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Edexcel Maths FP1 UNOFFICIAL Mark Scheme 19th May 2017

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    Sign change (-0.8888888 to 1.4149....)
    alpha = 6.45
    lambda = +- 4sqrt(5)
    a = 9/2 or -27/2
    Area (FXD) 15/16 a^2
    D: (-a , -3/2 a)
    Induction at the end (use assumption)
    k = 2/9 for sum
    other root: -3-2i
    a = 2
    b = -11 (or something like that)
    y=x/2 xy=16 (4sqrt(2), 2sqrt(2)) and (-4sqrt(2),-2sqrt(2))
    p = 20 somewhere

    Anyone else remember?
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    19th May. FP1 Edexcel Unofficial Mark Scheme.

    Here are the [speculative] answers with [speculative] full mark scheme breakdown:
    Question 1: Numerical Methods
    Question 2: Matrices
    Question 3: Coordinate Systems (hyperbola)
    Question 4: Complex Numbers
    Question 5: Matrices
    Question 6: Complex Numbers
    Question 7: Coordinate Systems (parabola)
    Question 8: Series
    Question 9: Proof

    Credit to X_IDE_sidf for providing the latex solutions for me.

    UMS Grade Boundaries are as follows:
    80=A
    70=B
    60=C
    50=D
    40=E

    Estimated Raw => UMS Conversion is as follows:
    Spoiler:
    Show

    Raw (/75) => UMS (/100)
    75=>100
    72=>95
    69=>91
    68=>89
    65=>85
    62=>80
    61=>79
    59=>76
    58=>74
    57=>73
    56=>72
    55=>71
    54=>70
    53=>69
    51=>66
    50=>64
    49=>63
    48=>61
    47=>60
    46=>59
    44=>56
    43=>55
    42=>54
    41=>53
    40=>51
    39=>50
    38=>49
    36=>46
    35=>45
    34=>44
    33=>43
    32=>40
    31=>39
    29=>36
    27=>33
    26=>32
    24=>29
    23=>28
    22=>27
    21=>26
    20=>24
    19=>23
    18=>22
    16=>19
    15=>18
    14=> 17
    13=> 16
    12=> 15
    11=> 14
    9=>11
    8=>10
    7=>9
    6=>8
    5=>6
    4=>5
    3=>4
    2=>3
    1=>2
    0=>0

    Methodology for nerds:
    Spoiler:
    Show
    So, I took similar questions from previous papers to use as a template for the mark scheme. I took the ums conversions for that paper, and multiplied each conversion value by the proportion of marks in this years paper which were from the similar questions. I then summed the values across all the different papers to get a set of results which look loosely correct. Some conversions (particularly the top 10) are lacking. This is because a lot of these UMS values do not have raw values attached to them on the old papers, and so it is difficult to anticipate where these will fall. Sometimes, if only one or two raw values are missing for that UMS value, I will bisect the interval between the two adjacent conversion values, to help get more conversions on my list. If anyone has any questions about my methods, ask me and I will send you my spreadsheet and you will have a nerd-gasm at how complex it looks.
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    Question 1: Numerical Methods
    f(x)=\frac{1}{3}x^2+\frac{4}{x^2  }-2x-1
    a.) Show that 6<a<7 [2]
    b.) Newton Raphson
    Answer = 6.45 [5]
    Full mark scheme below:
    Spoiler:
    Show
    1. a.) M1 for substitution and at least 1 correct value.

