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    My answers for Question 1:
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    A (a)
    B (c)
    C (d)
    D (b)
    E (d)
    F (a)
    G (d)
    H (b)
    I (b)
    J (b)
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    (Original post by souktik)
    The former.

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    Seriously? I am so. screwed.

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    (Original post by yl95)
    Seriously? I am so. screwed.

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    Yeah, I'm pretty sure. I believe they wanted to see if we could apply part (iv) and get the answer directly.
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    (Original post by souktik)
    Yeah, I'm pretty sure. I believe they wanted to see if we could apply part (iv) and get the answer directly.
    bro what were your anwers for the 2nd and 4th parts of the q2?
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    Looks like iv got 60 if im pessimistic 67 if im optimistic
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    (Original post by souktik)
    Yeah, I'm pretty sure. I believe they wanted to see if we could apply part (iv) and get the answer directly.
    That's annoying. Other people I know got the same answer as me. This paper has told me that I can't read properly.

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    (Original post by RichE)
    ...
    Do you know the lowest ever score an applicant has ever been shortlisted with in recent years?
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    can someone explain multiple choice part G??
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    Cannot believe I put the wrong answer down for the graph - I wrote down d) instead of b). l o l this is going downhill.

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    (Original post by kapur)
    bro what were your anwers for the 2nd and 4th parts of the q2?
    2.ii.a.
    f(t)-f(1-t)=t (for all t)
    implies: f(1-t)-f(t)=1-t (replacing t by 1-t)
    Add the two equations and you get t+(1-t)=0, that is 1=0.

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    2.ii.b. Condition: g(t)=-g(1-t)

    2.ii.c.
    I used f(t)=t.(2t-1)^3. In general, f(t)=t.g(t) is a valid example as long as the condition in 2.ii.b. is followed. A simpler example would have been f(t)={\frac{(2t-1)^3}{2}}, or f(t)={\frac{g(t)}{2}} in general.

    (Original post by yl95)
    That's annoying. Other people I know got the same answer as me. This paper has told me that I can't read properly.

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    Well, it was confusing. In an exam hall, it's easy to misinterpret "one digit at least 5" as "at least one digit 5".
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    (Original post by souktik)
    2.ii.a.
    f(t)-f(1-t)=t (for all t)
    implies: f(1-t)-f(t)=1-t (replacing t by 1-t)
    Add the two equations and you get t+(1-t)=0, that is 1=0.

    Spoiler:
    Show
    2.ii.b. Condition: g(t)=-g(1-t)

    2.ii.c.
    I used f(t)=t.(2t-1)^3. In general, f(t)=t.g(t) is a valid example as long as the condition in 2.ii.b. is followed. A simpler example would have been f(t)={\frac{(2t-1)^3}{2}}, or f(t)={\frac{g(t)}{2}} in general.



    Well, it was confusing. In an exam hall, it's easy to misinterpret "one digit at least 5" as "at least one digit 5".
    Do you have a copy of the paper?
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    (Original post by CD315)
    Do you have a copy of the paper?
    It's online on the MAT page alreadyn

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    (Original post by yl95)
    It's online on the MAT page alreadyn

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    Ok, here I go trying to count my scores.
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    Pretty sure I got 52 due to silly mistakes in the multiple choice more than anything. Jesus. Anyone think this will be enough to get me an Imperial offer?
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    (Original post by qwertyuiopg)
    can someone explain multiple choice part G??
    p_n(x) = nx-\frac{n(n+1)}{2} for all n.
    Therefore, p_{n-1}(x)=(n-1)x-\frac{(n-1)n}{2}
    Now some simple manipulation:
    p_n(x)=nx-\frac{n(n+1)}{2}
     = {\frac{n}{n-1}}{(n-1)x}-{\frac{n}{n-1}}{\frac{(n-1)(n+1)}{2}}
     = {\frac{n}{n-1}}{(n-1)x}-{\frac{n}{n-1}}{\frac{(n-1)n}{2}}-\frac{n}{2}
     = {\frac{n}{n-1}}p_{n-1}(x)-\frac{n}{2}

    So the answer is (d) -\frac{n}{2}.
    (Original post by CD315)
    Do you have a copy of the paper?
    https://www.maths.ox.ac.uk/system/fi...nts/test13.pdf
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    (Original post by CD315)
    Ok, here I go trying to count my scores.
    I'm relying on marks from my working...

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    (Original post by RichE)
    ..
    Have to say, I really don't like 1C. Wrong/right answers depending on whether you interpret  f''(2x) = \dfrac{d^2}{dt^2} f(t)\Big|_{t=2x} or f''(2x) = \dfrac{d^2}{dt^2} f(2t)\Big|_{t=x}. I confess I'm not sure what the correct interpretation is myself!
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    (Original post by DFranklin)
    Have to say, I really don't like 1C. Wrong/right answers depending on whether you interpret  f''(2x) = \dfrac{d^2}{dt^2} f(t)\Big|_{t=2x} or f''(2x) = \dfrac{d^2}{dt^2} f(2t)\Big|_{t=x}. I confess I'm not sure what the correct interpretation is myself!
    I know, right! I used the second interpretation and I'm all set to lose 4 marks.
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    (Original post by souktik)
    I know, right! I used the second interpretation and I'm all set to lose 4 marks.
    I think you'll be fine. There are people who got in the 50-60s who are worrying much more than you.

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    (Original post by DFranklin)
    Have to say, I really don't like 1C. Wrong/right answers depending on whether you interpret  f''(2x) = \dfrac{d^2}{dt^2} f(t)\Big|_{t=2x} or f''(2x) = \dfrac{d^2}{dt^2} f(2t)\Big|_{t=x}. I confess I'm not sure what the correct interpretation is myself!
    Does this mean they're likely to allow for 2 answers?
 
 
 
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