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Yes I agree in that top students can be differentiated luck due to whether their strong topic appears in the last question. However keep in mind that Q8 is not always the hardest. However I would argue that at the very top this isn't the case as much. Were you continuing your last line?

Sample Question:

The STEP papers are papers made my Cambridge that are taken by students (worldwide?) in order to get into Cambridge Mathematics. The format of the paper is: 13 Questions: 8 Pure Mathematics 3 Mechanics 2 Probability/Stats. You have the opportunity to pick 6 out of these 13 questions, and...

do it do it do it

Re: HSC 2013 2U Marathon Simply logarithm laws: a^{\log_a(b)} = b

Re: HSC 2013 2U Marathon $Consider a$ \ \ n \ \ $sided regular polygon$ $There exists a point in this regular polygon, that is equidistant from all other points in the polygon, call this point$ \ \ O $The distance from$ \ \ O \ \ $to any corner of the polygon is given by$ \ \ r $Let$ \ \...

Re: MX2 Integration Marathon Nice work. $First denote from the previous part$ I_k = \int_0^{\pi/2} \cos x \sin^{2k}x \ dx = \frac{1}{2k+1} $ii)$ J = \int_{0}^{\pi/2} \cos x \cos^{2n}x \ dx = \int_0^{\pi /2} \cos x (1-\sin^2 x)^n \ dx $By the binomial theorem$ J = \int_{0}^{\pi /2}...

Re: HSC 2013 4U Marathon \sin(3\theta) = \sin \theta \cos(2\theta) + \cos \theta \sin(2\theta) \frac{\sin 3\theta}{\sin \theta} = \cos 2\theta + 2\cos^2 \theta \therefore \ \ \frac{\sin 3\theta}{\sin \theta} = 1+ 2\cos 2\theta \ \ \dots \ \ (1) $Let$ \ \ \theta_k = \frac{\pi 3^k}{3^n+1}...

Is circumference of a triangle the circumfrence of the circumcircle or the perimeter of the triangle?

Re: MX2 Integration Marathon Maybe, sorry for being annoying but for this one I didn't do the question that gave out, I just look at it, determine what strategy should be used, deem if its possible etc etc, then post it. (only for integrals) In this case, I would decide to do u=e^x, since du...

I'm playing CS GO at the moment, at the moment I usually just play competitive matchmaking on and off.

By using the quadratic formula, we can find that: \alpha = \frac{-5 + \sqrt{25-24k}}{2k} \beta = \frac{-5 - \sqrt{25-24k}}{2k} What do we get when alpha - beta? We get an inequality in terms of k. What is the solution to this inequality? One of them is k > 1.

The only thing you need to actually show is that k > 1. The upper bound ( k < 25/24) comes much more simply than you might think.

I haven't done the test yet and I don't plan to in the near future. Please stay on topic.

I have an IQ of 62 Also I fixed the link now

Re: MX2 Integration Marathon Yes it is definitely correct, but note that what you used in your working out was sin(u) not sin(2u) like you have now.

Re: MX2 Integration Marathon dx \neq \frac{-2 \sin u}{\cos 2u} \ du oo i geddit now, clever

Re: MX2 Integration Marathon Yep, nice work.

http://www.politicaltest.net/ Do the test and post your results. I fit 'Conservative Cosmopolitan'

How do you obtain the roots of this quadratic equation?