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#### RuiAce

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##### Compilation of solutions to HARD past HSC papers (3U)
« on: June 18, 2017, 07:55:24 pm »
+14
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

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
« Last Edit: April 14, 2019, 09:32:28 am by RuiAce »

#### RuiAce

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##### 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

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##### 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

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##### Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #3 on: June 23, 2017, 01:20:09 pm »
+2
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

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##### 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

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##### 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

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##### Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #6 on: October 24, 2017, 09:52:47 pm »
+2
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*}
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$\text{Subbing in }\tan A \tan B = -1\text{ gives the equation}\\ \cos (A-B) = 0.$
$\text{Because }B < A < \pi\text{ we have }\boxed{0 < A-B < \pi - B}\\ \text{Because }B > 0, \boxed{\pi - B < \pi}\\ \text{So combining inequalities, }0 < A-
B < 0$

$\text{Over this domain, the only solution to }\cos(A-B)=0\\ \text{is }\boxed{A-B=\frac\pi2}$
_________________________________________________________________________________

This entire question uses $\tan x = \frac{\sin x}{\cos x}$ a lot
$\text{Firstly, from trigonometry (note our right-angled triangle)}\\ \tan \alpha = \frac{h}{d} \implies \boxed{h=d\tan \alpha}$
$\text{Also, subbing }t=\frac{x}{v\cos \theta}\text{ gives our usual Cartesian eqn of motion}\\ y = x\tan \theta -\frac{gx^2}{2v^2}\sec^2\theta$
\text{So if the basketball passes through }(d,h)\text{, i.e. }(d,d\tan \alpha),\\ \begin{align*}d\tan \alpha &= d\tan \theta - \frac{gd^2}{2v^2}{\sec^2\theta}\\ \tan\theta-\tan\alpha&=\frac{gd}{2v^2}\sec^2\theta \\ v^2&=\frac{gd}{2(\tan\theta -\tan \alpha)}\times \frac1{\cos^2\theta}\\ &= \frac{gd}{2(\cos\theta\sin\theta - \cos^2\theta\tan\alpha)}\end{align*}
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$\text{When }\theta\to \alpha,\text{ it's clear that the denominator gets affected. In fact,}\\ \cos\theta \sin \theta -\cos^2\theta \tan \alpha \to \cos\theta\sin\theta-\cos\theta\sin\theta =0\\ \text{So }v^2\to \infty\text{ and hence }v\to \infty$
$\text{When }\theta \to \alpha, \, \cos\theta\to 0\\ \text{so }\cos\theta\sin\theta-\cos^2\theta\tan\alpha\to 0\\ \text{and once again, }v\to \infty$
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$\text{Using double angle formulae, we may rewrite}\\ F(\theta) = \frac12\sin 2\theta -\frac12\left(1+\cos2\theta\right)\tan\alpha$
\text{So setting the derivative to 0}\\ \begin{align*}\cos 2\theta + \sin 2\theta \tan \alpha&=0\\ \sin 2\theta\tan \alpha &=\cos2\theta\\ \tan2\theta\tan\alpha &= -1\end{align*}
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$\text{Comparing what we had in a) and b) iii)}\\ \text{it makes sense to take the choice }\boxed{A = 2\theta, B = \alpha}$
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.)
$\text{We know that if }F^\prime(\theta) = 0,\text{ then }\tan 2\theta \tan \alpha = -1.\\ \text{When }\tan2\theta \tan \alpha = -1\text{, using a) ii) we have}2\theta - \alpha = \frac\pi2$
$\text{This gives }\boxed{\theta = \frac\alpha2 + \frac\pi4}$
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$\text{The main problem is that we had to realise that}\\ \boxed{v^2 = \frac{5d}{F(\theta)}}\\ \text{An important implication becomes necessary following this.}$
$\textit{Whenever }y=f(x)\textit{ is minimised}\\ y=\frac{1}{f(x)}\textit{ is MAXIMISED.}$
Note that the $5d$ bit doesn't impact this, as it's just a constant.
$\text{So we're interested in the }\textbf{maximum}\text{ value of }F(\theta).\\ \text{Effectively, we need to prove that }\theta = \frac\alpha2+\frac\pi4\text{ maximises }F(\theta)$
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.
$\text{Note that }\theta = \frac{\alpha}{2}+\frac\pi4\text{ lies}\\ \text{in between }\theta = \alpha\text{ and }\theta = \frac\pi2.\\ \text{Furthermore, we know that }F(\alpha) = 0\text{ and }F\left(\frac\pi2\right) = 0.$
$\text{So if we can check that }F\left(\frac\alpha2+\frac\pi4\right) > 0\text{, then}\\ \text{we ultimately can conclude that }\theta=\frac\alpha2+\frac\pi4\text{ gives the maximum.}$
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:
\text{Execute:}\\ \begin{align*}F\left(\frac\alpha2+\frac\pi4\right)&=\frac12 \sin \left(\alpha + \frac\pi2\right) - \frac12\left[1+\cos \left(\alpha+\frac\pi2\right)\right]\tan\alpha\\ &= \frac12\cos\alpha - \frac12 (1-\sin \alpha)\frac{\sin\alpha}{\cos\alpha}\\ &=\frac{1}{2\cos \alpha}\left(\cos^2\alpha-\sin\alpha\cos\alpha+\sin^2\alpha\right)\\ &= \frac1{2\cos\alpha}(1-\sin 2\alpha)\\ &> 0\end{align*}\\ \text{So regroup everything and then we are done.}
Note: Line * falls out from the range of the function $f(\alpha) = 1-\sin 2\alpha$, which is $0\le y \le 2$.

