Thoughts on CSSA Mathematics Extension 2 Paper (1 Viewer)

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Davo_01

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Ahh ok sorry a misunderstanding...

How did you show that 4p^3=qr^2 in the roots of multiplicity I think I got most of the way there..
Oh thats one i dont remember perfectly but when you differentiate, i remember getting or something similar. From original equation you wanna make it interms of for example
Subbing gives 4p^3=qr^2 i think
Thats the method but i cant remember the equation
 
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panakap

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Yes I do remember something along those lines but I don't remember if I made it in terms of x^2
I'm pretty sure I did but for some reason I don't think I ended up getting the required answer, oh well.
 

Davo_01

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Yes I do remember something along those lines but I don't remember if I made it in terms of x^2
I'm pretty sure I did but for some reason I don't think I ended up getting the required answer, oh well.
making it in terms of x^2 is not neccesary but neater since you can substitute the equation without having to square root which looks a bit messy with the plus or minus sign, you end up squaring anyway though
 

panakap

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That's what I did, I squared later haha something seemed familiar.
Yes originally making it in terms of x^2 is much neater
 

panakap

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Ahh thought of the other question I'm unsure about.
It's the combinations question (A's B's and C's in certain order)

Anyone?
 

Davo_01

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Ahh thought of the other question I'm unsure about.
It's the combinations question (A's B's and C's in certain order)

Anyone?
my answer was 27 and 56.
I thought of it as 3 seperate blocks, first block has 2B and 1C second has 1A and 2C then third has 2A and 1B, this is the only arrangement that we obey the restriction and total arrangements is


for part ii, if we take the case there was 1B and 2C we get 27 again same concept. this leaves with the case with 3B or 3C at front which gives a total of 2 arrangements so the total is 56
 

Davo_01

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lol the concept of telescoping sum isn't explicitly taught in high school, but in uni.
Yes but many students will probably be familiar with it without even knowing that its called a telescoping sum.

i learnt it when i saw it quoted on the forums so i researched it.
 
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panakap

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I got 27 for part (i) (yay me, I thought of it in the same way)

However for part (ii) I got a slightly larger number I believe.
The first 3 has 6!/3!3! ways = 20... and then its this cubed for the other 2 sections of 3?
Explain your method because I don't think this was right :/
 

Davo_01

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I got 27 for part (i) (yay me, I thought of it in the same way)

However for part (ii) I got a slightly larger number I believe.
The first 3 has 6!/3!3! ways = 20... and then its this cubed for the other 2 sections of 3?
Explain your method because I don't think this was right :/
how did u get 6! since it has only 3 spots. You need to take case or atleast i did. case 1 when only 1b and 2c are on block 1 case 2 when there is 2B and 1C then when either all 3B or 3C are on first block
 

panakap

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For some reason I thought of the 3B's and 3C's as 6 total and work from there.

Ahh such a blunder yeh your method makes sense now :/
 

panakap

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Anyone have questions I can answer from the paper? (can be some of the easier stuff, eg. volumes, complex, integration, conics)
 

wagig

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Just saying, I got like 20 million or something for the permutations question hahahaha
 

dunjaaa

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Pretty easy exam, we had all of harder 3u and circular motion taken out. Man I really wanted to attempt those harder 3u questions especially the last one haha :D. The only question that probably had me stumped for a bit was the Q16 volume; getting the area of the slice, other than that was all goods. Better exam than last year for sure.
 

panakap

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We weren't supposed to have the harder 3u in our paper our teacher only taught us the inequalities section.

So w/out mechanics and harder 3u I would've found this paper easy #100% lol.

I didn't see last year's paper so I wouldn't know however Q16 was VERY DIFFCULT if you didn't know some fancy tricks.
 

pheelx3

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so... does anyone actually have the paper?
if so... PM me please?
 

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