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# Edexcel A2 C4 Mathematics June 2016 - Official Thread watch

1. (Original post by cjlh)
My full solutions:

1) 1/8 - 15/16 x + 75/16 x^2 - 625/32 x^3

2) 0.6595, 1.083, 8/3 ln(2) - 7/9

3) (-2-4xy)/(2x^2 + 4 + pisin(0.5pi)), (3pi+62)/(pi+22)

4) x=e^(ln(60)-2.5t)), 633 mins

5) (-5root(3)/16), (4,5root3)

6) 3ln(3y+2)-2ln(y)+c, lambda=8, 4pi/3 - root3

7) k=17/2, 1696pi/5

8) {3,5,0}, r = {1,5,2} + lambda{-5,4,3}, k=2, cos(theta)=4/5, area=12/5, possible positions {3, 33/5, 16/5} and {-1, 17/5, 4/5}
Fairly sure E is actually (3, 17/5, 4/5) or (-1, 33/5, 16/5) or otherwise it doesnt fit on the line L2.
E = (1,5,2)+k(-5,4,3) so
1-5k=3 --> k= -2/5
5 + 4 x -2/5 = 17/5
2 + 3 x -2/5 = 4/5

You can do the same for the other value of E to work out if its on the line
2. (Original post by Vannzzoo)
I GOT EXACTLY THE SAME IM SOO HAPPY, hope they're right tho
Got all the same, except i got 634 minutes, or 10 hours 34 minutes
3. The exam went really well for except for the last 5 marker, could someone please explain to me how you were meant to do it as i literally spent about 25minutes on it and knew i had done one similar in the past. I was using the fact P was the midpoint and magnitudes were 2root2. But was proberly going off on a tangent
4. (Original post by Smithy1597)
How was it not 1024pi/5
thats volume for under graph, S would be volume of the rectangular part (becomes a cone) subtracted by the volume under the graph
5. (Original post by riffy123)
For the volume question, how was it 544pi? It had to be 576pi as u had to work out the whole cylinder and take it away by the integral surely!
It was 544π because radius is 8. 8.5 was the length
π x 8 x 8 x 8.5 =544π

Not rocket science

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6. any unofficial mark scheme yet?
7. (Original post by Woodford_97)
Fairly sure E is actually (3, 17/5, 4/5) or (-1, 33/5, 16/5) or otherwise it doesnt fit on the line L2.
E = (1,5,2)+k(-5,4,3) so
1-5k=3 --> k= -2/5
5 + 4 x -2/5 = 17/5
2 + 3 x -2/5 = 4/5

You can do the same for the other value of E to work out if its on the line
answers can't always be exactly right but you get the picture
9. (Original post by Zuzuvela)
It was 544π because radius is 8. 8.5 was the length
π x 8 x 8 x 8.5 =544π

Not rocket science

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you could integrate it if you wanted to
10. (Original post by Yua)
thats volume for under graph, S would be volume of the rectangular part (becomes a cone) subtracted by the volume under the graph
It becomes a cylinder no?
11. Can someone explain how you were supposed to do the last vector question (8e)?
12. (Original post by ryanroks1)
What did everyone get for the normal question where you had to find X? I remember 22 + pi being on the bottom?
Is that the a,b,c and are integers, find their values question?
I remember one of the bottom values was '1', I think it was the coefficient of pi.
13. Someone quote me when solutions are out please
14. (Original post by Zuzuvela)
It was 544π because radius is 8. 8.5 was the length
π x 8 x 8 x 8.5 =544π

Not rocket science

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Wasn't the height of cylinder 17/2 +1/2 = 9 so Volume = 576pi
15. (Original post by ombtom)
I got that
So did I 😊

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16. E was definitely: (-1, 33/5 , 16/5 ) or (3, 17/5 , 4/5 )
17. (Original post by Smithy1597)
How was it not 1024pi/5
I got that too, but I think there was a volume between x=0 and x=1/2 which we didn't account for, this I would expect to lose 2 marks, 1 for accuracy one for method
18. (Original post by Yua)
(becomes a cone)
Wait do you mean a cylinder? Cos other than that I completely agree x
19. (Original post by Yua)
thats volume for under graph, S would be volume of the rectangular part (becomes a cone) subtracted by the volume under the graph
Rectangles become Cylinders, Triangles become Cones.
20. (Original post by ombtom)
I got that
(Original post by thehollowcrown)
i think this is what i got
Even better, hopefully should be 75/75 in this then maybe a few marks lost in some places because I have a tendency to occasionally be an idiot in exams

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