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Question Number 2454 by Yozzi last updated on 20/Nov/15
Find smallest a>1 for which ((a+sinx)/(a+siny))≤e^(y−x) for ∀ x≤y.
Findsmallesta>1forwhicha+sinxa+siny⩽ey−xfor∀x⩽y.
Commented by Rasheed Soomro last updated on 21/Nov/15
Find smallest a>1 for which ((a+sinx)/(a+siny))≤e^(y−x) for ∀ x≤y. −−−−−−−−−−− ((a+sin x)/(a+sin y))≤e^(y−x) a+sin x≤e^(y−x) (a+sin y) ≤ae^(y−x) +e^(y−x) sin y a−ae^(y−x) ≤e^(y−x) sin y −sin x a(1−e^(y−x) )≤e^(y−x) sin y −sin x a≤((e^(y−x) sin y −sin x)/(1−e^(y−x) )) assuming y>x [for y=x see sec A] Since a>1 1<a≤((e^(y−x) sin y −sin x)/(1−e^(y−x) )) a=Smallest ((e^(y−x) sin y −sin x)/(1−e^(y−x) ))>1 We have to use such values of x and y for which_(−) ((e^(y−x) sin y −sin x)/(1−e^(y−x) )) is smallest but greater than 1_(−) For y=x............................section A ((a+sinx)/(a+siny))≤e^(y−x) ((a+sin x)/(a+sin x))≤e^(x−x) 1≤1 No help for a Continue
Findsmallesta>1forwhicha+sinxa+siny⩽ey−xfor∀x⩽y.−−−−−−−−−−−a+sinxa+siny⩽ey−xa+sinx⩽ey−x(a+siny)⩽aey−x+ey−xsinya−aey−x⩽ey−xsiny−sinxa(1−ey−x)⩽ey−xsiny−sinxa⩽ey−xsiny−sinx1−ey−xassumingy>x[fory=xseesecA]Sincea>11<a⩽ey−xsiny−sinx1−ey−xa=Smallestey−xsiny−sinx1−ey−x>1Wehavetousesuchvaluesofxandyforwhich−ey−xsiny−sinx1−ey−xissmallestbutgreaterthan1−Fory=x……………………….sectionAa+sinxa+siny⩽ey−xa+sinxa+sinx⩽ex−x1⩽1NohelpforaContinue
Commented by prakash jain last updated on 20/Nov/15
Since a>1, a+sin y>0 and e^x >0 we can rewrite the inequality as f(x,y)=ae^x +e^x sin x−(ae^y +e^y sin y) For f(x,y) ≤0, we need to chose a such that value at maxima is 0. f_x =ae^x +e^x sin x+e^x cos x f_y =−(ae^y +e^x sin y+e^x cos y) function f will attain maxima or minima when f_x and f_y are both 0. f_x =0⇒e^x (a+sin x+cos x)=0 a+(√2)sin (x+(π/4))=0 similarly a+(√2)sin (y+(π/4))=0 The critical points are given by: x=2πn+2tan^(−1) (((−(√(2−a^2 ))−1)/(a−1))), n∈Z y=2πm+2tan^(−1) (((−(√(2−a^2 ))−1)/(a−1))), m∈Z The problem statement is to chose a such that f(x,y)≤0 at critical points. To be continued in answer.
Sincea>1,a+siny>0andex>0wecanrewritetheinequalityasf(x,y)=aex+exsinx−(aey+eysiny)Forf(x,y)⩽0,weneedtochoseasuchthatvalueatmaximais0.fx=aex+exsinx+excosxfy=−(aey+exsiny+excosy)functionfwillattainmaximaorminimawhenfxandfyareboth0.fx=0⇒ex(a+sinx+cosx)=0a+2sin(x+π4)=0similarlya+2sin(y+π4)=0Thecriticalpointsaregivenby:x=2πn+2tan−1(−2−a2−1a−1),n∈Zy=2πm+2tan−1(−2−a2−1a−1),m∈ZTheproblemstatementistochoseasuchthatf(x,y)⩽0atcriticalpoints.Tobecontinuedinanswer.
Answered by prakash jain last updated on 20/Nov/15
At critical point of f(x,y) sinx=siny so the given inequality is satisfied by all a>1 (∵e^(y−x) ≥1) I am not 100% confident of the answer since I expected a value different than 1. comments? The answer a=1 is wrong (see comments below). Wrong conclusion was arrived as critical points for f(x,y) in the comments are only for LOCAL MAXIMA.
Atcriticalpointoff(x,y)sinx=sinysothegiveninequalityissatisfiedbyalla>1(∵ey−x⩾1)Iamnot100%confidentoftheanswersinceIexpectedavaluedifferentthan1.comments?Theanswera=1iswrong(seecommentsbelow).Wrongconclusionwasarrivedascriticalpointsforf(x,y)inthecommentsareonlyforLOCALMAXIMA.
