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✨ THE PAPER ✨ (I presume, I didn't actually do it) Credit: @a.tilla from Conquer server

What was the last question?

Yep all of that is correct. In terms of being asked to show working out, the method is to divide by all terms by the highest power and then show that terms of the form 1/x, 1/x^2 etc. all go to 0 as x goes to infinity, hence the only coefficients that remain are that were part of the highest...

Some part-by-part commentary for Q14 (a) (i) Done well by almost everyone, with only a few silly mistakes here and there. I was surprised that some students could just look at the question and then write the result, but good for you I guess and you did receive the mark. Keep in mind that if...

There's a mapping grid on the last 2 pages that tells you what band level each question was intended to be.

FWIW, there were only a few attempts that moved past 1 or 2 lines, and very, very few marks were awarded across the cohort.

Yep, that's what I meant. Now who wants to give it a go?

Actually this model is not correct. If resistance applies to a projectile with a magnitude of kv^n, then the horizontal and vertical resistance can be correctly set as \ddot{x}=-k\dot{x}^n and \ddot{y}=-k\dot{y}^n if and only if n=1. You'll notice in Cambridge and the few HSC past papers where...

Remember that if velocity is the derivative of displacement, then displacement is the result of integrating velocity. If a particle moves from a to b, then its total distance travelled is \int^b_a {v} dt, assuming that the particle does not change direction. What the integral actually provides...

Fort Street 2022 which you can find on THSC.

f(x) = a + bx + 4ax^2 approaches positive infinity on both sides (since a > 1 > 0), thus both of the limits are non-zero and \frac{1}{f(x)}= \frac{1}{a + bx + 4ax^2} has a horizontal asymptote.

More precisely, \frac{1}{f(x)} has a horizontal asymptote iff at least one of \lim_{x\to\infty} f(x) or \lim_{x\to-\infty} f(x) is non-zero.

Consider the sum and difference of z and its conjugate and work from there.

Integrate both sides with respect to x, and apply \int{0 dx} = C

So... half of all HSC students are mentally handicapped or have shit luck?

I just know already that the physics paper is going to be crazy. Highly recommend everyone to take part if you can.

It was a part of the old syllabus, so no you don't need it. But to test your understanding, it's possible to do it using the transformation of graphs that you would've learnt in prelim extension 1.