BoS Trials Maths and Business Studies 2024 (3 Viewers)

lqmoney · Oct 9, 2024

Canteen said:
Here's my attempt. I've probably over-complicated it or am just plain wrong:

Base case ( $n=0$ ) – Trivial

Inductive hypothesis – Assume true for some non-negative integer $k$ :
$\text{i.e. } \exists p, q \in \mathbb{Z} : w_{2k} = 11 p \rightarrow u_{2k} = 11q \qquad (1)$

Inductive step – RTP true for $n = k + 1$ :
$\text{i.e. } \exists r,s \in \mathbb{Z}: w_{2k+2} = 11r \rightarrow u_{2k+2} = 11s$

$\begin{aligned} \text{Consider, } w_{2k+2} - u_{2k+2} &= (10^{2k+1}a_{2k+1} + 10^{2k}a_{2k} + w_{2k}) - [(-1)^{2k+1}a_{2k+1} + (-1)^{2k}a_{2k} + u_{2k}], \quad \text{by definition} \\ &= (10^{2k+1} + 1) a_{2k+1} + (10^{2k} - 1) a_{2k} + (w_{2k} - u_{2k}) \\ &= (10+1)(\underbrace{1 + 10 + 10^2 + \dots + 10^{2k}}_{=N})a_{2k+1} + [(11 - 1)^{2k} - 1]a_{2k} + 11(p - q), \quad \text{from (1)} \\ &= 11Na_{2k+1} + \left[11^{2k} - \binom{2k}{1} 11^{2k-1} + \dots - \binom{2k}{2k-1}11 + 1 - 1\right]a_{2k} + 11(p - q) \\ &= 11Na_{2k+1} + 11\underbrace{\left[11^{2k-1} - \binom{2k}{1} 11^{2k-2} + \dots - \binom{2k}{2k-1}\right]}_{=M} a_{2k} + 11(p-q) \\ &= 11Na_{2k+1} + 11Ma_{2k} + 11(p-q) \\ \therefore u_{2k+2} &= 11(\underbrace{r - Na_{2k+1} - Ma_{2k} + q - p}_{=s}), \quad \text{as } w_{2k+2} = 11r \\ &= 11s, \quad \text{as required} \end{aligned}$

i did something similar, problem is I don’t think you can assume w_2k and u_2k are still divisible by 11 when proving for the case of n=k+1, since you are only assuming that there is an implication between the 2. doing it the way you did means the original statement 11| w2n => 11|u2n is only necessarily true if w2, w4 … w2n are all divisible by 11 as well.

Trebla · Oct 9, 2024

I think you've highlighted an interpretation of the question that was different to what was intended. I believe the intention was to suppose that w_2n was divisible by 11 for all n, not for the same value of n as u_2n. I will confirm with the original author if that was the intention. If confirmed as I suspected, the wording will be adjusted in the final release file to remove that potential ambiguity and taken into consideration in the marking (and I will provide a less complicated proof as well

).

nonya2000 · Oct 10, 2024

Canteen said:
Here's my attempt. I've probably over-complicated it or am just plain wrong:

Base case ( $n=0$ ) – Trivial

Inductive hypothesis – Assume true for some non-negative integer $k$ :
$\text{i.e. } \exists p, q \in \mathbb{Z} : w_{2k} = 11 p \rightarrow u_{2k} = 11q \qquad (1)$

Inductive step – RTP true for $n = k + 1$ :
$\text{i.e. } \exists r,s \in \mathbb{Z}: w_{2k+2} = 11r \rightarrow u_{2k+2} = 11s$

