Join TSR now and get all your revision questions answeredSign up now

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

    Offline

    0
    ReputationRep:
    (Original post by PatrickM)
    Ran out of time, hardest C4 paper ever. Honestly think it was harder than Jan 2013. What a ****ing joke.

    I've been getting high 80/low 90s in practice papers. Even with grade boundaries I'm probably going to miss out on an A by a few marks. Absolutely livid - that exam was absolutely horrific in comparison to past papers. I'm pretty sure I dropped marks on nearly every single question.
    Literally the same as you mate! In Jun11/12 I got 100 ums. Ill be shocked if I get 80
    Offline

    0
    ReputationRep:
    (Original post by onimusha370)
    easiest paper ever, that means high grade boundaries
    100UMS - 75
    90UMS - 69
    80UMS - 63
    etc

    very very easy, finished after 35 minutes and think i've got 74/75 at least.
    what did everyone else think?
    You're a nob
    Offline

    1
    ReputationRep:
    (Original post by bugsuper)
    i got 11/3 ln2 + 2 but there seems to be some dispute over this

    i thought if you integrated y=2 between 0 and -1, the area is just a rectangle of base 1 and height 2..., so positive 2
    it doesn't go under the x-axis at any point

    I probably sound like a total idiot
    I also got +2.

    The integral had 2x in it.

    The limits were x=0 and x=-1, so the integral would be (2(0) + ...) - (2(-1) + ...)

    = ... + 2
    Offline

    0
    ReputationRep:
    (Original post by fizzbizz)
    UNOFFICIAL MARK SCHEME

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

    1b) Integral =  \frac {11}{3} ln(2)

    1c) C = 2

    1d)  \frac {11}{3} ln(2) -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) Find vector AB (can't quite remember this)

    6b) Vector line equation through AB

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

    6d) 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
    1d. I think this may be +2 , not 100% sure
    Offline

    0
    ReputationRep:
    (Original post by fizzbizz)
    UNOFFICIAL MARK SCHEME

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

    1b) Integral =  \frac {11}{3} ln(2)

    1c) C = 2

    1d)  \frac {11}{3} ln(2) -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) Find vector AB (can't quite remember this)

    6b) Vector line equation through AB

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

    6d) 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 can deduce from this that I have almost certainly achieved less than 50%.... although I need 34% for an overall A. Gonna be tight! xD
    Offline

    0
    ReputationRep:
    (Original post by Qwob)
    I also got +2.

    The integral had 2x in it.

    The integral would be (2(0) + ...) - (2(-1) + ...)

    = ... + 2
    Indeed. It was minus minus 2.

    It's quite telling about exam nerves that I had to draw a little picture to justify minus minus 2 being positive 2 in my mind
    Offline

    1
    ReputationRep:
    Does there happen to be any legends out there who took pictures of the paper?
    Offline

    0
    ReputationRep:
    [QUOTE=fizzbizz;43070678]UNOFFICIAL MARK SCHEME


    1d)  \frac {11}{3} ln(2) -2

    Wasnt it +2 because its 11/3ln2 - (-2)?
    Offline

    2
    ReputationRep:
    I got every single answer on the unofficial mark scheme besides the position vector of E. I got to the stage where I had a quadratic equation with 17u^2 I think it was. So hopefully I picked up 1/2 marks
    I don't quite remember the signs of position vector of D but I remember mew=-1 and it involved (5,1,2) in that order.

    Oh and for the sin(alpha) question I drew the triangle but I forgot to give a value for cos(alpha). But everything else seems legit. Confidence is sky high now
    Offline

    0
    ReputationRep:
    (Original post by Qwob)
    Think I did too.

    Happy looking at that mark scheme. I think I missed out the part where you had to find coordinates of Q unless I'm being an idiot and can't remember doing it. I didn't recheck that question though as I thought I'd done it right so I may have done it, but probably not. If I didn't look for the coordinate of Q would that be around 2 marks lost?
    I think coordinate Q was 3 marks? I'm not too sure. Those marks shouldn't matter too much, don't worry.
    Offline

    2
    ReputationRep:
    Oh yeah and it was 11/2 ln2 + 2 because it was --2
    Offline

    0
    ReputationRep:
    (Original post by fizzbizz)
    UNOFFICIAL MARK SCHEME

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

    1b) Integral =  \frac {11}{3} ln(2)

    1c) C = 2

    1d)  \frac {11}{3} ln(2) -2 EDIT: I may be wrong on this, since the integral gives  [2x]^{0}_{-1} + \frac {11}{3} ln(2) which gives  \frac {11}{3} ln(2)+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) Find vector AB (can't quite remember this)

    6b) Vector line equation through AB

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

    6d) 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
    you don't happen to do FP3 do you?
    Offline

    0
    ReputationRep:
    Can someone please explain how they get to the co-ordinates of D for Question 6c on the Vectors question?
    Offline

    1
    ReputationRep:
    [QUOTE=theslav;43070855]
    (Original post by fizzbizz)
    UNOFFICIAL MARK SCHEME


    1d)  \frac {11}{3} ln(2) -2

    Wasnt it +2 because its 11/3ln2 - (-2)?

    (Original post by bugsuper)
    Indeed. It was minus minus 2.

    It's quite telling about exam nerves that I had to draw a little picture to justify minus minus 2 being positive 2 in my mind

    (Original post by DudeBoy)
    1d. I think this may be +2 , not 100% sure

    (Original post by Qwob)
    I also got +2.

    The integral had 2x in it.

    The limits were x=0 and x=-1, so the integral would be (2(0) + ...) - (2(-1) + ...)

    = ... + 2

    (Original post by Student Number 4)
    I also got 11/3 ln 2 + 2 and when I checked it on the calculator it was right, but someone who got 7/2 ln 2 +2 also said they checked it on the calculator and got it right?
    My bad, I have updated the mark scheme!
    Offline

    0
    ReputationRep:
    (Original post by kesupile)
    I got every single answer on the unofficial mark scheme besides the position vector of E. I got to the stage where I had a quadratic equation with 17u^2 I think it was. So hopefully I picked up 1/2 marks
    I don't quite remember the signs of position vector of D but I remember mew=-1 and it involved (5,1,2) in that order.

    Oh and for the sin(alpha) question I drew the triangle but I forgot to give a value for cos(alpha). But everything else seems legit. Confidence is sky high now
    Offline

    2
    ReputationRep:
    Oh no


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    I'm basically guessing here, but wasn't the vector AB (-2, -3, 2) or something like that?
    Just trying to fill in the gaps
    Offline

    1
    ReputationRep:
    (Original post by kesupile)
    I got every single answer on the unofficial mark scheme besides the position vector of E. I got to the stage where I had a quadratic equation with 17u^2 I think it was. So hopefully I picked up 1/2 marks
    I don't quite remember the signs of position vector of D but I remember mew=-1 and it involved (5,1,2) in that order.

    Oh and for the sin(alpha) question I drew the triangle but I forgot to give a value for cos(alpha). But everything else seems legit. Confidence is sky high now
    We seem to have done exactly the same on everything. :eek: That is scary.

    I got the quadratic containing 17u^2 and just solved it on my calculator, knowing it was wrong.
    Offline

    1
    ReputationRep:
    Barely attempted Q7, it seemed really misleading in my eyes, I think I just had a mental retard attack half way through the exam
    Offline

    0
    ReputationRep:
    can anyone remember the number of marks for each question?
 
 
 
Poll
Do you prefer hot dogs or burgers?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.