Print Page - Compilation of solutions to HARD past HSC papers (4U)

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: RuiAce on June 16, 2017, 01:23:27 pm

Title: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on June 16, 2017, 01:23:27 pm

Hi all,

As the thread title suggests, this thread will be reserved specifically for solutions to past BOSTES papers only. You should still post any other questions in the question thread, but please consider looking here if it is from a past BOSTES paper.

The main purpose of this thread is just so that if you have a really long answer, you won't need to go around asking for a solution but can just look for it here. Of course, you're always welcome to ask more about the solution if you're still confused :)

This is a work in progress and I've only decided to open it up now. I will most likely be opening up a thread for 2U and definitely one for 3U as well. I'll add more questions as more people ask them.
_____________________________________

2000
- Q8 a) i) Complex numbers and series

2001
- Q7 b) Hand-waving around polynomials and rational numbers
- Q8 b) Irrationality of $e$

2002
- Q7 b) Complex numbers, vectors and a pentagon

Parts i) and ii)

(https://i.imgur.com/5BKCPBb.png)
(https://i.imgur.com/F3rUGwg.png)
(https://i.imgur.com/5TZBdsX.png)

Parts iii) and iv

(https://i.imgur.com/nYUEP8S.png)
(https://i.imgur.com/eM805hL.png)

2005
- Q8 b) An arguably multifaceted question on a hyperbola

2008
- Q8 a) i) First of the urns question

2010
- Q4 c) Odd case of a quadratic in 4U

2011
- Q8 c) + d) Strange absolute values

2012
- Q15 b) i)-iv) Working around real roots
- Q16 c) i) First part of selecting integers

2013
- Q10 The hostel question

2014
- Q16 c) Surprising integral

2015
- Q16 b) Long winded complex numbers

"AN Final Revision 2017" contains:
- 2003 Q8b
- 2016 Q16b
- 2015 Q16b (Typo in the file)
- 2013 Q16b
____________________________________________

For papers from 2016 onwards, you can access handwritten solutions for the entire paper:
- 2016
- 2017
- 2018
- 2019
- 2020
- 2021

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: Wales on June 16, 2017, 01:25:46 pm

Would it be a good idea to start off with providing solutions to all the last questions for all 4u papers from 99-16? It seems they're usually the harder ones.

Just a thought.

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: brenden on June 16, 2017, 01:27:25 pm

Looks awesome Rui, you legend!

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on June 16, 2017, 01:29:21 pm

Quote from: Wales on June 16, 2017, 01:25:46 pm

Would it be a good idea to start off with providing solutions to all the last questions for all 4u papers from 99-16? It seems they're usually the harder ones.

Just a thought.

It takes time. Personally, I'm fine with people just posting their questions in the 4U questions thread and I'll (or Jake will) get to it afterwards.

If I have time after uni break I'll probably search up some solutions in the past as well. I'm sure there's been more than just that one derangement question that I've done before, which came out of the HSC and can be considered hard.

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: Wales on June 16, 2017, 01:31:11 pm

Quote from: RuiAce on June 16, 2017, 01:29:21 pm

It takes time. Personally, I'm fine with people just posting their questions in the 4U questions thread and I'll (or Jake will) get to it afterwards.

If I have time after uni break I'll probably search up some solutions in the past as well. I'm sure there's been more than just that one derangement question that I've done before, which came out of the HSC and can be considered hard.

Of course. That makes sense, sounds good.

This place is a blessing to us all.

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on June 18, 2017, 07:49:43 pm

Imported a few more and a 3U thread should be opening up soon.

I found it quite funny when three in a row were perms and combs.

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: Wales on June 19, 2017, 11:29:48 am

The "surprising integrals" solutions doesn't lead to anywhere.

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: jakesilove on June 19, 2017, 11:38:19 am

Rui mate this is the best idea ever. You will personally save the marks of hundreds of 4U students.

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on June 19, 2017, 10:43:56 pm

Quote from: Wales on June 19, 2017, 11:29:48 am

The "surprising integrals" solutions doesn't lead to anywhere.

