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

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Q1 Roots of Polynomials

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

(b) 7X^2-4X+7=0 [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 Complex Numbers

(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 Matrices

(a) matrix 4 0 [1 marks]
0 1
(b) stretch in x-direction SF4 [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
2. 4a 5b 6b all wrong. k=1/6 5b u whrre ment to get 3 linear factor and 6b an euality with greater and less than sign
3. do u remember marks for each question
4. (Original post by jojoj)
4a 5b 6b all wrong. k=1/6 5b u whrre ment to get 3 linear factor and 6b an euality with greater and less than sign
Nope. It was pi/k so k=6
5. (Original post by jojoj)
4a 5b 6b all wrong. k=1/6 5b u whrre ment to get 3 linear factor and 6b an euality with greater and less than sign
Yeah you are right about about 5 (b) i just couldnt remember the factors but I'll work it out now

Thanks man!
6. (Original post by henryhong)
Yeah you are right about about 5 (b) i just couldnt remember the factors but I'll work it out now

Thanks man!
yeah ur answer about k is right forgot it was pi/k thought it said pik
7. (Original post by -jordan-)
Nope. It was pi/k so k=6
Yep thats right!
8. (Original post by jojoj)
do u remember marks for each question
No but if you want I can find out for you?
9. (Original post by henryhong)
No but if you want I can find out for you?
10. (Original post by jojoj)
Okay Ill notify you once ive updated my answers
11. (Original post by henryhong)
Okay Ill notify you once ive updated my answers
I mean once ive updated the first post
12. (Original post by jojoj)
4a 5b 6b all wrong. k=1/6 5b u whrre ment to get 3 linear factor and 6b an euality with greater and less than sign
I knew something wasn't right about 5(b) what were your inquality signs?
and are the critical values atleast correct?
13. (Original post by henryhong)
I knew something wasn't right about 5(b) what were your inquality signs?
and are the critical values atleast correct?
5b should be n(2n+1)(2n-3)(n-1)? But definitely 4 linear factors, not 3.
14. (Original post by aoxa)
5b should be n(2n+1)(2n-3)(n-1)? But definitely 4 linear factors, not 3.
sorry I mean 6(b)
15. (Original post by henryhong)
I knew something wasn't right about 5(b) what were your inquality signs?
and are the critical values atleast correct?
sorry I mean 6(b)
16. (Original post by henryhong)
I mean once ive updated the first post
Its done
17. (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)(n-1)(4n+3) [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

Thank you so much for this!!
18. Are you sure about the minus signs on 8d? And 7bii) I think you have your imaginary and real components mixed up
19. (Original post by henryhong;[url="tel:65806509")

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)(n-1)(4n+3) [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
5b is n(2n+1)(2n-3)(n-1)
20. Question 8d is wrong, it's meant to be 2root3 2

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