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    (Original post by physicsmaths)
    Boundary thoughts?


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    Average (2002-2014) -: Distinction is 74 and Merit is 54

    I personally thought it was much easier than 2014 (and some of the other years), though i've seen online that many thought it was hard (preferably not as hard as 2014 but HARD).

    On that note; i will have a guess at (+- 2)
    Distinction: 71
    Merit: 51
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    (Original post by |-Solutions-|)
    Why wouldn't they put a logo...
    Some people might not want to blatantly tag their logo on a flagrant copyright violation.

    I mean, yeah, we all know that exam papers get copied left right and center, but actually removing copyright notices (*) and putting your own logo on just might (**) get you to the point where someone at Edexcel says "we really can't ignore this guy, call the lawyers".

    (*) I'm assuming this paper had the standard "This publication may be reproduced only in accordance with Pearson Education Ltd copyright policy. ©201X Pearson Education Ltd" notice that previous years have had.

    (**) Yes, I realise it's still very unlikely this will actually happen. Still.
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    (Original post by |-Solutions-|)
    Why wouldn't they put a logo...
    (Original post by DFranklin)
    Some people might not want to blatantly tag their logo on a flagrant copyright violation.

    I mean, yeah, we all know that exam papers get copied left right and center, but actually removing copyright notices (*) and putting your own logo on just might (**) get you to the point where someone at Edexcel says "we really can't ignore this guy, call the lawyers".

    (*) I'm assuming this paper had the standard "This publication may be reproduced only in accordance with Pearson Education Ltd copyright policy. ©201X Pearson Education Ltd" notice that previous years have had.

    (**) Yes, I realise it's still very unlikely this will actually happen. Still.
    Yes, the usual copyright bit was on the front page, but I-solutions didn't include that page in his posting.
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    Quick question.
    what does the mean value theorem show and how could I use it to solve problems. I know what it is but what can it show
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    (Original post by 11234)
    Quick question.
    what does the mean value theorem show and how could I use it to solve problems. I know what it is but what can it show
    There are lots of different ways of looking at it, but perhaps the most intuitive is the integral form:

    We can find \xi \in (a, b) such that \int_a^b f(x) \, dx = (b-a)f'(\xi).

    Since \frac{1}{b-a} \int_a^b f(x)\,dx is the mean value of f(x) between a and b, this is the same as saying that f(x) equals its mean value for some x between a and b.
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    (Original post by DFranklin)
    There are lots of different ways of looking at it, but perhaps the most intuitive is the integral form:

    We can find \xi \in (a, b) such that \int_a^b f(x) \, dx = (b-a)f'(\xi).

    Since \frac{1}{b-a} \int_a^b f(x)\,dx is the mean value of f(x) between a and b, this is the same as saying that f(x) equals its mean value for some x between a and b.
    My pure is not great but there is a differentiation mean value theorem which I remember pictorially and says if a function is continuous between 2 points A and B, the there is at least one point on the function with the same gradient as the gradient of the chord AB. (I think Rolle is a special case where the chord is horizontal)

    The one with the integral is used often by engineers to find the mean value of a function (sometimes the integrand is squared then the answer is square rooted).
    I think the f on the RHS is not dashed ...
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    (Original post by DFranklin)
    There are lots of different ways of looking at it, but perhaps the most intuitive is the integral form:

    We can find \xi \in (a, b) such that \int_a^b f(x) \, dx = (b-a)f'(\xi).

    Since \frac{1}{b-a} \int_a^b f(x)\,dx is the mean value of f(x) between a and b, this is the same as saying that f(x) equals its mean value for some x between a and b.
    many thanks
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    (Original post by DFranklin)
    There are lots of different ways of looking at it, but perhaps the most intuitive is the integral form:

    We can find \xi \in (a, b) such that \int_a^b f(x) \, dx = (b-a)f'(\xi).

    Since \frac{1}{b-a} \int_a^b f(x)\,dx is the mean value of f(x) between a and b, this is the same as saying that f(x) equals its mean value for some x between a and b.
    \int_a^b f(x) \, dx = (b-a)f(\xi), unless I'm getting confused in my sleep deprivation?