Fixed. Somehow my search link got pasted in and not the actual link to the thread

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on September 28, 2017, 09:56:38 am

Following the video release, I've attached my content block "handout" in the opening thread. I may convert them to screenshots later, but for now the PDFs should be enough :)

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: frog1944 on October 22, 2017, 12:02:49 pm

Hi RuiAce,

I was curious for Q 16 a) i), why does sin(theta)=sin(-alpha) mean that theta = - alpha? Why don't you have to consider it in terms of general solutions, with theta = 2*pi*k - (-1)^k alpha, and then some how work it out from there?

Thanks

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on October 22, 2017, 12:26:52 pm

Quote from: frog1944 on October 22, 2017, 12:02:49 pm

Hi RuiAce,

I was curious for Q 16 a) i), why does sin(theta)=sin(-alpha) mean that theta = - alpha? Why don't you have to consider it in terms of general solutions, with theta = 2*pi*k - (-1)^k alpha, and then some how work it out from there?

Thanks

What year are you talking about?

(Although, sometimes you can make assumptions about the domain of angles. Sometimes it makes no sense if theta falls outside the domains $0\le \theta \le \pi$ or $-\frac\pi2 \le \theta \le \frac\pi2$, and when that happens you can just jump straight to the answer without considering a general solution.)

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: frog1944 on October 22, 2017, 12:28:44 pm

Oh, sorry, HSC 2016 MX2.

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on October 22, 2017, 12:36:33 pm

Quote from: frog1944 on October 22, 2017, 12:28:44 pm

Oh, sorry, HSC 2016 MX2.

$\text{Yeah. Don't need general solution when they explicitly give you}\\ \boxed{-\pi < \theta \le \pi}\text{ and }\boxed{-\pi < \alpha \le \pi}$
Basically you just make the choices for alpha and theta in those specified domains, just like when you solved trig equations back in 2U. The general solution exists for when domains are unspecified.

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: frog1944 on October 22, 2017, 01:53:33 pm

I don't understand. If -pi < x <= pi, -pi < y < pi, and sin(x) = 1/2 and sin(y) = -1/2, then couldn't x = pi/6, 5pi/6 and y = -pi/6, -5pi/6 . Then, couldn't x = pi/6 and y = -5pi/6, then x does not equal -y?

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on October 22, 2017, 02:36:55 pm

$\text{That choice wouldn't have satisfied the other equation}\\ 1+\cos \theta + \cos \alpha = 0$
$\text{In fact }1+\cos \frac\pi6 + \cos \left(-\frac{5\pi}6\right) = 1.$
Looking back, I remember having an annoying time thinking about this question as well. The thing is, if we had only the one equation $\sin x = \sin (-\alpha)$ then we would actually have another case.

The whole point of the other equation was to eliminate the wrong case. When I think about it now, jumping straight to $ \theta = -\alpha$ whilst not rejecting the other case is a mistake. But it is certainly true that once you've rejected it, the only viable case is $\theta = -\alpha$

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: frog1944 on October 22, 2017, 05:41:04 pm

Oh, ok. That makes sense. Great! Thanks RuiAce :)

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on October 22, 2017, 05:44:32 pm

Pretty glad you asked that one though, because I finally resolved a confusion I had last year now because of it aha

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on December 13, 2017, 03:31:17 pm

$\textbf{2001 HSC MX2 Q8b}$
$\text{Because we know that }e^x\le e < 3\textbf{ for all }0\le x \le 1\\ \text{which consequently implies }x^{\alpha}e^x\le 3x^{\alpha} \\ \text{using the theorem in the AN final revision handout gives}\\ \begin{align*}\int_0^1 x^{\alpha}e^x\,dx &< \int_0^1 3x^\alpha\,dx\\ &= \frac{3}{\alpha+1}\end{align*}$
_________________________________________________________________
$\text{When }n=0\text{ we can just compute}\\ \int_0^1 e^x\,dx = -1+e\\ \text{so the choice }a_0 = -1\text{ and }b_0 = 1\text{ asserts the truth of the base case.}$
$\text{Assume that for }n=k\text{, there exists }\textbf{integers}\, a_k\text{ and }b_k\text{ such that}\\ \boxed{\int_0^1 x^k e^x\,dx = a_k + b_k e}\\ \text{and then consider the case }n=k+1.$
$\begin{align*}\int_0^1 x^{k+1}e^x\,dx &=\left[x^{k+1}e^x\right]_0^1-\int_0^1(k+1)x^k e^x\,dx\tag{integration by parts}\\ &= e - (k+1)(a_k + b_k e)\tag{inductive assumption}\\ &= -(k+1)a_k + \left[1-(k+1)b_k\right]e\tag{rearranging}\end{align*}$
$\text{so recalling that }k, a_k\text{ and }b_k\text{ are all }\textbf{integers},\\ \text{the choices }a_{k+1} = -(k+1)a_k\text{ and }b_{k+1}=1-(k+1)b_k\text{ assert our claim.}$
_________________________________________________________________

