BoS Maths Trials 2019 (1 Viewer)

Trebla · Sep 28, 2019

Sick of doing standard repetitive questions and want some good practice for the harder questions in the HSC maths exams?

Want to be challenged and tested on how well you really know your stuff?

Know someone at school who is a gun at HSC maths?

It is that time of year again where Bored of Studies will be holding trial exams for:
• Mathematics Extension 1
• Mathematics Extension 2

These are aimed at students seeking a challenging paper that will push their limits and test how well they really know their stuff. The exams have been designed by our team maintaining a high standard of challenging questions and proofs whilst still remaining within the bounds of the HSC syllabus. They will interest you, challenge you and show you ways of thinking for certain topics that you may not have considered before (which will be very handy for the actual HSC exam!). Students who have taken these trial exams in previous years can attest to this.

Students will sit the examination under supervised exam conditions and will be allowed to keep their copy of the exam paper to bring home. A team of markers will then mark the responses and post the results on this forum in due course, along with the worked solutions.

Note: Students who may be worried about losing motivation (due to results) are reminded that this is NOT intended to be an accurate reflection of the difficulty of the HSC. The questions are made to be as closely styled as possible to the HSC exams (see previous papers), but this will be your last chance to practice a full exam as close to HSC conditions as possible.

_________________________________________________

Location

University of Technology Sydney
Building 5 Block C
Room CB05C.01.029 (Level 1, Room 29)
1 Quay St, Haymarket (Sydney)

About a 10 minute walk from Central station

Date and times

Thursday 10th of October

Mathematics Extension 2
9:30am - 12:35pm

Mathematics Extension 1
2:00pm - 4:05pm

Target audience
It is highly recommended that students attempting the paper are comfortably in the Band E4 range, aiming for 95+ in any of the Mathematics Extension courses. The exams will very thoroughly examine your understanding of different topics, but more importantly your ability to apply this to solving problems under exam conditions. It is also highly recommended that those attempting the Mathematics Extension 2 paper also attempt the Mathematics Extension 1 paper, which I guarantee will also be sufficiently challenging.

What to bring
The usual materials allowed for the HSC exam, including pen and calculator. Exam writing booklets will be provided on the day.

Cost
We kindly ask that you each donate just $5 to partially cover our printing costs. If you are doing both papers you only need to pay once. The rest comes out of our own pockets. Contrary to what you might hear, we are actually NOT paid to put the time and resources into designing the paper and marking your responses. It is purely out of our passion to offer you guys something unique and valuable for your studies.

How to register

Please send me a private message (here or click on "Start a new conversation" on my profile) stating the following:
• Name
• School
• Which exam(s) you will be sitting (Ext1, Ext2)

If you wish to invite some friends (which I encourage provided they fit the recommended target audience described above), you may register on their behalf by providing their details as well. If you know anybody who is good at Mathematics or a 'beast' at it, I ask that you link them to this page. We want to put the best people in the state up against each other for this exam. Consider it a challenge from us.

If you do not have an account, I encourage you to make one.

Once you have registered, you will receive a student number to confirm your registration. Keep this student number with you on the day of the exam. You can use it to track your results when I put the marks online (only the student number and marks will be shown). You are only registered when I have replied with your student number.

Note we will cap registrations at 40 people for each exam, so get in quick!

Past exams and solutions
Here are some past papers since 2012 to give you an idea of what to expect. Feel free to use them as preparation for this year's paper or just practice in general.

Mathematics Advanced (2013-2015 only)
https://community.boredofstudies.or...hematics-2012-2015-exams-and-solutions.17177/

Mathematics Extension 1
https://community.boredofstudies.or...ension-1-2012-2018-exams-and-solutions.17484/

Mathematics Extension 2
https://community.boredofstudies.or...ension-2-2012-2018-exams-and-solutions.17485/

_________________________________________________

Look forward to seeing you there!

Drdusk · Sep 28, 2019

hahahahahahhaha good luck. Prepare to get destroyed

Trebla · Oct 3, 2019

Bump...this is just a week away!

Trebla · Oct 6, 2019

Bump! There are still places left and it's only a few days away...

Trebla · Oct 9, 2019

This is on tomorrow! Final registrations will be taken by 7pm today.

What do you reckon q16 and q14 of the 4u and 3u papers will be?

TheOnePheeph · Oct 9, 2019

Trebla said:
This is on tomorrow! Final registrations will be taken by 7pm today.

What do you reckon q16 and q14 of the 4u and 3u papers will be?

I'm willing to bet you will chuck a hard circle geo X conics question in there somewhere, since they are both gone from the syllabus next year. I have a feeling q16 will involve a really difficult polynomials question, possibly similar to 2001 hsc q 7b, but a lot harder. I'd love to see something like a proof for the fundamental theorem of algebra, but that may be too hard for x2.

