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    (Original post by niceguy95)
    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
    Hi,

    Could you copy out the new format, and add to that, and also put your additions in bold so I know what's new.

    Thanks!
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    I was one of the unfortunate ones who had to endure the June 2012 paper too. I got an E ...any maths paper in comparison to that one is easier. I wouldn't say this paper was easy, I flied through questions 1-5 then messed up the differentiation, log, series and trig questions so there go a ton of marks but I'm hoping to get a lot of follow through marks..fingers crossed!
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    (Original post by The H)
    It was 51 marks for an A, when it normally hovers around 61ish
    Oh wow.
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    (Original post by Gotzz)
    pretty sure it was a^c=b
    Quite possibly. It's all faded into the mists of my memory until the grades come out
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    (Original post by Dedicatedstudent)
    I was one of the unfortunate ones who had to endure the June 2012 paper too. I got an E ...any maths paper in comparison to that one is easier. I wouldn't say this paper was easy, I flied through questions 1-5 then messed up the differentiation, log, series and trig questions so there go a ton of marks but I'm hoping to get a lot of follow through marks..fingers crossed!
    I was aswell! These year12s don't know the true meaning of aqa torture hahaha. So yeah this paper.was alot easier! A lot of people from my year was resitting it

    Posted from TSR Mobile
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    (Original post by The H)
    Have you seen June 2012?
    I had to sit the June 12 one, got a D. Todays was honestly so much easier!
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    Core 1

    Coordinate Geometry Q1:
    p = -3 (2)
    k = 10 (1)
    m(grad) = 3/4 (1) OE i.e. accept 0.75 or 6/8 etc.
    Co-Ordinates D = (-7, -4) (4) M1 Normal m1xm2=-1, M1 grad=-4/3, M1 y=-4/3x+c OE, A1 -7, -4

    Surds (Q2):
    √48 = √16 x √3 = 4√3 (n=4) (1) CAO
    Solve x√12 = 7√3 - √48 --> 3√3/√12 --> x= 3/2 (3)


    Completing the Square (Q-):
    2(x+1.5)2 + 0.5 (2) A1 2(x+1.5)2 OE, A1 +0.5 OE
    min value = 0.5 (1) CAO
    Length AB min = 0.5(10)0.5. (?) (2)

    Polynomial (Q-)
    Show (x+3) is a factor --> f(-3) = 0 --> Is a factor (2) M1 f(-3) E1 Therefore is a factor
    (x+3)(x2 - 3x + 5) (2)

    Integration (Q-)
    Integral 31.5 or 63/2 (4)
    Shaded Region Area 27 (3) M1 (Area of Trapezium = 58.5), M1 58.5-31.5, A1 27

    Circles (Q-)
    (x-5)2 + (y+7)2 = 49 (3) A1 (x-5)2, A1 (x+7)2, A1 49 (accept 72)
    Centre: (5, -7) (2) A1 5, A1 -7
    r = √49 = 7 (1) CAO
    The point furthest away from the x-axis (5, -14) (2) A1 5, A1 -14
    Translation by the vector (6, -7). (3) B1 Translation, A1 6, A1 -7

    Inequalities (Last Question)
    Prove b2-4ac >= 0 and switch signs (4)
    The last question was 20/7 ( <=) k (<=) 4. (4) B2 20/7<= k NMS, B2 k <= 4 NMS



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

    Core 2

    Geometric Series (Q1):
    U3 = 20 (1) CAO
    Sum to Infinity = 160 (2) M1 a/1-r seen or used, A1 160
    Sum of first 12 terms = 159.96 (2dp) (2) M1 Use of formula, A1 159.96

    Circles and Radians: (Q-)
    S=r(theta) = 16cm (1) CAO
    A=1/2 x r2 x theta = 160cm2 (2) CAO
    Obtuse Angle = 1.87 radians (3) M1 Sine Rule M1 1.22r A1 1.87r

    Integration (Q3):
    Integral using Trapezium Rule: 7.12 (4)

    (Unknown Q)
    Integral = 39/2

    Logs (Q6):
    log(a)b = c in index form. - ac = b (1)
    Show that only one value of x satisfies 2log(2)(x+7) - log(2)(x-5) = 3
    (x+3)2 (6) M1 1 law of logs used correctly, M1 second law of logs used correctly, M1 x2+14x+49 M1 x2+6x+9, A1 (x+3)2 E1 Only one root?

