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AQA Core 4 - Monday 10th June 2013 (AM) - Official Thread Watch

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    how are your ums calculated? can you still get full ums but not full marks?
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    (Original post by Samtaylor47)
    Anyone got an Predictions on grade boundaries?
    Similar to Jan I'd say, bare in mind that Jan was mostly A2 resits and Further Maths, so June today would be a portion of resits and the usual cohort. So if the Jan one got low boundaries with the 'better candidates' <-- I only say that because of A2 resits and further then this one with the normal cohort should theoretically be lower.

    But I'm not sure if this one is harder because we actually sat this one for proper and not the Jan so perhaps Jan was harder. I did well in the Jan as a mock and thought this was a lot harder.

    I'd say 65 for an A* though, then decrease in 4-5 mark increments.
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    (Original post by Samtaylor47)
    Yes!
    Might just get myself a B if that was true
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    (Original post by Samtaylor47)
    Anyone got an Predictions on grade boundaries?
    im hoping they are similar to the grade boundaries on the jan paper!
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    (Original post by Kev.1995)
    When you form your own vector eqn of a line, I used alpha instead of meu, is that fine?
    Thats fine Kevin, its just a constant
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    (Original post by fizzbizz)
    UNOFFICIAL MARK SCHEME

    1a) Partial fractions: A = 3, B = 1

    1b) Integral =  \frac {11}{3} ln2

    1c) C = 2

    1d)  \frac {11}{3} ln2 -2 EDIT: I may be wrong on this, since the integral gives  [2x]^{0}_{-1} + \frac {11}{3} ln2 which gives  \frac {11}{3} ln2+2

    2a) Show  sin \alpha = \frac{2}{3} and hence find  cos \alpha = \frac {\sqrt{5}}{3}

    2b)  sin 2\alpha = \frac{4\sqrt{5}}{9}

    2c) \frac{2}{15}(5-\sqrt{5})

    3a) Binomial expansion:  1 -2x +8x^2

    3b)  (27-6x)^{-\frac{1}{3}} = [27(1-\frac{6}{27}x)]^{-\frac{1}{3}} = \frac{1}{3}[1-2(\frac{x}{27}) + 8(\frac{1}{27}x)^{2}] = \frac{1}{3} - \frac{2}{81}x + \frac{8}{2187}x^2

    3c)  \frac{2}{\sqrt[3]{28}} = 2[27+6(\frac{1}{6})]^{-\frac{1}{3}} which gave  \sqrt[3]{\frac{2}{7}} = 0.659... (cannot remember to 6 d.p.)

    4a) Show (2x + 3) is a factor of f(x)

    4b)  f(x) = (2x+3)(2x^2 -3x -1)

    4c) Show that function with  sin(\theta) and cos(2\theta) reduces to f(x) where  x = sin \theta

    4d) Solve for theta:  (2x+3)(2x^2 -3x -1) = 0

     sin\theta = \frac{-3}{2} has no solutions

    2x^2 -3x -1 = 0

     x = sin\theta = \frac {3 \pm\sqrt{17}}{4}

    \theta  = 196^{\circ}, 344^{\circ}

    5a) Parametric equations:  \frac {dy}{dx} = \frac {4e^{2t}}{-16e^{-2t}}

    5b) Gradient of tangent was -4

    5c) Co ordinates of P were (-2, 12)

    5d) Gradient of normal was  y-12=\frac{1}{4}(x+2)

    When y = 0, x = -50 and therefore Q (which was the point it cuts the x axis) was (-50,0)

    6a) Show C lies on line L

    6b) Find vector AB = \begin{pmatrix} -2 \\ -3 \\ 2 \end{pmatrix}

    6c) Vector line equation through AB

    6d) Point D was at (5, 1, 2)

    6e) For the two values of E I got  (\pm\frac{2}{3}, \pm 1, \mp\frac{2}{3})

    7)  \frac{dh}{dt} = a cos(kt) where  a=1.3 and k=\frac{\pi}{6}

    8a) Indefinite integral:  \int t cos(\frac{\pi}{4}t) dt

    By parts, this comes out to be:

     \frac {4}{\pi}t sin (\frac{\pi}{4}t) + \frac {16}{\pi^2}cos ({\pi}{4}t) + c

    8b) Differential equation (Courtesy of Nebula, post #600):

     16 \, {\rm x}^{2} = \dfrac{4 \, t \sin\left(\frac{1}{4} \, \pi t\right)}{\pi} + \dfrac{16 \, \cos\left(\frac{1}{4} \, \pi t\right)}{\pi^{2}} - \dfrac{16}{\pi^{2}} + 256

     x  = 3.65 m \quad (=3.64621 \ldots) or  365 cm


    Do feel free to correct me! I feel I have missed some questions out too
    I believe there was a question stating something like "show the cartesian equation is ???"
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    This paper actually almost made me cry- had been doing good on the past papers but got half way through this paper and pretty much gave up- didn't even attempt half the questions . So angry with myself- almost definatly a U which is a major pain in the bum considering I needed this grade to get into Uni.
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    I've noticed in all my maths and further maths exams this year that there seems to be a lot less working out space for answers... Timing has been super tight this year too, but I think I did alright in that exam this morning Not as bad as I, personally, thought it could've been, it just happened to involve a lot of working out.

    Did anyone else not think that last question was a pain to work out though? I got so confused because it just didn't look right...

    Anyone got a copy of the paper?
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    (Original post by nickroberts)
    how are your ums calculated? can you still get full ums but not full marks?
    Yeah, UMS are calculated based on the top and bottom percentage results across the year. The interval between each grade is the same, so say A is 54, A* is 58, so full UMS would be 62, as it's the interval added to the A* boundary.
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    Was 53 for an A in January! If that's the case then my grade might not be as bad as first thought!
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    (Original post by nickroberts)
    how are your ums calculated? can you still get full ums but not full marks?
    Yes you can it all depends on how everyone who took it did as a whole, I believe.
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    (Original post by Jamie T)
    Mine involves wrist exercises.
    Hahaha XD I wish I had more thumbs up for you.
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    (Original post by JTD77)
    I believe there was a question stating something like "show the cartesian equation is ???"
    yep there was - was it 2 or 3 marks - I seem to remember the value of k being 32
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    (Original post by Coconutter)
    This paper actually almost made me cry- had been doing good on the past papers but got half way through this paper and pretty much gave up- didn't even attempt half the questions . So angry with myself- almost definatly a U which is a major pain in the bum considering I needed this grade to get into Uni.
    I feel the pain dude. Still, could be worse, I only answered part a for question 1 ^^
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    (Original post by Anime Lover)
    Yeah, UMS are calculated based on the top and bottom percentage results across the year. The interval between each grade is the same, so say A is 54, A* is 58, so full UMS would be 62, as it's the interval added to the A* boundary.
    ah right thanks for clearing that up so i have nothing to worry about if i missed out 9 marks but think i got everything else right?
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    On the last question, I worked out C correctly and got the differential equation, however, didn't work out the height correctly as I'm not sure if it was because I didn't do the whole equations over pi or would it be because I had my calculator in radians? -_-
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    (Original post by nickroberts)
    how are your ums calculated? can you still get full ums but not full marks?
    yeah you can For example on the Jan 2013 Core 4 exam, you could get 63 raw marks and still get full UMS marks.

    http://www.aqa.org.uk/exams-administ...t-marks-to-ums

    ^ This website converts your raw marks to ums marks and you can see which raw marks get you which ums marks at the bottom.
    (Yes I know this well because I'm an obessived paranoid maniac...)
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    (Original post by mulac1)
    yep there was - was it 2 or 3 marks - I seem to remember the value of k being 32
    I believe it was 3 marks, I also think you're right with k=32. I put 16 as rushed it in the last minute.
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    (Original post by Student Number 4)
    yeah you can For example on the Jan 2013 Core 4 exam, you could get 63 raw marks and still get full UMS marks.

    http://www.aqa.org.uk/exams-administ...t-marks-to-ums

    ^ This website converts your raw marks to ums marks and you can see which raw marks get you which ums marks at the bottom.
    (Yes I know this well because I'm an obessived paranoid maniac...)
    thanks haha
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    Cant believe im gonna have 5 A's and a U... lol
 
 
 
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