This is where things get hard. It is not obvious at all that exhaustion is what we require to complete our proof. We will need to do some working backwards to figure out why.
$\text{Take }r = \frac{p}{q}\text{ as given. We keep in mind}\\ \text{that }p\text{ and }q\text{ are }\textbf{positive}\text{ integers}.$
$\text{Substituting this in,}\\ \begin{align*}|a+br| &= \left|a + \frac{bp}{q}\right|\\ &= \left| \frac{aq+bp}{q} \right|\\ &= \frac{1}{q}|aq+bp|\end{align*}\\ \text{recalling that }q > 0,\text{ so }|q| = q.$

Working backwards

$\text{If }|x|=0\text{ for any real (or in fact, complex) number }x,\\ \text{then }x=0.$
$\text{Using this theorem, if }|a+br| = 0\text{, then }a+br = 0\\ \text{so }a=-br.$
And that's that case sorted.
$\text{But instead, if }|a+br| \ge \frac{1}{q}\\ \text{an }\textbf{equivalent}\text{ proposition would be }\frac{1}{q}|aq+bp| \ge \frac{1}{q}\\ \text{i.e. }\boxed{|aq+bp| \ge 1}$
So the question now reduces to how do we prove that, GIVEN we've already eliminated the possibility of $ |aq+bp| = 0$?

As it turns out, we rely on what $a,b,p,q$ all are.

$\text{Since }a,b,p,q\text{ are integers, the expression }aq+bp\text{ must also therefore be an integer.}\\ \text{Hence, }|aq+bp|\text{ is also an integer.}\\ \text{In fact, }|aq+bp|\text{ is a non-negative integer, because of the absolute values.}$
$\text{Hence the only viable cases are that }|aq+bp|\text{ is zero, i.e.}\\\begin{align*} |aq+bp| &= 0\\ \frac1q|a+br| &= 0 \\ |a+br|& = 0\end{align*}$
$\text{or that its a strictly positive integer and hence}\\ \begin{align*}|aq+bp|&\ge 1\\ \frac{1}{q}|aq+bp|&\ge \frac{1}{q}\\ |a+br|&\ge \frac{1}{q}\end{align*}$
_________________________________________________________________
$\text{Given what we've proven in part iii)}\\ \text{it makes sense to deduce that we must find a contradiction for part iv).}$
$\text{For the sake of contradiction, suppose that }e\text{ is, in fact, rational.}\\ \text{Hence, taking }e=r\text{ in the above we have positive integers }p\text{ and }q\text{ such that}\\ \boxed{e=\frac{p}{q}}$
Note: The statement $e > 0$ is certainly safe to assume.
$\text{Keeping in mind part ii), we consider part i)}.\\ \text{The inequality can be continued to give}\\ \int_0^1 x^ne^x\,dx<\frac{3}{n+1} < \frac{3}{n}.$
$\text{So if we let }n=3q\text{, noting }q\text{ is an integer, we deduce that}\\ \boxed{\int_0^1 x^{3q}e^x\,dx < \frac{1}{q}}.$
$\text{But from part ii), we know that }\boxed{\int_0^1 x^{3q}e^x\,dx = a_{3q}+b_{3q}e}\\ \text{for some integers }a_{3q}\text{ and }b_{3q}.\\ \text{Therefore, we've found integers so that}\\ \boxed{a_{3q}+b_{3q}e < \frac{1}{q}}$
$\text{But also, }\boxed{\int_0^1x^{\alpha}e^x\,dx > 0}\\ \text{which can easily be deduced from a graph or basic facts.}\\ \text{Therefore }\boxed{a_{3q}+b_{3q}e > 0}\\ \text{so combining with the above, we arrive at }\boxed{0 < a_{3q}+b_{3q}e < \frac1q}.$
$\text{Of course, this now implies that}\\ \boxed{0 < |a_{3q}+b_{3q}e | < \frac1q}.\\ \text{But this }\textbf{literally contradicts what we said in part iii).}$
Remember that $r=e$ for our purposes.
In part iii), we have claimed that for any choice of integers $a$ and $b$, we must have $ |a+be| = 0 $ or $ |a+be| \ge \frac1q $. What we've just done, is shown that the integers $a_{3q}$ and $b_{3q}$ don't satisfy either of the above. This is because the above inequality can be dissected into $|a+be| \neq 0$ AND $ |a+be| < \frac{1}{q} $.
$\text{And hence, having found a contradiction,}\\ \text{our original assumption was wrong, and }e\text{ is thus irrational.}$
(Remark: This is my preferred way of doing this question. There is another alternate method very similar to this, but I find it hard to explain without using intuition beyond the MX2 student's capability. I will remark that the proof's conclusion is a bit nicer though - instead of contradiction part iii, we contradict the elementary statement $\text{a number cannot be smaller and greater than another number at the same time}$)

