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AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme] Watch

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    (Original post by datpr0)
    1. the marks for 1b were 6 not 5,
    2. the differential equation solution was x = 1/5[(-1/2(1+t)^-1 + 5/2)^2 - 4],
    3. the constant in the penultimate question was a not k, as stated in the question.
    Have changed marks but I'm fairly sure it was k?


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    (Original post by Jeff97)
    i got -10/root56 root14 from doing a.b=mod a mod b costheta
    Yeah, no; the scalar product came to 14 and mod a x mod b came to 2sqrt14 x sqrt 14 which equals 28.
    => cos(theta) = 14/18
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    (Original post by Alexandrite)
    On a slightly related note how harsh are examiners if you forget to mark the question reference but your work/answer is clearly for that question?

    I'm pretty sure I didn't make that mistake but at the same time I always have a nagging feeling maybe I did in one of my exams.
    They aren't bothered really; all that they are looking for is the answers. I forgot to put references in the margin many times last year, don't worry about it.
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    Grade Boundaries predictions?

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    (Original post by CD223)
    Have changed marks but I'm fairly sure it was k?


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    Hmm.
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    Can anyone show me the detailed solution of question 5? That one with sinq, not sure about how many marks i will lose
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    time to put c3/c4 book on ebay!

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    How many points will I lose for giving in 6a) the value of the angle in radians???

    I found cos=1/2 but then calculated angle in radians :/
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    guys i got x as -1/8 but i didnt use sine, i did it in quadratic form with x.... how many marks will i lose? (the sinp Q or whatever)
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    (Original post by datpr0)
    (let theta equal the acute angle)

    Therefore, it came from cos(theta) = 14/28
    => cos(theta) = 1/2
    => theta = 60 degrees
    Cos theta = 10/28 which is 69.1 :/
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    (Original post by majka)
    How many points will I lose for giving in 6a) the value of the angle in radians???

    I found cos=1/2 but then calculated angle in radians :/
    I doubt you'd lose any marks for it, the mark scheme would likely say "or/condone [insert the angle in radiains]".
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    My predictions:
    60 A*
    55 A
    50 B
    etc etc
    because it wasn't SOLID like we expected, but it was challenging and maybe a bit more difficult than last years but were supposedly getting 'smarter' so I think an increase of 1 isn't too unfair
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    Can anybody post predicted grade boundaries.
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    (Original post by Jed-Singh)
    My predictions:
    60 A*
    55 A
    50 B
    etc etc
    because it wasn't SOLID like we expected, but it was challenging and maybe a bit more difficult than last years but were supposedly getting 'smarter' so I think an increase of 1 isn't too unfair
    second this.

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    I think there by quite high the paper was quite easy to honest
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    (Original post by HennersPD)
    Cos theta = 10/28 which is 69.1 :/
    ... I'm still pretty sure the scalar product came to 14.
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    (Original post by Jed-Singh)
    My predictions:
    60 A*
    55 A
    50 B
    etc etc
    because it wasn't SOLID like we expected, but it was challenging and maybe a bit more difficult than last years but were supposedly getting 'smarter' so I think an increase of 1 isn't too unfair
    60 for an A* will be lovely <3 I do think that there will be less separation from the A* boundary to the C grade boundary due to the nature of this paper.
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    65 for an A*
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    (Original post by CD223)
    *** OFFICIAL AQA CORE 4 MPC4 JUNE 2015 EXAM DISCUSSION THREAD ***
    UNOFFICIAL MARK SCHEME
    1.
    A) Express f(x) = (19x-2)/(5-x)(1+6x) in the form:
    f(x) = A/(5-x) + B/(1+6x):

    A=3 B=-1 (3 marks)

    B) Find Integral between x = 4 and x = 0 of f(x). Show it equals pln5.

    = 8/3ln5 (6 marks)

    2.
    A) Express 2cosx - 5sinx in the form Rcos(x+alpha), R>0, 0<alpha<pi/2
    √29cos(x+1.19) (3 Marks)

    B) Find value of x in interval 0<x<2pi where 2cosx - 5sinx is at a maximum.
    (i). x=5.09 (2 Marks)
    (ii) Solve equation 2cosx - 5sinx + 1 = 0
    x=0.567, 3.34 (3 Marks)

    3.
    A) g(x) = 8x^3 - 12x^2 - 2x + d
    Given when g(x) is divided by (2x+1) the remainder is -2, find d.
    d=1 (2 Marks)

    B) Given x = -0.5 is a solution of g(x)=0, express g(x) as 3 linear factors.
    g(x) = (2x+1)(2x-1)(2x+3) (3 marks)

    C) h(x) is given by 4x^2 - 1 / f(x). Simplify h(x), showing it is a decreasing function.

    h(x) = 1/(2x+3), h'(x) = -2(2x+3)^-2 this is always negative, therefore it is a decreasing function (4 Marks)

    4.
    A) Find the binomial expansion of (1+5x)^(1.5):
    = 1 + x - 2(x^2) (2 Marks)
    B) Expand (8 + 3x)^(-2/3)
    (i). 1/4 - (1/16)x + (5/256)x^2 (3 Marks)
    (ii) Find an estimate for Cube root of 1/81.
    0.2313 (2 Marks)

    5. x = cos2t, y = sint
    A) Find the gradient at t = pi/6.
    dy/dx= -1/2 (3 Marks)

    B) Equation of normal at t=pi/6
    y=2x-1/2 (2 Marks)

    C) Normal cuts curve at Q(cos2q, sinq). Set up a quadratic in sinq to find where the normal cuts the curve again. Then find the x co-ordinate of Q.
    cos2q = -1/8 (5 Marks)

    6. A = (3,2,10), B = (5,-2, 4)
    Line l passes through A and has equation:
    r = [3,2,10] + lambda[x,y,z]

    A) Find angle between line l and AB.
    Theta = 60 degrees (4 Marks)
    B) C(15, 6, 2) (4 Marks)
    C) E(11, 0, 0) and (23, 3, 0) (4 Marks)

    7. y^3 + 2e^(-3x)y + x = k
    Find k given point P (ln2, 0.5) lies on curve.
    A) k=1/4 - ln2 (1 Mark)
    B). Find the gradient at P:
    dy/dx = 11/8 (6 Marks)

    8. dx/dt = (4+5x)^0.5/5(1+t)^2
    A) Given t=0 when x=0 find x in terms of t.
    x=9/20-1/2(t+1)-1/20(1+t)^2
    OR
    x = 1/5[(-1/2(1+t)^-1 + 5/2)^2 - 4]

    (7 marks)

    B) Another pond models a circle of radius r. Rate of change of radius is inversely proportional to pond surface area. Write a differential equation to model this situation.
    (i) dr/dt = k/(pir^2) (3 Marks)

    (ii) The rate of change of radius is 4.5 when the radius is 1. Find the radius when the rate is 0.5.
    r=3 metres (2 Marks)

    Date: 9th June 2015
    Time: 09:00am
    Duration: 1h 30m
    ______________________________
    CORE 3 EXAM DISCUSSION THREAD:
    http://www.thestudentroom.co.uk/showthread.php?t=3041463

    MECHANICS 2 EXAM DISCUSSION THREAD:
    http://www.thestudentroom.co.uk/showthread.php?t=3047357

    ______________________________

    RESOURCES:
    7a) i got k=1/4+ln(2)
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    (Original post by datpr0)
    ... I'm still pretty sure the scalar product came to 14.
    It did
 
 
 
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