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

1. (Original post by JRN-95)
For 8b)i) isnt it dr/dt= (pi*r^2)/k
No because you were told that dr/dt is inversely proportional to the area (therefore 1/pi*r^2).

Also, I am pretty sure it told you in the question the constant was a. So the actual answer should have been dr/dt = a/pi*r^2.
2. (Original post by RavinderKlair97)
how much ums would 55 marks get?
I think I got around 55 as well :// i hope very much for 80+
3. (Original post by ForgottenApple)
That's what a moron would say because it hasn't been marked yet
4. (Original post by amyrah)
coordinates for C were (-9, -2, 18) ?
I got this too! Does anyone know whether this is right/wrong with explanation?
5. Do you remember the actual diff equation they asked you to solve?
6. (Original post by datpr0)
No because you were told that dr/dt is inversely proportional to the area (therefore 1/pi*r^2).

Also, I am pretty sure it told you in the question the constant was a. So the actual answer should have been dr/dt = a/pi*r^2.
****!!! i did dr/dt= pi*r^2/k damnnn!!! could i still get some marks and follow through marks for the next part
7. Does anyone remember the question for the implicit differentiation? Just want to go over it again as I don't remember if I got 11/8 or not
8. is there an agreed upon mark scheme anywhere?
9. (Original post by HennersPD)
Do you remember the actual diff equation they asked you to solve?
Yes.

It was: dx/dt = (4+5x)^1/2 / 5(1+t)^2
10. Anyone got a worked solution for the vectors question part b and c?
11. (Original post by HennersPD)
Do you remember the actual diff equation they asked you to solve?
if think it was dx/dt= (4+5x)^1/2 divided by 5(t+1)^2
12. (Original post by jchap9776)
Potentially, did anybody else show x tending to infinity = 2x-3 tending to infinity = 1/(2x-3) tends to 0? Therefore is decreasing. (Idk if it was 2x-3 but was something like that)
Thats what i literally done.
13. for 2)b)ii) I got

-1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

Therefore √29cos(x+1.19) = √29

cos(x+1.19) = 1
x + 1.19 = 0
x = -1.19

So why is it 5.09 where as it could be both...
14. For the last question, I forgot to divide 0.5 by 60. I wrote a quick comment, and some working-out to correct the value, but then I realise that I forgot to square root the 60. Now I'm pissed >

Also I forgot the +C in the differential equation. How many marks do you think I'll lose
15. (Original post by Kennethm)
for 2)b)ii) I got

-1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

Therefore √29cos(x+1.19) = √29

cos(x+1.19) = 1
x + 1.19 = 0
x = -1.19

So why is it 5.09 where as it could be both...
It said between 0 < x <2pi I think
16. question 2 needs sorting and for vectors I swear it was E(11,0,0) and E(23,4,-8)
17. (Original post by Kennethm)
for 2)b)ii) I got

-1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

Therefore √29cos(x+1.19) = √29

cos(x+1.19) = 1
x + 1.19 = 0
x = -1.19

So why is it 5.09 where as it could be both...
because 0<x<2pi
18. (Original post by Kennethm)
for 2)b)ii) I got

-1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

Therefore √29cos(x+1.19) = √29

cos(x+1.19) = 1
x + 1.19 = 0
x = -1.19

So why is it 5.09 where as it could be both...
Because they told you to find the maximum value in the range 0 < x < 2pi !
19. (Original post by jamal_m96)
Nah love im pretty sure my core 3 exam had a question 9
There was deffo 8 questions, are you sure you're in the right thread?
20. (Original post by JAW-97)
There was deffo 8 questions, are you sure you're in the right thread?
He's trolling, ignore him.

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