Question-195511

Question Number 195511 by Calculusboy last updated on 04/Aug/23

Answered by mr W last updated on 04/Aug/23

there are infinite values for k+1. curve x^y +y^x =8 is symmetric about y=x. say it passes the point (a,a), then 2a^a =8 ⇒a=2. straight line x+y=k^2 intersects the curve x^y +y^x =8 if k^2 ≥a+a=4, i.e. if k≤−2 or k≥2. so we have k+1≤−1 or k+1≥3

$${there}\:{are}\:{infinite}\:{values}\:{for}\:{k}+\mathrm{1}. \\ $$$$ \\ $$$${curve}\:{x}^{{y}} +{y}^{{x}} =\mathrm{8}\:{is}\:{symmetric}\:{about}\: \\ $$$${y}={x}.\:{say}\:{it}\:{passes}\:{the}\:{point}\:\left({a},{a}\right), \\ $$$${then}\:\mathrm{2}{a}^{{a}} =\mathrm{8}\:\Rightarrow{a}=\mathrm{2}. \\ $$$${straight}\:{line}\:{x}+{y}={k}^{\mathrm{2}} \:{intersects}\:{the} \\ $$$${curve}\:{x}^{{y}} +{y}^{{x}} =\mathrm{8}\:{if}\:{k}^{\mathrm{2}} \geqslant{a}+{a}=\mathrm{4},\:{i}.{e}. \\ $$$${if}\:{k}\leqslant−\mathrm{2}\:{or}\:{k}\geqslant\mathrm{2}.\:{so}\:{we}\:{have} \\ $$$${k}+\mathrm{1}\leqslant−\mathrm{1}\:{or}\:{k}+\mathrm{1}\geqslant\mathrm{3} \\ $$

Commented by Calculusboy last updated on 04/Aug/23