December 15, 2019, 07:30:11 am

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#### Jefferson

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##### Re: 3U Maths Question Thread
« Reply #4140 on: July 05, 2019, 08:17:22 pm »
+2
THANKS A MILLION!

attached are some more questions!
q1: i got a negative discriminant so that's a bit weird. the answer is 63 deg 6 mins or 50 deg 52 mins
q2: i got around 8.6 seconds (7.22-0.58+2), but the answer's 2.7 seconds
q3: b) sorry, i think this one's kinda long. I found the cartesian equation for both and equated to get x=16.57 but the answer is x=11.5, y=8.2 and no, they pass that at different times.

See attachment for full working.
« Last Edit: July 06, 2019, 02:12:34 pm by Jefferson »

#### Jefferson

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##### Re: 3U Maths Question Thread
« Reply #4141 on: July 05, 2019, 08:19:39 pm »
+2
Answers to 25 (a) (b)

{maximum file size error}

#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4142 on: July 07, 2019, 03:52:44 pm »
0
thank you so much!!!

for the integration of 1/sqr(49-x^2), I got arcsinx/7 +c. the solution they have is -arccosx/7 +c, I was just wondering how to convert that?

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4143 on: July 07, 2019, 04:03:23 pm »
+1
thank you so much!!!

for the integration of 1/sqr(49-x^2), I got arcsinx/7 +c. the solution they have is -arccosx/7 +c, I was just wondering how to convert that?
They're both correct.
$\text{To go from }-\arccos \frac{x}{7}+C\text{ into yours however, recall the identity}\\ \boxed{\arcsin y + \arccos y = \frac\pi2}.$
$\text{Therefore }-\arccos \frac{x}{7}+C = \arcsin \frac{x}{7}-\frac\pi2 +C\\ \text{which gives the same answer.}$
In practice though, integrating a positive $(a^2-x^2)^{-1/2}$ into negative inverse cosine is pretty uncommon. It's not incorrect, but usually very redundant.

#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4144 on: July 07, 2019, 04:38:18 pm »
0
i see, thank you! pi/2 + c just becomes c, right?

I was wondering if anyone could help me with ii) - I don't understand the solutions given. This is 2012 Ascham Trials Q13d), by the way.

And for iii), the solutions say we have to sub in t=5, r=0. I guess I understand that r=0, but wouldn't the first thing someone think of is V=0?

« Last Edit: July 09, 2019, 07:43:13 pm by spnmox »

#### CompletelyNewUser

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##### Re: 3U Maths Question Thread
« Reply #4145 on: July 10, 2019, 06:45:21 pm »
+1
i see, thank you! pi/2 + c just becomes c, right?

I was wondering if anyone could help me with ii) - I don't understand the solutions given. This is 2012 Ascham Trials Q13d), by the way.

And for iii), the solutions say we have to sub in t=5, r=0. I guess I understand that r=0, but wouldn't the first thing someone think of is V=0?

$\text{13b): Using the fact that } \frac{dr}{dt} = p \\ r = \int \frac{dr}{dt} dt \text{ (anti-differentiation)} \\ r = \int p \, dt \\ r = pt + C \\ \text{When t = 0, r = 0.5 (initial radius)} \\ 0.5 = p(0) + C \\ C = 0.5 \\ r = pt + 0.5$
$\text{13c): When t = 5, r = 0 (tablet completely dissolves)} \\ r = pt + 0.5 \\ 0 = p(5) + 0.5 \\ 5p = -0.5 \\ p = -0.1$

In question 13b), you need to remember rates of change (when anti-differentiating a rate of change), and then plug in the values for the initial radius of the tablet. It also flows naturally from the proof in 13a).

For question 13c), I would be using the result from 13b), and try to find a value of t and r. As it mentions "The tablet dissolves completely
after 5 minutes" in the question, it would probably want you to use this information somewhere (t is given, and r would be 0). Also it's only 1 mark so it wouldn't need much working out.

Hopefully this helps!

#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4146 on: July 11, 2019, 12:46:03 pm »
0
1. Given t=112.5 degrees, show that 2t/1-t^2 =1. Hence show that tan 112.5 degrees = -sqr2 -1.

I'm working on the second part, not sure why tan112.5 degrees is the negative answer and not sqr2 - 1 ?

