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

Yes. The angle is always between the 2 vectors. B is the angle in the middle, made by AB and BC

Right, I got you :smile:

One last question, are you able to explain to me why in 8b they have positive 5? Because if you do OB.BA you get -5?

https://39a16d3272cfbc6efbbec4a18e06ad4b39d4616d.googledrive.com/host/0B1ZiqBksUHNYWlRPQ3B0dGFMc1k/January%202009%20QP%20-%20C4%20AQA.PDF

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Reply 1101

ubisoft

3

Original post by Jankie

Right, I got you :smile:

One last question, are you able to explain to me why in 8b they have positive 5? Because if you do OB.BA you get -5?

https://39a16d3272cfbc6efbbec4a18e06ad4b39d4616d.googledrive.com/host/0B1ZiqBksUHNYWlRPQ3B0dGFMc1k/January%202009%20QP%20-%20C4%20AQA.PDF

Because the angle is acute. If you use the negative then you get an obtuse angle.

Reply 1102

CD223

OP

10

Finally getting around to revise this after history today... Really hope the paper isn't as horrible as I think it will be.

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Reply 1103

lionelmessi19

3

Original post by CD223

Finally getting around to revise this after history today... Really hope the paper isn't as horrible as I think it will be.

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Hey man! How was history!?
How you feeling for tomorrow??

Can you do me a favour please and write a solution to June 2010 7c - its on vectors
and Jan 2011 - 7bi

Thank you!!! :biggrin:

Reply 1104

KAKAKA17

12

Does anybody know of any good websites or youtube videos to do vector questions? Definitely my biggest weakness along with exponential growth and decay.

With vectors I can only do the basics, which are usually the questions up to 3 marks, any more than that I don't know how to do them; so I'm really hoping the vectors section tomorrow isn't worth any more than 10 marks otherwise I'm screwed

Reply 1105

Jankie

3

Original post by ubisoft

Because the angle is acute. If you use the negative then you get an obtuse angle.

Argh sorry, how do you know when to do that then? :ashamed2:

Reply 1106

the_googly

7

Original post by the_googly

I struggle with forming differential equations from wordy questions. this usuallyll leads to messing the whole question up. any tips?

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help plzzz

Reply 1107

Jimmy20002012

4

Original post by Jimmy20002012

Any help on whole of part b! ImageUploadedByStudent Room1433771048.183159.jpg

binomial question

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Anyone?

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Reply 1108

SomeGuy96

1

Would really appreciate it if somebody could help me out. When i look at mark schemes i can see where I've went wrong or made stupid mistakes. But on June 2014 Q5(c)(iii) i am completely stumped. I managed to get the first mark but i don't understand the jump that is made to obtaining values of x.

Reply 1109

moinoi

5

Original post by SomeGuy96

Would really appreciate it if somebody could help me out. When i look at mark schemes i can see where I've went wrong or made stupid mistakes. But on June 2014 Q5(c)(iii) i am completely stumped. I managed to get the first mark but i don't understand the jump that is made to obtaining values of x.

can you post your workings? I'm about to do that paper if someone doesn't beat me to it I'll send you my workings, I think I got that one right when we did it as a mock.

Reply 1110

gloria0816

1

For this question june 12
How do you know the values of x and y to sub in to solve dy/dx =0 ImageUploadedByStudent Room1433774901.215632.jpg

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Reply 1111

datpr0

2

Original post by gloria0816

For this question june 12
How do you know the values of x and y to sub in to solve dy/dx =0 ImageUploadedByStudent Room1433774901.215632.jpg

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You never sub in values of x or y when solving dy/dx = 0.

In this question you differentiate implicitly, then either sub in 0 for each dy/dx term of rearrange to make dy/dx the subject and then sub in 0 for it (because stationary points occur when dy/dx = 0). You then rearrange the resultant equation to form two sets of solutions, one where x = a bunch of y's and one where y = a bunch of x's. What these tell you is that a stationary point occurs when x = a bunch of stuff to do with y and vice-versa. The next step is to substitute these solutions into the original equation to find a numerical coordinate of either x or y for each stationary point respectively.

