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    Would anyone who has the paper be willing to post the questions?
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    1/4 (e^2-1)

    correct!!!
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    how did u do threeee?
    p=128/289
    n yeash (e^2-1)/4 as well
    n the 'show' questions were pretty alrite as well..
    but 3 n 6 for me :s
    wat was the 'trick' for 3?
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    (Original post by hajira)
    hard paper
    the integrend in question 3
    This'll drive you insane but it collapsed to ∫sinxcosx dx with constants out front.
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    (Original post by Savioahang)
    i got 6/5 (Pi)a² for Q3
    and 128/289 for Q2b because Cosh2x = 8/17 , 17*17=289.
    the others were correct
    Sorry ye I was just going from memory !...
    2b) was either 128/289 or 128/259, you're all confusing me!
    3) was definately 6/5 (Pi)a²

    Keep em coming
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    Well, I thought the paper was difficult, too (not completely impossible, but v. tricky).

    It'd be great for someone to post the questions and then work out the answers by general consensus.

    I agree with things posted so far (6pia²/5 etc)...

    With question 8 c (tanh²x.... + sech x.....) did you get just one answer ? I did, something like ln (5/2 + √3/2) ; in that form, anyway...
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    (Original post by Daft Wader)
    This'll drive you insane but it collapsed to ∫sinxcosx dx with constants out front.
    :eek: nooooooo
    how?!
    it was all a mad rush of
    somethin like ∫ 3pi a² √[sin²x cos4x + cos²x sin4x] dx

    i think
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    i think it was ∫ 3pi a² √[sin^4x cos^2x + cos^4x sin²x] dx

    simplified down to cos^2(x)sin^2(x) or summin like that
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    √[sin²x cos4x + cos²x sin4x]
    =√[(sin²x cos²x)(cos²x + sin²x)]
    =√(sin²x cos²x)
    =sinxcosx

    i hope...
    well good luck with the UMS skewing everyone!
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    Q1 0.3 something
    Q2 128/289
    Q3 6/5 (Pi)a²
    Q4 (e^2-1)/4
    Q5 locus (25/16a,0) directrix x= - 25/16a
    Q6 proved
    Q7 radius of curvature was 2 root 3
    Q8 x= ln( (3+root 5)/2) and ln( (3-root 5)/2)

    Hope all correct....
    my written mistake for Q8 just now
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    (Original post by Yttrium)
    With question 8 c (tanh²x.... + sech x.....) did you get just one answer ? I did, something like ln (5/2 + √3/2) ; in that form, anyway...
    Yeah, I think that's what I got. Something like that.
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    sin^4x cos^2x + cos^4x sin²x
    = sin²xcos²x ( sin²x + cos²x )
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    (Original post by hajira)
    :eek: nooooooo
    how?!
    it was all a mad rush of
    somethin like ∫ 3pi a² √[sin²x cos4x + cos²x sin4x] dx

    i think
    It's along the lines of ∫ y 3pi a² √[sin²x cos4x + cos²x sin4x] dx
    The √[sin²x cos4x + cos²x sin4x] goes to √[sin²xcos²x(sin²x + cos²x) goes to √[sin²xcos²x = sinxcosx

    Multiply buy y = asin³ therefore ∫ sinxcosxsin³x = ∫ cosxsin4x = (1/5)sin5

    Hope that makes sense, I know its kinda hard to follow :tsr:
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    (Original post by Daft Wader)
    √[sin²x cos4x + cos²x sin4x]
    =√[(sin²x cos²x)(cos²x + sin²x)]
    =√(sin²x cos²x)
    =sinxcosx

    i hope...
    well good luck with the UMS skewing everyone!
    arghhhhhhh
    :mad: :mad: :mad:
    i feel quite blind now...was trying to get out a factor of (cos²x + sin²x)
    how many marks u think till that working out of 7?
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    (Original post by deity47)
    It's along the lines of ∫ y 3pi a² √[sin²x cos4x + cos²x sin4x] dx
    The √[sin²x cos4x + cos²x sin4x] goes to √[sin²xcos²x(sin²x + cos²x) goes to √[sin²xcos²x = sinxcosx

    Multiply buy y = asin³ therefore ∫ sinxcosxsin³x = ∫ cosxsin4x = (1/5)sin5

    Hope that makes sense, I know its kinda hard to follow :tsr:
    yeah the y was there alrite...but i never managed to spot that (cos²x + sin²x) :/
    thanx both of u!
    at least i can b at peace now..but thats probably a good 4-5marks gone!!
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    (Original post by Savioahang)
    Q1 0.3 something
    Q2 128/289
    Q3 6/5 (Pi)a²
    Q4 (e^2-1)/4
    Q5 locus (25/16a,0) directrix x= - 25/16a
    Q6 proved
    Q7 radius of curvature was 2 root 3
    Q8 x= ln (3+5^2)/2 and ln(3-5^2)/2

    Hope all correct....
    Agree with all of those, except almost (but didn't) prove q6, agree with q7 for sure, couldnt get q8 b) or c)
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    I didn't get that sin²x + cos²x factor in the exam but it came to me almost instantly just then. Looks like I really really suck under exam pressure (and the fact that to finish that paper you had to rush like nothing else).
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    it wasnt a hard paper (touch wood) but i mean it was all doable (bar qu 6) but it just took me a long time
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    (Original post by hihogyu)
    same. couldn't finish Q6, and just managed to finish Q8 part 1
    I did something similar: completed the integration in q 6, after splitting the integrand into 3 eventually, but ran out of time when applying the limits. I know I got Q3(?) - the exact fraction answer - wrong.

    I took ages to see q8 - I was probably running out of steam at this point.
    On the plus side, the reduction formula and the intrinsic coordinates question were not excessively difficult, and I've done some much harder conics questions recently.

    However, compared with the past Edexcel papers, and the Solomon P5s, all of which I've done at least once, I thought it was a quite tough, very time-consuming paper. I came close to finishing it, but partially because I had rehearsed some parts of the paper, like the reduction formulae, which I have regularly got stuck on.

    Aitch
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    (Original post by Savioahang)
    Q1 0.3 something
    Q2 128/289
    Q3 6/5 (Pi)a²
    Q4 (e^2-1)/4
    Q5 locus (25/16a,0) directrix x= - 25/16a
    Q6 proved
    Q7 radius of curvature was 2 root 3
    Q8 x= ln (3+5^2)/2 and ln(3-5^2)/2

    Hope all correct....
    i got a differn qu8 answer, i think mine had a log with a root in it
 
 
 
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