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

1. I did too - is this wrong?
2. I simply cannot believe I read maximum as minimum. I found it for -root29 instead of 29. That will undoubtedly be the difference between a grade FFS
3. (Original post by datpr0)
Relatively sure the differential equation question's answer given there is incorrect.
It's not what I got, This other users,
4. (Original post by George1231)
1) A=3 B=-1
1b) 8/3ln5

2)ai r=root 29 alpha=1.19 (3 marks)
2bi) max value root 29 x=5.09 (2 marks)
2bii) x=0.568 or 5..72

3a) d=1 (1 mark)
3b) (2x+1)(2x-1)(2x-3)
c) h(x)=(1=2x)^-1 differentiate shows its decreasing (4 marks)

4a) 1+x-2x^24 (2 marks)
4aii) 1/4 -x/16+5x^2/256 (3 marks)
bii) 0.2313

5a)gradient of normal =-1/2 (1 mark)
5b) equation of normal y=2x-1/2
5c) x cooridinate -1/8 ( 5 marks)

6a) 60 degrees (3 marks)
6b) (15,3,2) (4 marks)
6c) (23,3,0) and (11,0,0) (4 marks)

7a) 1/4-ln2 (1 mark)
b)11/8 (6 Marks)

8a) x=9/20-1/2(t+1)-1/20(1+t)^2 (7 marks)
b)dr/dt=k/pir^2 (3marks)
c) 3 (2 marks)
How confident are u that the last vectors question is right
5. how did people get -1/8 for 5c? I got -3/4 for some reason!
6. (Original post by majka)
I had 0,568 and 3,34 is that correct?
yep thats what i got
7. Does anyone rememebr if they told us to simplify/write in a certain format the solution to 8a?
8. (Original post by geeky_penguin)
how did people get -1/8 for 5c? I got -3/4 for some reason!
-3/4 was the y coordinate -1/8 was the x coordinate
9. i had g(x) factorised to (2x+1) (2x-1) (2x-3) but people are putting +3 which is right?
10. (Original post by geeky_penguin)
how did people get -1/8 for 5c? I got -3/4 for some reason!
me too!
11. (Original post by George1231)
1) A=3 B=-1
1b) 8/3ln5

2)ai r=root 29 alpha=1.19 (3 marks)
2bi) max value root 29 x=5.09 (2 marks)
2bii) x=0.568 or 5..72

3a) d=1 (1 mark)
3b) (2x+1)(2x-1)(2x-3)
c) h(x)=(1=2x)^-1 differentiate shows its decreasing (4 marks)

4a) 1+x-2x^24 (2 marks)
4aii) 1/4 -x/16+5x^2/256 (3 marks)
bii) 0.2313

5a)gradient of normal =-1/2 (1 mark)
5b) equation of normal y=2x-1/2
5c) x cooridinate -1/8 ( 5 marks)

6a) 60 degrees (3 marks)
6b) (15,3,2) (4 marks)
6c) (23,3,0) and (11,0,0) (4 marks)

7a) 1/4-ln2 (1 mark)
b)11/8 (6 Marks)

8a) x=9/20-1/2(t+1)-1/20(1+t)^2 (7 marks)
b)dr/dt=k/pir^2 (3marks)
c) 3 (2 marks)
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
12. Can someone check where I've got (OR...???) written in the first post? which is it? as I have so many different sources telling me differe,t
13. (Original post by JAW-97)
Reckon I lost 8 marks in the vectors question, but otherwise OK. Didn't finish finding the x coordinate for Q, got sinq= -3/4, 1/2 (Probably wrong). Probably lost a couple of marks on the decreasing function one too as I didn't differentiate.

A* - 62
A - 57
B - 52
C - 47

Etc.
I got the gradient as 8/11 ://
14. (Original post by Jonbazza1997)
-3/4 was the y coordinate -1/8 was the x coordinate
how many marks lost for writing -3/4 then?
15. (Original post by danniegee)
Does anyone rememebr if they told us to simplify/write in a certain format the solution to 8a?
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
16. (Original post by JackThorpe7)
Attachment 423177All My answers, really not sure on the coordinates for E though.
Your coordinates for E seem to be correct. The only problem I see here is that the final differential should be dr/dr = k/r^2 (rather than k/Pir^2), as 1/Pi is a constant and is therefore included within k.
17. Somebody make a poll?
18. (Original post by Jonbazza1997)
How confident are u that the last vectors question is right
not at all confident. i felt like i did it wrong but a few people have similar answers.
19. (Original post by CD223)
It's not what I got, This other users,
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
20. (Original post by geeky_penguin)
how did people get -1/8 for 5c? I got -3/4 for some reason!
I got coordinates (-3/4, -1/8) and (1/2, 3/4)

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