Question Number 217660 by ArshadS last updated on 17/Mar/25
$$\:{f}\left({x}\right)\:+\:{f}\left({y}\right)={f}\left({x}+{y}\right)+{xy}\: \\ $$$${f}\left({x}\right)=? \\ $$
Answered by vnm last updated on 19/Mar/25
$${f}\left({x}\right)+{f}\left(\mathrm{0}\right)={f}\left({x}+\mathrm{0}\right)+{x}\centerdot\mathrm{0}\:\Rightarrow\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)+{f}\left(−{x}\right)={f}\left({x}−{x}\right)+{x}\left(−{x}\right)=−{x}^{\mathrm{2}} \\ $$$${f}\left({x}\right)=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{g}\left({x}\right),\:{g}\left(−{x}\right)=−{g}\left({x}\right) \\ $$$${g}\left({x}\right)\:{is}\:{an}\:{odd}\:{function} \\ $$$${f}\left({x}\right)−{f}\left({x}+{a}\right)={ax}−{f}\left({a}\right) \\ $$$$\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{g}\left({x}\right)\right)−\left(−\frac{\left({x}+{a}\right)^{\mathrm{2}} }{\mathrm{2}}+{g}\left({x}+{a}\right)\right)={ax}−{f}\left({a}\right) \\ $$$${g}\left({x}\right)−{g}\left({x}+{a}\right)=−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}−{f}\left({a}\right)\:\forall{a},{x} \\ $$$${g}\left({x}\right)={kx} \\ $$$${g}\left({x}\right)\:{is}\:{a}\:{direct}\:{proportionality} \\ $$$${f}\left({x}\right)=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{kx} \\ $$
Commented by ArshadS last updated on 19/Mar/25
$${nice}\:{sir}! \\ $$