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#1 Posted 3:10pm 12-10-08 Some must see problems......Q1. L[ss]1[/ss] ≡ ax+by+c and L[ss]2[/ss] ≡ lx+my+n , where a,b,c,l,m,n are real. L[ss]1[/ss]+λL[ss]2[/ss]=0 represent all lines passing through intersection of L[ss]1[/ss]=0 and L[ss]2[/ss]=0 (True/False)? ______________________________________________________ Q2. Lim(x→0)(sin(1/x))/(sin(1/x)) = ? ______________________________________________________ Q.3 x = (2y±(2y-1-y[p]2[/p])[p]1/2[/p] )/2 is a second degree general curve here it represents ? ______________________________________________________ Q.4 The solution of f(x)=f[p]-1[/p](x) if exists lie on ? a) y=x b) y=-x c) None(Specify) ______________________________________________________
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#2 Posted 3:12pm 12-10-08 Re: Some must see problems......is the 2nd question Lim(x→0){(sin(1/x))/(sin(1/x))} = ? or {Lim(x→0)(sin(1/x))}/{Lim(x→0)(sin(1/x))} Both will have different answers! I think u meant the first one... btw both are worth trying :)
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#3 Posted 3:16pm 12-10-08 Re: Some must see problems......To add to question 1 (first part) If false, give a general form which will actually satisfy all lines criterion
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#4 Posted 6:36pm 12-10-08 Re: Some must see problems......I think no one is trying this!
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#5 Posted 6:38pm 12-10-08 Re: Some must see problems......To nishant Lim(x→0){(sin(1/x))/(sin(1/x))} = 1 or {Lim(x→0)(sin(1/x))}/{Lim(x→0)(sin(1/x))} =1 both are smae |
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#6 Posted 6:40pm 12-10-08 Re: Some must see problems......i think second part is easy.... other questions i cannoth think now.. in Q1 i think it will be only that it will use all lines |
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#7 Posted 6:49pm 12-10-08 Re: Some must see problems......Q4) f(f(x)) =x we have to solve... so it can be anywhere! |
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#8 Posted 7:01pm 12-10-08 Re: Some must see problems......i think the first one is true. isnt it?
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#9 Posted 7:02pm 12-10-08 Re: Some must see problems......NO :(
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#10 Posted 7:03pm 12-10-08 Re: Some must see problems......akshat all ur answers have mistakes :(
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#11 Posted 09:19am 13-10-08 Re: Some must see problems......Nishant , isn't the first one true ? let the point of intersection be ( h , k) L[ss]1[/ss] = ah + bk + c = 0 and L[ss]2[/ss] = lh + mk +n = 0 therefore any linear combination i.e L[ss]1[/ss] +λ L[ss]2[/ss] = 0 is true for all λ , as L[ss]1[/ss] and L[ss]2[/ss] are separately 0 .
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#12 Posted 09:23am 13-10-08 4 2nd question...Read the def of limits... lim(x→a)f(x)=b means as value of x approaches 'a' [i](always keeping itself in the domain of the function)[/i] f(x) approaches a fixed value 'b'.
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#13 Posted 09:25am 13-10-08 @Srinath...Does that linear combination represents the parent line L[ss]2[/ss] for any value of λ. More appropriate linear combination would be ...........
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#14 Posted 3:54pm 13-10-08 Re: Some must see problems......Actually the linear combination is λ[ss]1[/ss]L[ss]1[/ss]+λ[ss]2[/ss]L[ss]2[/ss]=0 and we divide entire eqn by λ[ss]1[/ss] and write λ[ss]1[/ss]/λ[ss]2[/ss] = λ assuming λ[ss]1[/ss] is not equal to 0 thus 1 line is lost in the family.
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#15 Posted 6:04pm 13-10-08 Re: Some must see problems......well abhishek there u are perfectly correct... btw ur answers will give a few repetitions... may be this form? λL[ss]1[/ss]+(1-λ)L[ss]2[/ss]=0 {0≤λ≤1}
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#16 Posted 6:05pm 13-10-08 Re: Some must see problems......In my opinion, this is a slightly better form of ur solution.. *only slightly :)
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#17 Posted 11:48pm 13-10-08 Re: Some must see problems......Gr8 i never took notice of that λ/1-λ captures whole R
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#18 Posted 11:50pm 13-10-08 Re: Some must see problems......only +ve R! why dont we need the negative side here?
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#19 Posted 00:04am 14-10-08 Re: Some must see problems......What a gud question i have touched!!! now i need time to think... may be 4 same reason we need only 0-pi angle 4 line's slope.. well may be..
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#20 Posted 00:19am 14-10-08 Re: Some must see problems......hey hang on.. i am getting confused :D wait give me 5 mins.. i will tell u exactly what :)
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#21 Posted 00:23am 14-10-08 Re: Some must see problems......0 to Pi was a guess
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#22 Posted 00:23am 14-10-08 Re: Some must see problems......yeah i am wrong.. it should have been -1<=λ<=1 sory :) y? because take an example x=0, y=0 will this give the set of all lines? no it will give lines only on +ve quadrant! It will not generate x-y=0!