    A1 for both values correct and sign change stated.

    b.) Differentiation (Fully Correct) M1 (A1)

    Application of N-R (Fully Correct) M1 (A1)

    Correct answer = 6.45
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    Question 2: Matrices
    A=\begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}
    a.) A^{-1} = \frac{1}{10}\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix}(2)
    b.) Given AB=P find B
    B = \begin{bmatrix} 2 & 1 \\ 1 & -4 \end{bmatrix}(3)


    Full mark scheme below:
    Spoiler:
    Show

    2. a.)  \frac{1}{det(A)}\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix} For M1



     \frac{1}{10}\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix} For A1

    b.) M1 for P=AB

    M1 for Applying “their A-1

    A1  \begin{bmatrix} 2 & 1 \\ 1 & -4 \end{bmatrix} cso
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    Question 3 - Coordinate systems (hyperbola)

    x=4t , x=\frac{4}{t}
    a) Find a line perpendicular to line between points t=\frac{1}{4} and t=2 that goes through origin. Ans: y=\frac{1}{2}x [3]

    b) Cartesian equation y=\frac{16}{x} or: xy=16[1]

    c) Points of intersection (4\sqrt{2},2\sqrt{2}) and (-4\sqrt{2},-2\sqrt{2}) [3]

    Full mark scheme below:
    Spoiler:
    Show
    3. a.) B1 for P: (1, 16); Q: (8, 2) seen or implied

    M1 for y=\frac{16-2}{1-8} or for y=\frac{2-16}{8-7}

    A1 for -1/“their PQ gradient”; y=\frac{1}{2}x

    b.) y=\frac{16}{x} or: xy=16 B1

    c.) M1 Attempt to solve 2 equations in x and y only.

    A1 for Correct Substitution or elimination of one variable to find: x= (\pm4\sqrt{2} or y= (\pm2\sqrt{2})

    A1 for Substitution to find: x= (*+/-4\sqrt{2} or y= (+/-2\sqrt{2})
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    Thank you for this
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    Yes got the same for all, decent exam- proof by induction a little harder than usual
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    (Original post by glad-he-ate-her)
    Yes got the same for all, decent exam- proof by induction a little harder than usual
    Remember any others? I got the induction.
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    I got the same things
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    Am i the only one who forgot the sum for 2 to the power of r-1 from c2 so manually wrote out all 12 terms :clap2:such mathematical elegance
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    Question 4: Complex Numbers
    i) w=\frac{p-4i}{2-3i}
    a) a+bi form : \frac{2p+12}{13}+\frac{3p-8}{13}i [3]
    arg(w) = \frac{\pi}{4}
    b) p=20 [2]
    ii) z=(1-\lambda i)(4+3i) , \mid z \mid = 45

    \lambda = \pm 4\sqrt{5} [3]
    Full mark scheme below:
    Spoiler:
    Show


    4.) i.) a.) M1 for multiplying by 3+2i

    B1 for 13 seen

    M1 for obtaining a numerator with 2 real and 2 imaginary parts.

    A1 for \frac{2p+12}{13}+\frac{3p-8}{13}i or for \frac{(2p+12)+(3p-8)i}{13}

    b.) M1 for \frac{(2p+12)+(3p-8)i}{13} = tan(pi/4) o.e.

    A1 for p = 20

    ii.) M1 for correct expansion of brackets

    M1 for sqrt{(4 + 3\lambda)+ (3 – 4\lambda)}=2025 o.e.

    A1 for \lambda = \pm 4\sqrt{5}

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    (Original post by Redcoats)
    Remember any others? I got the induction.
    matrix was 1/10 times something like 20 20 -40 and 10 but the point is they were all divisble by 10.
    proof by induction i got 27f(k) -19 times 3 to the something
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    Question 5 Matrices
    A= \begin{bmatrix} p & 2 \\ 3 & p \end{bmatrix} , B= \begin{bmatrix} -5 & 4 \\ 6 & -5 \end{bmatrix}
    (a) Find AB
    AB= \begin{bmatrix} (12-5p) & (4p-10) \\ (6p-15) & (12-5p) \end{bmatrix} (2)

    AB + 2A = kI
    find k and p
    (b) p=\frac{3}{2} , k = \frac{15}{2} (4)

    ii) a=\frac{9}{2} and \frac{-27}{2}[5]
    Full mark scheme below:
    Spoiler:
    Show
    5.)i.)a.) M1: Correct matrix multiplication method implied by one or two correct terms in correct positions.

