 June 03, 2020, 07:18:35 am

### AuthorTopic: 4U Maths Question Thread  (Read 337784 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#### louisaaa01

• MOTM: NOV 19
• Forum Regular
•  • Posts: 82
• Respect: +33 ##### Re: 4U Maths Question Thread
« Reply #2385 on: October 01, 2019, 11:48:56 am »
+1
Hey, would somebody be able to explain why the answer to this question is C? I feel myself going insane and hope somebody can assist me in my time of lunacy. Thanks heaps!

Hey not a mystery mark!

So basically the way we approach this question is to split our F and N vectors into vertical and horizontal components.

Firstly, we know that F is inclined at θ to the horizontal, thus the horizontal component of F is Fcosθ and vertical component is Fsinθ.

On the other hand, N is at an angle of θ to the vertical (applying geometry - N is perpendicular to F). So the horizontal component of N is Nsinθ and the vertical component is Ncosθ

Now, we consider the net horizontal and vertical forces on C.

Since the particle isn't going anywhere vertically, we can conclude that all vertical forces are balanced. Since F and N are both pointing upwards, this means that: Fsinθ + Ncosθ = mg.

Horizontally, the particle is undergoing uniform circular motion - the net force is mrw2. Since F and N are in opposite directions, labelling the direction towards the centre as positive, we can say that:

Fcosθ - Nsinθ = mrw2

Which gives us C.

Hope this helps - if you need any further clarification, let me know!
« Last Edit: October 01, 2019, 11:52:19 am by louisaaa01 »
2019 ATAR: 99.95

2020 - 2026

USYD BSci / Doctor of Medicine

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
•       • • Posts: 8737
• "All models are wrong, but some are useful."
• Respect: +2517 ##### Re: 4U Maths Question Thread
« Reply #2386 on: October 01, 2019, 11:54:31 am »
+1
Hey, would somebody be able to explain why the answer to this question is C? I feel myself going insane and hope somebody can assist me in my time of lunacy. Thanks heaps!
Above explanation should sum it up pretty much. Here's a diagram to go with it.   #### not a mystery mark ##### Re: 4U Maths Question Thread
« Reply #2387 on: October 01, 2019, 12:31:52 pm »
0
Above explanation should sum it up pretty much. Here's a diagram to go with it.

I don't know how, but you made that so much clearer. Thank you so much. I never understood it in class but this has diamond clarity now.
Thanks so much!
Class of 2019: Advanced English , Extension 1 English , Extension 1 Maths , Extension 2 Maths , Physics , Business Studies 
ATAR: 98.55
2020-2025:  B Science (Honours)/B Arts [UNSW], Specialisation Physics/Philosophy

#### duncand

• Fresh Poster
• • Posts: 3
• Respect: 0 ##### Re: 4U Maths Question Thread
« Reply #2388 on: October 02, 2019, 03:36:25 pm »
0
hi, I am not sure the answers for this trial paper for part ii)
went from the 4th last line to the last line
(or from the 2nd tick to the third tick)
thanks. #### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
•       • • Posts: 8737
• "All models are wrong, but some are useful."
• Respect: +2517 ##### Re: 4U Maths Question Thread
« Reply #2389 on: October 02, 2019, 06:37:13 pm »
+2
hi, I am not sure the answers for this trial paper for part ii)
went from the 4th last line to the last line
(or from the 2nd tick to the third tick)
thanks. This requires a somewhat combinatorial argument to justify.
$\text{Firstly, using the AM-GM inequality}$\begin{align*}
\sum_{1\leq i < j \leq n} 2\sqrt{\binom{n}{i}}\sqrt{\binom{n}{j}}  &= \sum_{1\leq i<j\leq n} 2\sqrt{\binom{n}i\binom{n}j}\\
&= \sum_{1\leq i < j \leq n} \left[\binom{n}{i} + \binom{n}{j} \right]
\end{align*}
$\text{When expanding this sum, we obtain}\\ \left[\binom{n}1+\binom{n}2 \right]+ \left[\binom{n}1+\binom{n}3 \right] + \left[\binom{n}1+\binom{n}4\right]+ \cdots + \left[ \binom{n}1+\binom{n}n\right]\\ \qquad \qquad\qquad+ \left[\binom{n}2+\binom{n}3 \right] + \left[\binom{n}2+\binom{n}4\right]+ \cdots + \left[ \binom{n}2+\binom{n}n\right] \\ \qquad \qquad\qquad\qquad \qquad\qquad+ \left[\binom{n}3+\binom{n}4\right]+ \cdots + \left[ \binom{n}3+\binom{n}n\right]\\ \qquad \qquad\qquad\qquad \qquad\qquad+\ddots \\ \qquad \qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad+ \left[\binom{n}{n-1}+ \binom{n}{n}\right]$
Basically, we had to sub (i,j) = (1,2), (1,3), (1,4), ..., (1,n),
and then (i,j) = (2,3), (2,4), ..., (2,n),
and then (i,j) = (3,4), ..., (3,n), and so on.
$\text{Now exactly how many times does each term appear?}\\ \text{It is hopefully clear that }\binom{n}{1}\text{ appears exactly }n-1\text{ times}\\ \text{because it's only ever in the first row, so we just count the number of pairs in the first row.}$
$\text{But then, observe how }\binom{n}{2}\text{ appears once in the first row}\\ \text{followed by }n-2\text{ times in the second row.}\\ \text{And then, observe how }\binom{n}3\text{ appears once in the first and second row}\\ \text{followed by }n-3\text{ times in the third row.}$
$\text{In general, every term }\binom{n}{k}\text{ will appear}\\ \text{once for each of the preceding }k-1\text{ rows}\\ \text{and then }n-k\text{ times in the }k\text{-th row.}\\ \text{(And never again).}$
$\text{The consequence:}\\ \textbf{Each }\binom{n}k\text{ term always appears }\textbf{exactly }n-1\text{ times.}$
This is the transition from the 4th last to the 3rd last line.
_________________________________________________________________________________
$\text{Then, following identity is used:}\\ \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots + \binom{n}{n} = 2^n.\\ \text{This is a classic MX1 question, proven by subbing }x=1\\\text{into the binomial theorem.}$
Note: At the MX2 level, some trial papers treat this as easy to prove. And hence there are no hints provided on when it can be used.
$\text{But in the sum we have from pulling out the }n-1,\\ \text{our sum actually starts at }\binom{n}{1}\text{ instead.}\\ \text{Hence it effectively evaluates to }2^n - \binom{n}{0}\\ \text{and it is well known that }\binom{n}{0}=1\text{ always.}$
And this is the transition from the 3rd last tot he 2nd last line.

