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$\noindent Well the group is $\{e, a, b\}$ (all \underline{distinct} elements). By closure of the group, $a^2$ must be one of $e, a$, or $b$. Now, what would happen if $a^{2} = e$ or if $a^{2} = a$?$

There is a way to do it without needing to do any calculations.

$\noindent So you essentially got out the first part?$ $\noindent For the second part, it's fine for $a^2$ to equal $b$. In that case, in fact, $a^2$ would be the same as $a^{-1}$. (For example, remember from complex numbers how a cube root of unity $w$ satisfies $w^{2} = w^{-1}$.)$

What was your progress on the questions so far? Also to use LaTeX on the forums here, you need to enclose TeX code in so-called "tex tags". (Using LaTeX here is a bit different to just using it on your own computer.) You need to enclose the TeX code in between: [tex.] [/tex.] (but leave out...

$\noindent This is a hyperbola. To get the asymptotes, make the RHS $0$. So its asymptotes are $x^{2} - 4y^{2} = 0$, that is, $y = \pm \frac{x}{2}$.$

$\noindent Your answer is equal to the book's answer. Multiply top and bottom of your answer by $e^{2}$ to see this.$

$\noindent By inspection,$ $$x^2 + 6x - y^2 - 8y - 7 = (x + y + 7)(x - y - 1).$$

$\noindent \textbf{Hint.} Since the base is the same ($(1+x)$ both times), we can combine the two factors:$ $$(1+x)^{2}(1+x)^{8} = (1+x)^{???}.$$ $\noindent See if you can fill in the ``???'' using your knowledge of index laws, and then make use of the binomial theorem.$

Re: HSC 2018 MX2 Marathon $\noindent Remember, adding $1$ (or any real number) to something won't change its imaginary part! (You can probably prove this fact easily if you're unsure of it.) So to compute$ $$\mathrm{Im}\left(z+z^{2} +\cdots + z^{n}\right),$$ $\noindent you can just compute$...

$\noindent Sketch of another way to do it via ``hence'': use the fact that the general solution for $z$ to $z^{n}= a^{n}$ (where $a$ is a given complex number) is $z= a\times \left\{n^{\text{th}}\text{ roots of unity}\right\}$. So you basically just need to compute the cube roots of unity and...

Oh yeah, I didn't even notice the "hence" (or look at part (i)).

$\noindent \textbf{Hint.} First solve using polar form (``mod-arg form''). Write $z = r\, \mathrm{cis}(\theta)$ (so our goal is to solve for $r$ and $\theta$ first. See what the equation $z^3 = 8i$ when we substitute $z = r\, \mathrm{cis}(\theta)$, and then compare modulus and argument of both...

Re: HSC 2018 MX2 Marathon One thing (probably not the only thing) you can try is searching up past years' HSC 4U marathons here on BOS (including the Advanced Level ones, though these are typically harder than HSC style Q's or would be broken into multiple steps if asked in the HSC exam).

Re: HSC 2018 MX2 Marathon $\noindent Well done! Incidentially, another method is to consider $ $$\sin( k\theta )\sin \frac{\theta}{2} = \ldots \quad (\text{use a product to difference formula})$$ $\noindent and then observe we get a \textit{telescoping sum} if we sum both sides from $k=1$ to...

Re: HSC 2018 MX2 Marathon Correct!

When you calculated the no. of ways to have roses adjacent and azaleas adjacent, you also made the block of roses adjacent with the block of azaleas. However, this is not necessary in order to have roses adjacent and azaleas adjacent. The meaning of "roses adjacent and azaleas adjacent" in this...

It seems like your question comes down to: "what is the number of ways of having the roses adjacent and also the azaleas adjacent?" (since you agree with the book on the other components) Label the plants as R1, R2, R3, R4, A1, A2, A3, X, Y, Z. (The Rj's stand for roses and Aj's for azaleas.)...

$\noindent The topic ``Harder 3U'' is actually a 4U topic (\underline{not} a 3U topic). As such, \color{blue}{the AM-GM inequality is \underline{only required for 4U} (not 3U)}\color{black}. (Inequalities are taught as a sub-topic in Harder 3U.) If you have a HSC 4U textbook, you should be able...

$\noindent Well if we want to show that$ $$x + \frac{1}{x}\geq 2 \quad \text{if }x > 0,$$ $\noindent then a perhaps more intuitive approach is to get everything on one side first. So it's equivalent to show that$ $$x + \frac{1}{x} -2 \geq 0 \quad \text{if }x > 0. $\noindent Now you...

$\noindent I assume you could do i) and ii). Here's some hints for iii). Note that if $y=0$ (note then $x \neq 0$ since $z$ can't be $0$), then $z=x$. So you just need to show that$ $$\left|x + \frac{1}{x}\right| \geq 2$$ $\noindent if $x$ is real$.$ $\noindent To show this, first show that...