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    (Original post by TheWolf)
    Indeed. Btw, is it just me or does paper A3's first q not show up?
    yeah, some of the questions are missing. I think that's the only one.
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    Im trying to figure out sin 3x = cos x now....
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    The normal to the curve at the point Q, with x coordinate q, passes through the origin. Show that x = q is a solution to the equation x² + ln3x = 0

    can anybody help me with this please?!?!
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    i've found a mistake on markscheme for c3 paper 4.
    It says 3² = 3, λ should be 5/3 not 5.
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    (Original post by luketennisboy)
    The normal to the curve at the point Q, with x coordinate q, passes through the origin. Show that x = q is a solution to the equation x² + ln3x = 0

    can anybody help me with this please?!?!
    Not enough information. Can you give the full question?
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    It doesnt show up.
    I need help with this.

    f(x) = 1/(x+1)(x-1)

    What is the range? The answers just say the range of f(x) is y where y>0
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    (Original post by sb1986)
    It doesnt show up.
    I need help with this.

    f(x) = 1/(x+1)(x-1)

    What is the range? The answers just say the range of f(x) is y where y>0
    f(x) = 1/(x²-1)
    f(x) = u/v
    => f'(x) = (v.du/dx - u.dv/dx)/v²
    f'(x) = [(x²-1) x 0 - 1(2x)]/(x²-1)²
    f'(x) = -2x/(x²-1)²

    As x -> infinity, f'(x) --> zero
    Therefore range is f(x) > 0

    You can also see from the equation that as x becomes very large and positive, f(x) > 0 so doesn't cross the x-axis.
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    ok now can someone explain how to find the range, i didnt know there was a specific method i wasnt taught. Iv been guessing all this time man.

    by the way thanks for sortin me out on that range question
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    (Original post by sb1986)
    It doesnt show up.
    I need help with this.

    f(x) = 1/(x+1)(x-1)

    What is the range? The answers just say the range of f(x) is y where y>0
    That's right because for all values of x, y will always be positive, because denominator is always positive.
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    (Original post by Seth)
    That's right because for all values of x, y will always be positive, because denominator is always positive.
    what about f(1/2) = -3/4
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    (Original post by mik1w)
    what about f(1/2) = -3/4
    If I remember correctly, the domain of the function is x>1
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    (Original post by mik1w)
    what about f(1/2) = -3/4
    Sorry, my mistake still half asleep im afraid.. :rolleyes: well spotted
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    I'v been tryin dat sin3x = cosx for 30 mins now lol, i ain gettin anywhere
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    Hi there,

    I'm having a bit of trouble with Question 6a) Page 52.

    f(x)≡e^0.8x - 1 / 3-2x

    a) Show that the equation f(x)=0 can be written as x = 1.5 - 0.5e^-0.8x

    Ive been trying it for a long time now but I can't do it. Maybe i'm missing something :confused:

    Thanks In Advance
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    (Original post by Widowmaker)
    If I remember correctly, the domain of the function is x>1
    In that case, what I said above applies...

    cheers
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    (Original post by PoseidonX)
    Hi there,

    I'm having a bit of trouble with Question 6a) Page 52.

    f(x)≡e^0.8x - 1 / 3-2x

    a) Show that the equation f(x)=0 can be written as x = 1.5 - 0.5e^-0.8x

    Ive been trying it for a long time now but I can't do it. Maybe i'm missing something :confused:

    Thanks In Advance
    If f(x) = 0
    e^0.8x - 1/3-2x = 0
    e^0.8x = 1/3-2x
    1/e^0.8x = 3 - 2x
    e^-0.8x = 3 - 2x
    3 - e^-0.8x = 2x
    3/2 - 0.5e^-0.8x = x

    so x = 1.5 - 0.5e^-0.8x
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    (Original post by PoseidonX)
    Hi there,

    I'm having a bit of trouble with Question 6a) Page 52.

    f(x)≡e^0.8x - 1 / 3-2x

    a) Show that the equation f(x)=0 can be written as x = 1.5 - 0.5e^-0.8x

    Ive been trying it for a long time now but I can't do it. Maybe i'm missing something :confused:

    Thanks In Advance
    e^0.8x - 1 / 3-2x = 0
    e^0.8x = 1 / 3-2x
    e^0.8x = (3-2x)^-1
    e^-0.8x = 3-2x (inverse of both sides)
    -3+e^0.8x = -2x
    x = 1.5 -0.5e^-0.8x

    yep featherflare got to it first...
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    (Original post by PoseidonX)
    Hi there,

    I'm having a bit of trouble with Question 6a) Page 52.

    f(x)≡e^0.8x - 1 / 3-2x

    a) Show that the equation f(x)=0 can be written as x = 1.5 - 0.5e^-0.8x

    Ive been trying it for a long time now but I can't do it. Maybe i'm missing something :confused:

    Thanks In Advance
    e^0.8x = 1 / 3-2x
    e^0.8x(3-2x) = 1
    3e^0.8x - 2xe^0.8x = 1
    3 - 2x = 1 / e^0.8x
    3 - e^-0.8x = 2x
    1.5 - 0.5e^-0.8x = x

    Beaten to the punch...
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    Can someone please help with this:
    [1-cos2x]/{[1-cos2x]/sin2x}

    Does this cancel into just sin2x?

    I always get confused with this, is there some easy way of remembering how things cancel? thanks
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    Yup
 
 
 
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