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#### classof2019

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##### Re: 4U Maths Question Thread
« Reply #2370 on: September 13, 2019, 08:57:19 pm »
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The mistake was in how you wrote it: xx5xx7xx9xx.

Consider the following. On one hand, we can do 6x58x7xx9xx. This would give us the number 65879.

On another hand, we can do x65x87xx9xx. This is a different arrangement in how you did 8P2, but observe that it also gives 65879! Hence the number 65879 has been double counted.

The flaw was that you focused too much on the possibility of something like 68579, which actually escapes any double counting. But it only does so because the even numbers got grouped between two odd numbers. It doesn’t work when the even numbers are split.

Instead, a better option would be the following. First arrange the 6 by picking one of the 4 crosses in: x5x7x9x.

Then arrange the 8. Note that now that the 6 has been arranged, we’d have something like x6x5x7x9x. Therefore we pick one out of 5 crosses for the 8.

In total, we therefore have $4\times5=20$ favourable arrangements, giving a probability of $\frac{20}{120}=6$.

Note: there is also a clever approach here. Recall the problem of arranging the letters ABCDEFG, if say the E must come before the F. But you can match each arrangement with E before F with another one with F before E. (For example, match BCDEFGA with BCDFEGA.) So exactly half of the arrangements have E before F, I.e. $\frac{7!}{2}$ arrangements.

Here, the idea is that we can group any configuration of the odd numbers among each other. Take for example 65879. We can match it with:
- 65897
- 67859
- 67895
- 69875
- 69857

So each number with 5,7,9 in order can be matched with exactly five other numbers, without changing the position of the 6 and the 8, but worn 5,7,9 not in order. Hence only one sixth of arrangements have 5,7,9 in order.

Nevertheless, that was a nice idea with the stars and bars like idea. Just executed with a flaw!

Thank you so much! This really helps.

Just a thought, and I'm not sure whether this is purely coincidental, but could we potentially say that all of these 5 digit numbers have some sort of arrangement of 5,7, and 9. Now, 5,7 and 9 can be arranged in 3! = 6 ways. Only one such arrangement is them in ascending order (5,7,9). So the probability is 1/6.

Is this a mathematically sound justification or a coincidence?
« Last Edit: September 13, 2019, 09:01:13 pm by classof2019 »

#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2371 on: September 13, 2019, 09:10:14 pm »
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Thank you so much! This really helps.

Just a thought, and I'm not sure whether this is purely coincidental, but could we potentially say that all of these 5 digit numbers have some sort of arrangement of 5,7, and 9. Now, 5,7 and 9 can be arranged in 3! = 6 ways. Only one such arrangement is them in ascending order (5,7,9). So the probability is 1/6.

Is this a mathematically sound justification or a coincidence?
That is perfectly mathematically sound. It’s just the more cleaner way of expressing my quicker way

#### Ollierobb1

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##### Re: 4U Maths Question Thread
« Reply #2372 on: September 15, 2019, 10:19:01 am »
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Given that  zn+z-n=2cosnx, find the solutions for 2z4+3z3+5z2+3z+2=0. Help would be much appreciated

#### fun_jirachi

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##### Re: 4U Maths Question Thread
« Reply #2373 on: September 15, 2019, 12:42:24 pm »
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Hey there!

In this sort of situation, we want to use the result given in some shape or form, so we try to manipulate any expressions to 'match up' with the given result.
For 2z4+3z3+5z2+3z+2=0, note that  z2(2z2+3z+5+3z-1+2z-2)=0.
Then, we have z2(4cos2x+6cosx+5)=0, using the result that zn+z-n=2cosnx where z=cisx. Clearly, from the original expression, z cannot equal zero, so we essentially just have a quadratic in cos x ie. 8cos2x+6cosx+1=0. Solving for cosx, we have that
$\cos x = \frac{-6\pm \sqrt{36-32}}{16} = \frac{-6\pm 2}{16} = -\frac{1}{2} \ \text{or} \ -\frac{1}{4} \\ \text{ie.} \ x = \frac{2\pi}{3}, \frac{4\pi}{3}, \cos^{-1} \left(-\frac{1}{4}\right), 2\pi-\cos^{-1} \left(-\frac{1}{4}\right) \\ \text{ie.} \ z=\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}, \cos \frac{4\pi}{3} + i\sin \frac{4\pi}{3} \\ \text{or} \ z= \cos \left(\cos^{-1} \left(-\frac{1}{4}\right)\right) + i\sin \left(\cos^{-1} \left(-\frac{1}{4}\right)\right), \cos \left(2\pi-\cos^{-1} \left(-\frac{1}{4}\right)\right) + i\sin \left(2\pi-\cos^{-1} \left(-\frac{1}{4}\right)\right)$

Hope this helps
« Last Edit: September 15, 2019, 12:44:28 pm by fun_jirachi »
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#### worldno1

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##### Re: 4U Maths Question Thread
« Reply #2374 on: September 17, 2019, 08:13:58 pm »
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Hey guys,

For this question:  An urn contains n red balls, n white balls and n blue balls. Three balls are drawn at random from the urn, one at a time, without replacement. What is the probability  that the three balls are all the same colour?

