HSC 2016 MX2 Marathon (archive) (2 Viewers)

Trebla · Nov 2, 2015

Post any questions within the scope and level of Mathematics Extension 2. Once a question is posted, it needs to be answered before the next question is raised.

This thread is mainly targeting Q1-15 difficulty in the HSC. Any questions of Q16 difficulty or harder should be posted in the Advanced thread.

I encourage all current HSC students in particular to participate in this marathon.

To start off:

$Find the values of $n$ such that $(1-i)^n$ is purely real$

Drsoccerball · Nov 2, 2015

Re: HSC 2016 4U Marathon

Trebla said:
Post any questions within the scope and level of Mathematics Extension 2. Once a question is posted, it needs to be answered before the next question is raised.

This thread is mainly targeting Q1-15 difficulty in the HSC. Any questions of Q16 difficulty or harder should be posted in the Advanced thread.

I encourage all current HSC students in particular to participate in this marathon.

To start off:

$Find the values of $n$ such that $(1-i)^n$ is purely real$

$(1-i)^n = \sqrt{2}^n cis(\frac{-\pi n}{4})$

$\therefore n=4p $ Where P is an integer $$

Drsoccerball · Nov 2, 2015

Re: HSC 2016 4U Marathon

$i) $ Prove that $ cos(n\theta)= \frac{z^n +\frac{1}{z^n}}{2} $ and $ \frac{z^n -\frac{1}{z^n}}{2i} = sin(n\theta) $ where $ z= cos\theta+ i sin\theta$

$ii) $ Hence find $ \int_0^{2 \pi}{sin(nx)sin(2x)}dx$

Trebla · Nov 2, 2015

Re: HSC 2016 4U Marathon

Not sure if that second part is in scope of HSC...

Carrotsticks · Nov 2, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
$(1-i)^n = \sqrt{2}^n cis(\frac{-\pi n}{4})$

$\therefore n=4p $ Where P is an integer $$

Trebla said:
I encourage all current HSC students in particular to participate in this marathon.

Trebla said:
I encourage all current HSC students in particular

Trebla said:
current HSC students

Drsoccerball, I'm afraid that it's time to move on.

You are no longer one of them.

InteGrand · Nov 2, 2015

Re: HSC 2016 4U Marathon

Carrotsticks said:
Drsoccerball, I'm afraid that it's time to move on.

You are no longer one of them.

Well he encouraged all current ones to participate – he didn't ban non-current ones from participating!

Drsoccerball · Nov 2, 2015

Re: HSC 2016 4U Marathon

Carrotsticks said:
Drsoccerball, I'm afraid that it's time to move on.

You are no longer one of them.

< / 3

Trebla said:
Not sure if that second part is in scope of HSC...

It's very easy

Carrotsticks · Nov 2, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
Well he encouraged all current ones to participate – he didn't ban non-current ones from participating!

Please Integrand, allow me to be mean completely arbitrarily.

Trebla · Nov 2, 2015

Re: HSC 2016 4U Marathon

This is probably a general note across all the marathon threads. Whilst it is great to for non-HSC students to participate and have discussions about mathematical concepts, please don't overdo it at the expense of not giving current HSC students a chance to participate.

Drsoccerball · Nov 2, 2015

Re: HSC 2016 4U Marathon

Trebla said:
This is probably a general note across all the marathon threads. Whilst it is great to for non-current HSC students participate and have discussions about mathematical concepts, please don't overdo it at the expense of not giving current HSC students a chance to participate.

The roast is real...

Trebla · Nov 2, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
It's very easy

Nvm, just realised when I actually tried doing it.

Ekman · Nov 3, 2015

Re: HSC 2016 4U Marathon

Carrotsticks said:
Please Integrand, allow me to be mean completely arbitrarily.

Carrot is clearly trying to make himself look like a big boy in front of the 2016ers. Please do continue...

(inb4ban)

qwert73 · Nov 3, 2015

Re: HSC 2016 4U Marathon

Any textbooks/pastpapers which help because I have no idea how to do some of these from looking at them

calamebe · Nov 3, 2015

Re: HSC 2016 4U Marathon

I haven't done much integration, so I may have done something wrong, though it more or less looks right: https://m.imgur.com/HVTvkKl

Drsoccerball · Nov 3, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
$i) $ Prove that $ cos(n\theta)= \frac{z^n +\frac{1}{z^n}}{2} $ and $ \frac{z^n -\frac{1}{z^n}}{2i} = sin(n\theta) $ where $ z= cos\theta+ i sin\theta$

$ii) $ Hence find $ \int_0^{2 \pi}{sin(nx)sin(2x)}dx$

$I= \int_0^{2\pi}{sin(nx)(sin(2x)}dx$

$I= \int_0^{2 \pi}{\frac{z^n -\frac{1}{z^n}}{2i} \cdot \frac{z^2-\frac{1}{z^2}}{2i}}dx$

$I= \frac{-1}{4}\int_0^{2\pi}{(z^{n+2} +\frac{1}{z^{n+2}} -(z^{n-2}+\frac{1}{z^{n-2}}}))dx$

$I= \frac{-1}{4} \int_0^{2 \pi}{cos(n+2)x -cos(n-2)x}dx$

The rest is straight forward...

calamebe said:
I haven't done much integration, so I may have done something wrong, though it more or less looks right: https://m.imgur.com/HVTvkKl

So yes yours seems right.

Paradoxica · Nov 3, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
$I= \int_0^{2\pi}{sin(nx)(sin(2x)}dx$

$I= \int_0^{2 \pi}{\frac{z^n -\frac{1}{z^n}}{2i} \cdot \frac{z^2-\frac{1}{z^2}}{2i}}dx$

$I= \frac{-1}{4}\int_0^{2\pi}{(z^{n+2} +\frac{1}{z^{n+2}} -(z^{n-2}+\frac{1}{z^{n-2}}}))dx$

$I= \frac{-1}{4} \int_0^{2 \pi}{cos(n+2)x -cos(n-2)x}dx$

The rest is straight forward...

So yes yours seems right.

$Alternatively, one could use the fact that $z = e^{ix}$, take the logarithm of both sides and differentiate to obtain a substitution for $x$. But you know, that works too.$

$However, I suspect that may be unjustifiable given that $\log{z}$ is multi-valued over $\mathbb C $ (Calling in Carrotsticks to comment on the situation here!!!)$

Paradoxica · Nov 3, 2015

Re: HSC 2016 4U Marathon

Are the Dürer's Shell Curves, Devil's Curve, Conchoid, and Bicorn sketchable using Extension 2 techniques?

Drsoccerball · Nov 3, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
$Alternatively, one could use the fact that $z = e^{ix}$, take the logarithm of both sides and differentiate to obtain a substitution for $x$. But you know, that works too.$

$However, I suspect that may be unjustifiable given that $\log{z}$ is multi-valued over $\mathbb C $ (Calling in Carrotsticks to comment on the situation here!!!)$

$e^{ix} =cisx$ Is not a HSC result.

Paradoxica · Nov 3, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
$e^{ix} =cisx$ Is not a HSC result.

My teacher said it's allowed....
.-.

KingOfActing · Nov 3, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
My teacher said it's allowed....
.-.

It used to be in the curriculum but they took it out at some point, I think. I don't really know why they would, since $e^{ix} = cis(x)$ is just a different way of writing the exact same thing, and to a lot of students (myself included) Euler's Formula was the "OMG MATH IS COOL" moment (especially seeing the Taylor series derivation of it). I wouldn't use it in exams, just to be on the safe side, though.

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