2. Suppose tan alpha =-1/3 and pi/2<alpha<pi. Find the exact value of tanalpha/2.

Not sure how to get exact value
« Last Edit: July 11, 2019, 01:01:05 pm by spnmox »

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4147 on: July 11, 2019, 06:06:41 pm »
0
1. Given t=112.5 degrees, show that 2t/1-t^2 =1. Hence show that tan 112.5 degrees = -sqr2 -1.

I'm working on the second part, not sure why tan112.5 degrees is the negative answer and not sqr2 - 1 ?

2. Suppose tan alpha =-1/3 and pi/2<alpha<pi. Find the exact value of tanalpha/2.

Not sure how to get exact value

Since you've done the first part, you've already recognised the need for the double angle ie. tan 225 = 1. t must be negative since 112.5 degrees is in the second quadrant, and tan is negative in the 2nd and 4th quadrants, hence we take the negative root.

For the second question, try a similar idea ie. 2t/(1-t^2)=-1/3, where t=tan (alpha/2). tan alpha/2 is one of the two roots, and since pi/2<alpha<pi, pi/4<alpha/2<pi/2 ie. alpha/2 is a first quadrant angle, ie tan alpha is positive. So basically just solve the quadratic, and take the positive root

Hope this helps
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#### terassy

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##### Re: 3U Maths Question Thread
« Reply #4148 on: July 19, 2019, 11:27:47 pm »
0
Hi, need help with this question:

Consider the function y=sin(cos-1x)
(i) Find the domain and range of the function.
(ii) Sketch the function.

I know how to do (ii) but for (i):

The domain is -1 ≤ x ≤ 1 because thats the domain of cos-1x
but I don't know how to find the range. In the solutions they wrote:
range 0 ≤ cos-1x ≤ pi               <---- how does this help find range?
therefore, range 0 ≤ y ≤ 1

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4149 on: July 20, 2019, 12:13:27 am »
+2
range 0 ≤ cos-1x ≤ pi               <---- how does this help find range?
therefore, range 0 ≤ y ≤ 1

This helps find range because we're taking the sine of the inverse cosine function, and thus the range of the whole function
$ie. \ \text{For any functions} \ f \ \text{and} \ g \\ \text{The range of} \ f(g(x)) \ \text{satisfies the following} \\ a \leq g(x) \leq b \ \text{(For real a and b)} \\ f(c) \leq f(g(x)) \leq f(d) \ \text{(where} \ a \leq c,d \leq b) \\ \text{Hence for} \ y=f(g(x)), m \leq y \leq n \ \text{(where} \ m=f(c), n=f(d))$

Now substitute in pi as b, and 0 as a, arccos x as g, and sin x as f. Have a think about it, and you'll probably see how this works
« Last Edit: July 20, 2019, 12:17:12 am by fun_jirachi »
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4150 on: July 20, 2019, 01:17:31 am »
0
hey guys, when studying for maths, would you recommend writing down the mistakes you make while doing past papers down somewhere? that's what i've been doing so far... and how long before exams should i attempt to redo them? should i read over them often so i don't forget where i made the mistake? and what do you guys think about silly mistakes - should write down, just try and be more careful, or practise with timing?

thankyou!

#### laura_

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##### Re: 3U Maths Question Thread
« Reply #4151 on: July 20, 2019, 05:52:50 am »
+1
hey guys, when studying for maths, would you recommend writing down the mistakes you make while doing past papers down somewhere? that's what i've been doing so far... and how long before exams should i attempt to redo them? should i read over them often so i don't forget where i made the mistake? and what do you guys think about silly mistakes - should write down, just try and be more careful, or practise with timing?

thankyou!

I am constantly re-doing all the questions I get incorrect for whatever reason, but I also keep a list that I do as I am revising for the exam. Once you get close to exam time and are beginning your solid preparation, perhaps then is a good time to do them so that they are fresh for the exam. Definitely, do practice papers and things with timing, but as for whether or not you include silly mistakes on that list, up to you.

For some subjects I write out all my silly mistakes as an encouragement to check my work more carefully, while as for others I don't so that I am being time-efficient. Doing a little extra practice may not be a bad thing, but if they are concepts you have a firm grasp on I guess there is no real need.
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#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4152 on: July 20, 2019, 10:42:49 am »
+3
hey guys, when studying for maths, would you recommend writing down the mistakes you make while doing past papers down somewhere? that's what i've been doing so far... and how long before exams should i attempt to redo them? should i read over them often so i don't forget where i made the mistake? and what do you guys think about silly mistakes - should write down, just try and be more careful, or practise with timing?

thankyou!
Your strategy will work provided you do go back on the mistakes. Which, judging by your post, you've already implied that you plan on doing so.

