I was told that they have been. You can't see them down at the bottom of the forum?
BoS trials Maths and Chemistry 2021 (2 Viewers)
- Thread starter Trebla
- Start date
The paper was challenging... and then some.
Also, some general thoughts / comments based on my observations in marking:
^2} \\ &= \cfrac{{V_T}^2}{g} \int \cfrac{dv}{{V_T}^2 - v^2} \\ &= \cfrac{{V_T}^2}{g} \times \cfrac{1}{2V_T}\int \cfrac{1}{V_T + v} + \cfrac{1}{V_T - v}\ dv \\ &= \cfrac{V_T}{2g}\bigg[\ln{\left|V_T + v\right|} - \ln{\left|V_T - v\right|}\bigg] + C, \quad \text{for some constant $C$} \\ \text{As the object is dropped from rest, at}\ t &= 0,\ v = 0\ \implies\ 0 = \cfrac{V_T}{2g}\ln{1} + C\ \implies\ C = 0 \\ \text{Thus,} \quad t &= \cfrac{V_T}{2g}\ln{\left(\cfrac{V_T + v}{V_T - v}\right)} \quad \text{noting that $0 \leqslant v \leqslant V_T$ for all $t \geqslant 0$} \\ \text{and hence,} \quad v &= V_T\left(\cfrac{e^{\frac{2gt}{V_T}} - 1}{e^{\frac{2gt}{V_T}} + 1}\right) \end{align*})
}}{\frac{V_T}{2g}\ln{\left(\frac{1 + \frac{v}{V_T}}{1 - \frac{v}{V_T}}\right)}} \qquad \text{with $v=rV_T$ and $0< r < 1$} \\ &= \cfrac{-2\ln{(1 - r)}}{\ln{\left(\frac{1 + r}{1 - r}\right)}} \\ &= \cfrac{2}{1 - \frac{\ln{(1 + r)}}{\ln{(1 - r)}}} \\ &\to 2^-\ \text{as}\ r \to 1^- \end{align*})
- On q11(c)(ii), the working can be simplified as follows:
- An alternative approach to q11(c)(iii):
- This shows that the time to
or
or
for the
particle approaches double the time taken to the same fraction of
under the
scenario, neglecting the fact that "coefficient"
- There were several approaches used in q12(c), several more efficient than what is show in the solutions so long as they were presented properly.
- On q12(d), quite a few students used the given expansion to establish that
and then replaced
with
, etc, which is perhaps more intuitively obvious.
- On q12(e)(ii), several students took as their counterexample
which gives
. Another example that was offered was
giving
and
. Both produce
without having
and
all being positive.
- The average (amongst those who attempted the questions) was 9.38 / 15 on question 11 and 10.66 / 15 on question 12. Only one student scored 15 / 15 for both questions 11 and 12.
- q11(a) was very difficult for most candidates. Only one student got the full answer and only one other made any merit-worthy progress beyond establishing that
. The average was 1.43 / 4 amongst the 21 who made an attempt.
- q11(b)(i) was done well, the average in 21 attempts being 1.86 / 2. Part (ii) was less well done, mostly because the answers were insufficient to establish that there were always two distinct pairs - too many noted two values of
and two values of
without proving that led to precisely two distinct
pairs. The average was 1. 4 / 2 in 20 attempts.
- q11(c) had averages of 1.76 / 2 for (i) in 21 attempts, 2.3 / 3 for (ii) in 20 attempts, and 1.06 /2 for (iii) in 16 attempts, so an average of 0.77 / 2 for (iii) if the zeroes from non-attempts are included. Some were given partial credit for error-carried work in part (iii) where the result from part (ii) was incorrect. A considerable number of students made the algebra worse by defining the resistance as
rather than
, and a few had
mysteriously become
between one line and the next.
- The mark for q12(a)(i) was earned by all of the 20 students who attempted the question. (ii) scored 0.83 / 1 amongst the 18 attempts, and (iii) scored 1.33 / 2 amongst 18 attempts. There were several examples of silly mistakes in part (iii).
- q12(b) was handled well by most, earning 1.56 / 2 on average amongst 18 attempts. I was surprised at the number of responses that did not make use of the conjugate root theorem, instead showing that
and then that
. There were some tries at following the first of these with "and similarly" and asserting the second.
- q12(c) attracted 7 non-responses and an average of 1.53 / 2 amongst the 15 who made an attempt.
