Turn on thread page Beta
    • Thread Starter
    Offline

    1
    ReputationRep:
    To the best of my knowledge these are the answers to the AQA MPC1 Paper (13/05/15)
    Feel free to discuss them and they may well be wrong
    Other than that enjoy discussing the exam


    (1)(a) Gradient = -3/5
    (1)(b)(i) -5x+3y-1=0
    (1)(c) A(9,-4)
    (2) 7+sqrt15 (n=7)
    (3)(a) y=-10x-4
    (3)(b)(i) 108/5
    (3)(b)(ii) 162/5
    (4)(a) (x+1)^2 + (y-3)^2 = 50
    (4)(b)(i) C(-1,3)
    (4)(b)(ii) radius = 2√5
    (4)(c) k= -8, 2
    (4)(d) minimum distance =7
    (5)(a) (x+3/2)^2 -1/4
    (5)(b)(i)(-3/2, -1/4)
    (5)(b)(ii) x=-3/2
    (5)(c) y = x^2 - x + 4
    (6)(a)(i) h= (48-r^2)/2r
    (6)(b)(ii)-12π therefore max (r=4)
    (7)(a) draw x^2(x-3)
    (7)(b)(i) R = 36
    (7)(b)(ii) R= 0 therefore x+2 is a root (of x^3 -3x^2 +20)
    (7)(b)(iii) (x+2)(x^2-5x+10)
    (7)(b)(iv) x=-2
    (8)(a) Show that x^2 + 3(k-2)x -13-k=0
    (8)(b) 9k^2 -32k -16 < 0
    (8)(c) -4/9 < k <4
    Offline

    1
    ReputationRep:
    1. Gradient: -3/5

    Equation of line: 5x – 3y +1=0
    Co-ordinates of A (9,-4)

    2. 7 + root15

    3. Tangent = y-6=-10(x+1) (no form required)

    Area under graph = 108/5 Trapezium = 54 Therefore Area Bound by curve and line =54-108/5 = 162/5 square units

    4. (x+1)2 + (y-3)2 = 50

    C = (-1,3)
    R = root50 = 5root2
    K=-2, K=8
    Shortest Distance from C to QR = root49 = 7

    5. x2 + 3x +2 = (x+3/2)2 -1/4

    Vertex = (-3/2,-1/4)
    LOS: x=-3/2
    Translated Graph = (x-1/2)2 +15/4
    This then expands to x2 –x +4

    6. 48π=πr2 +2πrh

    So h = 48π – πr2 / 2πr

    Hence h = 24/r - r/2

    V=24πr – πr3 / 2

    dV/dr = 24π – 3πr2 / 2

    Stationary Point: r=4

    D2V / Dr2 = -3πr = -12π = negative = maximum point



    7. Remainder = 36
    P(x) = (x+2)(x2 -5x+10)
    B2-4ac < 0 therefore no real solutions for quadratic.
    Only solution is when x = -2

    8. 9k2 – 32k – 16 < 0

    (9k+4)(k-4) < 0
    Critical values = -4/9 and 4
    Therefore -4/9 < k < 4

    Offline

    2
    ReputationRep:
    For 4bii I believe the radius was 5√2 (√50 = √25 * √2)
    Offline

    1
    ReputationRep:
    does anyone remember the equations used in question 1 to calculate the coordinates of A?
    Offline

    0
    ReputationRep:
    For 4c I think k = 8, -2 because the equation was 25+(k-3)^2=50 so k-3 must be equal to 5 or -5 which you get with 8 and -2
    Offline

    3
    ReputationRep:
    Got everything right except the cylinder question but may have picked up a mark or so from method marks to get me as close to 100ums as possible... :crossedf:
    Offline

    0
    ReputationRep:
    For the last past on the cylinder question (6). I got d^v2/dr2 = -16π which then gave a maximum. I am finding it hard to understand what the unofficial mark scheme says. is that correct?
    Offline

    3
    More comprehensive markscheme: http://www.thestudentroom.co.uk/show....php?t=3322639
    Offline

    2
    ReputationRep:
    Ahhh i don't even know what to think about it, i feel like i got an A but at the same time i feel like i got an E

    Does anyone thing the grade boundaries will be high
    Offline

    6
    ReputationRep:
    (Original post by rhoxc)
    Ahhh i don't even know what to think about it, i feel like i got an A but at the same time i feel like i got an E

