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

[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.

View attachment HSC 2018 Extension 2 Solutions.pdf

 

[Paradoxica]

Question 16 is a meme.

 

[NinjaDatHSC]

The entire paper was a fat meme.

 

[jacqt]

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

 

[Carrotsticks]

 

[Carrotsticks]

 

[peter ringout]

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

 

[budgetjackiechan]

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

 

[Carrotsticks]

 

[fan96]

I really didn't like that question...

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

 

[Q16slayer]

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

 

[bruh00]

Any guesses on the e4 cutoff?

 

[Paradoxica]

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]

For your solution to the last question of the paper, although not impactful to the proof, you mistakenly represented as when it really should be . The point I'm making is that, by definition, , 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

 

[fan96]

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]

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.



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

 

[Carrotsticks]

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 as when it really should be . The point I'm making is that, by definition, , 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.

 

[Paradoxica]

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]

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

 

[Q16slayer]

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