The final one should be obvious.  #### duncand

• Fresh Poster
• • Posts: 3
• Respect: 0 ##### Re: 4U Maths Question Thread
« Reply #2390 on: October 03, 2019, 03:54:14 pm »
0
.
wow, would of never thought of this
thanks!!   #### classof2019

• Trailblazer
• • Posts: 30
• Respect: 0 ##### Re: 4U Maths Question Thread
« Reply #2391 on: October 07, 2019, 03:54:58 pm »
0

A group of 12 people is to be divided into discussion groups.
(i) In how many ways can the discussion groups be formed if there are
8 people in one group, and 4 people in another?
(ii) In how many ways can the discussion groups be formed if there are
3 groups containing 4 people each?

More specifically, why do we need to divide our answer by 3! instead of just doing 12C8 x 8C4 x 4C4 + why doesn't a similar division apply for part (i)?

Cheers.

#### fun_jirachi

• MOTM: AUG 18
• Moderator
•     • • Posts: 641
• All doom and Gloom.
• Respect: +353 ##### Re: 4U Maths Question Thread
« Reply #2392 on: October 07, 2019, 04:05:50 pm »
+1
Hey there!

Consider 12 people A, B, C, D, E, F, G, H, I , J, K, L.
Dividing these people into a group of 8 and a group of 4 can only be done in one way ie. 12C8 x 4C4.
However, when we divide into 3 groups of 4, we need to account for the equal group sizes ie. we can't differentiate between say ABCD EFGH IJKL and EFGH ABCD IJKL. Essentially, dividing by 3! removes the ordering of identical groupings, as it doesn't matter whether we choose a certain four people in a group first, second or third. This doesn't occur with the first case because neither group is identical to the other.

Hope this makes sense Failing everything, but I'm still Flareon up.

HSC 2018: Modern History  | 2U Maths 
HSC 2019: Physics  | Chemistry  | English Advanced  | Maths Extension 1  | Maths Extension 2 
ATAR: 99.05

UCAT: 3310 - Verbal Reasoning  | Decision Making  | Quantitative Reasoning  | Abstract Reasoning 

#### louisaaa01

• MOTM: NOV 19
• Forum Regular
•  • Posts: 82
• Respect: +33 ##### Re: 4U Maths Question Thread
« Reply #2393 on: October 10, 2019, 04:47:09 pm »
0
Hi,

I'm wondering if anyone can provide a more efficient way to solve this question? (multiple choice Q10 from 2013 HSC)

A hostel has four vacant rooms. Each room can accommodate a maximum of four people.
In how many different ways can six people be accommodated in the four rooms?
The way I approached it was a little tedious. I basically listed the possible ways of separating 6 people into four groups (allowing groups of zero):

4, 1, 1, 0
4, 2, 0, 0
3, 1, 1, 1
3, 2, 1, 0
3, 3, 0, 0
2, 2, 2,0
2, 2, 1, 1

Then I determined the number of ways of selecting people to occupy these groups and 'arranging' these groups into the four different hotel rooms - I wound up with 4020 which is the correct answer.

