HSC 2012 MX2 Marathon (archive) (4 Viewers)

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

This means that as n approaches an infinitely large number, the stationary point and inflexion point will coincide (ignoring x = 0) and there will be a horizontal inflexion point. (is this right? lol)

Good argument. Indeed they tend to the same x value, and therefore the same y value as well.

There won't strictly be a horizontal inflexion point, though I can understand where you're coming from (point of inflexion + stationary point = HPOI).

Geometrically it means that as n approaches positive infinity the limiting ratio of the x-coordinates of the inflexion point to the stationary point of the curve y=x^n*log[e][x^n] is one?

You just repeated the question basically haha. A stronger argument is required.

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

Good argument. Indeed they tend to the same x value, and therefore the same y value as well.

There won't be a horizontal inflexion point, though I can understand where you're coming from (point of inflexion + stationary point = HPOI).



You just repeated the question basically haha. A stronger argument is required.

So just saying that they will tend to the same (x, y) values would be enough?

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

I realised lol.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

So just saying that they will tend to the same (x, y) values would be enough?

Yes.

Somebody post a new question (or answer seanieg's question).

 

[largarithmic]

Re: 2012 HSC MX2 Marathon

An easy polynomials question:

Consider R(x) = P(x) - Q(x): note that it is of degree at most n.

Now suppose R(x) is not the zero polynomial; in this case it has at most n roots (from unique factorisation, which is provable from the division algorithm). However it clearly has n+1 roots x1, x2, ..., xn+1. Therefore R(x) is the zero polynomial, so P(x) is identically equal to Q(x).

Here's a decently nice question: by considering its complex roots, find all polynomials P(x) such that:
for all complex x.

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

What happened to this thread?

 

[IamBread]

Re: 2012 HSC MX2 Marathon

What happened to this thread?

Woo new question XD

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Perhaps you can post up a question now.

 

[cutemouse]

Re: 2012 HSC MX2 Marathon

If 0, z_1, z_2 and z_3 lie on a circle then prove that the points 1/z_1, 1/z_2 and 1/z_3 are collinear. (z_1, z_2, z_3 are non zero complex numbers)

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

If 0, z_1, z_2 and z_3 lie on a circle then prove that the points 1/z_1, 1/z_2 and 1/z_3 are collinear. (z_1, z_2, z_3 are non zero complex numbers)

I've seen this question before... I don't quite remember when.

 

[IamBread]

Re: 2012 HSC MX2 Marathon

Perhaps you can post up a question now.

I couldn't think of any to post
But here's one now lol.

 

[math man]

Re: 2012 HSC MX2 Marathon

I've seen this question before... I don't quite remember when.

question is in patel and terry lee

 

[mnmaa]

Re: 2012 HSC MX2 Marathon

Easy questions
(a).If w is a complex root of z^5 -1=0 with the smallest positive argument, show that w^2 ,w^3 and 2^4 are the other complex roots.And hence prove that 1+w^2+w^3+w^4=0

(b).Find the quadratic equation whose roots are a=w+w^4 and B=w^2+w^3

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Easy questions
(a).If w is a complex root of z^5 -1=0 with the smallest positive argument, show that w^2 ,w^3 and 2^4 are the other complex roots.And hence prove that 1+w^2+w^3+w^4=0

(b).Find the quadratic equation whose roots are a=w+w^4 and B=w^2+w^3

Also, no "Hence find the exact value of":

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

Easy questions
(a).If w is a complex root of z^5 -1=0 with the smallest positive argument, show that w^2 ,w^3 and 2^4 are the other complex roots.And hence prove that 1+w^2+w^3+w^4=0

(b).Find the quadratic equation whose roots are a=w+w^4 and B=w^2+w^3

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha@plus;\beta=w@plus;w^2@plus;w^3@plus;w^4=-1~(\textrm{since}~1@plus;w@plus;w^2@plus;w^3@plus;w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w@plus;w^4)(w^2@plus;w^3)\\ =w^3@plus;w^4@plus;w^6@plus;w^7=w^3@plus;w^4@plus;w@plus;w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2@plus;x-1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha+\beta=w+w^2+w^3+w^4=-1~(\textrm{since}~1+w+w^2+w^3+w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w+w^4)(w^2+w^3)\\ =w^3+w^4+w^6+w^7=w^3+w^4+w+w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2+x-1" title="\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha+\beta=w+w^2+w^3+w^4=-1~(\textrm{since}~1+w+w^2+w^3+w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w+w^4)(w^2+w^3)\\ =w^3+w^4+w^6+w^7=w^3+w^4+w+w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2+x-1" /></a>

 

[AAEldar]

Re: 2012 HSC MX2 Marathon



And for something a little harder...



Interested in seeing different ways people solve it.

 

[zeebobDD]

Re: 2012 HSC MX2 Marathon

And for something a little harder...



Interested in seeing different ways people solve it.

that first one from cambridge polynomials?:/

 

[AAEldar]

Re: 2012 HSC MX2 Marathon

that first one from cambridge polynomials?:/

Yes it is hahaha. Got asked to do it last night and thought it was quite nice.

 

[mnmaa]

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha@plus;\beta=w@plus;w^2@plus;w^3@plus;w^4=-1~(\textrm{since}~1@plus;w@plus;w^2@plus;w^3@plus;w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w@plus;w^4)(w^2@plus;w^3)\\ =w^3@plus;w^4@plus;w^6@plus;w^7=w^3@plus;w^4@plus;w@plus;w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2@plus;x-1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha+\beta=w+w^2+w^3+w^4=-1~(\textrm{since}~1+w+w^2+w^3+w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w+w^4)(w^2+w^3)\\ =w^3+w^4+w^6+w^7=w^3+w^4+w+w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2+x-1" title="\\ z^5=1=cis0=cis2k\pi\\ z=(cis2k\pi)^{\frac{1}{5}}=cis\frac{2}{5}k\pi_{(k=0,1,2,3,4,)}~\textrm{(By De Moivre's theorem)}\\ ~\\ z_1=cis0=1~~~~~~ z_2=cis\frac{2}{5}\pi~~~~~ z_3=cis\frac{4}{5}\pi~~~~~ z_4=cis\frac{6}{5}\pi~~~~~ z_5=cis\frac{8}{5}\pi\\ ~\\ \therefore w=z_2=cis\frac{2}{5}\pi~\textrm{(Since it's the root with the smallest positive argument)}\\ z_3=(z_2)^2,~~~~~ z_4=(z_2)^3,~~~~~ z_5=(z_2)^4~\textrm{(By De Doivre's theorem)}\\ \therefore \textrm{the other complex roots are}~w^2,w^3, w^4 ~\\~\\~\\~\\ \frac{-b}{a}=\alpha+\beta=w+w^2+w^3+w^4=-1~(\textrm{since}~1+w+w^2+w^3+w^4=0)\\ \therefore b=1\\ ~\\ \frac{c}{a}=\alpha \beta=(w+w^4)(w^2+w^3)\\ =w^3+w^4+w^6+w^7=w^3+w^4+w+w^2=-1~(\textrm{since}~w^5=1)\\ \therefore c=-1\\~\\ \therefore \textrm{the required quadratic equation is}~x^2+x-1" /></a>

well done but you didnt prove that w+w^2+w^3+w^4=-1