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# Edexcel FP2 Official 2016 Exam Thread - 8th June 2016 watch

1. Can anyone help me understand how to do this?
Also can a question like this appear in the fp2 paper? bcuz this is from old spec
2. (Original post by Music With Rocks)

Attachment 540653First order differential question

Please tell me this makes no sense because it certainly makes no sense to me. They seem to have the x terms flipped for no reason and pulled some t terms out of thin air.

Could someone please check this for me to see if I still have my sanity

Jeez man, all these errors you're finding

Spoiler:
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Like what youre doing! Keep it up!
3. (Original post by rm761)
Can anyone help me understand how to do this?
Also can a question like this appear in the fp2 paper? bcuz this is from old spec
It can, indeed, come up on current spec FP2 papers. This question would not appear on its own, however. It would be preceded by a question involving proof of an identity.

I've found the full question for you. I've just done it, and note that there's a cheeky little thing you need to spot later on in part b, when it comes to using this identity.

4. (Original post by LikeASomebody)
Could someone explain how to do part B to this question please. I always trip up on these simple 1 markers. Thanks.
Attachment 540681
Hint: for x > -3, the inequality in (b) is the same as that in (a)
5. (Original post by P____P)
Jeez man, all these errors you're finding
Spoiler:
Show
Like what youre doing! Keep it up!
I seem to be a natural at finding mistakes. As they say takes one to know one
6. (Original post by PhysicsIP2016)
I cannot seem to upload my picture for some reason but I have a major arc going through the coordinates (2,0) (0,2) (-1,0) and (0, -1) and then I've added a rough point of where the centre should be. I then drew 4 lines, one from (0, -1) to the centre, one from (0, -1) to the point (0,2), one from (2,0) to the centre and another one from (2,0) to the point (0,2). You can see the two angles subtended at the centre and the point (0,2).
I am now trying to work on creating simultaneous equations from the diagram to calculate the coordinates of the centre, but hopefully you can see how it could work?
Thanks so much, this is what I've drawn so far but I don't think it's correct? If it is how do we go from here please to finding the coordinates of C? this question is really killing me! Thanks
7. (Original post by oinkk)
It can, indeed, come up on current spec FP2 papers. This question would not appear on its own, however. It would be preceded by a question involving proof of an identity.

I've found the full question for you. I've just done it, and note that there's a cheeky little thing you need to spot later on in part b, when it comes to using this identity.

Thanks for the help, i was confused because it didnt have the identity where i saw it.
8. (Original post by rm761)
Thanks for the help, i was confused because it didnt have the identity where i saw it.
No worries
9. Please could anyone explain question 47c here https://8fd9eafbb84fdb32c73d8e44d980...ZweHc/REV1.pdf why is the cot1/2 theta part ignored when sketching the graph? Thanks
10. (Original post by economicss)
Please could anyone explain question 47c here https://8fd9eafbb84fdb32c73d8e44d980...ZweHc/REV1.pdf why is the cot1/2 theta part ignored when sketching the graph? Thanks
It's not ignored. It's just that as theta ranges between 0 and 2pi, cot (theta/2) ranges between -infinity and infinity, whilst the real part is u=-1/2.
11. (Original post by economicss)
Thanks so much, this is what I've drawn so far but I don't think it's correct? If it is how do we go from here please to finding the coordinates of C? this question is really killing me! Thanks
Haha yeah it does take a while to get! That diagram looks correct You can now go about the coordinates of the centre either by using simultaneous equations generated from the angle rule and the idea that radii in a circle are of the same length (long method) or you could sub each of the coordinates into the general equation of the circle which is a lot less time consuming.

if you do the long method, you know that the distance between (2,0) and the centre is equal to the distance between (0,-1) and the centre, so from here you can find an equation in terms of x and y. From the angle rule, you can work out that the angle subtended at the centre is pi/2 radians, so the two lines that meet at the centre are perpendicular. Use this idea to form another equation using the gradients of the lines and then sub the first equation in. You'll get a quadratic equation but there is one solution which does not lie inside the circle.

