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# OCR C2 non-mei 23rd May 2018 Unofficial Markscheme watch

I managed to do the show that log question.

For 9 is a 2 or 4
2. I’ve been told by friends that a = 2. Didn’t get anywhere close to that myself but I spent more time crying about the paper than answering questions
3. (Original post by bethanngreen)
I’ve been told by friends that a = 2. Didn’t get anywhere close to that myself but I spent more time crying about the paper than answering questions
Yah I feel you.

I got a =4 for some weird reason...
4. When I was integrating I kept getting 1/2 a in my answers so might be something to do with that??
(Original post by Bulletzone)
Yah I feel you.

I got a =4 for some weird reason...
5. Pretty sure a = 2,
I used simultaneous equations with the f(x) and the integration of that to eliminate c at the very end.
Question 8 was very fiddly though so I imagine there would be quite a few method marks.
6. (Original post by humanteaparty)
Pretty sure a = 2,
I used simultaneous equations with the f(x) and the integration of that to eliminate c at the very end.
Question 8 was very fiddly though so I imagine there would be quite a few method marks.
Yep simultaneous equations and got a = 2 as well. Agree with method marks. Plenty of places to slip up on that question.

Also OP, pretty sure the question with a was question 8 not 9?
7. Also got a =2, integrate once and rearrange to make C the subject so you get something like c = a/2 + 3 then sub that in and integrate once more.
Sub the limits in and keep going till your down to a/3 + 20 or whatever it was then equate that to 30 and solve for a
8. I thought the paper was easier than previous papers so I expect grade boundaries to be high. The question I found most confusing was the sum to infite question for the geometric sequence. What did everyone get for that?
9. (Original post by tislam1)
I thought the paper was easier than previous papers so I expect grade boundaries to be high. The question I found most confusing was the sum to infite question for the geometric sequence. What did everyone get for that?
u1 =20
u2 = 20(0.8)
u3 = 20(0.8)^2
u4 = 20(0.8)^3

Therefore sum to infinity of Sn for u2n = a / (1-r)

a = u2 and r = u4 ÷ u2
10. (Original post by Bulletzone)

I managed to do the show that log question.

For 9 is a 2 or 4
Last answer I got x=2, x=4add2root10, x=4-2root10 or something like that. There was definitely a plus or minus 2root10
11. That was the worst paper I’ve every done
12. (Original post by Bulletzone)
u1 =20
u2 = 20(0.8)
u3 = 20(0.8)^2
u4 = 20(0.8)^3

Therefore sum to infinity of Sn for u2n = a / (1-r)

a = u2 and r = u4 ÷ u2
That’s exactly what I did!
u2 = 16
u4 = 10.24
r = 0.64
Sum to Infinity = 16/(1-0.64) = 4.44
13. (Original post by Finlaycollins)
That’s exactly what I did!
u2 = 16
u4 = 10.24
r = 0.64
Sum to Infinity = 16/(1-0.64) = 4.44
That's the answer have some rep.
14. a=1/2
15. Shouldn't it be 44.4

Also what was the answer to the area and r for the triangle and sector of a circle question?
16. I got sum to infinity to be 400/9As its u2N you get 20x(0.8)^2N-1 So if you put in n=1 you get a=16If you put in n=2 you get it as 10.24R=0.64 Do a sum to infinity you get 400/9Not sure if its right
17. Got a= -28. Brilliant
18. 9.40 for area
19. (Original post by aurora0914)
9.40 for area
I got the obtuse angle as something around 2.35?
and part iii) the arc length as 2.69 give or take
but not 100% sure on it
20. (Original post by fzzpop)
i got the obtuse angle as something around 2.35?
And part iii) the arc length as 2.69 give or take
but not 100% sure on it
ffs i forgot to do (pi) - angle

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