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    (Original post by tiny hobbit)
    Yes, but on someone else's paper! It really should just say "Edexcel" in big print.
    sure ... I do not think it is appropriate.
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    (Original post by tiny hobbit)
    Yes, but on someone else's paper! It really should just say "Edexcel" in big print.
    Meh. It's a watermark to indicate where you got the paper from. It's annoying but I guess as TeeEm says, it's good advertising for their site.
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    https://twitter.com/mathsdot/status/614495610943750144 solutions here
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    We take neg root for inverse of g as positive root obviously gives the other side of the inverse for x>0. Another mistake is not neglecting -1 as that comes from squaring in Q2 and upon checking 1/2 is the only solution. Other then that it is all correctz


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    Hopefully talking about sufficient conditions will get me an S+ mark in the convergence question .


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    (Original post by physicsmaths)
    Hopefully talking about sufficient conditions will get me an S+ mark in the convergence question .


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    I haven't looked at the solutions yet so ill keep all those things in mind when I do! Theyre not mine tho just some random teacher on twitter
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    Can someone explain these presentation marks and s+ marks? im a bit confused by them all
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    (Original post by tiny hobbit)
    Why do people who simply copy a paper belonging to someone else feel the need to put their own logo on it?
    Why wouldn't they put a logo...
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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
    _____


    ____
    Spoiler:
    Show
    Distinction : Merit
    Advanced Extension (June 2014) 68 48
    Advanced Extension (June 2013) 70 51
    Advanced Extension (June 2012) 75 56
    Advanced Extension (June 2011) 70 50
    Advanced Extension (June 2010) 81 63
    Advanced Extension (June 2009) 69 51
    Advanced Extension (June 2008) 72 54
    Advanced Extension (June 2007) 68 49
    Advanced Extension (June 2006) 67 48
    Advanced Extension (June 2005) 69 50
    Advanced Extension (June 2004) 70 48
    Advanced Extension (June 2003) 70 50
    Advanced Extension (June 2002) 74 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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