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Doing advanced math only (2U) : 2 units total Doing extension 1 math (3U): 3 units total (2 from advanced, 1 from extension 1) Doing extension 2 math (4U): 4 units total (2 from extension 1, 2 from extension 2, 0 from advanced - you also don't sit the 2U HSC exam, although its dependant on your...

Re: HSC 2018 MX2 Marathon $\noindent Given $ z = x + iy $ is a variable point and $ \alpha = a + ib $ is a fixed point on the Argand plane,$ $\noindent (i) Show that $ z\alpha -\bar{z}\bar{\alpha} = 0 $ represents a straight line through the origin $ O $\noindent (ii) Suppose that...

Re: HSC 2018 MX2 Marathon $For reference anyway: $ P(x) = 2x^4 + 7x^3 + 2x^2 -7x + 2 = x^2 (2x^2 + 7x + 2 - \frac{7}{x} + \frac{2}{x^2}) = x^2 \left(2(x^2 + \frac{1}{x^2}) + 7(x - \frac{1}{x}) + 2\right) $Using $ (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 = x^2 \left(2(x -...

Re: HSC 2018 MX2 Marathon Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too

School Rank: Usually between 200-220 English Advanced: 10/40 MX1: 1/20 MX2: 2/7 Physics: 2/18 Economics: 2/16

Re: HSC 2018 MX2 Marathon $\noindent If $ P(x) = 2x^4 + 7x^3 + 2x^2 -7x + 2 $, solve:$ $\noindent$ P(x) = 0 $ over C without using long division, sum/product of roots etc or the polynomial remainder theorem$

Re: HSC 2018 MX2 Marathon $Roots: $ cis\frac{\pi}{4} (\frac{1}{\sqrt2} + \frac{i}{\sqrt2}), cis\frac{-\pi}{4} (\frac{1}{\sqrt2} + \frac{-i}{\sqrt2}), cis\frac{3\pi}{4} (\frac{-1}{\sqrt2} + \frac{i}{\sqrt2}), cis\frac{-3\pi}{4} (\frac{-1}{\sqrt2} + \frac{-i}{\sqrt2}) $Now $ z^4 + 1 =...

Re: HSC 2018 MX2 Marathon My bad, yeah thats the correct equation

Re: HSC 2018 MX2 Marathon Correct! Here is another (I really like these questions) $a) Show that if n is divisible by 3 then:$ (1 + \sqrt{3}i)^{2n} + (1 - \sqrt{3}i)^{2n} = 2^{2n + 1} $b) Simplify the expression if n is \textbf NOT divisible by 3$

Re: HSC 2018 MX2 Marathon $Find the minimum value of the positive integer m for which $ (\sqrt3 + i)^m $ is: $ $a) real$ $b) purely imaginary$

Re: HSC 2018 MX2 Marathon Nice! $Alternatively $ (1+i)^m = (1-i)^m \noindent \frac{(1+i)^m}{(1-i)^m} = 1 \frac{m\sqrt{2}(cis \frac{m\pi}{4})}{m\sqrt{2}(cis \frac{-m\pi}{4})} = 1 cis \frac{m\pi}{2} = cis0 \frac{m\pi}{2} = 0 + 2k\pi $ where k = 0, -1, 1, -2, 2...$ m = 4k...

Re: HSC 2018 MX2 Marathon $New Question!$ $Find the smallest \textbf{positive} value of m if $(1 + i)^m = (1 - i)^m

Re: HSC 2018 MX2 Marathon Yeah that makes more sense now, its pretty creative. Thank you

Re: HSC 2018 MX2 Marathon Sorry if I'm missing something obvious, but just to clarify $\noindent$ S = \Im{\left( \sum_{k=0}^n z^k \right)} = \Im(z^0 + z^1 + z^2 + ... + z^{n+1}) = \Im(1 + z^1 + z^2 + ... + z^{n+1}) = \Im(\frac{(1)(z^{n+1} - 1)}{z - 1}) = $ ... $ = \frac{...

Re: HSC 2018 MX2 Marathon I'm not sure what you did on this line, I'm guessing its in regard to the fact that Im(1) = 0 \Im{\left( \frac{z^{n+1}-1}{z-1}\right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{z^{\frac{n+1}{2}} - \overline{z^{\frac{n+1}{2}}}}{\sqrt{z}-\overline{\sqrt{z}}} \right)}...

Re: HSC 2018 MX2 Marathon Ah I remember seeing telescopic stuff in an old MX2 marathon as a hard Q7 or Q8 Are there any resources in particular to practice Q7/Q8 level questions apart from past hsc papers?

Re: HSC 2018 MX2 Marathon $Let $ z = cosx + isinx$ sinx + sin2x +sin3x + ... + sin(nx)= \mathrm{Im}(z + z^2 + z^3 + ... + z^n) $GP with $a = z$ and $r = z$$ \frac{z(1-z^n)}{1-z} = \frac{(cosx + isinx)(1-(cosx + isinx)^n)}{1-cosx-isinx} = \frac{(cosx + isinx)(1- cosnx -...

Re: HSC 2018 MX2 Marathon Before I write up the latex, is the answer \frac{sin(\frac{nx}{2})(sin(x\frac{n+1}{2}))}{sin(\frac{x}{2})} $ ???$

Re: HSC 2018 MX2 Marathon $Roots $(\alpha $ and $ \beta): $$ 1 \pm i $Subbing in roots and $ y + 1 = \cot(x) $ into the LHS: LHS = \frac{(cotx + i)^n - (cotx - i)^n}{2i} = \frac{(\frac{cosx}{sinx} + i)^n - (\frac{cosx}{sinx} - i)^n}{2i} = \frac{\frac{1}{sin^nx}(cosx + isinx)^n...

Re: HSC 2018 MX2 Marathon I used all the stuff you gave me and went from lhs -> rhs but I end up with crappy trig inside the ()^n that I haven't been able to transform into cosx - isinx to use with de moivre ill write it up in latex later this afternoon