    A1: for all four terms correct and simplified and in correct position

    b.) M1 A1 for: \begin{bmatrix} (12-5p) & (4p-10) \\ (6p-15) & (12-5p) \end{bmatrix} + \begin{bmatrix} (2p) & (4) \\ (6) & (2p) \end{bmatrix} = \begin{bmatrix} (k) & (0) \\ (0) & (k) \end{bmatrix} o.e.

    (A1 for the above all correct.)

    M1 for equating 4p-6 = 0 or 6p-9 = 0; to reach p=…

    A1 for p=\frac{3}{2} , k = \frac{15}{2}

    ii.) M1 for finding det(M)

    A1 = 2a+9

    M1 for applying 270 = 15 x “their det(M)”

    A1ft for “their det(M)” = 18

    A1 for a=\frac{9}{2} and \frac{-27}{2}
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    Question 6: Complex Numbers
    a.) -3-2i [1]
    (x-4)(x+2-3i)(x+2+3i)
    b.) a=2, b=-11 [5]
    Full mark scheme below:
    Spoiler:
    Show




    6.) a.) B1 for -3-2i or -2i-3 o.e.

    b.) M1 for expansion of (x-4)(x+2-3i)(x+2+3i)

    M1 for -3i2=3

    M1 for (x-4)(x+2-3i)(x+2+3i) => (x-4)(x<sup>2</sup>+6x+13)

    A1 for a=3

    A1 for b=-11


    Question 7: Coordinate systems (parabola)
    a.) Show that [4]
    b.) D: (-a , -3/2 a) [4]
    c.)  \frac{15}{16}a^2[2]
    Full mark scheme below:
    Spoiler:
    Show

    7.)a.) M1 for dy/dx = a1/2x-1/2 or for 2y(dy/dx)=4a

    A1 for gradient = (1/q)

    M1 for correct use of (y-y1) = m (x-x1); or y=mx+c, where c is found as a constant.

    b.) M1 for substituting X into tangent to find q

    A1 for q=1/2

    B1 for x=-a

    A1ft for D: (-a , “-3/2 a”) using “their q”

    (Special case: if x is found x=/=-a then maximum is M1A1A0A0)

    c.) M1 for correct formula used: \frac{1}{2} ×  FX × \frac{-3a}{2}

    A1 for (15a2/16) or (15/16)a2
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    (Original post by glad-he-ate-her)
    matrix was 1/10 times something like 20 20 -40 and 10 but the point is they were all divisble by 10.
    proof by induction i got 27f(k) -19 times 3 to the something
    I got 8f(k) + 19(3^3k+blah). There are loads of different methods for it.
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    My Answer for FP1 2017,
    1. ST 6.45
    2. (3/10, 1/10
    -2/5, 1/5)
    (2,1
    1,-4)
    3. y=x/2 xy=16 (4sqrt(2), 2sqrt(2)) and (-4sqrt(2),-2sqrt(2))
    4. (2p+12)/13 + (3p-8)i/13 p=20 lambda=+-4sqrt(5)
    5. (12-5p, 4p-10
    6p-15, 12-5p)
    p=3/2 k=15/2 a=9/2 or -27/2
    6. -3-2i a=2 b=-11
    7. ST (-a, -3a/2) s=15a^2/16
    8. ST k=2/9
    9. ST ST
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    Question 8: Series
    a.) Show that [5]
    b.) k=\frac{2}{9}[4]

    Full mark scheme below:

    Spoiler:
    Show
    8.) a.) M1 for
    Attachment 650180
    M1 for application of standard results.

