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March 29, 2024, 03:18:28 am

Author Topic: 3U Maths Question Thread  (Read 1230307 times)  Share 

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katherine123

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Re: 3U Maths Question Thread
« Reply #300 on: July 14, 2016, 02:30:54 pm »
0
for part iv)  is it okay to say:
since <TAQ=<QCB (from part iii)
therefore converse of equal chord RQ subtend equal angle
therefore CAQR is a cyclic quad

given <AQT=90
therefore <AQT= <CRA=90 (equal chord subtend equal angle in cyclic quad CAQR)
therefore AR is perpendicular to CB

RuiAce

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Re: 3U Maths Question Thread
« Reply #301 on: July 14, 2016, 02:50:52 pm »
+1
for part iv)  is it okay to say:
since <TAQ=<QCB (from part iii)
therefore converse of equal chord RQ subtend equal angle
therefore CAQR is a cyclic quad

given <AQT=90
therefore <AQT= <CRA=90 (equal chord subtend equal angle in cyclic quad CAQR)
therefore AR is perpendicular to CB
Not written with as much sophistication as my picky self would prefer e.g. <AQT = <AQC = <CRA however enough has been covered to say it is most certainly correct.
« Last Edit: July 14, 2016, 03:06:53 pm by RuiAce »

jakesilove

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Re: 3U Maths Question Thread
« Reply #302 on: July 14, 2016, 02:51:31 pm »
+1
for part iv)  is it okay to say:
since <TAQ=<QCB (from part iii)
therefore converse of equal chord RQ subtend equal angle
therefore CAQR is a cyclic quad

given <AQT=90
therefore <AQT= <CRA=90 (equal chord subtend equal angle in cyclic quad CAQR)
therefore AR is perpendicular to CB

Definitely comprehensive enough for full marks! Nice one
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zoe_rammie

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Re: 3U Maths Question Thread
« Reply #303 on: July 15, 2016, 11:25:50 am »
0
Hi Jake (and whoever else reads this),

Just to get straight to the point,

I don't quite understand the answer to this binomial question (Taken from CSSA Trial 2012)
(Image attached, because I don't know how to paste it into this textbox LMAO much technology yay)

And alsoooo,
Isn't it so much easier to just belt out factorial notation, and slay LHS=RHS


jakesilove

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Re: 3U Maths Question Thread
« Reply #304 on: July 15, 2016, 12:52:20 pm »
+2
Hi Jake (and whoever else reads this),

Just to get straight to the point,

I don't quite understand the answer to this binomial question (Taken from CSSA Trial 2012)
(Image attached, because I don't know how to paste it into this textbox LMAO much technology yay)

And alsoooo,
Isn't it so much easier to just belt out factorial notation, and slay LHS=RHS

Hey Zoe,

First of all, your description of a potential method of answering this question is absolutely gold. Keep at it, love your work.

Your method, and the method in the answers, is actually the same. If you do what you were planning on doing, that's great, but you would have to equate coefficients as there isn't another way to compare the RHS to the LHS and prove that they are equal. So, you would equate coefficients, and get exactly the same answer as they did. They're answer probably isn't comprehensive enough anyway; if your method gets you the right relationship, you're absolutely doing the right thing.

If you wanted a more comprehensive proof, though, let us know, however as far as I can tell your "do shit get marks" methodology is spot on.

Jake
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RuiAce

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Re: 3U Maths Question Thread
« Reply #305 on: July 15, 2016, 04:09:36 pm »
+2
Hi Jake (and whoever else reads this),

Just to get straight to the point,

I don't quite understand the answer to this binomial question (Taken from CSSA Trial 2012)
(Image attached, because I don't know how to paste it into this textbox LMAO much technology yay)

And alsoooo,
Isn't it so much easier to just belt out factorial notation, and slay LHS=RHS
Firstly, exactly what Jake said.

Inspection of coefficients hastes the process (possibly without you realising it because it appears as more writing) because you just look at it and pull things out.

