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    Expecting something nice after the M1? 19/06/2018
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    (Original post by Ryan Right)
    Expecting something nice after the M1? 19/06/2018
    true
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    (Original post by Ryan Right)
    Expecting something nice after the M1? 19/06/2018
    This belongs in the Maths Exams subforum, in order to avoid clogging up the main forum. Notnek request move?
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    C3 is a generally well comprehended module, so even if the paper isn't too nice, I'd expect high grade boundaries.
    The last few C3 papers have been quite nice which is not a normal thing for MEI, so I wouldn't be surprised if this paper was quite challenging.

    After that horrid excuse of an M1 paper though, I do feel as though MEI owe us
    Regardless, good luck to everyone. All the best, and I'm sure you'll smash it!
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    what time is the exam again, I know its in the afternoon but forgot when :s

    (Nvm just checked its at 13:00)
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    how did people find that
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    Relatively easy except for stupidly doing 2ln(x-2) by parts in 9(iv) smdh.
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    Pretty good. Wait I got 0.05 for the half life question for K, did I make a stupid mistake?
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    (Original post by lowkeybuthighkey)
    Relatively easy except for stupidly doing 2ln(x-2) by parts in 9(iv) smdh.
    i did it by parts too, what was the other way of doing it?
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    (Original post by Rowanhuggins123)
    Pretty good. Wait I got 0.05 for the half life question for K, did I make a stupid mistake?
    Yes because it asked for 2 sig fig
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    this is what i got, the rest were like show that..
    • 1)(i) -3x/4y (ii)-6
    • 2) (i) odd

    (ii) periodic (pi/2)
    (iii) even
    • 3) (i)0%

    (ii)k=0.048
    • 4) (0.55, 0.514)
    • 5) (ii) -1/3
    • 6) (i) 5pi/6

    (ii) sin(0.5(x-pi/2))
    • 7) n^3-3n^2+2n. Prove it is divisible by 6 for all positive integers n. I factorized, they were 3 consecutive integers, so at least one divisible by 2 and one divisible by 3, thus it is divisible by 6
    • 8) (i) equation of the tangent: x-4y-3=0

    (iii) 23/8-4ln2
    • 9) (ii) find P and Q. P(3,0) Q(4, ln4)

    (iv) 4ln2-1
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    Yeah thought so ahah
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    (Original post by 2long)
    this is what i got, the rest were like show that..
    • 1)(i) -3x/4y (ii)-6
    • 2) (i) odd

    (ii) periodic (pi/2)
    (iii) even
    • 3) (i)0%

    (ii)k=0.048
    • 4) (0.55, 0.514)
    • 5) (ii) -1/3
    • 6) (i) 5pi/6

    (ii) sin(0.5(x-pi/2))
    • 7) n^3-3n^2+2n. Prove it is divisible by 6 for all positive integers n. I factorized, they were 3 consecutive integers, so at least one divisible by 2 and one divisible by 3, thus it is divisible by 6
    • 8) (i) equation of the tangent: x-4y-3=0

    (iii) 23/8-4ln2
    • 9) (ii) find P and Q. P(3,0) Q(4, ln4)

    (iv) 4ln2-1
    I think I got all the same apart from 25% of substance left after 28.8 years
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    (Original post by 2long)
    this is what i got, the rest were like show that..
    • 1)(i) -3x/4y (ii)-6
    • 2) (i) odd


    (ii) periodic (pi/2)
    (iii) even
    • 3) (i)0%


    (ii)k=0.048
    • 4) (0.55, 0.514)
    • 5) (ii) -1/3
    • 6) (i) 5pi/6


    (ii) sin(0.5(x-pi/2))
    • 7) n^3-3n^2+2n. Prove it is divisible by 6 for all positive integers n. I factorized, they were 3 consecutive integers, so at least one divisible by 2 and one divisible by 3, thus it is divisible by 6
    • 8) (i) equation of the tangent: x-4y-3=0


    (iii) 23/8-4ln2
    • 9) (ii) find P and Q. P(3,0) Q(4, ln4)


    (iv) 4ln2-1
    Yes although I think the period was pi
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    Wasn’t too bad, made a few stupid mistakes tho
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    (Original post by 2long)
    this is what i got, the rest were like show that..
    • 1)(i) -3x/4y (ii)-6
    • 2) (i) odd

    (ii) periodic (pi/2)
    (iii) even
    • 3) (i)0%

    (ii)k=0.048
    • 4) (0.55, 0.514)
    • 5) (ii) -1/3
    • 6) (i) 5pi/6

    (ii) sin(0.5(x-pi/2))
    • 7) n^3-3n^2+2n. Prove it is divisible by 6 for all positive integers n. I factorized, they were 3 consecutive integers, so at least one divisible by 2 and one divisible by 3, thus it is divisible by 6
    • 8) (i) equation of the tangent: x-4y-3=0

    (iii) 23/8-4ln2
    • 9) (ii) find P and Q. P(3,0) Q(4, ln4)

    (iv) 4ln2-1
    I got 8ln2 - 2 for the last part of last question... Can someone confirm this?
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    (Original post by uniftw)
    I think I got all the same apart from 25% of substance left after 28.8 years
    Your answer makes much more sense it was 1 mark so I kinda just guessed it lol
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    (Original post by a.b.a)
    I got 8ln2 - 2 for the last part of last question... Can someone confirm this?
    its 4ln2 -1
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    (Original post by 2long)
    this is what i got, the rest were like show that..
    • 1)(i) -3x/4y (ii)-6
    • 2) (i) odd

    (ii) periodic (pi/2)
    (iii) even
    • 3) (i)0.25%

    (ii)k=0.048
    • 4) (0.55, 0.514)
    • 5) (ii) -1/3
    • 6) (i) 5pi/6

    (ii) sin(0.5(x-pi/2))
    • 7) n^3-3n^2+2n. Prove it is divisible by 6 for all positive integers n. I factorized, they were 3 consecutive integers, so at least one divisible by 2 and one divisible by 3, thus it is divisible by 6
    • 8) (i) equation of the tangent: x-4y-3=0

    (iii) 23/8-4ln2
    • 9) (ii) find P and Q. P(3,0) Q(4, ln4)

    (iv) 4ln2-1

    period is pi not pi over 2
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    The final answer for the last question I got 4ln2-1? The one where you had to write it in the form mln2-n

    I got 25% for the half like one- but why the hell was there like half a page of answer space for that one mark? 😂
 
 
 
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