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    (Original post by morgan8002)
    What sid you get for AB? Off the top of my head it was r = \dfrac{4}{3\sin \theta}.You set this equal to r = 1 + \cos 2\theta to get a cubic in \sin \theta, which you solve for \sin \theta. One value is less than -1, so can be discarded, one is associated with A and the other gives you the given result.
    Yeah did that just couldn't solve the cubic
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    (Original post by Lau14)
    In 6b I think you're missing the end of the first bracket (should be after Bsin(lnx/2) maybe?), and for question 7 part (a) was 5 marks and (b)(i) 4 marks from what I remember. Can't believe how fast you put this together and how much you can remember, thanks!
    Updated all that now, thanks!
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    (Original post by Stepidermis)
    Yeah did that just couldn't solve the cubic
    You had to notice that one of the roots will correspond to A, so will be \sin \theta = \frac{1}{2}. Then it's just a case of algebraic polynomial division to get the other quadratic factor. Solving that with the quadratic formula will give you the other answers.
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    This unofficial mark scheme is a great help, really reassuring to see so many of my answers up there, yay! Any chance that an unofficial question paper has been put together? Only on the parts I got wrong I can't quite seem to remember the questions, thanks (:
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    Wait you had to state the validity of the series? I completely missed that hahaha...
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    (Original post by morgan8002)
    What sid you get for AB? Off the top of my head it was r = \dfrac{4}{3\sin \theta}.
    You set this equal to r = 1 + \cos 2\theta to get a cubic in \sin \theta, which you solve for \sin \theta. One value is less than -1, so can be discarded, one is associated with A and the other gives you the given result.
    It should be r = 3/(4sin theta).
    Carrying on give me sin theta = 1/2 and (-1 +/- √13)/4. I substituted them back into the AB equation but its not the right answer.
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    I don't think there was a question on the validity of the series.
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    It was in the Ln[(1+2x)(1-2x)] question, had to state the range of validity for values of x
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    so no one else got AB=1.44 for the last question then...
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    I would suggest a slight refinement - set sin theta = 3/4r

    You then can change 1 + cos 2 theta into 2 - 2 (3/4r)^2

    Rearranging you get 8r^3 -16r^2 + 9 = 0

    The key point then is that you know r = 3/2 is a solution, so 2r - 3 is a factor. Use that to find the quadratic factor and then solve with the formula or cts.
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    (Original post by Lordvulk)
    It was in the Ln[(1+2x)(1-2x)] question, had to state the range of validity for values of x
    Oh Damn! How many marks was it?
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    (Original post by SilenceOrNoise)
    Oh Damn! How many marks was it?
    The validity will probably just be worth 1 mark, can't imagine it would be worth any more than that.
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    (Original post by eddie1221)
    The validity will probably just be worth 1 mark, can't imagine it would be worth any more than that.
    thanks.
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    (Original post by Criggers)
    I would suggest a slight refinement - set sin theta = 3/4r

    You then can change 1 + cos 2 theta into 2 - 2 (3/4r)^2

    Rearranging you get 8r^3 -16r^2 + 9 = 0

    The key point then is that you know r = 3/2 is a solution, so 2r - 3 is a factor. Use that to find the quadratic factor and then solve with the formula or cts.
    Cheers mate it works.I should have thought about that earlier.
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    (Original post by onlinekute17)
    It should be r = 3/(4sin theta).
    Carrying on give me sin theta = 1/2 and (-1 +/- √13)/4. I substituted them back into the AB equation but its not the right answer.
    I did say it was off the top of my head.

    So you got the right answer to II?
    I used the calculator to find the value of theta at B(the question said not exact value). I then used cosine rule to find the length of AB. There are probably a few ways to do it.
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    (Original post by morgan8002)
    I did say it was off the top of my head.

    So you got the right answer to II?
    I used the calculator to find the value of theta at B(the question said not exact value). I then used cosine rule to find the length of AB. There are probably a few ways to do it.
    It was the right answer but not in the right form so I think I may lose 1 mark for that.
    I spent too much time doing ii so I ran out of time for the last one (
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    Well this is worrying is think I only got 48/75 :eek:
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    I don't really remember 4b but I don't think that was my answer...what was the question again?
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    Anyone posted the questions yet?
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    There's some bits missing/not sure on and the wording isn't right but these are the questions as far as I know
    Spoiler:
    Show
    1. f(x,y) = (y2 + x)/x
    f(2) = 5
    (a) Use Euler with h = 0.05 to find an approximation for y(2.05) [2]
    (b) Use the formula yr+1 = yr-1 + 2hf(xr, yr) to obtain an approximation for y(2.1), giving your answer to three significant figures [3]

    2. Use an integrating factor to solve dy/dx + (tanx)y = tan3xsecx, given that y = 2 when x = π/4. [9]

    3. (a)(i) Write down the expansion of ln(1 + 2x) up to and including terms in x4 [1]
    (a)(ii) Find the first two non-zero terms of ln[(1 + 2x)(1 – 2x)] and write down the range of validity for this expansion [3]
    (b) Find the value of Limx>0 [(3x – x√(9 + x))/ln[(1 + 2x)(1 – 2x)]] [4]

    4. (a) Why is the integral (range 0 – infinity) of (x-2)e-2x improper? [1]
    (b) Find the value of the integral, showing the limiting process used [6]

    5. (a) Find the general solution of d2y/dx2 + 6dy/dx + 9 = (-36?)sin2x [7]
    (b)(i) Given that y = f(x) is a solution of the equation, y(0) = 0 and y’(0) = 0, show that y’’(0) = 0 [1]
    (b)(ii) Find the first two non-zero terms of the expansion of f(x) [3]

    6. x = et, y is a function of x.
    (a) Show that the substitution x = et turns the differential equation (?) into d2y/dt2 -2dy/dt + 2y = 4t2 + 5e-t [7]
    (b) Hence find the general solution of the equation (?) [10]

    7. (a) Find the area of r = 1 + cos2Θ, range –π/2 ≤ Θ ≤ π/2 [5]
    (b)(i) The curve 1 + sinΘ intersects the curve 1 + cos2Θ at the origin and point A. Find the polar coordinates of A [4]
    (b)(ii) A horizontal line through A intersects the curve 1 + cos2Θ again at B. Show that the length of OB is (√13 + 1)/4 [6]
    (b)(iii) Find the length of AB [3]
 
 
 
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