Carrotsticks' Extension 2 HSC 2018 Solutions + Memes (1 Viewer)

Drdusk

π
Moderator
hey carrot what do you think e4 cutoff would be?

fan96

617 pages
If you expand by only keeping the imaginary part, then at least you essentially only have to write four terms for the expansion part.
The question said "Show that" - doesn't this mean we would basically have to perform the whole expansion in order to do the "showing"?

Honestly if we were just required to expand $\bg_white (\cos x + i \sin x)^8$ it wouldn't be that bad, but it is also necessary to express $\bg_white \cos^6 x$ and $\bg_white \cos^4 x$ in terms of $\bg_white \sin x$, which require even more expansions...

Edit: just read your other reply, that would have been a nice time saver. I wish I thought of that.

InteGrand

Well-Known Member
The question said "Show that" - doesn't this mean we would basically have to perform the whole expansion in order to do the "showing"?

Honestly if we were just required to expand $\bg_white (\cos x + i \sin x)^8$ it wouldn't be that bad, but it is also necessary to express $\bg_white \cos^6 x$ and $\bg_white \cos^4 x$ in terms of $\bg_white \sin x$, which require even more expansions...

Edit: just read your other reply, that would have been a nice time saver. I wish I thought of that.
$\bg_white \noindent I don't know what the HSC marking standard is, but in my opinion we shouldn't need to show the real parts (essentially because they are completely irrelevant to us and we (should!) know how to find imaginary parts of expansions like this without calculating the real part). We simply note that \sin 8\theta = \mathrm{Im} \left((c + is)^{8}\right), and then what I would do (if you wanted to show justifications) is write that 8-th power as \sum\limits_{k=0}^{8}\binom{8}{k}c^{8-k}i^{k}s^{k}. From this, we can see that the imaginary part comes from the terms with k=1,3,5,7 (since i^{k} is real if k is even and i^{k} is imaginary if k is odd), whence we can write down the imaginary part as the expression they gave us (which is obtained by taking just the k=1,3,5,7 terms from the sum).$

$\bg_white \noindent Anyway, don't worry too much now if you didn't spot these timesavers, it can be hard to spot them in exam conditions. Good luck with your remaining exams (everyone)!$

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kwonjiyong

New Member
I think there's an error with the trig integration
it should be 2 root3 not 2/root3
yeah i thought that too..

kwonjiyong

New Member
nah the resulting integral should be root 3 tan inverse root 3???? not one on root 3 tan inverse root 3?

kwonjiyong

New Member
what would be the e4 cutoff ???? carrotsticks? anyone?

stesoo

New Member
nah the resulting integral should be root 3 tan inverse root 3???? not one on root 3 tan inverse root 3?
if you're talking about the t sub question, im pretty sure carrotstick's solution is correct

kwonjiyong

New Member
if you're talking about the t sub question, im pretty sure carrotstick's solution is correct
ohhhh okay yep no worries LOL i was remembering wrong

1729

Active Member
lol
im also interested in what the cutoff would be
possibly anywhere between 68 and 74 based off the large majority of people finding the exam very difficult but who knows there could be a lot of quiet kids that may have found it easy

Carrotsticks

Retired
I've updated the solutions. Thanks everybody for your input.

Also please like this meme so I can feel validated as an oldie who can still connect with students.

Smash that like button, remember to subscribe, and click on the bell notification so you don't miss out.

Carrotsticks

Retired
Hey @Carrotsticks, because it was a 'find' question for Q16bii), I chucked one of these ones to find the ratio lmao

https://www.dropbox.com/s/thzsne7o0afhc0a/Solution to Q16b).png?dl=0

Surely they wouldn't penalise this?! The co-ordinate axes could quite easily be generalised via a linear transformation.
This is actually the best way of doing these. Regarding penalty, I actually cannot say for sure, but I am reasonably confident it will be okay. Amongst endless non-attempts, they see someone have a different solution. It will get given a lot of attention for sure, and hopefully those people know the maths behind it. I think it'll be accepted.

@other people reading this, in general any system that's been tilted you can 'untilt' and work the ratio from here. For more reading look up "Affine Transformation". Heaps of past HSC questions on things like this.

1039213

New Member
2 things,

Looking for a copy of the paper, is there anything anywhere?

With the rotation of the conic point thingo, i made it into a complex number and multiplied by i then converted it back. It got me the answer so I hope I'm not wrong.

raghib483

New Member
This is actually the best way of doing these. Regarding penalty, I actually cannot say for sure, but I am reasonably confident it will be okay. Amongst endless non-attempts, they see someone have a different solution. It will get given a lot of attention for sure, and hopefully those people know the maths behind it. I think it'll be accepted.

@other people reading this, in general any system that's been tilted you can 'untilt' and work the ratio from here. For more reading look up "Affine Transformation". Heaps of past HSC questions on things like this.
cheers bruv!

tbh I used to do stuff like this all the time in my HSC before I knew what affine transformations were. At SB the teachers were pretty chill about hack solutions, but I was often quite suspect if they were like that as markers. It was mostly out of sheer laziness, I'd just look at a length and be like 'Nah fk you imma let you be one unit' (3d trig <3). Or look at a point and go 'nah I don't like where you are, gtfover here'.

I was super psyched when I saw you put this idea to use in your 4U 2013 BOSTES transformation in Q16b)-d), but ofc you made em prove it lmao.

MatISrandom

New Member
Real talk; really disappointed that the difficulty of HSC papers (even this years one) have gone done drastically over the years :'(

This 4u paper was a meme

boredofstudiesuser1

Active Member
This is coming from someone who skipped anything that looked like mechanics, conics or any other question that required heaps of thought.
But I reckon there were a lot of easy marks in there. I haven't done past papers so I don't know if it's common, but yeah without studying I reckon you could get a solid 40 or 50 (not sure if this is usually the case)
But yeah, the hard stuff was REALLY hard

Lolihentai

New Member
u guys reckon a low 60 could make an e4? I need closure lol