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March 29, 2024, 11:10:09 pm

Author Topic: 3U Maths Question Thread  (Read 1230660 times)  Share 

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david.wang28

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Re: 3U Maths Question Thread
« Reply #3870 on: January 28, 2019, 11:09:41 am »
0
I'm not providing the full solution to this one yet because the way of thinking follows along the lines of the angle sum question from a while back you asked. You should revisit it.

Hint: The base case should be obvious - two lines can only intersect once, or they're parallel (and thus they never intersect).

For the inductive step, use a similar train of thought. Assume that every pair of \(k\) lines intersect at most \( \frac12 k (k-1) \) times. Then, given any \(k+1\) lines, you can split them into a group of \(k\) lines (which takes advantage of the inductive assumption), and that one extra line. But at most how many times can that last line intersect with each of the \(k\) other lines?
Hi Rui! It took me a while to do this question. Is this the right way to do it?
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3871 on: January 28, 2019, 12:05:31 pm »
+2
Hi Rui! It took me a while to do this question. Is this the right way to do it?
It's a bit fudgy the way you wrote it sadly because whilst your recurrence is correct, you haven't really convinced me why it's correct apart from subbing in a bunch of values. This is how I would've done it, and you can argue it similarly for recurrence relations.
\[ \text{Assume that the greatest number of intersection points of}\\ k\text{ lines is }\frac12 k(k-1). \]
\[ \text{Consider any }k+1\text{ lines that have the greatest number of intersection points.}\\ \text{Take }k\text{ of those lines. From the assumption these intersect}\\ \text{at most }\frac12 k(k-1)\text{ times.}\]
\[ \text{Then, the remaining line can intersect the other }k\text{ lines}\\ \text{at most }k\text{ times. This occurs essentially by ensuring}\\ \text{the remaining line intersects }\textbf{all}\text{ previous }k\text{ lines}.\]
\[ \text{Hence, the }k\text{ lines intersect at most}\\ \frac12 k(k-1) + k = \dots = \frac12 (k+1)k\text{ times.}\]
Whilst your approach follows the lines of what they examine in 2U with financial applications of series, for a 3U induction question it becomes insufficient. This is because the whole point of induction was to prove that the pattern exists, rather than just see something convenient by eye.

david.wang28

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Re: 3U Maths Question Thread
« Reply #3872 on: January 28, 2019, 02:32:02 pm »
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It's a bit fudgy the way you wrote it sadly because whilst your recurrence is correct, you haven't really convinced me why it's correct apart from subbing in a bunch of values. This is how I would've done it, and you can argue it similarly for recurrence relations.
\[ \text{Assume that the greatest number of intersection points of}\\ k\text{ lines is }\frac12 k(k-1). \]
\[ \text{Consider any }k+1\text{ lines that have the greatest number of intersection points.}\\ \text{Take }k\text{ of those lines. From the assumption these intersect}\\ \text{at most }\frac12 k(k-1)\text{ times.}\]
\[ \text{Then, the remaining line can intersect the other }k\text{ lines}\\ \text{at most }k\text{ times. This occurs essentially by ensuring}\\ \text{the remaining line intersects }\textbf{all}\text{ previous }k\text{ lines}.\]
\[ \text{Hence, the }k\text{ lines intersect at most}\\ \frac12 k(k-1) + k = \dots = \frac12 (k+1)k\text{ times.}\]
Whilst your approach follows the lines of what they examine in 2U with financial applications of series, for a 3U induction question it becomes insufficient. This is because the whole point of induction was to prove that the pattern exists, rather than just see something convenient by eye.
These type of questions really get me stuck in the mud. It's difficult to write up my own proof if someone understands it but I don't. Anyways, thanks for the insight! :)
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iktimal

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Re: 3U Maths Question Thread
« Reply #3873 on: January 28, 2019, 10:05:25 pm »
0
Hi, I have a question from a worksheet for extension where the answers they have given me has left me quite confused.
The question is:
A particle moves in a line so that its distance from the origin at time t is x.
a) Find v in terms of x.
b) Describe the motion.

The answers are:
a) d^2 x / d t^2 = dv/dt = dv/dx * dx/dt = v* dv/dx = d/dx (v^2/2)

b) d/dx(v^2/2) =3x - 2x^3
Intergrating: v^2/2 = 3x^2 / 2 - x^4/2 + C
When x=1, v = 0, giving C=-1
Therefore v^2 = 3x^2 - x^4 - 2

Can someone please explain to me how to do this question  :)
« Last Edit: January 28, 2019, 10:07:08 pm by iktimal »

Cali.doyle

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Re: 3U Maths Question Thread
« Reply #3874 on: January 29, 2019, 08:53:50 am »
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Holla,

I was having a look through past HSC papers, and found this binomial question from the 2017 one:
When expanded, which expression has a non-zero constant term?

a. (x+(1/x^2))^7
b. (x^2+(1/x^3))^7
c. (x^3+(1/x^4))^7
d. (x^4+(1/x^5))^7

Obviously since it's a multiple choice, I'm assuming actually expanding them is the last thing you want to do, so what would be the most efficient means of finding the answer?

