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    Heya, this is a question from the Edexcel Practice paper A4, I have never seen anything like this before (with logs and e's in) and also I think there's a bracket missing out of the question which makes it even more confusing!

    The curve C with equation y=p+qex , where p and q are constants, passes through the point (0,2). At the point P (ln2,p+2q on C, the gradient is 5.
    a) Find the value of p and the value of q
    The normal to C at P crosses the x-axis at L and the y-axis at M.
    b) Show that the area of triangle OLM, where O is the origin is approx 53.8

    I have the mark scheme in from of me and I dont understand it, the answer for a is:
    q=2.5
    p=-0.5

    Im not really sure am I meant to differentiate the curve - and if so, how so ..... :confused:

    Thanks !
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    a) dy/dx = qex

    At the point P, dy/dx = 5

    => 5 = q(eln2)
    5 = 2q
    q = 2.5

    We also know the curve passes through (0,2), so sub these values of x and y into the equation of the curve:

    2 = p + 2.5(e0)
    p = -0.5


    b) m1m2 = -1, where m1 = 5
    m2 = -1/5

    P(ln2 , 4.5)

    Equation of normal at P: y - 4.5 = (-1/5)(x-ln2)

    y + x/5 = 4.5 + (1/5)ln2

    At L, y=0 => x/5 = 4.5 + (1/5)ln2
    x = 22.5 + ln2
    => L(22.5 , 22.5 + ln2 )

    At M, x=0 => y = 4.5 + (1/5)ln2
    => M(0 , 4.5 + (1/5)ln2 )

    Now you have a triangle of height 4.5 + [(1/5)ln2] units, and base [22.5 + ln2] units, so the area is:

    (1/2)[4.5 + (1/5)ln2][22.5 + ln2] = 53.8 units2
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    a)
    y' = qex
    at P,
    x = ln2
    ex = 2
    =====
    at P,
    m = qex
    5 = q*2
    q = 2.5
    =====

    b)
    at A(0,2),
    y = 2 = p + qex
    2 = P + qe0
    2 = p + q
    2 = p + 2.5
    p = -0.5
    ======
    • Thread Starter
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    dy/dx = qex
    ok, yes I know im being stupid, its just the first line I dont get...
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    hm, I was never taught how to differentiate y=p+qe^x
    C2 exam next week too
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    and can you quickly explain what 'e' is and how it relates to ln and stuff, thank youuuu
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    hm, I was never taught how to differentiate y=p+qe^x
    C2 exam next week too
    edexcel? i will love you forever if you tell me I dont have to know it for the exam!
    lol although it would be nice to find out what it means... and what is goin on !
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    How would you differentitate y = ex?
    dy/dx = ex

    p and q are just constants, so the derivative of qex is just qex. And when you differentiate a constant on its own, you get 0, hence,
    d/dx[p + qex] = qex
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    How would you differentitate y = ex?
    dy/dx = ex
    why dont you just do it normally, and so get dy/dx =xe^(x-1) ?
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    (chants) times by the power, lower the power by 1 ...

    Im on edexcel franks, and Ive never come across differentiant y = a^x or anything
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    Im on edexcel franks, and Ive never come across differentiant y = a^x or anything
    I dont reckon we need to know it. Why on earth is it on this edexcel practice paper tho !!!?!

    Mockel - so is this just another rule you have to learn - like differentiation and integration, something you do when the equation is 'to the power of x' ?
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    Ive never come across differentiant y = a^x or anything
    You should know how to do it for P3 or higher.
    y=a^x
    lny=ln(a^x)
    lny=xlna
    (1/y)(dy/dx)=lna
    (dy/dx)=ylna
    (dy/dx)=(a^x)(lna)
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    Erm...yeah.
    Differentiate ex (with respect to x), and you get ex.
    You definitely had to know that for P2. Don't really know about the C's, but I'd think so too :confused:
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    oooook, I may just start another little thread and ask peeps.
    Thanks everybody for your help Stars the lot of you!
    franks x
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    I know you don't need to know it for C2 OCR
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    You learn this in C3 for all exam boards.

    e is a number, a constant 2.71... and basically it is the base of natural logarithms, which you'll learn about in C3. :rolleyes:
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    (Original post by franks)
    why dont you just do it normally, and so get dy/dx =xe^(x-1) ?
    You'd do this if you were differentiating with respect to e, but you are not, you have a constant to the power of x, not x to the power of a constant... so the normal method to diffrentiate doesnt work.
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    Its a C3 thing, but e is the exponential function. e^x is the curve (between 2^x and 3^x) where at (0,1) the gradient is 1.

    Essentially, at x=1, 3^x = 1 and 2^x = 1 etc... if you work out the gradients for these lines at this point, 3^x has a gradient of 1.099 and 2^x has a gradient 0.693... the gradient of e (2.71828...) at x=1 is 1.

    The inverse of this is ln|x| the natrual logarithm.

    Just to clear up any interests. But for now, just concentrate on the exams you have
 
 
 
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