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First off, I would sub in the point given (the origin). Next, you differentiate and realise that y = 2x is the tangent at x = 0 Then, use the fact that the maximum turning point is at x = 1. In these questions, make sure you use all the information provided - one for each variable you have to find.

Brain fart. I'll edit cause braintic is 100% correct.

For the first part, it looks like you use de Moivre's theorem after converting to polar form. Polar form then is the z=cis\theta=\cos\theta+i\sin\theta form. I believe the second part means that w is a root of unity ie w^n=1 but it's rather confusing. Are you sure you've got it down right...

I have a friend who practically self taught himself 3U and 4U off Cambridge by doing practically every question and that "it's the best textbook to learn from". Then again, he's pretty darn smart and topped maths. That being said, I myself like the formatting and questions from Cambridge. You...

Roots of unity are when z^n=1 So in the question, you would relate it to z^3-1=0 and thus find the values for z^2+z+1

I found the paper and there's a diagram which clears some ambiguity here (2002 q6c). This means that you should integrate with respect to y from 0 to 4 since you're rotating about the y axis. Then, you want the volume between the curve and y=4: V=\int_0^4{\pi x^2}\,dy=\pi \int_0^4{2\sqrt{y}}\,dy...

With rate of change questions, you usually have to use the given rate (dV/dt) and multiply it with another rate you derive to find another rate in terms of t. In this case, you probably have to find dh/dt and integrate (the -kt+C hints at this). Remember that you can treat dy/dx as a fraction.

It might help if you split up the inequality and use the z = x + iy form, then find the intersection of the two. 1 \leq |z| gives you everything outside the circle with centre the origin. |z| \leq |z-i| gives you a straight line from (0 \leq) \sqrt{x^2+y^2} \leq \sqrt{x^2+(y-1)^2} 0\leq...

yep, you can find them here: https://drive.google.com/folderview?id=0B4D3wNobxSuvVmpsY0tES251dlk&usp=sharing ps if you want math help don't feel bad to inbox me or whatever :)

I'll scan it up if someone hasn't already. edit: after dinner :/ .

I'm a dingus, absolutely correct. Wish I got it in the test though haha.

Carrot, another one: question 14div) (the radius one) there's a wee error...

Concensus is that you can, since it's a 1marker (I wrote by symmetry just in case). Carrot, 11c) says factorise, not solve.

I dunno, I know two of the NSB guys and after a bit of a chat one of them realised he lost a couple marks. I think it was just a comparatively easier test to other years.

you add 25x^2 to both sides and factorise. I'm sure there's a better way though.

Re: HSC 2013 4U Marathon Hrmm. So from iii) with c = 0: $Let the roots be $ 2, \alpha, mi, -mi d^2 = bd(0) - eb^2 = -eb^2 $but a, b, c, d, e all integers and therefore real, $ d = \pm b\sqrt{-e} $and $ e = 2\beta m^2 $ must be even (since e is an integer, $\beta $ is real, m is real)$...

Re: HSC 2013 4U Marathon It's terribly inelegant IMO. Here's one: $For a monic cubic, the discriminant is $ {(\alpha - \beta)^2 (\beta - \gamma)^2 (\gamma - \alpha)^2 $Prove that there are three distinct real zeroes iff $ \Delta > 0

Re: HSC 2013 4U Marathon Alright, I got a funky answer building on Lanxal and Sy's stuff. http://i.imgur.com/8tUqegy.jpg I was happy when the top part (light blue) cancelled, but less so when the integral (green) sucked. Again, how to solve x - \tan x = 2\pi :/

Re: MX2 Integration Marathon Derp. Good thing you don't need to memorise stuff for maths, right. Another: $Evaluate $I=\int_{2+\varepsilon}^4\frac{\mathrm{d}x}{x^2 \sqrt{x^2 -4}} $Hence, find I as $ \varepsilon \to 0^+

Re: MX2 Integration Marathon IBP to get \left [ x\ln^n{x} \right ]_1^e - \int_1^e{\frac{nx\ln^{n-1}x}{x}\mathrm{d}x} + \int_1^e{n\ln^{n-1}x\mathrm{d}x} = e\ln^n{e} - \ln^n{1} = e == also, we all overthought the one from before since \sqrt{a\sqrt{b\sqrt{x}}} =...