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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    (Original post by ryandaniels2015)
    Thanks for the reply, so the method does not have to be the same as the mark scheme?
    Not necessarily for a question like that, there may be different ways of achieving the same answer. These variations would normally be put down in the notes section under the question (examiners copy)
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    (Original post by Ayman!)
    Dude, what's a C3 triangle? Is that on the spec? :eek:

    That C3 proof triangle 45 degree, 45 degree + 90 right angle

    with 1, 1 and sqroot (2) lengths
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    (Original post by XxKingSniprxX)
    That C3 proof triangle 45 degree, 45 degree + 90 right angle

    with 1, 1 and sqroot (2) lengths
    Just remember it as a square cut by it's diagonal
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    (Original post by edothero)
    Not necessarily for a question like that, there may be different ways of achieving the same answer. These variations would normally be put down in the notes section under the question (examiners copy)
    I checked the notes section and my method isn't there, I went some long winded route, so u think I will still get the marks? Thanks for the help
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    (Original post by ryandaniels2015)
    I checked the notes section and my method isn't there, I went some long winded route, so u think I will still get the marks? Thanks for the help
    Mind showing me the question and your answer?
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    (Original post by edothero)
    Another interesting way of doing Question 10 on the IAL C34 Specimen Paper ( https://07a69ccf283966549a9350d1a669...%20Edexcel.pdf )
    Spoiler:
    Show
    x=tan\theta

y\ =\ sin\theta

    Find the Cartesian equation

    cos\theta = \sqrt{1-y^{2}}

    x=\dfrac{sin\theta}{cos\theta} = \dfrac{y}{\sqrt{1-y^{2}}}

    x\sqrt{1-y^{2}} = y

    x^{2}(1-y^{2}) = y^{2}

    x^{2} = y^{2} + y^{2}x^{2}

    \therefore \ y^{2} = \dfrac{x^{2}}{x^{2}+1}

    Volume of solid revolution equation : \pi \displaystyle \int y^{2}\ dx

    \pi \displaystyle \int_0^{\sqrt{3}} \dfrac{x^{2}}{x^{2}+1}\ dx\ \ = \ \ \pi \displaystyle \int_0^{\sqrt{3}} \dfrac{x^{2}+1-1}{x^{2}+1}\ dx\ \ =\ \ \pi \displaystyle \int_0^{\sqrt{3}} \dfrac{x^{2}+1}{x^{2}+1} - \dfrac{1}{x^{2}+1}\ dx

    \pi \displaystyle \int_0^{\sqrt{3}} 1 - \dfrac{1}{x^{2}+1}\ dx

    This bit is FP3 level:

     \displaystyle \int \dfrac{1}{x^{2}+1}\ dx\ =\ arctan(x)

    = \displaystyle \pi \left[x - arctan(x)\right]^{\sqrt{3}}_{0}

    \pi \displaystyle \left[(\sqrt{3}-arctan(\sqrt{3}))-(0-0)\right]

     = \pi\sqrt{3} - \dfrac{1}{3}\pi^{2}
    This is how I did it lmao, I wasn't thinking straight. Been doing a lot of FP3 lately :laugh:
    I would have done pi times integral of sin^2 theta x sec ^2 theta, which is integral of tan^2 theta... I'm stuck now, where did I go wrong


    Edit: oh wait that is sec^2 theta minus 1, so then it is pi [tan x - x] bounded by the theta value
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    (Original post by metrize)
    I would have done pi times integral of sin^2 theta x sec ^2 theta, which is integral of tan^2 theta... I'm stuck now, where did I go wrong
    Ignore my solution, its an alternative way and some methods used towards the end are beyond C4 level.

    \pi\displaystyle \int y^{2}\ \dfrac{dx}{d\theta}\ d\theta is the correct way of doing it.

    So \pi \displaystyle \int sin^{2}\theta \times sec^{2}\theta\ \ d\theta as you said.
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    (Original post by edothero)
    First you deal with the x

    So you would have

    0 < 2x < 720

    Then if you're adding/subtracting anything

    -60 < 2x-60 < 660

    Then once you have your values you do it in the opposite order

    so \dfrac{(x+60)}{2}
    Cheers mate. Is there a reason why it's done like that rather than the usual order when inside a bracket?


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    (Original post by thad33)
    Cheers mate. Is there a reason why it's done like that rather than the usual order when inside a bracket?


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    Don't quite get what you mean sorry, could you elaborate?
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    (Original post by edothero)
    Don't quite get what you mean sorry, could you elaborate?
    So sin(2x-60) for example is a transformation. So you would follow the rules of transformation inside the bracket. Why do you change the range of values the opposite way to that?


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    (Original post by edothero)
    Don't quite get what you mean sorry, could you elaborate?
    Wait, I think I got it. Is it because you need to get back to the original sin function in order to get the original values that you then need to change?


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    (Original post by thad33)
    So sin(2x-60) for example is a transformation. So you would follow the rules of transformation inside the bracket. Why do you change the range of values the opposite way to that?


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    Oh right, when you apply it to the ranges and do sin^{-1}=2x-60 you obtain values for 2x-60. Question asks for values of x, so you rearrange
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    (Original post by thad33)
    Wait, I think I got it. Is it because you need to get back to the original sin function in order to get the original values that you then need to change?


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    Take it like this, (sorry for double posting)

     -60 < 2x-60 < 660

    \therefore (2x-60) = sin^{-1}(1)

    (2x-60) = \theta_{1}\ ,\ \theta_{2}\ ,\ \cdots\ , \theta_{n}

    Where \theta is some value in the range
    Then you need to find the values of x.
    Therefore you would do

    x = \dfrac{(\theta_{1}+60)}{2}\ ,\  \dfrac{(\theta_{2}+60)}{2}\ ,\ \cdots\ ,  \dfrac{(\theta_{n}+60)}{2}

    Hence doing the opposite.
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    For those who dont know, Arsey's been banned from making model answers for edexcel, that means he wont be making them for c3 or c4, so brace yourselves for not being able to check your answers afterwards, unless someone else takes on the responsibility :adore:
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    (Original post by boyyo)
    For those who dont know, Arsey's been banned from making model answers for edexcel, that means he wont be making them for c3 or c4, so brace yourselves for not being able to check your answers afterwards, unless someone else takes on the responsibility :adore:
    Wow
    I guess this is good in some way


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    (Original post by Bloom77)
    Wow
    I guess this is good in some way


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    Darn it.
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    (Original post by Bloom77)
    Wow
    I guess this is good in some way


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    In what way is it good?
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    (Original post by Bloom77)
    Wow
    I guess this is good in some way


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    lol how??
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    (Original post by boyyo)
    unless someone else takes on the responsibility :adore:


    Won't be able to be as quick though.
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    (Original post by Zacken)


    Won't be able to be as quick though.
    Broo as long as its done!!! Honestly though I would appreciate it if you did
 
 
 
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