Commented by Yozzi last updated on 20/Nov/15
I′m unsure about how to tackle the problem, but how about this. If x=y the inequality becomes ((a+sinx)/(a+sinx))≤e^0 . ∵ a>1 ⇒∄x,a∈R∣a+sinx=0. So 1≤1 and hence the inequality holds. In this case there isn′t any smallest a∈R since one can infinitely set a to be a value above 1 and very close to 1. No highest lower bound exists for the set {a∈R∣a>1}. For y>x⇒y−x>0⇒e^(y−x) >e^0 =1 ∴1−e^(y−x) <0. Keeping note of this we have ((a+sinx)/(a+siny))≤e^(y−x) [a+sinx,a+siny>0 ∵ a>1] a+sinx≤ae^(y−x) +e^(y−x) siny a(1−e^(y−x) )≤e^(y−x) siny−sinx ⇒a≥((e^(y−x) siny−sinx)/(1−e^(y−x) )) [1−e^(y−x) <0] a≥((e^(y−x) siny)/(1−e^(y−x) ))−((sinx)/(1−e^(y−x) )) (∗) If the RHS of the inequality above is strictly greater than 1 then the smallest a possible would be found to be a=((e^(y−x) siny−sinx)/(1−e^(y−x) )). If this is true then ((e^(y−x) siny−sinx)/(1−e^(y−x) ))>1 e^(y−x) siny−sinx<1−e^(y−x) e^(y−x) (siny+1)<1+sinx If siny+1≠0 (otherwise siny+1>0) ∴e^(y−x) <((1+sinx)/(1+siny)) Checking that this is obeyed by the proposed value for a ((((e^(y−x) siny−sinx)/(1−e^(y−x) ))+sinx)/(((e^(y−x) siny−sinx)/(1−e^(y−x) ))+siny))≤e^(y−x) ((e^(y−x) siny−sinx+sinx−e^(y−x) )/(e^(y−x) siny−sinx+siny−e^(y−x) siny))≤e^(y−x) ((e^(y−x) (siny−1))/(siny−sinx))≤e^(y−x) ((siny−1)/(siny−sinx))≤1. y>x so siny≠sinx. If siny>sinx siny−1≤siny−sinx⇒sinx≤1 (possible) If siny<sinx⇒siny−1≥siny−sinx sinx≥1 (possible if x=nπ+(−1)^n (π/2)) Together the cases suggest that the inequality holds for all x<y, if siny+1≠0. If the RHS of (∗) is not greater than 1 I don′t know how one could acquire the necessary value for a from the information given. In this case we must have a>1.
I′munsureabouthowtotackletheproblem,buthowaboutthis.Ifx=ytheinequalitybecomesa+sinxa+sinx⩽e0.∵a>1⇒∄x,a∈R∣a+sinx=0.So1⩽1andhencetheinequalityholds.Inthiscasethereisn′tanysmallesta∈Rsinceonecaninfinitelysetatobeavalueabove1andverycloseto1.Nohighestlowerboundexistsfortheset{a∈R∣a>1}.Fory>x⇒y−x>0⇒ey−x>e0=1∴1−ey−x<0.Keepingnoteofthiswehavea+sinxa+siny⩽ey−x[a+sinx,a+siny>0∵a>1]a+sinx⩽aey−x+ey−xsinya(1−ey−x)⩽ey−xsiny−sinx⇒a⩾ey−xsiny−sinx1−ey−x[1−ey−x<0]a⩾ey−xsiny1−ey−x−sinx1−ey−x(∗)IftheRHSoftheinequalityaboveisstrictlygreaterthan1thenthesmallestapossiblewouldbefoundtobea=ey−xsiny−sinx1−ey−x.Ifthisistruetheney−xsiny−sinx1−ey−x>1ey−xsiny−sinx<1−ey−xey−x(siny+1)<1+sinxIfsiny+1≠0(otherwisesiny+1>0)∴ey−x<1+sinx1+sinyCheckingthatthisisobeyedbytheproposedvalueforaey−xsiny−sinx1−ey−x+sinxey−xsiny−sinx1−ey−x+siny⩽ey−xey−xsiny−sinx+sinx−ey−xey−xsiny−sinx+siny−ey−xsiny⩽ey−xey−x(siny−1)siny−sinx⩽ey−xsiny−1siny−sinx⩽1.y>xsosiny≠sinx.Ifsiny>sinxsiny−1⩽siny−sinx⇒sinx⩽1(possible)Ifsiny<sinx⇒siny−1⩾siny−sinxsinx⩾1(possibleifx=nπ+(−1)n(π/2))Togetherthecasessuggestthattheinequalityholdsforallx<y,ifsiny+1≠0.IftheRHSof(∗)isnotgreaterthan1Idon′tknowhowonecouldacquirethenecessaryvalueforafromtheinformationgiven.Inthiscasewemusthavea>1.
Commented by Yozzi last updated on 20/Nov/15
This question is from an admissions quiz and I think its purpose is to see how one breaks it down rather than obtaining just a value for a. It′s all about the process rather than the answer.
ThisquestionisfromanadmissionsquizandIthinkitspurposeistoseehowonebreaksitdownratherthanobtainingjustavaluefora.It′sallabouttheprocessratherthantheanswer.
Commented by prakash jain last updated on 20/Nov/15
Ok. The answer a>1 is wrong ((1.01+sin x)/(1.01+sin y))≤e^(y−x) x=(π/2), y=((3π)/2) ((2.01)/(0.01))=201≰e^π
Ok.Theanswera>1iswrong1.01+sinx1.01+siny⩽ey−xx=π2,y=3π22.010.01=201≰eπ
Commented by Yozzi last updated on 20/Nov/15
It probably was as simple as observing no such a exists.
Itprobablywasassimpleasobservingnosuchaexists.
Commented by prakash jain last updated on 20/Nov/15
I have updated answer on where the error is in the original arguments. If the answer is no such a exists then we need to prove that there is no global maxima for f(x,y) with y>x.
Ihaveupdatedansweronwheretheerrorisintheoriginalarguments.Iftheanswerisnosuchaexiststhenweneedtoprovethatthereisnoglobalmaximaforf(x,y)withy>x.
Commented by Yozzi last updated on 20/Nov/15
Yes.
Yes.

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