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    (Original post by 786)
    The last question where it asked you to show that AB was 3a(root3)/2. Did anyone manage this? The point of intersection was at theta=60, therefore r=3a(1 -cos(theta)) i.e 3a/2. Using simple trigonometry i get AB as twice the given answer. What did i do wrong?
    u can draw a triangle with vertices ABO and AB is 2rsin(theta) which gives the answer
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    (Original post by 786)
    The last question where it asked you to show that AB was 3a(root3)/2. Did anyone manage this? The point of intersection was at theta=60, therefore r=3a(1 -cos(theta)) i.e 3a/2. Using simple trigonometry i get AB as twice the given answer. What did i do wrong?
    AB = 2 times the y value at B (3a/2,Pi/3)
    y = r Sin (-)
    r=3a(1 - Cos(-))

    y = 3a(1 - Cos(-))Sin(-)

    Cos(Pi/3) = 0.5 and Sin(Pi/3) = Sqrt(3)/2

    =============

    y = 3a(1-0.5) X Sqrt(3)/2
    y = 3a/2 X Sqrt(3)/2
    y = 3aSqrt(3)/4

    Double y to get full distance from A to B.

    AB = 3aSqrt(3)/4 X 2
    AB = 3aSqrt(3)/2
    QED
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    A very pedantic question i know, but does anyone know if you lose a mark for omitting the cm^2 from an answer? (I did this for the last part of the last question)
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    (Original post by 786)
    Why is AB 2rsin(theta)?
    the y-coordinate of B is rsin(theta) and so AB is its double (A,B have the same x-coordinate)
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    (Original post by sam99)
    A very pedantic question i know, but does anyone know if you lose a mark for omitting the cm^2 from an answer? (I did this for the last part of the last question)
    yup
    a is actually root3 cm

    so the area has the unit of cm^2
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    (Original post by keisiuho)
    yup
    a is actually root3 cm

    so the area has the unit of cm^2
    I know it's cm^2, but will i lose a mark for not putting it?
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    These answers may not be entirely accurate, as i did p6 after p4 and have managed to forget half my answers to the first paper.

    Q1
    a) prove that SUM (r+1)(r+5) = 1/6n(2n+7)(n+7)
    b) sum it from r=10 to n=40

    I just used the standard results as it was only worth 3 or 4 marks.
    SUM (r+1)(r+5) = SUM r^2 + 6 SUM r + 5 SUM 1
    Then just put in the standard results, multiply out, and then refactorise to get the above result.

    Q2
    a) use linear interpolation on {1, 2}
    b) use newton raphson with a=1

    Just used the standard results for linear interpolation, and newton raphson is in the formula booklet.

    Q3
    a) so ab = 15
    b) find roots of z^2 + 6z -30i (not sure bout middle term)

    Substitute z = a + ib into the equation and multiply out, then equate coefficients of real and imaginary parts.

    Q4
    a) find solutions sets of MOD (x^2 - 2) = 2x

    X < 0.74 and X > 2.74 (something like that anyway, not sure about after the decimal point)

    Q5
    a) find w
    b) find arg w
    c) find mod w
    d) plot an argand diagram
    e) find distance AB in surd form

    Cant remember

    Q6
    a) find general solution
    b) find co-ordinates of minimum value
    c) plot a graph

    P.I. = e^2x (I think)

    Q7
    a) Find general solution
    b) find particular solution

    y = e^-t(2Sint - Cost + 2)

    Q8
    a) Find points of intersection
    b) Show the line = 3root3a/2
    c)Area of shaded region
    d)Area of badge given a=4.5cm

    See previous post.
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    (Original post by sam99)
    I know it's cm^2, but will i lose a mark for not putting it?
    I really hope not, cos i cant remember if i put it or not!
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    q7 was y = e-t (2sint-cost) +2e-t
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    i thought it went quite well.

    should be an A at least.
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    (Original post by rockindemon)
    q7 was y = e-t (2sint-cost) +2e-t
    i got 4sint ... forgot to divide by 2 !!! ...
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    hhmm, i got e-t (-2sint-cost) +2e-t, i probably forgot to get rid of a minus somewhere along the lines, how many marks you think i lose?
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    (Original post by mathematician)
    i personally thought it went very well.
    but it seems that everyone and i mean everyone was short of time and they all rekon they didnt do that great.
    I thought it went really well, had 15 mins to spare after checking it through. Definitely better than P5.
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    (Original post by Hovick)
    hhmm, i got e-t (-2sint-cost) +2e-t, i probably forgot to get rid of a minus somewhere along the lines, how many marks you think i lose?
    most likely you'll loose 1 mark only but if their being picky 2 for the final wrong answer
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    (Original post by rockindemon)
    q7 was y = e-t (2sint-cost) +2e-t
    You could simplify that to: y = e^-t(2Sint - Cost + 2)
    • Thread Starter
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    (Original post by rockindemon)
    most likely you'll loose 1 mark only but if their being picky 2 for the final wrong answer
    ACED!
    no prob.
    u know for the differential of
    a^x did everyone remember the standard differential of a^xlna, for the NR question
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    (Original post by mathematician)
    ACED!
    no prob.
    u know for the differential of
    a^x did everyone remember the standard differential of a^xlna, for the NR question
    was the area of the region something like
    -4pi + 8 or 9 root 3
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    think it was 9root 3 - pi
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    (Original post by [_Z_])
    think it was 9root 3 - pi
    nah it werent, it cant of been,
    your tellin me the area of the badge is 37.34.
    length AB was 4.5, no way is it 37.34
    my answer (9root3 - 4pi), gives an area of 9cm^2
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    yeh i concur
 
 
 
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