DSE 2008 Q55, Q56, Q57

[Dreyfus]

[Granpa Simpson]

57) S(alpha) is a point on the curve which means that it is an ordered pair in the coordinate system.
Now consider the following.
S1=(1/x)
S2=(1/x^2)
S3=(1/x^3)
.
.
.
.
S(n)=(1/x^n)
Now consider the points in S1={(x,y) s.t xER, y=(1/x)}
Similarly S2={(x,y) s.t. xER and y=(1/x^2)}.
Also S3={(x,y) s.t. xER and y=(1/x^3)}
.
.
.
for S(n)={(x,y) s.t. xER and y=(1/x^n)}
Now if u consider the set S1⋂S2⋂S3⋂...⋂S(n) the only common point is for sufficiently large x, y=0 which means that the intersection contains only one point which is the common point of convergence of each series i.e 0, thus the only elements are is (inf,0) and (-inf,0), however the answer given is a single elements...Plz do check whether the procedure seems alright or not and where am i wrong..??

[Dreyfus]

OK thanks subhayu I got it now....and what about q55...I m not getting interpretation of g(1,0/n)...does fn= n*1 if xEA and fn=n*0 if x not belongs to A...? Am I going right?

[Granpa Simpson]

Yes Vaibhav i think the procedure for 57 is ok..it will be a single point only because the question says "for positive real number x"...

[Granpa Simpson]

I think you are going in the right way...!!!!!

[kangkan]

 56..i am getting constant sequence(which is the key ans)

 for any arbitary integer k, fk (x)=k if   0<x<1/k
                                            = 0 otherwise

integration of fk over its domain is itgretate from 0 to 1/k of kdx=1.

so the sequence will be 1+1+1+1....a cosntant sequence


guys please help me with the working of question 57 of 2012 ..i am getting option a though the ans is c.

[The Villain]

Pls explain 55 ...m stuck

[kangkan]

http://discussion-forum.2150183.n2.nabble.com/DSE-2012-Q-57-td7591897.html#a7591969

Akshay solved question 55 here

[phelps.phan]

Q57: The different possible sets are 1/x, 1/x^2, 1/x^3,...
To find the intersection we equate 1/x = 1/x^2 which gives x=1. If we do it for any of the sets above we get x^n=1 ie x=1 (as the domain is defined for positive a). So that single point is x=1

[Ridhika]

Hey Subhayu.. I follow your logic for the rest, but how do you know this point is unique? How can we be certain these curves dont intersect elsewhere as well?

[Granpa Simpson]

Yes Ridhika dats a valid question..??infact i missed (1,1)..this will be common for all the sets..so by this logic it shud contain exactly two points, but not any other point except dat..because consider two curves y=(1/x^2) and y=(1/x^3)..except x=1, you cannot find any value for which the ordered pair will be same for both the curves..similarly if u consider for all the curves you will find the same..however according to my logic there will be exactly two points..!!!

[Dreyfus]

Hey Subhayu does (infinity, 0) is really a point? I think that single point mentioned by you is _(1,1)

[Granpa Simpson]

Yes Vaibhav actually (inf,0) cannot be a point in a rectangular ordered coordinate system..so (1,1) is the only point....

[Ridhika]

Okay guys.. I see why 1,1 is a point of intersection. But again how are we sure that no point corresponding to x E (1,infinity ) is also an intersection?

----- Reply message -----
From: "subhayu [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: DSE 2008 Q55, Q56, Q57
Date: Wed, Jun 11, 2014 12:39 AM

Yes Vaibhav actually (inf,0) cannot be a point in a rectangular ordered coordinate system..so (1,1) is the only point....

---

If you reply to this email, your message will be added to the discussion below:

http://discussion-forum.2150183.n2.nabble.com/DSE-2008-Q55-Q56-Q57-tp7591879p7592320.html

To unsubscribe from DSE 2008 Q55, Q56, Q57, click here.
NAML

[Granpa Simpson]

Ridhika even (1,infinity) is not a point in the coordinate system...!!!!

[Ridhika]

Yaya ! I know that ! I'm just asking how to show that (1,1) is unique?

----- Reply message -----
From: "subhayu [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: DSE 2008 Q55, Q56, Q57
Date: Wed, Jun 11, 2014 12:51 AM

Ridhika even (1,infinity) is not a point in the coordinate system...!!!!

---

If you reply to this email, your message will be added to the discussion below:

http://discussion-forum.2150183.n2.nabble.com/DSE-2008-Q55-Q56-Q57-tp7591879p7592325.html

To unsubscribe from DSE 2008 Q55, Q56, Q57, click here.
NAML

[Granpa Simpson]

Ridhika I have just used the logic that since the exponents of x are different you cannot get a point except (1,1) which is common to all the sets..may be a counter example can disprove this fact..but i dont think there exists one..however there must be a rigorous mathematical proof related to this fact which i am unaware of..but if you insist for such a proof i will surely look in to it and let u know (obviously if it is within the scope of my brain)...

[Ridhika]

Hehe .. no no I don't insist! I just kept thinking I'm missing something! Thanks a lot :)

----- Reply message -----
From: "subhayu [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: DSE 2008 Q55, Q56, Q57
Date: Wed, Jun 11, 2014 1:17 AM

Ridhika I have just used the logic that since the exponents of x are different you cannot get a point except (1,1) which is same..may be a counter example can disprove this fact..but i dont think there exists one..however there must be a rigorous proof related to this fact which i am unaware of..but if you insist for such a proof i will surely look in it and let u know...

---

If you reply to this email, your message will be added to the discussion below:

http://discussion-forum.2150183.n2.nabble.com/DSE-2008-Q55-Q56-Q57-tp7591879p7592329.html

To unsubscribe from DSE 2008 Q55, Q56, Q57, click here.
NAML

[mrittik]

Gracias @ Subhayu

[Granpa Simpson]

Mrittik: Si...