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

    0
    ReputationRep:
    Vector questions are always like me in bed

    Gentle and calm foreplay before excessive violent shafting
    Offline

    2
    ReputationRep:
    (Original post by CD223)
    Not very good! Prefer C4 haha! You?
    Which history do you do? I'm doing Edexcel Kaiser to Fuhrer?


    Posted from TSR Mobile
    Ah I'm on aqa, and cover the Cold War and international relations up to 2004! Really interesting. The thing I dislike about history tho is that I'll be mentioning the tiniest percentage of what I've learnt the whole year ... Feels like a waste (even tho it isnt).
    Offline

    2
    ReputationRep:
    I struggle with forming differential equations from wordy questions. this usuallyll leads to messing the whole question up. any tips?

    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    can somebody explain what I would do for 3c

    Posted from TSR Mobile
    Attached Images
     
    Offline

    3
    ReputationRep:
    (Original post by 2014_GCSE)
    Here you go:

    Attachment 421419

    Basically, B & C could be in either place. So you just solve it for both places.
    And as it's a parallelogram, it is two pairs of equal sides (like a rectangle).
    Hi, thanks for this. I'm just wondering if you are able to explain to me how and why the OC = OA+AC or something? I don't really understand that concept, thank you!!
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by nairline)
    Ah I'm on aqa, and cover the Cold War and international relations up to 2004! Really interesting. The thing I dislike about history tho is that I'll be mentioning the tiniest percentage of what I've learnt the whole year ... Feels like a waste (even tho it isnt).
    I couldn't agree more. Best of luck!!


    Posted from TSR Mobile
    Offline

    12
    ReputationRep:
    (Original post by datpr0)
    Yes, well, they certainly diminsh the point of learning certain graphs (i.e. inverse sin/cos/tan and the graphs for cosec/sec/cot). I assume, also, that their existence is the reasoning for having to explicitly state points of intersection with the axis to get the marks nowadays as anyone can draw a correctly shaped graph by copying it from their graphing calculator.

    They're also the reason why many papers deal with the graph of y = f(x), where f(x) is undefined, so that your graphing calculator won't help; which, in turn, begs the question of why their use is even permitted in the first place ^^.
    They tell you the points of intersection/min/max
    Offline

    0
    ReputationRep:
    Name:  3WMdVqX.png
Views: 151
Size:  54.5 KB we're looking at part 6)c)ii) boys and girls.

    I've got PS as
    \begin{bmatrix} 5 \\ -8 \\ 2 \end{bmatrix} + n \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}

    and we know the coordinates of R, so RS is
    \begin{bmatrix} 3 \\ -2 \\ 4 \end{bmatrix} + m \begin{bmatrix} a \\ b \\ c \end{bmatrix}

    and we also know RS is perpendicular to
    \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}

    and where PS and RS intersect, is S, which we're looking for.
    Where do I go from here ...?
    Offline

    12
    ReputationRep:
    (Original post by LetMeInPlox)
    Name:  3WMdVqX.png
Views: 151
Size:  54.5 KB we're looking at part 6)c)ii) boys and girls.

    I've got PS as
    \begin{bmatrix} 5 \\ -8 \\ 2 \end{bmatrix} + n \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}

    and we know the coordinates of R, so RS is
    \begin{bmatrix} 3 \\ -2 \\ 4 \end{bmatrix} + m \begin{bmatrix} a \\ b \\ c \end{bmatrix}

    and we also know RS is perpendicular to
    \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}

    and where PS and RS intersect, is S, which we're looking for.
    Where do I go from here ...?
    Its nice you read the question out lol
    Can't help you soz you're doing it the wizard way
    Offline

    2
    ReputationRep:
    I'm confident with all the concepts of the vectors in terms of distance between 2 vectors, parallel, intersecting, skew lines...blah blah.

    However, I always seem to be confused by what is often the final question of the section which most of the time asks you to apply your knowledge to a shape...

    Has anyone got any tips when tackling these styles of questions? :/
    Offline

    2
    ReputationRep:
    (Original post by LetMeInPlox)
    Name:  3WMdVqX.png
Views: 151
Size:  54.5 KB we're looking at part 6)c)ii) boys and girls.