Why is the sample space (3N)C3, instead of (3N)C1 * (3N-1)C1 * (3N-2)C1?

My logic was that, since each ball was taken out one at a time, the total number of balls available would decrease by one each time. Why is this not so?

Thanks in advance! Also, sorry, I don't know how LaTex works.
« Last Edit: September 17, 2019, 08:15:53 pm by worldno1 »

#### fun_jirachi

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##### Re: 4U Maths Question Thread
« Reply #2375 on: September 17, 2019, 10:55:59 pm »
+1
$\text{The sample space} \ \binom {3n} 1 \times \binom {3n-1} 1 \times \binom {3n-2} 1 \ \text{includes order.} \\ \text{This notation essentially means you're picking from one less ball each time,} \\ \text{while having one less colour to pick from each time, which is incorrect.} \\ \text{You can't really use the same idea because the balls are all in one bag} \\ \text{and you can't place restrictions within the bag itself.}$

Basically you have when you're picking for the second and third colour you're still allowing yourself to pick from colours that you can no longer pick from ie. with the 3n-1. You're still after all just picking 3 balls from 3n balls.

Hope this helps
« Last Edit: September 17, 2019, 10:59:21 pm by fun_jirachi »
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#### mani.s_

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##### Re: 4U Maths Question Thread
« Reply #2376 on: September 18, 2019, 09:17:36 pm »
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Will you guys make a new Maths Extension 2 course notes for the new syllabus???

#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2377 on: September 18, 2019, 09:19:37 pm »
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Will you guys make a new Maths Extension 2 course notes for the new syllabus???
Yep.

Sadly, they won't be out in time for the October head-start lectures due to other commitments I have with the company.

#### david.wang28

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##### Re: 4U Maths Question Thread
« Reply #2378 on: September 20, 2019, 05:19:31 pm »
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Hello, I have trouble working out this integral. My working out is attached, the answer is 4𝜋/27 - √3 /9. Can anyone please help me out? Thanks
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#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2379 on: September 20, 2019, 05:55:59 pm »
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Hello, I have trouble working out this integral. My working out is attached, the answer is 4𝜋/27 - √3 /9. Can anyone please help me out? Thanks
Your mistake was pretty deep in actually. It was in the transition from line 7 to line 8, next to the blue text.
$\text{Notice how you wrote}\\ \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{1+9x^2} = \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{\frac19 + x^2}\\ \text{without appropriately adjusting in the numerator as well!}$
$\text{The correct change is}\\ \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{1+9x^2} = \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{\frac19\, dx}{\frac19 + x^2}$

#### david.wang28

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##### Re: 4U Maths Question Thread
« Reply #2380 on: September 20, 2019, 06:40:31 pm »
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Your mistake was pretty deep in actually. It was in the transition from line 7 to line 8, next to the blue text.
$\text{Notice how you wrote}\\ \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{1+9x^2} = \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{\frac19 + x^2}\\ \text{without appropriately adjusting in the numerator as well!}$
$\text{The correct change is}\\ \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{1+9x^2} = \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{\frac19\, dx}{\frac19 + x^2}$
Ahhh I see, thank you
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#### louisaaa01

