Post by: RuiAce on June 25, 2017, 09:11:28 am
Similar to the other threads, this thread will be reserved specifically for solutions to past BOSTES papers only. You should still post any other questions in the question thread, but please consider looking here if it is from a past BOSTES paper.
The main purpose of this thread is just so that if you have a really long answer, you won't need to go around asking for a solution but can just look for it here. Of course, you're always welcome to ask more about the solution if you're still confused :)
Again, it's all just a work in progress and I'm sure more solutions will appear in time.
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2001
- Q10 b) The absolutely dreaded farmhouse question
2002
- Q10 b) Boats
2003
- Q10 b) iii) Stupidly long proof that we maximise the length
- Q9 c) ii) Easier max/min
2005
- Q10 b) Xuan and Yvette
2007
- Q9 a) ii) Growing annuity - Nice but tricky case
- Q10 a) iii) Overkilled explanation regarding the estimate
- Q10 a) iv) Guidelines for the sketch.
Sample sketch
(https://i.imgur.com/mxbvLqL.png)
2010
- Q10 b) The rotated hemispherical container
2011
- Q9 a) Some effort needed to dig up the ratio
- Q9 d) ii) Interesting telescoping sum
- Q10 b) A sector paddock
2012
- Q16 b) i) Strange tangent
- Q16 c) ii) Dealing with the weird circle and parabola. (Briefly touches on i).)
2013
- Q10 The particle's behaviour
- Q14 b) i) Working out the roads
- Q14 d) Integrals used as areas
- Q15 a) ii) Mysterious absolute value equation; gradient and number of solutions
2014
- Q7 Quick mention as to why the answer is not 3.
- Q15 b) Building up to a ratio of areas
- Q16 b) Growing annuity - Tedious case
2015
- Q9 Slightly unconventional motion question, overkilled explanation for a mere multiple choice
- Q16 c Classic similar triangles optimisation problem
2016
- Q15 b) ii) A bait - the reason why n = 12 and not n = 11
Solutions to Recent Papers
- 2016
- 2017
- 2019
- 2020
- 2021
Post by: georgiia on July 24, 2017, 08:33:00 pm
Just Q's that are a bit different from normal or have a twist or something. But not like the ones in your marathon post; hard but in line with whats expected
Thank You : :D !
Post by: RuiAce on July 25, 2017, 08:33:36 am
Hi, My trials are next week and I feel like I'm not using my time as efficiently as I'd like just by doing past trials. I was wondering if you have a list/collection of challenging exam style questions that I can do because I feel like I'm waisting my time with what I'm doing at the moment.Tbh - Anything that's Q16 (or Q10 for pre-2012 papers) of a past paper will give you that. Because maths papers are designed so that the questions go in ascending difficulty.
Just Q's that are a bit different from normal or have a twist or something. But not like the ones in your marathon post; hard but in line with whats expected
Thank You : :D !
Post by: georgiia on July 25, 2017, 12:39:57 pm
Tbh - Anything that's Q16 (or Q10 for pre-2012 papers) of a past paper will give you that. Because maths papers are designed so that the questions go in ascending difficulty.
Yeah I know, but usually where I loose my marks is in questions that are supposedly easy topics but have a hard/unexpected unusual thing about them. Are there any questions over the years that have stood out as different in any way?
Post by: RuiAce on July 25, 2017, 12:45:29 pm
Yeah I know, but usually where I loose my marks is in questions that are supposedly easy topics but have a hard/unexpected unusual thing about them. Are there any questions over the years that have stood out as different in any way?It's pretty hard telling what you mean by this. A suggestion is to look at Q10 of some BOSTES papers, 2012 onwards (i.e. last question of the multiple choice) as they usually have a twist to them, but otherwise mind posting an example to help clarify your question?
Post by: georgiia on July 25, 2017, 01:15:32 pm
It's pretty hard telling what you mean by this. A suggestion is to look at Q10 of some BOSTES papers, 2012 onwards (i.e. last question of the multiple choice) as they usually have a twist to them, but otherwise mind posting an example to help clarify your question?
I have this worksheet at home so I can't post a photo until tonight but it basically has a random assortment of all topics where the questions, although in the scope of 2u, they make you think. You can't just rote do them which I find you easily get away with for most of 2u. Ill upload it later but It's really good and I just want to do more like that because I always got them wrong the first time
Thanks!
Post by: RuiAce on July 26, 2017, 02:01:40 pm
I have this worksheet at home so I can't post a photo until tonight but it basically has a random assortment of all topics where the questions, although in the scope of 2u, they make you think. You can't just rote do them which I find you easily get away with for most of 2u. Ill upload it later but It's really good and I just want to do more like that because I always got them wrong the first timeBump. Still waiting :P
Thanks!
