Question 16 is a meme.
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
Last edited by Carrotsticks; 30 Oct 2018 at 9:25 AM.
Question 16 is a meme.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
The entire paper was a fat meme.
Does anyone by any chance know how many marks were allocated to each part of each question in the paper?
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 by peter ringout; 25 Oct 2018 at 11:06 PM.
I really didn't like that question...
What is the point of wasting our time by forcing us to perform so many expansions?
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 by Q16slayer; 25 Oct 2018 at 10:18 PM.
Any guesses on the e4 cutoff?
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
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 by aa180; 25 Oct 2018 at 10:30 PM.
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 by InteGrand; 25 Oct 2018 at 10:39 PM.
I think there's an error with the trig integration
it should be 2 root3 not 2/root3
hey carrot what do you think e4 cutoff would be?
2018 HSC!!
English | Maths Ext 1 | Maths Ext 2 | Physics | Software Design and Development | IPT
ATAR aim: 96+ (Advanced Science(Physics Honours) and Comp Sci)
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 it wouldn't be that bad, but it is also necessary to express and in terms of , 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.
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