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    Roots question: 2x²+3x-6 = 0 (A = alpha, B = beta)
    a) Solve A+B and AB
    -> A+B = -3/2, AB = -6/2 = -3
    b) Show A³+B³ = -135/8
    -> (A+B)³ - 3AB(A+B) = (-3/2)³-3(-3)(-3/2) = -135/8
    c) Find equation with roots A+A/B² and B+B/A²
    -> Sum of roots: A+B + (A/B² + B/A²)
    = -3/2 + (B³+A³/A²B²) = -3/2 + ((-135/8)/(-3)²) = -27/8
    -> Product of roots: (A+A/B²)(B+B/A²) = AB + (B/A+A/B) + (AB/A²B²)
    = -3 + (A²+B²/AB) + (-3/(-3)²) -> A²+B² = (A+B)² - 2AB = (-3/2)² - 2(-3) = 33/4. Thus product roots = -3 + ((33/4)/-3) + (-3/(-3)²) = -73/12
    So we get the equation: x²+27/8x-73/12 = 0, but we want INTEGER coefficients, so multiply by the lowest common multiple which is 24, hence 24x²+81x-146 = 0
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    (Original post by KBarney)
    I thought thats what mine shows, doesnt what you put show a stretch in the x direction? Can you explian how you got to that awnser of the roots then?
    Putting [1 0 0 3] means that the stretch is just 1 in the x-direction (i.e stays the same) and the 3 is the stretch in the y-direction as it was stretched 3x in y-direction and none across if that makes sense? and i put the answer to the roots question above.
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    (Original post by Annie_o)
    Yeah exactly, same. Not sure how I'll do in M1 sometimes I do really well, others not so good. M2 is horrible which is why I needed fp1 to go well...


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    Good luck on M2! M1 should be okay you just need to watch out for how they word stuff, I did the Jan 2013 as a mock and I hated how they worded that boat question but I got it in the end :P
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    anyone got the AQA D1 Jan 2013 paper?
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    (Original post by mathsRus)
    anyone got the AQA D1 Jan 2013 paper?
    No mark scheme though sorry.

    EDIT: oh, I thought I'd attached it. Hold on....
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  1. File Type: pdf AQA-MM1B-QP-Jan13.pdf (181.2 KB, 141 views)
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    (Original post by mathsRus)
    anyone got the AQA D1 Jan 2013 paper?
    (Original post by Vernish)
    No mark scheme though sorry.

    EDIT: oh, I thought I'd attached it. Hold on....
    You attached the M1 one lol. :P

    Here's the D1 one.
    https://docs.google.com/viewer?a=v&p...EzOTJlNGUzMzRl
    https://docs.google.com/viewer?a=v&p...I5N2NjMWQ3NTNj
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    :doh: Woops!
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    What really put me off was the amount of answering space there was, especially in the summations question. I was only given 1.5 pages for the lot of the sigma summation questions, which is ridiculous.

    During the exam, I had to ask for two extra booklets of answering space!
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    (Original post by AlanDu)
    What really put me off was the amount of answering space there was, especially in the summations question. I was only given 1.5 pages for the lot of the sigma summation questions, which is ridiculous.

    During the exam, I had to ask for two extra booklets of answering space!
    Yeah I had to ask for extra paper for the summation as well, I don't get how they expect us to use only 2 pages to do all those questions it's just insane...
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    (Original post by matt242o1552)
    Roots question: 2x²+3x-6 = 0 (A = alpha, B = beta)
    a) Solve A+B and AB
    -> A+B = -3/2, AB = -6/2 = -3
    b) Show A³+B³ = -135/8
    -> (A+B)³ - 3AB(A+B) = (-3/2)³-3(-3)(-3/2) = -135/8
    c) Find equation with roots A+A/B² and B+B/A²
    -> Sum of roots: A+B + (A/B² + B/A²)
    = -3/2 + (B³+A³/A²B²) = -3/2 + ((-135/8)/(-3)²) = -27/8
    -> Product of roots: (A+A/B²)(B+B/A²) = AB + (B/A+A/B) + (AB/A²B²)
    = -3 + (A²+B²/AB) + (-3/(-3)²) -> A²+B² = (A+B)² - 2AB = (-3/2)² - 2(-3) = 33/4. Thus product roots = -3 + ((33/4)/-3) + (-3/(-3)²) = -73/12
    So we get the equation: x²+27/8x-73/12 = 0, but we want INTEGER coefficients, so multiply by the lowest common multiple which is 24, hence 24x²+81x-146 = 0
    I don't know how but for the sum of the roots I got -9. I got what you got for the product of the roots though. Grrr I'm really annoyed because those questions are my favourite ( should get 4 of the 6 marks though I'm hoping
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    Has anyone found a mark scheme yet?
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    (Original post by kingmango)
    Has anyone found a mark scheme yet?
    The earliest anyone can have access to the official mark scheme is teachers via e-AQA 10 days after the exam takes place (according to AQA's website.) So 27th May for us!

