Carrotsticks' 2016 HSC MX2 Solutions (1 Viewer)

Carrotsticks · Oct 21, 2016

Solutions here

https://www.dropbox.com/s/7mw7k1nfdzknfzu/HSC 2016 Extension 2 Solutions.pdf?dl=0

Glyde · Oct 21, 2016

Carrotsticks said:
To be uploaded here. I will initially only do Q15 and 16, and then complete the remaining questions tonight when I return home from an event.

Am currently at Mittagong Maccas if anybody wants to drop by!

i work there haha

severus snake · Oct 21, 2016

Carrotsticks · Oct 21, 2016

Sorry all, I was only able to finish Q15. There's only so much I can do on my tiny laptop.

I will begin the rest of the solutions at around 11pm tonight when I return home, so a full set should be ready for you tomorrow morning. But I am sure somebody else would have done full solutions by then.

Here it is.

https://www.dropbox.com/s/xzwktjrjci6k7b4/HSC 2016 Extension 2 Solutions (Incomplete).pdf?dl=0

wu345 · Oct 21, 2016

no need to do the last part of Q16 because it was in the 2014 BOS Trial

InteGrand · Oct 21, 2016

wu345 said:
no need to do the last part of Q16 because it was in the 2014 BOS Trial

Also the Year 12 3U Pender Textbook had some extension Q's on derangements, but I can't remember whether they got you to derive the formula for D(n) too (I suspect they did, but they didn't guide you through it as much as this HSC paper).

raghib483 · Oct 21, 2016

the relevant Cambridge question about derangements: its from exercise 10E

Carrotsticks · Oct 22, 2016

Uploaded solutions in the original post!

InteGrand · Oct 22, 2016

Carrotsticks said:
Uploaded solutions in the original post!

$\noindent Thanks for taking the trouble to write these up!$

$\noindent Really minor thing, but for $16$ (c) (iv), you used an initial condition of $D(0) = 1$. But the question as written only ever defined $D(n)$ for $n\geq 1$, giving $D(1) = 0$ and $D(2) = 1$ as the initial conditions. (Of course, when derangements are defined starting from $0$, we do set $D(0) = 1$, but the Q. here hasn't defined a $D(0)$, so I think they wanted $n=1,2$ as the initial conditions. We could obtain a value of $D(0)$ using the recurrence though so that $D(2)$ would be $1$, and it'd be $D(0) = 1$. This is indeed why the convention is to set $D(0) = 1$ if we define a derangement at $0$.)$

$\noindent The proof without using $D(0)$ ends up being essentially the same though. From $T_{n+1} = -T_{n}$, we get $T_{n} = (-1)^{n-2}T_2 = (-1)^n T_2$, for $n\geq 2$ (i.e. $n>1$). Using the given initial conditions, $T_{2} = D(2) -2D(1) = 1-0 = 1$, so we have $T_n = (-1)^n$ again. The HSC might not have cared if $D(0)$ was used (not sure), but I guess a lot of students wouldn't have considered $D(0)$ in the first place.$

calamebe · Oct 22, 2016

Also for 14) a) iii) you forgot the squared on the pi, again fairly minor thing.

Carrotsticks · Oct 24, 2016

Have fixed the errors you have mentioned. Thank you for pointing them out. The updated dropbox link is now in the original post.

Carrotsticks' 2016 HSC MX2 Solutions (1 Viewer)

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