HSC 2012 MX2 Marathon (archive) (1 Viewer)

bleakarcher · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

It was pretty simple actually that last piece of info was very crucial though.

rolpsy · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Hint: Put the RHS into cis form somehow.

Oh and I forgot to type into the question for part (b):

$$Where $n=4k, k\in \mathbb{Z}$

Terribly sorry about that.

that makes it easier!

(a)
$\begin{align*}\sin(\theta) + i\cos(\theta) &= \frac{\left (\sin(\theta) + i\cos(\theta) \right ) \left ( 1 + \sin(\theta) - i\cos(\theta) \right )}{1 + \sin(\theta) - i\cos(\theta)}\\&= \frac{\sin(\theta) + \sin^2(\theta) - i\cos(\theta)\sin(\theta) + i\cos(\theta) + i\cos(\theta)\sin(\theta) + \cos^2(\theta)}{1 + \sin(\theta) - i\cos(\theta)}\\&= \frac{\sin^2(\theta) + \cos^2(\theta) + \sin(\theta) + i\cos(\theta)}{1 + \sin(\theta) - i\cos(\theta)}\\&= \frac{1 + \sin(\theta) + i\cos(\theta)}{1 + \sin(\theta) - i\cos(\theta)}\end{align*}$

(b)
$\begin{align*}\frac{1 + \sin\left (\frac{\pi}{n} \right ) + i\cos\left (\frac{\pi}{n} \right )}{1 + \sin\left (\frac{\pi}{n} \right ) - i\cos\left (\frac{\pi}{n} \right )} &= \sin\left (\frac{\pi}{n} \right ) + i\cos\left (\frac{\pi}{n} \right )\\&= \frac{\left (\sin\left (\frac{\pi}{n} \right ) + i\cos\left (\frac{\pi}{n} \right ) \right ) \left ( \cos\left (\frac{\pi}{n} \right ) + i\sin\left (\frac{\pi}{n} \right ) \right )}{\cos\left (\frac{\pi}{n} \right ) + i\sin\left (\frac{\pi}{n} \right )}\\&= \frac{i}{\cos\left (\frac{\pi}{n} \right ) + i\sin\left (\frac{\pi}{n} \right )}\\ \end{align*}$

$\begin{align*}\left (\frac{1 + \sin\left (\frac{\pi}{n} \right ) + i\cos\left (\frac{\pi}{n} \right )}{1 + \sin\left (\frac{\pi}{n} \right ) - i\cos\left (\frac{\pi}{n} \right )} \right )^n &= \left (\frac{i}{\cos\left (\frac{\pi}{n} \right ) + i\sin\left (\frac{\pi}{n} \right )} \right )^n\\&= \frac{i^n}{\cos\left ( \pi \right ) + i\sin\left ( \pi \right )}\\&= -i^n\\ &= -1\\\left (1 + \sin\left (\frac{\pi}{n} \right ) + i\cos\left (\frac{\pi}{n} \right )\right )^n &= -\left (1 + \sin\left (\frac{\pi}{n} \right ) - i\cos\left (\frac{\pi}{n} \right ) \right )^n\\\therefore \left (1 + \sin\left (\frac{\pi}{n} \right ) + i\cos\left (\frac{\pi}{n} \right )\right )^n &+ \left (1 + \sin\left (\frac{\pi}{n} \right ) - i\cos\left (\frac{\pi}{n} \right ) \right )^n = 0\end{align*}$

damn just realised it is supposed to be a plus

oh silly me, cis(pi) = -1 – fixed.

bleakarcher · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Why would you give him that question on an mx2 marathon? Are you studying university maths already? Also, Carrotsticks, Im scanning my solution. In part b)it should have been much shorter but i cbf rewriting.

SpiralFlex · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

On my iPhone guys so I can't do much. When I get home at around six or seven I will post something cool!

bleakarcher · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Here.

Carrotsticks · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

hayabusaboston said:
Consider T(subscript (a))(z)=z/(1+az). Show that T(subscr.(a))(z) constitutes a transformation group by examining the composition T(subcr.(b))(T(subscr.(a))(z))

Go Carrotsticks, try and prove this. I can give you some hints if you want./

Is this what you mean?

$$Consider $ T_a(z)=\frac{z}{1+az} \\\\ $Show that $ T_a(z) $ constitutes a transformation group by examining the composition $ T_b \circ T_a(z)$

hayabusaboston said:
It's conformal field theory. Want to see if Carrotsticks knows. He's at Uni after all.

No, it is not 'conformal field theory'.

It is higher-level algebra, specifically isometries within the Euclidean plane.

Please don't try to sound intelligent by posting up such questions that you most likely just pulled out from the internet. You really are making a fool of yourself.

Carrotsticks · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

hayabusaboston said:
Yes that is what I mean, if the little circle indicates a multiplication. It is part of conformal field theory, specifically the two dimensional conformal group. Physics is the application of mathematics. I find that quite offensive, you think I pulled this off the internet. Look it up. You wont find it.

You are mistaken.

The 'little circle' is the universal symbol for 'function composition'.

