HSC 2012 MX2 Marathon (archive) (5 Viewers)

lolcakes52 · Jan 9, 2012

Re: 2012 HSC MX2 Marathon

Yes, infinite geometric series. I think I am just confusing myself.
This is from the unsw school mathematics comp, it is interesting and I would consider it harder 3 unit.

Screen shot 2012-01-09 at 6.53.45 PM.png

tywebb · Jan 9, 2012

Re: 2012 HSC MX2 Marathon

----------------------------------------------------------------------------------------------

I already posted up a question before but maybe people forgot, so here it is again:

$\ \newline(a)\text{ If }y=\textstyle{1\over x}+\sum_{n=1}^\infty{2x\over x^2-n^2}\text{ for }0<x<1\text{ use the fact that }\pi^2=\sum_{n=1}^\infty{6\over n^2}\text{ to show }\newline\ \newline{\hskip0.25in} \pi^2+y'+y^2=0\newline\ \newline (b)\text{ Hence show }y=\pi\cot\pi x\text{ for }0<x<1\newline\ \newline{\hskip0.25in}\text{ and }\cot x=\textstyle{1\over x}+\sum_{n=1}^\infty{2x\over x^2-\pi^2n^2}\text{ for }0<x<\pi\newline\ \newline (c)\text{ Find }\int_0^x(\cot t-{1\over t})\ \!dt\text{ and hence show }\newline\ \newline{\hskip0.25in}\sin x=\textstyle x\prod_{n=1}^\infty(1-{x^2\over\pi^2n^2})\text{ for }0<x<\pi\newline\ \newline (d)\text{ Let }x=\textstyle{\pi\over k}\text{ to show }\prod_{n=1}^\infty{(kn)^2\over(kn)^2-1}={\pi/k\over\sin(\pi/k)}\text{ for }k>1\newline\ \newline (e)\text{ Let }k=2\text{ to prove Wallis' Product, }\prod_{n=1}^\infty{(2n)^2\over (2n)^2-1}={\pi\over2}$

Carrotsticks · Jan 9, 2012

Re: 2012 HSC MX2 Marathon

tywebb, I have already done it, but I didn't want to post a solution because I knew it would be a crazy amount of typing

I think somebody else should try posting a solution, this is a thread aimed at MX2 students after all.

tywebb · Jan 9, 2012

Re: 2012 HSC MX2 Marathon

No. If you've got a solution, post it up. Don't wait for others to do it or we might be waiting forever.

You don't need to type it. Just scan and upload will do.

Trebla · Jan 9, 2012

Re: 2012 HSC MX2 Marathon

Funny how this thread is titled "2012 HSC MX2 Marathon" yet there are hardly any current HSC students participating. How about toning down the difficulty and topic coverage of the questions to encourage participation from HSC students who've only done a term of the course? I'd imagine the questions above would be very intimidating for most of them.

SpiralFlex · Jan 9, 2012

Re: 2012 HSC MX2 Marathon

You are right.

Note: People who have posted in this thread are NOT ALLOWED to answer this. I want some fellow extension 2 students to feel welcomed.

$a) Solve in mod arg form: z^4+1=0$

$b) Plot the roots on an Argand diagram.$

$c) Find the area of the quadrilateral formed by these roots.$

Trebla · Jan 9, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
$$(ii) In the case when n is odd, \textbf{\underline{shower}} by considering $(\sqrt{a+1} + \sqrt{a})T_m$ where m is even, or otherwise, that Tn can be written in the form $C_n\sqrt{a+1} + D_n \sqrt{a}$ where Cn and Dn are integers depending on a and n and $(a+1)C_n^2 = aD_n^2 + 1$

lol

deswa1 · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

$a) z^4&=-1$
$&=cis(\pi +2k\pi)$
$&=cis(\pi/4), cis(3\pi/4), cis(-\pi/4), cis(-3\pi/4)$

I don't know how to plot them on here but the third part:
c) All diagonals are equal length (all have modulus 1). They are perpendicular to each other (separation of pi degrees). Therefore the quadrilateral is a square with diagonal length 2 units. Each side has length root 2 so the area of the quad is 2 units squared.

Hooray, first time using latex

.
A few questions: How do you type text without it becoming bunched up and how do you align your equals signs? I tried following the guide but it wasn't working...

tywebb · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
$a) z^4&=-1$
$&=cis(\pi +2k\pi)$
$&=cis(\pi/4), cis(3\pi/4), cis(-\pi/4), cis(-3\pi/4)$

How do you type text without it becoming bunched up and how do you align your equals signs?

