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    K was 65. To find the radius you used the length equation so: r^2 = 7^2 + 4^2 = 65 hence r = root 65. K is r^2 therefore K was 65.
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    (Original post by Zaydy33)
    Hi everyone, I'm pretty sure I've got almost everything correct but for the length of CT, I stupidly left it as root 81 instead of 9. Do you think they may allow it in the mark scheme as normally it says OE (or equivalent) and root 81 is equivalent to 9!

    Thanks
    Difficult to say, at most you've lost 1 mark and 74/75 will still get you 100 UMS so you shouldn't worry
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    Does anyone remember the formula we had to intergrate?
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    For q7bi and bii, if I got the right answer but forgot to simplify the fraction how many marks will I lose?
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    For question 4d shouldn't it be -14? (-2 * -14=28 +20=48)
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    (Original post by chloelaura27)
    For q7bi and bii, if I got the right answer but forgot to simplify the fraction how many marks will I lose?
    1 at most.
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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.
    I did the same but can't remember what value i actually got when i put it into the second derivative equation, and am now wondering if i differentiated it correctly. Do you remember what the formula was for d^2y/dx^2? And the coordinate that we were supposed to sub in as well? Thanks very much if you can.

    I hardly remember question 1 and forgot how to get coordinates of B as i don't remember getting (-3, 4) i I just seriously don't remember question one at all. I'm sure i checked it but if someone can tell me how they came to those coordinates?
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    (Original post by BlackthornRose)
    I did the same but can't remember what value i actually got when i put it into the second derivative equation, and am now wondering if i differentiated it correctly. Do you remember what the formula was for d^2y/dx^2? And the coordinate that we were supposed to sub in as well? Thanks very much if you can.

    I hardly remember question 1 and forgot how to get coordinates of B as i don't remember getting (-3, 4) i I just seriously don't remember question one at all. I'm sure i checked it but if someone can tell me how they came to those coordinates?

    I recall the x-coordinate being -1.5 , it was in mixed fraction form i think.
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    If I got the coordinates wrong for B on 5b) but did 5c) correctly (with the wrong coordinates) will I get follow through marks, if so would it be all 5 or would I have lost some for having the wrong coordinates?
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    (Original post by ExamsCauseStress)
    Difficult to say, at most you've lost 1 mark and 74/75 will still get you 100 UMS so you shouldn't worry

    Okay thank you! I hope i get 100 UMS cos core 2 is gonna be hard!
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    (Original post by Zaydy33)
    Hi everyone, I'm pretty sure I've got almost everything correct but for the length of CT, I stupidly left it as root 81 instead of 9. Do you think they may allow it in the mark scheme as normally it says OE (or equivalent) and root 81 is equivalent to 9!

    Thanks
    Hi, I was doing a past paper the other day and left a length as root 49 and the mark scheme said 'Do not accept root 49' so i hate to burst your bubble but i dont think so ://///
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    I think that getting m wrong for 1a wouldn't matter because 1b was simultaneous equations and 1c was just subbing in the coordinates into one of the equations (can't remember which)
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    (Original post by Ywna097)
    I recall the x-coordinate being -1.5 , it was in mixed fraction form i think.
    Thanks
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    The equation for the integral was 4 - x^2 - 3x^3 I think as
    I remember integrating it to give [4x - x^3/3 - (3x^4)/4] between -2 and 1
    Sub in -2 giving -52/3
    Sub in 1 giving 35/12

    So 35/12 - -52/3 gives 81/4 which is 20.25
    Then subtract from the area of the triangle
    20.25-9 = 11.25 or 45/4.
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    (Original post by Pentaquark)
    The equation for the integral was 4 - x^2 - 3x^3 I think as
    I remember integrating it to give [4x - x^3/3 - (3x^4)/4] between -2 and 1
    Sub in -2 giving -52/3
    Sub in 1 giving 35/12

    So 35/12 - -52/3 gives 81/4 which is 20.25
    Then subtract from the area of the triangle
    20.25-9 = 11.25 or 45/4.

    Thanks
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    (Original post by Parhomus)
    I think that getting m wrong for 1a wouldn't matter because 1b was simultaneous equations and 1c was just subbing in the coordinates into one of the equations (can't remember which)
    Do you happen to remember the simultaneous equations? I'd hate to lose marks on the first question and it's the one I remember least as I probably gave it the least thought.
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    I reckon like 72/75 will be 100UMS
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    (Original post by qwertyuiop1998)
    Any estimates for 100 UMS?
    I'm guessing about 72 for 100 UMS.

    60 for an A.

    56 for a B.

    52 for a C.
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    Unofficial Mark scheme for C1 AQA 2016

    It would help if you could link answers to questions as I can't remember them

    Questions:

    1)
    a)Asked to work out gradient of a tangent, m= -{5\over3} [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) 75-32√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 [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) I think there were 4 parts to this question I can't remember part a and b
    b)
    c) Definite integral of \int_{-2}^{1} 4-x^2-3x^3 = [ {{-3x^4}\over{4}} - {{x^3} \over {3}} + 4x + c] = 81 /4 [5]
    d) Area of shaded region = {84\over4} - {24\times2-{5\over4}\over2} = 45/4 [3]

    8) a) can't 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
    - d^2y/dx^2= - 2x - 9x^2 sub in x-coordinate of p to get 45, 45>0 therefore minimum

    Updated again.
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    If I got the wrong equation when completing the square for 8-4x-2x^2. Could i still get marks for the next part where it asked to calculate the roots by error carried forward

    Could i also get error carried forward on question 7 if i got the wrong equation in 7a leading to the wrong co-ordinate for Q and so the wrong area of the triangle
 
 
 
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