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    With all the answers I've seen on all comments, I believe I may have smacked d exam maybe 100%

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    (Original post by Fruitbasket)
    Given that sin(π/12) = (√3 - 1) / (2√2), you can take out the √2 and put it into a seperate fraction, then separate the remaining fraction to get:

    (1/√2)(√3/2 - 1/2)

    Which is the same as cos(π/4)(cos(π/6) + cos(2π/3)).
    What I did was to use Sin(π/12) = Cos(π/2 - π/12)
    Sin (
    π/12) = Cos(5π/12)

    But Cos(5
    π/12) = Cos(π/6 + π/4)

    But Cos(a+b) = Cos(a)Cos(b) - Sin(a)Sin(b)

    Therefore,
    Cos(
    π/6 + π/4) = Cos(π/4)Cos(π/6) - Sin(π/4)Sin(π/6)

    But Cos(
    π/4) = Sin(π/4) and Cos(π/3) = Sin(π/6)

    Cos(π/6 + π/4) = Cos(π/4)Cos(π/6) - Sin(π/4)Sin(π/6)

    Cos(
    π/6 + π/4) = Cos(π/4)Cos(π/6) - Cos(π/4)Cos(π/3)
    = Cos(
    π/4) [Cos(π/6) - Cos(π/3)]

    But -Cos(
    π/3) = Cos(2π/3)

    Therefore,
    Cos(
    π/6 + π/4) = Cos(π/4) [Cos(π/6) + Cos(2π/3)] = Cos(5π/12) = Sin(π+12)

    Therefore, a = 1/6 and b = 2/3.
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    (Original post by don_ayo)
    What I did was to use Sin(π/12) = Cos(π/2 - π/12)
    Sin (
    π/12) = Cos(5π/12)

    But Cos(5
    π/12) = Cos(π/6 + π/4)

    But Cos(a+b) = Cos(a)Cos(b) - Sin(a)Sin(b)

    Therefore,
    Cos(
    π/6 + π/4) = Cos(π/4)Cos(π/6) - Sin(π/4)Sin(π/6)

    But Cos(
    π/4) = Sin(π/4) and Cos(π/3) = Sin(π/6)

    Cos(π/6 + π/4) = Cos(π/4)Cos(π/6) - Sin(π/4)Sin(π/6)

    Cos(
    π/6 + π/4) = Cos(π/4)Cos(π/6) - Cos(π/4)Cos(π/3)
    = Cos(
    π/4) [Cos(π/6) - Cos(π/3)]

    But -Cos(
    π/3) = Cos(2π/3)

    Therefore,
    Cos(
    π/6 + π/4) = Cos(π/4) [Cos(π/6) + Cos(2π/3)] = Cos(5π/12) = Sin(π+12)

    Therefore, a = 1/6 and b = 2/3.
    Are you allowed to use Core 4 trig identities in FP1?
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    (Original post by fizzbizz)
    Are you allowed to use Core 4 trig identities in FP1?
    I guess so

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    (Original post by sohailkm96)
    Good luck on M2! M1 should be okay you just need to watch out for how they word stuff, I did the Jan 2013 as a mock and I hated how they worded that boat question but I got it in the end :P
    Thank you! Yeah I'm not overly worried about M1 it's normally just like one question that can be a bit nasty. M2 is horrible though


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    hey guys in the trigonometry i made a very silly mistake by using both degree and 2kpi and in the limit question i made a wrong sign -2/x + 5/2x^2 instead of -2/x - 5/2x^2 and it leads to wrong answer 9/2. So how many marks will i lose for each of the questions ?
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    (Original post by mathsRus)
    anyone got the AQA D1 Jan 2013 paper?
    I did the AQA D1 paper this January got 88UMS. Which is pretty good considering I didn't bother revising much coz decision isn't proper maths. Why do you ask
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    (Original post by sohailkm96)
    Yeah I had to ask for extra paper for the summation as well, I don't get how they expect us to use only 2 pages to do all those questions it's just insane...
    I finished the summation question with about 3/5ths of the second page left
 
 
 
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