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    (Original post by revelry26)
    Do you guys think this is correct ?
    Question two part twoAttachment 252683 Attachment 252684


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    To be honest, I can't really understand what you did. :-/
    It's probably because I'm using my phone and everything is really tiny, but I don't think our answers match, in any case. :-/

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    (Original post by revelry26)
    Do you guys think this is correct ?
    Question two part twoAttachment 252683 Attachment 252684


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    I think we use the same method.
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    (Original post by revelry26)
    Do you guys think this is correct ?
    Question two part twoAttachment 252683 Attachment 252684


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    I think you are getting lost here and trying to be too complex. The picture is small and not very clear.

    But I think you say f(t) = 1/(1-k) what is your reasoning behind this?
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    (Original post by revelry26)
    Do you guys think this is correct ?
    Question two part twoAttachment 252683 Attachment 252684


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    What is your answer of the last question of part 2?
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    (Original post by yxcai)
    I think we use the same method.
    Can you please explain how there's a k in part ii? I thought that the k was in part i alone. I'm pretty confused right now.

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    (Original post by souktik)
    Can you please explain how there's a k in part ii? I thought that the k was in part i alone. I'm pretty confused right now.

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    g(t)=t, the RHS only has t but LHS=m(t)+n(k), so n(k) must be 0
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    (Original post by souktik)
    To be honest, I can't really understand what you did. :-/
    It's probably because I'm using my phone and everything is really tiny, but I don't think our answers match, in any case. :-/

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    What are your answers?


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    (Original post by yxcai)
    g(t)=t, the RHS only has t but LHS=m(t)+n(k), so n(k) must be 0
    Where do you get m(t) and n(k)? I'm sorry, I'm not being able to follow at all. :confused:

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    (Original post by souktik)
    Where do you get m(t) and n(k)? I'm sorry, I'm not being able to follow at all. :confused:

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    You can see g(t) in two parts. One part has t and k(m(t)), one part does not have t but has k (n(k))
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    (Original post by souktik)
    Where do you get m(t) and n(k)? I'm sorry, I'm not being able to follow at all. :confused:

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    (Original post by revelry26)
    What are your answers?


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    I don't remember part i. For ii.a, I proved it by showing that if g(t)=t, g(1-t)=1-t, but 1 = t+(1-t) = g(t)+g(1-t) = f(t)-f(1-t)+f(1-t)-f(t) = 0. Clear contradiction, therefore g(t) can't be t.
    2.ii.b. Replacing t by 1-t in the equation and adding, you get the condition g(t)+g(1-t)=0.
    2.ii.b. f(t)=t.g(t) is a solution, in general, when the above condition is followed. So I got f(t)=t(2t-1)^3 as a valid solution.

    (Original post by yxcai)
    You can see g(t) in two parts. One part has t and k(m(t)), one part does not have t but has k (n(k))
    What on earth is k? There's no k in the second part! :O

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    (Original post by Oxford Computer Science Dept)
    I think this example is very misleading. The MAT is an extremely important part of the shortlisting process (ie deciding who we invite to interview) for Computer Science. But yes we do look at all the information we have about a candidate.

    When the test scores are in we draw a rough line in the sand. Last year that was 53. It's about the average score for those taking that set of questions that year. (We do also factor in contextualised GCSE scores here.) We interview pretty much everyone who gets over that line. The tutors also have the freedom to call for interview people who fall below that line, where they think there is good reason to. So, for example if someone was ill on the day of the test, or there was a family crisis, but their application otherwise looked strong, they could still be interviewed and ultimately offered a place.

    Most people have similar grades for achieved AS (or equivalent) or predicted A2s. We interview and get to meet people so the personal statement isn't as important as with other universities. The MAT and interviews are therefore extremely important part of the application process. The MAT is one of the few mechanisms we have to fairly compare all of the applicants, no matter which educational system they have come through.

    So yes, it is feasible that someone with a low MAT test score, who for reasons beyond their control performed badly on the day of the test, could be offered a place. But just because there is the odd case like this, it really doesn't mean that the MAT is unimportant for CompSci applications. It is.
    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?
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    (Original post by souktik)
    I don't remember part i. For ii.a, I proved it by showing that if g(t)=t, g(1-t)=1-t, but 1 = t+(1-t) = g(t)+g(1-t) = f(t)-f(1-t)+f(1-t)-f(t) = 0. Clear contradiction, therefore g(t) can't be t.
    2.ii.b. Replacing t by 1-t in the equation and adding, you get the condition g(t)+g(1-t)=0.
    2.ii.b. f(t)=t.g(t) is a solution, in general, when the above condition is followed. So I got f(t)=t(2t-1)^3 as a valid solution.



    What on earth is k? There's no k in the second part! :O

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    The following are the question.
    (i) Let k not equal to ±1. The function f(t) satisfies the identity
    f(t) − kf(1 − t) = t
    for all values of t. By replacing t with 1 − t, determine f(t).
    (ii) Consider the new identity
    f(t) − f(1 − t) = g(t). (∗)
    (a) Show that no function f(t) satisfies (∗) when g(t) = t.
    (b) What condition must the function g(t) satisfy for there to be a solution f(t) to (∗)?
    (c) Find a solution f(t) to (∗) when g(t) = (2t − 1)3
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    (Original post by yxcai)
    The following are the question.
    (i) Let k not equal to ±1. The function f(t) satisfies the identity
    f(t) − kf(1 − t) = t
    for all values of t. By replacing t with 1 − t, determine f(t).
    (ii) Consider the new identity
    f(t) − f(1 − t) = g(t). (∗)
    (a) Show that no function f(t) satisfies (∗) when g(t) = t.
    (b) What condition must the function g(t) satisfy for there to be a solution f(t) to (∗)?
    (c) Find a solution f(t) to (∗) when g(t) = (2t − 1)3
    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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    (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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    But I think "NEW identity" should be based on i
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    (Original post by yxcai)
    But I think "NEW identity" should be based on i
    It can't be based on i. You probably misinterpreted part ii, I think.
    Anyway, I have a statistics exam in 6 hours and I need to catch a few hours of sleep.

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    (Original post by souktik)
    It can't be based on i. You probably misinterpreted part ii, I think.
    Anyway, I have a statistics exam in 6 hours and I need to catch a few hours of sleep.

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    So I will get 0 . I want cry!
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    (Original post by souktik)
    It can't be based on i. You probably misinterpreted part ii, I think.
    Anyway, I have a statistics exam in 6 hours and I need to catch a few hours of sleep.

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    I will only get 70-75 for my mat.
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    (Original post by yxcai)
    So I will get 0 . I want cry!
    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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    (Original post by yxcai)
    I will only get 70-75 for my mat.
    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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