To align, type without spaces for [ tex ] and [ /tex ]:

[ tex ]\begin{aligned}a)\ z^4&=-1\\ &=\mathrm{c}i\mathrm{s}\; (\pi+2k\pi)\\ &=\mathrm{c}i\mathrm{s}\; (\pi/4),\ \mathrm{c}i\mathrm{s}\; (3\pi/4),\ \mathrm{c}i\mathrm{s}\; (-\pi/4),\ \mathrm{c}i\mathrm{s}\; (-3\pi/4)\end{aligned}[ /tex ]

to get

$\begin{aligned}a)\ z^4&=-1\\ &=\mathrm{c}i\mathrm{s}\; (\pi+2k\pi)\\ &=\mathrm{c}i\mathrm{s}\; (\pi/4),\ \mathrm{c}i\mathrm{s}\; (3\pi/4),\ \mathrm{c}i\mathrm{s}\; (-\pi/4),\ \mathrm{c}i\mathrm{s}\; (-3\pi/4)\end{aligned}$

Although your third line is actually incorrect and so it should be

$\begin{aligned}a)\ z^4&=-1\\ &=\mathrm{c}i\mathrm{s}\; (\pi+2k\pi)\\ {\color{BrickRed}{\therefore z}}&=\mathrm{c}i\mathrm{s}\; (\pi/4),\ \mathrm{c}i\mathrm{s}\; (3\pi/4),\ \mathrm{c}i\mathrm{s}\; (-\pi/4),\ \mathrm{c}i\mathrm{s}\; (-3\pi/4)\end{aligned}$

And suppose I want to type this sentence in latex without it being bunched up.

I'd do it like this:

[ tex ]\text{And suppose I want to type this sentence in latex without it being bunched up.}[ /tex ]

to get

$\text{And suppose I want to type this sentence in latex without it being bunched up.}$

deswa1 · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

Thanks for that

. How would you do this question:

I could do it but the second part took me a bit of time whilst in the answers they just wrote it down with no working (past school paper). Am I missing something ridiculously simple?

Carrotsticks · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
Thanks for that . How would you do this question:
View attachment 24089

I could do it but the second part took me a bit of time whilst in the answers they just wrote it down with no working (past school paper). Am I missing something ridiculously simple?

Show us the method that you used.

rawrence · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
Thanks for that . How would you do this question:
View attachment 24089

I could do it but the second part took me a bit of time whilst in the answers they just wrote it down with no working (past school paper). Am I missing something ridiculously simple?

Answer the second question using first

deswa1 · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

The answer is pi/4. The way I did it was find the gradient of the line passing through the origin that was a tangent to the parabola since that would have the minimum argument of z and then used the fact that the gradient of a line equals the tangent of the argument to get pi/4. There must be some easier way though isn't there?

lolcakes52 · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
The answer is pi/4. The way I did it was find the gradient of the line passing through the origin that was a tangent to the parabola since that would have the minimum argument of z and then used the fact that the gradient of a line equals the tangent of the argument to get pi/4. There must be some easier way though isn't there?

Did you sketch the graph? That sometimes gives the answer away especially in complex number questions. However I don't think there is a quicker algebraic method but I could be wrong.

bleakarcher · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

I considered that the arg(z)=tan^(-1)[y/x] and once I had the cartesian equation of the curve I just used calculus. Note that the curve is a parabola that is concave up and that the minimum clearly must occur in the first quadrant

Carrotsticks · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

The method I'm using below can be handy sometimes, because it gives me max AND min argument immediately. Of course there are other ways, but it's really just preference.:

$\text{The locus is defined to be } y=\frac{1}{4} (x^2+4) \\ \text{The line going through the origin and touching the curve, hence yielding the max/min argument, is in the form: }\\y=mx$

$\text{Substitute into one another to acquire a quadratic in terms of x and m: }\\ x^2-4mx+4=0$

$\text{Let the discriminant be equal to 0, since we are looking for tangents (which have 1 root with the curve) and solve for m:}$

$m=\pm 1\\ \\ \therefore tan\theta =\pm 1 \\ \\ \therefore \theta = \pm \frac{\pi}{4}$

$\text{Thus, minimum argument is } \frac{\pi}{4}$

EDIT:

bleakarcher said:
I considered that the arg(z)=tan^(-1)[y/x] and once I had the cartesian equation of the curve I just used calculus. Note that the curve is a parabola that is concave up and that the minimum clearly must occur in the first quadrant

How did you use calculus for this problem?

bleakarcher · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

arg(z)=tan^(-1)[y/x]=tan^(-1)[(x^2+4)/4x]. Let arg(z)=A.
Now, A=tan^(-1)[(x^2+4)/4x]
Differentiate with respect to x and set the derivative equal to 0 in order to find the point at which A is a minimum. You find that a minimum occurs at the point (2,2). Hence it follows that the minimum value for the arg(z)=tan^(-1)[2/2]=tan^(-1)[1]=pi/4

Carrotsticks · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
arg(z)=tan^(-1)[y/x]=tan^(-1)[(x^2+4)/4x]. Let arg(z)=A.
Now, A=tan^(-1)[(x^2+4)/4x]
Differentiate with respect to x and set the derivative equal to 0 in order to find the point at which A is a minimum. You find that a minimum occurs at the point (2,2). Hence it follows that the minimum value for the arg(z)=tan^(-1)[2/2]=tan^(-1)[1]=pi/4

It's good to see new methods pop up. I never thought of this. Nice work!

However, It is a wee bit slow. Another problem I can see coming is that you will need to actually show that it is in fact max/min, which can get quite tedious.

bleakarcher · Jan 10, 2012

Re: 2012 HSC MX2 Marathon

yeh lol, i couldnt think of anything else.

Carrotsticks · Jan 11, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
yeh lol, i couldnt think of anything else.

Better than being stuck and not thinking at all

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