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March 28, 2024, 10:23:01 pm

Author Topic: 3U Maths Question Thread  (Read 1230219 times)  Share 

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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #2790 on: September 20, 2017, 11:40:53 am »
+3
Thank you rui, so you know for the first question, its not like a normal 3u hsc question yeah?



It becomes 3U as soon as you get told that you should use the substitution \(x=2\sin{\theta}\), only 4U students are expected to come up with their own substitutions :)

So you'll be dealing with:



See if you can play with some trig identities from there ;D
« Last Edit: September 20, 2017, 11:44:06 am by jamonwindeyer »

pikachu975

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Re: 3U Maths Question Thread
« Reply #2791 on: September 20, 2017, 12:20:15 pm »
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As trigonometry is involved, you should be thinking of trigonometric methods first and not rush for quadratic methods.



Your first question is a 4U question, not 3U. You would have been given the substitution \( x = 2\sin \theta\) if that was the intended approach.

You can add 4 and minus 4 to the numerator rather than a substitution to make it 3u

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RuiAce

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Re: 3U Maths Question Thread
« Reply #2792 on: September 20, 2017, 12:29:35 pm »
+5
You can add 4 and minus 4 to the numerator rather than a substitution to make it 3u



It could be worked around if the upper boundary were 2. But the upper boundary would be 1 here.
« Last Edit: September 20, 2017, 12:35:26 pm by RuiAce »

davidss

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Re: 3U Maths Question Thread
« Reply #2793 on: September 23, 2017, 11:09:39 pm »
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Hi!!! can someone help with this 1 pls  :)


RuiAce

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Re: 3U Maths Question Thread
« Reply #2794 on: September 24, 2017, 12:54:37 am »
+7
Hi!!! can someone help with this 1 pls  :)

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winstondarmawan

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Re: 3U Maths Question Thread
« Reply #2795 on: September 24, 2017, 06:09:18 pm »
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Hello!
Not the typical exam question that gets posted here, but I've noticed that I have not been able to finish the old 3U papers (/84) on time - usually get up to the last few parts of the last question.
Is this a large problem? Considering I make a few silly mistakes throughout the paper - it seems like it will add up.
Basically, I just want to know what I should be aiming for for the older papers and if my worrying is warranted.
Thanks in advance.

RuiAce

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Re: 3U Maths Question Thread
« Reply #2796 on: September 24, 2017, 06:28:05 pm »
+4
Hello!
Not the typical exam question that gets posted here, but I've noticed that I have not been able to finish the old 3U papers (/84) on time - usually get up to the last few parts of the last question.
Is this a large problem? Considering I make a few silly mistakes throughout the paper - it seems like it will add up.
Basically, I just want to know what I should be aiming for for the older papers and if my worrying is warranted.
Thanks in advance.
"Worrying" is a completely relative word, because every person has different aims for their final mark. Which can be for many reasons; the paper is simply too much, aiming low just to have a safety net etc.

To get an idea of what you need to achieve your aim, you may consider using the raw marks database as a guideline. Me saying something is a large problem in itself is more or less a gateway to me being rude, because I cannot simply assume what you're aiming of regardless of the difficulty of the questions. It only gives me an indication as to where your abilities are at.

If you want advice with handling silly mistakes, try the following checklist:
1. Look at Jamon's guide on silly mistakes in maths to avoid
2. Write down those relevant to you
3. Dig up the papers you've done, and write down the silly mistakes you make in those as well (if you haven't already)
4. The next time you do a paper, whether or not you do it with a textbook beside you is up to you, but keep that silly mistake checklist nearby. When you're done, before you read the answers, check your answers based off the checklist. It's a grind, but it works.

winstondarmawan

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Re: 3U Maths Question Thread
« Reply #2797 on: September 24, 2017, 09:03:45 pm »
0
« Last Edit: September 24, 2017, 09:12:46 pm by winstondarmawan »

Dragomistress

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Re: 3U Maths Question Thread
« Reply #2798 on: September 25, 2017, 08:40:16 am »
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How does one solve this?

The point M divides the interval AB internally in the ratio 3:1. The point N divides the interval AB externally in the ratio 3:1. What is the value of NB:MB.

RuiAce

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Re: 3U Maths Question Thread
« Reply #2799 on: September 25, 2017, 10:30:33 am »
+4
How does one solve this?

The point M divides the interval AB internally in the ratio 3:1. The point N divides the interval AB externally in the ratio 3:1. What is the value of NB:MB.





RuiAce

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Re: 3U Maths Question Thread
« Reply #2800 on: September 25, 2017, 11:00:14 am »
+4
Would appreciate help with the following (both links are for the same Q) This was from 2009HSC last Q but I don't understand the provided solution.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21984023_1334137693378417_1111165516_n.png?oh=065a246eb62d7e0b274077d3b0122249&oe=59C9A7BE
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21984439_1334137783378408_2124958823_n.png?oh=8e2351abe5ad2855f9e7f499e35c6b88&oe=59C9BFCF
Parts (c). Thanks in advance!


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The BOSTES answers probably aren't the best, but ultimately this question then relies on intuition. I was hoping for something better but I couldn't find any.

Firstly, consider the case where P is to the left of T, as shown in the diagram. We will gradually move the point P towards the point T. Once the points P and T coincide, we no longer have \(\theta < \phi\), but rather \(\theta = \phi).

This happens because we must keep in mind that Q, P and S are collinear. As we gradually move P towards T, we are moving S towards T as well.This is why when P and T coincide, S coincides with it as well, which then imply that the angles \( \theta \) and \( \phi \) coincide.

GeoGebra simulation attached. You will find that as P gets dragged away from T, \( \theta\) gets smaller. And when P gets dragged towards T, \( \theta\) gets larger.
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Now that's one half of the story. The other half is when P is to the right of T instead.

You can repeat the same thing we did earlier. In a similar way, if P is to the right of T, we still have \( \theta < \phi \). And as we move P to the left, gradually coinciding with T again, eventually we will have \(\theta = \phi\) once they coincide.
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From there on, we need to recollect what we know.



Remark: Perhaps the most curious thing about this is that \( \phi \) is a fixed angle, which is what allows this argument to work. The reason for this is because of the circle geometry theorem "angles in the same segment are equal", aka. "angles standing on the same arc are equal". Here, the arc is just QR, and we're moving the angle along the major arc.

(Note, however, that phi doesn't necessarily have to be 28 degrees. That exact value was just a consequence of how I simulated it.)
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(May need to stare at the diagram really closely to see this.)
« Last Edit: September 26, 2017, 12:49:13 pm by RuiAce »

K9810

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Re: 3U Maths Question Thread
« Reply #2801 on: September 25, 2017, 05:09:55 pm »
0
Hey guys,

How do I do these questions?

RuiAce

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Re: 3U Maths Question Thread
« Reply #2802 on: September 25, 2017, 07:39:07 pm »
+4
Hey guys,

How do I do these questions?



The second question is cut off.

K9810

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Re: 3U Maths Question Thread
« Reply #2803 on: September 25, 2017, 09:54:30 pm »
0



The second question is cut off.

Thank you!
The second question was: Prove that PXQT is a cyclic quadrilateral

RuiAce

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Re: 3U Maths Question Thread
« Reply #2804 on: September 25, 2017, 10:03:59 pm »
+3