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Re: MX2 2016 Integration Marathon Haha wasn't it in a Carrotsticks trial as well? Or maybe one of the 'scrapped' questions.

Re: MX2 2016 Integration Marathon Was this in a Carrotsticks Trial recently?

Re: MX2 2016 Integration Marathon $\noindent If I am not mistaken, improper integrals aren't explicitly in the syllabus now. But they have asked Q's in the current syllabus asking for an area under a curve (or maybe it was a volume) over an infinite interval, by asking to find the area from...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $This is because for our original function $f(x)$, it's the function $x-\frac{1}{x}$ \textsl{restricted to }$x>0$. In other words, we only took one branch of the curve $y=x-\frac{1}{x}$, namely the branch to the right of the...

Re: MX2 2016 Integration Marathon Yeah, these aren't in the syllabus.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Actually, this is when we sub. in \emph{small} $y$-values, so that $\frac{1}{y}$ is large.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread It actually does do all these things. Observe that in the WolframAlpha plot, the "x"-axis is actually the vertical axis. So they simply plotted the same graph but switched the axes around. So as you can see, the horizontal...

Re: MX2 2016 Integration Marathon Yeah, there is definitely a question there about their derivatives' properties, but they don't mention that these functions are the hyperbolic functions. HSC students aren't expected to know the properties of these functions, or even what these functions...

What would happen if BOSTES around places where you should have lost marks (but didn't) and also places where you should have actually gained some extra marks. Would they just give you those extra marks you should have gained, or would they not give you those. I'm pretty sure if there are only...

Re: MX2 2016 Integration Marathon You're not mistaken. Some teachers might teach it as an 'extension'.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent We sketch the curve $y=f(x) = x - \frac{1}{x},x>0$ using subtraction of ordinates (i.e. subtract $y$-values of the second graph they told us to consider from the first one). Consider the graphs of $y=x$ and...

Re: MX2 2016 Integration Marathon $\textbf{NEW QUESTION}$ $\noindent Find $\int \frac{1}{x+\sqrt{x^2 +x + 1}}\text{ d}x$.$

Re: HSC 2016 2U Marathon $\noindent \textbf{ANOTHER QUESTION}$ $\noindent Let $a,b,c,d$ be numbers satisfying $\frac{a}{b} =\frac{c}{d}$, $a,b\neq 0$. Show that if $b+d \neq 0$, then both of $\frac{a}{b}$ and $\frac{c}{d}$ are equal to $\frac{a+c}{b+d}$.$

$\noindent Usually can be told from context. I think I've seen it used for both square of log or log of log. Best to just say what you mean by it before using it I suppose.$ $\noindent (Similar story with using something like $f^2$ etc. E.g. if writing about physics, it may be quite clear that...

Re: MX2 2016 Integration Marathon Good news, LaTeX appears to be working here again. Test: x

Re: MX2 2016 Integration Marathon Nice! The last integral can be quickly dealt with using the auxiliary angle method to transform the denominator into a single sinusoid or cosinusoid. Then have fun transforming everything back in terms of x (something you'd probably feel is the last thing...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Not sure why they omitted those points for b) though, since they do give imaginary results if you sub. them in to the expression.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread I haven't bothered doing the question, but notice that your answers match the answers given apart from some omissions of points in the answers. E.g. your x + y = 1 line is precisely the line through 1 (the point (1,0)) and i...

Re: MX2 2016 Integration Marathon And to compound matters, LaTeX doesn't seem to be working on the site right now. Test: $x^2$

We have two cases. Case 1. 2x – 6 ≥ 0 (i.e. x ≥ 3) In this Case, we have |2x – 6| = 2x – 6. So the inequation is 2x – 6 < x + 3, x ≥ 3 ⇒ x < 9, x ≥ 3. So 3 ≤ x < 9 is one part of the solution. Case 2. 2x – 6 < 0 (i.e. x < 3) In this Case, we have |2x – 6| = –(2x – 6) = 6 – 2x. So the...