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    How is
    For pic 1: 8C done
    Pic 2: 17 b done
    Pic 3: 3 a done
    Pic 4: 8 c and 8d done

    Thanks very much!

    Zacken
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    (Original post by Rkai01)
    How is
    For pic 1: 8C done
    Pic 2: 17 b done
    Pic 3: 3 a done
    Pic 4: 8 c and 8d done

    Thanks very much!
    8c is done the same way as 8b. Square root both sides to get +/- 1 on the right, take the inverse tan of both sides and get an expression for the argument. The locus should be two intersecting straight lines.
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    Does anyone have any clues as to different ways to go about this one?

    Please spoiler if so
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    (Original post by Euclidean)


    Does anyone have any clues as to different ways to go about this one?

    Please spoiler if so
    See below. Missing a bit of the question. Attachment 540701
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    (Original post by Zacken)
    See below. Missing a bit of the question. Attachment 540701
    I was wondering why it was so difficult :lol:

    Thanks prsom
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    (Original post by Zacken)
    See below. Missing a bit of the question. Attachment 540701
    I'm in awe of your ability to know questions. Just wow.
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    (Original post by Zacken)
    See below. Missing a bit of the question. Attachment 540701
    Was just going to say...it would be very surprising if they gave no help whatsoever on what function to consider.
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    Name:  fp2 capture 4.JPG
Views: 71
Size:  22.4 KBDoes anyone have an idea on how to go about this? I've tried rearranging for z, then using the conjugate to get the x and y values and then subbing them in but it got far too complicated so I thought that there may be an easier way. Thanks
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    (Original post by Alby1234)
    Name:  fp2 capture 4.JPG
Views: 71
Size:  22.4 KBDoes anyone have an idea on how to go about this? I've tried rearranging for z, then using the conjugate to get the x and y values and then subbing them in but it got far too complicated so I thought that there may be an easier way. Thanks
    Note that x^2 + y^2 = 1 is a circle. You can represent a circle as a complex number: |z|=1

    Does that help things?
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    (Original post by oinkk)
    Note that x^2 + y^2 = 1 is a circle. You can represent a circle as a complex number: |z|=1

    Does that help things?
    Ah yes it does, I'll give it a shot and let you know if there's a problem. Thanks!
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    (Original post by Alby1234)
    Ah yes it does, I'll give it a shot and let you know if there's a problem. Thanks!
    No worries
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    (Original post by A Slice of Pi)
    8c is done the same way as 8b. Square root both sides to get +/- 1 on the right, take the inverse tan of both sides and get an expression for the argument. The locus should be two intersecting straight lines.
    How about the other questions?
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    (Original post by Rkai01)
    How is
    For pic 1: 8C done
    Pic 2: 17 b done
    Pic 3: 3 a done
    Pic 4: 8 c and 8d done

    Thanks very much!

    Zacken
    For 17b (pic 2) I couldn't see any obvious method other than Polar coordinates. I use r = 2sin(theta) and use standard area finding by integrating r^2 / 2 with limits of pi/3 and pi/4. I'm not sure that is the polar coordinates for a circle of center off the origin but I think it is. Hope this helps.
    Can you send a link of the answers?
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    (Original post by Pyslocke)
    For 17b (pic 2) I couldn't see any obvious method other than Polar coordinates. I use r = 2sin(theta) and use standard area finding by integrating r^2 / 2 with limits of pi/3 and pi/4. I'm not sure that is the polar coordinates for a circle of center off the origin but I think it is. Hope this helps.
    Can you send a link of the answers?
    I agree with this method. I can't see any other way of approaching it either.

    Note that I really can't see this appearing on an examination. We do have to be able to convert loci into Cartesian form. And we do have to be able to convert Cartesian forms into polar forms. But this has never been tested together, and is very unlikely to be tested together in the future.

    Method and answer (not a worked solution solution) below if you need it. The question is actually quite easy once you can access the method.
    Spoiler:
    Show
    You can change the complex number |z-i| = 1 into Cartesian form (easy enough done), and then use x=rcos\theta and x=rsin\theta to transform your Cartesian equation into polar form. Then it is very simple to find the area bounded between that curve.

    Doing so gives you -\frac{\sqrt{3}}{4} + \frac{\pi}{12} + \frac{1}{2}.
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    Regarding the hint in the question attached, how would you determine to have t^3 in the P.I. instead of t^2 or t^1?
    What from the C.F. indicates that it should be this way?
    Thanks.
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    (Original post by paradoxequation)
    Regarding the hint in the question attached, how would you determine to have t^3 in the P.I. instead of t^2 or t^1?
    What from the C.F. indicates that it should be this way?
    Thanks.
    See this from the other day: http://www.thestudentroom.co.uk/show...9#post65317659

    Reply if you need any clarification. It may help reading Examples 10, 11 & 12 on page 95—98 of the Edexcel textbook first to understand where the first extra t comes from, and then reading that post to help understand where the second t comes from in this question.
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    (Original post by kennz)
    From part a(i) and a(ii) you get lambad =3 and mu=-sqrt(3)
    From the sketch you did in part b, you have 4 points that belong on the circle.
    So, you can subistute the cartesian equivlents of these points into the equation for a circle (x-a)^2 + (y-b)^2 = r^2 and then solve silmutaneously. Because the 4 points lie on either the x or y axis, this makes it a lot easier to solve.
    you should get centre (-0.5(1+sqrt(3)), 0.5(1+sqrt(3)))
    Thank you, I've done parts a and b now, please could you explain, I've found that the radius is root6 but not sure whether this is right? And how do I find the coordinates of the centre from here please? Thanks
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    (Original post by Pyslocke)
    For 17b (pic 2) I couldn't see any obvious method other than Polar coordinates. I use r = 2sin(theta) and use standard area finding by integrating r^2 / 2 with limits of pi/3 and pi/4. I'm not sure that is the polar coordinates for a circle of center off the origin but I think it is. Hope this helps.
    Can you send a link of the answers?
    (Original post by oinkk)
    I agree with this method. I can't see any other way of approaching it either.

    Note that I really can't see this appearing on an examination. We do have to be able to convert loci into Cartesian form. And we do have to be able to convert Cartesian forms into polar forms. But this has never been tested together, and is very unlikely to be tested together in the future.

    Method and answer (not a worked solution solution) below if you need it. The question is actually quite easy once you can access the method.
    Spoiler:
    Show
    You can change the complex number |z-i| = 1 into Cartesian form (easy enough done), and then use x=rcos\theta and x=rsin\theta to transform your Cartesian equation into polar form. Then it is very simple to find the area bounded between that curve.

    Doing so gives you -\frac{\sqrt{3}}{4} + \frac{\pi}{12} + \frac{1}{2}.
    Haven't actually tried it, but surely a sketch and some basic GCSE geometry should get you the answer?
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    (Original post by oinkk)
    See this from the other day: http://www.thestudentroom.co.uk/show...9#post65317659

    Reply if you need any clarification. It may help reading Examples 10, 11 & 12 on page 95—98 of the Edexcel textbook first to understand where the first extra t comes from, and then reading that post to help understand where the second t comes from in this question.
    Thank you! I didn't consider that the repeated roots would affect this, but I get it now.
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    Hey guys, just to clarify (or to be corrected), to get an A* in FM do you need to average 90 UMS across 3 of 4 A2 units, and then average 80 UMS across all the units in total? (6)

    Thanks
 
 
 
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