HSC 2014 MX2 Marathon ADVANCED (archive) (5 Viewers)

glittergal96 · Sep 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Does this question need clarification?

No, the question makes sense to me.

I am 99% sure the answer is yes, but a proof of this fact would need me to lay out some kind of "greedy algorithm" for doing this as well as a demonstration that the algorithm does what we require. I can't see a quick way of doing this (although I expect it is more cumbersome than technically difficult. the key idea is clearly induction and subdividing.)

dan964 · Sep 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

r

3ddementor said:
Can anyone give me a heads up as to how to do Q16 (iii) , it's really doing my head in :S (from Dan964's set of questions)

Thats kind of the point.
But I realised I forgot a constant term so I have to redo the question.

dan964 · Sep 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

edited questions of mine, discard old one.
errors fixed:
Q15 is now Q13
Q15(b) corrected now Q13 (b)
Q16 rewritten, some parts was possible but not mathematical legitimate (It assumed incorrectly that dt=dn which it wasn't)
Marks changed (some q were out of 16)

summary/difficulty rating:
Q11 - really easy (part (a) - do I even need to mention?, part (b) - standard 4U, snuck (iv) in though, (c) - almost standard 4U, good practice q, (d) - standard 4U)
Q12 - really easy (part (a) - some school trial, part (b) - standard 4U question, part (c) - easy)
Q13 (a) - moderate except (ii) and (iv) (same as old Q15(a))
Q13 (b) - moderate (adapted from old Q15(b) to eliminate Carmichael Numbers)
Q14 (a) - moderate (adapted from SGS trial - can't remember year, part (a)(i) adapted from 2009 HSC Q7(b)(i))
Q14 (b) - hard (adapted from Moriah 2001)
Q14 (c) - moderate, part (ii) is good (slightly standard 4U question, especially part (i)
Q15 (a) - moderate (adapted from James Ruse 2012 3U, except for part (ii) )
Q15 (b) - moderate/hard (adapted from past paper, probably HSC 2010 4U, especially part (c)
Q16 - red herring, it took some time to get this one right. In fact it is so difficult I didn't bother attempt part (vi)

ok

Q11-16 here >>>> View attachment dan964 q11-16.pdf

Errata (26-09-14):
Q13 (b)(ii), prove for prime p.
Q16 (a)(ii), use the fact to show the result.

TL1998 · Sep 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Glittergal96 u MUST go to the BOS trials 2014 event. Basically its where you sit this difficult mathematics extension two paper written by carrotsticks and Trebla. Normally if someone gets over 65% in the extension two exam they get a state rank. First in state last year attended the bos 2013 trials and even he couldnt break the 70% mark although he topped the exam anyways. You should most definitely consider!

mreditor16 · Sep 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
Glittergal96 u MUST go to the BOS trials 2014 event. Basically its where you sit this difficult mathematics extension two paper written by carrotsticks. Normally if someone gets over 65% in the extension two exam they get a state rank. First in state last year attended the bos 2013 trials and even he couldn't break the 70% mark although he topped the exam anyways. You should most definitely consider!

ftfy

dunjaaa · Sep 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

it'll be a honour to meet you

Trebla · Sep 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
Glittergal96 u MUST go to the BOS trials 2014 event. Basically its where you sit this difficult mathematic extrnsion two paper written by Carrotsticks and Trebla. Normall if someone gets over 65% in the extension two exam they get a stste rank. First in stste lsst year attended the bos 2013 trials and even he couldnt break the 70% mark although he topped the exam anyways. You should most definitely consider!

ftfy

mreditor16 · Sep 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Trebla said:
ftfy

yeh true - why does everyone forgot Trebla, when talking about BOS trials? :/

glittergal96 · Sep 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
Glittergal96 u MUST go to the BOS trials 2014 event. Basically its where you sit this difficult mathematic extrnsion two paper written by carrotsstiks. Normall if someone gets over 65% in the extension two exam they get a stste rank. First in stste lsst year attended the bos 2013 trials and even he couldnt break the 70% mark although he topped the exam anyways. You should most definitely consider!

