Actually, 81 is the highest mark a candidate has ever achieved in the BoS trials for Maths Ext2. The previous record was 80 achieved five years ago in 2019.
Will reserve saying anything for Ext1 until those marks are finalised.
You don't need to "learn" about the equations of planes per se, but you shouldn't be surprised if something like this pops up the in HSC. The only thing to "learn" is really to broaden your familiarity with applications of dot product.
Actually, the word "plane" is used everywhere in HSC Maths e.g. "complex plane", "x-y plane" or "inclined plane". It's even in one of the vector topic dot points:
Seems pretty obvious that students are expected to know what the word "plane" actually means.
Even if they don't explicitly use the...
That just means that you are not required to recall conceptual content specific to planes. However, it doesn't mean that it is ruled out from the "application" side of the syllabus.
It is possible for an exam question to guide you into doing a vectors question on planes by derivation. For...
Actually, there are often parts in Q16 that are easy marks. For example, suppose a question has two parts where part (i) is a hard proof but part (ii) is hence deduce this using part (i). In many cases, part (ii) is quite easy so you can just skip straight to that to get some free marks.
Also...
Here are the results and solutions for the 2024 Mathematics Extension 2 BoS Trials! :)
Please also note that the question paper has been updated (which these solutions are aligned to) with some minor adjustments to a few questions. The updated file is attached in my original post of the papers...
Pretty much in agreement with this. The image appears to have been doctored. The join date of that account was actually today and yet the post was dated “yesterday”.
General advice is to take these “leaks” with a grain of salt and at your own risk.
In part (i), you are asked to show that if X is true then Y follows.
In part (ii), you need to show that X is actually true for this specific scenario, so that you can then deduce that Y follows.
In this case, you would show that w2n = 0 which is an integer that is divisible by 11 and holds...
As soon as you show that w2n = 0 in general for any palindrome, regardless of the value of n (within the positive integers that is), then it must be true for all positive integers of n.
By definition, a state government agency like NESA are not run for profit - if anything they're probably severely underfunded by the government.
Even then, I doubt they would make much "profit" from this because very few people in the state actually bother to seek access to their HSC scripts...
Confirmed with author that the intended solution was to take as given that w2n is divisible by 11 for all n so going to adjust wording accordingly. Hence, the outline of the solution would be:
From the assumption n=k
w2k = 11P and u2k = 11Q
Given w2k+2 = 11R then
w2k+2 - w2k = 11(R-P)
=> a2k...
What if you were not clear headed in that one single day on externals? If that was weighted 100% as you suggested then you won’t have internal marks to save you.
At least with internals if you stuff up one assessment, you have an opportunity to make up for it in another assessment or the externals.
Um it doesn’t quite work like that?
You don’t get “dragged down” if your internal marks are higher than the most of the cohort. The only time you get “dragged down” is if your internal ranks/marks are similar to the middle/lower end of your cohort (in which case your moderated mark reflects...
I think you've highlighted an interpretation of the question that was different to what was intended. I believe the intention was to suppose that w2n was divisible by 11 for all n, not for the same value of n as u2n. I will confirm with the original author if that was the intention. If confirmed...