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    How many marks was 8ii and iii and does anyone have a pic of the paper? Or can post an unofficial mark scheme with the mark breakdown in each question
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    (Original post by fazee)
    UNOFFICIAL MARK SCHEME:

    1. Integrate 1 + cos(0.5x) between limits of 0 and pi/2. Answer is pi/2 + root2

    2. Solve for x. Answer is x = ln2

    3. Can't remember the context. Answer is 4ln4 - 4 - alternatively may be written as 4(ln4 -1) or ln256 - 4 or any equivalent form.

    4. |2x+1| = -x. First case, 2x+1 = -x, gives 3x+1 = 0, 3x = -1, x = -1/3
    Second case, -(2x+1) = -x, negate both sides to give 2x+1 = x, x+1 = 0, x = -1
    Alternative approach, square both sides, 4x2 + 4x + 1 = x2, 3x2 + 4x + 1 = 0, (x+1)(3x+1) = 0, x= -1, x = -1/3 as before

    5i. V = 4root(h3+1) - 4. dV/dh using chain rule gives 4*1/2(h3+1)-1/2 * 3h2 = 6h2 * (h3+1)-1/2. Substituting h=2 gives dV/dh = 8 (feel free to check this for yourself).
    5ii. dh/dt = dh/dV * dV/dt = 1/8 * 0.4 = 0.4/8 = 0.05

    6i. Can't remember the context. Answer is root3 /3
    6ii.Transformation is first translate 1 unit to the right, THEN stretch parallel to y-axis, scale factor 1/2

    7. x2n-1 = (xn+1)(xn-1). Substituting for x=2, 2n has to be even, so it's a product of two consecutive odd numbers, so one of those must be divisible by 3, so their product is divisible by 3. That's why I put anyway, I've seen someone say it's not right, so not completely sure about this one.

    8i. Differentiate using quotient rule, then multiply your result by 2root(x+4), to obtain dy/dx in the form given.
    8ii. Using equation for y, asymptote is x= -4, so x coordinate of P is -4. Use dy/dx for x=0, to show that tangent is y = 0.5x (y-intercept c is 0). Then y = 0.5 * -4 = -2, so P is (-4,-2)
    8iii.Area under the curve is found by integration to be 14/3. Area of triangle with vertices O,Q and (5,0) is 1/2 * 5 * 5/2 = 25/4
    So area of required region is 25/4 - 14/3 = 19/12

    9i. y = e2x + ke-2x. When x=0, y = 1 + k*1 = 1+k
    9ii. Answer is 2rootk
    9iii. Answer simplifies to (0.5k - 1/2) - (1/2 - 0.5k) = 0.5k - 1/2 - 1/2 + 0.5k = k - 1
    9ivA. Replace x with (x+ 0.25lnk), to give g(x) = rootk(e2x + e-2x).
    9ivB. g(-x) = g(x), so g(x) is an even function.
    9ivC. f(x) is symmetrical about the line x = 0.25lnk. This can be deduced because g(x) is f(x) translated 0.25lnk units to the left. g(x) is even, so is symmetrical about the y-axis i.e. symmetrical about line x=0. f(x) is g(x) translated 0.25lnk units to the right, therefore its line of symmetry is x = 0.25lnk. Full reasoning will probably be required for all 3 of these marks.

    These are the answers of my friend, who does Additional Further Maths, so I think they should all be right, or most of them anyway. Of course, feedback is more than welcome, if I've made a mistake please point it out.

    I think boundaries will be 88 or 89 for 100 UMS (out of 90, for exam and coursework). 90% should be 80 or 81, and A should be 72 or 73. Difference between each grade will be about 7 or 8 marks.

    Thanks for reading
    I had the exact same for every single answer. General consensus seems that these are the correct answers too.
    I got 15/18 for coursework + hopefully 65-72 out of 72 for the paper (assuming I made some errors) so hopefully that will put me close to 100ish ums (fingers crossed)
    overall I found it quite an easy paper - hope every else did well. no point stressing now as there is nothing that can be done
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    Also as response to some of the people debating the proof, this is what I put:

    (2^n -1)(2^n + 1)
    Since 2^n = 2,4,8,16,32,64.... so on
    therefore 2^n - 1= 1,3,(7),15,(31),63... every other number is divisible by three
    also 2^n + 1 = 3,(5),9,(17),33,(65)... every other number is divisible by three
    therefore the product of (2^n -1)(2^n +1) must be divisible by three as one of the factors is divisible by three
    the divisible factor alternates between every value of n

    that is what I put I'm not sure if that is exactly how you 'prove' the conjecture but makes sense to me
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    (Original post by Leechayy)
    That's the translated one isn't it?:smug::p:p:holmes:

    Posted from TSR Mobile
    :P well the un-translated one certainly isn't 2cosh2x
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    Out of interest what did people get for the coursework component seeing as that can cripple your grade.
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    thanks !
    (Original post by ComputerMaths97)
    Add the score from your coursework to your raw mark from the exam, giving a score out of 90, then you get corresponding ums out of 100.
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    (Original post by Aph)
    Out of interest what did people get for the coursework component seeing as that can cripple your grade.
    17/18
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    for question 9.ii i showed that the minimum point has x co ordinate of 1/4ln4 and then i subbed that back into f(x) to find out the y co ordinate but i got an incorrect y co ordinate. how many marks in the question will i lose because of this
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    (Original post by Aph)
    Out of interest what did people get for the coursework component seeing as that can cripple your grade.
    16/18
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    (Original post by Aph)
    Out of interest what did people get for the coursework component seeing as that can cripple your grade.
    18/18
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    For question 3 I didn't sketch the graph, how many marks is that out of 4?
    And 8iii how many marks is integration of just the curb work out of 9?
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    I think I got 46/72 in this and I got 15/18 in my coursework, what would I average out with?
    What does everyone think the grade boundaries for B will be?
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    Can anyone tell me that if i got 69/72 for c3 paper and 13/18 for c3 coursework, what is the UMS out of 100 for c3 please?
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    (Original post by fkjames)
    Can anyone tell me that if i got 69/72 for c3 paper and 13/18 for c3 coursework, what is the UMS out of 100 for c3 please?
    Since a lot of people are asking this: here's MEI's UMS conversion site. You can estimate based on previous years, but remember that it changes slightly based on the difficulty of the exam. http://www.ocr.org.uk/i-want-to/conv...-marks-to-ums/
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    (Original post by y0uSmartass)
    I think I got 46/72 in this and I got 15/18 in my coursework, what would I average out with?
    What does everyone think the grade boundaries for B will be?
    Based on previous years that would be a mid-C
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    (Original post by mrbazz12)
    Okay paper apart from 2 things:

    1. For the proof question, I wrote down the difference of two squares i.e. (X^n + 1)(X^n - 1) and then had a mind blank, so will I get 1 mark for that? (out of the 4)

    2. The modulus question I used the 'square both sides' method only to forget to square one of the sides (the X on the other side to the | 2x - 1 |) so got an answer of X = -1 and X = 1/4. How many marks will I lose for this?

    Apart from those lost marks^ I thought paper was alright.
    Can anyone answer this?
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    Can anyone remember question 3? What was the graph you had to draw?
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    (Original post by mrbazz12)
    Can anyone answer this?
    You would definitely get the first mark as it asked for that in the question.

    You would probably lose 2 marks on the second one, the final A1 for the wrong answer, and probably a M1 mark
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    can somebody please say the marks given to each question please haha
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    (Original post by mrbazz12)
    Can anyone remember question 3? What was the graph you had to draw?
    the graph was the modulus function and y=-x. You had to show the intersects.
 
 
 
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