Last set of 4u I'll be doing.
Multiple choice
(https://i.imgur.com/hHocnBG.png)
Q13 a) picture from tywebb below
Explanation: Basically consider the modulus and argument separately.
- Modulus: Since a+bi has modulus greater than 1, i.e. |a+bi| > 1, we also know that |a+bi|1/4 > 1, therefore they should lie outside the unit circle. However, the fourth root of a number greater than 1 will also be smaller than the original number. Therefore, the fourth roots would also be closer to the origin, than a+bi itself is.
- Argument: In general, if all values for \(\arg(a+bi)\) take the form \(\arg(a+bi) = \theta + 2k\pi\) for some integer \(k\), then all values for \( \arg((a+bi)^{1/4})\) take the form \( \arg((a+bi)^{1/4}) = \frac{\theta}{4} + \frac\pi2 k\). This can be proven using the standard algebraic method of explicitly computing fourth roots.
Therefore, we know that the first value for \( \arg((a+bi)^{1/4})\) will be one quarter of the argument of a+bi, which gives the first root. After that, we see that the other values require us to add \( \frac\pi2\) to the argument to obtain all of the other roots.
(https://i.imgur.com/flI1gMz.png)
Q14 b)
(https://i.imgur.com/8OYkDvc.png)
Q15
(https://i.imgur.com/UO5MYDz.png)
Q16 (a)+(c)
Some comments regarding the graph for part c) :
- Asymptotic behaviour occurs as Arg(z) approaches -pi/2 from below. Note that the entire ray corresponding to Arg(z)=-pi/2 belongs in the required region.
- With enough time spent brainstorming, you'll soon realise that none of the 2nd quadrant matters, and all of the 4th quadrant matters. These are the two easier quadrants to address in the graph. For example, in quadrant 2 specifically, Arg(z) is always positive whereas Re(z) is always negative, and hence none of the region will be included.
- The point of inflexion in the third quadrant is really not obvious. I would be so shocked if they penalised people for it. But the asymptote along the ray Arg(z)=-pi/2 is important.
- The ray Arg(z)=pi is NOT included.
(https://i.imgur.com/qRMguvZ.png)
Q16 (b)
(https://i.imgur.com/z4GWTLb.png)
I make no promises on every solution being accurate. Please point out mistakes and I'll get to them slowly.
Refer to fun_jirachi's solutions for questions I did not do.
11
(https://i.imgur.com/3ndQ9n6.jpg)
12
(https://i.imgur.com/q6xblVs.jpg)
(https://i.imgur.com/1ZPiFKQ.jpg)
13
(https://i.imgur.com/wVdDNlo.jpg)
14ac
(https://i.imgur.com/IeqEWxd.jpg)
Thoughts: Advanced was harder to compare 2020 and 2021. I felt neither year's paper was friendly.
This, however, has got to be the most draining MX2 paper I've done in a very long time. The only other time I recall of feeling so defeated was in 2018 or something where the chaotic Euclidean geometry question showed up.
- Have never in my life (including uni) seen anybody told to sketch Re(z) >= Arg(z) before. The sketch I did was freehand. I can't imagine how they're gonna mark that though, like what details do they want honestly?
Regarding that question, after talking to a friend...
Here's a more formal way of considering the problem that I was told later. If we want to sketch Re(z) = Arg(z), we can consider tan(Re(z)) = tan(Arg(z)). Then, if z=x+iy, the curves would be tan(x) = y/x. Hence, first make a sketch of \(y=x\tan x\), using multiplication of ordinates, which you learnt in Prelim MX1. Finally, you may wish to just use guess and check, to see what regions are involved.
- Very easy to make mistakes in projectile motion this year.
- I never even knew that the vector triangle inequality was permitted in the HSC. Not sure where that got pulled from.
- The alternate method for the Cauchy-Schwarz inequality felt like a cheap trick. I'm not sure if that was what they intended. It very may well be. But also, the method seemed a bit short for 3 marks, and the alternate method I had was another battle with algebra.
- Q15d can throw people off. Although I think I came up with a relatively simple method, it's very easy to get carried away into bizarre nonsense with it, just because of the nature of the final result itself.
Paper difficulty really felt like it was doing a \(y=a^x\) this year. It could have been the length instead, but we've had lengthy papers in the past that weren't too hard. This year was just questionable. No clue what NESA were thinking, especially with COVID struggles this year.
Alternatively, I'm just bad. But this is a stretch for me though...
Funnily enough though, Q10 of this paper was not as cruel as Q10 of Advanced.