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$\noindent The symbol $\phi$ (which can also be written as $\varphi$ (some Greek letter symbols have two versions)) is often used to represent a phase angle in a sinusoid. Here, $\theta$ is typically not used even though the $\theta$ symbol itself isn't used much for sinusoids (as $\omega t$ is...

Yeah, many-to-one just means a function that's not injective, i.e. not one-to-one.

Never lol. It's rare for White to move like that at the start, even if you're playing a beginner.

$\noindent Surjective is different, meaning that for every $b$ in the specified codomain of the function $f$, there exists an $a$ in its domain such that $f(a)=b$. The many-to-one property simply means that $f$ is a function that is not \textit{one-to-one} (another word for one-to-one is...

He was actually talking about functions (a circle graph doesn't represent a function).

$\noindent Ah right, so you meant that if there's more than $y$-value for a certain $x$-value, it's not a function? Yeah, that's correct by definition of function.$ $\noindent Edit: Oh wait, you meant that a function is allowed to have more than one $x$-value map to a given $y$-value. Yes, this...

$\noindent What do you mean by the \textit{vice versa} in this case?$

By the way, every function is a relation, but not every relation is a function.

$\noindent The angle one ($\phi$) is the Greek letter phi (this symbol is distinct from $\varnothing$). If $\phi$ is drawn a bit differently, it can refer to the empty set symbol (this symbol is the one you typed and is not the Greek letter phi): $\varnothing$. This comes from the Scandinavian...

$\noindent For this one, the only powers of $z$ present in $P(z)$ are even powers. Hence if $\alpha$ is a root, then so is $-\alpha$, since $P(\alpha) =P(-\alpha)$, since raising a negative of a number to an even power is the same as raising that number without the negative sign. In other words...

$\noindent To show that the L.H.S. and the R.H.S. are in the same branch of the tan graph, note that $0<\tan^{-1}\frac{3}{4}<\tan^{-1}1 = \frac{\pi}{4}$, so $0< \frac{1}{2}\tan^{-1}\frac{3}{4}< \frac{\pi}{8}$, which implies that $\frac{3\pi}{4}- \frac{\pi}{8}< LHS <\frac{3\pi}{4}$, i.e...

$\noindent Oh yeah, it might help to first multiply through the equation by 2 actually, so that the L.H.S. just has a $\tan^{-1}\frac{3}{4}$ without a $\frac{1}{2}$ in front, so it's easier to find the tan of this when taking the tan of the L.H.S. For the R.H.S., you can use the fact that $\tan...

$\noindent To prove this result, take the tan of both sides and show they are equal (use trig. identity for R.H.S.: $\tan (\pi -\alpha )=-\tan \alpha$. Use compound angle formula for L.H.S. and the fact that $\tan \frac{3\pi}{4} = -1$.). Then show that both the L.H.S. and R.H.S. lie in the same...

If there's no unusual circumstances preventing you from doing this task, why would you not do it? It's like just throwing away free marks. If you don't care about your marks at all though, then you can just skip it.

Does this assignment count towards your internal marks (it seems like it does)? If so, you should try and do it unless there are circumstances preventing you from doing so, in which case you should probably apply for some sort of alternative assessment.

$\noindent As long as it's of the form $A^2-B^2$, where $A$ and $B$ are some algebraic expressions, the answer will be similar, i.e $(A+B)(A-B)$.$

$\noindent Think of it as $a^2-b^2 =(a+b)(a-b)$, where $a= h+3i$ and $b=7j$.$

They just did difference of two squares.

Yeah, you should describe at least one negative, and then make an overall judgement at the end.

Re: MX2 2016 Integration Marathon $\noindent Complete the square, so the quantity inside the radical becomes $(x+1)^2 -2$. So the answer is $\ln \left(x+1 +\sqrt{(x+1)^2 -2}\right) + C = \ln \left(x+1 +\sqrt{x^2 + 2x-1\right) + C$, using the standard integrals sheet from past HSC papers.$