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Random answers:
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: 32i
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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Edexcel Maths FP1 UNOFFICIAL Mark Scheme 19th May 2017 watch

Redcoats
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 19052017 10:28
Last edited by Redcoats; 19052017 at 10:49. 
04MR17
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 19052017 10:32
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:ShowSo, 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 nerdgasm at how complex it looks.Last edited by 04MR17; 21052017 at 19:19. 
04MR17
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 19052017 10:33
Question 1: Numerical Methods
a.) Show that 6<a<7 [2]
b.) Newton Raphson
Answer = 6.45 [5]
Full mark scheme below:
Spoiler:Show1. 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 NR (Fully Correct) M1 (A1)
Correct answer = 6.45Last edited by 04MR17; 21052017 at 18:18. 
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 19052017 10:33
Last edited by 04MR17; 21052017 at 18:20. 
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 19052017 10:34
Question 3  Coordinate systems (hyperbola)
,
a) Find a line perpendicular to line between points and that goes through origin. Ans: [3]
b) Cartesian equation or: [1]
c) Points of intersection and [3]
Full mark scheme below:
Last edited by 04MR17; 21052017 at 18:22. 
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 19052017 10:36
Thank you for this

DwarfShortage
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 19052017 10:37
Yes got the same for all, decent exam proof by induction a little harder than usual

Redcoats
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 19052017 10:38
(Original post by gladheateher)
Yes got the same for all, decent exam proof by induction a little harder than usual 
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 19052017 10:38
I got the same things

DwarfShortage
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 19052017 10:38
Am i the only one who forgot the sum for 2 to the power of r1 from c2 so manually wrote out all 12 terms such mathematical elegance

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 19052017 10:39
Last edited by 04MR17; 21052017 at 20:09. 
DwarfShortage
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 19052017 10:39
(Original post by Redcoats)
Remember any others? I got the induction.
proof by induction i got 27f(k) 19 times 3 to the something 
04MR17
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 19052017 10:40
Question 5 Matrices
,
(a) Find AB
(2)
find and
(b) , (4)
ii) and [5]
Full mark scheme below:
Spoiler:Show5.)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: o.e.
(A1 for the above all correct.)
M1 for equating 4p6 = 0 or 6p9 = 0; to reach p=…
A1 for ,
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 andLast edited by 04MR17; 21052017 at 18:29. 
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 19052017 10:40
Question 6: Complex Numbers
a.) 32i [1]
b.) a=2, b=11 [5]
Full mark scheme below:
Spoiler:Show
Question 7: Coordinate systems (parabola)
a.) Show that [4]
b.) D: (a , 3/2 a) [4]
c.) [2]
Full mark scheme below:
Spoiler:Show
7.)a.) M1 for dy/dx = a^{1/2}x^{1/2} or for 2y(dy/dx)=4a
A1 for gradient = (1/q)
M1 for correct use of (yy_{1}) = m (xx_{1}); 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:
A1 for (15a^{2}/16) or (15/16)a^{2}
Last edited by 04MR17; 21052017 at 18:37. 
Redcoats
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 19052017 10:41
(Original post by gladheateher)
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 
KingsAlpaca
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 19052017 10:42
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 + (3p8)i/13 p=20 lambda=+4sqrt(5)
5. (125p, 4p10
6p15, 125p)
p=3/2 k=15/2 a=9/2 or 27/2
6. 32i 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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 19052017 10:44
Question 8: Series
a.) Show that [5]
b.) k=[4]
Full mark scheme below:
Spoiler:Show8.) 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 3term 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(2^{r1})
A1 for 2(12^{12})/12
(2^{nd} M1 1^{st} A1: These two marks can be implied by seeing 910 or 910k or 910/k)
A1 k=2/9Last edited by 04MR17; 21052017 at 18:51. 
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 19052017 10:45
Question 9 Proof
a.) Double recurrence [6]
b.) Divisibility [6]
Full mark scheme below:
Spoiler:Show
9.i.) B1 for u_{1}=3^{1}((1)+1) =6 o.e. and u_{2}=3^{2}((2)+1) =27 o.e.
M1 for attempting u_{k+2} in terms of u_{k+1} and u_{k}
A1 for correct expression (e.g. u_{k+2}=6(3^{k+1}(k+1+1)) – 9(3^{k}(k+1)) o.e.
M1 for Attempting u_{k+2}in terms of 3^{k+2}
A1 for correct expression: 3^{k+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 3^{1}+2^{4}=19 o.e.
M1 Attempts f(k+1)f(k) (see below for correct solution)
A1 Correct expression for f(k+1): [3^{3k+1}+2^{3k+4}] –[3^{3k2}+2^{3k+1}]
M1 Achieves an expression in 3^{3k2 }and 2^{3k+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 3^{1}+2^{4}=19 o.e.
M1 Attempts f(k+1) (see below for correct solution)
A1 Correct expression for f(k+1) (Can be unsimplified) [3^{3(k+1)2}+2^{3(k+1)+1}]
M1 Achieves an expression in 3^{3k2 }and 2^{3k+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 3^{1}+2^{4}=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) [3^{3(k+1)2}+2^{3(k+1)+1}]+[3^{3k2}+2^{3k+1}]
M1 Achieves an expression in 3^{3k2 }and 2^{3k+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.”
Last edited by 04MR17; 21052017 at 20:06. 
DwarfShortage
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 19052017 10:50
(Original post by Redcoats)
I got 8f(k) + 19(3^3k+blah). There are loads of different methods for it. 
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 19052017 10:51
(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 + (3p8)i/13 p=20 lambda=+4sqrt(5)
5. (125p, 4p10
6p15, 125p)
p=3/2 k=15/2 a=9/2 or 27/2
6. 32i 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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