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

5b is n(2n+1)(2n-3)(n-1)

Do you know where i went wrong?

It asked for the original coordinates before the transformation I got ((3^0.5)/2, 2)

I got (-root3/2 , -2)

Those are my workings

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.

Those are my workings

My answers for this morning's exam, please let me know if you agree/disagree on answers!

Q1 Roots of Polynomials

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

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

Q2 Gradients using h

(a) -h-3 [3 marks]

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

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

(b)gradient was -0.4 [1marks]

(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.

are you my teacher? lol I just got a *******ing. but yeah ill corect them now

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

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?😂😶


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