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    (Original post by Jasmine149)
    I got 1/2 :dontknow:

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    In fairness, the 9 marker and the last question were all one massive **** up for me so I have no clue xD
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    (Original post by sonron202)
    Anybody get 19/12 for the area question in the last part of question 8?
    I did


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    (Original post by sonron202)
    I got x as ln2
    So did I oh wow. I was nervous about that one


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    (Original post by Maligaha)
    I got 0.5k-0.5
    You forgot to do the 0 part of the substitution which gave (0.5-0.5k) so when you subtract that from your answer you get K-1
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    (Original post by Alexw1812)
    a was 1 and b was -1
    ?

    I got 0.5 and -0.5!! Oh God........
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    (Original post by Cadherin)
    ? Point P was on the asymptote I thought - given asymptote is x=-4 and y=1/2x was equation of tangent, I thought it was (-4, -2)... Oh dear, I really hope I haven't ****ed that one up as well...
    I thought the line was y=0.5x too. The majority are saying (-4,-2) I think we're safe :P
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    (Original post by daisyn97)
    4ln4 - 4?
    hmmm, i got -4ln4-4 (which i know realise should be 4ln+4...)
    not sure who's right tho
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    (Original post by Maligaha)
    I got 0.5k-0.5
    I got k-1
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    I thought it was a nice paper except question 9 was a bit weird. I definitely messed up on a few bits of that, i think i kept getting my es and ins confused
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    (Original post by treeporn)
    How the hell did you guys find out the equation of the tangent?
    The line was a tangent to the curve at the origin. So you sub x=0 into the dy/dx of the curve giving you a gradient=1/2

    The line went through the origin so C=0, so y=0.5x
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    (Original post by gemdarkstone)
    hmmm, i got -4ln4-4 (which i know realise should be 4ln+4...)
    not sure who's right tho
    It was 4ln4 - 4 I'm pretty certain - it evaluated to (4ln4-8)-(-4).

    Anyone else get 0.5k - 0.5 for ak + b?? I'm pretty sure that's right as you had to take a factor of 0.5 out before you made the substitution...
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    (Original post by ScalyJake)
    The line was a tangent to the curve at the origin. So you sub x=0 into the dy/dx of the curve giving you a gradient=1/2

    The line went through the origin so C=0, so y=0.5x
    Yeah, I was a bit rushed... Will I get full marks on that part for correct coordinates and writing down y=0.5x without actually showing how I got it, as I didn't have time?
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    (Original post by daisyn97)
    4ln4 - 4?
    Same, but I put it as ln256 - 4. Would that still get the marks?
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    very last q, about geometrical properties of f(x), what did you guys put
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    dr/dt 0.03?
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    (Original post by VokeA)
    So did I oh wow. I was nervous about that one


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    yea me to, for 5 marks i was worried....
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    You calculate dy/dx at the point x=0 (because it was at the origin) and then you use y-y1=m (x-x1) to find the equation of the tangent. X1 and y1 are (0,0) and m is the value of dy/dx.
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    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)
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    yo wtf was that paper man
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    (Original post by Alexw1812)
    You forgot to do the 0 part of the substitution which gave (0.5-0.5k) so when you subtract that from your answer you get K-1
    Im pretty sure you have substituted 0 twice....Name:  IMG_20160621_112701.jpg
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