AQA Core 4 14th June Unofficial Mark Scheme

Hello, guys. I've not seen an unofficial mark scheme yet, so I thought I'd make one and edit the answers and questions to what most people got. I have tried to do as much as I can off memory, but after the awfulness of it, my mind must be telling me "no more", and so I can't remember vast amounts.

If you remember any questions/answers, please reply, and help bulk it up! It's looking quite flimsy at the moment!

Question 1 (Partial Fractions):
a) A=2 B=3
b) P=-1, q=3, r=-5

Question 2 (R-addition):
a) Alpha=71.6
x=32, 291

Question 3 (Binomial Expansion):

Question 4 (the exponential thing with Mr Brown and Mrs White):
a) V=£1160
b) Years at which balance will exceed £2000, N=24
c) Year at which second balance will exceed first balance, T=28

Question 5 (trig question with cartesian):
a) Gradient of normal = 1/3
b) ?
c) p=9/4

Question 6 (implicit):
Stationary points((1/3),1) and ((-1/3),-1)

Question 7 (vectors):
a) Show q=4, find point of intersection
Find B if APB is the right angle of an isosceles triangle.

Question 8 (differential equations):
a) dh/dt = k(2-h)
b) t = 2/3 (2x-1)^3/2 + 2/5 (2x-1)^5/2 - 4
c) Between 30 and 40, lol.

(edited 11 years ago)

Reply 1

SomeoneIveNeverMet

I don't remember what I got for 8,b, but wouldn't you have to do 2/3 x 1/2 (because differenting 2x gives 2?).
Sorry if I'm wrong, like I said, I actually can't remember what I put!

Reply 2

piette

Your question 8b here is wrong. Integration by parts should have been used. Sub in the values which were given (x=1, t=0) and see if they work for your equation above.

Reply 3

joe900

Hi;

Could someone please explain to me how to get dy/dx a sinx+cosecx from x=2cosx and y=3sin2x
Many thanks :smile:

Reply 4

RickyW123

Can you put the marks for each question down too please??

Reply 5

jamil78613

1

Agree with the first 4 questions as well as question 6, the rest I do not remember. Wasn't 8c something like 37.4 / 38 or something along those lines.

Reply 6

lrbailey

Hi

for the last part on question 7 where you had to find the coordinates of B, you end up getting a quadratic equation with μ in it, which factorised to (3μ+1)(3μ+5) and then the values for μ were -1/3 and -5/3
then I subbed those back in to B to get the actual coordinates but they're the only numbers I remember, I can't remember what I got for B in the end but they were the μ values! :smile:

Reply 7

lrbailey

I think 8b is wrong,
when you intergrate 15x(2x-1)^1/2, you use by parts and the overall equation at the end is:
t= 5(2x-1)^3/2 - (2x-1)^5/2 - 4

Reply 8

Jozhogg

Original post by joe900

Hi;

Could someone please explain to me how to get dy/dx a sinx+cosecx from x=2cosx and y=3sin2x
Many thanks :smile:

your notation is a little bit problematic here (too many x's :P) If we call theta t then:

x = 2cos t and y = 3sin 2t

dx/dt = -2sin t dy/dt = 6cos 2t

and also we know dy/dx = (dy/dt)/(dx/dt)

so dy/dx = (-3cos 2t) / sin t then using cos 2t = 1- 2sin² t gives

dy/dx = (6 sin² t -3) / sin t = 6sin t -3cosec t

That's what I did at least, I hope it's right and that this helped!

Reply 9

jamil78613

1

Agree with the first 4 questions as well as question 6, the rest I do not remember. Wasn't 8c something like either 32.4 or 37.4 or something along those lines, any of these ring a bell?

Reply 10

lrbailey

sorry another thing, I keep remembering bits!
for 7a I remember that it was something like P = (6,-2,1), may be wrong though :smile:

and the other part of the question was find AP^2 and I got it equal to 20.
then you use that value to find B and you get this once you've expanded all brackets:
20 = 45μ^2 + 90μ +45
so then: 45μ^2 + 90μ +25 = 0
then divide by 5: 9μ^2 + 18μ + 5 = 0
factorises to: (3μ +1)(3μ+5) as I said before

Reply 11

APonderingMind

I am such an idiot..... for question 4, I think I put down N= 28 as my final answer when I should have written T=28 =_____=

(edited 11 years ago)

Reply 12

Appeal to reason

14

Really messed up on 8b I think, as I got 17.8 for 8c (Although at the time this seemed fine to me)
I'm so annoyed I left the 6 mark vector question for the end, as when I actually read it, I realised I'd done a practice question of a similar form last night and could've actually done it fairly quickly. I only managed to get to the point where I'd formed the quadratic though :frown:

My method for 8)b) is as follows:

