BoS trials Maths and Chemistry 2020 (1 Viewer)

Trebla

Just marked Q12 of Maths Ext1 and have some feedback on Q12e)i)

The question asked students to prove a given vector relationship between A, B and P given that

$\bg_white \dfrac{AP}{BP}=\dfrac{1-\mu}{\mu}\quad 0<\mu<1$

A very common response from students was to assert that

$\bg_white \overrightarrow{AP}=\left(\dfrac{1-\mu}{\mu}\right)\overrightarrow{BP}$

This is not correct because the given ratio above is purely related to vector magnitudes. To convert to vector equations, you need to take into consideration both the magnitude AND the direction of the vectors. Notice that vector AP does NOT point into the same direction as BP (as P lies in the interval AB) so they cannot be related by a positive scalar. Many students attempted to fudge the result when they realised the algebra did not work out from this assertion. Better responses carefully considered the directions that the vectors were pointing and often related the directions of AP and AB.
Also, what do you know....a similar question as above appeared in the HSC Maths Ext 2 paper - but with more guidance so it was not as easy to make the above mentioned error.

OnJob

New Member
Aiming for tomorrow. In the meantime, feel free to ask any specific questions here if you want some answers.
Hey! Is it possible to send the answers to q13c of the EXT 1 paper?

OnJob

New Member
Hey! Is it possible to send the answers to q13c of the EXT 1 paper?
Sorry I meant 11C not 13C

Trebla

I have also marked Q13 for Maths Ext1 and have some feedback for a)ii).

A common response to this question was that since f’(x) > g’(x) for x > 0 then therefore f(x) > g(x) in that same domain. This is insufficient because the result does not hold more generally. A trivial example to illustrate is f(x) = 3x-1000 and g(x) = x.

The gradient of one curve being steeper than another curve doesn’t tell you anything about whether one curve lies above another. Also, a derivative function can have multiple primitive functions which differ by a constant. This is why you cannot simply integrate both sides of an inequality, unless you use a definite integral.

Better responses recognised that f(0) = g(0), which allowed them to argue that since the curves start at the same common point then a steeper gradient implies a higher y-value for x > 0.

Trebla

Sorry I meant 11C not 13C
After applying the substitution you should get

$\bg_white \dfrac{2}{3}\displaystyle\int_0^{\infty} e^{-\frac{u^2}{2}}\,du$

Now recall that the standard normal probability density function is

$\bg_white f(x)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\,dx$

Hence

$\bg_white \displaystyle\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\,dx=1$

But the standard normal probability density function is an even function so

$\bg_white \displaystyle\int_{0}^{\infty}\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\,dx=\dfrac{1}{2}$

Hence

$\bg_white \displaystyle\int_{0}^{\infty}e^{-\frac{x^2}{2}}\,dx=\dfrac{\sqrt{2\pi}}{2}$

Use this result and it all follows from there

Sid386

New Member
Hi, I just wanted to ask, for Question 11 b) in the extension 1 paper, would it be valid to assume that 3/4 of the trajectory of the projectile is what defines whether the distance increases over time? (i.e. subbing in the time after the projectile has travelled 3/4 of its range)

Trebla

Hi, I just wanted to ask, for Question 11 b) in the extension 1 paper, would it be valid to assume that 3/4 of the trajectory of the projectile is what defines whether the distance increases over time? (i.e. subbing in the time after the projectile has travelled 3/4 of its range)
Not sure how you can justify that assumption? The question is asking IF the distance (from particle's current position to its projection point) is always increasing then what restrictions apply to the launch angle.

Sid386

New Member
Not sure how you can justify that assumption? The question is asking IF the distance (from particle's current position to its projection point) is always increasing then what restrictions apply to the launch angle.
Because what I was thinking was up to halfway, both the vertical and horizontal distances increase, meaning that D would obviously also increase.
So to prove for all times, I thought you could use a time after the maximum height to find the condition for the launch angle which would be always increasing. But I guess picking a specific point would not be sufficient.
Thanks!

-insert title here-
All jokes aside it is not me, but someone I know. He does not want people to know as he is a humble man. Keep working hard and maybe you can see him at NESA Party!
i'll put the dox in paradox for \$100

i won't pls don't harass me to reveal the name

Trebla

Boom.

