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

OP

Hey guys, hope it went well for everyone. I know quite a few people in my school thought it was a reasonably tricky paper but for those of you messaging me about grade boundaries, I really have no idea. I'm too tired to make an unofficial mark scheme on the original post and honestly I can't remember any of my answers, but if Mr M posts his own, I'll make sure I add the link to it to this thread :smile:

good luck with the rest of your exams!
PS for those of you doing S1, check out this thread.

Reply 301

Jmedi

6

Original post by Wolfram74

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?

If you calculate it you get 256 and this is incorrect as it should be higher than this;
some area gets subtracted as a small part of the area is below the x axis

Reply 302

Patsella

3

Can someone write out how the 2n question works ?? So confused !!!!

Reply 303

xsanda

2

Original post by Wolfram74

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?

You can check it by entering "integrate (x3-19x+30) from -5 to 3" into Wolfram Alpha. It's definitely positive, as the majority of the curve is above the x axis.

Reply 304

BradWerritt

Quite happy with how that exam went, although it probably was the hardest C2 paper I've ever done! Think I've got everything correct apart from 9. a. i) and ii), which were a total of 3 marks. Did the very last question specify for the answer to be in Radians? I gave mine in degrees...

Surely the grade boundaries will be low for this paper - any estimates on what you'd need for 100 UMS? January 2013 was 68/72, but I felt that this paper was harder than that one.

Reply 305

GM16

1

Original post by Maprunner

the first logarithm one

Thank you!

Posted from TSR Mobile

Reply 306

Nuvertion

6

Original post by BradWerritt

Quite happy with how that exam went, although it probably was the hardest C2 paper I've ever done! Think I've got everything correct apart from 9. a. i) and ii), which were a total of 3 marks. Did the very last question specify for the answer to be in Radians? I gave mine in degrees...

Surely the grade boundaries will be low for this paper - any estimates on what you'd need for 100 UMS? January 2013 was 68/72, but I felt that this paper was harder than that one.

It's better to compare this to June 2014. I reckon the grade/UMS boundaries will be pretty similar.

Reply 307

yuqiwang

1

Original post by AlfieH

Anyone remember how the marks were split up for Q9?

9（i）(a)1 mark
(b)2 marks
(ii)2 marks
(iii)around 5

Reply 308

xsanda

2

Original post by Patsella

Can someone write out how the 2n question works ?? So confused !!!!

Sum from N to 2N
= Sum to 2N - Sum to N-1

= 2N/2 [2a-(2N-1)d] - (N-1)/2 [2a-(N-2)d]
where a = 5, d =3

This gives a quadratic 9N²+13-5496 = 0, which solves to N=24 or -229/9, so N must be 24, positive integer.

Reply 309

Blessedness

3

Hi guys. I put all my answers for question 9 in degrees. Would you have an estimate of how many marks I might lose if all the answers were correct in degrees?

Posted from TSR Mobile

Reply 310

xsanda

2

Original post by Blessedness

Hi guys. I put all my answers for question 9 in degrees. Would you have an estimate of how many marks I might lose if all the answers were correct in degrees?

Posted from TSR Mobile

You'd lose the mark on 9i), and possibly one of the two on 9ii).

Reply 311

Charco63

1

Original post by xsanda

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

3i) shaded area: [4]
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

Wasn't question -7 not -3

Reply 312

Wolfram74

2

For the N = 24 question how many marks would you get for attempting to complete the square on an incorrect quadratic (9N^2 + 7N - 5500 = 0)?

Reply 313

SAMBAN

does anyone know what the log question was

Reply 314

terry_011

what will i get if i integrate the d2y/dx2 twice but put the (4,1) in instead, for log i get logx=5 and logy=8, didn't work out the x and y :/ how many marks will i lose?