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

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Original post by CD223
What I wrote on the mark scheme was what I put down in the paper so I'm hoping I got the majority right but probably missed some method marks - 71-72 maybe?:smile:

Core 4 will be hard :frown:


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thats great for you then! I'm just worried Bath won't let me in if I don't get that A* in Maths :frown: it's annoying as they've increased the grades I need from AAB to A*AA :frown: I know just hopefully not as hard as this exam! :/
Reply 821
Original post by UOE15
Yeah I do understand what you've done, it is logical but I just don't think I'd remember how to do it in the exam haha
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Haha just keep attempting past questions on differential equations until you're confident?
Reply 822
Original post by HennersPD
thats great for you then! I'm just worried Bath won't let me in if I don't get that A* in Maths :frown: it's annoying as they've increased the grades I need from AAB to A*AA :frown: I know just hopefully not as hard as this exam! :/


Seems harsh?!


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Reply 823
Original post by saad97
Haha just keep attempting past questions on differential equations until you're confident?


Yeah I am going to do that as I did that for vectors and can vaguely do them now!


I know that's why I'm panicking about getting an A* now as I'll need 100 UMS in core 4 as i only think I've got about 64/75 in core 3 :frown: can you suggest any tough vectors questions as thats what I'm worst on!
Reply 825
5b Jan 11 explanation anyone?
Reply 826
Original post by amyrah
5b Jan 11 explanation anyone?


Yeah, so it's just saying that when t=d, m = m0/16.

i.e. 10/16 = 10 x 2^(-1/8)d

=> 1/16 = 2^-d/8 (as the 10's cancel)
(edited 8 years ago)
Reply 827
Original post by datpr0
Yeah, so it-s just saying that when t=d, m = m0/16.

i.e. 10/16 = 10 x 2^(-1/8)d

=> 1/16 = 2^-d/8 (as the 10's cancel)


So basically I can put any value as m0?
Could someone explain q7 from June 11 for me please :/


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Reply 829
Original post by amyrah
So basically I can put any value as m0?


Well yes, because the values you enter would just cancel each other. I think, actually a better way of showing it mathematically would be to just indicate the m0's cancelling.
Could someone explain q7 from June 11 for me please :/


EDIT: also 8b of the same paper. Why is it + 1/(1-x) as I keep getting minus. Also how do you change to the form they want the answer in

Thanks!
Reply 831
Original post by EmilyC96
Could someone explain q7 from June 11 for me please :/


EDIT: also 8b of the same paper. Why is it + 1/(1-x) as I keep getting minus. Also how do you change to the form they want the answer in

Thanks!


image.jpg

This is 8b. I did it a few days ago too and was confused by it but apparently you multiply the -1 by (-x/1-x) to get (x/1-x). Well that's what my teacher said anyway.
Hope that helps :smile:
Reply 832
Original post by EmilyC96
Could someone explain q7 from June 11 for me please :/


EDIT: also 8b of the same paper. Why is it + 1/(1-x) as I keep getting minus. Also how do you change to the form they want the answer in

Thanks!


7
a) basically rewrite what they've told you in a differential equation.

They said surface area (A) is decreasing at a constant (-k) rate (t).
=> dA/dt = -k

... dA/dt is the rate of change of surface area. It is DECREASING at a CONSTANT rate, so -k.

b) i. Integrate your differential equation.

=> A = -kt +C

Its INITIAL (t=0) radius is 60cm so C = 4pi x 60^2 (as they tell you to assume surface area = 4pi x r^2).

With C calculated, input the values they give you and C into the integrated equation. This gives:

4pi x 30^2 = -9k + 4pi x 60^2

Solve to find k.

Substitute k & C into integrated equation and just take 1200pi out as a factor to get it in the correct form.

b) ii. A = 0, => -t + 12 = 0...
(edited 8 years ago)
Reply 833
Original post by EmilyC96
Could someone explain q7 from June 11 for me please :/


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image.jpg
Original post by saad97
image.jpg

This is 8b. I did it a few days ago too and was confused by it but apparently you multiply the -1 by (-x/1-x) to get (x/1-x). Well that's what my teacher said anyway.
Hope that helps :smile:


Thanks :smile:


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Anyone fancy explaining the necessary and sufficient notations?

According to the spec we need to be able to use and understand them.
Guys I will be really grateful , if someone helps me with a question, I attempted it like 10 times and don't know what to do.
Its a question from JUNE 2010 QUESTION 3(b)
Thanks !
Could someone help me with 3c from jan 12 please


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Reply 839
Original post by EmilyC96
Could someone help me with 3c from jan 12 please


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It's pretty simple actually.

Cubed-root of 100 = 10^2/3

Therefore just consider what value of x is required for 8 + 6x = 10 to be true. Then substitute this value of x into your answer to b and simplify into the form a/b.
(edited 8 years ago)

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