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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
    Thank you! What were the marks for each question?
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    (Original post by klefki)
    Attachment 554089

    I got 55/12 for the area in q8, can someone tell me where I went wrong please? (sorry it's sideways)
    When you expanded the bracket in the last step the calculations are wrong... 2/3 of 27 is 18, and 8x3 is 24
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    (Original post by Ddgv12)
    I think i got 57, anyone reckon that'll be an A?
    Do you know your coursework mark?

    (Original post by Shreya Randev)
    How many marks was the integration question where you had to minus stuff?
    That was the least descriptive question ever, any more detail please? xD There was quite a few integration questions.
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    [QUOTE=ComputerMaths97;65979885]Do you know your coursework mark?

    Yeh i got 15
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    Last years paper was similar to this years and that was relatively more difficult than normal. So IMO:

    86/90 = 100 UMS
    79/90 = 90 UMS
    72/73 = 80 UMS
    65/90 = 70 UMS
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    (Original post by i<3emmawatson)
    even my teachers couldn't do parts of Q9
    ...
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    Can we colab to get the marks per question please? From what I can remember:
    3) 5m?
    7) 3m? or 2m?
    9)i) 1m
    iii) 4m?
    iv)B) 2m?
    C) 3m

    My memory isn't great with this sort of thing
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    Was it just me who thought this paper was really hard 😩
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    (Original post by Herthen)
    Can we colab to get the marks per question please? From what I can remember:
    3) 5m?
    7) 3m? or 2m?
    9)i) 1m
    iii) 4m?
    iv)B) 2m?
    C) 3m

    My memory isn't great with this sort of thing
    for q8, one part was 4 marks and another was 5 marks.. i think
    9 IV A was 2 or 3
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    Anyone know what question 3 was? Answer was 4ln(4)-4
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    (Original post by Goulden_J)
    Last years paper was similar to this years and that was relatively more difficult than normal. So IMO:

    86/90 = 100 UMS
    79/90 = 90 UMS
    72/73 = 80 UMS
    65/90 = 70 UMS
    I'd agree although maybe 85/90 for 100 ums (same as last year) think that proof question might have caused a problem as a lot of people I know just used the odd X odd method which in this case wasn't enough
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    (Original post by groovy_q)
    5 and 7 are consecutive odd numbers but their product isnt divisible by 3..
    Yeah i realised that i missed part of my explanation:
    Posted since somewhere: I
    missed something in the proof (q7) when writing my unnoficial markscheme:
    http://www.thestudentroom.co.uk/show...3930177&page=8
    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 is not a multiple of 3 as its prime factorisation is 2^n so the brackets are consecutive odd numbers. Therefore one is a multiple of 3 as we have 3 numbers and the middle one is not a multiple of 3 so either the one above or the one below must be. Therefore the product of the two brackets is divisible by 3.
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    In question 8 when working out P I got the gradient of tangent at 0 to come out as 1? So the tangent equation was y=X and then I used this to get the co-ordinates (-4,-4) for P... How have I got the gradient wrong at the origin from just subbing in 0 ?!
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    I thought last years paper was easier. This years wasn't terrible, but I thought question 9 was quite challenging. Personally I think the grade boundaries will be similar to last years, maybe even a little lower?
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    (Original post by ComputerMaths97)
    Change the answer to question 7 already, don't give people false hope. That is incorrect. 11 and 13 are consecutive odd numbers, but neither divisible by 3. It proves nothing, maybe get 1 or 2/4 for that.

    ALTHOUGH CAN CONFIRM IS IT CORRECT OTHERWISE.
    Agreed, please change the answer for 7. I don't get this odd and even number hype for this question, it is not relevant at all. Loads of my friends did this too.

    It's just that 2^n-1, 2^n, 2^n+1 are 3 consecutive numbers and hence one of them must be a multiple of 3. 2^n is not a multiple of 3, therefore it's one of 2^n-1 and 2^n+1 which is a multiple of 3. So (2^n-1)*(2^n+1) must be divisible by 3. There is absolutely no need to mention even or odd numbers, although if you did you shouldn't get penalised as long as you did the actual proof correctly.

    Stuff like "it is divisible by 3 because its two consecutive odd numbers multiplied..." That's absolute rubbish lol.
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    I found it weird there weren't any indefinite integrals in this paper lol, and if you got 14/3 as the area but forgot to subtract how many marks would you get overall? I feel pretty good seeing the answers people got, feel like I dropped only 6-7 marks which is pretty solid for me.
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    on the unofficial mark scheme, which answer is related to the question which told you to draw sketch of graphs and hence show it had two roots?

    question 4 is, my bad
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    (Original post by ZoeWoah)
    How did everyone get 19/12 for the area? I got between 30 and 40, I found the area beneath the curve then subtracted the triangle from it, I feel like that was completely wrong and I misinterpreted the graph wrong hahahha
    I did this aswell! Can't quite remember what I got but I think that's the right method
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    (Original post by Hannahlcxx)
    I did this aswell! Can't quite remember what I got but I think that's the right method
    You were supposed to subtract the area from the triangle not the other way round
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    I think I got 60-62 marks

    and I have 17/18 in the CW... what ums should be roughly ( as a range please) /100

    Thanks.
 
 
 
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