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    f(x) = (e^2x).sin2x , 0 < x < 180

    a)Find the values of x for which f(x) = 0 in terms of pie

    b) use calculus to find the coords of the turning points of the graph y = f(x)


    2)

    f(x) = p - 2x/x+q x not equal to -q

    where p and q are constant. The curve, C, with equation y = f(x) has an asymptote with equation x = 2 and passes through the point with coords (3,2)

    a) write down value of q (I got -2)

    b) show that p = 8 (Did this)

    c) Write down an equation for the second asymptote to C (cant do this)

    Cheers
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    I get the feeling that I want to do this Edexcel P3 Module just to see how it is.
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    (Original post by imasillynarb)
    f(x) = (e^2x).sin2x , 0 < x < 180

    a)Find the values of x for which f(x) = 0 in terms of pie
    b) use calculus to find the coords of the turning points of the graph y = f(x)



    2)

    f(x) = p - 2x/x+q x not equal to -q

    where p and q are constant. The curve, C, with equation y = f(x) has an asymptote with equation x = 2 and passes through the point with coords (3,2)

    a) write down value of q (I got -2)

    b) show that p = 8 (Did this)

    c) Write down an equation for the second asymptote to C (cant do this)

    Cheers
    1a) I get 0, pie/2
    b) I get 3/8pie and 7/8pie
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    (Original post by bono)
    I get the feeling that I want to do this Edexcel P3 Module just to see how it is.
    What do you mean?
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    (Original post by imasillynarb)
    What do you mean?
    Seems like the majority of Maths Questions on here are Edexcel P3 ones! How hard can it be!
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    f(x) = (e^2x).sin2x , 0 < x < 180

    a)Find the values of x for which f(x) = 0 in terms of pie

    e^2x =/= 0 for any x => sin2x = 0 => x = 0, pi/2 = steak and kidney pie

    b) use calculus to find the coords of the turning points of the graph y = f(x)

    f'(x) = 2(e^2x)sin2x + 2(e^2x)cos2x = 0 for stat point

    sin2x + cos2x = 0

    sin2x = -cos2x

    2x = 3pi/4, 7pi/4

    x = 3pi/8, 7pi/8

    y = √2/2 e^(3pi/4), -√2/2 e^(7pi/4) respectively
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    2)

    f(x) = p - 2x/x+q x not equal to -q

    where p and q are constant. The curve, C, with equation y = f(x) has an asymptote with equation x = 2 and passes through the point with coords (3,2)

    a) write down value of q (I got -2)

    b) show that p = 8 (Did this)

    c) Write down an equation for the second asymptote to C (cant do this)

    f(x) = 8 - 2x/(x-2) = 8 - 2/(1-2/x) = 6 as x-> +/- infinity

    => f(x) = 6 is the horizontal asymptote
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    (Original post by elpaw)


    sin2x = -cos2x

    2x = 3pi/4, 7pi/4

    x = 3pi/8, 7pi/8

    y = √2/2 e^(3pi/4), -√2/2 e^(7pi/4) respectively
    How can you just jump from the first step to the second?

    Surely there's more simplification possible as well?
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    What was with the steak and kidney pie thing?
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    (Original post by XTinaA)
    How can you just jump from the first step to the second?

    Surely there's more simplification possible as well?
    Just to clarify my position I would've gone on to say tan2X = -1 and so on.
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    (Original post by XTinaA)
    How can you just jump from the first step to the second?

    Surely there's more simplification possible as well?
    yes there is, but i know that the only time sin/cos values are equal in magnitude are when they are √2/2 (i.e. (45 + 90n) degrees for natural n), so cos45==sin45, cos 135 == -sin135 etc. its just logic (you dont have to do everything in a strict mathematical formulation). but if you wanted to do the simplification, this is what you'd get: (eg)

    sinx = cosx
    => sin²x = cos²x
    => sin²x = 1 - sin²x
    => 2sin²x = 1
    => sin²x = 1/2
    => sinx = +/- 1/√2 = +/- √2/2

    then get the corresponding values of x (remembering they have to satisfy sinx = cosx, because some will satisfy sinx = -cosx because of the squaring)
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    (Original post by bono)
    What was with the steak and kidney pie thing?
    sillynarb wanted the answer in terms of pie, so thats what i gave him.
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    (Original post by elpaw)
    yes there is, but i know that the only time sin/cos values are equal in magnitude are when they are √2/2 (i.e. (45 + 90n) degrees for natural n), so cos45==sin45, cos 135 == -sin135 etc. its just logic (you dont have to do everything in a strict mathematical formulation). but if you wanted to do the simplification, this is what you'd get: (eg)

    sinx = cosx
    => sin²x = cos²x
    => sin²x = 1 - sin²x
    => 2sin²x = 1
    => sin²x = 1/2
    => sinx = +/- 1/√2 = +/- √2/2

    then get the corresponding values of x (remembering they have to satisfy sinx = cosx, because some will satisfy sinx = -cosx because of the squaring)
    I think I'll stick to turning it into tan
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    (Original post by XTinaA)
    I think I'll stick to turning it into tan
    yeah, thats probably easier, but i hate tans.
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    elpaw:

    c) Write down an equation for the second asymptote to C (cant do this)

    f(x) = 8 - 2x/(x-2) = 8 - 2/(1-2/x) = 6 as x-> +/- infinity

    => f(x) = 6 is the horizontal asymptote



    sorry mate but i think the horizontal asymptote is f(x)= -2
    cause you didnt factorise the top correctly or
    its really late for me!!!!

    cheers
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    (Original post by elpaw)
    yeah, thats probably easier, but i hate tans.
    Stay indoors then <bu-boom, crash!>
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    are you saying the equation is (8-2x)/(x-2)? :mad: brackets would help!

    i took f(x) = 8 - [2x/(x-2)], where the asymptote is f(x)=6, but if f(x)=(8-2x)/(x-2), the asymptote would be f(x)=-2, as you said.
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    (Original post by elpaw)
    are you saying the equation is (8-2x)/(x-2)? :mad: brackets would help!

    i took f(x) = 8 - [2x/(x-2)], where the asymptote is f(x)=6, but if f(x)=(8-2x)/(x-2), the asymptote would be f(x)=-2, as you said.
    From the looks of the question Paw's right.
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    I'm sorry, i didnt read it properly. if it is p - [2x/x-2] then f(x) = 6.

    sorry again I said its too late for me!!!!
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    Sorry, my bad for not using brackets..(8-2x)/(x-2)

    Why would it be -2 sorry?
 
 
 

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