Question Number 2300 by Syaka last updated on 14/Nov/15
$${if}\:{f}\left(\mathrm{4}{xy}\right)\:=\:\mathrm{2}{y}\left({f}\left({x}\:−\:{y}\right)\:+\:{f}\left({x}\:+\:{y}\right)\right)\:{and}\:{f}\left(\mathrm{5}\right)\:=\:\mathrm{3} \\ $$$${what}\:{is}\:{the}\:{value}\:{of}\:{f}\left(\mathrm{2015}\right)\:=\:? \\ $$
Commented by Rasheed Soomro last updated on 17/Nov/15
$$\mathcal{EXCELLENT}\:! \\ $$
Commented by prakash jain last updated on 14/Nov/15
$${y}=\mathrm{0}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$
Commented by prakash jain last updated on 14/Nov/15
$${y}={x},{x}={y} \\ $$$${f}\left(\mathrm{4}{xy}\right)=\mathrm{2}{x}\left({f}\left({y}−{x}\right)+{f}\left({x}+{y}\right)\right) \\ $$$${yf}\left({x}−{y}\right)+{yf}\left({x}+{y}\right)={xf}\left({y}−{x}\right)+{xf}\left({x}+{y}\right)\:\:\:...\left(\mathrm{1}\right) \\ $$$${y}=−{y} \\ $$$${f}\left(−\mathrm{4}{xy}\right)=−\mathrm{2}{y}\left({f}\left({x}+{y}\right)+{f}\left({x}−{y}\right)\right) \\ $$$${f}\left(\mathrm{4}{xy}\right)=−{f}\left(−\mathrm{4}{xy}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{odd}\:\mathrm{function} \\ $$$${from}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right) \\ $$$${f}\left({x}+{y}\right)\left({x}−{y}\right)=\left({y}+{x}\right){f}\left({x}−{y}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\left(\mathrm{3}\right) \\ $$$${y}={x} \\ $$$${f}\left(\mathrm{4}{x}^{\mathrm{2}} \right)=\mathrm{2}{xf}\left(\mathrm{2}{x}\right)\Rightarrow{f}\left({x}^{\mathrm{2}} \right)={xf}\left({x}\right) \\ $$
Answered by prakash jain last updated on 14/Nov/15
$$\left({x}+{y}\right){f}\left({x}−{y}\right)=\left({x}−{y}\right){f}\left({x}+{y}\right) \\ $$$${x}=\mathrm{1010} \\ $$$${y}=\mathrm{1005} \\ $$$$\mathrm{2015}×{f}\left(\mathrm{5}\right)=\mathrm{5}×{f}\left(\mathrm{2015}\right) \\ $$$${f}\left(\mathrm{2015}\right)=\mathrm{403}×\mathrm{3}=\mathrm{1209} \\ $$