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Here's what I got based on skimming through them: $\noindent 1 (C); 2 (C); 3 (A); 4 (D); 5 (B); 6 (C); 7 (D); 8 (C); 9 (A); 10 (B)$

There's Q's about it in an extension section of the 3U Year 12 Pender textbook. Also, the question explained everything needed to do it (i.e. no prior knowledge of derangements was prerequisite). $\noindent Incidentally, the last three parts of that Q. had essentially nothing to do with...

2014 BOS Trial last page. Also, as Carrotsticks pointed out on the other thread, Question 15(c) was very similar to a question from this year's BOS 3U 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).

Yeah - 2014 BOS Trials (last batch of Q's in that paper iirc). Also, I think the Year 12 3U Pender textbook had a couple of extension Q's on derangements, but they didn't guide you through it like this paper did.

$\noindent For 13(d), $p(x) = ax^{3} + bx^2 + cx + d$, $p'(x) = 3ax^2 + 2bx + c$. Roots of $p'$ are $\frac{-2b\pm \sqrt{4b^2 - 4 \times 3ac}}{2\times 3a} = \frac{-b \pm \sqrt{b^2 - 3ac}}{3a}$. So if $b^{2} -3ac = 0$ and $p\left(x_{0}\right) = 0$, where $x_{0} = - \frac{b}{3a}$, we have that...

$\noindent Yeah, 10 is (B). Note that if we let $x = \mathrm{cis}\left(\theta\right)$ (by symmetry it suffices to let $0 \leq \theta \leq \pi$), we obtain $2\cos \theta = -1 \Rightarrow \cos \theta = - \frac{1}{2}$. So $\theta = \frac{2\pi}{3}$. It follows that $x^{2016} + \frac{1}{x^{2016}} = 2...

Yep it's y = 100. All it amounts to is that the parabola y(200 - y) (which is concave down obviously since the leading coefficient is negative) has its vertex (hence maximum) at y = 100 (midpoint of the roots).

14 a(ii): cos(x) is odd about x = pi/2, so raising cos(x) to an odd power (or more generally, applying any odd function to it) will maintain this property. In other words, (cos(x))^(2n-1) is also odd about pi/2 for any positive integer n, so integrates to 0 over 0 to pi (integrating a function...

7 is (D) $\noindent The result is that $x^{2} - y^{2} =k$ and $xy = \frac{1}{2}k$ ($k > 0$) are $45^\circ$ rotations of each other.$

One reason they ask Q's like that I think is because they want to try and ensure that everyone gets at least some marks.

tywebb posted it here: http://community.boredofstudies.org/1189/mathematics-extension-2/355617/ext-2-paper.html .

They're probably not written up yet, but once they are, they should come up under the Extension 2 part of this page: http://www.hsccoaching.com/documents/35.html .

Ah yeah, you're right. Took a look at that ellipse Q. and the marks there are essentially all standard. But I think one reason they ask Q's like that is to try and make sure everyone can at least get some marks.

Do you think there was a bit too much guidance given in some places? (It seems like that in some places to me, made some of the questions much easier.)

Is the general consensus then that it was easy this year but just had Q's involving lots of tedious algebra?

Yeah, that's right.

$\noindent The domain is $t\geq 0$, so the range is $\left[10,200 \right)$. This is because the function $1 + 19e^{-0.5t}$ ($t \geq 0$) has range $\left(1,20 \right]$ (it starts at $20$ and then decreases (monotonically and asymptotically) to $1$). Reciprocating makes the range become $\left[...

Yeah looking at that question I was also thinking I'd seen it in some paper (was also thinking a BOS Trial) somewhere recently.

Wow last Q. was derangements, something covered in a past 4U BOS Trial (the 2014 one I think) from memory!