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    does anyone remember the number of marks for each question? would help to see how many marks i roughly got
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    (Original post by Jacobisswaggy)
    For the one where you had to prove it was a minimum value, I believe you first need to find that it is a stationary point by subbing in to dy/dx, then subbing in to d^2y/dx^2 to show it is a minimum.

    I base this on the fact the question was 4 marks, and it wouldn't be 4 marks for only subbing in to d^2y/dx^2.
    yes u are right but do you remember what question that was?
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    I got root 65 for the radius as well. How did everyone find the paper and what do you guys think about grade boundaries?
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    (Original post by Parhomus)
    That method only works when you are given the equation in the way that was already shown. You HAD to use the fact that it said the point was on the circle to find the radius. I almost made that mistake for k=34 but if you plug in the coordinates for the point on the circle then you see it doesn't make sense.
    ohhhhhh well i ****ed up on that entire question how much was it worth?
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    for Q8 where you had to show point P was a minimum, i failed at solving 6x^2-27x-54 to find the stationary points. i was really stressed with 10 minutes left and so i forgot to simplify by dividing by 3 and so made things really hard on myself. Instead i just put 3/2 into the second derivative and got 45>0 therefore point P was a minimum. Would i get any marks for this? it was worth 4 marks.
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    (Original post by Andrew Dainty)
    for Q8 where you had to show point P was a minimum, i failed at solving 6x^2-27x-54 to find the stationary points. i was really stressed with 10 minutes left and so i forgot to simplify by dividing by 3 and so made things really hard on myself. Instead i just put 3/2 into the second derivative and got 45>0 therefore point P was a minimum. Would i get any marks for this? it was worth 4 marks.
    all you had to do was make dy/dx=0
    and enter the value of P i believe it was -3/2 as X and you should get 0 anyway
    that proves it is a stationary point
    then use that same x value and put into d2y/dx2 and you should get 45
    hence since dy/dx=0 when x=-3/2 it is stationary
    also since d2y/dx2>0 it is a minimum point
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    Pretty sure the last question was that:
    b) y=k(x+4) intercepts the line at y=x^2 +4x-8 show that it equals...
    c) y=k(x+4) is the tangent of y=x^2+4x-8 and to find the values of k when it has real roots. I got k= 12+/- (8root3)
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    (Original post by timtjtim)
    Okay, I've added a little bit.

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

    2) a) simplify (3√5)^2 = 45
    b) Put ((3√5)^2 + √5) / (7 + 3 √5) into m + n√5= 75 - 325

    3) a) y=(x-7/2)^2 - 41/4
    b) Translation of (1/2 , 41/4)

    4) a) show that (x+3) was a factor of x^3 - 5x^2 -8x + 48. p(-3) = -3^3 - 5 * 3-^2 - 8 * -3 + 48 = 0. As p(-3) = 0, (x+3) is a factor of p(x).

    b) Three linear factors goes to (x+3)(x-4)(x-4)
    c) find the remainder when x^3 - 5x^2 -8x + 48 was divided by (x+2) R=20
    d) factorise p(x) using your answer to part c) as R = (x-2)(x^2-3x+14) + 20

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

    6) a)

    7) a) I think there were 4 parts to this question I can't remember part a and b
    b)
    c) Definite integral (from 1 to -2) = 81/4 (20.25)
    d) Area of shaded region (integral - triangle area) = 45/4 (11.25)

    8) a) can't remember anything except the answers were k>6 and k<-3/2 (NOT 6 > k > -3/2 )

    Answers that need a question to be assigned to
    - Q(-5/4), y=-32x-40, upside down positive graph passing through y axis at 8
    - k=4 and k=20
    - x = -1±√5
    - d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
    3b) Threw me out a bit with 41/4, until I realised it was just 10.25 which isn't too abstract I suppose
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    (Original post by Andrew Dainty)
    for Q8 where you had to show point P was a minimum, i failed at solving 6x^2-27x-54 to find the stationary points. i was really stressed with 10 minutes left and so i forgot to simplify by dividing by 3 and so made things really hard on myself. Instead i just put 3/2 into the second derivative and got 45>0 therefore point P was a minimum. Would i get any marks for this? it was worth 4 marks.
    I did the same thing as you, up until they said "Please stop writing", managed to make dy/dx equal to zero and get in a few calculations. Hopefully workings won't need to be too thorough in the mark scheme.
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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
    b)Asked to find co-ordinates of B B( - 3,4)
    c)Asked to find K K= - 30

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

    3) a) y=(x-7/2)^2 - 41/4
    b) min value = -41/4
    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

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

    6) a) y=-32x-40,
    b)Q(-5/4)
    c) upside down positive graph passing through y axis at 8
    d)

    7) a) i think there was 4 parts to this question i cant remember part a and b
    b)
    c) definite integral = 81/4
    d) area of shaded region = 45/4

    8) a) 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
    - x = -1±√5
    - d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
    x = -1±√5 was for question 6, it was the x coordinates where the graph passed through the x axis
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    (Original post by UbaidS)
    Exactly i keep seeing 65 making me worried and that
    Centre was (x-5)^2 + (y+3)^2 or something but it also had to go through point (-2,1) hence (-2-5)^2 + (1+3)^2 = 49+16 = 65
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    (Original post by sasha1699)
    x = -1±√5 was for question 6, it was the x coordinates where the graph passed through the x axis
    thanks
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    (Original post by beanigger)
    yes u are right but do you remember what question that was?
    I don't, sorry I'm terrible at remembering tests, haha
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    Can anyone remember the intergral - completly messed up the numbers (definite integral)
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    Could you include how many marks each question is work. I think Question 1) 2
    2) 5
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    d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
    This was 7a right?
    or 8a?
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    For the remainder question I got 76, what have I done wrong?
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    Yh thats what i got. Thanks god lol. I think it was q8 as q7 was the intergration section
    (Original post by DrPepper22)
    d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
    This was 7a right?
    or 8a?
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    Wait, are we 100% positive the first question asked for a parallel line? I could have sworn it said perpendicular and gave 3/5 as my gradient...

    Edit: Sorry, said 5/3, meant 3/5. Or something. I don't even know anymore.
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    (Original post by Muronaldo777)
    For the remainder question I got 76, what have I done wrong?
    48 - 28, not 48 + 28. This is likely where you went wrong.
 
 
 
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