Your originally posted question was "composition T(subcr.(b))(T(subscr.(a))(z))"

Then you assert that the 'little circle indicates a multiplication'.

This is what the little circle (in the definition of composition) means:

$f \left (g(x) \right) = f \circ g(x)$

Going back on track, good work bleakarcher and rolpsy!

However bleakarcher, your method was quite long (and admittedly very illegible as you reached the bottom of the page).

rolpsy · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

hayabusaboston said:
How long does this take you in LaTex?

not long

rolpsy said:
$\begin{align*}\textup{(a)} &\textup{ It is given that } A,B>0 \textup{ and } n \textup{ is a positive integer.}\\i. &\textup{ Find } \frac{A^{n+1} - A^{n}B + B^{n+1} - B^{n}A}{A-B}, \textup{ and deduce }\\& A^{n+1} + B^{n+1} \geq A^{n}B + B^{n}A.\\ii. &\textup{ Using } i, \textup{ show by mathematical induction that}\\&\left ( \frac{A+B}{2} \right )^n \leq \frac{A^n + B^n}{2}.\end{align*}$

this will help with part i.

$\\A,B>0\\\textup{Note: } x^n \textup{ is increasing for } x>0, \textup{ i.e. if } A>B \textup{ then } A^n > B^n.\\\begin{align*}A>B &\Rightarrow A - B > 0, A^n - B^n > 0 \\&\Rightarrow (A-B)(A^n - B^n) > 0\\\\A<B &\Rightarrow A - B < 0, A^n - B^n < 0 \\&\Rightarrow (A-B)(A^n - B^n) > 0\\\\A=B &\Rightarrow (A-B)(A^n - B^n) = 0\end{align*}\\\therefore (A-B)(A^n - B^n) \geq 0$

bleakarcher · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
You are mistaken.

The 'little circle' is the universal symbol for 'function composition'.

Your originally posted question was "composition T(subcr.(b))(T(subscr.(a))(z))"

Then you assert that the 'little circle indicates a multiplication'.

This is what the little circle (in the definition of composition) means:

$f \left (g(x) \right) = f \circ g(x)$

Going back on track, good work bleakarcher and rolpsy!

However bleakarcher, your method was quite long (and admittedly very illegible as you reached the bottom of the page).

Yeh, I had a lot of unnecessary working at the bottom of the page since i didnt know what to do until you told me that n=4k, k integral.

Carrotsticks · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
Yeh, I had a lot of unnecessary working at the bottom of the page since i didnt know what to do until you told me that n=4k, k integral.

Again, my apologies for causing you grief by forgetting about that little bit of crucial information.

SpiralFlex · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:
not long

this will help with part i.

$\\A,B>0\\\textup{Note: } x^n \textup{ is increasing for } x>0, \textup{ i.e. if } A>B \textup{ then } A^n > B^n.\\\begin{align*}A>B &\Rightarrow A - B > 0, A^n - B^n > 0 \\&\Rightarrow (A-B)(A^n - B^n) > 0\\\\A[B] 0\\\\A=B &\Rightarrow (A-B)(A^n - B^n) = 0\end{align*}\\\therefore (A-B)(A^n - B^n) \geq 0$

I loved this question!

I recall this question from a HSC paper. One of the earlier years.

rolpsy · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
I loved this question!

I recall this question from a HSC paper. One of the earlier years.

1984 I think – first year with standard integrals

bleakarcher · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

no worries carrotsticks.

Carrotsticks · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Find the radius and centre of the circle that describes the locus of:

$Arg(z+2)-Arg(z-4)=\frac{\pi}{3}$

rolpsy · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

centre: (1,√3)
radius: 2√3.

$\\\textup{Let }z_{1}=3\left (\cos\left (\frac{\pi}{3} \right )+i\sin\left (\frac{\pi}{3} \right )\right ) \textup{ and } z_{2}=4\left (\cos\left (\frac{\pi}{6} \right )+i\sin\left (\frac{\pi}{6} \right )\right ).\\\\\textup{Determine the smallest integral } n \textup{ for which } \left ( \frac{z_{1}}{z_{2}} \right )^{n} \textup{ is real}.$

deswa1 · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

n=6 (assuming n=0 isn't allowed).

Given that w is the complex root of z^3=1 with smallest positive argument, evaluate:
(1-w)(1-w^2)(1-w^4)(1-w^8)

SpiralFlex · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Also smallest positive integral value.

SpiralFlex · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Level I

Level II

Level III

TBA tomorrow when I am bothered.

Carrotsticks · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Level I

Level II

Level III

TBA tomorrow when I am bothered.

This is a good idea. Having difficulty 'levels' when posting up questions is a good way to cater for different skill levels on the forums.

However, I suggest you co-ordinate the colours with the level of difficulty.

ie: Green = Easy, Blue = Medium, Red = Hard.

Black = Destroy.

SpiralFlex · Jan 14, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
This is a good idea. Having difficulty 'levels' when posting up questions is a good way to cater for different skill levels on the forums.

However, I suggest you co-ordinate the colours with the level of difficulty.

ie: Green = Easy, Blue = Medium, Red = Hard.

Black = Destroy.

It's always black when it comes to negative. =.=

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