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AQA A2 MFP2 Further Pure 2 – 24th June [Exam Discussion Thread] Watch

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    (Original post by Jm098)
    yeah I'd agree! some questions are repetitive though... hoping for some nice integrals
    Some are but nothing like FP3 where there was basically only 3 topics in the entire unit. Hope so too

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    (Original post by PiTheta97)
    Some are but nothing like FP3 where there was basically only 3 topics in the entire unit. Hope so too

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    can't wait to see what they throw M2 for the last question
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    Anyone mind explaining some of the harder questions to me if I post the ones I'm struggling with?
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    Hey can some one help me with the proof by induction question on the 2015 paper please ?
    I've attached what I've managed to do so far, but don't understand how to simplify Name:  fp2.jpg
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    (Original post by Kedric123)
    Anyone mind explaining some of the harder questions to me if I post the ones I'm struggling with?
    yeah i will
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    (Original post by Roxanne18)
    Hey can some one help me with the proof by induction question on the 2015 paper please ?
    I've attached what I've managed to do so far, but don't understand how to simplify Name:  fp2.jpg
Views: 132
Size:  95.6 KB
    16 is 2^4 so the 2^4k+7 s cancel
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    (Original post by Jm098)
    16 is 2^4 so the 2^4k+7 s cancel
    Oh wow can't believe I didn't see that, thank you !
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    (Original post by girlruinedme)
    Someone please explain the factorials in the 2015 paper for
    (r+1)! can be expressed as (r+1)r!, it can be expressed as that because for example 4! = 4*3! because 4! = 4*3*2*1 = 4*3!. (r+2)! can be expressed as (r+2)(r+1)r! just like how 4! = 4*3*2!.

    Here's some working for the question.
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    Could someone kindly explain june 11 q4) aiv), bi) and bii) - I don't understand the method shown in the solutions

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    Thanks in advance!
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    Where did they get the results in the mark scheme for June 14 4bii?
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    (Original post by C0balt)
    Where did they get the results in the mark scheme for June 14 4bii?
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    Honestly I have no idea. I just used the method shown to the right.
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    (Original post by Buntar)
    Could someone kindly explain june 11 q4) aiv), bi) and bii) - I don't understand the method shown in the solutions

    Name:  Screen Shot 2016-06-23 at 00.01.01.png
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    Thanks in advance!
    \alpha satisfies the equation since it is a root. But so are \beta and \gamma so they also satisfy the equation.
    You want the sum of the cubes of the roots, so clearly you want to add the equations \alpha^3-2\alpha^2+k=0, \beta^3-2\beta^2+k=0,\gamma^3-2\gamma^2+k=0 to get \Sigma\alpha^3-2\Sigma\alpha^2+3k=0 and the result follows.

    For the next two cases you want the sum of fourth powers and fifth powers, so it makes sense to multiply the equation by x and x^2 respectively and do the same thing.
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    (Original post by IrrationalRoot)
    Honestly I have no idea. I just used the method shown to the right.
    Do you mind explaining your method
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    (Original post by Hjyu1)
    Do you mind explaining your method
    There's no method to explain, the whole method is shown in two lines to the right on the mark scheme.
    If there's something specific you don't understand within it, let me know and I'll explain it.
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    (Original post by C0balt)
    Where did they get the results in the mark scheme for June 14 4bii?
    Name:  image.jpg
Views: 110
Size:  47.4 KBAttachment 555315555317
    (Original post by IrrationalRoot)
    Honestly I have no idea. I just used the method shown to the right.
    (Original post by Hjyu1)
    Do you mind explaining your method
    This is how I did it:

    (\alpha + \beta)(\beta + \gamma)(\gamma + \alpha)

= (\alpha + \beta)(\alpha\beta + \beta\gamma + \gamma\alpha +\gamma^2)

= \alpha^2\beta + \gamma\alpha^2 + \alpha\beta^2 + \beta^2\gamma + \beta\gamma^2 + \gamma^2\alpha + 2\alpha\beta\gamma 

= (\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) - 3\alpha\beta\gamma + 2\alpha\beta\gamma

= (\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma

= \Sigma\alpha\Sigma\alpha\beta - \alpha\beta\gamma

    Expanding the brackets on line 4 gives us an extra 3\alpha\beta\gamma which we don't want, hence the -3\alpha\beta\gamma.
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    (Original post by IrrationalRoot)
    \alpha satisfies the equation since it is a root. But so are \beta and \gamma so they also satisfy the equation.
    You want the sum of the cubes of the roots, so clearly you want to add the equations \alpha^3-2\alpha^2+k=0, \beta^3-2\beta^2+k=0,\gamma^3-2\gamma^2+k=0 to get \Sigma\alpha^3-2\Sigma\alpha^2+3k=0 and the result follows.

    For the next two cases you want the sum of fourth powers and fifth powers, so it makes sense to multiply the equation by x and x^2 respectively and do the same thing.
    Thank you!
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    (Original post by sam_97)
    This is how I did it:

    (\alpha + \beta)(\beta + \gamma)(\gamma + \alpha)

= (\alpha + \beta)(\alpha\beta + \beta\gamma + \gamma\alpha +\gamma^2)

= \alpha^2\beta + \gamma\alpha^2 + \alpha\beta^2 + \beta^2\gamma + \beta\gamma^2 + \gamma^2\alpha + 2\alpha\beta\gamma 

= (\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) - 3\alpha\beta\gamma + 2\alpha\beta\gamma

= (\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma

= \Sigma\alpha\Sigma\alpha\beta - \alpha\beta\gamma

    Expanding the brackets on line 4 gives us an extra 3\alpha\beta\gamma which we don't want, hence the -3\alpha\beta\gamma.
    Thanks, I see it but how did you spot going from line 3 to 4 without seeing the ms
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    (Original post by C0balt)
    Thanks, I see it but how did you spot going from line 3 to 4 without seeing the ms
    No problem. I haven't done the paper for a while so I can't remember too clearly, I think I just presumed that the terms in line 3 might factorise somehow with all the squared terms knocking around.
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    Do we have to know to derive surface area revolution thing
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    (Original post by C0balt)
    Do we have to know to derive surface area revolution thing
    No
 
 
 
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