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# MAT Prep Thread - 2nd November 2016

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1. (Original post by lewman99)
It's been going quite well - I'm doing roughly a paper a week, and I'm doing the 2011 paper tomorrow. I haven't been focusing too much on scores so far, moreso on tackling and understanding the questions, but I feel I understand most of it. Let's just say I'm very happy to do the logic question rather than geometry. I feel I've been rushing through the papers a bit, though - I usually have quite a bit of time left, and I'm unsure whether going a bit more slowly would help me or just cause me to doubt myself.

Also, can somebody explain 1H from the 2009 paper? I'm in Scotland, and the trapezium rule isn't taught here at all, not even as Advanced Higher. I know a little bit about it (approximating an integral with trapeziums, if the graph curves up it'll be an overestimate otherwise it's an underestimate), and that's been enough for some of the other trapezium rule questions so far, but I just don't get this one.
Ahh fair, are you getting help from your college? I'm all alone haha! I would recommend originally going through without to much emphasis on time making sure you understand as much as you can and appreciate what's going on- i find if i rush through i generally dont realise what they want you to see! I am exactly the same with preference to logic haha!
As for 1 H, assuming you know the formulae for the trapizium rule, if not you'll need to learn it! Write it out with the subinterval terms and try to form a partial geometric sequence and use the sum formula to evaluate it, you should be able to simplify using index laws and obtain one of the answers.
If you still get stuck let me know and i'll try and explain more explicitly but, would recommend trying to do it yourself to start with, hope i helped
2. (Original post by lewman99)
x
http://i.imgur.com/djWpe0V.png

Here are the notes from our (UK edexcel) textbook on the trapezium rule. Hope it helps

