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lol they use a excel spreadsheet i've been told. So they're going to get your two papers and see that one of the Q6 papers has already been typed in, and then just give the other one a look and go, fuck this kid, he gets 0. Nah you will get whatever you get /12 (prob 1 or 2 im guessing)

that's ok bro :) thanks anyway. Do you know anyone who does have a scanned version? Goodluck with the rest of all your exams too

THIS + Portal @ MathLinks Buchanan's post contains a plethora of trial and HSC exam papers. There is almost no better prep than the REAL thing

Namu, do you have a copy/link of the 2U 2009 paper? just wanna see it. thanks bro

Can someone scan the paper? i just wanna see it/

What do you want to study? and don't dwell on it bro

dude, i bet you surprise yourself. 95 is still very good man. I'm hoping for 100+, but i dunno, i might have been a bit vague in some explanations and lost a mark here and there. 95 will scale pretty high bro, the majority of state would have found it quite difficult

P'(x)=24x^2-24x+6 P'(x)>0 for increasing values 24x^2-24x+6>0 4x^2-6x+1>0 x=[6+-sqrt(36-16)]/8 (quadratic formula) =(6+-2rt5)/8 So they are the values (sorry i dont have a calculator in hand) then set up on a number line: <-------A-------B-------> Test points for each section ie. <A...

1e) int(pi/4-->0) cos@sin^2@ d@ Let u=sin@ du=cos@ d@ Change limits: when @=pi/4, u=1/rt2 when @=0, u=0 int(1/rt2-->0) u^2 du |(1/3)u^3| (1/rt2-->0) |(1/3)(1/2rt2)|-0 (1/6)rt2 Alternatively: This can just be done via inspection, |(1/3)sin^3@| (pi/4-->0) get the same answer Q3aii)...

lim(x-->0) sin3x/x x 3/3 (which is the same as multiplying by 1) therefore lim(x-->0) sin3x/3x x 3/1 Therfore 1 x 3 =3 -Solution isn't 1/3

nice work daryl, couldn't be a team player and post the paper could ya? Just itching to see what all the fuss is about lol

Could someone post the paper? Just curious

Have a look at the y=xarcsin(x) graph, it will help you understand the domain and range.

hey, the answer is no, it doesn't require proof, might benefit you seeing the proof once though just to help with understanding, but it's not testable to my understanding.

It should be "1+tan^2@" on the numeratorthen 2tan@/sec^2@2(sin@/cos@).cos^2@2sin@cos@=sin2@

Get a writer, it can be a student from year 11. Find your teachers number or somehow contact school authority and get them to help you out.

"A man shooter usually hits a bulls-eye 80% of the time. how many times must he fire his pistol in order that the probability of getting at least one bulls-eye exceeds 0.999?"P(bullseye)=0.8P(atleast one bullseye) = 1-P(no bullseye)therefore 1-(0.2)^n=0.999(0.2)^n=(1-0.999)n=ln(1-0.999)/ln(0.2)...

Hmm.. maybe 50/120??Band 6 is approx 96, so get an estimate off that

lol good one namu!Yeah i would agree with what namu said, i would say 2003 being the hardest. I think Q 10b) of 2000 was pretty hard also, put works out nicely once you get the idea of the q.2007 was also a pretty easy exam.

sent:)