Author Topic: 4U Maths Question Thread (Read 667865 times) Tweet Share

RuiAce · « **Reply #555 on:** October 07, 2016, 09:50:01 am »

Quote from: katherine123 on October 07, 2016, 01:34:23 am

how to do part iii)

$\text{Firstly, briefly note that the ordering mattered for part (ii)}\\ \text{The ordering doesn't matter for part (iii) as }WWLL\text{ is the same as }LLWW = 2$
$\text{List out all the possibilities for the home team to have more points than the away team.}\\ \begin{align*}WWWW\\ WWWD\\ WWWL\\ WWDD\\ WWDL\\ WDDD \end{align*}$
$\text{Probability of 4 wins is easy; that's just }0.2^4$
$\text{The probability of 3 wins and a draw is a }\textit{binomial probability}\\ \text{This is your classic }k\text{ successes in }n\text{ tries}\\ \text{This can be achieved in }\binom{4}{3}(0.2)^3(0.6)$
$\text{Similarly, 3 wins 1 loss is }\binom{4}{3}(0.2)^3(0.2)\\ \text{and 3 draws 1 win is }\binom{4}{3}(0.6)^3(0.2)$
$\text{It should be clear now that 2 draws 2 wins is also binomial. This one has probability }\binom{4}{2}(0.6)^2(0.2)^2$
$\text{Hence, noting that all cases are different, if we tear apart the binomial coefficient into factorials we have}\\ 0.2^4+\frac{4!}{3!}(0.2)^3(0.6)+\frac{4!}{3!}(0.2)^2(0.2)^2+\frac{4!}{3!}(0.6)^3(0.2)+\frac{4!}{2!2!}(0.6)^2(0.6)^2+Pr(WWDL)$
$WWDL\text{ is a bit trickier. I purposely decomposed the binomial coefficients because}\\ \text{Notice that }\frac{4!}{3!}\text{ is just the number of ways we can make words out of }WWWD!\\ \text{And how }\frac{4!}{4!}\text{ is the number of ways we can make words out of }WWDD$
$\text{The number of ways we can make words out of }WWDL\text{ is }\frac{4!}{2!}\\ \text{Hence this choice has a probability of }\frac{4!}{2!}(0.2)^2(0.6)(0.2)$
$\text{Add these probabilities up and we are done.}$

kiwiberry · « **Reply #556 on:** October 07, 2016, 06:35:22 pm »

Hey, I'm new here so I'm not sure how this works haha
this is probably really simple but how would you graph
i) 1/f(x)
ii) [f(x)]^2

thank you

jakesilove · « **Reply #557 on:** October 07, 2016, 06:42:35 pm »

Quote from: kiwiberry on October 07, 2016, 06:35:22 pm

Hey, I'm new here so I'm not sure how this works haha
this is probably really simple but how would you graph
i) 1/f(x)
ii) [f(x)]^2

thank you

Hey! Welcome to the forums! You've done the exact right thing; post up any questions you've got, and we're here to answer them! You can also answer others' questions if you feel up to it

For these kinds of questions, you just need to think about what happens at critical points on the graph. So, let's start with 1/f(x).

We know that at x approaching infinity, y will approach 2. That means that, for 1/f(x), as x approaches infinity, y will approach 1/2. We also know that y equals -1 and x=2, therefore for 1/f(x), when x=2, y=1/(-1)=-1.

Now, we need to recognise that if we divide by zero, there will be an asymptote at that point. We know that at x=0, y=0. Therefore, on our new graph, there will be an asymptote at x=0!

On the left hand side, we know that as x reaches negative infinity, y reaches infinity. As 1/infinity=0, our new graph will reach zero as x reaches negative infinity, and reach infinity as x reaches zero from the left.

This is super hard to explain. I have to run, but hopefully someone can give a better explanation/show you the second graph before I come back!

RuiAce · « **Reply #558 on:** October 07, 2016, 06:46:08 pm »

Quote from: kiwiberry on October 07, 2016, 06:35:22 pm

Hey, I'm new here so I'm not sure how this works haha
this is probably really simple but how would you graph
i) 1/f(x)
ii) [f(x)]^2

