33I think no one is trying this!
Q1. L1 ≡ ax+by+c and L2 ≡ lx+my+n , where a,b,c,l,m,n are real.
L1+λL2=0 represent all lines passing through intersection of L1=0 and L2=0
(True/False)?
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Q2. Lim(x→0)(sin(1/x))/(sin(1/x)) = ?
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Q.3 x = (2y±(2y-1-y2)1/2 )/2 is a second degree general curve here it represents ?
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Q.4 The solution of f(x)=f-1(x) if exists lie on ?
a) y=x
b) y=-x
c) None(Specify)
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89 Answers
1i 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
1To 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
62To add to question 1 (first part)
If false, give a general form which will actually satisfy all lines criterion
62As 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..) !
62I 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 :)
62and 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 !
62hmm.. 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...
1hehe :)
I am reading... good god...
but i think it is reflection along x=y... so they cant meet anywhere else!
33But 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
62if 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!)
62is 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 :)
33Last answer is not x=y, consider f(x)=-x+sin(x)..................
1last 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 !
14th 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
1hmm.. good question.. i not think that way..
good abhishek .. i like your answer
1oh.. yes i think.. indeterminate by indeterminate? right?
so be indeterminate na?
33Oh 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. but i'm not sure......
1good 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.
62yup.. unfortunately not too many active users on the forum...
they come.. they see.. they go :D