Search results

Re: HSC 2016 3U Marathon I think Flop21's thread just got merged with this one too. (Or got moved somewhere...can't find it right now, but when I click on it, it directs me to this thread.) Edit: Ah, there's a post related to it at the top of this page.

Re: HSC 2016 3U Marathon Also, realised that Shuuya's SOS thread got merged with the HSC 3U Marathon.

Re: Shuuya's endless maths questions SOS thread $\noindent (i) Since $f(x) = 2x^4 - 3x^2 +1$, we have $f^\prime (x) = 8x^3 - 6x = 2x(4x^3 -3) = 2x\cdot 4\left(x^2 - \frac{3}{4}\right)$. So to analyse the sign of $f^\prime (x)$, we need to analyse the sign of the factors $x$ and $x^2 -...

HSC Biology is pointless generally speaking. Only do it if you think it'll help your ATAR, not because you think it's needed for tertiary studies (it's not, generally speaking). USyd and Griffith Medicine don't need UMAT (USyd requires 99.95 ATAR). Most places do require UMAT, however.

Re: MX2 2016 Integration Marathon RealiseNothing was referring to what leehuan was saying was beyond MX2 scope, namely the Taylor polynomial bounds for sin / cos etc. So he was saying that this was in a past HSC (getting you to prove those inequalities). The 2010 HSC Integration Question 8 was...

Yeah, this is just simultaneous equations.

Re: Shuuya's endless maths questions SOS thread $\noindent We have the system$ $$\left\{\begin{matrix} p+2q-r=5& (1) & \\2p-q+r=3 & (2)& \\ p+3q-2r = 8& (3) & \end{matrix}\right.$$ $\noindent The goal is to first eliminate one of the variables completely, so we are left with only two...

$\noindent We have the system$ $$\left\{\begin{matrix} p+2q-r=5& (1) & \\2p-q+r=3 & (2)& \\ p+3q-2r = 8& (3) & \end{matrix}\right.$$ $\noindent The goal is to first eliminate one of the variables completely, so we are left with only two variables.$ $$(1) + (2): 3p + q = 8\quad...

It is very possible to do well in the HSC by memorising (sometimes you need to memorise).

Re: Flop math question thread $\noindent So the solutions are of the form $x=-\frac{\pi}{4} + (-1)^m \frac{\pi}{6}+m\pi$ (and any number of this form is a solution). If we only want solutions in the range $\left(0,2\pi\right)$, we just take the integers $m$ that give solutions in this range...

Re: Flop math question thread $\noindent Oh yeah, sorry, forgot to simplify in my original answer that $\sin^{-1} \frac{1}{2}=\frac{\pi}{6}$.$

Re: MX2 2016 Integration Marathon This is why seanieg89 said he was sure you'd seen them before.

Re: Flop math question thread $\noindent We have$ $$\begin{align*}\sqrt{2}\sin \left(x+\frac{\pi}{4}\right) &= \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \\ \Longleftrightarrow \sin \left(x+\frac{\pi}{4}\right) &= \frac{1}{2}.\end{align*}$$ $\noindent Assuming you're after general solutions, the...

Re: Flop math question thread $\noindent For interests's sake, in general, $a\cos \theta + b\sin \theta = R\cos \left(\theta - \varphi \right)$ for all $\theta \in \mathbb{R}$, where $R=\sqrt{a^2 + b^2}$ and $\varphi$ is the angle to the positive $x$-axis made by the vector $(a,b)$ (with $a,b...

Re: Flop math question thread $\noindent It should become $\sqrt{2} \cos \left(x-\frac{\pi}{4} \right)$. If you want to use $\sin$, it's $\sqrt{2} \sin \left(x+\frac{\pi}{4} \right)$. $

Re: MX2 2016 Integration Marathon Maybe referring to the Taylor polynomial bounds? E.g. things like $$x - \frac{x^3}{3!}\leq \sin x \leq x,\, x\geq 0$.$

$\noindent Since the coefficients are real, another root is $-i$ by the conjugate root theorem. Note that there will be four roots in total. The constant term is 3, and $P$ is monic, so it is of the form $P(x) = x^4 + ax^3 +bx^2 + cx + 3$ (note that the coefficient of $x^3$ is negative as the...

Re: Shuuya's endless maths questions SOS thread $\noindent i) It's an even function, so it fails the Horizontal Line Test, so is not invertible.$ $\noindent ii) Restricted to $x\geq 0$, clearly $x^2$ is monotone, so $\mathrm{e}^{x^2}$ is monotone (as the exponential function is monotone and a...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent It's in the answers because $2\pi$ just falls inside the domain, as the domain had $0\leq x \leq 2\pi$.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Is that $\left(\sqrt{9-x}\right)^2$?$