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Question Number 14747 by 433 last updated on 04/Jun/17

ε>0 6−ε≤xy≤6+ε 5−ε≤x+y≤5+ε Find x & y

ϵ>0 6−ϵ⩽xy⩽6+ϵ 5−ϵ⩽x+y⩽5+ϵ Findx&y

Answered by ajfour last updated on 04/Jun/17

Let if ε=0 x and y are roots of the equation: z^2 −(x+y)z+xy=0 or z^2 −5z+6=0 z^2 −3z−2z+6=0 (z−3)(z−2)=0 ⇒ z=3, or z=2. let us choose x=3, y=2 (when ε=0) Now let − h+3≤x≤3+h −k+2≤y≤2+k then (x+h)+(y+k)−(x+y)=ε or h+k=𝛆 .....(i) (x+h)(y+k)−xy= ε ⇒ (3+h)(2+k)−6=ε or 3k+2h+hk=𝛆 ...(ii) ⇒ k+2ε+hk=ε k+hk+ε=0 as k=ε−h ε−h+h(ε−h)+ε=0 −h^2 −h(1−ε)+2ε=0 h^2 +(1−ε)h−2ε=0 h,k= ((−(1−ε)±(√((1−ε)^2 +8ε)))/2) −h+3≤x≤3+h −k+2≤y≤2+k .

Letifϵ=0 xandyarerootsoftheequation: z2−(x+y)z+xy=0 orz2−5z+6=0 z2−3z−2z+6=0 (z−3)(z−2)=0 ⇒z=3,orz=2. letuschoosex=3,y=2(whenϵ=0) Nowlet−h+3⩽x⩽3+h −k+2⩽y⩽2+k then(x+h)+(y+k)−(x+y)=ϵ orh+k=ϵ.....(i) (x+h)(y+k)−xy=ϵ ⇒(3+h)(2+k)−6=ϵ or3k+2h+hk=ϵ...(ii) ⇒k+2ϵ+hk=ϵ k+hk+ϵ=0 ask=ϵ−h ϵ−h+h(ϵ−h)+ϵ=0 −h2−h(1−ϵ)+2ϵ=0 h2+(1−ϵ)h−2ϵ=0 h,k=−(1−ϵ)±(1−ϵ)2+8ϵ2 −h+3⩽x⩽3+h −k+2⩽y⩽2+k.

Commented byajfour last updated on 04/Jun/17

there might be some 𝛆rror.. someone please confirm..

theremightbesomeϵrror.. someonepleaseconfirm..

Commented bymrW1 last updated on 04/Jun/17

intuitively it should be something like 3−h≤x≤3+h 2−k≤y≤2+k

intuitivelyitshouldbesomethinglike 3−h⩽x⩽3+h 2−k⩽y⩽2+k

Commented byajfour last updated on 04/Jun/17

thank you sir, God bless you sir. but then the 𝛆rror had to be there.

thankyousir,Godblessyousir. butthentheϵrrorhadtobethere.

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