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AQA A2 Mathematics Pure Core 4 MPC4 - 17th June 2016

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Original post by -jordan-
You loose the same percentage in the same time period.


Not of the original value. That would mean an element would only ever have 2 half lives..
Reply 441
Can anyone remember how many marks q3 was and q8?
Original post by df97
partial fractions and binomial expansion, i think answers are as follows:
A=-5/2 B=9/2
-1+7x-22/3x^2
not sure on limit, think some people have something like mod x < 4/3


wasnt there a smaller range mod x< 1/2
Reply 443
Original post by C0balt
What he said


So what form was the answer to the last question?
Original post by df97
partial fractions and binomial expansion, i think answers are as follows:
A=-5/2 B=9/2
-1+7x-22/3x^2
not sure on limit, think some people have something like mod x < 4/3


mod x <1/2 i believe , since the range of 1/2 falls within the range of 3/4
Original post by ihsanulhaq101
13/35 for me also.


SAME
Can we please verify if 1/5 or 13/35?
How did everyone find it? I thought it was alright, couldnt do last vectors or last page though
Original post by Pingu7
So what form was the answer to the last question?

3sin3y - 3cos3y =1/6 tan^-1(3x/2)+pi
Or sth
Original post by 1/2(p^2-p)
wasnt there a smaller range mod x< 1/2

yes thats right
Reply 449
Original post by tanyapotter
does anyone remember the two parametric equations that you had to find dy/dx for?


x = (4-4e^2-6t)4 y=e^3t/33t
Original post by m00ngrk
Can anyone remember how many marks q3 was and q8?

last one was 10 marks
3 was the 5 marker i think? Trig quadratic
Whoops, did anyone get lamda=3 and mu=2?:confused:
Original post by mickel_w
Not of the original value. That would mean an element would only ever have 2 half lives..


After 100 days you have k^100 lots of the original value left. After 200 days you have (k^100)^2 of the original value. So therefore you loose (0.8)^3 over 300 days, given that k^100 is approximately 0.8 because you had roughly 8=(k^100)*10
(edited 7 years ago)
Original post by hi-zen-berg
SAME
Can we please verify if 1/5 or 13/35?


I got 13/35
Original post by jtebbbs
Yeah sounds right. Just made me realise I put |x|<1/4 cause I was looking at (3-4x) and not (1-4/3x), there goes a mark lol

Edit: think the right answer would be |x|<3/4, as the |x| in |x|<1 is replaced by |4/3x|, so |4/3x|<1, so |x|<3/4


I got -1/2<x<1/2
What did everyone do for p(5-5^0.5)?
Original post by Jm098
x = (4-4e^2-6t)4 y=e^3t/33t


what was the gradient when you use these to find dy/dx
Original post by AQslAyy
I got -1/2<x<1/2


Me too, as this was the narrowest of the two ranges
Original post by tanyapotter
does anyone remember the two parametric equations that you had to find dy/dx for?


Attachment not found
Original post by AQslAyy
I got -1/2<x<1/2


Yeah don't listen to me mate, you're right, didn't even consider the (1+2x) part until just now, oh well it's only one mark

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