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Good fun.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $We have $$\begin{align*} z^3 + 1 &= 0 \\ \Rightarrow z^3 &= -1 \\ \Rightarrow z^3 &= \mathrm{cis}\left(\left(2k+1\right)\pi \right), \quad k\in \mathbb{Z} \\ \Rightarrow z &= \mathrm{cis}\left(\left(2k+1\right)\frac{\pi}{3}...

And most of these integrals generally aren't asked in the HSC (the integrals that get asked in the HSC are generally easy for the computer to compute).

It depends on what level of maths you did for your HSC (like 2U etc.). Did you do General? I think you had a thread before but I can't remember what level of maths you said you did (if you said it).

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $In general, if we have the graph of $y=f(x)$, then the graph of $y=-f(x)$ is the graph of $y=f(x)$ reflected about the $x$-axis. This is explained as follows. Consider any point $P\left(a,b\right)$ that lies on the graph of...

$This is because our method for doing this Q. is first arrange the `groups', and for each group-arrangement, we have $2!$ sub-arrangements for the ordering of the couple. So when we first order the groups, it's $\frac{5!}{5}$ (as there are 5 groups), and for each of these $\frac{5!}{5}$...

Re: HSC 2016 4U Marathon That question came with a diagram in the HSC paper if I recall correctly.

Others will know more, but this is roughly what I get. $\noindent (1) There was a man (say Bob) ($\exists$ means ``there exists'').$ $\noindent (2) He worked hard (in Physics, $ \bold{F} \cdot \bold{d}$ is the work done by a force $\bold{F}$ displacing a particle by $\bold{d}$ ).$...

Or more simply, just state that the square of any real number is non-negative; this would be sufficient in an exam.

Re: MX2 2016 Integration Marathon $Note that we don't need the graph of the integrand to see that it is not continuous over the interval in question, since we can tell that the function $f(x)=\frac{1}{1+2^{\frac{1}{x}}}$ is not continuous here as it is undefined at $x=0$. So the fact that...

Re: MX2 2016 Integration Marathon $The answer should be $\frac{2}{3}$; also, you don't need to actually differentiate it, because the integrand is continuous in the split integrals when making the bounds replaced by $\epsilon$ and then taking the limit, so we can just use the fundamental...

$We are given that $f(0)=0$. We also have $f(4)=0$, since the net change in function value between $x=0$ and $4$ is 0. From $x=4$ to 6, the function decreases at a constant rate of $3$ per unit increase in $x$. So$ $$\begin{align*} f(6) &= f(4) - 3\times (6-4) \quad (\text{as the change in }x...

$The incorrect thing with your reasoning is dividing by 6. Instead, you should have divided by 5, because there are 5 ``groups'' here. It's the number of groups that is relevant, rather than the number of chairs. So even though there's 6 chairs in total, it's as though only 5 chairs are there...

$The answer is $2!\cdot 4!=48$; there are $2!$ ways to arrange the couple (either the hostess is on the \textsl{left} of the host, or on the \textsl{right} of the host), and for each of these arrangements of the couple, we can arrange the remaining four people in $4!$ ways.$

$Here's a start. Let the other (non-hypotenuse) side be length $y$ metres, then we have the hypotenuse being $\sqrt{x^2 + y^2}$ (Pythagoras' Theorem). So we must have $x+y+\sqrt{x^2 + y^2} = 5$ (as the total length of the wire is 5 metres). This is our constraint equation. The function to...

Re: MX2 2016 Integration Marathon Yes. Although of course, such a question would be unlikely to be asked directly in the HSC, and if it were asked, students would probably be guided through the steps.

Re: MX2 2016 Integration Marathon $\textbf{NEW QUESTION} $Find $\int _{-1} ^{1} \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{1 + 2^{\frac{1}{x}}}\right)\text{ d}x$.$

$Let the double root (which is rational) be $\alpha$. By the rational roots theorem, since the polynomial is monic (leading coefficient is 1) $\alpha$ must be a factor of the constant term, so the test values for $\alpha$ are $\pm 1,\pm 7$ (as these are the factors of the constant term $-7$).$...

Re: MX2 2016 Integration Marathon Oh yes, I realise I forgot to square-root a hypotenuse for one of them.

Re: MX2 2016 Integration Marathon $\noindent It's the good method for doing these ones where it's a fraction with a linear combination of $\sin x$ and $\cos x$ on both the numerator and denominator. You rewrite the numerator as a linear combination of the denominator and the denominator's...