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    (Original post by TeddyKC)
    Could someone help with this beast of a question?
    (Part b)
    Attachment 425815


    Posted from TSR Mobile
    Which paper is it from?
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    (Original post by frozo123)
    I wish I knew how to use a calculator...
    The struggle is real
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    (Original post by Bealzibub)
    Do I need to learn these? Do they come up in papers?

    they are quoted in the formula book;
    you could be asked to prove one of them but they would probably give you a hint - let P=A+B and Q= A-B, then using the addition formulae show that sinP-sinQ =....
    I do think this is unlikely to be asked though.
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    (Original post by TeeEm)
    question with solution posted earlier (original post 1335)

    well done to those who did it correctly ...
    I did it with 5 of my students today, and they all got "eaten" ...
    Ty for the worked method,

    3 lines before the answer, where you took e to one side, then simplified it, and turnt it into one e^ to the power of ln's , how did you get 1/10 ln4.
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    Why has 3sinx from the question changed to 3cosx, how did they get to this stage ?? the model answer matches the mark scheme
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    Has anyone got an easy way to remember transforming graphs? Like what to do when it's af(x) or f(ax)
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    Can some one explain how -1/5ln(11/20)= 1/5ln(20/11) please.
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    (Original post by nayilgervinho)
    How many sig fig should I put down for solutions to contextual rcostheta/rsintheta questions.
    they normally tell you how they want it, but if they dont, 3sf
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    (Original post by Gilo98)
    it is five. but look below at the conventional approach using similar triangles. Apply a similar method for f(5) on the RHS branch.
    Hi, could you please explain June 2013 withdrawn 8c and d, I'm confused by the format of H since the sign is the opposite
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    (Original post by BioAgent)
    Can some one explain how -1/5ln(11/20)= 1/5ln(20/11) please.
    -ln11/20=-(ln11-ln20)=ln20-ln11=ln20/11
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    (Original post by MrBowcat)
    Them asking to draw inverse trig graphs is very very very very very very very very very very very very unlikely.
    That does not mean it's impossible, and a two or three marker based on that could be the difference between two grades for someone. People might as well skim them
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    (Original post by BioAgent)
    Can some one explain how -1/5ln(11/20)= 1/5ln(20/11) please.
    Do you agree that 2ln(x)=ln(x^2)? And do you agree that (\frac{11}{20})^{-1}=\frac{20}{11}?
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    (Original post by suyoof123)
    Ty for the worked method,

    3 lines before the answer, where you took e to one side, then simplified it, and turnt it into one e^ to the power of ln's , how did you get 1/10 ln4.
    1/5(....) - 1/10(....) = 1/10(....)
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    (Original post by AbubakarB)
    Can you accurately draw the sin, cos and tan graphs?

    From there everywhere where the graph crosses the x-axis (when y=0) draw an asymptote.

    1) for the cosec graph...
    every maximum point on the sin graph - draw a happy face lol
    every minimum point draw a sad face.

    2) same concept with the sec graph

    3) the tan graph is a bit trickier to grasp but still the same concept. y=0 ... asmyptote then flip the graph
    yes I can
    what do you mean flip? reflection in the asymptote?
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    Predictions for 90 ums.
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    (Original post by TeeEm)
    1/5(....) - 1/10(....) = 1/10(....)
    Thank you! cant believe I missed that....
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    (Original post by toddle1)
    Has anyone got an easy way to remember transforming graphs? Like what to do when it's af(x) or f(ax)
    af(x) = multiply y coords by a
    f(ax) = divide x coords by a
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    But how do you solve the quadratic?

    I got x^2(a) +x(k) +a = 0

    How would you solve a quadratic like this???
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    (Original post by TeeEm)
    1/5(....) - 1/10(....) = 1/10(....)
    Last minute advice/predictions?

    (Original post by adorablegirl1202)
    Hi, could you please explain June 2013 withdrawn 8c and d, I'm confused by the format of H since the sign is the opposite
    check out Arsey's model answers, haven't done it myself and chilling out now
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    (Original post by TeddyKC)
    Could someone help with this beast of a question?
    (Part b)
    Attachment 425815


    Posted from TSR Mobile
    set the two functions equal to each other to get an equation for their points of intersection. It should be a quadratic in x. Since they curves are tangential, you know this equation only has one solution. This means the discriminant of the equation is zero. You'll then get an equation you can solve for k.
 
 
 
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