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Edexcel Maths FP1 UNOFFICIAL Mark Scheme 19th May 2017

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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: -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?
(edited 6 years ago)

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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


Methodology for nerds:

Spoiler

(edited 6 years ago)
Question 1: Numerical Methods
f(x)=13x2+4x2−2x−1f(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

(edited 6 years ago)
Question 2: Matrices
A=[2−143]A=\begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}
a.) A−1=110[31−42]A^{-1} = \frac{1}{10}\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix}(2)
b.) Given AB=P find B
B=[211−4]B = \begin{bmatrix} 2 & 1 \\ 1 & -4 \end{bmatrix}(3)


Full mark scheme below:

Spoiler

(edited 6 years ago)
Question 3 - Coordinate systems (hyperbola)

x=4tx=4t , x=4tx=\frac{4}{t}
a) Find a line perpendicular to line between points t=14t=\frac{1}{4} and t=2t=2 that goes through origin. Ans: y=12xy=\frac{1}{2}x [3]

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

c) Points of intersection (42,22)(4\sqrt{2},2\sqrt{2}) and (−42,−22)(-4\sqrt{2},-2\sqrt{2}) [3]

Full mark scheme below:

Spoiler

(edited 6 years ago)
Thank you for this :biggrin:
Yes got the same for all, decent exam- proof by induction a little harder than usual
Reply 7
Avatar for Redcoats
Redcoats
OP
10
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.
Reply 8
Avatar for Ace2233
Ace2233
7
I got the same things :biggrin:
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
Question 4: Complex Numbers
i) w=p−4i2−3iw=\frac{p-4i}{2-3i}
a) a+bi form : 2p+1213+3p−813i\frac{2p+12}{13}+\frac{3p-8}{13}i [3]
arg(w)=π4arg(w) = \frac{\pi}{4}
b) p=20 [2]
ii) z=(1−λi)(4+3i)z=(1-\lambda i)(4+3i) , ∣z∣=45\mid z \mid = 45

λ=±45\lambda = \pm 4\sqrt{5} [3]
Full mark scheme below:

Spoiler

(edited 6 years ago)
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
Question 5 Matrices
A=[p23p]A= \begin{bmatrix} p & 2 \\ 3 & p \end{bmatrix} , B=[−546−5]B= \begin{bmatrix} -5 & 4 \\ 6 & -5 \end{bmatrix}
(a) Find AB
AB=[(12−5p)(4p−10)(6p−15)(12−5p)]AB= \begin{bmatrix} (12-5p) & (4p-10) \\ (6p-15) & (12-5p) \end{bmatrix} (2)

AB+2A=kIAB + 2A = kI
find kk and pp
(b) p=32p=\frac{3}{2} , k=152k = \frac{15}{2} (4)

ii) a=92a=\frac{9}{2} and −272\frac{-27}{2}[5]
Full mark scheme below:

Spoiler

(edited 6 years ago)
Question 6: Complex Numbers
a.) -3-2i [1]
(x−4)(x+2−3i)(x+2+3i)(x-4)(x+2-3i)(x+2+3i)
b.) a=2, b=-11 [5]
Full mark scheme below:

Spoiler



Question 7: Coordinate systems (parabola)
a.) Show that [4]
b.) D: (-a , -3/2 a) [4]
c.) 1516a2 \frac{15}{16}a^2[2]
Full mark scheme below:

Spoiler

(edited 6 years ago)
Reply 14
Avatar for Redcoats
Redcoats
OP
10
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.
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
Question 8: Series
a.) Show that [5]
b.) k=29\frac{2}{9}[4]

Full mark scheme below:

Spoiler

(edited 6 years ago)
Question 9 Proof
a.) Double recurrence [6]
b.) Divisibility [6]

Full mark scheme below:

Spoiler



(edited 6 years ago)
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.
Reply 19
Avatar for Redcoats
Redcoats
OP
10
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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