The 2004 HSC - Mathematics Extension 2 Paper (1 Viewer)

[Dougie]

for those of us who really can't get close to 120 physically, these guys make it sound easy, hey! wish it was for an average person!

 

[BillyMak]

I looked at the induction question with about 30 seconds left (thought I'd come back to it cos it would be too much writing), then quickly proved true for n = 1 to try and scab a mark

 

[fantasia]

can someone explain the conics question? i completely left that out..

also what'd u guys get for the question about n students in two rooms?

 

[BillyMak]

Anyone care to estimate the state average, based on the previous state averages of past papers?

 

[derek_]

for those of us who really can't get close to 120 physically, these guys make it sound easy, hey! wish it was for an average person!

we can only aspire to these individuals, until then you and i are merely nobodies :>

 

[Rorix]

easy question 8
but I didn't or couldn't do Q2 (d) (complex number), Q4 c (conics), Q5 (b) (ii) (probability, and I dont know if I get part (i) right), Q6 b (ii)-(vi), Q7 (a) (i), Q 8 (a) (iii)-(v), Q8 (b) (i), (iii).
Noo.. that adds up to 32 marks!!!

in Q6 (b) (ii), is it meant to be verify B satisfies the equation?
coz I couldnt do it...

2d use either 2 eqi triagles or (the shitty way which I did) use that OBCA is a parallelogram
4c is just realising that everything isa straight line gradient +- e and e<1 therefore angle < pi/2
bi) is 2^(n-1), bii i got wrong im almost certain,

6b was just maniupulation, the trickiest bit was ii which needs u to use product rule and realise that dv/dt = x''.

7ai.....i assume u mean ii see above

8a is pretty much just basic geometry, the similar triangles you need a construction

8b is sec^2 for i and for iii the terms on the RHS from the sigma cancel

 

[Dougie]

we can only aspire to these individuals, until then you and i are merely nobodies :>

i'm just not going to read any suggeted answers or anything like that, i'm here to chat, not remember that exam!!

 

[BillyMak]

also what'd u guys get for the question about n students in two rooms?

I said that for the two rooms not to be empty, take 2 students and place 1 in each. There are n(n-1) ways of randomly selecting 2 students and 2! ways of putting them into the 2 rooms so neither is empty.

There are (n-2) students left, and they can choose any room since none is empty. The number of ways (n-2) students can choose 1 of 2 rooms is 2^n-2

:. total number of ways = n(n-1)2!2^n-2
= n(n-1)2^n-1

Correct me if I'm wrong....

 

[Shuter]

can someone explain the conics question? i completely left that out..

also what'd u guys get for the question about n students in two rooms?

I just guessed the first by putting (n+1)Cn.

The second one I drew a big table and I got 5 ways.

 

[Danny11]

i got 116 after counting. i think. should get 98 at least.

 

[mathock]

i thought it was a damn hard test... im not too confident with all these proofs, and i was hoping for some more marks on conics, polynomials etc.. there wasnt any factor theorem, no ellipses or hyperbolae to draw, minimal de moivre's theorem... i anticipated there to be a lot more marks that we had to use methods to work out, not all these f@#ken proofs... my aspirations for a 80% raw score were shattered within the 5 minutes reading time

 

[ben_ratus]

i got 2^(n-2)
i only left out 3 marks in the whole paper

 

[Dougie]

THose guy getting in the 100's arnt freaks they just study and do like 5 practice papers. Me only 2 got about 90 reckon out of 120. Hopefully e4. Do you reckon.

no worries.

 

[Danny11]

for the prop question i said that each person could choose either room. ie 2 arrangements. Since n people than using multiplication law 2^n arragements. But rooms cant be empty. 2 arrangements with a room emty. therefore answer is 2^n-2. I think

same here. part ii) from memory is 3^5 - (3 + 3*2^5) = 144 or something.

 

[mic_k]

what does a raw mark 80 - 90 go to?

 

[johnamos123]

can someone explain the conics question? i completely left that out..

also what'd u guys get for the question about n students in two rooms?

i said that there are n-1 combinations the rooms can be filled. 1:n-1, 2 :n-2 .... n-1: 1
= n-1 ways
and the students can be arranged in n! ways therfore arrangements are (n-1)n!

prob wrong but who cares cant change it now

 

[BillyMak]

I didn't study for 4-unit. Won't be studying for 3-unit.

Meh... I better learn that time payment stuff, I was away when we covered it and never caught up.

 

[mathock]

i said that there are n-1 combinations the rooms can be filled. 1:n-1, 2 :n-2 .... n-1: 1
= n-1 ways
and the students can be arranged in n! ways therfore arrangements are (n-1)n!

prob wrong but who cares cant change it now

yeah i got that too!! it seemed liek the logical answer, but now im thinking that their actual positions might be relevant... like sally tom and jeff in room 1 and josh and sophie in room 2 is a DIFFERENT way to sally m and SOPHIE in room 1 and josh and tom in room 2... if u get what i mean :-S

 

[Danny11]

i said that there are n-1 combinations the rooms can be filled. 1:n-1, 2 :n-2 .... n-1: 1
= n-1 ways
and the students can be arranged in n! ways therfore arrangements are (n-1)n!

prob wrong but who cares cant change it now

lol. thats a 3u problem. its permutation with replacement. formula should be in your books. a^n. a = total number choices, n = how many times you allowed to choose. its 2^n - 2, because there are two ways of having one room emptied, and since both rooms can't be emptied at the same time its -2.

 

[Archman]

THose guy getting in the 100's arnt freaks they just study and do like 5 practice papers. Me only 2 got about 90 reckon out of 120. Hopefully e4. Do you reckon.

i only did 2 as well.