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Edexcel Maths FP1 UNOFFICIAL Mark Scheme 19th May 2017

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Original post by erawein
Ffs, i thought the k series one was kr(2^(r-1)) not k(2^(r-1))


How did you solve this one, tbh I just added it 12 times
Reply 81
Original post by Charles Omer
Hopefully this helps.

IMG_3029.jpg


Oh... We weren't taught that you could do multiple f(k) :/
Original post by Faznaz55
U have to prove true for n=k+2 by induction and assume true for n=k and n=k+1


I believe this is correct too.
IMG_3030.jpg
Can anyone remember the actual question, 5, part ii?
Original post by Charles Omer
I believe this is correct too.
IMG_3030.jpg

Yes it is
(edited 6 years ago)
Original post by __Will__
Oh... We weren't taught that you could do multiple f(k) :/


I think there are different ways to do it, it would have been better if you were taught all the methods... I'm sorry to hear that.
Reply 86
Original post by glad-he-ate-her
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


Fully did the same thing then reverse engineered the 4095 into the sum of a geometric sequence formula to get the marks 😅😂
Reply 87
What do you think grade boundaries for 90% will be?
Original post by X_IDE_sidf
How did you solve this one, tbh I just added it 12 times


have to use geometric series formula from c2.
Also in the determinant question there is another solution, a = -27/2 as well as -9/2 as both of these satisfy mod(detM) = 18. Great job for putting up this mark scheme with the latex and stuff
Reply 89
Original post by k.russell
You have to test n=1 and n=2, assume it's true for n = k and n = k + 1 then test if it's true for n = k + 2


Use the assumption of k with k-1 to get answer and factorise it into 3^k+1((k+1)+1)
Original post by ~K1
Use the assumption of k with k-1 to get answer and factorise it into 3^k+1((k+1)+1)


it's the same damn thing b xxx
Reply 91
Must a Cartesian equation be written with y as the subject, as I had written xy=16 and not y=16/x.
Original post by k.russell
have to use geometric series formula from c2.
Also in the determinant question there is another solution, a = -27/2 as well as -9/2 as both of these satisfy mod(detM) = 18. Great job for putting up this mark scheme with the latex and stuff


ffs, that was a sneaky one. I always seem to forget to consider the negative case.
Original post by X_IDE_sidf
ffs, that was a sneaky one. I always seem to forget to consider the negative case.


yeah I have made that mistake before. luckily I noticed that it was asking for values of a and a wasn't squared so sort of guessed it must be the negative one
Original post by X_IDE_sidf
Sorry, I'll use English,

The question was to prove that 19 divides 33n223n+13^{3n-2}-2^{3n+1} where n is a natural number (positive, non zero integer), which I'm sure you can easily do yourself.

I cannot remember the other one, but it was a sequence of some sort.


Shame you couldn't have typeset it properly so that text was a consistent size.
Original post by TBoy11
Must a Cartesian equation be written with y as the subject, as I had written xy=16 and not y=16/x.


Both are correct, btw c=4 and use formula book to get ans directly lol
Original post by TBoy11
Must a Cartesian equation be written with y as the subject, as I had written xy=16 and not y=16/x.


this is fine
Original post by TBoy11
Must a Cartesian equation be written with y as the subject, as I had written xy=16 and not y=16/x.

Either way is fine but xy=16 is probably better.
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)

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} (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} and 272\frac{-27}{2} (4 marks)

9) two proofs questions, each worth 6 marks


Thanks To:
- @k.russell for 8 b second answer


the second answer should be going in 5ii) btw
Original post by Mathsislive
Shame you couldn't have typeset it properly so that text was a consistent size.


If you care so much, feel free to typeset it properly. I'd recommend doing it in latex with the exam document class. Post the PDF after, would be pretty.

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