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

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

    c=0 => 2ab = ad
    => 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
    => b = Answer given
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    I was skirting around that answer the whole time, thanks :-D
    (Original post by britishtf2)
    2ab = ad - cd,
    a^2 - b^2 = ac + bd

    c=0 => 2ab = ad
    => 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
    => b = Answer given
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    (Original post by Maths_Hector)
    I was skirting around that answer the whole time, thanks :-D
    No problem.
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    (Original post by Maths_Hector)
    Has anyone done a walkthrough for question 10 ii).?
    z^2 = z^*w

    (a+ib)^2 = (a - ib)(c + id)

    a^2 - b^2 + 2abi = ac + bd + i(ad-bc)

    Now look at imaginary parts ...

    Edit: Sorry I thought you were asking about part (i) but others have answered in the meantime.
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    Does anyone know what it was that had to be simplified in question 1??
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    (Original post by Kara_h)
    Does anyone know what it was that had to be simplified in question 1??
    \displaystyle \sum_{r=1}^{n} (r^2 - r - 8)
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    (Original post by Mr M)
    \displaystyle \sum_{r=1}^{n} (r^2 - r - 8)
    Thank you!
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    (Original post by Mr M)
    Mr M's OCR (not OCR MEI) Further Pure 1 answers May 2017



    (ii) \sqrt{18} and \frac{7\pi}{20} (5 marks)

    (iii) Perpendicular bisector of line segment joining z_1 and z_2 (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
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    (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.
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    (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.
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    (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?
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    (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!
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    (Original post by shreys)
    Dear Mr. M, Have you got any idea about how the marks will be allocated for the final question?
    2ab = ad - cd,
    a^2 - b^2 = ac + bd [1]

    c=0 => 2ab = ad
    => 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.
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    (Original post by britishtf2)
    2ab = ad - cd,
    a^2 - b^2 = ac + bd [1]

    c=0 => 2ab = ad
    => 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.
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    (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!
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    (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.
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    (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.
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    (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
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    (Original post by shreys)
    Don't think many people found it easier lol
    I hope not
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    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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