Question Number 160246 by mathlove last updated on 26/Nov/21
Answered by Rasheed.Sindhi last updated on 26/Nov/21
![af(x)+bf((1/x))=(1/x).............(i) Replacing x by (1/x) af((1/x))+bf(x)=x................(ii) (i)×a : a^2 f(x)+abf((1/x))=(a/x).....(iii) (ii)×b: b^2 f(x)+abf((1/x))=bx......(iv) (iii)−(iv): (a^2 −b^2 )f(x)=(a/x)−bx f(x)=((a−bx^2 )/((a^2 −b^2 )x)) [∵a≠b]](https://www.tinkutara.com/question/Q160247.png)
$$\mathrm{af}\left(\mathrm{x}\right)+\mathrm{bf}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{x}}………….\left(\mathrm{i}\right) \\ $$$$\mathrm{Replacing}\:\mathrm{x}\:\mathrm{by}\:\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{af}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{bf}\left(\mathrm{x}\right)=\mathrm{x}…………….\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)×\mathrm{a}\::\:\mathrm{a}^{\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)+\mathrm{abf}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\frac{\mathrm{a}}{\mathrm{x}}…..\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{ii}\right)×\mathrm{b}:\:\mathrm{b}^{\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)+\mathrm{abf}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\mathrm{bx}……\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{iii}\right)−\left(\mathrm{iv}\right):\:\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}−\mathrm{bx} \\ $$$$\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}−\mathrm{bx}^{\mathrm{2}} }{\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)\mathrm{x}}\:\:\:\:\:\:\:\:\:\:\:\left[\because\mathrm{a}\neq\mathrm{b}\right] \\ $$$$ \\ $$
Commented by mathlove last updated on 26/Nov/21
$${thanks}\:{thear} \\ $$