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    (Original post by TeeEm)
    this is an old one of mine
    How do you do this?
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    (Original post by TeeEm)
    P and Q are distinct
    May you check this please? 2014 Q6 first part. Because the ms goes the other way. I'm sure my way is fine but not 100%. It is essentially the same thing except suggesting u is a dummy in replacement of x. Instead of using the sub in the expression.
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    (Original post by fatart123)
    Sketch for a = -1, b = 0 then? That seems to work.
    ..

    excuse the bad graphs
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    (Original post by imusti96)
    How do you do this?
    look at post 243
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    (Original post by TeeEm)
    look at post 243
    Thats a different question
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    (Original post by physicsmaths)
    May you check this please? 2014 Q6 first part. Because the ms goes the other way. I'm sure my way is fine but not 100%. It is essentially the same thing except suggesting u is a dummy in replacement of x. Instead of using the sub in the expression.
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    that is the correct answer. Isn.t it?
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    (Original post by TeeEm)
    ..

    excuse the bad graphs
    Should it not be -a^2 = a on the y-axis? So a = -1 is the only value that'll work as a = 0 gives P and Q as the same point?
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    (Original post by imusti96)
    Thats a different question
    was it this one?
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  2. File Type: pdf solutionn.pdf (116.2 KB, 31 views)
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    (Original post by TeeEm)
    that is the correct answer. Isn.t it?
    Yes but I they just said let u=x-h into the expression needed. I used a dummy variable in the expressions given then subbed it into the expression.


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    (Original post by TeeEm)
    was it this one?
    Yes thank you
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    (Original post by fatart123)
    Should it not be -a^2 = a on the y-axis? So a = -1 is the only value that'll work as a = 0 gives P and Q as the same point?
    there is no actual numbers ... the configuration could be slid anywhere on the xy plane. You just have to prove that the two curves intersect at the stationary points of one another
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    (Original post by physicsmaths)
    Yes but I they just said let u=x-h into the expression needed. I used a dummy variable in the expressions given then subbed it into the expression.


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    Are you asking is the method ok?
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    (Original post by TeeEm)
    Are you asking is the method ok?
    Yes.


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    (Original post by physicsmaths)
    Yes.


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    I cannot see why not ... in most high end problems there are several clever ways of manipulating and no marking scheme could cater for all.

    these paper are marked by mathematicians
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    (Original post by TeeEm)
    I cannot see why not ... in most high end problems there are several clever ways of manipulating and no marking scheme could cater for all.

    these paper are marked by mathematicians
    Cheers. I hope its not like some of my teachers who mark for edexcel and couldnt see why the + square root in Q1 C3 was wrong for some x .........


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    (Original post by TeeEm)
    there is no actual numbers ... the configuration could be slid anywhere on the xy plane. You just have to prove that the two curves intersect at the stationary points of one another
    Could you tell me where I've gone wrong? Don't really understand how that can work

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    (Original post by physicsmaths)
    Cheers. I hope its not like some of my teachers who mark for edexcel and couldnt see why the + square root in Q1 C3 was wrong for some x .........


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    I said mathematicians, not maths teachers.
    very few are both!
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    (Original post by fatart123)
    Could you tell me where I've gone wrong? Don't really understand how that can work

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    is this part b?
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    (Original post by TeeEm)
    is this part b?
    Mhm
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    (Original post by fatart123)
    Mhm
    if that is a yes then that is fine

    (-1,0) and (0,-1)
 
 
 
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