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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
    3.37%
    31-50
    11
    6.18%
    51-60
    17
    9.55%
    61-70
    16
    8.99%
    71-80
    25
    14.04%
    81-90
    29
    16.29%
    91-100
    53
    29.78%
    101-105
    21
    11.80%

    • Thread Starter
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    1a. 30 (and -30 is probably allowed) Marks=3?
    b. -1.5 Marks=?

    2a. \begin{pmatrix} -13 & 2 \\ 6 & 1 \end{pmatrix} Marks=?
    b.  \frac{11}{8} Marks=?
    c. Because the number of columns of B does not equal the number of rows of A Marks=1

    3a. Since 1365 = 3*5*7*13 , choose any two factors, multiply those together, and write the other two factors: e.g 3, 5, 91 Marks=?
    b. Factorised, ab-11b = b(a-11) so b is an arbitrary square e.g 4, and a=36 Marks=?

    4.  2744 Marks=?

    5. \frac{4}{5} Marks=?

    6. 12.31 Marks=?

    7. 16 Marks=?

    8. 106.1 Marks=?

    9. x=-2 Marks=? www.desmos.com/calculator/um4a23ffwt .
    Otherwise it is clear that  \{-2, -1, 0, 1\} \cap \{-2, -3, -4, ...\} = \{-2\} , the first two sets being the integer solutions to the two inequalities

    10. 15.2 Marks=?

    11a. Draw the thing. Marks=?
    11b.  -5 <= f(x) <= 7 Marks=?

    12a. 3(5-x)(5+x) Marks=?
    b. 12n Marks=?

    13. \frac{50a^2}{3} Marks=5?

    14. a=4, b=10 Marks=?

    15. y=\frac{8x}{w+1} Marks=3?

    16a. 3^{-2b} Marks=1
    b. 5^{x+2} Marks=1
    c. 2^{3m} Marks=1

    17a. (4, -25) Marks=?
    c. The roots are x=5.79 and x=1.21 Marks=?
    18. \frac{32y^2}{5} Marks=?

    19a. k. Marks=1 Think of 19a. and 19b. as transformations: https://www.desmos.com/calculator/4kphkki8vl
    b. -k Marks=1
    c.  \sqrt{1-k^2}. This is rearranging of  \sin^2 x + \cos^2 x = 1 with  \sin^2 x = k^2 Marks=2

    20a. Angle at circumference is half angle at centre Marks=1
    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 Marks=5?
    There are obviously many other ways to do this. I think we all know if we got or did not get marks on this question

    21a. (0, 8) Marks=?
    b. -2x+2 Marks=?
    c. Normal equation y= \frac{-1}{2}x + 8
    So intersects x-axis at (16, 0)
    The rest is easy (show one length is twice the other) Marks=?

    22.  (-6/5, -7/5) and (2, 5) were the pairs of solutions Marks=?

    23. -1 is the constant, because

     \frac{1}{\tan^2 x} - \frac{1}{\sin^2 x}

     = \frac{1}{(\frac{\sin x}{\cos x})^2} - \frac{1}{\sin^2 x}

     = \frac{1}{\frac{\sin^2 x}{\cos^2 x}} - \frac{1}{\sin^2 x}

     = \frac{\cos^2 x}{\sin^2 x} - \frac{1}{\sin^2 x}

     = \frac{\cos^2 x - 1}{\sin^2 x}

     = \frac{- (-\cos^2 x + 1)}{\sin^2 x}

     = \frac{- \sin^2 x}{\sin^2 x}

     = -1 Marks=4?

    24. 3(2x-5)^2 - 70 Marks=5
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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
    "Because number of columns of B does not equal rows of A"... Wut?
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    How did you do the cos a question that was worth 2 marks
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    How much do you think 148 would be
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    (Original post by Shanpatel2000)
    How much do you think 148 would be
    Probably an A^, but at least an A*.
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    (Original post by Frankiitch)
    How did you do the cos a question that was worth 2 marks
    Square it so cos^2(a)= 1-sin^2(a)
    Sina=k so Sin^2(a)=k^2
    Thus cos^2(a)=1-k^2
    Square Root it so cosa= root(1-k^2)
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    Because number of columns of B does not equal rows of A

    ^ For this question I put there are fewer columns in B than A. Mark or no mark?
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    Q1 difference between 20th and 8th terms = 20th term - 8th term = -30?
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    (Original post by _gcx)
    Probably an A^, but at least an A*.
    Hoping for A^; thanks
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    (Original post by LUSer21)
    Q1 difference between 20th and 8th terms = 20th term - 8th term = -30?
    Yeah but the values of the 20th term nd the 8th term were both negative numbers so you don't need to write -30 as a difference, just 30 is correct
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    Can someone describe question 9 and 10 for me since I forgot?
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    I got around 87 in this and 55 in the other.... is this an A*
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    (Original post by Mermaidqueen)
    Yeah but the values of the 20th term nd the 8th term were both negative numbers so you don't need to write -30 as a difference, just 30 is correct
    Will I still get the mark if I write -30?
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    What was the answer for the question where you had to work out the area of OAB i got 24 but I don't see it in the ms
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    (Original post by LUSer21)
    Will I still get the mark if I write -30?
    Im not sure, i don't remember how many marks it was but if it was out of 2 marks
    you'd probably get one mark for calculating -30 and the second for writing 30
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    (Original post by Oliver_2001)
    What was the answer for the question where you had to work out the area of OAB i got 24 but I don't see it in the ms
    It was the area of OPB (which was 16) not OAB
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    (Original post by Oliver_2001)
    What was the answer for the question where you had to work out the area of OAB i got 24 but I don't see it in the ms
    What did u do for it? I don't remember my answer but what i did was:
    calculate the coordinates of A and B, then found the distance AB
    then i did 2/5 multiple by distance AB bc the ratio was 2;3 and the length we needed for the little triangle was the 2 part.
    Anyways so i did 2/5 (distance AB) multiplied by 8 then divide by 2
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    (Original post by Mermaidqueen)
    Im not sure, i don't remember how many marks it was but if it was out of 2 marks
    you'd probably get one mark for calculating -30 and the second for writing 30
    Idk, but I think the first mark is going to be substituting 20 and 9 into the nth term and the second mark is going to be the answer. So I think -30 will be in the accept column.
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    Could you put the marks in please?
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    what was the first question again?
 
 
 
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