    Arithmetic Series: (Q-)
    Show p=3/2
    72=96p+q
    24=24p+q
    48=72p
    48/72= 2/3 --- (Or simultaneous equations solving q then p) (4)
    U3 = 56 (1) CAO

    Transformations (Q-):
    Stretch SF 2 x-direction (2) B1 Stretch, A1 SF 2 X-Direction
    The transformation of [2 -0.7] was √((x-2)3 +1) -0.7
    g(4) = 2.33 (3)
    k = 6/5


    x-2 and x-6 cancel - left with 16 + 12x-4. for binomial question

    Trig (Last Question)
    tan x = -1 --> x = 135, 315 (2) A1 135, A1 315
    Show identities equal one another (3)
    x = 16, 104 and 136 (6) M1 (3cosx-2), M1 cos x=2/3, M1 3x = 48, A3 16, 104, 136

    Marks are a guide.. I have no clue, but it is an indication
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    (Original post by Hart1995)
    I was aswell! These year12s don't know the true meaning of aqa torture hahaha. So yeah this paper.was alot easier! A lot of people from my year was resitting it

    Posted from TSR Mobile
    Hahaha I agree with you there! I retook core 2 again in January but I had some circumstances so I couldn't revise for it properly and ended up with another awful grade. So this is third time lucky! But I know I messed up a lot of the big questions :/ ...how do you think you did?
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    you people complain too much... i took june12 core2 paper and got a B
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    Right, made an account for this.

    I used a new calculator for this exam and you can check back through the previous 99 calculations (I cleared it for the exam obviously) and these are my answers, workings and what I think the question was (the question part references are most likely incorrect):

    1. The first term of a geometric series is 80 and the common ratio is 0.5.
    (a) Find the value of u3.
    A: un=ar^(n-1)
    u3=80x0.5^(3-1)=20

    (b) Find the sum to infinity of this series.
    A: s=a/(1-r)=80/(1-0.5)=160

    (c) Find the sum of the first 12 terms of this series, giving your answer to two decimal places.(??)
    A: sn=(a(1-r^n))/(1-r)
    s12=(80(1-0.5^12))/(1-0.5)=159.96

    2. Sector AOB has radius 20cm and angle AOB=0.8 radians.
    (a) Find the arc length of the sector.
    A: arc length=rθ=20cm x0.8=16cm

    (b)Find the area of the sector.
    A: area=0.5r^2
    θ=0.5x(20cm)^2x0.8=160cm^2
    (On my calculator I have 0.5x16^2x0.8=102.4, I'm guessing this is a mistake?)

    (c)OB meets a line from B at point D. Line BD is 15cm. Find the size of obtuse angle ODB to 3 significant figures.
    A: sinA/a=sinB/b
    =>sinA=asinB/b
    sinODB=20cmsin0.8/15cm=0.95647...
    α=1.27...
    ODB=
    π-α=π-1.27...=1.87

    3. (a)(i) Using binomial expansion, or otherwise, expand (2+x^-2)^3.
    A: 2^3+ncr(3,1) 2^2 x^-2+ncr(3,2) 2 (x^-2)^2 +(x^-2)^3
    =8+12x^-2+6x^-4+x^-6

    (ii) Hence show
    (2+x^-2)^3+(2-x^-2)^3 can be written in the form p+qx^-4.
    A:
    (2+x^-2)^3+(2-x^-2)^3
    =
    8+8+12x^-2-12x^-2+6x^-4+6x^-4+x^-6-x^-6
    =16+12x^-4

    (b)(i) Integrate
    (2+x^-2)^3+(2-x^-2)^3 dx.
    A:
    (2+x^-2)^3+(2-x^-2)^3 dx
    =
    ∫ (16+12x^-4) dx
    =16x+(12x^-3)/-3+c
    =16x-4x^-3+c

    (b)(ii) Hence find the value of
    ∫2:1 (2+x^-2)^3+(2-x^-2)^3 dx.
    A:
    ∫2:1 (2+x^-2)^3+(2-x^-2)^3 dx
    =[16x-4x^-3]2:1
    =32-4/8-(16-4)
    =19.5

    4. [Can't remember the exact order of this question. I've only got a log calculation in my calculator]
    (a) Sketch the graph of y=9^x, indicating the points where the curve crosses the co-ordinate axis.
    A: This question comes up every year, it's a exponential shaped graph with the point (0,1) clearly marked.