thank you

$\text{Consider what is going on. What exactly happens when you reciprocate a function.}\\ \text{All the }y\text{-coordinates that aren't }0\text{ get reciprocated for starters.}\\ \text{So your new horizontal asymptote would be }y=\frac{1}{2}$
$\text{What happens at the }x\text{-intercepts? }\frac{1}{0}\text{ should be undefined.}\\ \text{And rightfully so. As you divide something getting as small as possible to 0 e.g. }\frac{1}{0.001}\\ \text{You get something really large: }1000\\ \text{So we have a vertical asymptote replacing the }x\text{-intercepts.}$
$\text{Does it work in reverse? Do vertical asymptotes become }x\text{-intercepts?}\\ \text{Not quite. You get closer and closer back down to 0, but you don't know if you ever reach 0.}\\ \text{Instead, we have discontinuities (holes in the graph) on the }x\text{-axis.}$
$\text{Also, }\frac{1}{f(x)}\text{ has the exact same sign as }f(x)\\ \text{So whatever was above the }x\text{-axis stays there, and whatever was below stays as well.}$
$\text{Lastly, consider the derivative:}\\ \frac{d}{dx}\frac{1}{f(x)}=-\frac{f^\prime(x)}{[f(x)]^2}\\ \text{Unlike the above, the sign of the derivative is flipped! Hence if }f(x)\text{ was increasing, }\frac{1}{f(x)}\text{ decreases}\\ \text{and if }f(x)\text{ was decreasing, }\frac{1}{f(x)}\text{ increases.}\\ \text{Note also that stationary points are preserved, but }\textit{flipped.}\text{ Local max become local min and etc.}$
$\text{Have an attempt now at the first graph and post up your solution.}$
Jamon went through the effort of making a guide for 4U curve sketching. Consider reading it.

znaser · « **Reply #559 on:** October 09, 2016, 10:03:04 am »

Hi. I'm not sure of the working out for this question. The answer is A.

RuiAce · « **Reply #560 on:** October 09, 2016, 10:17:09 am »

Quote from: znaser on October 09, 2016, 10:03:04 am

Hi. I'm not sure of the working out for this question. The answer is A.

$\text{Counting technique: Consider the complementary event.}\\ \text{The complement is the case of there being more than 4 in a room.}\\ \text{i.e. There are 5, or 6 in a room.}$
$\text{The total possible outcomes is just }4^6\text{ as we have 6 people and 4 rooms}$
$\text{The }\textit{non-favourable}\text{ outcomes (complementary events) are:}\\ \text{If all six are in a room, all we need to do is choose 1 room to put them in. This can be done in }4\text{ ways.}$
$\text{If five are in a room, we choose the first room out of 4 and then allocate 5 of the 6 people into them.}\\ \text{Note that the order we allocate the people into the rooms is irrelevant, so this is done in }4\times \binom{6}{5}\text{ways}$
$\text{And then the last guy goes in any of the 3 leftover rooms.}\\ \text{So if 5 are in a room, this can be done in }4\binom{6}{5}\times 3\text{ ways}$
$\text{Hence, our answer is }4^6 - 4 - 12\binom{6}{5}$
A more brute force approach that would work is a case by case analysis. Note, however, that this requires significantly more effort (just because of how many cases there are), and thus we choose the complement for convenience.

znaser · « **Reply #561 on:** October 09, 2016, 10:31:17 am »

Quote from: RuiAce on October 09, 2016, 10:17:09 am

$\text{Counting technique: Consider the complementary event.}\\ \text{The complement is the case of there being more than 4 in a room.}\\ \text{i.e. There are 5, or 6 in a room.}$
$\text{The total possible outcomes is just }4^6\text{ as we have 6 people and 4 rooms}$
$\text{The }\textit{non-favourable}\text{ outcomes (complementary events) are:}\\ \text{If all six are in a room, all we need to do is choose 1 room to put them in. This can be done in }4\text{ ways.}$
$\text{If five are in a room, we choose the first room out of 4 and then allocate 5 of the 6 people into them.}\\ \text{Note that the order we allocate the people into the rooms is irrelevant, so this is done in }4\times \binom{6}{5}\text{ways}$
$\text{And then the last guy goes in any of the 3 leftover rooms.}\\ \text{So if 5 are in a room, this can be done in }4\binom{6}{5}\times 3\text{ ways}$
$\text{Hence, our answer is }4^6 - 4 - 12\binom{6}{5}$
A more brute force approach that would work is a case by case analysis.

Thanks!

birdwing341 · « **Reply #562 on:** October 09, 2016, 09:26:00 pm »

Hello!

Another out-of-left-field question, but I was wondering how you would order the HSC Extension 2 papers in difficultly (from 2001 to 2015) or in rough equivalents?

Thanks again!

RuiAce · « **Reply #563 on:** October 09, 2016, 09:44:29 pm »

Quote from: birdwing341 on October 09, 2016, 09:26:00 pm

Hello!

Another out-of-left-field question, but I was wondering how you would order the HSC Extension 2 papers in difficultly (from 2001 to 2015) or in rough equivalents?

Thanks again!

Quote from: RuiAce on October 09, 2016, 09:27:40 pm

That's hard. I need to look at every single paper to give an answer to this one.

