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Marking guidelines are strict about what is required in an answer but much more flexible about the content used to meet the requirements. An "assess" question that does not make a judgement, based on discussed evidence, will not score highly. However, there are many questions where what the...

I showed that \left(r^3 + 3rs^2d - 3r - 1\right) + s\sqrt{d}\left(3r^2 + s^2d - 3\right) = 0. Recognise that this expression is the sum of a rational and an irrational term, and yet is also zero... \begin{align*} \text{Let} \qquad \left(r^3 + 3rs^2d - 3r - 1\right) + s\sqrt{d}\left(3r^2 +...

I was asked to expand on my comments with a more detailed answer, so here goes: Part (a) \begin{align*} x^3 - 3x - 1 &= 0 \qquad \qquad \text{. . . . . (*)} \\ \\ \text{Let $x = \frac{p}{q}$, where $p$ and $q$ are coprime and $pq \neq 0$, be a root of} \qquad ax^3 - 3x + b &= 0 \qquad...

This I wouldn't give 2 marks for, because Fermat's Last Theorem only establishes the result for integers n > 2... but yes, a succinct version that shows n = 2 separately should be sufficient.

My mistake, I mis-remembered the year. :)

@tywebb, the cross-product solution is mathematically flawless. The question did not include any direction that the cross-product method contravenes. I struggle to see any reason why it should be penalised - and especially if the solution included a statement as to what the cross-product is...

They could have just come from Victoria, the VCE specialist maths course includes vector cross products!

Ultimately, past papers are the best practice resource as the questions come with content crossing topics and without being pre-sorted to tell you where to start.

The solution has been posted, but on the strategy / approach front, looking at the result gives a few clues: first, you start with sums and end up with a product, which suggests the sums-to-products formulae second, the terms will need to be taken in pairs, so which of the three possible...

This thread actually illustrates how identity problems allow for multiple solutions, some quicker than others: \textbf{Theorem:} \qquad \sin{3x} - \sin{x} = 2\sin{x} - 4\sin^3{x} Proof 1: Conventional LHS / RHS approach \begin{align*} \text{LHS} &= \sin{3x} - \sin{x} \\ &= \sin{(2x + x)} -...

These concepts are spread between modules 2 and 3, technically. Net ionic equations are covered in the module 2 worksheets for KISS, for example. The reactions above are in Module 3 topic tests from Hegarty, rather than module 2. I wouldn't expect a question this hard until perhaps a Yr 11...

Standard reaction in general chemistry: acid (aq) + metal carbonate (s or aq) ----> salt (aq if soluble) + carbon dioxide (g) + water (l) For example, neutral species equation: 2HCl (aq) + Na2CO3 (s) -----> 2NaCl (aq) + CO2 (g) + H2O (l} As a net ionic equation: 2H+ (aq) + Na2CO3 (s) ----->...

It looks to me like you are trying to project onto the vector i + j to get onto the xy-plane. Any plane can be defined in terms of a vector normal to it - in the case of the xy-plane, one such unit normal vector is k. To project a vector v onto that plane, the projection of v onto the normal...

Part (b) is a LHs = ... and RHS = ... \begin{align*} \text{By definition:} \qquad \text{LHS} &= T_n \\ &= a^n + b^n \qquad \text{where $a$ and $b$ are the roots of $x^2 + x + 1 = 0$} \\ \\ \text{RHS} &= -T_{n-1} - T_{n-2} \qquad \text{provided $n > 2$ and $n \in \mathbb{Z}$} \\ &=...

To start, for Q 17... Put x = p / q into the second equation given and eliminate fractions. You should get that ap^3 - 3pq^2 + bq^3 = 0 where a, b, p, and q are all integers. Expressing this as ap^3 = 3pq^2 - bq^3 = q^2(3p - bq), what can we deduce about a, recalling that p and q share no...

For Q16(a)... \text{Suppose $A = 1$. Then, at the common root $x = \alpha$, $g(\alpha) = \alpha^2 + \alpha + 1 = 0$ . . . (*)} \begin{align*} \text{Hence,} \quad f(\alpha) = \alpha^3 + \alpha^2 + \alpha + B &= 0 \\ \alpha\left(\alpha^2 + \alpha + 1\right) + B &= 0 \\ \alpha \times 0 + B &= 0...

Very few HSC students will have heard of the uranyl ion, nor the precipitation described. It's a challenging question because it tests if you can deduce such things from the information given. The product given has nine acetate anions with a total 9- charge, so the cations must add to 9+...

Excess HCl liberates CO2, so C and O are two of the four elements in the mineral... and they are present as carbonate ions, CO32-, or as hydrogencarbonate ions, HCO3- (or possibly both)... but the latter options are only possible if hydrogen could be one of the four elements in the mineral. The...

Here is a more exam-like structuring of question 1, modified so that the result to prove is actually correct: \text{Let}\ \ z = 1 + \cos\theta + i\sin\theta \text{(a)} \qquad \text{Show that } |z| = 2\left|\cos{\frac{\theta}{2}}\right|. \text{(b)} \qquad \text{Express $z$ in modulus-argument...

Your teacher is wrong. Reliability is about consistency of results. Precision is about measurements, but also about reporting results. If you have data to 3 sig fig and you report your result to 6 sig fig, that is unjustified / false precision.