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What did you get '3/7' in? In principle it's possible to come third in your trials but still sixth overall.

$\noindent (Of course, these are just MCQ's worth 1 mark, so when in the exam it's best to do these quickly if possible, you don't need full working etc. It's pretty clear using the given solution's method for example what option has to be the right one (the other ones won't always satisfy...

$\noindent Yeah, technically the answer as written just shows that if $|z_1| = 1$, then $|z_2| = \sqrt{11}$. One way to see that the whole circle is obtained is as follows. Let $w = \sqrt{2} - 3i = R\,\mathrm{cis}\left(\phi\right)$, where $R = \sqrt{11}$ and $\phi = \mathrm{Arg}\left(w\right)$...

The topics in Physics that might be helpful for MX2 would mainly be circular motion (and that's from Prelim Physics). But it's definitely not necessary to Physics in order to do MX2.

I simply wrote S for the given sum so that we have a variable assigned to it if we need to refer to it (so it's more convenient to have to keep referring to it as "the sum" and so we can write equations involving it etc.).

$\noindent For Q.4, let the desired sum be $S$, i.e. $S = \sum _{k =0}^{\infty} (-1)^{k} \frac{\sin ((2k+1)\theta)}{9^k}$. This is the imaginary part of the geometric series $\widetilde{S} = \sum_{k=0}^{\infty} \left(-\frac{1}{9}\right)^{k} \mathrm{cis}\left((2k+1)\theta\right)$. This has common...

$\noindent Method for Q.6: let $w = \frac{\alpha \overline{\beta}-1}{\alpha -\beta}$, so $\overline{w} = \frac{\overline{\alpha}\beta -1}{\overline{\alpha} -\overline{\beta}}$. Expand out the expression for $w \overline{w}$ to show that $w \overline{w} = 1$, noting that $\alpha \overline{\alpha}...

Once we have the candidate points, yeah, just evaluate the function at them to find the maximum and minimum. (In this case we're optimising a continuous function over a compact domain, so we're guaranteed a global maximum and a minimum value on this domain.)

Depends a lot on how you go externally.

Yes, it's possible.

Did Ekman make any solutions?

You seem to be first in your subjects. In that case, your cohort is essentially irrelevant (in terms of whether they can drag you down). Just make sure to do well externally too.

Re: HSC 2016 3U Marathon (Note that due to the convexity of a parabola, its tangents will never go 'inside' it, so there's no way two tangents could intersect somewhere inside the parabola. With normals though, they go both in and outside the parabola, so a priori they could well meet outside...

Re: HSC 2016 3U Marathon $\noindent No. Consider the parabola $x^2 = 4y \iff y = \frac{x^2}{4}$. The normal at a point $\left(2t,t^2\right)$ is $y = t^2 -\frac{1}{t}\left(x -2t\right)$. So the normal at the point on the parabola where $t = 1$ (the point $(2,1)$) is $y = 1 -(x-2)\iff y = -x+3$...

You flipped the two around. The "hypotenuse" is the mg and the tangential force is the "adjacent". Basically this (zoom in if it's a bit unclear initially): http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0011/pendulum_FBD.gif .

$\noindent The acute angle in the picture between the red weight vector and the tangent is $90^\circ - \theta$, since the angle between the radius and tangent is $90^\circ$ and using co-interior angles in parallel lines. So the component of the weight force in the tangential direction (pointing...

Part (c) was done here: http://community.boredofstudies.org/1003/maths/352034/math1231-1241-1251-sos-thread-2.html#post7184625 . (For part b), remember sum of roots is 0, since the z^{n-1} coefficient in the polynomial is 0.)

Re: HSC 2017 Maths (Advanced) Marathon Any (real-valued) function (of a real variable) defined on a symmetric interval (about 0) that is both odd and even must satisfy f(x) = f(-x) = -f(-x) for all x in the domain, which implies f(-x) = 0 for all x in the domain, whence f(x) = 0 for all x in...

$\noindent It's easier to spot by keeping Paradoxica's factorisation in mind: since $f(x,y) = (x+3y)(x+y)$, we can easily find lines of the form $x = ky$ such that $f$ will be made positive or made negative. E.g. taking $x = -2y$, we have $f(x,y) = (-2y+3y)(-2y+y) = -y^2$. Etc.$

No, you don't need to know it for Year 12. You could (and are expected to) just find the second derivative of the function normally (i.e. find the first derivative first, then differentiate again). It's just a bit more tedious to do it that way I think.