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Author Topic: 3U Maths Question Thread  (Read 1230284 times)  Share 

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RuiAce

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Re: 3U Maths Question Thread
« Reply #675 on: September 23, 2016, 02:55:58 pm »
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Hi! This is a silly question but I never understood how to do these types of polynomials :/

P(x)=(x+1)(x-3)Q(x)+(-2x+6)
Where Q(x) is a polynomial.
Find the remainder when P(x) is divided by (x+1)(x-3)

Sorry if this is dumb D: Thank you!

Neutron
The answer is just (-2x+6)

Because this is just the division transformation P(x) = A(x)Q(x) + R(x)
Where: P(x) is your original polynomial
A(x) is your divisor (x+1)(x-3)
Q(x) is your quotient which doesn't matter
R(x) is your remainder.


If instead you wanted the remainder upon division by x-3 then you would need to use the remainder theorem and find P(3)
« Last Edit: September 23, 2016, 02:58:29 pm by RuiAce »

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #676 on: September 23, 2016, 03:04:21 pm »
+1
Hi! This is a silly question but I never understood how to do these types of polynomials :/

P(x)=(x+1)(x-3)Q(x)+(-2x+6)
Where Q(x) is a polynomial.
Find the remainder when P(x) is divided by (x+1)(x-3)

Sorry if this is dumb D: Thank you!

Neutron

Don't worry, I still hate these, always makes me do a double take, not sure why ;)

Just to add a bit more to Rui's explanation. Go back to basics (and make it a little more colloquial than normal). Remember that when we divide something, we are really determining how many times the divisor goes (fully) into the dividend, that number being the quotient. The number leftover is the remainder. So really, what we are doing is separating the dividend into [quotient] groups of the divisor, plus the remainder. That's the division transformation:



As Rui said, in your case the question is easy. Our divisor is \((x+1)(x-3)\), so our remainder is actually in the question itself!!


lha

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Re: 3U Maths Question Thread
« Reply #677 on: September 25, 2016, 03:53:26 pm »
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And finally, 5 minutes is enough time to say 'I think I can' about 150 times. At the end of the day, no matter what you do in the 5 minutes reading time, the important thing is to be in the right mindset. If finding tricky questions, and feeling safe in knowing you can figure out how to do them, is the way to make you feel comfortable, go for that. If reading the formula sheet, or listing pneumonic devices, is your thing; do that. Reading time isn't a huge factor in exam success (although, if you use it in a way best suited to you, it can be very effective)

This helps a lot, thank you!

nimasha.w

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Re: 3U Maths Question Thread
« Reply #678 on: September 25, 2016, 04:31:42 pm »
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hi! can someone please explain this question for me :)

RuiAce

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Re: 3U Maths Question Thread
« Reply #679 on: September 25, 2016, 04:39:02 pm »
+1
hi! can someone please explain this question for me :)




massive

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Re: 3U Maths Question Thread
« Reply #680 on: September 26, 2016, 12:20:57 am »
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Guys how on earth would you do the q attached in an exam?? The answer is C but I don't get how you can get without using something like wolfram alpha. Any tips would be greeeatly appreciated.

massive

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Re: 3U Maths Question Thread
« Reply #681 on: September 26, 2016, 07:58:14 am »
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another one, how do you do part ii guysss??

RuiAce

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Re: 3U Maths Question Thread
« Reply #682 on: September 26, 2016, 08:39:34 am »
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Guys how on earth would you do the q attached in an exam?? The answer is C but I don't get how you can get without using something like wolfram alpha. Any tips would be greeeatly appreciated.
That question is kinda mean.

Draw a graph, but draw it to SCALE.


The only solution you can guarantee is x=0. The rest you cannot solve via algebra; you can only do so by graph.

RuiAce

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Re: 3U Maths Question Thread
« Reply #683 on: September 26, 2016, 08:41:28 am »
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another one, how do you do part ii guysss??


jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #684 on: September 26, 2016, 03:55:19 pm »
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Guys how on earth would you do the q attached in an exam?? The answer is C but I don't get how you can get without using something like wolfram alpha. Any tips would be greeeatly appreciated.

To clarify above, we are drawing two separate graphs:



The points of intersection of this graph will be solutions to the equation (can you see why?) ;D

To emphasise something even further from Rui's explanation, definitely draw to scale. Like, crazy to scale. You'll get this graph:



It is very easy to miss those two solutions on the right there. There are 3 solutions in the domain ;D

lha

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Re: 3U Maths Question Thread
« Reply #685 on: September 26, 2016, 07:39:46 pm »
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Hi can anyone help me with question 12b)i) and ii) in the 2015 hsc 3u exam? Its not letting me attach a pic of the question so if someone could look it up that would be great! I have no idea how to do these types of questions so any help is extremely appreciated!

RuiAce

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Re: 3U Maths Question Thread
« Reply #686 on: September 26, 2016, 08:05:52 pm »
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Hi can anyone help me with question 12b)i) and ii) in the 2015 hsc 3u exam? Its not letting me attach a pic of the question so if someone could look it up that would be great! I have no idea how to do these types of questions so any help is extremely appreciated!
These are one mark circle geometry questions. Tips:

(i) <ACD is right next to it and labelled 30 degrees. That's something I immediately keep in mind. And then just read the question. They explicitly tell you BD is a diameter, so what does that say about <BCD?

(ii) I see the angle formed by a chord (AD) and a tangent (XD). I immediately think about alternate segment theorem. There are two triangles to use alternate segment theorem on, so just look at both of them and find a useful once.

Remember - You're only looking for like one thing for a one marker. Don't overanalyse in these scenarios.
(Exclusion: That 1 marker is the last question of the paper. Maybe think a bit there)
« Last Edit: September 26, 2016, 08:13:08 pm by RuiAce »

lha

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Re: 3U Maths Question Thread
« Reply #687 on: September 26, 2016, 08:46:56 pm »
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These are one mark circle geometry questions. Tips:

(i) <ACD is right next to it and labelled 30 degrees. That's something I immediately keep in mind. And then just read the question. They explicitly tell you BD is a diameter, so what does that say about <BCD?

(ii) I see the angle formed by a chord (AD) and a tangent (XD). I immediately think about alternate segment theorem. There are two triangles to use alternate segment theorem on, so just look at both of them and find a useful once.

Remember - You're only looking for like one thing for a one marker. Don't overanalyse in these scenarios.
(Exclusion: That 1 marker is the last question of the paper. Maybe think a bit there)

Thank you that helps but i think you got the questions mixed up, I need help on 12b) not 12a). Sorry!

RuiAce

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Re: 3U Maths Question Thread
« Reply #688 on: September 26, 2016, 09:24:27 pm »
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Damn it lol thought something was weird!

For the correct question

(i) This is one of the more fundamental proofs that are included in the textbook. The derivation isn't hard - a chord being "focal" means it runs through the focus. For that parabola, the focus is (0,a). So all you do is just sub that in.

(ii) This now requires you to call in a bit of knowledge. Look closely at the coordinates of P. The coordinates of P are (2ap, ap^2) and (8a, 16a) simultaneously! That means all you have to do is equate.

Regardless of whether you equate x or y coordinates, you're going to get p=4.

So if PQ is a focal chord, you can safely use part (i) and sub back in.

lha

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Re: 3U Maths Question Thread
« Reply #689 on: September 26, 2016, 09:27:39 pm »
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I just realised how to edit my posts hahaha

Well I have another question. How do you do q14a)ii) from the 2015 hsc paper?
« Last Edit: September 26, 2016, 10:05:29 pm by lha »