However, I'm aware that a multiple-choice question should take about 1.5 minutes, so I was wondering if there's a better way to go about this question?

Thanks!
« Last Edit: October 10, 2019, 04:49:06 pm by louisaaa01 »
2019 ATAR: 99.95

2020 - 2026

USYD BSci / Doctor of Medicine

#### fun_jirachi

• MOTM: AUG 18
• Moderator
•     • • Posts: 641
• All doom and Gloom.
• Respect: +353 ##### Re: 4U Maths Question Thread
« Reply #2394 on: October 10, 2019, 05:11:49 pm »
0
Hey there!

You're right in that there is a quicker way; usually, if you're listing all these possibilities and the exhaustive list gets a bit long, using the complement is a better idea.

What the restriction actually does is ensure you can't have 5 or 6 people in a single room ie. the combinations 5/1/0/0 and 6/0/0/0 are invalid. Noting that the total number of combinations is just 46, we subtract (4C1x6C5x3C1x1C1 + 4C1x6C6) (which are just the total number of ways of the above combinations respectively) to find the total with the restriction.

Hope this helps Failing everything, but I'm still Flareon up.

HSC 2018: Modern History  | 2U Maths 
HSC 2019: Physics  | Chemistry  | English Advanced  | Maths Extension 1  | Maths Extension 2 
ATAR: 99.05

UCAT: 3310 - Verbal Reasoning  | Decision Making  | Quantitative Reasoning  | Abstract Reasoning 

#### louisaaa01

• MOTM: NOV 19
• Forum Regular
•  • Posts: 82
• Respect: +33 ##### Re: 4U Maths Question Thread
« Reply #2395 on: October 10, 2019, 05:48:50 pm »
0
Hey there!

You're right in that there is a quicker way; usually, if you're listing all these possibilities and the exhaustive list gets a bit long, using the complement is a better idea.

What the restriction actually does is ensure you can't have 5 or 6 people in a single room ie. the combinations 5/1/0/0 and 6/0/0/0 are invalid. Noting that the total number of combinations is just 46, we subtract (4C1x6C5x3C1x1C1 + 4C1x6C6) (which are just the total number of ways of the above combinations respectively) to find the total with the restriction.

Hope this helps Thanks so much!
2019 ATAR: 99.95

2020 - 2026

USYD BSci / Doctor of Medicine

#### classof2019

• Trailblazer
• • Posts: 30
• Respect: 0 ##### Re: 4U Maths Question Thread
« Reply #2396 on: October 19, 2019, 01:43:26 pm »
0
Hi, does anyone have any sort of tips they could offer when attempting the harder/ more abstract questions in the 4U paper (e.g. in Q16)? Especially those involving Harder 3U topics. Cheers.

#### fun_jirachi

• MOTM: AUG 18
• Moderator
•     • • Posts: 641
• All doom and Gloom.
• Respect: +353 ##### Re: 4U Maths Question Thread
« Reply #2397 on: October 19, 2019, 02:31:35 pm »
+2
Hey there!

I think people by the time they get to Q16 will have a decent amount of time to weigh up how to attack each question in Q16. Therefore, there should be time to make a few considerations 1. Try and identify what the question is asking you to do
A tip I've restated here on the forums a lot and that I've found really handy is to think about what you're working towards. This usually works for when you have to prove a result of some sort, and this can be done in many ways, whether it's working backwards, or converting something into a more familiar form.

If you're stuck on circle geo, the best thing you can do sometimes is just mess around and chase angles. A similar principle applies; like have a go and see where an approach takes you, and with the time you have you should have ample opportunity to try a few things.

Outside of this, it's best to just get in loads of practice for end questions, and perhaps take a look at them during reading time instead of solving all the MC in your head. Identifying what you need to do relatively early then seeing it again makes things a lot easier more often than not. As well as this, you don't actually lose much time for not solving MC in reading time, since MC (especially Q1-6/7) are usually quite quick Hope this helps Failing everything, but I'm still Flareon up.

HSC 2018: Modern History  | 2U Maths 
HSC 2019: Physics  | Chemistry  | English Advanced  | Maths Extension 1  | Maths Extension 2 
ATAR: 99.05

UCAT: 3310 - Verbal Reasoning  | Decision Making  | Quantitative Reasoning  | Abstract Reasoning 

#### louisaaa01

• MOTM: NOV 19
• Forum Regular
•  • Posts: 82
• Respect: +33 ##### Re: 4U Maths Question Thread
« Reply #2398 on: October 20, 2019, 02:07:45 pm »
0
Hi!

