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    (Original post by tsr1)
    c3- jan 2007 -question 8iia)

    given that: y= arcosx

    -1 < x ≤ 1 and 0 ≤ y ≤ pi

    express arcsin in terms of y

    can anyone help with this? really stuck.. thanks
    The exam solutions link will show you the answer, but I did it differently.

    arccos(x) = y
    so cos(y) = x

    now, assuming you know that cos(x) = sin(90-x) [and also that sin(x)=cos(90-x), though not relevant in this case] you know that:

    sin(90-y) = x

    and hence arccsin(x) = 90-y
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    (Original post by justinawe)
    Generally, the domain would be everything except for the value that makes the denominator equal 0, and the range would be anything other than 0.

    Of course, it can depend on additional information given in the question as well.


    Is there not an algebraic method?
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    (Original post by überambitious_ox)
    Find the exact value of x such that

    3arctan (x-2) + pi =0

    Please can someone explain how this would be solved. Thank you
    Name:  1371074324132.jpg
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Size:  26.1 KB

    Hope this is right... :ninja:

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    (Original post by überambitious_ox)
    Find the exact value of x such that

    3arctan (x-2) + pi =0

    Please can someone explain how this would be solved. Thank you
    I'll use degree instead of radians for convenience.

    3arctan(x-2) + 180 = 0
    3arctan(x-2) = -180
    acrtan(x-2) = -60
    x-2 = tan(-60)
    x-2 = -root3
    x = 2 - root3


    That's what I got anyway, not sure of the actual answer.

    Edit: Looks like I've been beaten to it
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    (Original post by MathsNerd1)
    It will be due but just remember they are reflections in line y=x so it's not too hard really
    Yer I think theyre ok, but not the nicest questions in the world, especially back in 2007! Apart fro that, theres nothing in the content that strikes me as something 'bad'.
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    (Original post by suncake)
    Name:  1371074324132.jpg
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Size:  26.1 KB

    Hope this is right... :ninja:

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    Thank you; yes it is
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    Apparently c3 was leaked and the replacement is way harder !!!


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    (Original post by Flounder1)
    Apparently c3 was leaked and the replacement is way harder !!!


    Posted from TSR Mobile
    Why would it be harder? Why not just a different paper?
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    (Original post by Flounder1)
    Apparently c3 was leaked and the replacement is way harder !!!


    Posted from TSR Mobile
    I call bs
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    (Original post by Flounder1)
    Apparently c3 was leaked and the replacement is way harder !!!


    Posted from TSR Mobile
    Source?
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    (Original post by ApplePearOrange5297)
    I call bs
    No, the papers were stolent/leaked/lost, it's on the edexcel website. It affected C3, C4 and some others. I don't see why they'd be harder, though. Just a different paper.

    Edit: https://fbcdn-sphotos-a-a.akamaihd.n...77688065_n.jpg
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    Can anyone help me with this quick problem?

    The question states

    f: ln(x+k)
    g: |2x-k|

    Evaluate fg(k/4)


    I got that fg = ln(2x-k+k) = ln2x = ln2k/4

    Simplifies to lnk/2



    HOWEVER, the markscheme says that the answer is ln3k/2?
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    Hows everyone feeling about tommorow ? :/ did anyone else make loads of silly mistakes in jan that made them miss the A by 1 mark :/ errh retakes
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    (Original post by Flounder1)
    Apparently c3 was leaked and the replacement is way harder !!!


    Posted from TSR Mobile
    Yeah i heard about that too. Just hope theyre nice to us. The c4 paper got leaked too
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    (Original post by physicso)
    Can anyone help me with this quick problem?

    The question states

    f: ln(x+k)
    g: |2x-k|

    Evaluate fg(k/4)


    I got that fg = ln(2x-k+k) = ln2x = ln2k/4

    Simplifies to lnk/2



    HOWEVER, the markscheme says that the answer is ln3k/2?
    What papers this from ?
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    (Original post by physicso)
    Can anyone help me with this quick problem?

    The question states

    f: ln(x+k)
    g: |2x-k|

    Evaluate fg(k/4)


    I got that fg = ln(2x-k+k) = ln2x = ln2k/4

    Simplifies to lnk/2



    HOWEVER, the markscheme says that the answer is ln3k/2?
    f(k/4) = |-k/2| = k/2 sub this into f(x) and BOSH!
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    (Original post by BearOfStown)
    Source?
    http://www.thestudentroom.co.uk/show....php?t=2362486

    And I'm guessing the source of that is the edexcel website.

    Yes, there's a replacement paper but this doesn't mean it will be harder. As edexcel have said, the replacement papers have gone through the same processes as the originals, and are no harder/easier than the original papers.

    So there really isn't anything to worry about.

    Edit:

    And if you don't believe edexcel, and still think they'll give a harder paper, the FP1 replacement paper was done on Monday, and most candidates found it (fairly) easy. So I doubt they'll screw you guys over

    Good luck!
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    (Original post by kingpro88)
    Hows everyone feeling about tommorow ? :/ did anyone else make loads of silly mistakes in jan that made them miss the A by 1 mark :/ errh retakes
    that is gutting! first time ive taken the paper and feeling confident
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    C3 is going to be so much Banter. I can't wait for it.
 
 
 
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