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Re: HSC 2013 3U Marathon Thread I believe him.

Re: HSC 2013 4U Marathon Yeah I know, I was just responding to you saying you think arctan(cos x) can't be expressed in terms of elementary functions. Anyway, all good :).

Re: HSC 2013 4U Marathon Arctan and cos are both elementary functions themselves lol. I still don't think it can be done though.

Re: HSC 2013 4U Marathon Pretty sure that integral can't be expressed in terms of elementary functions.

Re: HSC 2013 4U Marathon Semiclassical analysis, microlocal analysis and harmonic analysis mostly.

Re: HSC 2013 4U Marathon Differences to products implies that cos(nx) = 2*cos((n-1)x)*cos(x) + cos((n-2)x) applying this formula inductively with x = 1 under the assumption that cos(1 degree) is rational implies the assertion that cos(m) is rational for EVERY integer m. But we know that...

Re: HSC 2013 4U Marathon 1. You are on the right track. Forget about Euler's formula though. From the perspective of pretty much any MX2 level question it is just an alternative way of writing cis, and one that will cost you marks / potentially confuse you. 2. I didn't but you may find it...

Fun little exercise for those with some familiarity with convergence tests for integrals. (A first year university course in calculus should suffice). $For which real $\alpha,\beta\in\mathbb{R}$ with $\beta>0$ does the improper integral: \\ \\ $\int_0^\infty x^\alpha \sin(x^\beta)\, dx$\\...

Re: HSC 2013 4U Marathon $To sum the cosine series, take the real part of a suitable geometric series. You definitely don't need anything like Euler's formula. I will let you do this yourself, but I shall provide an evaluation of the integral that does not require it.\\ \\To evaluate the...

Nah, they wouldn't want that. The question is just wrong, missing a piece of information or something. (They clearly asked for the equation of the hyperbola.)

The locus of P? P is the fixed point (15,16), it does not vary. Remember, a locus is a set of points, not a set of cuves.

There have also been a couple of errors in his HSC solutions in recent years.

$ $\mathbb{Z}$ is the set of INTEGERS, not complex numbers.\\ You should end up with the answer to 3 being an infinitude of equally spaced collinear points on the line $y=2x/3$, the answer to 5 being the whole line $y=2x/3$, and the answer to 6 being the whole complex plane. $

The locus of what? The best one could do is eliminate either a or b and get a one-parameter family of hyperbolas passing through the given point.

Re: HSC 2013 4U Marathon I won't spoil the question yet for those who are working on it, but there are likely several approaches. A hint to doing it the way that I did is dividing the poly by x^2, (observing first that 0 cannot be a root of the poly). Having at least one real root does...

The key word is rectangular. There is only one rectangular hyperbola (up to reflection, translation, dilation, rotation etc), and its eccentricity is e=\sqrt{2}.

Re: HSC 2013 3U Marathon Thread Yes. Check the number of terms on the LHS, there are n+1, not n. A minor change needs to be made to your formula for the sum of a finite geometric series.

Re: HSC 2013 4U Marathon Yep although not directly relevant to the HSC syllabus they are good practice for problem solving and rigorous argument, especially for those who want to pursue mathematics further. You certainly don't have to be of modern olympiad standard to solve most of them.

Re: HSC 2013 3U Marathon Thread Err, not true. The last term on the LHS alone is greater than the RHS...

Re: HSC 2013 4U Marathon From the early 70s. Those problems are often doable by good mx2 students with no prior olympiad training and this one is no exception.