- q12(d) had several "otherwise" proofs that used the
case of the AM-GM inequality to establish the required result. Amongst these, some did not start by proving the
- q12(e) was well done in parts (i) and (iv), but with some weaknesses in between. (i) averaged 0.95 / 1 amongst 20 attempts and all 20 attempts at (iv) scored 1 / 1. Amongst 19 attempts at (ii), the average was 1.4 / 2. This was mostly due to inadequate handling of the counterexample, but there were some errors in stating the converse. In part (iii), the average was 0.74 / 1 amongst 20 attempts. The statement of the contrapositive gave some difficulties, with a number of students stating that at least one of the four roots must be negative and not recognising that "not positive" and "negative" are not synonymous.
- As a more general comment, there were several places where marking took quite a bit of time to decide whether working was correct when setting out was poor, steps were omitted, and the initial setting up of solutions was scant or careless. I took the time to give the benefit of the doubt when that was justified, but expecting markers to deduce what is meant or to look for an interpretation that warrants credit is unrealistic at times - and especially when reading what is written is unclear. I have noted on some papers areas that could cause problems and careless mistakes (some penalised, some not, depending on the situation).
MX2 Multi-choice. Percentages are the proportion of respondents selecting each option. Correct answer bolded and in green.
Options | Q1 | Q2 | Q3 | Q4 | Q5 | Q6 | Q7 | Q8 | Q9 | Q10 |
|---|---|---|---|---|---|---|---|---|---|---|
A | 27% | 23% | 5% | 59% | 14% | 45% | 0% | 23% | 9% | 9% |
B | 59% | 9% | 0% | 5% | 55% | 9% | 95% | 14% | 5% | 68% |
C | 5% | 68% | 5% | 36% | 32% | 41% | 5% | 59% | 82% | 18% |
D | 9% | 0% | 91% | 0% | 0% | 5% | 0% | 5% | 5% | 5% |
Paradoxica
-insert title here-
For MX2, part 14 b iv, the question itself gave you all the information needed to solve that part. It was effectively free marks.
Similarly, part 16 c ii was completely free, as the question itself tells you what the reduced locus is (the interior of a disk).
Similarly, part 16 c ii was completely free, as the question itself tells you what the reduced locus is (the interior of a disk).
Paradoxica
-insert title here-
For MX1, part 14 d iv was free marks, as the previous parts gave you all the information needed to perform the inductive proof.
Paradoxica
-insert title here-
I will stress that you are always allowed to freely assume the information required by a later part in earlier parts, even if you have not actually shown those results. (At least when I did my HSC) You do not get penalised for not showing the previous parts if the question itself states them.
I am pointing this out, as there were no attempts for any of these three sub-questions, when there should have been plenty.
I am pointing this out, as there were no attempts for any of these three sub-questions, when there should have been plenty.
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Here are the solutions and student results for the 2021 Mathematics Extension 1 BoS trials. Thanks to CM_Tutor, 1039213 and Paradoxica for helping with solutions and marking!
Once, again a reminder that whilst the questions are made to be as closely styled to the HSC exams as possible, this is NOT intended to be an accurate reflection of the difficulty of the HSC exam. Please don't get too hung up on the mark you got. The whole point of the trial exam is to give you good practice in exercising your problem solving skills in preparation for those nasty harder questions in the HSC (ahem... Advanced and Extension 2 HSC).
Please note that the question paper has also been updated (in the same post I originally attached it to) to include changes to some questions to remove some ambiguity and/or make them more aligned with the original intent.
The marking of the paper was based on the original version released, whereas the solutions reflect the latest version, so please be aware of these minor differences if you participated in the exam event.
If you find any typos in the solutions, please flag that for our consideration.
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I was marking Q13 and found a recurring theme in part (a), that I should probably highlight here. The question was effectively asking to identify the sign of the rate of change of OP, as its length evolved over time travelling along the curve. All students who attempted this question ended up analysing d(OP)/dx, rather than d(OP)/dt.
In the original version, it turned out that the sign of d(OP)/dx was the same as d(OP)/dt because x was increasing over time. So whilst the conclusions were valid, the link to the rate of change was missing. This question has now been updated so that d(OP)/dx is NOT the same sign as d(OP)/dt so you need to analyse the sign of d(OP)/dt when x is decreasing over time. I've also added the words "over time" in the question to make that more explicit.
In the original version, it turned out that the sign of d(OP)/dx was the same as d(OP)/dt because x was increasing over time. So whilst the conclusions were valid, the link to the rate of change was missing. This question has now been updated so that d(OP)/dx is NOT the same sign as d(OP)/dt so you need to analyse the sign of d(OP)/dt when x is decreasing over time. I've also added the words "over time" in the question to make that more explicit.
