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# FP1 AQA 15th June 2016 UNOFFICIAL MARK SCHEME (what you have been looking for) Watch

1. (Original post by Sameechoudhury;[url="tel:65832083")
65832083[/url]]Question 8d is wrong, it's meant to be 2root3 2
It asked for the original coordinates before the transformation I got ((3^0.5)/2, 2)
2. (Original post by JackLloyd98)
5b is n(2n+1)(2n-3)(n-1)
Do you know where i went wrong?
3. (Original post by henryhong)

Do you know where i went wrong?
You forgot to to take out the n with the 3n
4. (Original post by henryhong;[url="tel:65832655")
65832655[/url]]

Do you know where i went wrong?
Those are my workings
5. (Original post by JackLloyd98)
It asked for the original coordinates before the transformation I got ((3^0.5)/2, 2)
I got (-root3/2 , -2)
6. (Original post by moodyloo1998)
I got (-root3/2 , -2)
Maybe I read it wrong, I thought the question was stretch followed by reflection.
7. (Original post by JackLloyd98)

Those are my workings
Thank you man, ill update the Unofficial MS
8. (Original post by JackLloyd98)
Maybe I read it wrong, I thought the question was stretch followed by reflection.
You are correct about the question, it did say stretch followed by reflection so
reflection matrix in front and stretch matrix behind. I think the problem might be caused by the simultaneous equation.
9. (Original post by moodyloo1998)
You are correct about the question, it did say stretch followed by reflection so
reflection matrix in front and stretch matrix behind. I think the problem might be caused by the simultaneous equation.
Those are my workings
10. (Original post by JackLloyd98)
Those are my workings
I'm pretty sure the question told us that the point transformed to (0, 4), not (0, -4). Well actually, the answer I told you is the one I just worked out by memory now, my answer might be the same as yours during the exam if it is (0,-4), I just can't really remember anything right now. All the calculation seems correct though.
11. (Original post by henryhong)

Q1 Roots of Polynomials

(a) alpha+beta=6 [2marks]
alpha*beta=14

(b) 7X^2-4X+7 [5 marks]

(a) -h-3 [3 marks]

(b) as h-->0 gradient-->-3 [2 marks]

Q3 Log graph
(a) show linear relationship log(Y)= Xlog(B) + Log(A) [2 marks]

(c) a= 316 [4 marks]
b=0.398

Q4 General Solutions

(a) k=6 [1 marks]

(b) x= πn+ π/2 [4marks]
x= πn+ π/3

(c) root 3 [2 marks]

Q5 summations

(a) show it equal to 3n(4n^2-1) [5 marks]

(b) n(2n+1)(2n-3)(n-1) [4marks]

Q6 parobola

(a) show a=2 [3 marks]

(b) k= 1/2 [6 marks]
k= -2
k=0

Q7 imaginary

(a) -2+4i [3 marks]
-2-4i

(bi) i+qi=0 therefore q must beequal to -1 [2 marks]

(bii) p=4 q= -12-i [5marks]
p=-4 q= 4-i

Q8 matrix

(a) matrix 4 0 [1 marks]
0 1
(b) stretch in x-directionSF4 [1 marks]

(c)matrix 0.5 -root3/2 [2marks]
-root3/2 -0.5

(d) matrix p= -2root3 [6marks]
-2

Q9 graph

(a) Asymptotes: x=0.5 x=2 y=0 [2 marks]

(b) Intersections: x=5/2 x=1 x=0 [3 marks]

(c) a sketch [3 marks]

(d) x<equal to 0 [3marks]
0.5< to x<equal to1
2<x<equal to 2.5
Q1 - part a seems about right. But part b, i'm pretty sure it asked for an equation, so in that case, you need to put 7x^2-4x+7=0, the =0 matters you see. Stop being so sloppy.

Q2, Q3, Q4 & Q5 - looks about right

Q6 - (a) seems familiar. Pretty sure (b) asked for a range, or did it not?

Q7 - the answers seem right. But the question title "imaginary", none of these answers were imaginary, -2+4i is complex, -2-4i is complex, -12-i is complex, 4-i is complex, so where the hell did you get "imaginary" from?!?

Q8 - parts (a) through to (c) seem familiar, but I think part d, the COORDINATES, not matrix, P were (2rt3, 2). Also, the question should probably be titled "matrices" since we were asked about two of them, not just one.

Q9 - all good.
12. Can someone post the question for the general solution? I can't remember my answer so I want to try it again to see if I got it right
13. How did you do Q7bii? I spent ages on it in the exam and couldn't work out how to do it at all
14. The only thing I got wrong by the looks of it was Q4 part c, I thought it was 1 anyone mind clearing up with it was root 3?
15. (Original post by rektangle)
Q1 - part a seems about right. But part b, i'm pretty sure it asked for an equation, so in that case, you need to put 7x^2-4x+7=0, the =0 matters you see. Stop being so sloppy.

Q2, Q3, Q4 & Q5 - looks about right

Q6 - (a) seems familiar. Pretty sure (b) asked for a range, or did it not?

Q7 - the answers seem right. But the question title "imaginary", none of these answers were imaginary, -2+4i is complex, -2-4i is complex, -12-i is complex, 4-i is complex, so where the hell did you get "imaginary" from?!?

Q8 - parts (a) through to (c) seem familiar, but I think part d, the COORDINATES, not matrix, P were (2rt3, 2). Also, the question should probably be titled "matrices" since we were asked about two of them, not just one.

Q9 - all good.
are you my teacher? lol I just got a *******ing. but yeah ill corect them now
16. (Original post by henryhong)
are you my teacher? lol I just got a *******ing. but yeah ill corect them now
b o ll o c king was what I wrote initially but student room didn't allow it
17. (Original post by Chickenslayer69)
Can someone post the question for the general solution? I can't remember my answer so I want to try it again to see if I got it right
help pls thx
18. I did product or roots and sum of roots for complex numbers and got p=+-4 (from POR= (p+2i)(p-2i)= p^2-4i^2 = 20 so p^2=16....) then SOR ( -(4+i+iq)= p+2i + p-2i) so -4-i-qi = 2p, sub in p= 4 and -4, so qi+i= 4 so I did q = (4-i)/i to get 4/i -1 and the other q= -12/i -1) anyone get this or do this?😂😶

Posted from TSR Mobile
19. (Original post by Olsmarto)
I did product or roots and sum of roots for complex numbers and got p=+-4 (from POR= (p+2i)(p-2i)= p^2-4i^2 = 20 so p^2=16....) then SOR ( -(4+i+iq)= p+2i + p-2i) so -4-i-qi = 2p, sub in p= 4 and -4, so qi+i= 4 so I did q = (4-i)/i to get 4/i -1 and the other q= -12/i -1) anyone get this or do this?😂😶

Posted from TSR Mobile
I didn't do that (I was completely stuck) but seems like a really good way of doing it haha
20. Wht are the steps for q)7b)ii)

I'm really confused

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