As for the when factor, that will vary from person to person. You can choose to go redo them as soon as you feel prepared to, or towards the very end, or anywhere in-between. You can also choose to redo a small bundle of those questions you got wrong over multiple days, rather than just bombard all of them in the one go. There is no real 'optimal' time to do them because redoing them will always benefit somehow regardless.

To help minimise the amount of your mistakes, you should try completing all the questions from a textbook exercise/past paper and then get into the habit of checking your work before you start turning to the answers/solutions. It's going to be a hard habit to develop on the spot in the exam, because students are generally out of it once they've completed the paper. But usually there will be a point in time where they should check over what they've written, because almost everyone is very likely to make some minor error along the way. By practicing that now, you'll significantly reduce how prone you are to this in the exam.
« Last Edit: July 20, 2019, 10:44:38 am by RuiAce »

#### terassy

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##### Re: 3U Maths Question Thread
« Reply #4153 on: July 20, 2019, 05:40:51 pm »
0
Stuck with part (ii) of this question, could someone explain how they got the answer:

(b) Bill and Ben play a game in which a fair die is rolled. Bill wins the game if the die shows 1, 2, 3 or 4. Ben wins the game if the die shows 5 or 6.
(i) If they play the game 6 times, find in simplest fraction form the probability that Bill wins two games more than Ben. (answer: 80/243)
(ii) If they continue playing until one wins two games more than the other, find in simplest fraction form the probability that Bill is the eventual winner. (answer: 4/5)

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4154 on: July 20, 2019, 08:43:53 pm »
+3
Stuck with part (ii) of this question, could someone explain how they got the answer:

(b) Bill and Ben play a game in which a fair die is rolled. Bill wins the game if the die shows 1, 2, 3 or 4. Ben wins the game if the die shows 5 or 6.
(i) If they play the game 6 times, find in simplest fraction form the probability that Bill wins two games more than Ben. (answer: 80/243)
(ii) If they continue playing until one wins two games more than the other, find in simplest fraction form the probability that Bill is the eventual winner. (answer: 4/5)
Note that the first part was a binomial probability but the second part is NOT. This is because in the first part, we don't care about the ordering of the wins; we just care that Bill wins 4 out of the 6. The second part imposes a restriction that actually impacts ordering.
$\text{Note obviously that there must be more than one game}\\ \text{for one person to win twice more than the other.}$
Denote $L$ for Bill and $N$ for Ben.
$\text{If only two games are played and Bill wins,}\\ \text{the only possibility is }LL\\ \text{which occurs with probability }\left( \frac23 \right)^2.$
$\text{Now observe that there cannot have been only three games played either. }$
See that NONE of the following configurations are possible:
- $LLL$ - can't happen because Bill would've won on the second game and hence the third game doesn't happen
- $LLN$ - as above
- $LNL$ - score becomes 2 to 1, so the games must continue
- $NLL$ - as above
- $LNN$ - score becomes 1 to 2, so the games must continue
- $NLN$ - as above
- $NNL$ - can't happen because Ben would've won on the second game and hence the third game doesn't happen
- $NNN$ - as above
$\text{So the next possibility is when Bill wins on game }\textbf{four}.\\ \text{Whilst not really necessary, we illustrate all 16 cases to make it extremely obvious.}$
- $LLLL$ - can't happen because Bill would've already won on the second game
- $LLLN$ - as above
- $LLNL$ - as above
- $LLNN$ - as above
- $LNLL$ - a favourable outcome!
- $LNLN$ - score becomes 2 to 2, so the games must continue
- $LNNL$ - as above
- $LNNN$ - Ben wins here because the score is 1 to 3
- $NLLL$ - a favourable outcome!
- $NLLN$ - score becomes 2 to 2 as well
- $NLNL$ - as above
- $NLNN$ - Ben wins here as well on game 4
- $NNLL$ - Ben will already have won on game 2, so this cannot happen
- $NNLN$ - as above
- $NNNL$ - as above
- $NNNN$ - as above
$\text{So the possibilities are }LNLL\text{ and }NLLL\\ \text{with probability }2\times \left(\frac13\right)\left(\frac23\right) \times \left( \frac23\right)^2.$
___________________________________________________________________________________________

Now this will become annoying to deal with for 5 games, because then I'd have to write out 32 cases. So let's see if we can spot the pattern.

