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

1. Not sure what I've done wrong, but I've just done june 12 and on 8 b)I) I got C=4/15 and in the mark scheme it's -4/15. I collected the constants from both sides over to one. One side comes out as 0+C so the other side was equal to C. Are you meant to put C to the other side if it equals 0 or something?
2. can someone help me with June 11 question 8b? really confused
3. (Original post by datpr0)
The funky vectors questions are basically an application of a bunch of the standard stuff. In this case, if you're unsure just draw yourself a labelled diagram of what's going on and you should be able to solve it.

As for shapes you just need to remember their basic properties, generally what sides are of equal length/ parallel/ perpendicular.
Yeah I did this paper the other day and I only got on coordinate, wasn't sure how to get the other. And it takes so long for 4 marks it's just awful haha
4. (Original post by 2014_GCSE)
Someone posted this list on a previous page:

guys what shapes could they ask?~~~[vectors]
Rhombus
isosceles triangle
trapeziums
kites
rectangles
parallelograms
squares
perpendicular a.b=0
parallel=a.b=-1
cut put circles in there
maybe use cosine rule/sine rule also given that we find the magnitudes out???
circles????
5. (Original post by JackNorman)
Not sure what I've done wrong, but I've just done june 12 and on 8 b)I) I got C=4/15 and in the mark scheme it's -4/15. I collected the constants from both sides over to one. One side comes out as 0+C so the other side was equal to C. Are you meant to put C to the other side if it equals 0 or something?
I did the same thing! I just ignored it because you end up with the same answer in the end, and I think the examiner would be smart enough to see that you've just put the constant on the other side because it's the exact same number, just negative.
6. (Original post by madmenace)
circles????
Yeah... I don't quite get that either. I'm guessing they might have a vector going across a circle and you might have to work out the circumference by the vector length and then work out the area of the circle? Something like that.
7. (Original post by datpr0)
The scalar product = 0 if they're perpendicular and as such if you're told that AP-> is perpendicular to CD-> you know you'll probably need to use the scalar product to find the answer they want.
I understand this, but sometimes we dot a vector with another vector and sometimes we have to dot a vector with a direction vector of another line, how do we know? Is it if we do not have the direction vector of the line we have to dot two vectors?

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8. (Original post by 2014_GCSE)
I did the same thing! I just ignored it because you end up with the same answer in the end, and I think the examiner would be smart enough to see that you've just put the constant on the other side because it's the exact same number, just negative.
Ah right cheers. Think in future I'll just put the C over to the other side to be safe 😂
9. (Original post by 2014_GCSE)
I did the same thing! I just ignored it because you end up with the same answer in the end, and I think the examiner would be smart enough to see that you've just put the constant on the other side because it's the exact same number, just negative.
Also did you manage the vectors question? I just don't see where the squaring comes into it, never did that in lessons :s
10. (Original post by madmenace)
circles????
Yeah, that's unlikely. The only things I can imagine they'd be able to ask is if you find the magnitude of a line which you're told is the radius of a circle so then you could find area/circunfrence I guess... Or finding the angle between two lines of equal length and finding the arc length/ area of a segment using r(theta)/ r^2(theta).
11. (Original post by JackNorman)
Also did you manage the vectors question? I just don't see where the squaring comes into it, never did that in lessons :s
Which paper was it again, haha?
12. (Original post by 2014_GCSE)
Yeah... I don't quite get that either. I'm guessing they might have a vector going across a circle and you might have to work out the circumference by the vector length and then work out the area of the circle? Something like that.
oh so i guess that isnt too bad?? vector length is magnitude right?
13. (Original post by RhyaCeri)
can someone help me with June 11 question 8b? really confused
Sure. I'll go do it now and then take a picture of how it's done.
14. (Original post by JackNorman)
Also did you manage the vectors question? I just don't see where the squaring comes into it, never did that in lessons :s
Oh, it's really simple . AP is the magnitude of vector AP->, which you find by the sqrt(x^2 + y^2 + z^2). So AP^2 just cancels out the sqrt.
15. (Original post by madmenace)
oh so i guess that isnt too bad?? vector length is magnitude right?
Yes.
16. (Original post by 2014_GCSE)
It doesn't say in the mark scheme you're not allowed to do that, it's just not one of the listed methods.
I think as long as you leave it in the: "y=x^2/9 + 9/x^2" form at the end, it doesn't matter how you do it.

I literally just did that paper and I subbed in the values.
Yeah I normally just sub them in but then I read this is the Examiner's Report.
17. (Original post by 2014_GCSE)
Sure. I'll go do it now and then take a picture of how it's done.
thank you so much
18. (Original post by Tibooster)
Yeah I normally just sub them in but then I read this is the Examiner's Report.
That says you can sub the numbers in!! As long as you leave a final equation at the end, you can't just "verify" it works by subbing them in.

You need to leave the final "y=x^2/9-9/x^2".
19. (Original post by RhyaCeri)
thank you so much
I hope you can follow what I did because I know it's a bit messy at times. Feel free to ask me about any parts where you don't understand what I did!

20. (Original post by 2014_GCSE)
That says you can sub the numbers in!! As long as you leave a final equation at the end, you can't just "verify" it works by subbing them in.

You need to leave the final "y=x^2/9-9/x^2".
Oh sorry my bad, that's a relief because I thought I was going to have to learn a new method

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