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C4 in January? Post here watch

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    I thought we could find out which tsrians are doing c4 and then create a revision thread soon.
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    i did it last june.
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    (Original post by Undry1)
    i did it last june.
    Thanks for that. :rolleyes:
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    I am doing it in June which is kind of a waste since I've finished sylabuss and am at my peak now ,6months down line probably would have forgotton everything. Dont mind revising with you though name a topic
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    (Original post by Malik)
    I am doing it in June which is kind of a waste since I've finished sylabuss and am at my peak now ,6months down line probably would have forgotton everything. Dont mind revising with you though name a topic
    vectors.
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    (Original post by Undry1)
    vectors.
    A vector is a quantity that has both magnitude and direction :cool: :rolleyes:
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    full marks!!
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    I saw this in a P3 paper, or something similar.

    Split into partial fractions.

    4x+5/(1+x²)(1-x)(2+4x)

    Never seen one like this before. Is it just the same method?
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    4x+5/(1+x²)(1-x)(2+4x) = A/(1-x) + B/(2+4x) + C/(1+x²)
    4x + 5 = 2A(1+2x)(1+x²) + B(1-x)(1+x²) + 2C(1-x)(1+2x)

    now let x=1,

    9 = 2A(3)(2)
    A = 9/12 = 3/4

    let x=-1/2

    3 = B(3/2)(5/4)
    B = 8/5

    let x=0

    5 = 2A(1)(1) + B(1)(1) + 2C(1)(1)
    5 = 2A + B + 2C
    C = (5 - 2A - B)/2
    C = 19/20

    So 4x+5/(1+x²)(1-x)(2+4x) = 3/ 4(1-x) + 4/ 5(1+2x) + 19/ 20(1+x²)

    I hope..
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    (Original post by john!!)
    4x+5/(1+x²)(1-x)(2+4x) = A/(1-x) + B/(2+4x) + C/(1+x²)
    4x + 5 = 2A(1+2x)(1+x²) + B(1-x)(1+x²) + 2C(1-x)(1+2x)

    now let x=1,

    9 = 2A(3)(2)
    A = 9/12 = 3/4

    let x=-1/2

    3 = B(3/2)(5/4)
    B = 8/5

    let x=0

    5 = 2A(1)(1) + B(1)(1) + 2C(1)(1)
    5 = 2A + B + 2C
    C = (5 - 2A - B)/2
    C = 19/20

    So 4x+5/(1+x²)(1-x)(2+4x) = 3/ 4(1-x) + 4/ 5(1+2x) + 19/ 20(1+x²)

    I hope..
    Why 2A and 2C?
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    (Original post by john!!)
    4x+5/(1+x²)(1-x)(2+4x) = A/(1-x) + B/(2+4x) + C/(1+x²)
    4x + 5 = 2A(1+2x)(1+x²) + B(1-x)(1+x²) + 2C(1-x)(1+2x)

    now let x=1,

    9 = 2A(3)(2)
    A = 9/12 = 3/4

    let x=-1/2

    3 = B(3/2)(5/4)
    B = 8/5

    let x=0

    5 = 2A(1)(1) + B(1)(1) + 2C(1)(1)
    5 = 2A + B + 2C
    C = (5 - 2A - B)/2
    C = 19/20

    So 4x+5/(1+x²)(1-x)(2+4x) = 3/ 4(1-x) + 4/ 5(1+2x) + 19/ 20(1+x²)

    I hope..
    Ah okay. my teacher said there is a completely different method for these ones. :confused:
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    maybe there is, because what I just wrote down is completely wrong.
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    (Original post by Widowmaker)
    I saw this in a P3 paper, or something similar.

    Split into partial fractions.

    4x+5/(1+x²)(1-x)(2+4x)

    Never seen one like this before. Is it just the same method?
    (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] ≡ A/(1 + x²) + B/(1 - x) + C/(2 + 4x)

    4x + 5 ≡ A(1 - x)(2 + 4x) + B(1 + x²)(2 + 4x) + C(1 + x²)(1 - x)

    x = 1 gives:

    9 = B(2)(6)

    12B = 9

    B = 3/4

    x = -1/2 gives:

    3 = C(5/4)(3/2)

    15C/8 = 3

    C = 8/5

    x = i gives:

    4i + 5 = A(1 - i)(2 + 4i)

    4i + 5 = A(6 + 2i)

    A = (4i + 5)/(2i + 6)

    Therefore:

    (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] = (4i + 5)/[(2i + 6)(1 + x²)] + 3/[4(1 - x)] + 8/[10(1 + 2x)]

    I'm not quite sure how the imaginary bits got in there :p: ! Not done any of these questions before, thought I'd give it a stab for the hell of it :p: !
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    (Original post by Widowmaker)
    I saw this in a P3 paper, or something similar.

    Split into partial fractions.

    4x+5/(1+x²)(1-x)(2+4x)

    Never seen one like this before. Is it just the same method?
    (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] ≡ A/(1 + x²) + B/(1 - x) + C/(2 + 4x)

    4x + 5 ≡ A(1 - x)(2 + 4x) + B(1 + x²)(2 + 4x) + C(1 + x²)(1 - x)

    x = 1 gives:

    9 = B(2)(6)

    12B = 9

    B = 3/4

    x = -1/2 gives:

    3 = C(5/4)(3/2)

    15C/8 = 3

    C = 8/5

    x = 2 gives:

    13 = A(-1)(10) + 0.75(5)(10) + 1.6(5)(-1)

    13 = -10A + 37.5 - 8

    10A = 16.5

    A = 1.65 = 33/20

    Therefore:

    (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] = 33/[20(1 + x²)] + 3/[4(1 - x)] + 8/[5(2 + 4x)]

    Hehe, I hope this is right ! Not done any questions like this before, so might be wrong sorry ...

    Doing my C4 exam in June, sorry I can't join you in January !
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    Isn't the general form Ax + B for the numerator of a quadratic factor?

    i.e, (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] ≡ (Ax+B)/(1 + x²) + C/(1 - x) + D/(2 + 4x)

    ...and then proceed as usual.
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    (Original post by mockel)
    Isn't the general form Ax + B for the numerator of a quadratic factor?

    i.e, (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] ≡ (Ax+B)/(1 + x²) + C/(1 - x) + D/(2 + 4x)

    ...and then proceed as usual.
    I thought as much. I think you are right
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    (Original post by mockel)
    Isn't the general form Ax + B for the numerator of a quadratic factor?

    i.e, (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] ≡ (Ax+B)/(1 + x²) + C/(1 - x) + D/(2 + 4x)

    ...and then proceed as usual.
    It could well be ... I'll leave it for someone else to have fun with ...
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    Partial fractions are NOT my friend. Whenever I'm doing any kind of integration question, I try to avoid partial fractions at all cost. They're just uncool.
    So I'll leave it for someone else to do as well... :p:
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    im taking c1 to c4 in jan
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    (Original post by john!!)
    So 4x+5/(1+x²)(1-x)(2+4x) = 3/ 4(1-x) + 4/ 5(1+2x) + 19/ 20(1+x²)

    I hope..
    Incorrect.

    --------------

    (Original post by Simba)
    (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] = 33/[20(1 + x²)] + 3/[4(1 - x)] + 8/[5(2 + 4x)]
    Correct.
 
 
 
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