HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

Status
Not open for further replies.

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004
Re: HSC 2014 4U Marathon - Advanced Level

ok here are the details. My previous calculation is for the case X, Y and theire wives all in one boat.



They can be in any of the five boates, which is why 5 times; 8 C 2 squared is for two men and two women chosen for the first boat of the remaining four boats; similarly for all other boats.

Now calculation for X and his wife in one boat and Y and his wife in the other boat (explanation quite similar so omitted)



Therefore, total number of ways =31752000+508032000=539784000
 
Last edited:

abecina

Member
Joined
Aug 24, 2008
Messages
44
Gender
Male
HSC
N/A
Re: HSC 2014 4U Marathon - Advanced Level

I think what's going on here is that you are assuming that different boats means different arrangement. I think the question is asking how many different ways can they be separated. Sort of life "how many ways can 10 people be divided into two teams" doesn't assume which team is Team A and which team is Team B
 

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004
Re: HSC 2014 4U Marathon - Advanced Level

I think what's going on here is that you are assuming that different boats means different arrangement. I think the question is asking how many different ways can they be separated. Sort of life "how many ways can 10 people be divided into two teams" doesn't assume which team is Team A and which team is Team B
but boats are different, aren't they?

Anyway, my answer divided by 5! will be the same as yours
 
Last edited:

glittergal96

Active Member
Joined
Jul 25, 2014
Messages
418
Gender
Female
HSC
2014
Re: HSC 2014 4U Marathon - Advanced Level

This one should be a good one:

Nice question :).

Let

Then by repeated integration by parts:



Each of the summands in the second term vanish, because the polynomials are all odd, so we are left with



But for , we can integrate by parts to obtain a recurrence for .







Hence the expression we need to evaluate is equal to





 
Last edited:

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004
Re: HSC 2014 4U Marathon - Advanced Level

Nice question :).

Let

Then by repeated integration by parts:



Each of the summands in the second term vanish, because the polynomials are all odd, so we are left with



But for , we can integrate by parts to obtain a recurrence for .







Hence the expression we need to evaluate is equal to





well done. nice question and nicer solution. lol
 
Last edited:

glittergal96

Active Member
Joined
Jul 25, 2014
Messages
418
Gender
Female
HSC
2014
Re: HSC 2014 4U Marathon - Advanced Level

Nice question :).

Let

Then by repeated integration by parts:



Each of the summands in the second term vanish, because the polynomials are all odd, so we are left with



But for , we can integrate by parts to obtain a recurrence for .







Hence the expression we need to evaluate is equal to





Actually, thats not a correct explanation of why the sum vanishes, I will post again after lunch fixing that part.
 

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004
Re: HSC 2014 4U Marathon - Advanced Level

maybe it is good, one odd and the other even, always like that
 

glittergal96

Active Member
Joined
Jul 25, 2014
Messages
418
Gender
Female
HSC
2014
Re: HSC 2014 4U Marathon - Advanced Level

Actually, thats not a correct explanation of why the sum vanishes, I will post again after lunch fixing that part.
Ahh simple, because each of these terms has a factor that comes from hitting an (x^2-1)^n with less than n derivatives and evaluating at the roots of this polynomial. As this polynomials roots are of multiplicity n, these factors all vanish, and hence the whole sum does.
 
Last edited:

glittergal96

Active Member
Joined
Jul 25, 2014
Messages
418
Gender
Female
HSC
2014
Re: HSC 2014 4U Marathon - Advanced Level

maybe it is good, one odd and the other even, always like that
No its not, because odd is not what we actually need there. But my above post tidies up this inaccuracy.
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2014 4U Marathon - Advanced Level

(HSC Methods only)



---

 
Last edited:

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004
Re: HSC 2014 4U Marathon - Advanced Level

(HSC Methods only)



---

Part (a) Consider the area under the curve y=1/x between 1 and n. Divide the interval into (n-1) equal parts. If we use upper rectangles to approximate the area, we have



While we use lower rectangles to approximate the area, we obtain



Conbine these two inequalities, we arrive at



Therefore, if the limit exists, it must be finite.

NOTE: We assume the limit exists. To prove this, it seems to me we need monotone convergence which is beyond HSC scope.
 

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004
Re: HSC 2014 4U Marathon - Advanced Level

For part (b) (i):



(since odd numbers go away, even numbers k=2m get doubled)

Part (b)(ii):



From which it follows



 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2014 4U Marathon - Advanced Level

NOTE: We assume the limit exists. To prove this, it seems to me we need monotone convergence which is beyond HSC scope.
Right. My bad.
 

SilentWaters

Member
Joined
Mar 20, 2014
Messages
55
Gender
Male
HSC
2014
Re: HSC 2014 4U Marathon - Advanced Level

NOTE: We assume the limit exists. To prove this, it seems to me we need monotone convergence which is beyond HSC scope.
Not really. Investigating monotonic behaviour in functions and looking at boundaries is well within the scope of Extension 2.

Consider . So:



We know by its positive square second derivative that is concave down, and that by its first derivative, is tangential at (thus exceeding the former function for all non-zero ). Popping in its place, where we can conclude that (*) is negative. Hence our sequence is monotonic decreasing.

Additionally, looking at the relationship between the area beneath the curve to the right of , and the sums of lower-bound rectangles of unit width, we can say or But due to the monotonic increasing nature of the logarithmic curve. Hence

In summary, we have a decreasing sequence bounded below by zero. So as it converges to a finite value.
 
Last edited:

SilentWaters

Member
Joined
Mar 20, 2014
Messages
55
Gender
Male
HSC
2014
Re: HSC 2014 4U Marathon - Advanced Level

Part (a) Consider the area under the curve y=1/x between 1 and n. Divide the interval into (n-1) equal parts. If we use upper rectangles to approximate the area, we have



While we use lower rectangles to approximate the area, we obtain



Conbine these two inequalities, we arrive at



Therefore, if the limit exists, it must be finite.
Also I think the third line should read
 
Last edited:

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Re: HSC 2014 4U Marathon - Advanced Level

In summary, we have a decreasing sequence bounded below by zero. So as it converges to a finite value.
Sure, but this fact cannot be proven within the scope of MX2.

One would need to rigorously define the reals to prove such a statement as it is reliant on completeness or an equivalent axiom.
 
Last edited:
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top