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    (Original post by Rit101)
    This means that the gradient of AB was -3/5
    Hence the parallel line should have the same gradient correct?
    That's what I did
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    (Original post by Ywna097)
    Thanks
    It may be wrong
    My friends said something like 4 - 8x - 2x^2 but I feel like that was for another question Argh I normally remember nearly all the questions but today was just a blur tbf.
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    (Original post by timtjtim)
    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.

    R = (x-2)(x^2-3x+14) + 20
    For this question, shouldnt it be -14
    So that when you multiply it out you'd get +28+20 which would be +48 which was the original number in the equation.
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    How do you work out ques 1. It was find the coordinate of B ... lies on diameter???

    Posted from TSR Mobile
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    (Original post by Rit101)
    Why was it not 6>k>-3/2
    because it was equation>0, so it was all the values above the x axis. Meaning it was anything under -3/2 (x<-3/2) and anything above 6 (x>6)
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    For 6d
    I used the quadratic formula, I got -1+-squareroot|80

    How did you get -1+-squareroot|5 ?
    Sorry for poor quality post, I'm on my **** phone.
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    Would error carried forward be possible Q7 for the triangle area and co ordinate of Q if 7a was done wrong to get the line.
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    smh the only marks i'm counting on are method marks.
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    (Original post by kiiten)
    How do you work out ques 1. It was find the coordinate of B ... lies on diameter???

    Posted from TSR Mobile
    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)
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    messed up the tangent to circle question (5marks) think I used different coordinates to what the point actually was. If I did the working out correct but did this how many marks do you think ill gain?
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    (Original post by Pavansanga01)
    For 6d
    I used the quadratic formula, I got -1+-squareroot|80

    How did you get -1+-squareroot|5 ?
    Sorry for poor quality post, I'm on my **** phone.
    It worked out as (-4 +- sqrt80)/(4) which turns into (-4 +- sqrt16 * sqrt5)/(4) which turns into (-4 +- 4*sqrt5)/(4) which turns into -1 +- sqrt5
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    (Original post by vyyzx)
    What was the question for k=4 and k=20?
    i think it was the one about if line blank is a tangent to the circle what is k and you had to do a discriminant or some shi
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    (Original post by timtjtim)
    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.
    Thanks this is really useful! Was 5d definitely only 2 marks?? I completely missed it and only saw it when I was closing the paper and checking through and I think I remember it being 3 marks? Would rather it was 2 obviously aha
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    (x-5)^2-25+(y+3)^2-9=0
    = (x-5)^2+(y+3)^2=34 therefore radius=Square root of 34
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    (Original post by Pavansanga01)
    For 6d
    I used the quadratic formula, I got -1+-squareroot|80

    How did you get -1+-squareroot|5 ?
    Sorry for poor quality post, I'm on my **** phone.
    thats what i did :'(
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    (Original post by beanigger)

    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
    How many marks was 6d??? 3/4?
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    (Original post by Applebananaaaa)
    (x-5)^2-25+(y+3)^2-9=0
    = (x-5)^2+(y+3)^2=34 therefore radius=Square root of 34
    Thats what I got!
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    for Q4 d) i swear the answer is (x-2)(x^2-3x+28) + 20

    because 28+20= 48 ...
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    (Original post by money-for-all)
    for Q4 d) i swear the answer is (x-2)(x^2-3x+28) + 20

    because 28+20= 48 ...
    I think when you times the whole thing out you had to get +48
    In your case when you times it all out then you'd get -56+20
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    (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)
    this is like the only question i agree with you tim tim my man
 
 
 
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