#### RuiAce

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##### Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #7 on: December 28, 2017, 09:05:25 pm »
+1
$\textbf{2007 Q7 b)}$
$\text{The first part is very standard.}\\ \text{Rearranging the first equation gives }t=\frac{x}{14\cos \theta}.$
\text{So subbing into the second gives}\\ \begin{align*}y&= \frac{14\sin\theta}{14\cos\theta}x - 4.9 \left(\frac{x}{\cos\theta}\right)^2\\ &= x\tan\theta -\left(\frac{1+\tan^2\theta}{40}\right)x^2\\ &= mx - \left(\frac{1+m^2}{40}\right)x^2\end{align*}
having recalled that $\frac1{\cos^2\theta} = \sec^2\theta = 1+\tan^2\theta$.
____________________________________________
\text{The barrier is at }x=10\\ \text{so if }y=h\text{ when }x=10,\text{ we have}\\ \begin{align*}h&=10m - 2.5(1+m^2)\\ 0&=2.5m^2-10m+(h-2.5)\\ m&=\frac{10\pm \sqrt{100 - 10(h-2.5)}}{5}\tag{quadratic formula}\\ &= \frac{10\pm \sqrt{75 - 10h}}{5}\end{align*}
\text{Exploiting the property }\sqrt{ab} = \sqrt{a}\sqrt{b}\text{ we have}\\ \begin{align*}m &= \frac{10 \pm \sqrt{25 \left(3 - \frac{10h}{25}\right)}}{5}\\ &= \frac{10\pm 5\sqrt{3-0.4h}}{5}\\ &= 2\pm \sqrt{3-0.4h}\end{align*}\\ \text{as required.}
\text{Now, this imposes a restriction on }h\\ \text{because anything under a square root cannot be negative! Hence,}\\ \begin{align*}3-0.4h &\ge 0\\ h&\le 7.5\end{align*}
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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.
\text{All we can do is sub the upper and lower heights of the hole in.}\\ \text{When }h=3.9\text{ we first have}\\ \begin{align*}m& = 2\pm \sqrt{3-0.4\times 3.9}\\ &=0.8, \, 3.2\end{align*}
\text{Then, when }h=5.9,\\ \begin{align*}m&=2\pm \sqrt{3-0.4\times 5.9}\\ &= 1.2, \, 2.8\end{align*}
$\text{So the other interval must be the interval between 0.8 and 1.2}\\ \text{i.e. }0.8\le m \le 1.2$
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$\text{Now, if the wall blocks the ball, then the range is obviously 10.}\\ \text{When that does }\textbf{not}\text{ happen, we really have a classic again,}\\ \text{because the range of flight will be determined by when the ball lands,}\\ \text{i.e. }y=0.$
Of course, since we have the Cartesian equation of motion, we can just directly use that instead of go back to the time equations.
\begin{align*}mx - \left(\frac{1+m^2}{40}\right)x^2&=0\\ m - \left(\frac{1+m^2}{40}\right)x&=0\\ x&= \frac{40m}{1+m^2}\end{align*}
having divided out by $x$ since $x=0$ is the initial starting point, which we're not interested in.
\text{Now, this last part involves a bait.}\\ \text{We first observe the correspondences between }m\text{ and }\theta:\\ \begin{align*}m=0.8&\implies \theta\approx 38.7^\circ\\ m=1.2&\implies \theta \approx 50.2^\circ\\ m=2.8&\implies \theta \approx 70.3^\circ\\ m=3.2&\implies \theta \approx 72.6^\circ\end{align*}
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.
$\text{In the first interval }2.8\le m\le 3.2\\ \text{since }45^\circ\text{ does not lie within the corresponding angles,}\\ \text{we can just compute the values at the endpoints:}\\ m=2.8\implies R\approx12.67\\ m=3.2\implies R\approx11.39$
$\text{Hence the width of the interval for the range will be approx }12.67-11.39=1.28$
\text{In the second interval }0.8\le m\le 3.2\\ \text{since }45^\circ\text{ DOES lie within the corresponding intervals}\\ \text{we need the values at the endpoints AND at }\theta=45^\circ.\\\begin{align*} m=0.8&\implies R\approx 19.512\\ m=1.2&\implies R\approx 19.672\\ m=1&\implies R\approx 20\end{align*}
$\text{Hence the width of the interval for the range will be approx }20 - 19.512 = 0.488$
« Last Edit: December 29, 2017, 03:56:07 pm by RuiAce »