$\begin{aligned} \text{Consider, } w_{2k+2} - u_{2k+2} &= (10^{2k+1}a_{2k+1} + 10^{2k}a_{2k} + w_{2k}) - [(-1)^{2k+1}a_{2k+1} + (-1)^{2k}a_{2k} + u_{2k}], \quad \text{by definition} \\ &= (10^{2k+1} + 1) a_{2k+1} + (10^{2k} - 1) a_{2k} + (w_{2k} - u_{2k}) \\ &= (10+1)(\underbrace{1 + 10 + 10^2 + \dots + 10^{2k}}_{=N})a_{2k+1} + [(11 - 1)^{2k} - 1]a_{2k} + 11(p - q), \quad \text{from (1)} \\ &= 11Na_{2k+1} + \left[11^{2k} - \binom{2k}{1} 11^{2k-1} + \dots - \binom{2k}{2k-1}11 + 1 - 1\right]a_{2k} + 11(p - q) \\ &= 11Na_{2k+1} + 11\underbrace{\left[11^{2k-1} - \binom{2k}{1} 11^{2k-2} + \dots - \binom{2k}{2k-1}\right]}_{=M} a_{2k} + 11(p-q) \\ &= 11Na_{2k+1} + 11Ma_{2k} + 11(p-q) \\ \therefore u_{2k+2} &= 11(\underbrace{r - Na_{2k+1} - Ma_{2k} + q - p}_{=s}), \quad \text{as } w_{2k+2} = 11r \\ &= 11s, \quad \text{as required} \end{aligned}$

how does one write in latex

lqmoney · Oct 10, 2024

lqmoney said:
i did something similar, problem is I don’t think you can assume w_2k and u_2k are still divisible by 11 when proving for the case of n=k+1, since you are only assuming that there is an implication between the 2. doing it the way you did means the original statement 11| w2n => 11|u2n is only necessarily true if w2, w4 … w2n are all divisible by 11 as well.

A proof for this interpretation

carrotsss · Oct 10, 2024

nonya2000 said:
how does one write in latex

[TEX]normal latex code [/TEX]

Trebla · Oct 10, 2024

Confirmed with author that the intended solution was to take as given that w_2n is divisible by 11 for all n so going to adjust wording accordingly. Hence, the outline of the solution would be:

From the assumption n=k
w_2k = 11P and u_2k = 11Q

Given w_2k+2 = 11R then
w_2k+2 - w_2k = 11(R-P)
=> a_2k - a_2k+1 = 11(R-P)

Consider
u_2k+2
= u_2k + 10^2ka_2k + 10^2k+1a_2k+1
= 11Q + 10^2k(a_2k+10a_2k+1) by assumption
= 11Q + 10^2k(11(R-P)+11a_2k+1)
= 11S

lqmoney · Oct 11, 2024

do you have to account for this in ii)? because if w_2n is divisble by 11 for all n then a0=a1, a2=a3 etc which wouldnt allow the implication in i) to be applied to palindromic numbers without some manipulation atleast

yanujw · Oct 12, 2024

lqmoney said:
do you have to account for this in ii)? because if w_2n is divisble by 11 for all n then a0=a1, a2=a3 etc which wouldnt allow the implication in i) to be applied to palindromic numbers without some manipulation atleast

I think you've confused a few implications. Firstly, don't proceed by assuming w_2n is divisible by 11, but apply the palindromic nature of the coefficients to w_2n first. If u_2n is palindromic, a_k = a_(2n-1-k) for all k, then w_2n = 0, and then you can make the argument u_2n is divisible by 11.

lqmoney · Oct 12, 2024

yanujw said:
I think you've confused a few implications. Firstly, don't proceed by assuming w_2n is divisible by 11, but apply the palindromic nature of the coefficients to w_2n first. If u_2n is palindromic, a_k = a_(2n-1-k) for all k, then w_2n = 0, and then you can make the argument u_2n is divisible by 11.

But if the proof in i) relies on w_2k being divisble by 11 for all integers up to n, then showing w_2n is divisible by 11 is not satisfactory to show that u_2n is also divisible by 11, since to apply the proof from i) you must also show w_2, w_4 … are also dividble by 11. For 11|w2, the only solution is w2 = 0 since w2 is at most 9 and at-least -9, so a1 = a2. then for 11|w4 similarly a3 = a4 and so on.

Trebla · Oct 12, 2024

lqmoney said:
But if the proof in i) relies on w_2k being divisble by 11 for all integers up to n, then showing w_2n is divisible by 11 is not satisfactory to show that u_2n is also divisible by 11, since to apply the proof from i) you must also show w_2, w_4 … are also dividble by 11. For 11|w2, the only solution is w2 = 0 since w2 is at most 9 and at-least -9, so a1 = a2. then for 11|w4 similarly a3 = a4 and so on.

