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    (Original post by xyz9856)
    Oh I'm so glad, I was thinking this was maybe written before the specification change? By the way I had a look at the January 2016 IAL F1 paper, there is a question there I just don't understand. Is F1 different to FP1 or is it significantly harder but the same content?.

    http://imgur.com/Nu0pRS5
    It's the same difficulty but with a teensy bit of different content. Which is precisely what the question you've linked me to is about - that topic is not on spec for FP1, every other questions except this "alpha, beta" stuff is though.
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    (Original post by Zacken)
    I would have expected it in exact form, no?
    oh yeah, my bad. I got the same as you for a . And for b i got square root of (-0.5 + 1/(2^0.5) )


    Sorry for lack of proper notation.
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    (Original post by alfmeister)
    oh yeah, my bad. I got the same as you for a . And for b i got square root of (-0.5 + 1/(2^0.5) )


    Sorry for lack of proper notation.
    Nah, no worries - it's just massively ugly and not a proper question.
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    (Original post by Zacken)
    Nah, no worries - it's just massively ugly and not a proper question.
    Now looking back I can see where you are coming from. However I wanted a bit of a challenge so thought I would try it.
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    (Original post by Zacken)
    It's the same difficulty but with a teensy bit of different content. Which is precisely what the question you've linked me to is about - that topic is not on spec for FP1, every other questions except this "alpha, beta" stuff is though.
    Oh I see, going to finish the rest of that paper now. I was a bit disheartened at first and stopped in search of the solution. Thanks again.
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    (Original post by alfmeister)
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    Sure, I did a different example so that you can try and attempt the question in the paper again.

    Attachment 529671
    Thanks so much! What did you get as your answers for questions 3 and 7?
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    (Original post by economicss)
    Thanks so much! What did you get as your answers for questions 3 and 7?
    No worries, for 7 I got it was an anticlockwise rotation 45 degrees and an enlargement by a scale factor of 3.

    For 3 I think I got -4n^2 +12n +8.

    What did you get?
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    (Original post by alfmeister)
    No worries, for 7 I got it was an anticlockwise rotation 45 degrees and an enlargement by a scale factor of 3.

    For 3 I think I got -4n^2 +12n +8.

    What did you get?
    I got the same as you for q7! I struggled with question 3, I got 4n(n+1) so I'm probably wrong! haha
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    I don't have time now but tommorow I'll post my working if you wanted to check it out?

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    (Original post by alfmeister)
    I don't have time now but tommorow I'll post my working if you wanted to check it out?

    Posted from TSR Mobile
    Yeah that would be good thanks, I'll post mine in a bit
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    (Original post by alfmeister)
    No worries, for 7 I got it was an anticlockwise rotation 45 degrees and an enlargement by a scale factor of 3.

    For 3 I think I got -4n^2 +12n +8.

    What did you get?
    (Original post by economicss)
    Yeah that would be good thanks, I'll post mine in a bit
    Bleh, might as well:

    Note that (3-r)^2 - (r+1)^2 = 9 - 6r + r^2 - r^2 - 2r - 1 = 8 - 8r

    So:

    \displaystyle

\begin{align*} \sum_{r=0}^{n} 8 - 8r &= 8 - 8(0) + \sum_{r=1}^n (8-8r) \\&= 8 + 8n - 8 \times \frac{n}{2}(n+1) \\&= 8(n+1) - 4n(n+1) \\&= 4(n+1)(2 - n) \\ & = 4(2n - n^2 +2 -n) \\& = -4n^2 +4n + 8\end{align*}

    Did you remember that the sum starts from r=0 and hence the sum of \sum_{r=0}^n 8 = 8(n+1) \neq 8n?
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    (Original post by Zacken)
    Bleh, might as well:

    Note that (3-r)^2 - (r+1)^2 = 9 - 6r + r^2 - r^2 - 2r - 1 = 8 - 8r

    So:

    \displaystyle

\begin{align*} \sum_{r=0}^{n} 8 - 8r &= 8 - 8(0) + \sum_{r=1}^n (8-8r) \\&= 8 + 8n - 8 \times \frac{n}{2}(n+1) \\&= 8(n+1) - 4n(n+1) \\&= 4(n+1)(2 - n) \\ & = 4(2n - n^2 +2 -n) \\& = -4n^2 +4n + 8\end{align*}

    Did you remember that the sum starts from r=0 and hence the sum of \sum_{r=0}^n 8 = 8(n+1) \neq 8n?
    Damn it, silly error. Got the right working out but made a simple calculation error of 8n + 4n instead of 8n - 4n. Should have checked over my working again. Thanks
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    (Original post by alfmeister)
    Damn it, silly error. Got the right working out but made a simple calculation error of 8n + 4n instead of 8n - 4n. Should have checked over my working again. Thanks
    Ah, that's okay then - as long as your concepts are solid.
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    (Original post by Zacken)
    Bleh, might as well:

    Note that (3-r)^2 - (r+1)^2 = 9 - 6r + r^2 - r^2 - 2r - 1 = 8 - 8r

    So:

    \displaystyle

\begin{align*} \sum_{r=0}^{n} 8 - 8r &= 8 - 8(0) + \sum_{r=1}^n (8-8r) \\&= 8 + 8n - 8 \times \frac{n}{2}(n+1) \\&= 8(n+1) - 4n(n+1) \\&= 4(n+1)(2 - n) \\ & = 4(2n - n^2 +2 -n) \\& = -4n^2 +4n + 8\end{align*}

    Did you remember that the sum starts from r=0 and hence the sum of \sum_{r=0}^n 8 = 8(n+1) \neq 8n?
    Thanks for showing the working, made the silly mistake of not working the r=0 term out separately!
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    (Original post by economicss)
    Thanks for showing the working, made the silly mistake of not working the r=0 term out separately!
    No worries!
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    Anybody got any further qns on conics - the ones listed by kingaaran don't have answers...
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    Hii guys
    Question; under proof by induction, is it necessary to write the whole essay conclusion like the one in the FP1 Edexcel text book or is there a summarised one that could still get the marks for the conclusion?
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    (Original post by Reeyap)
    Hii guys
    Question; under proof by induction, is it necessary to write the whole essay conclusion like the one in the FP1 Edexcel text book or is there a summarised one that could still get the marks for the conclusion?
    "So if the statement is true for n, then it is true for n + 1; as it is true for n = 1 it is true for all n by induction" sounds sufficient.
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    (Original post by 13 1 20 8 42)
    "So if the statement is true for n, then it is true for n + 1; as it is true for n = 1 it is true for all n by induction" sounds sufficient.
    Ohh alright
    thank youu
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    (Original post by Reeyap)
    Ohh alright
    thank youu
    Though I doubt what I said wouldn't be enough, I'd advise looking at what kind of conclusions are on the mark schemes to be on the safe side. In general markers apply the scheme "positively", but always good to check...
 
 
 
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