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    (Original post by jf1994)
    Did anyone else get that there was no asymptote for the second transformation graph? plz plz plz
    Nah. I got y is 5 for the asymptote. It was just reflected on the y axis.


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    (Original post by Chazley123)
    tan = sin / cos
    therefor cos = sin/tan so sin/p
    I put this at first, but its not correct. Cause they wanted it with respect to p so you had to use the identity tan^2x =1+sec^2x (or whatever it is)

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    (Original post by sj97)
    Wasn't the range for g supposed to have an e value in it?
    I had an e value in my answer. Something like 4e^-4?
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    (Original post by PaulCJH)
    My Answers:
    1. 2p/(1-p^2), 1/((p^2+1)^1/2), (1+p)/(p-1)
    2. x> or equal to ln5/2, Solutions: ln3/2, ln7/2.
    3. 26.57, 2 root 5, 51.8 and -25.3, k<2 root 5 and k>2 root 5
    4. 20, ln2/40, 93
    5. 4pi^2, 1/3pi
    7. -4e^-4<g(x)<1/(8e)
    8. 2.820, 5.962,
    9. increasing, -3k/(x-2k)^2

    Some might be wrong please bear that in mind.
    Yup same
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    (Original post by sj97)
    Wasn't the range for g supposed to have an e value in it?
    4e and -2.5e
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    (Original post by taichingkan)
    Was it an increasing or decreasing function?
    Increasing, due to the positive gradient.
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    (Original post by jf1994)
    Did anyone else get that there was no asymptote for the second transformation graph? plz plz plz
    Well, I think there is one. When x<ln(5/2), the asymptote would be y=5.
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    (Original post by phoenixsilver)
    Why are people getting values with e in them for the range of g(x)?
    The only hard questions in this paper was the trig prove one and (although easier) the range one.

    I don't understand why people are getting e values for the range though.
    The question gave the domain of g(x), so this must be equal to the range of the inverse function. I equated this to inverse g(x), getting a quadratic with x taken out of it.
    How do you get e values..?
    Because the exact answer had e values in
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    POLL ----> http://strawpoll.me/4609428
    POLL ----> http://strawpoll.me/4609428
    POLL ----> http://strawpoll.me/4609428
    POLL ----> http://strawpoll.me/4609428
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    (Original post by PaulCJH)
    My Answers:
    1. 2p/(1-p^2), 1/((p^2+1)^1/2), (1+p)/(p-1)
    2. x> or equal to ln5/2, Solutions: ln3/2, ln7/2.
    3. 26.57, 2 root 5, 51.8 and -25.3, k<2 root 5 and k>2 root 5
    4. 20, ln2/40, 93
    5. 4pi^2, 1/3pi
    7. -4e^-4<g(x)<1/(8e)
    8. 2.820, 5.962,
    9. increasing, -3k/(x-2k)^2

    Some might be wrong please bear that in mind.
    Exactly the same as mine.
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    Asymptote for second graph was y=5, I think?
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    8a)

    1/sin2a + 2sinacosa/cos^2a - sin^2 a
    1 = cos^2a + sin^2

    therefore = cos^2a + sin^2a + 2sinacosa/ cos^2a - sin^2a

    the top forms a quadratic and the bottom is a difference of two squares thus leading to the final answer
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    Please vote on poll:

    http://strawpoll.me/4609429
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    (Original post by HannahSan)
    Well, I think there is one. When x<ln(5/2), the asymptote would be y=5.
    *x=5 you mean
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    (Original post by MCMXCVIII)
    Nah. I got y is 5 for the asymptote. It was just reflected on the y axis.


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    But it was the mod of f(x). right? So anything under the x axis would be reflected up, with anything over the x axis staying where it is. So there can't have been an asymptote at y=5 because the portion of graph that remains where it is clearly cuts through y = 5
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    Can someone post an unofficial mark scheme?
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    (Original post by phoenixsilver)
    Increasing, due to the positive gradient.
    Good then. Just making sure. Thanks!
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    (Original post by RThornton)
    Yeah, then you have to use sin^2x + cos^2x = 1, to get rid of the sin
    oh did it have to be only in terms of tan then? thats shame
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    (Original post by Neilg99)
    The paper was quite easy.. I think I'll get above 70/75.. Any solutions you need?

    Do all
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    (Original post by PaulCJH)
    My Answers:
    3. 26.57, 2 root 5, 51.8 and -25.3, k<2 root 5 and k>2 root 5
    I got k<-4 root 5, k>4 root 5 for that
 
 
 
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