# Thread: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

1. ## Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

I hated this paper. Typing it was a massive bitch to do compared to previous years. I'm also going to post some memes here too.

Usual stuff here everybody. Comment if you see errors but I think we should mostly be good. I've taken care to explain things as much as I can to aid understanding where needed.

HSC 2018 Extension 2 Solutions.pdf

2. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Question 16 is a meme.

3. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

The entire paper was a fat meme.

4. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Does anyone by any chance know how many marks were allocated to each part of each question in the paper?

7. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by Carrotsticks
I hated this paper. Typing it was a massive bitch to do compared to previous years. I'm also going to post some memes here too.

Usual stuff here everybody. Comment if you see errors but I think we should mostly be good. I've taken care to explain things as much as I can to aid understanding where needed.

https://www.dropbox.com/s/obrxl4z3gz...tions.pdf?dl=0
Hi all
In Q16 you attempt to use the converse of the equal intercept theorem. This converse does not hold. Just a few sketches will convince you that equal ratios of intercepts does NOT imply that the three lines are parallel. Fortunately it is given that the lines are parallel
Cheers

8. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by Carrotsticks
that fking question wasted so much of my fkin time i coulda done q16 if it wasnt for that fukin stupid long fk

10. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

I really didn't like that question...

What is the point of wasting our time by forcing us to perform so many expansions?

11. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by Carrotsticks
I hated this paper. Typing it was a massive bitch to do compared to previous years. I'm also going to post some memes here too.

Usual stuff here everybody. Comment if you see errors but I think we should mostly be good. I've taken care to explain things as much as I can to aid understanding where needed.

https://www.dropbox.com/s/obrxl4z3gz...tions.pdf?dl=0
Not really much of an error, but just an inquiry about Q16 bii
Let DE = x
DE/BC = BF/BC = 1/root2 (because BF/BC = FG/CA = 1/root2 by corresponding sides of similar triangles in same ratio)
therefore, DE = BF = x and so BC = root2 x
FC = BC - BF = (root2 - 1)x
ZE = FC = (root2 - 1)x (opposite sides of a parallelogram are equal)
so we get, YZ = DE - (DY+ZE) = DE - 2ZE = x - 2(root2 -1)x = (3-2root2)x
so YZ/BC = (3-2root2)x/(root2) x = (3 - 2root2)/root2
which i think is a quicker and better solution haha

12. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Any guesses on the e4 cutoff?

13. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by fan96
I really didn't like that question...

What is the point of wasting our time by forcing us to perform so many expansions?
they've finally run out of idea so they just pulled a giant expansion question

14. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

For your solution to the last question of the paper, although not impactful to the proof, you mistakenly represented $\overline{\alpha} - \alpha$ as $-2\text{Im}(\alpha)$ when it really should be $-2i\text{Im}(\alpha)$. The point I'm making is that, by definition, $\text{Im}(\alpha)\in\mathbb{R}$, so the Im function doesn't include i and so you have to add it explicitly if you want it to be included.

EDIT: Ignore what I said about it being negative lol

15. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

they've finally run out of idea so they just pulled a giant expansion question
Did it really have to be to the eighth power though?

Wouldn't three or four have been enough...

16. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by fan96
Did it really have to be to the eighth power though?

Wouldn't three or four have been enough...
If you expand by only keeping the imaginary part, then at least you essentially only have to write four terms for the expansion part.

$\noindent (Also for brevity, it would be wise to do something like define c\equiv \cos \theta and s \equiv \sin \theta.)$

Edit: just had a proper look at the question. Yeah it's a bit annoying, looks like it was done to make it worth as many marks as it was. I think though once you get that

$\frac{\sin 8\theta}{\sin 2\theta} = a_{1}c^{6} + a_{2} c^{4} s^{2} + a_{3} c^{2} s^{4} + a_{4} s^{6}$

$\noindent where the constants a_{1}, a_{2}, a_{3}, a_{4} have been kept track off, then sub. in c^{2} = 1-s^{2}, c^{4} = 1-2s^{2} + s^{4} and c^{6} = 1-3s^{2} + 3s^{4} - s^{6}, you can effectively go straight to the answer (since it's given to you) and say it was by grouping like terms.$

17. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by Q16slayer
Not really much of an error, but just an inquiry about Q16 bii
Let DE = x
DE/BC = BF/BC = 1/root2 (because BF/BC = FG/CA = 1/root2 by corresponding sides of similar triangles in same ratio)
therefore, DE = BF = x and so BC = root2 x
FC = BC - BF = (root2 - 1)x
ZE = FC = (root2 - 1)x (opposite sides of a parallelogram are equal)
so we get, YZ = DE - (DY+ZE) = DE - 2ZE = x - 2(root2 -1)x = (3-2root2)x
so YZ/BC = (3-2root2)x/(root2) x = (3 - 2root2)/root2
which i think is a quicker and better solution haha
I like this answer. I knew as I was doing it that I was going a bit round-about. I rushed a bit too much and ended with a longer solution. Would you be okay with me adding your answer in as an alternative?

Originally Posted by aa180
For your solution to the last question of the paper, although not impactful to the proof, you mistakenly represented $\overline{\alpha} - \alpha$ as $-2\text{Im}(\alpha)$ when it really should be $-2i\text{Im}(\alpha)$. The point I'm making is that, by definition, $\text{Im}(\alpha)\in\mathbb{R}$, so the Im function doesn't include i and so you have to add it explicitly if you want it to be included.

EDIT: Ignore what I said about it being negative lol
Nice spotting I will fix this.

18. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by fan96
Did it really have to be to the eighth power though?

Wouldn't three or four have been enough...
general expansion question xd

idk

19. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

I think there's an error with the trig integration
it should be 2 root3 not 2/root3

20. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by Carrotsticks
I like this answer. I knew as I was doing it that I was going a bit round-about. I rushed a bit too much and ended with a longer solution. Would you be okay with me adding your answer in as an alternative?

Nice spotting I will fix this.
thats completely fine! im happy you liked it, although i didnt manage to pull off this answer during the actual exam so it really is a shame. The time- limit is a complete bummer to conceptual questions :c

21. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

hey carrot what do you think e4 cutoff would be?

22. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by InteGrand
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 $(\cos x + i \sin x)^8$ it wouldn't be that bad, but it is also necessary to express $\cos^6 x$ and $\cos^4 x$ in terms of $\sin x$, 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.

23. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by fan96
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 $(\cos x + i \sin x)^8$ it wouldn't be that bad, but it is also necessary to express $\cos^6 x$ and $\cos^4 x$ in terms of $\sin x$, 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.
$\noindent I don't know what the HSC marking standard is, but in my opinion we shouldn't need to show the real parts (essentially because they are completely irrelevant to us and we (should!) know how to find imaginary parts of expansions like this without calculating the real part). We simply note that \sin 8\theta = \mathrm{Im} \left((c + is)^{8}\right), and then what I would do (if you wanted to show justifications) is write that 8-th power as \sum\limits_{k=0}^{8}\binom{8}{k}c^{8-k}i^{k}s^{k}. From this, we can see that the imaginary part comes from the terms with k=1,3,5,7 (since i^{k} is real if k is even and i^{k} is imaginary if k is odd), whence we can write down the imaginary part as the expression they gave us (which is obtained by taking just the k=1,3,5,7 terms from the sum).$

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

24. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

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

25. ## Re: Carrotsticks' Extension 2 HSC 2018 Solutions + Memes

Originally Posted by kwonjiyong
yeah i thought that too..
he just didnt rationalise the denom?

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