Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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!

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.

See below. Missing a bit of the question.

See below. Missing a bit of the question.

Does 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?

Ah yes it does, I'll give it a shot and let you know if there's a problem. Thanks!

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 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?

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.

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)))

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.

See this from the other day: http://www.thestudentroom.co.uk/showthread.php?t=3696793&page=29&p=65317659#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.