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    Do you know what either the 1st or 2nd derivative was?
    (Original post by beanigger)
    well the co ordinate u had was x=-1.5 and u had to prove that was a minimum by first proving it was a stationary point (showing that the dy/dx of that x was zero) then subbing into second derivative to get a positive value which showed it was a minimum as second derivative was greater than 0
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    Can someone confirm that m= -5/3 for 1a?
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    (Original post by Igzzy__)
    that makes sense thanks a lot for clearing that up for me. So if I wrote, translation by vector (-1/2,41/4) as my answer. would I get two marks for that ?
    I wrote that too, are we wrong ?
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    (Original post by yasaminO_o)
    I think 7a was finding the tangent of the curve at P and part b was finding the x-coordinate of Q but I'm not entirely sure
    wasnt that six thought seven a and b was a differentiation then also 2nd derivative to find out if there was a minimum point
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    (Original post by Thequickspark)
    Can someone confirm that m= -5/3 for 1a?
    Correct.
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    (Original post by GabbytheGreek_48)
    wasnt that six thought seven a and b was a differentiation then also 2nd derivative to find out if there was a minimum point
    That was 8 :P
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    MARK SCHEME UPDATE:

    Mark scheme is attached. Some answers are missing - please comment below if you know them!
    Attached Images
  1. File Type: pdf MarkSchemeV1.pdf (99.8 KB, 294 views)
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    (Original post by timtjtim)
    That was 8 :P
    lol ohh well it was somewhere too many questions
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    (Original post by GabbytheGreek_48)
    wasnt that six thought seven a and b was a differentiation then also 2nd derivative to find out if there was a minimum point
    I think you are talking about question 8 for the dy/dx stuff. It was right at the end of the paper.
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    I found this paper kinda hard do you think the grade boundaries will be lower?
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    It was
    (2k^2-9k+18)>0
    I guess :yes:
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    (Original post by beanigger)
    Unoffical Mark scheme for C1 AQA 2016

    It would help if you could link answers to questions as i cant remember them

    Questions:

    1) a)Asked to work out gradient of a tangent, m= - 5/3 [2]
    b)Asked to find co-ordinates of B B( - 3,4) [3]
    c)Asked to find K K= - 30 [2]

    2) a) simplify (3√5)^2 = 45 [1]
    b) i cant remember the question but the answer was75 - 32√5 [4]

    3) a) y=(x-7/2)^2 - 41/4
    b) min value = -41/4 [1]
    c) Translation of (1/2 , 41/4)

    4) a) show that (x+3) was a factor of x^3 - 5x^2 -8x + 48, p(-3) = 0, (x+3) is a factor of p(x).
    b) three linear factors of p(x) = (x+3)(x-4)(x-4)
    c) find remainder when p(x) was divided by (x+2) R=20
    d) Factorise p(x) using (x+2) R = (x-2)(x^2-3x-14) + 20 [3]

    5) a) asked to find equation of circle (x-5)^2+(y+3)^2=65 (note 65 = r^2) [3]
    b) asked to find co-ordinates of B (AB is diameter) B ( 12, -7) [2]
    c) asked to find equation of tangent at A 7x-4y+18=0 [5]
    d) asked to find length of CT = 9 [2]

    6) a) y=-32x-40
    b)Q(-5/4) [1]
    c) upside down positive graph passing through y axis at 8 [2]
    d) x = -1±√5

    7) a)
    b)
    c) definite integral = = 81 /4 [5]
    d) area of shaded region = = 45/4 [3]


    8) a) d^2y/dx^2= - 2x - 9x^2 [2]
    b) verify that P was a minimum point sub x co-ordinate of P into dy/dx (given in the question) to prove it was a stationary point, then sub in x-coord of p into d^2y/dx^2 to get +45, d^2y/dx^2>0 therefore minimum point [4]
    c)show that y is decreasing, so make your dy/dx<0 and rearrange to get into the form they wanted (note when you multiply by -1 the inequality sign flips)
    d) cant remember anything except the answers were k>6 and k<-3/2