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on September 28, 2018, 05:32:39 pm

$\textbf{2005 HSC MX2 Q8 b)}$
$\text{Since }AP\text{ and }PB\text{ are both vertical line segments, we only need to}\\ \text{consider the difference in their }y\text{-coodinates for the distance.}\\ \begin{align*}\therefore AP \times PB &= |b \sec \theta - b \tan \theta| |b\sec \theta + b \tan \theta|\\ &= |b|^2 |\sec^2\theta - \tan^2\theta|\\ &= b^2.\end{align*}$
Can of course, still be done with the usual distance formula if you wish.
_______________________________________________________________

This part more or less requires a lot of angle-chasing. It's not immediately obvious that $ \angle ACP = \alpha - \beta$, but firstly the presence of $AC$ and $AP$ should've hinted at narrowing our focus to $ \triangle ACP$, and also the weird expressions should've hinted at trigonometry.
$\text{First note that }\angle QPB = \frac\pi2 - \alpha\text{ from the angle sum of the right-angled triangle containing }\alpha.\\ \text{Then also, }\angle OAB = \frac\pi2 - \beta\text{ from the angle sum of the right-angled triangle containing }\beta.$
$\text{Therefore }\boxed{\angle CAP = \frac\pi2 + \beta}\text{ from the straight angle }\angle UAC\\ \text{and thus }\boxed{\angle ACP = \alpha - \beta}\text{ from the angle sum of }\triangle ACP$
$\text{It follows that by the sine rule on }\triangle CAP,\\ \frac{CP}{\sin \left( \frac\pi2 + \beta \right)} = \frac{AP}{\sin (\alpha - \beta)} \implies \boxed{CP = \frac{AP \cos \beta}{\sin (\alpha - \beta)} }.\\ \text{Note that }\sin \left( \frac\pi2 + \beta\right) = \sin \frac\pi2 \cos \beta + \cos \frac\pi2 \sin \beta = \cos \beta.$
$\text{Then, note that the angle 'below' the marked }\beta\text{ is also }\beta\\ \text{just because it's the angle subtended by the asymptote.}\\ \text{Also, write down the vertically opposite angle at }\alpha.$
$\text{Considering the exterior angle of a triangle, this shows that }\boxed{\angle PDB = \alpha+\beta}.$
(I didn't feel like adding labels because it would look complicated, but apologies in that I wasn't bothered to copy out the diagram either.)
$\text{Then by the angle sum of }\triangle PDB, \boxed{\angle PBD = \frac\pi2 - \beta}\\ \text{so a similar sine rule argument shows that }\boxed{PD = \frac{PB \cos \beta}{\sin (\alpha+\beta)}}$
_______________________________________________________________
$\text{We see that }CP \times PD = \frac{AP \times PB \cos^2\beta}{\sin (\alpha-\beta) \sin (\alpha+\beta)} = \frac{b^2\cos^2\beta}{\sin (\alpha-\beta)\sin (\alpha+\beta)}.$
$\text{This is 'dependent' on }b, \beta\text{ and }a,\\ \text{however }b\text{ and }\beta \text{ are fixed constants from the hyperbola itself.}\\ \text{Therefore the only dependence is on }\alpha\text{, and consequently not }\theta\text{, which reflects the position of }P.$
It may be worth noting that $CP$ does depend on the position of $P$ due to the problem of $AP$. It's just that upon multiplying, we can take advantage of the fact that $ AP\times PB$ is equal to the constant value of $b^2$. Also, the reason behind why $ \beta$ is constant is because it's uniquely determined by the asymptotes, which are also determined by the hyperbola itself.
_______________________________________________________________
$\text{The significance in part c) is that the dependence of }\alpha\\ \text{allows us to view the point }P\text{, and the point }Q\text{ in a similar way.}\\ \text{Basically, we can repeat the method used in parts a) and b) to show that}\\ \boxed{CQ\times QD}\text{ also depends on only }\alpha.\\ \text{This follows from that }Q\text{ is other point of intersection between the line and the hyperbola.}$
$\text{In fact, exploiting that they }\textbf{do}\text{ lie on the same line, we can anticipate that}\\ \text{they are in fact, equal: }\boxed{CP \times PD = CQ \times QD}.$
$\text{So decomposing }PD\text{ and }CQ,\\ \begin{align*}CP (PQ + QD) &= (CP + PQ)QD\\ p(r+q) &= (p+r)q\\ pq&=qr\\ p&=q \end{align*}$
_______________________________________________________________