Please note: if I don't sound very coherent it's because I'm REALLY tired!

Reply 1112

Tom^

1

Original post by gloria0816

For this question june 12
How do you know the values of x and y to sub in to solve dy/dx =0 ImageUploadedByStudent Room1433774901.215632.jpg

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For the Question you get the differentiation for the equation and you know it = 0
therefore 0=6y -18x this means 3x=y, plug this back into the original equation to find the stationary points of the line which you will end up with a quadratic,

9X^2-6X(3X)+4(3X)^2=3

giving you 27X^2=3 therefore X=+ or - 1/3

Hope this helps

(edited 8 years ago)

Reply 1113

Hexaneandheels

4

Considering I was getting 98/99 and 100UMS on all previous mock papers, I'd say something went wrong in the exam considering I must have dropped at least ten marks. It is not necessarily unpreparedness when not doing well in exam, sometimes there can be other factors, such as health, or the ACTUAL EXAM QUESTIONS!!!

Original post by ubisoft

How is you not being prepared for the paper being robbed?

Reply 1114

problemq

1

Original post by Jimmy20002012

Anyone?

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You want to find the expansion for (1+kx)^-2 - (1+x)^n up to x^3. Then compare coefficients with the given result. You obtain n = -2k by looking at the x coefficient (Here this coefficient is 0). Then compare the x^2 term to find k (replace n by -2k here) you should get a quadratic in k, reject positive k since n<0 (given). Then complete by looking at x^3 term and subbing in k and n to obtain p.

Reply 1115

datpr0

2

Original post by SomeGuy96

Would really appreciate it if somebody could help me out. When i look at mark schemes i can see where I've went wrong or made stupid mistakes. But on June 2014 Q5(c)(iii) i am completely stumped. I managed to get the first mark but i don't understand the jump that is made to obtaining values of x.

Okay well the first thing to remember is that it's c.iii. and that when part of a question is split further into sub-questions it will likely mean these "sub-questions" will be linked together in some manner. In this case, your told explicitly to "use the results from (c)(i) and (c)(ii) to show...", so I'm not really sure why you can't see the 'jump' :/.

They tell you 4cos(2theta) x cos(theta) + 1 = 0, and in (c)(ii) you just proved that could be written as 8x³- 4x + 1 (= 0), where x = cos(theta). Also, in (c)(i) you proved that (2x - 1) was a factor of 8x³ - 4x + 1, which means you can go a step further to rewrite it as (2x - 1)(4x² + 2x - 1) = 0.

The quadratic doesn't factorise and consequently you use the quadratic formula, giving you two values for x on top of the one from 2x - 1 = 0. You know x = cos(theta) and you've been told theta = 72 degrees in the question. Therefore cos72 = the value for x equivalent to that they asked you to show.

Again, please note: if I don't sound very coherent it's because I'm REALLY tired!

(edited 8 years ago)

Reply 1116

Jimmy20002012

4

Original post by problemq

You want to find the expansion for (1+kx)^-2 - (1+x)^n up to x^3. Then compare coefficients with the given result. You obtain n = -2k by looking at the x coefficient (Here this coefficient is 0). Then compare the x^2 term to find k (replace n by -2k here) you should get a quadratic in k, reject positive k since n<0 (given). Then complete by looking at x^3 term and subbing in k and n to obtain p.

Still don't seem how you get n= -2k??

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Reply 1117

AaronA330

1

In the tables book, there are two formulae for binomial expansion. When do you know which method to use?

Reply 1118

problemq

1

Original post by Jimmy20002012

Still don't seem how you get n= -2k??

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What did you get for the expansion of (1+kx)^-2 - (1+x)^n? To obtain this use your answer from part (a) and then subtract the expansion of (1+x)^n from this (expansion of (1+x)^n is in formula book). Then compare coefficients. The coefficient of x in 6x^2 + px + ... is 0 (as this is equal to 0 + 0*x + 6x^2 + px +....). so set your coefficient of x in your expansion you found to 0 so you can obtain an equation in k and n which solves to n=-2k