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Edexcel Maths FP1 UNOFFICIAL Mark Scheme 19th May 2017

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@donnaseemchandra
How are you feeling?
For the record, I noticed on an IAL paper that if you don't use the geometric progression formula from C2, you immediately lose 2 marks, they didn't accept simply plugging in numbers from 1 to 12
Reply 42
What were the 2 starting matrices in question 2 and how much was that question worth per part?
(edited 6 years ago)
Also for the matrix question where it was AB +2A= kI, me and my friends got p=3/2, k=15/2
Original post by Redcoats
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?


agree with all :smile:) hope we're on for 75/75
Original post by arniethepie
For the record, I noticed on an IAL paper that if you don't use the geometric progression formula from C2, you immediately lose 2 marks, they didn't accept simply plugging in numbers from 1 to 12


link?
anyone else get a=5 , or a=-13
@Redcoats, I have done questions separately if you want to order the OP.:wink:

I should be able to access the paper tomorrow, I've seen all the mark schemes so should be able to guess how the marks are structured once I've had a good luck.:smile:
Reply 48
Original post by k.russell
-27/2 because -27/2 x 2 + 9 = -18. The scale factor is absolute value of det m, so you need to find a such that det m = +/-18


ahh do you think you would get 3/5 for getting one solution?
These are my answers, looking for mild confirmation, here also are the marks:

1) This was a newton r f(x)=13x2+4x22x1f(x)=\frac{1}{3}x^2+\frac{4}{x^2}-2x-1

a) Prove root [6,7] Change of sign (2 marks)
b) Find first newton approx starting at 6 to 2dp , 6.45 (5 marks)

2) some matrix question:
A=[2143]A=\begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}
(a) A1=110[3142]A^{-1} = \frac{1}{10}\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix} (2 marks)

P=[36118]P=\begin{bmatrix} 3 & 6 \\ 11 & -8 \end{bmatrix}

(b) Given AB=P find B B=[2114]B = \begin{bmatrix} 2 & 1 \\ 1 & -4 \end{bmatrix} (3 marks)

3)
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 marks)
b) cart equation y=16xy=\frac{16}{x}

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

4) Complex Numbers, horra.
i) w=p4i23iw=\frac{p-4i}{2-3i}
a) a+bi form : 2p+1213+3p813i\frac{2p+12}{13}+\frac{3p-8}{13}i (3 marks)
arg(w)=π4arg(w) = \frac{\pi}{4}
b) p=20 (2 marks)
ii) z=(1λi)(4+3i)z=(1-\lambda i)(4+3i) , z=45\mid z \mid = 45

λ=±45\lambda = \pm 4\sqrt{5} (2 marks)

5) i)
A=[p23p]A= \begin{bmatrix} p & 2 \\ 3 & p \end{bmatrix} , B=[5465]B= \begin{bmatrix} -5 & 4 \\ 6 & -5 \end{bmatrix}
(a) Find AB (2)
AB+2A=kIAB + 2A = kI
find kk and pp
(b) p=32p=\frac{3}{2} , k=152k = \frac{15}{2}

ii) I forget what the question was but a=92a=\frac{9}{2} and 272\frac{-27}{2}(5 marks)

6 (a) -2-3i (1)
(x4)(x+23i)(x+2+3i)(x-4)(x+2-3i)(x+2+3i)
(b) a=2 b=-11 (5 marks)
7 (a) (4 marks)
(b) (4 marks)
(c) 1516a2 \frac{15}{16}a^2 (2 marks)

8) Proof involving sums:
(a) (5 marks)
(b) 29\frac{2}{9} (4 marks)

9) two proofs questions, each worth 6 marks


Thanks To:
- @k.russell for 5 ii second answer
- @DystopiaisReal for 5 a, I put down the wrong sign
(edited 6 years ago)
a=4.5,-13.5
k=2/9
p=1.5
k=7.5
Original post by Nik298
ahh do you think you would get 3/5 for getting one solution?


tbh I am not sure, I reckon that sounds fair though
Original post by X_IDE_sidf
These are my answers, looking for mild confirmation, here also are the marks:

1) This was a newton r f(x)=13x2+4x22x1f(x)=\frac{1}{3}x^2+\frac{4}{x^2}-2x-1

a) Prove root [6,7] Change of sign (2 marks)
b) Find first newton approx starting at 6 to 2dp , 6.45 (5 marks)

2) some matrix question:
A=[2143]A=\begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}
(a) A1=110[3142]A^-1 = \frac{1}{10}\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix} (2 marks)

P=[36118]P=\begin{bmatrix} 3 & 6 \\ 11 & -8 \end{bmatrix}

(b) Given AB=P find B B=[2114]B = \begin{bmatrix} 2 & 1 \\ 1 & -4 \end{bmatrix} (3 marks)

3)
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 marks)
b) cart equation y=16xy=\frac{16}{x}

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

4) Complex Numbers, horra.
i) w=p4i23iw=\frac{p-4i}{2-3i}
a) a+bi form : 2p+1213+3p+813i\frac{2p+12}{13}+\frac{3p+8}{13}i (3 marks)
arg(w)=π4arg(w) = \frac{\pi}{4}
b) p=20 (2 marks)
ii) z=(1λi)(4+3i)z=(1-\lambda i)(4+3i) , z=45\mid z \mid = 45

λ=±45\lambda = \pm 4\sqrt{5} (2 marks)



what question was the p=20 one?
I ****ed up all the induction, could i still get some marks?
Reply 54
I had a wicked nosebleed round solving question 10 and it broke my concentration
Sooooo disappointing...
Do you guys remember the question about summations that led to k = 2/9.
Question 9i solution/answer.
IMG_3030.jpg
Can you please do it for the k series question?
Question 9ii solution/answer.
IMG_3029.jpg
Original post by 04MR17
Question 2: Matrices
a.) 1 .(3. 1)
....10 (-4 2) [2]
b.) (2. 1)
.....(1 -4) [3]


The determinant was 10??? Oh ffs

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