#### RuiAce

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##### Re: Compilation of solutions to HARD past HSC papers (3U)
« Reply #8 on: September 29, 2018, 12:14:18 pm »
+2
$\textbf{2010 HSC MX1 Q7c)}$
$\text{Essentially, we can have 0, 1, 2, 3, }\dots\\ \text{up to and including }r\text{ red balls in our selection of }r\text{ balls.}\\ \text{Note by default that all the remaining balls are automatically blue.}$
$\text{So the number of possibilities is just }\boxed{r+1}\\ \text{(we're literally counting how many numbers are there from 0 to }r\text{ inclusive).}$
________________________________________________________________________
$\text{This part is essentially equivalent to asking}\\ \text{in how many ways can we choose }n-r\text{ numbers}\\ \text{out of }n\text{ different numbers.}$
$\text{Which is obviously }\binom{n}{n-r}.\\ \text{But which is then }\binom{n}{r}\text{ because of the symmetry property.}$
In case you may have forgotten: $\binom{n}{n-r} = \frac{n!}{(n-r)! [n-(n-r)]!} = \frac{n!}{r!(n-r)!} = \binom{n}{r}$
________________________________________________________________________
$\text{The question presents the scenario where we choose }n\text{ balls}\\ \text{out of any of the }\textbf{three}\text{ colours mentioned in the previous parts.}\\ \text{(Namely red, blue and white)}$
$\text{We can }\textbf{decompose}\text{ this into the cases,}\\ \text{where we choose }r\text{ that are either red or blue,}\\ \text{and }n-r\text{ that are white, and consequently have numberings.}$
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.
$\text{The cases for }r\text{ are that }r = 0, 1, 2, \dots, n\\ \text{and we stop at }n\text{ essentially because}\\ \text{we draw at most }n\text{ balls.}$
$\text{If }r = 0\text{, then there is only }r+1 = 1\text{ way}\\ \text{of drawing the red/blue balls.}\\ \text{Then, there are }\binom{n}{r} = \binom{n}{0}\text{ ways}\\ \text{of drawing the white balls.}\\ \text{So this case yields }\binom{n}{0} \text{ different selections.}$
$\text{If }r = 1\text{, then there are now }r+1 = 2\text{ ways}\\ \text{of drawing the red/blue balls.}\\ \text{Then, there are }\binom{n}{r} = \binom{n}{1}\text{ ways}\\ \text{ of drawing the white balls.}\\ \text{So this case yields }2\binom{n}{1}\text{ different selections.}$
$\text{If }r = 2\text{, then there are now }r+1 = 3\text{ ways}\\ \text{of drawing the red/blue balls.}\\ \text{Then, there are }\binom{n}{r} = \binom{n}{2}\text{ ways}\\ \text{of drawing the white balls.}\\ \text{So this case yields }3\binom{n}{2}\text{ different selections.}$
$\text{Essentially, this continues until we reach }r = n\\ \text{in which there are now }r+1 = n+1\text{ ways of drawing the red/blue balls}\\ \text{and }\binom{n}{r} = \binom{n}{n}\text{ ways of drawing the white balls.}\\ \text{So this case yields }(n+1)\binom{n}{n}\text{ different selections.}$
$\text{Regrouping the cases, the total number of selections is}\\ \binom{n}{0} + 2\binom{n}{1} + 3\binom{n}{2} +\dots + (n+1) \binom{n}{n} = \boxed{\sum_{k=0}^n (k+1) \binom{n}{k}}$
This is the generic method:
$\text{For some fixed value of }r\text{ (between 0 and }n),\\ \text{part i) tells us that there are }r+1\text{ ways}\\ \text{of drawing the red/blue balls.}$
$\text{Part ii) tells us there are }\binom{n}{r}\text{ ways}\\ \text{of drawing the remaining }n-r\text{ white balls on top of them.}\\ \text{So the }n\text{ balls can therefore be drawn in }(r+1) \binom{n}{r}\text{ ways.}$
$\text{Therefore, by considering all cases of }r\text{, i.e. }r = 0, 1, 2, \dots, n,\\ \text{the total number of selections is therefore }\boxed{\sum_{r=0}^n (r+1) \binom{n}{r}}$
$\text{From here, it's just a small battle with algebra.}\\ \text{Firstly upon expanding, the sum splits into }\sum_{r=0}^n r \binom{n}{r} + \sum_{r=0}^n \binom{n}{r}$
$\text{In 7b i), we've already proven that }\sum_{r=0}^n \binom{n}{r} = 2^n.\\ \text{But furthermore, it is true that }\boxed{\sum_{r=0}^n r \binom{n}{r} = \sum_{r=1}^n r \binom{n}{r}}.\\ \text{This should be clear if you try expanding the entire sum out.}$
$\text{So using 7b iii), we know that }\sum_{r=0}^n r \binom{n}{r} = n 2^{n-1}.\\ \text{Therefore the final answer is }n2^{n-1} + 2^n\text{, which upon factorising becomes }(n+2)2^{n-1}.$
« Last Edit: September 29, 2018, 12:17:31 pm by RuiAce »