Alternatively you can use (x-x1)^2 + (y-y1)^2 = r^2 to create two equations (one in terms of x and the other in terms of y) by using the coordinates where the circle meets each of the axes, but that's not as much fun!
12. For the question:
it says in the textbook use the P.I.

Now I get the Complimentary function

So why is the PI not just ?
13. (Original post by Zacken)
It's not ignored. It's just that as theta ranges between 0 and 2pi, cot (theta/2) ranges between -infinity and infinity, whilst the real part is u=-1/2.
Ah I see, thanks for your help
14. (Original post by PhysicsIP2016)
Haha yeah it does take a while to get! That diagram looks correct You can now go about the coordinates of the centre either by using simultaneous equations generated from the angle rule and the idea that radii in a circle are of the same length (long method) or you could sub each of the coordinates into the general equation of the circle which is a lot less time consuming.

if you do the long method, you know that the distance between (2,0) and the centre is equal to the distance between (0,-1) and the centre, so from here you can find an equation in terms of x and y. From the angle rule, you can work out that the angle subtended at the centre is pi/2 radians, so the two lines that meet at the centre are perpendicular. Use this idea to form another equation using the gradients of the lines and then sub the first equation in. You'll get a quadratic equation but there is one solution which does not lie inside the circle.

Alternatively you can use (x-x1)^2 + (y-y1)^2 = r^2 to create two equations (one in terms of x and the other in terms of y) by using the coordinates where the circle meets each of the axes, but that's not as much fun!
Thank you! I think I'm getting there, for the quicker method is the general equation of the circle I'm subbing in to (x-x1)^2 +(y-y1)^2=r^2? If so I tried it but I couldn't seem to get an answer, not sure where I'm going wrong! Thanks
15. (Original post by Music With Rocks)
For the question:
it says in the textbook use the P.I.

Now I get the Complimentary function

So why is the PI not just ?
Might be just to make it a little bit more challenging.
16. Question: https://gyazo.com/3a02dd69a1250ddf22c5f7b51a4946aa
2nd ODEs
Basically I've used the transformation y=xv to make the DE go from y wrt. x to v wrt. x and found the GS of v wrt.x. When I sub the v back in since v y/x , which means everything on the RHS is to be times by x. I'm wondering why in the mark scheme only the (Asin3x+Bsin3x) is times by x but the (x^2/9)-(2/81) isn't? Is there a mistake?
17. (Original post by Music With Rocks)
For the question:
it says in the textbook use the P.I.

Now I get the Complimentary function

So why is the PI not just ?
I've seen them do this before as well. I think it's so that when you differentiate to get , and then when you differentiate again to get , t is always in your differential value, (if you used , then by the time you work out , the t term would have disappeared).

I don't see why it's necessary though, but I think that's their logic.
18. (Original post by economicss)
Thank you! I think I'm getting there, for the quicker method is the general equation of the circle I'm subbing in to (x-x1)^2 +(y-y1)^2=r^2? If so I tried it but I couldn't seem to get an answer, not sure where I'm going wrong! Thanks
Yes that's right
To find the value of x1, sub only the points where C meets the x-axis into the general equation to get (2-x1)^2 + (0-y1)^2 = r^2 and (-1-x1)^2 + (0-y1)^2 = r^2. If you then equate them, you do not need to find the value for r^2 and the y1^2 terms cancel out.
Do a separate one using the points where C meets the y-axis in order to find the value of y1 and then x1 and y1 will be your centre coordinates.
19. (Original post by Music With Rocks)
For the question:
it says in the textbook use the P.I.

Now I get the Complimentary function

So why is the PI not just ?
That's a weird one. For what its worth, I get with the particular integral .

I'd also be appreciative of anyone who could weigh in on this... since doesn't form part of our complimentary function?
20. (Original post by Music With Rocks)
For the question:
it says in the textbook use the P.I.

Now I get the Complimentary function

So why is the PI not just ?
Just had a thought... I think it's to do with the fact that appears in our complimentary function, but also our auxiliary equation has repeated roots, so we multiply it by t to give to account for it appearing in our complimentary function, then t again to give to account for the repeated roots.

Or that could be crazy talk.

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