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Edexcel FP3 - 27th June, 2016 watch

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    This is part c im not sure why ive got the wrong answer


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    (Original post by itsalexs)
    Applying for Computer Science at Southampton, received an offer for AADistinction in Maths/FM/BTEC Comp
    OMG I totally applied for Comp Sci at Southampton too. Ended up firming Durham, who gave me an offer of AB because I had D*d* in BTEC IT.
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    ok so maybe this is really simple but how do you do part a for this question - I assumed that the normal to the plane would be the direction vector of the line so (-i + 2j + 3k) but the mark scheme says its (-i + 5j + 3k)
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    (Original post by Zacken)
    I know the feel, I need approximately 0 UMS to get an A* in Further Maths in this. :lol:
    Same! Although I'm pretty sure I haven't got grade 1s in step 2 and 3 so looks like I'm going to Warwick now not Cambridge

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    (Original post by the-anonymous-me)
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    ok so maybe this is really simple but how do you do part a for this question - I assumed that the normal to the plane would be the direction vector of the line so (-i + 2j + 3k) but the mark scheme says its (-i + 5j + 3k)
    Mistake in the question
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    (Original post by Hineshtailor)
    Mistake in the question
    Ok thanks x - I thought i was missing something
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    (Original post by oinkk)
    OMG I totally applied for Comp Sci at Southampton too. Ended up firming Durham, who gave me an offer of AB because I had D*d* in BTEC IT.
    I got an offer from Imperial in CS and rejected them for Durham.. Dno if I regret it yet lol
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    reduction formula is starting to trigger the **** out of me. ok so question 6 a) https://3a14597dd5c7aa2363f067571766...%20Edexcel.pdf

    why can't you just use integration by parts with u=x^n and v'=(16-x^2)^(1/2)? why does the mark scheme only allow rewriting the initial expression as x^(n-1)*x*(16-x^2)^(1/2)?

    and then if you did rewrite it that way, how would you use integration by parts on that expression, since there's three x terms? I know it's probably something simple, but I cant think straight atm
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    (Original post by Anon-)
    reduction formula is starting to trigger the **** out of me. ok so question 6 a) https://3a14597dd5c7aa2363f067571766...%20Edexcel.pdf

    why can't you just use integration by parts with u=x^n and v'=(16-x^2)^(1/2)? why does the mark scheme only allow rewriting the initial expression as x^(n-1)*x*(16-x^2)^(1/2)?

    and then if you did rewrite it that way, how would you use integration by parts on that expression, since there's three x terms? I know it's probably something simple, but I cant think straight atm
    So that you can integrate the square root bit using the chain rule. The x term on the outside allows you to do that because if you differentiate what is inside of the square root it would give you an x term on the outside. I've attached a picture of the solution.
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    (Original post by Zacken)
    I know the feel, I need approximately 0 UMS to get an A* in Further Maths in this. :lol:
    Have you already done 3 A2 units?
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    (Original post by kelvin1338)
    I got 90 UMS in D2 last year

    This year's FP2 I think i got like 67 raw marks. In my M2 I got about 64-66 raw marks.

    Is that enough for an A*, you guys think? If not, how much UMS should I aim for in FP3 to get A* overall in Further Maths?
    It will be close. I think you would have to get full ums (or just below) in fp3 in order to have a chance I think. Then again no one knows the grade boundaries so we've just got to hope they're lower than average.
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    Don't really understand part d. Could someone explain it to me please?
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    (Original post by cjlh)
    I got an offer from Imperial in CS and rejected them for Durham.. Dno if I regret it yet lol
    Oh so you're hopefully off to Durham in October too?
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    (Original post by oinkk)
    Oh so you're hopefully off to Durham in October too?
    Yeah I will be indeed St. Chad's College

    My offer's reduced from A*AA to AAB so I'll definitely be going whether I get my A*s in maths or not but I'm hoping to do as well as possible in FP3 to ensure I get an A* in FM. How about you?
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    (Original post by cjlh)
    Yeah I will be indeed St. Chad's College

    My offer's reduced from A*AA to AAB so I'll definitely be going whether I get my A*s in maths or not but I'm hoping to do as well as possible in FP3 to ensure I get an A* in FM. How about you?
    Ah exciting! I'm St. Aidans. I'm basically the same, I only need 2 A-levels but still want to absolutely smash all three

    I actually live in Durham too, so when you come here I can show you around
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    (Original post by oinkk)
    Ah exciting! I'm St. Aidans. I'm basically the same, I only need 2 A-levels but still want to absolutely smash all three

    I actually live in Durham too, so when you come here I can show you around
    Sweet man, what do you think of the department? I'm really excited for next year tbh
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    Zacken, Tom, could you guys help with 8(c) here? I can see how implicit differentiation works, but I'm not sure I understand the main mark scheme method of using summation of roots (which is supposedly not even in the UK spec?)
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    (Original post by target21859)
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    So that you can integrate the square root bit using the chain rule. The x term on the outside allows you to do that because if you differentiate what is inside of the square root it would give you an x term on the outside. I've attached a picture of the solution.
    ah right, makes sense. but I still don't get why you have to split the initial equation into x^(n-1)*x*(16-x^2)^(1/2)?
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    (Original post by Ayman!)
    Zacken, could you help with 8(c) here? I can see how implicit differentiation works, but I'm not sure I understand the main mark scheme method of using summation of roots (which is not even in the UK spec?)
    you're looking for the midpoint right? that's the sum of the first x coordinate and the second x coordinate, i.e (x1+x2)/2
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    (Original post by Anon-)
    ah right, makes sense. but I still don't get why you have to split the initial equation into x^(n-1)*x*(16-x^2)^(1/2)?
    When you differentiate (-1/3)(16-x^2)6(3/2) you get x(16-x^2)^(1/2) so you can see that there is an x on the outside. This x is needed so that you can integrate the expression exactly into what I just differentiated.
 
 
 
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