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##### Re: 4U Maths Question Thread
« Reply #2381 on: September 27, 2019, 10:18:36 pm »
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#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2382 on: September 28, 2019, 07:09:23 am »
+1
Edit: At later glance I realised I slightly messed up the base.
$\text{When }n=0,\\ \text{the only configurations for }(r,s)\text{ are }(1,0)\text{ and }(0,1).\\ \text{So we can just check each individually.}$
$\text{For the }(0,1)\text{ case, we have }P(0,1)=0.\\ \text{Here }s=1\text{, so we verify that}\\ \frac{1}{2^1}\left[ \binom10 + \binom11 \right] =\frac22 = 1$
$\text{For the }(1,0)\text{ case, we have }P(1,0) = 0.\\ \text{Here }s=0\text{, which is outside the valid range }s\geq 1.\\ \text{Hence this does not contradict the inductive assumption anyhow.}$
So in both cases, the base case $n=0$ is valid.
_________________________________________________________________________________
$\text{Now assume the statement holds }\textbf{for all}\text{ configurations }(r,s)\\ \text{such that }r+s-1=k. \\ \text{We then need to prove the statement holds for all configurations }(r,s)\\ \text{such that } r+s-1=k+1.$
That was perhaps the most confusing part of the induction - figuring out the goal of the inductive step.
$\text{Firstly, from the definition of the recurrence,}\\ P(r,s) = \frac12 P(r-1, s) + \frac12 P(r,s-1).$
$\text{Now if }r+s-1 = k+1\text{, we know that}\\ (r-1)+s-1 = k\text{ and } r + (s-1) - 1= k.\\ \text{Hence we may use the inductive assumption on }\textit{both}\text{ terms:}$
\begin{align*}
P(r,s) &=  \frac12 P(r-1, s) + \frac12 P(r,s-1)\\
&= \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] + \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right] \\
&= \frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] +\frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right]
\end{align*}
Ensure that you understand why the first sum terminates at $s-1$, whilst the second terminates at $s-2$.
$\text{Now to finish, the tricks here are similar to the induction proof of the binomial theorem.}\\ \text{We need to recall that firstly, }\binom{k}{0} = \binom{k+1}{0} = 1.\\ \text{Also, we need Pascal's identity }\boxed{\binom{N+1}{K+1} = \binom{N}{K} + \binom{N}{K+1}}.\\ \text{With }\textit{both}\text{ properties in mind, we proceed to cleverly rearrange the terms.}$
\begin{align*}
P(r,s) &= \frac1{2^{n+1}} \left[ \binom{k}{0} + \left[ \binom{k}0 + \binom{k}1\right] + \left[ \binom{k}1+ \binom{k}2 \right]+ \dots + \left[\binom{k}{s-2}+\binom{k}{s-1} \right] \right]\\
&= \frac1{2^{n+1}} \left[ \binom{k+1}{0} + \binom{k+1}{1}+\binom{k+1}{2} + \dots + \binom{k+1}{s-1} \right].
\end{align*}
And this is exactly what we wished to prove, so we are done.

*I did this in a bit of a rush so I may have given too little clarification. Let me know if there's any bit you want me to elaborate on
« Last Edit: January 15, 2020, 12:17:33 am by RuiAce »

#### louisaaa01

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##### Re: 4U Maths Question Thread
« Reply #2383 on: September 28, 2019, 11:56:42 am »
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$\text{When }n=1,\\ \text{the only configurations for }(r,s)\text{ are }(1,0)\text{ and }(0,1).\\ \text{In both cases, we know that }P(r,s) = 1\text{ from the question.}$
$\text{And also, }\\ \frac{1}{2^1}\left[ \binom10 + \binom11\right] = \frac22 = 1.\\ \text{Hence the statement holds for }n=1.$
_________________________________________________________________________________
$\text{Now assume the statement holds }\textbf{for all}\text{ configurations }(r,s)\\ \text{such that }r+s-1=k. \\ \text{We then need to prove the statement holds for all configurations }(r,s)\\ \text{such that } r+s-1=k+1.$
That was perhaps the most confusing part of the induction - figuring out the goal of the inductive step.
$\text{Firstly, from the definition of the recurrence,}\\ P(r,s) = \frac12 P(r-1, s) + \frac12 P(r,s-1).$
$\text{Now if }r+s-1 = k+1\text{, we know that}\\ (r-1)+s-1 = k\text{ and } r + (s-1) - 1= k.\\ \text{Hence we may use the inductive assumption on }\textit{both}\text{ terms:}$
\begin{align*}
P(r,s) &=  \frac12 P(r-1, s) + \frac12 P(r,s-1)\\
&= \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] + \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right] \\
&= \frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] +\frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right]
\end{align*}
Ensure that you understand why the first sum terminates at $s-1$, whilst the second terminates at $s-2$.
$\text{Now to finish, the tricks here are similar to the induction proof of the binomial theorem.}\\ \text{We need to recall that firstly, }\binom{k}{0} = \binom{k+1}{0} = 1.\\ \text{Also, we need Pascal's identity }\boxed{\binom{N+1}{K+1} = \binom{N}{K} + \binom{N}{K+1}}.\\ \text{With }\textit{both}\text{ properties in mind, we proceed to cleverly rearrange the terms.}$
\begin{align*}
P(r,s) &= \frac1{2^{n+1}} \left[ \binom{k}{0} + \left[ \binom{k}0 + \binom{k}1\right] + \left[ \binom{k}1+ \binom{k}2 \right]+ \dots + \left[\binom{k}{s-2}+\binom{k}{s-1} \right] \right]\\
&= \frac1{2^{n+1}} \left[ \binom{k+1}{0} + \binom{k+1}{1}+\binom{k+1}{2} + \dots + \binom{k+1}{s-1} \right].
\end{align*}
And this is exactly what we wished to prove, so we are done.

*I did this in a bit of a rush so I may have given too little clarification. Let me know if there's any bit you want me to elaborate on

Thank you!
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#### not a mystery mark

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##### Re: 4U Maths Question Thread
« Reply #2384 on: October 01, 2019, 11:36:18 am »
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Hey, would somebody be able to explain why the answer to this question is C? I feel myself going insane and hope somebody can assist me in my time of lunacy. Thanks heaps!
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