Question Number 46478 by rahul 19 last updated on 27/Oct/18
$${If}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{1}.\:{Find}\:{range}\:{of}\: \\ $$$${E}\:=\:{x}^{\mathrm{3}} {y}+{xy}^{\mathrm{3}} +\mathrm{4}\:. \\ $$
Commented by rahul 19 last updated on 27/Oct/18
$${Sir},\:{ans}.\:{is}\: \\ $$$$\mathrm{2}\leqslant{E}\leqslant\frac{\mathrm{38}}{\mathrm{9}}\:. \\ $$$${Is}\:{it}\:{wrong}\:{as}\:{i}\:{cannot}\:{find}\:{any}\:{mistake} \\ $$$${in}\:{your}\:{solution}….?? \\ $$
Answered by MJS last updated on 27/Oct/18
![x^2 +yx+y^2 −1=0 this is an ellipse with center ((0),(0) ) and θ=−45° y=−(x/2)±((√(4−3x^2 ))/2) ⇒ −((2(√3))/3)≤x≤((2(√3))/3) ⇒ −((2(√3))/3)≤y≤((2(√3))/3) [because of symmetry] E(x)=x^3 (−(x/2)±((√(4−3x^2 ))/2))+x(−(x/2)±((√(4−3x^2 ))/2))^3 +4= =(x^4 /2)−((3x^2 )/2)+4±((x(x^2 +1)(√(4−3x^2 )))/2) solving E′(x)=0 we get minima and maxima at x=±1 and x=±((√3)/3) ⇒ range(E)=[2; ((38)/9)]](https://www.tinkutara.com/question/Q46493.png)
$${x}^{\mathrm{2}} +{yx}+{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{an}\:\mathrm{ellipse}\:\mathrm{with}\:\mathrm{center}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\mathrm{and}\:\theta=−\mathrm{45}° \\ $$$${y}=−\frac{{x}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}}\:\Rightarrow\:−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\leqslant{x}\leqslant\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\Rightarrow\:−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\leqslant{y}\leqslant\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\left[\mathrm{because}\:\mathrm{of}\:\mathrm{symmetry}\right] \\ $$$$ \\ $$$${E}\left({x}\right)={x}^{\mathrm{3}} \left(−\frac{{x}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}}\right)+{x}\left(−\frac{{x}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}}\right)^{\mathrm{3}} +\mathrm{4}= \\ $$$$=\frac{{x}^{\mathrm{4}} }{\mathrm{2}}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{4}\pm\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{4}−\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\mathrm{solving}\:{E}'\left({x}\right)=\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{minima}\:\mathrm{and}\:\mathrm{maxima} \\ $$$$\mathrm{at}\:{x}=\pm\mathrm{1}\:\mathrm{and}\:{x}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:\Rightarrow\:\mathrm{range}\left({E}\right)=\left[\mathrm{2};\:\frac{\mathrm{38}}{\mathrm{9}}\right] \\ $$
Commented by rahul 19 last updated on 27/Oct/18
thanks sir! ����