Welcome, Guest. Please login or register.

December 07, 2019, 02:33:07 pm

Author Topic: Compilation of solutions to HARD past HSC papers (3U)  (Read 8703 times)  Share 

0 Members and 1 Guest are viewing this topic.

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8613
  • "All models are wrong, but some are useful."
  • Respect: +2411
Compilation of solutions to HARD past HSC papers (3U)
« on: June 18, 2017, 07:55:24 pm »
+15
Hey all,

Similar to the 4U thread, 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.
______________________________

2001
- Q5 b) + c) Weird binomial probability
- Q7 b) iii) Finishing off the trig with a max/min problem

2003
- Q7 b) iii) Chris, Sandy and some painful ball and ceiling

2004
- Q7 b) ii)+iii) Working with the binomial coefficients after dealing with geometric series
- Q7 b) iv) Finishing it off

2005
- Q7 a) ii) Oil and related rates
- Q7 b) i)+ii) Cubics, S.P.s and zeroes (Scroll up a bit for part i)

2006
- Q6) a) iii) What it means to be ascending

2007
- Q7 b) A hole in the wall

2008
- Q7 d)-f) The later parts of the projectile with angles all over the place
  - e) redone

2009
- Q6 b) Points on a grid, albeit all three parts here here
- Q7 c) The billboard and moving the point P around

2010
- Q6 a) + b) The basketball
- Q7 c) Three colours

2012
- Q11 f) ii) Follows on from part i), constant term explanation

2013
- Q10 The absolute value inequality
- Q14 b) iii) The dreadful binomial theorem question
- Q14 c) ii) Some approximated common tangent

2014
- Q13 c) iii) Brief explanation to deducing the circle bit (Requires a little scrolling down)
- Q14 b) Spinners and spinners

2015
- Q13 c) iii) Short explanation to the SHM graph question
- Q14 c) iii) Long binomial probability

2016
- Q9 Reading the graph
« Last Edit: September 30, 2019, 04:48:14 pm by RuiAce »

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8613
  • "All models are wrong, but some are useful."
  • Respect: +2411
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #1 on: June 19, 2017, 11:09:01 pm »
+3
Added a few more. 2U thread might be coming up in the next few days.

jingyi.ren1999

  • Adventurer
  • *
  • Posts: 12
  • Respect: 0
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #2 on: June 23, 2017, 01:07:59 pm »
0
anyone have tips for have tips for 3u (inequalities)

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8613
  • "All models are wrong, but some are useful."
  • Respect: +2411
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #3 on: June 23, 2017, 01:20:09 pm »
+3
3U inequalities are very similar to 2U. The only difference is that they can modify a few things to appear as though they are no longer relevant anymore.

As in the question thread, please provide more details to your questions. Arbitrary questions mean that we do not know where your problems are exactly, and there's no reason for us to type a long-winded response that might not even address where your troubles are.
« Last Edit: June 23, 2017, 03:11:44 pm by RuiAce »

jingyi.ren1999

  • Adventurer
  • *
  • Posts: 12
  • Respect: 0
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #4 on: June 27, 2017, 10:44:41 am »
0
3U inequalities are very similar to 2U. The only difference is that they can modify a few things to appear as though they are no longer relevant anymore.

As in the question thread, please provide more details to your questions. Arbitrary questions mean that we do not know where your problems are exactly, and there's no reason for us to type a long-winded response that might not even address where your troubles are.


Thanks RuiAce, i'll find some questions but I feel as if I've posted on the wrong thread.. as it was or harder 3u inequalities for 4 unit. But that's ok.
Thanks anyway

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8613
  • "All models are wrong, but some are useful."
  • Respect: +2411
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #5 on: June 27, 2017, 03:45:59 pm »
0
Oh. Those are definitely harder but yeah, perhaps post those ones in the 4U section.