One may be tempted to jump straight into bashing it (finding the equation of the tangent, finding any intercepts and explicitly using the midpoint formula). But given that it was only a 1 mark question, it had to relate to part iv) somehow.
$\text{Essentially, visualise what happens as we progressively move the point }P\text{across}\\ \text{to eventually }\textbf{coincide}\text{ with the point }T.$
$\text{As }P \text{ approaches }T,\, Q\text{ must approach }T\text{ as well.}\\ \text{That is to say, as }PT\to 0, \, QT \to 0.$
Note that this is equivalent to saying that the points $P$ and $Q$ must coincide...
$\text{Therefore, the secant }PQ\text{ must tend towards the tangent at }T.\\ \text{This means that furthermore, }C\text{ must be approaching }U\\ \text{and also }D\text{ must be approaching }V.$
$\text{So since }\boxed{CP = DQ}\text{, by limiting }P\text{ and }Q\text{ towards the point }T,\\ \text{we see that }\boxed{UT = VT}.\\ \text{Therefore }T\text{ must be the midpoint of }UV.$

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on November 21, 2018, 01:35:21 pm

I have completed the solutions to the 2020 sample paper uploaded by NESA recently to the new syllabus resources. This was obviously unnecessary as they already had solutions at the end of the paper. It is only here for the sake of completeness.

(The images zoom in heaps if you tap on them.)

Multiple Choice

(https://i.imgur.com/Ld2borQ.png)

Short Answers

(https://i.imgur.com/zTFB0MY.png)

All NESA resources are accessible here for the new MX2. Links to the other maths syllabus materials are also available. Note that the formula sheet has also been updated and whilst less well formatted, it covers a larger range of material.

Note that the sample paper is not a full paper. Only 30 marks of short response questions have been provided, however you will be given 90 marks worth of those in the actual exam.

Note that HSC-2019 students do not need to worry about this post. However there are some questions that are still relevant to your current syllabus, if you want that extra practice.

Title: Re: Compilation of solutions to HARD past HSC papers (4U)
Post by: RuiAce on December 18, 2019, 11:30:02 pm

I had a look at my solution to 2001 Q7b and I'm kinda disappointed by it. It looks like i went down a needlessly convoluted path somewhere. I'm swapping out for the following set of handwritten solutions.

Click me

(https://i.imgur.com/u1Mj1HV.png)
(https://i.imgur.com/PXvBxy8.png)
(https://i.imgur.com/OEJzCgX.png)
(https://i.imgur.com/F3AaalY.png)

ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: RuiAce on June 16, 2017, 01:23:27 pm