Your proof using the definition of the binomial coefficient is in no way whatsoever flawed. The method of equating coefficients is just manipulation of an identity, whereas the definition just uses subtraction.

Of course, such a question would most likely not be asked in the HSC because it will be too offputting. EVERYONE will think the definition is faster to use.
« Last Edit: July 15, 2016, 04:23:01 pm by RuiAce »

ATWalk

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Re: 3U Maths Question Thread
« Reply #306 on: July 15, 2016, 05:26:04 pm »
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Hi,

Could someone please attempt this question and tell me which answer they get? I got:



I was just wondering whether or not I got hopelessly lost somewhere... It's a super hard question.  :-\

Thanks in advance!

P.S. Sorry if the way I wrote my solution is confusing. Not that good with the LaTex stuff.
« Last Edit: July 15, 2016, 05:27:46 pm by ATWalk »

Happy Physics Land

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Re: 3U Maths Question Thread
« Reply #307 on: July 15, 2016, 06:31:09 pm »
+3
Hi,

Could someone please attempt this question and tell me which answer they get? I got:



I was just wondering whether or not I got hopelessly lost somewhere... It's a super hard question.  :-\

Thanks in advance!

P.S. Sorry if the way I wrote my solution is confusing. Not that good with the LaTex stuff.

Hey ATWalk!

I did the question and I think our answers are out by a factor of 1/a. It was, after all, a very tedious question. I really could have been wrong. I will stick my solution here so that you can see any mistakes you have made. If you dont see anything wrong with yours, chances are I could have made a mistake.



Best Regards
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conic curve

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Re: 3U Maths Question Thread
« Reply #308 on: July 15, 2016, 06:52:16 pm »
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Hey guys

Just curious but how would you tackle 3D triginometry?

RuiAce

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Re: 3U Maths Question Thread
« Reply #309 on: July 15, 2016, 07:06:31 pm »
+2
Hey guys

Just curious but how would you tackle 3D triginometry?
Jamon gave some tips which will be on his lecture slides when they're released.

One tip for 3D trig is to draw your 3D diagram, then draw your triangles separately. This reduces the ease of getting lost in your diagram

RuiAce

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3U Maths Question Thread
« Reply #310 on: July 15, 2016, 07:10:21 pm »
+3

I get what HPL got.

The answers are most likely wrong.

Edit: in fact so does WolframAlpha
« Last Edit: July 15, 2016, 07:12:28 pm by RuiAce »

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Re: 3U Maths Question Thread
« Reply #311 on: July 15, 2016, 07:32:47 pm »
+2
Hey guys

Just curious but how would you tackle 3D triginometry?

As well as Rui's tips, I mainly spoke about working backwards. I need this tangent ratio in my answer, okay, I need this side then, oh, that means I need to use this triangle. Etc, etc.

Extension HSC 3D Trig questions will commonly lead you through the process of getting the answer, the trick is just spotting the right triangles to use that give you the correct answer. Always keep your eye on the answer you need to guide your working.

I also leave 3D Trig questions till last, just to make sure they don't waste time I need for other questions  ;D

amandali

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Re: 3U Maths Question Thread
« Reply #312 on: July 16, 2016, 10:07:53 am »
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how to do part iv )

RuiAce

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Re: 3U Maths Question Thread
« Reply #313 on: July 16, 2016, 01:05:08 pm »
+1
(Image removed from quote.)
how to do part iv )
Do you have a solution to this question, or the source of it? I requested a solution of this question and it used concepts WELL beyond that of the HSC.

jakesilove

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Re: 3U Maths Question Thread
« Reply #314 on: July 16, 2016, 01:15:38 pm »
+2
Do you have a solution to this question, or the source of it? I requested a solution of this question and it used concepts WELL beyond that of the HSC.

Totally agree: I can prove that it equals e, but only using university level maths. I'm sure you can prove the relationship somehow using iii) but honestly it doesn't fall out as easily as I would have expected.
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