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

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
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
 
Last edited:

jacqt

New Member
Joined
Feb 6, 2017
Messages
10
Gender
Undisclosed
HSC
2019
Does anyone by any chance know how many marks were allocated to each part of each question in the paper?
 

peter ringout

New Member
Joined
Sep 5, 2017
Messages
9
Gender
Undisclosed
HSC
N/A
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
 
Last edited:

fan96

617 pages
Joined
May 25, 2017
Messages
543
Location
NSW
Gender
Male
HSC
2018
Uni Grad
2024
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
Joined
Sep 23, 2018
Messages
5
Location
HSCland
Gender
Male
HSC
2018
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
 
Last edited:

aa180

Member
Joined
Sep 21, 2016
Messages
55
Location
Euclidean 3-space
Gender
Male
HSC
2012
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
 
Last edited:

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
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



 
Last edited:

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
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.
 

thewildfrog

New Member
Joined
Aug 17, 2018
Messages
1
Gender
Male
HSC
2018
I think there's an error with the trig integration
it should be 2 root3 not 2/root3
 

Q16slayer

New Member
Joined
Sep 23, 2018
Messages
5
Location
HSCland
Gender
Male
HSC
2018
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
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top