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

#### Carrotsticks

##### Retired
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.

View attachment HSC 2018 Extension 2 Solutions.pdf

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##### -insert title here-
Question 16 is a meme.

##### Member
The entire paper was a fat meme.

#### jacqt

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

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#### peter ringout

##### New Member
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/obrxl4z3gz7pebo/HSC 2018 Extension 2 Solutions.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

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#### budgetjackiechan

##### New Member
that fking question wasted so much of my fkin time i coulda done q16 if it wasnt for that fukin stupid long fk

#### fan96

##### 617 pages
I really didn't like that question...

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

#### Q16slayer

##### New Member
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/obrxl4z3gz7pebo/HSC 2018 Extension 2 Solutions.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

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#### bruh00

##### New Member
Any guesses on the e4 cutoff?

##### -insert title here-
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

#### aa180

##### Member
For your solution to the last question of the paper, although not impactful to the proof, you mistakenly represented $\bg_white \overline{\alpha} - \alpha$ as $\bg_white -2\text{Im}(\alpha)$ when it really should be $\bg_white -2i\text{Im}(\alpha)$. The point I'm making is that, by definition, $\bg_white \text{Im}(\alpha)\in\mathbb{R}$, 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

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#### fan96

##### 617 pages
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...

#### InteGrand

##### Well-Known Member
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.

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

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

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

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

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#### Carrotsticks

##### Retired
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?

For your solution to the last question of the paper, although not impactful to the proof, you mistakenly represented $\bg_white \overline{\alpha} - \alpha$ as $\bg_white -2\text{Im}(\alpha)$ when it really should be $\bg_white -2i\text{Im}(\alpha)$. The point I'm making is that, by definition, $\bg_white \text{Im}(\alpha)\in\mathbb{R}$, 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.

##### -insert title here-
Did it really have to be to the eighth power though?

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

idk

#### thewildfrog

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

#### Q16slayer

##### New Member
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