Thanks,
Cali

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Re: 3U Maths Question Thread
« Reply #3875 on: January 29, 2019, 09:43:10 am »
+1
Holla,

I was having a look through past HSC papers, and found this binomial question from the 2017 one:
When expanded, which expression has a non-zero constant term?

a. (x+(1/x^2))^7
b. (x^2+(1/x^3))^7
c. (x^3+(1/x^4))^7
d. (x^4+(1/x^5))^7

Obviously since it's a multiple choice, I'm assuming actually expanding them is the last thing you want to do, so what would be the most efficient means of finding the answer?

Thanks,
Cali

Hello! :)

If you're looking for a non-zero constant term, you're looking for where the x's cancel out. Considering the first term of option A, you would have x7. The next term would have the coefficient multiplied by x6 multiplied by 1/x^2. This equates to the coefficient by x4. The logic here is that by lowering the power of x by one, and increasing the power of 1/x^2 by one, you're effectively making the 'power differential' 3. So you would get down to x^4, then x^1, and so on in an arithmetic series until the last term.

The same logic follows for the other options. For B, you start with the first term x^14, and count the powers down by fives (because 2+3=5). For C, you start with first term x^21, and count down by sevens. This will get to an x^0 term, and therefore C is the answer.
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InnererSchweinehund

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Re: 3U Maths Question Thread
« Reply #3876 on: January 29, 2019, 10:11:53 am »
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Hi Guys,
I am currently taking Ext. 1 Math into year 12 and truth be known, I'm not loving it.
I finished accelerated 2u with a mark a little lower than I was hoping for, but still a band 6.
A lot of people are saying that it's better to stick with Ext because it will be good for the ATAR in the long run, however I'm wondering if I would be better off dropping it, seeing that I'm not really enjoying it, and using my time to focus on getting even better marks in my other subjects??
I was curious to see what you guys, who probably enjoy Ext 1. way more than me (and especially those who might have already finished), think??
Thanks in advance!!

charlottemchenry

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Re: 3U Maths Question Thread
« Reply #3877 on: January 29, 2019, 10:34:38 am »
+1
I was curious to see what you guys, who probably enjoy Ext 1. way more than me (and especially those who might have already finished), think??

I dropped extension maths at the start of year 12 as I wasn't enjoying it, my marks had dropped and I wanted to spend more time on 2u. As you're not doing 2u then this wouldn't be a factor however by dropping I got 3 extra studies. Whilst this doesn't seem like a lot you also get extra time at home as 3u was a lot of work including a lot of homework.

If you're not enjoying it and it's not a prerequisite for your university course I'd probably drop it, however this decision also depends on your number of units. Talk to your teacher as soon as you get back as this was super helpful towards my own decision.

Best of luck
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meerae

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Re: 3U Maths Question Thread
« Reply #3878 on: January 29, 2019, 10:56:14 am »
+1
Hi Guys,
I am currently taking Ext. 1 Math into year 12 and truth be known, I'm not loving it.
I finished accelerated 2u with a mark a little lower than I was hoping for, but still a band 6.
A lot of people are saying that it's better to stick with Ext because it will be good for the ATAR in the long run, however I'm wondering if I would be better off dropping it, seeing that I'm not really enjoying it, and using my time to focus on getting even better marks in my other subjects??
I was curious to see what you guys, who probably enjoy Ext 1. way more than me (and especially those who might have already finished), think??
Thanks in advance!!

Hi Megan!
I also completed 2u last year in the same situation, however, I am keeping 3u. What I would ask is, if you are not enjoying it, and you drop it, are you going to use the time to focus on other subjects. As you would know, maths is incredibly time consuming and when you complete 2u, unless you're enjoying maths/need it for your course, i would probably drop it. As charlottemchenry said, definitely talk to your maths teacher/any teacher you trust for advice on what to do!

Best of Luck!
meerae :)
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InnererSchweinehund

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Re: 3U Maths Question Thread
« Reply #3879 on: January 29, 2019, 11:57:22 am »
+1
I dropped extension maths at the start of year 12 as I wasn't enjoying it, my marks had dropped and I wanted to spend more time on 2u. As you're not doing 2u then this wouldn't be a factor however by dropping I got 3 extra studies. Whilst this doesn't seem like a lot you also get extra time at home as 3u was a lot of work including a lot of homework.

If you're not enjoying it and it's not a prerequisite for your university course I'd probably drop it, however this decision also depends on your number of units. Talk to your teacher as soon as you get back as this was super helpful towards my own decision.

Best of luck
Hi Megan!
I also completed 2u last year in the same situation, however, I am keeping 3u. What I would ask is, if you are not enjoying it, and you drop it, are you going to use the time to focus on other subjects. As you would know, maths is incredibly time consuming and when you complete 2u, unless you're enjoying maths/need it for your course, i would probably drop it. As charlottemchenry said, definitely talk to your maths teacher/any teacher you trust for advice on what to do!

