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    (Original post by souktik)
    You still have part i, and there's also the chance that I'M the one who is wrong. I don't want to entertain that thought in my head, but it's possible. Maybe one of the seniors will clarify if they notice...

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    I really hope I can receive the interview
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    (Original post by souktik)
    70-75 is above average for admitted candidates, so you don't have much to worry about. Cheer up! And as I said, maybe I'm wrong.

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    thank you~
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    (Original post by yxcai)
    I really hope I can receive the interview
    -_-
    If you're scoring 70+, it's pretty obvious that you'll get the interview. Anyone above 55 is sure to get an interview, I think.

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    Here's my solution to Q2 FWIW:

    (i) f(t) - k f(1-t) = t, so f(1-t) - k f(t) = 1-t, so kf(1-t) -k^2 f(t) = k(1-t). Adding the first and last equation we get:
    f(t) - k^2 f(t) = t - k(1-t)
    So (1-k^2)f(t) = k + (1+k)t, so f(t) = \dfrac{k + (1+k)t}{1-k^2}.

    (ii) (a) Whatever the function f, if we set t = 0.5 the LHS = 0 but the RHS = 1/2.
    (b) Since the LHS changes sign on applying the map t -> 1-t, the RHS must also, so g(t) = -g(1-t) is necessary.
    (c) Note that if g(t) = -g(1-t), then if f(t) = g(t) / 2 then f(t) - f(1-t) = (g(t) + g(t))/2 = g(t). So in this particular case, f(t) = \frac{1}{2} (2t-1)^3 works.
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    (Original post by souktik)
    Yeah, I remember the questions. Please note that 2.i deals with a k, but you're working with a NEW identity for 2.ii. There's no k in 2.ii, I'm sorry.

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    i used k in part 2 (ii) a. for 2(i) by solving the 2 equatiosn simultaneously you get f(t)= (t +k(1-t) )/ (1-k^2).
    so i took part 2(ii) as a special case of part 2(i) where k=1 . and then you get 1/0 .
    do you think this is fine?
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    (Original post by DFranklin)
    Here's my solution to Q2 FWIW:

    (i) f(t) - k f(1-t) = t, so f(1-t) - k f(t) = 1-t, so kf(1-t) -k^2 f(t) = k(1-t). Adding the first and last equation we get:
    f(t) - k^2 f(t) = t - k(1-t)
    So (1-k^2)f(t) = k + (1+k)t, so f(t) = \dfrac{k + (1+k)t}{1-k^2}.

    (ii) (a) Whatever the function f, if we set t = 0.5 the LHS = 0 but the RHS = 1/2.
    (b) Since the LHS changes sign on applying the map t -> 1-t, the RHS must also, so g(t) = -g(1-t) is necessary.
    (c) Note that if g(t) = -g(1-t), then if f(t) = g(t) / 2 then f(t) - f(1-t) = (g(t) + g(t))/2 = g(t). So in this particular case, f(t) = \frac{1}{2} (2t-1)^3 works.
    Thanks! I feel relieved now.
    There's probably a minor calculation mistake in part i of your solution, though.
    (i) f(t) - k f(1-t) = t, so f(1-t) - k f(t) = 1-t, so kf(1-t) -k^2 f(t) = k(1-t). Adding the first and last equation we get:
    f(t) - k^2 f(t) = t + k(1-t)
    So (1-k^2)f(t) = k + (1-k)t, so f(t) = \dfrac{k + (1-k)t}{1-k^2}.

    (Original post by kapur)
    i used k in part 2 (ii) a. for 2(i) by solving the 2 equatiosn simultaneously you get f(t)= (t +k(1-t) )/ (1-k^2).
    so i took part 2(ii) as a special case of part 2(i) where k=1 . and then you get 1/0 .
    do you think this is fine?
    No. I think that's faulty reasoning, I'll explain as soon as I can.

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    (Original post by kapur)
    i used k in part 2 (ii) a. for 2(i) by solving the 2 equatiosn simultaneously you get f(t)= (t +k(1-t) )/ (1-k^2).
    so i took part 2(ii) as a special case of part 2(i) where k=1 . and then you get 1/0 .
    do you think this is fine?
    No. Dividing by zero leads to a singularity and therefore is just not allowed. It doesn't mean that something is not true. I have been trying to think of an example and I think the best I can do at the moment is x^2 - x = 0 when x = 0 is a valid solution but where dividing by x 'loses' this as a solution.