Yeah, I read that thread the other day. I'll definitely consider it.

Even if I don't come to the event though, I will still have an attempt at the paper if it gets put online.

faisalabdul16 · Sep 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

mreditor16 said:
yeh true - why does everyone forget Trebla, when talking about BOS trials? :/

I wonder as well!

Sy123 · Oct 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find all roots with modulus 1 of the polynomial$ \\ x^{n}-x^{n-2} -x + 2 = 0 \\ $where$ \ n \geq 2$

dunjaaa · Oct 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Can you give us a hint? I'm not that good with these types of questions

Carrotsticks · Oct 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dunjaaa said:
Can you give us a hint? I'm not that good with these types of questions

Think about what kinds of polynomials you know that involve its roots having modulus 1 (ie: All lying on the unit circle).

Also, as we're dealing with the moduli, that's hinting at the use of the triangle inequality at some point.

Sy123 · Oct 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find all roots with modulus 1 of the polynomial$ \\ x^{n}-x^{n-1} -x + 2 = 0 \\ $where$ \ n \geq 2$

I just realised I made a small typo, it should be

$x^n - x^{n-2} - x + 2 = 0$

Edited the original post, apologies for any inconvenience (though as far as I can see the method I had in mind can be used for the original question as well as the typo'd one)

dunjaaa · Oct 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

I don't know if this is correct, but here's what I did:

Screen shot 2014-10-02 at 6.10.10 PM.png

Sy123 · Oct 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dunjaaa said:
I don't know if this is correct, but here's what I did: View attachment 30983

Not all those solutions are valid since they give a non-zero Re(z)!

For instance, try your first one, $z = cis \frac{\pi}{3(n-1)}$ (when k=0)

If you try putting that into Re(z) you won't get 0

dunjaaa · Oct 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

oh yes, I forgot to test that

glittergal96 · Oct 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find all roots with modulus 1 of the polynomial$ \\ x^{n}-x^{n-2} -x + 2 = 0 \\ $where$ \ n \geq 2$

Let

$\alpha=\textrm{cis}(\theta)$

be a root. Then

$P_n(\alpha)=0 \Rightarrow |\alpha^2-1|=|\alpha^n-\alpha^{n-2}|=|\alpha-2|$

$\Rightarrow 4\sin^2(\theta)=(\cos(\theta)-2)^2+\sin^2(\theta).$

Using the Pythagorean identity, this becomes a quadratic in $\cos(\alpha)$ , with unique solution $1/2$ .

So the only possible solution on the unit circle is $\beta=\textrm{cis}(\frac{\pi}{3})$ (and its conjugate comes with it of course). It remains to determine the n for which this is actually a solution.

Noting that $\beta^2-\beta+1=0$ , we get

$P_n(\beta)=\beta^n-\beta^{n-2}-\beta+2=\beta^n-\beta^{n-2}-\beta^2+1=(\beta^{n-2}-1)(\beta^2-1).$

This quantity vanishes if and only if $n$ is such that $\beta$ is an $(n-2)$ -th root of unity.

That is, there must exist k with

$\frac{2k\pi}{n-2}=\frac{\pi}{3}$ .

Solving this for $n$ allows us to conclude that the only possible solutions to the polynomial of modulus $1$ are $\textrm{cis}(\pm\frac{\pi}{3})$ , and these are solutions if and only if $n=6k+2$ for some non-negative integer $k$ .

mreditor16 · Oct 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

OMG glittergal, you didn't got to BOS trials!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! why?? :/

glittergal96 · Oct 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

mreditor16 said:
OMG glittergal, you didn't got to BOS trials!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! why?? :/

To be honest, I actually forgot to register lol. For some reason I had it in my head that it was next week :/.

No biggie though, I will just do the exam at home later whenever it gets uploaded.

HSC 2014 MX2 Marathon ADVANCED (archive) (5 Viewers)

Active Member

what

what

Member

Well-Known Member

Active Member

Administrator

Well-Known Member

Active Member

Wot

This too shall pass

Active Member

Retired

This too shall pass

Active Member

This too shall pass

Active Member

Active Member

Well-Known Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 5)