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    Four problems that I hope you'll like. A level knowledge (inc FM where appropriate) is required.

    Problem 1

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    Determine whether or not there is a rational number r that satisfies

    \tan(r\pi) = \sqrt{7}

    Hint 1: a rational number is any number that can be expressed in the form p/q where p and q are coprime (i.e they have a gcd of 1).

    Hint 2: Use complex numbers.

    Hint 3: An algebraic integer is a complex root of a polynomial with integer coefficients that has leading coefficient (that is the non-zero coefficient of the highest power) 1. The sum, difference and product of two algebraic integers is also an algebraic integer. If a number is both rational and an algebraic integer then it is an integer.





































    Problem 2

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    Find all functions f that satisfy

    f((a-b)^2) = (f(a))^2 - 2af(b) + b^2

    for all real a and b































    Problem 3

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    Evaluate

    \displaystyle \int_0^1 [x(1-x)]^{-1/2} \ln \left(\dfrac{3+x}{3-x}\right) \ dx














    Problem 4

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    Evaluate

    \displaystyle \int \dfrac{dx}{\sqrt{4x^2 - x + 1}+2x}













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    I have a funny feeling none of these are actually appropriate for "A level students"...
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    (Original post by DFranklin)
    I have a funny feeling none of these are actually appropriate for "A level students"...
    Well, definitely inappropriate for an A-level paper

    Whenever I've done this in the past I've always received a decent response. We'll see what happens :lol:
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    (Original post by DFranklin)
    I have a funny feeling none of these are actually appropriate for "A level students"...
    Having done A-Level maths + FM just last summer, I can safely say I have never seen anything like this unless I missed something... :lol:
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    (Original post by Indeterminate)
    Well, definitely inappropriate for an A-level paper

    Whenever I've done this in the past I've always received a decent response. We'll see what happens :lol:
    Since there's a set of 10 problems I posted only yesterday with little response (and although some of them were hard, some were reasonably accessible) I'm not expecting a great deal...
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    (Original post by RDKGames)
    Having done A-Level maths + FM just last summer, I can safely say I have never seen anything like this unless I missed something... :lol:
    :lol:

    Well here's a start. For problem 4 consider the quadratic in the denominator and try a substitution.

    (Original post by DFranklin)
    Since there's a set of 10 problems I posted only yesterday with little response (and although some of them were hard, some were reasonably accessible) I'm not expecting a great deal...
    Oh.

    Well, erm, at least we try. :lol:
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    I think I may be able to do 4...
    I'll have a crack at it tomorrow

    Edit: Couldn't hack it
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    Can you confirm whether or not Problem 2 is just f(x) = x. If so, do you want us to prove it?
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    (Original post by BobBobson)
    Can you confirm whether or not Problem 2 is just f(x) = x. If so, do you want us to prove it?
    I think f(x) = 1+x works as well. Plug a,b = 0 in to get f(0)=0 or 1. Then let a=0, and see what you get. Also, with functional equations substitute back in to see if what you get works.
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    (Original post by BobBobson)
    Can you confirm whether or not Problem 2 is just f(x) = x. If so, do you want us to prove it?
    (Original post by A02)
    I think f(x) = 1+x works as well. Plug a,b = 0 in to get f(0)=0 or 1. Then let a=0, and see what you get. Also, with functional equations substitute back in to see if what you get works.
    If you think you've found one (or more than one) solution, can't you find any others? Or can you show that there aren't any others?
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    (Original post by Indeterminate)
    If you think you've found one (or more than one) solution, can't you find any others? Or can you show that there aren't any others?
    Yes.
    I think this is right:
    Answer

    a=0;b=0
    f(0)=f(0)^2
    So f(0)= 0 or 1
    Case 1: f(0)=0
    Let a=0
    f(b^2)=b^2
    so f(x) = x
    Case 2: f(0)=1
    Let a=0
    f(b^2)=1+b^2
    so f(x) =1+x
    Substitute back into check, and both of these work.
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    Problem 4 is just ugly. Problem 3 isn't so bad, on the other hand (doubt an A-Level student would manage it, though):

    Use x \mapsto (\sin x)^2 to get

    \displaystyle 

\begin{align*}\int_0^1 \frac{\log \frac{3+x}{3-x}}{\sqrt{x(1-x)}} \, \mathrm{d}x &=\int_0^{\pi/2} \log\left(\frac{3 + \sin^2 x }{3-\sin^2 x}\right)\frac{2\sin x\cos x \, \mathrm{d}x}{\sqrt{\sin^2 x(1-\sin^2 x)}} \\ &= 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin^2 x}{3-\sin^2 x}\right) \, \mathrm{d}x \end{align*}

    Now use x \mapsto \frac{x}{2} and \sin^2 x = \frac{1-\cos \frac{x}{2}}{2} to get

    \displaystyle \int_0^1 \frac{\log \frac{3+x}{3-x}}{\sqrt{x(1-x)}} \, \mathrm{d}x = \int_0^{\pi} \log \frac{7 - \cos x}{5 + \cos x} \, \mathrm{d}x

    Now use \displaystyle \int_0^{\pi} \log(\kappa \pm \cos \theta) \, \mathrm{d}\theta =\pi \text{arccosh}\, \kappa on our integral (pretty nifty identity, try proving it) to get:

    \displaystyle I = \pi (\text{arccosh} \,7 - \text{arccosh} \,5)
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    "An algebraic integer is a root of some polynomial with leading coefficient (that is the non-zero coefficient of the highest power) 1."

    Really?
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    (Original post by Zacken)
    Problem 4 is just ugly. Problem 3 isn't so bad, on the other hand (doubt an A-Level student would manage it, though):

    Use x \mapsto (\sin x)^2 to get

    \displaystyle 

\begin{align*}\int_0^1 \frac{\log \frac{3+x}{3-x}}{\sqrt{x(1-x)}} \, \mathrm{d}x &=\int_0^{\pi/2} \log\left(\frac{3 + \sin^2 x }{3-\sin^2 x}\right)\frac{2\sin x\cos x \, \mathrm{d}x}{\sqrt{\sin^2 x(1-\sin^2 x)}} \\ &= 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin^2 x}{3-\sin^2 x}\right) \, \mathrm{d}x \end{align*}

    Now use x \mapsto \frac{x}{2} and \sin^2 x = \frac{1-\cos \frac{x}{2}}{2} to get

    \displaystyle \int_0^1 \frac{\log \frac{3+x}{3-x}}{\sqrt{x(1-x)}} \, \mathrm{d}x = \int_0^{\pi} \log \frac{7 - \cos x}{5 + \cos x} \, \mathrm{d}x

    Now use \displaystyle \int_0^{\pi} \log(\kappa \pm \cos \theta) \, \mathrm{d}\theta =\pi \text{arccosh}\, \kappa on our integral (pretty nifty identity, try proving it) to get:

    \displaystyle I = \pi (\text{arccosh} \,7 - \text{arccosh} \,5)
    I expected you to chip in


    (Original post by Zacken)
    "An algebraic integer is a root of some polynomial with leading coefficient (that is the non-zero coefficient of the highest power) 1." Really?
    Well spotted :lol:
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    (Original post by Indeterminate)
    I expected you to chip in
    Might as well before lectures start (tomorrow!!). ;lol:

    When do your lectures start?
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    (Original post by Zacken)
    Might as well before lectures start (tomorrow!!). ;lol:

    When do your lectures start?
    Fair enough :lol:

    Day 1 for me too, but I know a few who haven't got any lectures until Monday.

    But yeah looking forward to it of course!
 
 
 
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