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Oxford MAT Test - 5th Nov 2014

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Can anyone help with this question?

I don't know how to prove the square of an arithmetic progression... I can only do it in a roundabout sort of way which takes a while.

Original post by donutellme
Can anyone help with this question?

I don't know how to prove the square of an arithmetic progression... I can only do it in a roundabout sort of way which takes a while.



Is there a way for you to rewrite the series in terms of an arthrithmetic progress (Hint, consider the difference of two squares)
Original post by Slowbro93
Is there a way for you to rewrite the series in terms of an arthrithmetic progress (Hint, consider the difference of two squares)


So I've split it into n(n+1)/2 and two arithmetic progressions (+(1+3+5...) and -(2+4+6...)).

But I end up with -n^2(n+1)/2...
Reply 183
I'm stuck on the last multiple choice question on the 2013 paper..
I'm not sure if I understand it, does it have a 'stair' shaped graph? I just did the geometric sum and the answers all look alot more complicated -.-'
Original post by xigi
I'm stuck on the last multiple choice question on the 2013 paper..
I'm not sure if I understand it, does it have a 'stair' shaped graph? I just did the geometric sum and the answers all look alot more complicated -.-'


Yes, the graph is stair-shaped.

Think about what type of function will x have to be for 2^x to be an integer. Then sketch a graph and think about how you would find the total area of that graph.
Original post by donutellme
So I've split it into n(n+1)/2 and two arithmetic progressions (+(1+3+5...) and -(2+4+6...)).

But I end up with -n^2(n+1)/2...


Well for example,

you see that 1222=3,3242=7,5262=111^2-2^2=-3, 3^2-4^2=-7, 5^2-6^2=-11 etc.

You then see that you have a sequence of -3, -7, -11 etc. where you will have n terms.

Can you see what you can do from here and the other terms? :smile:
Original post by Slowbro93
Well for example,

you see that 1222=3,3242=7,5262=111^2-2^2=-3, 3^2-4^2=-7, 5^2-6^2=-11 etc.

You then see that you have a sequence of -3, -7, -11 etc. where you will have n terms.

Can you see what you can do from here and the other terms? :smile:


Yeah that's pretty clear now!

I could see that they were coming in pairs, but I went for factoring out the two series (which is completely wrong). I learnt something new today! I can't believe I didn't spot this...:P
Reply 187
Would PAT papers help MAT?
Original post by Bob Ni
Would PAT papers help MAT?


Probably not much. Maybe they'd help with question 1. And one question or so from the PAT can't hurt, choose one of the harder 7+ markers to do. I wouldn't recommend doing a whole paper, though.
Please can someone explain why the area isn't maximised when Q and R are very close to P so that nearly all the circle is shaded?

Original post by EconFan_73
Please can someone explain why the area isn't maximised when Q and R are very close to P so that nearly all the circle is shaded?



Theta is fixed.
Original post by EconFan_73
Please can someone explain why the area isn't maximised when Q and R are very close to P so that nearly all the circle is shaded?



THe "largest" bit is about varying the position of P while keeping theta fixed.
For 1j) this was my working. Not sure whether there is a faster method?
I think i am right. (sort had to guess that last identity lol)
Original post by iPixelBlue
For 1j) this was my working. Not sure whether there is a faster method?"Dirty" method:

When theta = pi/2, QR is a diameter, and the area of triangle QPR is clearly maximized when the triangle is isoceles, at which point we see it's area is 1 (right angled triangle with base = height = sqrt(2)). The remaining area (the curved bit) is the area of a semicircle (radius 1), so equals pi/2.

So any correct answer must = 1 + pi/2 when theta = pi/2. The only option that matches is B.
Original post by DFranklin
"Dirty" method:

When theta = pi/2, QR is a diameter, and the area of triangle QPR is clearly maximized when the triangle is isoceles, at which point we see it's area is 1 (right angled triangle with base = height = sqrt(2)). The remaining area (the curved bit) is the area of a semicircle (radius 1), so equals pi/2.

So any correct answer must = 1 + pi/2 when theta = pi/2. The only option that matches is B.


why would theta being 0.5pi imply QR is the diameter?
Original post by iPixelBlue
why would theta being 0.5pi imply QR is the diameter?
GCSE geometry:

http://www.bbc.co.uk/schools/gcsebitesize/maths/geometry/circles2hirev5.shtml



LOOOOL okay now i understand what you're saying. I thought you were referring to the angle between the diameter and x. Anyway I got the right answer so hehe
Does anyone know are we allowed to use extra paper for the MAT if we use all the space? The instructions can kinda suggest that we can't but i'm not sure - "Answers to questions 2-7 should be written in the space provided, continuing on to the blank pages at the end of this booklet if necessary. Each of Questions 2-7 is worth 15 marks." I REALLY hope we can use extra paper or i'll have to be much more careful with how much space I use :/
Original post by zarifina
Does anyone know are we allowed to use extra paper for the MAT if we use all the space? The instructions can kinda suggest that we can't but i'm not sure - "Answers to questions 2-7 should be written in the space provided, continuing on to the blank pages at the end of this booklet if necessary. Each of Questions 2-7 is worth 15 marks." I REALLY hope we can use extra paper or i'll have to be much more careful with how much space I use :/


No extra paper. They will only mark what is in the answer book.
Original post by tiny hobbit
No extra paper. They will only mark what is in the answer book.



Oh.... great. are we allowed to use extra paper for rough working if we think we're going to need it? e.g. for the multi-choice as we're not being marked for working out anyway? :smile:

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