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    Hopefully A will be like 52 marks, hardest paper I've seen
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    What's everyone thinking for the cap?
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    (Original post by The Monk)
    Hopefully A will be like 52 marks, hardest paper I've seen
    Have you seen June 2012?
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    (Original post by dannyb2012)
    I got 135 but how did u get 315 it was interval 0 to 540 werent it ? i got 415
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    (Original post by Declan_Brennan)
    What's everyone thinking for the cap?
    No it was 136 as it was interval of 0 to 1Pi radian and 3x so 3Pi was a cos graph ?
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    Andyt, it wasn't 3*pi, we were working in degress so it was simply 180*3 !
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    (Original post by The H)
    Have you seen June 2012?
    Thought it was easier than this I ****ed up the integration, series and logarithms
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    (Original post by 394130)
    first term 80

    24=24p+q
    72=96p+q
    48=72p
    p=48/72=3/4

    24=2/3(24)+q
    24-16=q
    q=8

    I messed up the 12th term thing pfft... think this was a calc error. I type it in now and I come out with 159.96... oops

    arc length = r theta = 20*0.8=16cm
    sector area a r^2theta/2 (20)^2*0.8/2 = 160cm^2



    equation of the normal y-11=-4/3(x-4) (sounds like this might be wrong from other answers)

    Trapezium rule (8x^3+1)^1/2 between 0 and 2
    0 = 1
    0.5 = root 2
    1.0 = 3
    1.5 = 2 root 7
    2.0 = root 65

    0.25(28...) = 7.12 to 3sf (4)

    sqrt(8x^3+1) to sqrt (x^3+1) was (1/2)^3 before the x, so a stretch of scale factor 2 in the x-direction (2)

    Plot a graph of y=9^x, showing the intercept on the y-axis (1)
    Loga(b) = c. Find b in terms of c and a

    c^a = b

    9^x = 5 (I think?) and so log9(5) = x = 0.732 to 3sf (think my memory might deceive me on this one)



    y=9^x > y=9^-x reflection in y-axis


    2log2(x+7)-log2(x+5)=3
    log2(x+7)^2-log2(x+5)=3
    log2(x+7)^2/log2(x+5)=3log2(2)
    log2(x+7)^2/log2(x+5)=log2(8)
    (x+7)^2/(x+5)=8
    x^2+14x+59=8x+40
    x^2+6x+9=0
    (x+3)^2=0
    x= -3 (6)


    Plot a graph of tan(x) (3)
    tanx = -1 from 0 to 360 (2)
    Trig identity (3)
    Trig solutions: 16, 104, 136 (6)
    pretty sure it was a^c=b
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    (Original post by Gotzz)
    Are you sure this is right cus my maths teacher said no matter what you do the inverse of the positive value ???
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    (Original post by The Monk)
    Thought it was easier than this I ****ed up the integration, series and logarithms
    1st person i've known to say that
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    (Original post by The H)
    Have you seen June 2012?
    What was so hard about that? I got 71 marks in it.
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    Hi everyone. This is the official thread for the following exams

    Maths Pure Core 1 JUN13/MPC1
    Maths Pure Core 2 JUN13/MPC2


    Guys, if you want to make corrections or anything, please copy and paste the whole thing, and get the answers in the right order. I won't be online for now, most of you should be able to get most of this mark scheme done. Please make corrections |etc, and at the end, I will update my own post to reflect this.

    THANKS


    ******************************** ******************************** ******************************** ******************************** **********************
    Core 1


    question 2)a)ii) the question asked x√12 = 7√3 - √48
    x= 3/2

    For Q1, p was -3, k = 10, D = (-7, -4).
    For the surds question, x = 1.5 and for part b the answer they were looking for was 8 - (15)^0.5.
    The complete the square question was 2(x+1.5)^2 + 0.5 and the length of AB was 0.5(10)^0.5.
    Dividing the polynomial by x+3 gave x^2 - 3x + 5.
    The integral was 31.5 or 63/2 and the area of the shaded region was 27.
    The point furthest away from the x-axis was (5, -14) and the translation was by vector (6, -7).
    The last question was 20/7 ( < or =) k (< or =) 4.


