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    (Original post by louisforrest)
    Is this an acceptable answer for 2015 2 part iv or have I made a wrong assumption in line 4
    Wrong assumption in line 4. FWIW I don't think this is the right (or salvageable) approach.
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    This is important. The MAT will be sat at different times around the world tomorrow. Do not discuss how the test went immediately afterwards: you will be giving an unfair advantage to those still to sit the test. It's best to say nothing: it's a small step from "I found Q3 hard" to "I found Q3 on ... hard". You can resume discussions at 1pm the day after the test, i.e. Thursday.

    Good luck everyone!

    Gavin
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    (Original post by louisforrest)
    Is this an acceptable answer for 2015 2 part iv or have I made a wrong assumption in line 4
    The error probably arises from the fact that the factorization is not necessarily unique.

    Your argument assumes that if a number C can be written as AB where A and B are both positive integers greater than 1, then B=A^2 (or A=B^2) is a necessary condition for C to be a cube number. However, this is not the case (it is actually a sufficient condition instead).

    Consider 216=6^3. 216 could be written as 3(72) but 72=/=3^2.

    Therefore showing that x^2+x+1=/=(x+1)^2 is not enough to prove that (x+1)(x^2+x+1) is not a cube.
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    (Original post by gavinlowe)
    This is important. The MAT will be sat at different times around the world tomorrow. Do not discuss how the test went immediately afterwards: you will be giving an unfair advantage to those still to sit the test. It's best to say nothing: it's a small step from "I found Q3 hard" to "I found Q3 on ... hard". You can resume discussions at 1pm the day after the test, i.e. Thursday.

    Good luck everyone!

    Gavin
    When do you normally release the mark scheme?
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    I've been trying to find software to draw it but I can't find any- but in light of last year, would anyone know what the graph of tan(y)= tan(x) would look like?
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    (Original post by louisforrest)
    I'm going to have a day off, watch a movie tonight. No point in doing lots of work today.

    Posted from TSR Mobile
    You know, just want to be well rested and relaxed
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    (Original post by KloppOClock)
    When do you normally release the mark scheme?
    I forget. My colleagues in Mathematics normally do this. But certainly not before the marking is finished (the end of next week).

    The questions are put online a couple of days after the test.

    Gavin
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    (Original post by QueenOfCaffeine)
    I've been trying to find software to draw it but I can't find any- but in light of last year, would anyone know what the graph of tan(y)= tan(x) would look like?
    Try wolfram alpha
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    (Original post by gavinlowe)
    I forget. My colleagues in Mathematics normally do this. But certainly not before the marking is finished (the end of next week).

    The questions are put online a couple of days after the test.

    Gavin
    That's fast, thanks.
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    (Original post by QueenOfCaffeine)
    I've been trying to find software to draw it but I can't find any- but in light of last year, would anyone know what the graph of tan(y)= tan(x) would look like?
    As tan has period pi we have infinitely many lines of the form y=x+k(pi) where k is an integer. So y=x, y=x+pi, y=x-pi etc. Visually it looks like a series of regularly spaced, diagonal (bottom left to upper right) lines.
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    Thanks- I could be completely wrong but is that not y=tan(x)?? I've just plotted tan(y)=tan(x) on wolfram alpha and it looks like lots of right-angled triangles?
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    (Original post by redsquared)
    Try wolfram alpha
    thank you!
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    (Original post by QueenOfCaffeine)
    Thanks- I could be completely wrong but is that not y=tan(x)?? I've just plotted tan(y)=tan(x) on wolfram alpha and it looks like lots of right-angled triangles?
    I'm not sure what wolfram alpha displays but what I described is definitely not y=tanx. y=tanx looks like a horizontal series of infinitely many asymptotical curves.
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    (Original post by QueenOfCaffeine)
    thank you!
    No worries its a great website
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    I heard if you score above a certain mark, you will be auto accepted for interview? Is this correct? gavinlowe


    If so, what was that threshold for last year?
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    (Original post by ttva)
    I heard if you score above a certain mark, you will be auto accepted for interview? Is this correct? gavinlowe


    If so, what was that threshold for last year?
    Close, but not quite right.

    We use a "provisional shortlisting indicator" (PSI), which is a linear combination of the MAT score and the proportion of A*s at GCSE. A threshold is then set such that everyone with a PSI above that threshold is shortlisted by default. A college may choose to go against that default, but must justify their decisions; this happens rarely, e.g. when a candidate has already done part of a degree, so could have been expected to get a higher mark.

    There's a second, lower, threshold: everyone below that lower threshold is, by default, not shortlisted; again colleges can go against the default. We look carefully at candidates between the two thresholds, particularly taking contextual information into account. We use our academic judgement as to whom to interview.

    Last year, every Computer Science candidate with a MAT score of at least 62 was interviewed.

    Gavin
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    (Original post by QueenOfCaffeine)
    I've been trying to find software to draw it but I can't find any- but in light of last year, would anyone know what the graph of tan(y)= tan(x) would look like?
    Ignoring the points where tan x or tan y are undefined, It's going to be the set of diagonal lines

    y = x + n\pi (n \in \mathbb{Z})

    (I'm confident the horizontal / vertical lines Wolfram Alpha is drawing are spurious - probably caused by its treatment of the undefined points).
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    For question 4 for 2014 for part (v) i understand swapping alpha and beta but why does the area of ABC still remain as 1/2sin2alpha shouldn't it be 1/2sin2beta since we swapped it?
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    (Original post by gavinlowe)
    Close, but not quite right.

    We use a "provisional shortlisting indicator" (PSI), which is a linear combination of the MAT score and the proportion of A*s at GCSE. A threshold is then set such that everyone with a PSI above that threshold is shortlisted by default. A college may choose to go against that default, but must justify their decisions; this happens rarely, e.g. when a candidate has already done part of a degree, so could have been expected to get a higher mark.

    There's a second, lower, threshold: everyone below that lower threshold is, by default, not shortlisted; again colleges can go against the default. We look carefully at candidates between the two thresholds, particularly taking contextual information into account. We use our academic judgement as to whom to interview.

    Last year, every Computer Science candidate with a MAT score of at least 62 was interviewed.

    Gavin
    Do you know what that threshold would be for Maths last year. Would it be a few points lower than CS as CS has a lower acceptance rate,
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    (Original post by ttva)
    Do you know what that threshold would be for Maths last year.
    There's a graph at the end of the feedback report for 2015:

    https://www.maths.ox.ac.uk/system/fi...ack%202015.pdf

    I don't know how reliable it is when discerning "1 person failed to get an interview in this bracket" v.s. "no people failed in this bracket", but looking at the 70-74 bracket v.s. the 75-79 bracket strongly indicates someone was rejected in that bracket (75-79)

    I'd be pretty confident that anyone scoring that high but rejected was rejected for reasons unrelated to the MAT. (E.g. someone who already had a degree in maths and wanting to get a 2nd degree from Oxford).
 
 
 
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