Here are the solutions and student results for the 2020 Mathematics Extension 1 BoS trials. Thanks to everyone who participated and for your patience!

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. So don't get too hung up on the mark you got. It is designed to give you good practice in exercising your problem solving skills in preparation for those harder questions in the exam. Some easier versions of the BoS trial questions have appeared in this year's HSC and previous years as well, so hopefully the HSC exam itself will look easy after doing that.

Please note that the question paper has also been updated (in the same post I originally posted it) to include minor edits to wordings, revised mark allocations and correcting some typos. If you find any typos in the solutions, please flag that for our consideration.

Best of luck for the HSC exam on Friday!

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laraknight

New Member
Hey, I was just wondering if you could post the solution to question 27 of the chemistry paper??

Many thanks!

jazz519

Moderator
Moderator
Hey, I was just wondering if you could post the solution to question 27 of the chemistry paper??

Many thanks!
Structure A is 2,2-dimethylpropanal
Structure B is 2,2-dimethylpropanoic acid

Reactivity ==> A is an aldehyde or primary alcohol as it is oxidised to produce B which is a carboxylic acid as it gives off CO2 when Na2CO3 is added.

IR ==> Strong sharp peak at 1700 cm^-1 is a C=O, peaks at 3000 cm-1 are C-H. Therefore A is an aldehyde

MS ==> 86 m/z molecular ion peak so formula is C5H10O as 5x12 + 10x1 + 16 = 86

13C NMR ==> 3 peaks so 3 unique carbon environments. 210 ppm is a C=O (aldehyde), 50 ppm R-C-C=O, 28 ppm C-C

1H NMR ==> 2 peaks so 2 unique hydrogen environments. 9.5 ppm - singlet (no H neighbours) and 1H integration likely a C-H. 1.3 ppm - singlet (no H neighbours) and 9H integration likely a 3xCH3

rp28

New Member
Hey, would u be able to post the chem trials solutions pls??

YonOra

Well-Known Member
For 13 b) ii) on the Ext 1 paper, how are the roots alpha and beta interchangeable when you've previously states that alpha > beta?

Trebla

Hey, would u be able to post the chem trials solutions pls??
Should be out in a few days. If you have any specific questions you want a solution for, ask it here.
For 13 b) ii) on the Ext 1 paper, how are the roots alpha and beta interchangeable when you've previously states that alpha > beta?
Good pick-up. There's actually a typo in that alpha and beta should not be equal (so the denominator is non-zero), rather than one being greater than the other which is a stronger condition (notice how the inequality alpha > beta is never used to construct the relation in part (i)). I have fixed that and uploaded the updated question paper in the post.

Terminator21

New Member
Hi I was just wondering if you could post the solution to question 25.c and 26 of the chemistry paper?
Thanks

Trebla

Hi I was just wondering if you could post the solution to question 25.c and 26 of the chemistry paper?
Thanks
Waiting for jazz519 to comment, but for Q25c) you should recognise that a condensation polymer is formed by linking N monomer chains which releases (N-1) water molecules in the process. Compute total polymer molecular weight based on the molecular weight of the N monomers and subtracting out the (N-1) water. Equate that to the given value polymer molecular weight in the question and solve for N.

For Q26 you need to recognise that the negative charge of the soap will be attracted to the Mg2+ and Ca2+ cations rather than the oil whilst it is the opposite case for the cationic detergent. Your answer should refer to the intermolecular forces and include diagrams.

MrGresh

Active Member
Any news on the chem solutions? The exam is creeping up haha

Trebla

Any news on the chem solutions? The exam is creeping up haha
Hoping to release them later tonight

Trebla

Sorry for the delay. Thanks again for your patience.

Here are the results and the marking guidelines/comments with sample answers for the 2020 Chemistry BoS trial. Thanks to everyone who participated in our very first Chemistry BoS trial!

Massive thanks to @jazz519 for his efforts in writing the paper, marking the responses and writing the sample answers!

If you find any typos in the sample answers, please flag that for our consideration.

Best of luck for the HSC exam on Monday!

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