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

1. (Original post by ubisoft)
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

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://39a16d3272cfbc6efbbec4a18e06...20C4%20AQA.PDF
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2. (Original post by Jankie)
Right, I got you

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://39a16d3272cfbc6efbbec4a18e06...20C4%20AQA.PDF
Because the angle is acute. If you use the negative then you get an obtuse angle.
3. 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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4. (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!!!
5. 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
6. (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?
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
8. (Original post by Jimmy20002012)
Any help on whole of part b! Attachment 422239 binomial question

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Anyone?

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9. 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.
10. (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.
11. For this question june 12
How do you know the values of x and y to sub in to solve dy/dx =0

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12. (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

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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!
13. (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

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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
14. 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?
15. (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.
16. (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!
17. (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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18. In the tables book, there are two formulae for binomial expansion. When do you know which method to use?
19. (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
20. 3/2Ln3 - 2Ln2 becomes what? Im not sure what you do with the numbers infront of the Ln?

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