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I think it depends on what ATAR range you're looking at. Relatively low ATARs would probably (I'm guessing) have higher "variance" in their estimates compared to high ones.

Re: HSC 2018 MX2 Marathon $\noindent By the way, you can do this with any function, not just $e^x$ (provided that it makes sense to input both $x$ and $-x$ into your function). We can write a given $f(x)$ as a sum of an \color{red}{even }\color{black} function and an \color{blue}{odd...

The slope of the line element at a point (x, y) gives you the value of the slope of a solution (i.e. dy/dx) at the point (x, y). So, which equation for dy/dx could be represented by the given graph? Examine things like where the line elements on the graph have positive slope, zero slope...

Try looking for the places where the graph cuts the x-axis.

Look for the places the graph cuts the x-axis (y = 0).

$\noindent By the way, the answer to ii) can't be $n(n-1)\cdot 2^{n-2}$. If this was the case, then substituting $n=1$ into your answer gives $0$, so the series would have to equal $0$ when $n=1$. But clearly when $n=1$, the series equals $\binom{1}{1}1^{2} = 1 \color{red}\neq\color{black} 0$.$

$\noindent To deal with such an integral, we have to split it up at the discontinuity and separately consider limits over the continuous parts. That is, we must consider$ $$\lim\limits_{a\to 0^{-}}\int_{-1}^a \frac{1}{x}\,dx +\lim\limits_{b\to 0^{+}}\int_{b}^{1}\frac{1}{x}\, dx.$$...

$\noindent Here's a way to do the first one. Hopefully you can then try the second one.$ $\noindent We have $w = \frac{z-2i}{1-z}$, so $w-wz = z - 2i \Rightarrow z= \frac{w+2i}{1+w}$. Therefore,$ $$\begin{align*}|z|&= 2 \\ \Rightarrow \left|\frac{w+2i}{1+w} \right| &= 2 \\ \Rightarrow |w +...

$\noindent Were you given a domain to use? Otherwise, since that's an odd-degree polynomial, it won't have a maximum value on $\mathbb{R}$ (in fact it goes to $\infty$ as $x\to-\infty$). If you want to find a \emph{local} maximum, you can use the second derivative test.$

$\noindent By the way, to check your answer, you don't need to redo the whole question. You can just check by substituting your answer of $\color{red}n=9$ into your formulas for $T_6$ and $T_7$ and seeing whether $\frac{T_6}{T_7}$ equals $4/9$. This is a much faster way to check the answer...

$\noindent The error was after your line $T_{6}:T_{7} = 4:9$, you forgot to include the ratio $\frac{4}{9}$ when equating the expressions. That is, you wrote $\color{red}\binom{n}{5}2^{n-5}3^{5} = \binom{n}{6}2^{n-6}3^{6}$, but should have written $\binom{n}{5}2^{n-5}3^{5} =...

$\noindent In general, the domain of $f\circ g$ would be the set of all $x$ values such that $g(x)$ is in the domain of $f$. In this case, what is the domain of $f$ and which values $x$ in the domain of $g$ make $g(x)$ land in the domain of $f$?$

You don't need to do anything, because the domain is already given to us.

$\noindent For any positive real number $x$, the symbol $\sqrt{x}$ refers to the \emph{positive} square root of $x$.$

Just 1/2.

Have you found and simplified f(g(x))? Once you do this, you should be able to do the question.

I haven't seen the question wording, but did it say it's a GP? Non-GP sums can also have limiting sums. Without having seen the question, I think that if you know (either via proof or by being given it) that it's a GP, and the question tells you the sum converges, then you shouldn't need to...

I would be quite surprised if you had to write out the actual number for 14 b) ii).

$\noindent To type mathematical formulas here, we use something known as \emph{\LaTeX}. To type maths formulas in $\LaTeX$ on these forums, you need to know how to write $\LaTeX$ formulas, and also use what are known as ``TeX tags'' when you type things into the message box. TeX tags are like...

$\noindent If you have the first derivative $y'=\frac{dy}{dx}$ in terms of $t$, to get the second derivative, you have to differentiate the first derivative $y'$ with respect to $t$ \emph{but then also \color{blue}\underline{multiply by $\frac{dt}{dx}$}}\color{black}. In other words, we have...