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OCR (non mei) C4 Friday 24th June 2016

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1. (Original post by TheNicholas)
For the +c I got 3??
Which part of the question are you talking about, I did get c=3 for something but I don't remember what it was
2. (Original post by Ohagan07)
Which part of the question are you talking about, I did get c=3 for something but I don't remember what it was
For the +c after you integrate by subbing in the values of x and y
3. Anyone remember the actual question for 6? Like what we had to integrate + what we were meant to substitute in
4. Unofficial mark scheme anywhere?
5. (Original post by kathrynloops)
Unofficial mark scheme anywhere?
6. For 9 (ii) I got C=3, but then I got (1/944875)e^2.
This worked when I subbed it in to check, but perhaps I copied the partial fractions wrong from previous page. Any ideas for how many marks?
7. pretty sure the unofficial mark scheme is wrong it should be

3/(x+1)^2 -4/(x+4) +2/(x+1) NOT 3/(x+1)^2 +2/(x+4) -4/(x+1)

when you integrate you get

lny+C=2ln(x+1)-4ln(x+4)-3/(x+1)

sub in y=1/256 and x=0

ln(1/256) + C = -4ln4 -3
-ln256 + C=-ln256 -3
C=-3
therefore

lny =2ln(x+1)-4ln(x+4)-3/(x+1)+3
when x=2

lny=2ln3-4ln6+2
lny=ln9-ln1296 +2
ln(144y)=2
y=(e^2)/144
??
8. Do you know how 10ii was worked out?

edit: nvm I noticed from another post that I got it completely wrong.
9. (Original post by Jacquea)
pretty sure the unofficial mark scheme is wrong it should be

3/(x+1)^2 -4/(x+4) +2/(x+1) NOT 3/(x+1)^2 +2/(x+4) -4/(x+1)

when you integrate you get

lny+C=2ln(x+1)-4ln(x+4)-3/(x+1)

sub in y=1/256 and x=0

ln(1/256) + C = -4ln4 -3
-ln256 + C=-ln256 -3
C=-3
therefore

lny =2ln(x+1)-4ln(x+4)-3/(x+1)+3
when x=2

lny=2ln3-4ln6+2
lny=ln9-ln1296 +2
ln(144y)=2
y=(e^2)/144
??
That's literally what it says as the answer?
10. (Original post by TheNicholas)
That's literally what it says as the answer?
yeh part ii) is right the partial fractions are wrong i was just showing why the partial fractions have to be the other thing I suggested
11. (Original post by Jacquea)
pretty sure the unofficial mark scheme is wrong it should be

3/(x+1)^2 -4/(x+4) +2/(x+1) NOT 3/(x+1)^2 +2/(x+4) -4/(x+1)

when you integrate you get

lny+C=2ln(x+1)-4ln(x+4)-3/(x+1)

sub in y=1/256 and x=0

ln(1/256) + C = -4ln4 -3
-ln256 + C=-ln256 -3
C=-3
therefore

lny =2ln(x+1)-4ln(x+4)-3/(x+1)+3
when x=2

lny=2ln3-4ln6+2
lny=ln9-ln1296 +2
ln(144y)=2
y=(e^2)/144
??
I agree with you
12. (Original post by Jacquea)
yeh part ii) is right the partial fractions are wrong i was just showing why the partial fractions have to be the other thing I suggested
Oh okay, my mistake
13. 58 on this paper and 64 on C3, I'm thinking this won't quite be enough to get an A* which is annoying because I have been very ill this past week
14. for the vector question would it have been okay to show that both lines pass through the same two points? and do you think 57 on this and 72 on C3 is enough for an A*
15. Hi, i may be wrong here, but with the graph question, i think it had a range of just 0 to 2 because x = 1 - cos theta. this can only ever range from 0 to 2.
16. For the last question I did not get c=3 but I did get the correct answer will I lose any marks?
17. (Original post by Sid1234)
for the vector question would it have been okay to show that both lines pass through the same two points? and do you think 57 on this and 72 on C3 is enough for an A*
Well last year I got 70/72 in C3 and 56/72 in C4 and I got the exact amount of UMS needed for an A* (99 in C3 and 81 in C4) so you have probably just made it.
18. (Original post by marioman)
Well last year I got 70/72 in C3 and 56/72 in C4 and I got the exact amount of UMS needed for an A* (99 in C3 and 81 in C4) so you have probably just made it.
19. (Original post by Sid1234)
No idea, I've not seen the paper.
20. (Original post by Ohagan07)
The answer to the last question was: 1/144 e^2
I got either 144e^3 or 1/144e^3 ... Not e^2...

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