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# AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme] watch

1. (Original post by ThePanoramicGame)
I did i rounded to 3.33 and another solution
I got something like 0.567 and 3.34
2. On question 5 I got part a as just 1/2 instead of -1/2. I've then done the rest correctly but go slightly wrong answers because of my part a). Is this gona penalise me for all of them?
3. Is this a correct answer for diff eqn?

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4. (Original post by CD223)
Dude I couldn't find E. I got a horrid differential equation and can't even remember it? I got r=3 and drew the graph for h.

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I did the same for h and got te gradient as 8/11
5. (Original post by danniegee)
Does anyone rememebr if they told us to simplify/write in a certain format the solution to 8a?
They didn't. I didn't have time to simplify it so I really hope it doesn't cost me a mark. It would be kind of ****ty of them to do that when they didn't mention it but at the same time you are taught to always simplify so I don't know.
6. (Original post by ThePanoramicGame)
yep apparently they did i didnt although i did work out c to hoping if my differential is right i wont lose much for not putting in x=t form
Oh no I just left it as it was! I messed up this paper quite badly :'(
7. The answer to 8)a) isn't correct I don't think
8. (Original post by majka)
I got coordinates (-3/4, -1/8) and (1/2, 3/4)
I believe the X-coordinate was -1/8 and the y coordinate was -3/4.

Does anyone know whether we needed to state both the X and Y coordinate of Q or just the X?
9. (Original post by CD223)
Is this a correct answer for diff eqn?

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That's right! I got that
10. how much ums would 55 marks get?
11. (Original post by CD223)
Is this a correct answer for diff eqn?

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That's exactly what I got mate
12. (Original post by MoHoosen9167)
That's right! I got that
Same!
13. (Original post by MoHoosen9167)
That's right! I got that
Phew.

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14. (Original post by datpr0)
I know , but it remains incorrect.

I just did it again and the answer I came to (same as in the exam) is:

x = 1/5[(-1/2(1+t)^-1 + 5/2)^2 - 4]

EDIT: the question gave dx/dt = (4+5x)^1/2 / 5(1+t)^2
THIS

also I included the constant, not sure if I lose any marks for that. So if the constant was included it will just be:
x = 1/5[(-1/2(1+t)^-1 + 5C/2)^2 - 4]
15. (Original post by datpr0)
I know , but it remains incorrect.

I just did it again and the answer I came to (same as in the exam) is:

x = 1/5[(-1/2(1+t)^-1 + 5/2)^2 - 4]
I GOT SOMETHING SIMILAR TO THAT! defo that form, cant quite remember inside the brack but yeh defo 1/5 and -4 outside
with the gradient of the parametric function, would you get marks for differentiating and then subbing in x and y, then rearranging for dy/dx got either 10/11 or 11/10

also for the vector question I got a fraction mu, and did all the other parts with my wrong mu.. will I still get all the marks?

damnn!! the differential equation q efed me over......... got 1/3 for r

hoping I get 85+ums so baaadly
i dont understand what you have done for the parametric. Did you do (dy/dt)/dx/dt. I assume you get one method mark for differentiating x and on method mark for differentiating y and then one for the answer. For the vector question I'm still unsure if i got it right myself so can't answer. And the last part of the differential question do you mean dr/dt or 7a.
17. (Original post by CD223)
Is this a correct answer for diff eqn?

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Looks similar to mine but i got it as x=
18. (Original post by sam_97)
The x values in question 2)c) were 0.57 and 3.44
what was the question for this? think i may have accidentally missed this question. nooooo
19. (Original post by sam_97)
That's exactly what I got mate
Thanks man - I'm still so disappointed in myself
20. did no one get 69.1 for the angle where does 60 come from?

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