Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    17
    ReputationRep:
    (Original post by timtjtim)
    You know that C is the midpoint of the line AB.

    To get from A to C, you go down by 4 and across by 7. So from C to B, you do the same. C(5,-3) so B(5+7,-3-4) = B(12,-7)
    :facepalm: only question I didn't answer - glad it was only 2 marks
    Offline

    0
    ReputationRep:
    It was root 65
    Offline

    0
    ReputationRep:
    I'm positive 4d was(x-2)(x^2-3x-14) + 20 [
    Offline

    0
    ReputationRep:
    Perfection
    Offline

    9
    ReputationRep:
    (Original post by kiiten)
    :facepalm: only question I didn't answer - glad it was only 2 marks
    same here i was like well its only 2 marks oh well
    Offline

    1
    ReputationRep:
    (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 [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 was 4 parts to this question i cant remember part a and b
    b)
    c) definite integral = 81/4 [5]
    d) area of shaded region = 45/4 [3]

    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
    - d^2y/dx^2= - 2x - 9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
    How could make this mark scheme?? Are these answers 100% right??
    Offline

    2
    ReputationRep:
    (Original post by LouisRobins)
    it was 25 not 65

    65 aint a square number! must have been a typo
    Oh dear. You do know that r^2 doesn't have to be a square of an integer right? 65 is correct.
    Offline

    2
    ReputationRep:
    Anyone know that the final answer was deffo k>6 and k<-3/2

    Because I got k= -1 and k= 9, would be helpful if someone could remember the inequality we had to solve!
    Offline

    2
    ReputationRep:
    (Original post by Alvi98)
    Translation got (1/2, 33/4) anyone?
    Nah man got 41/4
    Offline

    14
    ReputationRep:
    (Original post by Alvi98)
    Translation got (1/2, 33/4) anyone?
    the 1/2 is correct, but the 33/4 probs came from outer space mate
    Offline

    0
    ReputationRep:
    (Original post by Rit101)
    Why was it not 6>k>-3/2
    the quadratic was set as greater than 0 so must be the values above the x axis. so k is greater than 6 OR less than -3/2
    Offline

    2
    ReputationRep:
    what do you think the grade boundaries will be like? did you guys find the paper hard, like on a scale of 1-10...
    Offline

    2
    ReputationRep:
    (Original post by omarharoun3)
    what do you think the grade boundaries will be like? Did you guys find the paper hard, like on a scale of 1-10...
    59/61 a
    54/55 b
    48/49 c
    Offline

    0
    ReputationRep:
    8a was when you had to divide by -3 and so swap the inequality to get the equation to solve in part b, wasn't it? so your answers to 8a are actually 8b I think
    Offline

    1
    ReputationRep:
    (Original post by Jin99)
    Do you know for the area of shaded region, do you subtract the triangle from the area under the curve?

    Posted from TSR Mobile
    yep
    Offline

    5
    ReputationRep:
    almost certain the remainder question worked out to be
    (x-2)(x^2-3x-14) + 20

    the '-14' of the quadratic had to be multiplied by -2 to get 28.
    28 then added to the +20 (your r value)
    to get the +48 (constant of p(x))
    Offline

    2
    ReputationRep:
    I didn't simplify my fractions but got the same answers, think I would lose around 2 marks for both integration questions?
    Offline

    7
    ReputationRep:
    (Original post by _wsburn)
    almost certain the remainder question worked out to be
    (x-2)(x^2-3x-14) + 20

    the '-14' of the quadratic had to be multiplied by -2 to get 28.
    28 then added to the +20 (your r value)
    to get the +48 (constant of p(x))
    Yeah, that's how you do it, I didn't even bother with algebraic division for any of them, I just used (x-2)(ax^2+bx+c) and found a,b,c just with inspection :P.It's how you do it in c4 anyways.
    Offline

    7
    ReputationRep:
    The one about the coordinates of the point of the circle I used vectors to figure it out :P. I think they'll give me the marks. It gave me (12,-7) for B or something.
    Offline

    2
    ReputationRep:
    (Original post by Adam998)
    I didn't simplify my fractions but got the same answers, think I would lose around 2 marks for both integration questions?
    i did the same! what were your fractions?!
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

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.