Post by: georgiia on July 26, 2017, 06:36:46 pm
Bump. Still waiting :P(https://uploads.tapatalk-cdn.com/20170726/16c8e9c917f9ee6ba3e386fc92165b7c.jpg)
Post by: RuiAce on July 26, 2017, 07:44:39 pm
Something like Q2 is just an example of geometry going haywire and that has appeared in past HSC papers before (e.g. BOSTES 2013, independents - copyrighted).
Something like Q4a) is probably a bit weirder though. That stuff is more likely to appear in CSSA papers (again, unfortunately copyrighted) because CSSA likes twisting with your brain.
Post by: georgiia on July 26, 2017, 08:15:58 pm
Most of the questions in that picture seem fairly reasonable to me and don't really involve any tricks.
Something like Q2 is just an example of geometry going haywire and that has appeared in past HSC papers before (e.g. BOSTES 2013, independents - copyrighted).
Something like Q4a) is probably a bit weirder though. That stuff is more likely to appear in CSSA papers (again, unfortunately copyrighted) because CSSA likes twisting with your brain.
Ahh oops forgot to say Q2 was the one I meant, not so much the others.
Post by: RuiAce on July 26, 2017, 08:23:16 pm
Ahh oops forgot to say Q2 was the one I meant, not so much the others.It's pretty much just geometry (and on occasion questions like these may involve trig). If this is what you have trouble with, just keep digging through papers and attempt all the questions that involve geometry. Some copyrighted papers do have a fair amount of it but it's not that common in the HSC to be fair (again, 2013 is an example of where it did appear).
Off the top of my head I'm not sure which papers on THSC involve nice geometry questions though. At least, those that require thorough geometry.in combination with trig. The Xuan and Yvette one also involves it but the focus is somewhere completely different
Post by: georgiia on July 26, 2017, 09:12:03 pm
Post by: sophiegmaher on October 03, 2017, 11:35:12 am
Post by: Eric11267 on October 03, 2017, 11:40:56 am
I don't understand how to do part (iii) of the attached question!By integrating f'(x) between 0 and 6 you know that f(6)-f(0)=A1-A2-6
And since you know f(0) and the values of A1 and A2 you can extrapolate f(6)
Though I'm not too sure of this, so someone correct me if I'm wrong
Post by: sophiegmaher on October 22, 2017, 10:45:00 am
Post by: RuiAce on October 22, 2017, 10:55:05 am
Hey I'm confused as to where I've gone wrong for question 16 part (iv) of the 2016 HSC paper. I've attached my working, and I thought I've differentiated correctly by using the product rule but I'm not getting the same result as Jamon in his worked solutions :/
The computation mistake is responsible for your derivative not matching up. But keep in mind that if your working does not logically flow you can still be penalised.
Ironically enough, I actually think Jamon took a convoluted path. I think the best way of doing that question is to just sketch dy/dt against y (which is a parabola), and then just read off that. Alternatively, from prelim, we know that the axis of symmetry (and hence the x-coordinate of the turning point) of \(y=ax^2+bx+c\) is at \(x=-\frac{b}{2a}\), which can also be used to our advantage.
(I forgot if that minus sign should be there or not.)
Post by: sophiegmaher on October 22, 2017, 11:10:30 am
The computation mistake is responsible for your derivative not matching up. But keep in mind that if your working does not logically flow you can still be penalised.
Ironically enough, I actually think Jamon took a convoluted path. I think the best way of doing that question is to just sketch dy/dt against y (which is a parabola), and then just read off that. Alternatively, from prelim, we know that the axis of symmetry (and hence the x-coordinate of the turning point) of \(y=ax^2+bx+c\) is at \(x=-\frac{b}{2a}\), which can also be used to our advantage.
(I forgot if that minus sign should be there or not.)
That makes so much more sense, thank you so much! And yes, the minus sign should be there :)
Post by: Prerna Kumar on October 22, 2017, 06:09:56 pm
Thanks :)
Post by: Mandynguyenn on February 11, 2018, 03:20:49 pm
from a past paper it mentions "finite region", in the question: "find the area of the finite region bounded by C and L", to confirm does this just want us to find the area enclosed by C and L?
Post by: RuiAce on February 12, 2018, 05:44:03 pm
hi,Yeah
from a past paper it mentions "finite region", in the question: "find the area of the finite region bounded by C and L", to confirm does this just want us to find the area enclosed by C and L?
