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OCR (not MEI) C2 - Wednesday 20th May 2015 Watch

1. Maximum of 40 marks... woo...
2. Couldnt find N. Made a quadratic that didnt work but then managed to fumble the right answer using trial and error. N was 24 for anyone wondering
3. What did a = for the coefficient q
4. It was a really weird exam, they pulled questions that Hadn't been used in the whole spec

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5. Xsanda's solutions:
1i) common ratio:[1]
r=-2
1ii) 11th term: [2]
ar¹⁰ = 3072
1iii) sum to 20: [2]
Sum to 20 = -1048575

2i) trapezium rule: [4]
21.4
2ii) improvement: [1]
more/narrower strips

32.1
3ii) perimeter: [5]
22.7

4i) Binomial expansion: [4]
64+192ax+240a²x²+…
4ii) Find a: [5]
a=2/3

5) Double antidifferential: [7]
y=4x^(3/2)-7x-3

6i) Factorise: [4]
(x-2)(x+5)(x-3)
6ii) Integrate: [4]
256
6iii) Problem with integral: [2]
negative section of integration counts negatively

7i) u₂₀: [2]
62
7ii) Σ 10 to 20 show that: [3]
517
7iii) Σ N to 2N: [6]
N=24

8a) log 2^(n-3) = 18000: [4]
n=17.1
8b) simultaneous logs: [5]
x=32, y=8

9ia) Next solution of cos: [1]
6π-α
9ib) Solution with negative: [2]
3π-a
9ii) Draw graph: [2]
graph of sin, half height
9iii) Solve graphs: [4]
3.32, 12.7
6. (Original post by ben5312)
What did a = for the coefficient q
2/3

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7. In the time it'd been running

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8. (Original post by xsanda)
Xsanda's solutions:
R=-2
ar¹⁰ = 3072
Sum to 20 = -1048575

trapezium: 21.4
narrower strips

triangle area: 32.1
perimeter: 20.7

expansion: 64+192ax+240a²x²+…
a=2/3

y=4x^(3/2)-7x-3
(x-2)(x+5)(x-3)
256
negative section of integration counts negatively

u₂₀ = 62
Σ = 517 (show that)

N=24
n=17.1
x=32, y=8

6π-α
3π-a
graph of sin, half height
3.32, 12.7
Speedy x_x
9. That has just given me so much confidence! Thanks xandra

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10. Very last part of the last section I messed up straight away by dviding 2cos1/3x by cos1/3x.. Carried on though to get two solutions correct to what I was doing.... Any marks at all? How marks was it worth?
11. (Original post by xsanda)
Xsanda's solutions:
R=-2
ar¹⁰ = 3072
Sum to 20 = -1048575

trapezium: 21.4
narrower strips

triangle area: 32.1
perimeter: 20.7

expansion: 64+192ax+240a²x²+…
a=2/3

y=4x^(3/2)-7x-3
(x-2)(x+5)(x-3)
256
negative section of integration counts negatively

u₂₀ = 62
Σ = 517 (show that)

N=24
n=17.1
x=32, y=8

6π-α
3π-a
graph of sin, half height
3.32, 12.7
I've gotta know... How the heck do you remember all of those answers?!
12. (Original post by xsanda)
Xsanda's solutions:
R=-2
ar¹⁰ = 3072
Sum to 20 = -1048575

trapezium: 21.4
narrower strips

triangle area: 32.1
perimeter: 20.7

expansion: 64+192ax+240a²x²+…
a=2/3

y=4x^(3/2)-7x-3
(x-2)(x+5)(x-3)
256
negative section of integration counts negatively

u₂₀ = 62
Σ = 517 (show that)

N=24
n=17.1
x=32, y=8

6π-α
3π-a
graph of sin, half height
3.32, 12.7
For the sum to 20 one did you but the negative number in brackets? I got the same as you but put it in brackets and got a more sensible andwer if i remember correctly!
13. (Original post by xsanda)
Xsanda's solutions:
R=-2
ar¹⁰ = 3072
Sum to 20 = -1048575

trapezium: 21.4
narrower strips

triangle area: 32.1
perimeter: 20.7

expansion: 64+192ax+240a²x²+…
a=2/3

y=4x^(3/2)-7x-3
(x-2)(x+5)(x-3)
256
negative section of integration counts negatively

u₂₀ = 62
Σ = 517 (show that)

N=24
n=17.1
x=32, y=8

6π-α
3π-a
graph of sin, half height
3.32, 12.7
For the expansion I put 240(ax)^2, will I lose a mark for this because it said to simplify so I didn't know if I had to expand out the brackets fml
14. (Original post by xsanda)
Xsanda's solutions:
R=-2
ar¹⁰ = 3072
Sum to 20 = -1048575

trapezium: 21.4
narrower strips

triangle area: 32.1
perimeter: 20.7

expansion: 64+192ax+240a²x²+…
a=2/3

y=4x^(3/2)-7x-3
(x-2)(x+5)(x-3)
256
negative section of integration counts negatively

u₂₀ = 62
Σ = 517 (show that)

N=24
n=17.1
x=32, y=8

6π-α
3π-a
graph of sin, half height
3.32, 12.7
May just be me, but 12.7 isn't in the range of 0 to 2pi
15. (Original post by eddy29)
May just be me, but 12.7 isn't in the range of 0 to 2pi
The range is 0 — 6π, because it was sin(1/3 x) so the range had to be tripled.
16. (Original post by Zallin98)
For the sum to 20 one did you but the negative number in brackets? I got the same as you but put it in brackets and got a more sensible andwer if i remember correctly!
I checked it manually, which was stupid, as I ran out of time by q9, but I got the same as he/she did
17. (Original post by eddy29)
May just be me, but 12.7 isn't in the range of 0 to 2pi
You had to find the points of intersection on the graph, which covered a greater range than 0 to .
18. (Original post by fayecaravelli)
For the expansion I put 240(ax)^2, will I lose a mark for this because it said to simplify so I didn't know if I had to expand out the brackets fml
I left mine in the same form you left yours in- should get the marks because that is simpler.
19. Anyone reckon 60 will be an a :/
20. I got the perimeter as 22.7

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