    A1 for correct unsimplified expression e.g.:
    3[(n/6)(n+1)(2n+1)]+8[(n/2)(n+1)]+3(n) or better
    M1 for factorising out (n/2)or equivalent to leave a 3-term quadratic e.g.:

    A1 for (n/2)(2n+5)(n+3) * given answer cso

    b.) M1 for application of previous answer, this alone scores M0

    M1 for attempt to supply the sum to n terms of k(2r-1)

    A1 for 2(1-212)/1-2

    (2nd M1 1st A1: These two marks can be implied by seeing 910 or 910k or 910/k)

    A1 k=2/9
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    Question 9 Proof
    a.) Double recurrence [6]
    b.) Divisibility [6]

    Full mark scheme below:
    Spoiler:
    Show

    9.i.) B1 for u1=31((1)+1) =6 o.e. and u2=32((2)+1) =27 o.e.

    M1 for attempting uk+2 in terms of uk+1 and uk

    A1 for correct expression (e.g. uk+2=6(3k+1(k+1+1)) – 9(3k(k+1)) o.e.

    M1 for Attempting uk+2in terms of 3k+2

    A1 for correct expression: 3k+2((k+2)+1)

    A1 for correct statements seen anywhere in the solution. cso.

    “If the result is true for n = k and n = k + 1 then it has been shown true for n = k + 2. As it is true for n = 1and n = 2 then it is true for all n (positive integers.)”

    (Do not award final A if n defined incorrectly e.g. ‘n is an integer’ award A0)

    ii.) B1 for 31+24=19 o.e.

    M1 Attempts f(k+1)-f(k) (see below for correct solution)

    A1 Correct expression for f(k+1): [33k+1+23k+4] –[33k-2+23k+1]

    M1 Achieves an expression in 33k-2 and 23k+1

    A1 Achieves a correct expression for f(k+1) which is divisible by 19

    A1 Correct conclusion at the end, at least as given, and all previous marks scored. cso.

    “If the result is true for n = k, then it is now true for n = k + 1. As the result has shown to be true for n = 1, then the result is true for all n.”

    Way 2:

    B1 for 31+24=19 o.e.

    M1 Attempts f(k+1) (see below for correct solution)

    A1 Correct expression for f(k+1) (Can be unsimplified) [33(k+1)-2+23(k+1)+1]

    M1 Achieves an expression in 33k-2 and 23k+1

    A1 Achieves a correct expression for f(k+1) which is divisible by 19

    A1 Correct conclusion at the end, at least as given, and all previous marks scored. cso.

    “If the result is true for n = k, then it is now true for n = k + 1. As the result has shown to be true for n = 1, then the result is true for all n.”

    Way 3:

    B1 for 31+24=19 o.e.

    M1 Attempts f(k+1)+f(k) (see below for correct solution)

    A1 Correct expression for f(k+1) (Can be unsimplified) [33(k+1)-2+23(k+1)+1]+[33k-2+23k+1]



    M1 Achieves an expression in 33k-2 and 23k+1

    A1 Achieves a correct expression for f(k+1) which is divisible by 19

    A1 Correct conclusion at the end, at least as given, and all previous marks scored. cso.

    “If the result is true for n = k, then it is now true for n = k + 1. As the result has shown to be true for n = 1, then the result is true for all n.”


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    (Original post by Redcoats)
    I got 8f(k) + 19(3^3k+blah). There are loads of different methods for it.
    That is right as well- i did it twice and got yours and my answer but went with mine.
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    (Original post by kingsalpaca)
    my answer for fp1 2017,
    1. St 6.45
    2. (3/10, 1/10
    -2/5, 1/5)
    (2,1
    1,-4)
    3. Y=x/2 xy=16 (4sqrt(2), 2sqrt(2)) and (-4sqrt(2),-2sqrt(2))
    4. (2p+12)/13 + (3p-8)i/13 p=20 lambda=+-4sqrt(5)
    5. (12-5p, 4p-10
    6p-15, 12-5p)
    p=3/2 k=15/2 a=9/2 or -27/2
    6. -3-2i a=2 b=-11
    7. St (-a, -3a/2) s=15a^2/16
    8. St k=2/9
    9. St st
    exact same as what i got
 
 
 
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