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# Mr M's OCR (not OCR MEI) Further Pure 1 Answers May 2017 watch

1. (Original post by Maths_Hector)
Has anyone done a walkthrough for question 10 ii).?
a^2 - b^2 = ac + bd

=> 2b = d
c=0, d=2b => a^2 - b^2 = bd
=> a^2 - b^2 = b(2b)
=> a^2 = 3b^2
=> (a^2)/3 = b^2
2. I was skirting around that answer the whole time, thanks :-D
(Original post by britishtf2)
a^2 - b^2 = ac + bd

=> 2b = d
c=0, d=2b => a^2 - b^2 = bd
=> a^2 - b^2 = b(2b)
=> a^2 = 3b^2
=> (a^2)/3 = b^2
3. (Original post by Maths_Hector)
I was skirting around that answer the whole time, thanks :-D
No problem.
4. (Original post by Maths_Hector)
Has anyone done a walkthrough for question 10 ii).?

Now look at imaginary parts ...

Edit: Sorry I thought you were asking about part (i) but others have answered in the meantime.
5. Does anyone know what it was that had to be simplified in question 1??
6. (Original post by Kara_h)
Does anyone know what it was that had to be simplified in question 1??
7. (Original post by Mr M)
Thank you!
8. (Original post by Mr M)
Mr M's OCR (not OCR MEI) Further Pure 1 answers May 2017

(ii) and (5 marks)

(iii) Perpendicular bisector of line segment joining and (2 marks)
for part ii arg(z1) = 3pi/5, arg(z2) = -9pi/10, |z1|=|z2|=3, wouldn't that mean the angle between them is pi/2 thus making the arg(z1-z2) = 17pi/20 not 7pi/20

unless i remember/read the question differently
9. (Original post by RalphDG)
for part ii arg(z1) = 3pi/5, arg(z2) = -9pi/10, |z1|=|z2|=3, wouldn't that mean the angle between them is pi/2 thus making the arg(z1-z2) = 17pi/20 not 7pi/20

unless i remember/read the question differently
Angle between them is pi/2, you are correct.
However, arg(z1-z2) is 7pi/20. There is a diagram on the previous (I think) page which shows what z1-z2 looks like and you'll so it is around 7pi/20. If not, I could write a proof/show an answer.
10. (Original post by RalphDG)
for part ii arg(z1) = 3pi/5, arg(z2) = -9pi/10, |z1|=|z2|=3, wouldn't that mean the angle between them is pi/2 thus making the arg(z1-z2) = 17pi/20 not 7pi/20

unless i remember/read the question differently
Have a look at the diagram on the previous page.
11. (Original post by Mr M)
Have a look at the diagram on the previous page.
Dear Mr. M, Have you got any idea about how the marks will be allocated for the final question?
12. (Original post by shreys)
Dear Mr. M, Have you got any idea about how the marks will be allocated for the final question?
Not really. 6 marks seems a lot for this relatively small piece of work!
13. (Original post by shreys)
Dear Mr. M, Have you got any idea about how the marks will be allocated for the final question?
a^2 - b^2 = ac + bd [1]

=> 2b = d [2]
c=0, d=2b => a^2 - b^2 = bd [3]
=> a^2 - b^2 = b(2b) [4]
=> a^2 = 3b^2
=> (a^2)/3 = b^2 [5]
=> b = [Correct] Answer given [6]

Something along the lines of that I would imagine. Not 6 marks of work imo though, so just a guess.
14. (Original post by britishtf2)
a^2 - b^2 = ac + bd [1]

=> 2b = d [2]
c=0, d=2b => a^2 - b^2 = bd [3]
=> a^2 - b^2 = b(2b) [4]
=> a^2 = 3b^2
=> (a^2)/3 = b^2 [5]
=> b = [Correct] Answer given [6]

Something along the lines of that I would imagine. Not 6 marks of work imo though, so just a guess.
Do you think I'd get 3 out of 6 for getting 2b=d and also rearranging a bunch of other stuff but somehow managing to get a cube root? It was quite close to the answer btw - i think a cube root instead of a square root or something similar - also forgot a plus/minus.
15. (Original post by shreys)
Do you think I'd get 3 out of 6 for getting 2b=d and also rearranging a bunch of other stuff but somehow managing to get a cube root? It was quite close to the answer btw - i think a cube root instead of a square root or something similar.
Will really heavily depend on the markscheme. Not easy to tell. You might, you might not. You might get 3/6 for getting d = 2b outright, you might just get 1 for it. Sorry I can't help more :/

EDIT: Forgetting +/- might not do anything to your mark as you got something different, but there might be a whole mark for putting +/-. Again, sorry I can't help more!
16. (Original post by britishtf2)
Will really heavily depend on the markscheme. Not easy to tell. You might, you might not. You might get 3/6 for getting d = 2b outright, you might just get 1 for it. Sorry I can't help more :/

EDIT: Forgetting +/- might not do anything to your mark as you got something different, but there might be a whole mark for putting +/-. Again, sorry I can't help more!
That's fine, thanks. I think it'll be around 56/57 for an A.
17. (Original post by shreys)
That's fine, thanks. I think it'll be around 56/57 for an A.
Honestly no idea to be honest. I got like 64-66 just cos I'm bad at exams tbh :/ Seemed easier than last year's paper to me, but I've had another year so Idk. We'll see.
18. (Original post by britishtf2)
Honestly no idea to be honest. I got like 64-66 just cos I'm bad at exams tbh :/ Seemed easier than last year's paper to me, but I've had another year so Idk. We'll see.
Don't think many people found it easier lol
19. (Original post by shreys)
Don't think many people found it easier lol
I hope not
20. I feel really thick, for the 3x3 matrix one i got two values for a, one being 5 and the other being 2, will i lose marks for this? Really wasn't my exam today :/

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