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

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    (Original post by amyrah)
    I got up to the equation of the normal but don't understand what to do after? Do you mind explaining?
    All he did was create a cartesian equation for the curve from the two parametric ones and then solved it simultaneously with the equation of the normal at (9, 1) to find the x-coordinates of the points of intersection. This gives two points of intersection, one of which being 9 which was already known; thus, you then simply substitute the other, previously unknown, point of intersection in to the parametric equation where x is the subject to find the respective value for t at that point.
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    Really stuck on question 4 a) ii) http://filestore.aqa.org.uk/subjects...4-QP-JAN13.PDF

    It looks like it should be really easy? But I can't figure out what to do after forming the two equations.
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    (Original post by 2014_GCSE)
    Really stuck on question 4 a) ii) http://filestore.aqa.org.uk/subjects...4-QP-JAN13.PDF

    It looks like it should be really easy? But I can't figure out what to do after forming the two equations.
    Well surely you simply solve them simultaneously to find their point of intersection, as you know the x-axis occurs when y=0, if the y-coordinate of their point of intersection is 0, that indicates they intersect on the x-axis?
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    (Original post by amyrah)
    I got up to the equation of the normal but don't understand what to do after? Do you mind explaining?
    Found the Cartesian equation of the curve and equated the normal and curve to find intersection points.


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    (Original post by datpr0)
    Well surely you simply solve them simultaneously to find their point of intersection, as you know the x-axis occurs when y=0, if the y-coordinate of their point of intersection is 0, that indicates they intersect on the x-axis?
    But when you put them equal to each other, they just completely cancel down due to being effectively mirror equations, so you just get 0 = 0.
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    In the Jan 2013 paper Q3bii there are 3 values for x given in the mark scheme of 90°, 45°, 135°. I've solved cos2x = 0 in the range 0° < x < 180° and come out with 45° and 135°. Can someone point out what I've forgotten :/?

    I'm pretty sure cos(2x90) =/= 0...
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    (Original post by datpr0)
    In the Jan 2013 paper Q3bii there are 3 values for x given in the mark scheme of 90°, 45°, 135°. I've solved cos2x = 0 in the range 0° < x < 180° and come out with 45° and 135°. Can someone point out what I've forgotten :/?

    I'm pretty sure cos(2x90) =/= 0...
    Write cot theta as cos/sin


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    Can someone please help on 4bii)
    I feel so stupid for missing what to do... I keep re-reading the mark scheme and just can't grasp it.

    http://filestore.aqa.org.uk/subjects...4-QP-JAN13.PDF
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    (Original post by 2014_GCSE)
    Can someone please help on 4bii)
    I feel so stupid for missing what to do... I keep re-reading the mark scheme and just can't grasp it.

    http://filestore.aqa.org.uk/subjects...4-QP-JAN13.PDF
    If you mean 4aii, it's the same as I said previously - you just solve the two equations simultaneously, by adding them together, leaving you with 2y = 0 => y = 0. Then you just make a statement like: x-axis occurs when y = 0 *therefore* intersection occurs on x-axis.
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    (Original post by datpr0)
    If you mean 4aii, it's the same as I said previously - you just solve the two equations simultaneously, by adding them together, leaving you with 2y = 0 => y = 0. Then you just make a statement like: x-axis occurs when y = 0 *therefore* intersection occurs on x-axis.
    Ahhh!! Of course. Adding them solves it.
    I was getting two y= equations and then putting them equals to each other which just made 0=0, not an x or y value. Thanks.
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    (Original post by 2014_GCSE)
    Ahhh!! Of course. Adding them solves it.
    I was getting two y= equations and then putting them equals to each other which just made 0=0, not an x or y value. Thanks.
    No problem . Gotta remember those three GCSE methods for solving simultaneous equations: addition, subtraction and substitution.
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    Can anyone help me with how to do Qu.7c on June 2010 Core 4?
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    (Original post by HennersPD)
    Can anyone help me with how to do Qu.7c on June 2010 Core 4?
    Here you go:

    Name:  IMG_0615.jpg
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    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).
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    (Original post by 2014_GCSE)
    Here you go:

    Name:  IMG_0615.jpg
Views: 205
Size:  507.0 KB

    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).
    wouldn't the vector AB be a diagonal though? thats what confuses me!? :/ could you explain how to do it step by step please?
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    Can someone explain 4ci for June 14 please :/


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    (Original post by EmilyC96)
    Can someone explain 4ci for June 14 please :/


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    Sure here you go let me know if you don't understand anything from my working Name:  June 14 4c.jpg
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    (Original post by HennersPD)
    wouldn't the vector AB be a diagonal though? thats what confuses me!? :/ could you explain how to do it step by step please?
    The vector would be diagonal yes, because it is a parallelogram, but it is still a straight line so the fact it is diagonal doesn't really matter. The main point they're making with the fact it is a parallelogram is that the shape has two pairs of equal sides.

    I'm going to write out a full explanation on it to try and help you. Give me 5-10 mins.
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    (Original post by 2014_GCSE)
    The vector would be diagonal yes, because it is a parallelogram, but it is still a straight line so the fact it is diagonal doesn't really matter. The main point they're making with the fact it is a parallelogram is that the shape has two pairs of equal sides.

    I'm going to write out a full explanation on it to try and help you. Give me 5-10 mins.
    Thank you i really appreciate your help are there any questions you'd like help with as I'd like to return the favour
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    (Original post by HennersPD)
    Sure here you go let me know if you don't understand anything from my working Name:  June 14 4c.jpg
Views: 89
Size:  262.0 KB
    Yes!!! Thank you so much! Been sat staring at the markscheme like what?! Hahaa thanks ☺️


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    (Original post by EmilyC96)
    Yes!!! Thank you so much! Been sat staring at the markscheme like what?! Hahaa thanks ☺️


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    you're welcome could you help me with the vectors question on Jan 2012 at all? any other questions you'd like help on?
 
 
 
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