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    (Original post by cbnz)
    Hi, I'm applying for maths and philosophy and I'm devastated about this test, I think I probably got in the 40s (Max 55) I panicked on question four and so only answered the first two parts of 5... I got 9 A*s at GCSE and have a 4 A* prediction, do you think they'll still interview me?
    With a score in the 40s, it's unlikely you will get called for interview, but not impossible. You have to remember that a lot of Oxford applicants have academic backgrounds as impressive as yours, and the MAT is used to distinguish between these students
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    (Original post by dutchmaths)
    What did people get on the first question? I was fairly sure is was a>2 but now I'm not sure anymore.
    Could we set up a document of all multiple choice questions and the answers that everybody agrees on? I'm sure our collective memory is good enough to remember all the questions!
    it was a not equal to 2 for sure . option A
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    (Original post by IceKidd)
    ..
    Seen you around this thread quite a bit. How did you find it? I know you were doing well in the past papers, so I thought I'd ask
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    (Original post by kapur)
    it was a not equal to 2 for sure . option A
    I realize that now after working it out... *stabs self*
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    (Original post by Jooooshy)
    Below average would probably be sub-50-ish. I'm not going to say you definitely won't get an interview with that score (because they can 'rescue' students who fall under the cut-off), but you'll have a lot to prove if you do!
    Thanks, I'm just hoping I did better than I think I did!
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    (Original post by Jooooshy)
    Seen you around this thread quite a bit. How did you find it? I know you were doing well in the past papers, so I thought I'd ask
    haha not as good as past papers. contrary to everyone else i thought the multiple choice was fairly easy and the longer questions hard, though i know i did better in the longer questions than the multiple choice!

    Im fairly sure i have a decent mark( above 50), just annoyed that exam pressure got the best of me because i was getting a bit better in past papers.

    i know thats good enough for an interview, mind

    thanks for asking btw
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    (Original post by ziggyl)
    Thanks, I'm just hoping I did better than I think I did!
    I'm sure you have done better than you think. Reading answers on here will do nothing for your confidence!

    Best of luck with the rest of your application process!
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    if i remember correctly, the question was:
    x^2+ax+a=1, or
    x^2+ax+a-1=0
    if you want distinct solutions then by doing the discriminant you get:
    a^2-4a+4 is equal or greater than 0....
    or simpler:
    (a-2)^2 is equal or greater than 0...
    correct me if i'm wrong but this is always equal or greater than 0 so it is true for every real a...
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    (Original post by Jooooshy)
    I'm sure you have done better than you think. Reading answers on here will do nothing for your confidence!

    Best of luck with the rest of your application process!
    Thankyou, I hope so!
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    (Original post by Papazachariou)
    if i remember correctly, the question was:
    x^2+ax+a=1, or
    x^2+ax+a-1=0
    if you want distinct solutions then by doing the discriminant you get:
    a^2-4a+4 is equal or greater than 0....
    or simpler:
    (a-2)^2 is equal or greater than 0...
    correct me if i'm wrong but this is always equal or greater than 0 so it is true for every real a...
    distinct means the roots can't be equal
    sorry
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    (Original post by IceKidd)
    haha not as good as past papers. contrary to everyone else i thought the multiple choice was fairly easy and the longer questions hard, though i know i did better in the longer questions than the multiple choice!

    Just annoyed that exam pressure got the best of me because i was getting a bit better in past papers.

    i know thats good enough for an interview, but any ideas on how well id have to do in the interview for an offer?

    thanks for asking btw
    In my opinion, longer questions are always harder, since they require more thought (and a lot of people give up on the latter halves of the long questions!).

    Yeah, you're looking at an interview. To be honest, there aren't many ways you can prepare for an interview. You're probably going to be introduced to some problems you've never seen before and be asked to talk through how you think you might go about solving them (with some very helpful hints)!

    MAKE SURE YOU HAVE READ ANY BOOKS IN YOUR PS. A friend of mine mentioned a book he read about infinity and they love that **** here. Every interview he was asked about it.

    Which college did you apply for?
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    (Original post by sun_tzu)
    distinct means the roots can't be equal
    sorry
    do you remember what you got, or the exact problem, because i really can't
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    Anybody got a copy of the paper? Or a rough idea of what the questions were?

    If you do, I'll have a word with the other UGs here in compsci/maths and we'll scrap up a rough mark scheme for y'all.
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    (Original post by Jooooshy)
    Anybody got a copy of the paper? Or a rough idea of what the questions were?

    If you do, I'll have a word with the other UGs here in compsci/maths and we'll scrap up a rough mark scheme for y'all.
    I'll give you a sketch of question 4, which is going to be incomplete and possibly incorrect. My hope is that someone will see this, remember the entire question correctly.

