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    That was really hard

    The first part want absolutely fine (q1 & q2)

    The rest was really complicated trig stuff, -C3 shouldn't be this hard
    The questions were nothing like those in the textbook and were much harder than past paper questions

    soooo much trig and rearranging!

    worst thing is I completely forgot to put my calculator in radian mode for the iteration question -that's 4 easy marks down the drain
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    What were the marks for each question?
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    (Original post by ComputerMaths97)
    No that's actually the best possible answer. There's no way of proving it in less than that, in such a way that wouldn't earn 4 marks. Congrats The only way you could improve it would be by saying "2 is the only prime factor of 2^n by definition, so 2^n is not divisible by 3" but I'm assuming what you've written will be considered sufficient. I expect it to be.
    I was just so worried writing that because I didn't think a wordy proof was what they were looking for!!! I almost said by definition on the exam paper but I wasn't sure if it was the right thing to do

    I know I've messed but on a few things but hopefully I've got a high A...

    Thanks
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    For the question asking you to draw cos^-1(x) and a quadratic and show they only had one root... I got 2 points of intersection :-(
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    (Original post by Rawsonj)
    Answers that I got:
    1)integrating gives area = pi/2+sqrt2
    2) ln(2+e^x)=2x is a quadratic in e^x with roots 2 and -1. e^x is always greater than 0 so e^x=2. x=ln2
    3)Area = ln(256) -4 or 4(ln(4) -1) by parts
    4)draw the graphs, intersect at x=-1 and x=-1/3
    5)dv/dh=8 dh/dt=0.4/8=0.05 m per minute
    6)dy/dx=1/2cos2y gradient at x=1.5 y =pi/12 is sqrt(3) /3 For second part y=1/2 arcsin(x-1) a translation of 1 in the positive x direction followed by an enlargement of scale factor 1/2 parallel to the y axis.
    7)x^2n-1 = (x^n+1)(x^n-1) (2^n+1)(2^n-1) as n is greater than but not equal to 0, 2^n is even and so the brackets are consecutive odd numbers. Therefore one is a multiple of 3. Therefore the product of the two brackets is divisible by 3.
    8)first part was a show that. gradient was 1/2 so point was (-4,-2) third part was integration by substition which gave 19/12.
    9) p=1+k show that minimum is at x=1/4ln(k) and y value at this point is 2*sqrt(k). Integrating gives area =k-1. Next part was to show what g(x) is equal to. Show it is even g(x)=g(-x). Last part: line of symetry in x=1/4ln(k). explanation is that as when translated by 1/4ln(k) in the negative x direction, we see that the origin is a line of symetry (line of reflection) as it is even. Therefore when we translate g(x) back to f(x) we move the line of symetry to x=1/4ln(k)
    5 and 7 are consecutive odd numbers but their product isnt divisible by 3..
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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
    Thanks alot for this.
    According to this mark scheme. I think i've about 66/67 out of 72 plus 16 in the coursework. Hopefully c4 is a lot easier and the grade boundaries, lower than last year.
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    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 drew a 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^ (9 at most if I assume that I scored a fat old 0 on both questions above) I thought paper was alright, think grade boundaries will be very similar to last year.
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    (Original post by sullyking)
    Thanks alot for this.
    According to this mark scheme. I think i've about 66/67 out of 72 plus 16 in the coursework. Hopefully c4 is a lot easier and the grade boundaries, lower than last year.
    coursework? -Am I on the right thread for OCR C3 maths A2
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    (Original post by Aph)
    I was just so worried writing that because I didn't think a wordy proof was what they were looking for!!! I almost said by definition on the exam paper but I wasn't sure if it was the right thing to do

    I know I've messed but on a few things but hopefully I've got a high A...

    Thanks
    Ahaha no worries. A proof is a proof, an exam board cannot decline marks for a perfect proof if it not in the form they wanted. Especially if they didn't specify in the question that it must not have any words or something ridiculous like that.

    People always over complicate the proof questions. For me they're the easiest marks in the paper.
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    (Original post by BongoBuddha)
    coursework? -Am I on the right thread for OCR C3 maths A2
    this is MEI m8
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    (Original post by mrk1357)
    GO97: I assume you meant "2 to the power of an integer cannot make 12".

    I agree.

    I am saying that in your solution, you can't just say "odd*odd is always divisible by three and therefore the expression is divisible by three" because odd*odd is NOT ALWAYS divisible by three.

    You need to do what I did in my previous post or what ComputerMaths97 did to show that for any pair of odd numbers similar to the pair 11 and 13 in the way that none of the 2 numbers is divisible by three, the number between them is never perfect square. There are a lot of such pairs besides 11 and 13:
    19 and 21, 141 and 143, 23 and 25...
    ah fair enough, my brains fried
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    (Original post by BongoBuddha)
    coursework? -Am I on the right thread for OCR C3 maths A2
    This is the thread for OCR MEI C3, OCR C3 will probably have a different thread
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    What are the method marks for each question?
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    (Original post by fazee)
    This is the thread for OCR MEI C3, OCR C3 will probably have a different thread
    ooooooooooo thanks!
    was getting really confused -silly me
    I had an exam this morning C3 but not MEI -doh!
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    (Original post by ComputerMaths97)
    If anyone has any questions about the paper I'll answer them, fairly certain I got full marks so I'm happy to help out

    How many marks was the integration question where you had to minus stuff?
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    question 9(ii)-- how did people get 2rootk? I couldn't cancel it down for the life of me
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    I think i got 57, anyone reckon that'll be an A?
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    (Original post by Shreya Randev)
    How many marks was the integration question where you had to minus stuff?
    What marks were each question?
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    (Original post by Aph)
    For 7 my argument was
    2^n will never have a multiple of 3 in it, as by definition it is made only of 2's multiplied together.
    In any 3 consecutive integers there will be one which has to be divisible by 3.
    :. As 2^n won't be that number it is necessary for either
    2^n-1
    or
    2^n+1
    to be divisible by 3...

    It feels really clunky but it was the best I could do.
    yeah thats sound exactly what i did, i didnt feel like it was very elegant but nothing else i could do lol
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    What will a grade B and C be?
 
 
 
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