Unparseable latex formula:

dx/dt = 1/(15x\sqrt(2x-1))[br]\int x\sqrt(2x-1)dx=\int 1/15 dt[br]LHS:[br]x=u[br]\sqrt(2x-1)=dv/dx[br]v=1/3 ((2x-1)^3^/^2)[br][br]=x/3 (2x-1)^3^/^2 - \int 1/3((2x-1)^3^/^2)dx[br][br]=x/3 (2x-1)^3^/^2 - (2/over15) ((2x-1)^5^/^2)[br][br]RHS: [br]\int 1/15 dt = t/15+c[br][br] t/15+c=x/3 (2x-1)^3^/^2 - (2/over15) ((2x-1)^5^/^2)[br]t+c=5x(2x-1)^3^/^2 - 2((2x-1)^5^/^2) (multiplied throughout by 15)[br]t=0 x=1[br]c=5-2 = 3[br]t=5x(2x-1)^3^/^2 - 2((2x-1)^5^/^2) - 3[br]x=2[br]t=30\sqrt3 - 18\sqrt3 - 3[br]t=-3+12\sqrt3 = 17.8 minutes.[br]

Where is the mistake in this, as it is completely different to what everyone else I know got.

Reply 13

Appeal to reason

14

Original post by lrbailey

sorry another thing, I keep remembering bits!
for 7a I remember that it was something like P = (6,-2,1), may be wrong though :smile:

and the other part of the question was find AP^2 and I got it equal to 20.
then you use that value to find B and you get this once you've expanded all brackets:
20 = 45μ^2 + 90μ +45
so then: 45μ^2 + 90μ +25 = 0
then divide by 5: 9μ^2 + 18μ + 5 = 0
factorises to: (3μ +1)(3μ+5) as I said before

I got all of these, but didn't have time to factorise the quadratic :frown:

Reply 14

lrbailey

Original post by Appeal to reason

Really messed up on 8b I think, as I got 17.8 for 8c (Although at the time this seemed fine to me)
I'm so annoyed I left the 6 mark vector question for the end, as when I actually read it, I realised I'd done a practice question of a similar form last night and could've actually done it fairly quickly. I only managed to get to the point where I'd formed the quadratic though :frown:

My method for 8)b) is as follows:

Unparseable latex formula:

dx/dt = 1/(15x\sqrt(2x-1))[br]\int x\sqrt(2x-1)dx=\int 1/15 dt[br]LHS:[br]x=u[br]\sqrt(2x-1)=dv/dx[br]v=1/3 ((2x-1)^3^/^2)[br][br]=x/3 (2x-1)^3^/^2 - \int 1/3((2x-1)^3^/^2)dx[br][br]=x/3 (2x-1)^3^/^2 - (2/over15) ((2x-1)^5^/^2)[br][br]RHS: [br]\int 1/15 dt = t/15+c[br][br] t/15+c=x/3 (2x-1)^3^/^2 - (2/over15) ((2x-1)^5^/^2)[br]t+c=5x(2x-1)^3^/^2 - 2((2x-1)^5^/^2) (multiplied throughout by 15)[br]t=0 x=1[br]c=5-2 = 3[br]t=5x(2x-1)^3^/^2 - 2((2x-1)^5^/^2) - 3[br]x=2[br]t=30\sqrt3 - 18\sqrt3 - 3[br]t=-3+12\sqrt3 = 17.8 minutes.[br]

Where is the mistake in this, as it is completely different to what everyone else I know got.

when you've integrated the 1/3(2x-1)^3/2 on the by parts bit you should get:
1/over15(2x-1)^5/2 wherease you've put 2/over15(2x-1)^5/2

Reply 15

Appeal to reason

14

Original post by lrbailey

when you've integrated the 1/3(2x-1)^3/2 on the by parts bit you should get:
1/over15(2x-1)^5/2 wherease you've put 2/over15(2x-1)^5/2

Ahh I see, I forgot to divide by the differential of (2x-1) :angry:

Well thats me lost a lot of marks. -.-

Reply 16

canI?

Who did find the paper hard?

Reply 17

lrbailey

Original post by Appeal to reason

Ahh I see, I forgot to divide by the differential of (2x-1) :angry:

Well thats me lost a lot of marks. -.-

I don't think you would've lost too many, you probably will get method marks :smile:

Reply 18

username896004

Original post by lrbailey

I think 8b is wrong,
when you intergrate 15x(2x-1)^1/2, you use by parts and the overall equation at the end is:
t= 5(2x-1)^3/2 - (2x-1)^5/2 - 4

its funny because thats actually equal to what the original poster has written :')

Reply 19

lrbailey

Original post by DavidMRoper

its funny because thats actually equal to what the original poster has written :')

No it isn't the coefficients are different

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AQA Core 4 14th June Unofficial Mark Scheme

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