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

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    (Original post by Stepidermis)
    There must be other ways of showing that y=0, however in this instance I don't think your way works because in the question they have told you that q cannot equal 0 but they haven't given you a restriction for p (correct me if I'm wrong).

    So the fact that you've found two equivalent expressions for x in this case is nullified by the fact that they both contain (q/p), as if p=0 then they don't intersect as this would instead be an asymptote, does that make sense?
    I see! Argh how I hate some of these questions that contradict everything I thought I knew... 😭


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    (Original post by CD223)
    I see! Argh how I hate some of these questions that contradict everything I thought I knew... 😭


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    Well actually I think I'm being a bit overscrupulous here as I doubt the mark scheme will be THAT picky, so you'll probably get the marks anyway, just thought I'd show you a more efficient approach is all...

    *walks away*

    :danceboy:
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    (Original post by Stepidermis)
    Well actually I think I'm being a bit overscrupulous here as I doubt the mark scheme will be THAT picky, so you'll probably get the marks anyway, just thought I'd show you a more efficient approach is all...

    *walks away*

    :danceboy:
    Your approach makes more sense! It's just frustrating that what I thought was right could be wrong if p was zero :L


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    (Original post by CD223)
    Ever so sorry, I thought you needed help with 3c and not 3b. Here's my method for both parts.

    Attachment 408151

    Is this correct?


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    yep that is correct!

    but what I don't understand is how finding the dot product of pq and qr will help me find angle q?

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    (Original post by the_googly)
    yep that is correct!

    but what I don't understand is how finding the dot product of pq and qr will help me find angle q?

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    The dot product helps you find the angle between two lines.

    Hopefully this clears it up

    (https://m.youtube.com/watch?v=5p5gZL6FW28)


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    (Original post by CD223)
    Oh right thanks!

    Is this an acceptable way of showing the two tangents meet on the x axis?

    (Jan 13 Q4 a ii)

    Attachment 410827


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    I would have thought so. But I'm not too sure.
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    (Original post by a123a)
    I would have thought so. But I'm not too sure.
    Cool! Thanks for your help


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    Hi can anybody tell me why on Jan 11 Q8 b(ii), when you get two values for mu, you're supposed to choose mu=5/7 and not mu=1 ? I've been trying for a while now and i can't understand why Sorry if this has already been covered; here's the paper/markscheme, thanks

    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
    http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF
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    Can somebody check my working for question six. can't seem to get around it

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    (Original post by CD223)
    The dot product helps you find the angle between two lines.

    Hopefully this clears it up

    (https://m.youtube.com/watch?v=5p5gZL6FW28)


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    thanks

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    Think I've done it properly

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    (Original post by the_googly)
    Think I've done it properly

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    Here's my workings for the same questions.

    I believe our answers agree, even though they're expressed slightly differently

    Name:  ImageUploadedByStudent Room1432849810.025160.jpg
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    (Original post by CD223)
    Here's my workings for the same questions.

    I believe our answers agree, even though they're expressed slightly differently

    Name:  ImageUploadedByStudent Room1432849810.025160.jpg
Views: 143
Size:  190.3 KB


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    your way seems much better and more convincing. :-)

    how are you revising now for this exam?

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    (Original post by the_googly)
    your way seems much better and more convincing. :-)

    how are you revising now for this exam?

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    Just purely past papers! How about you?


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    (Original post by CD223)
    Just purely past papers! How about you?


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    will be starting past papers tomorrow!

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    (Original post by CD223)
    Could someone explain where I've gone wrong on June 2010 Q7?

    Attachment 409213

    I've set up a quadratic in terms of \alpha like so:

    5 \alpha^{2} - 12 \alpha + 7 = 0

    To which the solutions are:
    \alpha = 1 or \alpha = \frac{7}{5}

    This gives co ordinates of C as:
    (3, -1, 3) which is correct, and (\frac{19}{5}, -1, \frac{17}{5}) which is incorrect.

    How can I get to the other possibility of (-1, -1, 1)?

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    Hey I don't know if anyone gave a solution for this but I think it may be because in your second parallelogram BP isn't parallel to AC due to C being in a different position so the equation is only valid for one solution if that makes sense? Sorry if this is wrong or someone already answered it!
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    (Original post by the_googly)
    will be starting past papers tomorrow!

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    Oh nice have you not done any yet?


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    (Original post by Baloney26)
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    Hey I don't know if anyone gave a solution for this but I think it may be because in your second parallelogram BP isn't parallel to AC due to C being in a different position so the equation is only valid for one solution if that makes sense? Sorry if this is wrong or someone already answered it!
    Brilliant! Thank you. No one had answered haha.


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    (Original post by CD223)
    Brilliant! Thank you. No one had answered haha.


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    No problem 😊.
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    (Original post by Baloney26)
    No problem 😊.
    How's your prep going?


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