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Here's my attempt. I've probably over-complicated it or am just plain wrong: Base case (n=0) – Trivial Inductive hypothesis – Assume true for some non-negative integer k: \text{i.e. } \exists p, q \in \mathbb{Z} : w_{2k} = 11 p \rightarrow u_{2k} = 11q \qquad (1) Inductive step – RTP true...

The above proof relies on the extension of calculus to complex numbers, which is not within the syllabus (although I'm sure most students won't hesitate in accepting it). The Cambridge ME2 Textbook provides a lengthier proof that is more within the confines of the syllabus. At any rate...

Your comment and a few recent articles by Derek Buchanan at 4unitmaths.com have both mentioned the beta and gamma functions as alternative, more efficient, approaches to certain questions. Furthermore, in his most recent article 'Alternative solutions to 2023 HSC Extension 1 Q14ci' Derek states...

To get this thread going again I thought I'd give one of these a go. Letting u = \tfrac{\pi}{2} - x: \begin{aligned}I &= \int_{0}^{\pi}\left(3\pi-\cos x-2x\right)\left(\frac{x}{1+\sin x}\right)^{2}dx \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(2\pi-\sin u + 2u...

Thanks for that, I might properly add the edits in later so it looks more polished

Yes you are quite right, I somehow thought 3^2 was 6 o_O

Here are my typeset solutions to this year's MX1 HSC exam. I wrote these in a great rush so there are probably mistakes lol

Lol. I can't type into my calculator. I did: 252\times4\times4=4032 Should be: 252\times24\times24=145152 as I had above.

oops forgot to arrange them around the tables! That would make it: 252\times (5-1)!\times(5-1)!=4032 I think that should be correct.

Choose 5 people to sit at the mahogany then let the rest sit at the oak: \binom{10}{5}=252

Technically they are not the "cube roots of unity" but simply the "cube roots".

2. |z|^{\frac13}=(\sqrt{1^2+1^2})^{\frac13}=\sqrt[6]{2} \frac13\textrm{arg}z=\frac13\arctan(1)=\frac\pi{12} So the "primary root" will be z=\sqrt[6]{2}\textrm{cis}\frac{\pi}{12} 3. Note that 1/3arg(z) can also take other values namely: \frac13\textrm{arg}z=\frac{\pi}{12}+\frac{2\pi}{3}\textrm{...

I vaguely remember this being a property of cubic polynomials with three distinct roots.... I think its because the vertex of a parabola (the first derivative) occurs at an x value which is the average of the x intercepts (which will correspond to the max and mins on the cubic). As the vertex of...

I think this is what it means. 1. 14\ddot{x} =- \frac65v \frac{dv}{dt}=-\frac3{35}v \int_{21}^V\frac{dv}{v} =-\int_0^{11}\frac3{35}dt\ \ (\textrm{where }V\textrm{ is the final velocity}) [\ln|x|]_{21}^V=-\left[\frac3{35}t\right]_0^{11} \ln\left(\frac{V}{21}\right)=-\frac{33}{35} \therefore...

\frac{F}{\ell}=k\frac{I^2}{d} \implies I=\sqrt{\frac{Fd}{\ell k}}\ \ \left(\textrm{where } k\textrm{ is } \frac{\mu_0}{2\pi}\right) So we want something that looks like y=\sqrt{x} making the answer D

My teacher told us that whenever a question requires you to provide a personal response/reading a strong personal voice should be adopted i.e. use "I" not "we". This really emphasizes that it is you who is making an informed response to the text you are studying.

This question relies on too many variables. It is impossible to say. I dont really see how saying "aim for rank 3 in English" will drive you to do any better. Just do your best!

Before every one of my maths exams I write a list of all the silly mistakes I have made in previous exams, class tests and even homework. This helps you to know what to look out for and make a conscious effort to avoid. This has helped me a lot and now I consistently get over 95% in 4U.

I think you should keep it. Only consider dropping it if, after a few assessments, you are doing badly in it and you think that the time spend studying it could be better used to study other stronger subjects. It is probably good having a backup in case one you stuff up one of your other...

One of my friends at school did the test. He received 87% and got into A2. He said that you need at least 90% for A1. Topics included polynomials, combinatorics and physical applications of calculus. I don't think there was any integration from what he told me.