1039213
New Member
Good morning,
From being a BoS student to now marking the exam 3 years later makes me really honoured to be a part of this community. Thank you to Trebla for helping organise this. I'll post my marker's comments in the next two replies.
From being a BoS student to now marking the exam 3 years later makes me really honoured to be a part of this community. Thank you to Trebla for helping organise this. I'll post my marker's comments in the next two replies.
1039213
New Member
QUESTION 11:
Throughout the marking period, I found the answers tended to vary in style and methodology. With an average of 7.23, and a maximum of 14, it was clear that individuals struggled in certain aspects.
(a) Students who successfully used 10^n = (7+3)^n and provided clear logical reasoning were able to reach the answer without any problems. The majority of the candidature performed well on this part.
(b) Very few candidates scored full marks on this part. Squaring both sides, but not removing the incorrect solutions was a common error in the method, as well as not accounting for the domain of the square root function. Checking and testing your solutions is imperative in these inequality questions.
(c)(i) Students must remember not to exclude |x| as the question doesn't define a condition on x. Those who didn't remove |x| were highly successful in obtaining full marks.
(ii) Carry forward errors were given to those who answered (i) incorrectly. This part was performed well.
(iii) The majority of students obtained full marks on this part, well done!
(d)(i) Sufficient setting out was required to show the given answer, as opposed to stating it as true. Around half the candidature were able to successfully answer this part.
(ii) Only a handful of students were able to answer this part correctly. Sound geometrical understanding as well as using the earlier result, and in some cases, the sine rule, deemed most appropriate for obtaining the full marks.
Throughout the marking period, I found the answers tended to vary in style and methodology. With an average of 7.23, and a maximum of 14, it was clear that individuals struggled in certain aspects.
(a) Students who successfully used 10^n = (7+3)^n and provided clear logical reasoning were able to reach the answer without any problems. The majority of the candidature performed well on this part.
(b) Very few candidates scored full marks on this part. Squaring both sides, but not removing the incorrect solutions was a common error in the method, as well as not accounting for the domain of the square root function. Checking and testing your solutions is imperative in these inequality questions.
(c)(i) Students must remember not to exclude |x| as the question doesn't define a condition on x. Those who didn't remove |x| were highly successful in obtaining full marks.
(ii) Carry forward errors were given to those who answered (i) incorrectly. This part was performed well.
(iii) The majority of students obtained full marks on this part, well done!
(d)(i) Sufficient setting out was required to show the given answer, as opposed to stating it as true. Around half the candidature were able to successfully answer this part.
(ii) Only a handful of students were able to answer this part correctly. Sound geometrical understanding as well as using the earlier result, and in some cases, the sine rule, deemed most appropriate for obtaining the full marks.
1039213
New Member
QUESTION 12:
A decent effort on a tough question. With an average mark of roughly 4.5/15, and a top score of 13/15, it is crucial to pick up any marks where necessary. The variety in topics combined with time restraints made this question difficult overall.
Q12a could only progress if the expression was multiplied by a multiple of sin(pi/2n). Students who failed to do so were unsuccessful, and those who did were likely to obtain the full marks. Well done to the student who used sigma notation, saved lots of time and was concise.
Q12b made students very easily fall into the trap of expanding the brackets and simplifying a long and difficult algebraic expression. To circumvent this, using (sin(theta))^2 + (cos(theta))^2 = 1 was the most ideal method to take, as opposed to using a right-angled triangle or otherwise.
Q12ci could have been done simply by using the result of Q12b, as the result was given. Make sure not to skip any latter parts as the question simplified nicely when using the correct components of each vector. Q12cii used the definition of the derivative so those who applied the limit were very successful in their answers. The integral in Q12ciii wasn't too difficult once making the substitution, but students must make sure to leave their answers in terms of T instead of t. No candidate received full marks in Q12civ as the path of the Cartesian equation is in fact a circle with centre (0,a) and radius a. The limiting value was performed successfully by those that attempted it.
Finally, Q12di was straightforward for most, if not all candidates, whereby some made minor computational errors. Q12dii deemed difficult for most however as the majority of candidates attempted to substitute tan(theta) into the cubic, instead of the factored quadratic result. Only very few candidates had perfect justification to which solution of theta was to match with each root; this was justified via the respective quadrants. A handful of candidates were awarded full marks for this question.
