((x−1)/x))+ f ((1/(1−x))) ]= 1+x +((2x−1)/x) + ((2−x)/x) substitute equation (2) 2 [ f(x) + ((2x−1)/x) ] = ((x^2 +2x+1)/x) f(x) = ((x^2 +2x+1)/(2x)) − ((2x−1)/x) f(x) = ((x^2 −2x+3)/(2x)) ⇒f(2)= ((4−4+3)/4) ⇒f(x) = (3/4)](https://www.tinkutara.com/question/Q118543.png)
$$\:{Given}\:\left(\mathrm{1}\right)\:{f}\left({x}\right)\:+\:{f}\:\left(\frac{{x}−\mathrm{1}}{{x}}\right)\:=\:\mathrm{1}+{x} \\ $$$${replacing}\:{x}\:{by}\:\frac{{x}−\mathrm{1}}{{x}} \\ $$$$\left(\mathrm{2}\right)\:{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)\:+{f}\:\left(\frac{\frac{{x}−\mathrm{1}}{{x}}−\mathrm{1}}{\frac{{x}−\mathrm{1}}{{x}}}\right)\:=\:\mathrm{1}+\frac{{x}−\mathrm{1}}{{x}} \\ $$$$\Rightarrow\:{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)\:+\:{f}\:\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:=\:\frac{\mathrm{2}{x}−\mathrm{1}}{{x}} \\ $$$${replacing}\:\frac{{x}−\mathrm{1}}{{x}}\:{by}\:{x} \\ $$$$\left(\mathrm{3}\right)\:{f}\:\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:+\:{f}\left({x}\right)\:=\:\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow{f}\:\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:+\:{f}\left({x}\right)\:=\:\frac{\mathrm{2}−{x}}{{x}}\: \\ $$$${adding}\:\left(\mathrm{1}\right),\left(\mathrm{2}\right)\:{and}\:\left(\mathrm{3}\right)\:;\:{gives}\: \\ $$$$\mathrm{2}\:\left[\:{f}\left({x}\right)\:+\:{f}\:\left(\frac{{x}−\mathrm{1}}{{x}}\right)+\:{f}\:\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:\right]=\:\mathrm{1}+{x}\:+\frac{\mathrm{2}{x}−\mathrm{1}}{{x}}\:+\:\frac{\mathrm{2}−{x}}{{x}} \\ $$$${substitute}\:{equation}\:\left(\mathrm{2}\right) \\ $$$$\:\mathrm{2}\:\left[\:{f}\left({x}\right)\:+\:\frac{\mathrm{2}{x}−\mathrm{1}}{{x}}\:\right]\:=\:\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{{x}} \\ $$$$\:{f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}}\:−\:\frac{\mathrm{2}{x}−\mathrm{1}}{{x}} \\ $$$$\:{f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}}{\mathrm{2}{x}}\:\Rightarrow{f}\left(\mathrm{2}\right)=\:\frac{\mathrm{4}−\mathrm{4}+\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{f}\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by I want to learn more last updated on 18/Oct/20
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$