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    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...
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    Hardest C3 paper so far, even my teachers couldn't do parts of Q9, so hopefully the boundaries for a B should be pretty low. :p:U:cool:
    Still pretty frustrated though, way beyond the C3 syllabus. Lets hope C4 will be easier to compensate.
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    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
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    (2016 i think)
    full ums = 70/72
    90 ums = 64/72
    80 ums = 59/72
    70 ums = 54/72

    (2015)
    full ums = ? (i think it was 69/72)
    90 ums = 62/72
    80 ums = 56/72
    70 ums = 51/72

    (2014)
    full ums = 71/72 (i think)
    90 ums = 65/72
    80 ums = 58/72
    70 ums = 52/72
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    (Original post by sullyking)
    (2016 i think)
    full ums = 70/72
    90 ums = 64/72
    80 ums = 59/72
    70 ums = 54/72

    (2015)
    full ums = ? (i think it was 69/72)
    90 ums = 62/72
    80 ums = 56/72
    70 ums = 51/72

    (2014)
    full ums = 71/72 (i think)
    90 ums = 65/72
    80 ums = 58/72
    70 ums = 52/72
    m8 the differences should be the same between the boundaries, you ok fam???????:albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein:
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    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
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    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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    (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
    You needed to do as you said, but subtract the area under the curve from the triangle, not the other way around.

    Surely you would have got a negative value from doing it that way?
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    (Original post by i<3emmawatson)
    m8 the differences should be the same between the boundaries, you ok fam???????:albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein::albertein:
    It was a good paper compared to the 2015 paper which was much harder. Everyone I spoke to found it good. There was only two 3 marker's I couldn't get the answer to. But I'll probably get method marks. The rest of the paper was calm.
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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). A should be 80 or 81, and B should be 72 or 73. Difference between each grade will be about 7 or 8 marks.

    Thanks for reading
    That looks good.
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    (Original post by WhiteBison)
    You needed to do as you said, but subtract the area under the curve from the triangle, not the other way around.

    Surely you would have got a negative value from doing it that way?
    Yeah, maybe I did it that way round. I think I got the wrong number for the area under the curve though! Oh well, should get method marks!
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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). A should be 80 or 81, and B should be 72 or 73. Difference between each grade will be about 7 or 8 marks.

    Thanks for reading
    Let's hope your friend is correct as I've got exactly the same answers for every question, probably dropped a couple of marks on the explanation of proof and symmetry though
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    (Original post by Bealzibub)
    Did you get k-1?
    Yep for the ak+b one
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    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.
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    What was q9ii and how many marks was it worth
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    http://www.thestudentroom.co.uk/show...9#post65978509
    Thread for the unofficial mark scheme

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    Two consecutive odd numbers are not always divisible by 3

    i.e 11 and 13.

    Therefore that way of "proof" is not sufficient.
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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.
    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.
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    Yes, i have edited it now thank you.
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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
    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.
 
 
 
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