for the MAT question just carefully think about what each y value will be, then do some tidying up and as non-euclidean said you should look to use the geometric series formula
3. so is everyone generally finding the logic 15 markers easier and the geometry 15 markers harder?
4. for 2007 question 4, i don't really understand the solution. i used a different method to answer the A(thetha)=B(*) and A(pi/4) etc whereby i essentially found the area of both A(theta) and B(theta) by finding area of triangle-area of sector for both and checked that B(pi/2-theta)=A(theta). Is that a valid method of doing it? Also can someone explain to me the official solution to that question from the A(theta)=B(pi/2-theta) part.
Thanks!
5. (Original post by danielhx)
for 2007 question 4, i don't really understand the solution. i used a different method to answer the A(thetha)=B(*) and A(pi/4) etc whereby i essentially found the area of both A(theta) and B(theta) by finding area of triangle-area of sector for both and checked that B(pi/2-theta)=A(theta). Is that a valid method of doing it? Also can someone explain to me the official solution to that question from the A(theta)=B(pi/2-theta) part.
Thanks!
It seems they're just looking for an intuitive answer: it's sort of obvious when you look at the picture that that result should follow, and what they've written is basically a long-winded way of saying "symmetry". I doubt you'd be penalised for concretely working it out. But in general if something says "explain" it's license to just use words.
Not sure I can explain better than they have but I'll try. Imagine I fix some arbitrary value for theta. Then consider the corresponding diagram with pi/2 - theta in theta's place, so the areas are A(pi/2 - theta) and B(pi/2 - theta) Now envision this diagram being reflected in the line y = x. The circle stays the same, because we don't care which of x and y is which right? (remember the equation is (x - 1)^2 + (y - 1)^2 = 1 ). But the line changes, swapping theta and pi/2 - theta. This is just our original diagram with theta. You can envision the section with area B(pi/2 - theta) being reflected where it becomes A(theta) in said original diagram.
6. (Original post by 13 1 20 8 42)
It seems they're just looking for an intuitive answer: it's sort of obvious when you look at the picture that that result should follow, and what they've written is basically a long-winded way of saying "symmetry". I doubt you'd be penalised for concretely working it out. But in general if something says "explain" it's license to just use words.
Not sure I can explain better than they have but I'll try. Imagine I fix some arbitrary value for theta. Then consider the corresponding diagram with pi/2 - theta in theta's place, so the areas are A(pi/2 - theta) and B(pi/2 - theta) Now envision this diagram being reflected in the line y = x. The circle stays the same, because we don't care which of x and y is which right? (remember the equation is (x - 1)^2 + (y - 1)^2 = 1 ). But the line changes, swapping theta and pi/2 - theta. This is just our original diagram with theta. You can envision the section with area B(pi/2 - theta) being reflected where it becomes A(theta) in said original diagram.
Thank you very much!
7. im going to my first private lesson for mat tonight, apart from just going over MAT questions, should I be asking anything in particular in order to get the most value out of my money?
8. (Original post by KloppOClock)
should I be asking anything in particular in order to get the most value out of my money?
Yep! "When can I quit?" - there's no substitute for getting better at MAT than just working at it. A private lesson is going to be a waste of money in my opinion, but each to his own...
9. (Original post by Zacken)
Yep! "When can I quit?" - there's no substitute for getting better at MAT than just working at it. A private lesson is going to be a waste of money in my opinion, but each to his own...
m8 im failing atm
10. (Original post by Non-Euclidean)
Ahh fair, are you getting help from your college? I'm all alone haha! I would recommend originally going through without to much emphasis on time making sure you understand as much as you can and appreciate what's going on- i find if i rush through i generally dont realise what they want you to see! I am exactly the same with preference to logic haha!
As for 1 H, assuming you know the formulae for the trapizium rule, if not you'll need to learn it! Write it out with the subinterval terms and try to form a partial geometric sequence and use the sum formula to evaluate it, you should be able to simplify using index laws and obtain one of the answers.
If you still get stuck let me know and i'll try and explain more explicitly but, would recommend trying to do it yourself to start with, hope i helped
I'm not doing it entirely alone - I've got a maths teacher who's willing to help with questions and the like - but I'm the first person to apply to Oxford in my school (definitely for maths, and possibly ever). It makes forums like this pretty damn useful!
I sat the 2011 paper today, without as much emphasis on time - I was still within time, but not ridiculously so. Handwriting's better as well . I haven't had time to mark it properly yet, but having a skim through the marking scheme it seems I've done well on it. Logic questions aren't too difficult, but they can be pretty annoying with the sheer amount of writing you've got to do.
So I think I've got 1H, thanks to your explanation, trying it out and looking over the marking scheme. By the looks of it, I might also be able to try it with a single trapezium (which would be of area 3/2), then sub N=1 into each option. That's perhaps cheesing it, though, and I don't want to rely on it in case it's not multiple choice.

11. i tried geometric series, wot i do
12. (Original post by KloppOClock)

i tried geometric series, wot i do
Not sure what you're asking. Yes it is a geometric series, and that term is its sum.
13. (Original post by RichE)
Not sure what you're asking. Yes it is a geometric series, and that term is its sum.
how is it a geometric series, it doesnt have a common ratio
14. (Original post by KloppOClock)
how is it a geometric series, it doesnt have a common ratio
1 + 4 + 4^2 + 4^3 + ... + 4^(k-1)

common ratio of 4
15. (Original post by Zacken)
1 + 4 + 4^2 + 4^3 + ... + 4^(k-1)

common ratio of 4
but when you apply f1(x) the first three terms are
4x+1
16x+5
64x+21

how does the common ratio of 4 apply there
16. (Original post by KloppOClock)
but when you apply f1(x) the first three terms are
4x+1
16x+5
64x+21

how does the common ratio of 4 apply there
It's

4x + 1

16x + (1 + 4)

64x + (1 + 4 + 4^2)

Can you not see that the bracket terms are a geometric series???????
17. (Original post by Zacken)
It's

4x + 1

16x + (1 + 4)

64x + (1 + 4 + 4^2)

Can you not see that the bracket terms are a geometric series???????
.. yeah... of course... just testing you fam
18. (Original post by KloppOClock)
just testing you fam
What?
19. (Original post by Zacken)
What?
what do you mean what?
20. (Original post by KloppOClock)
what do you mean what?
Are you being serious about "just testing me"?

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