    Equation of normal
    X=4 Y=20 normal -4/9 So Y-20=-4/9(x-4)

    3. (Circle question)
    (a) (x-5)^2 + (y+7)^2 = 49

    Radius was 7 so to draw the circle, you plot the middle point (5, -7) and go out 7 from the centre.

    This implies a shift to the right of 6 units in x and 7 down in y. This is represented by the vector [6, -7].

    Also, for question 5)b)ii), it asked you to use the coordinates that we got from part a)ii) (-3,-13) or something I think,
    and find a minimum value putting it into 1/2√n

    any done that one? And did you get 1/2√100? - DID ANYONE ELSE GET THIS BECAUSE I DID?

    ******************************** ******************************** ******************************** ******************************** *******************************

    Core 2

    Trapezium Rule Question - 7.12

    1.

    The first term of a geometric series in 80 and the common ratio is 0.5
    a) work out the third term in the series - 20
    b) work out the sum to infinity - 160
    c) work out the sum of the first 12 terms in the series to 2 decimal places - 159.96

    2. Work out Arc --> 20 x 0.8 = 16
    ii) work out area of sector ----> 0.5 x 20^2 x 0.8 = 160

    b) work out obtuse angle ABD

    Ad was 15 I think, and rad was 0.8

    Other length was 20

    Sin0.8/15 = SinD/20

    20sin0.8 = 15sinD

    SinD= 20sin0.8/15

    D= sin^-1 ( 20sin0.8/15) in rads

    D= 1.27 rads

    Since its obtuse, it's Pi -1.27

    D= 1.87

    3.
    a) (2+Y)^3 = 8 + 12y + 6y^2 + Y^3

    B) 16 +12x^-4

    (c) Integral = 39/2

    6.
    a. Write log(a)b = c in index form. - a^c = b
    b. Show that only one value of x satisfies 2log(2)(x+7) - log(2)(x-5) = 3

    log2(x+7)2 - log2(x+5) - 3 = 0

    →log2(x2+14x+49) = 3 + log2(x+5)
    →log2(x2+14x+49) = log223 + log2(x+5)
    →log2(x2+14x+49) = log2(8x+40)
    →x2+14x+49 = 8x+40
    →x2+6x+9 = 0
    →(x+3)2

    7.

    a) Show p=3/2

    72=p96+q
    24=p24+q

    48=p72

    48/72= 2/3

    b) Find u3

    72= 2/3*96 +q
    72=64 + q
    Q=8

    u3= 2/3*72 +q
    U3= 48 +q

    u3= 48 + 8
    = 56

    The transformation of [2 -0.7] was √((x-2)3 +1) -0.7

    2k question k = 6/5



    x = 16, 104 and 136 for the last one

    also.. tan x = -1
    solves to
    x = 135, 315
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    (Original post by The Monk)
    Hopefully A will be like 52 marks, hardest paper I've seen
    Agreed, my maths teacher also agrees!
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    (Original post by dannyb2012)
    Are you sure this is right cus my maths teacher said no matter what you do the inverse of the positive value ???
    Yeah your maths teacher is wrong, you can see where it crosses on the graph yourself.
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    (Original post by AunShah)
    My answer was, transformation (6,-7)
    Not sure if i got it right :-/
    I'm not sure if its the same for Core 1, but in Core 2 you have to specify TRANSLATION to get the mark rather than transformation. I could be wrong about Core 1 but in Core 2 you would most probably lose one mark for that.
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    Is there a chance C1 might be like 62 for A and 68-70 for 100 UMS, I literally need a 100 UMS to make up for C2 and get an overall A
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    Has anyone got the actual paper?
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    (Original post by Gotzz)
    Yeah your maths teacher is wrong, you can see where it crosses on the graph yourself.
    [email protected] Ill kill him lol this is discrimination ! haha
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    (Original post by Gotzz)
    What was so hard about that? I got 71 marks in it.
    It was 51 marks for an A, when it normally hovers around 61ish
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    In this question "The transformation of [2 -0.7] was √((x-2)3 +1) -0.7" I made a stupid mistake and put "-1" rather than "+1". How many marks will I lose and will I get follow through marks when working out G(4)?
 
 
 
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