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Old MX1 had circle geometry (now gone), simple harmonic motion (now MX2), projectiles but without vectors, and lacked vectors, slope fields and some parts of DEs. New MX1 also has some curve sketching that was previously MX2.

Sums to products rule: \begin{align*} \cos\alpha + \cos\beta &= 2\cos{\left(\cfrac{\alpha + \beta}{2}\right)}\cos{\left(\cfrac{\alpha - \beta}{2}\right)} \\ \text{Put $\alpha = 50^\circ$ and $\beta = 70^\circ$:} \quad \cos{50^\circ} + \cos{70^\circ} &= 2\cos{\left(\cfrac{50^\circ +...

I am sad to see anyone who has actively contributed to this community leave. I know that it is inevitable that many will leave after the HSC, this site no longer being as valuable for support and their immediate goals. Some will stay and offer to others support similar to what they have...

PS: On the original question, whilst I agree that the Cambridge Advance books are excellent and certainly more than adequate, the Cambridge MX1 books do cover all of the Advanced content and the exercises on that content do include extra questions that further explore the Advanced content. For...

If an exam is well written, it will be structured to preclude the use of more advanced methods if that is the examiner's intent. One simple way is to specify a method, or to create a connection to a prior part with "Hence". If a method uses theory beyond the syllabus of a particular course, an...

Under the HSC timetable as it used to be, the MX2 and Advanced exams were held simultaneously.

Trebla has been sorting out the BoS MX2 Trial, coordinating the marking (and doing some of it too) and now we need to sort MX1. Site activity is high, with the extra work that brings, and he has a full-time job. There's even been some quite vile trolling. In that spare hour he has from 2 to 3...

He's being paid $3.50 and paying out $2.85 on average, so he is profiting

The expected payout on each game is \$1 \times 0.25 + \$2 \times 0.25 + \$3 \times 0.2 + \$4 \times 0.1 + \$5 \times 0.1 + \$6 \times 0.1 = E(X) = \$2.85 So, the profit on each game should average $3.50 - $2.85 = $0.65 So, expected profit on 200 games = 200 x $0.65 = $130

It's more an illustration of the difficulty that can be worked into a question by mixing topics, IMO.

How? At the half equivalence point, half of the acid or base has been converted to its conjugate, so the ratio of the two forms is 1:1. Put this into the formula for Ka and the two conjugate forms cancel, leaving Ka = [H3O+].

Plot the pH curve. If Vequiv is the volume required to reach the equivalence point, then at 0.5Vequiv, pH = pKa.

Solution posted above. You need to use the condition that was established in part (a) that links \sin\theta and \cos\theta and allows you to find the specific value of \theta.

For an AP, we need 2\cos\theta - \sin\theta = 2\sin\theta - 2\cos\theta\ \implies\ 4\cos\theta = 3\sin\theta\ \implies\ \tan\theta = \cfrac{4}{3} from which we can find \theta. For (b), we have that the common difference is d = 2\cos\theta - \sin\theta and so the next term after 2\sin\theta...

The 9th post has had a note added to note that there is a fault in it, as is posted above. The function should be F(x) = \int \cfrac{x + 1.5}{x^2 + 3x}\ dx The 10th post, M and N are two of the three points of intersection of the parabloa y=2x^2 and the circle x^2 + (y-1)^2 = 1. The points...

If you have specific examples of errors, post them and I will check.

If the middle term in the AP is a, and we take d > 0, then the three sides of the triangle are, in increasing order, a-d,\ a,\ \text{and}\ a+d. Now, which of these must be opposite the angle given as 120o? Draw a diagram and recognise that you can use the cosine rule...

You are correct for the integral given as written. For the question to work, however, the integral should be: F(x) = \int \cfrac{x + 1.5}{x^2 + 3x}\ dx

FYI, I've edited to add the 2015 paper as well.

Yes. The six 6th roots will form a regular hexagon. Further, (C) and (D) have modulus of 4, so can't be roots of i, while (A) and (B) both have modulus of 1 as required. arg(A) is 3pi / 4, so arg(A^6) = 9pi / 2, equiv to pi / 2. arg(B) is 5pi / 4, so arg(B^6) = 15pi / 2, equiv to 3pi / 2...