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    • Thread Starter

    Trying to work through some past papers and got stuck on a couple of questions, would really appreciate it if anyone could help out

    1) first part of the question is to find the normal to the curve y = 2x - 3ln(2x + 5) at the point (-2,-4)... i got this to be... y = (1/4)x - 7/2

    The normal to the curve at P intersects the curve again at the point Q with x-coordinate q. Show that 1 < q < 2... just don't know where to go - thought about simultaneous equations but doesn't look too nice

    other question is...

    2) f(x) = x^2 + 5x - 2secx -(pi)/2 < x < (pi)/2
    a) show that the equation f(x)=0 has a root in the interval [1,1.5]

    A more accurate estimate of this root is to be found using iterations of the form

    x (n+1) = arccos g(xn)

    b) find a suitable form for g(x)

    i don't even understand what a suitable form is so no idea where to start on that!

    Thanks for any help, Jack

    1. Simultaneous equations are the way to go. Re-arrange to get:
    x - (12/7)ln(2x+5) + 2 = 0, and remember that q is a root of this equation.

    Let f(x) = x - (12/7)ln(2x+5) + 2, and we find that:
    f(1) < 0
    f(q) = 0
    f(2) > 0

    So q must lie between 1 and 2.

    a) f(1) > 0, f(1.5) < 0. So the roots must lie somewhere in between.

    b) Let's re-arrange f(x)=0:
    x^2 + 5x - 2sec(x) = 0
    x^2 + 5x = 2/cos(x)
    cos(x) = 2/(x^2 + 5x)
    x = arccos(2/(x^2 + 5x))
    • Thread Starter

    That's great thanks, half of this exam lark is just getting used to the type of question! got one final C3 question which would appreciate mucho if someone could help me out... looks cunningly like differential equations but it isn't - still C3 - and has got me stuck...

    T = 5 + Ae^(-kt) where A and k are constants
    Given that T = 18 when t = 10 and T = 12 when t = 60,
    show that k = 0.0124 to 3SF and find value of A

    Thanks, Jack
    • Thread Starter

    Any ideas anyone? you can use the information to give the equations...

    Ae^(-10k) = 13
    Ae^(-60k) = 7

    With differential equations like that you normally get a t=0, making e^0=1 and therefore you can work out the A constant, however the question doesnt give that so am stuck on where to go now!

    Thanks for any help, Jack

    You have a pair of simultaneous equations:
    18 = 5 + Ae^(-10k)
    12 = 5 + Ae(-60k)

    So solve them. If you can't see how, then maybe re-writing them as:
    13/e^(-10k) = A
    6/e^(-60k) = A
    will help.
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Updated: April 10, 2006
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