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8613
  • "All models are wrong, but some are useful."
  • Respect: +2411
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #6 on: October 24, 2017, 09:52:47 pm »
+3
I wasn't particularly pleased with the old solutions for the 2010 basketball question, so I'm redoing them here.
\begin{align*}\cos A \cos B (1 + \tan A \tan B) &= \cos A \cos B \left(1+\frac{\sin A \sin B}{\cos A \cos B}\right)\\ &= \cos A \cos B + \sin A \sin B\\ &= \cos(A-B)\end{align*}
_______________________



_________________________________________________________________________________

This entire question uses \(\tan x = \frac{\sin x}{\cos x}\) a lot



_______________________


_______________________


_______________________

Note that this choice is permitted.
Previously, we had \( 0 < B < \frac\pi2 \), and here we have \( 0 < \alpha < \theta <\frac\pi2 \), so \( 0 < \alpha < \frac\pi2 \) is satisfied.
Previously, we had \( B < A < \frac\pi2 \), and here we have \(\alpha < \boxed{2\alpha < 2\theta < \pi}\), so \(\alpha < 2\theta < \pi \) is satisfied.

(With those inequalities above, you should read them one step at a time. Not all at once.)


_______________________


Note that the \(5d\) bit doesn't impact this, as it's just a constant.

This can be done by simply computing the second derivative of \(F(\theta)\) and going about it the old way, if you wish to. NESA did something similar to testing both sides. But they tested both sides for the original function \(F(\theta)\) instead of \(F^\prime(\theta)\), and it was in a very ambiguous way.


You may want to sketch a diagram to figure out why this is the case. The idea is that if \(F\left(\frac\alpha2+\frac\pi4\right) < 0\) then we would've had a minimum.

You may be able to fudge around it and argue that "obviously" \(F(\theta) \ge 0\) for all \(\theta\). Here's one way of not doing that:

Note: Line * falls out from the range of the function \( f(\alpha) = 1-\sin 2\alpha \), which is \(0\le y \le 2 \).

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8613
  • "All models are wrong, but some are useful."
  • Respect: +2411
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #7 on: December 28, 2017, 09:05:25 pm »
+3



having recalled that \( \frac1{\cos^2\theta} = \sec^2\theta = 1+\tan^2\theta\).
____________________________________________



____________________________________________
To understand what's going on here
There's two ways the ball can enter the hole. The ball can either be thrown at a very shallow angle and it just goes right through during its ascent (when it goes up).

OR, it can be thrown at a very steep angle, but still make its way through during its DESCENT instead (when it goes down)!

What we're actually given is basically the second case. We need to find the conditions for the first.



____________________________________________

Of course, since we have the Cartesian equation of motion, we can just directly use that instead of go back to the time equations.

having divided out by \(x\) since \(x=0\) is the initial starting point, which we're not interested in.

Explaining the bait
Recall that \(\boxed{m=\tan\theta} \).

The cause of the problem: What we're after is the range of the particle. Recall that the range of the particle is maximised when \(\theta = 45^\circ\). In other words, as you increase \(\theta\) from \(0^\circ\) up to \(45^\circ\), the range increases. But as you increase \(\theta\) from \(45^\circ\) to \(90^\circ\), the range decreases.

What I will do now is sketch \( R = \frac{40\tan\theta}{1+\tan^2\theta} \) on GeoGebra, with the four angles above included. (You can actually prove that in fact, \(R = 20\sin 2\theta\)).

The problem: If we just plug \(m=0.8\) and \(m=1.2\) blindly into \( \frac{40m}{1+m^2} \), we will miss out on the fact that there's a stationary point at \(45^\circ\). Plugging \( m = 0.8\) and \(m = 1.2\) will give our ranges \(R = 19.512\) and \(R = 19.672\). This would make us misbelieve that the width of the interval is just 19.672 - 19.512 = 0.160.

But in reality, when \(\theta=45^\circ\) and hence \(m=1\), we see \(R = 20\). This is actually a longer range! Hence the width of the interval is actually 20 - 19.512 = 0.488

Moral of the story: If you're given an interval, you need to consider any STATIONARY POINTS as well as the endpoints.

