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    (Original post by kprime2)
    92ish I reckon
    thankss, wanted a bit less pressure on C4 but oh well
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    How many marks for 100 ums do you think? I think i got full raw but i will probably slip up with accuracy mark somewhere
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    What's 45 marks typically I messed up badly

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    Could someone link me the past C3 grade boundaries table?
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    (Original post by Craig1998)
    I used x=0 and x=1 and they both lead to a=3 and b=4..

    I really cant afford to lose any more marks, just realised I lost some on the differentiation for forgetting d/dx (cosx) = -sinx.
    Yeah I was initially going to use 0 and 1 too but then I read x>2 and decided not to just to be safe. On second thought though, 0 and 1 works just fine too. The question just happens to specify the domain as x>2. However using x=0 and x=1 just to simplify f(x) doesn't really violate the function. I can't imagine you losing marks.
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    (Original post by kprime2)
    Wasn't too bad
    Most questions were standard.
    Lots would have done 6(a) the incredibly long and tedious way.
    9(b)(c) would probably have confused students.

    I reckon 62ish for an A

    EDIT: IGNORE THE PDF ATTACHMENT IT HAS A QUESTION MISSING

    MODEL ANSWERS: https://drive.google.com/open?id=0B3...mlERk5GUG1DNDg
    Thank you! Quick question. I idiotically differentiated the sin^2(2y) to be 8cos(2y)sin(2y) and contiuned on, getting the last answer to be 0.25cosec(4y). Do yiu have any idea how many marks would i have lost? Thank you!xx
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    For 8b i did:
    3cot x = cosec^2 x - 2

    Times by sin^2 x to get

    3sin x cos x = 1- 2 sin^2 x
    3/2(sin 2x)=cos2x
    Tan 2x = 2/3

    Then you can get the same solutions.

    Would i lose marks for not doing quadratic formula way
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    (Original post by metrize)
    For 8b i did:
    3cot x = cosec^2 x - 2

    Times by sin^2 x to get

    3sin x cos x = 1- 2 sin^2 x
    3/2(sin 2x)=cos2x
    Tan 2x = 2/3

    Then you can get the same solutions.

    Would i lose marks for not doing quadratic formula way
    Yes. As a general rule of thumb, if you use a different method and still get the correct solutions, you still get full marks. The mark scheme tends to have several alternative ways of answering each question.
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    (Original post by kprime2)
    seems fine, you get the correct answer
    Also, sorry to bother, as you would imagine; my method for part C was also a little odd:
    Spoiler:
    Show
    7.5 = (5.181916.. + 15)e^{-0.2T}

    \dfrac{7.5}{5.181916.. + 15} = e^{-0.2T}

    ln(0.365529..) = -0.2T

     -5ln(0.365529..) = T

     T = 5ln[(0.365529..)^{-1}] = 5ln(2.73575...)

     b + \dfrac{b}{e} = 2.73575...

    \therefore b = \dfrac{2.73575e}{e+1} = 2

    I used ANS on my calculator so all my values were as accurate as they could be, and it canceled down to exactly 2

    In hindsight, it would've been better for me to use 15e^{-1} instead of 5.181916.. but exam pressure kinda got to me I guess lol
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    Yesssss Arse- I mean kprime is on it loool


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    (Original post by kprime2)
    .
    if on the last part of the last q i had -5(0.5+...), would i get 1-2 marks? The inside wasn't tidy or what it was supposed to be but i got -5 on outside and half on inside so cause of ln rules it's the same as having 5 and 2 ? (my inside wasn't correct but i showed working to get a)
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    (Original post by coolguy123456)
    if on the last part of the last q i had -5(0.5+...), would i get 1-2 marks? The inside wasn't tidy or what it was supposed to be but i got -5 on outside and half on inside so cause of ln rules it's the same as having 5 and 2 ? (my inside wasn't correct but i showed working to get a)
    Question says a and b are integers. Probably -1
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    (Original post by edothero)
    Question says a and b are integers. Probably -1
    For some reason my final answer was -5ln(1/2+.../e^1.4), how many marks would this get me?
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    (Original post by edothero)
    Also, sorry to bother, as you would imagine; my method for part C was also a little odd:
    Spoiler:
    Show
    7.5 = (5.181916.. + 15)e^{-0.2T}

    \dfrac{7.5}{5.181916.. + 15} = e^{-0.2T}

    ln(0.365529..) = -0.2T

     -5ln(0.365529..) = T

     T = 5ln[(0.365529..)^{-1}] = 5ln(2.73575...)

     b + \dfrac{b}{e} = 2.73575...

    \therefore b = \dfrac{2.73575e}{e+1} = 2

    I used ANS on my calculator so all my values were as accurate as they could be, and it canceled down to exactly 2

    In hindsight, it would've been better for me to use 15e^{-1} instead of 5.181916.. but exam pressure kinda got to me I guess lol
    Hmm I'm not sure. I think the purpose of the question is to test your skills of manipulating logs and exponentials to maintain an exact answer throughout your working, which is why they want you to show that T=a ln(b+b/e) rather than just finding the value of T. You've set b+b/e = 2.7375... and just let the calculator do the work for you. I can't answer for sure whether you would or wouldn't get marks but I reckon the examiner wants to see your skills in maintaining exactness throughout workings. Perhaps your answers will be sent to review and you will get marks. A lot of unconventional methods do get marks provided you get the right answer.
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    Anyone have any idea what ums 58 would be? I messed that paper up. Missed whole question 6.
    Any help appreciated
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    (Original post by kprime2)
    Hmm I'm not sure. I think the purpose of the question is to test your skills of manipulating logs and exponentials to maintain an exact answer throughout your working, which is why they want you to show that T=a ln(b+b/e) rather than just finding the value of T. You've set b+b/e = 2.7375... and just let the calculator do the work for you. I can't answer for sure whether you would or wouldn't get marks but I reckon the examiner wants to see your skills in maintaining exactness throughout workings. Perhaps your answers will be sent to review and you will get marks. A lot of unconventional methods do get marks provided you get the right answer.
    Yeah I know, this was very sloppy from me. I had 5 minutes remaining so I just wanted to solve the question via brute force.
    If they are as picky as they say, would you see me getting at least 2 marks?
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    (Original post by edothero)
    Yeah I know, this was very sloppy from me. I had 5 minutes remaining so I just wanted to solve the question via brute force.
    If they are as picky as they say, would you see me getting at least 2 marks?
    I don't think they'd be too harsh. You might get full marks but if not then you'll probably only lose max 2 marks
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    (Original post by kprime2)
    I don't think they'd be too harsh. You might get full marks but if not then you'll probably only lose max 2 marks
    Cheers for your help, I'll double check with my teacher :yy:
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    I messed up so much on this paper
    How many ums will 45 be?
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    70-71/75
 
 
 
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