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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Are you talking about for the first question?

$\noindent Observe from a sketch that the graph of $y= 5-x^2$ is the above that of $y=(x-1)^2$ for the region whose area we wish to calculate. We need to find the $x$-values of the points of intersection of the two curves. To do this, let $(x-1)^2 = 5-x^2$, so$ $$\begin{align*}x^2 - 2x + 1 &=...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread What is the question?

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Another way to do your part (b) of first Question (proving that $PQ = PT$) is to just show that $\theta = 45^\circ$, where $\theta = \angle PQT$. This can be done by using the angle between two lines formula: $\tan...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent I think because they're both just tedious algebra. For the first one, let $P$ be like $\left(2p,p^2\right)$, where $p\in \mathbb{R}$ and $p\neq 0$ ($p$ can't be 0, because then $P$ is the origin, and so the normal at...

$\noindent Vol. of cylinder is $\pi r^2 h$, where $r=3$, $h = \ln 3$.$ $\noindent Then subtract off the volume from rotating area between $y$-axis and graph, i.e. Subtract off: $\pi \int _c ^d x^2 \text{ d}y = \pi \int _0 ^{\ln 3} e^{2y}\text{ d}y$ (as $x =e^y$). Are these the integrals you...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent The base case of $n=13$ clearly holds, as $2^{13} = 8192 > 4394 = 2 \cdot 12^3$. Now suppose the result holds for some particular integer $k\geq 13$, i.e. $2^k \geq 2k^3$. We need to show that $2^{k+1}\geq...

$\noindent Here's a proof by induction for 2.$ $\noindent The result is clearly true for $n=1$, since then, $LHS =(2-1)! = 1! =1$, and $RHS = 1^1 = 1 = LHS$.$ $\noindent Now, suppose that the result holds for some particular positive integer $n$, so that we have $(2n-1)! \geq n^n$. We need to...

There's no complex numbers in the Prelim. or HSC Physics syllabi, though the mechanics stuff in these syllabi can be a bit helpful for HSC 4U Maths mechanics (like for understanding circular motion in 4U a bit better etc.).

Oh right. Thought you meant "space-like" as in motion-based things haha. Yeah, Cosmic Engine is quite pointless.

Re: MX2 2016 Integration Marathon Correct!

Re: MX2 2016 Integration Marathon $\noindent \textbf{NEW FOR-FUN QUESTION (NOT HSC STYLE)} $ $\noindent Find $\int _{\frac{1}{4}} ^{\frac{1}{2}} \left \lfloor \ln \left \lfloor \frac{1}{x}\right \rfloor \right \rfloor \text{ d}x$.$ $\noindent \textbf{Note.} The symbol $\lfloor \cdot \rfloor$...

That topic in Prelim. is "Moving About". It's slightly useful in HSC Physics, like knowing what conservation of momentum is, but you won't actually need to calculations with those stuff for HSC Physics most likely.

How can you avoid memorisation? A lot of the stuff in the HSC Science subjects require or are amenable to rote learning. I know a lot of state-rankers etc. who memorised and did well.

I guess that indeed was a good Q., if only all the Q's in the HSC papers were good Q's :). (Here's the link to that paper for anyone interested: http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-physics.pdf .)

The problem is, while some things can indeed be understood with extra effort (outside-syllabus reading etc.), the exams are designed in such a way that such understanding is neither necessary nor rewarded. Hence people can rote learn their way to high marks, provided they've mastered how to...

$\noindent What do you mean? I did do it assuming $z= (1-\cos x)+i\sin x$.$

Year 12 HSC Science is mostly like humanities anyway (in the sense that there's not much science in it and is mostly based on impacts of science on society, history of science, etc.).

$\noindent (Btw, you've drawn your $\lambda$'s the wrong way round :$\mathcal{P}$. Probably wouldn't matter much since it's not a common symbol in the HSC, and they wouldn't detract marks for that probably.)$

$\noindent Yeah this was similar to my method. The $2^m$ combination of which mouse drinks and does not drink a given bottle (where $m$ is the number of mice) is more easily seen by noting that for a given bottle $i$, we can see that each mouse either drinks it, or does not. Hence $2^m$, as...