Remark: This problem does not occur for the other interval, since we don't bump into any more stationary points.



« Last Edit: December 29, 2017, 03:56:07 pm by RuiAce »

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8613
  • "All models are wrong, but some are useful."
  • Respect: +2411
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #8 on: September 29, 2018, 12:14:18 pm »
+2



________________________________________________________________________


In case you may have forgotten: \( \binom{n}{n-r} = \frac{n!}{(n-r)! [n-(n-r)]!} = \frac{n!}{r!(n-r)!} = \binom{n}{r} \)
________________________________________________________________________


I write down explicitly how the cases work here in case you don't understand what comes next. Basically the next part just generalises all the cases, rather than does them one at a time. The method is the same.






This is the generic method:





« Last Edit: September 29, 2018, 12:17:31 pm by RuiAce »

Rabi

  • Adventurer
  • *
  • Posts: 9
  • Respect: 0
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #9 on: January 24, 2019, 11:20:19 am »
0
Quote from: Rabi on January 23, 2019, 07:40:56 pm
Hey I really need help with this question plz plz help.
"A water trough is 200 cm long and has a cross section of a right angled isosceles triangle. Show that when the depth of the water is x cm, the volume of water in the tank is 200x^2 cm^3. Water is poured in at a constant rate of 5 litres per minute. Find the rate at which the water level is rising when the depth is 30cm.

diggity

  • Trailblazer
  • *
  • Posts: 26
  • "Somebody once told me,the world is gonna roll me"
  • Respect: 0
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #10 on: September 30, 2019, 03:57:23 pm »
0
Hey Rui, I'd like to provide an alternate solution that I used when I originally did 2008 7) e), as yours looked (at a glance) to be more confusing than mine (imo). It may be useful to some. Here it is:

https://discordapp.com/invite/zEWnJnC

HSC 2019 discord. Minimal work is done, but more people to do that work would be nice. Feel free to join.

Aiming for an E4 in MX1 and MX2, and a band 6 in IT (Multimedia). Band 5 in English would require a miracle. Need a 90+ atar to actually recieve my scholarship at WSU (which is why I'm here, haha)

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8613
  • "All models are wrong, but some are useful."
  • Respect: +2411
Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #11 on: September 30, 2019, 04:46:53 pm »
+1
I'm actually not too sure what I did either there. I looked at how I did it at TuteSmart and it was far more elegant. I'm gonna type that up and then just link to both solutions here.
\begin{align*}
\tan\phi &= \text{gradient}\\
&= \frac{dy}{dx}\\
&= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\
&= \frac{-gt_1 + V\sin\theta}{V\cos\theta}.
\end{align*}
(This was definitely how you started the question as well. Haven't properly checked the rest.)
\[ \text{Recalling that }t_1+t_2 = \frac{2V}{g}\sin\theta\\ \text{we have }\boxed{t_2 = \frac{2V}{g}\sin\theta - t_1} .\\ \text{Hence consider:}\] \begin{align*}
\tan \alpha - \tan \beta &= \frac{h}{Vt_1\cos\theta} - \frac{h}{Vt_2\cos\theta}\\
&= \frac{h}{V\cos\theta}\left( \frac1{t_1}-\frac1{t_2}\right)\\
&= \frac{h}{V\cos\theta}\left( \frac{t_2-t_1}{t_1t_2}\right)\\
&= \frac{h}{V\cos\theta}\left( \frac{t_2-t_1}{\frac{2h}{g}}\right)\\
&= \frac{g}{2V\cos\theta}(t_2-t_1)\\
&= \frac{g}{2V\cos\theta}\left( \frac{2V\sin\theta}{g} - 2t_1 \right)\\
&= \frac{V\sin\theta-gt_1}{V\cos\theta}.
\end{align*}
\[ \therefore \tan\phi=\tan\alpha-\tan\beta. \]
(Possibly subject to typos)