    Answers that need a question to be assigned to
    - k=4 and k=20
    What are we thinking of grade boundaries people? 2015 = 64 for an A, and i feel this year the integral question/many fractions made it a bit harder so i personally feel it will be between 59 and 63, probably 61/2 though. Thoughts?
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    (Original post by timtjtim)
    MARK SCHEME UPDATE:

    Mark scheme is attached. Some answers are missing - please comment below if you know them!
    8) a) d^2y/dx^2= - 2x - 9x^2 [2]

    b) verify that P was a minimum point sub x co-ordinate of P into dy/dx (given in the question) to prove it was a stationary point, then sub in x-coord of p into d^2y/dx^2 to get +45, d^2y/dx^2>0 therefore minimum point [4]

    c)show that y is decreasing, so make your dy/dx<0 and rearrange to get into the form they wanted (note when you multiply by -1 the inequality sign flips)

    d) cant remember anything except the answers were k>6 and k<-3/2 "

    I believe these are correct?
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    (Original post by Thequickspark)
    Can someone confirm that m= -5/3 for 1a?
    Yes , AB was 5x+3y+3=0 & the line with the equation y=mx +(whatever it was) was parallel
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    (Original post by Chickenslayer69)
    8) a) d^2y/dx^2= - 2x - 9x^2 [2]

    b) verify that P was a minimum point sub x co-ordinate of P into dy/dx (given in the question) to prove it was a stationary point, then sub in x-coord of p into d^2y/dx^2 to get +45, d^2y/dx^2>0 therefore minimum point [4]

    c)show that y is decreasing, so make your dy/dx<0 and rearrange to get into the form they wanted (note when you multiply by -1 the inequality sign flips)

    d) cant remember anything except the answers were k>6 and k<-3/2 "

    I believe these are correct?
    Correct
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    Does anyone think the grade boundaries might be the lowest they've ever been this year? I've done every single past paper and This was just vile. There's an old paper (2013?) where it was like 56 for an A or something and that was way easier than this.
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    (Original post by RueXO)
    Does anyone think the grade boundaries might be the lowest they've ever been this year? I've done every single past paper and This was just vile. There's an old paper (2013?) where it was like 56 for an A or something and that was way easier than this.
    It's very likely for the grade boundaries to be lower rather than higher, but if an A is less than 60 It'll be a pleasant suprise
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    My brain stopped working on question 7 😠😢😠😢 because i made a mistake somewhere so i spent about 20 minutes on it(mainly staring blankly...and crying inside 😥) so I'm pretty sure it was as follows:

    a) answer was y=-32x - 40

    b)find x coordinate of axis intersect (of Q i think?)

    c) the 5 marker integration question: 81/4

    d) shaded area (I believe you used the x coordinate you found of on part b to find the area of the triangle) .: area= 45/4

    The point is, i somehow got y=-32x - 26 for part a so i got part b and c wrong too. Anybody know how many marks was it for QUESTION 7 PART A, B & D?!?

    Thank you 😄
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    Mark Scheme with some workings etc.
    Attached Images
  2. File Type: pdf MarkSchemeV2.pdf (112.5 KB, 135 views)
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    (Original post by Chickenslayer69)
    8) a) d^2y/dx^2= - 2x - 9x^2 [2]

    b) verify that P was a minimum point sub x co-ordinate of P into dy/dx (given in the question) to prove it was a stationary point, then sub in x-coord of p into d^2y/dx^2 to get +45, d^2y/dx^2>0 therefore minimum point [4]

    c)show that y is decreasing, so make your dy/dx<0 and rearrange to get into the form they wanted (note when you multiply by -1 the inequality sign flips)

    d) cant remember anything except the answers were k>6 and k<-3/2 "

    I believe these are correct?
    Boy it's late. I'll add these into V3, tomorrow - thankyou!
 
 
 
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