A decent effort on a tough question. With an average mark of roughly 4.5/15, and a top score of 13/15, it is crucial to pick up any marks where necessary. The variety in topics combined with time restraints made this question difficult overall.
Q12a could only progress if the expression was multiplied by a multiple of sin(pi/2n). Students who failed to do so were unsuccessful, and those who did were likely to obtain the full marks. Well done to the student who used sigma notation, saved lots of time and was concise.
Q12b made students very easily fall into the trap of expanding the brackets and simplifying a long and difficult algebraic expression. To circumvent this, using (sin(theta))^2 + (cos(theta))^2 = 1 was the most ideal method to take, as opposed to using a right-angled triangle or otherwise.
Q12ci could have been done simply by using the result of Q12b, as the result was given. Make sure not to skip any latter parts as the question simplified nicely when using the correct components of each vector. Q12cii used the definition of the derivative so those who applied the limit were very successful in their answers. The integral in Q12ciii wasn't too difficult once making the substitution, but students must make sure to leave their answers in terms of T instead of t. No candidate received full marks in Q12civ as the path of the Cartesian equation is in fact a circle with centre (0,a) and radius a. The limiting value was performed successfully by those that attempted it.
Finally, Q12di was straightforward for most, if not all candidates, whereby some made minor computational errors. Q12dii deemed difficult for most however as the majority of candidates attempted to substitute tan(theta) into the cubic, instead of the factored quadratic result. Only very few candidates had perfect justification to which solution of theta was to match with each root; this was justified via the respective quadrants. A handful of candidates were awarded full marks for this question.
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Hi all,
It is with regret that I have to inform you that we will not be able to release a solutions document in time before the HSC Chemistry exam.
Unfortunately, some things have gone wrong and/or taken longer than expected behind the scenes. We will aim to ensure this does not occur again in future years.
However, we do have the marks and student results for those who participated in the event in the attachment below. Thanks to CM_Tutor who has marked the paper and has kindly returned scanned copies of the papers back to those students in their submission threads. A reminder that whilst the questions are made to be as closely styled to the HSC exams as possible, this is NOT intended to be an accurate reflection of the difficulty of the HSC exam. Please don't get too hung up on the mark you got. The whole point of the trial exam is to give you good practice in exercising your problem solving skills in preparation for those nasty harder questions in the HSC.
As the next best thing, we also have some drafted sample answers to selected questions. Thanks to jazz519 for writing this up.
Also, CM_Tutor has drafted some marker's comments and answers to some of the other questions in the thread below:
boredofstudies.org
If you have any specific questions that you want the answer for, just ask in this thread. Apologies again for not having a full solution file at this point in time.
Best of luck for your HSC exam!
It is with regret that I have to inform you that we will not be able to release a solutions document in time before the HSC Chemistry exam.
Unfortunately, some things have gone wrong and/or taken longer than expected behind the scenes. We will aim to ensure this does not occur again in future years.
However, we do have the marks and student results for those who participated in the event in the attachment below. Thanks to CM_Tutor who has marked the paper and has kindly returned scanned copies of the papers back to those students in their submission threads. A reminder that whilst the questions are made to be as closely styled to the HSC exams as possible, this is NOT intended to be an accurate reflection of the difficulty of the HSC exam. Please don't get too hung up on the mark you got. The whole point of the trial exam is to give you good practice in exercising your problem solving skills in preparation for those nasty harder questions in the HSC.
As the next best thing, we also have some drafted sample answers to selected questions. Thanks to jazz519 for writing this up.
(a) Functional group isomers
(b) and (c)
(b) and (c)
Also, CM_Tutor has drafted some marker's comments and answers to some of the other questions in the thread below:
2021 BoS Chemistry Trial - Marking and Answers
First, let me start with an apology that this process has taken much longer than expected. Second, here are the distributions of answers for the MCQ: The most difficult questions, based on the rate at which the question was answered correctly, were questions 5, 14, 11, and 2. The easiest...
boredofstudies.org
If you have any specific questions that you want the answer for, just ask in this thread. Apologies again for not having a full solution file at this point in time.
Best of luck for your HSC exam!
This thread, to which Trebla referred, contains discussion of every part of the short answer section of the paper including common mistakes.
This post in another thread, and subsequent discussion, covers all of the MCQ on the paper.
I regret that a full solution file is not completed but every question and its solution is covered above and invite any questions that anyone may have.
This post in another thread, and subsequent discussion, covers all of the MCQ on the paper.
I regret that a full solution file is not completed but every question and its solution is covered above and invite any questions that anyone may have.