    We define a function fk(x) = x(x-2)(x-k). (This is obviously a cubic which has roots at x=0,k,2. k was drawn as being between 0 and 2 in the accompanying graph, but this wasn't explicitly stated)
    A(k) is either drawn or stated as being the area/integral of fk between 0 and 2.
    (i)Define A(k) in two integrals.

    As far as I could tell this was just the integral from 0 to k plus the integral from k to 2, which is equal to the area from 0 to 2, so I'm not sure why the two integrals? Maybe to put k into the equation?

    (ii)Explain why A(k) is a polynomial in k with a degree no greater than 4.

    The degree seemed pretty obvious; an integral of a cubic can't be of a higher degree than 4. How to prove the fact that it's a polynomial in k I wasn't sure. I definitely felt like x wasn't a variable in the same way that k was, but I was uncertain as to how to put this into words.

    (iii) Show that A(k)=A(2-k)

    Again, this seemed intuitive, but hard to put into words. I don't quite remember if/whether I showed this rigorously.

    (iv) Show that A(k) can be written as: a(k-1)4 + b(k-1)2 + c with real a,b,c

    I deduced from A(k)=A(2-k) that A(k) was symmetric in the line k=1, and since it was a polynomial in k it was a polynomial in k-1 as well. However, since it was symmetric in k=1, the polynomial in k-1 had to be symmetric as well. And a polynomial graph is only symmetric when it only has components of an even degree. Earlier we showed that the maximal degree of A(k) was 4, so A(k) can be written as a polynomial in k-1 with only components of degrees 0,2,4, i.e. as A(k)= a(k-1)4 + b(k-1)2 + c

    There was another part of this question I don't quite remember, I'm hoping someone will fill me in on this.
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    (Original post by Papazachariou)
    do you remember what you got, or the exact problem, because i really can't
    you got the problem exactly right, it's just rather than
    a^2-4a+4 is equal or greater than 0....
    it should be: a^2-4a+4 is greater than 0....

    therefore it's true for all values of a except a=2
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    I can't remember the entire question, but for the multiple choice question about loga(b) something etc., was the answer 'for a unique value of a' ?
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    (Original post by dutchmaths)
    (i)Define A(k) in two integrals.

    As far as I could tell this was just the integral from 0 to k plus the integral from k to 2, which is equal to the area from 0 to 2, so I'm not sure why the two integrals? Maybe to put k into the equation?
    No, that's wrong. The area is the integral from 0 to k MINUS the area from k to 2. If you do a sketch, you will see that the area between k and 2 is below the x axis, so integrating would give you a negative value.
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    (Original post by Jooooshy)
    Anybody got a copy of the paper? Or a rough idea of what the questions were?

    If you do, I'll have a word with the other UGs here in compsci/maths and we'll scrap up a rough mark scheme for y'all.
    While I'm at it, here's a sketch of question 2. Again, I'm hoping people will reply to this comment and expand on it/correct me.

    Q2:
    (i) For k=/=±1, we have a function f(x) that satisfies:
    f(x)-k*f(1-x)=x
    Find f(x) by replacing x with 1-x

    I was absolutely clueless on this one. The answer I ended up giving was k=0 and f(x)=x, which technically is correct, but probably not what they were looking for.

    (ii) We redefine the function f(x) as follows:
    f(x)-f(1-x)=g(x)
    Show that there is no solution for f(x) if g(x)=x

    I solved this by replacing x with 1-x, and proving some paradoxical property of x. I think it was x=-(1-x)=x-1. I don't quite remember, though, and my answer might not have been correct.

    (iii) What condition does g(x) have to satisfy for there to be a valid f(x)?

    The answer I ended up giving was: g(x) has to be a polynomial with components of an even degree. I figured this was the only way this weird type of symmetry would crop up. Again, I probably had this wrong.

    (iv) What is f(x) if g(x)=(2x+1)3?

    I had no idea. Also, the g(x) I give here is probably incorrect, but it was something similar to this.
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    (Original post by Tarquin Digby)
    No, that's wrong. The area is the integral from 0 to k MINUS the area from k to 2. If you do a sketch, you will see that the area between k and 2 is below the x axis, so integrating would give you a negative value.
    But wasn't that the point? They never said that the area had to be positive. Also, I'm fairly sure I forgot one part of this question, would you mind filling me in? Something about fk-2(x)=-fk(x)
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    (Original post by Tarquin Digby)
    No, that's wrong. The area is the integral from 0 to k MINUS the area from k to 2. If you do a sketch, you will see that the area between k and 2 is below the x axis, so integrating would give you a negative value.
    instead of a minus you can simply take the absolute value of the integral from k to 2 (which will always be positive, as the area has to e positive)
    that's what we did in my Math HL class... Still could be wrong but in the end gives out the same result
 
 
 
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