Best of Luck!
meerae :)

Thank you for the advice!!

jfole05

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Re: 3U Maths Question Thread
« Reply #3880 on: January 29, 2019, 06:46:23 pm »
0
Hi Everyone,
I am new here and i was just wondering if anyone had advice or any notes on how to
summarise the circle geometry laws. As i am finding it very hard to memorise
them, i have tried looking in the notes section but saddly couldn't find anything.
Any help would be appreciated even attachments that i can print out. Thank you
in advance to anyone who can help. :)

From Joshua.

meerae

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Re: 3U Maths Question Thread
« Reply #3881 on: January 29, 2019, 07:15:58 pm »
+1
Hi Everyone,
I am new here and i was just wondering if anyone had advice or any notes on how to
summarise the circle geometry laws. As i am finding it very hard to memorise
them, i have tried looking in the notes section but saddly couldn't find anything.
Any help would be appreciated even attachments that i can print out. Thank you
in advance to anyone who can help. :)

From Joshua.

Hi Joshua!
Welcome to the forums!
I find the best way to master circle geo is to do a bunch of questions. I would do questions with the theorems in front of me, just as a list, and write out my proofs fully (with no shortcuts) so I would memorise them by writing them out.

Hope this helped!
meerae :)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3882 on: January 30, 2019, 12:26:52 am »
+4
Hi Everyone,
I am new here and i was just wondering if anyone had advice or any notes on how to
summarise the circle geometry laws. As i am finding it very hard to memorise
them, i have tried looking in the notes section but saddly couldn't find anything.
Any help would be appreciated even attachments that i can print out. Thank you
in advance to anyone who can help. :)

From Joshua.
When I first learnt circle geometry I had to take a while to drill in all those theorems.

Over time however, I managed to memorise what the circle geometry theorems "looked like" and why the words "made sense". Take two examples:
- Angles in the same segment (AKA angles standing on the same arc) are equal. They're literally angles on the same arc because the arc is essentially subtending both of the angles. But even though the theorem says they look the same, I always looked at the interesting inverted shape pattern it has. I eventually just associated it with that theorem.
- Alternate segment theorem. Sometimes called the surfboard theorem because of how it looks, but honestly I literally just memorised a triangle and a tangent. That was it.

Then I moved onto the key features. I thought about which theorems involved the centre of the circle/diameters, or which theorem involved tangents. And of course which theorems involved nothing special (i.e. only chords and secants).

Over time I had the words basically just associated to what it "looked" like. Last piece of the puzzle was remembering where everything actually went. With stuff like alternate segment theorem and the weirder cases of angle at centre = 2x angle at circumference this takes a bit more effort.

Still though, sitting down and focusing on them is not gonna do anything - practice problems are not avoidable. But because I knew what it all "looked" like. when I looked at a complicated diagram I just chose to not focus on the diagram in its entirety. Instead, I just looked at what I wanted to prove (or find), and "saw" what the theorems looked like to navigate my way across.

And then there's also other little things to watch out for e.g. angle chasing

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Re: 3U Maths Question Thread
« Reply #3883 on: February 01, 2019, 04:32:37 pm »
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Hi all,
This is just out of my own curiousity, but
if the second derivative, y" = 0, is true for a point (x,y) on a function f(x), and y' ≠ 0, i.e. it's not a stationary point, then do we know for sure that it's an inflextion point without testing for any change in concavity on either side?
In other words, is it possible to not have an inflexion point when y" = 0, given that the point is not a stationary point? If so, then could you provide me with an example of such case?
Thanks!
« Last Edit: February 01, 2019, 04:36:15 pm by Jefferson »

RuiAce

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Re: 3U Maths Question Thread
« Reply #3884 on: February 01, 2019, 04:35:02 pm »
+1
Hi all,
This is just out of my own curiousity, but
if the second derivative, y" = 0, is true for a point (x,y) on a function f(x), and y' ≠ 0, i.e. it's not a stationary point, then do we know for sure that it's an inflextion point without testing for any change in gradient/concavity on either side?
In other words, is it possible to not have an inflexion point when y" = 0 when it is not a stationary point? if so, then can you provide me with an example of such case?
Thanks!
Yeah.

The special case of \(y^\prime = 0\) and \( y^{\prime\prime} = 0\) is called a horizontal point of inflexion. This is because the tangent to the curve at that point also has gradient 0, on top of the concavity change. This is what we observe for curves such as \(y=x^3\).

If we want just a (not special) point of inflexion, we just want to eliminate that stationary nature. An example is \(y = \sin x\), at \(x=0\).

Another example is \(y = x(x-1)(x+1)\), at \(x=0\).
« Last Edit: February 01, 2019, 04:36:45 pm by RuiAce »