And now that I had a quick glance (I'm not spending 45 hours of my life doing all those papers...) this is what I reckon

From easiest to hardest

2014
2015
2012
2013
2010
2011
2007
2009
2004
2006
2005
2002
2003
2001

Reminder: This is very easily debatable/arguable.

kiwiberry · « **Reply #564 on:** October 09, 2016, 09:55:11 pm »

Quote from: RuiAce on October 07, 2016, 06:46:08 pm

$\text{Consider what is going on. What exactly happens when you reciprocate a function.}\\ \text{All the }y\text{-coordinates that aren't }0\text{ get reciprocated for starters.}\\ \text{So your new horizontal asymptote would be }y=\frac{1}{2}$
$\text{What happens at the }x\text{-intercepts? }\frac{1}{0}\text{ should be undefined.}\\ \text{And rightfully so. As you divide something getting as small as possible to 0 e.g. }\frac{1}{0.001}\\ \text{You get something really large: }1000\\ \text{So we have a vertical asymptote replacing the }x\text{-intercepts.}$
$\text{Does it work in reverse? Do vertical asymptotes become }x\text{-intercepts?}\\ \text{Not quite. You get closer and closer back down to 0, but you don't know if you ever reach 0.}\\ \text{Instead, we have discontinuities (holes in the graph) on the }x\text{-axis.}$
$\text{Also, }\frac{1}{f(x)}\text{ has the exact same sign as }f(x)\\ \text{So whatever was above the }x\text{-axis stays there, and whatever was below stays as well.}$
$\text{Lastly, consider the derivative:}\\ \frac{d}{dx}\frac{1}{f(x)}=-\frac{f^\prime(x)}{[f(x)]^2}\\ \text{Unlike the above, the sign of the derivative is flipped! Hence if }f(x)\text{ was increasing, }\frac{1}{f(x)}\text{ decreases}\\ \text{and if }f(x)\text{ was decreasing, }\frac{1}{f(x)}\text{ increases.}\\ \text{Note also that stationary points are preserved, but }\textit{flipped.}\text{ Local max become local min and etc.}$
$\text{Have an attempt now at the first graph and post up your solution.}$
Jamon went through the effort of making a guide for 4U curve sketching. Consider reading it.

Quote from: jakesilove on October 07, 2016, 06:42:35 pm

Hey! Welcome to the forums! You've done the exact right thing; post up any questions you've got, and we're here to answer them! You can also answer others' questions if you feel up to it

For these kinds of questions, you just need to think about what happens at critical points on the graph. So, let's start with 1/f(x).

We know that at x approaching infinity, y will approach 2. That means that, for 1/f(x), as x approaches infinity, y will approach 1/2. We also know that y equals -1 and x=2, therefore for 1/f(x), when x=2, y=1/(-1)=-1.

Now, we need to recognise that if we divide by zero, there will be an asymptote at that point. We know that at x=0, y=0. Therefore, on our new graph, there will be an asymptote at x=0!

On the left hand side, we know that as x reaches negative infinity, y reaches infinity. As 1/infinity=0, our new graph will reach zero as x reaches negative infinity, and reach infinity as x reaches zero from the left.

This is super hard to explain. I have to run, but hopefully someone can give a better explanation/show you the second graph before I come back!

Thanks so much guys

I had a read of Jamon's guide and it was really helpful!!
Here's my attempt at the questions

RuiAce · « **Reply #565 on:** October 09, 2016, 10:01:29 pm »

Quote from: kiwiberry on October 09, 2016, 09:55:11 pm

Thanks so much guys
I had a read of Jamon's guide and it was really helpful!!
Here's my attempt at the questions

Looking good. Well done

Justina Shehata · « **Reply #566 on:** October 09, 2016, 10:05:33 pm »

Do you lose marks in the HSC for not putting working out in your answers?

RuiAce · « **Reply #567 on:** October 09, 2016, 10:06:32 pm »

Quote from: Justina Shehata on October 09, 2016, 10:05:33 pm

Do you lose marks in the HSC for not putting working out in your answers?

Yes.

That's why some questions are worth 3 marks, not 1 mark each. You're actually AWARDED marks for working out.

If you magically write down an answer, people will think ok where did you get it from

birdwing341 · « **Reply #568 on:** October 09, 2016, 10:20:13 pm »

Quote from: RuiAce on October 09, 2016, 09:44:29 pm

And now that I had a quick glance (I'm not spending 45 hours of my life doing all those papers...) this is what I reckon

From easiest to hardest

2014
2015
2012
2013
2010
2011
2007
2009
2004
2006
2005
2002
2003
2001

Reminder: This is very easily debatable/arguable.

Man you are a legend, thanks rui! Just did one of the past papers so wanted to see where it sat in terms of difficulty

Justina Shehata · « **Reply #569 on:** October 09, 2016, 11:49:10 pm »

Quote from: RuiAce on October 09, 2016, 10:06:32 pm

Yes.

That's why some questions are worth 3 marks, not 1 mark each. You're actually AWARDED marks for working out.

If you magically write down an answer, people will think ok where did you get it from

would the working out in the sample answers be a good indication to how much they require?

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