Search results

Re: HSC 2016 4U Marathon - Advanced Level $This isn't sufficient to prove it's monotone decreasing on $[1,\infty)$.$

Yeah, since the Key Word is "assess", which means you need to make a judgement. Assess: "Make a judgement of value, quality, outcomes, results or size" ( The BOSTES definitions of key words is found here, and you should be familiar with them...

Re: Announcement from BOSTES - significant change to calculus courses What's the main reason some people aren't happy with the formula sheet?

Here's a playlist of videos from MIT OpenCourseWare which is from a course called Mathematics for Computer Science: https://www.youtube.com/watch?v=L3LMbpZIKhQ&list=PLB7540DEDD482705B

Re: HSC 2016 4U Marathon $Technically $y=x$ is both a concave and convex function (which is why I didn't use it as the example). However, it is not a \textsl{strictly convex} function.$

Re: HSC 2016 2U Marathon Are you sure you meant sum, or did you just mean limit of that expression? Because that sum clearly will not converge, as can be seen by using the integral test ( https://en.wikipedia.org/wiki/Integral_test_for_convergence ).

Re: Announcement from BOSTES - significant change to calculus courses Doesn't make a difference except they'll probably stop asking Q's like 'calculate the angle between these two lines', since those were just a test of memory and substitution (if they ask it now, it'll just be a test of...

Re: HSC 2016 4U Marathon $Simple complex numbers Q:$ $Let $z$ be a non-zero complex number. Show that $\frac{1}{z}$ is real if and only if $z$ is real.$

Re: HSC 2016 4U Marathon $This part is not true. If $y=\mathrm{e}^{f(x)}$ and $f$ is twice differentiable, then $y^\prime = f^\prime (x)\mathrm{e}^{f(x)}$, and so$ $$\begin{align*}y'' &= f'' (x)\mathrm{e}^{f(x)} + f^\prime (x) \times f^\prime (x)\mathrm{e}^{f(x)} \\ &= f'' (x)...

Re: HSC 2016 2U Marathon This is not first principles.

Re: HSC 2016 2U Marathon $$\frac{1}{1048577}x^{1048577}+c$$

Re: HSC 2016 MX2 Combinatorics Marathon You also apparently got a probability greater than 1 – a sign there's something wrong somewhere in the working.

Re: HSC 2016 3U Marathon It's this one (Cauchy's forward-backward induction proof): https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Proof_by_Cauchy_using_forward.E2.80.93backward_induction This proof was also an Extension Q in a 3U Pender (Cambridge) textbook...

$Have you learnt the cosine rule? If you have, you don't need to draw up right-angled triangles as this rule applies to arbitrary triangles, and you can do this question straightaway. Applying this rule, we have$ $$\begin{align*}BC^2 &= AB^2 + CA^2 - 2\cdot AB \cdot CA \cdot \cos A \\ &= 13^2...

Re: HSC 2016 MX2 Combinatorics Marathon $Note that for (a), the arrangements can be like this too:$ $ $\text{MATHS}\, \text{MATHS}\, *\, *$$ $ $* \,\text{MATHS} \, * \text{MATHS}$$ $and other arrangements, where * represents an arbitrary letter from the alphabet.$

Re: HSC 2016 2U Marathon $Sky is advantaged slightly by going first.$ $Let $P$ be the probability that Sky wins. Then $P=\frac{1}{36}+\frac{35}{36}\times \frac{35}{36}\times P$ (use tree diagram if you want). Solving for $P$ yields $P=\frac{36}{71}$. An alternative method is to use an...

$One of your answers is $\frac{1}{3}\ln \left| 3x\right|+C_1$. Another is $\frac{1}{3}\ln \left| x\right|+C_2$ ($C_1,C_2$ arbitrary constants). Note that the first answer an be simplified to $\frac{1}{3}\left(\ln \left|x\right| + \ln 3\right)+C_1 = \frac{1}{3} \ln |x| + \frac{1}{3}\ln 3 + C_1 =...

Re: MX2 2016 Integration Marathon $Here are two currently unanswered ones from the user juantheron:$

Re: MX2 2016 Integration Marathon The 2015 one still has some unanswered ones.

Re: HSC 2016 4U Marathon Yeah, integral is another word for integer (as an 'adjective').