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$They should have made it clearer, maybe by writing $\mu(v)$ and $\mu (V)$ and stating it's a function of the speed. But somehow I'm guessing most people weren't confused, since no-one's mentioned it yet I think. Maybe because most people don't pay much attention to the physics behind it.$...

$It's looking like in the banked turn question, $\mu$ is to be treated as a variable (a function of the speed), which confused me for a bit, because normally $\mu$ is a constant (a coefficient of friction).$

Q.11 is missing from the WhoStanLeee one.

re: HSC Physics Marathon Archive Hard to see from that document what the extension courses would actually be like (i.e. whether they are real science or just more of the same as now).

None of these are working for me. The second one just lags and the first one says to download an app to view the PDF. Did anyone manage to download from here?

Complex numbers in HSC 4U generally isn't too hard once you get used to them (obviously there exist hard questions in every topic, but the average Q. isn't too hard). And there's complex numbers appears in HSC 4U as part of Complex Numbers, and also in Polynomials.

Yes.

Lol.

re: HSC Physics Marathon Archive Calculus in HSC Physics, I wish.

I thought the only time someone got 100 raw in the HSC 4U paper was in 1993 (Anthony Henderson). People usually lose a mark or two from silly mistakes or something.

$This is the general method of finding the limit as $x\to \infty$ of a ratio of equal degree polynomials. I.e. $\lim _{x\to \infty}\frac{a_N x^N +a_{N-1}x^{N-1} \cdots + a_0}{b_N x^N +b_{N-1}x^{N-1}+ \cdots + b_0}=\frac{a_N}{b_N}$. This is seen by dividing top and bottom of the quotient by...

$I'm assuming $p$ is a fixed positive integer. Then $\frac{n^p}{\binom{np}{p}}=\frac{n^p}{\frac{(np)(np-1)\cdots (np -(p -1))}{p!}}=\frac{n^p}{\underbrace{(np)(np-1)\cdots (np -(p-1))}_{p\text{ terms}}}\times p!$. Hence as $n\to \infty$, this quantity approaches $\frac{p!}{p^p}$, since the...

They probably wanted to minimise the amount of wasted pages.

Maybe in the HSC Maths Extension 1 paper.

Unsurprising.

In general, there's no fixed way to go about tricky proofs. Sometimes you just need to try a lot of things until something works. It's usually a good idea to get the expressions in the RTP statement involved, since you're trying to prove something about them.

Basically, you should try doing as much as possible to somehow come up with expressions involved in the RTP statement. In other words, work towards what you're trying to prove, and then everything should hopefully fall out. So they added those things together as they knew that'd get them one of...

$Since the region is \textsl{below} the line $y=x-2$, the region must satisfy $y\leq x-2$, so the only options to consider now are (A) and (C). Note that the point $(x_0,y_0)=(3,0)$ lies in the region, and $y_0^2 = 0^2 = 0$, $4-x_0 = 4-3 = 1$, so $y_0^2 < 4-x_0$ (as $0<1$), so the region...

$First get it into the form $(x-h)^2 = 4a(y-k)$ or $(x-h)^2 = -4a(y-k)$:$ $Completing the square, we have $y=(x+3)^2 -2$$ $$\Rightarrow (x+3)^2 = y+2 \Longleftrightarrow (x-(-3))^2 = 4\times \frac{1}{4}\left(y-(-2) \right)$.$ $So the vertex is $(-3,-2)$, and the focal length is $\frac{1}{4}$...

$That was basically to obtain $XY\left(\frac{1}{BC}+\frac{1}{AD} \right)$, which is one of expressions in the identity that is required to be proved.$