    To show that something isn't true you assume that it is and then show that this leads to a contradiction such as 0 = 1.
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    (Original post by jadoreétudier)
    The MAT is supposed to test for raw talent/ability, but I'd expect that fee-paying schools will be able to prepare their pupils better for it than state schools?
    There's a lot of truth in that (although there are also non-fee-paying schools that prepare their students very well, and some fee-paying schools that prepare theirs less well). That's why we consider contextual information very carefully, both in short-listing and in making final decisions.
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    (Original post by souktik)
    Thanks! I feel relieved now.
    There's probably a minor calculation mistake in part i of your solution, though.
    (i) f(t) - k f(1-t) = t, so f(1-t) - k f(t) = 1-t, so kf(1-t) -k^2 f(t) = k(1-t). Adding the first and last equation we get:
    f(t) - k^2 f(t) = t + k(1-t)
    Yes, thanks. I'm never as good when going straight into TSR as opposed to pen and paper.
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    wow I really messed up Q2
    I think I was relying on 2(ii) to be a case of the first identity for k=1
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    (Original post by sun_tzu)
    wow I really messed up Q2
    I think I was relying on 2(ii) to be a case of the first identity for k=1
    Same, ugh. I can't believe I missed k=-1 but I didn't do part i) properly so everything just cascaded

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    (Original post by jadoreétudier)
    The MAT is supposed to test for raw talent/ability, but I'd expect that fee-paying schools will be able to prepare their pupils better for it than state schools?
    This is a huge generalisation. I'd like to think that my (state) sixth form college does a reasonable job in helping students to prepare for the MAT (and the STEP). It does come down to their efforts at trying questions however, We can give them hints when they are stuck, but unless they are prepared to put the time in to trying questions, telling them how to do them does not have much value.
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    Which college are you applying to? Hope we won't crash...
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    (Original post by superhjd)
    Which college are you applying to? Hope we won't crash...
    st hugh's college
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    (Original post by yxcai)
    st hugh's college
    Me Worcester... Btw, why you said you will only get 70-75? According to your answers, it won't be below 80 which is really good
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    (Original post by superhjd)
    Me Worcester... Btw, why you said you will only get 70-75? According to your answers, it won't be below 80 which is really good
    My part 2 is wrong~
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    (Original post by yxcai)
    My part 2 is wrong~
    Only loss 12 since part i of question 2 seems correct
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    I'm not sure if this has been asked before, but I was wondering exactly what aspects of the MAT are used in the admissions process. Will the tutors only know the score, will they get a more detailed mark distribution, or will they see the paper itself? For example, will they see that I had the right idea on the last part of q4, and simply made a mistake with a trig identity? Or will they only see that I got 12/15 or 70 or whatever? The point of the test is to measure aptitude, but does anyone know whether that's based only on the score or on a more detailed analysis of a candidate's work and mistakes? I understand that it would be very time-consuming to evaluate the candidates in this way, and there are other pros and cons to doing it like that, which is why I'm not really sure which way the MAT is used.

    I know I can't do anything to change it now, but I am curious to know the role of the MAT in the admissions decision. Thanks!
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    (Original post by GMU10)
    I'm not sure if this has been asked before, but I was wondering exactly what aspects of the MAT are used in the admissions process. Will the tutors only know the score, will they get a more detailed mark distribution, or will they see the paper itself? For example, will they see that I had the right idea on the last part of q4, and simply made a mistake with a trig identity? Or will they only see that I got 12/15 or 70 or whatever? The point of the test is to measure aptitude, but does anyone know whether that's based only on the score or on a more detailed analysis of a candidate's work and mistakes? I understand that it would be very time-consuming to evaluate the candidates in this way, and there are other pros and cons to doing it like that, which is why I'm not really sure which way the MAT is used.

    I know I can't do anything to change it now, but I am curious to know the role of the MAT in the admissions decision. Thanks!
    Tutors see all of the above. The breakdown of marks by question will be on the database that we use, so very easily available. The tutors at the first-choice college receive the paper (and other tutors can ask to see it). As you say, it's time-consuming to study every script in detail, but often scripts of border-line candidates will be looked at.
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    any1 applied to merton ?
 
 
 
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