    I've got PS as
    \begin{bmatrix} 5 \\ -8 \\ 2 \end{bmatrix} + n \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}

    and we know the coordinates of R, so RS is
    \begin{bmatrix} 3 \\ -2 \\ 4 \end{bmatrix} + m \begin{bmatrix} a \\ b \\ c \end{bmatrix}

    and we also know RS is perpendicular to
    \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}

    and where PS and RS intersect, is S, which we're looking for.
    Where do I go from here ...?
    Just did this myself

    Sub PS into where you have m(abc) then dot product

    A.B = Cos 90 (which as you know, is 0)

    My problem with this is that I did the line SR when the mark scheme says I should have done line RS, why is this? Aren't they both supposed to be going outwards from the angle? PQ is. I came to this thread to ask why this is, can anyone help?
    • Study Helper
    • Welcome Squad
    Offline

    18
    ReputationRep:
    Just wondering how people do these sorts of question? :beard:
    Spoiler:
    Show
    Offline

    2
    ReputationRep:
    (Original post by 8472)
    Just wondering how people do these sorts of question? :beard:
    Spoiler:
    Show
    Mulitply everything by the (2x^2 -x +2) and you have that 4x^3 = 2AX^3 so you can see that A would be 2, then you use this value for A in 16x= 2Ax +4Bx to get the value for B which is 3
    Offline

    2
    ReputationRep:
    (Original post by 8472)
    Just wondering how people do these sorts of question? :beard:
    Spoiler:
    Show

    I would use algebraic long division first. The result becomes the Ax and the remainder is the B(4x-1).

    There's apparently a quicker/easier way according to the mark scheme though.
    Offline

    2
    ReputationRep:
    (Original post by 8472)
    Just wondering how people do these sorts of question? :beard:
    Spoiler:
    Show
    Polynomial long division :P
    • Study Helper
    • Welcome Squad
    Offline

    18
    ReputationRep:
    (Original post by Alexandrite)
    I would use algebraic long division first. The result becomes the Ax and the remainder is the B(4x-1).

    There's apparently a quicker/easier way according to the mark scheme though.
    :woo:

    No idea how I did not see that my result was Ax. :lol: Awesome thanks.
    Spoiler:
    Show
    Repped you all.
    Offline

    2
    ReputationRep:
    (Original post by datpr0)
    Yes ;D! I completely agree with that rant about Q2, like, I know we have graphing calculators but I was still waking up at that point - my brain was not yet functioning properly xD...
    LOOOL IKR, at the end "I'll end up doing Dance at Slough with college dropouts" :ahee:
    Offline

    2
    ReputationRep:
    (Original post by the_googly)
    can somebody explain what I would do for 3c

    Posted from TSR Mobile
    You just state the stricter of the two validity ranges from the two expansions you did in b. .
    Offline

    3
    ReputationRep:
    (Original post by Alexandrite)
    Just did this myself

    Sub PS into where you have m(abc) then dot product

    A.B = Cos 90 (which as you know, is 0)

    My problem with this is that I did the line SR when the mark scheme says I should have done line RS, why is this? Aren't they both supposed to be going outwards from the angle? PQ is. I came to this thread to ask why this is, can anyone help?
    your PS is ACTUALLY OS, RS is then OS - OR
    Offline

    3
    ReputationRep:
    (Original post by LetMeInPlox)
    Name:  3WMdVqX.png
Views: 151
Size:  54.5 KB we're looking at part 6)c)ii) boys and girls.

    I've got PS as
    \begin{bmatrix} 5 \\ -8 \\ 2 \end{bmatrix} + n \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}

    and we know the coordinates of R, so RS is
    \begin{bmatrix} 3 \\ -2 \\ 4 \end{bmatrix} + m \begin{bmatrix} a \\ b \\ c \end{bmatrix}

    and we also know RS is perpendicular to
    \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}

    and where PS and RS intersect, is S, which we're looking for.
    Where do I go from here ...?
    your PS is ACTUALLY OS, RS is then OS - OR

    I think?
 
 
 
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.