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    Question 1

    3x^2 - 2y^2 + 2x - 3y = 0
    6x - 4ydy/dx +2 -3dy/dx = 0
    4ydy/dx + 3dy/dx = 6x + 2
    dy/dx(4y+3) = 6x + 2
    dy/dx = (6x + 2)/(4y+3)


    dy/dx = 2/7
    Gradient of Normal = -7/2
    y-y1 = m(x-x1)
    y-1 = (-7/2)(x-0)
    y-1 = (-7/2)x
    2y - 2= -7x

    7x + 2y - 2 = 0


    y = (x-1)lnx

    a) Complete table:
    x = 1.5, y = 0.5ln1.5
    x = 2.5, y = 1.5ln2.5

    b) i 1.792
    ii 1.684

    c) Because the shape of the trapezium more closely follows the curve of the graph with an increased number of trapeziums

    d) Integrate between 1 and 3 (x-1)lnx dx
    One way is to expand the brackets:
    => xlnx - 1lnx

    Integrate each by parts:

    xlnx => u = lnx dv/dx = x v = (1/2)x^2 du/dx = 1/x
    => [(1/2)x^2][lnx] - (integral)[((1/2)x^2)(1/x)]
    => [(1/2)x^2][lnx] - (integral)[(1/2)x]
    => [(1/2)x^2][lnx] - (1/4)x^2

    1lnx => u = lnx dv/dx = 1 v = x du/dx = 1/x
    => xlnx - (integral)[1]
    => xlnx - x

    Combine both integration by parts:
    => [ [(1/2)x^2][lnx] - (1/4)x^2 - xlnx - x ] between 3 and 1
    => ((1/2)(9)ln3 - (1/4)(9) - 3ln3 - 3) - ((1/2)(ln1) - (1/4)(1) - 1ln1 - 1)
    => (9/2)ln3 - (9/4) - 3ln3 - 3 -(0 - (1/4) - 0 -1)
    => (9/2)ln3 - 3ln3 - (9/4) + (1/4) - 3 + 1
    => (27/6)ln3 - (18/6)ln3 -2 + 2
    => (9/6)ln3
    => (3/2)ln3


    a) Surface area, S=6x^2
    ds/dx = 12x

    dx/dt = ds/dt x dx/ds
    dx/dt = 8 x 1/12x
    dx/dt = 8/(12x) =2/(3x) Therefore k=2/3

    b) dv/dt = dv/dx x dx/dt
    V = x^3
    dv/dx = 3x^2
    dv/dt = 3x^2 x 2/(3x) (from part a)
    dv/dt = 2x
    As V = x^3
    x = V^(1/3)
    So: dv/dt = 2V^(1/3)

    c) t=0 V=8

    dv/dt = 2V^(1/3)
    integral 1/(V^(1/3)) dv = integral 2 dt
    integral V^-(1/3) dv = integral 2 dt
    So: (3/2)V^(2/3) = 2t + c

    Insert values of V and t, C = 6

    (3/2)V^(2/3) = 2t + 6

    V=16root2, t=unknown

    (3/2)(16root2)^(2/3) = 2t + 6
    2t = (3/2)(16root2)^(2/3) - 6
    2t = 6
    t = 3

    Anyone remember questions 2-5? Or want to provide solutions?

    (Original post by AhmadMujtaba)
    Here r some answers:

    The gradient for the normal was -7/2

    The binomial expanded as -1-x+4x^3 (A was -1.5 and B was 0.5)

    The value of the area of the graph (the one in the shape of a hump) was 12
    The volume was 9pi^2

    the ratio AP:PB was 2:3 or 3:2 (i dont remember correcctly)

    In the last question t was 3.

    (Original post by Smashingdude)
    well depends how u preceive. i found c3 way hard so thats why maybe i found this easier. anyway how many marks was the first question? i just wrote the wrong sign and ended up with y=x+1 .. the gradient came to -1 of the tangent. all screwed up cause one one damn sign..........
    how many marks will i loose?
    first question was 7 marks!

    (Original post by matmike)
    Brainstorm any answers you got guys:
    1. 2y - 2 + 7x = 0
    2. -1 - x + 4x^3

    3. Area = 12 units2
    Volume = 9 pi squared

    7c) t = 3

    Got same answers than you.. found this paper quite easy.. couldn't revise that much because of my Physics and i did better than in phy lol!!

    (Original post by Nasha_k2)
    Hi, I'm from Spain and I'm sitting Biology edexcel, maths edexcel, spanish cambridge and media AQA in a British school.
    I thought C3 was an evil paper but was quite pleased with C4 (comparatively!). Everyone in my class came out gloomy though and several people reckon if they got all they answered correct they can only get 35% maximum. I'm glad to hear everyone found C3 so hard though! Hopefully grade boundaries will be VERY VERY VERY low!

    que tal? yo soy español tambien y estudio en colegio britanico en las palmas.... pues en mi clase el examen lo encontramos bastante dificil tambien... aunq los que aspiramos a sacar A pues damos por hecho que mas de 80% sacamos... a mi no me salio nada mal... era dificila primera vista pero a la segunda pasada ya me salian las preguntas. De dnd eres tu?

    (Original post by bigbadb1319)
    first question was 7 marks!
    I thought I was off to such a good start when I powered my way through such an easy question worth a whole 7 marks! :p: Shame it went downhill from there.

    why was the area of the humpy thing 12 and not 6 - i dont remember. Was it cos(2pi) or cos(pi/2) for the limits? I dont remember, i mustve thought it was cos(pi/2) to get 6?

    the limits were 0 and 2pi.

    (Original post by AhmadMujtaba)
    the limits were 0 and 2pi.
    ok thats why i got 6, lol! how silly of me...! oh well... i used pi/2 to get 0 instead of -1 for cos 2pi - how many marks lost is that? It was 6 right, for the whole thing.

    i guess it was 6 marks.....

    (Original post by AhmadMujtaba)
    i guess it was 6 marks.....
    do you think all the marks would be lost for getting the limits mixed up?

    I think i got the limits right, it was just the last calc because i got the volume of revolution right.

    Also, is it just me or did you hardly use a calculator for that test? I'm getting really worred now.

    i better get some marks for methods, even though all my answers were wrong.
    horrible paper.
    c3 was a **** aswell.
    no A for me then...

    The last question was the one with the differential equations in relation to the cube. :hmmmm: I think you had to write everything out in terms of x (length of the sides), and use that in the calculations.
    Of course I figured that out a bit too late for it to be of any use :banghead:

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