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#23 Posted 00:25am 14-10-08 Re: Some must see problems......okie.. that 0-pi was a good guess... i mean the line sweeps the whole plane I though the same thing wud work here as well.. but it does not.. So this correction :)
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#24 Posted 00:28am 14-10-08 Re: Some must see problems......Now here comes a complete soln of a question but lets see if others try other questions....
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#25 Posted 00:33am 14-10-08 Re: Some must see problems......yup.. unfortunately not too many active users on the forum... they come.. they see.. they go :D
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#26 Posted 1:27pm 14-10-08 Re: Some must see problems......good everyone.. i like this all questions.. i will try to answer... Q2. Lim(x→0)(sin(1/x))/(sin(1/x)) = ? answer will be 1 . because it is lim(x-0) {1} the other thing. when in same question there was another case... answer will be 1 again.
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#27 Posted 2:24pm 14-10-08 Re: Some must see problems......Oh no first Lim(x→0)(sin(1/x))/(sin(1/x)) , according 2 me limit should not exist as limit exists if the limit approach a fixed quantity no matter from which path we approach it but here we can't approach 0 with path i.e. x=1/n*Pi and a very infinetisimal interval across zero will contain numerous points not in domain. [u]but i'm not sure......[/u]
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#28 Posted 2:24pm 14-10-08 Re: Some must see problems......oh.. yes i think.. indeterminate by indeterminate? right? so be indeterminate na?
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#29 Posted 2:26pm 14-10-08 Re: Some must see problems......yes.. priyam u are perfectly right.. my wrong :D
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#30 Posted 2:28pm 14-10-08 Re: Some must see problems......hmm.. good question.. i not think that way.. good abhishek .. i like your answer
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#31 Posted 2:30pm 14-10-08 Re: Some must see problems......4th part is it x=y? i think if not wrong.. it is reflection on x=y but will this is correct answer i dont know.. my guess only
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#32 Posted 6:07pm 14-10-08 Re: Some must see problems......last question is x=y.. this is because of reflection property of inverse functions good discussion b/w abhishek and nishant .. i liked it and learned from it as well ! |
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#33 Posted 6:15pm 14-10-08 Re: Some must see problems......it is x=y..
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#34 Posted 9:10pm 14-10-08 Re: Some must see problems......Last answer is not x=y, consider f(x)=-x+sin(x)..................
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#35 Posted 9:19pm 14-10-08 Re: Some must see problems......As long as inverses are functions.. they will meet at x=y the condition for them not to meet is that the "inverse" are not functions but relations (sorry for putting this relation crap in between..) !
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#36 Posted 9:30pm 14-10-08 Re: Some must see problems......What about f(x)=-x+sin(x)
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#37 Posted 9:32pm 14-10-08 Re: Some must see problems......you two are interesting :)
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#38 Posted 9:33pm 14-10-08 Re: Some must see problems......and for that what is the point which is not on x=y?
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#39 Posted 9:36pm 14-10-08 Re: Some must see problems......if u are not convinced... i will try to find some good reference to support my case :) Or try to prove this thing myself (which seems unlikely!)
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#40 Posted 9:37pm 14-10-08 Re: Some must see problems......But if we plot graph of f(x)=-x+sin(x) and reflect it about y=x and the reflected graph and original graph intersect at many points other than y=x
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#41 Posted 9:37pm 14-10-08 Re: Some must see problems......hehe :) I am reading... good god... but i think it is reflection along x=y... so they cant meet anywhere else!
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#42 Posted 9:39pm 14-10-08 Re: Some must see problems......hmm.. i can see where u got this from.. ok i will try to make sure i am right.. and then tell u why this is happening... because u could have givne a simpler function x=-y!! their inverse will meet everywhere.... but just let me check once more...
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#43 Posted 9:40pm 14-10-08 Re: Some must see problems......and after seeing this function x=-y.. even i have started to believe that i am wrong! these two meet at all points.. and they seem to be invertible !
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#44 Posted 9:40pm 14-10-08 Re: Some must see problems......interesting! i never think that way!
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#45 Posted 9:50pm 14-10-08 Re: Some must see problems......I searched the net.. somehow i am getting 2-3 sites that say what i am saying.. But there is no denying that u are right... absolutely. I remember my fiitjee teacher ofcourse many years back having taught this same thing :D And he was an IIT Kanpur Gold Medalist Mathematics(Rank 1!) So he coulnt have been wrong :D .. alas :)) This one is going in the next news letter with credits to you :)
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#46 Posted 9:50pm 14-10-08 Re: Some must see problems......Oh, my teacher asked me this and we were blank so he gave us that example so i never thought of simpler example y=-x
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#47 Posted 9:52pm 14-10-08 Re: Some must see problems......Oh my fiitjee teacher also has taught that last year.....
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#48 Posted 9:55pm 14-10-08 Re: Some must see problems......Now 3 questions have lead 2 good conclusions now last question of the topic is left but an easy one...
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#49 Posted 11:32pm 14-10-08 Re: Some must see problems......q3-- 2 lines.. 3y-1=x and y+1=x... is this the answers?
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#50 Posted 11:38pm 14-10-08 Re: Some must see problems......Check question once more.......
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