Question Number 211365 by JuniorKepler last updated on 06/Sep/24
Answered by mr W last updated on 07/Sep/24
$$\frac{\left({x}+{y}\right)^{{p}} \left({x}+{y}\right)^{{q}} }{{x}^{{p}} {y}^{{q}} }=\mathrm{1} \\ $$$$\left(\mathrm{1}+\frac{{y}}{{x}}\right)^{{p}} \left(\mathrm{1}+\frac{{x}}{{y}}\right)^{{q}} =\mathrm{1} \\ $$$${say}\:{y}={kx} \\ $$$$\left(\mathrm{1}+{k}\right)^{{p}} \left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)^{{q}} =\mathrm{1}\:\Rightarrow\left(\mathrm{1}+{k}\right)^{{p}+{q}} −{k}^{{q}} \:=\mathrm{0} \\ $$$$\Rightarrow{k}={constant}\:{w}.{r}.{t}.\:{x} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={k}\:\Rightarrow\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{answer}\:\left({a}\right) \\ $$
Answered by MATHEMATICSAM last updated on 07/Sep/24
$$\mathrm{0} \\ $$