Unfortunately I'm not able to attend this, but I'm really excited to see the paper. Good luck everyone who is!

Drdusk · Oct 9, 2019

Trebla said:
What do you reckon q16 and q14 of the 4u and 3u papers will be?

Do I really wanna guess.. haha

But if I had to I would say it's probably some hard as proof by contradiction mixed with some integrals, maybe inequalities.

psmao · Oct 9, 2019

Should we get their 30 minutes earlier?

CellerySticks · Oct 9, 2019

Do you know what would be funny, if they made us draw the graph of

y=\left|\sin x\right|+5e^{-x^{100}}\cdot\cos x (just copy paste this into desmos)
or my personal favourite
0=2.8x^{2}(x^{2}(2.5x^{2}+y^{2}-2)+1.2y^{2}(y(3y-0.75)-6.0311)+3.09)+0.98y^{2}((y^{2}-3.01)y^{2}+3)-1.005 (copy paste into desmos)

EDIT: Possibly NSFW

Drdusk · Oct 9, 2019

CellerySticks said:
Do you know what would be funny, if they made us draw the graph of View attachment 27276
y=\left|\sin x\right|+5e^{-x^{100}}\cdot\cos x (just copy paste this into desmos)
or my personal favourite
0=2.8x^{2}(x^{2}(2.5x^{2}+y^{2}-2)+1.2y^{2}(y(3y-0.75)-6.0311)+3.09)+0.98y^{2}((y^{2}-3.01)y^{2}+3)-1.005 (copy paste into desmos)

I would cry seeing that in an exam...

mathsbrain · Oct 9, 2019

As a quick warm up for tomorrow, can anyone answer with reasons
Given 50 cards with the integers 1, 2, 3, ... 50 printed on them, how many ways are there to select 9 distinct cards, such that no two cards have consecutive numbers printed on them?

yaboinan · Oct 9, 2019

hi, im currently doing maths 4 unit and in my trial i had received 98/100. My other subjects are 3 unit 97/100, maths advanced 97/100, english advanced 93/100, chemistry 98/100 and physics 96/100. Hope these marks stay for the hsc. Also i have no studied yet. RIP

yaboinan · Oct 9, 2019

mathsbrain said:
As a quick warm up for tomorrow, can anyone answer with reasons
Given 50 cards with the integers 1, 2, 3, ... 50 printed on them, how many ways are there to select 9 distinct cards, such that no two cards have consecutive numbers printed on them?

By jumping of a building.

Ilovecarrots · Oct 9, 2019

yaboinan said:
By jumping of a building.

off* are you sure you got 93 in English?

InteGrand · Oct 9, 2019

mathsbrain said:
As a quick warm up for tomorrow, can anyone answer with reasons
Given 50 cards with the integers 1, 2, 3, ... 50 printed on them, how many ways are there to select 9 distinct cards, such that no two cards have consecutive numbers printed on them?

There's probably a simpler way to do it, but here's a sketch for one way.

$\noindent In general we can ask the same question but with $K$ integers chosen from $\{1,\ldots,N\}$.$

$\noindent Assume the integers are $a_1 < a_2 < \cdots < a_K$ ($K=9$ in your example). We need $a_k \ge a_{k-1}+2$ for all $k =2,\ldots,K$. So we can write $a_2 = a_1+ b_1$, $a_3 = a_2+b_2 = a_1+b_1+b_2,\ldots, a_K = a_1 + b_1 + b_2+\cdots + b_{K-1}$, where $b_k \ge 2$ and $b_k\in \Bbb{Z}$ for all $k=1,\ldots K-1$. Since $a_K \le N$ (remember $N = 50$ in your example), we have $b_1+b_2+\cdots+ b_{K-1} \le N-a_1$

$\noindent Equivalently, we can write $b_k = 2+c_k$, where $c_k\in\Bbb{N}$. Then $c_1+\cdots+c_{K-1} \le N - 2(K-1) - a_1$.$

$\noindent Note we must have $a_1 \le N - 2(K-1)$ to have a valid choice. Now, find the number of solutions $(c_1,\ldots,c_{K-1})$ to $c_1+\cdots+c_{K-1} \le N - 2(K-1) - a_1$ (call this $f(a_1; K,N)$), and then your answer is given by $\sum\limits_{a_1=1}^{N-2(K-1)}f(a_1;K,N)$.$

$\noindent Some further helpful facts:$

$\noindent 1) the number of non-negative integer solutions $(c_1,\ldots,c_{K-1})$ to $c_1+\cdots+c_{K-1} \, \color{red}{ = } \color{black}\,j$ is $\binom{j+K-2}{K-2}$ (using Stars and Bars)$

$\noindent 2) $\sum\limits_{t=k}^{n}\binom{t}{k} = \binom{n+1}{k+1}$ (``Hockey-Stick Identity'').$

mathsbrain · Oct 9, 2019

InteGrand said:
There's probably a simpler way to do it, but here's a sketch for one way.

