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the trick is to not be intimidated because of my traumata i rarely get intimidated which massively helped with blitzing through maths

look at the sum and product of roots of the quadratic to show that they are both positive

1995 is more than a few years....

For a non-binomial method, divide both sides by 3^n to obtain (2/3)^n + 1 = (5/3)^n The left hand side has a base smaller than 1 and is a strictly decreasing function of n, and the right hand side is likewise strictly increasing. Thus, there can only be at most one solution, which is...

specifying the branches of the function is a pain so i'm leaving that as an exercise to the reader

I can't say. I know that for proving the irrationality of the nth root of 2, the proof of fermat's last theorem does in fact require some elementary irrationality results (including this very thing) so it would be circular reasoning. This would need the consultation of an expert in algebraic...

Meme Solution for 15. (d). If the result were true, this would be a non-trivial solution to Fermat's Last Theorem for every integer n>2. Hence, the result must be false for all n>2. The result is clearly false for n=2 by manually checking. QED ⬛

Here is the region for which Re(z) ≥ Arg(z)

Suppose T(2n) = H(m) for some integer m. 2n(2n+1)/2 = m(2m-1) 2n²+n = 2m²-m m+n = 2m²-2n² = 2(m-n)(m+n) Since m and n are positive integers, we can divide. 1 = 2(m-n) This is impossible, as integers cannot have a difference of ½. Therefore, the even triangular numbers are never equal to...

Only in the HSC. Try the BoS Trials Integrals :P

I will stress that you are always allowed to freely assume the information required by a later part in earlier parts, even if you have not actually shown those results. (At least when I did my HSC) You do not get penalised for not showing the previous parts if the question itself states them. I...

For MX1, part 14 d iv was free marks, as the previous parts gave you all the information needed to perform the inductive proof.

For MX2, part 14 b iv, the question itself gave you all the information needed to solve that part. It was effectively free marks. Similarly, part 16 c ii was completely free, as the question itself tells you what the reduced locus is (the interior of a disk).

yeah I would have taken more marking but idk I wasn’t assigned everything I was willing to take on

well it does help i got question 16 which only had 7 submissions, and most of question 13 were attempts consisting of less than one complete part (on average)

honestly we could have taken at least 50% more people due to no physical space requirements and digital marking is a breeze compared to marking a physical booklet

The grind never stops. (We have the drafting process running in the background behind our lives and all question ideas we come up with are thrown into the basket, which gives us a lot to work with in the months leading up to the exam)

lmao no questions asked this year

k metres per second squared. just slightly bad formatting

Wouldn't you like to know :P