The idea is that the issues stemming from winning on the 3rd game will occur again on winning on the 5th game. Basically, either the game will already have been won, or the score will be 3-to-2 or 2-to-3 instead. Similarly, for the 7th game, either it will already have won or the score becomes 4-to-3 or 3-to-4. So in general, we never have the possibility of winning on an odd number of games, so we may discard it.

As for the evens, let's try figuring out the go with Bill winning on the 6th game without mapping out all the cases for this one. (Which would suck because there are 64 of them.) Basically we recycle what's above.
$\text{For Bill to win on game }\textbf{six},\\ \text{firstly we require the game has progressed that far.}\\ \text{This would mean that after the fourth game,}\\ \text{the match's outcome has }\textbf{still}\text{ not been decided.}$
$\text{If we look above, we see that there are 4 cases where the game continues on game four:}\\ LNLN,\, LNNL,\, NLNL,\, NLLN.$
$\text{But also, if the game continues into game four, we required it to not have been finished by game two.}\\ \text{Hence at the end of game 2, we must've had }LN\text{ or }NL.$
___________________________________________________________________________________________
$\text{Indeed there is a pattern.}\\ \text{The idea is that we always require the game to be unfinished on the }(2n-2)\text{th turn}\\ \text{for it to finish on the }2n\text{th turn.}$
To formally prove this we'd require induction, so let's not.
$\text{More importantly however is this.}\\ \text{Note that at each }2n\text{th stage, the previous sequences must've involved}\\ \text{a sequence of }n\,\,\boxed{NL}\text{'s or }\boxed{LN}\text{'s!}$
So for example, if instead the game were finished on the 8th game, it must've been unfinished after the 6th. But that would require three $NL$'s or $LN$'s appearing. In fact, the possibilities are
- $LNLNLN$
- $LNLNNL$
- $LNNLLN$
- $LNNLNL$
- $NLLNLN$
- $NLLNNL$
- $NLNLLN$
- $NLNLNL$
$\text{The idea is that if the game ends on the }2n\text{th turn,}\\ \text{it must be that we've had a sequence of }n-1\, \boxed{LN}\text{'s}\text{ or }\boxed{NL}\text{'s.}\\ \text{Note also that they're independent of one another - one being }LN\text{ does NOT mean the other must be }LN\text{ or }NL.$
$\text{Therefore we realise that to win on the }2n\text{th game, we have }\boxed{2^{n-1}}\text{ choices}\text{ for the ordering of the previous }LN\text{ and }NL\text{ pairs.}\\ \text{So the probability that the game is unfinished on the }2(n-1)\text{th turn is}\\ 2^{n-1} \left(\frac13 \right)^{n-1} \left( \frac23\right)^{n-1}$
$\text{And of course, the icing on the cake is that the last two games must always go to Bill.}\\ \text{(Should be obvious, but make sure you understand why!)}\\ \text{This, of course, occurs with probability }\left( \frac23 \right)^2 .$
$\text{So the probability of winning on exactly the }2n\text{th turn is}\\ 2^{n-1} \left(\frac13 \right)^{n-1} \left( \frac23\right)^{n-1} \left( \frac23\right)^2$
___________________________________________________________________________________________
\text{Hence as we require the probability of winning on the 2nd, 4th, 6th, 8th or so on-th turn,}\\ \text{the probability we wish to compute is}\\ \begin{align*} &\quad\sum_{n=1}^\infty 2^{n-1} \left(\frac13 \right)^{n-1} \left( \frac23\right)^{n-1} \left( \frac23\right)^2\\&= \frac49 \sum_{n=1}^\infty \left( \frac49 \right)^{n-1} \tag{simplifying}\\ &= \frac49 \left(1+\frac49 + \left( \frac49 \right)^2 + \dots \right) \\ &= \frac49 \times \frac{1}{1-\frac49} \tag{recognising geometric series}\\ &= \frac45\end{align*}
« Last Edit: July 20, 2019, 08:53:06 pm by RuiAce »