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Re: HSC 2016 2U Marathon $\noindent In general, we multiply by the derivative of the power. The reason why we so often just multiply by the coefficient of $x$ is that the exponential is in the form $\mathrm{e}^{kx}$ ($\mathrm{e}^{2x}$ for instance). In this case, the derivative of the power is...

Re: HSC 2016 2U Marathon $\noindent Restrict them as in, as part of your solution to the inequation, you need to ensure that the quantities inside the logs originally will always be positive. Generally speaking, whenever you get inequations with logs, you will need to do this. So for...

Re: HSC 2016 2U Marathon $\noindent Yes, this is true. But the statement ``if $\log _a A > \log_a B$ ($A,B>0$), then $A>B$'' is correct if and only if $a>1$. If $a<1$ (and of course $a>0$ since log bases are positive), then the correct statement is ``if $\log _a A > \log_a B$ ($A,B>0$), then...

Re: HSC 2016 2U Marathon $\noindent Here's the proof that $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2x}\right)=2x^{2x}\left(\ln x +1 \right)$. In order to do the differentiation here, we need to make the base into something we can differentiate with, so we convert it to base e. So we have...

Re: HSC 2016 2U Marathon $\noindent To do this integral, we use the fact that $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2x}\right) = 2x^{2x} \left(\ln x + 1 \right)$ (in the HSC, they would probably ask you to show this as a sub-part before asking the integral; we'll show this below). So...

Re: HSC 2016 2U Marathon $\noindent Hey, why did you multiply both sides of the inequality by $\frac{1}{2}$ at the start? Also, Drsoccerball's question had a trick in it! Remember that $y=\log_a x$ is a \textit{decreasing} function if $a<1$. So you need to flip the inequality sign when...

Re: HSC 2016 3U Marathon $\noindent This ``stars and bars'' trick is a classic trick for doing such combinatorial problems. I suggest you read about it online, maybe here for starters:$ https://www.quora.com/How-do-I-use-the-stars-and-bars-combinatorics $\noindent It's a method that allows...

Re: HSC 2016 3U Marathon $\noindent 26. (a) Label the colours 1,2,3,4,5,6. We can put Colour 1 anywhere on the cube by symmetry. Now fix Colour 1's position, then the face opposite Colour 1 can have its colour chosen in \boxed{5} ways (any of the remaining colours), and each of these choices...

Don't worry, HSC maths papers have 10 MCQ's only, so it'd take less time to complete I presume :p.

Re: MX2 2016 Integration Marathon $\noindent For $a\neq 0$, we can also do it as follows. Let $a\in \mathbb{R}$ and $a\neq 0$. Let $I:= \int \frac{1}{\left(\sqrt{x^2 + ax}+x\right)^2}\text{ d}x$. Make the substitution $z=\sqrt{x^2 + ax}+x$, so$ $$\begin{align*}\sqrt{x^2 +ax}&= z-x \quad...

Re: MX2 2016 Integration Marathon Yeah.

Re: MX2 2016 Integration Marathon $\noindent Note that this assumes $a\neq 0$. For the case of $a=0$, we have $I_0:=\int \frac{1}{\left(\sqrt{x^2}+x\right)^2}\text{ d}x$. Note that $x\neq 0$ as then the integrand has 0 denominator. In fact, $x\nless 0$ also, since if $x<0$, the denominator...

Imagine if this something like this came up as a BOS Trial Multiple Choice question set.

He says in his first line about the aether model: "That is aether is the medium in which light travels"

$\noindent Here are some common transformations of the graph of $y=f(x)$:$ $$\bullet$ $y=f(x+k)$, $k>0$: shift the graph of $y=f(x)$ \textsl{left} by $k$ units$ $$\bullet$ $y=f(x-k)$, $k>0$: shift the graph of $y=f(x)$ \textsl{right} by $k$ units$ $$\bullet$ $y=f(kx)$, $k>1$: contract the...

$\noindent No. In actuality, the transformation $f(a-x)$ consists of flipping the graph of $y=f(x)$ about the vertical line $x=\frac{a}{2}$. To see why this is true, set $g(x) \equiv f(a-x)$, then we just need to show that $g\left(\frac{a}{2}-\xi\right)=f\left(\frac{a}{2}+\xi\right)$ for any...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Note that as $\alpha$ is acute, $\sin \alpha,\cos \alpha >0$. Since $\tan \alpha = \frac{2}{5}$, we can easily see by constructing a right-angled triangle that $\sin \alpha = \frac{2}{\sqrt{29}}$ and $\cos \alpha =...

Depends how good your memory is. If you're relatively fast at rote learning things or have a photographic memory, there's probably no need to start memorising HSC Chemistry stuff heaps early.

Those modules are largely irrelevant for HSC Chemistry content. Some things are useful though, like polar bonds etc.

Re: MX2 2016 Integration Marathon $\noindent In the radical in the second line, it should actually be $x^2 + 2+\frac{1}{x^2}\textbf{-1}$, not $+1$.$