    (b) Using logarithms, find x to three decimal places given 15=9^x.
    A: 15=9^x
    => x=log9(15)=1.23

    (c) The curve y=9^x is reflected in the y axis to give y=f(x). Find an expression for f(x).
    A: f(x)=9^-x

    5.(a) Using the trapezium rule and 5 ordinates (4 strips) find an approximation for
    ∫2:0 √(8x^3+1) dx, giving your answer to 3 decimal places.
    A: h=0.5
    f(x)=√(8x^3+1)
    ∫2:0 √(8x^3+1) dx
    ≃0.5h(f(x0)+f(xn)+2(f(x1)+f(x2 )+...f(xn-1)))
    =
    0.5x0.5(1+√65+2(√2+3+2√7))
    =7.12

    (b) Describe fully the transformation that maps the curve y=
    √(8x^3+1) to y=√(x^3+1).
    A: Stretch in the x direction, scale factor 2.

    (c)
    y=√(x^3+1) is translated via vector [2, -0.7] to give curve g(x). Find g(4).
    A: g(x)=
    √((x-2)^3+1)-0.7
    g(4)=
    √((4-2)^3+1)-0.7=2.3

    ...more coming once i decipher my remaining calculations...
    By the way, these are my calculations and answers, please validate them before taking them as the right answers.
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    (Original post by dannyb2012)
    U know for the 9th question of tanx=-1 my teacher said do the inverse of the positive value only so i did and got 45degrees so i got 180 - 45 and 360 + 45, X = 135, 405 Anyone else remember this and then for the very last one i got summit like 14degrees and cant remember the other 2 values U had to factorise i got summit like (3cos(theta)-2) (2cos(theta)+3) Anyone else confirm or better ?
    You had to find the inverse tan of -1 which was -45, but that wasn't in the range so you had to add 180 to that to get 135, and add 180 to that to get 315, which were the two answers I got.
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    (Original post by Dzintra)
    I had to sit the June 12 one, got a D. Todays was honestly so much easier!
    Oh yeah
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    (Original post by solosdk)
    Hi,

    Could you copy out the new format, and add to that, and also put your additions in bold so I know what's new.

    Thanks!
    I Believe the geometric series bit to be wrong if the common ratio is 0.5 then the sum of the 12 terms is 80(1-0.5^12) / 1-0.5 = 79.48 ???? the first bit is right but not sure about this last bit you state as 2 answers surely wont be the same as if it 159.96 then it would be 80(1-0.5^12) / 0.5 only and forget about the 1 ??
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    What was the actual question for the 2^k?
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    Dannyb, what I have is correct, and you must be wrong, it isn't 79.48.

    Also you could easily work 3rd term out by doing 80 x 0.5 x 0.5
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    Well.. I lost 2 or 3 on the prove the identity question
    lost 4 or so on the log question haha
    didn't get the 2^k one...
    nor did I get the translations fully...

    but all in all, hopefully only lost 12 or 13... :P
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    I know I lost 1 mark for saying scale factor of 1/2 instead of 2, and up to 2 marks on the last question as I only got two solutions. Hopefully that's all though!
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    Carry on get the answers coming guys I need them!
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    Does anyone remember how much the transformation question was worth all together. Question 5 that is. I got g(4) as 7. 34 and was wondering how many marks I would lose.
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    For the polynomial on core one...
    there was a question of prove only one real root

    so.. b^2 - 4ac < 0
    (-3)^2 - 4 x 1 x5 = 9 - 20
    = -11
    which is < 0
    therefore, no real roots for the quadratic
    only x = -3 is a root
 
 
 
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