If there isn't a response in the next little while I might have to concede there was a typo. It came from a past paper from the 90's of a reputable school.
Ive tried to guess the typo, but am interested to know if anyone has solved it.
Thanks
Re: HSC 2016 4U Marathon - Advanced Level
\large \textrm{The shaded (parallel) cross section is an isosceles trapezium with height } h \textrm{ and parallel sides equal to } y \textrm{ and } z \textrm{ as shown}. \\\textrm{The rear face } ABCD \textrm{ is a square with side length of } a...
Re: HSC 2015 4U Marathon - Advanced Level
Hi seanieg89. I can get some sort of induction argument going, but was wondering if there is a proof by contradiction. It seems just right for one!?
Re: HSC 2015 4U Marathon - Advanced Level
$A function $f(x) = ax^2 +bx+c$ has real coefficients and satisfies $|f(x)|\leq 1$ for all $x\in[0,1]$. \\ Find the maximal value of $|a|+|b|+|c|$ $
Re: HSC 2014 4U Marathon - Advanced Level
$Case I: \textbf{X and his wife and Y and his wife are all in the same boat}. \\\\This means that the remaining eight men can be distributed amongst the other boats in $ \frac{8!}{4!(2!)^4} $ ways. Here I am assuming the boats themselves are...
Re: HSC 2014 4U Marathon - Advanced Level
I think what's going on here is that you are assuming that different boats means different arrangement. I think the question is asking how many different ways can they be separated. Sort of life "how many ways can 10 people be divided into two teams"...
Re: HSC 2014 4U Marathon - Advanced Level
yep both cases. i.e. Case 1. X and Wife and Y and Wife all in the same boat. Case 2: X and Wife in one boat. Y and wife in another boat