p-x-xp-x-x-2-1-faind-p-2-

Question Number 164135 by mathlove last updated on 14/Jan/22

$${p}\left({x}\right)+{xp}\left(−{x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${faind}\:\:\:{p}\left(\mathrm{2}\right)=? \\ $$

Answered by mr W last updated on 14/Jan/22

p(x)+xp(−x)=x^2 +1 ...(i) p(−x)−xp(x)=x^2 +1 xp(−x)−x^2 p(x)=(x^2 +1)x ...(ii) (i)−(ii): (1+x^2 )p(x)=(x^2 +1)(1−x) ⇒p(x)=1−x ⇒p(2)=1−2=−1

$${p}\left({x}\right)+{xp}\left(−{x}\right)={x}^{\mathrm{2}} +\mathrm{1}\:\:\:…\left({i}\right) \\ $$$${p}\left(−{x}\right)−{xp}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${xp}\left(−{x}\right)−{x}^{\mathrm{2}} {p}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right){x}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right){p}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}−{x}\right) \\ $$$$\Rightarrow{p}\left({x}\right)=\mathrm{1}−{x} \\ $$$$\Rightarrow{p}\left(\mathrm{2}\right)=\mathrm{1}−\mathrm{2}=−\mathrm{1} \\ $$

Commented by mathlove last updated on 14/Jan/22