Question-109569

Question Number 109569 by bemath last updated on 24/Aug/20

Answered by john santu last updated on 24/Aug/20

((⇋JS⇌)/♠) let ((x−1)/x)=t →x=(1/(1−t)) (1)f((1/(1−t)))+f(t) = (1/(1−t))+1=((2−t)/(1−t)) set x = ((t−1)/t) (2) f(((t−1)/t))+f(((((t−1)/t)−1)/((t−1)/t)))=((t−1)/t)+1 ⇒f(((t−1)/t))+f((1/(1−t)))=((2t−1)/t) (3)f(t)+f(((t−1)/t))=t+1 (2)−(3)⇒f((1/(1−t)))−f(t)=((2t−1)/t)−t−1 f((1/(1−t)))−f(t)=((t−1−t^2 )/t) ...(4) (1)−(4)⇒ f((1/(1−t)))+f(t)=((2−t)/(1−t)) ⇒f((1/(1−t)))−f(t)=((t−1−t^2 )/t) ____________________ − 2f(t) = ((2−t)/(1−t))−((t−1−t^2 )/t) 2f(t)= ((2t−t^2 −(1−t)(t−1−t^2 ))/(t−t^2 )) 2f(t)=((2t−t^2 −(t−1−t^2 −t^2 +t+t^3 ))/(t−t^2 )) 2f(t)=((2t−t^2 −(2t−2t^2 −1+t^3 ))/(t−t^2 )) 2f(t)=((t^2 +1−t^3 )/(t−t^2 ))⇒f(x)=((−x^3 +x^2 +1)/(2x−2x^2 )) ∴ f(x)=((x^3 −x^2 −1)/(2x(x−1))). [answer : B ]

$$\:\:\:\frac{\leftrightharpoons{JS}\rightleftharpoons}{\spadesuit} \\ $$$${let}\:\:\frac{{x}−\mathrm{1}}{{x}}={t}\:\rightarrow{x}=\frac{\mathrm{1}}{\mathrm{1}−{t}} \\ $$$$\left(\mathrm{1}\right){f}\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\right)+{f}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{t}}+\mathrm{1}=\frac{\mathrm{2}−{t}}{\mathrm{1}−{t}} \\ $$$${set}\:{x}\:=\:\frac{{t}−\mathrm{1}}{{t}} \\ $$$$\left(\mathrm{2}\right)\:{f}\left(\frac{{t}−\mathrm{1}}{{t}}\right)+{f}\left(\frac{\frac{{t}−\mathrm{1}}{{t}}−\mathrm{1}}{\frac{{t}−\mathrm{1}}{{t}}}\right)=\frac{{t}−\mathrm{1}}{{t}}+\mathrm{1} \\ $$$$\Rightarrow{f}\left(\frac{{t}−\mathrm{1}}{{t}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\right)=\frac{\mathrm{2}{t}−\mathrm{1}}{{t}} \\ $$$$\left(\mathrm{3}\right){f}\left({t}\right)+{f}\left(\frac{{t}−\mathrm{1}}{{t}}\right)={t}+\mathrm{1} \\ $$$$\left(\mathrm{2}\right)−\left(\mathrm{3}\right)\Rightarrow{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\right)−{f}\left({t}\right)=\frac{\mathrm{2}{t}−\mathrm{1}}{{t}}−{t}−\mathrm{1} \\ $$$$\:\:\:{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\right)−{f}\left({t}\right)=\frac{{t}−\mathrm{1}−{t}^{\mathrm{2}} }{{t}}\:\:…\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{4}\right)\Rightarrow\:{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\right)+{f}\left({t}\right)=\frac{\mathrm{2}−{t}}{\mathrm{1}−{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\right)−{f}\left({t}\right)=\frac{{t}−\mathrm{1}−{t}^{\mathrm{2}} }{{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\:− \\ $$$$\mathrm{2}{f}\left({t}\right)\:=\:\frac{\mathrm{2}−{t}}{\mathrm{1}−{t}}−\frac{{t}−\mathrm{1}−{t}^{\mathrm{2}} }{{t}} \\ $$$$\mathrm{2}{f}\left({t}\right)=\:\frac{\mathrm{2}{t}−{t}^{\mathrm{2}} −\left(\mathrm{1}−{t}\right)\left({t}−\mathrm{1}−{t}^{\mathrm{2}} \right)}{{t}−{t}^{\mathrm{2}} } \\ $$$$\mathrm{2}{f}\left({t}\right)=\frac{\mathrm{2}{t}−{t}^{\mathrm{2}} −\left({t}−\mathrm{1}−{t}^{\mathrm{2}} −{t}^{\mathrm{2}} +{t}+{t}^{\mathrm{3}} \right)}{{t}−{t}^{\mathrm{2}} } \\ $$$$\mathrm{2}{f}\left({t}\right)=\frac{\mathrm{2}{t}−{t}^{\mathrm{2}} −\left(\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}+{t}^{\mathrm{3}} \right)}{{t}−{t}^{\mathrm{2}} } \\ $$$$\mathrm{2}{f}\left({t}\right)=\frac{{t}^{\mathrm{2}} +\mathrm{1}−{t}^{\mathrm{3}} }{{t}−{t}^{\mathrm{2}} }\Rightarrow{f}\left({x}\right)=\frac{−{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\therefore\:{f}\left({x}\right)=\frac{{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}\left({x}−\mathrm{1}\right)}.\:\:\left[{answer}\::\:{B}\:\right] \\ $$

Commented by mnjuly1970 last updated on 24/Aug/20