2018 Unofficial Markscheme Edexcel AS Further Maths Core Pure Paper 1 Watch

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camfanclash
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3a (black lines aren't part of the half-line)
Name:  Argand diagram.JPG
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3b
lw-1-1il = 3
(x-1)^2 + (y-1)^2 = 9
tan (pi/4) = y / (x-2)
y = x-2
(x-1)^2 + (x-3)^2 = 9
2x^2 - 8x + 1 = 0
x = 7.5 + 2 * 14^0.5
x^2 = 19.5 + 4 * 14^0.5
y = 0.5 * 14^0.5
y^2 = 3.5
x^2 + y^2 = 11 + 2 * 14^0.5

4b












7
We know that the real root 3 must be accompanied by two imaginary roots or a triangle won't be formed in the diagram.
The coefficient of z^2 is 1, so the sum of roots is -1.
3 + x + iy + x - iy = -1
2x = -4
x = -2
So the two other roots have a real part of -2.
3 - (-2) gives a base of 5.
The triangle has area 35.
So it's height is 14.
This means the imaginary part of the two other roots is 7 and -7 respectively.
Hence the roots are -2 + 7i, -2 -7i and 3.

If anyone remembers any other questions post the solution and I'll paste it into the OP.
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username2684465
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Lemur14
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Note to self to come back and add what I remember later... Could probably manage the prove by induction questions (if I can remember the matrix)!
The first two questions were nice, unfortunately I can't remember the numbers for them
Questions ~~(will check when I complete them tomorrow)
5 - 8 ^n 4n - 8n
2 1 ? 1-4n


F(n)=4^{n+1}+5^{2n-1}
Is divisible by 21



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Tbarker1
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for 3b you forgot to square the absolute value of the complex number
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susanbthapa
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(Original post by IWantIPods)








I got the hard part of the question right then got the easy part wrong , the last clearing up step
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nyxnko_
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Disclaimer:
These are my own answers. Whether they are right or not, I have no way to tell. Also, I don’t particularly remember the order of the questions so it’s going to be weird. I may have messed up the parts as well.
1
(a) This was a find the inverse q. I can’t remember this exactly but the determinant was -69 and the numbers were very very weird.
(b) \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}
(c) The 3 planes meet at a single point.

Questions 2-6
Sorry. I can’t remember the order of the questions at all but I do have some idea of what the answers are… :dontknow:

Summations q:
(a) Show that q. Hope everyone got this
(b) n=29

Transformations q:
(a) Rotation 120 degrees anticlockwise about the origin
(b) \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} or something like that. I think it was a reflection in the line y=-x?
(c) You had to multiply the matrices from (b) and (a) together. It had to be that order.
(d) k was something like 2 + \sqrt 3 It was something like that. It had an integer and an irrational number.

Vectors q:
(a) Can’t remember
(b) angle: 7 point something degrees, shortest distance: 3 hundred and something cm

Complex Numbers q:
(a) It was a shade the region q
(b) |w| was 11+2\sqrt14 I remember it was something like that.

7
p = 41, q = -159

8
Proof by Induction q. I’m hoping everyone was able to do this.

9
(a) a=-\frac{4}{15}, [tex]b=\frac{274}{15}
(b) 268pi
(c) 1 mark q about something insignificant.
(d) Something along the lines of there was a 12% difference so the model is unreliable.

PS. Again, these were just my answers. Whether they are right, I don’t know. If you guys have anything you disagree with, please comment.
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Lemur14
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The vectors question was 7.58 if you used sine, and 7.60 if you used cosine. I imagine they'll be looking for something which rounds to 7.6 :yep:
I got 375cm for the pipe, but I know there were a few other answers floating around
Part c) in question 9 was a limitation on the model (eg. assumes bottle is full/has no air in etc.)

