Question Number 206746 by sonukgindia last updated on 23/Apr/24
Answered by A5T last updated on 23/Apr/24
$${f}\left(\mathrm{2}\right)+\mathrm{2}{f}\left(−\mathrm{1}\right)=\mathrm{2}…\left({i}\right) \\ $$$${f}\left(−\mathrm{1}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{1}…\left({ii}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2}{f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}…\left({iii}\right) \\ $$$$\mathrm{2}×\left({iii}\right)−\left({ii}\right):\:\mathrm{4}{f}\left(\mathrm{2}\right)−{f}\left(−\mathrm{1}\right)=\mathrm{2}…\left({iv}\right) \\ $$$$\mathrm{2}×\left({iv}\right)+\left({i}\right):\:\mathrm{9}{f}\left(\mathrm{2}\right)=\mathrm{6}\Rightarrow{f}\left(\mathrm{2}\right)=\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Answered by mr W last updated on 24/Apr/24
![f(x)+2f((1/(1−x)))=x ...(i) f((1/(1−x)))+2f(((x−1)/x))=(1/(1−x)) ...(ii) f(((x−1)/x))+2f(x)=((x−1)/x) ...(iii) (iii) into (ii): f((1/(1−x)))+2[((x−1)/x)−2f(x)]=(1/(1−x)) f((1/(1−x)))=4f(x)+(1/(1−x))−((2(x−1))/x) this into (i): f(x)+2[4f(x)+(1/(1−x))−((2(x−1))/x)]=x ⇒f(x)=(1/9)[x−(2/(1−x))+((4(x−1))/x)] f(2)=(1/9)[2−(2/(1−2))+((4(2−1))/2)]=(2/3) ✓](https://www.tinkutara.com/question/Q206766.png)
$${f}\left({x}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)={x}\:\:\:…\left({i}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+\mathrm{2}{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:\:…\left({ii}\right) \\ $$$${f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)+\mathrm{2}{f}\left({x}\right)=\frac{{x}−\mathrm{1}}{{x}}\:\:\:…\left({iii}\right) \\ $$$$\left({iii}\right)\:{into}\:\left({ii}\right): \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+\mathrm{2}\left[\frac{{x}−\mathrm{1}}{{x}}−\mathrm{2}{f}\left({x}\right)\right]=\frac{\mathrm{1}}{\mathrm{1}−{x}}\: \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)=\mathrm{4}{f}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{2}\left({x}−\mathrm{1}\right)}{{x}} \\ $$$${this}\:{into}\:\left({i}\right): \\ $$$${f}\left({x}\right)+\mathrm{2}\left[\mathrm{4}{f}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{2}\left({x}−\mathrm{1}\right)}{{x}}\right]={x} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{9}}\left[{x}−\frac{\mathrm{2}}{\mathrm{1}−{x}}+\frac{\mathrm{4}\left({x}−\mathrm{1}\right)}{{x}}\right] \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{9}}\left[\mathrm{2}−\frac{\mathrm{2}}{\mathrm{1}−\mathrm{2}}+\frac{\mathrm{4}\left(\mathrm{2}−\mathrm{1}\right)}{\mathrm{2}}\right]=\frac{\mathrm{2}}{\mathrm{3}}\:\checkmark \\ $$