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2017 AQA Level 2 Further Maths Paper 2 Unofficial Mark Scheme Watch

  • View Poll Results: How many marks do you think you got?
    0-30
    6
    3.33%
    31-50
    11
    6.11%
    51-60
    17
    9.44%
    61-70
    16
    8.89%
    71-80
    25
    13.89%
    81-90
    29
    16.11%
    91-100
    54
    30.00%
    101-105
    22
    12.22%

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    8
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    do you guys still remember what q23 was?
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    2
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    (Original post by gcse0)
    I got 19.2 units squared for the area of the triangle - is that right?
    No it was 24
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    8
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    (Original post by darude_sand)
    No it was 24
    i swear it was 16 square units
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    8
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    (Original post by Nudelauflauf)
    I swear OPB was 24 cause it was 10 across and 4.2 up so 10*4.2*1/2 is 24?
    no because 4 was the height, 8 was the base and therefore area o the triangle was 16
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    9
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    (Original post by darude_sand)
    No it was 24
    How?
    Could you show your working?
    I thought it was 4.8*0.5*8 = 19.2
    8-3.2 = 4.8 which was the y-coordinate
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    11
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    What do you think between 151 and 165 will be? Do you think the boundaries will be higher or lower than usual?
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    5
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    (Original post by Npt512)
    i swear it was 16 square units
    It was definitely 16 square units yes
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    5
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    (Original post by Loci Pi)
    What do you think between 151 and 165 will be? Do you think the boundaries will be higher or lower than usual?
    I'd say lower than usual, those two papers combined were harder than the 2015 papers, and that was 140 for an A^. Either way, 151 and 165 will both almost definitely be an A^
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    5
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    (Original post by Loci Pi)
    What do you think between 151 and 165 will be? Do you think the boundaries will be higher or lower than usual?
    I'd say lower than usual, those two papers combined were harder than the 2015 papers, and that was 140 for an A^. Either way, 151 to 165 will almost definitely be an A^
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    3
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    (Original post by Npt512)
    do you guys still remember what q23 was?
    Prove that 1/tan^2 x -1/cos^2 x(maybe sin^2 x idk) is a constant or something and you get -1 as an answer when you simplify it
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    1
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    Does anyone remember that last question? I think I messed it up
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    2
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    (Original post by gcse0)
    How?
    Could you show your working?
    I thought it was 4.8*0.5*8 = 19.2
    8-3.2 = 4.8 which was the y-coordinate
    You had to split the area into to triangles, as you can't directly work out the area from (4.8*0.5*8). I don't know how you got 8, agree with you on 4.8 though. Did you work out the x coord of the point C, or whatever it was (the point in the middle)? And are you sure it was 8 for the x axis, and not 10? Because I remember C being at (4, 4.8) (I think that was it), and so as the total length was 10, you had two triangles. So then I did (6*0.5*4.8) + (4*4.8*0.5) = 24. I could be wrong I guess but I'm pretty sure that was on the right lines at least.



    (Original post by Npt512)
    i swear it was 16 square units
    What was your working out?
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    11
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    (Original post by Tomazza123)
    I'd say lower than usual, those two papers combined were harder than the 2015 papers, and that was 140 for an A^. Either way, 151 to 165 will almost definitely be an A^
    Yay, thanks
    I think the boundary for A^ will be between 138 and 142. I found these papers really hard but I didn't find the 2015 paper too bad.
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    2
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    (Original post by gcse0)
    How?
    Could you show your working?
    I thought it was 4.8*0.5*8 = 19.2
    8-3.2 = 4.8 which was the y-coordinate
    Yeah 8 might have been the base not 10 actually, still not 100% sure though
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    8
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    What were the questions/working outs for these answers?:

    8. 106.7
    9. x=-2
    10. 15.2
    13. 50a^2 / 3
    18. (32/5) y^2
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    (Original post by etothepiiplusone)
    1a. 30
    b. -1.5
    2a. 13 -2
    6 1
    b. 11/8
    c. Because number of columns of B does not equal rows of A
    3a. e.g 3, 5, 91
    b. Factorised ab-11b = b(a-11) so b is arbitrary square e.g 4, a=36
    4. 2744
    5. 4/5
    6. 12.31
    7. 16
    8. 106.7
    9. x=-2
    10. 15.2
    11b. -5 <= f(x) <= 7
    12a. 3(5-x)(5+x)
    b. 12n
    13. 50a^2 / 3
    14. a=4, b=10
    15. y=(8x)/(w+1)
    16a.3^(-2b)
    b. 5^(x+2)
    c. 2^(3m)
    17a. (4, -25)
    c. 5.79 and 1.21
    18. (32/5) y^2
    19a.k
    b. -k
    c. sqrt(1-k^2)
    20a. Angle at circumference is half angle at centre
    b. e.g Angle in triangle 90-x
    Angle on straight line 90+x
    Opposite angle in cyclic quadrilateral 90-x
    Other base angle 90-x
    Other angle in triangle 180-2(90-x) = 2x QED
    21a. (0, 8)
    b. -2x+2
    c. Normal equation y=-1/2 x + 8
    So intersects at (16, 0)
    2*6 = 16-4
    22. (-6/5, -7/5) and (2, 5)
    23. -1
    24. 3(2x-5)^2 - 70
    Can you please put the marks in, I don't care if they are only a guestimate


    Please someone make a list of marks, I want to see how I did, +rep to whoever wants to
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    What do you think 140 would be? A*? I really wanted that A^... oh well
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    13
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    Not sure about the others, but for 13. 50a^2 / 3 it was a 3 mark question where you had to simplify three fractions (one multiplied and one divided) that had powers of a.
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    8
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    (Original post by darude_sand)
    You had to split the area into to triangles, as you can't directly work out the area from (4.8*0.5*8). I don't know how you got 8, agree with you on 4.8 though. Did you work out the x coord of the point C, or whatever it was (the point in the middle)? And are you sure it was 8 for the x axis, and not 10? Because I remember C being at (4, 4.8) (I think that was it), and so as the total length was 10, you had two triangles. So then I did (6*0.5*4.8) + (4*4.8*0.5) = 24. I could be wrong I guess but I'm pretty sure that was on the right lines at least.





    What was your working out?
    you can see from the diagram that OBP = 8 x 4 x 0.5 = 16

    Attachment 665876
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    0
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    (Original post by LUSer21)
    Will I still get the mark if I write -30?
    that's what I thought too I wrote 30 though I think you will only lose 1 mark if any
 
 
 
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