Join TSR now and get all your revision questions answeredSign up now

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

    Offline

    1
    ReputationRep:
    (Original post by datpr0)
    It's fine. That's equivalent to 1/2 x y^-1/2, so you just integrate from there (it becomes 1/4 x y^1/2).
    don't you need to divide by 1/2 not mulitply? so 1/2 * 1/2 =1/4 is wrong, should be 1/2 divided by 1/2 giving 1. Idk if I've just got that very wrong tho
    Offline

    2
    ReputationRep:
    (Original post by 2014_GCSE)
    You've done 1/2 x 1/2. It's 1/2 / 1/2

    so it's y^1/2

    Ay, yes, haha :P... I just realised and edited it. Don't worry, though, I will go to sleep earlier tonight so I am actually awake in the exam ;D!
    Offline

    1
    ReputationRep:
    (Original post by moinoi)
    1/2rootY is equal to 1/2(y^-1/2)
    So integrates to y^1/2 or root Y
    Thanks


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by moinoi)
    don't you need to divide by 1/2 not mulitply? so 1/2 * 1/2 =1/4 is wrong, should be 1/2 divided by 1/2 giving 1. Idk if I've just got that very wrong tho
    Damn, everyone was too fast for my editing skills ! Yes, you're correct in the sense that it was 1/2 divided by 1/2 (i.e. 1/2 x 2) .
    Offline

    1
    ReputationRep:
    (Original post by student0042)
    I commented earlier on this on this post. It is a trapezium, therefore you need to smaller value because if you use the larger value, the shape will be a parallelogram. (As the lengths will be the same CD, i believe, but the shape wrong). If you still don't understand let me know and I will draw a diagram.
    A diagram would be very useful, thank you!
    Offline

    2
    ReputationRep:
    Are you guys ready for this? I cant wait to get it overw with.
    Offline

    1
    ReputationRep:
    (Original post by saeed97)
    Are you guys ready for this? I cant wait to get it overw with.
    Im scared they will make it different to a degree of difficulty
    Offline

    2
    ReputationRep:
    8a) is ln(y) = -cos(t) + c but in the mark scheme they changed the -cos(t) to A and this gave the final mark. Why did they do this? And would we have to do it in all questions of this type?

    Thanks!
    Attached Images
     
    Offline

    2
    ReputationRep:
    Does anybody know if we need to know these formulae?:

     sinA + sinB = 2sin(\frac{A +B}{2})cos(\frac{A - B}{2})
     sinA - sinB = 2cos(\frac{A + B}{2})sin(\frac{A - B}{2})
     cosA + cosB = 2cos(\frac{A + B}{2})cos(\frac{A - B}{2})
     cosA - cosB = 2sin(\frac{A + B}{2})sin(\frac{A - B}{2})
    Offline

    3
    ReputationRep:
    (Original post by Tiwa)
    Does anybody know if we need to know these formulae?:

     sinA + sinB = 2sin(\frac{A +B}{2})cos(\frac{A - B}{2})
     sinA - sinB = 2cos(\frac{A + B}{2})sin(\frac{A - B}{2})
     cosA + cosB = 2cos(\frac{A + B}{2})cos(\frac{A - B}{2})
     cosA - cosB = 2sin(\frac{A + B}{2})sin(\frac{A - B}{2})
    Nope we don't need to know them
    Offline

    1
    ReputationRep:
    (Original post by CD223)
    Plot twist when vectors doesn't appear because AQA forgot to put them in.


    Posted from TSR Mobile

    AKA The dream
    Offline

    2
    ReputationRep:
    (Original post by 2014_GCSE)
    Nope we don't need to know them
    Thanks!
    Offline

    3
    ReputationRep:
    With vectors, how do you use the AB = OA+OB? Like how does that work? Thanks!
    Offline

    1
    ReputationRep:
    (Original post by 2014_GCSE)
    How do you manipulate

    3sin4x

    My teacher swears that we'll get 4x's and stuff where we have to manipulate the double angle formulae to do this.
    How does it work?

    Does it become 3Sin2xCos2x?
    Would also like to know this.

    Would something like Cos(4x) be equivalent to
    2Cos^2(2x) - 1
    1 - 2Sin^2(2x)
    cos^2(2x) - sin^2(2x)
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Tiwa)
    Does anybody know if we need to know these formulae?:

     sinA + sinB = 2sin(\frac{A +B}{2})cos(\frac{A - B}{2})
     sinA - sinB = 2cos(\frac{A + B}{2})sin(\frac{A - B}{2})
     cosA + cosB = 2cos(\frac{A + B}{2})cos(\frac{A - B}{2})
     cosA - cosB = 2sin(\frac{A + B}{2})sin(\frac{A - B}{2})
    That's Edexcel.


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by Jankie)
    With vectors, how do you use the AB = OA+OB? Like how does that work? Thanks!
    The rule is AB = OB - OA.

    This is because vector AB is unknown. However, travelling from vector a to vector b is the same as travelling from vector a to the origin and the from the origin to vector b (i.e. AO + OB). You then proceed to write AO as -OA (AO = -OA), giving AB = OB - OA.
    Offline

    2
    ReputationRep:
    (Original post by Haza100)
    Would also like to know this.

    Would something like Cos(4x) be equivalent to
    2Cos^2(2x) - 1
    1 - 2Sin^2(2x)
    cos^2(2x) - sin^2(2x)
    No, they're not equivalent. You would have to write cos(4x) as cos(2x + 2x) and go from there.
    Offline

    2
    ReputationRep:
    (Original post by CD223)
    That's Edexcel.


    Posted from TSR Mobile
    Ah, I understand. A bit weird it's on the AQA formula booklet. :hmmmm2:
    Offline

    2
    ReputationRep:
    (Original post by Tiwa)
    Ah, I understand. A bit weird it's on the AQA formula booklet. :hmmmm2:
    Yeah but that formula booklet is full of stuff you won't need, at least for the core exams.
    Offline

    2
    ReputationRep:
    (Original post by datpr0)
    Yeah but that formula booklet is full of stuff you won't need, at least for the core exams.
    That is true.
 
 
 
Poll
Which web browser do you use?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.