Question 2 was the roots question, I can't remember what the numbers were though...I think the squared coefficient was -19, but I might be misremembering.
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SigmaSigma
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Ques 1. M=
(2 1 -3
4 -2 1
3 5 -2)
and i found the inverse by finding the cofactors of minors. Changing the signs. Transpose and dividing by determinant (although I know a shortcut but couldnt risk the marks).
(1/69) ( 1 13 5
-11 -5 14
-26 7 8)
part b had matrix M as three equations which had to be solved simultaneously.
2x+y-3z = -4
4x-2y+z = 1
3x+5y-2z = 5
I just multiplied the answer to part a by (-4,1,5) to get (2,1,3).
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Kadeo
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Does anyone remember how much marks each question was worth?
I remembered some of them and I've tried estimating the ones I couldn't, which gave me something like this:

Q1 Matrices (5 marks)
Q2 Roots of Polynomials (5 marks)
Q3 Complex Number / Argand Diagram (12 marks)
Q4 Summations proof and the Cosine Summation (12 marks)
Q5 Vectors (12 marks)
Q6 Matrix transformations (5 marks)
Q7 Roots on Argand diagram create a triangle with are of 35 (6 marks)
Q8 Proof by induction (12 marks)
Q9 Volumes of Revolution (11 marks)

P.S. The order of the questions is probably wrong, can't really remember much about questions 3-7 :dontknow:
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username3284896
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How is the proof by induction done? Year 12 student who will take the AS next year. I can get down to f(k+1) -f(k) but do not know where to go from there. Is it really 21?
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Lemur14
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(Original post by TurkishMathsHelp)
How is the proof by induction done? Year 12 student who will take the AS next year. I can get down to f(k+1) -f(k) but do not know where to go from there. Is it really 21?
Yes it is If I'm remembering right it was actually f(k+1)-4f(k)

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username3284896
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(Original post by Lemur14)
Yes it is If I'm remembering right it was actually f(k+1)-4f(k)

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So it would be f(k+1)-4f(k)= 16(4^k) + 5(5^2k) - 16 (4^k) + 4/5 (5^2k)

That gives 21(5^k-1) oh okay thanks.

Also if anyone can explain it please, the summation formula but the cos part of it. I do not understand it and my teachers only give the textbook for work. I used the classwiz to get the number but would like to know how to do the expanation. Sorry, probably sound stupid
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Lemur14
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(Original post by TurkishMathsHelp)
So it would be f(k+1)-4f(k)= 16(4^k) + 5(5^2k) - 16 (4^k) + 4/5 (5^2k)

That gives 21(5^k-1) oh okay thanks.

Also if anyone can explain it please, the summation formula but the cos part of it. I do not understand it and my teachers only give the textbook for work. I used the classwiz to get the number but would like to know how to do the expanation. Sorry, probably sound stupid
I can't remember what it was but that sounds about right. If it simplifies to that I'm sure it's fine

I don't think there was a summation formula for the cos part. You had to go through and substitute the values in and work out the total like that as you did.

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StugglinAlevels
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I don’t understand the cosine part of the summations question, why is it 130(2 2 2 2 2 2 2)
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Sirrup
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(Original post by StugglinAlevels)
I don’t understand the cosine part of the summations question, why is it 130(2 2 2 2 2 2 2)
Hey sorry this is really late for everyone but the summations with the cosine was a weird questions. The TLDR is that for the sigma rcos(rpi/2) you subbed in 1,2,3,4,5,6,7,8....28 in for the r's. Now if you started this you can notice a pattern (the cos is in radians by the way) it goes , 0,-4,0,6,0,-8,0,10... you notice that after each set of 4 the difference between the sum of the numbers is 2 (0+-4+0+6=2). Because you know it goes up until 28 you know there will be 28/4 sets=7. Because the sum of each set of 4 is 2. the total is 14 because there is 2x7 two's in the brackets.
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Henry james 896
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Hi there, do you guys have an unofficial mark scheme for the further applied as Edexcel 2018 paper?
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Henry james 896
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Hi there, do you have an unofficial mark scheme for the further applied paper? Thanks - Henry
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kellymay1
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could anyone show how you got from the third line to the fourth line on the series question where n=29 please
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