Question Number 44350 by Necxx last updated on 27/Sep/18
$${If}\:{f}\left({x}+{y}\right)={f}\left({x}\right).{f}\left({y}\right)\:{for}\:{real}\:{x},{y} \\ $$$${and}\:{f}\left(\mathrm{0}\right)\neq\mathrm{0}.{Let}\:{F}\left({x}\right)=\frac{{f}\left({x}\right)}{\mathrm{1}+\left({f}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$${then}\:{F}\left({x}\right)\:{is} \\ $$$$\left.{a}\left.\right)\left.{even}\:{b}\right){odd}\:{c}\right){neither}\:{even}\:{nor}\:{odd} \\ $$
Commented by maxmathsup by imad last updated on 27/Sep/18
$$\Rightarrow{f}\left({x}\right).{f}\left(−{x}\right)={f}\left(\mathrm{0}\right)\:\Rightarrow{f}\left(\mathrm{0}\right)^{\mathrm{2}} ={f}\left(\mathrm{0}\right)\:\Rightarrow{f}\left(\mathrm{0}\right)\:=\mathrm{1}\:\:\Rightarrow{F}\left(\mathrm{0}\right)=\frac{{f}\left(\mathrm{0}\right)}{\mathrm{1}+\left({f}\left(\mathrm{0}\right)\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${due}\:{to}\:{F}\left(\mathrm{0}\right)\neq\mathrm{0}\:{F}\:{can}\:{t}\:{be}\:{odd}\:{let}\:{see}\:{if}\:{F}\:{is}\:{even} \\ $$$${F}\left(−{x}\right)\:=\frac{{f}\left(−{x}\right)}{\mathrm{1}+\left({f}\left(−{x}\right)\right)^{\mathrm{2}} }\:=\frac{\frac{\mathrm{1}}{{f}\left({x}\right)}}{\mathrm{1}+\left(\frac{\mathrm{1}}{{f}\left({x}\right)}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{f}\left({x}\right)\left\{\frac{{f}^{\mathrm{2}} \left({x}\right)+\mathrm{1}}{{f}^{\mathrm{2}} \left({x}\right)}\right\}}\:=\frac{{f}\left({x}\right)}{\mathrm{1}+{f}^{\mathrm{2}} \left({x}\right)}={F}\left({x}\right) \\ $$$${so}\:\:{F}\:{is}\:{even}. \\ $$
Commented by Necxx last updated on 27/Sep/18
$${thank}\:{you}\:{sir} \\ $$
Commented by Necxx last updated on 27/Sep/18
$${that}'{s}\:{it} \\ $$