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    Revise c3 and do more papers today.
    Is everyone ready?
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    does any1 have the makscheme for c3 paper L the soloman ? cant find it anywhere
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    Guys how do you work out the range or domain of a graph/equation?
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    I'm feeling pretty upbeat for C3! Although there's always lots of little things to trip you up, which is what I'm terrified of. Tis a resit for me so I HAVE to do well :eek:

    Is anybody else finding C4 a complete and utter nightmare? The thought of it makes me feel sick
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    c4 is all about integration T_T
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    how did u guys find the jan 2010 paper? was it hard?
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    it was ticky
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    I am screwed for this exam...if I lose my uni place because of Maths I'm going to bomb Edexcel, then kill myself!
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    (Original post by miss_maddy)
    I am screwed for this exam...if I lose my uni place because of Maths I'm going to bomb Edexcel, then kill myself!
    Count me in. I'll help you! :yep:
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    It looks like every chapter in the C3 book will be assessed. Algebriac fractions/long division are easy marks, as are Iteration Numerical method questions The trig identities are'nt given in the formuale book, so make sure you know them well and there is lots of differentation as well in most of the past papers...It won't be an easy paper, January is usually very similar to June, unless January was ridicuously hard, remember there are A* questions in C3! so it's expected to be a tough and challenging paper.
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    I did January's paper and scrapped an A. I remember there was a few annoying questions on log and lnx, which I completely screwed up on. But it did have some wonderfully nice drawing the graph questions xD, simple transformations. I don't think there was any major exponential questions and no finding the equation of tangents/normal questions came up in that paper so it may come up this time.

    Well most trig identities you can make from the ones given in the formula book i.e. sin2x.

    To my knowledge arcsin..etc questions are rare. There was one in January 2007 Q8ii).

    Probably a few Solomon papers to face those A* style questions.
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    Hi can any of you do this question for me please.. Im feeling a bit stupid =\
    ai) Using the substitution y= 2^x to express 8^x + 2^x+2 -2 = 0
    aii) Hence find an approximate solution to the equation 8^x + 2^x+2 -2 = 0, giving your answers to 2 sig fig.
    The answers are:
    ai) y^3 + 4y - 2 =0
    aii) -1.1

    I got part ai) but Im not gettin part 'aii' right? Can anyone help .. thanks
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    doing a c3 paper is actually quite easily after struggling so much with c4 :|
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    Would be ever so grateful if someone just could just explain the last step in this question to me, its taken from January 2006 Edexcel, and its question 4b:
    1/8
    Given that x = 4 sin (2y + 6), find dy/dx IN TERMS OF X. I differentiated to find dy.dx in terms of y easily enough which was 1/8cos(2y+6). However I have no idea how to put this into terms of x ? The answer given is

    1/2(16-x^2)^1/2. Ok its hard to read but if anyone could explain how to put it into terms of x after differentiating I'd be grateful!
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    Hint... you know that sine the angle is x over 4, just draw a little triangle to find out the cosine in terms of x, and there you go.
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    (Original post by game well and truly over)
    Hint... you know that sine the angle is x over 4, just draw a little triangle to find out the cosine in terms of x, and there you go.
    im stuck on this as well and dont understand your point.
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    First re-arrange equation to in terms of y:

    y = [ arcsin(x/4) - 6 ] / 2

    Then sub y into the dy/dx:

    dy/dx= 1 / 8cos(2y+6) =>

    = 1 / 8cos{2[arcsin(x/4)-6]/2 +6} times 2 and divide 2 => 1 / cancels out
    -6 and +6 => 0

    = 1 / 8cos[arcsin(x/4)]

    Now arcsin(x/4) is the same as sin feta = x/4 ... draw triangle with opposite as 4 and hypotenuse as x

    Find adjacent which is √4² - x² => √16 - x²

    Cos feta is √16 - x² / 4 (ajacent over hypothenuse)

    = 1 / 8cos[arcsin(x/4)] = 1 / 8 (√16 - x² / 4 )

    => 1 / 2 √16 - x²

    Sorry hard to explain! Hope it helps!
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    Easier way

    Get to your dy/dx

    Rearrange x=4sin(2y+6)
    x^2 = 16sin^2(2y+6)
    x^2 = 16 - 16cos^2(2y+6)
    16cos^2(2y+6) = 16 - x^2
    4cos(2y+6) = (16-x^2)^1/2
    8cos(2y+6) = 2(16-x^2)^1/2
    Sub in bottom of dy/dx
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    in the original equation rearrange to make 2y+6 the subject

    therefore sin inverse (x/4) also known as arcsin (x/4)

    then go to your dy/dx : 1 over 8 (cos 2y+6)

    replace the 2y+6 with arcsin(x/4) so its now 1 over 8cos (arcsin x/4) i think they should accept this ans
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    (Original post by houseo2)
    im stuck on this as well and dont understand your point.
    The sine of 2y+6 is x over 4. Doesnt matter what you call the angle, in this case, 2y plus 6. Call it theta or whatever. Draw a triangle whose opposite side is length x, with hypotenuse 4. Then the length of the adjacent side is the sq root of (16 minus x squared). The cosine is therefore root(16-x squared) over 4. The dreivative is therefore the inverse. See now? Much easier than all those complex explanations involving arcsin etc.
 
 
 
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