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    (Original post by constellarknight)
    Exactly. Like I said in my previous post, Indeterminate is not making much sense, which is apt given his username.
    We all make silly mistakes from time to time. He's deleted his post a few minutes back. Don't harp on about it.
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    (Original post by Zacken)
    We all make silly mistakes from time to time. He's deleted his post a few minutes back. Don't harp on about it.
    Alright whatever.
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    (Original post by constellarknight)
    Exactly. Like I said in my previous post, Indeterminate is not making much sense, which is apt given his username.
    :facepalm:
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    How do you get the spanking smiley?
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    (Original post by Indeterminate)
    Enjoy

    Problem 1

    Let

    p and q be positive integers (not necessarily distinct)

    Prove that

    p^q - 1 is divisible by (p-1)

    without using the remainder theorem




    Problem 2

    Find

    \displaystyle \int \dfrac{\sin t - \cos t}{(\sin t + \cos t)\sqrt{\sin^2 t \cos^2 t + \sin t \cos t}} \ dt




    Problem 3

    Let f be a function satisfying

    f + f' = |x|

    for real x

    You are given that f(-1)=0

    Find the value of f(1)




    Problem 4

    Find

    \displaystyle \int \dfrac{x\cos x + 1}{\sqrt{x^2 + 2x^3 e^{\sin x}}} \ dx




    Problem 5

    A rational function g(x) is defined as follows

    g(x) = \displaystyle \int \dfrac{f(x)}{x^2(x+1)^3}\ dx

    where f(x) is a quadratic satisfying f(0)=1

    Find the value of f'(0)

    oops I screwed my constants up lol, think this one is right now..
    Is Q2 something like..
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    -arccos(1/(sint + cost)^2)
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    Problem 3
    \displaystyle 2(e^{-1}-e^{-2})
    The function \displaystyle f(x) = x - 2(e^{-1}-1)e^{-x} - 1 for  \displaystyle x \ge 0, and  \displaystyle f(x) = -x -2e^{-x-1} +1 for  \displaystyle x < 0.


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    This is a really lovely question.
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    (Original post by Indeterminate)
    Enjoy

    Problem 5

    A rational function g(x) is defined as follows

    g(x) = \displaystyle \int \dfrac{f(x)}{x^2(x+1)^3}\ dx

    where f(x) is a quadratic satisfying f(0)=1

    Find the value of f'(0)

    Is question 5 correctly stated? Because I'm not enjoying it too much at the moment.
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    5 is nice. Currently thinking about 3.
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    (Original post by morgan8002)
    5 is nice. Currently thinking about 3.
    Did you get 3 for problem 5?
    I just went looking for a function g(x), I think \displaystyle \frac{-1}{x(x+1)} works.
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    (Original post by 13 1 20 8 42)
    oops I screwed my constants up lol, think this one is right now..
    Is Q2 something like..
    Spoiler:
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    -arccos(1/(sint + cost)^2)
    Or equivalently...
    Spoiler:
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    -\arctan(\sqrt{(\sin t+\cos t)^4 - 1} )
    ...but yours is much neater
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    (Original post by EricPiphany)
    Did you get 3 for problem 5?
    I just went looking for a function g(x), I think \displaystyle \frac{-1}{x(x+1)} works.
    Yeah. I just wrote f(x) = ax^2 + bx + 1, since f(0) = 1, then integrated. You get some logarithmic terms that only go away if b = 3 and since g(x) is rational, b must be 3.
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    (Original post by morgan8002)
    Yeah. I just wrote f(x) = ax^2 + bx + 1, since f(0) = 1, then integrated. You get some logarithmic terms that only go away if b = 3 and since g(x) is rational, b must be 3.
    Very well reasoned.
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    (Original post by morgan8002)
    Yeah. I just wrote f(x) = ax^2 + bx + 1, since f(0) = 1, then integrated. You get some logarithmic terms that only go away if b = 3 and since g(x) is rational, b must be 3.
    I'm an idiot - I didn't even try to integrate - I assumed that there would be some very slick theoretical approach.
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    (Original post by atsruser)
    I'm an idiot - I didn't even try to integrate - I assumed that there would be some very slick theoretical approach.
    same, by the time i was getting around to integrating i lost motivation, really wished i had stuck with it now
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    (Original post by drandy76)
    same, by the time i was getting around to integrating i lost motivation, really wished i had stuck with it now
    In fact, I'd got as far as writing out the integrals ready to integrate and didn't bother - I went to watch TV instead. I've been led astray by the complexity of Indeterminate's other problems, I think. Still, it's a useful lesson - you need persistence as much as ability to solve problems, and if you don't follow a lead, you'll never know if it was the right one.
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    Impressed by the response to this thread!

    I see that #4 is the only one that remains, so here's a hint
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    u = 2x e^{\sin x}
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    Looks like I'll have to wrap things up

    Solution 4


    Let I denote our integral.

    Multiplying through by e^{\sin x} > 0 gives us

    I = \displaystyle \int \dfrac{x\cos x e^{\sin x} + e^{\sin x}}{e^{\sin x}\sqrt{x^2 + 2x^3 e^{\sin x}}} \ dx

    = \displaystyle \int \dfrac{x\cos x e^{\sin x} + e^{\sin x}}{xe^{\sin x}\sqrt{1+ 2x e^{\sin x}}} \ dx

    Now let  u = 2x e^{\sin x} so that

     \dfrac{1}{2} \ du = (e^{\sin x} + x\cos x e^{\sin x}) \ dx

    and this leads to

     I = \displaystyle \int \dfrac{\ du}{u \sqrt{u+1}}

    Now v^2 = u+1 \Rightarrow 2  \ dv = \dfrac{du}{v} = \dfrac{\ du}{\sqrt{u+1}}

    and so

    I = \displaystyle \int \dfrac{2}{v^2 - 1}  \ dv = -2 \tanh^{-1} v + \text{constant}

    Now back in terms of x

    I = -2 \tanh^{-1} \left(\sqrt{1+ 2xe^{\sin x}}\right) + \text{constant}



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    (Original post by Indeterminate)
    Looks like I'll have to wrap things up

    Solution 4

    :drool:
 
 
 
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