Also, just a question, I've heard (but I'm not entirely sure) that the question types in the MX2 HSC changed after 2001 - what exactly does this mean? I know they introduced MC in 2012 but I heard something about a change in 2001 as well.

Thanks « Last Edit: October 20, 2019, 02:11:06 pm by louisaaa01 »
2019 ATAR: 99.95

2020 - 2026

USYD BSci / Doctor of Medicine

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
•       • • Posts: 8737
• "All models are wrong, but some are useful."
• Respect: +2517 ##### Re: 4U Maths Question Thread
« Reply #2399 on: October 20, 2019, 08:08:25 pm »
+2
Hi!

Also, just a question, I've heard (but I'm not entirely sure) that the question types in the MX2 HSC changed after 2001 - what exactly does this mean? I know they introduced MC in 2012 but I heard something about a change in 2001 as well.

Thanks The main thing is that from 2001 onwards, mark allocation for each part of a question was promised.

But apart from that, it's especially true with 4U that many past papers liked to test some perhaps common concepts, but in absurdly weird ways in the final exam. (And of course, they were notably more time consuming.) For a question like this, the notation is somewhat peculiar, and the use of graphing in ii) and iii) seems a bit out of place to me. It turns out that part v) is just a recursive argument, but I'm not sure if I've ever seen that kind of pattern in recent papers.
$\textbf{1993 HSC 4U Additional}$
Note that $n$ is a fixed integer for this question.
$\text{By immediately using our existence assumption that }\\ x_n = \cot \theta_n\text{ for some }0<\theta_n<\pi,\text{ we obtain} \\ \boxed{\cot \theta_1 = \cot \theta_{n+1}}.$
$\text{Now upon applying the recursive relationship once, we obtain}\\ \boxed{\cot\theta_1 = \cot 2\theta_n}.\\ \text{But applying it again, we obtain}\\ \boxed{\cot \theta_1 = \cot 2^2\theta_{n-1}}.\\ \text{And just to make it clearer, explicitly applying it once more gives}\\ \boxed{\cot \theta_1= \cot 2^3 \theta_{n-2}}.$
$\text{The idea is that upon further repeated use of this recursion,}\\ \text{we will eventually arrive at }\boxed{\cot \theta_1 = \cot 2^n\theta_1 }.$
One way of tracking the required power of 2 is to observe that the power plus the index on $\theta$ must equal to $n+1$.
$\text{At this point, it is now perfectly fine for us to use the general solution}\\ \text{in an attempt to track down all solutions.}\\ \text{Solving }\cot (2^n \theta_1) = \cot\theta_1 \text{ gives}\\ 2^n\theta_1 = k\pi + \theta_1 \implies \boxed{\theta_1 = \frac{k\pi}{2^n-1}}\\ \text{where }k\text{ is an integer.}$
Note: If you put the $k\pi$ on the other side, you'd end up with $\theta_1 = \frac{-k\pi}{2^n-1}$, But that's all good - just sub out $\ell = -k$ and work with $\ell$ instead.
$\text{But we need to track down which values of }k\text{ give values for }\theta_1\text{ that are not out of bounds.}\\ \text{Recall that we still require }0< \theta_1 < \pi.$
$\text{Subbing into that expression gives}\\ 0 < \frac{k\pi}{2^n-1} < \pi \implies \boxed{0 < k < 2^n -1}$
Note: Of course, we're assuming that $2^n -1 > 0$, since $n \geq 1$. This assumption on $n$ is not explicitly stated in the question, but is implicitly assumed to be the lowest index.
$\text{Hence our valid values for }k\text{ are}\\ k = 1, 2, 3, \dots, 2^n - 2.\\ \text{Which means our final answers are}\\ x_1 = \cot \left( \frac{\pi}{2^n - 1} \right), \, \cot \left( \frac{2\pi}{2^n - 1}\right), \dots, \cot \left( \frac{ (2^n-2)\pi}{2^n -1} \right).$

Aside: For the interested reader, this question basically shows one special class of consequences when Newton's method is applied on a function that has no roots to begin with. Newton's method is designed with the intention of converging to a root once used often enough, so if we don't have a root, well where does it converge to?

More often than not, at each iteration we just get some debatably randomised value appearing. In other words, it definitely does not converge. But this question basically asks - is it possible for the values to start cycling at some point? That is, could Newton's method somehow exhibit periodic behaviour? The question more or less answers yes to this question, but challenges us by asking what initial values must we start with, to exhibit a period of $n$.
« Last Edit: October 20, 2019, 08:24:29 pm by RuiAce »  