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

1. what's the question where the answer is N=24??
2. (Original post by czj1997)
what's the question where the answer is N=24??
it was find the value of N if 2750 = 2N above sigma and n = N below sigma.
3. (Original post by czj1997)
what's the question where the answer is N=24??
The one about summations from 2n to n where it equaled 2570 or something

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4. (Original post by ella_chloe)
This would be lovely boundaries

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These are the grade boundaries all the way back to 2005
Attached Images

5. For the last question, i worked in degrees as i find it easier, then converted to radians at the end and got the same answers as the unofficial mark scheme, would that still be full marks?

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6. Does anyone remember how many marks the binomial expansion question was worth?
7. How many marks will I drop on the last one for giving 3.33 rad and 12.75 rad?
9. Hi can someone check through this working for the Sum from N to 2N being 2750 question? I can't understand why I don't get the same answer... :/
Working:0.5(N+1)[2N+3N] = 2750N+1[5N] = 55005N2 + 5N - 5500 = 0
Gave me an answer of like 33.6??I'm wondering where I went wrong...!Thanks!
10. (Original post by ella_chloe)
For the last question, i worked in degrees as i find it easier, then converted to radians at the end and got the same answers as the unofficial mark scheme, would that still be full marks?

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yes full marks
11. (Original post by shizhang)
Does anyone remember how many marks the binomial expansion question was worth?
No sorry
12. (Original post by jampot98)
These are the grade boundaries all the way back to 2005
this gives me hope thank you

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13. (Original post by shizhang)
Does anyone remember how many marks the binomial expansion question was worth?
7 in total

4 for the first
3 for the second
14. (Original post by ella_chloe)
For the last question, i worked in degrees as i find it easier, then converted to radians at the end and got the same answers as the unofficial mark scheme, would that still be full marks?

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Yeah it would be

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15. (Original post by gabby07)
Hi can someone check through this working for the Sum from N to 2N being 2750 question? I can't understand why I don't get the same answer... :/
Working:0.5(N+1)[2N+3N] = 2750N+1[5N] = 55005N2 + 5N - 5500 = 0
Gave me an answer of like 33.6??I'm wondering where I went wrong...!Thanks!
I did exactly the same thing, how many marks will we get?
16. 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
17. (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: 22.7

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
What question is the n =17.1 from?

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18. (Original post by GM16)
What question is the n =17.1 from?

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the first logarithm one
19. (Original post by GM16)
What question is the n =17.1 from?

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See post above this, questions and marks added in
20. (Original post by xsanda)
Xsanda's solutions:1i) common ratio:[1] r=-21ii) 11th term: [2] ar¹⁰ = 30721iii) sum to 20: [2] Sum to 20 = -10485752i) trapezium rule: [4] 21.42ii) improvement: [1] more/narrower strips3i) shaded area: [4] 32.13ii) perimeter: [5] 22.74i) Binomial expansion: [4] 64+192ax+240a²x²+…4ii) Find a: [5] a=2/35) Double antidifferential: [7] y=4x^(3/2)-7x-36i) Factorise: [4] (x-2)(x+5)(x-3)6ii) Integrate: [4] 2566iii) Problem with integral: [2] negative section of integration counts negatively7i) u₂₀: [2] 627ii) Σ 10 to 20 show that: [3] 5177iii) Σ N to 2N: [6] N=248a) log 2^(n-3) = 18000: [4] n=17.18b) simultaneous logs: [5] x=32, y=89ia) Next solution of cos: [1] 6π-α9ib) Solution with negative: [2] 3π-a9ii) Draw graph: [2] graph of sin, half height9iii) Solve graphs: [4] 3.32, 12.7
Wouldn't the integration one be -206.5 or -207.5 (can't remember which) because the next question asked why this was not the area under the curve between -5 and 3?

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