Post by: talitha_h on June 20, 2018, 07:59:00 pm
Post by: itssona on June 20, 2018, 09:42:37 pm
Could someone please explain question 16b)iv) from the 2013 hsc? Thanksso we want rate of increase of carp to equal rate of decrease of trout.
rate of increase of carp is given by dN/dt which is -e^0.04t x 0.04
for rate of decrease of carp, we differentiate P and make sure to put a negative sign since we want decrease ;)
so P=P0e^0.02t (this is from the standard formula)
where we know P0 is 10 since 10 carp are introduced
and differentiate this so dP/dt = 0.2e^0.02t
rate of decrease of carp is therefore -0.2e^0.02t
now we can equate such that 0.2e^0.02t = 0.04e^0.04t
5e^0.02t=e^0.04t
ln(5)+ln(e^0.02t)=ln(e^0.04t)
ln(5)=0.02t
t=ln(5)/0.02 = 80.5 approx
Post by: RuiAce on October 10, 2018, 01:23:11 pm
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Now for the hard part. To obtain some intuition behind solving this part we can do some working backwards.
Observe that if we rearrange \( r < \frac{P}{2} \) we obtain \( P > 2r\). We ask ourselves - is there any good reason for that inequality to be true? And the idea is that yes we can because we know from part i) that \( \boxed{P = r(\theta + 2)} \). Note that \( \theta > 0\), because \(\theta\) is the angle that the sector makes.
Similarly, if we rearrange \( \frac{P}{2(\pi + 1)} < r \) we obtain \( P < r(2\pi + 2) \). And again, there is a good reason for this. It essentially relies on the fact that if \( \theta > 2\pi\), we get something nonsense. That nonsense, is essentially an angle that can't be the angle in the sector we are given. Because how can it make sense to go beyond a full revolution and still have a sector? (In fact, if you go beyond a full revolution, you get the entire circle plus a little bit more.)
With some intuition built up, we prepare to answer the whole thing.
Post by: emilijab on October 23, 2018, 05:25:46 pm
I was having difficulty with Q9 B from the 2010 HSC paper, parts ii and iii. Derivative graphs in general really confuse me and i was wondering, in this question, how do i find the coordinates of the points without an equation to sub the x value into? The sample answers are of no help.
If you could provide some general advice on derivative graphs i'd greatly appreciate it :)
Thanks!!
Post by: fun_jirachi on October 23, 2018, 05:39:08 pm
So with b)ii), you can see that the definite integral of f'(x) has a maximum value of 4, remembering that area above the x-axis is positive and that area under the x-axis is negative. So basically you're looking for when the area on top of the graph is greatest, and its value when it is the greatest. Fortunately for you, it's given in the question :)
A similar thing happens with part iii), the rectangle next to A2 can be calculated to be 6 units squared (3x2), and since A1 and A2 cancel, you get -6.
Because you know that at x=6, f(x)=-6 (from part iii) and that f(x) has a maximum at x=2, and f(x)=4, and that f(x)=0 at x=4 (from the total signed area of f'(x) between 0 and 4), you can accurately draw the graph given in part iv, which is a parabola :)
I guess general tip here is to notice that when there's no equation, theres usually some other way to solve the question that's right in front of you that most times you won't even notice. Sometimes it doesnt have to be as easy as bringing out your integrals and your dxs. When you're given the derivative graph, I guess look for the area under the graph, and not at the graph itself, because more likely than not you're gonna find f(x) anyway!
hope this helps :)
Post by: emilijab on October 23, 2018, 05:48:14 pm
Remember that to get y=f'(x) from y=f(x) you differentiate the function. In the same way, you integrate y=f'(x) to get y=f(x). What this essentially means is you dont actually care what the graph y=f'(x) actually is, you care a hell of a lot more about the area under the graph, which for the most part in the question is given to you or is easily calculated (for 4<=x<=6).
So with b)ii), you can see that the definite integral of f'(x) has a maximum value of 4, remembering that area above the x-axis is positive and that area under the x-axis is negative. So basically you're looking for when the area on top of the graph is greatest, and its value when it is the greatest. Fortunately for you, it's given in the question :)
A similar thing happens with part iii), the rectangle next to A2 can be calculated to be 6 units squared (3x2), and since A1 and A2 cancel, you get -6.
Because you know that at x=6, f(x)=-6 (from part iii) and that f(x) has a maximum at x=2, and f(x)=4, and that f(x)=0 at x=4 (from the total signed area of f'(x) between 0 and 4), you can accurately draw the graph given in part iv, which is a parabola :)
I guess general tip here is to notice that when there's no equation, theres usually some other way to solve the question that's right in front of you that most times you won't even notice. Sometimes it doesnt have to be as easy as bringing out your integrals and your dxs. When you're given the derivative graph, I guess look for the area under the graph, and not at the graph itself, because more likely than not you're gonna find f(x) anyway!
hope this helps :)
ahhhh okay, I didn't realise that the area could indicate the value of y like that. Thanks heaps!! :D Glad I came across that sort of question because I wouldn't have had a clue how to do it in the exam.