    EDIT: TeeEm is ahead of me . The Wikipedia page has a nice diagram explaining the "derivative" version.
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    (Original post by TeeEm)
    My pure is not great but there is a differentiation mean value theorem which I remember pictorially and says if a function is continuous between 2 points A and B, the there is at least one point on the function with the same gradient as the gradient of the chord AB. (I think Rolle is a special case where the chord is horizontal)
    Yes, this is the version I'm most familiar with, but the correct conditions on f are a a bit finicky: if a function f is continuous on the closed interval [a,b], and differentiable on the open interval (a, b), then there is at least one point \xi \in (a, b) with f'(\xi)(b-a) = f(b) - f(a).

    (I think Rolle is a special case where the chord is horizontal.
    You usually prove Rolle first, because we know that a cts function on a closed interval is bounded and attains it's bounds, and it's reasonably easy to show that the derivative at a maximum point must be zero. It's then easy to go from there to the general MVT.

    I think the f on the RHS is not dashed ...
    (Original post by shamika)
    \int_a^b f(x) \, dx = (b-a)f(\xi), unless I'm getting confused in my sleep deprivation?
    Yes, thanks. Freudian slip on my part, since I'm so used to the non-integral formulation that I reflexively put the dash in, even though I knew I shouldn't.
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    (Original post by TeeEm)
    My pure is not great but there is a differentiation mean value theorem which I remember pictorially and says if a function is continuous between 2 points A and B, the there is at least one point on the function with the same gradient as the gradient of the chord AB. (I think Rolle is a special case where the chord is horizontal)

    The one with the integral is used often by engineers to find the mean value of a function (sometimes the integrand is squared then the answer is square rooted).
    I think the f on the RHS is not dashed ...
    could you recommend any further reading for maths over the holidays. Many thanks
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    (Original post by 11234)
    could you recommend any further reading for maths over the holidays. Many thanks
    I assume you've applied to warwick as you sat the aea. If so: http://www2.warwick.ac.uk/fac/sci/ma...ug/read_list2/

    Or, https://www0.maths.ox.ac.uk/courses/material , click on a topic and select course synopsis, there are reading lists there
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    (Original post by 11234)
    could you recommend any further reading for maths over the holidays. Many thanks
    (Original post by Gome44)
    I assume you've applied to warwick as you sat the aea. If so: http://www2.warwick.ac.uk/fac/sci/ma...ug/read_list2/
    Or, https://www0.maths.ox.ac.uk/courses/material , click on a topic and select course synopsis, there are reading lists there

    Morally I could not help anybody who has an intention of going to Warwick, supports Arsenal or dislikes cats ...
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    (Original post by TeeEm)
    Morally I could not help anybody who has an intention of going to Warwick, supports Arsenal or dislikes cats ...
    I dont like cats and have an intention of goin warwick as an insurance. Luckily i dont support arsenal ....


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    (Original post by TeeEm)
    Morally I could not help anybody who has an intention of going to Warwick, supports Arsenal or dislikes cats ...
    Why is that?
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    (Original post by Jai Sandhu)
    Why is that?
    Clearly I strongly indicate my personal likes and dislikes.
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    (Original post by TeeEm)
    Clearly I strongly indicate my personal likes and dislikes.
    That really does not answer my question.
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    (Original post by TeeEm)
    Clearly I strongly indicate my personal likes and dislikes.
    Why do hate warwick? We know they are better then UCL for maths 👀🌝.


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    (Original post by Jai Sandhu)
    That really does not answer my question.
    I would love to re-explain my stance/opinion which many find biased but today I am very busy so I will have to postpone my explanation.
    I find the thread which exactly the same debate took place I will post it.
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    (Original post by physicsmaths)
    Why do hate warwick? We know they are better then UCL for maths 👀🌝.


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    Only for pure and simply I have no interest in Pure.

    However UCL in terms of institutional prestige makes mince meat of Warwick.
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    (Original post by TeeEm)
    Only for pure and simply I have no interest in Pure.

    However UCL in terms of institutional prestige makes mince meat of Warwick.
    Maybe, i rejected ucl rather go warwick and ucl is in the city lol.


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