As soon as you show that w_2n = 0 in general for any palindrome, regardless of the value of n (within the positive integers that is), then it must be true for all positive integers of n.

lqmoney · Oct 12, 2024

Trebla said:
Confirmed with author that the intended solution was to take as given that w_2n is divisible by 11 for all n

i think im misunderstanding something. with this change to the q, in part i. it is proven that u_2n is divisble by 11 if w_2k is divisble by 11 for all integer k from 1 to n, right? so even if w_2n is 0, it is not necessarily true that u_2n is divisbile by 11? (it is true but only working from the proof from i. we cannot assume it is true from this step).

Trebla · Oct 12, 2024

In part (i), you are asked to show that if X is true then Y follows.

In part (ii), you need to show that X is actually true for this specific scenario, so that you can then deduce that Y follows.

In this case, you would show that w_2n = 0 which is an integer that is divisible by 11 and holds for all values of n. This means your "if" condition in part (i) has been satisfied so the deduction can be made.

svrw · Oct 17, 2024

Hey guys,
I don't know if this has been brought up already, but I think there might be a typo in BoS 2024 3u for Q12 c). I have attached my working for it below. Feel free to tell me if there is a mistake in my working or in fact the question itself.
Thanks.

loading.... · Oct 17, 2024

svrw said:
Hey guys,
I don't know if this has been brought up already, but I think there might be a typo in BoS 2024 3u for Q12 c). I have attached my working for it below. Feel free to tell me if there is a mistake in my working or in fact the question itself.
Thanks.

3rd step you can’t take cos out the front

C_master · Oct 17, 2024

loading.... said:
3rd step you can’t take cos out the front

yea u can

loading.... · Oct 17, 2024

oh shit mb its wrt x not theta

Trebla · Oct 17, 2024

svrw said:
Hey guys,
I don't know if this has been brought up already, but I think there might be a typo in BoS 2024 3u for Q12 c). I have attached my working for it below. Feel free to tell me if there is a mistake in my working or in fact the question itself.
Thanks.

Yes, I agree there is a typo. It will be considered in the marking and will be corrected in the final version of the paper.

Paradoxica · Oct 17, 2024

General remark for MX2 Q11d: Most students were unable to decipher and extract the information from the question as presented in order to get the correct roots. All of the information in the polynomial is written into the question (and the polynomial itself), but you have to pull it out.

Quite a few students assumed the square was centred on 0 (the origin) of the complex plane, which is not stated anywhere in the question.

One student found the insidious solution of completing the binomial and solving the trivial quartic equation from there, which does get full marks as the question merely asks you to find the roots, not how.

Trebla · Oct 19, 2024

Here are the results and solutions for the 2024 Mathematics Extension 2 BoS Trials!

Please also note that the question paper has been updated (which these solutions are aligned to) with some minor adjustments to a few questions. The updated file is attached in my original post of the papers which replaces the older version.

In terms of results, massive congrats to the the student who scored the top mark of 81 followed by two students who tied on 66!

Should go without saying, but once again to remind everyone:

Note: These are not your average trial papers. These papers are heavily skewed towards the more challenging questions (i.e. filtered out most of the boring/repetitive easy stuff in a typical trial paper). Students who may be worried about losing motivation after attempting the paper are reminded that this is NOT intended to be an accurate reflection of the difficulty of the HSC exam itself.

We are continuing to work through the remaining subjects, so please be patient...

lmkae · Oct 19, 2024

Trebla said:
Here are the results and solutions for the 2024 Mathematics Extension 2 BoS Trials!

Please also note that the question paper has been updated (which these solutions are aligned to) with some minor adjustments to a few questions. The updated file is attached in my original post of the papers which replaces the older version.

In terms of results, massive congrats to the the student who scored the top mark of 81 followed by two students who tied on 66!

Should go without saying, but once again to remind everyone:

We are continuing to work through the remaining subjects, so please be patient...

81 is insane congrats

BoS Trials Maths and Business Studies 2024 (3 Viewers)

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