    Does anyone thing the grade boundaries will be high
    They should be about the same as last year, an A being 60±1. (As people will have found the cylinder question quite challenging.)
    Offline

    1
    ReputationRep:
    (Original post by LiesandAlibis)
    To the best of my knowledge these are the answers to the AQA MPC1 Paper (13/05/15)
    Feel free to discuss them and they may well be wrong
    Other than that enjoy discussing the exam


    (1)(a) Gradient = -3/5
    (1)(b)(i) -5x+3y-1=0
    (1)(c) A(9,-4)
    (2) 7+sqrt15 (n=7)
    (3)(a) y=-10x-4
    (3)(b)(i) 108/5
    (3)(b)(ii) 162/5
    (4)(a) (x+1)^2 + (y-3)^2 = 50
    (4)(b)(i) C(-1,3)
    (4)(b)(ii) radius = 2√5
    (4)(c) k= -8, 2
    (4)(d) minimum distance =7
    (5)(a) (x+3/2)^2 -1/4
    (5)(b)(i)(-3/2, -1/4)
    (5)(b)(ii) x=-3/2
    (5)(c) y = x^2 - x + 4
    (6)(a)(i) h= (48-r^2)/2r
    (6)(b)(ii)-12π therefore max (r=4)
    (7)(a) draw x^2(x-3)
    (7)(b)(i) R = 36
    (7)(b)(ii) R= 0 therefore x+2 is a root (of x^3 -3x^2 +20)
    (7)(b)(iii) (x+2)(x^2-5x+10)
    (7)(b)(iv) x=-2
    (8)(a) Show that x^2 + 3(k-2)x -13-k=0
    (8)(b) 9k^2 -32k -16 < 0
    (8)(c) -4/9 < k <4

    I feel as if i got all of those questions right except for 6a and 7a but how did everyone else find the exam?
    Offline

    0
    ReputationRep:
    Think I read the last question wrong so I got -4 and 4/9
    Do you know how many marks I would've lost for this?
    Offline

    0
    ReputationRep:
    How many marks do you think would give a B for this paper?
    Offline

    3
    ReputationRep:
    I believe there are some errors here.

    (4) (c) 8 and -2
    (4) (d) Is 3. How can the distance from the centre of a circle to the midpoint of a chord be greater than the radius?
    (6) (b) (ii) Wasn't it +12pi? And so it's a minimum not a maximum? (not sure about this)

    @ihatehistory The last question was worth 4 marks so I'm guessing probably most of them. Factorisation (1) Diagram (1) Correct inequalities (2) As a guess.
    Offline

    1
    ReputationRep:
    (Original post by ihatehistoryy)
    Think I read the last question wrong so I got -4 and 4/9
    Do you know how many marks I would've lost for this?
    usually those questions are 5 marks, so you will get definitely get a mark for drawing the graph, if you factorised correctly then i suspect that is two marks for both of the brackets and you may get another mark if your inequality signs are correct even if the numbers are wrong but i'm not sure. I think that you should get at least 3 marks for that question
    Offline

    1
    ReputationRep:
    (Original post by InaneRambler)

    (4) (d) Is 3. How can the distance from the centre of a circle to the midpoint of a chord be greater than the radius?
    Good point. Radius was rt50 though wasnt it? Slightly greater than 7
    Offline

    0
    ReputationRep:
    i agree with all of it except the cylinder question which you have to have wrong purely because you cannot have a negative radius.
    Offline

    3
    ReputationRep:
    (Original post by eekeek)
    Good point. Radius was rt50 though wasnt it? Slightly greater than 7
    Ye... I got that one wrong. Whoops! Still 70+ I hope :/
    Offline

    6
    ReputationRep:
    (Original post by 2014_GCSE)
    Got everything right except the cylinder question but may have picked up a mark or so from method marks to get me as close to 100ums as possible... :crossedf:
    Not being able to do a whole question is going to crush that UMS
    Offline

    3
    ReputationRep:
    (Original post by JamieOH)
    Not being able to do a whole question is going to crush that UMS
    I did the second half of it, and got it correct! You could still do the differential/stationary point stuff because the previous question was a show that so you could still use it.

    I only lost 5 marks (and may get like 1 method mark for my various attempts). So 70-71/75, which I think should be a decent UMS, no?
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 11, 2016

1,686

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should universities take a stronger line on drugs?

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

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