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# OCR C2 exam Wed 25th May watch

1. Did it say give the period in radians??
I know the rest of the Q was in radians but for part i I just wrote 360/a
2. (Original post by Heyitsmerhi)
I did -1 to 0 and 0 to 3 as otherwise it wouldn't give the exact area would it? Not 100% sure but I got something like 179/5
it should give u the same answer if you split it, but really you are supposed to do it from -1 to 3.
you only have to split it when the part of the area is under the *X* axis, which in this case it wasnt
3. Does anyone remember roughly what the questions were for these answers?5bi) -6/a+2/a^2+4 (4m) 8iv) x=6.73 (3m)
4. (Original post by morrissies)
Did it say give the period in radians??
I know the rest of the Q was in radians but for part i I just wrote 360/a
Check the mark scheme last year for 9i and ii.
5. (Original post by AlfieH)
My teacher has confirmed that the answer to 9iii is Pi/3a and 4Pi/3a.
But 4Pi/3a gives a negative solution when substituted into the original equation, so it cannot be a right answer.

sin(a*4Pi/3a)=sqrt3*cos(a*4Pi/3a)=-sqrt3/2
6. (Original post by abrahammurciano)
But 4Pi/3a gives a negative solution when substituted into the original equation, so it cannot be a right answer.

sin(a*4Pi/3a)=sqrt3*cos(a*4Pi/3a)=-sqrt3/2
Maybe it meant the first two positive solutions in respect to the x axis.
7. (Original post by student.1997)
Does anyone remember roughly what the questions were for these answers?5bi) -6/a+2/a^2+4 (4m) 8iv) x=6.73 (3m)
5bi was something like integrate 6x^-2+4x^-3 between A and 1 (or -1).

8iv was 180=3^(n-2); solve for n.
8. Anyone got an unofficial mark scheme ??
9. For the last question how many marks do you think I would get,
I squared everything to get sin^2ax=3cos^2ax
Then used the identies to get 1-cos^2ax=3cos^2ax
Then rearranged to get 4cos^2ax=1 ,then cos^2ax=1/4
then square rooted to get cosax=+ or - 1/2
Then somehow I ended up with pi/3a and 2pi/3a as my final answer.
Thanks very much
10. (Original post by Sabreak10)
Anyone got an unofficial mark scheme ??
1i) 10cm

1ii) 5.04cm

2i) 3/10π

2ii) 20.4cm

3i) 27+27kx+9k^2x^2+k^3x^3

3ii) +/-√3

4i) log base 3 x^2/x+4

4ii) x=12

5a) x^4/2-x^3+2x^2-6x

5bi) -6/a+2/a^2+4

5bii) 4

6i) k=91

6ii) Sum of terms = 978

6iii) N=38

7i) R=0, quotient = x^2-4x+3

7ii) (x+1), (x-1), (x-3)

7iii) Prove

7iv) 512/15

8i) Translation 2 units in positive x direction

8ii) Stretch scale factor 1/9 in y direction

8iii) Sketch, crosses y at (0,1/9)

8iv) x=6.73

8v) Area =9.60

9i) 2pi/a

9ii) a=5/3 and k=√3/2

Someone else came up with this one
11. (Original post by micycle)
5bi was something like integrate 6x^-2+4x^-3 between A and 1 (or -1).

8iv was 180=3^(n-2); solve for n.
Thank you!!
12. On this mark scheme they haven't included the +c on the end of 5a, don't u need it for full marks
13. What were the questions 8)iv and 8)v
14. (Original post by Sabreak10)
Anyone got an unofficial mark scheme ??
15. I got 1/9 and 0 for question 3)ii but the mark scheme says sqrt3
Maybe I misread the question... What did everyone else get?

Edit: Never mind, turns out I didn't read the question properly -_- I had so much time to check as well! Oh well
16. Wow awesome, know I got between 65 and 69/72- definitely an A . Hopefully that will cancel out my mediocre performance in C1.
17. For the one where k is +/-root3, people say they equated k to 27 - where's the 27 come from and how is it the constant
18. (Original post by Bobrey)
For the one where k is +/-root3, people say they equated k to 27 - where's the 27 come from and how is it the constant
I think it's a constant as it stays the same whatever the k value is and is unaffected by it and it's got no x's with it, I may be wrong though
19. (Original post by morrissies)
not word for word but it was:
sum of first 20 terms of AP > sum to infinity of GP

(for the AP a=5 d=1.5) (for the GP a=1.2 (?) r=0.9)
Wasn't A for the g.p. 120
20. (Original post by Bobby21231)
I think it's a constant as it stays the same whatever the k value is and is unaffected by it and it's got no x's with it, I may be wrong though
You are correct.

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