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    Wait, guys, is 1J) really B? Still means I got 7/10 then from my lucky guess.
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    (Original post by yl95)
    Wait, guys, is 1J) really B? Still means I got 7/10 then from my lucky guess.
    yes it is
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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".
    I left the last part of q2 man. Was it harder / longer than the other parts? I'm screwed if the last part was worth 6 marks!
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    (Original post by qwertyuiopg)
    yes it is
    Well, I thought it was one with a log in it so uh....lol....
    Looking back at the test, I guess the difficulty was slightly increased but it wasn't as hard as I originally said it was. However, there were plenty of traps where applicants could easily fall into like I did.
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    (Original post by souktik)
    My answers for Question 1:
    Spoiler:
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    A (a)
    B (c)
    C (d)
    D (b)
    E (d)
    F (a)
    G (d)
    H (b)
    I (b)
    J (b)
    I is (c)

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    (Original post by journeyinwards)
    I is (c)

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    Really? How did you get that? :confused:

    (Original post by yl95)
    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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    I'm not worried, just a little annoyed. Meh.

    (Original post by CD315)
    Does this mean they're likely to allow for 2 answers?
    Unlikely. The first interpretation seems to be correct. They'd put those options on purpose, possibly to check if we're familiar with the notation. I find that a little unfair, but there's no point in cribbing.

    (Original post by kapur)
    I left the last part of q2 man. Was it harder / longer than the other parts? I'm screwed if the last part was worth 6 marks!
    What do you mean by the last part? 2.ii.c, right? What did you get for 2.i, 2.ii.a and 2.ii.b? I don't think not doing 2.ii.c will cost you 6 marks if you got the other parts correctly. 4+3+4+4 might be the division of marks.
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    Just compared the multiple choice questions from this year with those of the other years and I still feel like they were way harder this year. Does anyone feel the same?
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    I hate the uncertainty surrounding how Imperial will use this. Damn.
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    (Original post by souktik)
    Really? How did you get that? :confused:



    I'm not worried, just a little annoyed. Meh.



    Unlikely. The first interpretation seems to be correct. They'd put those options on purpose, possibly to check if we're familiar with the notation. I find that a little unfair, but there's no point in cribbing.



    What do you mean by the last part? 2.ii.c, right? What did you get for 2.i, 2.ii.a and 2.ii.b? I don't think not doing 2.ii.c will cost you 6 marks if you got the other parts correctly. 4+3+4+4 might be the division of marks.
    Yea I mean 2 ii c. I got the same as you in the other parts. And part 1 I multiplied the 2nd eqn with k and addded. That's why I'm hoping the last part wasn't worth 6 :P I was too excpeting above 80 but it turns out I markwd 2 mcq's incorrectly and misread one part of the 5th q 70 + it is then
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    (Original post by dutchmaths)
    Just compared the multiple choice questions from this year with those of the other years and I still feel like they were way harder this year. Does anyone feel the same?
    Yes.
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    (Original post by dutchmaths)
    Just compared the multiple choice questions from this year with those of the other years and I still feel like they were way harder this year. Does anyone feel the same?
    No way bro I found 2012 paper really hard for some reason. Especially the multiple choice and q5. In this paper other than the last multiple choice all were fine.
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    (Original post by kapur)
    No way bro I found 2012 paper really hard for some reason. Especially the multiple choice and q5. In this paper other than the last multiple choice all were fine.
    2012? Hard?...


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    Don't worry. Only time will tell whether you will get shortlisted or not. Have hope!
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    (Original post by Tarquin Digby)
    https://www.maths.ox.ac.uk/system/fi...nts/test13.pdf

    Paper's up. I got BCDDDBDBBB for multiple choice, might be wrong tho!
    u tr0ll

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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".
    Another possibility was to notice that

    f(\frac{1}{2}+t)-f(\frac{1}{2}-t))=8t^3

    ,thus

    f(t)=4(t-\frac{1}{2})^3

    is a solution (although I forgot to undo the t \rightarrow t+\frac{1}{2} transformation: I'm happy that part wasn't multiple choice!).

    The stars and bars method killed question 5 very quickly.

    I was under the impression that f'(u)=\frac{df(u)}{dx}. I dislike Leibniz notation.
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    (Original post by Tarquin Digby)
    https://www.maths.ox.ac.uk/system/fi...nts/test13.pdf

    Paper's up. I got BCDDDBDBBB for multiple choice, might be wrong tho!
    Not might be - will be. At least 3 wrong possibly 4 depending on interpretation of (c).
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    (Original post by CD315)
    Does this mean they're likely to allow for 2 answers?
    I doubt it, since the fact they gave both "possible" answers implies they wanted you to choose the "correct" one.

    I'm just arrogant enough to think that if I'm not sure what the correct answer is, it's not a very fair question. [Although I could just be being stupid or forgetful about a standard definition].

    I'm interested to see what RichE has to say about it.
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    (Original post by Tarquin Digby)
    I don't always troll!
    pestis! furcifer!


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    (Original post by henpen)
    ...
    (Original post by msmith2512)
    ...
    I just did the first 5 multiple choice questions and got BCCBA.
    Could you guys compare your answers against mine?
    (Also I didn't do this paper, I'm just curious)

    Edit, I just did G and got D as my answer.
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    (Original post by DFranklin)
    I doubt it, since the fact they gave both "possible" answers implies they wanted you to choose the "correct" one.

    I'm just arrogant enough to think that if I'm not sure what the correct answer is, it's not a very fair question. [Although I could just be being stupid or forgetful about a standard definition].

    I'm interested to see what RichE has to say about it.
    I take it as meaning the second in your original post (i.e. to differentiate it using the chain rule on the 2x as opposed to just on some t and subbing in t=2x at the end). But you're right, the question is ambiguous.
 
 
 
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