$\noindent In general we can ask the same question but with $K$ integers chosen from $\{1,\ldots,N\}$.$

$\noindent Assume the integers are $a_1 < a_2 < \cdots < a_K$ ($K=9$ in your example). We need $a_k \ge a_{k-1}+2$ for all $k =2,\ldots,K$. So we can write $a_2 = a_1+ b_1$, $a_3 = a_2+b_2 = a_1+b_1+b_2,\ldots, a_K = a_1 + b_1 + b_2+\cdots + b_{K-1}$, where $b_k \ge 2$ and $b_k\in \Bbb{Z}$ for all $k=1,\ldots K-1$. Since $a_K \le N$ (remember $N = 50$ in your example), we have $b_1+b_2+\cdots+ b_{K-1} \le N-a_1$

$\noindent Equivalently, we can write $b_k = 2+c_k$, where $c_k\in\Bbb{N}$. Then $c_1+\cdots+c_{K-1} \le N - 2(K-1) - a_1$.$

$\noindent Note we must have $a_1 \le N - 2(K-1)$ to have a valid choice. Now, find the number of solutions $(c_1,\ldots,c_{K-1})$ to $c_1+\cdots+c_{K-1} \le N - 2(K-1) - a_1$ (call this $f(a_1; K,N)$), and then your answer is given by $\sum\limits_{a_1=1}^{N-2(K-1)}f(a_1;K,N)$.$

$\noindent Some further helpful facts:$

$\noindent 1) the number of non-negative integer solutions $(c_1,\ldots,c_{K-1})$ to $c_1+\cdots+c_{K-1} \, \color{red}{ = } \color{black}\,j$ is $\binom{j+K-2}{K-2}$ (using Stars and Bars)$

$\noindent 2) $\sum\limits_{t=k}^{n}\binom{t}{k} = \binom{n+1}{k+1}$ (``Hockey-Stick Identity'').$

Thanks, but i think this is way over the top for 4unit.
I saw a way a teacher did it through binary. The analogy goes something like this:

1 0 1 1 1 1 0 0 0 is a binary string of length 10. In a binary string of length 50, how many ways are there to have a string with exactly nine lots of ones and that no ones are next to each other?

I fail to see how these two are the same question/where is the one-to-one correspondence.
Maybe trebla can help?

mathsbrain · Oct 9, 2019

So obviously the answer is 42C9

StudyOnly · Oct 9, 2019

InteGrand said:
There's probably a simpler way to do it, but here's a sketch for one way.

$\noindent In general we can ask the same question but with $K$ integers chosen from $\{1,\ldots,N\}$.$

$\noindent Assume the integers are $a_1 < a_2 < \cdots < a_K$ ($K=9$ in your example). We need $a_k \ge a_{k-1}+2$ for all $k =2,\ldots,K$. So we can write $a_2 = a_1+ b_1$, $a_3 = a_2+b_2 = a_1+b_1+b_2,\ldots, a_K = a_1 + b_1 + b_2+\cdots + b_{K-1}$, where $b_k \ge 2$ and $b_k\in \Bbb{Z}$ for all $k=1,\ldots K-1$. Since $a_K \le N$ (remember $N = 50$ in your example), we have $b_1+b_2+\cdots+ b_{K-1} \le N-a_1$

$\noindent Equivalently, we can write $b_k = 2+c_k$, where $c_k\in\Bbb{N}$. Then $c_1+\cdots+c_{K-1} \le N - 2(K-1) - a_1$.$

$\noindent Note we must have $a_1 \le N - 2(K-1)$ to have a valid choice. Now, find the number of solutions $(c_1,\ldots,c_{K-1})$ to $c_1+\cdots+c_{K-1} \le N - 2(K-1) - a_1$ (call this $f(a_1; K,N)$), and then your answer is given by $\sum\limits_{a_1=1}^{N-2(K-1)}f(a_1;K,N)$.$

$\noindent Some further helpful facts:$

$\noindent 1) the number of non-negative integer solutions $(c_1,\ldots,c_{K-1})$ to $c_1+\cdots+c_{K-1} \, \color{red}{ = } \color{black}\,j$ is $\binom{j+K-2}{K-2}$ (using Stars and Bars)$

$\noindent 2) $\sum\limits_{t=k}^{n}\binom{t}{k} = \binom{n+1}{k+1}$ (``Hockey-Stick Identity'').$

oh ye, that makes sense

mathsbrain · Oct 9, 2019

StudyOnly said:
oh ye, that makes sense

lol

aa180 · Oct 9, 2019

Is my solution not correct or something?:
$\sum\limits_{k=0}^{N-2n+1}(N-2n+2-k)\cdot\binom{n-2+k}{k}$

BoS Maths Trials 2019 (1 Viewer)

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