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olympiad questions might require quite a bit of knowledge outside HSC (hence alot of irrelevant material), and the average time to solve each is around 40 or so minutes? I don't recommend that for practice.

Precisely, the general formula isn't that long

it's probably some model for price of something and income mouse: where's the question?

2^n - 1 are mersenne primes and 2^n +1 are fermat primes. reminds me of a question (B session question for anyone who knows what that is): find all primes in the form 2^n + 1 where n is a fibonnaci number.

quartics aren't that scary, I think turtle meant half a page :P Cool poster turtle, where did you get it ?

depends on the school though, if you're at an average comprehensive highschool, these marks quite marginal. It's still early, but you need to lift these marks

part a probably wants you to derive the acceleration

Hmm.. none of above.. just have a habit of checking up message boards before I sleep

the bash: the asymtyopes are: y = x/2 and y = -x/2 the gradient at p is 1/(2sinA) the tangent will therefore be (x-2secA)/(2sinA) = y - tan(A) x - 2secA = 2ysinA - 2*(sinA)^2/cos(A) x - 2[(1-(sinA)^2)/(cosA)] = 2ysinA x - 2cosA = 2ysinA solving this for x simultaneously...

11!/(2!)^3 using all the letters.. that's not telling you the Ms are different

My advice: Never look at solutions before you finish a question. 7 integration drills + 3 differentiation question + 10 algebra drills per day. (skip if you are good) Don't be obsessed with papers but instead, do textbook questions -> cambridge and coroneous. work through them carefully...

should've? he can do it any way he likes. And no.. not all of us to go/use to go to james ruse or selectives e.g. me: parramatta high.

pcx: *cough*... back in the old days when I learnt this, you had 3 trig tables, 2 log tables and 2 anti log tables. you should be content with the fact that you have a calculator.

omg. 3cos(x) + sqrt(3)sin(x) = (2*sqrt(3))*[ (sqrt(3)/2)*cos(x) + (1/2)sin(x) ] = 2sqrt(3)*[cos(pi/6)cos(x) + sin(pi/6)sin(x)] = 2sqrt(3)*[cos(x - pi/6)] = 2sqrt(3)*[cos(x + 11pi/6)]

ryan: using demoivre for this is superfluous.

you can pick any thing not less than N^2 to be M, N^2 is the smallest one which works though

You missed 'trivial'

hmm dunno.. this does though: to prove as (x-> inf) sqrt(x) -> inf we only need to show that for any given N>0, there exist M such that whenever x>M, sqrt(x) > N and M = N^2 fulfills this.

One size fits all g(x) = (1/2)[ f(x) + f(-x) ] h(x) = (1/2)[ f(x) - f(-x) ] works for any f(x)

q1.) Q-> circle, centered at 3+i, radius 2k. To help you see this consider the locus of R = w = 2z then since Q = R + 3+i.. it's just r translated. q3.) as Z describe the y axis, z=ki for some real k. ik(w+1) = (w-1) 1+ik = (1-ik)w w= (1+ik)/(1-ik) w= (1-k^2 + 2ik) /(1+k^2)...