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    I got stuck in last part
    what I have done so far is substitute x and y in 9x-4y-15 and
    expand it and got
    4t^3 - 9t^2 -54t +135=0
    I don't know how to factorize it further....
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    (Original post by Shaanv)
    May help in 2 seconds

    I posted now
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    (Original post by Qer)
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    I got stuck in last part
    what I have done so far is substitute x and y in 9x-4y-15 and
    expand it and got
    4t^3 - 9t^2 -54t +135=0
    I don't know how to factorize it further....
    Work out the value of t at B. This will be a factor of the equation you have. Then u can do some division to get a quadratic in terms of t.
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    i tried factorising it and got t as -15/4 and 3.
    try subing in=t into eqaution for x to try and find x.
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    (Original post by Shaanv)
    Work out the value of t at B. This will be a factor of the equation you have. Then u can do some division to get a quadratic in terms of t.

    got right answer now

    can you tell me how you know that t-3 would be a factor(at b t=3) or how can I remember that thing?
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    (Original post by Qer)
    got right answer now

    can you tell me how you know that t-3 would be a factor(at b t=3) or how can I remember that thing?
    Hmm, not sure i can explain it very well but ill have a go.

    So the tangent has to pass through B as that is the condition given in the question. So u know when x=15 and y=0 the equation for the tangent is satisfied. But as u have the equation in terms of the parameter t, the t value corresponding to the point B will satisfy the equation, so that t value is a factor.

    Sorry my explanations lacking, perhaps someone else could explain it more clearly.

    Anyways hope i helped.

    Which exam board are you with?
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    (Original post by Shaanv)
    Hmm, not sure i can explain it very well but ill have a go.

    So the tangent has to pass through B as that is the condition given in the question. So u know when x=15 and y=0 the equation for the tangent is satisfied. But as u have the equation in terms of the parameter t, the t value corresponding to the point B will satisfy the equation, so that t value is a factor.

    Sorry my explanations lacking, perhaps someone else could explain it more clearly.

    Anyways hope i helped.

    Which exam board are you with?
    edexcel
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    (Original post by Qer)
    edexcel
    With them too, i am also on that past paper grind.

    The grind never ends😂
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    (Original post by Shaanv)
    With them too, i am also on that past paper grind.

    The grind never ends😂
    are you doing M1?
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    (Original post by Qer)
    are you doing M1?
    I sat it last year. Managed to get 100 ums after doing sh** loads of past papers
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    (Original post by Shaanv)
    I sat it last year. Managed to get 100 ums after doing sh** loads of past papers
    GREAT

    what u got in c1 &c2
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    (Original post by Shaanv)
    I sat it last year. Managed to get 100 ums after doing sh** loads of past papers
    100 ums!! thats amazing!! wow @Shaanv!! i did loads of paper as well but ended up with 82 ums which is pretty good but was expecting above 90.
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    (Original post by Qer)
    GREAT

    what u got in c1 &c2
    C1 i got 95 and c2 i got 97. I am doing further maths as well and i feel like the more math i do the better i get at the stuff ive already done so that probably helped.

    (Original post by Brudor2000)
    100 ums!! thats amazing!! wow @Shaanv!! i did loads of paper as well but ended up with 82 ums which is pretty good but was expecting above 90.
    Thanks, ur making me blush😂. 82 is still an awesome score and will help u get the grades u need.
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    (Original post by Shaanv)
    Thanks, ur making me blush😂. 82 is still an awesome score and will help u get the grades u need.
    Yh but 100 is just awesome!!! How r u finding A2 stuff. Do u do further maths??
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    I need little more help with some questions, have a look, please

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    (Original post by Qer)
    ...
    You're not working with the entire range. Yes you got one part of the inequality, then look for the upper bound for which a straight horizontal line would intersect that graph twice.
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    (Original post by Qer)
    I need little more help with some questions, have a look, please
    What do you mean...?

    \sin 2 \theta - \tan \theta = \sqrt{3} \cos 2\theta

    \Leftrightarrow \tan \theta \cos 2 \theta = \sqrt{3} \cos 2 \theta

    \Leftrightarrow \cos 2\theta (\tan \theta - \sqrt{3}) = 0

    So you solve \cos 2 \theta = 0 and \tan \theta = \sqrt{3} in the required range.
 
 
 
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