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    (Original post by 1TrueMeaning)
    Hi guys,
    For the polynomial/long division question where we asked to use the factor theorem to show (x+2) is a factor of p (x) or did we just have to put it in the form specified and then show that (×+2) is the only factor?
    Thanks.
    Isn't question 5 wrong? if you complete the square you get +1/2 don't you intstead of -1/4?
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    do you know if you had to state n=7 for question 2. I only said 7+sqrt15 and didn't restate it was n=7
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    I think the question asked you to state the value of n.
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    (Original post by JamieOH)
    I think the question asked you to state the value of n.
    **** me i probably lost one mark then
    this is really stupid though, losing mark for not stating the obvious
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    (Original post by C0balt)
    **** me i probably lost one mark then
    this is really stupid though, losing mark for not stating the obvious
    I think the only mark I lost in the paper was that mark -.-
    I don't even remember it saying 'state the value of n' so annoying!!!! >:O
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    (Original post by Heffalump .)
    I think the only mark I lost in the paper was that mark -.-
    I don't even remember it saying 'state the value of n' so annoying!!!! >:O
    I'm pretty sure it will be less than 75/75 for 100UMS... (HOPEFULLY)
    So whatever... Unless there was some other question with the stupid "state qerqwer" lol

    I don't remember either tbh. I thought it said express in the form n+sqrt15 where n is an integer or something lol But my friends said it said to state n lmao
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    (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 left mine as: 48pi-pir^2/2pir, you simplified by taking out the 'pi's'
    But you (or me) didn't simplify to: 24/r - r/2.. do you think I'll still get 3 marks? or do you think I'll lose one..
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    (Original post by C0balt)
    I'm pretty sure it will be less than 75/75 for 100UMS... (HOPEFULLY)So whatever... Unless there was some other question with the stupid "state qerqwer" lolI don't remember either tbh. I thought it said express in the form n+sqrt15 where n is an integer or something lol But my friends said it said to state n lmao
    Yeah hopefully, I hate dumbass questions like that -.- I'm an even bigger dumbass for not doing it - it's not even maths! >:O haha nevermind, hopefully it will be around 73/74 for 100UMS (due to that cylinder question)
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    (Original post by Heffalump .)
    Yeah hopefully, I hate dumbass questions like that -.- I'm an even bigger dumbass for not doing it - it's not even maths! >:O haha nevermind, hopefully it will be around 73/74 for 100UMS (due to that cylinder question)
    You know what, I hate exams only because of stupid MARK SCHEMES. It is ****ing obvious I am able to do maths from all the workings, yet I lose mark for not doing what mark scheme wants WTF

    Yeah cylinder was tricky and also the transformation was harder than anything i've ever seen before. So it should be fine lol
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    (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
    Do you know how to draw the graph for 7a?
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    (Original post by sciencebiology1)
    Do you know how to draw the graph for 7a?
    Repeated root is at origin so graph should have positive correlation, turn at origin and then turn again and pass through (3,0)
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    Do you think it matters that instead of putting 108/5 and 162/5 I put
    21 3/5 and 32 2/5?
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    (Original post by Amelia.D.)
    Do you think it matters that instead of putting 108/5 and 162/5 I put
    21 3/5 and 32 2/5?
    I second this I put 648/30 and 972/30 which are equivalents would I still get the marks?
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    (Original post by chancehaycock)
    Repeated root is at origin so graph should have positive correlation, turn at origin and then turn again and pass through (3,0)
    Thank you! I think I drew that so phew
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    Is 55 pushing it for a B?
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    For the cylinder question i said h=V/(pi)r^2 , how many marks dropped in the first two parts for using this?
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    (Original post by chancehaycock)
    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

    Hey r u sure q2 is 7+ and nt 7-
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    (Original post by girl :D)
    Hey r u sure q2 is 7+ and nt 7-
    It's 7+

    Posted from TSR Mobile
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    (Original post by girl :D)
    Hey r u sure q2 is 7+ and nt 7-
    Do you know how many marks it was for Q1c?
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    (Original post by sciencebiology1)
    Do you know how many marks it was for Q1c?
    Was it simultaneous equation? I don't remember tbh

    Posted from TSR Mobile
 
 
 
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